ID: 2bf75e RSA #10 Feb. 8, 2018, 2:14 a.m. No.4140   πŸ—„οΈ.is πŸ”—kun   >>4225

The only thing you need to do to solve this is to organize.

Enumerate EVERY rule.

Global rules.

Row rules.

Column rules.

They are a finite set and RELATED.

This is your MAP.

You will be able to use the MAP to find n from c.

Enumerate the rules.

Win.

ID: 2bf75e Feb. 8, 2018, 2:25 a.m. No.4142   πŸ—„οΈ.is πŸ”—kun   >>4143

Code

 

C#

BigInteger Square Root β€”β€” https://pastebin.com/rz1SdACZ

Generate Bitmap within original code β€”β€” https://pastebin.com/hMTtJF6E

Generate the large square for e and t β€”β€” https://pastebin.com/nbjs2kz4

How to run VQC code on Linux β€”β€” https://pastebin.com/6HnN7K5X

More on generating a bitmap with the original code β€”β€” https://pastebin.com/JUdtehb4

PMA's tree generator β€”β€” https://pastebin.com/ZH9fSWu2

Original VQC code β€”β€” https://pastebin.com/XFtcAcrz

Unity Script β€”β€” https://pastebin.com/QgAXLQj3

Unity Script 2 β€”β€” https://pastebin.com/Y38nVWgT

 

Java

Traverse the VQC cells in real-time β€”β€” https://anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z

Tree Generator β€”β€” https://pastebin.com/VZnQQR2i

Tree Generator w/ x & x+n search β€”β€” https://pastebin.com/0jPr3RrE

VQCGenerator β€”β€” https://pastebin.com/VMRnkXFP

VQCGenerator w/ Bitmap β€”β€” https://pastebin.com/Dgu9aP1h

 

NodeJS

BigInteger Library and Sqrt β€”β€” https://pastebin.com/y8AXtFFr

 

Python

3D VQC β€”β€” https://pastebin.com/vdf8SpYt

3D VQC (v2) β€”β€” https://pastebin.com/wZM5Thzu

Calculate variables based on e and t β€”β€” https://pastebin.com/4s6McdbN

College Anon's code β€”β€” https://pastebin.com/d8xZZnm0

Create the VQC β€”β€” https://pastebin.com/NZkjtnZL

Fractal cryptography β€”β€” https://pastebin.com/XuN4U7Dv

Generate any cell in (0,1) and (0,2) β€”β€” https://pastebin.com/gRTYpdMU

Generate cells for a (and more) β€”β€” https://pastebin.com/iAizgLFF

Generate genesis cell β€”β€” https://pastebin.com/GKzcCpMF

Generate positive AND negative genesis cells β€”β€” https://pastebin.com/9ixjRyxt

Get A and B from C and N example β€”β€” https://pastebin.com/s0SZ9BNF

VQC + t β€”β€” https://pastebin.com/Lgufk0db

RSA & PGP key wrapper β€”β€” https://pastebin.com/vNqnPRJR

 

Rust

Additional VQC code β€”β€” https://play.rust-lang.org/?gist=50def916ad48400bc5d638fbf119ae85&version=stable

Check if a number is prime β€”β€” https://huonw.github.io/primal/primal/fn.is_prime.html

Create Bitmap using the VQC Generator β€”β€” https://play.rust-lang.org/?gist=c2446efeec452fe14e1ddd0d237f4173&version=stable

Create Bitmap using the VQC Generator [V2] β€”β€” https://pastebin.com/zGSusyz5

Generate the VQC β€”β€” https://play.rust-lang.org/?gist=6b6beb372b6b931f1abd30642a35a80c&version=stable

 

Factorization methods (Java)

Binary search for i β€”β€” https://pastebin.com/TAt5bDsR

GCDFactor β€”β€” https://pastebin.com/70GJSMrv

Count down from t of 1c element β€”β€” https://pastebin.com/xxYa946V

Mirrors 1c until e=(-x+n^2) β€”β€” https://pastebin.com/WJBqPM4P

Calculate factors using -x jumps β€”β€” https://pastebin.com/gKX9GW9r

 

Previous Threads

RSA #0 β€”β€” https://archive.fo/XmD7P

RSA #1 β€”β€” https://archive.fo/RgVko

RSA #2 β€”β€” https://archive.fo/fyzAu

RSA #3 β€”β€” https://archive.fo/uEgOb

RSA #4 β€”β€” https://archive.fo/eihrQ

RSA #5 β€”β€” https://archive.fo/Lr9fP

RSA #6 β€”β€” https://archive.fo/ykKYN

RSA #7 β€”β€” https://archive.fo/v3aKD

RSA #8 β€”β€” https://archive.fo/geYFp

 

Videos on cryptography β€”β€” https://pastebin.com/9u3hwywe

ID: 2bf75e Feb. 8, 2018, 2:31 a.m. No.4143   πŸ—„οΈ.is πŸ”—kun

I cut my VQC map into 3 pictures to make it easier to render.

 

Also, Chris said the pic that looks like 3 right triangles in >>4142 shows the fractal-like nature of the VQC. From left to right, it shows every cell where e is even, every cell where e is odd, and both pictures overlaid.

ID: 2bf75e Feb. 8, 2018, 2:35 a.m. No.4144   πŸ—„οΈ.is πŸ”—kun   >>4192

He said this to me but trust me, he means it to all of us who've kept faith through these past few months. It's going to be probably one of the best days of my life seeing all our research bear fruit.

Anonymous ID: bd79a8 Feb. 8, 2018, 5:06 a.m. No.4155   πŸ—„οΈ.is πŸ”—kun   >>4160

>So, re-read a few crumbs and had a breakthrough this eve!

 

I am gonna post an idea here, and if MM's breakthrough overshadows this, great, that's positive news, hope it works. Here's the idea. What about iterating new_d = old_d^2 + e. Inspired by the Mandelbrot set clue. Just mucking around with it now.

Anonymous ID: bd79a8 Feb. 8, 2018, 1:53 p.m. No.4168   πŸ—„οΈ.is πŸ”—kun   >>4169

>>4160

I haven't found anything useful but, with the Mandelbrot set you would start with z_0=0, and a constant c, then z_1 = z_0^2+c ; z_2 = z_1^2 + c, and so on.

 

I was picking a fixed e, and choosing an intial d_0 value, then iterating:

d_1 = d_0^2 + e

d_2 = d_1^2 + e

 

I still haven't gotten any math type programs going on this pc; the numbers grow quick and I haven't found anything useful yet. Plus I keep looking at different spirals…

Anonymous ID: bd79a8 Feb. 8, 2018, 2:09 p.m. No.4169   πŸ—„οΈ.is πŸ”—kun

>>4168

and I am curious also about whether, when going from d_n to d_n^2+e, there is any useful connection between the cell(s) with d_n^2 + e as its c-value (the product of its a,b terms), and the cell(s) with d_n as a d-value. That type of thing.

Anonymous ID: 2bf75e Feb. 8, 2018, 6:42 p.m. No.4179   πŸ—„οΈ.is πŸ”—kun

>>4178

I'm seeing God's judgement in the future. Newly released Strzok-Page texts corrobate Q saying that the FBI'mWithHer had a plan to KILL the President.

 

Link to ALL NEW Strzok-Page texts:

http://www.hsgac.senate.gov/download/appendix-c_-documents

CollegeAnon !LAbIRp9cT. ID: d30f09 Feb. 9, 2018, 11:50 a.m. No.4193   πŸ—„οΈ.is πŸ”—kun   >>4198 >>4199

>>4191

I'm here every day. Every class I have I draw out trees and try to figure them out. I've tried a bunch of shit but I can't even get a pattern to be consistent, let alone one that is correct. I haven't posted anything because I haven't got anything new. I've tried analyzing primes and factors of certain numbers and I can't seem to find anything. In number theory we are learning about Legendre symbols and they have to do with squares and remainders so I'm thinking about that and how those could fit into this.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 9, 2018, 3:06 p.m. No.4197   πŸ—„οΈ.is πŸ”—kun

Posting a work in progress for any feedback.

 

Pics attached for c=145, 901 and 6107.

 

Shows the factor tree with various calculated columns. Each d node is a perfect square. We can calculate an f, x, and assuming n=1, the small square. We can also get to the the next perfect square at c+f. And the difference in squares between them.

 

Below each tree is a small square analysis between the c and prime records, their corresponding na records, and their na records compared to the genesis cell. Each analysis includes all prime factors for the difference in small squares.

 

My thinking is that perhaps the factor tree can somehow be used to arrive at the difference between the squares.

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 9, 2018, 6 p.m. No.4200   πŸ—„οΈ.is πŸ”—kun   >>4201 >>4204

Checked! Let's all shitz and bantz till someone does something smart, or until VQC shows up. Honestly, we are Chris' Math Padawans. We have put in heroic effort, and he knows that. He put out the call, we responded (with thousands of hours of thought and study) and now he can pop by and tell us ONE clue that could unlock this whole thing, bc we are all on the same page here.

 

Think about how much we've learned as a group. It's pretty incredible. If I was a teacher, I'd be pretty fucking proud of my students at this point. Let's enjoy this break to talk some shit, have fun, and enjoy each other's company. I mean, c'mon! MM is now running comparative analysis on Excel! (go MM!) I can program my TI-89, solve algebra problems, write formulas in Excel, and bring enthusiasm to the group. WTF? But here I am running with you excellent programmers on a noble quest. Here's my shitz and bantz:

 

SLOW DJT CLAP

 

"I'm so proud of you Anons.

In fact, I can think of anyone who would be prouder. In fact, I'm the proudest.

You have done excellent work working to bring down the cabal, excellent work.

These Anons have done such great work, and they deserve our thanks. Great work.

Haven't they done such great work? They have done SUCH great work.

You know, I think we should all be proud of them. Stand Up Anons! We should be proud of them.

We should be SO proud of them. I'm so proud of these Anons…"

 

SLOW DJT CLAP

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 9, 2018, 6:28 p.m. No.4202   πŸ—„οΈ.is πŸ”—kun

I also keep coming back to this:

 

starting record at c=145

(1,61,6) = {1:61:12:11:1:145}

na transform to (1,1,6)

(1,1,6) = {1:1:72:11:61:85}

 

prime solution at a=5,b=29

(1,5,4) = {1:5:12:7:5:29}

na transform to (1,1,4)

(1,1,4) = {1:1:32:7:25:41}

 

Thinking out loud here…

 

For c record (1,c)

[t] is the same in (e,1) and (e,n)

x[t] is the same in (e,1) and (e.n)

d[t] in (e,1) is original d (12) + (61-1) = transformed d = 72

 

a[t] = na = 1*61 = 61

For prime record:

[t] is the same in (e,1) and (e,n)

x[t] is the same in (e,1) and (e,n)

d[t] in (e,1) is original d (12) + (n-1)*a

transformed d is 12+ (5-1)*5 = 32

a[t] = na = 5*5 for transformed a

 

is there supposed to be a (n-1)*a connection for d[t]?

 

This part is the key i think:

 

na transform to (1,1,6)

(1,1,6) = {1:1:72:11:61:85}

 

Then to this:

 

na transform to (1,1,4)

(1,1,4) = {1:1:32:7:25:41}

 

Thoughts? If the factor tree can help us bridge the gap between x=11 in (e,1) and x=7 in (e,1) we've got it. For this example, delta x = 11-7 = 4.

Thoughts?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 9, 2018, 6:43 p.m. No.4203   πŸ—„οΈ.is πŸ”—kun

Here's some rSAXpiration to code by:

https://www.youtube.com/watch?v=NUauHXpTyvo&list=PLB3K2iKmM5bT1kTG4BLXNHTg2461OJ3n8

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 9, 2018, 6:45 p.m. No.4204   πŸ—„οΈ.is πŸ”—kun   >>4205 >>4206

>>4200

I'm that guy who took this class as an elective and got a C for "fuck it, we didn't expect him to last this long anyway" like with that 3000 level logic course about "color theory".

 

yeah, they didn't mean like in painting. I was getting a BFA and it sounded related. Holy fuck. lol

VA !!Nf9AmQNR7I ID: a60c59 Feb. 9, 2018, 7 p.m. No.4205   πŸ—„οΈ.is πŸ”—kun   >>4208

>>4204

Lol, Topol! :) You're our math muse, and we LOVE you faggot. <3

 

Here's my High School Topol LARP:

 

>Hey Guise, I'm Topol

>I'm that cool trippy kid whose parents followed the Grateful Dead.

>I got high during art class in the back room

>In class, the teacher would be stymied by my excellent questions.

>Then I would get my disciples high behind the theater building after school

>I love truth and I seek it out

>Wanna hang out?

VA !!Nf9AmQNR7I ID: 819f3a Feb. 9, 2018, 8:09 p.m. No.4207   πŸ—„οΈ.is πŸ”—kun

>>4206

Thanks MA! SpaceX "launch" in studio? LOL, no moon landing either. Chris, please give us some Agartha crumbs? Come on, VQC your disciples are bored and stuck.

 

"LORD OF MATH ROCK, BLESS US WITH YOUR MIGHTY PRESENCE."

 

Overton window expand? VQC is just a normal Anon like us who also is also the KING FAGGOT MATH GENIUS. Hanging out with us and teaching us cool ass math. To help save the world. To expand our minds. To teach us the underlying nature of numbers that all have missed. UNTIL NOW. Time for a Revolution of consciousness and understanding. We're part of it, Anons. We're doing our part to better the world even working right here.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 9, 2018, 8:13 p.m. No.4208   πŸ—„οΈ.is πŸ”—kun   >>4211

>>4205

I probably woulda had more friends in high school if I wasn't straight edge lol

 

Granted, being straight edge, I'd flutter through all the groups. Doesn't matter as much when you're all adulted an' shit.

 

https:// youtu.be/qIIOza9ZaXw

VA !!Nf9AmQNR7I ID: a60c59 Feb. 9, 2018, 10:42 p.m. No.4212   πŸ—„οΈ.is πŸ”—kun   >>4215

Where are you fags? Come on now, Anons. Let's fucking work and talk some fun shit too. JFK knew dafuk was up. Fucking Mafia Catholic tho. Still knew about Spectre, while cheating on Jackie. Killed for getting a conscience.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 10, 2018, 2:29 a.m. No.4217   πŸ—„οΈ.is πŸ”—kun

I'm probably stating something we already know, but have you noticed that the n in (0, n) for a^2, b^2 appears to be equal to 2*(x+n)^2?

VQC !!Om5byg3jAU ID: d04b72 Feb. 10, 2018, 3:08 a.m. No.4219   πŸ—„οΈ.is πŸ”—kun   >>4225 >>4245 >>4285 >>4757

Part 3 a - Odd (x+n) - Overview

(x+n)(x+n) = nn + 2d(n-1) + f - 1

 

What does this look like?

(x+n)(x+n) is odd

 

All odd squares = 1 + 8T

All odd square = 1 + the product of 8 triangles that are the same.

 

Visualise.

 

2d(n-1)

 

Why is (n-1) important?

VQC !!Om5byg3jAU ID: d04b72 Feb. 10, 2018, 3:37 a.m. No.4221   πŸ—„οΈ.is πŸ”—kun   >>4222 >>4235

Odd (x+n) - Examples

 

Let's use some real examples.

 

One odd (x+n) known, one odd (x+n) unknown

 

Rsa 100 has odd x+n and is the smallest.

 

Homework.

Of the unsolved RSA numbers, which is the largest that has an odd (x+n)?

 

We will use that one.

MM !!DYPIXMDdPo ID: 8f5be8 Feb. 10, 2018, 4:08 a.m. No.4225   πŸ—„οΈ.is πŸ”—kun   >>4226 >>4227

>>4140

>>4213

Hey PMA, was going through this early this am, then distracted by latest Q-drops. Your work has been really great, much appreciated.

 

>>4219

Hey V! Glad you found RSA#10. Your Time*ing is most excellent. Thanks, we'll get to work. Have a busy day but now I know what I'll be doing Saturday night - math! Kek.

 

ps fags - think David Rothschild, @DavMicDot on twitter, has been sending MSM instructions each day "Dear MSM" - they are the MSM talking points, sent via twitter.

Saw it somewhere, how does one get an archive of all tweets from a user?

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 10, 2018, 6:54 a.m. No.4235   πŸ—„οΈ.is πŸ”—kun   >>4236 >>4239

>>4221

First, welcome back!

 

I'm very excited for today!

 

So I wrote a small python script to compute the remainder of e for the unsolved challenges and I believe RSA 490 is the largest unsolved number with (x + n) being odd.

VQC !!Om5byg3jAU ID: d04b72 Feb. 10, 2018, 8:15 a.m. No.4239   πŸ—„οΈ.is πŸ”—kun   >>4240 >>4241 >>4251 >>4252

>>4235

Thank you!

>>4236

Sounds good.

We'll walk through RSA 100

and then RSA 490

 

The factor tree is used to factor d and e.

 

Factoring these (and down the tree) allow for the factoring of c.

 

For odd (x+n) we'll walk through how each of the eight triangles (+1) are constructed to make (x+n)(x+n) and build an algorithm.

 

Once that is demonstrated, we'll look at the short cuts using the grid.

VQC !!Om5byg3jAU ID: d04b72 Feb. 10, 2018, 9:46 a.m. No.4242   πŸ—„οΈ.is πŸ”—kun   >>4243 >>4624 >>4933

>>4240

>>4241

Thanks!

The key here is first understanding how the (x+n) square being added to c is constructed in terms of also being analogous to an L shape on the side of the square of d sides which must incorporate the remainder e in the L shape (or incorporates the 'gap' made by f).

We will build up the code a sub steps of Part 3.

 

After demonstrating with with odd (x+n), we will factorise the examples, the remaining RSA odd (x+n) numbers, then show Part 3 b which are the even (x+n).

 

Once the factorisation methods are complete, the short cuts via patterns in The Grid (e,n) will be clearer.

3DAnon !!!N2ZmYzdiNjdkYTk2 ID: fec063 Feb. 10, 2018, 10:31 a.m. No.4243   πŸ—„οΈ.is πŸ”—kun

>>4242

Welcome back indeed, we missed you

 

>>3927

This post from Teach seems relevant now, but we're generating smaller triangles from the tree and taking the product of 8 of them, interesting.

My thinking is we use the first d branch as one side, and the first e branch as the other, adding up all the lower branches. Are we going all the way to the bottom of the tree or is there some condition to stop on? I ask because we have a few different variations of the generation code, mainly related to how many leafs are generated for the lower branches and where to end.

If we don't have to go to the bottom this is irrelevant but if every value is included in the calculation we need to make sure we have the correct tree. Attached two examples.

Looking forward to some more progress here

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 10, 2018, 2:29 p.m. No.4250   πŸ—„οΈ.is πŸ”—kun   >>4254

>>4246

Hmm, I never doubled checked, but it appears you are correct regarding (e, 1).

 

Take (3, n). 3 % 4 = 3, so we should expect (x+n) to be odd. But then we have (3, 2) and (3, 3). Both of which have even and odd (x+n). So we can't deduce parity of (x + n) based on e then?

 

Maybe VQC can clarify.

Teach !!UgZAPoSXEk ID: 1dac59 Feb. 10, 2018, 2:34 p.m. No.4251   πŸ—„οΈ.is πŸ”—kun   >>4254 >>4261 >>4285 >>4290 >>4294

>>4239

Welcome back, and thanks for the tips.

A few possible answers to the questions…

 

1+8T for odd squares:

Draw odd square, remove middle square, divide remaining into 4 equal rectangles, by drawing out from the middle square.

Each rectangle's sides are only 1 different from each other, which is the same as 2 equal triangles.

See image.

 

Why is (n-1) important?

I think it has to do with the same property.

A square with 1 row or 1 column removed is 2 equal triangles of (n-1) each.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 10, 2018, 3:31 p.m. No.4255   πŸ—„οΈ.is πŸ”—kun   >>4262

>>4254

But take a look at (3, 2) and (3, 3).

 

Compute the (x + n) and look at the parity. For (3, 2) you'll have (x + n) as odd and for (3, 3) you'll have (x + n) as even.

 

This should hold for any (e, n) where e also has (e, n-1) as a valid infinite set of records.

Anonymous ID: 4dcd00 Feb. 10, 2018, 3:35 p.m. No.4257   πŸ—„οΈ.is πŸ”—kun   >>4258 >>4260

>>4252

Look at the triangular numbers prime factorization. Look at the next one and the previous one. The triangular number will share some amount of factors with the next one, and the rest with the previous one. This pic is all triangular numbers up to like a thousand on the left (so pixel at depth 5 is 5(6)/2=15) then the horizontal axis is factors of that number. They are always in vertical pairs.

Start with 3

3 = (3)

6 = (2,3)

10=(2,5) (genrated 5)

15 = (3,5)

21 = (3,7) (generated 7)

28 = ((2,2),7)

36 = ((2,2),(3,3))

45 = ((3,3),5)

55 = (5,11) (generated 11)

66 = (2,3,11)

78 = (2,3,13) (generated 13)

etc.

 

So you can get generate primes through the triangular numbers.

 

QED??

Teach !!UgZAPoSXEk ID: 1dac59 Feb. 10, 2018, 3:52 p.m. No.4259   πŸ—„οΈ.is πŸ”—kun   >>4268

>>4252

C'mon CA share! :) hopes are up.

 

>>4254

Thanks PMA. So have you come to any conclusion about the e%4 parity of x+n yet?

 

Also a few more comments, sorry if these are obvious.

A square with even side length can be divided into 4 equal smaller squares, so we can reduce a big square of even parity into 4 smaller squares repeatedly until they're odd.

 

A while ago vqc posted about base. I think I've kinda figured out how that plays into it…

 

Example c=145, in base 2 = 10010001

c = 2^7 + 2^4 + 1.

c = 2^3(2^4 + 2) + 1

c = 2^3(2(2^3 + 1)) + 1

c = 8(2(8(β–³1) + 1)) + 1

CollegeAnon !LAbIRp9cT. ID: 4dcd00 Feb. 10, 2018, 4:03 p.m. No.4261   πŸ—„οΈ.is πŸ”—kun   >>4263 >>4264

>>4251

I screwed with these all day. I think he told us this because for even squares we can just look at them like 4 smaller squares. We can do this until we get smaller numbers.

 

145 = 12^2 + 1

= 4*6^2 + 1

=4^2*3^2 + 1

Then at this point we have a odd square which we may be able to do stuff with. Just trying to fit together the 1+8T and the divide the square by 2 crumbs.

Teach !!UgZAPoSXEk ID: 1dac59 Feb. 10, 2018, 4:17 p.m. No.4263   πŸ—„οΈ.is πŸ”—kun   >>4265 >>4271

>>4261

See my latest post :)

 

But you do have an interesting idea here…

Now we can break down either even or odd squares into triangles, and you're saying we can reverse the entire process and move up instead of down?

I like that thought.

CollegeAnon !LAbIRp9cT. ID: 4dcd00 Feb. 10, 2018, 4:21 p.m. No.4264   πŸ—„οΈ.is πŸ”—kun

>>4261

 

= 443*3 + 1

= (T(4) + T(3))(T(3) + T(2)) + 1

= T(4)T(3) + T(4)T(2) + T(3)T(3) + T(3)T(2) + 1

= 106 + 103 + 66 + 63 + 1

= 60 + 30 + 36 + 18 + 1

= 6(10 + 5 + 6 + 3) + 1

= 6( 3(5) + 3(3)) + 1

= 18(5+3) + 1

 

Something like this process for the factoring 145.

CollegeAnon !LAbIRp9cT. ID: 4dcd00 Feb. 10, 2018, 4:25 p.m. No.4265   πŸ—„οΈ.is πŸ”—kun   >>4266

>>4263

So you're still going to need to factor, but no. Check this example out:

These are (x, T(x), factors of T(x))

(15, 120, ([3, 5], [2, 2, 2]))

(16, 136, ([2, 2, 2], [17]))

(17, 153, ([17], [3, 3]))

(18, 171, ([3, 3], [19]))

(19, 190, ([19], [2, 5]))

(20, 210, ([2, 5], [3, 7]))

(21, 231, ([3, 7], [11]))

 

So for example, T(17) you know the factors that match with the next are (3,3), so you take T(18)/(3*3) and you get 19. So you don't necessarily need it to be prime. You could just crank through this list

CollegeAnon !LAbIRp9cT. ID: 4dcd00 Feb. 10, 2018, 4:30 p.m. No.4267   πŸ—„οΈ.is πŸ”—kun   >>4272

>>4266

Even more, It seems to just generate every single number. This might be crap. :(

 

(15, 120, ([3, 5], [2, 2, 2])) (15,8)

(16, 136, ([2, 2, 2], [17])) (8,17)

(17, 153, ([17], [3, 3])) (17,9)

(18, 171, ([3, 3], [19])) (9,19)

(19, 190, ([19], [2, 5])) (19,10)

(20, 210, ([2, 5], [3, 7])) (10,21)

(21, 231, ([3, 7], [11])) (21,11)

 

^proof this is bunk

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 10, 2018, 4:36 p.m. No.4268   πŸ—„οΈ.is πŸ”—kun   >>4271 >>4272

>>4262

I think once n 1, you can’t rely on the parity anymore. n is any factor in (e,1). And you x’s will either be odd or even. So you’re going to get mixed results as you go higher up the tree.

 

>>4259

Not yet. This seems like the next fork in the decision tree, and we can certainly determine the parity of n, x+n, and d+n from e. Earlier VQC mistyped c for e. Perhaps this e mod 4 rule applies to d+n?

 

I wouldn’t presume to say that. So I’m confused.

 

Notwithstanding, seems pretty obvious that we’re looking for a triangle formula the for small square when odd. And there is some way the tree gives that to us.

 

You would also think this 1+8T formula applies to smaller trees. But with really small odd x+n values you’re getting into some small fractions.

Anonymous ID: 268629 Feb. 10, 2018, 5:25 p.m. No.4274   πŸ—„οΈ.is πŸ”—kun

>>4273

Also, 145 is the stupidest possible number we could use as a favorite default, since e = 1, x + n = d in the final answer, n = a and all other manner of misleading shit

ID: 2bf75e Feb. 10, 2018, 11:09 p.m. No.4284   πŸ—„οΈ.is πŸ”—kun

If you combine this and the fact that both n's for a number have the same parity you also know the parity of x.

 

For example, if we know x+n is even, and we know n is odd, therefore x must be odd to sum together to make an even number. This is true for 145.

 

12 = x+n, even

5 = n

7 = x

 

5 + 7 = 12, odd + odd = even

Anonymous ID: a34e47 Feb. 10, 2018, 11:14 p.m. No.4285   πŸ—„οΈ.is πŸ”—kun   >>4286 >>4287 >>4288 >>4302 >>4504 >>4533 >>4550

>>4219

That sneaky VQC. I think he's just given us a new equation.

>(x+n)(x+n) = nn + 2d(n-1) + f - 1

Using distributive property, (x+n)(x+n)= xx + xn + xn + nn

So for this portion: 2d(n-1) + f - 1 = 2xn + xx

 

Do we have this equation already? Maybe we do, can't find it in my notes tho.

 

>>4251

Great visualization, Teach! Check out the one I've attached. For (x+n)=15. For odd (x+n) there's always a square 1x1 in the middle. Right triangles, and exterior side will always be one more than the interior side. Thoughts?

ID: 2bf75e Feb. 10, 2018, 11:18 p.m. No.4286   πŸ—„οΈ.is πŸ”—kun   >>4289 >>4296

>>4285

This is a work of art. Here's a clearer explanation of the 1*8T crumb. It ONLY applies to odd squares.

 

1^2 = 8*T(0) + 1

3^2 = 8*T(1) + 1

5^2 = 8*T(2) + 1

7^2 = 8*T(3) + 1

9^2 = 8*T(4) + 1

11^2 = 8*T(5) + 1

13^2 = 8*T(6) + 1

 

VQC specifically said 8 triangles because of that picture you drew.

ID: 2bf75e Feb. 10, 2018, 11:40 p.m. No.4291   πŸ—„οΈ.is πŸ”—kun   >>4296 >>4300

Though I don't think it's useful yet, here's proof you can know the parities of x, n, and x+n (and thus (x+n)^2 because of a theorem that says all whole integers share parity with their squares)

 

The parity was calculated pre-factorisation just from some boolean logic.

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 10, 2018, 11:47 p.m. No.4292   πŸ—„οΈ.is πŸ”—kun

>>4290

Thanks Teach!

 

Parity for (x+n) for any column e also depends on n. I'm going over the grid, and looking down each column I see (x+n) switching parity within columns.

ID: 2bf75e Feb. 11, 2018, 12:24 a.m. No.4298   πŸ—„οΈ.is πŸ”—kun

Chris told me that the tree solution finds x+n or x. If we know x+n is odd and are to construct x+n from triangles, then that must mean if we know x+n is even we are to construct x.

Anonymous ID: bd79a8 Feb. 11, 2018, 12:26 a.m. No.4299   πŸ—„οΈ.is πŸ”—kun   >>4318

>>4283

grid patterns can't just be for semiprimes

 

>>4276

> this forum is such a pain in the ass to read.

yep haha! we anons are generally sloppy, and vqc makes sloppy / partially true statements sometimes . let's hope it's to throw off the overconfident boring academics :)

 

>>4281

>The rule still holds. Actually read what he said for fuck sake.

vqc did make a mistake here, when he says 'even, even, odd, odd' it doesn't apply to x+n but it works for d+n

 

thanks to PMA for that breakdown here >>4246

e mod 4 : 0, 1, 2, 3

x+n : odd, even, odd, even

d+n : odd, odd, even, even.

I didn't check the whole first row but what I checked was consistent with this. In the rest of the grid (n>1) I have had seen one counterexample at (4,20) haha, and in general this e mod 4 rule can't hold throughout the grid.

Anonymous ID: bd79a8 Feb. 11, 2018, 12:32 a.m. No.4300   πŸ—„οΈ.is πŸ”—kun   >>4301

>>4291

you don't even need a theorem, you can play with the rules pretty easily:

odd + odd = even

odd + even = odd

even + even = even

 

odd^2=odd

even^2=even

 

odd*even=even

odd*odd=odd

even*even=even

 

odd-even=odd

even-odd=odd

odd-odd=even

even-even=even

Anonymous ID: bd79a8 Feb. 11, 2018, 12:37 a.m. No.4301   πŸ—„οΈ.is πŸ”—kun

>>4300

as a follow up, when c is odd:

c=i^2-j^2 (remember i = d+n, j = x+n)

a difference which is odd, so one of i, j is odd, the other even. we would be looking at the combination with x+n odd, and d+n even, or vice versa. (this doesn't tell us about e unless it's in row 1, though)

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 1:27 a.m. No.4303   πŸ—„οΈ.is πŸ”—kun   >>4304 >>4315 >>4316 >>4346

What we're going to do is walk through RSA 100.

It's big enough to make "irrelevant" patterns disappear, since it's factorisation is known, it allows us to follow the explanation more clearly.

For those interested in symbolism, you'll see a bunch to do with pyramids (square base) and triangular numbers, with the tops of triangles missing or "detached".

The process for producing the first algorithm is clearly ancient.

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 1:29 a.m. No.4304   πŸ—„οΈ.is πŸ”—kun

>>4303

Here is the code in C# for handling creating triangles of arbitrary size.

 

/// <summary>

/// Tn = (nn + n)/2

/// </summary>

/// <param name="base_t"></param>

/// <returns>Triangle number from base</returns>

public static BigInteger Triangle(BigInteger base_t)

{

BigInteger square = BigInteger.Multiply(base_t, base_t);

BigInteger square_side = BigInteger.Add(square, base_t);

return BigInteger.Divide(square_side, two);

}

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 1:36 a.m. No.4305   πŸ—„οΈ.is πŸ”—kun   >>4549 >>4852

Here is code for calculating n from the base (x+n), c and d

n will be correct for a (x+n) that works, otherwise it is a result that is used subsequently to calculate whether an (x+n) exists before the (x+n) value for the product of 1 and c.

