I knew this would make a swastika eventually.
Code
C#
BigInteger Square Root —— https://pastebin.com/rz1SdACZ
Generate Bitmap within original code —— https://pastebin.com/hMTtJF6E
Generate the large square for e and t —— https://pastebin.com/nbjs2kz4
How to run VQC code on Linux —— https://pastebin.com/6HnN7K5X
More on generating a bitmap with the original code —— https://pastebin.com/JUdtehb4
PMA's tree generator —— https://pastebin.com/ZH9fSWu2
Original VQC code —— https://pastebin.com/XFtcAcrz
Unity Script —— https://pastebin.com/QgAXLQj3
Unity Script 2 —— https://pastebin.com/Y38nVWgT
Java
Traverse the VQC cells in real-time —— https://anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z
Tree Generator —— https://pastebin.com/VZnQQR2i
Tree Generator w/ x & x+n search —— https://pastebin.com/0jPr3RrE
VQCGenerator —— https://pastebin.com/VMRnkXFP
VQCGenerator w/ Bitmap —— https://pastebin.com/Dgu9aP1h
VQC Triangle Number Methods —— https://pastebin.com/NCQ3HK2K
NodeJS
BigInteger Library and Sqrt —— https://pastebin.com/y8AXtFFr
Python
3D VQC —— https://pastebin.com/vdf8SpYt
3D VQC (v2) —— https://pastebin.com/wZM5Thzu
Calculate variables based on e and t —— https://pastebin.com/4s6McdbN
College Anon's code —— https://pastebin.com/d8xZZnm0
Create the VQC —— https://pastebin.com/NZkjtnZL
Fractal cryptography —— https://pastebin.com/XuN4U7Dv
Generate any cell in (0,1) and (0,2) —— https://pastebin.com/gRTYpdMU
Generate cells for a (and more) —— https://pastebin.com/iAizgLFF
Generate genesis cell —— https://pastebin.com/GKzcCpMF
Generate positive AND negative genesis cells —— https://pastebin.com/9ixjRyxt
Get A and B from C and N example —— https://pastebin.com/s0SZ9BNF
VQC + t —— https://pastebin.com/Lgufk0db
RSA & PGP key wrapper —— https://pastebin.com/vNqnPRJR
Rust
Additional VQC code —— https://play.rust-lang.org/?gist=50def916ad48400bc5d638fbf119ae85&version=stable
Check if a number is prime —— https://huonw.github.io/primal/primal/fn.is_prime.html
Create Bitmap using the VQC Generator —— https://play.rust-lang.org/?gist=c2446efeec452fe14e1ddd0d237f4173&version=stable
Create Bitmap using the VQC Generator [V2] —— https://pastebin.com/zGSusyz5
Generate the VQC —— https://play.rust-lang.org/?gist=6b6beb372b6b931f1abd30642a35a80c&version=stable
Static Java/C# class with all RSA numbers —— https://pastebin.com/XYFpsDWE
Factorization methods (Java)
Binary search for i —— https://pastebin.com/TAt5bDsR
GCDFactor —— https://pastebin.com/70GJSMrv
Count down from t of 1c element —— https://pastebin.com/xxYa946V
Mirrors 1c until e=(-x+n^2) —— https://pastebin.com/WJBqPM4P
Calculate factors using -x jumps —— https://pastebin.com/gKX9GW9r
Previous Threads
RSA #0 —— https://archive.fo/XmD7P
RSA #1 —— https://archive.fo/RgVko
RSA #2 —— https://archive.fo/fyzAu
RSA #3 —— https://archive.fo/uEgOb
RSA #4 —— https://archive.fo/eihrQ
RSA #5 —— https://archive.fo/Lr9fP
RSA #6 —— https://archive.fo/ykKYN
RSA #7 —— https://archive.fo/v3aKD
RSA #8 —— https://archive.fo/geYFp
RSA #9 —— https://archive.fo/jog81
Videos on cryptography —— https://pastebin.com/9u3hwywe
Thanks Baker! Delicious bread! Appreciate the new graphics for DM's, and for the new crumb maps. Great work.
WHERE WE GO {1},
WE GO (RS)A[11]!!!
Also:
I E Sanctus Discurre, { >BOOM! }
Dona Eis Prandium. { >BOOM! }
But if you'd like me to actually answer your question:
There was a conversation today that might should maybe be added to the bread.
https://8ch.net/qresearch/res/567922.html#q568733
https://8ch.net/qresearch/res/568699.html#q568803
Get ready for a Math BOOM ;)
>>5075
Could…. could you reformat that? lol
Even I'M having difficulty following that…
ty. And context for this, posted by same user:
Hope he is ok!?
See you guessed:
thought the same reading:
>Hmmm, is the pope a believer in hollow earth theory?
>Nice find anon.
..and ya pissed the BV off:
(left trip on few posts)
ty!
Oh, did I switch over and it loaded into thingy?
My B. I try to be conscious off that…
It's why I derp and post as Anon over here every now and then.
Tell BV he has my sincerest apologies.
Also… >>>/qresearch/568733…
I… hmm…. missed something the first time…
Wife and son?
Work is good, but slightly overrated. Nice to pop in here and afterwards and see Topol breaking news about VQC sightings and posting awesome memes. Where is that fag VQC? Did he really lose a wife and son? Wtf?
Not sure if anyone noticed, but Q has been talking about The End recently.
Also, recent Q post linked (a few steps removed) to a Haiti photographer in bed with Comcast who had been working on blockchain debt products.
Hrm. Also, where is VQC twitter, he seems to have been disappeared?
Been away from VQC board for a while. What'd I miss?
Who dat be? Q is talking about The End? Sauce? Block chain and all encryption will suffer a loss of confidence when we solve this math challenge. VQC twatter has been down since last week, part of the purge. You missed a bunch of cool ideas and shitposting. Same thing you'd miss on any other board. Welcome back, Anon!
wow, just caught up. incredibly sad if true.
I know, but it's been a while and its frequency increased in span.
>>568863
https://wikileaks.org/clinton-emails/emailid/3672
Who are we taught to trust the most?
This will not be easy.
The END.
Q
https://8ch. net/qresearch/res/568699.html#568863
Is the stage set for a drop of HRC +++ + +++++(raw vid 5:5). EX-rvid5774.
We have it all.
Re_read re: stage.
The nail in many coffins [liberal undo].
[Impossible to defend].
[Toxic to those connected].
WE must work TOGETHER.
WE are only as strong as your VOICE.
YOU must organize and BE HEARD.
THIS is why they keep you DIVIDED and in the DARK.
WEAK.
We are here to UNITE and provide TRUTH.
Dark to LIGHT.
EVIL surrounds us.
WE are FIGHTING for you.
Where we go one, we go ALL.
The choice, to KNOW, will be yours [end].
Q
https://8ch. net/qresearch/res/563179.html#563806
Working through n0 and this piece:
>>The objective is to find a base larger than n and smaller than x+n at this stage.
>So 5 is a "base". The objective I think is analogous to Newton's method; you need an initial guess to get the process started. The process will transform our initial guess to be n < HERE < x + n
I made that. That is not completely related to the grid but its neat. So missing to the left in that pic is a column of 1's. Then each cell is determined by the sum of the cell above and the cell to the left for the rest of the grid. If we index this by (x,y) where x is the column and y is the row, it would be a triangle with side length y in the xth dimension. Maybe we can use this to help us out. 1st column is integers, 2nd is triangular numbers (which we can use to construct the squares), 3rd is pyramidal numbers etc.
Now if you change the column that is missing on the left to all 2's, this results in odd numbers in the next column. Then, the next column would be successive odd numbers. Then the next column would be a sum of successive odds which is also the square numbers, which can also be used in our square calculation stuff. The next column is cube numbers.
Essentially, if you have the left column as x and the top column as 1, then an entry in row r column c would be a (x+2)-gon of side length r in the cth dimension. Idk maybe we can use some geometry for this. Also, maybe since everything is derived from the 2 initial seeds, (x and 1), we could reduce everything to multiples of 2 and multiples of 1 and use that. I'm not sure.
The more interesting thing I discovered recently was a pattern in the continued fractions. Last week my Professor taught us continued fractions and mentioned that there were patterns of the continued fraction for sqrt(2) or something carved onto Babylonian tablets or something. He also mentioned that nobody knew anything about continued fractions. So I discovered a pattern for continued fractions that oddly relates to this whole dd+e = c thing.
So for pic related that continued fraction would be notated as [a0, a1, a2…]. Also for this, I will write this [a0, (a1, a2)] to say that the thing starts with a0, then the pattern inside the parenthesis repeats forever.
So if you have a continued fraction calculated (also if you do the calculation of these by hand, everything always conveniently cancels out, which might be good for computing) for this number sqrt(dd+e), then if (2*d) is divisible by e, then the continued fraction would be [d, (2d/e, 2d)].
(look in pic related for some good examples). If we extrapolate from there, and say ab=c, then we would know that there exists a fraction [c, (a, 2c)] which is equivalent to the square root of (c^2 + 2b). He was telling us to look at c^2 so maybe this is it?? idk. Also you could do root(c^2 + b) = [c,(2a,2c)]. These numbers would all be like the parent number but idk I've been cranking at it algebraically to solve for a and I can't poke any holes in it. Also if you do the continued fraction of [(1)] it is equivalent to the golden ratio. Also, you can calculate it out to a certain amount of terms, and you will notice that the numerator and the denominator are always fibonacci numbers. These numbers are where I was talking about the 'cancellation' because it happens with most repeated continued fractions.
Also I didn't post it in that thread, somebody else did
The theorem for calculating the terms out is this:
let N(t) be the numerator of the tth fraction. Guess what D(t) is. This is for [a0,a1,a2,…]
N(0) = a0
N(1) = a0*a1 + 1
D(0) = 1
D(1) = a1
Then, N(t) = atN(t-1) + N(t-2), and D(t) = atD(t-1) + D(t-2).
These formulae make it easy to see how [1,1,1,1,..] is fibonacci. This could be easier to program cuz you can cache the stuff
Also basically the higher your t value, the more precise a fraction you will get to approximate the true value of the continued fraction. N(0)/D(0) will be a shit approximation, but N(100)/D(100) will be pretty damn close to the number you want.
This just keeps getting weirder… lol
http://investmentwatchblog.com/i-am-amazed-you-will-be-too-just-put-together/
Hate to be a fag but was the “have succumbed to alcohol, can’t handle my son’s death, see you on the other side - vqc” /qresearch/ post shown to be false in any way?
I hope so, obviously, but I haven’t noted any VQC posts here since.
Lemme try to dig it up. I thought it was an expertly-crafted shill/troll sce it was around the time of the Revelations post from VQC.
Plus, it had no trip and was signed lowercase vqc. If VQC is around, he could clear this up quickly…….
When it happened I almost posted it here but, honestly, PMA was kicking so much ass and it seemed like such a predictable shill tactic that I didn’t out of shill propagation caution. However, now that a number of weeks have elapsed, I’m more concerned.
Yes, I recall the threatening post back then.
Also the suicide exchange that Topo references.
I believe VQC was in todays thread referenced earlier >>5082 and also the following one.
Took these 2 posts to be VQC, 2 different threads:
The first is ODD. Seems a compartmentalization of personality, or perhaps it's multiple individuals working together.
Well, I'm not gonna believe that shit. Chris is fine, and he disappears periodically as we all know. He beat alcohol with nothing to return to, and now he has us and an amazing truth to reveal. He has no reason to do any of that.
Fake.
Read through all that and it sounds like someone Is "driving" him. Idk. Something is off. Not sure what but something.
>>>5114
>qresearch posts you posted!!!
>I remember the topo posts because back then I was insisting he be banned (sorry bro)
What the absolute fuck is going on?
VQC please post with trip and verify, even if it is the worst, we can handle it.
I believe so.
To clear the air, the "threatening post" was this (at bottom). Including prior posts for context, same ID, same thread:
on 2018.01.02
https://8ch.net/cbts/res/228142.html - #270
▶Anonymous 01/02/18 (Tue) 09:47:46 153378 No.228325>>228412
I wondered when this card would be played.
North Korea is going for the 'nuclear option'.
"Threatening to release classified information on the Moon landings."
What other NASA secrets is NK threatening?
▶Anonymous 01/02/18 (Tue) 09:49:41 224158 No.228335>>228353
>>228321
Who controls the gold??
Peter Munk
▶Anonymous 01/02/18 (Tue) 09:54:01 153378 No.228353
>>228335
Controlled
▶Anonymous 01/02/18 (Tue) 10:10:24 153378 No.228445>>228450 >>228459 >>228461 >>228463 >>228491
>>228412
It's not the first time.
KYU's dad did it previously and every time the US backed off.
Google for the sauce.
Site has decent info and disinfo.
Choose wisely.
Do you think NASA has things to hide?
Why no unshopped pictures of the North Pole.
Let's not pretend there is nothing in that.
Let's not assume what is or isn't there.
All roads lead to the revelation about the poles, what (all) the pyramids really are, what is in Kentucky, Tibet, etc.
Critical thinking.
Why are satellite pictures of the North Pole above 40,000ft classified?
Why on Earth would this be?
It is DEFINITELY connected.
▶Anonymous 01/02/18 (Tue) 10:14:26 153378 No.228471>>228478 >>228484 >>228562
>>228450
Kek
When all this is done, the reality is pretty special.
Critical thinking.
No one needs to leap to Hollow Earth.
Just post a single unshopped image of the North Pole from 40,000ft or above (satellite preferred) and I will never come back.
▶Anonymous 01/02/18 (Tue) 10:17:31 153378 No.228487>>228492 >>228497
>>228461
I've been here every single day.
Sometimes you recognise me, sometimes you don't.
We're on the same side.
We all want the Cabal locked up or executed.
Some of us want the Revelation.
Personally, I'm just about done waiting.
Fortunately, this wait is almost over.
It's about Time.
▶Anonymous 01/02/18 (Tue) 10:18:57 153378 No.228493
>>228463
I will leave (well, I'll not post) for good if you show me a single unadulterated photo (preferably satellite) image of the North Pole from 40,000ft or above.
▶Anonymous 01/02/18 (Tue) 10:19:47 153378 No.228498>>228511
>>228484
What did I lie about?
▶Anonymous 01/02/18 (Tue) 10:21:20 153378 No.228509
>>228501
Expand your thinking.
It's all connected.
The 40,000ft that is classified.
THAT is what it has to do with Q.
▶Anonymous 01/02/18 (Tue) 10:27:15 153378 No.228537>>228543 >>228545 >>228546 >>228550 >>228557 >>228558 >>228561 >>228569
>>228511
Fair enough.
Some answers then.
The round up has started in the EU.
Assange is about to drop bombs.
Iran will spill evidence everywhere.
The Cabal will flood with NASA truth bombs as their very last card. It's their only card left.
Everything is about to get very real, very quickly.
It's been a pleasure anons.
I'll go back to anon posting.
BUT we will meet up for a picnic inside the Earth at some point.
I'll be the one on the unicorn.
I'm not even joking.
▶Anonymous 01/02/18 (Tue) 10:30:38 153378 No.228558
>>228537
One more thing.
The next time (((you))) send someone to my home, or even anywhere near it, I will take care of it personally and there will be the worst kind of Hell to pay.
You know who you are and you know EXACTLY what I mean.
▶Anonymous 01/02/18 (Tue) 10:32:04 153378 No.228565
>>228562
>>228561
>>228550
<3
▶Anonymous 01/02/18 (Tue) 10:32:54 153378 No.228570
>>228545
XXX
Well… judging by the images the person they're responding to would post, and considering that it's super likely that it's who gave me the "(e,n) : (4,1) : a[t] where t = 2,3,4"…
Yeah, I can't figure out who they're talking to either.
I absolutely hate saying this: could also be an AI bot trained on /vqc/ as well as /qresearch/ or /cbts/. A muddy-the-waters type of thing.
Yes, I've figured out you were one of the people in the thread, ya fag.
CBTS #270 ID: 153378
https://archive.fo/jzRqB
This is definitely Chris.
I think he's right about the pole stuff. It's so fucking easy to shop a hole out of the poles, they're covered in white.
Most of the conversation is with me on both threads. Yes.
As for what you linked to…
"You have a certain… bickleness"
Know where I've seen THAT before?
https://ia802300.us.archive.org/8/items/rofschildv1/IAmARofschildAxeMeAQuestion.html
Topol is certainly the easiest to spot of all. Though there are a few ponies that are give-aways too, ha.
Topol, you couldn't help yourself yesterday, taking the CDC plane and putting the little DaWae on the tail. imo, that was distracting to anons trying to work and dig, just one example - this is not a game. Trying to change the course of the future of humanity, let alone what this entire operations costs (yuge), and you can't help shitposting. Anyway, enough preaching from me.
While we're clearing the air, I don't think in retrospect this was VQC, but this is the suicide context. And the one that Topol got bent wrt my reaching out directly (which took about 3minutes of searching) .
ID: 40e61e in:
https://8ch.net/cbts/res/148453.html
After that, there were more and more 'kill myself' scenarios over time, and it appears to have been a new shill tactic, possibly with some real cases mixed in.
>Let's have fun and get to work!
Yes, let's get to work!
CA, one thing that's been lingering is the reference to "pyramids", specifically with square base (versus triangular, or tetrahedron pyramid).
The Johnson solid J_1 is the square-based pyramid with equilateral faces. Numbers that are simultaneously Triangular and square pyramidal satisfy the Diophantine Equation.
Also, there is the concept of a "Pyramidal Frustrum". A Frustrum is defined as the portion of a solid which lies between two parallel planes cutting the solid. (e.g. our 5-layer thick blue zone!).
Cool, looking back, only (3) posts by that user id in the thread, so couldn't have been you. j/k!
Here's the CDC plane, check the tail (this was posted short bit later in bread after first):
You're so not living that down!
MY POOR MAMMY!
:P I had to say you were some professor's assistant and it was regarding some yob interview.
That being said…doooon't hate me buuuut:
A person with a history of alcoholism who has a wife and son to worry about: Quinn Michaels.
That or Assange, who the current VQC is not, but if anything happened to his wife and kid… I feel like that'd be frontpage news.
Aight, I'm going back to the EZ Bake to have a Russian Tea Party with my ponies.
So, this is what I had focused on earlier:
>The objective is to find a base larger than n and smaller than x+n at this stage.
So, that gives us a window of "x" for the start. Is that an 'x' for that iteration, or the final result? Not sure.
Also, the start. Have been starting with the (1,c). I don't think that's correct, better to start using the f-2 method, or even using (d+1) as 'a' and 'b'. We need to move TOWARD the 1,c, and if we get there before finding a factor pair, then c is prime. This says the (1,c) should be the destination of the iteration, no?
>We will add a method to calculate what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d.
>That gap will allow us to use more geometry of triangular numbers to establish what multiple of 2d could be added to make our eight triangles.
>That geometry with multiples of 2d will give us the solution, if it exists (which in our example we know it does). If doubling the width of the chunks of (f-2) didn't give us a solution, the number c would be prime, the worst case (doubling the width of (f-2) chunks very quickly determines this as the number of times it doubles is logarithmic in the length of (f-2)).
I wish I could help more with the math but I'm both overrun with other things to do and some pretty intense non-physical developments in my life lately. Shit's absolutely crazy, but I don't want to derail the thread. Still here daily to mod if anyone needed reassurance of that. It always makes me happy to see how many of you are still here too.
Keep in mind, there was that person who kept posting about Chris saying he was going to kill himself and that his only evidence was Topolanon being there. Topol said here >>3121 that it was a lie. Another thing to keep in mind is that drunk people barely ever spell things properly, so even if drinking would make it more likely that he'd spell vqc in lowercase, what are the chances that being drunk enough to do that wouldn't be too drunk to be able to spell every other word in that post properly? If it's true, obviously it's going to be significantly harder to do this unless someone actually is going to take over, but it obviously isn't impossible, and it's also obviously going to be easier with several people working on it. If it isn't true, which I think is the case, all we have to do is keep working and he'll pop in at some point.
Yeah those would be sums of successive squares, which would appear in the last pic on the second graph in the third column. I think my hypothesis about the n-gon of x-dimensions may be false but it is true for the 1,1 grid on the top left.
Hello lads! Working and thinking over here. I have a few good new ideas, just working on testing them first. Thinking on the (x+n) square and this:
>Then fill the square with n0^2 and multiples of 2d.
I'm thinking of n0^2 as little squares inside the bigger (x+n) square. Like doing a block puzzle with n0^2 pieces and 2d sized pieces. Or maybe 2d(n0-1) pieces?
VA - Still going around in circles with the iterative approach.
MM - we are iterating towards (1,c).
Revised c=259 pictures attached at n0=2, n0=3, n0=4 and n=6. The previous pictures didn't display the fact that both the (x+n)(x+n) and (d+n)(d+n) squares were changing sizes on each iteration. These pictures are more accurate. x+n square is outlined in red. d+n square is dark grey. I've also highlighted the "r" blocks, which is the question I'm looking to pose with this post.
We start with a triangle base u. Calculate an estimated small square est_XPN from 1 + 8*T(u) and maybe some remainders.
Next calculate the n0 = sqrt( c + est_XPN ) - d. This may or may not be the n we are looking for.
To determine this, we use the Get_Remainder_2dnm1 formula to calculate the difference between the estimated XPN, and the n,d,f formula. Simplified, this is essentially:
remainder = est_XPN - ((n0 * n0 - 1) + 2d(n0 - 1) + f)
If the remainder = 0, we have a match.
Otherwise, according to VQC:
>That gap will allow us to use more geometry of triangular numbers to establish what multiple of 2d could be added to make our eight triangles.
> That geometry with multiples of 2d will give us the solution, if it exists (which in our example we know it does).
Unless I am missing something, that gap is the remainder from Get_Remainder_2dnm1.
And in order to be useful, it would have to tell us more than simply how to fill the square we already know.
So what could it be?
Recently in the qresearch Q posted USE LOGIC. Could this mean we are supposed to utilize logical functions? Such as AND, OR, NAND, or XOR. I don't know if there are any proofs about these functions and the integers but we might get something
Not that it means dick all to me, but a little birdy mentioned something about:
2 steps, one of them being related to:
"Modulo log-1".
Kept coming up in a conversation where dude kept saying someone in Cicada told them how to crack RSA since they did it years ago.
Granted, i don't know if that's the most solid method or just what they'd worked out at the time of the conversation.
And I could be way off as usual since I have no I have no earthly I where to plug that in if it's even applicable.
Hello PMA! Your diagrams are great, and exactly what I was working on, increasing n0^2. Don't forget that f = 2d + 1 - e. Would that help remove the green squares? Or said in formula, e has already been removed in the formula for f…
((n0 * n0 - 1) + 2d(n0 - 1) + 2d +1 - e)
That would effectively take away the 3 green squares and make a match for the correct (x+n). Am I missing something here? Looks like you've successfully increased values of n0^2 and 2d(n-1) until finding a match.
Also, having n0^2 in the upper corner really helps observe the increase in n0^2.
>Looks like you've successfully increased values of n0^2 and 2d(n-1) until finding a match.
For small numbers, I have a process that works, but it essentially just increments u until it finds a match. When taking larger jumps of u, it's hit and miss. Trying to analyze a possible next step.
Went back to the grid (we've been away far too long!).
Attached pics are for e=0, e=2, e=4, and e=6. Each record now includes a value of u for odd x+n.
Notice how the u increments by n, and in some cases by both 1 and n.
Ok, I'm looking at the u values. This is the base of 1Tu, correct? Have you noticed that the other side of the 1Tu is always u+1? If you draw the straight lines from the center square, for (x+n)=15 you always get Tu with base of 7 and side of 8. How can we connect this with n0 base estimation?
Ok, so if we find u, we can solve (x+n). How does n0 work into this? For smaller examples we can iterate by 1. Then, once f is big enough to divide (f-2)/8 we can iterate by n0. Then when we’re dealing with RSA sized numbers we do (f-2)/40 and iterate by n0. What are the problems you’re currently having with the iteration process for larger numbers? It seems like we have a good working process for smaller numbers. The key is tying the verification calc to the formula ((n0 * n0 - 1) + 2d(n0 - 1) + 2d +1 - e).
I think this is what VQC was getting at. n0^2 and 2d(n-1) combine to form the (x+n) square. I could be stating the obvious, but this is news to me.
>((n0 * n0 - 1) + 2d(n0 - 1) + 2d +1 - e)
These formulas are all equivalent.
(x+n)(x+n) 1 + 8T(u) (nn-1) + 2d(n-1) + f
VQC refactored into:
8T(u) == nn +2d(n-1) + f - 2
T(u) == (nn +2d(n-1) + f - 2) / 8
And then reduced size by 1/5 as an estimated starting position. That's where the (f-2) / 40 came from.
What's missing in the iteration process, is a way to appropriately handle the (nn + 2d(n-1)) / 40 piece and it's mod. This is where inaccuracies come into play as the numbers get larger. We know d and are estimating n as n0. And for RSA size numbers, we fly right on bye extremely quickly with this approach.
So something else has to come into play.
PMA, you just made a triangle joke! And you made your first shitpost last week. Nice!
I miss you all, triangular anons!
Sorry - I became very busy with work.
I'm trying to catch up a bit, and still trying to understand the triangular number formulas - if anyone has any specific posts they recommend I read first, I'm all ears.
I've been thinking a bit about the relationship between triangles and squares, so here's a list of things I've been thinking about regarding that topic (not sure if my whitespace formatting is going to work as intended):
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We know that the series of all odd squares are described as 8 triangles + 1. (8T(u) + 1).
series of odd squares:
1, 9, 25, 49, 81, 121, 169, 225, 289, 361
minus 1, divide 8:
0, 1, 3, 6, 10, 15, 21, 28, 36, 45
Interesting to note the parity of the numbers here. Similar to the columns of e.
-
We know that the series of all even squares can be divided by 4.
series of even squares:
0, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400
divide by 4:
0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
So we have a way to break a square into either a smaller square or a triangular number. I've been thinking about how a triangular number can be expressed as a square, so my own 3 observations (although I think others have already expressed maybe some of these).
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Any triangular number can be expressed as a square minus the triangle of number -1.
T(u) = u^2 - T(u-1)
I'm not sure if I like this expression, because in algorithmic terms the movement of u is linear.
Conversely, a square can be expressed as the sum of 2 consecutive triangular numbers.
u^2 = T(u) + T(u-1)
-
An even triangle number can be expressed as the largest square that fits under it (u/2) plus 2 smaller triangles of size (u/2).
T(u) = (u/2)^2 + 2T(u/2)
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An odd triangle number can be expressed as the largest square that fits under it ((u+1)/2) plus 2 smaller triangles of size (u-1)/2.
T(u) = ((u+1)/2) + 2T((u-1)/2)
What I like about these last two is the u/2 in the recursive part of the algorithmic interpretation.
Ok, so another couple of known relationships:
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The series of differences of 2 consecutive squares is the series of odd numbers:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144
-
0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121
=1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23
-
The series of differences of 2 consecutive triangle numbers is the series of whole numbers:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91
-
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78
=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
-
The relationship between the two series above is 2n - 1.
So I'm not quite sure what to conclude of all this. I can imagine growth in several aspects of the relationship. I'm going to continue to get caught up. I hope this helps someone!
>Do you have a method yet to check each iteration?
Only thing we have at this point is a check on the result from Get_Remainder_2dnm1. And still not seeing how that helps us determine the "more geometry of triangular numbers" hint.
I was under the impression in the last thread that VQC dropped enough clues to solve the odd number RSA values. And he would stay away until that was accomplished. (Although it would be AWESOME if he came back soon!)
Going to once again go back and review previous posts to see if anything jumps out.
Working on improving the quadratics to help verify (x+n) iteration path. Check this out, found a major (but very basic) improvement for the formula. Staet with the difference of squares equation, and sub in (Tu XPN Est) for the small square. Then begin to solve the equation for n (isolate n). When taking SQRT of c and (Tu XPN Est), we need to include the remainders of both c and (Tu XPN). c = dd + e. For clarity, I'll call remainder of SQRT(Tu XPN) var g.
c = (d+n)^2 - (Tu XPN Est)^2
c + (Tu XPN Est)^2 = (d+n)^2
Floor SQRT(c) + e + Floor SQRT(Tu XPN Est) + g = d + n
dd + e + (Tu XPN Est) + g = d + n
dd + e + (Tu XPN Est) + g - d = n
We're looking for a match where n is a whole integer. Should be easy to spot in a list of calcs.
I was bored and had nothing useful to contribute to the odd squares thing so I decided to have a look at how triangle numbers relate to even squares purely out of interest. If we are meant to use triangle numbers to figure out (x+n) when it comes to even squares, it looks like it'll be confusing finding a relationship between (x+n) and f but there are already some obvious patterns.
