I'm only going to shill this once since it isn't RSA-related, but I know some of you might be interested.
This is why it's good that those of us here on this board (as far as I'm aware) don't want to be known for this (we're not famefags, as one you said). I have a feeling since most of the /cbts/ people came from other places on the internet and haven't always necessarily been part of anonymous imageboard culture that they got in over their heads and their egos took over when they became part of a 'super awesome secret club for special people'. That might be why some of them talked to Infowars etc, and it's obviously why so many 50+ year olds are making YouTube videos about it constantly. I have a feeling it's part of why the mods are being dicks. Either they're overwhelmed or they're compromised. I wouldn't want to run a board like that.
Why doesn't Q just make his own board? That way he could lock and sticky all of his own posts and everyone would know it could only be him.
>VQC said we only need 3 variables to solve for c
The only variables they could be are e, n, d and f, right? If we knew any others we'd already be able to do the factorization, and that would be O(1).
>x = floor(sqrt(abs(f)))
Am I missing something here? Does it only work for some cases? Pic related.
I don't know if this is useful but I haven't really contributed to this thread yet.
Where n = 1, x = floor(sqrt(abs(f)))
Where n = 1, i - d = 1, which means i = d+1
Where n = 1, x + n = j, which means j = floor(sqrt(abs(f))) + 1
Where n = 1, i scales up linearly with d and j scales up linearly with floor(sqrt(abs(f))).
When was the last time Chris posted here?
Has anyone figured out how O(log n) factors into the grid navigation? I noticed you're mostly all working on calculations and things. Surely if we knew how the idea of this being done in O(log n) time makes sense it would help immensely.
What are you talking about? O(1) means it's instant. Also what does that have to do with anything? VQC said it would be in O(log n) time. I'm not aware of many ways in which an algorithm can work in O(log n) time. Wouldn't that be a constructive thing to think about? It might more clearly show what we should be looking for in these relationships.
How many different O(log n) algorithms can you list? What n is is irrelevant to what I'm trying to say. Even if we don't apply any existing O(log n) algorithms to the grid, what we do with the grid is meant to fit the criteria of an O(log n) algorithm according to Chris. Does that mean there has to be some variable or relationship between variables that allows us to halve n's search pool every time? Does that mean doing something weird with powers? If we know what criteria are necessary for something to be O(log n), we'll know more about what kind of mathematical relationship we're looking for, so that whenever we find an interesting relationship, we can figure out if we're going in the right direction.
I don't say this to be a demoralization shill but I've really lost enthusiasm for this lately. Chris hasn't even posted for two weeks. I know it's real, and I haven't contributed since I gave up looking for a linear relationship to use in binary search so I have no idea if any progress has really been made. I just wish Chris hadn't kept saying "oh I'm going to tell you how it works now, oh wait I'll tell you in a week, oh wait I'll tell you in a month" etc.
For someone who seems to know a lot of obscure information and have a "free mind", I'm surprised you'd use gmail.
>Try multiplying pi and golden mean together and find the square of 5, and 79, and start peaking around
3.1415926 * 1.618 = 5.083096827, but I have no idea what you mean about using that to find the square of 5 and 79
>THE PUPPET MASTERS HAVE BEEN REMOVED.
Well, if those aren't the most encouraging words Q has ever posted. Seems like great circumstances for the VQC to be figured out.
Of course we would.
Factoring known subprimes is O(n) currently, so it isn't a polynomial. You haven't posted much, which means I don't know how much you know about computer science, so I may as well explain it anyway (even if you already know, some lurkers probably don't).
Polynomial time means the number of calculations the computer has to make for a set of code can be calculated by n^k. When I mentioned O(n) earlier, I was referring to the same concept of the number of calculations a computer has to make. In O(n) time, there might be a big list of numbers or words or something n objects long, and you might iterate through each element of that list of objects to find the right one. That means it takes at most n calculations (if the list is 5 long, it's very quick, but if it's 10^39523 long, it'll take far too long for it to work). In polynomial time, there might be a list of objects n objects long, but the number of times you have to look through that list is n^k, where k is some constant number. This isn't a good example for the P/NP thing, but, for the sake of explaining polynomial time, generating the grid takes polynomial time, because the inner loop runs however many times as specified by the outer loop (meaning it's O(n^2), k = 2). If you had a loop within a loop within a loop and the outer loops determined the number of calculations done in the inner loops like with the grid, it would be O(n^3), k = 3. As n gets higher in any polynomial problem, the number of calculations necessary (and therefore the time taken to complete the code) goes up at an increasing rate (it isn't just a straight linear line).