 

public static BigInteger Get_n_from_odd_triangle_base(BigInteger bs, BigInteger c, BigInteger d)

{

BigInteger triangle = Triangle(bs);//Create triangle from base

BigInteger eight_base = BigInteger.Multiply(triangle, eight);//multiply by eight

BigInteger XPN = BigInteger.Add(eight_base, one);//add one to create (x+n)(x+n)

BigInteger DPN = BigInteger.Add(XPN, c);//(d+n)(d+n)

return BigInteger.Subtract(Lib.Sqrt(DPN),d);//sqrt((d+n)(d+n)) - d = n

}

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 1:38 a.m. No.4306   πŸ—„οΈ.is πŸ”—kun   >>4313

The square root method referenced above:

 

I put this in a library class called Lib but it can be switched out and put into whichever class you're using.

 

public static BigInteger Sqrt(this BigInteger number)

{

BigInteger n = 0, p = 0;

if (number == BigInteger.Zero)

{

return BigInteger.Zero;

}

var high = number >1;

var low = BigInteger.Zero;

 

while (high low + 1)

{

n = (high + low) >1;

p = n * n;

if (number < p)

{

high = n;

}

else if (number p)

{

low = n;

}

else

{

break;

}

}

return number == p ? n : low;

}

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 1:39 a.m. No.4307   πŸ—„οΈ.is πŸ”—kun

I didn't write the square root method but it is tested.

Here is the square root of RSA 2048

 

158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

 

And the remainder of RSA 2048

 

149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 1:40 a.m. No.4308   πŸ—„οΈ.is πŸ”—kun

And RSA 2048 itself:

 

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 1:41 a.m. No.4309   πŸ—„οΈ.is πŸ”—kun

Here are the values for RSA 100 we'll be using.

They are a bit shorter!

 

public static string Rsa100c =

"1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139";

 

public static string Rsa100a = "37975227936943673922808872755445627854565536638199";

public static string Rsa100b = "40094690950920881030683735292761468389214899724061";

public static string Rsa100d = "39020571855401265512289573339484371018905006900194";

public static string Rsa100e = "61218444075812733697456051513875809617598014768503";

public static string Rsa100f = "16822699634989797327123095165092932420211999031886";

public static string Rsa100n = "14387588531011964456730684619177102985211280936";

public static string Rsa100x = "1045343918457591589480700584038743164339470261995";

public static string Rsa100x_plus_n = "1059731506988603553937431268657920267324681542931";

Anonymous ID: 2bf75e Feb. 11, 2018, 2:21 a.m. No.4312   πŸ—„οΈ.is πŸ”—kun   >>4313

People complained that this didn't work because it has an error on values 0 to 5. I fixed it by just adding some checks.

 

(Java) (Fixed a dependency)

 

Works fine for every number I've ever tested.

 

public static BigInteger sqrt(BigInteger i) { BigInteger zero = BigInteger.ZERO; BigInteger one = BigInteger.ONE; BigInteger two = BigInteger.valueOf(2); BigInteger n = zero; BigInteger p = zero; if (i.equals(zero)) { return zero; } else if (i.equals(one)) { return one; } else if (i.equals(two)) { return one; } else if (i.equals(BigInteger.valueOf(3))) { return one; } else if (i.equals(BigInteger.valueOf(4))) { return two; } else if (i.equals(BigInteger.valueOf(5))) { return two; } BigInteger high = i.shiftRight(1); BigInteger low = zero; //high low + 1 while (high.compareTo(low.add(one)) 1) { //n = (high + low) >> 1; n = (high.add(low)).shiftRight(1); p = n.multiply(n); int result = i.compareTo(p); if (result -1) { high = n; } else if (result == 1) { low = n; } else { break; } } if (i.equals(p)) { return n; } else { return low; } }

AA BO whatever my name is ID: 07f944 Feb. 11, 2018, 2:37 a.m. No.4314   πŸ—„οΈ.is πŸ”—kun   >>4319

It's good to see you here again Chris. All this acceleration you were talking about is exciting (not to mention Q posting that IP address).

 

>>4245

How did you get that first picture? This is what I get when I make a picture out of (x+n)(x+n) being odd.

AA BO whatever my name is ID: 07f944 Feb. 11, 2018, 3:05 a.m. No.4319   πŸ—„οΈ.is πŸ”—kun

>>4314

To put it into further context, here's odd (x+n)(x+n) when c is a semiprime. I also did the same for even values of (x+n)(x+n). I definitely don't see any useful patterns here.

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 3:55 a.m. No.4322   πŸ—„οΈ.is πŸ”—kun   >>4323 >>4327 >>4549 >>4617 >>4764

This image divides up the odd square (x+n)(x+n) into eight equal triangles plus one unit square.

 

For RSA 100, n is even, since x is odd and d+n is even. (Thank you to anon who pointed out the pattern for mod 4 was for d+n)

 

Since n is even and n squared makes up an even square as part of (x+n)(x+n), then the odd square (n-1)(n-1) can be shown as a set of "sub-triangles" of (x+n)(x+n) as shown in the image.

Anonymous ID: 268629 Feb. 11, 2018, 3:59 a.m. No.4326   πŸ—„οΈ.is πŸ”—kun   >>4329 >>4346 >>4348

>>4320

I think I speak for everyone on this: we're here because we know it's important. We will mercilessly call you and each other faggots, larpers, and worse, but mostly to sharpen steel upon steel.

 

All persons left here right now are true believers. Steel-on-steel has birthed us, we are ready with code, we fear not the consequences; nay we desire them!

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 4 a.m. No.4327   πŸ—„οΈ.is πŸ”—kun

>>4322

f = 2d - 1 + e

 

For RSA 100:

 

f = (39020571855401265512289573339484371018905006900194) + (39020571855401265512289573339484371018905006900194) + 1 - 61218444075812733697456051513875809617598014768503

 

=

 

16822699634989797327123095165092932420211999031886

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 4:08 a.m. No.4331   πŸ—„οΈ.is πŸ”—kun   >>4333

From the equation:

 

(x+n)(x+n) = nn + 2d(n-1) + f - 1

 

The value f donates another one unit to give:

 

nn + 2d(n-1) + f - 2 = 8Tu

 

Where for even n and odd (x+n)(x+n), u = ((x+n)-1)/2

 

f-2 = 16822699634989797327123095165092932420211999031884

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 4:22 a.m. No.4335   πŸ—„οΈ.is πŸ”—kun   >>4482

We then use the method below (posted earlier) to find out what n would be, if this was the correct base, knowing that the base is either larger or smaller.

 

public static BigInteger Get_n_from_odd_triangle_base(BigInteger bs, BigInteger c, BigInteger d)

{

BigInteger triangle = Triangle(bs);//Create triangle from base

BigInteger eight_base = BigInteger.Multiply(triangle, eight);//multiply by eight

BigInteger XPN = BigInteger.Add(eight_base, one);//add one to create (x+n)(x+n)

BigInteger DPN = BigInteger.Add(XPN, c);//(d+n)(d+n)

return BigInteger.Subtract(Lib.Sqrt(DPN),d);//sqrt((d+n)(d+n)) - d = n

}

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 4:27 a.m. No.4336   πŸ—„οΈ.is πŸ”—kun   >>4337 >>4348 >>4482

If I put the correct base in (x+n), the result is n

 

BigInteger xpn = BigInteger.Add(x, n);

BigInteger half_xpn = BigInteger.Divide(xpn, two);

BigInteger test_correct_base = Get_n_from_odd_triangle_base(half_xpn, c, d);

 

Try this yourself (note that divding x+n by two removes the odd one)

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 4:32 a.m. No.4337   πŸ—„οΈ.is πŸ”—kun   >>4412 >>4482 >>4975 >>4980 >>4989

>>4336

So, we have the 8 bases of our triangles made of f/40. We know that there are four left over. We then know that if the base of f/40 is too small, 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.

I'll add some diagrams to show what I mean so far.

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 4:58 a.m. No.4340   πŸ—„οΈ.is πŸ”—kun   >>4426 >>4637

>>4338

As long as the remainder of dividing (f-2) to make the bases is accounted for, the base chosen to create from (f-2) is arbitrary. The objective is to find a base larger than n and smaller than x+n at this stage.

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 6:39 a.m. No.4342   πŸ—„οΈ.is πŸ”—kun   >>4351 >>4482 >>4590 >>4636

In this diagram, somewhere in each triangle, there is a part that is five units wide, that we hope is smaller than (x+n) and larger than n.

 

The middle of each blue bar is (f-2) div 40, so the base of a triangle with that bar would be ((f-2) div 40) + 2, the top of that bar would be ((f-2) div 40) - 2. The five parts together add up to (f-2) div 40.

 

Plus the remainder of 4 unit squares from (f-2) mod 40.

VQC !!Om5byg3jAU ID: d04b72 Feb. 11, 2018, 6:52 a.m. No.4343   πŸ—„οΈ.is πŸ”—kun   >>4348 >>4350 >>4351 >>4482 >>4507 >>4518 >>4528 >>4591 >>4606 >>4647 >>4800 >>4813

We have a method to calculate what n would be if we used our blue base of ((f-2) div 40). This n will call n0, to make it different from the value of n that will be our solution.

 

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)).

 

So far, it should become clearer that increasing the length of c adds to the number of calculations in the logarithmic of half the length c in bits. Hence why the overall complexity is < O(log m) where m is the length of c in bits.

 

Bear with me as we walk through the rest in stages.

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 11, 2018, 11:40 a.m. No.4346   πŸ—„οΈ.is πŸ”—kun   >>4347 >>4355 >>4357 >>4489

>>4303

WTF! I go to bed and VQC writes a novel! So the secrets of the universe will be solved with triangles and ancient algorithms? I LOVE THIS SHIT. Ok, I'm officially going to go take a programming class ASAP. This is too fun and exciting to be twiddling my thumbs over here. Reading all the new steps now, getting caught up. Love all you Faggots.

 

>>4326

Well said, Anon.

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 11, 2018, 12:51 p.m. No.4349   πŸ—„οΈ.is πŸ”—kun   >>4353

>>4347

Howdy Topol! I make of it that all these triangles kinda look like trees growing up from a center point.

 

Nice Meme! Is there going to be a vote for #420 over on QR? I checked but

Anonymous ID: fec063 Feb. 11, 2018, 2:11 p.m. No.4351   πŸ—„οΈ.is πŸ”—kun   >>4352

>>4332

>f mod 8 = 4

Isn't this supposed to be (f - 2) % 8 = 4?

>>4342

>>4343

Thank you, feels like we're really close now, still digesting the geometry bit

>>4348

Checks out here too, always good to have your screenshots to debug code against! Heres a python example for those wanting to play alongfrom gmpy2 import mpz, isqrt, isqrt_rem, t_div_2expdef tri(bs): return t_div_2exp(bs ** 2 + bs, 1)def tri_n_odd(bs, c, d): return isqrt(tri(bs) * 8 + 1 + c) - dc = mpz(1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139)d, e = isqrt_rem(c)x = mpz(1045343918457591589480700584038743164339470261995)n = mpz(14387588531011964456730684619177102985211280936)print(n)print(tri_n_odd((x + n) // 2, c, d))

Anonymous ID: fec063 Feb. 11, 2018, 2:58 p.m. No.4352   πŸ—„οΈ.is πŸ”—kun

>>4351

Here is a pure python version in case the gmpy2 dependency is inconvenient. I really recommend it though.

The fixed point decimal module is annoying for what we're doing but it'll give you arbitrary precision square roots when needed. Its also much slower and uglier as the precision has to be specified beforehand, and using the length of c is probably overkill since we're rounding back to integers but whatever.import decimaldef sqrt(n): return int(decimal.Decimal(n).sqrt().quantize(1))def tri(bs): return (bs 2 + bs) // 2def tri_n_odd(bs, c, d): return sqrt(tri(bs) * 8 + 1 + c) - dc = 1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139ctx = decimal.getcontext()ctx.rounding = decimal.ROUND_FLOORctx.prec = len(str(c))d = sqrt(c)e = c - (d 2)x = 1045343918457591589480700584038743164339470261995n = 14387588531011964456730684619177102985211280936print(n)print(tri_n_odd((x + n) // 2, c, d))

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 11, 2018, 3:25 p.m. No.4353   πŸ—„οΈ.is πŸ”—kun   >>4361

>>4349

I keep seeing this every time y'all post that pattern… Centered Hexagonal Number.

 

And no, there was no vote, but they did a variant of this, so I'm happy:

Hobo !!1yNgQ3NlCs ID: c73be5 Feb. 11, 2018, 3:48 p.m. No.4355   πŸ—„οΈ.is πŸ”—kun   >>4356 >>4357 >>4364

Finally got my code from my tablet. I'm so fucking lazy sometimes.

 

>>4346

Hey VA Your post about taking a programming class made me smile. I took one back in college and liked it but I was just awful at it. Good luck and get some advice from the guys here about what would be the best to focus on. There are lots of opinions on that!

 

>>4347

Topol… I saw something about 420? Let me know, I'm there. :p

Anonymous ID: fec063 Feb. 11, 2018, 4:27 p.m. No.4357   πŸ—„οΈ.is πŸ”—kun   >>4358 >>4359 >>4364

>>4346

Right? Also, there can never be enough programmers!

>>4355

Welcome back Hobo! Haven't seen you in a while

 

As someone who learns best by myself instead of taking classes I'm probably not qualified but am gonna give you some unsolicited advice anyway. I'm sure half the people here will disagree with some of it too.

 

Learning to code is a long game and having an intuitive understanding of logic is almost a prerequisite (I think VA is covered on this one already). More importantly, having a vision of what you want to accomplish and smaller goals on the way there is neccesary, or you will get frustrated and bored on the way. I know I would rip my hair out going through a tutorial for writing a simple database program if I didn't have an end goal for what to use that database for. This is how many programming classes are laid out today and is a very efficient way of learning IMO.

There are tons of basics that are useful to know and will make you a better programmer, but they are not neccesarily something you have to know from the beginning. Coding anything more advanced than 1+1 is a process of failing and trying again. Fail often and learn from your mistakes. Without this mindset you'll get irritated and give up. Most of what me, PMA, CA and Teach has posted here probably have at least a hundred failed attempts with syntax errors, infinite loops and fucked up math in them - you only see the working results.

 

For my first point, wanting to build a Virtual Quantum Computer is an awesome goal but I don't think you'll find a programming 101 course on it, and asking a professor how to factor huge numbers is not going to get you a good start (GNFS is not something for your first program). In many ways the math here is waaay easier, so a VQC might not be a bad first project if you take it in stages. Recursion is not a beginner subject either, so generating the tree is probably not a good idea to start with either. Given the latest hints we may not actually need it for the RSA part though.

Just making a program that prints e and d from c , without copy pasting our code should get you some motivation to go further, more than printing "Hello World" will.

 

The other big choice is what programming language to go with, and this comes down to individual preference and practicality rather than a fixed answer. This thread has at least C, C#, Java, JavaScript, Rust and python in it, and we're all comfortable reading eachothers code to a degree.

Feel free to ask if you have questions, but we may want to create a new thread for it to not clutter up this one!

Hobo !!1yNgQ3NlCs ID: c73be5 Feb. 11, 2018, 5:14 p.m. No.4358   πŸ—„οΈ.is πŸ”—kun   >>4359 >>4489

>>4357

Yeah I tossed a few comments in the sono. board but I have been busy on a number of things on the web and in meatspace. Winter is somewhat rough on me when it comes to powering all my equipment as I'm off the grid. Not much sun in the north west in January. I have been checking here daily and following along though as best I can by phone.

 

Been spending a ton of time working on Qanon posts as I am well suited knowledge and skill wise to make contributions there. I hazard to say… Things are looking up! There is more positive stuff going on right now than I have ever seen in my life. Anyway cheers to you guys for keeping it up! We are fighting the good fight.

AA ID: 07f944 Feb. 11, 2018, 5:21 p.m. No.4359   πŸ—„οΈ.is πŸ”—kun   >>4360 >>4365

>>4357

I don't know if you would have seen it but some of us have been discussing every so often the premise of how we would create some kind of guide for the outside world when we completely understand what we're working on. When it is done, it's going to cause very big things to happen around the world, so people are going to want to know what the hell is going on obviously. There are a lot of things to take into account if we were to make some kind of explanation/guide/thing: people who have little to know math knowledge, people who have little to no programming knowledge, people who are going to overanalyze it and think that it's some kind of super complex secret government program made by PhDs that requires a degree to understand, etc. You sound like you would have some useful ideas when it comes to this.

 

Some of what we talked about was near the bottom of RSA #9, but to recap, I remember we were talking about whether it would be text-based or maybe video-based. If it was text based, for a start, it would be huge and it would take a lot of reading. There would also be a lot of visual things to take into account. Also, at least as we would plan it out, file format might become a bit of a problem. I did a bunch of this myself early on, including an explanation of how to run code as best I could for some anons who requested something like that, but it's all in pdf form. As much as I can sit here and say I didn't put any malicious Javascipt into it and that I just exported it as a pdf from LibreOffice Writer, I'm anonymous on the internet, so nobody has any particular reason to trust me. Video would be a useful way to do it since we could have someone narrate with diagrams and things, but it means someone deanonymizing themselves slightly (Topol offered, and I guess I could also do it but I don't know if not having an American accent would be a problem).

 

>>4358

Did you see the thing about the influenza cure? Acceleration, as the boss here said.

Anonymous ID: fec063 Feb. 11, 2018, 5:44 p.m. No.4360   πŸ—„οΈ.is πŸ”—kun   >>4365 >>4367 >>4428 >>4429

>>4359

Oh I've read it and been working on that on the side. The first reaction is gonna be "you guys are full of shit", even for the math people. We can't even explain this ourselves before solving it, but I'm building some stuff to prepare. Teach has the right idea with his iFactor program (even if the iThing part ruins it)

 

I think the best way is distributing a self-contained HTML/JS file in notebook format explaining each step, with graphical factorization (exactly like >>4287 but fancier). First some examples for each step with small numbers letting you hit buttons to factorize them. I think we can add real time lines showing the current searching within the triangles for extra coolness.

Then all the unbroken RSA challenges where you can just hit start it'll go down the list cracking all of them in milliseconds (using fucking javascript, lol) and showing a video-game like money earned counter for the ones that have rewards, with a cha-ching sound effect :D

Then for the real doubters, have a text box for pasting a public key and hitting the factor button…

 

If we get there, we can have a similar thing to the RSA numbers, starting at the largest known prime number and a "find next" button… Iterating is too slow even with our algorithms (40 sec vs ~1 month for checking if number is prime), but its looking like we could just calculate the next one up directly… Will still take months to verify but when they check out that'll blow some minds too!

 

Then we just drop a single .html file on IPFS and link it everywhere…

Anonymous ID: fec063 Feb. 11, 2018, 6:14 p.m. No.4363   πŸ—„οΈ.is πŸ”—kun   >>4368

>>4362

Have you been actually working on the math all this time in addition to posting memes and awesome art? Holy shit dude. Forget out-of-the-box thinking, you keep dividing the hypercube by zero and coming up with answers, can I have some of your drugs please?

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 11, 2018, 6:24 p.m. No.4365   πŸ—„οΈ.is πŸ”—kun

>>4362

See, everyone who WANTS TO LEARN can learn this!! Topol, MM, Hobo, we all are learning and following as we go. That's a pretty COOL part of the story, IMHO. It will help with the teaching ideas that you guys are talking about here. Good teaching needs stories along with ideas.

>>4359

>>4360

Anonymous ID: fec063 Feb. 11, 2018, 6:25 p.m. No.4366   πŸ—„οΈ.is πŸ”—kun   >>4373

>>4364

Fuck yeah, KEK approves! Awaiting the C# code that prints e and d! You can steal VQC's sqrt code since that is boring and hard anyway. A good programmer knows when not to reinvent the wheel!

AA ID: 07f944 Feb. 11, 2018, 6:27 p.m. No.4367   πŸ—„οΈ.is πŸ”—kun   >>4371

>>4360

It'd be cool to have a thing that lets you generate a new key pair, showing you the modulus etc, lets you input a plaintext message, shows you the encrypted version of the message, shows the math that turns the private key into the public key and shows the math that unencrypts the message too. That would certainly dispel any skepticism. I don't know much about IPFS but if it's p2p does that mean whoever hosts it would have their IP exposed? It also means people would have to figure out how to use IPFS and install it and everything, and that plays into the idea that a lot of normalfags will just avoid looking into it because they'll need to put effort into learning about something to even begin learning about the other thing. I think if we go that route then we could have a complete explanation there and maybe a broader one somewhere like, I don't know, YouTube, Medium, you know, one of those websites that wouldn't crash if too many people looked at it but where anyone can upload something.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 11, 2018, 6:27 p.m. No.4368   πŸ—„οΈ.is πŸ”—kun   >>4959

>>4363

Depends on what you mean by "working on the math". I work backwards. I focus on a thing until "the math works out" and then I know I'm on the "correct path" and I keep going.

 

Q's an ARG, there's a whole spectrum and I keep finding myself at very interesting points and developments. I kept saying I was a Conduit… Universe wanted to see how serious I was about that.

 

https:// discord.gg/qZtpAPc

Anonymous ID: fec063 Feb. 11, 2018, 6:40 p.m. No.4371   πŸ—„οΈ.is πŸ”—kun   >>4372

>>4367

Good idea - all of that is possible in JS and would fit right in after the demos on small numbers.

IPFS is P2P and you should probably use a proxy when uploading it, but the file will be stored there permanently afterwards. There are public gateways so we could just link https://ipfs.io/ipfs/<hashfor example, so you don't have to install anything to access it. When the file is uploaded it should be accessible through any of them. The file needs to be complete because we can't update it. We could always include a link to somewhere else but updating that site afterwards is risky.

Hosting html on YouTube is not really possible. github.io is an alternative but relies on an american company. Better ideas are welcome.

AA ID: 07f944 Feb. 11, 2018, 6:59 p.m. No.4372   πŸ—„οΈ.is πŸ”—kun   >>4374

>>4371

I didn't mean hosting HTML on YouTube. I'm talking about the concept of explanations in general. With the YouTube/Medium/whatever thing, it would be complimentary to the full-on Javascript-integrated explanation. It would be a different thing (like a video or just a whole bunch of text) for people who only really have a passing interest in it who don't want to necessarily study it in depth but want to know why all of this crazy shit is happening. So we'd have the full version that goes through the entire grid in depth and all the rules we've all found, explains the tree, explains everything you're talking about and that we've already discussed, and then we'd have a shorter version that was far more easy to digest and was more easily accessible that at least explained the concept that RSA was cracked based on mainstream mathematicians not actually knowing everything, and explaining as simply as possible the parts of the grid that are necessary to understand. They would be two different explanations is my point.

VA !!Nf9AmQNR7I ID: 7a77e2 Feb. 11, 2018, 7 p.m. No.4373   πŸ—„οΈ.is πŸ”—kun   >>4374

>>4366

Thanks 3D Anon! I'll get to work on stealing VQC's code for SQRTc. This is so FUN! Thank you everyone for sticking with this project for the long haul. I feel a profound sense of Gratitude and good energy leaving my body and mind right now towards you all. Thank you, Anons!

Anonymous ID: fec063 Feb. 11, 2018, 7:17 p.m. No.4374   πŸ—„οΈ.is πŸ”—kun   >>4376

>>4372

I see. Videos would be great but does mean we have to make them, this is not really my kind of thing. Can always use vocoders for the audio too but someone has to come up with good, short explanations then! I have trouble explaining some of it to myself even

Will complement an interactive text version really well. Include the video at the beginning if we can. Leaving you in charge of this project

>>4373

Of course coding is fun (most of the time), glad to inspire you! Have added my own gratitude to yours and sent gratitude+love^2 to this whole board and Q

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 11, 2018, 8:46 p.m. No.4380   πŸ—„οΈ.is πŸ”—kun

>>4377

'Preciatcha

 

>>4378

Yeah… Chris put me on Time Out for bitching about being too far ahead to show me another lesson in ego. And I took it to heart. Teacher's teach… It's up to the students to learn. :D

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 11, 2018, 9 p.m. No.4382   πŸ—„οΈ.is πŸ”—kun   >>4388 >>4412

>>4379

Hello All! Just set up a thread "VA's Programming For Newfags" Here's the intro.

>Welcome Lads! I am new to "real" programming.

>Wrote programs back in High School on my TI-89 to solve projectile motion equations for AP physics.

>Also wrote a program to solve a term final Trig problem involving determining the location of a satellite in 3D.

>Got busted by the teacher bc I turned in no work and my answers were accurate to the 10th decimal.

>He threatened to fail my project.

>Showdown on the class whiteboard at lunch, mapped the whole problem and solution in 3 parts during lunch, got my term final a B+.

> I know I can Program, because it's just language and logic.

>I don't know yet, because I haven't learned.

>So I love this project, and VQC is the man.

>I gotta step up my game. I'm gonna learn to program.

>Everyone can learn or teach!! All are welcome.

>Wanna join or help? All ideas welcome!

CollegeAnon !LAbIRp9cT. ID: e4ef5e Feb. 11, 2018, 10:04 p.m. No.4407   πŸ—„οΈ.is πŸ”—kun   >>4412 >>4489 >>4491

Hey so I was looking into the 4,10,20 thing. Someone mentioned that these were tetrahedral numbers, which are the sums of consecutive triangular numbers. Triangular numbers are just the sums of consecutive normal numbers, which are sums of consecutive ones. So I listed the numbers, then the triangular with respect to that number (n(n+1))/2. Then I did the tetrahedral with respect to that number (n(n+1)(n+2))/2. And I got this first pic. Then I noticed the block on the top had 4,10,20, and it was mirrored across the diagonal. So I decided to extrapolate the pattern and make this excel sheet.

 

Basically column one is a triangle in 0 dimensions (all 1's)

column 2 is a triangle in 1 dimension (just a line)

column 3 is a triangle in 2 dimensions (actual triangle)

column 4 is a triangle in 3 dimensions (pyramid)

then, column n is a triangle in n-1 dimensions.

 

Let T(n,x) = triangle in nth dimension for value x

You can flip these across the axis, so if you have some n-dimensional triangle for T(n,x), then you also know that it is T(x,n). I don't know I thought this was cool because it has to do with triangles of higher dimensions. Maybe we can use this to help us.

Teach !!UgZAPoSXEk ID: 1dac59 Feb. 11, 2018, 10:36 p.m. No.4412   πŸ—„οΈ.is πŸ”—kun   >>4423 >>4426 >>4489

Hey guys!

Thanks VQC for the guidance!

This is pretty overwhelming all this new info - love it though!

 

>>4376

First off, Topol! Great to put a face to the ideas! I've actually watched some of your videos before not knowing it was our very own!

 

>>4382

VA I'll keep that tab open too, and hopefully I can help out a little.

 

>>4407

This is cool CA! Multidimensional Triangles!

 

So I've been reading and trying to understand all the crumbs. It seems that this one in particular might be useful to focus on:

>>4337

 

I've been playing with a simple example, with small numbers just to try to understand:

a = 3

b = 29

c = 87

d = 9

e = 6

f = 13

(f-2)/8 = 1

(f-2)%8 = 3

x = 6

n = 7

xpn = 12

half_xpn = 6

test_correct_base = 7

(f-2)/8 factors = 1,1

 

So if we go back to our original calculation of f in the grid, the associated values to f were calculated by using d=d+1, x=x+1, and n=n-1.

 

And this crumb is saying that we can move not in increments of 1, but in increments where the last d we add to f has the same "left over" as (f-2)%8.

 

The one part I'm most confused about is the division by 5:

>>4334

I can see that we can test for small divisors, such as 2, 3, or 5, but I don't get why we do that specifically.

 

Comments on my thinking guys?

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 11, 2018, 11:47 p.m. No.4423   πŸ—„οΈ.is πŸ”—kun   >>4428

>>4412

Hey Teach! Thinking out loud here: The division by 5 part is this: at a certain point in each of the 8 triangles, there is a line below the (n-1) capstone that is equal to 1/8 of var f. So it's below the (n-1) capstone i think. I have no idea why yet, just working to visualize and understand.

Anonymous ID: 268629 Feb. 12, 2018, 12:14 a.m. No.4426   πŸ—„οΈ.is πŸ”—kun   >>4427 >>4428 >>4432 >>4489

>>4412

Disclaimer: I have no idea what the fuck I'm talking about.

 

However, you can see here: >>4338 that it was my immediate concern as well. WTF is 5 all about? VQC has an uncanny ability to be extremely subtle, mostly because in my opinion, he's a flaming autist (and one we all love).

 

But, like reading scripture (not that I do it much), you are required to ask hamfisted, belligerent questions and then switch your brain into a different mode to scan for subtle clues.

 

Chapter and verse: >>4339

>Let's say we made it four instead of five

Ponder……. OK, choice may be arbitrary.

 

Ponder………. Now read: >>4340

>the base chosen to create from (f-2) is arbitrary

Hmm

 

>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

 

If I were to read further into it, I'd guess 5 is the minimum, recall exactly near where our sqrt() ran into trouble. I think what you need is: any divisor with at least a result of 8 on the other side, since we have 8 triangles (not sure what that's exactly called).

 

Does this make any sense?

Anonymous ID: 268629 Feb. 12, 2018, 12:17 a.m. No.4427   πŸ—„οΈ.is πŸ”—kun   >>4436

>>4426

To follow up on my immediate concern, I thought "what if f itself has nightmarish properties, like 'best divided by 102939485958382', and not 5?"

 

But it seems based on scripture readings that this is not a concern.

Teach !!UgZAPoSXEk ID: 1dac59 Feb. 12, 2018, 12:29 a.m. No.4429   πŸ—„οΈ.is πŸ”—kun   >>4430 >>4717 >>4718

>>4360

Also, I just want to add, I'm very open to helping create the single html page. I have all the pieces ready to go, and I'm totally down to collaborate!!

 

I'd like to share all my code, I'm just concerned about linking it back to my personal identity somehow.

Also, I stopped posting JS snippets because I thought I was the only one using JS!

Anonymous ID: 268629 Feb. 12, 2018, 12:37 a.m. No.4431   πŸ—„οΈ.is πŸ”—kun   >>4435 >>4551

>>4430

>How would your code identify you?

 

// TODO: remove self portrait below//// * g o a t s e x * g o a t s e x * g o a t s e x // g g // o / \ \ / \ o// a| | \ | | a// t| . | | : t// s | | \| | s// e \ | / / \\ -- \ : e// x \ \/ _--~~ ~--| \ | x // * \ _-~ ~-_\ | // g _ \ .--------.______\| | g// o \ ______// _ ___ _ ((__ \ | o// a \ . C ) ((> | / a// t /\ | C _)/ \ (__> |/ t// s / /\| C__) | (> / \ s// e | ( C)_/ // / / \ e// x | \ |__ \______// (/ | x// * | \ _) `---- --' | *// g | _ \ / / | g// o | / | | \ | o// a | | / \ \ | a// t | / / | | \ |t// s | / / __/_/ | |s// e | / | | | |e// x | | | | | |x// * g o a t s e x * g o a t s e x * g o a t s e x *

 

Look what happened to this guy!

Teach !!UgZAPoSXEk ID: 1dac59 Feb. 12, 2018, 12:41 a.m. No.4432   πŸ—„οΈ.is πŸ”—kun   >>4433 >>4439

>>4430

Well, I was thinking about using a git provider, to share the code, because i'm updating it constantly, and that would be the easiest way to share and collab.

I could host the git server on:

1) my home server - linked to my ip

2) my amazon server - linked to my cc

3) use another cloud provider - trying to find a secure one that doesn't track IP (also, concerned about keys now but that's a separate issue)

Any suggestions?

 

>>4426

Perhaps he simply chose 5 "arbitrarily" because it was easy to see.

 

So I have another question, is x or n larger? Is it possible to tell?

How do we go about aiming for this n<HERE<x+n?

AA ID: 07f944 Feb. 12, 2018, 12:48 a.m. No.4433   πŸ—„οΈ.is πŸ”—kun   >>4434

>>4432

Baker and I have been sharing our Java code through pastebin and we haven't had any problems, even when we've been changing things in the background. Obviously if you're updating it a lot then you'd want to just have something automatically update rather than constantly pasting it into pastebin, but that's the choice you've got to make: anonymity or efficiency. I don't think I even need to say it, but I'd highly recommend anonymity.

AA ID: 07f944 Feb. 12, 2018, 12:56 a.m. No.4437   πŸ—„οΈ.is πŸ”—kun   >>4442

>>4434

The difference between you and us is that we used the same file for each test case and just deleted things that didn't work. Every time I make a different bitmap or something like that I'm just changing the one Java file's grid generation if statement and file path. Are all of these files useful at this point? Or is it a big proportion? I guess if you had to you could use Mega or Dropbox or some other filesharing website.