4 0 4 0 4 0 4 0 4 0 4 0 4 0
Any triangle number plus the previous is the square of the longest side of the first triangle number
During his explanation for RSA 100, VQC shifted from using (f/40) to (f-2)/40 as the triangle base.
>Since f can be divided by 5, we can make a base for each triangle which is made of (f/40), still with 4 left over.
And then f-2 took over.
Tried both with and without mod handling, and the results with larger numbers were similar.
I am posting 2 examples for my latest iterative process for a matching record at c=259, and a failed result at c=14596111.
The current process uses (f-2) / 40 as the incremental u, does not included either the (f-2 mod 40) or (d mod 40) in the totals, although they are displayed in the grid.
The previous u is carried forward from the last calculation and is the starting position for the next f-2 chunk to test for est u. Also showing estimated values for T(u), 1+8T(u), n0, the remainder from 2d(n-1), and x+n values.
Finally, checking those results again to determine if a match and the difference to the solution x+n.
Reason for the double checking of the 2d(n-1) remainder, is that I am unsure where the mods need to be included in the calculations, and this gave a way to tinker with different combinations to see what was happening.
Each iteration is also checking +/- 2 u values for possible matches.
Not sure if the -20 x+n diff for c=14596111 is coincidental or related to the same value from (d mod 40).
Just looking to see if anyone has any other ideas to try out here.
Hey PMA.
I too am confused with the description of the last steps in the process.
I'm confused about the triangle stuff.
There are many cases when n-1 is larger than (x+n-1)/2, and therefore 1+8T(n-1) is larger than (x+n)^2.
Also, there many cases where (f-2)/8 doesn't lie between n-1 and (x+n-1)/2.
I'm very confused indeed.
I've been focused on understanding this post:
In particular how the multiples of 2d can be turned into a square, or parts of the triangles.
This last series of posts in RSA 10, were super messy, and contained many errors.
Very unlike Chris.
And now his twitter is gone.
He has deleted his Twitter before you know.
Teach - in looking at that >>4678 post, and VQC's previous question "Why is (n-1) important?".
Doesn't it seem that n-1 gives the width of one side of the nn-1 square?
With regards to the 2d(n-1) contribution to the triangle, I posted a few pictures in the previous thread that connected the (x+n)(x+n) square for the prime solution to the starting c small square.
This idea, and subsequent posts, was the closest I've come to understanding how the triangles work. The red blocks represent the 2d(n-1) contribution. When coupled with the f and (nn-1) pieces in the center prime small square, always form an outlining square or "capstone".
Those formulas work everywhere, but I wasn't able to figure out a way to arrive at that "capstone" square.
Same iteration problem that we're running into now with f-2.
I'm trying to understand now.
What about this image that VQC posted:
Does your solution explain this post? I can't make the connection.
Based on that image, I was assuming that the 2 halves of (f-2) would be a band of the triangle, or maybe a band of the square, as in the image attached.
I like what you've done with the nn-1 on the outside, but originally I was thinking that the nn-1 + f is basically a recursive version of the problem.
I went about the problem a bit differently than VQC explained, so I may have gone down a wrong path. Who knows?
The image VQC posted shows the relationships between small and big squares of a single record.
I combined the nn-1 + 2d(n-1) + f formulas from both the c and prime records into a single image. Think of them as one square layered on top of the other, sharing the f value and portions of the nn-1 and 2d(n-1) values.
To me this represented a nice growth perspective. Fractal even.
The positioning of the f value didn’t really matter.
What was really interesting was the appearance of the 2d(n-1) as a square every time. And then the ability to represent the inner nn-1 algebraically.
The relationship definitely exists. But don’t know how to connect the dots so to speak.
ENCOURAGEMEEEEEEEENT!!!!
>You're saying that 2d(n-1) is a square? In which cases?
Put together another test to try and explain a bit clearer what I'm seeing regarding the "capstones".
The tests are for odd x+n only.
The "capstone area", represents the red square in >>4758. I am arriving at that number by calculating a triangle base (difference between nn-1 in c and p records), and then subtracting that from c (x+n)(x+n). (nn-1 is the purple area.)
Therefore, the formula for the "capstone area" is 2d(n-1) from the c starting position, plus f, plus the nn-1 portion of the prime solution we are looking for.
This is always a square that can be defined using the 1+8T(u) formula, plus some remainder. Where the remainder is always divisible by 8. In order words, 1 + 8 triangles + an equal extra portion for each triangle.
Love the new images. The world is a confusing place.
Tetrahedral numbers.
Seeing as how I haven't checked in in awhile, hello everything. Special welcome back to Teach. So I've been working on some patterns, and have reduced my work into one, fairly underwhelming formula.
N = (D-A)^2+E / (2*A).
I'll be extra salty if this was stated somewhere else, other than the known "X^2+E = 2NA". I'm using the form D-A instead of X, because this is the fruit of working on the X+N square, and my observation that it seems like an overly complex way of iterating over A.
Here in this formula, if (2A) evenly divides (D-A)^2+E, A is valid, thus X, and you get N. Which isn't useful except, finding the intercept of these two equations, (where (D-A)^2+E < 2A) gives us the upper bound for A, (i.e, the lower bound for X). It may also be possible to push X into the negative (where B and A become swapped) and give us the upper bound for B.
There's some more patterns with this I'm exploring, and I could do a visualization, but what's happening here is fairly complex to show without coloring it manually.
So a visual of what I've been playing with, every single cell comes from the origin cells in N0. So N0, F0, F1, F4, F9, F16, etc etc. The square root of these original cell's F, is X. The color coding guide for this is green is A=1. Red is A=2. Magenta A=3. Orange A=4. Again, all these blocks share the same X, with the origin at F(X^2), N0.
So there are an infinite amount of these, and their spacing is what controls if a cell exists at a certain E,N or not. This is the key to unlocking T as well. What we understand as an X-chain, (where the cell's B becomes the next cell's A), is a deep relationship to where these objects overlap, and multiple X chains are where multiples of THAT pattern overlap.
Going deeper, this first picture is the X=1 object in Red. In the second picture, we add the X=3 object in Green. The third, the X=5 object in Blue. Lastly, the X=7 object in Yellow.
X=1 in Red, X = -1 in light blue, VS. X=12 in Red, X = -11 in light blue. This is part of my tangent that the original VQC code only generates a part of the grid, which obscures some patterns like this one.
Hello. Some math stuff has showed up in
>>>/truthlegion/
He says hes related to Q. Also has some stuff to say about Phi, Phi, Fibonacci, etc. Could be insane though. I believe him
Hey lads! Still working on tying the n0 iterations together with the quadratics to verify when we reach the correct n value. Just checking in.
>Again, take with a massive grain of salt. I've recently had some experiences that lead me to believe this type of stuff though.
It's good anon, we are open minded. Be comfy.
Regarding crop circles, yes there are "real" and "fake" ones. The 'real' are likely created using a space-based microwave beam technology, painted onto fields (think x-y galvanometer such as in confocal microscopes or stereolithography systems). We don't know the exact purpose of these demonstrations, nor who is creating them.
Could be a star-wars weapon from Reagan days, protecting EU countries, and the crop-circles are demonstrations of 'operational readiness' to serve the mission (therefore, "pay up for your protection, we're operational!").
Could be blackmail - "pay up or else!"
The geometries are fascinating though.
>Can't come off as too crazy
Too crazy? Our senpai claims insight into all millennial challenges, as the Lamb of God. You are fine.
Familiar with the Arecibo Message And the Chilbolton Reply. Carl Sagan was likely in an elite circle of scientists, likely a hidden group, just a theory.
His first wife, Lynn Margulis, is very interesting. She did some very novel work regarding bacteria and species merging. Think - how did mitochondria get into our cells? Is this 'alien' life within us?
Also some interesting views on 9/11:
https:/ /hooktube.com/watch?v=SS3fz3yfVbU
Pic attached for c=87 is an attempt to explain the "capstone area", and how it could be used as a basis for the iterative search.
It seems to me that there are 2 ways to iteratively find a solution. Both have been problematic.
The first is to start at a sufficiently small triangle base and iterate in chunks of (f-2)/40 towards the c (x+n)(x+n) square. Using the remainder 2d(n-1) = 0 along the way to determine if an n match has been found.
The second, that I pursued in the previous thread, is to iterate towards the "capstone square" which always lies somewhere between the prime and c small squares. The problem here was determining if a solution was found.
Both of these approaches share a problem of handling mods.
In >>5177, for odd (x+n) test cases, the "capstone square" can be defined as 1+8T(u) plus a remainder (if any) that is divisible by 8.
In the picture, the "capstone square" is the middle red square plus the additional remainder of 48. (6 allocated to each triangle). Note the symmetry. This "capstone square" + remainder includes the f, nn-1 and 2d(n-1) portions of the prime square we are looking for, plus the additional 2d(n-1) portion from our starting position.
The legend and math breakdown on the right hand side shows how the different squares are calculated from shared pieces.
With this approach, we are searching for the total area of the capstone. See the "Capstone Info" box.
A reasonable starting position for the search can be calculated as 2d(n-1) + f from our initial c record. The only piece we are missing from the "capstone square" is the nn-1 prime portion we are looking for.
The point of determining this "capstone square" is to calculate the "8 Triangle bases" from the c (x+n)(x+n) square - the dark purple area. Recall that VQC split his triangle into a base and capstone. >>4322
From there, the prime solution is trivial.
In order to take this further, we may need some way to incorporate the Get_n_from_odd_triangle_base and Get_Remainder_2dnm1 to relate the "capstone square" and remainder to our original c value.
What's also interesting about this approach is that outside the prime (x+n)(x+n) square, we're only trying to determine "how many lots of 2d would be needed to fill in the gap" that VQC mentioned.
I looked this up the other night. We are in the midst of first contact. Have been for a while. Powerful people are trying to control that influence. If they Do we are ruined. We have to wake up. Good luck everyone.
There is more geometry we can bring to bear to find our solution.
Attached is a partial picture for the same c=87 with labels indicating different u and remainder values.
On the right hand side is an explanation of how the differences between each triangle of the (x+n)(x+n) squares can be calculated as a trapezoid, following the basic formula Area=h(b1+b2)/2.
The first shows an example calculating the triangle base from the prime (x+n)(x+n) without any remainder.
The second shows an example calculating the triangle base using the capstone with a remainder.
Obviously both calculations rely on our ability to iterate to a correct u value (either for the prime or capstone squares). But perhaps this can assist along the way.
Hello PMA! Great work as usual, thanks man. I'm a bit stuck where you are on the iterative approach.
>It seems to me that there are 2 ways to iteratively find a solution. Both have been problematic.
Yup.
>The first is to start at a sufficiently small triangle base and iterate in chunks of (f-2)/40 towards the c (x+n)(x+n) square. Using the remainder 2d(n-1) = 0 along the way to determine if an n match has been found.
Yup.
>The second, that I pursued in the previous thread, is to iterate towards the "capstone square" which always lies somewhere between the prime and c small squares. The problem here was determining if a solution was found.
See pic attached, I got 0 at the correct (x+n) value. Also, very interesting to note that it was n0[99] - 4 = 7227 - 4 = 7223. What was the part about the mod being 4? Found these related crumbs:
Seems like the mod for each Tu is 4 in VQC's example? Thoughts, Anons? In my sheet I got n0[99] - 4 = 7223 = correct (x+n) value. However, mod for my example is 28.
I'm working through all of RSA#10 re-reading VQC's crumbs. Every time I read them I understand more.
VA - think we're getting close.
The key to the problem with iterating, I believe, lies in understanding how the remainder works.
In the "capstone square", there is a remainder of 48 (6 per triangle). This value is different from Get_Remainder_2dnm1. Each iteration (unless by chance we happen upon a perfect match), is going to be off by a remainder evenly divisibly by 8.
In >>5206 it is labeled as "capstone remainder".
We need a formula to calculate this.
So is this just a matter of running tons of different tests and analyzing results or is this a matter of thinking outside the box and just happening to come up with the correct ideas? I have been following along this whole time and I can tell how close we are, but I haven't been able to put a great deal of time towards doing math. VQC went through a full example and gave the answer, so we've obviously got all the tools we need. I just can't tell if it's a matter of taking a whole bunch of examples of each of these new functions with various a and b values and their corresponding (x+n) squares and analyzing the relationships as finely as possible to figure out the f/40 thing etc or if there's a fundamental rule behind these concepts that we haven't picked up on yet.
The tons of different tests I've already tried. Which is why I've switched to spreadsheets and diagrams.
The "capstone square" that I keep mentioning is a known point between the c and p records. We can calculate it directly. Which means it can tell us something about how the squares behave when they're not "perfect" - so to speak.
When there is a remainder, that remainder will fall in the u+1 portion just outside of the u square. So for the c=87 example, the 48 remainder falls in the 4*(2(u+1)+1) range. This is the perimeter of the square surrounding u. So if u=12, the surrounding perimeter is 108. 48 units of that is our remainder, and the rest goes towards balancing out the difference of squares formula.
The remainder is always divisible by 8, so there are a number of potential remainders within this range. What makes the 48 so special?
Attached are 2 spreadsheets to explore this idea for c=87 and c=259. The green highlighted row represents the "capstone square" that provides the nn-1 solution.
I'm trying to figure out what determines the "capstone square" remainder from these list of choices. Perhaps understanding this will lead to a more accurate iteration algorithm.
Can't really do much much math right now but I can be a Jesus-fag in the meantime (As you so eloquently put it)
I'll just put the link for that crazy DNA shit here
https://archive.fo/1z3rH
TRANGLES!
I think our key could be a certain number. I remember reading something along the lines of "the key is 5 and 5 your hands are a reminder". Now 55 is in a lot of stuff.
Triangular Numbers
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, [55], 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, …
Square Pyramidal
1, 5, 14, 30, [55], 91, 140, 204, 285, 385, 506, 650,…
Fibonacci
1,1,2,3,5,8,13,21,34,[55],89,144, …
Heptagonal Numbers [(5n^2-3n)/2]
1, 7, 18, 34, [55], 81, 112, 148, 189, …
Trianglar numbers are cool. The sum of any 2 consecutive ones is a square. Square pyramidals are also cool. The sum of squares up to n is a square pyramidal. Then fibonacci is the sum of previous terms. Heptagonal can be broken into the three thing in pic related. Also heptagonal numbers have to do with 7 sides, which is a holy number.
Another thing I recently learned is that if you take any 2 numbers, a and b, and do the fibonacci sequence with them, the sum of the first 10 numbers will be equal to the 7th number times 11.
term sum
1: a a
2: b a+b
3: a+b 2a+2b
4: a+2b 3a+4b
5: 2a+3b 5a+7b
6: 3a+5b 8a+12b
7: 5a+8b 13a+20b
8: 8a+13b 21a+33b
9: 13a+21b 34a+54b
10: 21a+34b 55a+88b
And it is obvious that (5a+8b)*11 = 55a+88b.
Maybe we can use this stuff to help us find de wey
More Ancient Mathematics. Magic square. My teacher said we would be going over "ancient indian math" next class (monday). Now I don't know what it is, but he said it had to do with the Pell Equation (x-ny^2=1) and he mentioned that Pell didn't even create it (maybe THEY wanted to hide its origin). Then, as a complete COINCIDENCE, I stumbled on these magic square concepts, which are related to the Pell Equation somehow (saw on wikipedia, ask me monday how they actually are).
You have an nxn square, and fill the square in such a way with the values 1,2,…,n^2 such that each row and column and diagonal have the same sum. The universe named these Magic squares so they might mean something. Each n, here are the column numbers:
n
3 45 15 (3*5)
4 136 34 (2*17)
5 325 65 (5*13)
6 666 111 (3*37)
7 1225 175 (557)
8 2080 260 (225*13)
9 3321 369 (13*13)
10 5050 500 (3341)
11 7381 671 (11*61)
12 10440 870 (235*29)
13 14365 1105 (51317)
14 19306 1379 (7*197)
A mathematician that worked on the Pell equations also came up with another thing, a seperate fibonacci type guy.
f(0) = f(1) = f(2) = 1
f(n) = f(n-1) + f(n-3)
[fib is f(n) = f(n-1) + f(n-2), so this is a small modification]
But then look at our numbers for this sequence:
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425,…
I don't want to get all nazi-ie because I actually think they were a Rothchild scheme, but this simple modification gives us 88, which in (>>5219) is related to the fibonacci sequence (55a+88b). 88 is also the 14th number in the sequence, so go at it nazi larpers.
Hello AA! I think there's a fundamental idea we haven't quite grasped yet. It will take calculations to verify, but we need the idea first. Remember this crumb?
>the fractal nature of integers. They are related into families
>Our current understanding of math and numbers is lacking
>New Math Kangz
Also, VQC hinted that this iteration process could be done (shortcut?) by using the grid. PMA has been working on tying together the (1,c) and (prime) records using a capstone calculation.
Hello PMA! Would you please give a quick explanation to us of how and why the capstone idea works? Does it connect (1,c) and (prime) records in the grid? We know we're looking for a method to verify the correct answer.
Hello Baker! Love these verses, ideas, and images. How about this verse for a Sunday?
“If you abide in my word, you are truly my disciples, and you will know the truth, and the truth will set you free.”
That's some cool ass triangle math Topol! Inspiration to step our thinking up a few levels.
Whoa! Great images and math. More mind expanding ideas!! Thanks Anon!
>Nine Nine Nine!!!
I seriously LOLd
Hello CA! Magic Squares? That is some cool shit. I'll study this in detail later today.
Lads, "The Kingdom Of Heaven Is Within Us" so let's raise our thoughts and emotions up a few quantum levels and visualize our success with happiness, joy, and excitement. We don't have to know the answer yet, we need to know that we WILL find it.
Also I read something from book 2 of Corpus Hermeticum (early christianity/the actual religion. Is the same as other ancient religions) it says:
"21. And he learning diligently and understanding their Essence, and partaking their nature, resolved to pierce and break through the Circumference of the Circles, and to understand the Power of him that sits upon the Fire."
Which has to do with pi. Also there are also mentions of the 7 holy circles in book 4. There are 7 notes in music. 7 Colors (ROYGBIV). There are 7 chakras. Someone in an earlier thread said 7 is important because it is 111 in binary (3 horns in book of revelations).
Whoa.
>to understand the Power of him that sits upon the Fire.
We are definitely searching for the underlying numerical order of the universe, which would be related to pi, music, color, sound, DNA, everything, created by the Power of him who sits on the fire.
Your idea about 7 being 111 in binary brought this crumb to mind that I've attached. VQC said that the connection gap between (1,c) and (prime) records in the grid is easier to see in Binary. Have we explored this yet, Anons? Maybe the missing link is easier to see when look at the gap in binary. (e,1) is supposed to be the key to the whole grid?
>Corpus Hermeticum (early christianity/the actual religion. Is the same as other ancient religions)
Are you saying the Corpus Hermeticum is from early Christianity? Hermeticism is its own thing. Maybe you just worded that confusingly.
Maybe a particular power of 2 comes up throughout all of the primes or something. That seems a little too obvious but at this point it could be anything.
Oh I thought that was Christianity my bad. I remember learning about it as an issue in the Catholic church with people trying to revert to hermeticism, so I just assumed it was a previous iteration of Christianity so to say. Thanks for the clarification
Catholicism and papacy are a stain on all of Christianity. It's sad
It's totally different. They don't worship YHVH, for a start. If you're curious about what Hermeticism is, read this. A lot of my beliefs are based around it.
I think I understand how to fill in the missing pieces in the iteration process that were giving inconsistent results in previous tests.
This is a work in progress related to >>5215 for c=87, and a continuation of trying to understand the "capstone square" remainder in >>5204 and >>5206.
Pic attached for c=87 shows a simple iteration of u. For each u, I am then iterating a remainder factor from 0 to u. (see the remainder offset column (r off)).
The formula for each estimated (x+n)(x+n) is: 1+8T(u) + 8*(r off).
For each u, there are 0 to u remainders possible.
This formula produces a difference of 8 between each iteration. And hits both the primary, capstone, and starting c squares properly. Something I was previously unable to do just by manipulating the u.
Next step is to add back in the (f-2)/40 for bigger jumps.
What if the integers themselves are a type of tagged pointer? https://en.wikipedia.org/wiki/Tagged_pointer
On 64 bit architectures, pointers to word-aligned objects have the lower 3 bits clear. What if an implementation detail of our universe is integers are tagged pointers that can be partially decoded to 3 bit detail before being dereferenced?
Jesus Christ some of the ideas spawned from these threads are truly heretical…
I've thought something like this. Like for an NP problem, say backpack, it would point to a record, then you read in the next possible item for the backpack and it points you somewhere else, then , the next and the next and finally your last pointer is pointing at the solution, so no real calculations were done, the optimization is implicit in the directing.
Great work PMA! So it is a recursive adding back in of remainders? I studied your output closely and understand. This example is the prime, but it also works for capstone and starting c as well?
So to clarify, we iterate by n0, but check each iteration against all possible remainders?
I never thought remainders would be so important again! Who would have known, lol :)
Read the article, and I think that's what VQC is hinting at! Integers are grouped into families somehow.
Seconded. PMA is our MVP for sure!
Nomnomnomnomnom
Alright guys, I've taken the iteration a bit further and am now including the (f-2)/40 for bigger jumps.
Please bear with me as I try to explain and forgive the small pics attached.
There are 2 variables that control the iteration process. The first, fm2_factor, is the number of (f-2)/40 chunks to include in the triangle base (u) (see the "#" column), and the second, rm_factor, is the number of chunks of (f-2)/40 to include in the remainder calculation (see the "r off" column).
The estimated (x+n)(x+n) square (est_XPN), is calculated as follows:
est_XPN = 1 + 8 * est_tu + ((f-2)%40) + (rm total)
where est_tu = T(fm2_factor * (f-2)/40)
and rm_total = 8 * (rm_factor * (f-2)/40) + ((f-2)%40)
The rm_total is only included if the rm_factor != 0.
Conceptually, we are creating a square from triangles and then adding chunks of (f-2)/40 around the square, checking for a match at each step.
In my previous example for c=87, the area around the square to check was simply u+1 and fit within the 4*(2(u+1)+1) perimeter. "For each u, there are 0 to u remainders possible."
For larger gaps in u, however, we need to expand the possible area for the remainder search to include up to the next possible u: (fm2_factor + 1) * (f-2)/40). See column "rm u" in the examples. Not accounting for this gap is what lead to inconsistent results in previous attempts.
The area that could possibly contain the remainder can be defined as follows:
remainder_max = 8 * (T(rm_u) - T(est_u))
This maximum remainder is essentially the "triangle base", and is necessary to know when to switch from iterating through a remainder to the next potential u. The "rm next" column is a calculation for the next total remainder.
rm_next = 8 * ( rm_factor + 1 ) * _fm2_chunk
When the next remainder is greater than the max possible remainder, the rm_factor is reset, and the fm2_factor is increased.
The pictures attached are various test cases from small to medium sized numbers, and are an attempt to illustrate that the process works, and that there are a couple of pitfalls that still need to be addressed. Number of iterations and elapsed time for each test are included.
Tests for 363, 6107, 7463 just show the process working for small numbers.
For 208975787, 208997671, and 209194627 note the additional iterations and processing time required for small values of f.
For 9872452111 and 9872801899, the starting fm2_factor of 0 enables searching of smaller values of n.
For 9874400051, the rm 2d(n-1) didn't equal zero, but the n0 value does equal the prime result. The concern here is that a minor adjustment to this algorithm needs to be made to tackle larger numbers and/or an additional validation method needs to be written.
This may or may not work for RSA sized numbers. The number of iterations required is still too large to effectively test. Most likely, we need another breakthrough with regards to the geometry of numbers to make this more efficient.
[MoT-{i-(n-no)-vation}]!
Stated another way: you are making (f-2)/40 jumps but why that small? For what specific, articulatable reason? Why are we doing 5 * 8 rather than 6543277645 * 8 to really speed things up?
Note that at the end of the day I have no fucking clue what I’m talking about, I’m just trying to help by pointing out what I don’t understand about your generally excellent posts and program output.
Here's the whole grid in binary:
https://files.catbox.moe/cm95wo.csv
And here's the grid when a and b are primes in binary:
https://files.catbox.moe/1ihylr.csv
I have no idea if malicious code can be injected into a csv file so I don't know how paranoid you should all be about downloading these from a stranger on the internet. At least they aren't pdfs. I don't really see any patterns myself but I haven't looked over them too deeply. I just put them together a couple minutes ago since nobody else had yet. It might be more useful if a and b were base 10 and the other variables were base 2.
Further from >>5243
Here's the grid with e, n, d and x in base 2 and a and b in base 10
https://files.catbox.moe/dzbpoe.csv
And here's the same but with a and b as the x and y axes instead of e and n (maybe it'll help us see patterns in cells with the same a or b values if there are any)
https://files.catbox.moe/qasjuc.csv
I had gathered that (f-2)%40 was specific to the example VQC used because (f-2) from that RSA number could be divided by 5 and there are 8 triangles. I don't have any useful suggestions but it seems weird to use it universally.
>Non-Euclidean Riemann Jackoffery
Kek
Super duper busy, and it's bumming me out, but I thought I'd come procrastinate here a little.
Alright guys I may have a new strategy. VQC said it only took a couple steps for a c=145 right? So I am making an exhaustive search for a certain depth of steps using functions we know. I have generated a ton of pathways doing this
def AB(A,B):
C = A * B
D = int(math.sqrt(C))
E = C - D * D
X = D - A
N = int(((X * X)+E)/(2 * A))
return (int(E),int(N),int(D),int(X),int(A),int(B))
def EDX(E,D,X):
C = D*D + E
A = D - X
B = int(C/A)
N = int((XX+E)/(2A))
return (E,N,D,X,A,B)
def F(rec):
newE = rec[e] - ((2*rec[d])+1)
newN = rec[n] - 1
newD = rec[d] + 1
newX = rec[x] + 1
return (newE, newN, newD, newX, rec[a], rec[b])
def FI(rec): #F inverse
N = rec[n] + 1
D = rec[d] - 1
X = rec[x] - 1
E = rec[e] + (2*D) + 1
return (E,N,D,X, rec[a], rec[b])
def C(C):
return AB(1,C)
def nextT(rec):
A = rec[b]
B = 2*rec[b] - rec[a] + 4
X = rec[x] + 2
D = rec[b] + rec[x] + 2
return (rec[e], rec[n], D, X, A, B)
def search(c, depth):
for i in range(2,c):
if(c%i==0):
if(i<c/i):
goal = AB(i,int(c/i))
else:
goal = AB(int(c/i),i)
start = AB(1,c)
path = "S/"
inSearch(start, goal, depth, path)
def checkRec(a,b, path):
for i in range(len(a)):
if(a[i]!=b[i]):
return
print("True", path)
def l(i):
if(i==0):return "e"
if(i==1):return "n"
if(i==2):return "d"
if(i==3):return "x"
if(i==4):return "a"
if(i==5):return "b"
return "?"+str(i)
def inSearch(rec,goal, count, path):
if(rec==None): return
checkRec(rec,goal, path)
if(count==0): return
inSearch(F(rec), goal, count-1, path + "F/")
inSearch(FI(rec), goal, count-1, path + "Fi/")
inSearch(nextT(rec), goal, count-1, path + "+t/")
for i in range(len(rec)):
if(rec[i]<0):continue
inSearch(C(rec[i]), goal, count-1, path + "C("+l(i)+")/")
for i in range(len(rec)):
for j in range(len(rec)):
if(rec[i]*rec[j]<0):continue
if(rec[i]0 or rec[j]0):continue
inSearch(AB(rec[i],rec[j]), goal, count-1, path + "AB("+l(i)+","+l(j)+")/")
for i in range(len(rec)):
for j in range(len(rec)):
for k in range(len(rec)):
if(rec[j]-rec[k]==0):continue
inSearch(EDX(rec[i],rec[j],rec[k]), goal, count-1, path+"EDX("+l(i)+","+l(j)+","+l(k)+")/")
Here is all the python code.
Here is some output:
>>> search(145,2)
True S/C(n)/EDX(a,e,d)/
True S/AB(e,n)/EDX(a,e,d)/
True S/AB(n,e)/EDX(b,e,d)/
True S/AB(n,a)/EDX(b,e,d)/
True S/AB(a,n)/EDX(a,e,d)/
>>> search(145,3)
True S/+t/C(n)/EDX(a,e,d)/
True S/+t/AB(e,n)/EDX(a,e,d)/
True S/+t/AB(n,e)/EDX(b,e,d)/
True S/C(n)/C(b)/EDX(a,e,d)/
True S/C(n)/AB(a,b)/EDX(a,e,d)/
True S/C(n)/AB(b,a)/EDX(b,e,d)/
True S/C(n)/EDX(e,n,d)/EDX(n,e,x)/
True S/C(n)/EDX(e,d,x)/EDX(a,e,d)/
True S/C(n)/EDX(e,d,a)/EDX(n,e,d)/
True S/C(n)/EDX(e,d,a)/EDX(x,e,d)/
True S/C(n)/EDX(e,a,d)/EDX(d,e,x)/
True S/C(n)/EDX(n,d,a)/EDX(x,b,d)/
True S/C(n)/EDX(d,e,a)/EDX(x,d,e)/
True S/C(n)/EDX(d,a,e)/EDX(d,x,e)/
True S/C(n)/EDX(d,b,e)/EDX(n,x,e)/
output for depth of 3.
