ID: a2017c RSA #6 Dec. 21, 2017, 1:52 a.m. No.848   πŸ—„οΈ.is πŸ”—kun

Our task is simple. We are going to completely break the entire RSA cryptosystem!

And how are we going to do that?! With the VQC! Quick rundown to follow.

 

The main mathematical task that must be achieved to break RSA is to be able to factorize any integer instantly. The VQC is a promising way to do that.

 

What is the VQC?

The virtual quantum computer (VQC) is a grid made of infinite yet constructable sets that follow a known pattern. Like a quantum spreadsheet.

The grid is the superposition. The collapse of that superposition will be two input parameters, d and e which can be calculated easily for all integers, c, where c is the difference of two squares.

 

When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows calculation instead of searching is possible.

 

Legend

The map's legend is {e:n:d:x:a:b}, where d is the result of removing the largest square from c AKA the square root,

e is the remainder,

n is what you add to d to be exactly halfway between a and b,

and x is what you add to a to make d.

c is any number that is the difference of two squares, so odd numbers are included.

na and nb for any c can be found n places apart in the cell at (e,1).

 

Rules of the grid: global rules

Each cell of the grid (e,n) has infinite elements or ZERO elements.

Each cell with one value has infinite elements, since every element can make a new one.

By induction, a cell only needs one value to make infinite values, that's part of the power of this and is why it is a virtual quantum computer as a whole.

The t variable is what will allow you to walk across these infinite elements.

If a grid cell has elements, all elements are constructable from a finite set of root elements.

The grid is indexed using e, n, and t, where e is the rows, n is the columns, and t is the specific element in the cell-group.

Thus, only three variables are required to identify an element: e, n and t.

All products of odd numbers and all products of pairs of even numbers are the difference of two squares.

The x-intercept of the line that goes through the point containing the factors of c is (a + 1).

 

(1, 1) - the key

The values of a and b at 1,1 are related to the length of the longest side in right angled triangles.

The values here can be used to create the entire grid.

The values here determine the values of the rows to the left and right, which determine the values of the whole column.

 

Columns

Each cell at n=1 contains the roots of products in the column.

If c is a prime number, it will appear in one column exactly once.

If c is the product of two prime numbers that do not equal eachother, c will appear in two cells of one column.

All products (integers) c that are the sum of two squares appear (only) in columns where e=0,1,4,9,16,25…

All factors in a column are factors of the elements of the first cell in their column.

All Fermat primes (except) 3 appear in column one.

 

(e, 1)

If a number at position t has a factor s, then s is a factor at (t+s), (t+2s) and so on for a at (e,1).

Also, if a number at position t has a factor s at (e+1), then s is a factor at (s+1-t), (2s+1-t), etc for a at (e,1).

 

Rows

(1, n)

The cells in row one where n=1 have a relationship with the cells 2n to the right and 2n to the left.

Each "a" from the first row equals na because xx+e = 2na and na is half of that. That's BIG part of the KEY

Each element in a cell can be generated by moving up (t-1 = x-2) or down (t+1 = x+2). Other variables can be generated from x.

Every single factor of any value of a in the first row will be referred to as s.

 

Want an even better quick rundown? Read VQC's introduction.

>>19

 

Useful Equations and Notation

ab = c

dd + e = c

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

a + 2x + 2n = b

a = d - x

d = a + x

d = floor_sqrt(c)

e = c - (dd)

b = c / a

n = the difference between the square root d and the larger of the two squares

n = ((a + b) / 2) - d

d + n = number that is exactly halfway between a and b

d + n = i

x = d - a

x = (floor_sqrt(( (d+n)*(d+n) - c))) - n

x + n = j

f = e - 2d + 1

t = the variable that lets you traverse the infinite elements in for a given (e, n) that has values.

if (e is even) t = (x + 2) / 2

if (e is odd) t = (x + 1) / 2

ID: a2017c Dec. 21, 2017, 1:53 a.m. No.849   πŸ—„οΈ.is πŸ”—kun

CodePost Guide

Because of the tendency for post links to disappear I will now be using Pastebin.