NP (non-deterministic polynomial) is a name that refers to a group of problems that we can solve with code with some special properties. The solution to an NP problem can be verified in polynomial time. That means, for one of these problems, if you think you know the answer, you can figure out if it's correct or incorrect in polynomial time. The problem is that you can't actually necessarily figure out an NP problem's solution in polynomial time. This abbreviation, NP, does also cover all P problems (polynomial problems) that can be solved in polynomial time. NP covers any problem that can have a potential answer verified in polynomial time but it also covers problems that both can and can't be solved in polynomial time. Pic related.
Then there's this other class of problems called NP-complete problems. A problem is NP-complete if it's an NP problem and every other NP problem can be rephrased as a special case of it. An algorithm that solves an NP-complete problem can solve every other NP problem (those really time-consuming ones). That means, if we find a really quick way to solve an NP-complete problem, we can solve every other NP problem in a reasonable amount of time. There are a few NP-complete problems, such as the Hamiltonian Cycle problem. Say you had a map with a bunch of towns on it and you wanted to visit each town once in the shortest possible path. If you had a computer figure that out, it would need to go through each possible path starting from each possible city, which could take an extremely long time depending on how many cities there are. If you had an algorithm that could solve this problem quickly, you'd be able to solve all the others really quickly.
The P=NP problem refers to this picture. We know that all P problems fit the criteria of an NP problem. What we don't actually know (at least in mainstream mathematics) is whether all NP problems actually are all P problems. If there was a way to solve all NP problems in P time, that would mean P=NP. If there wasn't, and there were some problems we really couldn't solve in a reasonable amount of time, that would mean P!=NP.
If there's anything anyone wants me (or anyone other than me obviously) to explain any better, or if you want more information (or more context in relation to RSA), don't be afraid to ask. I only stop there because I have to go somewhere.
Hey there. You mentioned earlier that you hadn't read any of the previous posts. I don't know if you've caught up, but if you're planning to stick around and help out, even if it isn't necessarily with the RSA stuff, I'd highly recommend getting far more context than you seem to have. You can find links to archived threads (in chronological order) at the bottom of this post >>2556. Speaking of which, if you click the number at the end of each post header, it allows you to reply directly to someone, rather than just posting text (and then people have to figure out who it was directed towards).
Also, someone reported one of your posts. The last thing I want to do is ban you, but you've got to understand: this whole board is for discussing the entire scope of possibilities that arise from what we're all talking about (including what you're talking about), but this particular thread is for discussing RSA. We like to keep organized here, so if you want to discuss another topic (such as ways in which to use this kind of mathematics unrelated to RSA), you need to do it in another thread. One of the people here was kind enough to create one for you, which is right here >>2943. You can clearly help a great deal, which I'm sure we'll all be thankful for, but even if you don't like the idea of RSA-cracking technology being open source and available to anyone, that's what this particular thread is about. You're more than welcome to explain other things; we'd just ask that you do it in another thread is all.
Hey Topol, would you mind responding to this conversation starting here >>3008 in the Sonoluminescence thread? Someone's saying that VQC said some stuff about killing himself if the VQC wasn't public by Jan 1st and their only claim of evidence is that you probably saw it too (i.e. if you don't know what they're talking about it's probably bullshit).
Wild stab in the dark by someone who hasn't contributed for a week so maybe doesn't know what they're talking about: maybe 2 is that offset Chris mentioned and b for every e follows some pattern of d + x in some way with a similar offset.
Part of why I wish Chris gave us a bit more detail.
For when you need to crack RSA numbers on the go huh?
I've been going through Chris' old hints and found this
>I am going to show how to do it in a complexity of less than O(log t) where log t is the natural log of the length of c in bits.
>less than O(log t)
I wasted an entire week on that binary search wild goose chase when all I had to do is read the hints again properly. Nice. Anyway, that means that this algorithm's running time probably isn't something that will lead us to the solution, in case anyone still thought that might be the case.
Aside from it being something to do when I'm not busy and having the company of anonymous internet people, it was a bit of a waste of time in my opinion. I still can't tell if we're just waiting around for Chris to decide the time is right again or if we really are just going to keep going around in circles like this until it magically happens several months or years from now. No lurkers with malicious intent seem to have solved it, or we'd have heard about it by now. That means either there are no selfish people watching (that's a bit farfetched considering the recognition you'd get if you claimed it was you), Chris hasn't given us enough information to solve it (at this point I trust his judgement so I don't think it's that; plus, he figured all of it out himself, albeit, if memory serves, it took him years), or we just happen not to have figured it out yet (which Chris would be aware of and could solve by making it way more obvious than he has so far).