I am going to make a thing that generates these strings and matches them up against other correct pathways for other coprimes and I'm going to see if we can brute force this thing.
>The concern here is that a minor adjustment to this algorithm needs to be made
Think I've found the minor adjustment.
Revised tests for 211073351 and 9874400051 attached. The change is that the remainder 2d(n-1) needs to be subtracted from the estimated XPN.
XPN = est XPN - rm 2d(n-1)
The XPN column is the final value to check. Notice now that the "x+n rm" column is zero. We have a perfect square.
This works for EVERY one of my test cases.
I have tried many different denominators in multiples of 8. Understanding how to handle the remainders is the key. Not how the f-2 chunks are divided.
In order to make this a very efficient solution, we need a way to sub-divide the remainder area to determine if the solution could fall within that area, or simply skip to the next one. Perhaps this is where the "more geometry of triangular numbers" comes into play.
Here are some paths that worked for this 3 depth.
So for the products of the primes [3,5,7,11,13,17,19] I did a depth search where I went at most 3 steps deep to see if I could solve for the correct record. Each step deeper could either be F, F inverse, next T (didn't do previous t), or AB(1,y), AB(x,y), EDX(x,y,z), where x,y,z could be any 3 entries in (e,n,d,x,a,b).
S/EDX(e,a,d)/EDX(x,e,a)/EDX(d,e,n)/ 7 27
S/EDX(n,e,d)/EDX(x,d,n)/EDX(d,e,n)/ 6 27
S/EDX(d,b,e)/EDX(b,x,e)/AB(a,n)/ 6 27
S/EDX(x,e,d)/AB(a,b)/EDX(d,x,e)/ 9 27
S/EDX(x,d,e)/AB(a,b)/EDX(x,d,e)/ 6 27
S/EDX(d,e,n)/EDX(d,e,n)/EDX(e,d,b)/ 6 27
Numbers at the end are amount of times this path worked, then the total times tried. Someone want to take a crack at analyzing these? It doesn't work for every one, but maybe one step (or a couple) need to be iterated.
Here is output:
https://pastebin.com/M17sJM8A
It works a good amount of times. Next I'll break down each of these individual steps and do an individual search just with those. The depth things took wicked long but I might run it in the background for depth=4 for more primes while I write a paper.
I broke it down into 13 moves
"EDX(e,a,d)/", "EDX(x,e,a)/", "EDX(d,e,n)/", "EDX(n,e,d)/", "EDX(x,d,n)/", "EDX(d,b,e)/", "EDX(b,d,e)/", "AB(a,n)/", "EDX(x,e,d)/", "AB(a,b)/", "EDX(d,x,e)/", "EDX(x,d,e)/", "EDX(e,d,b)/"
Then went for all the products of these primes
[3, 5, 7, 11, 13, 17, 19, 23, 29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
this is a total of 184 coprimes. I found 3 patterns that work 41/184 times:
41 184 C(c)/EDX(x,e,d)/AB(a,b)/EDX(d,x,e)/
41 184 C(c)/EDX(d,x,e)/EDX(d,x,e)/AB(a,b)/EDX(d,x,e)/
Also I found some other notable ones:
23 184 C(c)/EDX(x,d,e)/AB(a,b)/EDX(x,e,a)/EDX(e,a,d)/
26 184 C(c)/EDX(x,e,d)/AB(a,b)/EDX(x,e,d)/EDX(x,e,d)/
28 184 C(c)/EDX(x,d,e)/AB(a,b)/EDX(x,d,e)/
Some of these may overlap or be redundant. But not all the AB(a,b) are worthless, because sometimes you can have same a,b with different other values.
Hello PMA! Studied your output closely. So the (x+n) and u values are the same for both (1,c) and (prime). This the essence of your capstone idea, correct?
Also, I understand and follow your idea about dealing with the remainders, especially the large remainders in the (1,c) calcs.
For this improved method, we are now essentially running an iterative search that also accounts for the remainder at each iteration and searches for a value where rm (x+n) = 0 and rm 2d(n-1) = 0. Have I understood your idea correctly?
VA - I'll try to summarize my recent posts and the current iterative approach using f-2 chunks of (f-2)/40.
1) Starting from a 0 factor, create an initial square as 1+8T(u) + remainder. Where u is a multiple of chunks and the remainder is the mod.
2) Calculate the next u (I'm calling this rm u) as u + (f-2 chunk).
3) Within the area between u and rm u, a solution may be found. Iterate through this section by adding 8 times incremental f-2 chunks to the initial square, plus another remainder.
4) At each iteration, calculate the n0, and remainder for 2d(n-1). Subtract this remainder from the estimated XPN to get a final XPN.
5) This final XPN is a perfect square with no remainder. This is the x+n rm column and is only a sanity check. Test the final XPN against the rm 2d(n-1) to see if we have a match.
The recent work on the "capstone square", lead to an understanding of how the remainders are always evenly divisible by 8. Which in turn lead to a more accurate iterative approach.
The slow part of this process is checking within the u and rm u area.
>still looking for a way to verify when we arrive at the correct solution
No. This part is done.
I'll post a bit later some thoughts on how to tackle speeding up the remainder search.
In trying to understand a bit more about the remainders, I stumbled upon something that seems too coincidental not to share.
This has not yet been verified against other test cases.
The pics attached are for the last iteration that solves the prime for c=6107. A full output of the execution can be found at pastebin.com/EnNfXudK if anyone is interested. These pictures show only 1 of the 8 triangles that will eventually make up the final (x+n)(x+n) square.
The first picture (c6107_triangle_base_triangles) shows how the triangle base can be represented as the sum of two triangles.
The triangle base is T(42) - T(39) = 123, which is equal to T(12) + T(9).
And T(12) = 78 just happens to be the remainder portion that solves our prime problem. (!!!!!) The math breakdown is on the right hand side.
The second picture (c6107_triangle_iteration_mapping) shows how the T(39) and T(12) triangles are mapped into the T(41) solution triangle.
Pay particular attention to the 3 dark green squares on the bottom right side of the triangle. These 3 remainder squares represent a portion of the original (f-2) mod 40.
VQC hinted at this in the previous thread:
>then the number of 2d to add must have the same left over on the last d to create the final base of each triangle. THIS is key.
In this example, (f-2) mod 40 = 12. We add 12 to the initial 1+8T(u) triangle, and then again in the remainder iterations. So the total mod to be added is 24.
In my picture, 3 dark green squares, 8 triangles. 24.
If this breakdown of the triangle base into smaller triangles holds for other test cases, then we have a way to speed up the iteration process.
I've been briefly looking for some patterns in the binary versions of various things since nobody else is. Maybe if we find that particular patterns in powers of 2 come up that we can find a way to go from iteration to calculation. VQC has hinted as such, after all, however vaguely.
Here are the first 100 triangle numbers in base 2:
0000000000001
0000000000011
0000000000110
0000000001010
0000000001111
0000000010101
0000000011100
0000000100100
0000000101101
0000000110111
0000001000010
0000001001110
0000001011011
0000001101001
0000001111000
0000010001000
0000010011001
0000010101011
0000010111110
0000011010010
0000011100111
0000011111101
0000100010100
0000100101100
0000101000101
0000101011111
0000101111010
0000110010110
0000110110011
0000111010001
0000111110000
0001000010000
0001000110001
0001001010011
0001001110110
0001010011010
0001010111111
0001011100101
0001100001100
0001100110100
0001101011101
0001110000111
0001110110010
0001111011110
0010000001011
0010000111001
0010001101000
0010010011000
0010011001001
0010011111011
0010100101110
0010101100010
0010110010111
0010111001101
0011000000100
0011000111100
0011001110101
0011010101111
0011011101010
0011100100110
0011101100011
0011110100001
0011111100000
0100000100000
0100001100001
0100010100011
0100011100110
0100100101010
0100101101111
0100110110101
0100111111100
0101001000100
0101010001101
0101011010111
0101100100010
0101101101110
0101110111011
0110000001001
0110001011000
0110010101000
0110011111001
0110101001011
0110110011110
0110111110010
0111001000111
0111010011101
0111011110100
0111101001100
0111110100101
0111111111111
1000001011010
1000010110110
1000100010011
1000101110001
1000111010000
1001000110000
1001010010001
1001011110011
1001101010110
1001110111010
There are a few complicated visual patterns but there are some simple things I noticed that I'll point out first. Whether these patterns have any relevance, I don't know. The point is to find patterns and then figure out if they're relevant. What I thought was that maybe some of these patterns also occur in prime numbers or something, as in maybe they're made up of similar patterns of powers of 2. Either way, some of them definitely don't seem like visual coincidence.
For anyone who doesn't know much about how binary works, from right to left, it you want to convert it to decimal, you add each column based on it being 2 to the power of however many digits in you are minus 1. So, if there's a 1 in the fourth column across and the second column (i.e. 1010), the rightmost 0 represents 2^0 (but it's a 0 so you don't add it), the next one represents 2^1 (so add 2), the next is 2^2 (but it's 0), and the next is 2^3 (so add 8). That means 1010 in binary is 10 in decimal.
So, patterns. Firstly, in the 2^0 column from top to bottom (far right, 2^0 = 1 to translate to decimal), the pattern is 1100 repeated. In the 2^1 column (next one to the left, 2^1 = 2), it's 01111000 repeated. From then on the patterns get more complicated (so in the 2^2 column, it's 0010111111010000). It seems like the patterns double in size each time, but I don't know how patterns from 2^2 onwards come about. If there is some way of figuring that out, there's a way of calculating which powers of 2 are in which triangle numbers.
Secondly, you can see how many triangle numbers there are with a particular highest power of 2 (based on the leftmost 1), and this has a pattern too. It seems like a weird pattern so far.
1 1 1 2 3 5 7 9 13 19 27
That's how many triangle numbers there are with each ascending power of 2 as its highest. So there are 27 triangle numbers with 2^10 as their highest power of 2. I don't know about the earlier numbers, and I've only looked at this for the first 100 triangle numbers so far, but from 2^6 onwards, it's +2, +4, +6 and +8. So maybe it keeps going like that. That might means there are 37 triangle numbers with 2^11 as their highest power of 2.
Thirdly, there are some things I've noticed visually so I'll have to draw them in Paint or something.
So here's a picture of the visual thing I mentioned. Now, where the triangles are light blue, I felt like I might have been seeing triangles because I wanted to see triangles. I can admit that. There definitely are triangular shapes, but they aren't perfectly triangular. That said, the green ones are definitely a pattern. In the earlier powers of 2 downward through the triangle numbers, there are triangles of varying size within the zeroes. There is almost definitely a pattern as to the size of the triangle. As I said, I have no idea what if anything we're meant to do with this information, nor how we'd extrapolate to 8Tu + 1. I'm just trying to follow leads that nobody seems to have followed yet. I think it would be useful to do the same thing with values from endxab and see if any patterns arise that are the same in more than one variable.
This is fucking awesome.
A couple stupid questions from your pastebin output: what happens if you keep going past the solution? Does x+n diff become negative? Just looking for stably-sorted things to binary search. If x+n diff did become negative and we could jump around, it would be an attractive binary search target.
Also, I don't really understand u but it's weird in your pastebin output that final u is x+n diff/2.
Weird, you're right. Not sure if it means anything but interesting. I took your list and re-did it with only LSB (least significant bits) and I'm thinking mirror rather than triangle… Triangle often has an odd element at the center, but these don't. Also, I boxed in with red zero values that I felt to be unexpected.
They're two of the same triangle number mirrored, yeah. I meant triangle in terms of the shape I guess. So at least based on your image, it's going 1, 3, 1, 6, 1, 3, 1, 10, 1, 3, 1, 6, 1, 3, 1, 15, 1, 3, 1, 6, 1, 3, 1, 10. The peaks of the triangles are always predictable since the 2^0 pattern is 1100 over and over and wherever there are zeroes in the 2^0 column there are mirrored triangles, so the peak of the mirrored triangle number shapes is always on the 3rd and 4th triangle numbers. That means the 115th and 116th triangle numbers should make a mirrored triangle of 21 out of the zeroes. I really can't tell if I'm taking this too seriously but he did say that patterns would be more obvious in binary, and this is a pretty obvious fucking pattern.
My headcanon wants these triangles and pyramids to turn into conicals.
I don't mean to be a fag, but there are no straight lines on curved earth.
BOOM. PMA you are a fabulous faggot (!!!!)
>And T(12) = 78 just happens to be the remainder portion that solves our prime problem. (!!!!!) The math breakdown is on the right hand side.
So you can verify solution because the remainder is accounted for? if so, then we may have this ==>>
So then we solve for n:
SQRT( c + (x+n)^2) - d = n
SQRT(6107 + (83^2)) - 78 = 36 (prime n)
BAKER=
RSA #10 —- http://archive.fo/anGD7
I've got all VQC maps and archives saved offline. Can upload to anonfile if anyone wants to save offline as well.
>Does x+n diff become negative
The x+n diff column is a sanity test against the known prime solution x+n value. We won't know this value when searching through unsolved RSA numbers. I just use it as an indicator.
u is a triangle base. For odd x+n values, it is defined as u = (x+n-1)/2. And the area for any triangle base is T(u) = u*(u+1)/2.
I tried converting the linear search into something that will handle T(u) values.
See pastebin.com/qEgSaq4r for the the full iterative output for c=6107.
The main difference here is that instead of computing the remainder as a multiple of (f-2)/40, I am using the multiple of (f-2)/40 as the triangle base to the T(u) formula. This has the effect of both dramatically improving performance, and at the same time creating inconsistent results for different test cases. For the c=6107 example, it has reduced iterations from 287 to 53. For other test cases, the results are hit and miss.
Recall that the iterative solution presented previously leads a prime solution for ANY odd x+n (given enough processing time). And so we need a faster way to iterate.
VQC mentioned previously in a DM that the factor tree was one way to understand the recursive nature of solutions.
To that end, I think we need a different approach to solving this problem. And I think the factor tree we worked on previously now needs to come into play.
See the attach pic for a "triangle" analysis of c=6107 and c=7463.
The c=6107 example is just another breakdown of the previously presented final iteration.
For c=7463, however, the last iteration doesn't fit as cleanly into 2 triangles. I am attempting to show in this spreadsheet how we could conceptually recurse into the correct x+n value using the factor tree. Notice that there are 5 separate triangles, plus the 2 mod values that when added together equal the prime solution.
The solution for c=7463 was produced by trial and error just to illustrate an idea.
Also recall that in the previous thread CA found an interesting connection between the x+n and d values as numbers scaled higher.
I'm not quite sure yet how to incorporate these different ideas into a unified solution. But certainly something to consider further.
>>5273
Question (3 points):
This anon has no posting history on this board. Why would they pose a question we've already discussed in great detail?
Answers:
a) Because they didn't read the thread.
b) For demoralization purposes because they're that one person I keep having to ban who just keeps coming here and spamming the threads telling us to give up.
If the answer is (a), I don't intend to be patronizing, but we've already discussed this and everyone seems to have come to the conclusion that it doesn't matter and that we should keep working on it. It would be a bit of a waste of time to repeat ourselves so read the thread and post something a little more constructive. If the answer is (b), fuck off.
Very cool indeed. I don't think they are triangles though. I did the same output and got what seems to be normal distributions. Also on this picture to the left of the normal distribution seems to be another symetrical distribution which is flipped over the 1's and 0's. This happens repeatedly through the output.
>What about VQC?
>Has anyone heard from that side of the pond lately?
>EU in progress?
>I'm here in almost every thread, every day, just no namefagging.
>I've got to put my dog down this evening. He's got cancer and it's sad.
>Good to be amongst fine anons.
>Much loves.
What I meant when I said they were triangles was that they were reflected triangle numbers, not that they were visual triangles. Although, that non-linear triangle you pointed out (the red one) makes me think that maybe the whole thing really is the triangle, if this does actually mean something. Partially because once it isn't linear (one less 0 the lower the column) the number of 0s until the next column is 1, 1, 1, 1, 2, 2, 3, 5. Aside from the first one, that's the same sequence as here >>5261 in terms of rows where the highest power of 2 is. 1, 1, 1, 2, 2, 3, 5. So I'm just blindly extrapolating here but I would think bigger triangles might follow that pattern and that the next would be 7, then 9, then 13 and so on. That's if it isn't the other explanations I was thinking, which is that it is reflected triangle numbers and that the fact that its a big triangle means it overlaps with other 0 patterns that we haven't figured out yet.
If that wasn't a larper and you're reading this VQC, I had to have my dog put down about a month ago for similar reasons. You aren't alone. It would also be good to have confirmation with trip that you didn't an hero, but obviously that's not a priority when you're going through whatever you're going through, so whenever you're ready.
I may have found an alternate way to iterate. Not sure how effective it will be, but it's interesting nonetheless.
Attached pics are for c=209325931.
Using the current linear approach of incrementing f-2 div 40 chunks, it takes 51699 iterations to solve. The solution is found in the 12th increment of the T(est u) triangle, and at the 4886th iteration in remainder portion T(rm u).
We can calculate the "triangle base" for the remainder portion by T(rm u) - T(est u).
Here's the interesting part.
If you divide this "triangle base" by (f-2 div 40), you get a u that is exactly half way between the rm u and est u values. And the "triangle base" mod (f-2 div 40) is 1/2 the starting (f-2 div 40).
If you repeat that calculation, you basically find a way to subdivide the triangle base between T(rm u) and T(est u) by valid chunks of (f-2 div 40), but using triangle values instead of increments.
For example, at iteration 12:
(f-2) div 40 = 650
(T(8450) - T(7800)) / 650 = 8125 with remainder 325
The next iterations:
(T(8125) - T(7800)) / 325 = 7963
(T(7963) - T(7800)) / 162 = 7930
and so on.
The rm_u_tree pic attached is the partial result of a parse tree that starts from the mid point between est u and rm u, and splits u values in both the positive and negative direction. So the triangle base is divided into 1/2, 1/4, 1/8, 1/16, 1/32, etc in a tree structure until the leaf nodes are just 1 u apart.
There are only 255 entries for this parse tree, so we're down to 3060 total calculations to solve. But most likely a few less than that. Notice from the picture that T(8196) is only 5 levels down in this tree, and is only 1 away from the prime solution of 8197.
It has been many years that we have seen these kind of numbers.
id: c4ad45
45
Hex #c4ad45
Gold
Coincidence?
I think not.
I don't know if it's just that information has a way of clustering into self-similar patterns or it may very well be that our POTUS has just addressed this board. What a trip (pun).
Now, what to do.
I'm an abstract thinker.
Keep working on market(s)/comms?
Numbers?
What a time to be alive.
MAGA
Hey guys, long time lurker. I've really admired the effort you guys are putting into this and have wanted to help. But I've been hesitant because I don't know a lick of coding and didn't want to drag you guys down. But as I see the world changing and changing I really want to give this a try.
So I guess I'll ask this dumb question but how do I start and how do I catch up to what you guys are working on now?
First Chris gave us the equations and the code to generate the VQC, an example of which you see here.
Then he gave us many many special tips at using the VQC to change numbers around.
Then he gave us the tree.
Then he gave us the triangle number connection.
Read the maps, then skim the threads for what jumps out at you.
I say skim because each one is 750 posts.. so don't feel like you have to read it all.
If you want to learn some basic coding and create the grid for yourself, see what you think of this thread >>4379 here. If you're wondering what the grid is and what it's used for, you might like to read the pdf here >>5290 if you aren't paranoid about pdfs (I wrote it so if you'd prefer a text dump I or someone else could do that), read this guy's (the baker's) other post above it too, and maybe read through the images in the OP of each of these threads if you don't want to have to read 11 entire threads. Don't worry about being a burden or anything. At the moment it seems like PMA is the only one who has any idea where to take the mathematic side of this (I mean, just personally, I understand most of the current stuff but presently have no idea where to take it next or what other things to test). Don't worry about needing a name, either. Most of the time these names come from necessity.
I should mention: the thing about f in the pdf is wrong. I wrote that based on an earlier version of the Java code that had some errors in it. I haven't actually edited (or looked at) that pdf since probably November or December. You might not understand this until you read to that section but f is kind of like a negative e value. It isn't fbaxdn (that's what the error was); it's still endxab but e = f and n = n-1. It's used to make the side of the grid for a negative e value. e is the difference between c and the highest square number below it. f (from my understanding) is the difference between c and the lowest square above it.
>c4ad45
Resident Artfag chiming in.
HexCOLOR
And check this shit… it's the same one they use for OnePunchMan in the Reddit style sheet.
Now THAT'S when you say "What a coincidence! Right? We still believe in those?"
https://www.reddit.com/r/OnePunchMan/wiki/config/stylesheet
>Coincidence? I think not.
Very cool!
Welcome, Anon! Glad to have you on board.
Howdy Topol! Thanks for the snacks.
Hello PMA! Thanks for the new ideas and output. I've studied them closely and understand. So at this point you're working to speed up the iteration process by looking for quicker ways to deal with the remainders, correct? Remember these crumbs? Handling remainders is key, base selection is arbitrary. We're on the right track, lads!
>The objective is to find a base larger than n and smaller than x+n at this stage.
Hello AA! Thanks for keeping shills off the board, and thanks for your work on the binary ideas. I've checked out your output and there are definitely patterns there. I have no clue yet how to decode them, but thanks for chasing down new ideas.
VQC was saying the connection between (1,c) and (prime) records is easier to see in binary. We got very close to connecting them a few breads back. PMA's output had us one step away from a connection. Takes a look at the numbers I've put a box around in binary??
Hey homos!
You know what would make this go faster?
Talking to each other.
Are you on Discord? If not, get it.
Put this into a browser without the space:
https:// discord.gg/qZtpAPc
That's an invite to my server.
Send me explicit verification of who you are, meaning an image in your style of output that hasn't been posted on the boards or better yet, take that image, cut it in half, post one half in the EZ Bake and DM me the other half and I'll put all the Pathgawds together.
I already got Minecraft and Sheeeeitbaked ;)
Ever notice my tone changed a bit and now I talk about soooomewhat more intentionally relevant things?
I'm the only member of the G-Unit.
I could set up a private chat or we could have a whole server dedicated to this, if y'all like, and that would be our back up bakery. We can privately send that invite to each other on Discord.
Savvy?
Just posting one more example to explain the u midpoint and tree connections. This might make it a bit easier to see what I was talking about.
Pics attached are for c=9874400051. And include the iterative solution, the d and e factor tree including some f properties, and an annotated u div 2 parse tree.
Because of the low n value in the prime solution, the iteration process happens entirely in the "remainder" portion. Before any triangle can be made from a factor of (f-2) div 40.
The iterative solution here is reasonably quick at 103 calculations.
The u/2 parse tree calculated from the midpoint between the starting est u of 0 and the next rm u of 4889 results in 4095 nodes.
Only the path to the solution u of 996 is included. And would require only 12 iterations, assuming we could determine the correct path to take.
Which leads me back to thinking that we really need to understand the original parse tree to finish this iterative approach.
VA - Hope this clarifies. I'm considering the "remainders" as falling within the gap between two triangle bases, f-2 div 40 units apart.
Hmm I wonder if the correct path ends up being some monkey business in the original “the end”
Funny thing happens when you play with numbers (right guys?)…
51.99% of my screen boils down to:
1331x749
1331=11^3
749=107*7, both primes.
(also works with 52% but I entered the dimensions by constraining to 1331.)
Fuckin' suck it, nerds!
Are we figuring out the dag blast pyramidz?!
And, fuck it, since I just so happen to be here:
5303 – Sophie Germain prime, balanced prime[5]
And shits'n'giggles:
5329 – 73^2, centered octagonal number[2]
5333 – Sophie Germain prime
5335 – magic constant of n × n normal magic square and n-queens problem for n = 22.
4
Thanks for the binary output AA! I looked it over closely and see some possible patterns. I guess I'll need to learn binary and examine some other examples for na to verify!
Hey Topol, I'm open to the idea, just wondering bc lots of people on /GA/ keep saying Discord is a honeypot. Is it compromised in some way?? Honest question here.
Nice digits PMA! I studied your output and understand your idea. So do we actually have a way to know when we're getting Closer or Farther away from a solution? Do the calcs for x+n rm and rm 2d(n-1) show us when we're approaching a solution?
Already got Komrade Baker, Minecraft, PMA, and Sheeeeeeitbaked. Think about what you've been doing on an open forum already. Think of all the doubt being seeded about 8chan, let alone a tool that lets talk with people in real time. Why wouldn't (((they))) try to dissuade you from utilizing it.
And again, think of what you've been doing, publicly, for the last 5 months. If you or any of us were boned, we'd be boned by now. If you're comfy with your OPSec here, then you'll be fine on Discord. Otherwise, you might as well get off the internet. The whole thing's #COMPED, anyway.
The work's been going so much faster on the Discord server. Do the verification and join us! There are no "bots" in the server that would inject screwball activity. I've never trusted those things and some of them WILL fuck with the servers so I keep mine clean.
Savvy?
Discord is definitely a honeypot. I saw a detailed explanation on /pol/ a while ago. I swear I'm not saying this because I'm the BO here and I'm on some kind of stupid power trip, but I really don't see any reason why we should be using Discord.
>Why wouldn't (((they))) try to dissuade you from utilizing it
It could be argued that (((they))) would actually try to persuade you to use it.
>it'll be all the same people with the same names from here, so it just means moving to a different platform that does the same basic thing but with less useful features
>people can impersonate each other because there aren't tripcodes
>there's less shill protection because you won't have someone who can see IP hashes
>we can't have multiple threads about different topics
>this is meant to be a completely public disclosure
That's my opinion. So why do you think we should use Discord?
>Discord is definitely a honeypot. I saw a detailed explanation on /pol/ a while ago. I swear I'm not saying this because I'm the BO here and I'm on some kind of stupid power trip, but I really don't see any reason why we should be using Discord.
Get off the internet.
>>Why wouldn't (((they))) try to dissuade you from utilizing it
>It could be argued that (((they))) would actually try to persuade you to use it.
Can they magically read more than they are here? If you're playing the VPN game… what's the difference?
>>it'll be all the same people with the same names from here, so it just means moving to a different platform that does the same basic thing but with less useful features
Which is why I'm verifying people and not posting the invite to the actual VQC server. Literally no one can be in there without my permission or the permission of one of the VQCrew. Hell, I even send self-destructing invites.
>>people can impersonate each other because there aren't tripcodes
See above.
>>there's less shill protection because you won't have someone who can see IP hashes
Unless they're logging in AS YOU ON YOUR ACCOUNT, we are literally the only ones in there. If someone adds someone we don't know who is always set to invisible, I'd see it in the audit logs.
>>we can't have multiple threads about different topics
We already do. Catch up. I can add as many as I like, with voice capability.
>>this is meant to be a completely public disclosure
Think of this as another one of my side ovens. Where do you think I bake when I'm not here? We still have to present breakthroughs to VQCreators/"Chris". We can't not post here, I don't have his Discord name anymore and he never added me. He used to pop into bakeries.
>That's my opinion. So why do you think we should use Discord?
Because we're getting so much more done so much faster due to the difference in platform and usage. Because we're not concerned with working up a solid post. So that this isn't a bunch of individuals working in the loosest possible tandem by themselves.
It's time to come together and wrap this shit up.
The NSA/Palantir System is already fully aware of who you are and no one's shown up at your door for trying to break the Rothschilds Sociatial Authoritarianism.
Or stay here. Maybe that'll get the others to post the findings more often since you're not seeing the partial steps and quick conversations about tweaks and outputs that aren't complete enough to "officially put on the board".
Long time lurker from the start. This place works. Yall do good work. Don't leave a public forum for a private one. If this thing you guys are making is made in private it will be captured and used for evil. If it is public and free it can't be. Don't fix what's not broke.
>they have access to your information anyway, just give it to them even easier
There's a difference between having all of your metadata funneled into NSA warehouses and having it sold to advertising companies. One you can't prevent and the other you can avoid by not going on those websites.
>Can they magically read more than they are here?
I don't understand the relevance.
>CAPITAL LETTERS FOR EMPHASIS
I hope you're not getting mad over this. I'm just trying to have a conversation with you.
>with voice capability
So it's less anonymous? That makes me want to use it less.
>Because we're getting so much more done so much faster due to the difference in platform and usage
How is being on a slightly different website getting anything done any faster?
>Because we're not concerned with working up a solid post.
What, do people actually care about whether their posts seem like they had effort put into them? This is an anonymous imageboard. Nobody's ever cared when people have made useless posts here in the past. Why would they start caring now?