 

C#

BigInteger Square Root: pastebin.com/rz1SdACZ

Create Bitmap in: pastebin.com/hMTtJF6E

Generate Bitmap within original code: pastebin.com/JUdtehb4

Generate the large square for e and t: pastebin.com/nbjs2kz4

Original VQC code: pastebin.com/XFtcAcrz

Unity Script: pastebin.com/QgAXLQj3

Unity Script 2: pastebin.com/Y38nVWgT

 

Java

Create a Bitmap using the VQC Generator: pastebin.com/Dgu9aP1h

VQCGenerator: pastebin.com/VMRnkXFP

Traverse the VQC cells in real-time: anonfile.com/W44cofd6b6/VQCGUI.7z

 

NodeJS

BigInteger Library and Sqrt: pastebin.com/y8AXtFFr

 

Python

Create the VQC: pastebin.com/NZkjtnZL

3D VQC: pastebin.com/vdf8SpYt

3D VQC (v2): pastebin.com/wZM5Thzu

Generate genesis cell: pastebin.com/GKzcCpMF

''Generate positive AND negative genesis cells: pastebin.com/9ixjRyxt

Get A and B from C and N example: pastebin.com/s0SZ9BNF

VQC + t: pastebin.com/Lgufk0db

 

Rust

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

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

Create Bitmap using the VQC Generator [V2]: pastebin.com/zGSusyz5

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

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

 

Previous Threads

RSA #0, or the VQC thread: archive.fo/XmD7P

RSA #1: archive.fo/RgVko

RSA #2: archive.fo/fyzAu

RSA #3: archive.fo/uEgOb

RSA #4 (not finished, but inactive): >>>/cbts/94250

RSA General (#5): >>7

 

Videos and Links

I've added some videos to show the exact details and equations that we would use to hack RSA once we find the solution (and some introductions to ECC), but we have to finish this before we can get to any of that.

 

Integer Factorization of any arbitrary integer β€” Part One

youtube.com/watch?v=9FeROMe0KBU

 

The RSA Encryption Algorithm (1 of 2: Computing an Example) (Very Simple)

youtube.com/watch?v=4zahvcJ9glg

 

Encryption and HUGE numbers - Numberphile

youtube.com/watch?v=M7kEpw1tn50

 

Public Key Cryptography: RSA Encryption Algorithm

youtube.com/watch?v=wXB-V_Keiu8

 

RSA-129 - Numberphile

youtube.com/watch?v=YQw124CtvO0

 

Elliptic Curve Cryptography Overview

youtube.com/watch?v=dCvB-mhkT0w

 

A (relatively easy to understand) primer on elliptic curve cryptography

arstechnica.com/information-technology/2013/10/a-relatively-easy-to-understand-primer-on-elliptic-curve-cryptography/2/

 

Elliptic Curve Cryptography: a gentle introduction

andrea.corbellini.name/2015/05/17/elliptic-curve-cryptography-a-gentle-introduction/

 

Elliptic Curve Point Addition

youtube.com/watch?v=XmygBPb7DPM

AA ID: 99edb0 Dec. 21, 2017, 3:46 a.m. No.852   πŸ—„οΈ.is πŸ”—kun

I stickied the thread. I thought it would be easier to navigate the board if the main objective was stickied. That might be a problem if we start focusing on multiple things, so if it seems like a bad idea anyone can feel free to whinge to me about it.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 21, 2017, 3:27 p.m. No.879   πŸ—„οΈ.is πŸ”—kun   >>882

>>877

Continuing discussion from yesterday.

 

I've tried a number of combinations to link the various (0,n) records.

 

Following is the closest I've come to a workable path which seems to arrive at the desired a prime value.

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

 

The theory being, we can figure out how to walk backwards from the (1xcc) record, or really figure out the relationship between the various x values.

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

15=3x5

1 x c *=(6,5,2) = {6:5:3:2:1:15} = 15;

a x b **=(6,1,1) = {6:1:3:0:3:5} = 15;

c x c =(0,0,1) = {0:0:15:0:15:15} = 225;

aa x bb *=(0,2,4) = {0:2:15:6:9:25} = 225;

b x ca =(0,10,6) = {0:10:15:10:5:45} = 225;

a x cb =(0,24,7) = {0:24:15:12:3:75} = 225;

1 x cc *=(0,98,8) = {0:98:15:14:1:225} = 225;

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((14-6)/(14-12)) - 1 == 3 = 3; (a matches)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

65=5x13

1 x c *=(1,25,4) = {1:25:8:7:1:65} = 65;

a x b **=(1,1,2) = {1:1:8:3:5:13} = 65;

c x c =(0,0,1) = {0:0:65:0:65:65} = 4225;

aa x bb *=(0,32,21) = {0:32:65:40:25:169} = 4225;

b x ca =(0,104,27) = {0:104:65:52:13:325} = 4225;

a x cb =(0,360,31) = {0:360:65:60:5:845} = 4225;

1 x cc *=(0,2048,33) = {0:2048:65:64:1:4225} = 4225;