Here's some pointless c = c algebra I messed around with for a while. Maybe something magical will pop out if you stare at it for a really long time.
c = d^2 + e
c = (d+n)(d+n)-(x+n)(x+n)
Therefore
d^2 + e = (d+n)(d+n) - (x+n)(x+n)
d^2 + e = d^2 + n^2 + 2dn - x^2 - 2nx - n^2
d^2 + e = d^2 + 2dn - x^2 - 2nx
e = 2dn - x^2 - 2nx
x = d - a
Therefore
e = 2dn - (d-a)(d-a) - 2n(d-a)
e = 2dn - (d^2 - 2ad + a^2) - 2dn + 2an
e = 2an - d^2 + 2ad - a^2
n = ((a + b) / 2) - d
Therefore
e = 2a(((a+b)/2)-d) - d^2 + 2ad - a^2
e + d^2 = 2a(((a+b)/2)-d) + 2ad - a^2
c = 2a(((a+b)/2)-d) + 2ad - a^2
c = 2a(((a+(c/a))/2)-d) + 2ad - a^2
c = 2a(0.5(a+(c/a)) - d) + 2ad - a^2
c = 2a(0.5a + 0.5(c/a) - d) + 2ad - a^2
c = a^2 + c - 2ad + 2ad - a^2
c = c
If it was c^3 and c = 91
(1, 753571) (1, c^3)
(7, 107653) (a, a^2*b^3)
(13, 57967) (b, b^2*a^3)
(49, 15379) (a^2, b^3*a)
(91, 8281) (c, c^2)
(169, 4459) (b^2, a^3*b)
(343, 2197) (a^3, b^3)
(637, 1183) (c^2/b, c^2/a) or in other words c^4/c
Each factor is just every list of possible combinations of a, b and c that give c^k. I was trying to look for relationships between the cells I mentioned here >>3243 for a lower value of iMax than is necessary for c = 91 so it doesn't take hours to generate the grid, but for some reason it cuts off around n = 511 with setSize = 1 (I'm assuming it would be at t = 1 because VQC pointed out a while ago that in e=0's column, for subsequent t cells in a particular (e, n), a and b increment upwards in particular patterns. I'll find the post if you want. I read it in the pdf I posted in the last thread). With c^2 for c = 35, (a, b) = (1, 1225), i should equal 613, so that cell should show up at (e, n) = (0, 578). Can you think of why that isn't happening?
I know that. I think my code is wrong because even if I change setSize to 18 or more it doesn't show up. That's weird though. I thought it would have been at t = 1. Have you looked at the patterns as t increments always starting from lower numbers and occasionally a = one of the bs from before? What does the pattern of a and b look like before (0, 578, 18)?
You're using the Java version too, right? I'm trying to figure out how these cells are actually put into the grid. We know t = 18 for the cell we want based on math, but I'm looking at outputGrid and I'm not seeing how it would put it in at t = 18 instead of just the first instance of (e, n), which, for this cell, since those other ones aren't valid, would be t = 1.
>changing d and e without changing a or b
That doesn't make any sense. d = floor(sqrt(a*b)) and e = c - dd. They don't change if a and b don't change. What are you talking about? Are you using d and e to describe different variables? Because if you are it explains why I also don't understand this >>3258 stuff and you should pick different letters.
>>3272
But you haven't explained the logic by which you can create multiple records with the same a and b values but with changing e and d values when e and d are calculated based on a and b. You can't just change the formula to e-2*d-1. e = c - dd. That's what I don't understand.
You've also avoided answering my questions over and over. Jesus Christ anon. If you're right then all you have to do is explain your logic, and if you're wrong maybe I'll save you the time sink of a wild goose chase.
>you have specific values for a and b, say, a = 3, b = 5
>c = a*b (in this case 15), so if a and b don't change, c doesn't change
>d = floor(sqrt(c)) (in this case 3), so if c doesn't change, d doesn't change (also meaning if a and b don't change, d doesn't change)
>e = c - dd (in this case 6), so if c and d (and by extension a and b) don't change, e doesn't change
>if a and b don't change, d and e don't change
Why is it that whenever I ask where the logic in changing the formula is, all you say is "test them"?
So let me get this straight. Is this what you're suggesting?