>So that this isn't a bunch of individuals working in the loosest possible tandem by themselves.
How does 8ch force us to work as individuals? People work together on the same things all the time here.
>It's time to come together and wrap this shit up.
That isn't going to happen specifically because we started using Discord. In fact, splintering us and creating a space where we can forget about anons who potentially have useful ideas could quite easily slow us down.
This is the first I've heard of anyone using Discord as a supplement to this board so the fact that everyone thinks some of their posts aren't "complete enough" to post here is news to me. And it doesn't make any sense to me either. This entire time people have continually posted ideas they hadn't fleshed out yet and that nobody understood for a while. I still seriously don't see any reason why using Discord is a good idea that doesn't already apply to 8ch where we're already firmly established, where every post is public, and where people who don't pay attention very often know where to find us. Being secretive also really doesn't seem like a good idea to me. I won't and obviously can't really stop anyone from using it but you're yet to convince me.
I'm glad someone agrees with me.
Yep for sure. I, and another have been backing up all the work here. You know for a rainy day. I just don't understand the sudden heavy handed push to leave here. I think it is a bad idea at best and a trap at worst. Don't do it.
I'll be keeping my home base here, but I'm willing to check out Topol's "side oven" on Discord. My vote is for this board to remain the main HQ where ideas are being presented, discussed, and improved. We've accomplished a lot here, and I enjoy my daily routine of checking in and catching up on everyone's posts. One day we're all gonna meet up at Tony Stark's lab anyways, right? :)
No one's suggesting we leave here.
Again, we have to turn in our work for review.
Instead of everyone working on this in their isolated corners, there's now an option to spitball, if you will.
There's voice and text… comments and questions can be made and answered on the fly instead of waiting for days to check out the boards, as some people were falling victim to the discouragement shilling that was going on here.
Think community, not division.
I honestly don't see the upload rate changing.
And no one has to join the server. If y'all wanna act as anchor to make sure we don't get tunnel visioned on the discord, which can also be backed up, that's fine too.
Here's the thing… it's also a bunker if the boards get comped and we can't get back here. Not the first time that idea's been floated.
But here's the thing…
Why are you trying to hard to sway the board, Anon? You're taking all the work and don't want to be cut off… trying to sell as in our interests instead of yours?
Perhaps that's a reason to go to the boards…
Who are you? ;)
Triangles, Triangles, and more Triangles
>>5239 (f-2) div 40 iterations
>>5251 rm 2d(n-1) fine tuning
>>5257 overview
>>5260 woah another triangle?
>>5272 more than 1 triangle??
This post is an attempt to further explain the iteration process. How the numbers "click" into place, that everything can be represented as triangles, a possible connection to the factor tree that needs more research, and perhaps even a connection to the grid using x+n values.
First off, thanks to Topol for posting the attached pic (topol_triangles.png) in the EZ Bake >>5304. This pic shows how I believe the "remainder" portions come together as separate triangles to form the solution square.
I have created a new "triangle" prime solution output for all my odd x+n test cases, which you can find at pastebin.com/qanuvycg (again if anyone is interested).
From that list, I've chosen c=208932019 for the walkthough, but the same approach applies to all numbers. The full explanation is in the pic (c208932019_formula_breakdown.png).
The final XPN value we are searching for is comprised of:
XPN = 1 + 8T(u) + mod + 8 * (remainder) + mod - (rm 2d(n-1))
XPN = 1 + 8T(7875) + 4 + 8 * 2440500 + 4 - 1128
The (remainder) portion can be broken down further into a series of 0 or more triangle numbers. By removing the largest possible triangle, then the next, and so on until the remainder is zero.
2440500 = T(2208) + T(58) + T(9) + T(3) + T(1) + T(1)
The 2 mod values can be added together, and then subtract the 2d(n-1) remainder. This ALWAYS leaves you with a number that is divisible by 8. I'm referring to this as the "balancing total" and is the "bal total" column in the sample output.
balancing total = (4 + 4) - 1128 == -1120
The balancing total can be positive or negative. To visualize this, see the "dark green" remainder blocks in the triangle iteration mapping picture at >>5260. It is the adjustment from this "balancing total" that enables the estimated XPN from the iteration process to "click" into the exact final XPN value. This number can also be converted into triangle representation.
balancing total = -( T(16) + T(2) + T(1) )
Therefore, we can represent the ENTIRE final XPN formula as a series of triangular numbers.
final XPN = 1 + 8 * ( T(7875) + T(2208) + T(58) + T(9) + T(3) + T(1) + T(1) - T(16) - T(2) - T(1) )
See the pic (c208932019_triangle_breakdown.png) for a further breakdown of how these triangles fit together to form the (x+n)(x+n) solution square. It also shows how the (x+n) = 2u + 1 relationship can be used to solve for (x+n)(x+n). The color coding for the respective triangles matches pictures at >>5260. So the u and (x+n) formulas are really interchangeable.
Also notice how the triangle numbers grow to form the solution. There are most likely many combinations of triangle numbers that can be pieced together to arrive at a solution. My methodolgy is just an attempt to illustrate the point.
I believe that this growth in u values is what the factor tree is meant to provide. And might very well be the last remaining key to an extremely fast solution.
>There's a difference between having all of your metadata funneled into NSA warehouses and having it sold to advertising companies. One you can't prevent and the other you can avoid by not going on those websites.
Again, if you're using a VPN, what do you care? Otherwise… it already is. Get off the internet. Palantir, nigga, AT LEAST. And you can't access the code on this open source platform, or have you not been following the shillery closely enough?
>>Can they magically read more than they are here?
>I don't understand the relevance.
Could the people you're afraid of getting your information get more info from you using Discord on a website or as an app? Yes, No? Prove it?
>I hope you're not getting mad over this. I'm just trying to have a conversation with you.
nigga I just don't know the markup
>So it's less anonymous? That makes me want to use it less.
Great, you do you, booboo.
Or modulate your voice, you suddenly important motherfucker, you! I can't tell you dick about anyone here, but y'all know me. Who are you talking to? lol
>How is being on a slightly different website getting anything done any faster?
Because the Boards aren't for chatting and derping around or half baked thoughts. We even set up an EZ Bake that no one utilized. This is just an extension of the EZ Bake. Now with voice and faster responses for to get this done quicker. Some folks weren't posting on the boards because they didn't trust it. So… whatever camp you fall into… you do you.
>What, do people actually care about whether their posts seem like they had effort put into them? This is an anonymous imageboard. Nobody's ever cared when people have made useless posts here in the past. Why would they start caring now?
I've had a thread created dedicated to chewing me out because people care. Who are you talking to?
>How does 8ch force us to work as individuals? People work together on the same things all the time here.
Think of it as a conference call verses mailing something. You don't mail half assed letters, and reputations/trip/namefagging has been built up at this point, so the Discord is like… a break room? Informal social while also getting work done and throwing ideas "out there" more loosely.
>That isn't going to happen specifically because we started using Discord. In fact, splintering us and creating a space where we can forget about anons who potentially have useful ideas could quite easily slow us down.
Sweetheart, you're the only one talking about splintering. You and Anon. Like how the left goes around shooting up schools and gay nightclubs, and then try to take guns from the right.
>This is the first I've heard of anyone using Discord as a supplement to this board so the fact that everyone thinks some of their posts aren't "complete enough" to post here is news to me. And it doesn't make any sense to me either. This entire time people have continually posted ideas they hadn't fleshed out yet and that nobody understood for a while. I still seriously don't see any reason why using Discord is a good idea that doesn't already apply to 8ch where we're already firmly established, where every post is public, and where people who don't pay attention very often know where to find us. Being secretive also really doesn't seem like a good idea to me. I won't and obviously can't really stop anyone from using it but you're yet to convince me.
There's nothing "secretive" about it. Check it out for yourself. There's a lot of shit posting that doesn't belong in the Work Board and it's not like anyone but me really utilizes the EZ Bake. People stopped posting because 8chan, you're afraid of discord, and some random fuckwit is upset because they won't have the same flow that they once did.
If you want to be divisive and call it splintering, go right ahead. Everyone's waiting for the rest of y'all to join us cuz… hoooooow many of us are here? Half are on the discord and enjoying interacting. You wanna stay super secret secret squirrel… you go right ahead…
>>>5312
>I'm glad someone agrees with me.
Y'all keep each other company, and things will come in at, like I said, probably the same frequency they were coming in in the first place… now with friends.
Would you mind weighing in on this Discord thing? He said you were part of it. Excellent work, by the way. If it wasn't for you I don't think we'd be getting anywhere until Chris comes back.
I really don't see why we can't spitball here where everything we say is public. That's the entire point of this. Even a tiny bit of a closed door on what we're doing and we're going against the meaning of disclosure. Do you want 40/60 from Q or do you want the entire 40,000ft view? Because people are relying on us and VQC for that same thing (although obviously to a lesser degree), so if we don't do everything publicly, how are people meant to know we aren't hiding something? I'd like >>5315 your opinion on this too, VA.
>Why are you trying to hard to sway the board, Anon? You're taking all the work and don't want to be cut off… trying to sell as in our interests instead of yours?
You're doing what you're accusing that anon of doing, Topol. You're trying to sway the board to use an extra platform where you have to let them in.
>the Boards aren't for chatting and derping around or half baked thoughts
Says who? Why can't we do that here? In fact we've been doing that this whole time. This binary triangle number stuff (e.g. >>5277 here) is a good example. It hasn't gotten us anywhere. That doesn't mean we can't post about it. If you're all so worried about cluttering the RSA general, make a "half-baked ideas thread" or something. The boards are for chatting and derping around and posting half baked thoughts. In my opinion (and I had assumed everyone else's until now), they're for everything we do that is related to this.
>We even set up an EZ Bake that no one utilized.
That thread was set up specifically for you because you weren't posting anything related to cracking RSA and people were getting annoyed and reporting you. See >>356 this thread.
>There's nothing "secretive" about it
There's nothing secretive about having a separate place to talk about this where only people you allow can see it and that you don't post the link to? That doesn't sound like the definition of the word "secretive" to you?
>You wanna stay super secret secret squirrel… you go right ahead…
You're the one doing that (although you seem to disagree). What are you talking about?
>Freakin' out over nothin'.
You might not see this as a big deal but I can see it turning into a major problem very quickly. What's to stop the people in the Discord from figuring it out and then never posting it here? Nothing. The fact that that's even a possibility defeats literally the entire purpose of doing this in a public way on 8chan, and the fact that we even need to have this discussion honestly worries me a little about your intentions here. Not to mention, patronizing shit like this
>Great, you do you, booboo.
>Sweetheart
>you suddenly important motherfucker, you!
makes you needlessly sound like an asshole and makes me wonder if you're just being stubborn for no good reason.
Your concerns have been noted.
And there's never been anything stopping anyone from "figuring it out and not posting it".
C'mon.
sorry to hear VQC has lost his friend (peace to you VQC if you read this). Dogs are special.
thanks for invite, but finding this nice and comfy here. VQC with DM's splintered off was bad enough, perhaps good that twitter is gone. WWG1WGA
ty
welcome lurkerAnons
good.
interesting to take a different view on this. Viewed but not analyzed.
Would be good if a cell is valid, to be able to list more t's, they drop off when the variables hit maximums. Isn't this a parameter in the code, such that we could get more elements in a cell if that cell were valid?
hey CA, hope all is well. Any further progress?
excellent post and progress, thank you. Looks like worthwhile to replicate in my own excel models. Too bad we can't share excel files, but don't think there's a good anon way with excel. Did some graphs, have been intending to post a couple screencaps.
>re: Padovan sequence
That is an interesting triangular form! Note the sequence: take the sum of any two triangles adjacent on outermost node, say 7+9, skip one, and the next triangle is that value (16 here).
Have reviewed post and will take another look and try to get the model going.
Solid post, PMA! I'm loving all the triangles, and they make understanding of how to deal with remainders straightforward. As usual, thanks for your work!
General Note: I'm gonna check out Topol's Discord chat, AND my home base will remain here. Also, we shouldn't have kicked Topol out of the RSA general. I've visited a few times to EZ bake, but no many others have. I enjoy his posts. Nuff Said.
Let's keep working and having fun, faggots!
Hey AA! Your analysis is accurate, and your foresight and experience kept us solid when CBTS gave us the boot. My personal conclusion is that this board is the best place to get work done, AND if people want to supplement that with chatting on Discord, then great.
At this point we’ve all been working together so long, it would be a waste to disrupt our harmonious frequencies with discord among the brethren (lol, couldn’t resist lads). Seriously tho, let’s not make this a big issue. A team needs harmony to thrive. Thanks to both AA and Topol all they have contributed. Let’s keep going!
41/184 C(c)/EDX(x,e,d)/AB(a,b)/EDX(d,x,e)/
41/184 C(c)/EDX(d,x,e)/EDX(d,x,e)/AB(a,b)/EDX(d,x,e)/
23/184 C(c)/EDX(x,d,e)/AB(a,b)/EDX(x,e,a)/EDX(e,a,d)/
26/184 C(c)/EDX(x,e,d)/AB(a,b)/EDX(x,e,d)/EDX(x,e,d)/
28/184 C(c)/EDX(x,d,e)/AB(a,b)/EDX(x,d,e)/
These are the algorithms and the relative frequencies of success for the first however many coprimes. Nothing rock solid yet.
All I wanted was a bar graph or even a line graph or a fuck load of plot points of Prime Numbers and this is what the searches give me.
Interestingly, with the circle chart,
Notice the pure white, 3-rows each, sections?
and there's a stripe of whit between lines with primes…
if we base the "graph" on something like that …. and expand it.. I feel like we could exclude any number that falls into the "pure white" regions.
Marked a lot of the semiprimes. Didn't feel like finishing it.
So neither of you are concerned that we're having non-public discussion about this? That's the point I'm trying to get across and it doesn't seem like either of you addressed it.
What's that last image from? It looks like the same triangle sequence created by the zeros in the binary triangle numbers.
Having to give your e-mail?
Do I a setting fuckety somewhere?
When did that pop up?
Being conscious of reporting back to the board is a topic that comes up but honestly there's a lost of derp and quick messages. If what was happening on the discord were happening on the discord were happening on the board, it'd be hell.
That being said, I have no fucking idea where that comes from. I was looking for a bar graph consisting of Prim Plot Points. Which… doesn't even sound like it makes sense in a strict understanding, but I hope you know what I mean.
Here's a sample of what you're "missing"
Might just be a discord thing… I don't have any verification settings checked on either channel and I ran into that once and thought it was a server setting for that particular one.
Might be a thing that got implemented after I started using discord. Either way… throw-away-e-mails can always be created. It's not an issue for ProgrammingMathGawds.
This highlights all the cells with same a values. Stumbled on this website but its eerily similar
http://www.divisorplot.com/5.html
>So neither of you are concerned that we're having non-public discussion about this?
I appreciate your concern. But our individual work and effort is all private until we choose to publish here. Which I will continue to do because I believe in the importance of this project and I want us all to succeed together. This knowledge is too profound to be kept in the dark.
And if I get stuck in my thinking and am able to reach out directly to a fellow anon on an anonymous chat platform for a quick question, why wouldn’t I take advantage of that to try and speed up this process?
Hello Lads! Just checking in over here to see what's going on. I've been turning over all the ideas in my brain and studying our recent posts, along with VQC crumbs. It's pretty Mellow over here.
Hey AA! Eh, Discord is cool for chatting, but I prefer working here.
Mmmmmm…. I think I found something that partially confirms my "Scissor Theory"…
If C=33015879=19406579 and the square root is about 4405.3
4405.3/3301=1.33453
5879/4405.3=1.334528…
8783500069=4392106027, with the square root being 66272.96
66272.96/8783=7.45595…
500069/66272.96=7.545596
so it's like…
(√C)/p=X
P/(√C)≈X
… did i just sorta back up my scissors theory? O_o
…where the decimal places being slightly off relate to prime numbers not being purely sequential?
lol, holy shit if i did ^_^
Aaaaaand then when you have an EXTREME difference between P and p, your relationship would be found via the Extreme Scissors.
P=10000000343
p=3
C=30000001029
√C≈173205.084
P/(√C)=57735.0278
(√C)/p=57735.028
So, as stated, thiiiiiis one woulda been caught by the Extreme Scissors faster than the Mid Scissors
Updated:
holy formatting, batsmang!
I'll stop with the * and switch to x since i'm not using it as a variable right now so that it's more legible:
When you Floor √C, the values for "d" sync generally to the 100ths, sometime 1000ths
so….
p=8783
P=500069
C=8783 x 500069 = 4392106027
√C≈66272.96
Floor it to 66272…
P/(√C)=500069/66272=7.54570557701593
(√C)/p=66272/8783=7.54548559717636
d≈7.545
Hopefully that reads better.
I'm not really sure what the scissors context is, but the approximations you're seeing are actually exact equalities, give or take your calculator's precision:
If C = p * P, then √C = √p * √P,
so P / √C = (P / √P) / √p = √P / √p
and √C / p = √P * (√p / p) = √P / √p, the same.
Scissors:
You have two Scissors of Iteration, or if you want to imagine Compasses because masonry, go for it. The primes can be the Square. Anyway.
Green Scissors start at the extremes, so 2 and C/2 and close inward.
Blue Scissors start closed at √C and then open outward.
From what I'm seeing, the Blue Scissors will hit the primes before the Green Extreme.
Also, I've been confusing CoPrimes and SemiPrimes. But on my diagram I just picked variables anyway.
So, the extremes will block off large amounts of numbers.
So… C/3 C/5 C/7 on the low end Green, but every time you do that, you can cross off MASSIVE amounts of primes at the Upper Extreme.
The Blue Scissors seem like they'll be faster at getting to the large, complex C's (Semiprimes).
Constrained by the relationship we're talking about.
Also:
Lower Range:
{2<p<√C}
Upper Range:
{√C<P<C/2}
Lllllllllimits!
A bit of a performance improvement.
While further reviewing the data for the iterative search, I came across a performance improvement relating to the "balancing total" in my previous post.
Attached pic for c=7463 shows the last few iterations for the prime solution. Notice how the n0 and XPN values remain the same for different values of est XPN. This is a result of the "balancing total" adjustment explained previously that enables relatively close results to "click" into place.
So why do we need to test the same n0 value multiple times?
Turns out we don't.
The range over which the XPN remains the same for various est XPN values can be determined.
By calling the Get_n_from_odd_triangle_base method with the est XPN and next n0 (i.e. n0+1), and adding the absolute value of the result to the current rm 2d(n-1), we can determine the exact spread of est XPN values that will result in the same XPN. This spread can then be used to adjust the iterative search factors to skip relatively large chunks of f-2 div 40 that would otherwise result in duplicate calculations.
In addition, because the parity of the prime n and starting n values are always the same, we can increase the next n0 to n0+2 in cases where the parities match.
The remaining pics attached show the original and revised iteration results for various odd x+n values. Full revised results are posted at https://pastebin.com/imn7X5M8.
The most significant performance improvement can been seen for c=208975787, where the number of iterations has been reduced from 11,153,430 to 4274. Another noteable result is for c=9874400051 where iterations have been reduced from 103 to 11.
And in all except the smallest test cases (where the gcd would apply), the iteration process is now faster than directly iterating n values.
W00T MUFFIIIIIIIINS!!!!!
So it turns out, I was looking for g, not d.
Funny how that works out.
See the opening and closing?
I'm unsure if this is trivial, but I've never come across it before.
With the above definition of g, then the following relationship is true.
g/a = b/g
g = sqrt(145) = ~12.04159457879229548013
g/5 = ~2.40831891575845909603
29/g = ~2.40831891575845909603
g = sqrt(7463) = ~86.38865666278183307262
g/17 = ~5.08168568604599018074
439/g = ~5.08168568604599018074
g = sqrt(93801) = 306.26948917579106446528
g/3 = ~102.08982972526368815509
31267/g = ~102.08982972526368815510
This is awesome. How big of a number can you factor with about an hour’s worth of CPU time? I assume you’re single threaded right now and using a heavier-weight or scripting language. To me it appears to be the time to test the limits.
ON DUBZ!!!!!
Magical even.
Good work anon.
That was me asking a question about whether it was worth continuing the fight against alcoholism, staying sober, a massive personal struggle, and also whether I had messed up by participating in Discord, I saw some good in them at the time.
I think the response from Q was general to all anons who get down and/or struggle.
Start in any way you can.
Everything good begins like that.
Momentum.
Good to have you aboard anon.
I diiiid try to friend you when you were on back in the day… but I think you'd sidestepped discord before seeing my friend request.
Or you just didn't do friend requests.
I saw ya back then.
Not sure where you are in the reading but…
It's a straight VQC Server :D
Beautiful
There were things to be taken care of.
That has been done.
Thank you for your kinds thoughts.
Minecraft only explained the gist of coprimes and RSA like… a month ago.
"Original" youtube video?
"en"… I'm thinking of limits…
as in….
0, e (the .000…1 appearing) but… inverse of n?
.999…, n, 1
does e=n^-1, by any chance?
Fantastic if I'm off…
I'm just glad we could resurrect you through the powers of autism.
There's a reason I'm back now.
I've got to go through a fair bit of setup as I've wiped and replaced my primary devices and restore to this temporary one.
I believe in the plan of Q, POTUS and WW support.
This is truly BIGger than imagined.
Hi, you're new here.
We've been solving not only the Secrets of the Pyramids, the G of Freemasonry, and the Sacred Geometry that describes the Divine Math of the Great Architect's construction of Existence.
Also something about Tyler Durden, Albino Morpheus's Cables hidden in the Blockchain (but his internet is "cut off" right now, "officially"), bye bye Linux, bye bye RSA Rothschilds' Stranglehold of Aarde….
So like… Hoooooooooow much bigger you need me to imagine?
Cuz…
Topologies of Existence…
What's "up"?
And by Topologies, plural, I mean I don't remember the last upderp you got on my weirdness.
Glad you made it through the Stargate.
Welcome Home.
and by "first" I mean "chronological first" as in an understanding of "Original".
Have you ever posted a video?
I know I post a bunch….
That's a good question.
I will stick here to the process of you all discovering how the VQC works and the subsequent consequences of P = NP and tech, etc.
We've come a long way together.
Through the hard times and the good.
…
We're still in the foothills.
The view at the top will be worth it.
It gets suuuuuuuuper weird.
(Don't worry, this business is no longer a thing.)
I either give sympathy for your son or…
You mean Son in like… Korean?
https:// wikileaks.org/clinton-emails/?q=&mfrom=Harold%20Hongju%20Koh
Dafuq?
#VirtualQuantumComputer = TechKnowLogos reaching out for warm fuzzies back into play!
#Gametheory #QuantumAutisticIntuition #CommsLearning
Thanks
Welcome back Senpai!! What a nice surprise to wake up and find your new posts. What’s the best way to check iterations to know when you’re getting close to a (x+n) match? Is there a way, or do we just iterate until a match shows no remainder?
>>5401 6th or 7th, welcome back.
>>5355 k, will forgive you based on the dubs, you're on a roll.
>>5366 May Harold R.I.P.
>>5382 2nd!
>>5400 Checked! 99&00
>>5377 perhaps as you go through any personal / local setup, work here in parallel, descriptively? Also, if there are specifics to your setup, please lay it out here so we are ready.
>>5353 good question, apart from VQC's input now, that would have been a great next step as a checkpoint. >>5348 Some nice progress there PMA, way to stick to it!
>>5364 3rd that after VA, PMA has been forging ahead.
>>5325 me.
>>5361 juicy q post, drops will come fast. Acceleration. Also, STAY TOGETHER (e.g. avoid discord). ty
We're together on Discord no matter how much you stay away. You're welcome to join us.
I'd rather you talked to me directly instead of calling the house, y'know?
Plus, we've posted all our findings so far.
We're MORE together now that it's not everyone working on their parts on their own AND we have faster iterations right now.
And we got to such a point that Chris Resurrected, so you can take your divisive concern trolling elsewhere if you don't see that the Discord has paid off. :P
Code is c# running single threaded.
My largest test cases are the products of 5 and 6 digit primes. I can certainly put together larger test cases if you think it worth while.
A guesstimate of iterations required is roughly n/2. So RSA numbers are safe for a little bit.
There is one more efficiency I am looking into with regards to removing large chunks at the higher end of the remainder portion.
>>5408 cool.
Was thinking would be good to do test cases where the ratio of a:b is very low, as well as when it approaches 1, with 1 being a perfect square. The (1,c) are as low as you can go where a=1, and they go lower and lower to zero as c goes to infinity. Those are some of the graphs I was alluding to.
>>5407 chillax brah. ty for the standing invite.
Gladly.
No more of your bitching about the Discord.
You can't accuse US of being divisive when YOU refuse to join. Since you refuse to join, you're just going to have to trust when se say that you're seeing what's going on on the discord without the random side comments.
HOORAY! THAT'S OVER!
That trangle come from her':
https://en.wikipedia.org/wiki/Fano_plane
Here's a Fano 3Space… Look familiar?
Fano n-space didn't pop any results, but 4space did:
https://www.researchgate.net/publication/256844851_Fano_Hypersurfaces_in_Weighted_Projective_4-Spaces
Also:
https://en.wikipedia.org/wiki/Fano_surface
And:
https://en.wikipedia.org/wiki/Fano_variety
Someone followed me to the discord from the boards, but they live in my personal server, not in the VQC server. That's the reason I did that. They, or Christopher Bouzy, were the troll/shill here.
Speaking of which… I don't think I ever turned the screenshot into a png sans info… check this shit out:
You are closer than you think.
It is efficiencies.
From O(EXP) to O(log t).
From exponential complexity class (well, sub exponential with the general number field sieve) to the natural log of the length of the integer c in bits.
Breaking the problem down.
First by roots.
Then by triangular numbers.
I'll clean up my set up, then walk through a clean version.
Perhaps one day I'll pastebin the whole dump of the last seven years and show you the tortured/enlightened path through all this.
The probability of it all happening by accident seems remote. Just needed someone like Trump/Q et al to come along.
NOTE: You don't need to know the general number field sieve. It is a horrible approach in terms of understanding. Just know that it is based on making the Fermat factori(z)ation method more efficient.
FUN and IMPORTANT later: any coincidences in the Mandelbrot Set? E.g. one of them relates to Pi.
I wonder if were supposed to blend the Compass, the AlmondBrot Set, and… I'm seeing fractals.
Is the coprime integer (that we seem to be calling "c") the "zoom" of the set and the two primes we're working for are on theeeee… edge of the zoom?
Do we zoom to √C knowing that P and p are equidistance? So at that zoom, theeee….
Which points on the Almond are we rubbing?
@[C{0√(D)}]
(After typing that out and thinking better of something else)
Dafuq's a 0-root? lol
https://en.wikipedia.org/wiki/Zero_of_a_function
And does it relate? We're making some sort of Tree or something, right? Need roots, I'd imagine…
If you were Munder it'd be a white faux-telecomm van
So… a conversation that happened in an unrelated server…. someone was talking about the "Sevens" Cicada puzzle and that it had something to do with an RSA crack and something hidden in an image through layers of steno… but it was never solved by that particular guy
But… buuuuuut… also someone of high intelligence…
"Fermat factori(z)ation"
Z.
Well… don't get distracted by puzzles and A.R.G.s…
But that's a gosh darn what the fuck!
Math of Existence…
Because I have declared that P=NP, proof me wrong, then I'm not surprised that I'm on points I'm just making up that happen to exist, apparently.
Get on my n-level.
(Someone explain to me what "n" is. In my headcanon, it's that "1" that shows up "immediately" after 0, and fills the same gap/jump between ".999…" and 1.)
Finished a couple of tests on the products of some random 9 digit prime numbers. Stopped testing a bit "early" as I didn't see any benefit in the code running for days.
Output posted at pastebin.com/Y0tvNJHu and the summary is in the attached pic.
The performance is pretty close to n/2. And it works.
>Breaking the problem down.
>First by roots.
>Then by triangular numbers.
Thank you for this.
I don't believe removing the higher end of the remainder portion or the u/2 parse tree >>5299 are going to get us the required efficiencies.
So back to trying to integrate the factor tree.
I've got all threads saved offline, along with all Maps. Here's links for this bread (easy to update) and RSA #10, which hasn't been added to the dough yet.
BAKER
RSA #11 —- http://archive.is/YERWc
RSA #10 —- http://archive.fo/anGD7
Hello VQC! Good to see you. To your knowledge, are Q or POTUS aware of this mission? We're not gonna screw anything up are we? POTUS always talks about how the old ways of keeping secrets were better… Almost like he's waiting for a P=NP habbening.
Hello PMA! I'll start working on smaller examples of this:
>Breaking the problem down.
>First by roots.
>Then by triangular numbers.