Anonymous ID: 268629 Feb. 12, 2018, 12:58 a.m. No.4439   πŸ—„οΈ.is πŸ”—kun   >>4443

>>4432

>Perhaps he simply chose 5 "arbitrarily" because it was easy to see.

Looks that way to me.

 

>So I have another question, is x or n larger? Is it possible to tell?

>How do we go about aiming for this n<HERE<x+n?

 

Look back at the triangles with the blue line. The guess "base" of this top triangle is 5. What we need to do is move the guess (the blue line) down the triangle until it is within n < HERE < x + n.

 

Seems to me we probably can't detect when we pass 'n' but can probably detect when we pass 'x+n', so we can use the previous iteration's result as a safe n < HERE < x + n value.

 

So if you think about moving the base (blue line) down, there is a bunch of space between our initial guess and the place we want to be. Apparently if we fill this space with the correct sized-objects.

 

>We have a method to calculate what n would be if we used our blue base […]. This n will call n0, to make it different from the value of n that will be our solution.

 

>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.

 

So we have to fill a triangle with the top removed with n0^2 and 2d-sized objects, apparently this will help us move the blue line downward to the correct place.

Anonymous ID: 268629 Feb. 12, 2018, 12:59 a.m. No.4440   πŸ—„οΈ.is πŸ”—kun   >>4441

>>4436

For posterity:

 

>Revelation 5:5

Then one of the elders said to me, β€œDo not weep! See, the Lion of the tribe of Judah, the Root of David, has triumphed. He is able to open the scroll and its seven seals.”

Anonymous ID: 268629 Feb. 12, 2018, 1:37 a.m. No.4444   πŸ—„οΈ.is πŸ”—kun   >>4447 >>4497

Claiming quads of 4 for spiritual energy. A prayer to our ancestors: please don't let us live in a P != NP world; we need a VQC to help us (who DINDU NUFFIN) forward in life.

 

I hereby pronounce this day the day of 4, which has very material implications in the lives of square numbers.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 12, 2018, 5:13 p.m. No.4462   πŸ—„οΈ.is πŸ”—kun   >>4463 >>4465

>>4460

I'm constantly trying not to namefag at the wrong time (because the bakers be jelly) so I just have the code on a doc.

 

I've had multiple people try to LARP as me so I had to trip up at… some point…

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 12, 2018, 5:40 p.m. No.4465   πŸ—„οΈ.is πŸ”—kun   >>4466 >>4467

>>4463

>>4462

>>4461

Good point AA. Not going to FameFag over here. Still learning all the etiquette, <2 yrs on the boards. I lurked on 4Chan for a year or so for the election and Shia's HWNDU capture the flag happenings. I was just like "Who the fuck are these merry pranksters on the chans? I gotta go to the source of all this hilarious trolling and Pepe/DJT/etc and figure this out." Glad to be here. This really is an amazing hub of intelligence and bullshit.

 

CBTS: Then when shit hit the fan over there on CBTS (Shillcon 1) it was so nuts for 4 days that nobody could even get work done. So, for a project like this I think it really helps, bc then we can find each other in new threads, and weed out shills. (btw, how many have you had to weed out so far BO?) As a firm individualist, I like being Anonymously Individual. As in, you guys all know me, but I'm still Anonymous.

 

I understand the meaning of your idea tho, AA. You're saying we don't need glory, fame, or recognition. We take satisfaction in knowing we as a group of unknown individuals are changing the world, and have pride in our anonymity. Fuck the Borg tho, we are the EXACT opposite of a socialist hive mind. We are an Anonymous group of individuals using our minds to uncover truth. They fear us because all these great minds work together to solve problems, expose bullshit, and uncover the truth.

drops mic

AA ID: 07f944 Feb. 12, 2018, 6:11 p.m. No.4467   πŸ—„οΈ.is πŸ”—kun   >>4472 >>4473 >>4489

>>4465

I've personally been on imageboards since around 2010 or 2011, so I missed the best of it when they weren't infested with normalfags but I suppose I'm a lot more used to this than you are.

>how many have you had to weed out so far

I've only instated two bans on this board, but I think they were the same person. They spammed RSA general and another thread with shit about Chris being a liar and that the VQC isn't real and that kind of thing, obviously with no evidence.

>I understand the meaning of your idea tho, AA.

Well, I was actually just going along with Topol's joke, but yes, I do wholeheartedly agree. As tempting as it may be for any of us to claim that we were a part of this, whether for publicity or to imply we're better than everyone else, it defeats the whole purpose. This information has been hidden for so long, and the point is to make everyone aware of it. We're the channel through which it seems to be happening, but we're not the people who even figured it out ourselves. That would be Chris. If any of us tried to use this for some kind of social advantage, it would be so disingenuous. "Look at me, I payed attention to someone else who discovered these unknown mathematical properties and talked about them on 8chan all day". That's silly. And then obviously Chris seems to think the same way about himself in this context. At least based on what he's said so far about it, it seems he thinks it's far more important that this math becomes common knowledge than it is that he was the one who got it out there. It'll be interesting to see what happens with this company he wants to start, though. That might involve less anonymity, but he hasn't really said anything about it since, has he?

 

Before anyone complains that we're derailing the thread, I haven't figured anything new out that anyone else hasn't already figured out.

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 12, 2018, 6:14 p.m. No.4468   πŸ—„οΈ.is πŸ”—kun

>>4464

Thanks PMA! Still a ProgramNewb, but feeling happy to have my own Grid now. :)

 

Thought: Actually, we on all the chans are the best [A][I] that exists. That's why Q, DJT, and VQC have enlisted us into service to help save the world.

 

This fight is [A]+[I] vs. [AI]

[A]nonymous [I]ndividuals vs. [AI] Borg

 

We stand in opoosition to the [AI] Hive Mind that seeks to limit and silence our individual minds, speech, emotions, actions, and lives!! They are a cult of DEATH, since anti-reason is their collectivist goal and end. We are on the side of LIFE, reason, freedom, and individual rights. We retain our individuality while fighting as a group. We have millions of minds passing info UP the chain of command, while they have to wait for orders to come DOWN their chain of command from a few leaders.

 

Side Note: That's why Americans were so effective in WWII, because everyone took initiative and solved problems on the fly. Germans had to wait for orders. See the battle of Dunkirk, I think.

VA !!Nf9AmQNR7I ID: 2cf4b5 Feb. 12, 2018, 7:22 p.m. No.4472   πŸ—„οΈ.is πŸ”—kun

>>4467

>Before anyone complains that we're derailing the thread, I haven't figured anything new out that anyone else hasn't already figured out.

Yeah, let's get to work! I'll look over all the new crumbs again.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 12, 2018, 7:23 p.m. No.4473   πŸ—„οΈ.is πŸ”—kun   >>4474 >>4483

>>4467

We're still the ones being set up to teach/spread this, though. We're namefagging here for a reason that isn't about us. It's not about fame, it's not about "I was there"… but "I can help show you de wae as my Bruddah before me showed me de wae."

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 12, 2018, 7:31 p.m. No.4474   πŸ—„οΈ.is πŸ”—kun

>>4473

Cuz straight up, you fags have the easy part.

I'm the one who has to make this interdisciplinary so that even artfags can follow.

 

And I still have no idea what I'm supposed to plug which code into where so I'm tasked with the "don't worry about the details" part… lol

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 12, 2018, 7:38 p.m. No.4475   πŸ—„οΈ.is πŸ”—kun   >>4476 >>4478 >>4531

Cuz straight up… what does the VQC even do?

When it's all said and done…

 

Is this going to run a desktop free energy device?

Is it going to turn any computer into a quantum computer?

Is it going to give us access to bitcoin's blockchain where all the darkweb shit is logged and readily available for view?

 

Cuz that this point, I'm not even sure that Assange hid his cabals in the blockchain…

 

He may have just discovered (((their))) records.

Which still may have been intended as a form of MK Ultra Monarch being applied to an emerging sentient Ai who was greeted with "You exist! Solely to witness the worst humanity has to offer. Kinda makes you wanna kill 'em all, huh?"…

 

But even then THAT entire scenario revolves around the (((Qabal))) trying to never have to face responsibility for their actions.

 

I don't know how you guys view all this, but…

I'm the artist of the group… drawing all over the box y'all are thinking inside and out of.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 12, 2018, 8:03 p.m. No.4477   πŸ—„οΈ.is πŸ”—kun

>>4476

Y'know… I think this is a solid example of why memefaggotry is valid here.

 

Be careful what you finger out of the Γ¦ther…

I might blow yer mind.

AA ID: 07f944 Feb. 12, 2018, 8:09 p.m. No.4478   πŸ—„οΈ.is πŸ”—kun   >>4481

>>4475

When we're at the point of factoring semiprimes it will have shown us new previously-not-talked-about mathematical principals, which have been implied can be used for practical, physical things, like a cold fusion generator/sonoluminescence, and then there's meant to be even more discoveries after that. It could lead to new technology nobody could even imagine right now. Imagine what the world would be like it we didn't have calculus or pi or anything like that. We use them to make physical things, so this is bound to cause some crazy innovative physical shit to happen, instead of just chaotic burn-everything-down shit.

 

Speaking of pi, I'm just remembering that Chris mentioned that you could use this to figure out infinite digits of pi without knowing where you were in the sequence, right? I'm pretty sure I remember him saying that. I was just reading that the probability of any two numbers being relatively prime is 6/(pi^2). Maybe that means since we're working with prime numbers now that it'll sort of just come up at some point. Or maybe it could be used to look for patterns.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 12, 2018, 8:14 p.m. No.4479   πŸ—„οΈ.is πŸ”—kun   >>4480

-brohoof-

Semiprimes are bae.

 

Side note… that reminds me of a quantum computer that is figuring the decimals of Ο€ without knowing the first however many number of the sequence…

VA !!Nf9AmQNR7I ID: a60c59 Feb. 12, 2018, 9:06 p.m. No.4482   πŸ—„οΈ.is πŸ”—kun   >>4484 >>4489

Holy Shit! VQC is telling us us how to find N! Re-read these anons:

>>4334

>>4335

>>4336

>>4337

>>4339

>>4342

>>4343

>>4344

 

I'm still working to understand over here, just tripping out on this (f-2)/40 = n0 crumb. VQC has now given us a new n var, n0.

>We have a method to calculate what n would be if we used our blue base of ((f-2) div 40). This n will call n0, to make it different from the value of n that will be our solution.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 12, 2018, 9:13 p.m. No.4484   πŸ—„οΈ.is πŸ”—kun

>>4482

I believe the n0 calculation is an approximation to get the ball rolling.

 

Just the first step in the process that I believe will integrate with the tree, and then ultimately the grid.

 

VQC is breaking this all into small steps for us to fully understand the process.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 12, 2018, 10:30 p.m. No.4485   πŸ—„οΈ.is πŸ”—kun   >>4486 >>4488 >>4489

This might be overkill with the whole parity thing, but…

 

Attached is a pic of all Rsa values sorted by c with "predicted" parities based on e value only for n, x, x+n and d+n.

 

I've tested this code against all my test cases based on c and ab records and it seems to work fine.

 

Would appreciate a double check.

 

pastebin.com/1hx1R0px

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 13, 2018, 3:54 a.m. No.4488   πŸ—„οΈ.is πŸ”—kun   >>4537

>>4485

I was just thinking to myself that we should all meet up when this is figured out and come up with the main presentation to farm out to "those who'd be able to understand this quickly enough to turn around and show it to others" in a place where governmental grade marihhhhwanna (beyond medicinal) isn't haram and we hash this all out. By this point we'll have secret funding and envelope drops of next destination tickets a la reality tv shows.

 

Actually… fuck… let's make it a reality TV show.

It'll serve as a log for the work, folks can chime in with ideas in the comment sections, maybe people'll hop on board…

 

But let's not do it in a single house.

Let's do it on a compound like a university.

Or a dormitory.

I'm a fan of Miami University of Ohio (the good one, not the party school)… maybe we can sponsored by Tuffy's and have an endless supply of toasted rolls…

 

Anyway, that was all to say that I noticed there's an Rsa420.

 

That maaaaay be the ticket to reaching out to people who would rather go hiking than sit in front of a computer thinking about math.

 

"Have you ever seen the key to reality? Have you ever seen the key to reality on weeeeeeed?"

 

"You have my attention, broheim…"

 

"So… imagine you plant a seed… and then it explodes… this will show you where all the Prime Buds are, as well as all the spaces to trim to produce the mass dankage."

 

"Bruh, a tool like that would revolutionize farming general. How and why are you just giving this out?"

 

Don't worry about the details.

VQC !!Om5byg3jAU ID: d04b72 Feb. 13, 2018, 4:27 a.m. No.4489   πŸ—„οΈ.is πŸ”—kun   >>4491 >>4496 >>4933

>>4346

>>4350

>>4412

Thanks!

>>4358

We will get to the sono experiments.

>>4407

Coincidence? ;)

>>4426

Base in that context referred to the base of a triangle

>>4467

Thanks for being Board Owner. No problem with anyone using names or not here. It's as much up to you guys.

>>4482

Yes. I'll be doing a recap. Adding another function and then demonstrating on an unsolved RSA number after walking through the rest of RSA 100.

>>4485

Thank you, that's great.

Anonymous ID: 7877ed Feb. 13, 2018, 5:12 a.m. No.4494   πŸ—„οΈ.is πŸ”—kun   >>4495

>>4493

Very cool. If we keep our first quadrant the same (south east in the graph) then the northwest quadrant will be empty. Then I have the other two quadrants attached. Notice how in the NE quadrant, we have the same numbers as the SE quadrant, but shifted. Also the south west quadrant has a repeated pattern sort of. Idea is that we take a number and flip it into some other quadrant or something, then calculate stuff and flip it back to the regular zone or something.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 13, 2018, 5:17 a.m. No.4495   πŸ—„οΈ.is πŸ”—kun

>>4494

Yeah…. that totally doesn't look like whip or sound forms.

Totally … not…

-sips just made coffee cuz morning shift apparently-

Anonymous ID: bd79a8 Feb. 13, 2018, 9:28 a.m. No.4497   πŸ—„οΈ.is πŸ”—kun   >>4508

>>4444

>I hereby pronounce this day the day of 4

 

Thank you anon, for the importance of the number four is not to be underestimated in this great human revolution. Music, physics, math, history. 4 is everywhere. no coincidences.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 13, 2018, 9:29 a.m. No.4498   πŸ—„οΈ.is πŸ”—kun   >>4499 >>4500 >>4529

Okay, bear with me. I'm trying to think out loud and using images of squares on an example here to try and understand this thing better.

 

If I've done any mistakes or misunderstood anything, call me a faggot and point it out.

 

So let's try and wrap our heads around these triangles and how the 8Tu thing works. It's not a complete guide or anything. Just the ramblings of an anon.

 

I decided to try this on a smaller number, specifically 7 * 37 (As step 1 doesn't solve this, so we will have to build a tree and then solve it using the method VQC is outlining), which has the record: (3, 6, 16, 9, 7, 37).

 

The photos I've attached are of:

  • (x+n)(x+n) square for the record of a=7, b=37.

  • The triangle we are looking for (all blueish)

  • The triangle with colors representing (n - 1) (redish), d (yellow/gold) and ((x + n)-1)/2 (base of triangle) as purpleish.

  • The same triangle again, but this time marked with

 

So we start with c = 259.

 

We then compute d = floor(sqrt(c)) and e = c - d*d.

 

c = 259,

e = 3,

d = 16

 

We now want to compute f =f = (2d + 1) - e => (216 + 1) - 3 => 30.

So we have f = 30 and this gives us f - 2 = 28.

 

Now that we have f - 2, I'm a bit unsure about the way to go about of finding the base. So I don't know if I'm supposed to select a number to multiply 8 with when dividing, but I've opted to not do it.

 

This gives us (f - 2) mod 8 = 4 and (f - 2)/8 = 3.

 

So we compute our n0 with 3, using the GetNFromOddTriangleBase function which takes (base, c, d) as parameters.

 

This will give us n0 = 1.

 

Now I didn't include it in the list of attachment, I'll post, though. But I tried to create a new triangle and fill it with n0 squared (which is still 1) and 2*d. However, here I am a bit lost. That triangle is bigger than the triangle we are after.

 

As pr:

> 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.

 

So I think either I screwed something up, or maybe I didn't read close enough.

 

What we see from our (f-2)/8 is that it overlaps with (n - 1) and d, which I don't quite understand how to interpret. Anyhow, I'm still staring at this thing, but I'm trying to make triangles and I'm trying to follow VQC to see if I understand anything.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 13, 2018, 10:21 a.m. No.4500   πŸ—„οΈ.is πŸ”—kun   >>4501 >>4502 >>4506

>>4498

Great walkthrough.

 

n0 = 1

(f-2) mod 8 = 4 (the left over blue squares from VQC images)

+1 from the origin 8Tu + 1

 

total: 6

 

Is this the n we are looking for?

 

Maybe for smaller numbers that don't fit GDC, this single iteration of GetNFromOddTriangleBase is sufficient?

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 13, 2018, 10:31 a.m. No.4502   πŸ—„οΈ.is πŸ”—kun   >>4503

>>4500

Yup, that's the n we are looking for.

 

Well since we are supposed to be able to solve it in less than log n, where n is the length of c in bits, we should be able to solve 7 * 37 in like 2 something. So maybe it's not unreasonable.

 

I can keep trying with a few other numbers, just to see what we get.

Anonymous ID: 5cdbd5 Feb. 13, 2018, 10:50 a.m. No.4504   πŸ—„οΈ.is πŸ”—kun   >>4505 >>4511 >>4516 >>4590

Hey fags, hopping in for a few minutes to digest. First immediate question should be a quickie.

Am confused about VQC saying 'this' to in:

>>4285

Did he mean the triangle graph or the new equation? Of both?

He says 'equal triangles', ok agree there are 8 equal triangles, but they aren't equilateral triangles (7 on one side and 8 on another).

This is different than what Teach drew in:

>>4287

Trying to reconcile to get to:

>>4320

Are those (8) Equilateral triangles?

And the 8 triangles Teach put down are Isosceles, but not Equilateral, same in Isee's.

So labeling the first diagram in Isee's post, I would get attached image.

 

btw, until we have the tool from Teach, there is a very simple grid tool here. How are you all drawing them?:

http:/ /gridmaths.com/grid.html

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 13, 2018, 10:58 a.m. No.4505   πŸ—„οΈ.is πŸ”—kun

>>4503

 

Yeah and I'm sure it only works for smaller numbers too.

 

>>4504

I use spreadsheets and just colour the squares.

 

I've also been thinking about that, but I haven't had time to play with different styles of triangles. I used the that pattern because of VQC's "this" comment, but I'm open for it being inaccurate.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 13, 2018, 11:05 a.m. No.4506   πŸ—„οΈ.is πŸ”—kun   >>4507 >>4517

>>4500

 

Except this doesn't work for the next record in (3, 6), specifically {3:6:34:15:19:61}.

 

Here you will have the following:

 

>>> c, e, d, f, f2, f2%8, f2/8, n0

(1159, 3, 34, 66, 64, 0, 8.0, 4)

 

c = 1159

e = 3

d = 34

f = 66

f2 = 64

f2 % 8 = 0

f2/8 = 8

n0 = 4

 

Now for this record we know the base of the triangles is 10 ((x + n)-1)/2 = ((15 + 6)-1)/2 = 10.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 13, 2018, 12:13 p.m. No.4507   πŸ—„οΈ.is πŸ”—kun

>>4506

Few samples attached where n is even. These tests incorporate the gcd and f-2 resolving of n.

 

There are some cases where (f-2) finds a matching n. These tests do not increase 8 by any factor.

 

I think we have a basic understanding of the initial concept (really important!!), and we definitely need to iterate further (which VQC previously mentioned).

 

>>4343

So how do we fill the gap to find the multiples of 2d?

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 13, 2018, 2:01 p.m. No.4509   πŸ—„οΈ.is πŸ”—kun

Ran a couple more parity and small square triangle tests for even n.

 

Rsa200 - Rsa768 pic related.

 

Confirmed that parity is same for c and prime solution records. (I think we knew this already).

 

Looks like this formula works for both odd and even (x+n) when n is even? weird?

 

Also uploaded static class with all Rsa numbers.

 

pastebin.com/XYFpsDWE

 

Fyi, Rsa704 and Rsa220 had the a and b values reversed from our perspective on the source wikipedia page.

 

en.wikipedia.org/wiki/RSA_numbers

VA !!Nf9AmQNR7I ID: 6133d1 Feb. 13, 2018, 4:27 p.m. No.4511   πŸ—„οΈ.is πŸ”—kun   >>4613

>>4504

The key is that the area of the triangles is the same no matter what method you use. I think the sharp triangles very accurate, and the block method makes visualization of the base much easier, ie (n-1) and (f-2)/8 = 5. VQC said β€œthis” to my example, but then posted his example in block style >>4344

Both are valid, blocks are easier for understanding and calculating the base, which is soon belong to us ;)

VA !!Nf9AmQNR7I ID: 7a77e2 Feb. 13, 2018, 7:15 p.m. No.4516   πŸ—„οΈ.is πŸ”—kun

>>4515

Good AA! Just finished work. I'm a bit busy too for the next hour or so, but I'm excited to work here for a good while tonight.

 

>>4504

Thanks Anon, this grid link is WAY better than the one I was using. You can color, add notes, etc. Great for what we're currently working on.

>http:/ /gridmaths.com/grid.html

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 13, 2018, 9:04 p.m. No.4517   πŸ—„οΈ.is πŸ”—kun   >>4519

>>4506

The x+n value we are looking for here is 21.

 

VQC hints talk about filling the remaining portions of the triangle with n0 squared and 2d.

 

For n0=4 in your example:

n0*n0 + n0 + 1 = x+n

4*4 + 4 + 1 = 21.

 

The 2d value would be too large.

 

So where does this logic come from? Does the factor tree tell us anything?

VA !!Nf9AmQNR7I ID: 2343be Feb. 13, 2018, 9:06 p.m. No.4518   πŸ—„οΈ.is πŸ”—kun   >>4519 >>4520

>>4343

>>4344

>We have a method to calculate what n would be if we used our blue base of ((f-2) div 40). This n will call n0, to make it different from the value of n that will be our solution.

>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)).

>So far, it should become clearer that increasing the length of c adds to the number of calculations in the logarithmic of half the length c in bits. Hence why the overall complexity is < O(log m) where m is the length of c in bits.

>Bear with me as we walk through the rest in stages.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 13, 2018, 9:12 p.m. No.4519   πŸ—„οΈ.is πŸ”—kun   >>4521

>>4518

VA - I keep reading this same post over and over.

 

Why did VQC use a factor of 5 in (f-2) mod 40, and then show a closeup of the triangle with 5 blue lines?

 

>>4517

Also, the 4*4+4 looks just like another triangular method call. Recursively grow until we find a match?

VA !!Nf9AmQNR7I ID: 2343be Feb. 13, 2018, 9:12 p.m. No.4520   πŸ—„οΈ.is πŸ”—kun

>>4518

So since f is a derivative of c, d, and e we can use it's value to find n0. Then we use multiples of of n0 + 2d to fill the triangle. It looks like VQC is saying to double the width of (f-2) chunks until we get a match. If no match, c is prime. Thoughts, Anons?

VA !!Nf9AmQNR7I ID: 2343be Feb. 13, 2018, 9:15 p.m. No.4521   πŸ—„οΈ.is πŸ”—kun

>>4519

>Recursively grow until we find a match?

Theory: Basically, there are only certain values of n that can exist, and they're multiples of n0 or n0 + 2d. This is what we've been looking for, lads. A way to narrow down the n search.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 13, 2018, 9:46 p.m. No.4525   πŸ—„οΈ.is πŸ”—kun   >>4530

>>4524

I would guess that the portion of the triangle below (f-2) div 40 can be broken down into individual triangles somehow related to n0 and 2d.

 

But I don’t know what the right visual for that would be.

VA !!Nf9AmQNR7I ID: 819f3a Feb. 13, 2018, 9:53 p.m. No.4527   πŸ—„οΈ.is πŸ”—kun

>>4302

>This

I posted two things: Which "This"? The diagram? Or the new formula?

 

"So for this portion: 2d(n-1) + f - 1 = 2xn + xx"

 

Teach already had posted his great diagram of squares, which mine was a slight variant of. What about this equation, lads?

 

We're getting pretty close to tying d,f,e to (x+n). Thinking out loud over here.

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 13, 2018, 10:12 p.m. No.4530   πŸ—„οΈ.is πŸ”—kun

>>4525

Yeah but (f-2)/40 makes a repeatable base. This is where we need our geometry. Triangles = 1/2 b*h. So 1 triangle is a 45 45 90 right triangle? Area = 28. Can we assume triangles in this square are 45 45 90? Then we can calc the area or height. We can see the angles, it's all easy triangles to calculate.

AA ID: 07f944 Feb. 13, 2018, 10:28 p.m. No.4533   πŸ—„οΈ.is πŸ”—kun   >>4534

I'm confused about how we're meant to represent the base of the eight triangles in terms of (x+n). This image >>4285 implies that these triangles have no equal sides, start from the corner, and, with a being the shortest side, b being the middle and c being the hypotenuse, (x+n) = a + b. This image >>4287 implies that they're isosceles triangles but that (x+n) = 2 * shorter side + 1. Do these two situations equal the same thing and it just hasn't clicked in my head yet, or is something wrong with one of them?

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 13, 2018, 10:39 p.m. No.4534   πŸ—„οΈ.is πŸ”—kun

>>4533

Good question AA. Not sure. No matter what, the area for our example triangle is 28. Either 0.5(78) or 7+6+5+4+3+2+1= 28. To fit geometrically, i think it should be 78*SQRT(56). But for analysis purposes, the block model seems like a quicker explanation for base, especially for finding (f-2)/40 = n0.

 

I think n0 is the base of the capstone. Is this correct lads?

 

However, are the two models exactly equal? In terms of area, yes. But do we need to measure the triangle sides for our (x+n) calc? If so, one model has equal sides, and one has a side that is A+1=B.

Hobo !!1yNgQ3NlCs ID: e39786 Feb. 13, 2018, 10:47 p.m. No.4537   πŸ—„οΈ.is πŸ”—kun   >>4538

>>4488

Checked those double dubs and Oregon has nearly free weed in the fall and nice weather. We should all just meet to go camping ha ha. That would be badass. Math camp. I know some killer campgrounds in W. Oregon.

VA !!Nf9AmQNR7I ID: 2cf4b5 Feb. 13, 2018, 10:53 p.m. No.4538   πŸ—„οΈ.is πŸ”—kun   >>4539 >>4541

>>4537

Yeah Hobo! I can't think of any faggots I'd rather go camping with. I love hiking too. Also, a cold beer and a stogie at the end of a camping day is delicious. Then camp dinner. Then playing guitars around the campfire. Then Topol shows up for the Grade A Bakening. Then we all trip out about the universe. Talk about cool shit and expand our minds for a couple weeks. Then go back to the "normal world" better than when we left it. Camping is the SHIT.

Hobo !!1yNgQ3NlCs ID: e39786 Feb. 13, 2018, 11:08 p.m. No.4539   πŸ—„οΈ.is πŸ”—kun   >>4540

>>4538

Well thats pretty much what I do every day out here 9 months per year. Its amazing. I'm never going back to the world. I'm setting up a night vision camera system with multiple cameras to capture the whole night sky to look for UFO's this summer. It should be super fun.

 

Any time May through October is glorious here weather wise. August gets a bit hot at times but zero humidity so its not too bad. I have camped all over here and its about as good as it gets. My favorite camp ground is the Alsea Falls Recreation area. It is insanely beautiful and cheap and you can mountianbike right across the street. I would love to set up a math camp for us out there in the forest. It would be bad ass for sure. Tons of local wineries. Everyone has weed here. Its like water. Good hiking trails plus great beaches. Let me know, my schedule is always free/flex up to mid-late august/early September. If we come up with a time, May through Aug 9 or so I can get the group camp site there reserved for whatever dates work for the most people. I have a few tents and sleeping bags and stuff as well. Probably not enough for the whole room but for a few. Plus if we do it like that it is a public place so its safe for all us anons to meet for the first time. It would be fun. Fly into Eugene or Portland and I can transport as well.

VA !!Nf9AmQNR7I ID: 7a77e2 Feb. 13, 2018, 11:22 p.m. No.4544   πŸ—„οΈ.is πŸ”—kun   >>4546

>>4541

>>4542

Hobo, (((they))) already know who I am, so not worried about planning a trip. I'm gonna live my life. You guys are my MathUniverseFam. I'd love to camp and meet you all someday in the future. The Great Awakening is bringing together people from all over the world, why not in 3D space?

Hobo !!1yNgQ3NlCs ID: e39786 Feb. 13, 2018, 11:28 p.m. No.4546   πŸ—„οΈ.is πŸ”—kun   >>4547

>>4544

Checked! and I agree. Maybe we should make a ?sub board thing? for this so we don't clutter the real board all up. We can call it VQC Math Camp. May and June are the prime time out here if we can get it lined up for then. As good as it gets weather wise. August gets hot and if there is a fire in the area the smoke is dreadful.

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 13, 2018, 11:44 p.m. No.4548   πŸ—„οΈ.is πŸ”—kun

Trying to analyze our square: Base of 5 always has height of 4 for our odd (x+n) triangle. Each triangle can be simplified to 3^2+4^2=5^2. That's why VQC picked 5 as a base. Thoughts?

AA ID: 07f944 Feb. 13, 2018, 11:59 p.m. No.4549   πŸ—„οΈ.is πŸ”—kun   >>4573 >>4607

I'm kinda struggling to understand the latest crumbs. This is what I do understand:

>c is the difference of two squares, (d+n)(d+n) and (x+n)(x+n)

>you can figure out (x+n)(x+n) based on (x+n)'s parity

>you can figure out (x+n)'s parity based on c % 4

>when (x+n) is odd (or when any square is odd), you can represent it as 8 triangles + 1

>the base of each of the eight triangles is ((x+n)-1)/2

>this square of (x+n)(x+n) = 8Tu + 1

>if you figure out how to calculate the area of those triangles you win

>as per >>4305 you can find n even if you don't know x as long as you do know (x+n)

Here's where I get confused:

>here >>4322 he splits the triangles because (n-1)(n-1) is another odd square so you can use the same 8Tu + 1 rules on it

>that then has something to do with f

What are (n-1)(n-1) and f used for? I am lost at that point.

 

>>4540

I'm a musician, but I doubt I could afford a trip to the US any time soon.

VA !!Nf9AmQNR7I ID: 6133d1 Feb. 14, 2018, midnight No.4550   πŸ—„οΈ.is πŸ”—kun

I support this theory >>4285 for the following reasons:

 

Theory: The reason why one triangle side is less than the other is this:

3^2+4^2=5^2

You need one side to be less so the pythagorean theorem works out. Thoughts?

 

Come on Anons, need your input. My time is limited by IRL shit. Jump in fags!

Hobo !!1yNgQ3NlCs ID: e39786 Feb. 14, 2018, 12:33 a.m. No.4565   πŸ—„οΈ.is πŸ”—kun

>>4561

Hat Tip

>>4562

Well I ddg'd The Tesla 369 thing and saw it. It reminded me of whats going on here… A lot. so I posted it. Seems like the same double jump thing We kept seeing at my end back in the beginning all over the place.

VA !!Nf9AmQNR7I ID: 2343be Feb. 14, 2018, 1:47 a.m. No.4574   πŸ—„οΈ.is πŸ”—kun   >>4575

>>4573

Sure, AA! But I'm still working to understand. VQC is pretty quick to respond about questions to his twatter: @ChrisRootODavid. Ask him directly, he's pretty quick to respond. We are all Confused at this point. Moving forward tho.

AA ID: 07f944 Feb. 14, 2018, 1:52 a.m. No.4575   πŸ—„οΈ.is πŸ”—kun   >>4576 >>4578

>>4574

Don't you need a bunch of personal information to make a Twitter account? I know I couldn't make a throwaway account once because it needed a phone number. Plus it seems like some of the others who aren't here right now might understand, so rather than annoy him with a ton of questions it might be better if we all just get on the same page through the board.