>there are infinite f values for a given (a, b)
>the first is e - 2d + 1
>if you use a value of f that isn't the original one, you're changing the formula to calculate f
>that means you also need to change the formula for every other variable
>this works because f is what you add to c to get a square and each successive f creates the next square higher than the larger square
>this means the changed formulae for the other variables relates to the smaller square
>as the squares increase or decrease (whatever's going on), they calculate c correctly all the same, just based on the changed formulae that come from how far into the infinite set of fs you are
This doesn't seem like what VQC had in mind, and I can't tell how this relates to finding n, but if that's what you're saying and it makes mathematical sense then I guess there might be something we can do with it maybe.
It seems like for each extension of these variables you'd need a separate dimension of the grid, though, right? Otherwise you'd have regular ds in the same side (either negative or non-negative) of the grid as any of the infinite other ds, and I can imagine a lot of them would overlap with other cells.
Oh okay. I thought you were just using greentext wrong.
I've been looking into this >>3242 >>3246 stuff about finding c^2 in (0, n). I set up some grids and calculated it that way so I didn't have to wait a year for the grid to generate with a really high iMax. The idea with these grids is to see if there's a relationship between each possible c = c^2 record in the 0 column. For c^2, there are 5 cells in the 0 column: (1, c^2), (a, c^2/a), (b, c^2/b), (a^2, b^2) and (c, c). If c is a semiprime it won't be in column 0 unless it's the square of a prime (which we aren't looking for in RSA), but for all we know there might be a mathematical relationship in the c^2 cells that brings us to the answer.
The first image is (in the grid on the right) each possible combination of (a, b) on the x axis for a bunch of different c^2s on the y axis. I'll post the {e, n, d, x, a, b} and t values generated by each of these sets of (a, b) for a given c^2 in the next post.
So these are each of the {e, n, d, x, a, b} and t values for a given c^2 based on each possible (a, b). We know c and we're looking for a and b (meaning we could also find n or x). Can we can use (a, b) = either (1, c^2) or (c, c) and mathematically get to one of the cells with one of a and b actually equaling a or b? Can anyone see any relationship?
>Why are people like you more qualified to solve this problem than mathematicians?
We aren't all autistic by the way.
>(0,n) one row to rule them all
>x marks the spot
>the square in 0,1 is the basis for all patterns in n=1
So you're saying the solution has something to do with these posts?
Or are you saying something else (because nobody replied to those posts with anything)?
I'm looking at the 0 column and the relationship between c and c^2. I found a few weird things. For the square of a semiprime (probably every square but I'm only paying attention to that table of squared semiprimes), for (a, b) = (1, c^2), which will always be the first cell of the infinite set, t increases by the same number throughout the infinite set (e.g. 11, 21, 31, 41, 51, always +10, but that 10 changes set to set), a is the list of squares from 1 to infinity (the same 1, 4, 9, 16 each time in every set), b increases by a linearly increasing amount (e.g. 441, 484, 529, 576, 625, so it's +43, +45, +47, +49 etc), d also increases by a linearly increasing amount (e.g. 21, 44, 69, 96, 125, so it's +23, +25, +27, +29 etc), and x is linear with a gradient somehow related to t (e.g. 20, 40, 60, 80, 100, which is a gradient of t's increase * 2). Obviously e and n stay the same (0, 200) as you'll see in that table. The example I used for those numbers was c^2 = 441, c = 21.
Here's another example with c^2 = 225, c = 15. (e, n) = (0, 98). t is 8, 15, 22, 29 (+7 each time), a is 1, 4, 9, 16 etc like before, b is 225, 256, 289, 324, 361 (+31, +33, +35, +37), d is 15, 32, 51, 72, 95 (+17, +19, +21, +23), and x is 14, 28, 42, 56 etc (linear with a gradient of t's increase * 1 instead of *2 for some reason).
What that means is, if we use c^2 as our c, every variable follows a predictible pattern, although I don't know why each of the numbers starts where they are. Maybe it has something in common with (1, c). I looked at the (1, c) cells for c = 21 and c = 15 but the relationships are pretty weird and I can't figure out how they work or how they relate to their corresponding c^2 cell, but there's bound to be at least one if not several relationships there. It might just be how late early it is right now but I really hope someone else looks at this and finds something, because it definitely seems like an important relationship and Chris seems to have alluded to it.
When did he say that? Link?
>What's the third decision?
At the top of the decision tree, if e = 0, it's a square, so it's not an RSA number (right?). If e != 0, we need to find n or x, which apparently relies on whether it's odd or even. In the second decision, whether it's even or odd determines what t is, but then to calculate t you need to know x, and if you knew x you'd know a. So do we know any other calculations that require knowing if e is even or odd? Otherwise, should we be using this to find relationships?
Last post so I can sticky the new one already