I think that the "levels" of each root and remainder could be helpful in eliminating unnecessary calcs. Maybe our desired (x+n) is 1) a multiple of the lowest levels of the factor tree (or GCD) and 2) a triangular number. Process of elimination. Maybe the factor tree aids in more efficient searching, not providing an answer up front.
Some of the plots done a while back for visualization
Sea of c
a & b prime.
a along x-axis, b along y
a <= b (45 degree line = perfect squares, a=b)
pic 1 table
pic 2 zoomed out
pic 3 axes normalized for a & b
pic 4 veins/ribbons that show up with digit changes
pic 5 larger visual shifts from decimal to exp format
Fermat's factorization algorithm
https://archive.fo/e4crP
https://archive.fo/tAjTb
https://web.archive.org/web/20160604033254/https://trans4mind.com/personal_development/mathematics/numberTheory/divisibilityFermat.htm
Fermat's factorization is pretty much just iterating every number around d and seeing if it is one of the nonadjacent squares (d+n)^2 or (x+n)^2. It's pretty basic compared to some things we have here. But it's the hint Chris gave.
Eh. NSA knows all. “And by all I mean all.”
Google is about to suck some DJT NSA dick. If what Q is saying is true, the good guys are about to mop up 400 years of Evil Bastards and foil their ill-begotten plans. Google can suck our dicks.
That and he was "waiting for the Compass and "g" to show up". Why?
Because of the relationship or because Masons love math'n'artchitecture plus pyramids and I'd be totally into that and it was just "expected"?
Then Mandelbrot and Pi for later…
Do the Square of Primes (fixed) and Compass of Equality (measuring the equidistance from √C) come in at the same time?
Like… Did I miss it?
Iiiiiis "g" a thing here?
Or was that skipped of "for now/for reasons"?
I'z just curious.
I guess what I mean is that they will be forced to serve patriots instead of the Cabal. God, we dodged a bullet. We were about to go full totalitarian regime.
The demiurge and the archons.
>From exponential complexity class (well, sub exponential with the general number field sieve) to the natural log of the length of the integer c in bits.
How would even/odd decision tree result in natural log? It would result in log2.
Wall Confirmed.
And a few trangley pyramidals.
Also, we're getting out of the woods… or at least to a clearing or something…
A somethin' dank happeneded…
Could the (z) be singling out the z. What could z mean? Riemann Zeta function? N just sideways in our end record?
Also Fermat factorization method is just testing squares out to see if its a difference of them right? Basically the equivalent of doing the F function (incrementing d) repeatedly until e is a square. Then if e is a square (say of radius r) then you would satisfy the equation:
c = d^2 - r^2 which would imply that c = (d+r)(d-r)
This would mean that a = d-r, b = d+r, e = r^2
Also the pi thing is basically the amount of iterations it takes to go greater than 2 while just outside the cups of the mandelbrot set approaches pi to an increasing power. Looks like a cornucopia.
So we would want our e to be a negative square. Interesting because negative squares have entries in every column and they are easy to generate (i think).
If d^2-r^2 = c then d^2 + e = c where e = -r^2 would be a solution. This would imply that (d-r)(d+r) = c which means d-r = a, d+r = b.
If d-r=a then x = d-a = d-(d-r) = r
Then, since (d+n)^2 - (x+n)^2 = c and we already know that x = r and d^2 - x^2 = c so n must be zero. If n is zero then xx+e must be zero also.
Generally in talking about algorithm complexity, constant factors are ignored, so log2, ln, log10, etc. are all considered essentially the same.
That said, an even-odd decision tree or a normal binary search (like the square root code) would end up as log of the value c, since you get about one bit of the answer per iteration. Something that scales as the log of the number of bits of c is like log(log c), which is… shockingly efficient.
I hadn't really thought about this before, but if at most it takes O(log(c's length in bits)), that implies that we're doing some kind of logarithmic operation on the bit string of c, and that in some cases we'll need to iterate the logarithm of the whole bit string. The bigger factorization process is just meant to be a calculation, and not involve iterating through possible a and b values, so that refers to all of these pieces of information that we've been figuring out. We're using the tree to figure out information related to c and the parity of its respective variables.We're then jumping to figuring out the triangular numbers that make up the d+n and x+n squares based on their parity. We're using the specific numbers we find from doing those things to calculate a and b. If it's O(log(c's length in bits)), meaning at some point we need to do a logarithmic iteration, but to actually find the factors we're doing a calculation that just involves plugging in some specific numbers, doesn't that imply that rather than being part of finding the factors, the logarithmic iteration is actually done on the bit string version of c? That would mean one of those specific numbers we're using to calculate a and b has something to do with the powers of 2 that make up c, wouldn't it?
I’m aware of that, which is why I found it odd that it was specifically “natural log” instead of just “log”. Either that was a typo and he meant log2, or I really have no idea what’s going on because I can’t think of anything that would specifically scale ln here.
Yeah, that’s exactly what I’m thinking too. I could be way off here, but it’s my understanding that the reason this is called VQC in the first place is because a pre-existing virtual “data structure” kind of already exists in the universe that will factor any c. I’m thinking like a fractal of decision trees. You have to know where to start and what decisions to make, and walking this tree will lead you to the answer in log c time.
Sometimes when I’m talking to people, thry’re like “You can program? You must be so good at math.” And it drives me nuts because to me it’s not even remotely true. To me, math handicaps itself with its seeming insistence on having the entire program be one giant single expression, then having academic circlejerks about it for a few hundred years. When this whole thing started, it kind of resonated with me that there is unexplored structure in the numbers themselves and also a certain inherent decision-tree-ness that traditional approaches to math might not be interested in. Ok end of rant.
>pre-existing virtual "data structure"
That's what a lot of us have assumed this whole time. It's just a matter of even comprehending what that means. I'm currently trying to see if there's any correlation between binary c values and the binary of the triangle numbers used in x+n squares when x+n is odd and n is even.
Not seeing anything so far. I've sort of picked the variables I've looked at arbitrarily, and I don't even know what we're looking for, so I dunno.
I'd just like to point out that I was in a conversation with Defango a forever ago, and I'm pretty sure I posted it here… where he was saying that Cicada showed him the RSA crack and that it invoooooolved…. log something….>>5144
Yup, the most recent use of mod in our quest is solving the odd (x+n) square. We divide the square into 8 Triangles -1. The mod is the remainder left over after dividing into 8 matching triangles. Example:
(x+n)^2= 15^2 = 225
225 - 1 = 224
224 / 8 = 28
224 mod 8 = 0 (bc there is no remainder in this case)
Your cicada friend seems to be hinting that knowing how to properly deal with the remainders is a big key. PMA has done a great job on explaining our current process ==>>>>
>the remainder is the mod
>>5317. PMA's excellent explanation.
>XPN = 1 + 8T(u) + mod + 8 * (remainder) + mod - (rm 2d(n-1))
>XPN = 1 + 8T(7875) + 4 + 8 * 2440500 + 4 - 1128
I think we should be looking for a D value. Specifically the one where n = 0. Also where -x^2 = e Then, if we have that D value we can look at (d+n)^2-(x+n)^2 = c to get just d^2-x^2 = c. The thing is that for our first factors (1,c) we will not have n=0 here it will be something else. Another thing I noticed is that if you make a record (e,n,d,x,a,b) then make a new record with the same a,b but make the new d value equal to the old n value, then your new n value will be the same as your old d value, so in a way you can switch d and n but the x and the e will change. Then another thing is that VQC mentioned the record c^2. This is cool because if you shift it to n=0, then you can get d^2 - x^2 = c^2, a pythagorean triple (learned in number theory today you can generate these from a different triple. Could maybe just swap c and x in this). This would also mean that (d+x)(d-x) = c^2. This could only be true if x = 0 in which case d=c. Or this could make the scenario where (assuming b>a)
(d+x) = ab^2 and (d-x) = a
(d+x) = ba^2 and (d-x) = b
(d+x) = b^2 and (d-x) = a^2 [I like this one best]
then
d + x = b^2
d - x = a^2
2d = a^2 + b^2
We know the correct D for our original record is the one where D = (b+a)/2. 2D = b+a
Also for this the correct X would be equal to (b-a)/2.
(b+a)(b+a) = 4D^2
b^2 + 2c + a^2 = 4D^2
2c + 2d = 4D^2 =c + d = 2D^2
b^2 + a^2 = 4D^2 - 2c
b^2 + a^2 = 2D^2 + 2(D^2 - c)
b^2 + a^2 = 2D^2 - 2E = 2D^2 + X^2 = 2(D^2 +X^2)
b^2 + a^2 = 2(D^2 + X^2) = 2d
d = D^2 + X^2
from this we can infer that the correct d value for the record where (d-x)=a^2 and (d+x)=b^2 is also a square. Lets see if we can figure it out from here:
( * is for square records, capital letters are the goal record.)
d* = D^2 + X^2
a* = a^2 = (D-X)^2 = D^2 - 2DX - X^2
x = d - a = D^2 + X^2 - (D^2 - 2DX - X^2)
= 2X^2 + 2DX
b = b^2 = (a+2x+2n)^2 =(D^2 - 2DX - X^2 + 2(2X^2 + 2DX))^2
= ( D^2 - 2DX - X^2 + 4X^2 + 4DX )^2 = (D^2 + 2DX + 3X^2)^2
c* = c^2
n* = 0
e = c - d*^2 = c^2 - (D^2 + X^2)^2
= (D^2 + E)^2 - (D^2 + X^2)^2 = (D^2 - X^2)^2 - (D^2 + X^2)^2
= D^4 - 2D^2X^2 + X^4 - D^4 - 2D^2X^2 - X^4
= - 4D^2X^2
That was beautiful man.
>So back to trying to integrate the factor tree.
>Breaking the problem down.
>First by roots.
>Then by triangular numbers.
I have been working through an idea of how the factor tree could potentially integrate with the iterative solution.
Pics attached are for c=6107 and show the full factor tree (c6107_full_factor_tree.png) and a trimmed down version with squares (c6107_factor_tree_and_squares.png) that I will explain further.
The large number test results in >>5428 confirmed that the iterative formula works for any odd x+n. And it is EXTREMELY fast for small n values.
If there was some way we could turn every operation into a small n value search, then I think we would have our solution. (Reminder that in RSA #8 >>3264, an anon showed that there are multiple grid entries for c with prime a and b at various n values. A few samples for c=115 posted at >>3270.)
First thing to note is that the parsing of the e values has been removed. Could be wrong here, but I don't see the point of further factoring remainders.
There are three groups of nodes to the c=6107 parse tree. The first group corresponds to how the iterative search is currently working. The c, e, d, and f values get us to two iterative results where rm 2d(n-1)=0. First for the prime solution, and second for the starting c value.
The second group is for the 39 (d/2) node and is a result of 78/2, it's d and e values. And the third group is for the 3 (d/2) node, from 6/2 with it's d and e values.
By not further factoring the e values, we get a linear path of d values to the root of the tree. (First by root?)
So here's the working theory:
At the bottom of the tree, there is a way to iteratively search for two n0 values that satisfy an rm 2d(n-1)=0 constraint. (This is the "Then by triangular numbers" piece.)
These n0 values won't be the n values we are looking for. But from that operation (should be really quick), we will have something that can be subsequently used.
One of the n0 values will correspond to a path back up the tree in the "Prime Solution" column, while the other will lead us up the "Starting C" column.
This process will repeat at each level until we have the "final" result back at the top of the tree.
Still need to work out what the calculations look like at each level in terms of inputs and outputs. And what the relationship between the squares at different levels could be.
This seems to me to be exceedingly reasonable. I too was never clear on why to attempt to further break down e.
This root has two stalks.
>If there was some way we could turn every operation into a small n value search, then I think we would have our solution
This is where the (e,1) connection has to come in. Row 1 is the “one row to rule them all.”
Also for the good square record we can get:
Square record: (-x^2, 0, d, x, d-x, d+x) [where (d-x) = a^2, d+x = b^2, prod = c^2]
Goal record: (-X^2, 0, D, X, D-X, D+X) [conventionally D-X = a]
d^2 - x^2 = c^2 =d^2 - c^2 = x^2 => x^2 = (d-c)(d+c)
Moreover, when you look at these (d+c) (d-c) values, you see that
d+c = 2D
d-c = 2X
So you can say that 4DX = (d-c)(d+c) = x^2
Started to integrate the factor tree into the iterative search process.
Currently using the c and f values from the original starting position, and substituting in d from the node being evaluated.
Unfortunately, doesn't look like rm 2d(n-1)=0 is usable with this combination of values. So previous theory won't work.
However, the c=6107 pic attached does show something interesting in the very first calculation. Notice the (x+n) and u values relative to our prime solution. I'm seeing similar sized jumps on a number of other test cases, but not all.
For subsequent nodes, I'm calculating estimated factors based on the final XPN in the previous node.
I'm thinking that we need to generate a pythagorean triple with a^2+b^2=c^2 where a or b is our original number squared. Then we can use this to get:
c^2 - b^2 = a^2. We can then set our D and X vaues as c and b respectively, then we get:
D^2 - X^2 = a^2
Then we can use this to get the correct little d value = 2D-c and correct little x = 2X + c.
Thanks for sharing.
I'm trying to look at D,X,C where D^2-X^2=C^2. I've done it for a bunch of primes (mod 60) [60 = 345]. and I've generalized it into a few cases:
(D,X,C) MOD 60:
i) (1,0,c) c in {1,11,19,29,31,41,49,59}
ii) (25,24,c) c in {7,13,17,23,37,43,47,53}
iii) (25,36,c) c in {7,13,17,23,37,43,47,53}
iv)(49,0,c) c in {1,11,19,29,31,41,49,59}
So if c is in {1,11,19,29,31,41,49,59} we know that X is divisible by 60. If C is in {7,13,17,23,37,43,47,53} then we know that D-25 is divisible by 60. If we can get this D and X then we can use algebra to solve for d^2-x^2=c. We get this triple somehow, then we have a big D. Then we know that our D value can be split up into a sum of two squares because D = a^2 + b^2 and then get the squares by Fermat.
Continuing with the factor tree integration.
Pic attached is for test cases c=6107 and c=7463 showing iterative search results now satisfying rm 2d(n-1)=0 constraints for each node in the factor tree. Full output available at pastebin.com/bDv31qjY.
Main change here is that I am using node specific c, d, e and f values instead of my previous example that attempted to use the starting c and f values. For the c=6107 example, the c values for each node are 3, 39, and 6107.
So perhaps we do have valid small squares at each level of the factor tree after all.
Random side note I autismed:
+++Q=VQC+++
Same voice.
Not Q, Not VQC.
Hi VA, and everyone else, you make a good point.
I think the release and impact on integer factorisation has been anticipated.
I also see nothing that prevents anyone from progress in mathematics.
This is simply the solution of a mathematical problem after all, at heart, but in reality it is so MUCH more.
Thanks for your patience. I am waiting on new equipment at the moment.
To answer other questions, after the next big step, the RSA numbers will be solvable on the laptop I owned ten years ago, before even.
Going from a search to a calculation is an almost inconceivable step.
Think of a Rubik's cube.
Trying to solve it without a plan.
Then remembers those books on how to solve it?
You have to know which side to start with and which steps to take.
All we will end up doing with the next step is figuring out a pretty pattern with odd squares with a single unit hole in the middle.
The symmetry is beautiful in the solution.
When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.
The improvements on this method became the sieves, ending up with the general number field sieves.
The start of the general number field sieve talks about selecting a polynomial and that there is no known way to do this. This is critical. You don't have to know about all that, but in essence, what we are doing is bypassing that random step.
ty, and thanks for popping in. Hope you are well today. Had the last of leftover lamb from Easter in a sandwich the other day, and thought of Harold for a moment w/ his last supper.
The plots would look so much better as color gradients, but even just the numbers themselves show the pattern.
Also shows getting between the n, and x+n for some ranges is tighter than others.
The thought was perhaps by moving within a gradient (like along the same topological line on a mountain map, staying at the same height), maybe it would make a difference.
Most of this is spot on.
The main point to make about O(log t) is that the amount of effort in the calculation doesn't increase by ten if an extra decimal digit is added, it increase by the natural log.
The natural log of 100,000 is only two more than 10,000 as an example.
1000,000 is only two more than that (approx 13 : floor)
In reality, the process will be optimised to be AT MOST O(log t). It's basically dependent on how efficient your square root function is. Anyone got the Big Oh for the square root function?
Thank you so much, very kind.
On exactly the right lines.
The 'heart' of the problem is that breaking the problem down involves two sets of triangle solving (otherwise it is prime), with one set of triangles (say half of the eight) working on a triangular problem that is one unit longer than the other set. This is one reason why no current approach I have seen works.
Then we'll show how the grid (The End) solves/shows the solution.
I should have new equipment by the end of the week. If not, I'll get admin access to install what I need on the current borrowed equipment.
Then there is a week before my time comes under pressure, so hopefully next week is going to be big!
What about instead of a mountain… Mandelbrot zooms. Looking at it as a topological map, then each line/notation would relate to the zoom plane. Interesting, whether what I'm describing is possible or not.
Could also look at the mountain getting wider as the Compass opening, where the topological lines are the relationship:
√C=g, n=prime, N=Composite or Prime Number
N/g=d ("distance" from g)
g/n=d
I don't know if that's the "d" you nerds are using, but I'll settle on a variable eventually.
Haven't been able to follow your notation, doesn't mean it doesn't make perfect sense to you. Incorporating the Mandelbrot and fractal aspect is going to be very interesting.
The "d" we are using represent a square root. So for 64, d=8. If a number isn't a perfect square (say 65, instead of 64), then it would be 8.xxx. Instead of using some decimal square root, we say d=8, and e=1 (bring the 65 down to 64 first, a difference of 1, so e=1, giving you the closest integer d root).
So to find a root we could use newtons method which is pick a random x0 starting point, then x(n-1) is NOT x times (n-1) it is the (n-1)th x value
xn = x(n-1) + f(x(n-1))/f'(x(n-1))
Want to solve root(c) is solving the equation x^2-c for roots and the derivative of this is 2x.
xn = x(n-1) + (x(n-1)^2 - c)/(2x(n-1))
From stack overflow, some guy just posted "from c decimal places, you will need log(c) iterations" which seems to be the exact complexity of this.
Another way of finding roots is through the continued fractions. With these you find your best (good enough) approximation after you hit the first value 2*d or right when it starts repeating. This (I think) has an upper bound of log(n) or sqrt(n) not totally sure. The thing is this way you still need to calculate the continued fraction in the first place which would take some time so I bet you go with the first one above.
I posted some code for sqrt of big integers earlier on.
My understanding is that the complexity of finding square roots is at most O(log t) where t is the length or size of the number whose root we are finding (remember we don't mind what the remainder is as a fraction or decimal, just the integer part, once the integer part is found, the remainder is simple to find).
A note on Big Oh; what you're looking for in any process is the part that grows the fastest.
E.g.
x squared plus a billion x + a trillion
x^2 + 10^9x + 10^12 would still be dominated by x^2 at scale. From x=1 increasing to x=one million and up, the total of the equation begins to be dominated by x^2 and as x becomes a quadrillion, the other parts become less significant and so on.
One way to look at complexity or Big Oh is to think what if the size of operand (in our case right now, it is c) in the operations of the process tends towards infinity. What dominates? That's your Big Oh. For the process we are looking at, the square root function would dominate. I.e. there is nothing that would dominate MORE than that.
Hello VQC! Ah, this is why why you said "this" to the division of the (x+n) square that had the sharp lines. We end up with triangle side measurements of u and u+1 for each one of the 8Tu. Primes are often created by adding or removing 1, so our u and u+1 triangles can help generate primes??
I don't suppose it has anything to do with binary? I can see a pattern in the 2^1 and 2^2 columns (they're always 0). I don't see anything to do with symmetry though.
When I mentioned that we might have to do something with the logarithm of c in binary form, it was because you mentioned that we're meant to be looking for binary patterns. If the only reason it isn't O(1) is because of square roots, where should we be looking in binary?
Oh shiiiiiit I just noticed that it's following the same pattern as the triangle numbers from here >>5263 with the length of the 0s being 2, 3, 2, 4, 2, 3, 2, 5, 2, 3, 2. It's the same pattern but with a different difference each time. Here it's always a difference of 1. In the triangle numbers, it's always the next triangle number (so ^#2, ^#3, ^#2, ^#4 so on). "Coincidences", am I right? If that's not what you wanted us to find, that's still pretty crazy.
>The highest knowledge man can attain is the yearning for peace, for the union of his will with an infinite will, his human will with God's will.
https://www.youtube.com/watch?v=NoRjwZomUK0
AUTIZUUUUUUUUUUM!!!!
Extreme Compass hits first: 47 * 3301
Middle Compass his first: 2251 * 3301
In waaaaay less time than doing every prime from 2 to 2251
This was all by hand and covfefe.
Here's more binary stuff. I haven't looked for patterns in any of this yet but I thought I'd put it here in case anyone else wanted to. This is all for odd x+n squares with even ns. Small triangle and big triangle are the two triangle numbers you add together to get the x+n square. I didn't do the eight triangles since they don't seem to usually be whole numbers. I would have looked for patterns before posting but I really should be spending time on other things. If anyone wants anything turned to binary and they can't be bothered doing it themselves, let me know and I'll get to it when I can.
>patterns
I guess I meant correlations. Things that appear in the binary of two or more of these things that link them together in weird ways to do with powers of 2 that we weren't already aware of.
I apparently forgot 43, which gives you 3608.07
But that's beside the point.
47x3301
Extreme: {2}<p and P>(C/2)
47 is the 15th sequential prime
~vs~
*Middle: p>(√C)<P *
106th available prime from √C
2251x3301
Extreme: {2}<p and P>(C/2)
2251 is the 335th sequential prime
~vs~
*Middle: p>(√C)<P *
84th available prime from √C
So for the larger semiprime, C,
I savedja 'bout 251 steps.
Hey C-hris!
Is it just a "coincidence" that they talk about "C" when talking about Ologs?
https://en.wikipedia.org/wiki/Olog
Hey Topol,
Here are the grid records for your spreadsheets. c, prime solution, P, and p.
c = 155147
(698,77181,197) = {698:77181:393:392:1:155147} = 155147; f=89; (x+n)=77573; u=38786; (d+n)=77574
(698,1281,174) = {698:1281:393:346:47:3301} = 155147; f=89; (x+n)=1627; u=813; (d+n)=1674
(52,1594,29) = {52:1594:57:56:1:3301} = 3301; f=63; (x+n)=1650; (d+n)=1651
(11,18,3) = {11:18:6:5:1:47} = 47; f=2; (x+n)=23; u=11; (d+n)=24
c = 7430551
(4926,3712551,1363) = {4926:3712551:2725:2724:1:7430551} = 7430551; f=525; (x+n)=3715275; u=1857637; (d+n)=3715276
(4926,51,238) = {4926:51:2725:474:2251:3301} = 7430551; f=525; (x+n)=525; u=262; (d+n)=2776
(52,1594,29) = {52:1594:57:56:1:3301} = 3301; f=63; (x+n)=1650; (d+n)=1651
(42,1079,24) = {42:1079:47:46:1:2251} = 2251; f=53; (x+n)=1125; u=562; (d+n)=1126
AND, I just noticed something quite remarkable.
Take the various (x+n) values from the c=7430551 as an example.
a=2251 (x+n)=1125
b=3301 (x+n)=1650
(x+n) diff = 1650 - 1125 = 525
(x+n) from the prime solution record = 525!!!!!!
Holy cow:
here's c = 6107
(23,2976,39) = {23:2976:78:77:1:6107} = 6107; f=134; (x+n)=3053; u=1526; (d+n)=3054
(23,36,24) = {23:36:78:47:31:197} = 6107; f=134; (x+n)=83; u=41; (d+n)=114
(1,85,7) = {1:85:14:13:1:197} = 197; f=28; (x+n)=98; (d+n)=99
(6,11,3) = {6:11:5:4:1:31} = 31; f=5; (x+n)=15; u=7; (d+n)=16
a=31 (x+n) = 15
b=197 (x+n) = 98
(x+n) diff = 98 - 15 = 83
(x+n) from the prime solution record = 83!!!!!!
>The 'heart' of the problem is that breaking the problem down involves two sets of triangle solving (otherwise it is prime), with one set of triangles (say half of the eight) working on a triangular problem that is one unit longer than the other set.
Is this the 2 triangle iteration?!?!?!?!?!?!
So one of the prime numbers has an odd (x+n), the other prime number has an even (x+n), but an odd (d+n). We can iterative search on both records at the same time. And the difference between the 2 small squares is our prime solution!
Only if you already know a and b, heh heh.
Trying something different with the split triangles.
Attached pics are for c=6107 and c=7463 and show an attempt to improve the iterative search process based on VQC's split triangle hint.
Previous number of iterations for these examples were 23 and 83 for c=6107 and c=7463 respectively.
As you can see, these spreadsheets match the prime solution XPN in just 2 calculations.
The revised formula for testing is:
1 + 4T(u1) + 4T(u2) + 2*mod - (rm 2d(n-1))
Where u1 and u2 are each based on different factors of (f-2) div 40 chunks.
The efficiency comes in from the "step factor" and "u2 factor offset" variables. I arrived at these by trial and error. But it appears that the "step factor" variable may be somehow related to the factor tree d value.
u1 = step * "step factor" * (f-2 div 40 chunks).
u2 = u1 - "u2 factor offset"
I haven't gotten this to work for small values of n or yet figured out larger test cases.
But it is certainly interesting enough to share at this point.
One more example for c=14904371.
Previous iterative search solved this in 2166 steps.
For this example, the u2 formula needed to be adjusted slightly to:
u2 = (u1 factor - "u2 factor offset") * (f-2 div 40) - 1
previous examples used the formula:
u2 = (u1 factor - "u2 factor offset") * (f-2 div 40)
The "step factor" column is pretty irrelevant and just indicates the total number of iterations required to get to the correct solution. So in this example, the total number of iterations would be 60 instead of 2166. Still a pretty good performance improvement.
The main issues with this approach are:
1) determining where that "u2 factor offset" value comes from.
2) when the u2 needs the -1 additional offset.
3) How to make this approach work where the small n values are lower than the first f-2 div 40 triangle.
Hello PMA! This two triangle idea has my brain spinning in overdrive. What if the x+n square can only fit two triangles and the remainder, where one triangle is 1 unit smaller than the larger one? It’s Topol’s “scissors” idea on steroids. The two equations must balance. This also gives us more algebra opportunities to create a new equation…
nn-1 + 2d(n-1) + f = 8Tu + mod etc.
I’ll post later with correct formulas, out and about at the moment. Thoughts, Anons?
Alright sooooooo….
Using my method… if the two compasses cross before a hit, if they don't hit at the same time… then I thiiiiink the number is prime.
I did a test by hand because I'm not a programmer and even if I was following along with that side of it and up to as speed as I could be, I don't know how to program in a list of every known prime as a database/The Grid to reference from, let alone tell it how to skip to the next available prime if you don't get a hit.
So for randomly generated 59357 to start with, I did my method. Didn't find a factor.
Oddly, the median point seemed to be 113.
113 is the 30th sequential prime.
30 steps in and you get to 525.2… * 113
From there, anything below that, the "Opening Compass" would have already been tested by the Extreme Compass… so you can stop looking. It's prime.
So, there's some "mid-point" floating between the two methods that if you pass it, you're dealing with a prime number. That's my current hypothesis.
Pics are sequential, plus the original drawing so you can see what I mean without scrolling up.
I have them coming to a 2 points on the line-of-primes…. for any number that isn't prime, wherever the 2 compasses "collide" is the last possible position of where the factors are if neither compass "Hit" already.
The compasses would have collided at 30 "steps". but they didn't >, they X'ed.
Continuing with work on a revised iterative approach from 2 triangles.
Pics attached are for c=6107 and c=209785063.
These examples show multiple pathways to a solution using different combinations of factors and offsets.
The two triangles are calculated as follows:
u1 = "u1 factor" * (f-2 div 40)
u2 = ("u1 factor" - "u2 factor offset") * (f-2 div 40) - "u2 offset"
I'm not quite sure yet how to pick appropriate "u2 factor offset" or "u2 offset" values. These tests just run through multiple combinations and show only prime solution matches. (u2 factor offsets from 1 to 100, and u2 offsets from 0 to 5)
The "u1 factor" column + 1 is the number of iterations required for the "u2 factor offset" and "u2 offset" values.
For c=209785063, my current best case scenario is 289 iterations, compared to 3705 in the previous iterative search solution.
For c=6107, there are a number of cases where even negative u2 values find the correct solution. This may equate to the -x records we previously explored in the grid. Best scenario here is 1 calculation. (cool)
Haven't gotten this to work for all examples yet.
Hmmm….
I'd have to "run" another test but…
113 is the first prime right below (√59357)/2
Which is where the search would have run into each other…
Sooooo…. if someone can program whatever method I'm using, this'd go way faster lol
I'm guessing I'm not the only one who didn't expect anything resembling an "output" to come outta me.