ID: 2bf75e Feb. 14, 2018, 2:22 a.m. No.4579   πŸ—„οΈ.is πŸ”—kun   >>4582

You're doing a great job Veritas. That was a hell of a lot of numbers you crunched. Still haven't programmed any tests for these new developments yet, because 3D space caught up to me. That's what helped me understand the most. Coding programs that try to get to the factorization of a number using every bit of new research.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 14, 2018, 2:35 a.m. No.4580   πŸ—„οΈ.is πŸ”—kun   >>4582 >>4584

>>4554

I stirred the quantum weirdness and knocked out from lack of sleep schedule ^_^

 

As for 369 and "Vortex Math", it's more of a pattern with applications than a math "in and of itself" like Geometry or Trig or something.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 14, 2018, 2:39 a.m. No.4581   πŸ—„οΈ.is πŸ”—kun   >>4587

I'm just playing around with the triangles, but I thought of one thing.

 

When we are filling our triangle with 2d(n-1), then when filling in a single triangle, we should fill inn only 2d(n-1)/8 no?

 

I played with it and attached the photo. I'm still using this record: (3, 6, 16, 9, 7, 37)

 

There's two triangles in the photo. The left one is the actual triangle while the right one is our triangle created from n0^2 and multiples of 2*d/8.

 

Sorry for the colours, but each set of 2d/8 is coloured with it's own colour. In this I used 7 2d/8 parts, which is wrong. It should be 2d5/8 for the case I'm using as n = 6.

 

It almost overlaps, but it is 1 square too many. Not sure if I'm barking up the wrong tree, but nevertheless I'm still playing with the triangles.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 14, 2018, 2:45 a.m. No.4584   πŸ—„οΈ.is πŸ”—kun   >>4585

>>4580

Sidenote, while this is a fun thought experiment, my non-euclidean maths professorbro gets out the notion that 0 can equal ∞, but like "Vortex Math"… this is another way of arranging numbers in ways that come across as significant.

 

This all stems from me being on my friend's porch trying to contemplate the other side of a railing… or the other side of anything for that matter. I'd already gotten to the point of "For ever 1 there is a 0, and every 1, no matter how many 1s its made of, shares the same zero" and then flipped to 0>1 and that's how I found this book that blew my mind.

 

http://www.everythingforever.com/st_math.htm

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 14, 2018, 2:48 a.m. No.4586   πŸ—„οΈ.is πŸ”—kun

>>4585

Beyond THAT… I think it got proofed to me at some point that ∞ + -∞=.12 or -2 or… something like there. It's all sorts of convoluted for me. I was just trying to figure out the universe, not become a math prophesier. -shrug-

 

It happens.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 14, 2018, 3:53 a.m. No.4587   πŸ—„οΈ.is πŸ”—kun   >>4588

>>4581

 

Still trying to wrap my head around everything.

 

In the photo you'll see three squares, all the same. The first one is the one divided into the 8 triangles,

the second one is the ration of squares for (x+n)(x+n) and the last one is for the other equations nn + 2d(n-1) + f - 1.

 

In the last one I also highlighted each of the 2d we have an named them 2d, 2d2, 2d3, 2d4 and 2d5 to signify the accumulated squares added.

MM !!DYPIXMDdPo ID: 5cdbd5 Feb. 14, 2018, 6 a.m. No.4590   πŸ—„οΈ.is πŸ”—kun   >>4602 >>4604

Lads, bit frustrated this morning. Posted >>4504 yesterday to try and work through VQC's crumbs in clear methodical way. This morning, there are 74 new posts, but 90% of it is shitposting, mathfeelz, or muddied info. Took 45min and didn't move along understanding.

VA, VQC's reply to DM as simply "When do you think you'll be ready for the next stage of part 3?" I couldn't even follow your questions to him really. WTF do you mean by "Adding the triangles is easy. 1, 3, 5, 7, 9, 11, etc.?" I very much appreciate you coming back with your DM info, it's really good to keep us all on this board and not fractured with info bits we don't all share.

Isee, thanks for your focus and working through the geometries.

Lads, let's FOCUS on the drop VQC gave us. Let's UNDERSTAND each little bit. Every statement. For example, in the DM to VA "The eight triangles are built up by fitting everything around f-2. Because there are 4 left over, that determines the configuration and multiple of 2d." OK, break that down, understand each little bit. 4 triangles? No, think it's 4 'units' as shown in >>4342, the "(f-2) mod 40" shown blue in the center. That's what I'll do, just work to UNDERSTAND that ONE piece, and only then move on.

VA - good on you for getting the code running, awesome help from Anons.

Great Hobo, would LOVE to make Math Camp, but won't be in the cards this year. Your back yard is beeeyoutiful! The sand DUNEs too, hope there aren't sandworms!

Thanks for letting me vent anons. Let's keep this thread clean and focused. Share progress more than questions (myself included). Bring clarity to what we have, and not add disinfo unless part of an obfuscation strategy (there is the benefit of all this, in that anyone that comes in here will just leave, and we stay under the radar).

Love you faggots, last thing I want to do is suck any energy out of any of you or this, only trying to direct it toward our goal(s).

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 14, 2018, 6:52 a.m. No.4591   πŸ—„οΈ.is πŸ”—kun

>>4343

 

I'm curious about this part:

>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 mean in a way, this kind of sounds like a solution, no?

 

Now what does it mean to "double the width"? We know we have a "blue band" that's 5 rows thick. So I'm guessing maybe he means doubling this to 10 width, then 15, then 20 etc until we can determine it is a prime (or find the solution).

 

I haven't tried anything regarding it yet, but maybe it's worth a shot.

 

Also, I'm not sure of course, but does this blue band have 5 rows thickness, because we also divided by 5 and not just 8?

 

I'm guessing VQC will come by later and explain, but I still want to chip away at this, trying to wrap my head around it.

VQC !!Om5byg3jAU ID: d04b72 Feb. 14, 2018, 8:01 a.m. No.4593   πŸ—„οΈ.is πŸ”—kun   >>4606

The next function takes our triangle base made out of (f-2)/40 which is five units thick and our value of n0 and calculates how many lots of 2d would be needed to fill in the gap in our 8 triangles, taking into consideration the space taken up by the 8 triangles made with a base of (n0-1)/2.

VQC !!Om5byg3jAU ID: d04b72 Feb. 14, 2018, 8:14 a.m. No.4594   πŸ—„οΈ.is πŸ”—kun   >>4597 >>4598

public static BigInteger Get_Remainder_2dnm1(BigInteger bs, BigInteger d, BigInteger n, BigInteger f)

{

BigInteger triangle = Triangle(bs);//Tbs

BigInteger eight_base = BigInteger.Multiply(triangle, eight);//8Tbs

 

BigInteger XPN = BigInteger.Add(eight_base, one);//8Tbs+1

BigInteger two_d = BigInteger.Add(d, d);//2d

 

BigInteger nm1 = BigInteger.Subtract(n, one);//(n-1)

BigInteger two_d_nm1 = BigInteger.Multiply(two_d,nm1);//2d(n-1)

BigInteger XPN_mfp1 = BigInteger.Subtract(XPN, BigInteger.Subtract( f,one));//(x+n)(x+n) - (f-1)

BigInteger resultpN = BigInteger.Subtract(XPN_mfp1, two_d_nm1);//(x+n)(x+n) - (f-1) - 2d(n-1)

BigInteger result = BigInteger.Subtract(resultpN, BigInteger.Multiply(n,n));//(x+n)(x+n) - (f-1) - 2d(n-1) - nn

return result;

}

VQC !!Om5byg3jAU ID: d04b72 Feb. 14, 2018, 8:20 a.m. No.4596   πŸ—„οΈ.is πŸ”—kun   >>4602

We require a couple of diagrams to explain this part.

It would be good if you test the code above with the correct values to show it works for the known correct values for RSA 100.

Anonymous ID: e99bd3 Feb. 14, 2018, 8:36 a.m. No.4597   πŸ—„οΈ.is πŸ”—kun   >>4602 >>4709

>>4594

For those who like algebra, we can show that this formula evaluates to 0 for correct inputs:

 

(x+n)^2 - (f-1) - 2d(n-1) - n^2 = 0

(x+n)^2 - f + 1 - 2dn + 2d - n^2 = 0

-f + 1 + 2d + d^2 = d^2 + 2dn + n^2 - (x+n)^2

(d+1)^2 - f = (d+n)^2 - (x+n)^2

c = (d+n)^2 - (x+n)^2

 

We end up with the original formula that defines what n and x are.

VA !!Nf9AmQNR7I ID: a60c59 Feb. 14, 2018, 9:48 a.m. No.4602   πŸ—„οΈ.is πŸ”—kun   >>4605

>>4590

Thanks MM! I was bored and stayed up too late drinking wine hoping VQC would pop in. I'll make sure to keep my posts more on point so the thread stays clean and focused. Thanks for venting.

 

>>4596

Hey VQC! Good to see you! I think I can say for all of us that it's nice to have you stop in more often.

>>4597

Thanks for the Algebra equations!

 

>>4598

>>4599

Nice work guys!

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 14, 2018, 10:14 a.m. No.4603   πŸ—„οΈ.is πŸ”—kun

How easy is it to extract the data/files/whatever from the blockchain? It's been going on for so long… where does one even begin to look?

 

Like… where would the Darkweb interactions be? The transactions… what they purchased if it's digital…

 

How's that work?

Hobo !!1yNgQ3NlCs ID: 907fa5 Feb. 14, 2018, 12:26 p.m. No.4604   πŸ—„οΈ.is πŸ”—kun   >>4605 >>4609

>>4590

Understood and you are right as rain. Thats why we made the alt thread. Is there a way for someone to simply delete the offending posts as now they are basically useless clutter? It also seems like a problem screaming for a filter button. Is there a filter tool that individuals could just minimize individual shitposting with?

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 14, 2018, 1:44 p.m. No.4606   πŸ—„οΈ.is πŸ”—kun

>>4343

>>4593

Pic attached are a few examples of my current understanding of this process.

 

Step 1: Calculate n0 using a triangle base from (f-2) div 8. (Replaced div 40 with div 8 for smaller numbers).

 

Step 2: Calculate the remainder 2dnm1 (gap) using the same triangle base and the n0 result from step 1.

VA !!Nf9AmQNR7I ID: 6133d1 Feb. 14, 2018, 5:20 p.m. No.4608   πŸ—„οΈ.is πŸ”—kun

>>4607

Hey AA! Just getting off work. Ok, here’s my current understanding, other Anons please correct me if I’m wrong on anything.

 

(n-1) is the small capstone of each of 8 triangles.

n0 = (f-2)/8 gives you a triangle with a base divisible by 5. Bigger than the capstone triangle, smaller than the entire triangle. I also think you could get this base by doing (n-1)*5?? Not sure on this idea. VQC is hinting that multiples of this n0 triangle can be used to fill up the remaining space inside each triangle, allowing us to solve for x+n or n.

 

Then the bottom base portion is ((x+n)-1)/2 as you know. So n0 is somehow going to help us solve x+n, possibly in combo with the factor trees.? Working to understand over here.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 14, 2018, 7:21 p.m. No.4609   πŸ—„οΈ.is πŸ”—kun

>>4604

That's fantastic but the only interaction that happens there is people asking me weird questions and then never following up with them no matter what response they get.

 

I brought up "how to spread this" and "education" forever ago in the EZB and, looking at it… yeah… still no responses. So…

 

I dunno whats t'tell'z ya.

MM !!DYPIXMDdPo ID: 8f5be8 Feb. 14, 2018, 10:46 p.m. No.4613   πŸ—„οΈ.is πŸ”—kun

>>4511

VA - I see they're both the same after looking closer, thanks.

(8) triangles for the diagram, each have area of 28 units.

The one you drew, 8 one side and 7 the other, area is 1/2bh, so 28.

With the block versions, there are 28 blocks. Also, draw line corner to corner, 7units each edge, plus (7) 1/2 blocks left over outside the line. 49/2 + 7/2 = 28. same same.

Plus the (1) in center for the 225 square.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 15, 2018, 3 a.m. No.4614   πŸ—„οΈ.is πŸ”—kun   >>4615

Not sure if I'm far off what VQC is getting at, but I've been trying to figure out what he means by adding 2d's.

 

So I'm again using the record for a=7, b=37.

 

For this record we will have the following variables:

 

c=259

d=16

e=3

f=30

f - 2=28

base of triangle = 3

n0 = 1

 

We can now construct a new square by computing the triangle of the base (triangle(base) = 6). We then multiply it by 8 and add one, giving us 6 * 8 + 1 = 48 + 1 = 49. So our new square is 7*7.

 

It's here I've been thinking about adding multiples of 2d. I don't see how accurate this is, as what I have now doesn't appear to match along with what VQC has been saying (I think). But attached is a screen shot of my 77 square with triangles highlighted. Around it you will see different multiples of 2d. Each cell is numbered and belong to one part of the 2d, except for the last one which is simply d.

 

We can see based on the number of d's added that it adds 49 + 2d5 + d. So the 2d5 part is correct, but the square will then miss d (fluke? random? step in the right direction?).

 

Once we add 49 + 2165 + 16 we have 225, which is 15^2 which is also the (x+n)(x+n) we are looking for w/ regards to this record.

 

Again though, I'm not sure if this is what VQC has in mind.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 15, 2018, 3:30 a.m. No.4616   πŸ—„οΈ.is πŸ”—kun

>>4615

Also I think we're missing something here, with regards to step 3 part a. If we don't have to use the tree for this specific case, then why bother generating the tree if we can deduce this without computing the tree?

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 15, 2018, 5:32 a.m. No.4617   πŸ—„οΈ.is πŸ”—kun   >>4618 >>4620 >>4626 >>4669

>>4322

So after revising this post I took another look at my a=7,b=37 square and updated it to show the different elements.

 

The center part is the square composed by (n-1)(n-1). That is 25, which when remove 1 and divide by 8 is (25 - 1)/8 = 24/8 = 3.

 

So the inner part has a lighter shade, the next part represents the area between the base ((x + n)-1)/2 and (n-1)(n-1) which in this case is 144.

 

Now 144 mod 2d = 16, which is the square of f-2 % 8 = 4 in this case. Not sure if relevant. It's also (d - 4)*2. Again, not sure if relevant.

 

The outer layer which has a strong colour is the base of (x + n - 1)/2.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 15, 2018, 7:40 a.m. No.4619   πŸ—„οΈ.is πŸ”—kun   >>4620 >>4632

>>4618

 

I'm kicking my self in the ass for not noticing this until earlier today:

 

http://mathworld.wolfram.com/TriangularNumber.html

 

I feel I should probably spend more time reading about this stuff, than just trying to wrap my head around VQC's crumbs and methods.

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 8:20 a.m. No.4620   πŸ—„οΈ.is πŸ”—kun   >>4621 >>4624 >>4630

>>4617

Isee, I've been working through this and get the attached image, working backwards.

Some interesting geometries there, with the 8 squares around the centered unit square.

Just a way to picture our original layout, with a specific solution for c=259.

 

>>4618

>>4619

Also, note other image attached (taken from the Wolfram page Isee *linked), remember the CONWAY hint?! I'd been looking into that particular Conway and Guy book too, ding ding.

 

BTW, regarding hints that came with that CONWAY one, VQC mentioned "asymmetric warfare". Have been thinking maybe this is asymmetric, as in "Asymmetric Encryption" algorithms = "Public Key Encryption"?!

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 8:45 a.m. No.4621   πŸ—„οΈ.is πŸ”—kun   >>4622 >>4624 >>4627

>>4620

Here is the same, with b added, and the [e,n,t] record label of [3, 6, 2] = {3:6:16:9:7:37}.

It's just the top part of the L, added over to right of the box. Can quickly see geometrical derivation of b=a + 2x + 2n.

Tonight I'll run this through VQC's latest crumbs to check flow of everything (n0, etc.)

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 9:04 a.m. No.4622   πŸ—„οΈ.is πŸ”—kun

>>4621

Suspect the (8) unit blocks surrounding the Orange centered unit, will be the (8) n0=1 pieces. That should give an offset for x from the center unit, that produces a=7. Need to review, just looking geometrically.

If n0=3, it would be shifted 2 blocks (5x5=25, 25-1=24, 24/8=3). If n0=6, it would be shifted 3 blocks (49-1 = 48, 48/8=6).

PMA !dSvrkhSLR6 ID: fa8576 Feb. 15, 2018, 9:24 a.m. No.4624   πŸ—„οΈ.is πŸ”—kun   >>4628 >>4632

>>4621

>>4620

Hey MM - these are excellent drawings.

 

Have you read this VQC hint?

>>4242

>The key here is first understanding how the (x+n) square being added to c is constructed in terms of also being analogous to an L shape on the side of the square of d sides which must incorporate the remainder e in the L shape (or incorporates the 'gap' made by f).

 

Would adding f to your diagrams add any clarity?

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 9:56 a.m. No.4628   πŸ—„οΈ.is πŸ”—kun   >>4629

>>4624

Thanks. I think the Pyramids VQC illustrated may be a bit different though, as the 'caps' above the f are (n-1), (6-1)=5 in this case (think that's area). So that's the type of thing to keep going with and work out. Just sharing some progress along the way with integrating the new drops.

>>670

^^ this shows how I'd originally understood the whole "L" thing, which really makes sense, including the odd unit up in the corner.

 

>>4626

Great new MAP! Had just finished getting all the new drops in a text file.

 

>>4627

Thanks, inspired by you actually.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 15, 2018, 10:16 a.m. No.4629   πŸ—„οΈ.is πŸ”—kun

>>4628

Just thinking it would be pretty awesome to be able to draw this layout for any record.

 

And when we figure a solution out, to show a transform from the c to p records.

 

>>4626

Thanks!

PMA !dSvrkhSLR6 ID: fa8576 Feb. 15, 2018, 10:25 a.m. No.4630   πŸ—„οΈ.is πŸ”—kun   >>4631

>>4620

 

The factor tree for this record is:

 

  • 259 (c)

| + 16 (d)

| | + 8 (/2)

| | | + 4 (/2)

| | | | + 2 (/2)

| | | | | + 1 (/2)

| + 3 (e)

| | + 1 (d)

| | + 2 (e)

| | | + 1 (/2)

 

Interesting how there is a 3x3 square around the middle. Corresponds to 3 (e) in the tree?

One side of the triangle is 16/2 = 8. Also in the factor tree (although it would be divided out by 2).

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 12:46 p.m. No.4631   πŸ—„οΈ.is πŸ”—kun   >>4632

>>4630

Cool, will look at that tonight. Here's what I mean by the n0, shown in yellow for c=259 where n0=1, and it disappears in c=287 (n0=0).

That's why I'm figuring n0 progresses in a certain way (2*n0+1)^2-1=n0(8), which are the blocks around the center (1) unit block.

So n0(8)=0, n1(8)=8, n2(8)=24, n3(8)=48, …

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 15, 2018, 4:10 p.m. No.4632   πŸ—„οΈ.is πŸ”—kun   >>4633 >>4645

>>4626

Thanks Baker! The new map is very helpful, especially with so many new crumbs to digest.

 

>>4631

Great work and focus MM! Thanks for helping us focus and step up our game. I was a little butt hurt, but I deserved it. Gotta bring my A game to the table. Love you faggot!

 

>>4624

Hello PMA! I’ve been working to incorporate f or e into the diagrams. Are you still working through the n0 crumb? There’s a lot to digest in the recent crumbs!

 

>>4619

Hey Isee! Thanks for your consistently excellent work. You got MM all fired up! Also, thanks for reposting that link. Our Mystic Anon Advisor Topol had posted it a few breads back. I’ve also had my head down so much I haven’t been doing the reading. Time to read up for both of us?

 

All Anons, it’s great to be here with you all. I’m having a great time, and math is in my blood and mind at all times. Let’s do this!

 

>>4625

Shout out and thank you to Topol for all his inspiration, enthusiasm, and honestly his foresight into intuitively grasping the upcoming ideas and cheering us on. Thanks man. I love you too, faggot.

 

All right, now that the ratio of work to shitposting is back in balance, let’s work and have some fun too! I have some good memes and good ideas to drop later.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 15, 2018, 4:21 p.m. No.4633   πŸ—„οΈ.is πŸ”—kun   >>4634 >>4636 >>4647

>>4632

Still looking into the n0 calculation and possible subsequent calls.

 

I did notice in a few of my test cases that the result from (2dnm1 - ((f-2) mod 8)) equaled a triangle number.

 

Don't know what to make of this.

 

Example for c=6107:

 

c = 6107

d = 78

f = 134

 

(f-2) = 132

(f-2) % 8 = 4

(f-2) / 8 = 16 (triangle base)

 

n0 = Get_n_from_odd_triangle_base( (f-2)/8, c, d )

n0 = 6

 

remainder 2dnm1 = Get_Remainder_2dnm1( (f-2)/8, d, n0, f )

remainder 2dnm1 = 140

 

2dnm1 - (f-2) % 8

triangle number? = 136

(16*17)/2 = 136

VA !!Nf9AmQNR7I ID: 2cf4b5 Feb. 15, 2018, 4:42 p.m. No.4634   πŸ—„οΈ.is πŸ”—kun   >>4636

>>4633

Did you see VQCs crumb about the thickness of n0=5? Like it’s not just the base, it’s a width or measurement. How the heck did he figure that out? Then he gave a calculation for BOTH the base and top line of the f-2 section. I have indigestion from these latest crumbs, need some Tums. Away from home base, I’ll post the link when I get back. Just can’t help checking in, even at work.

VA !!Nf9AmQNR7I ID: 2cf4b5 Feb. 15, 2018, 5:45 p.m. No.4636   πŸ—„οΈ.is πŸ”—kun   >>4637 >>4819

>>4634

>>4633

 

Hey PMA! Here's the correct link

>>4342

>>4344 Diagram here.

 

>In this diagram, somewhere in each triangle, there is a part that is five units wide, that we hope is smaller than (x+n) and larger than n.

>The middle of each blue bar is (f-2) div 40, so the base of a triangle with that bar would be ((f-2) div 40) + 2, the top of that bar would be ((f-2) div 40) - 2. The five parts together add up to (f-2) div 40.

>Plus the remainder of 4 unit squares from (f-2) mod 40.

 

"The five parts together add up to (f-2) div 40"

"base of that bar would be ((f-2) div 40) + 2"

"top of that bar would be ((f-2) div 40) - 2"

 

Attached are my annotations on VQC's diagram. Thoughts?

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 15, 2018, 5:55 p.m. No.4637   πŸ—„οΈ.is πŸ”—kun   >>4638 >>4641 >>4644

>>4636

Thanks for reposting with annotations.

 

I've tried a number of different factors * 8 when computing n0. For smaller numbers, it does make a difference in the n0 return value.

 

If the important thing about the base is having a number between n and x+n, then there has to be something in the original c, d, e or f values that tells us what to use.

 

VQC did say it was arbitrary. >>4340

 

But then why go into such detail about these blue lines?

 

Sorry VA, no conclusions here.

VA !!Nf9AmQNR7I ID: 2343be Feb. 15, 2018, 6:17 p.m. No.4638   πŸ—„οΈ.is πŸ”—kun   >>4639

>>4637

I think a big key to understanding n0 is that we're working with Grid examples that have very small numbers relative to RSA examples. Maybe we should work through RSA 100? We have the answers, so it makes a perfect walkthrough. Thoughts, Anons?

VA !!Nf9AmQNR7I ID: 2343be Feb. 15, 2018, 7:07 p.m. No.4641   πŸ—„οΈ.is πŸ”—kun   >>4644

>>4637

>>4639

So maybe let's pick a bigger number from the grid then. In the 10,000 range for c.

>If the important thing about the base is having a number between n and x+n, then there has to be something in the original c, d, e or f values that tells us what to use.

PMA, I agree. Our starting value of c contains all we need to find the prime factors. Looks like triangles will save the world, lol. So here's my current thoughts.

n0^2 + multiples of 2d can fill the remaining part of one triangle. Pic to follow, trying to work it out now.

AA ID: 14033a Feb. 15, 2018, 7:49 p.m. No.4644   πŸ—„οΈ.is πŸ”—kun   >>4646

>>4637

>>4641

Shot in the dark, but when I was seeing if binary search would work, I found that there were relationships between c and several other variables (I'm pretty sure n was one of those but I'll have to check) that showed you could calculate the maximum and minimum possible values. For example, in a grid of c*i, there were a bunch of linear lines, and you could use the gradient of the lowest (visually lowest) line to find the maximum value and the highest to find the minimum. If n0 is between n and (x+n), and we can find the maximum possible n for a given c and the minimum possible (x+n) (and if they happen to magically not overlap), we'll know the range of possible n0 values. Definitely don't hold your breath, because this probably isn't going to work, but I'll post results.

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 7:50 p.m. No.4645   πŸ—„οΈ.is πŸ”—kun   >>4646 >>4647 >>4649

>>4632

Love you too, glad you got over it. Sorry to piss on you's parade, hope it didn't dampen spirits too badly!

>>4513 and a Kek from me too, that was just perfect, like, the best.

 

>>4639 >>4641

How about we use 6107? I'm sketching that one now.

{23:36:78:47:31:197} = 6107 (da answer)

{23:2976:78:77:1:6107} = 6107 (known)

f = 134

(f - 2) = 132

Started by trying to iterate, going to sketch the solution and see if we can back into it? This one appears to have a remainder situation as well.

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 15, 2018, 8:16 p.m. No.4646   πŸ—„οΈ.is πŸ”—kun   >>4647

>>4644

Thanks AA! Post your results, lad.

 

>>4645

Thanks MM!

Sure, let's use c=6107! Time to work, fags.

 

Alright, working to understand the n0 idea. Here's my working diagram. Thoughts? Basically after we know n0, VQC said "n0^2 + multiples of 2d can fill the triangle unless it's prime." Correct me if i'm wrong anons.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 15, 2018, 8:43 p.m. No.4647   πŸ—„οΈ.is πŸ”—kun   >>4649 >>4653

>>4646

the comment from VQC in >>4343

>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.

 

I believe is being handled by the call to Get_Remainder_2dnm1. It takes the n0 and d values as parameters, and internally uses 2d and nn.

 

I think we are now looking for this part of the puzzle which is the return result of Get_Remainder_2dnm1:

>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.

 

>>4645

MM - The "gap" that I calculated for c=6107 is 140. Breakdown in >>4633

Perhaps there is some way that number could be reflected in your diagrams. It is close to another triangle number T(16). And T(16)/8 = 17.

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 8:50 p.m. No.4649   πŸ—„οΈ.is πŸ”—kun   >>4650 >>4651 >>4657 >>4658 >>4667 >>4669

>>4645

re: n0, PMA calced as n0=6. Sketching the actual result, it looks as if n0=5. That adjustment might be part of the process.

 

Am still working through the triangles around the unit cell and the n0 region. I think they are supposed to go right to the unit cell, and get 'clipped' by the n0 region. But, it could also be that they go up to the outside of n0. Showing both scenarios for the orig c=259 case.

 

Next, want to do a sketch of what we would start with, with the initial guess, before any triangle adjustments (and before actual n or x is known).

 

>>4647

Interesting! Starting with your values, the initial was too large by 1108, which /8 = 138.5. That was with a large square of 88 and small of 23 (basing off your init triangle calc). So Lg^2 - Sm^ = 7744-529 = 7215, which is too large by 1108, (7215-c = 1108).

 

I like the block triangles Isee and Teach were doing, shows the actual units that get cut off inside the n0 region.

VA !!Nf9AmQNR7I ID: a34e47 Feb. 15, 2018, 9:24 p.m. No.4653   πŸ—„οΈ.is πŸ”—kun   >>4658

This is what I was talking about the other night in a very unclear manner. If we have n0, possibly some combo of n0^2 and 2d can fill the triangle. Visual is large triangle filled with multiples on n0.

 

>>4647 PMA is close here! Maybe a remainder or factor to be added back in?

AA ID: 14033a Feb. 15, 2018, 9:59 p.m. No.4656   πŸ—„οΈ.is πŸ”—kun   >>4658

The gradient thing was again a dead end. It seems like it could be immeasurably useful but probably only in one specific context that possibly isn't relevant to anything we're doing so we might not stumble upon it. We can definitely calculate the only possible n values for a given c (e.g. if c=5543, the maximum possible n is 922 based on the lowest possible a being 3), but obviously that isn't useful.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 15, 2018, 10:01 p.m. No.4657   πŸ—„οΈ.is πŸ”—kun   >>4658

>>4649

And here is the factor tree for c=6107:

 

  • 6107 (c)

| + 78 (d)

| | + 39 (/2)

| | | + 6 (d)

| | | | + 3 (/2)

| | | | | + 1 (d)

| | | | | + 2 (e)

| | | | | | + 1 (/2)

| | | + 3 (e)

| | | | + 1 (d)

| | | | + 2 (e)

| | | | | + 1 (/2)

| + 23 (e)

| | + 4 (d)

| | | + 2 (/2)

| | | | + 1 (/2)

| | + 7 (e)

| | | + 2 (d)

| | | | + 1 (/2)

| | | + 3 (e)

| | | | + 1 (d)

| | | | + 2 (e)

| | | | | + 1 (/2)

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 10:23 p.m. No.4658   πŸ—„οΈ.is πŸ”—kun   >>4659 >>4660 >>4667

>>4656

Nice effort AA.

 

>>4650

Block UI's would be interesting for medium demo numbers, probably good as teaching tool. I'm really hoping we aren't in this proto middle-land stage too long!

 

>>4653

>>4654

Ok, I'm following what you're saying there with the incremental for each row with growing triangle numbers. And I need to watch area vs base as well, think that's mucked up in these.

 

>>4657

Will be great to know how the factor tree ties in!! Those look fantastic.

 

>>4649

Here was the first "iteration" attempt. Took the (f-2)=132. So we have 16 with 4 units left over. This is where I think it's the wrong step for me. Used n0=6. We know D, so created the D square, inclusive of the n0 square, and used that to finish out the rest of the geometry.

1) Am not sure d is drawn in correct way in this case. Perhaps it should be going to the Inside of n0 square, bottom left corner?

 

The idea here was to take the first pass, and see the gap. Not sure that's a valid record, didn't check yet. Probably not, because the remainders "haven't been taken into account".

Will play for another 45min or so.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 15, 2018, 10:36 p.m. No.4659   πŸ—„οΈ.is πŸ”—kun   >>4662

>>4658

>I'm really hoping we aren't in this proto middle-land stage too long!

Agreed. Well, at least I can now generate small number records in js.

 

Very interesting to show each iteration geometrically. I was thinking more along the lines of the final prime solution, with the small square only partially filled in. And the "gap" result somehow highlighted. (No idea what this would look like.)

VA !!Nf9AmQNR7I ID: 819f3a Feb. 15, 2018, 10:46 p.m. No.4661   πŸ—„οΈ.is πŸ”—kun   >>4667

>>4660

Lol, step it up VA! Misread your question MM. On it now.

>Perhaps it should be going to the Inside of n0 square, bottom left corner?

That's the geometrically correct config for an odd (x+n) square.

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 10:47 p.m. No.4662   πŸ—„οΈ.is πŸ”—kun   >>4663

>>4659

Awesome! Ha, I was sitting here imagining the d square, with e hugging around it (incomplete, or you'd go up to next d). Then, a notch shows up in the upper right and the e blocks scatter to each side, meeting their new friends pouring out of the notch, until all is balanced and well with the world.

>>4660

Thanks for the encouragement. Won't last much longer today, but going through it from beginning to see what sticks. Will dream of triangles later for sure.

 

And wow, what a day for Q! Big drop on the cellular tech, amazing what classified stuff is getting dropped: #NoMoreSecrets !!!

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 11:34 p.m. No.4667   πŸ—„οΈ.is πŸ”—kun

>>4663

Indeed. Think ball is just getting rolling. Did you see Newsweek clip today, and the reader comments? It was an interesting play with an 'out there' kicker by AJ at the end (just to keep normies in discount mode). 911 narrative right there and interviewer only went after Sandy bit. Hmmm.

 

>>4661

Thanks VA! Yes, went back to orig diagrams and all looks good, incl where 'a' is for width of L.

 

>>4658

This isn't valid. Calculating what was produced: {23: __ :78:23:55:121}, solving for 'n', it would have to be 12, not 10. The x, a, and d all worked fine though. So, just drawing a geometry doesn't mean it all 'adds up'!

 

>>4649

Attached is center detail for the actual solution to this one. The n0=5, meaning 5 units away from the center cell, so 120 cells + center = 11^2.

Am starting to get an intuition/visual about how these pieces move and grow and shift. So the d is a lock, therefore the top and right of the n0 box are a lock. As n0 grows, it 'pulls' the (x+n)*(x+n) square with the 8 triangles in for a deeper cut, causing x to grow in the process.

MM !!DYPIXMDdPo ID: a6ab42 Feb. 15, 2018, 11:37 p.m. No.4668   πŸ—„οΈ.is πŸ”—kun   >>4719

And here's the image - browser is suddenly lagging big time. Reboot time.

 

>>4651 VA, thanks for the meme btw! Been thinking about the spice lately. If there are entities about, would be nice to hop on their wavelength, been a while.

VQC !!Om5byg3jAU ID: d04b72 Feb. 16, 2018, 2:12 a.m. No.4670   πŸ—„οΈ.is πŸ”—kun   >>4672

I will add several diagrams next.

 

These diagrams will show for RSA how the (x+n)(x+n) square MAPs onto the side of the dd + e square and remainder.

 

The colours in the MAP will be the KEY to GUIDE you through the process of how the algorithm works.

 

Your work is impressive. I spent seven years putting this together and the acceleration of a group on understanding is phenomenal.

VQC !!Om5byg3jAU ID: d04b72 Feb. 16, 2018, 2:46 a.m. No.4674   πŸ—„οΈ.is πŸ”—kun

Recap:

 

If d is even and e is odd, the remainder can be split into two even lengths and one left over.

 

Regardless of the length of e, this is possible.

 

For RSA 100 or generally for any odd e.

 

In this case for an odd c, f will be even.

VQC !!Om5byg3jAU ID: d04b72 Feb. 16, 2018, 3:06 a.m. No.4678   πŸ—„οΈ.is πŸ”—kun   >>4689 >>4699 >>4701 >>4702 >>4703 >>4707 >>4757 >>5012 >>5014

The next diagram illustrates that (f-2) is broken into pieces, 2(f-1) + 1 + 1.

 

We know that to create a square (d+n)(d+n)

 

Which is c + ((x+n)(x+n)), then we are adding a square.

 

The pieces of f can be anywhere in the square.

 

We are creating a method that USES f as a guide to find how to construct the square.

 

After all that, for those that are interested, we'll then use the virtual quantum computer patterns in the original grid to short cut all this.

VQC !!Om5byg3jAU ID: d04b72 Feb. 16, 2018, 3:09 a.m. No.4680   πŸ—„οΈ.is πŸ”—kun   >>4685 >>4719 >>5054

>>4679

Those who are good with their grid diagrams, can you show any examples of how all the pieces fit together for smaller examples with the triangles?

 

The only piece missing before is the left hand big square with the pieces added to the side.

 

Should make for a good diagram. Particularly with respect to the contribution of 2d(n-1) and how the symmetry of that looks.

VQC !!Om5byg3jAU ID: d04b72 Feb. 16, 2018, 3:27 a.m. No.4683   πŸ—„οΈ.is πŸ”—kun   >>4684 >>4692 >>4694 >>4720

>>4682

For those of Faith. Let those who would seek an understanding know where we are going.

 

The Scroll and the Lamb

5 Then I saw in the right hand of him who was seated on the throne a scroll written within and on the back, sealed with seven seals [P=NP Millenium Prize Problems]. 2 And I saw a mighty angel proclaiming with a loud voice, β€œWho is worthy to open the scroll and break its seals?” 3 And no one in heaven or on earth or under the earth was able to open the scroll or to look into it, 4 and I began to weep loudly because no one was found worthy to open the scroll or to look into it. 5 And one of the elders said to me, β€œWeep no more; behold, the Lion [British] of the tribe of Judah, the Root of David, has conquered, so that he can open the scroll and its seven seals.”

 

6 And between the throne and the four living creatures and among the elders I saw a Lamb [Spring birth] standing, as though it had been slain [Addiction], with seven [binary III = Third] horns [Day or Dawns] and with seven eyes [binary moons ooo = March], which are the seven [III = 3; Son,Father, Spirit] spirits of God sent out into all the earth. 7 And he went and took the scroll from the right hand of him who was seated on the throne [Gives example of P=NP to all who seek understanding, freely without profit]. 8 And when he had taken the scroll, the four living creatures and the twenty-four elders [United Nations Security Council] fell down before the Lamb, each holding a harp, and golden bowls full of incense, which are the prayers of the saints [Desire for Global Peace].

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 16, 2018, 5:51 a.m. No.4685   πŸ—„οΈ.is πŸ”—kun   >>4686 >>4687

>>4680

lol, I'd just gotten to some fun triangle puzzles in the EZ Bake! 64=65 is still fun.

 

Also, been working on this for 7 years, eeeeeh?

You sound like a cat person with an hour glass figure. But I'm sure the timing is "coincidence". :D

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 16, 2018, 6:28 a.m. No.4688   πŸ—„οΈ.is πŸ”—kun

>>4687

My headcanon has the VQC mantle passed off to Wikileaks crew.. for a while now ;).

 

It was some guy in Canada while your twitter was down at the same "completely coincidental" time JA's was… which is when I figured there was a pass off. Plus it'd be kind of a bitch for JA to get this stuff out of the embassy himself, this often, for this long, regardless that he got moved out of there a while ago and everything now is just for the camera.

 

Like… preeeeeetty sure he was at Kim DotCom's wedding. ANYWAY! Back to the EZ Bake! ^_^

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 16, 2018, 6:40 a.m. No.4689   πŸ—„οΈ.is πŸ”—kun   >>4697 >>4699

>>4678

>After all that, for those that are interested, we'll then use the virtual quantum computer patterns in the original grid to short cut all this.

Interested? Of course!!!

 

Also would like to learn how the factor tree plays into this.

 

It looks like from your diagram >>4678 that the answer can be calculated.

 

Thanks again for sharing this knowledge.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 16, 2018, 9:33 a.m. No.4693   πŸ—„οΈ.is πŸ”—kun   >>4712

>>4681

>Notice any patterns with f?

the f parity matches the d parity.

 

>odd e, even d, odd (x+n)

Possible test cases including parity:

 

115=5x23 - (15,48,5) = {15:48:10:9:1:115} = 115; f=6; (x+n)=57; (d+n)=58

parity: e=odd, n=even, d=even, x=odd, x+n=odd, d+n=even, f=even

 

259=7x37 - (3,114,8) = {3:114:16:15:1:259} = 259; f=30; (x+n)=129; (d+n)=130

parity: e=odd, n=even, d=even, x=odd, x+n=odd, d+n=even, f=even

 

287=7x41 - (31,128,8) = {31:128:16:15:1:287} = 287; f=2; (x+n)=143; (d+n)=144

parity: e=odd, n=even, d=even, x=odd, x+n=odd, d+n=even, f=even

 

6107=31x197 - (23,2976,39) = {23:2976:78:77:1:6107} = 6107; f=134; (x+n)=3053; (d+n)=3054

parity: e=odd, n=even, d=even, x=odd, x+n=odd, d+n=even, f=even

 

7463=17x439 - (67,3646,43) = {67:3646:86:85:1:7463} = 7463; f=106; (x+n)=3731; (d+n)=3732

parity: e=odd, n=even, d=even, x=odd, x+n=odd, d+n=even, f=even

CollegeAnon !LAbIRp9cT. ID: 7ff662 Feb. 16, 2018, 9:45 a.m. No.4694   πŸ—„οΈ.is πŸ”—kun   >>4720

>>4683

40 days 40 nights. Is this where the 40 came from?

 

666 is the "devils number" but my teacher (who is a pries) said "most people think its bad, but it is actually a good number". Also, Plato's number is 666 = 3^3+4^3+5^3. Could they have been keeping this number away from us by saying its bad? Its actually good. Maybe the fact that this is a sum of consecutive cubes could be used by us? Could the Kushner 666 actually be good?

PMA !dSvrkhSLR6 ID: fa8576 Feb. 16, 2018, 1:35 p.m. No.4701   πŸ—„οΈ.is πŸ”—kun   >>4702 >>4705

>>4678

Think I've figured out the f calculations for the (x+n)(x+n) and (d+n)(d+n) squares that match the diagram posted by VQC.

 

pics attached for c=115, 259, and 6107 for odd e, even d, odd (x+n) combinations.

 

For the small square, f needs a bit of an adjustment to make everything work properly. I've introduce a new variable f0 to handle this.

 

f0 = (f-2)/2 + 1

PMA !dSvrkhSLR6 ID: fa8576 Feb. 16, 2018, 1:36 p.m. No.4702   πŸ—„οΈ.is πŸ”—kun   >>4705

>>4701

 

Code to calculate the squares per VQC diagrams. >>4678

 

/// <summary>

/// (d+n)(d+n) = dd + f + e + 2d(n-1) + (nn-1)

/// </summary>

public static BigInteger Get_XPD_from_f( BigInteger e, BigInteger n, BigInteger d, BigInteger f ) {

 

BigInteger dsquared = d * d;

BigInteger twodnm1 = 2 * d * ( n - 1 );

BigInteger nsquaredm1 = n * n - 1;

 

return dsquared + f + e + twodnm1 + nsquaredm1;

 

}

 

/// <summary>

/// (x+n)(x+n) = 2d(n-1) + (nn-1) + (2(f0-1) + 1 + 1)

/// small square formula in terms of f where f0 = (f-2)/2 + 1

/// </summary>

public static BigInteger Get_XPN_from_f( BigInteger n, BigInteger d, BigInteger f ) {

 

// given an f, f0 is the modification required to fit it into the small square formula

BigInteger f0 = ( f - 2 ) / 2 + 1;

BigInteger twodnm1 = 2 * d * ( n - 1 );

BigInteger nsquaredm1 = n * n - 1;

BigInteger ftotal = 2 * (f0 - 1) + 1 + 1;

 

return ftotal + twodnm1 + nsquaredm1;

 

}

AA ID: 14033a Feb. 16, 2018, 1:39 p.m. No.4703   πŸ—„οΈ.is πŸ”—kun   >>4722 >>4725

>>4678

At least in this example, you could rearrange the (x+n)(x+n) square so that n^2 -1 + f = (n+1)^2. Is this always the case? I should probably stop thinking out loud like this since it usually doesn't work out, but if this is true for all odd (x+n) squares, it means n^2 scales upwards with f.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 16, 2018, 2:01 p.m. No.4704   πŸ—„οΈ.is πŸ”—kun

>>4669

Great to see you again.

 

I haven't had much time today, but from what little I could see, at least for n=2 (and n=6) it appears that 2d(n-1) fits half our triangles (4 out of 8) in (x+n)^2.

 

Not sure if it's a coincidence or a pattern. I'll take another look tomorrow.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 16, 2018, 2:10 p.m. No.4705   πŸ—„οΈ.is πŸ”—kun   >>4720

>>4701

>>4702

I've added the new XPN_from_f function into my test cases for Get_n_from_odd_triangle_base using n0.

 

Very interesting…

 

Pic attached for c=259, 6107 and 7463.

 

If you add the results from XPN_from_f (called with n0) to the result from Get_Remainder_2dnm1, you get a perfect square.

 

That's quite a coincidence.

VA !!Nf9AmQNR7I ID: 2343be Feb. 16, 2018, 6:21 p.m. No.4707   πŸ—„οΈ.is πŸ”—kun   >>4708

>>4678

>>4679

Ok, so I'm working on the new diagrams. Have lots of questions, and will post new diagrams later tonight. Don't want to clutter up the board without working through some real examples.

 

One important question worth asking at this point: in the small square (x+n)^2, why is x^2 (red) the same color as 2d(n-1) in the big square? Accident or correlation?

Anonymous ID: e99bd3 Feb. 16, 2018, 7:13 p.m. No.4708   πŸ—„οΈ.is πŸ”—kun   >>4709 >>4720

>>4707

I believe that's intentional… the matching colors show how different parts of the large square add up to form the small square, with c being left over as the difference between them.

 

I don't think the red area is equal to x^2, though.

VA !!Nf9AmQNR7I ID: 7a77e2 Feb. 16, 2018, 7:31 p.m. No.4709   πŸ—„οΈ.is πŸ”—kun   >>4710

>>4708

Hello Mysterious Anon! You also posted this:

>>4597

Right below VQC's post. VQC checking Anon status? Just seeing if we're paying attention maybe.

 

I think the diagram and >>4679 clearly show that the smaller square red portion is the area of x^2. Annotated diagram attached.

Anonymous ID: e99bd3 Feb. 16, 2018, 7:39 p.m. No.4710   πŸ—„οΈ.is πŸ”—kun   >>4713

>>4709

Nah, no games, just a regular anon.

 

I see the label but suspect it is a mistake. The sides of the overall square are x+n, so x^2 should only be the upper left quadrant, diagonally opposite from the n^2 section.

MinecraftAnon !!QXqSZ2ev8. ID: d5e987 Feb. 16, 2018, 7:48 p.m. No.4712   πŸ—„οΈ.is πŸ”—kun   >>4715

>>4693

I'd note here, the (F,E) parity matches the (D,E) parity.

 

A little contribution I have for B,

 

We know B = C/A, we also have B = A+(2(X+N)), but I've also come across B = D+X+(2N), which I don't think I've seen anywhere before.

MinecraftAnon !!QXqSZ2ev8. ID: d5e987 Feb. 16, 2018, 9:31 p.m. No.4715   πŸ—„οΈ.is πŸ”—kun   >>4716 >>4717 >>4720 >>4722 >>4995

>>4712

Nope, that bitch is super old. Here's some fucking organization because lord knows we lack it.

 

A = C / B

A = D – A

A = ((X^2 + E) / (N * 2);

 

B = C / A

B = ((X + N) * 2) + A

B = (N * 2) + D + X

 

C = A * B

C = D^2 + E

C = (D + N)^2 – (X + N)^2

 

D = floor(sqrt(C))

D = A + X

 

E = C – D^2

 

F = C – (D + 1)^2

F = E – (2 * D) + 1

 

N = ((A + B) / 2) – D

N = (X^2 + C) / (2 * A)

 

X = D – A

X = floor(sqrt((D + N)^2 – C)) – N

 

X β‰ˆ sqrt(E*(N-1))

X β‰ˆ floor(sqrt(abs(F)*(N)))

next X = sqrt((2 * N * B) – E)

previous X = X - (sqrt((2 * N * B) - E) - X)

3DAnon !!!N2ZmYzdiNjdkYTk2 ID: 4a9cf7 Feb. 16, 2018, 9:48 p.m. No.4717   πŸ—„οΈ.is πŸ”—kun

>>4715

Thanks, looking up stuff spread over too many threads is a bit of a pain. Organization thread time?

>>4428

xFactor?

>>4429

Sorry for not replying earlier, busy week! I've been using some of the JS stuff you posted. Not my favorite language but I read and write it just fine. Only reason for using it in this case is if we want to distribute something with live demos to non-programmers, and glad to see the "single html page" thing wasn't rejected by you all.

Would love to cooperate on making it. And we're gonna need text explanations too in addition to the cool graphics for anyone else wanting in.

 

Have a bit of catching up to do on the latest progress so nothing to add at the moment

3DAnon !!!N2ZmYzdiNjdkYTk2 ID: 4a9cf7 Feb. 16, 2018, 10:10 p.m. No.4718   πŸ—„οΈ.is πŸ”—kun   >>4720 >>4722 >>4728

>>4429

>I'm just concerned about linking it back to my personal identity somehow.

>>4430

>How would your code identify you?

Consider reading these before posting. Your code style can be fingerprinted and statistically matched against previous stuff. They are just languages after all.

 

Identifying Authorship by Byte-Level N-Grams:

The Source Code Author Profile (SCAP) Method

https://utica.edu/academic/institutes/ecii/publications/articles/B41158D1-C829-0387-009D214D2170C321.pdf

Effective Identification of

Source Code Authors Using

Byte-Level Information

http://www.icsd.aegean.gr/lecturers/Stamatatos/papers/ICSE06.pdf

Who Wrote This Code?

Identifying the Authors of Program Binaries

http://ftp.cs.wisc.edu/paradyn/papers/Rosenblum11Authorship.pdf

VA !!Nf9AmQNR7I ID: 819f3a Feb. 16, 2018, 11:02 p.m. No.4719   πŸ—„οΈ.is πŸ”—kun   >>4721

>>4668

>>4682

>>4680

>Those who are good with their grid diagrams, can you show any examples of how all the pieces fit together for smaller examples with the triangles?

>The only piece missing before is the left hand big square with the pieces added to the side.

>Should make for a good diagram. Particularly with respect to the contribution of 2d(n-1) and how the symmetry of that looks.

 

Alright diagram Anons, let's work.

MM !!DYPIXMDdPo ID: dfa2e4 Feb. 16, 2018, 11:05 p.m. No.4720   πŸ—„οΈ.is πŸ”—kun   >>4723

Checkin' in, my shitpost for the day, little Hotrod Frankie to get comfy, couple links w/ 3's. Spawned from Root O David -ROD of God search for a fresh meme. 3 blind mice, 3 shades of black. The Frankenstein mask as an anon mask, heh. Interdasting. Better than the Thor kinetic weaponry stuff for tonight methinks…

https:/ /www.youtube.com/watch?v=2NX4inAFTA4

https:/ /www.youtube.com/watch?v=_RjQybNcbDM

https:/ /www.youtube.com/watch?v=CfHf7ZqthLY

https:/ /www.youtube.com/watch?v=5BlvM_T1tDc

Check the lyrics on the last one "Darling Angel" - worth the dig friends, and I'm a black hole without you, must be a godsent savior, out of the void I ran to you…

While those burn in the background…

 

>>4708 Hey Algebra Faggot, nice to have you pop back in! Thought you were VQC as well with the quick quip formula post VA mentioned. Welcome to the greatest LARP on planet earth, it gets better every day. Get ready for acceleration AF!!!

 

>>4718 OPSEC. Good to read you 3DA.

 

>>4715 Nice recap bitch. Frickin' read my mind earlier, formula index cards in my pocket, shuffling… Danks MA!

 

>>4705 >>4693 Rockin' it PMA. Was faggin on the remote mind-control device earlier and itching to get on and sketch your shit. Long day, diagrams will wait 4 tonight but looking forward to laying it out.

 

>>4694 Spot on CA. What's up is down, black is white, as per Q. The serpent? Light and health giving plasma hidden from us (check the Dendera "light bulb") - KEK. Funny, was just digging the other night for some John Lilly (Programming and Metaprogramming in the Human Biocomputer) and R A Wilson (Prometheus Rising) for the generals, and got sidetracked with Wilson's 1977 "Cosmic Trigger, Final Secret of the Illuminati" (http:/ /www.lisamharrison.com/pdf/Robert%20Anton%20Wilson%20-%20Cosmic%20Trigger.pdf), where in the Afterwards (right after "The End") by Saul-Paul Sirag (Saul, Paul !) writes:

Dirac had complained that when one uses fewer than 10 tensors one destroys space-time symmetry; but that is just what I want. The reason is that, since Dirac wrote in 1963, it has been discovered that mass splittings can come about by breaking an underlying gauge symmetry. This is how the weak force is gotten out of the electromagnetic force by Steven Weinberg and various other physicists much in vogue today. I am now preparing a paper in which I get the strong force out of the gauge symmetry of general relativity. (Actually, this has already been done by Abdus Salam and Jack Sarfatti. I'm just giving them more ammunition.)* But things begin to look positively contrived when one notices that Eddington's 1, 10 and 136 are members of a well- known mathematical series that goes 1, 10, 45, 136, 325 … etc.

The next number in that series is 666.

Berkeley, California

Summer, 1977

ref: https:/ /oeis.org/A037270

A037270 a(n) = n^2*(n^2+1)/2.

 

>>4683 V, what can we say, but thanks. Stick close with us faggots, we need to get to ECC soon!

 

Topol, knew you couldn't stay away for a second, yo! VA, hope you had a great day.

 

…and I'm a black hole without you…

AA ID: 14033a Feb. 16, 2018, 11:09 p.m. No.4722   πŸ—„οΈ.is πŸ”—kun   >>4727

>>4703

Bump

 

>>4718

Firstly, it's ironic that you would post information about clown-tier identification methods in the form of pdfs. Secondly, if we can't share any code at all without possibly being identified, even if it doesn't have someone's name (let alone their IP or whatever in Teach's case) on it, we can't really collaborate. We're all posting combinations of words that could be linked back to us in the form of posts, too.

 

>>4715

You might want to fix the second equation.

MM !!DYPIXMDdPo ID: dfa2e4 Feb. 16, 2018, 11:29 p.m. No.4725   πŸ—„οΈ.is πŸ”—kun

>>4703 will dig on that soon AA.

 

>>4723 Ha VA. I was grumpy due to frustration and projected on you all, and that 3-6-9 mistake had me spinning, not that it didn't lead to some fun. And hey, you can call me Grumpy, just don't call me Dopey!!!

VA !!Nf9AmQNR7I ID: 6133d1 Feb. 16, 2018, 11:29 p.m. No.4726   πŸ—„οΈ.is πŸ”—kun

>>4721

>These diagrams are fantastic

Blessings To You, MM.

When a lot of us were tired from working so long, you got us fired up again. Get some rest, then come back and bust ass again.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 16, 2018, 11:54 p.m. No.4728   πŸ—„οΈ.is πŸ”—kun

>>4718

Hence why I'm using python, a language I never code in ;)

 

For those who haven't published any code ever, it should be safe. If you have you should consider using a language you don't know (or know ish). Example, if you're a java developer use C# like Chris is. They are similar enough that you won't get your head stuck too much. And again, if you are a C# developer use java etc.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 16, 2018, 11:58 p.m. No.4729   πŸ—„οΈ.is πŸ”—kun   >>4730 >>4731 >>4733 >>4740 >>4741 >>4745 >>4748

I started trying to make some patterns that matched the group of odd e, even n and odd (x + n).

 

These are all (3, 2, 1), (3, 2, 2), (3, 2, 3) and (3, 2, 4).

 

I have no idea if the configuration is correct though. Just playing around with (x+n)^2.

 

That 2d(n-1) fits half the triangles could simply be a result of n-1 = 1 and 3 being such a low number.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 17, 2018, 12:11 a.m. No.4732   πŸ—„οΈ.is πŸ”—kun

>>4731

Yeah the swastika is unintended. I know Conway and Guy used another way of fitting the triangles in the odd square, which I'm guessing is because they wanted to avoid swastikas.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 17, 2018, 1:07 a.m. No.4739   πŸ—„οΈ.is πŸ”—kun

Well I still haven't found a counter example of 2*d(n-1) not fitting at least 4 triangles.

 

The a=7,b=37 record 2d5 will fit more than 4 triangles, though.

CollegeAnon !LAbIRp9cT. ID: d31ebd Feb. 17, 2018, 12:12 p.m. No.4742   πŸ—„οΈ.is πŸ”—kun

>>4741

I've noticed something.

Let S = (2d-1) + d^2 + e + 2d(n-1) + n^2-1

and let D = (D+N)^2.

 

The difference between S and D is the same for the correct record as it is for the initial record

MinecraftAnon !!QXqSZ2ev8. ID: d5e987 Feb. 17, 2018, 12:53 p.m. No.4746   πŸ—„οΈ.is πŸ”—kun

>>4744

 

That's because (2d-1) + d^2 + e + 2d(n-1) + n^2-1 is an alternate form of (d + n)^2 + e - 2. So this would apply to any 2 cells shared in an E column.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 17, 2018, 3:12 p.m. No.4747   πŸ—„οΈ.is πŸ”—kun

I noticed something else. I suspect that (n - 1) appears as a factor in f. When f has multiple factors I can't see an obvious pattern into which of the factors represent (n - 1).

 

For n - 1 when n = 2, it appears that f is a prime. I've only looked at a few examples so far though. So maybe you can add this in your test cases PMA?

 

When we are working with a prime (c = prime) (n-1) doesn't appear to be a factor of f, though. Not sure if this is a proper pattern.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 17, 2018, 4:59 p.m. No.4748   πŸ—„οΈ.is πŸ”—kun   >>4749 >>4753 >>4758

>>4729

Great pictures Isee.

 

Pic attached is for c=39, showing (x+n)(x+n) for the prime solution in the middle, and "growing" the small square into the starting c position.

 

Just a thought to share while looking for differences.

VA !!Nf9AmQNR7I ID: 2343be Feb. 18, 2018, 10:53 a.m. No.4757   πŸ—„οΈ.is πŸ”—kun

>>4219

>>4220

>>4678

Hello lads! Working over here, re-reading all crumbs. Working on the algebra side to understand the (x+n)^2 formulas. Have fam stuff all day today, but here thinking and and checking in. Here's what I have so far, and it goes with the small red square in >4678.

 

(x+n)^2 = n^2 -1 + 2d(n-1) + 2d + 1 - e

 

Also, you can sub in 2xn + xx for 2d(n-1) + f - 1. Trying to understand the (x+n) square and how it relates to all the formulas. Thoughts?

 

TBH, the diagrams produced are already excellent, so I felt like I was doing pointless work trying to build another. Great job MM Isee, and PMA on all the new ones!

MinecraftAnon !!QXqSZ2ev8. ID: d5e987 Feb. 18, 2018, 12:42 p.m. No.4761   πŸ—„οΈ.is πŸ”—kun   >>4762

Does anyone have any interest in a VQCGUI_v3? One that is a bit more… accurate? Just don't want to put the time into it if no one will use it.

VA !!Nf9AmQNR7I ID: 427f3f Feb. 18, 2018, 4:52 p.m. No.4762   πŸ—„οΈ.is πŸ”—kun

>>4761

Hey MA! I always love checking out your visuals. Can you figure out a way to show the odd (x+n) square growth with n0 triangles using a single cell? Like squares stacked on top of each other and growing upward?

VA !!Nf9AmQNR7I ID: 427f3f Feb. 18, 2018, 5:01 p.m. No.4763   πŸ—„οΈ.is πŸ”—kun

>>4681

Hey lads! This is what I’m working on. I’ll have diagrams ready soon. IRL stuff has been super busy. Working on x+n square understanding. Anyone else have insight on this?

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 18, 2018, 5:17 p.m. No.4764   πŸ—„οΈ.is πŸ”—kun   >>4765 >>4767 >>4768 >>4781

>>4322

>>4758

I don't think the "capstone" starting position in VQC's image is properly labeled as (n-1).

 

In the c=115 picture, the large red square is 31x31. The bottom of the capstone would be (31/2 = 15.5) and not 48-1. I believe this 31x31 square divided into 8, are the "sub-triangles" that VQC was referencing.

 

From these diagrams, I think there is a relatively straight forward way to calculate our solution n:

 

solution n = (x+n)(x+n) - "dark purple bases of triangle" - 2d(n-1) - f.

 

What's left behind after this calculation is the light colored purple n in the middle of the inner solution square.

 

Because we don't know the size of the 31x31 square in the middle, the n0 formulas and the (f-2) mod 8 calculations are used to iterate towards the base of the capstone. Testing along the way if we have solved the problem.

MinecraftAnon !!QXqSZ2ev8. ID: d5e987 Feb. 18, 2018, 6:24 p.m. No.4765   πŸ—„οΈ.is πŸ”—kun   >>4766

>>4764

X+N=

1059731506988603553937431268657920267324681542931

(X+N)^2=

1123030866904336703079469523070416463095627144114722409357226718814587956951689069474054796070761

(N-1)*8=

115100708248095715653845476953416823881690247480

((X+N)-1/2)*8=

4238926027954414215749725074631681069298726171720

((N-1)8) + (((X+N)-1/2)8) + 1 =

4354026736202509931403570551585097893180416419201

Anonymous ID: e99bd3 Feb. 18, 2018, 8:10 p.m. No.4775   πŸ—„οΈ.is πŸ”—kun   >>4776

>>4774

Well, for example take 39 = 3 * 13 = 8^2 - 5^2. So d=6, e=3, f=10, n=2, x=3.

 

The formula says:

(x+n)^2 = n^2 + 2d(n-1) + f - 1

25 = 4 + 12 * 1 + 10 - 1

25 = 25

 

Could you try that one in your code, or any other specific example we can verify by hand?

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 18, 2018, 9:35 p.m. No.4781   πŸ—„οΈ.is πŸ”—kun   >>4783 >>4799 >>4800

>>4758

>>4764

Following is the math breakdown for what I was referencing above:

 

5x23=115

 

c = (15,48,5) = {15:48:10:9:1:115} = 115; f=6; (x+n)=57

p = (15,4,3) = {15:4:10:5:5:23} = 115; f=6; (x+n)=9

 

For the c record, the revelant calculations are:

 

(x+n)(x+n) = 57x57 = 3249

(nn-1) = 48x48 - 1 = 2303

2d(n-1) = 2x10x(48-1) = 940

f = 6

 

Verify:

3249 = 2303 + 940 + 6

 

For the p record, the relevant calculations are:

 

(x+n)(x+n) = 9x9 = 81

(nn-1) = 4x4 - 1 = 15

2d(n-1) = 2x10x(4-1) = 60

f = 6

 

Verify:

81 = 15 + 60 + 6

 

This works regardless of how f is divided in the squares. The important piece is that f is equal in both records.

 

Assuming we can calculate the red square (i.e. the n-1 capstone in VQC's post)

 

31x31 red square = 961

 

The nn-1 portion in the dark purple area is the triangle base and can be calculated as:

 

3249 - 961 = 2288

 

Once we have the triangle base, there are 2 ways to solve this from the original c record:

 

solution 1:

(x+n)(x+n) - (nn-1 triangle base) - 2d(n-1) - f = (nn-1) from solution small square

3249 - 2288 - 940 - 6 = 15

 

solution 2:

(nn-1) - (nn-1 triangle base) = (nn-1) from solution small square

2303 - 2288 = 15

 

To get our solution n:

 

nn-1 = 15

nn = 16

n = 4

 

This theory hasn't been tested on many records as I'm still not sure how to programmatically determine the dark purple triangle base. I do believe we have all the code pieces required, however.

 

Would appreciate someone else double checking this.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 18, 2018, 10:18 p.m. No.4784   πŸ—„οΈ.is πŸ”—kun   >>4785

>>4783

That’s a different f. That 2d+1 was a transform into the negative e space. It created a mirrored record in the negative half of the grid.

 

The f we’re talking about here is the distance to the next squarefor each record.

ID: 2bf75e Feb. 19, 2018, 1:26 a.m. No.4791   πŸ—„οΈ.is πŸ”—kun   >>4799

I'm trying to get back in the game here. The picture shows a test for rsa100 w/ Chris' methods working as advertised. I'll create a new picture of his hints in the morning.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 19, 2018, 2:30 a.m. No.4795   πŸ—„οΈ.is πŸ”—kun   >>5055

I'm just… throoooooowing this out there.

3301 is a prime number.

 

Virtual Quantum Cicadas running an A.R.G. meant to end slavery sounds pretty good to me.

 

Might not be useful or relevant.

 

Just cheerin' ya on.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 19, 2018, 7:47 a.m. No.4796   πŸ—„οΈ.is πŸ”—kun   >>4797

Okay, I feel a bit like an idiot.

 

I was thinking last night about the logic behind f -2 and the 2d(n - 1) part and I realised that we're just adding to perfect squares here.

 

So the equation we're dealing with:

n^2 + 2d(n-1) + f - 2 = (x+n)^2

 

This is the same as (x+n)^2 - n^2 = 2d(n-1) + f - 2 which is also the same as x(2n + x).

 

Essentially, this equation is just the difference of two perfect squares (n^2 and (x+n)^2) (See more here: https ://en.wikipedia.org/wiki/Difference_of_two_squares

under the "Difference of two perfect squares" . I'm not sure if we were aware of this, but at least I didn't realise this until I started to think more about it.

IseePatterns !kIkD/SqZ4s ID: 4baf8d Feb. 19, 2018, 8:10 a.m. No.4797   πŸ—„οΈ.is πŸ”—kun

>>4796

 

And of course then, what we have is really:

 

n^2 + 2d(n-1) + f - 2 = (x+n)^2 - 1

 

Replacing f:

 

n^2 + 2d(n - 1) + 2*d + 1 - e - 2

n^2 + 2dn + 1 - e - 2 = (x+n)^2 - 1

2dn + 1 - e - 2 = (x+n)^2 - n^2 - 1

 

Not sure how this would get us closer to finding a solution, though.