I think the reduction in calcs will come from iterating the two triangles against each other. Solution can be found by iterating with the +1 and (x+n) remainder. Triangle numbers scale quickly because +1 on the base adds (new base) +1 to the other side.
There’s also anumber tree path, and a path in the grid.
We still haven’t discovered this beautiful underlying pattern. We will tho!
Long time lurker, not a math anon but have been following VQC general concepts, purpose and progress since Chris' initial posts.
I remember back in the beginning, there were many discussions about purpose, application & causes of concern once the solution would be discovered. Among several theories proposed by anons of what could be done with VQC, I believe Chris mentioned one application that always stuck in my mind: that one could use VQC to compare complex original data with altered versions, specifically pictures & video. That it could be used to prove/disprove whether a certain video was authentic. At that time, the hint/theory was that it could be used to prove/disprove authenticity of ISIS videos and other "suspect" videos. But now, months later, in the light of many Q drops since, we are told the HRC video drop will be coming soon (5:5 could also mean 5/5) which correlates now to Chris' reappearance and hints this week that
a) there's a reason he's come back now and
b) he/VQC anons have a few weeks to wrap this all up?
Future proves past. If this is what I think it is, and Chris was sent here by Q Team to share VQC formula with autists via the same socratic method Q is using to expose everything else to other anons, then a timely solve of VQC would be mission critical.
I have no idea if I'm right but this has been bugging me for days and I can't get it out of my head, feel strongly I am being "told" to post this. Haven't followed VQC posts much over the past month or so due to lots of personal business that needed my attn so if this has already been suggested/posted then my apologies.
Carry on VQC anons - your genius & persistence have been an inspiration to me for months. I pray you find your solution at the right time and I thank all of you for your dedication and hard work these last several months. You are all heroes in my book!
Nice digits, and great post, lurker Anon!
>Future proves past. If this is what I think it is, and Chris was sent here by Q Team to share VQC formula with autists via the same socratic method Q is using to expose everything else to other anons, then a timely solve of VQC would be mission critical.
I agree. This is tied into Q’s objectives, and Chris mentioned that the impact of solving this has been taken into account… >>5481
Further on "Is this Prime?"
If it's prime, it'll cross/meet at a midpoint without hitting any co-prime or composite pairing.
It has to do with the Available Prime steps and the Sequential Prime steps.
Still looking for something concrete that can relate the 2 triangles together.
Attached pics are for c=115 and show various iterative results and a few diagrams of how the different sized triangles may come together around the solution small square.
Included in the diagram are the grid entries for the starting c, solution p, and prime a and b records, together with some observations about how the squares relate to each other.
Hello PMA! I'm thinking on this from VQC:
>All we will end up doing with the next step is figuring out a pretty pattern with odd squares with a single unit hole in the middle.
>The symmetry is beautiful in the solution.
>When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.
Last hint is pretty cool. Let's tie together triangle numbers and square numbers with primes. Can we create a method of generating primes within the x+n square?
We know that f is our starting point for n0. We know that u and u+1 play a role, bc VQC hinted about solving a triangle problem where one of the triangles is one unit larger. Can primes be generated as the difference between these two triangles and the mod/remainder? Thoughts, anons?
You're welcome here anytime, even if you don't have much to contribute. The thing about timing would make sense, but I would be surprised. He hasn't to my knowledge ever hinted at having personally known the Q team themselves or having talked to them (despite obviously knowing a great deal about the things Q mentions), and no comms are private. I don't know though, obviously.
>b) he/VQC anons have a few weeks to wrap this all up?
I think he was meaning he'll be busy the week after next. I mean, if it is meant to be done soon, that's pretty exciting. We've all been waiting here for nearly half a year. I hope you're right.
>When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.
That implies that there are several kinds of symmetry we might be looking at and that one of them can only exist in one way. So what ways can you think of that a square made of 4 asymmetrical rectangles with a one-unit hole in the middle can be symmetrical?
Thank you.
fyi, today I decided to search "spot fake video" and coincidentally there's breaking news that just came out in the last 24 hrs…
https:// nation.com.pk/13-Apr-2018/google-engineer-developing-tool-to-spot-fake-video
https:// phys.org/news/2018-04-lying-eyes-google-tool-fake.html
https:// www.businesstimes.com.sg/technology/google-engineer-developing-tool-to-make-and-spot-fake-videos
I'm just leaving this here in anticipation.
Hah. It's gotta be connected to that idiot Dnimeerf and his Ray Kurzweil Cronies. Kurzweil is a brilliant mind who has given up hope for humanity and is now in service to EVIL. He's like 'Singularity Now!! Merge with the Machines. Kill the humans!" No doubt Kurzweil Cronies or Googlefags are lurking here too.
The shifting by 2n has too much similarity to quadratic equations: (a,b,c) for ax^2 + bxy + cy^2. d is probably the determinant.
Thanks for waiting.
Pick up new equipment tomorrow after 3pm UK time.
Will be here everyday for seven days after that as well as tomorrow.
Thanks again for waiting.
Nice digits.
What we will move onto quickly later is using the Mandelbrot Set on both spatial and temporal transforms for signal input.
That will make your head spin.
Someone takes an illegal picture? Rotate the view.
Want to see what happened in previous 'frame'? Etc.
A perfect resonator like the application of the Mandelbrot Set on signal data is remarkable in a (growing) array of solutions. I'm surprised it is not in the public domain yet.
derped. lol now with trip
Ready!
All these 555's are making me frisky. I won't lewd the ponies tho, Topol. I'll take a Masonic oath if you want ;)
>"I swear by my life and my love of it, that I will never live for another man's sake, nor ask another man to live for mine. Furthermore, I will never lewd a pony, nor ask a pony to lewd herself before me. Should I break this Oath, may the Compass and Scissors cut my tongue out, and may I be buried in the sand at low tide for Pelicans to peck my desiccated carcass, and may my soul burn in Hades with all the nasty Pedos and members of the CF."
This I solemnly swear. May Topol, keeper of ponies and waifus hold me to this Oath.
HOLY SHIT 5555. You deserve it, man. Thanks for all your love and inspiration on this Quest. Can I Get An AMEN for TOPOL's epic Quad 5 Get? Topol, you have kept us inspired on many a despondent day. Including my favorite Dash Meme of all time which you posted on the fly back in RSA #4, right before Teach showed up. Where is that smarty pants faggot, anyhow? Miss him. Love you Topol, xoxoxoxo.
A simple thanks would have sufficed. Topol has a hard time accepting compliments? Please, no false humility here, faggot. You got skillz.
Shoulda worked in (P) and (p) to the Oath…
(P)elicans and (p)iranhas
(P)irates and (p)ussies
(P)imps and (p)rosties
(P)residents and (p)risoners
(P) = (np) ?
?
>>5559
Got it
>>5559
Roger that.
Recap and exercise.
Starting with odd squares.
(x+n) mod 2 = 1
Starting with the smallest odd squares.
(x+n) = 1,3,5,..,
For each odd square what are the permutations of possible values of f,d and (n-1)?
What are the patterns?
What do these look like?
VQC I was thinking could we do something with the equation x^2 + e as a function of x. Also I learned in Number Theory about quadratic forms, which are this:
ax^2 + bxy + cy^2 and you can denote them like this (a,b,c). This can be used to represent integers
(3,1,2) would be equivalent to (2,1,3).
In addition, another odd equivalence is (2,0,3) and (2,4,5). A way to generate these equivalent forms is through a unimodular substitution such as x = pX+qY and y=rX+sY. Even though these may generate different equations they are equivalent statements. But due to some stuff, you need ps-rq =+/- 1. Therefore different forms can be used to represent the same integers.
Def: Integer is properly represented by (a,b,c) if n=ax^2 + bxy + cy^2 where gcd(x,y)=1.
Def: The Discriminant d = b^2-4ac (also what I suspect d to be). If two forms are equivalent they have the same determinant (doesn't go the other way). If two forms have the same discriminant but aren't equivalent then they are disjoint. Also, discriminants must be congruent to 0 or 1 (mod 4).
If d<0 it is called definite. If definite, a>0 represents positive integers, a<0 represents negative integers. d>0 is definite, sometimes positive sometimes negative.
Given a form (a,b,c) we want to know if we can represent n by the form, where gcd(x,y) = 1.
Method: Supposes (a,b,c) represents
n = ap^2 + bpr + cr^2 gcd(p,r)=1
Want to find q,s such that ps-qr = 1 (easy to do euclids algorithm for this)
Then substitute x=pX+qY, y=rX+sY
then, you have
AX^2 + BXY + CY^2 = nX^2 + BXY + CY^2
also -A = ap^2 + bpr + cr^2 = n
So, if (a,b,c) properly represents n, then there is a form (n,h,L) (h,L not known) that also is equivalent. Also, if (a,b,c) represents n properly, then since discriminants are the same you know b^2-4ac = h^2-4nL. Also h^2 is congruent to d (mod 4n ) [n is absolute value if it was negative]. If we pick h, then L is fixed. We can test the congruence to see if n is or is not represented by this. Also, due to the congruence, we can assume that 0<h<2n. Watch this now. This is where I think it connects to the grid.
nx^2 + hxy + Ly^2
x = X + Y , y = Y
=n(X+Y)^2 + h(X+Y)(Y) + L(Y)^2
=nX^2 + 2nXY + hXY + hY^2 + lY^2
=nX^2 + XY(h+2n) + (h+L)Y^2
Now look at the middle h term, which we were picking. We can just add 2n to this to generate a new form. Teacher said to shift it until 0 < h < 2n. Remember, if we add 2n to the middle term, then since we have the same discriminant the next L would be different, but its easily calculable.
So to try and connect it here, the equation we would be solving would be (1,0,-1) for x=d+n, y=x+n because that would be a difference of two squares. Moreover, it could be (0,1,0) for x = a, y = b. My thinking is that since we can shift by 2n in this equation, then if we are aligning it with the grid, the middle term h would be the e value for (e,n,d,x,a,b) because that is the one you would be shifting by 2n.
>>5567
a is odd, so if x is even, then d is odd. If x is odd, then d is even.
(x+n) = 2j+1
(n-1) = 2j-x
Suppose x is even, this means that d is odd and n is odd also. Then, (d+n) can take values 2k so
2k = (d+n) = (d+ 2j-x + 1)
= a + 2j-1
= a + (x+n)-2
(d+n) = 2(k-1) = a + (x+n)
Suppose x is odd, then d is even, and n is even. Then, (d+n) takes values 2k also actually the same thing, but I think this is valuable
(d+n) = 2(k-1) = a + (x+n)
Attempt like 7 on this
Suppose
x = 2j+1,
(x+n) odd -n even
n = 2k
n-1 = 2k-1
x+n = 2(j+k) + 1
(d+n) even and n even =d even
d = 2p
f = 2d+1 = 2(2p) + 1
f = 4p+1
d - x = a -2p - 2j - 1 = a
a = 2(p-j) - 1
b = a + 2x + 2n
b = 2(p-j) - 1 + 2(2j+1) + 2(2k)
b = 2p - 2j - 1 + 4j + 2 + 4k
b = 4k + 2p + 2j + 1
Suppose x = 2j
(x+n) odd -n odd
n = 2k+1
n-1 = 2k
(x+n) = 2(j+k) + 1
(d+n) even n odd =d odd
d = 2p+1
f = 2(2p+1) + 1
f = 4p+3
a = d-x = 2p+1 - 2j = 2(p-j) + 1
b = a + 2x + 2n = 2(p-j) + 1 + 2(2j) + 2(2k+1)
b = 2p + 2j + 4k + 3
You forgot f = 2*d + 1 - e. So the equations for f aren't technically correct.
Just did some naive checking against some numbers. I haven't looked into any permutations or patterns, just parity:
Assume x = odd, n = even:
(a = 7, b = 21)
f = even
d = even
n - 1 = odd
x = even, n = odd:
(a = 31, b = 85)
f = odd
d = odd
n - 1 = even
Woah. If you think it's completely serious and you think continuing will get you and others killed, but you still want to continue, probably obvious logic would suggest that if you continue to dump crumbs and something happens to you, we'll be stuck for the next however many years. You might have to just dump the whole thing immediately. Obviously I don't know the details of your situation though.
I guess I'm a bit late to this but here's a visual thing for (x+n) = 1. (x+n) = 1 with primes whenever b = a + 2.
Good to see you VQC! Whoa, death threats are not good. (((They're scared.))) Q team should send you a security detail! If you can't post anything else at the moment, we'll work on your Recap and exercise.
Hey Lads, cell (2,1) has these exact ascending (x+n) examples to work through. They're also in(4,1) (6,1) etc as var e increases. I can see that as e ascends by 2's, d ascends by 1. Grid snapshot attached. Cool patterns everywhere. Let's delve into these (x+n) patterns.
(2,1,1) x+n = 1
(2,1,2) x+n = 3
(2,1,3) x+n = 5
(2,1,4) x+n = 7
etc.
VQC's questions:
>For each odd square what are the permutations of possible values of f,d and (n-1)?
>What are the patterns?
>What do these look like?
I also attached the graphic VQC was referencing, and my working spreadsheet for the exercise. Working up triangle diagrams now. I've got famalam stuff today, and I'll stop in as often as possible.
Question: For the example I've posted, is (n-1) just 0 for every example? Or is (n-1) a reference to something else I'm missing or not understanding yet?
From my current understanding, the "primes" that make up an RSA number are "randomly generated" and not even necessarily coprimes.
so these "randomly generated primes"… could they be controlled?
Couldn't we just "best guesstimate" the primes using the same methods between C/2 and 2?
The lower prime and upper prime will be "equidistant" from √C no matter what soooo
if you could chart it out on a line and have the scissors run until they hit 2 whole numbers which are "equidstant" from √C
Boooound by the lower region search… hmmmm…
That would also mean that each side is a fixed 50% of available numbers
{2, √C}=50%
{√C, (C/2)}=50%
one has more numbers than the other… so I wonder…
if {2, √C}=50%=P
{√C, (C/2)}=50%=NP
Then….
d/P=g
NP/d=g
And then P=NP, cuz I say so.
So basically, the upper set will have an equal amount of "available" primes to the lower set's sequential primes.
And if you're doing this "graphically", just reflect the lower prime range into the upper and just run it with a constrained g value until both lines point at whole numbers with no remainders/decimals.
So how to program only to look for those whole numbers… those only two "of course it's obviously these numbers" numbers… Cuz doing my method by hand, I eventually just started notating it as ###.x because it didn't matter WHAT the decimals were… just that they were there.
Whoa. Another cool reminder of why we can solve for all of row one using SQRT(f). Check it out. (x+n) = SQRT(f) = SQRT(2d+1-e) for row one.
Oops, here's the screenshot.
Pic attached is a summary of relevant odd x+n squares showing parities and values. Additional odd x+n records from e=1 to 24, n=1 to 10, and t=1 to 10 can be downloaded via pastebin.com/XJagiXg0.
Interesting that n-1 parity always matches e parity.
Also, noticed some weird behaviour in d parities for e values at e mod 8 - 1. (i.e. e=7, 15, 23, etc.)
Sample of this odd x+n data is posted at pastebin.com/AvSzBeFQ for e=7, n=1 to 50, and t=1 to 50.
Specifically, where n mod 4 = 0, the odd and even parities cycle, whereas other n values they look to always be odd.
For each (x+n) = 1, 3, 5, etc, there is an infinite set of possible {e,n,d,x,a,b}s. In by far the majority of these cases (i.e. there are a finite number that don’t follow this and an infinite number that do), n = 1. I found this through outputting {e,n,d,x,a,b} when x+n = several different incremental odd numbers. I found something pretty crazy so I hope someone reads through this.
There are some obvious patterns that I noticed and I don’t think I saw anyone else notice, so I’ll go over them, but first, here are the outputs for the first 5 odd numbers 1, 3, 5, 7 and 9. As you can probably see, for each odd (x+n), you can map out every possible value just by incrementing a and b. So although it’s a confusing infinite set, all you need to do is print out the stuff that goes in the grid to see it. Every incrementing a value has infinite odd (x+n) values based on incrementing b by 2 (i.e. a=1 has infinite odd (x+n)s, as do a=2, a=3 etc, and for a=1, one of those odd (x+n)s comes from b=3, another comes from b=5, b=7 etc).
So firstly we need to think about what x and n could possibly equal for (x+n) to be odd. n will only ever be 0 if c is a square (i.e. n = ((a+b)/2)-d = ((4+4)/2)-4 = 0). If that’s the case, x = 0 too (i.e. x = d-a = 4-4 = 0). So if (x+n) is odd, the lowest possible n value will be 1. In the case of (x+n) = 1 (the first picture), that means the only possible way to make (x+n) is with n=1 and x=0. For every other (x+n) value, e.g. (x+n)=5, there are more possibilities. (x+n)=5 when x=0 and n=5, x=1 and n=4, x=2 and n=3, x=3 and n=2, and x=4 and n=1. So if we were to try to find mathematical patterns, we would have to do so for every different possible combination of x and n values, right?
Well, as you’ll see in the pictures for every (x+n) above 1, there are a finite number of cells for which n=/=1, followed by an infinite set of cells for which n=1. I’m going to focus on the infinite sets for now in my explanation.
The main points of comparison are f, d and (n-1), as VQC said. Within the infinite sets where n=1, f will always equal (x+n)^2 and (n-1) will equal 0. This will be true for all odd values of (x+n) where n=1 (which is an infinite set). In the case of nn+2d(n-1)+f-1, the value of d is irrelevant, because 2d(1-1) is 0. Also, nn-1 will be 0. That means, for the infinite set for which (x+n) equals an odd number and n=1 (which will begin after a certain increment of a and b), (x+n)(x+n) = f. That means we can very easily calculate x and n, since, if f=(x+n)(x+n) and n=1, x will be sqrt(f)-1. This is true for an infinite set of numbers. Either I’m somehow wrong (I don’t see how I could be) or this has been staring us in the face the whole time.
Now, the obvious flaw in this finding is that it doesn’t represent the values of (x+n) where n=/=1. For each odd (x+n), this is a finite set, so we should be able to play around with them and find a pattern. It does look very noisy though (if you look at f in the first several outputs for 3, 5, 7 and 9, you'll see it doesn't seem very obvious how to straight calculate it). I only just put this all together, so I haven't messed with it myself. Could anyone double check my working? Because it seems too easy for (x+n)(x+n) to equal f for the vast majority of cases for it to be entirely true or useful.
Excellent approach. Looking into this further.
Initial observation is that perhaps there is an upper bound on n values for each possible x+n combination, which would enable us to make much larger jumps during iteration.
Is there a formula to calculate that maximum (n-1) from an (x+n)?
I don't know about a maximum possible (n-1) value but I have been working on figuring out where the finite set where n=/=1 ends and the infinite set where n=1 begins. Each universal set begins with the finite set, which begins with a=1, and a just increments by 1 the whole time. That means the length of the finite set is also the highest a value for which n=/=1. The finite sets end like this:
>(x+n)=3, finite set length is 1
>(x+n)=5, finite set length is 7
>(x+n)=7, finite set length is 17
>(x+n)=9, finite set length is 31
>(x+n)=11, finite set length is 49
This is going to look a bit convoluted, and I'll try to figure out how to explain myself, but you can calculate the highest a for n=/=1 / the length of the finite set from (x+n) as follows:
((((x+n)-1)/2)-1)*2 + 4(the ((((x+n)-1)/2)-1)th triangle number) + 1
Here are some examples so it maybe doesn't look so complex.
Say (x+n) = 5
(((5-1)/2)-1)*2 + 4(the (((5-1)/2)-1)th triangle number) + 1 = 2 + 4 + 1 = 7
Say (x+n) = 7
(((7-1)/2)-1)*2 + 4(the (((7-1)/2)-1)th triangle number) + 1 = 4 + 12 + 1 = 17
Say (x+n) = 9
(((9-1)/2)-1)*2 + 4(the (((9-1)/2)-1)th triangle number) + 1 = 6 + 24 + 1 = 31
So if you have a particular odd (x+n) value, if you plug it into that formula, all of the infinite a values above that point that creates your given odd (x+n) value will have n=1 and f=(x+n)(x+n).
I did try to see if there was an obvious way to calculate the maximum n value for a given (x+n) based on each of these universal sets beginning with a=1. b in every case for a given odd (x+n) is b=a+2(x+n). n is on the right end of this spreadsheet. It's obviously incrementing, but it isn't linear. Any ideas?
Here it is updated with the maximum n value's corresponding x value (which is obviously going to be the minimum x since it has to add with the maximum n to make (x+n)). It's incrementing by 1 in groups of two. Each group is one number long (i.e. 4, 4, 4, 5, 5, 5), and the next is one longer than that (i.e. 6, 6, 6, 6, 7, 7, 7, 7). That's the only obvious pattern I can see.
I just noticed d also has that pattern but it's 1 more than x since a is always 1 when n is max and x is min. Not that it changes anything.
Not sure, it was quite subtle.
Life Insurance offer that could have been spam but knew my details to a point too far and talked specifically about my family in a pointed way.
Let me put some more measures in place.
There is plenty to work on to come up to speed in the exercise as it is VERY relevant for all the subsequent pieces.
Brilliant.
If you colo(u)r the values of f in the grids that show the triangles, patterns will be added, ESPECIALLY if you divide f up into 8. The remainder of f when divided by 8 is a key to what values of d and (n-1) can be used.
Playing with these patterns will organically lead you exactly where we will be going.
Colored by F mod 8. Hmmmmm.
This is a great explanation, AA.
>The main points of comparison are f, d and (n-1), as VQC said. Within the infinite sets where n=1, f will always equal (x+n)^2 and (n-1) will equal 0. This will be true for all odd values of (x+n) where n=1 (which is an infinite set). In the case of nn+2d(n-1)+f-1, the value of d is irrelevant, because 2d(1-1) is 0. Also, nn-1 will be 0. That means, for the infinite set for which (x+n) equals an odd number and n=1 (which will begin after a certain increment of a and b), (x+n)(x+n) = f. That means we can very easily calculate x and n, since, if f=(x+n)(x+n) and n=1, x will be sqrt(f)-1. This is true for an infinite set of numbers. Either I’m somehow wrong (I don’t see how I could be) or this has been staring us in the face the whole time.
SNACKY TREEEEEEATS!
>>5602
Actually this
>If f%8 = 0, then:
(n-1) is an even number.
d is even
>If f%8 = 1, then:
(n-1) is divisible by 4.
d is odd
>If f%8 = 2, then:
(n-1) is an odd number.
d is even
>If f%8 = 3, then:
(n-1) is odd
(n-1) is prime (or 9 ?)
d is odd
>If f%8 = 4, then:
(n-1) is even
d is even
>If f%8 = 5, then:
(n-1) % 4 = 2
d is odd
>If f%8 = 6, then:
(n-1) is odd
d is even
>If f%8 = 7, then:
(n-1) is odd
d is odd
Pics attached are small odd x+n tests that cover the possible f mod 8 remainders of 0, 1, 2, 4, 5, 6. My tests indicate that remainders of 3 and 7 aren't possible for odd x+n, but someone should please verify.
These tests are grouped by x+n and all sorted by x+n ascending, n-1 ascending, and d descending.
The reason for the sorting is that the largest n-1 value always coincides with the smallest d value. And the smallest n-1 value coincides with the largest d value.
And so determining that range could be extremely beneficial performance wise, but I'd settle for just finding the top n-1 value!
Hey Lads! I think we've been banging our heads on the wall when VQC gave us everything we needed. Check this out.
-
We use f for two things: first, 8Tu + remainder gives us our starting n0. This is column G on my spreadsheet.
-
We also use f to begin filling the (x+n) square. For the first example (2,3,1) all we have is f mod 8 = 5. So we begin by filling the square with 5 units.
-
Now, we iterate n0 starting at n=1. We plug this into nn+2d(n-1)+f-1.
-
at each iteration, we check to see if our answer is a perfect square. If yes, we have reached the correct n. If there is a remainder, we are not at the correct n0.
-
So f serves two purposes, helping us find a starting n0 to iterate, and also to begin filling our (x+n) square, looking for a perfect match.
This is pretty much what PMA has been getting at about how to use rm 2d(n-1). I finally get it! Hopefully this is clear and helpful to other anons as well.
Another cool advantage of this method is that we solve for the correct n and the correct x at the same time! That will save some calc time.
When there is no remainder, we know we have the correct n. Then we calc SQRT((x+n)^2) - n = x
For example (2,3,3) this is:
SQRT(121)- 3 = x = 8
Pics attached are f mod 8 summary tables for odd x+n up to 99, showing min, max, spread, spread div 8, spread mod 8 for n-1 and d values. Last column shows all possible n-1 values for each x+n.
Only including where f mod 8 is 0 and 1, as it appears that f mod 8 behavior can be grouped as follows:
Odd n-1
f mod 8 = 0
f mod 8 = 2
f mod 8 = 4
f mod 8 = 6
Even n-1
f mod 8 = 1
f mod 8 = 5
The n-1 values between mod values in these groups aren't exactly the same, but they may be similar enough.
If there are 6 days left to finish this stage (per VQC's post >>5564 here), it might pay to get this conversation out of the way: none of us are going to be dicks and go on Infowars or whatever are we? There were some people who did that back when /cbts/ was still on 4chan.
I suppose I should be using this trip more often.
>THEY’RE TURNING OUR FRICKIN’ COPRIMES GAY
Fortunately there’s no shortage of number theory whackjobs on the Internet, so until we factor a RSA challenge number, it’s probably not a huge concern. Fuck if somebody did manage to get on infowars it would probably be ne of the top two funniest things ever. Posting w/ trip for posterity.
Then don’t let the doorknob hit ‘ya where the good Lord split ‘ya. Best of luck, but don’t expect replies.
Just throwing this out there:
If anyone's going on on InfoWars, it should be me, cuz you all sound too normal.
Just switch out "InfoWars" for "Louder with Crowder" since Jones might legitimately get turned on by the Ponies, but Not Gay Jared would potentially dress up as a couple of them.
In relation to what? Chris needing a dead man's switch or (x+n) = 1?
My point was that I hoped we didn't become famefags and didn't go on Infowars or whoever might ask us, you know? Do you not remember those people from /cbts/ who started acting like spokesmen for Q and talking to all the e-celebs? If any of those places wanted to interview us or anything all they would have to do is post here. I'm not that worried about this if we absolutely have to and we all agree about who it is, but I'd personally prefer just talking to them through here in text form. I have to admit, it would be pretty entertaining if you talked to them. You could talk about Padovan triangles and Dan Winter golden ratio fractal shit and mess with their heads. I have a pretty specific accent that would narrow down my location quite a bit, so I don't know if it would be a good idea for me to do it, but then again that might even be an advantage since I'm far away from most of the rich and powerful people who are going to be affected the most/pissed off the most. Or at least an annoying distance away.
That seems to be a pretty good explanation of what PMA has been doing. I'm not 100% sure why you're asking me though. As much as I have been following along as best I can, and I have been understanding the results of the tests each time, they have always been a tiny bit convoluted, so I honestly don't understand 100% of what PMA is doing. Either way, thanks for the helpful post VA.
Heya! If you're here for THAT, got to the EZ Bake.
Becaaaaause,
VQC+++ is not our VQC
It's the same guy who was going around as:
+++
Q
Also, didn't we all discuss at some point, since I'm already doxxed, that I'll happily be the face of this when it goes mainstream? I want Tucker over Hannity, but I'll take Hannity if he's all I can get.
Sure, Anon! More diagrams to follow after work today. Attaching VQC's diagram to help explain.
-
We start with c.
-
Then we get d, e, and f using known formulas.
-
Then we use f for two tasks related to constructing and searching for the correct (x+n) square.
-
First, we use f to estimate n0. This "Starting n" will allow us to iterate through possible values of n to look for a match.
-
Second, we use f to begin filling the (x+n) square. We don't yet know how big the square is. The formula for the Area of the (x+n) square is nn+2d(n-1)+f-1, so we plug in our full f value.
-
Now we begin to iterate n0. As we increase n0, we get new values for nn + 2d(n-1). f -1 remains constant in the formula. This is the bottom part of my spreadsheet.
-
We know that (x+n) is a perfect square, so each time we iterate, we check to see if we have a perfect square for the Area of (x+n)(x+n). We do this by taking the square root of our new (x+n)^2 value we've created. If it has a remainder, it's not a match. If it has no remainder, it's our match!
-
We now have the correct (x+n) and the correct n. We can also solve for x easily by taking SQRT(correct (x+n)^2) - correct n0 = x
-
We can now solve for a and b.
-
BOOM. Done. Simple and clean.
-
It works for all the small examples I've tested, but we need to test on many more examples.
-
The key idea I was missing is that f is also used to CONSTRUCT the (x+n) square. In fact the size of f fits perfectly with a multiple of nn+2d(n-1) -1.
-
VQC gave us all the pieces and formulas, but there was some assembly required. He kept hinting to build out all the (x+n) squares and play with the pieces to see how things fit together. Try it out for yourselves, Anons. When we iterate n0, the part that's increasing is nn+2d(n-1). f -1 remains constant.
-
I think we're about ready to move on to part 3b.
-
Can some of you excellent program Anons give this a full testing out? Let's run this bitch on the RSA 100 numbers.
Thankyou! My curiosity and confusion has been satiated.
Oooh! Oooh!
When I'm asked specifics on RSBN (couldn't get Hannity, let alone Tucker), I can be all cheeky and lead in with:
DAMN IT, JIM! I'M AN ARTIST, NOT A MATHEMATICIAN!!!
So that's another plus to having me be the face beyond already being doxxed. It's not like there's anything of value (math puns) that they can get out of me. :D
Here's some more examples from a new cell (7,8) I picked because it has odd (x+n) squares. The top two parts are just calculating out all variables for clarity.
The bottom part has the iteration process for the first element in (7,8).
You can see the solution at (x+n)^2 = 121 when n0=8. So (x+n) = 11
11 - 8 = x = 3
Then you can solve for a and b vars.
Not sure if relevant, but wanted to share some observations regarding the f mod 8 patterns.
Back to the c=6107 test, where f=134, and f mod 8 = 6.
Attached pics include:
1) c=6107 iterative search results based on f-2 div 40; (c6107_fm2_iterative_search)
2) valid n-1 and d values for x+n=17 where f mod 8 = 6; (c6107_x_plus_n_17)
3) x+n records where f=134 and f mod 8 = 6 up to the x+n=83 prime solution record. (c6107_f134)
The x+n=17 sample was chosen because the iterative search hits that record very early in the process.
It appears that there are a couple of ways to proceed.
One way is to figure out what the maximum n-1 value is for x+n=17, and use that to jump ahead. In this case, max n-1=5, n=6, and that would place us at least at x+n=30 as the next iteration.
The other, might be using the common f=134 value and various related records in the grid. Not quite sure what the path would be through these various records, but did notice that several chains exist between a and b values.
For example, working backwards from x+n=83:
x+n=83 a=403, b=569
x+n=69 a=265, b=403
x+n=55 a=155, b=265
x+n=41 a=73, b=155
Perhaps, notwithstanding the various e values, this shared f value across multiple records could point the way to the solution.
Good evening faggots. Our meme department is severely lacking in Star Trek memes. Working to correct this now.
>For each odd square what are the permutations of possible values of f,d and (n-1)?
I believe the following describes the valid possible values f, d, and n-1 for odd x+n. Would appreciate someone double checking this.
Attached pics show sample output for x+n=9.
Groups are defined as follows:
————-
Odd n-1
————-
f mod 8 = 0 n: even; d: odd; f: even;
f mod 8 = 2 n: even; d: even; f: even;
f mod 8 = 4 n: even; d: odd; f: even;
f mod 8 = 6 n: even; d: even; f: even;
n = n +/- 2
d = d +/- 4
f = f +/- (n-1) * 8
————-
Even n-1
————-
f mod 8 = 1 n: odd; d: alternates even/odd if (n-1) mod 4 == 0, otherwise even; f: odd;
f mod 8 = 5 n: odd; d: odd; f: odd;
n = n +/- 2 (for f mod 8 = 1)
n = n +/- 4 (for f mod 8 = 5)
d = d +/- 2
f = f +/- (n-1)/2 * 8
d and f work in opposite directions, if d increases then f decreases and vice versa.
Not sure yet what the adjustments are when moving between different n values. But the above should accurately describe how to adjust d and f values within a particular n and maintain the same (x+n)(x+n).
Things got a little more intricate in determining the n, d, and f values while doing further testing for odd x+n.
Pic attached is the relevant source code.
Based on f mod 8, n and d each have various start and step (increment) values. And in some cases, it appears that the d values can only be determined once n is known.
Likewise, I have so far only managed to define the incremental f changes in terms of n. (See the GetFStep method). The initial f value is determined once n and d are calculated via f = (x+n)(x+n) - (nn-1 + 2d(n-1)).
That being said, given any odd x+n, these formulas can be used to iterate through all valid n, d, and f combinations.
Good discussion going on on Topol's second Bake. Here it is anons.
I hope VQC hasn't suddenly gotten super busy. He hasn't been here for a few days now, and he definitely won't be here much in a week.
I hate to continue to be annoying but that conversation could have taken place here and nothing bad or inconvenient would have happened
AA, you're right. But we all (VQC Anons) wish (you) could just join the Chat. Vault 7. Wikileaks. We're all compromised at the chip and boot level. Maybe you're right, but we're fighting for the freedom of the world. It doesn't matter at this point. Join Or Die.
JFC why does every single human group endeavor inevitably become so incredibly fucking gay
I’m so deleriously tired of human beings. I mean I love you all, but this species has severe and debilitating issues that probably cannot be overcome without splitting into a new species altogether. It’s a miracle (and I mean literal miracle, as in God himself forcefully intervening) that anything at all ever gets accomplished, ever.
Sometimes you think, just for one second, that perhaps just this once this isn’t the case, some stupid human-caused derailment spontaneously appears out of the ether like matter/antimatter virtual particles in a vacuum.
I AM SO FUCKING SICK OF HUMANITY IN GENERAL REEEEREEEEEEEEEEEEEEE
Are you complaining about something we're doing?
I wouldn't have a problem with it if it was just for the sake of being social with you guys. We've spent the last 6 months working together anonymously on what's meant to be world-altering math, so it'd be cool to talk about regular shit of course. I suppose joining would also give me a chance to screenshot anything you guys haven't screenshotted and post it here if I think it needs to be here. Maybe I'll join at some point, but it still really doesn't sit right with me having a less public communication channel than this one being used to discuss what's meant to be for the sake of fully public disclosure (i.e. you just type in a url and all this information is here). Let's see what happens by the end of VQC's 8 days first (under the hope that he hasn't been murdered by someone masquerading as a life insurance emailing person, and under the hope that he also hasn't gotten busy again and he isn't having too hard of a time emotionally with the other things).
>I wouldn't have a problem with it if [things not the case]
>it still really doesn't sit right with me having a less public communication channel [that already exists in wide usage]
>VQC's 8 days
>hope that he hasn't been murdered
>a life insurance emailing person
>the hope that he also hasn't gotten busy again
>he isn't having too hard of a time emotionally
I suppose this is why phrases like "darkest hour" exist. I'm not sure humans really deserve redemption at this point, honestly. Maybe God has other plans.
Let me speak plainly: perhaps this is what niggers are always agitating about. The unspoken premise of society in general is "you can't really trust niggers", and this is for mostly good reasons. Hence, niggers should just be grateful for the drippings of fat that perchance fall into their mouths at random.
Here, we're the stupid niggers. VQC is the fat-dripper. I don't mean this in any fundamental negative way, but I'm calling out the truth as I see it. I see nothing to contradict my main thesis point. Feel free to educate me.
If VQC were acting in good faith, all he would have to do is keep to his public statements. However, it is a fact that he has not in fact kept to his public statements. Because I personally like him, and I have corresponded with him back and forth, it actually pains me that I have to actually say this. Especially since I am the "nigger" here.
The way I see it, VQC has some choices to make. If he is, in fact, Christian man du jour as claimed by himself in his Revelations analogies, avoidance of death is merely logistical and not otherwise fundamentally consequential. The Spirit's will would be the whole of the consequential. If that is in fact not the case, I for one would appreciate an update.
Also, from my nigger perspective, I see a VQC not always technically accurate– i.e., messing up f formulas, misspeaking about big-O complexity, and in general being sloppy with details to the extent that it seems to me to deliberately waste people's time. At this point I think PMA VQC in leading the charge to where this group is going. Plus PMA writes much better code than VQC.
OK, guys, let me have it. I'm sorry, I had to put this out there. The only thing you can't say about me is I haven't been here since the very beginning.
Or just call it what it is.
VQC is an A.R.G.
Tell me… what does Sonoluminescence have to do with any of this? That's your answer in a nutshell.
COME ON IN! We've been on a Jesus kick lately for whatever reason. :D But I'd rather be Christ-like than Christian, no matter how people try to spin it.
Oh man I totally understand you when all you do is quote me and not explain the relevance of any of it to your point. Of course. What the hell are you talking about?
Are you saying he's lying to us, or are you saying he's deliberately taking a long time for no reason, or is it something else? No offense, but you're very bad at explaining things, at least so far.
Pics attached are sample data for c=6107, c=14596111, and c=9874400051 showing iteration results using the formulas posted previously.
These test cases use the prime solution (x+n)(x+n) values and show the different iterations assuming f or d are treated as constants.
Interesting how holding f constant requires fewer jumps.
Wow, these results look good PMA. So You're holding f constant, and finding multiple matches for a given (x+n) square? Am I reading the results correctly? So different c and d values for each example, but f and (x+n) stay the same for the purpose of the test?
Trip change needed.
changed.
Thanks for clarifying. Interesting that holding d constant produces answers with large -e values, but then pops out the correct solution! On your second example, with such a large f value, how did you decide where to begin iterating?
God giving humans what they don't deserve, i.e mercy and forgiveness is called grace. From the beginning of the OT, God's plans were to redeem you.
https://archive.fo/mcvv4
Thinking out loud here. Could the remainder of f-2 mod 8 limit our search to certain multiples of n? Or said another way, does the mod have a connection to the correct n, or even the parity of n?
Bantz and Shitz! Nice to see you. It’s still the best LARP ever. ;)
You got banned because you started spamming it everywhere, and all of your accusations against VQC were false or had no evidence. If you don't want to get banned, don't be an asshole.
Hey everyone. Sorry I was being a complete faggot a couple of nights ago. Sometimes humans and associated human chaos drive me nuts, particularly when I've been forced to deal with a lot of people-related bullshit in my own life, as I have recently. I'm sorry famalams. I love you all, I truly do.
We're all humans here. Don't worry about it.
Have an idea about what we are searching for, but not quite sure how the math works out. Perhaps another anon can assist.
Pics attached are sample odd x+n f,d,n output for x+n=17 at f mod 8 = 0. One image shows all possible values for n=2 and n=4, the other shows the minimum d values for all possible n.
Full pastebin of these pictures for all f mod 8 combinations are available at pastebin.com/gaxUD1ex (all possible values) and pastebin.com/5FpEffS9 (min d values only).
The iterative search gives us n0 and XPN values at each step. And the common f value tells us which f mod 8 rules apply.
Per the formulas at >>5632, we can directly calculate the minimum d for each possible n.
And because there is an inverse relationship between the 2d(n-1) and f values (see the (2dnm1 + f) column) we should have everything we need to determine the first row in each n grouping of possible values.
That first row, I believe, is important because the list of available "d" values decreases as n increases. Notice how the max d for n=2 is 143, but the max d for n=4 is 43. This min d value, however, is available for every n.
The second picture shows only the possible n values using the min d=3 value.
And so when we search for a record with (x+n)=17 and f mod 8 = 0, we already know that the max n that can be used is 14. (This is the formula that needs to be determined. It could be related to the d step and/or f step values.)
Applying this logic at each step of the iterative search should provide a bit of a performance improvement.
No worries, Anon! I was posting dumb shit about "Join Or Die" after drinking wine. I'm probably the one who sent you over the edge with my post to AA. Luckily we still get plenty of work done and have fun too! A little human craziness keeps things interesting.
Recap and exercise for odd (x+n) squares.
I'm going to work on this, in addition to the (x+n) exercises. This part is cool:
>The remainder of f when divided by 8 is a key to what values of d and (n-1) can be used. Playing with these patterns will organically lead you exactly where we will be going.
Seriously guise, I forgot how much I love solving math problems before I found this forum. A part of my soul has come back to life here. Thank you all! And big thanks to Senapi VQC for inviting us to join on this merry Math quest :) Now if we can just get another hit/drip/crumb that would be great, mmmmmkay?? ;)
>What do these look like?
Pic attached shows some ideas of how the n,d, and f values may grow within the same odd x+n.
Example is for f mod 8 = 0 and x+n=13. I have attempted to retain as much symmetry as possible and moved the nn-1 portion to the edges.
The point here is to try and understand how n grows in relation to a constant f value. The green arrows indicate possible paths. And constant n with growth in f values are indicated by blue arrows.
Hello PMA! On your examples with 3 nn-1 squares, the symmetry would be perfect if you used just nn.
Hear me out.
In nn+2d(n-1)+f-1 there are two 1's that cancel. Expanded formula is this:
nn+2d(n-1)+2d+1-e-1
so the +1 and -1 cancel.
Is this correct, anons?
>666
666 trouble has arrived in the form of f stops ;) Hmmm. so f is an aperture opening for n to flow through? That is cool as shit. Topol, you enlighten the quest yet again. Thanks man!
Hey AA! I remember, and have been keeping an eye out. I'm honestly not too concerned about VQC, I think the timelines just keep getting bumped just like Q does. I can't spend any energy worrying about it TBH. It would be great if he could check in and let us know he's OK tho.
Still searching for something to enable n jumps greater than 2.
Pic attached shows a filtered search in odd x+n=83, where f mod 8 = 6 and f values are between 126 and 142. f=134 is the target value for c=6107 and the prime solution record is the last entry in the table.
Note from the table that valid n values for f=134 are 2, 6, 8 and 36. Whereas f=142 appears at n = 2, 4, and 14.
Perhaps the valid values of n are determined by f ?
While we wait for VQC to turn up again, and since I have no ideas re PMA's current workings, does anyone want to speculate about >>5481 with me?
>You have to know which side to start with and which step to take.
>All we will end up doing with the next step is figuring out a pretty pattern with odd squares with a single unit hole in the middle.
>The symmetry is beautiful in the solution.
>When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.
The solution lies in taking what we currently have and somehow making a pattern out of the odd (x+n) squares that has something to do with symmetry. It also has something to do with f, d and (n-1). It's a kind of symmetry that can only exist in one way for the correct (x+n) in terms of its corresponding f, d and (n-1) values. What are we looking for here? Could it be related to >>5664 this?
AA - I think we're on the same page here. VQC is teaching us the steps to take with the f mod 8 work.
Those relationships seem to be defined pretty well here >>5632. Only recent change I made was to adjust the n_start for f_mod_8 != 0 to 1 instead of 3. This enable me to integrate these n step values into the iterative search process. (only n value not contemplated is 0, but that's a special case anyway).
It would be helpful if someone else could verify these formulas.
So based on these values, iterative search is still n/2 performance, except where f mod 8 = 5. Then it's n/4. Obviously too slow.
The fixing of the odd x+n does allow deeper analysis into the movements of n.
From my latest post >>5670, there is more to it than meets the eye.
Regarding the symmetry, I've only posted pictures for f mod 8=0. There are different layouts for each of the other mod values that are also pretty interesting. All place f in the middle, the nn-1 portion around the edges, and 2d(n-1) filling the space in between. The patterns with f certainly do grow in predictable ways. (again the formulas posted define this quite well).
Possibly helpful, possibly not: I just realized that since f = 2d - 1 + e and we're looking for 2d(n-1), we're also technically looking for (f+1-e)(n-1).
I know, but I'm talking more specifically about visual symmetry, rather than the links between (x+n) squares based on the finite set of non-(x+n)^2=f values. The way VQC said it made it sound like it was far more specific about symmetry than in PMA's picture. If it was obvious and directly related to what we're currently paying attention to, we'd have it by now.
>It would be helpful if someone else could verify these formulas.
If you're talking about the formulae here >>5631 then sorry I didn't reply to this earlier considering I did do a reasonable amount of work on the concept (I guess I mustn't have been paying enough attention last week). It doesn't correlate with what I found in these tests >>5581 here. I found valid values of everything within a particular (x+n) set for every value of n iterating from 1 upwards by 1 each time up to some kind of maximum n value, whereas you have that n is +/- 2 (unless you're referring to something else or I'm reading it wrong). Here's another example in pic related, but I have more detail in the terminal screencaps in my original post. And here's the code I used to generate those outputs (I was lazy and manually changed the if statement's value each time): https://pastebin.com/0gF6rgiR
Different visualization approach:
Using the interior (n) to find where the triangles will be {1+8T(u)} or however that's said.
Or, using the Columella to generate where the outer shell will be; and where the spike points will be in the case of the technical drawing.
Little bit of progress to report.
Managed to integrate the n patterns for f mod 8 into the f-2 div 40 iterative search with reasonable performance improvements.
Pics attached show detailed output for c=6107, a portion of all test cases (with full output available at pastebin.com/3DvFwQmu), and an updated code snippet showing start and step values for n that now also includes patterns for d mod 8.
On the performance side, these revised tests have about 50% fewer iterations than the previous f-2 div 40. A few examples of the iteration improvements:
c=6107 from 23 to 10
c=208364311 from 3706 to 1853
c=9874400051 from 11 to 6
These improvements are a result of a few things:
1) Certain f mod 8 and d mod 8 values enable consistent n+4 jumps between records.
2) The fm2 factor and rm factors are now calculated directly from n0 and the (nn-1 + 2d(n-1) + f) formula.
3) This enables the initial n0 calculation to match the valid known n start.
4) Once in sync with starting n, triangles are created using the methods described previously. (1 + 8 * T(fm2_factor * fm2_chunk) + fm2_mod + 8 * (rm_factor * fm2_chunk) + fm2_mod)
5) The next valid n is then determined from the current n0 and f and d mods using the GetNIterateSettings method.
6) A side benefit is that f-2 div 40 chunks are no longer necessary and have been replaced with f-2 div 8 chunks.
For f mod 8 values of 0, 4, 5 and in some cases 1, the performance is still stuck at n/2. Unless we can figure out a few more patterns in how f and d mods affect valid n values.
Tried to incorporate everything we know so far about relationships between d, f and n.
Very possible I missed something.
Your f formula is incorrect though. Should be f = 2d + 1 - e.
The f that VQC included in the original grid code represented an "F movement" to the negative e side of the grid.
What I was trying to point out is that you're incrementing n by 2 each time when I've found that every value of n from 1 upwards to some specific point (incrementing by 1) is valid for any given (x+n). I can't tell if you're taking every n into account considering you have n = n+/-2 in >>5631 this post. Maybe I'm reading it wrong. The f thing in my quick algebra is a mistake but that's not what I'm talking about.
I think I understand what you’re saying.
My latest tests are a bit further along. Now at n+4 movements at best. n+2 at worst. And am in a position to make even larger jumps if we can figure out more patterns.
Wow! 50% improvement is great. That's a huge step forward, PMA.
I've been thinking about this work
Great find. I found another way to visualize this. If you notice with each successive c for this you are adding up another odd number, much like constructing squares. Think of the value t which is always a+b-1 [think of ab like a rectangle and count the squares on 2 adjacent sides. side a = a, side b = b and they share one so -1]. This is always odd if a and b are the same parity. Since it's odd, we can represent it as 2n-1. If we solve for t=2i-1 we get i = (d+n). Then if we take this value t and remove it, then do the same until we have a one dimensional line, the length of this line is equal to 2(x+n)+1.
Example: 5*29 = 145
t = 5 + 29 - 1 = 33
33 = 2*i - 1 =34 = 2*i => i=17 = (d+n)
5*29 = 145
145 - t = 145 - 33 = 112
4*28 = 112
t = 4 + 28 - 1 = 33 - 2 = 31
112 - t = 112 - 31 = 81
3*27 = 81
t = 2 + 27 - 1 = 31 - 2 = 29
81 - t = 81 - 29 = 52
2*26 = 52
t = 2 + 26 - 1 = 29 - 2 = 27
52 - 27 = 25
1*25 = 25
2*j+1 = 25
2j = 24
j = 12
17^2 - 12^2 = 145
Found a pattern that can be used to improve performance for f mod 8 = 5.
Pic attached shows the original and revised test cases with similar 50% less iterations.
This improvement comes from verifying that the d for the current n0 and XPN is a valid entry.
From nn-1 + 2d(n-1) + f = XPN, and given n0, f and XPN, we can test if (XPN - (nn-1 + f)) mod 2(n-1) == 0. If so, then the n+4 jumping works properly.
Very cool, PMA!
The early bird catches the security worm.
Blessed.
Anons, there are good reasons why I will never be the public face of this. These are not the same reasons why I would like you all to take credit when those organic moments happen.
I will simply be here during to help steer to the next level, then the next level and so on.
If any anons make anything from this in any way shape or form, be a return on investment for faith and effort or anything, then you have my blessing and you don't even need that but it is there.
Anyone invested in this effort to further the knowledge of beautiful mathematics has an opportunity to enjoy it and reap any reward they choose.
It is only my own experience that means money doesn't drive me and I fear celebrity, if I'm honest, though in the past I would have embraced it. That doesn't mean anyone else will have the same experience.
It would be more reward to help plant seeds and stand shoulder to shoulder in the crowd and admire the fruit.
The journey is always looked back on with fond memory, even when that journey involves a great struggle and suffering.
I'm about as far as as 1605 = 505. But it gets fairly tricky past that.
Ok, thank you for your patience.
It is clear that timelines have changed.
I choose caution, that doesn't mean I don't think about you all waiting patiently.
Truth is, at this stage you don't need me.
That will be obvious later when you look back.
Once the other happenings take off, and I'm 100% on my security, if things haven't moved forward here (which I think you are on the cusp of now), I'll come back to accelerate.
Hints:
For any idea, ensure that it scales.
Remember, the solution sets (for odd and even (x+n) will never be of a complexity above O(log t) where t is the length of c in bits.
For odd (x+n) where:
nn + 2d(n-1) + f - 1
This is eight triangles surrounding a single unit square.
Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)
This might help visualise and I hinted at this in an earlier diagram.
https://en.wikipedia.org/wiki/Polite_number
This will REALLY help if you visualise the smaller (x+n) squares and throw in a few much larger (x+n) squares. Focus on the patterns in the way f is distributed in each of the 8 triangles. Look at symmetries and how to construct the triangles when you know you need the two different portion sizes of f. This affects the values of 2d(n-1)
There is a pattern to these that matches the grid which is why later you'll look back at the grid (The End) and understand all the puns and analogies of using that term.
Also, another beauty of P=NP is that one approach is to start with the assumption that there is a solution and work backwards from how it works. I think this will be a massive shift in design methodology in the near future.
If you show some grid diagrams like some of the brilliant ones so far, I can point you in the right direction.
Minecraft anon, can you help visualise this if I guide you? Pyramids can be made with layers of odd and even squares?
Your pattern finding is absolutely excellent.
All what you're doing her will help hone your approach across a range of engineering problems in the real world and with what we'll do next.
It did for me and I ain't that special, just obsessive and with few friends :)
Thanks again for your input anon.
To re-iterate what I said earlier and to clear.
I will never ask you for money.
I will never ask you to take part in anything to profit financially from it.
Q mentioned something about this on qresearch after I posted earlier.
I have no problem with anyone here who has aspirations of using anything here to do so.
I love maths, friendly anons and technical progress.
I can offer advice, what i know, and support of efforts in maths and then design but I can never benefit financially with respect to anything here.
It's about the love of discovery.
Be back soon.
A: Will money even be a thing when this is finished?
B: You already have the code. Y'all woulda used it already if that was your intent.
C: Pic Related
D: Having an income would be nice though.
E: Oooh! Or friends. I miss those.
Hello VQC! Thanks for stopping by, glad you were over on /qr/ to see us fishing for ya! :)
I'm pretty sure he was just talking about YouTubers and Alex Jones types. We all know by now that you're not in this for malicious reasons. It's a big reason why I'm still here personally: because I want the same thing as you.
Out of curiosity, how long did it take you to solve it after the point we're up to?
CA Here, like usual doing my own thing. Have we tried this algorithm?
Take your start value c. Then make values e,n,d,x,a,b. Make a new value (c-n). Repeat with (c-n). It works a decent amount. But the problem is dealing with the parity because to make a record you need (a,b) to have the same parity. I've so far just taken away the largest power of 2 that I can from the number and I make the next record b-n.
should be 11 instead of 13
This part is cool! It also matches an earlier PMA diagram!
>each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)
Start with e,n,d,x,a=1,b=c
Then I make a new value from H = b-n.
Then from here I need to factor H (recursion) and get m (a factor of H)
Then, we make a new record e,n,d,x,a,b = AB(1,H/m)
Then make new value H=b-n
Factor H to get m
rec = AB(1,H/m)
etc..
At each step, if you check the gcd of the original number with each of H, n, and b one of these will have the result. I tried with the first 3741 coprimes and it worked
nvm you just need to check the gcd with the b value.
nvm false alarm
Here's a quick draw. Each inner rectangle has dimensions of u by u+1. This is just for visualizing the idea, and for small n it would be different. But I think for large n the pieces of the squares are ordered correctly. Thoughts, Anons?
For c=6107
f-2/8 = (134-2)8 = 16
f-2 mod 8 = 4
Interesting that the correct n=36 is a multiple of our mod, which is 4
Hey, welcome back.
That's an interesting way of visualizing it. Obviously we'd need to know a and b already but I can see that potentially leading to something.
Is this an accurate diagram in terms of the numbers, or did you just do that to visually show where everything goes with arbitrary values? Because if it's accurate, just thinking in terms of symmetry, the diagonal from the outer corner of f to the corner of the center square (so not dividing the f box in two, just a diagonal through the squares) is 7 boxes and so is from the outer corner of nn to the inner corner of 2d(n-1). And 7 is prime. I have a feeling this is just arbitrary values though.
Hello CA! Good to see you.
LOL! Thanks for this Topol :D
Hey AA! Yeah, the diagram is just arbitrary, to show the big picture. I'll be making some that are accurate soon! The key to this is that 8 triangles of u and u+1 make 4 rectangles that fit perfectly around the center square.
Diagram time? I need a set of colored tiles so I can play blocks with these ideas.
>This will REALLY help if you visualise the smaller (x+n) squares and throw in a few much larger (x+n) squares.
>Focus on the patterns in the way f is distributed in each of the 8 triangles. Look at symmetries and how to construct the triangles when you know you need the two different portion sizes of f. This affects the values of 2d(n-1)
look at the values for
nn + 2d(n-1) + f - 1
They are always square. The square root is either equal to (b-a)/2 or f
Also (b-a)/2 is x+n
Pic attached for c=6107 is an attempt to understand the staircase number breakdown for values of f that could be used to speed up the iteration process.
Both images are for the x+n=83 small square where f=134.
The left square follows my previous diagrams >>5664 with f in the center and nn around the edges. The main difference is that the -1 from the nn + 2d(n-1) + f - 1 formula has been moved from the nn portion into the f portion in the center.
The light blue squares indicate the f mod 8 = 6 value. And the total of blue, light blue, and the orange square in the middle is now f-1 = 133.
In the right square, is the idea for splitting the f-1 portion into 2 separate groups of 4 triangles each. The dark blue represents 4 triangles made up of the staircase number 4+5+6 = 15 for a total of 60. And the blue and light blue triangles are 4 triangles of 5+6+7 = 18 for a total of 72.
60 + 72 + the orange square in the middle gets us back to the f - 1 = 133 number.
Interesting to note that the 60 + 72 staircase numbers equal our f - 2 iteration chunk of 132.
Lads, VQC said we are really close at this point. He basically said we can do it from here.
Let's expand our minds, and feel the emotion of success!
Visualize.
Conceptualize.
Realize.
Our minds are our own Personal VQC's.
Space Jesus Says "You will know the Truth, and the Truth will set you free."
Hey CA! I'm following you well. This is a solid and interesting idea. Can you post your output to show and explain?
In row one, (x+n)^2 is always equal to f.
You're also saying that (b-a)/2 = x+n for all (e,n)?
Can you clarify Pls?
>
This is a uh…
Look into Vixie.
We're taking down Linux AND Unix now.
;)
http://dotcomeon.com/
Because when we crack RSA we'll be able to sign as the people who push code/updates… from my understanding.
I'm just messing with stuff so I don't know what this means but it looks nice so I thought I'd post it here.
So I'm thinking when f = d, we have a = 6k+1.
If f is a square, (x+n) is the square root of f.
If the gcd(f,x+n) = f, then (x+n) is fj where j is a factor of the original (x+n) = (c-1)/2
There is probably other stuff but I'm sleepy
Alright guys, I think I just found something really cool. Need your help to verify Pls!
So, for c = 6107
f = 134
f-2 = 132
(f-2)/8 = 16
(f-2) mod 8 = 6
Based on VQC's latest hints, there are two staircase numbers that make up (f-2)/8 which give the correct n0 inside each one of the 8Tu. I think I found out how to find them:
4+3+2+1 = 10
3+2+1 = 6
= 16
The correct f shape inside each of the 8Tu could be a triangle?
Key idea is that the overlay of (1+2+3) and (1+2+3+4) = 16
Overlay doesn't work for filling the square.