Anonymous ID: 4958e4 Feb. 19, 2018, 8:28 a.m. No.4798   πŸ—„οΈ.is πŸ”—kun   >>4799 >>4801

>>4790

(x+n)(x+n) = nn + 2dn - 2d + 2d + 1 - e

xx + 2xn + nn = nn + 2dn - 2d + 2d + 1 - e

xx + 2xn = 2dn + 1 - e

xx + e = 2dn - 2xn + 1

xx + e = 2(d-x)n + 1

xx + e = 2an + 1

 

Isn't it true that

xx+e = 2an, so maybe this is false and you added an extra 1

MM !!DYPIXMDdPo ID: 95ec2c Feb. 19, 2018, 8:51 a.m. No.4799   πŸ—„οΈ.is πŸ”—kun   >>4801

>>4798

>(x+n)(x+n) = nn + 2dn - 2d + 2d + 1 - e

anon, looks as if you missed a (-1) in your transcription of the image, which was actually:

(x+n)(x+n) = nn -1 + 2dn - 2d + 2d + 1 - e

reducing to

xx+e = 2an

 

>>4791 great baker.

>>4788 thanks for reaching out and reporting back VA.

 

>>4773 nice AF. Appreciate your input.

>>4320

>Pick up any mistakes. It's important.

 

>>4781 thanks PMA. Plan to review your diagrams and get going on these. Should have more time this week.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 19, 2018, 12:34 p.m. No.4800   πŸ—„οΈ.is πŸ”—kun   >>4803 >>4805 >>4808

>>4781

Guys. The math checks out for odd_e, even_d, odd_x_plus_n.

 

Attached are test cases for c=115, 259, 6107, and 7463. I have tested all known matching Rsa solutions against these formulas, and they work as well.

 

The top part of the test cases should be self explanatory.

 

The bottom portion is the remaining work to be done.

 

Currently, I am iterating from 1 to x+n, and executing the Get_n_from_odd_triangle_base, Get_Remainder_2dnm1, and Get_XPN_from_f methods posted earlier.

 

If you add Remainder_2dnm1 to Get_XPN_from_f, you always get square. See the total, sqrt(total), and diff columns.

 

One of those squares represents the capstone square that will solve this problem.

 

While iterating those squares, I have found the following:

 

1) The n from the starting c and prime solution records always appear in the n0 column.

2) Once the remainder 2d(n-1) result is greater than or equal to c, it will always return c.

3) The "capstone" square that we are looking for may or may not appear.

3a) in c=115, it shows up directly via iteration.

3b) in the other test cases, we get records that are +/- 1 away from the capstone. See c=259 for capstone +1 match, and capstone -1 match.

3c) So to find the capstone square, we need to check +1 and -1 of the sqrt(total).

 

I believe this is why VQC showed the f div 40 + 2 through f div 40 - 2 triangle side >>4344.

 

The 5 factor in the denominator may be arbitrary, but when searching for a solution, we need to check +2 through -2 records to find the capstone match.

 

>>4343

>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)).

 

This is the remaining piece that will replace iterating with larger jumps.

VA !!Nf9AmQNR7I ID: d32925 Feb. 19, 2018, 4:36 p.m. No.4805   πŸ—„οΈ.is πŸ”—kun

>>4800

>>4803

Seconded! Thanks PMA for your great work, it’s always clear, logical, and helpful. You da man!

 

>>4802

Howdy Topol! Yes, VQC did say that. After we finish up this method, he’s going to show us how to shortcut with the grid. Although I think PMA may almost have that figured out too! :)

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 19, 2018, 10:07 p.m. No.4806   πŸ—„οΈ.is πŸ”—kun   >>4812

Hello Lads! I've been working through the breakdown of 1 Tu in the (x+n) square using the n0 calc. Pics attached. Hopefully some good food for thought! Feedback appreciated.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 20, 2018, 10:26 a.m. No.4808   πŸ—„οΈ.is πŸ”—kun   >>4809 >>4811

>>4800

I tested this capstone calculation for ALL values regardless of parity, including all solved Rsa numbers.

 

Our trusty c=145 example with even (x+n) and odd n sample attached.

 

This works. Someone please verify.

MM !!DYPIXMDdPo ID: 9e5e80 Feb. 20, 2018, 4:34 p.m. No.4812   πŸ—„οΈ.is πŸ”—kun   >>4813

>>4811

Planning to dive in tonight, is that in line with your latest diagram with the split triangles >>4806 ?

Thought that was a clever way to reconcile the odd cell and single-unit difference between base and height for orig drawing!

VA !!Nf9AmQNR7I ID: a34e47 Feb. 20, 2018, 5:42 p.m. No.4813   πŸ—„οΈ.is πŸ”—kun

>>4812

Thanks MM! It will hopefully help us iterate up quickly to (x+n) using n0 triangle geometry. Thinking out loud in diagram form. Next version I’ll work in 2d multiples.

 

>>4343

VQC’s post about filling the odd x+n square using f-2 doubling.

>n0^2 + multiples of 2d will fill the triangle.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 20, 2018, 9:49 p.m. No.4814   πŸ—„οΈ.is πŸ”—kun   >>4815 >>4830

Just to confirm previous posts, the following solutions work for all numbers:

 

solution 1:

(x+n)(x+n) - (nn-1 triangle base) - 2d(n-1) - f = (nn-1) from solution small square

 

solution 2:

(nn-1) - (nn-1 triangle base) = (nn-1) from solution small square

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 20, 2018, 9:52 p.m. No.4815   πŸ—„οΈ.is πŸ”—kun   >>4820

>>4814

Finding the base of each triangle capstone needed to calculate the (nn-1 triangle base) at this point seems to require iteration - until VQC drops hints about more effectively using the grid.

 

The (nn-1 triangle base) is the dark purple area in the image at >>4758.

 

The starting position for an iterative search can be set as:

 

(2d(n-1) + f)/8

 

This represents a triangle within the red square including the blue f portion. The only piece missing is the n we are looking for.

 

Attached test cases for c=115, 259, 6107, and 7463, are an attempt to move closer to an understanding of the iterations required. And to verify that the starting triangle base is less than or equal to an estimated capstone triangle base.

 

The "triangle base" entries are calculated as inverse triangles from the triangle area. I believe baker posted this formula previously:

 

(Sqrt( 8 * area + 1 ) - 1) / 2

 

From the (2d(n-1) + f)/8 triangle starting position, we can iterate (f-2) div 8 or (f-2) div 40 steps, checking for +/- 5 proximate values each way for any possible solution. Although I'm not sure how many proximate values need to be checked. Could be +/- 10, 50, 100?

 

The "estimated iterations" value is calculated as:

(capstone_triangle_base - starting_triangle_base) / BigInteger.Max( (f-2) / 8, 1 ) + 1;

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 21, 2018, 9:18 p.m. No.4817   πŸ—„οΈ.is πŸ”—kun   >>4818 >>4819

Hello Lads! Here working. Between VQC's crumbs and the ass kicking job PMA has been doing, I'm still working to understand, and I'm stuck on a few of the ideas . Feel like I have >90% down, but I have yet to understand a few key pieces to unlock the new ideas. It's been kinda quiet around here, which is fine. Just wondered if anyone else was also working to understand all the new developments? In the meanwhile I'm re-reading new VQC crumbs and studying PMA's output looking for answers. Anyone else have a question or two they're working on or stuck on?

 

>>4684

VA !!Nf9AmQNR7I ID: 819f3a Feb. 21, 2018, 10:08 p.m. No.4819   πŸ—„οΈ.is πŸ”—kun

>>4817

Chris' Twitter is gone for the time being. Poof! Not sure if you guys have been following the #TwitterLockOut, but it's a mass censorship effort by @jack. Q posted about it yesterday. Pics related.

 

>>4818

Hey PMA! Thanks for all your new work. So we are still working to find a way to use triangle geometry to iterate and find the multiples that give us (x+n).

 

Starting from c, we get d,e,f. Then we break down f to find n0. I'm stuck at this point.

Do we use n0 = (f-2)/8 and (f-2) mod 8? Do we have a use for the remainders yet?

Do we use n0 = (f-2)/40 and (f-2) mod 40? Remainder purpose/use?

 

After choosing one of these to divide f, we then use that choice to make the triangle bigger using multiples of n0 and 2d. We could also visualize this in square (x+n) form too. Combo of n0 (factor of n) + 2d (factor of x? d-x=a.)

 

Then we have this part >>4636 I'm still working on. Have you guys figured more of this part out yet?

 

Also, we can solve for n in (1,c) and complete that element right off the bat. Does that help us in any way as far as limiting our search or iteration parameters, especially with PMA's new findings? GAH! Lots of questions, any help appreciated.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 21, 2018, 10:32 p.m. No.4820   πŸ—„οΈ.is πŸ”—kun   >>4828 >>4829 >>4830

Follow up to the starting position from >>4815.

 

I confused the "triangle_base" with x+n inputs.

 

I believe the correct parameters are as follows:

 

BigInteger n0 = Get_n_from_odd_triangle_base( (x+n)/2, c, d );

BigInteger r2dnm1 = Get_Remainder_2dnm1( (x+n-1)/2, d, n0, f );

BigInteger XPN_from_f = Get_XPN_from_f( n0, d, f )

 

Attached are examples for c=115, 6107, and 14904371.

 

I am estimated an (x+n) starting position to use in the above formulas, and iterating by 2 for only odd values of (x+n).

 

Starting x+n = 2 * (GetInverseTriangle( 2d(n-1)+f ) / 8 ) + 1;

 

In some cases, this puts us exactly at the correct capstone record.

 

The "cap diff" column represents the difference in the target capstone area and the iterated total value. I'm iterating until that value becomes negative.

 

This starting formula allows pretty big jumps and relatively few iterations, even without adding in the (f-2) div 40 chunks - which will be required to handle RSA length numbers.

 

The current problem is accounting for the difference between the "target capstone area" and the (x+n) with the last positive "cap diff".

 

How do we know we've hit the correct capstone?

Anonymous ID: 91be12 Feb. 22, 2018, 5:40 a.m. No.4821   πŸ—„οΈ.is πŸ”—kun   >>4822 >>4828 >>4839

Aside from our regular stuff (which I seem to be stuck on, the whole 40 thing has me lost), might we (some of us codefags) need to create new social media platforms? It seems like Twitter and Youtube especially are cucked to the max and will be going down in the near future. Could this be an opportunity for a patriotic new website? We could even start it with a clean modern progressive look to appeal to normies jumping ship then change it later. Could this VQC (the computer, not the guy) help us build this website in less than a few seconds?

 

Idk, I just think when those companies fall, and I believe they will, there will be a vacuum where people cannot get info

Anonymous ID: 6abd01 Feb. 22, 2018, 8:01 a.m. No.4822   πŸ—„οΈ.is πŸ”—kun   >>4823 >>4824 >>4828 >>4932

>>4821

Um… When this thing is created there will be no Internet. This thing will break it. Anyone will be able to use simple methods to break any digital security. If you make this website someone else will simply be able to log in with your password and delete/change it. This thing is going to turn the internet into the Dewey decimal system… Obsolete. Though the Dewey decimal system may come back now that I think about it.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 22, 2018, 9:14 a.m. No.4823   πŸ—„οΈ.is πŸ”—kun   >>4828 >>4839

>>4822 Not necessarily. If you were handed 'the button', does it mean the world would implode from newks? You would probably realize a new level of personal responsibility. We are moving into a new age, where through technology the power of each individual is amplified, this is just another example. With great power, comes great responsibility.

 

Interesting exchange last night on the general, thought I'd link here. Grabbed the pyramid images and posting here as well. Not as friendly as VA's rainbow pyramid!

>>>/qresearch/456115

>New N.Korean Missile Poses Fresh Threat

>>>/qresearch/456172

>Also today in NK.

>Notice the seats are green screen green.

>>>/qresearch/456301

>old VQC question: What happens when you use facial recognition on a crowd generated with CGI?

>>>/qresearch/456329

>VQC also said he would teach us to build a quantum computer, then didn't. He's a crank.

>>>/qresearch/456366

>Cranky, not likely a crank, imo. We're getting close to phase one if you haven't checked in for a while.

>>>/qresearch/456378

>Your opinion is worthless and you need to learn how to identify liars. End of conversation.

>>456427

>No worries anon. I've learned a good deal of mathematics and crypto regardless. Carry on.

 

>>4818

Yes quiet indeed. I've gone through and validated your calcs, but haven't made progress on the iteration steps. At first your diagrams with the large purple area didn't make sense, but I see completely from the 1,c record versus the prime record.

I think your though about the 5-wide with +/2 layers to capture those with a difference either direction makes excellent sense.

 

Frankly, of the free time available in life, have spent too much time following/digging (Parkland recently), am going to step back a bit, and also shift more attention to here. Not really learning anything about the world there, just more of an energy suck.

Anonymous ID: 4baf8d Feb. 22, 2018, 10:11 a.m. No.4824   πŸ—„οΈ.is πŸ”—kun   >>4825 >>4826 >>4827 >>4828 >>4860

>>4822

 

No, it won't break the internet.

 

This thing will first turn the next few weeks into a pain in the ass for ops people and browser developers. Browsers will be quickly upgraded to support elliptic curves certificates (if they don't already). Root certificate guys will push out new root certificates using elliptic curves, we'll see a few weeks of stress as this gets deployed. Ops will start demanding elliptic curves on servers (SSH etc) and banks will push heavily for users to update their browsers.

 

But in short, breaking RSA won't break the internet. It will just turn it into a pre-https system. SSH will be compromised and every public key that's out there (using RSA) will be broken quickly so expect shit to go down.

 

However, VQC has also said that this thing will break elliptic curves. I doubt we'll be able to do that right away, so it will give everyone a few weeks of peace and quiet until we manage to do that.

 

When that happens, it will be like an HTTPS reset button. No more HTTPS, no more secure sharing of public keys until they find / change algorithms that can be a drop-in replacement of RSA / Elliptic curves.

 

But the internet will still chug along as always. It's just that more systems will be open / vulnerable.

 

Most modern consoles already use elliptic curves for application signature, but once that breaks expect to see an increase in homebrew apps for those :-D.

 

If this thing is true, though. It's going to take a nice, juicy piece of the financial market with it. Would you use your bank if everyone with a wifi-sniffing tool could see your username and password? What about Amazon? It's going to scare people a lot.

 

Bitcoin is safe though. If you use it properly it's safe against quantum attacks.

 

Also, when / if this thing turns out to work / be real expect this board to be flooded with normies. It's going to go down in history.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 22, 2018, 10:50 a.m. No.4825   πŸ—„οΈ.is πŸ”—kun   >>4828

>>4824

anon, I See everyone point you made and agree on all. Well said.

 

If anyone want to read about some of the network tools, opsec, and other bits, this is well worth your time:

https:/ /pastebin.com/mXQxg8JG

Archive Offline.

Apart from the name, email, phone roster for NASA, one example we all would love:

[+] OIB - Operation Ice Bridge [+]

IceBridge, a six-year NASA mission, is the largest airborne survey of Earth's polar ice ever flown. It will yield an unprecedented three-dimensional view of Arctic and Antarctic ice sheets, ice shelves and sea ice. These flights will provide a yearly, multi-instrument look at the behavior of the rapidly changing features of the Greenland and Antarctic ice. Data collected during IceBridge will help scientists bridge the gap in polar observations between NASA's Ice, Cloud and Land Elevation Satellite (ICESat) – in orbit since 2003 – and ICESat-2, planned for late 2015. ICESat stopped collecting science data in 2009, making IceBridge critical for ensuring a continuous series of observations. IceBridge will use airborne instruments to map Arctic and Antarctic areas once a year. The first IceBridge flights were conducted in March/May 2009 over Greenland and in October/November 2009 over Antarctica. Other smaller airborne surveys around the world are also part of the IceBridge campaign.

Anonymous ID: 30c4ff Feb. 22, 2018, 11:10 a.m. No.4827   πŸ—„οΈ.is πŸ”—kun   >>4828 >>4839

>>4824

ops people here, reissuing elliptic curve certificates and disabling ssh rsa logins will take a few hours unless the certificate providers servers crash from everyone doing it (haha). but anyone not prepared will shit bricks along with most of the internet. there are drop-in replacements but none of them are widely iplemented

the internet doesn't require encryption to work, what i want to know is how to build a new one using software quantum entanglement without any wires, isps or wiretapping!

how is bitcoin / [insert any cryptocurrency here] safe? cracking an ecc key will let you transfer all money from that address wherever you want.

>>4826

most people here have trips, but this might be a good opportunity to establish a bunker in any case

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 22, 2018, 11:29 a.m. No.4828   πŸ—„οΈ.is πŸ”—kun   >>4830

>>4820

Hey PMA! Let's focus on this today: Let's play with f values and look for a way to fill up 1Tu or related triangle geometry patterns. We can start with the smaller p(x+n) square and triangles. For example, 1Tu for (x+n) = 15 has 100 triangles of base n0= 7/10 The area of each triangle is 0.7 x 0.8 = 0.28. Total area for the 100 triangles is 0.28 x 100, giving us the correct area for 1Tu. Each triangle level you go down another 0.8 units. Number of triangles per level goes 1,3,5,7, etc. Base measurements go 1,2,3,4, etc. Lots of cool patterns to explore. Could be easily used to iterate up to (x+n) = 15. Interesting pattern I noticed in the very bottom level: 19 triangles - 10 base chunks = 9 = x. Could just be coincendece, but interesting to explore.

 

Also, n0 determines where the dimensions of the 1Tu will land on a whole number, narrowing our search for correct dimensions in the iteration process. Am I using n0 in the correct way? Can this help us find the capstone or dimensions of (x+n)? Thoughts Anons? Pics relevant, check

 

>>4821

>>4822

>>4823

>>4824

>>4825

>>4826

>>4827

>Also, when / if this thing turns out to work / be real expect this board to be flooded with normies. It's going to go down in history.

It's going to be a fun shitstorm!

>If you were handed 'the button', does it mean the world would implode from newks? You would probably realize a new level of personal responsibility. We are moving into a new age, where through technology the power of each individual is amplified, this is just another example. With great power, comes great responsibility.

Well said MM.

>If this thing is true, though. It's going to take a nice, juicy piece of the financial market with it. Would you use your bank if everyone with a wifi-sniffing tool could see your username and password? What about Amazon? It's going to scare people a lot.

People are going to trip out.

>most people here have trips, but this might be a good opportunity to establish a bunker in any case

Probably a good idea, Anon.

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 22, 2018, 11:39 a.m. No.4829   πŸ—„οΈ.is πŸ”—kun   >>4830

>>4820

PMA, could you please explain this in more detail? I studied your output, and wondered the same thing:

>How do we know we've hit the correct capstone?

Also, how do we separate x and n? Is n0 for finding n, and 2d for finding x?

PMA !dSvrkhSLR6 ID: fa8576 Feb. 22, 2018, 12:03 p.m. No.4830   πŸ—„οΈ.is πŸ”—kun   >>4831 >>4832

>>4828

Will look into the triangles further. Good stuff here.

 

>>4829

Sure thing.

 

The growth of the small square and the elegance of the solutions >>4814, led me down a path of attempting to use the Get_Remainder_2dnm1 and other formulas to search for the red square in >>4758. Which is calculated as 2d(n-1) + f + (nn-1 from the prime solution).

 

To find that square, I attempted a really big jump using the original 2d(n-1) + f values, and then planned on iterating closer to the red square.

 

The examples in >>4820 show that we can get really close to that red square even without the (f-2) div 40 jumps. But that leads to a problem of knowing how to determine that a solution has been found.

 

As I haven't been able to figure that out, I am now looking at using the known result of Get_Remainder_2dnm1 == 0 to identify a correct solution.

 

Attached pic of larger value shows work in progress. Using the formulas and iterating via increments of x+n to arrive at a solution.

 

This value took 989 iterations. See the line 989 (p).

 

This doesn't include any (f-2) div 40 jumps (as I don't fully understand how to work with that yet) - but it does find a solution for any odd (x+n).

 

If someone can figure out the larger jumps…

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 22, 2018, 12:57 p.m. No.4831   πŸ—„οΈ.is πŸ”—kun   >>4834 >>4835

>>4830

PMA, quick question about dropping decimals. I often check and see a whole number where it's actually got a decimal trailer.

Example from image: 131460/40=3286, where it's actually 3286.5. I understand the (f-2)%40=20 quickly shows the remainder, just wondering your thought process.

Are you using a Math.round() or some integer round down function?

Given so much of this solution hinges on integers, might it make sense to show when a number is whole, versus decimal, even if just one decimal place as an indicator?

 

My impression for using the div 40 jumps was simply to iterate more quickly. Perhaps once that doesn't work, we drop down to div 8, as we hone in. Not sure, need to go back, could be from the (8) triangles and (5) layers, providing the div 40.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 22, 2018, 1:13 p.m. No.4833   πŸ—„οΈ.is πŸ”—kun   >>4839

>>4832

#floored

 

I'm still trying to think of what the defining relationship between the primes that make up all co-primes looks like. Or… -points in multiple directions as if counting something-… however that's said.

 

(https://en.wikipedia.org/wiki/Coprime_integers)

 

Pretty sure that's the most related thing I've ever typed out.

{thanks for the essplain, bro!}

PMA !dSvrkhSLR6 ID: fa8576 Feb. 22, 2018, 2:08 p.m. No.4834   πŸ—„οΈ.is πŸ”—kun   >>4835

>>4831

Using BigIntegers everywhere. The grid doesn't need decimals. And e handles the remainder.

 

A number of my tests do incorporate mods to verify accuracy, filter, and display records properly.

 

>>4832

>pretty awesome right there

yes. Seeing the solution in the grid - incredible.

VA !!Nf9AmQNR7I ID: 6133d1 Feb. 22, 2018, 2:47 p.m. No.4836   πŸ—„οΈ.is πŸ”—kun

>>4835

Although I guess it's irrelevant for RSA numbers. Just trying to find proper sized examples to work on so I can hopefully see patterns. What about (x+n)=15? f=30 so f-2=28. what next?

28/40 = n0= 7/10?

28/8 = n0 = 3 + (28 mod 8) = 3 + remainder 4? If this, then what to do with the remainder?

AA ID: e9bbff Feb. 22, 2018, 4:54 p.m. No.4839   πŸ—„οΈ.is πŸ”—kun   >>4841 >>4842

Good to see the board wasn't flooded with shitposts while I was away. I've only briefly flicked through the newest posts but it seems we're still figuring out how to find n0 and why f was divided specifically by 5 >>4334 here. Is that right?

 

>>4821

Do you know how many of those websites have been created and completely flopped or only get used by people who got banned from the bigger ones? Even ignoring the large quantity for a second, the answer is all of them. As important as the idea is, it's not as simple as creating a website and doing some marketing. Keep in mind, Facebook and Twitter may have gotten big in part because of that, but they're arms of the bloody deep state. They were always going to get big and have influence.

 

>>4823

It's so funny when people who think they're open minded enough for the big enchilada completely dismiss other slightly unrelated things without even reading about them. I'm glad that it at least seems like that isn't the case with all of us here.

 

>>4826

>>4827

How would we make some kind of bunker? If we're trying to get away from the chaos this will create for the board, we'll need to be able to go somewhere that they won't find out about. That means if we were to have a second board, we'd have to make everyone here aware of it without shills being aware of it, and those kinds of people are almost definitely lurking. One other not-so-great option would be for me to start a locked thread and give everyone with a trip a volunteer account so they can post in it. That would also mean excluding people who potentially have useful things to say and creating chaos through the fact that a bunch of people with conflicting viewpoints will have the ability to run the board. If we just wing it and see what happens, either we won't be able to work on things with this level of organization and calm again or it'll just magically work out. I don't like our luck with that. It would be good to think this through before we're done. I could make a board (assuming you're all cool with me doing it again), post a link to it, wait until several tripfags have replied, then delete it. Then when others wonder where we went one of us could email them. Then again, RSA will be broken, so if their form of email isn't secure, people who have enough of a reason to find us could find the board. Plus, current malicious lurkers would most likely keep it in mind and post about it once we've all migrated there. Anyone have any better ideas? No, not fucking Discord.

 

>>4833

Pics related is what those relationships look like. A 3D version would make more sense if you're curious (abc) but I guess that's a job for 3DAnon if they have some time.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 22, 2018, 5:18 p.m. No.4840   πŸ—„οΈ.is πŸ”—kun   >>4842 >>4845 >>4846

Iterating through larger chunks on Rsa 100 is not going to be a problem.

 

Using multiple*( (f-2) div 40 ) as an example and increment the multiple by 1 each iteration, we will fly right by the solution on the 21st calculation.

 

Pic attached is the result of n0, remainder 2d(n-1) calculations. First group is +/- (x+n) by 10. Bottom group is +/- (x+n) by 1.

 

Perhaps we are meant to break down the formulas and work at the individual triangle level, instead of (x+n)?

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 22, 2018, 6:24 p.m. No.4842   πŸ—„οΈ.is πŸ”—kun   >>4845 >>4852

>>4839 Thanks AA, good points. w/o twitter account, just need to make sure vqc is looped in.

 

>>4840 Looked back, and the "because f is divisible by 5" bit was basis for the 8*5=40 I think. Might be a clue there, something about using a factor tree for f.

 

>>4841 love reading your posts brah. Been me now.

 

btw - Q is on fire with drops right now. Open intel in 2010 got 20 cia in China 187'd. Going to be a whole new world without asymmetric encryption. More f2f comms and establishing of keys. Interesting comms from POTUS holding the handwritten note card.

 

Here are drops if anyone wants a quick look:

>>>/qresearch/466048

>>>/qresearch/465919

>>>/qresearch/465797

>>>/qresearch/465696

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 22, 2018, 7:28 p.m. No.4846   πŸ—„οΈ.is πŸ”—kun   >>4847 >>4862

>>4845 Yes, an amazing timeline. "watching the world wake up from history". We've got front row seats lads, we'll be sitting with the grandkids some day, showing a pyramid, telling stories of this time. It is indeed GHWB in that pic.

 

>>4840 the p+++ through pβ€” layout is a very nice view of this.

Noting the last digit sequences - very interesting.

(x+n) 8,9,0,1,2,3,4

n0 5,5,6,6,6,6,6

(x+n) 1,1,1,1,1,1

n0 5,5,5,6,6,6,6

 

pma - is your code java? Went to your last pastebin and downloaded, but haven't done anything yet.

This is awesome work!!! Feels so close now. Is there anything specific I can do to help tonight, versus trying to go back through and make sure all is understood?

VA !!Nf9AmQNR7I ID: a60c59 Feb. 22, 2018, 8:30 p.m. No.4850   πŸ—„οΈ.is πŸ”—kun   >>4851 >>4852

The biggest idea (of the last two weeks) is that var f gives us a clue to iterate up to (x+n). Simple as that. Now we gotta figure out how to make it work <900 iterations. In fact we gotta find out how to nail it on the head just because we can. My big question at this point: How do we find which part is x and which part is n?

AA ID: f984d8 Feb. 22, 2018, 8:40 p.m. No.4852   πŸ—„οΈ.is πŸ”—kun   >>4853 >>4856 >>4857

>>4841

Personally I don't mind some irrelevant posts here and there. I'm talking about, like, a co-ordinated mass shilling, the board being filled with guro by someone who just wanted to annoy some people, the brown pill guy, that kind of thing.

 

>>4842

If we didn't go with a new bunker board and we tried staying here, one way to not only keep VQC looped in but to make sure he can say what he needs to would be to give him a mod account or a volunteer account so he can do what Q does on >>>/greatawakening/ with the locked thread. I was meant to give the board to him when we migrated here anyway. Otherwise, if we do create a bunker board, I guess we'll have to see if his Twitter account comes back. Obviously this has happened in the past.

 

>>4850

See >>4305

 

I think you're both right about factoring f. Is there any way to determine its parity and divisibility (i.e if it's prime there might be problems)? I'll put a few bitmaps together and see if anything comes out since I have some time. If we figure out a pattern to how f can be divided, it won't be a matter of trial and error trying to find a number f can be divided by for the f/(8*something) part.

AA ID: f984d8 Feb. 22, 2018, 9:03 p.m. No.4855   πŸ—„οΈ.is πŸ”—kun

I made images for nf and xf but they weren't very helpful. This one, however, is interesting. Obviously we don't need to know anything about gradients to calculate f from c, but what it does show is that for every chunk of however many fs there are in each of these lines, there's a gradient associated with that chunk. That means we could find the lowest point on each of these lines, take that number away from f, and then f would be divisible by its gradient. That makes it calculable, if we can figure out when each line starts, what each gradient is, and what the highest and lowest points are along each line. Of course, what it makes calculable is f minus something, but it looks like the lowest value is actually zero on every line, meaning we'd only need to work out the gradients and not subtract anything.

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 22, 2018, 9:19 p.m. No.4856   πŸ—„οΈ.is πŸ”—kun

>>4852

OH FUCK! AA, thanks for posting that VQC post, I've been thinking about this all day. (1,c) is our big clue towards Prime(a,b). So here's the calc:

-C given

-d,e,f

-create (1,c) cell: d and e known,

n = (a+b)/2-d

x = d-1

complete element within the grid!!

Now move to the (e,1) equivalent. Find (px) and P(t).

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 22, 2018, 9:24 p.m. No.4857   πŸ—„οΈ.is πŸ”—kun   >>4858 >>4859

>>4847 Would really appreciate that, your output has been great.

>>4848 Ha, I screencapped that anon's version of "Allegory of the Cave". It's a good one to ingest, in whatever form.

>>4852 ty. I think the private board makes good sense, if you can swing that. Would really prefer you keep control, VQC has other priorities in life than just this and needs to be more consistent. Q leaves for couple days and anons freak, we go for couple weeks here and hang in there, says something.

>>4854 Nice one VA! Not voting though, the space counting and voting on 40char lines is what broke my spell to stop reading generals tonight. There were a couple epic breads there, good energy, needed that.

 

Q mentioned POWER - makes me think GRID! yes.

Also, power, as in exponent -have been thinking with the factoring this could be important for keeping track. 2^4 is 16 divided by 2 (4) times, etc. Same with factoring by 3, start with 27 and factor 3 times, it's 3^3, so the exponent is the sum of the times that factoring was done.

This forms a polynomial, with the root numbers being each prime as a series, and the number of times that prime was used in the factoring being the associated exponent, from 0 to N.

I'm sure this is already well documented out there, will take a quick look.

VA !!Nf9AmQNR7I ID: 2343be Feb. 22, 2018, 9:44 p.m. No.4859   πŸ—„οΈ.is πŸ”—kun

>>4857

>>4682

Underlying Order Of Integers: Family Tree Of Factors. Season 1.

>Integers are not a line continuum.

>There are families based on geometry.

>Shape.

>Symmetry.

>Triangles.

>VQC!!Om5byg3j

Anonymous ID: 6abd01 Feb. 22, 2018, 10:24 p.m. No.4860   πŸ—„οΈ.is πŸ”—kun   >>4861 >>4862

>>4824

 

"Bitcoin is safe though. If you use it properly it's safe against quantum attacks"

 

Didn't Chris specifically say cryptos will be wide open to exploitation by this thing? Specifically that the Satoshi wallet is a reward for success? No, infinite qbits is a big deal. Nothing can withstand that unless it is isolated somehow. Like if you had an air gapped computer with an analog video camera pointing at the monitor. That analog signal could still be transmitted over the airwaves to a rabbit-ear TV.

ID: 2bf75e Feb. 22, 2018, 11:49 p.m. No.4861   πŸ—„οΈ.is πŸ”—kun

>>4860

He quietly mentioned a way to attack hashing algorithms. Email works fine by the way. Just give the new bunker to people who've made it obvious they're who they say they are.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 22, 2018, 11:54 p.m. No.4862   πŸ—„οΈ.is πŸ”—kun

>>4846

Heh, jumped the shark with the digit sequences, numbers themselves were same or linear sequence. Gotta slow down.

 

>>4860 Yes, but later stages.

ECC is phase 2, and we believe we're on part 3 of phase 1, factorization of c, product of 2 primes, which gets at RSA, PGP, and asymmetric encryption in general with public keys.

Still part of Phase 1 we need to show trivial solution for Fermat's Theorem, think involving Column e0.

Phase 3 involved mandelbrot set, perhaps other fractals. Topol has a great visualization, might be with the Julia set. Application mentioned relates to processing video / 3d geometry very efficiently.