But making a triangle along the straight side of each Tu works. Thoughts, Anons?
Excellent work, VA!
First pic attached is a revised layout of the f-2 portion of the c=6017 small square originally posted at >>5716.
The f-2 value of 132 has been divided into 8 symmetrical triangles with 4 remainder squares in light blue. The remainder of 4 coming from f-2 mod 8.
The inner blue square represents the smaller staircase number of 6 (or T(3)), while the outside dark blue is the larger staircase number of 10 (or T(4)). And again, the single orange square in the middle. Total of all squares is still f-1 = 133.
The second image is a quick calculation based on using these two staircase numbers as the triangular bases for the calculations originally explored in >>5525.
It appears that we can get very close to the solution in 5 steps.
Still need to determine a formula for splitting f into two staircase numbers and if this methodology applies to different test cases.
Thanks PMA! More verification needed, but hopefully this turns out to be a good step forward. Can we find the two triangular numbers that create (f-2)/8 ? Is that the formula you're talking about?
In trying to find a relationship between (x+n) and c, I ran into gradients again. I don't know if anyone remembers what happened last time (4 or 5 months ago) but this probably isn't going to lead anywhere useful.
Since we all seem to be coming towards this from completely different angles, I'll explain my current thinking. If you look at the last pictures in my post >>5581 here, you can visually see where the cutoff is from the infinite set of values for which (x+n)^2 = f to the finite set we actually need to figure out the math for. I have a feeling we'll find the answer in some way that will be analogous to figuring out a mathematical ruleset that would be used to construct the finite set from a given c value. So obviously it's far more likely that we'll figure it out with pictures and symmetry, but thinking of it in the context of series math might show us more relationships. I mean, you can see that all the n, d, x, a and b values follow incremental patterns. It's just e and f are all over the place.
I think that the gcd of (x+n,f) for the starting record and the gcd of (x+n,f) for the correct record are always related somehow. Usually the correct one is a factor of the starting one. Sometimes the correct is a multiple of the starting one. We could do recursive like this:
get e,n,d,x,a,b
get gcd(x+n,f)
factor that:
go through the factors (or multiples of the factors) as (x+n) to test if (x+n)^2 + c is square.
If nothing works, start iterating the original gcd.
???
Profit!
Right, AA. CA, we can't know a and b starting from only c.
From c we only have d, e, and f.
VQC is hinting that we can solve the (x+n) square using f.
CA, check out these recent VQC posts.
>Going from a search to a calculation is an almost inconceivable step.
>Think of a Rubik's cube.
>Trying to solve it without a plan.
>Then remembers those books on how to solve it?
>You have to know which side to start with and which steps to take.
>All we will end up doing with the next step is figuring out a pretty pattern with odd squares with a single unit hole in the middle.
>The symmetry is beautiful in the solution.
>When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.
>Recap and exercise.
>Starting with odd squares.
>(x+n) mod 2 = 1
>Starting with the smallest odd squares.
>(x+n) = 1,3,5,..,
>For each odd square what are the permutations of possible values of f,d and (n-1)?
>What are the patterns?
>What do these look like?
>For odd (x+n) where:
>nn + 2d(n-1) + f - 1
>This is eight triangles surrounding a single unit square.
>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)
>This might help visualise and I hinted at this in an earlier diagram.
>https://en.wikipedia.org/wiki/Polite_number
>This will REALLY help if you visualise the smaller (x+n) squares and throw in a few much larger (x+n) squares. Focus on the patterns in the way f is distributed in each of the 8 triangles. Look at symmetries and how to construct the triangles when you know you need the two different portion sizes of f. This affects the values of 2d(n-1)
>There is a pattern to these that matches the grid which is why later you'll look back at the grid (The End) and understand all the puns and analogies of using that term.
Yeah I know. We can solve (x+n) from f if f is a square number. If f is square, then (x+n) is the root of f. This leads me to believe that in order to factor c we need to factor f (recursive solution). This is because for our first step if we had a case where e = 0 then we'd have a solution. For this if we (for our f) have e=0, then we have a solution. I think we need to do this because the correct (x+n) often seems to be a multiple of a factor of f (often a multiple of the gcd of the original (x+n) and f). If f is prime, not sure what to do. I've noticed that sometimes the correct (x+n) can be found by chopping off the last bit of e or f. Sometimes you take f*2 +/- 1 to get the correct answer. Still a lot to think about.
CA, check out row 1!
Sqrt(f) is ALWAYS (x+n)
f is also (x+n)^2 always.
That's why (e,1) row 1 is our reference to the correct solution.
Somehow row 1 gives us a clue to the (prime) solution.
Thoughts, Anons?
Good points, CA! I'm examining them closely to make sure I understand your ideas. Here we go.
>We can solve (x+n) from f if f is a square number.
>If f is square, then (x+n) is the root of f.
>this leads me to believe that in order to factor c we need to factor f (recursive solution). This is because for our first step if we had a case where e = 0 then we'd have a solution. For this if we (for our f) have e=0, then we have a solution.
>I think we need to do this because the correct (x+n) often seems to be a multiple of a factor of f.
(often a multiple of the gcd of the original (x+n) and f).
>If f is prime, not sure what to do.
>Still a lot to think about.
Now we're talking. (e,1) is supposed to be our big clue. We still haven't figured that out. After 6-7 months. How does (e,1) play into factoring and given c or x+n value?
Is there a reason why c + f forms a square number?
Good explanation, MA.
Have been focusing on the following hints from VQC's recent post.
>For any idea, ensure that it scales.
>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two … configurations in each triangle.
>The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other.
The quickest iterative search so far has been using 2 different sized triangles. (see >>5522, >>5522, >>5525 for some background)
The problem here has been determining the different u values to use, and perhaps we now have a bit of a clue.
First pic attached is a spreadsheet calculation for c=14904371 showing the solution in 18 steps using the dual triangle approach. Previous best result using f-2 iterative search and valid n jumps was 1083 iterations. So a pretty good performance bump.
In this test case, the u1 and u2 values of 271 and 84 were chosen by trial and error. But there may be a way to get to those values - which is the reason for the post.
The second pic is the square layout for f-1. It is broken into 8 triangles of 368, 4 remainder squares, and 1 square in the middle. 1 + 8*368 + 4 = 2949 == f - 1.
One of the triangles has been subdivided into f-2 div 40 chunks of 73, with 2 separate remainder areas of 12 and 64, to try and figure out where these two configurations come from.
For the working solution:
u1 = 271 == (1+2) * 73 + 52 ???
u2 = 84 == (0+1) * 73 + 11 ???
The staircase numbers may be the factors that are applied to the arbitrary divisor of f (40 in this case). (0+1) and (1+2) in my guestimate.
But then where do the differences of 52 and 11 come from?
And why is the solution found with only 4 chunks of 73, when there are 5 available in the 368 triangle?
This may all be coincidental, of course, as these large u values would prevent this solution from working with small values of n.
So I did my visual arts thing and looked for patterns and my first wondering was:
"Does the Yellow complete the individual Blue and Green trangles?"
Answer: You get an extra unit. 1 leftover square.
Aight so… how many units short (of a perfect trangle) is the combination trangle?
Answer: 10 units.
So I axed:
"Wha happen if'n's ya lay it out in the other orientation? Vertical instead of Horizontal?"
Since the original trangle was 27 rows deep, I stuck with that.
Those missing 10?
Is a pretty capstone, is what i'm seeing.
I woooooonder if it'll work like that every time or if this was just happenstantial…
A few visuals of what the full squares look like for the f-1 portion of c=14904371.
Color coding matches the single triangle breakdown posted in >>5747.
The darker center portions of the u2 and u1 images represent the "starting" values based on the staircase number * (f-2 div 40) chunks. The lighter area represents the "remainder" that I questioned earlier.
The "combined" image shows how the u1 and u2 squares mostly fit the starting f-1 original square.
It's possible that these 2 squares need to be searched for independently and then used to find the final answer.
f mod 8 pics attached for odd x+n = 13 showing (hopefully) more consistent symmetry with n, d, and f values.
The f mod 8 = 0 has been updated to incorporate VA's idea of not adjusting the nn portion. >>5665
Additional images have been created for f mod 8 = 1, 2, 4, 5, and 6.
In most cases, it seemed to make sense to fill the center square with 1 taken from (f-1). See the yellow center square in images for 0, 2, 4 and 6.
In the cases of f mod 8 = 1 and 5, however, the symmetry looked better taking the 1 center square from nn. See the green center square in those images.
Based on these adjustments, a few of these mod cases start to look increasingly similar. 1 and 2 have some overlaps. As does 5 and 6.
And pics for f mod 8 = 4, 5, 6.
>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two … configurations in each triangle.
Maybe that's the symmetry he's referring to.
>The symmetry is beautiful in the solution.
>When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.
If the arbitrary divisor for f gives one of two configurations, it exists in more than one way. I might be stating the obvious. I only came in here briefly and I'm not actually applying this to the math in my head. But we should be looking for a value of f that gives only one configuration in each triangle. So I think we'd find the answer if we took one triangle, found the correct value of f to give only one configuration in the triangles, and then compared it to other arbitrary values of f that give more than one configuration and looked for the differences between the singular f and the pluralur (I know that's not a word) f. That's if I'm reading into this right.
Exactly like this on one side and the (x+n) square constructed on the other side. This is all about patterns.
I would like to start a thread in parallel to this as we're reaching the point where an anon or group of anons is going to have a Eureka moment and solve this for one of the four types of solution (odd x+n with either even or odd e), then quickly for another and then the other two.
The other thread will describe how to take the work done to solve Fermat Last theorem relates to our approach and why two objects that are identical types in number theory had to be proved so in a difficult to under proof, when in reality, with the right number system or model, this should have been obvious or a tautology.
Brilliant.
Beautiful
This is amazing. Truly excellent.
This step edges towards the Eureka moment.
Love this thinking.
This is beginning to visualise the symmetric and asymmetry (still symmetric though) involved when numbers are not prime.
Important!
Love this.
This is also verging on the first solution (that quickly becomes the four types).
I think without ruining the last piece of the journey where you find out the first solution yourself, it looks like anons have made it onto the home stretch.
Brilliant work, especially with how your approach was so quick and organic.
The next piece that uses the same approach is not going to go into the details of elliptic curves and modular forms, it will be more about why these two identical things looked different, because realising they were the same thing solved a 350 year old math problem called Fermat's Last Theorem.
What does it tell us about a language like maths when you need a virtually incomprehensible proof JUST to show two things are identical?
How are we going to use this to our advantage? How have we been using this to our advantage to train in a new approach to the problem the solution is spring out in the RSA problem above? This relates back to the patterns in the grid.
Same way, same method. We begin to ask the right questions and point out the glaring thing that no one was talking about. We start to look at patterns which we anons are above anything else, experts at.
Here is an example of a paradox called the Liar's Paradox:
"This statement is false."
Thinkers, some great thinkers, said this statement was true when it was false, and false when it was true. This conclusion ruined progress in an entire branch of mathematics. A very important branch. Critically so.
Here is the statement above written in a computer language, resolving the dysfunctional application of the equality operator between two things which are not the same and removing the statement's ambiguous self reference:
bool b_statement = false; //this whole line is called a statement
The assignment of a value to a Boolean variable called "statement" was being held to be identical to the concept of a whole line or sentence being a "statement". Two different things, no paradox. A computer will compile the code, it woouldn't be able to compile something non-deterministic (where a choice is made that it doesn't understand enough about).
If we define properly the equality operator, i.e. when two objects are identical, AND WHEN THEY ARE NOT, and we do it properly, we classify numbers like integers differently and also other things, and then a whole bunch of things happen. In fact a whole bunch or category of problems become trivial to solve once we figure out what we were not seeing in terms of patterns.
That's where we'll be going next with elliptic curve encryption and then further using the Mandelbrot Set as a computer.
Hello VQC, nice to wake up on a Saturday and see your posts! Studying now. The adjacent polite triangle numbers are the key. We're working on how to make the remainders and "leftovers" fit. I think that's what Topol's new drawings are hinting at. Thanks for the new crumbs, back to work!
Oops! Here's a shot from Topol's VQC Fishing Expedition last night. Highly relevant.
Trying to work this out in a real example.
c=6107
f=134
(f-2)=132
(f-2)/8=16
(f-2) mod 8 = 4
(x+n) = 83
(d+n) = 114
u1 = 41
u2 (u+1) =42
correct n value = 36
Here's the formula: nn + 2d(n-1) + f - 1
f-1 = 134-1 = 133
2d(n-1) = 2*78(36-1) = 5460
nn = 1296
1296 + 5460 + 133 = 6889
SQRT(6889) = 83 = correct (x+n)
Thoughts, Anons?
I started a thread and stickied it here. >>5774
>one of the four types of solution (odd x+n with either even or odd e), then quickly for another and then the other two.
I've graphed four different graphs for the different parities of (x+n) and e (so (odd, even), (even, odd), (odd, odd) and (even, even)). They're the first four pictures here. (x+n) is the x axis and e is the y axis. The numbers in these grids are first the co-ordinates ((x+n), e) and then the third number is f.
As you can see in the (x+n) odd e even and (x+n) even e odd pictures, most of the f values are squares. So these are our infinite sets where (x+n)(x+n) = f. They only ever occur when (x+n) and e have opposite parity. As you'll also probably see (sorry if one or two of them are blurry), there are some other non-linear f values scattered around in weird places. I can't tell right away why that is. You'll also see they just make a big (visual) square and can infinitely and uniformly stretch downwards or sideways. It's an infinite set of infinite sets, each with an increasing amount of noise. If you figure out the noise, you win.
Looking at the pictures where (x+n) and e have the same parity, these contain the finite sets. Some parts of the finite sets are contained within the grids where (x+n) and e have different parity like I explained above as those "f values scattered around in weird places", so they aren't singled out completely, but none of the f values are square/infinite unless (x+n) and e have different parity. So that's quite a useful piece of information. Another thing from these is that since they show the finite series you can see them increasing in length as y increases. I've shown that a little more in the fifth picture. It increases at an increasing rate (so from (x+n)=3 it's 1, 5, 11, 19, 29 etc, meaning +4, +6, +8, +10). As you might also see if the numbers aren't too small is that in both cases where (x+n) and e have the same parity, most of the values of f follow strict patterns, but there's also noise here. So if there wasn't noise, this would be the pattern: from the (visually) top value of e downwards (in even even it's from 0, and in odd odd it's from 3 since when e=1 (x+n) is even according to the other grid where that's the case), the value of f decreases by 2 whenever e increases by 4, and as (x+n) increases by 2 and e stays the same, f increases by an increasing amount (so if you look at the even even picture, at {4, 0, 7} and {6, 0, 17}, as (x+n) increases by 2, f first increases by 10, then, if there wasn't noise, it would be 31 next, meaning an increase of 14, then an increase of 18 to 49, so it's increasing by 10, 14, 18, 22 etc, meaning the increase increases by 4 each time).
Hopefully that wasn't too much ramble. If there's anything to take away from it it's that (x+n)(x+n) = f when (x+n) and e have opposite parity.
Have you done this on any other examples or any other divisor fs? It looks like some degree of symmetry could come from that.
Nice digits, AA! I'm just building out the c=6107 example to try and complete a medium sized problem IRL. I'll post updates, trying to fit everything together like a block puzzle.
Here's 2(f-2) + 2(mod 4).
(f-2)+(mod 4) is a few posts above. Don't want to keep referencing myself. Baker is gonna shit his pants. Native American symbols everywhere. They must have had a few mathematicians in the tribe!
Hey Lads, bear in mind: My diagrams are an exploration of three things.
-
Using f to fill the square
-
Using f to find an iterative match.
-
Using the polite triangle numbers to understand f.
Just trying to work out the ideas!
The first two are potential ways to view the (d+n) square. The square seems to always have the length that is half of the square that you can construct with the a*b rectangles. The third pic is just me testing triangular numbers for e and f and making a rectangle out of them. I'm thinking like an aperture or something
Within(between) and Around….
Here's layer 4.
c=6107
f=134
(f-2)=132
(f-2)/8 = 16 (green triangles = 6 + yellow triangles =10)
(f-2) mod 8 = 4 (blue squares)
Current Values for layer 4:
(f-2)/8 value = 128 18 = 2304 = 24 * 24 4
(f-2) mod = 4 * 18 = 72
2304 + 72 = 2376
Check 132 * 18 = 2376,
mods + (f-2) values currently matching.
I think the mods are going to form a new border to create the +1 differential need for u and u+1.
Here's layer 6:
(f-2) div 8 = 128 * 50 = 6400
(f-2) mod 8 = 4 * 50 = 200
Check Math: 6400 + 200 + 1 = 6601
Distance remaining to correct (x+n)^2 = 6889 - 6601 = 288
PMA, is that close enough for rm 2d(n-1) to lock on it?
ALSO: The remaining 288 can be divided by our polite (f-2) div 8 triangle of 16 ( 6 green + 10 yellow)
288 / 16 = 18
This could just be chance, but still pretty cool!!
Look at these examples. I tried to construct the (x+n) square doing this weird way. I noticed something with these two records.
We can set (x+n)(x+n) = f + d + (d+n)
You fags have really kicked it up a notch. Wish I had more time to dive in atm, but enjoying following along for now.
>>5690 Thanks for the new crumbs! Good think we've been polite!
>>5763 Looking forward to it!.
>>5755 So nice PMA. Rocking it as usual.
>>5760 Topol joining the fray, nice inputs.
>>5791 Trippy evolution as you've built this idea out. Nice work.
>>5768 Enlighten us PMA?
Only input I have is 2 months old, so a slice of stale bread here from March.
Was playing around with patterns. One approach was was to do the n0 in the center, and then fill in with the n (the n^2-1) both around the outside and around the center n0.
Was putting 1 f unit in the center, and then the remainder of the f around the outside, then adding the rest as rings.
The interesting thing, is there were combinations where everything worked out evenly as integers. Hopefully the attached images give a sense of this approach.
Well, that was fun! I didn't find a perfect match, but got very close. Main point is the the BASES of the polite triangle number are 3+4=7 (our correct potential n0), and our correct u1 and u2 values are 41 + 42 = 83. So we could iterate n0 = 7 * 6 = 42, and then go -1 and +1 to test which one is our other u value.
For this last diagram, I added back in the (f-2), and we get another chunk.
2 * 50 = 100
6601 + 100 = 6701
correct (x+n) = 6889
6889 - 6701 = 188 remaining
Close enough to lock with PMA's formulas???
I think so.
Thoughts, Anons?
I know what Baker is thinking ;)
MM - nothing doing. Just responding to CA. >>5770
Attached spreadsheet explodes the pyramid numbers into left and right side polite numbers from 1 to 10 (and 73 just for fun).
XPN columns follow the formula 1+4T(right polite number) + 4T(left polite number).
Some interesting numbers are showing up.
Wow! Great job on these diagrams, CA. Studying now.
For all you nerdy fags ;)
I started on column to cuz Derp so the first one steps out from 19
Right 1, down 1:
19, 30, 42, 55, 69, 84, 100, 117, 135
19+11+12+13+14+15+16+17+18
Increment+1
Right 1, Down 2
19, 33, 50, 70, 93, 119
19+14+17+20+23+26
Increment+3
Right 1, Down 3
19, 36, 58, 85, 117, 154
19+17+22+27+32+37
Increment+5
Right 1, Down 4
19, 39, 66, 100, 141
19+20+27+34+41
Increment+7
Then I I realized… I'm a derp!
And it turns out… these relationships are the same no matter where ya start buuut
Starting form the First Column… you can see what's happening more clearly.
Right 1
10, 19, 27, 34, 40, 45
10+9+8+7+6+5
Inc-1
Right 2, down 1… from the base… row*2=inc
10, 30, 50, 70, 90
10+20+20+20+20
Increment = Same (the "Plus Self" when you start in Column 2)
Right 1, Down 1
10, 21, 33, 46, 60, 75
10+11+12+13+14+15
Inc+1
Right 1, Down 2
10, 23, 39, 58, 80 ,105
10+13+16+19+22+25
Inc+3
Right 1, Down 3:
10, 25, 45, 70, 100
10+15+20+25+30
Inc+5
Right 1, Down 4
10, 27, 51, 82, 120
10+17+24+31+38
Inc+7
Some Observations:
only the first and second rows seems to matter, oving in increments of 1's and 2's. LET'S TRY 3's!
Whatever number you're starting with can always be tracked back Up and Left.
Moving Up/Right or Down/Left makes things get weird. (For me, at least)
—————————————————
"right 3, down 2" is an increment of +3 after you find the base value of the increment ({Row# * 3}+2).
If you're starting with the first column,
Right 3, Down 2…
You base increment is ({Row# * 3}+2)
And then it increases in +3 per step
Example:
15*3=45, 45+2 = 47
Right 3, down 2 from 15:
62
now add +3 to the increment and that's your next step
62+50=112
+53
165
Also worked when I did it for 7, 21, and 23
Try it out fer yer own bedazzlement!
Preeeeeetty :D
New tab didn't work, but downloading it and opening it did. Thx!
Ok, lads. I have a match. Still don't quite understand how it all works. But additional iterations of 16 filled the gap. Take a look, can't confirm it's 100% correct, but the symmetry / asymmetry is pretty cool.
The border needed corners and midpoints of 40/8=5. So each corner and midpoint has and odd number. The 8 remaining gaps to be filled were all 36 units. Found a way to make them all fit.
Yellow = 6
Green = 10
They all fit.
But shouldn't there be extra mod units to add in? I'm stoked but at the same time confused. Any ideas welcome. Posting a work in progress for group discussion.
Thanks CA!
Update:
Properly filled borders.
10 * 18 = 180
6 * 18 = 108
= 288, our remaining units to be filled.
Still confused about additional mods related to the 288, but here is the diagram for your consideration and analysis.
Thoughts, Anons?
Based on VA's new square layout, and the pyramid numbers posted earlier, attached is another test case for c6107 that attempts to bring these ideas together.
The solution takes a few more iterations than posted in >>5728, and includes the following changes:
1) u1 and u2 are now incremented by pyramid staircase values offset by 4 (meaning each iteration is descending the pyramid).
2) There are 4 squares made up of 2 triangles each that are themselves 1 staircase unit apart.
3) The estimated XPN formula has been adjusted to:
1 + 4*(T(u1) + T(u2) + (2d + 1)) - (2 * f mod 8)
4) the 2d+1 portion represents the border around the 2 adjoining triangles and fills the remainder portion. The d value in each iteration is calculated as (u1+u2)/2.
5) Subtracting the (2 * f mod 8) resulted in each iteration returning a perfect square.
This method hasn't been tested on any other cases and is highly depended on identifying the correct starting u1, u2, and the "polite" way to descend the pyramid.
Pics attached show two possible layouts for the first iteration of the c6107 example previously posted.
Left square is the f-1 starting position.
Right squares represent the T(u1), T(u2), 2d+1 and f mod 8 portions of the formula.
new baker necessary
Ooh la laaaaa…
Timing and events…
Baker?
Jan?
Other unaccounted Id?
;)
We're doing this.
Nice following along with you all.
Found this old crumb by mr. curtis. Might be relevant now! (see pic)
Also, no breakthrough patterns, but just another dig back from couple months ago. Did those plots of EvenOdd x+n, D, n, etc in excel.
Attached is a matrix of (f Mod 8) for the product of the first 751 primes (5701 is the 751st prime). So the matrix is from 1x1 to 5701x5701.
Set color f=0 Red, 3=Yellow, 7=Green
The realisation that something is missing in our mathematics, and can be found starts to arrive doesn't it!
Thanks for being part of this.
Love this.
Love it.
OK, so looking at the case of (x+n) is odd.
If the unit square is thought of as the middle and then there are eight triangles.
Remember that (f-1) will be distributed among those eight triangles UNEVENLY.
Some triangles will have more of f than others.
How does this work?
Especially at scale? (Getting very large)
The size of f and d restrict the values of n and n-1.
If you draw out some odd (x+n) squares, relate them graphically side by side with the grid and see any patterns that f make and each 8 triangle, especially if portions of f are outside the (n-1) square in the middle.
Summary:
(x+n) is odd.
f is divided amonst the 8 triangles (some have a bit more than others).
f is within each triangle but OUTSIDE of (n-1), where the (n-1) square should go in the middle.
This will join up with what you were doing before.
This is for cases where f/8 n
Good to see you too anon!
Yes. The final lock and key construction steps.
Happy to give it but it is the Eureka moment and anons are close.
In hindsight, it shows exactly why this problem has existed for so long.
You are solving two problems at once in this method of constructing the answer.
Also in hindsight the steps give the pattern back in the grid (The End).
Essentially, as I was showing with RSA100, you take a stab at the number by creating a triangle base out of f div 8. The remainder (f mod 8) is divided among the eight triangles. This restricts the multiples of 2d that can be used because they must always be in a number that supplies the gap in f and satisfy another property. That property can be broken down.
It is easier to break down the less factors there are as there are less solutions.
The answer is easier to come to the more factors there are as there are more solutions.
You get the best of both worlds, which is the ideal for this problem.
I'll get less and less cryptic over time until I EITHER give it away, OR you guys solve it, OR WE MEET IN THE MIDDLE.
The excluded middle.
>The remainder (f mod 8) is divided among the eight triangles.
Like this?
I'm not sure about the 5th in the centers…
But… maybe?
Also, VA is out'n'about and can't 8chan at the moment.
"Ask him this: what about the (f-2) portion? Do we add that back in?"
Eastern Time
new baker on standby
Novice Baker here, but have my cookbook from /qr/ handy. Topol haz Dough? I'll bake in 2-3 hours if nobody more qualified steps up.
Holy shit that first key seems HUGE.
This equation works for A,B and D values.
A/B/D(t) = T(t)4 + tconst + start
where T(t) =(t*(t+1))/2
I figured out a way earlier to calculate the constant and start for each (e,1) record I just don't have it on hand. These are known values.
an = A(t) = T(t)4 + tconst + start
= 2t(t+1) + t*const + start
bn = A(t+n) = T(t+n)4 + (t+n)const + start
= 2(t+n)(t+n+1) + (t+n)*const + start
b = a + 2n + 2x
so
bn = an + 2nn + 2xn
2(t+n)(t+n+1) + (t+n)const + start = 2t(t+1) + tconst + start + 2nn + 2xn
(2t+2n)(t+n+1) + tconst + nconst + start = 2t(t+1) + t*const + start + 2nn + 2xn
[2tt] + 2tn + [2t] + 2tn + [2nn] + 2n + [tconst] + nconst + [start] = [2tt] + [2t] + [t*const] + [start] + [2nn] + 2xn
4tn + 2n + n*const = 2xn
4t + 2 + const = 2x
4t + 2 + const = 2d - 2a
Idk if this helps but I think the algebra is correct.
I forgot what the calculation for t was.
If x is odd (d even), then t = (x-1)/2,
so 4(x-1)/2 + 2 + const = 2x
2x - 2 + 2 + const = 2x
const = 0
FUCK
>(x+n) is odd.
>f is divided amonst the 8 triangles (some have a bit more than others).
May have an idea of how to proceed.
Pics attached are the (x+n)=83 small squares for c6107 and c27707 arranged according to the nn + 2d(n-1) + f - 1 formula.
nn around the edge, then 2d(n-1), and finally the f-1 portion in the middle including the center square.
c27707 is a new test case, so for reference, here are the relevant details from the grid representing the prime solutions.
{23:36:78:47:31:197} = 6107; f=134; (x+n)=83; u=41; (d+n)=114
{151:20:166:63:103:269} = 27707; f=182; (x+n)=83; u=41; (d+n)=186;
Notice the f-1 distribution for both cases is uneven with respect to the 8 triangles.
(1/2)
And here is (I think) the find.
Per VQC,
>f is within each triangle but OUTSIDE of (n-1), where the (n-1) square should go in the middle.
By taking one of the (n-1) portions out of the 2d(n-1) piece of the nn + 2d(n-1) + f - 1 equation, we are left with an equation that looks like:
nn + (2d-1)(n-1) + (n-1) + f - 1
This new (n-1) + f - 1 piece turns an "out of balance" f portion around the middle into something that is equally distributed among the 8 triangles.
The revised full square pics attached for the same c6107 and c27707 tests show how this new layout comes together. Also attached are close ups of the (n-1) + f - 1 in the centers.
The green squares in the middle represent the single (n-1) portion that has been removed from 2d(n-1).
The yellow squares are the portions of (f-1) that have been "displaced" from their original positions.
The red square is a single unit taken from (2d-1)(n-1).
And this is most likely why (n-1) is so important as a factor to all of the transforms that we worked on previously. (d[t]-d)/(n-1), a(n-1) na transform into (e,1).
(n-1) balances out the f center square.
(2/2)
Damn. Nice work PMA! Studying in detail now.
Also, every Anon archive everything offline. I've got most of it, but need to save the maps first. Baker did a great job with that.
Filling bread