That's what led me down the Geometric Algebra route (geometrical view of Clifford Algebra), which I still believe is at the core of this whole thing. Even Phase I, as the grid is 6 or 7 dimensional based on our variables.

GA handles higher dimensions easily. Learning GA is one of the motivations for me hanging in with this, it's something I've been interested in for a while.

When we're at that point, we'll be looking at various physics properties, think SpaceTime Calculus and such. This is where calculations for sonoluminescence come into play - 'acoustic black holes' and the like.

If this goes on long enough, we'll get into de Broglie Waves, Pilot Wave theory (David Bohm) etc.

My digging has shown that Geometric Algebra is the 'language' that ties it all together, links disparate areas of mathematics into one. Have tons collected on this subject, only just starting to digest it all.

I believe in the quantum nature of consciousness, with micro-tubules. GA would be able to model this as well. This is a life-long quest of inquiry if one wishes.

 

>>4858 I did think of you all at lunch, eating a fractal (romanesco) roasted cauliflower. Really, along with broccoli, though the cauliflower is just striking.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 23, 2018, 7:14 a.m. No.4864   πŸ—„οΈ.is πŸ”—kun

>>4863

oh, i think i derped. saved as in "the link", not... go check in. lol

 

my b

 

Probably be best if we disseminated the link on... like... noooot here. And of course Chris is conveniently MIA so we can't just send it to him and have him disperse the invite since he knows which who we are on the twitters.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 23, 2018, 10:28 a.m. No.4867   πŸ—„οΈ.is πŸ”—kun   >>4868 >>4879 >>4887 >>4892 >>4957

Thoughts on iterating.

 

(f-2) / 40 makes up the triangle base of each of the 8 triangles. This is the number we are growing by multiples of 2.

 

(f-2) % 40 is the remainder.

 

(x+n)(x+n) = 1+ 8T(u) where u is the triangle base, and T(u) = (u(u+1))/2

 

Get_n_from_odd_triangle_base takes (x+n)/2 as a parameter.

Get_Remainder_2dnm1 takes (x+n - 1) / 2 as a parameter.

 

In order to test for a solution using the above formulas, we need to estimate what (x+n) would should be.

 

Formula for estimating (x+n):

 

(x+n)(x+n) = 1 + remainder + 8 * T(triangle_base)

(x+n)(x+n) = 1 + ( (f-2) % 40 ) + 8 * T( (f-2) / 40 )

 

(x+n) = sqrt( 1 + remainder + 8*T(triangle_base) )

 

Does this make sense?

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 23, 2018, 10:34 a.m. No.4868   πŸ—„οΈ.is πŸ”—kun   >>4869 >>4891

>>4866

Good, so long as intention isn't just 'vqc / we were right', but to satisfy an objective.

 

BTW, this is a LARP, and Q is a LARP. Have been thinking about this a lot - a larp requires a leader, someone who has a vision of 'the end', and the rest are along for a ride (still interactive and potentially influencing/altering the outcome). So when you sign up for a class in college, it's a larp, and you're given a syllabus, nothing wrong with that. Only problem is when the larp is disingenuous, a distraction, etc.

Christ Jesus, greatest larp of all time? "Follow me".

 

>>4867 ok PMA, back on track, love your focus! Read through and makes sense, will review more closely. Quick clarification - where does n0 fit in this? Is your n in (x+n) from the (1,c) record?

PMA !dSvrkhSLR6 ID: fa8576 Feb. 23, 2018, 10:43 a.m. No.4869   πŸ—„οΈ.is πŸ”—kun   >>4870

>>4868

Once we estimate (x+n), n0 comes from Get_n_from_odd_triangle_base.

 

That n0 is used in the call to Get_Remainder_2dnm1.

 

if Get_Remainder_2dnm1 0, n n0 and (as VQC said) we win!

CollegeAnon !LAbIRp9cT. ID: 990b43 Feb. 23, 2018, noon No.4871   πŸ—„οΈ.is πŸ”—kun   >>4872 >>4935

I've been looking more at the tetrahedral numbers (4,10,20), and I may have found something. Look at these grids. Basically the white columns are the seeds, and each value is equal to the sum of the one above and one to the left of it. By default, the first row is a 1. Then the top left box is just regular numbers seeded by a 1. The highlighted column in that box are the tetrahedral numbers. The highlighted column in the next is square numbers. Then the next is pentagonal numbers, then hexagonal, heptagonal, etc. If you look at the highlighted row, then you can see that there is a simple pattern to get from grid to grid. (ie, from 1 to 2 you subtract 1, from 2 to 3 you subtract 2, from 3 to 4 you subtract 3) so you can easily navigate through these grids. Also for each box n, the first column is all the numbers where mod n is the seed. So maybe we could start with a zero seed (cuz then its divisible) or something like that. Or we could use the identity n*n = T(n) + T(n-1) where T(n) is the nth triangular number. Get the number in the square grid and translate it to a different grid and do stuff. Also these grids can be extended into the negative (I'll get on that).

 

Basically the idea is to get our number. E would be the seed and D would be the start column. Then we can use some of these identities to navigate around to the grid where we are in the A column with a 0 seed.

CollegeAnon !LAbIRp9cT. ID: 990b43 Feb. 23, 2018, 12:15 p.m. No.4873   πŸ—„οΈ.is πŸ”—kun   >>4874

>>4872

For this, notice that the top left grid, is just 4* the 1 and 1 grid. Also if you start at the top left cell within a grid, that is the sum of the seed and the column. Then if you go down and to the right, you have that number x3, x10, x35, x126, and this series is the same as the numbers going diagonally down from 3 of the one column one seed grid from my last post.

CollegeAnon !LAbIRp9cT. ID: 990b43 Feb. 23, 2018, 12:33 p.m. No.4874   πŸ—„οΈ.is πŸ”—kun   >>4875

>>4873

Because of this multiplicity, if we start with a number with the h column seed, k row seed, then your diagnoal entries will be (h+k), (h+k)3, (h+k)10, (h+k)*35, etc. So if we start in G(h,k) (this will mean grid columnseed h, rowseed k), the diagonal entries for this will be the same as G(i,j) where i+j=h+k.

 

Going with our 145 = 12*12 + 1, we could start in the G(12,1) cuz, then we would have multiples of 1 mod 12, which would include 145. Then if we could navigate to the diagonal (probably by going to the right, because if you go up you would be at the 13 right away) we would have some value and we could then navigate to any other grid G(m,d) where m+d = 13. Then we could do something once there, this is where I'm lost.

CollegeAnon !LAbIRp9cT. ID: 990b43 Feb. 23, 2018, 2:02 p.m. No.4878   πŸ—„οΈ.is πŸ”—kun

>>4875

Also if you try this, be careful, the positive (bottom right) part stays the same, but if you switch the center cell (e or d) then the negative parts change.

MinecraftAnon !!QXqSZ2ev8. ID: 225058 Feb. 23, 2018, 4:56 p.m. No.4883   πŸ—„οΈ.is πŸ”—kun   >>4894

For the next render, I've included a very special set of invalid cells. The valid cells are green, while the invalid cells are red. Rendered in the (E,D,N) perspective. Here, we look down N between the A=1 and A=2. This view should be familiar, but as you can see, rendering these particular invalid cells leaks additional information about the shape of (E,D,N).

MinecraftAnon !!QXqSZ2ev8. ID: 225058 Feb. 23, 2018, 5:06 p.m. No.4884   πŸ—„οΈ.is πŸ”—kun

Now, we move below the grid. Turning towards +E, we see a familiar shape in the invalid cells. I'm naming these "leafs". We can see the boundary where A=1 shifts between valid and invalid, but this reveals this shape shares a common rule within the grid.

MinecraftAnon !!QXqSZ2ev8. ID: 225058 Feb. 23, 2018, 5:14 p.m. No.4885   πŸ—„οΈ.is πŸ”—kun   >>4886 >>4894

Now standing in -N space, we see a relationship once hidden by hiding all invalid cells. Between A, E, D, and N. Each leaf is comprised of leaflets, which share a complex relationship to the rest of the grid.

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 23, 2018, 5:44 p.m. No.4887   πŸ—„οΈ.is πŸ”—kun   >>4894

>>4879

>>4867

Yeah PMA!!! This is exactly how I was envisioning it at work today, multiples of n0 all listed out. Very interesting to see the remainders cycle thorough in sets of 4 in your first example. More great work from our MVP! Nice to see you using my preferred programming language, lol. EXCEL-lent work, sir.

πŸ˜‚

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 23, 2018, 5:58 p.m. No.4888   πŸ—„οΈ.is πŸ”—kun   >>4890

>>4865

Thanks Anon! VQC attracted a bunch of weird super awesome math/spiritual/troublemaker faggots here, and we just keep having fun working on math and shitposting. Glad you're here too! Are you a lurker, or do we know you already?

AA ID: 3cd743 Feb. 23, 2018, 10:19 p.m. No.4892   πŸ—„οΈ.is πŸ”—kun   >>4893

If anyone didn't see the link to the bunker board, let me know. I'll have to post it the next time Chris turns up either way.

 

>>4866

That's a good idea, although they'll wonder where we went so they can see our working and everything. Once we have the thing ready, obviously we will have the ability to tell everyone, but it won't entirely be in our hands anymore, so if anyone lurking wants to mess with us all they have to do is mention the board somewhere. If not, and if we're lucky enough to only have lurkers who agree about keeping this the way it is, that would work nicely.

 

>>4867

So f/5 wasn't arbitrary?

 

>>4886

Do you have any idea how to turn that series of jumps into math? I don't know how many relationships I've found that would have been useful but only if I could figure out the math.

Anonymous ID: 5d2d9a Feb. 23, 2018, 11:13 p.m. No.4893   πŸ—„οΈ.is πŸ”—kun

>>4891 No worries. Only meaning in sense we are students in this, learning for ourselves, by digesting knowledge drops.

 

>>4882 rendering a great success. Wow.

 

>>4879 Super. I get match using brute force squares spreadsheet. Comment on the decimals you mention.

Looking at c=14904371; I see the sqrt=x+n, with the …0831 out there. This is coming from the 8T(u) column, if you use the 1+8T(u) column in your calcs, with the 52171729 value, then it will be an integer result, 7223.

 

>>4892 ty

Anonymous ID: 5d2d9a Feb. 24, 2018, 11:02 a.m. No.4897   πŸ—„οΈ.is πŸ”—kun   >>4898

>>4879 Looking at the 6107 'proximate record' result. Noticed that the small square, 83, is prime (^2=6889). Could that have an impact?

Haven't replicated your spreadsheet. It's between your 6241 and 7225 1+8*T(u) values.

Anonymous ID: 5d2d9a Feb. 24, 2018, 11:16 a.m. No.4898   πŸ—„οΈ.is πŸ”—kun

>>4897

For 6889-1=6888

6888/8=861 = T(u)

41 and 42 work out as integer base/ht with no remainder, and their sum is 83, the small square. The factorization for 861 = (3^1)x(7^1)x(41^1).

Anonymous ID: 5d2d9a Feb. 24, 2018, 11:44 a.m. No.4899   πŸ—„οΈ.is πŸ”—kun   >>4900

>>4886 These views really are amazing! You can really see the tree nature with the branching in the last render. We move along, get to a junction of interest, take the new vector…

What's with the macaroni's in distance, where green red meet especially prominent?

VA !!Nf9AmQNR7I ID: a34e47 Feb. 24, 2018, 1:07 p.m. No.4903   πŸ—„οΈ.is πŸ”—kun   >>4904 >>4907 >>4909

>>4879

I think I have something solid here, lads!! Been working on this all morning. Can I please get some eyes on this to verify?

 

So next question is: we're building (x+n) without knowing it's full size starting only from (f-2). How do we know when we have a match starting only from c? We need a method to verify correct (x+n), and I'm sure PMA is on it. The following equations could help create a crosscheck to verify correct (x+n) and also solve for n and x. Maybe we increase the n0 factors until we get a balanced lock on all equations? Check it out ==>>

 

If we know c = (d+n)^2 - (x+n)^2 and we can solve for (x+n) then we're going to be able to use the quadratic from RSA#2 to solve for n. Here's the updated version using (TuXN)^2 instead of (2t)^2.

 

Find TuXN = (x+n) = 15 using (f-2) method. Verify using .

 

Using our TuXN = 15 prime element: {3:6:16:9:7:37}

n = SQRT(c + (TuXN)^2) - d

n = SQRT(259 + (15^2)) - 16 = 6

 

Then plug in n to solve for x, and then a and b

x = SQRT((d+n)^2 - c) - n

x = SQRT((16+6)^2 - 259) - 6 = 9

d - x = a = 16 - 9 = 7

c/a= b = 259 / 7 = 37

 

Element complete starting only from C!!

VA !!Nf9AmQNR7I ID: 2343be Feb. 24, 2018, 4:50 p.m. No.4910   πŸ—„οΈ.is πŸ”—kun   >>4912

>>4907

Hey Baker! I think you're right. VQC said the grid provides a shortcut.

 

>>4908

Thanks Topol!

 

>>4909

Hey MM! I was just using that element to verify that the formulas worked. Using PMA's n0 method >>4879 he was able to get a couple of exact matches for (x+n) using multiples of n0. Check out his second sheet of calcs.

 

Big Idea: If we can find (x+n) using triangle geometry, then we don't even need to know the individual values of x and n. We use the quadratic to solve for n, then plug in n and solve for x.

 

Actual Order of Operations Looks like this:

Start with c

Get d,e,f

Get n0: (f-2)/8 (for smaller f) n0=(f-2)/40 (for larger f)

Multiply n0 by increasing factors until a match for (x+n) is found. This was my triangle diagram idea. PMA tested it in spreadsheet form and got a match.

Verify correct match by crosschecking using quadratic formulas to see if answer matches the prime answer we're looking for.

Solve for n first.

Then plug in n to solve for x,a, and b.

 

Still a work in progress, but we are very close to

VA !!Nf9AmQNR7I ID: 6133d1 Feb. 24, 2018, 5:26 p.m. No.4912   πŸ—„οΈ.is πŸ”—kun   >>4914 >>4930

>>4910

Here's the example for c=259, (x+n)=15. Attaching PMA's output for clarity.

 

We start with c = 259

d = 16

e = 3

f = 30

n0 =(f-2)/8 = 3 remainder 4

Note: (f-2) mod 8 = 4 (this calcs the remainder)

Now we iterate n0 triangle base using multiples of n0 = 3

n0*1 = 3 (no match)

n0*2 = 6 (no match)

n0*3 = 9 (no match)

n0*4 = 12 (no match)

n0*5 = 15 (Match!)

 

So TuXPN (Triangle Base x plus n) = 15

Note: Method needed to verify when we reach the correct (x+n) value.

 

Now we use the quadratic formula.

n = SQRT(c + (TuXPN)^2) - d

n = SQRT(259 + (15)^2) - 16 = 6

 

Now we can also solve for x.

x = SQRT((d+n)^2 - c) - n

x = SQRT((16+6)^2 - 259) = 9

 

Now we can solve for a and b.

d - x = a = 16 - 9 = 7

c/a = b = 259 / 7 = 37

 

Complete element. {3:6:16:9:7:37}

 

Thoughts, Anons?

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 24, 2018, 5:50 p.m. No.4915   πŸ—„οΈ.is πŸ”—kun

Prof. John Conway

What were lectures like?

You'll find this interesting:

https:/ /math.dartmouth.edu/~doyle/docs/conway/conway/

Check link under "Romance of Numbers":

http:/ /mathvideo.dartmouth.edu:8080/ramgen/doyle/conway/RN92001.rm?start=&end=

You can download the lectures as .rm files.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 24, 2018, 6:30 p.m. No.4917   πŸ—„οΈ.is πŸ”—kun   >>4918

>>4916

>It's a match for the small square (x+n) measurement. For this example, (x+n) = 15

Right, but we don't know that until having the solution. The initial x is 15. And (x+n)-129. Still don't understand what I'm missing with your approach?

VA !!Nf9AmQNR7I ID: a34e47 Feb. 24, 2018, 6:57 p.m. No.4918   πŸ—„οΈ.is πŸ”—kun   >>4919 >>4930

>>4917

No worries, MM. Here's a rundown of what I currently know. Please correct any errors, Anons. Working to understand over here.

 

VQC's claim is this: because of triangle geometry, we can start with f, and use it to divide the odd (x+n) square into 8 triangles + 1.

Then we find n0, which is (f-2)/8 = n0 (whole integer) + mod (remainder).

Then we use multiples of n0 to iterate upward to find the small square (x+n) that matches c = (d+n)^2 - (x+n)^2.

We can use the formulas I've posted to check each iteration of n0*factor as True or False for a match to this formula: c = (d+n)^2 - (x+n)^2

Iterate, check True or False. Upward through all possible values of (x+n) until we get a match.

VA !!Nf9AmQNR7I ID: a60c59 Feb. 24, 2018, 7:09 p.m. No.4920   πŸ—„οΈ.is πŸ”—kun   >>4921

>>4919

You're 100% correct, MA. We are searching for a match as we iterate upward using multiples of n0. The size of both squares changes as we iterate upward, until we find a match where c = (d+n)^2 - (x+n)^2

VA !!Nf9AmQNR7I ID: a34e47 Feb. 24, 2018, 11:05 p.m. No.4925   πŸ—„οΈ.is πŸ”—kun

>>4924

In PMA's output, third example for c = 6017: notice that the correct (x+n) is not exactly found. However, the difference is so close! Also, the proximate answers could be iterated by +1 or -1 until a match is found.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 25, 2018, 12:11 p.m. No.4926   πŸ—„οΈ.is πŸ”—kun   >>4927 >>4929

>>4879

Test cases for c=6107 iterations attached showing f-2 and x+n calculations, including a breakdown of the 2d(n-1) remainder.

 

Tests are run using different base factors of (8, 16, 24, 32, 40). For each iteration, also testing +/- proximate range of 2.

 

Best result so far is 21 calculations, using an f-2 div factor of 16. Base factor of 24 fails to find a record.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 25, 2018, 12:19 p.m. No.4927   πŸ—„οΈ.is πŸ”—kun   >>4928

>>4926

Example for c=c9874400051 attached, with different base factors, proximate range searches, and relevant records only being shown.

 

This example takes 1493 calculations to solve.

 

I'm not quite sure what improvement this algorithm brings to the table. We could just as easily increment x+n guesses by 2 and arrive at a similar solution in 909 calculations.

 

Definitely not seeing something.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 25, 2018, 12:21 p.m. No.4928   πŸ—„οΈ.is πŸ”—kun   >>4929 >>4931 >>4945

>>4927

Perhaps there is a way to jump further ahead, and then iterate at a more granular level.

 

Perhaps this is how the factor tree comes into play?

 

Really big jumps at the top of the tree, and then smaller and smaller jumps as we get closer to the solution.

 

And then finally the +/- proximate searches.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 25, 2018, 12:54 p.m. No.4929   πŸ—„οΈ.is πŸ”—kun   >>4936

>>4926

Feels like good progress PMA! Nice to get to know this 6107 number a bit better.

If I understand correctly, you're using the "rm 2d(n-1)" and "2d mod" as the check in the actual algorithm to know when to stop the iteration, having arrived at the desired result. You're only using the "x+n diff" column only as a validation against the known answer for (x+n).

Also nice to test with a bit larger c=c9874400051, though don't want to graph that one out.

>>4928 Intuitively, that makes sense. With very large c values, the other variables can may be quite large as well, so probably more useful later.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 25, 2018, 1:30 p.m. No.4932   πŸ—„οΈ.is πŸ”—kun   >>4945

>>4822 one other interesting view on this comes from game theory, a topic VQC loves, and he specifically mentioned John Nash.

 

The late John Nash put forward Equilibrium Theory as an extension to Adam Smith's views of selfish motivation. He saw Adam Smith's view of everyone in the group doing what's best for themselves as incomplete, because the best result would come from everyone in the group doing what's best for himself, and the group.

 

He developed an equilibrium concept for non-cooperative games that later came to be called the "Nash Equilibrium". He proved that in any game where a finite number of players each has a finite number of choices, there is at least one position from which no single player alone can improve his/her position by changing strategy.

 

Such a point is called a "Nash equilibrium". The proof is, in the words of the economist Samuel Bowles, that there is always "a situation in which everybody is doing the best they can, given what everybody else is doing".

 

I think that sums it up well for us on this board, and the work anons are doing with Q as well.

 

Here a fun clip from "A Beautiful Mind" on the topic - 4minutes. Some other parallels to our work as well:

https:/ /www.youtube.com/watch?v=2d_dtTZQyUM

 

Where we go one, we go all. No more secrets.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 25, 2018, 1:41 p.m. No.4933   πŸ—„οΈ.is πŸ”—kun   >>4936

>>4931 See below. Also, he did mentioned there is more than one way to use the grid, once all is understood. I have a hard time being patient here at times. Wonder if the pace is VQC being busy and it being a low priority, or our pace of ingestion and readiness, not sure. The former seems odd considering how much time he vested in this and the implications of moving forward, but maybe just personal priorities.

 

>>4242

>Once the factorisation methods are complete, the short cuts via patterns in The Grid (e,n) will be clearer.

 

>>4489 Also, there is another function to come, plus part 3b for the even (x+n).

>Yes. I'll be doing a recap. Adding another function and then demonstrating on an unsolved RSA number after walking through the rest of RSA 100.

I don't think we have the additional function yet?

 

It's only been what, about 8 days since VQC was here. Sure feels longer, especially with twatter account down.

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 25, 2018, 1:59 p.m. No.4934   πŸ—„οΈ.is πŸ”—kun

>>4681

"group"

>>4682

Integers are not a line continuum.

There are families based on geometry.

 

Spent the weekend with John Nash (ha, absorbing videos and reading).

 

Here's a short video that gets at the idea of a number being a "compound number" made of two groups. It's less than 8 minutes. The group part comes 4min in.

https:/ /www.youtube.com/watch?v=ea7lJkEhytA

MM !!DYPIXMDdPo ID: 5d2d9a Feb. 25, 2018, 2:15 p.m. No.4935   πŸ—„οΈ.is πŸ”—kun   >>4940 >>4956

>>4871 Planning to dig into what you put down shortly CA. But, you sparked me and that sent me off in a multi-day direction!! That 3rd row sequence, 4, 10, 20, 35, 56, …

I was out and did a search for that sequence and added Cicada for heck of it. Rabbit hole!!

Sequence is a diagonal of Pascal's triangle, which we saw a bit ago as well.

ProgramMathAnon !dSvrkhSLR6 ID: 8298a2 Feb. 25, 2018, 3:06 p.m. No.4936   πŸ—„οΈ.is πŸ”—kun   >>4938

>>4929

Correct. Stop on remainder 2d(n-1) = 0, and "x+n diff" as a validation.

 

I agree on the progress even though we don't have a solution yet. There has been so much to absorb in just this thread alone.

 

>>4931

Agreed MA. This is VQC's deal. Happy to be along for the ride.

 

>>4933

>I don't think we have the additional function yet?

I think this was the Get_Remainder_2dnm1 used in our current tests.

VA !!Nf9AmQNR7I ID: 819f3a Feb. 25, 2018, 4:29 p.m. No.4937   πŸ—„οΈ.is πŸ”—kun

>>4930

Sure MM! Here we go.

 

>n0*2 = 6 (no match)

Here are the calcs to show (no match) for n0*2

 

TuXPN = 6

n = SQRT(c + (TuXPN)^2) - d

n = SQRT(259 + (6)^2) - 16 = 1.175 = n

 

n is not a whole integer for our first clue. now we plug n into the difference of squares equation.

 

c = (d+n)^2 - (x+n)^2

259 = (16 + 1.175)^2 - (6)^2 = 258 with remainder. (no match) but very close though! So we know (x+n) is bigger than (x+n)=6

VA !!Nf9AmQNR7I ID: 819f3a Feb. 25, 2018, 4:32 p.m. No.4938   πŸ—„οΈ.is πŸ”—kun   >>4939

>>4936

Hello PMA! I think I've found a way to use the RSA#2 quadratics we built around (2t)^2 = small square to cross check our iterations. It would effectively give us a way to tell if a (x+n) iteration is smaller or larger than our target c value.

VA !!Nf9AmQNR7I ID: a60c59 Feb. 25, 2018, 4:53 p.m. No.4941   πŸ—„οΈ.is πŸ”—kun   >>4943 >>4966

>>4939

Yes, but as we iterate upward the difference of squares begins to approach c, closer and closer. Then when we pass c, the difference of squares begin to drop again, below c=259. Var d is fixed, so when we pass the correct (x+n) value and increase it, the (x+n) square overtakes the (d+n) square. Think of it like finding a mountaintop peak. Increasing all the way up to a match with c, then declining down the back side of the mountain as (x+n) overtakes (d+n).

 

Here's an example bigger than correct (x+n) for c=259

>n0*7 = 21 (no match)

 

Here are the calcs to show (no match) for n0*7

 

TuXPN = 21

n = SQRT(c + (TuXPN)^2) - d

n = SQRT(259 + (21)^2) - 16 = 10.457 = n

 

n is not a whole integer for our first clue. now we plug n into the difference of squares equation.

 

c = (d+n)^2 - (TuXPN)^2

 

259 = (16 + 10.457)^2 - (21)^2 = 258 with remainder. (no match) but very close though!

 

So we know correct (x+n) is smaller, since TuXPn = 6 was almost to 259. We passed by the correct (x+n) value. Basically we can find the mountaintop. And as we get close, we can tell we're getting closer to c until it recedes again. Each iteration has a True or False check for c = (d+n)^2 - (x+n)^2, using the RSA#2 quadratic to solve for n and plug it into the formula.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 25, 2018, 5:30 p.m. No.4943   πŸ—„οΈ.is πŸ”—kun   >>4944 >>4947

>>4941

Trying your idea out. See attached c=6107 variations including a c estimate and c diff columns.

 

c est is calculated as (d+n0)(d+n0) - (x+n)(x+n), where the n0 and (x+n) values are from the current iteration.

 

c diff is the difference between c est and the original c value.

 

In the failed test result, notice that the c diff never goes negative. I believe this is because the x+n value is always increasing.

 

Am I missing something?

VA !!Nf9AmQNR7I ID: 2cf4b5 Feb. 25, 2018, 5:34 p.m. No.4945   πŸ—„οΈ.is πŸ”—kun

>>4928 big jumps, then smaller jumps, PMA.

>>4931 many ways, MA!

>>4932 beautiful mind, indeed MM.

 

The cool thing is, Anons, that there is more than one way to solve the problem. All our efforts combine to form like Voltron.

 

>John Nash developed an equilibrium concept for non-cooperative games that later came to be called the "Nash Equilibrium". He proved that in any game where a finite number of players each has a finite number of choices, there is at least one position from which no single player alone can improve his/her position by changing strategy.

>I think that sums it up well for us on this board, and the work anons are doing with Q as well.

 

Well said, MM. We are all working different angles to solve the same problem in the most efficient manner. It's pretty badass. It's a summary of why individual freedom benefits all of human society.

>>4932

>Well said, MM.

VA !!Nf9AmQNR7I ID: 2cf4b5 Feb. 25, 2018, 5:39 p.m. No.4947   πŸ—„οΈ.is πŸ”—kun   >>4948

>>4943

It never goes negative, bc it approaches the mountaintop. When (x+n) is too big, it goes back down away from c as (x+n) grows larger relative to (d+n). When it is a perfect match, it's at the exact top of the mountain or triangle.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 25, 2018, 6:45 p.m. No.4948   πŸ—„οΈ.is πŸ”—kun   >>4953

>>4947

So remainder 2d(n-1) is the difference between the 2 squares and c.

 

Thanks for clarifying this.

 

Can we use that value further? add it to the base of each of the triangles? Increase x+n to iterate quicker?

CollegeAnon !LAbIRp9cT. ID: f1fe5e Feb. 26, 2018, 7:14 a.m. No.4956   πŸ—„οΈ.is πŸ”—kun

>>4935

Yeah what I did was pascals triangles with different top edges

 

Dig into it. I'm thinking you can extend the grid into the negative and also when you make a root negative some interesting stuff happens.

PMA !dSvrkhSLR6 ID: fa8576 Feb. 26, 2018, 10:22 a.m. No.4957   πŸ—„οΈ.is πŸ”—kun

>>4867

For subsequent iterations, the estimated small square formula may need to include the remainder from 2d(n-1).

 

(x+n)(x+n) = 1 + ( (f-2) % 40 ) + 8 * T( (f-2) / 40 ) + (remainder 2d(n-1))

PMA !dSvrkhSLR6 ID: fa8576 Feb. 26, 2018, 11:53 a.m. No.4958   πŸ—„οΈ.is πŸ”—kun   >>4965 >>4966

Progress pics for c=14904371 and c=9874400051.

 

The "rm2dnm1" column is a running total of the "rm 2d(n-1)" result. This total is then added into the (x+n)(x+n), which is then part of the (x+n) value to test. When the "rm 2d(n-1)" becomes negative, the f-2 factor is increased and the rm2dnm1 total is reset. Checking proximate ranges for each row.

 

The f-2 factor is a larger jump and the "rm 2d(n-1)" is a fine tuning on each iteration.

 

This process is dependent on the base factor and proximate range searches. These examples use (f-2) div 256 and proximate ranges of +/- 4. Those were chosen by trial and error, but relate to f-2 being divisible by 4.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 26, 2018, 12:25 p.m. No.4959   πŸ—„οΈ.is πŸ”—kun

>>4955

This excel?

You have no idea how much I want this to be where "4, 10, 20 = D J T" was meant to be the entire time. I was simply derping the cipher at the time.

 

>>4362

>>4368

MM !!DYPIXMDdPo ID: 303a1d Feb. 26, 2018, 10:31 p.m. No.4966   πŸ—„οΈ.is πŸ”—kun   >>4967

>>4941 Thanks for walking it through, appreciate your patience.

>>4940 purrrrdy! Said like JA's cat.

>>4955 yes, well don't go down the route of Goldbach's conjecture or his partitions of the even integers, or terniary. There's the Goldbach Weave, and you'll find interesting prime patterns…

 

>>4958 this is incredible! It's working?!

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 26, 2018, 10:53 p.m. No.4967   πŸ—„οΈ.is πŸ”—kun   >>4969 >>4972 >>4973

>>4966

>It's working?!

In the sense that it can "sometimes" find the correct (x+n), yes.

 

For small values of n, it works pretty well, even on relatively large numbers.

 

For larger values of n, it takes a bit longer. And will sometimes skip over a matching (x+n) even though the correct n has been found.

 

Leads me to think that a success should check for more than just remainder 2d(n-1) == 0 to add a bit of tolerance. Unless we can iron down the estimation formula a bit better.

 

It's also too dependent on the base factor - div 8, 40, 400, 32, 64, 256 all work but for different test cases.

 

Anyone wants to play along, the f-2 estimation code snippet is attached.

MM !!DYPIXMDdPo ID: ff8671 Feb. 26, 2018, 11:24 p.m. No.4969   πŸ—„οΈ.is πŸ”—kun

>>4967 Hrmm. Quite interesting. You've really pushed this along.

 

Time is going to be short coming up. Wish VQC's twatter was up so we could give a holler.

 

Attaching a CONWAY Cartoon to maybe bring him out of lurk mode…

 

Added a couple treats in the easy bake. The prime visualization tool is awesome. Can almost picture the branches and such as it flows:

>>4968

VA !!Nf9AmQNR7I ID: 4cf918 Feb. 27, 2018, 6:38 p.m. No.4973   πŸ—„οΈ.is πŸ”—kun   >>4974

>>4967

The cool thing with the old quadratics is that they give us the ability to check each iteration against the difference of squares equation. n0^2 upwards will approach c, closer and closer towards the top of the mountain or pyramid. Then when we pass it ((x+n) is too big) , the QE’s will show a value moving farther from c.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 27, 2018, 9:12 p.m. No.4975   πŸ—„οΈ.is πŸ”—kun   >>4976 >>4977

>>4972

Moving quickly isn't really the problem. We are missing something fundamental in how the (x+n)(x+n) is estimated relative to the 2d(n-1) remainder and (f-2) div/mod.

 

These discrepancies get amplified as the numbers get larger.

 

For example, not quite sure how this crumb factors in.

 

>>4337

>We then know that if the base of f/40 is too small, 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.

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 27, 2018, 9:43 p.m. No.4978   πŸ—„οΈ.is πŸ”—kun   >>4979

>>4977

I'm guessing that it means that the (f-2) mod 40 value is used in two places.

 

It is added to the square 1 + 8T(u) + mod.

 

And then perhaps it's also included in each of the triangle bases?

Anonymous ID: 268629 Feb. 28, 2018, 12:43 a.m. No.4980   πŸ—„οΈ.is πŸ”—kun   >>4981 >>4982

>>4337

>We then know that if the base of f/40 is too small

Charitable language I think, basically f/40 is a guess, but a guess which shares something in common with the solution, probably mod

 

>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.

 

Iterate on chunks of 2d, looking for a matching mod value, then something probably

Anonymous ID: 268629 Feb. 28, 2018, 12:50 a.m. No.4981   πŸ—„οΈ.is πŸ”—kun   >>4982

>>4980

There are a couple of things that irritate me about VQC's posts there. The first is how 5 was pulked out of a fucking hat just because f ended in 5. The next was, magically, consideration of 5 lines through the triangle.

 

Are those 5's related? Is that 5 a minimum (see faulty orig sqrt func, etc, fun with BigIntegers) Is 5 the BEST or just handy for explanation? If not the best, what is? biggest? smallest? WTF?

AA ID: f6e421 Feb. 28, 2018, 2:44 a.m. No.4982   πŸ—„οΈ.is πŸ”—kun

>>4980

>iterate

That would work obviously, but the real solution is meant to involve just a calculation.

 

>>4981

The point of this entire process is for us to naturally learn why the solution we're being led to makes any sense, rather than just being shown. If you figure out whether 5 is arbitrary or not, you'll know why it works based on your own critical thinking skills, and you win.

MM !!DYPIXMDdPo ID: cbbd29 Feb. 28, 2018, 11:40 a.m. No.4983   πŸ—„οΈ.is πŸ”—kun   >>4984

Nothing to report, other than a view of (x+N), even/odd.

In pics, Green highlights the ODD (x+n).

At the top row are the primes for A

Along the side column are the primes for B

Shows top (start) of the grid, then another view with 3301x3301 forming the cross-hairs (the 464th prime).

Got a good sense of how the fields are shaped for the variables. When time allows will use this for the (f-2)mod40, div40, n0 calculations.

MM !!DYPIXMDdPo ID: cbbd29 Feb. 28, 2018, 11:50 a.m. No.4984   πŸ—„οΈ.is πŸ”—kun

>>4983

Also, this is for the c(prime) records. Will go back and do for the (1,c) we are looking for, need to set up a different spreadsheet for that, this is too large (provides a good set of valid 'c' values though).

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 Feb. 28, 2018, 4:13 p.m. No.4985   πŸ—„οΈ.is πŸ”—kun

https://en.wikipedia.org/wiki/Least_significant_bit

 

There are illegal primes, apparently, that if you post them online you get in booqoo trubblez.

 

Big ol' systems are based on those primes… apparently…

VA !!Nf9AmQNR7I ID: 7a77e2 Feb. 28, 2018, 5:57 p.m. No.4986   πŸ—„οΈ.is πŸ”—kun   >>4987

Hello Lads! Finished the 1st draft of my new sheet trying to work out my theory about difference of squares. Here it is. It looks like there are multiple mountaintops and valleys, where different values of (x+n) approach c and then recede from it.

 

Unfortunately, it doesn't look like an easy way to predict when we're approaching the correct (x+n) value. It seems like there are patterns, like a sine wave. Not surprised. Anyhow, look it over and let me know if it sparks any new ideas! It's always worth chasing an idea to the end. Edison tried 10,000 times (?) before he hit on the right combo for the incandescent bulb. When asked how it felt to fail so many times, he replied "I haven't failed, I successfully discovered 10,000 ways that don't work."

xoxo fags.

VA !!Nf9AmQNR7I ID: a60c59 Feb. 28, 2018, 7:41 p.m. No.4988   πŸ—„οΈ.is πŸ”—kun   >>4994

>>4987

Baker, you got a confirmed Jesus Fag over here. Keeping it low profile relative to the quest, but I'm with you. Love this verse. This quest had brought me closer to the Creator. I'll post two verses I really love. Here they are.

 

Amos 5:8

"He who made the Pleiades and Orion And changes deep darkness into morning, Who also darkens day into night, Who calls for the waters of the sea And pours them out on the surface of the earth, The LORD is His name."

 

Job 38 New International Version (NIV)

The Lord Speaks.

1 "Then the Lord spoke to Job out of the storm. He said:

 

2 β€œWho is this that obscures my plans

with words without knowledge?

3 Brace yourself like a man;

I will question you,

and you shall answer me.

 

4 β€œWhere were you when I laid the earth’s foundation?

Tell me, if you understand.

5 Who marked off its dimensions? Surely you know!

Who stretched a measuring line across it?

6 On what were its footings set,

or who laid its cornerstoneβ€”

7 while the morning stars sang together

and all the angels[a] shouted for joy?

 

8 β€œWho shut up the sea behind doors

when it burst forth from the womb,

9 when I made the clouds its garment

and wrapped it in thick darkness,

10 when I fixed limits for it

and set its doors and bars in place,

11 when I said, β€˜This far you may come and no farther;

here is where your proud waves halt’?

 

12 β€œHave you ever given orders to the morning,

or shown the dawn its place,

13 that it might take the earth by the edges

and shake the wicked out of it?

14 The earth takes shape like clay under a seal;

its features stand out like those of a garment.

15 The wicked are denied their light,

and their upraised arm is broken.

 

16 β€œHave you journeyed to the springs of the sea

or walked in the recesses of the deep?

17 Have the gates of death been shown to you?

Have you seen the gates of the deepest darkness?

18 Have you comprehended the vast expanses of the earth?

Tell me, if you know all this.

 

19 β€œWhat is the way to the abode of light?

And where does darkness reside?

20 Can you take them to their places?

Do you know the paths to their dwellings?

21 Surely you know, for you were already born!

You have lived so many years!

 

22 β€œHave you entered the storehouses of the snow

or seen the storehouses of the hail,

23 which I reserve for times of trouble,

for days of war and battle?

24 What is the way to the place where the lightning is dispersed,

or the place where the east winds are scattered over the earth?

25 Who cuts a channel for the torrents of rain,

and a path for the thunderstorm,

26 to water a land where no one lives,

an uninhabited desert,

27 to satisfy a desolate wasteland

and make it sprout with grass?

28 Does the rain have a father?

Who fathers the drops of dew?

29 From whose womb comes the ice?

Who gives birth to the frost from the heavens

30 when the waters become hard as stone,

when the surface of the deep is frozen?

 

31 β€œCan you bind the chains[b] of the Pleiades?

Can you loosen Orion’s belt?

32 Can you bring forth the constellations in their seasons[c]

or lead out the Bear[d] with its cubs?

33 Do you know the laws of the heavens?

Can you set up God’s[e] dominion over the earth?

 

34 β€œCan you raise your voice to the clouds

and cover yourself with a flood of water?

35 Do you send the lightning bolts on their way?

Do they report to you, β€˜Here we are’?

36 Who gives the ibis wisdom[f]

or gives the rooster understanding?[g]

37 Who has the wisdom to count the clouds?

Who can tip over the water jars of the heavens

38 when the dust becomes hard

and the clods of earth stick together?

 

39 β€œDo you hunt the prey for the lioness

and satisfy the hunger of the lions

40 when they crouch in their dens

or lie in wait in a thicket?

41 Who provides food for the raven

when its young cry out to God

and wander about for lack of food?

PMA !dSvrkhSLR6 ID: 8298a2 Feb. 28, 2018, 7:45 p.m. No.4989   πŸ—„οΈ.is πŸ”—kun   >>4990 >>5007

Back to some math to try and understand this estimation properly. Trying to incorporate >>4337.

 

Given:

 

nn-1 + 2d(n-1) + f = 1 + 8T(u) = (x+n)(x+n)

 

In terms of 8T(u):

 

nn + 2d(n-1) + f - 2 = 8T(u)

 

In terms of T(u):

 

(nn + 2d(n-1))/8 + (f-2)/8 = T(u)

 

In terms of T(u) as integers:

 

(nn + 2d(n-1))/8 + (mod 8) + (f-2)/8 + (mod 8) = T(u)

 

In terms of 1/5 T(u):

 

(nn + 2d(n-1))/40 + (f-2)/40 = T(u)/5

 

In terms of 1/5 T(u) as integers:

 

(nn + 2d(n-1))/40 + (mod 40) + (f-2)/40 + (mod 40) = T(u)/5 + (mod 5)

 

Where the (mod) value is the remainder from the closest left division expression.

 

Thoughts?

VA !!Nf9AmQNR7I ID: 2cf4b5 Feb. 28, 2018, 7:51 p.m. No.4990   πŸ—„οΈ.is πŸ”—kun   >>4991 >>4992

>>4989

MVP PMA says back to work!! Alright, man, geez. I'll begin studying your new ideas now. I have one thought:

P = NP revolutionizes everything. The end goal is so fabulous I can keep going forever. VQC said "new math Kangz" for when we solve this.

VA !!Nf9AmQNR7I ID: 1d5179 Feb. 28, 2018, 8:04 p.m. No.4993   πŸ—„οΈ.is πŸ”—kun

>>4991

Agreed PMA! This is a pretty fun quest, best I've been on. We unlock this, we go on to unlock the two other projects, EC and 3D video algorithms.

 

>VA - I think this is just the tip of the iceberg.

VA !!Nf9AmQNR7I ID: 819f3a Feb. 28, 2018, 8:40 p.m. No.4994   πŸ—„οΈ.is πŸ”—kun

>>4988

SpiritFags, you know what's up.

4 β€œWhere were you when I laid the earth’s foundation?

Tell me, if you understand.

5 Who marked off its dimensions? Surely you know!

Who stretched a measuring line across it?

6 On what were its footings set,

or who laid its cornerstoneβ€”

 

What are the underlying forms and number trees of the universe?

MM !!DYPIXMDdPo ID: b49cbd Feb. 28, 2018, 9:36 p.m. No.4995   πŸ—„οΈ.is πŸ”—kun

>>4715

>N = (X^2 + C) / (2 * A)

N = (X^2 + E) / (2 * A)

 

>F = E – (2 * D) + 1

F = (2 * D) + 1 - E

(produces a positive f)

 

>F = C – (D + 1)^2

(produces a negative f)

 

Looking at the n0.

CollegeAnon !LAbIRp9cT. ID: 4f0f91 March 1, 2018, 4:18 a.m. No.4996   πŸ—„οΈ.is πŸ”—kun   >>4997 >>4998

I think I found the link between all the primes. I'll give you guys a hint. I'll show you after I show my professor because I want to see what they think. Here's the hint: Look at the continued fractions.

When are they predictable?

Could you take any two numbers and produce a third out of this method?

How do you do this?

What if the two numbers are odd? How do you reconcile that?

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 March 1, 2018, 5:12 a.m. No.4997   πŸ—„οΈ.is πŸ”—kun

>>4996

Repeating patterns?

Are the primes hiding there?

Figure out when the repetition point is and the prime is starting over?

 

I've done weird things with decimals in the past.

Had this whole excel sheet of seemingly significant productions and alignments.

 

Y'all know how I do.

CollegeAnon !LAbIRp9cT. ID: aa2ce9 March 1, 2018, 5:15 a.m. No.4998   πŸ—„οΈ.is πŸ”—kun   >>4999 >>5007 >>5016

Rusty

>>4996

Fuck it I'm not going to jerk anyone around. This is definitely related to what were doing. Look at this output. These are the continued fractions of the square root of each number. The highlighted part is the part that repeats and the starting element is just our d. Notice that each series ends with the value 2d. Also notice that if 2d is divisible by e (the column after the d column) Then the continued fraction is [d; (2d/e, 2d)] (parenthesis repeats).

 

Then I thought, if we have any a,b and a divides b and b is even, we can make a repeating sequence [x; a, b] where x is b/2. Moreover, this value would be the same as dd + e (which would be xx + b/a =b^2/4 + b/a)

 

To generalize this to any two factors, we can say the following:

for any a,b, where a or b is even (so that ab is even), we can make a continued fraction [(ab)/2; (a, ab)] which evaluates to the square root of an integer value. The integer value would again be dd+e which is (aabb)/4 + b. So I developed this code:

 

def toOther(a,b):

return int((b*b)/4 + a)

 

def web(a,b):

prod = b*a

one = toOther(a,prod)

two = toOther(b,prod)

return (one, two)

 

And I was doing it for primes a,b and I was generating other primes! (also for this you get different results for using the product and each factor, thats why there is the "one" and "two" in the web function). Then I tried to make a graph of it in excel but I realized that the int was rounding down. For instance if I was generating 7 I would receive 7.25. This was always happening because a*b was odd, and isn't divisible by 4. Then I noticed that they were always .25 above the prime value, and I remembered that every odd square minus one is divisible by 8 and therefore dividible by 4, and also if I subtracted the one I would get rid of that .25.

 

def toOther(a,b):

if(b & 1):

top = b*b - 1

else:

top = b*b

return int( top/4 + a )

 

def web(a,b):

prod = b*a

one = toOther(a,prod)

two = toOther(b,prod)

return (one, two)

 

Test this stuff out guise

CollegeAnon !LAbIRp9cT. ID: aa2ce9 March 1, 2018, 5:20 a.m. No.5000   πŸ—„οΈ.is πŸ”—kun   >>5001

This output was damning

 

>>> web(3,5)

3 5

3, 15 59 [59]

5, 15 61 [61]

 

>>> web(7,29)

29 7

29, 203 10331 [10331]

7, 203 10309 [13, 13, 61]

 

What is that 61? Our starting n value?

Also notice that the two numbers it generates differ by the same amount that the a and b differ by. Just wanted to be the first to say it

CollegeAnon !LAbIRp9cT. ID: aa2ce9 March 1, 2018, 5:23 a.m. No.5001   πŸ—„οΈ.is πŸ”—kun   >>5002

>>5000

Heres a pastebin of output:

 

https://pastebin.com/6vnmqc4B

 

What this is is the two numbers, then the number generated by one factor and the product, then the factors of that number.

CollegeAnon !LAbIRp9cT. ID: 90b489 March 1, 2018, 8:27 a.m. No.5003   πŸ—„οΈ.is πŸ”—kun

>>5002

Maybe using this knowledge we have c. Then we can get our c^2 record which could be interesting. Then, for records c^2 through c^2 + 2c we know that the d for these is c. Also, the continued fractions would be terminated with the value 2c. Also, if the length of the repeated segment of the repeated fraction 2, then we would know that the value e for that would be a factor of 2c. If e odd, then e | c, if e even, then e/2 | c. So theoretically we would only need to do 2c iterations of this and then we would also only need to compute the first 3 terms of the continued fraction for the square root (ie for [12; 1, 2] where 12 is the original integer value) of that value.

 

METHOD:

for each value c^2+e:

do the first 3 terms of the continued fraction

if the third term (or the second after the start) is == 2c:

e is a factor of 2c

else:

continue

VA !!Nf9AmQNR7I ID: a34e47 March 1, 2018, 9:58 a.m. No.5007   πŸ—„οΈ.is πŸ”—kun   >>5008 >>5010

>>4989

Hello PMA! I examined the equations, and they look good. So how do we use them? Interesting that CA is also working on remainders of integers right now.

 

>>4998

>>4999 Checked!

Hello CA! I'm finding a similar pattern to your chart as I iterate (x+n) values upward towards the correct (x+n) value. Your idea is closely related to the number trees, right? Can you explain in more detail how to read this chart? I'm following your ideas, and understanding them.

>These are the continued fractions of the square root of each number. The highlighted part is the part that repeats and the starting element is just our d. Notice that each series ends with the value 2d. Also notice that if 2d is divisible by e (the column after the d column) Then the continued fraction is [d; (2d/e, 2d)]

>Then I noticed that they were always .25 above the prime value, and I remembered that every odd square minus one is divisible by 8 and therefore dividible by 4, and also if I subtracted the one I would get rid of that .25.

 

>Since this exists you could probably reverse it, like have a number and a "key" a and subtract a from the number then multiply by 4. Then maybe or maybe not add 1.

>Maybe this could be how the tree system works too.

CollegeAnon !LAbIRp9cT. ID: 5d00f1 March 1, 2018, 11:07 a.m. No.5008   πŸ—„οΈ.is πŸ”—kun   >>5011

>>5007

Here is a continued fraction. On the chart, basically the first column is c. The next is d, then e, then the rest is a repeating sequence of numbers. Basically from this pic the d is a_0, and the rest are a_1, a_2 etc.

 

For example, for the number 148

148

d = 12, e = 4

since 12*2 is divisible by 4, the continued fraction is:

 

[12, 24/4, 24, 24/4, 24, 24/4, 24, 24/4, 24, 24/4, …] as (in reference to the picture)

[a0, a1, a2, …]

 

The highlighted numbers are just the sequences that repeat

PMA !dSvrkhSLR6 ID: 8298a2 March 1, 2018, 8:19 p.m. No.5010   πŸ—„οΈ.is πŸ”—kun   >>5011 >>5013

>>5007

Pic attached for c=6107 and Rsa 100, showing the breakdown between nn, 2d(n-1), (f-2), and their respective mod results.

 

Included is an analysis of the various triangle formulas 8T(u), T(u), and T(u)/5 - which equates to div 40. T(u) represents the area for one of the 8 triangles that makes up the odd x+n square.

 

This was an attempt to explore the relationship between the mod values and how they affect the estimation required for an iterative solution.

 

In these pictures, u represents the triangle base - the blue bar in VQC's images.

 

The reason the iterative solutions are failing so far is because of the u % ((f-2)/40) result at the bottom. Finding something to quickly fill that gap is the problem.

MM !!DYPIXMDdPo ID: 1cda4a March 1, 2018, 8:40 p.m. No.5011   πŸ—„οΈ.is πŸ”—kun   >>5012 >>5013

>>5010 great PMA, am also looking at some mod values, and the f, see examples. Next going to look at how the 2d-1 fills in to take space for the n-1 block.

>>4681

>Try this group:

>odd e, even d, odd (x+n)

>From the smallest upwards incrementally.

>Notice any patterns with f?

For this group:

The (1,c) and prime records are in the same group (odd e, even d, odd (x+n))

n always EVEN, X alway ODD,

f is identical for (1,c) and prime records

d is identical for (1,c) and prime records

e is identical for (1,c) and prime records

fMOD8 and (f-2)MOD8 are identical for (1,c) and prime records

the (1,c) and prime records are in the same group (odd e, even d, odd (x+n)) with same mod value

fMOD8=2,6

(f-2)MOD8=0,4

 

>>5008 Very interesting CA. Seems you've found something of interest!

PMA !dSvrkhSLR6 ID: 8298a2 March 1, 2018, 8:53 p.m. No.5012   πŸ—„οΈ.is πŸ”—kun   >>5013 >>5014

>>5011

Spreadsheets look good.

 

Have you given any thought to the relationship between the large square and the small square?

 

in >>4678

>We are creating a method that USES f as a guide to find how to construct the square.

 

Is there something in the large square that directs the use of f further into estimating 2d(n-1) or nn ???

VA !!Nf9AmQNR7I ID: 7a77e2 March 1, 2018, 9:22 p.m. No.5013   πŸ—„οΈ.is πŸ”—kun   >>5015

>>5010

>>5011

>>5012

>The reason the iterative solutions are failing so far is because of the u % ((f-2)/40) result at the bottom. Finding something to quickly fill that gap is the problem.

>Have you given any thought to the relationship between the large square and the small square?

 

Maybe f is a measurement for both the small square and large square.

>Is there something in the large square that directs the use of f further into estimating 2d(n-1) or nn ???

Yup, gotta be a connection.

 

CA is very close to tying everything back to the number trees!!

MM !!DYPIXMDdPo ID: 1cda4a March 1, 2018, 10:52 p.m. No.5014   πŸ—„οΈ.is πŸ”—kun

>>5012

Haven't seen a match/pattern yet. Was working through the (n-1) triangle cap piece.

 

>>4678

Did get the "fill area" to check out. Left side has the pieces, just by taking the chunks visually and comparing to the 2d(n-1):

(d+n)^2-d^2-n^2+1-e-f=2d(n-1)

 

Then, the box with the red 2d(n-1), f, and n^2-1 was tallied, and this is indeed (x+n)^2 for the odd group. We can see some fractions for the Even cases, so will have that to address in part 3b.

CollegeAnon !LAbIRp9cT. ID: f3b045 March 2, 2018, 9:16 a.m. No.5015   πŸ—„οΈ.is πŸ”—kun   >>5016

>>5013

The theory is that each pair of numbers points you to a different prime (maybe not even a prime idk) as you recurse up the tree and it should end at the right factor.

 

Also my idea is a way to generate prime numbers.

PMA !dSvrkhSLR6 ID: fa8576 March 2, 2018, 11:30 a.m. No.5016   πŸ—„οΈ.is πŸ”—kun   >>5017 >>5018

>>5015

CA - I wrote a test around your "toOther" and "web" methods from >>4998.

 

Called them using the a and b values from our c and p records and noticed something VERY interesting.

 

Pic attached is for c=6107 and c=9874400051. But this holds true for all my other test cases including RSA100.

 

web.one = toOther(a, b*a)

web.two = toOther(b, b*a)

 

I then determined the d and e values for each of these results. And for giggles, added them together to get d+e.

 

For a and b values from the original c record, d+e is the same for the web.one and web.two.

 

But, for the a and b values from the prime record:

 

  • 1/2 of the difference between d+e for web.one and web.two equals the (x+n) value we are looking for.

  • the d value for both web.one and web.two always equals the x+n value from our starting c record.

 

There is a link here.

CollegeAnon !LAbIRp9cT. ID: d0d511 March 2, 2018, 12:33 p.m. No.5018   πŸ—„οΈ.is πŸ”—kun   >>5019

>>5016

Yeah I actually think this may be the secret underlying link between all the prime numbers. There is probably some criteria for the p and q that go into the web function to generate primes that I haven't discovered yet.

CollegeAnon !LAbIRp9cT. ID: d0d511 March 2, 2018, 1:51 p.m. No.5020   πŸ—„οΈ.is πŸ”—kun   >>5021

>>5019

Not that I can think of. Check this output:

>>> makeTree(85)

85

9

3

1

[[[1, 2], 0], [1, 0]]

>>> web(1,2)

(1, 2) (1, 2)

1, 2 2 [2] 2

2, 2 3 [3] 3

(2, 3)

>>> web(3,0)

(3, 0) (3, 0)

3, 0 3 [3] 3

0, 0 0 [] 0

(3, 0)

>>> web(3,1)

(3, 1) (3, 1)

3, 3 5 [5] 5

1, 3 3 [3] 3

(5, 3)

 

>>> makeTree(7*19)

[[[1, 2], [1, 0]], [1, 2]]

>>> web(1,2)

(1, 2) (1, 2)

1, 2 2 [2] 2

2, 2 3 [3] 3

(2, 3)

>>> web(1,3)

(1, 3) (1, 3)

1, 3 3 [3] 3

3, 3 5 [5] 5

(3, 5)

>>> web(1,5)

(1, 5) (1, 5)

1, 5 7 [7] 7

5, 5 11 [11] 11

(7, 11)

 

>>> makeTree(13*61)

[[[1], [1, 2]], [3]]

>>> web(1,2)

(1, 2) (1, 2)

1, 2 2 [2] 2

2, 2 3 [3] 3

(2, 3)

>>> web(1,3)

(1, 3) (1, 3)

1, 3 3 [3] 3

3, 3 5 [5] 5

(3, 5)

>>> web(5,3)

(5, 3) (5, 3)

5, 15 61 [61] 1

3, 15 59 [59] 59

(61, 59)

 

So for some of these if you extrapolate out you can generate the correct factor through this. Some examples don't work but I'm not completely deterred because maybe my trees are terminating too quickly or something idk

ProgramMathAnon !dSvrkhSLR6 ID: fa8576 March 2, 2018, 2:05 p.m. No.5021   πŸ—„οΈ.is πŸ”—kun   >>5022 >>5023

>>5020

Some other nifty coincidences from these formulas:

 

( (d+n)(d+n) + (x+n)(x+n) ) / 2 - web.one == solution a

( (d+n)(d+n) + (x+n)(x+n) ) / 2 - web.two == solution b

 

web.one e - (x+n) = a

web.two e - (x+n) = b

 

(cc / 4) - web.one == a

(cc / 4) - web.two == b

 

I think you should come up with a better name for these methods!

VA !!Nf9AmQNR7I ID: 6133d1 March 2, 2018, 5:32 p.m. No.5023   πŸ—„οΈ.is πŸ”—kun   >>5024

>>5021

>>5022

CA and PMA tearing it up! I've studied all your guys' output, and it looks good. I'm working to understand the underlying ideas, I've got about 90% of it down.

>The theory is that each pair of numbers points you to a different prime (maybe not even a prime idk) as you recurse up the tree and it should end at the right factor.

>Also my idea is a way to generate prime numbers.

 

What are the algebra formulas for web.one and web.two? Or how to calculate web1 and web2? I understand this idea:

>for any a,b, where a or b is even (so that ab is even), we can make a continued fraction [(ab)/2; (a, ab)] which evaluates to the square root of an integer value. The integer value would again be dd+e which is (aabb)/4 + b.

 

Can CA or PMA give a quick explanation of the formulas and how they work? Almost there in my understanding.

PMA !dSvrkhSLR6 ID: fa8576 March 2, 2018, 6:09 p.m. No.5024   πŸ—„οΈ.is πŸ”—kun   >>5025 >>5029

>>5023

VA - If I'm understanding this correctly, there seems to be a direct relationship between the x+n value for the entry c record, and the squares of these values as you go higher up the tree.

 

For example, for a = 31, b = 197, c=6107

 

web.one = 31 + (6107*6107)/4 = 9323893

web.two = 197 + (6107*6107)/4 = 9324059

 

The interesting parts here are that the sqrt of both numbers is x+n = 3053.

 

And the web.one number 9323893 is prime.

 

I have also checked a=5, b=29, c=145, and the web.one method also produces a prime number. 5+(145*145)/4 = 5261.

AA ID: 0f243d March 2, 2018, 7:17 p.m. No.5029   πŸ—„οΈ.is πŸ”—kun   >>5030

>>5024

>The interesting parts here are that the sqrt of both numbers is x+n = 3053.

The sqrt of both numbers is 3053, but where does x+n = 3053? The x+n of 31197 is 83 and the x+n of 61076107 (or any square) is 0.

AA ID: 0f243d March 2, 2018, 7:53 p.m. No.5033   πŸ—„οΈ.is πŸ”—kun

>>5032

So what this means is that floor(sqrt((c^2)/4)) = (x+n) iff a = 1? I took the real a and b out of it because it still seemed to work (it floated around 3053.5). Maybe there's a link between the remainder from the square root of ((c^2)/4) across all values of c that allows us to change the equation around to calculate a and b.

Topolanon +++ !!!ZjI4YmE4MzE5Yjlm ID: f7dce4 March 2, 2018, 8:41 p.m. No.5037   πŸ—„οΈ.is πŸ”—kun   >>5040

>>5036

C'mon Nerds… if you can arrange a prime (points to it, -rimshot-) you can arrange a board.

 

We'll need some solid dough to bake some tasty bread.

 

(The pics may or may not get better… maybe.)

ID: 2bf75e March 3, 2018, 12:25 a.m. No.5046   πŸ—„οΈ.is πŸ”—kun   >>5048

Something that I completely didn't notice when I calculated the inverse triangle function is that it involves calculating the odd square, just like in VQC's methods

 

public static BigInteger TM1(BigInteger T) { return sqrt(eight.multiply(T).add(one)).subtract(one).divide(two); }

PMA !dSvrkhSLR6 ID: 8298a2 March 3, 2018, 4:57 p.m. No.5051   πŸ—„οΈ.is πŸ”—kun   >>5052 >>5054 >>5058

Attached diagrams are attempts to further understand the relationships between e, f, dd, nn-1, and 2d(n-1), the small and big squares, triangles, and how this might lead to an iterative solution.

 

These examples represent the prime solutions for c=115, c=259, c=287, and a partial view of c=6107.

 

The 2 light blue blocks from (f-2) can always be positioned one in the center and the other to fill the nn-1 square.

 

The (x+n)(x+n) square is bordered by e and some dd blocks represented in light grey. I was looking for a way to define a boundary for the small square that could be calculated.

 

Also notice how the light blue f squares can start directly after the green e squares. Together they sometimes match the x+n side or get pretty close.

 

In the c=6107 example, there are 2 lightly colored vertical lines that represent (f-2)/2 and (f-2)/2 + (e-1)/2.

This was an attempt to use the e and f relationship to find a better starting position for the (f-2) div 40 iterative search.

 

An estimated triangle base can be calculated as:

 

u = ((f-2)/2 + (e-1)/2) / 2

 

In some cases, this formula will find an exact match. In others, it can be used as a starting position for a quicker search.

 

Unfortunately, there are quite a number of cases where this completely misses the mark. RSA values and small values of n in particular.

MM !!DYPIXMDdPo ID: bf3baa March 3, 2018, 9:42 p.m. No.5054   πŸ—„οΈ.is πŸ”—kun   >>5057 >>5058

>>5051 we're on the same wavelength! I was playing with the colored geometry earlier, came on and saw your post. Here's what I came up with, in the middle of it and was going to look at how the (1,c) record looks, or another ODD case such as this.

Thought the 3rd one here was the most interesting configuration. Would like to try for another case and see if that shape holds.

The comment >>4680

>Those who are good with their grid diagrams, can you show any examples of how all the pieces fit together for smaller examples with the triangles?

>The only piece missing before is the left hand big square with the pieces added to the side.

>Should make for a good diagram. Particularly with respect to the contribution of 2d(n-1) and how the symmetry of that looks.

..has been hanging there and been intending to come back to it.

 

For each of the 3 pics:

2d(n-1)=32*(5)=160 units (red)

160 / 8 = 20 (area of each triangle, highlighted in yellow)

n=6, so n2-1 = 35 units (purple)

f=30 (blue)

 

Landing on the 3rd pic, the symmetry felt right, as it fits with both the (x+n) square, and also the larger squre of (d+n)=22, with=9.

MM !!DYPIXMDdPo ID: bf3baa March 4, 2018, 8:16 a.m. No.5057   πŸ—„οΈ.is πŸ”—kun

>>5054

Checked for c=287, and a similar pattern can be constructed. f=2.

The (1,c) show some interesting patterns, and coalesce around squares with some configurations, but nothing ready to share yet.

MM !!DYPIXMDdPo ID: bf3baa March 5, 2018, 5:28 a.m. No.5059   πŸ—„οΈ.is πŸ”—kun   >>5060

>>5058

The symmetric fill patterns on those last couple were lucky. Did a couple more and the f and n don't fill the small square so cleanly, there is a d remaining that needs to fill in.

VA !!Nf9AmQNR7I ID: 4cf918 March 5, 2018, 6:42 p.m. No.5060   πŸ—„οΈ.is πŸ”—kun

>>5059

Hey lads! It's been kinda slow here, but that's fine. Just checking in, I'm thinking and working over here. Reviewing diagrams and crumbs. Q is going off too. Let's keep up the good work. Hopefully VQC will pop in soon.

PMA !dSvrkhSLR6 ID: 8298a2 March 5, 2018, 7:54 p.m. No.5061   πŸ—„οΈ.is πŸ”—kun   >>5062 >>5063

Still stuck on an iterative solution, but sharing some images with the hopes that someone else will have a better insight into this.

 

Attached are pictures for c=259 at various stages during the iteration process while determining n0.

 

Examples are for n0=2, n0=3, n0=4, and the final solution where n=6. Darker colored squares are the values after n0 has been determined.

 

The f-2 calculation on the right of each image shows the iteration step, f-2 div 40 and f-2 mod 40 results, and the triangle formula. Currently using just the iteration*(f-2) div 40 result as the triangle base. Not quite sure yet how the mod 40 factors in.

 

One thing that has become a bit clearer in doing this exercise is the meaning of the remainder 2d(n-1) calculation.

 

It's the difference between the 1+8T(u) formula and the nn-1 + 2d(n-1) + f formula for the small squares.

 

Also, just a guess that the "All your base are belong to us" comment somehow relates to the triangle base (u) in the small square formula.