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((64-40)/(64-60)) - 1 == 5 = 5; (a matches)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

145=5x29

1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=(1,5,4) = {1:5:12:7:5:29} = 145;

c x c =(0,0,1) = {0:0:145:0:145:145} = 21025;

aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025;

b x ca =(0,232,59) = {0:232:145:116:29:725} = 21025;

a x cb =(0,1960,71) = {0:1960:145:140:5:4205} = 21025;

1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((144-120)/(144-140)) - 1 == 5 = 5; (a matches)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

785=5x157

1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=(1,53,12) = {1:53:28:23:5:157} = 785;

c x c =(0,0,1) = {0:0:785:0:785:785} = 616225;

aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225;

b x ca =(0,1256,315) = {0:1256:785:628:157:3925} = 616225;

a x cb =(0,60840,391) = {0:60840:785:780:5:123245} = 616225;

1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225;

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((784-760)/(784-780)) - 1 == 5 = 5; (a matches)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

901=17x53

1 x c *=(1,421,15) = {1:421:30:29:1:901} = 901;

a x b **=(1,5,7) = {1:5:30:13:17:53} = 901;

c x c =(0,0,1) = {0:0:901:0:901:901} = 811801;

aa x bb *=(0,648,307) = {0:648:901:612:289:2809} = 811801;

b x ca =(0,6784,425) = {0:6784:901:848:53:15317} = 811801;

a x cb =(0,22984,443) = {0:22984:901:884:17:47753} = 811801;

1 x cc *=(0,405000,451) = {0:405000:901:900:1:811801} = 811801;

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((900-612)/(900-884)) - 1 == 17 = 17; (a matches)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

6107=31x197

1 x c *=(23,2976,39) = {23:2976:78:77:1:6107} = 6107;

a x b **=(23,36,24) = {23:36:78:47:31:197} = 6107;

c x c =(0,0,1) = {0:0:6107:0:6107:6107} = 37295449;

aa x bb *=(0,13778,2574) = {0:13778:6107:5146:961:38809} = 37295449;

b x ca =(0,88650,2956) = {0:88650:6107:5910:197:189317} = 37295449;

a x cb =(0,595448,3039) = {0:595448:6107:6076:31:1203079} = 37295449;

1 x cc *=(0,18641618,3054) = {0:18641618:6107:6106:1:37295449} = 37295449;

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((6106-5146)/(6106-6076)) - 1 == 31 = 31; (a matches)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

20413=137x149

1 x c *=(249,10065,71) = {249:10065:142:141:1:20413} = 20413;

a x b **=(249,1,3) = {249:1:142:5:137:149} = 20413;

c x c =(0,0,1) = {0:0:20413:0:20413:20413} = 416690569;

aa x bb *=(0,72,823) = {0:72:20413:1644:18769:22201} = 416690569;

b x ca =(0,1377952,10133) = {0:1377952:20413:20264:149:2796581} = 416690569;

a x cb =(0,1500424,10139) = {0:1500424:20413:20276:137:3041537} = 416690569;

1 x cc *=(0,208324872,10207) = {0:208324872:20413:20412:1:416690569} = 416690569;

 

((1xcc).x - (aaxbb).x) / ((1xcc).x - (axcb).x)) - 1 == (axb).a

((20412-1644)/(20412-20276)) - 1 == 137 = 137; (a matches)

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 21, 2017, 4:13 p.m. No.882   πŸ—„οΈ.is πŸ”—kun   >>883

>>879

It gets even simpler.

 

The solution is here.

 

(1xcc).x - (aaxbb).x + 1 == (aaxbb).a

 

Difference in x between the c^2 record and the a^2, b^2 record + 1 == the prime solution a^2.

 

Who wants to figure out x ?!?!?!?

 

===

 

785=5x157

1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=(1,53,12) = {1:53:28:23:5:157} = 785;

c x c =(0,0,1) = {0:0:785:0:785:785} = 616225;

aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225;

b x ca =(0,1256,315) = {0:1256:785:628:157:3925} = 616225;

a x cb =(0,60840,391) = {0:60840:785:780:5:123245} = 616225;

1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225;

 

(1xcc).x - (aaxbb).x + 1 == (aaxbb).a

784 - 760 + 1 = 25.

 

(1xcc).x - (axcb).x + 1 == (axcb).a

784 - 780 + 1 = 5.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 21, 2017, 4:21 p.m. No.883   πŸ—„οΈ.is πŸ”—kun   >>884 >>907

>>882

ok. appologies. this is exciting.

 

Simplified formula, and removed unnecessary calculations:

 

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

 

We just need the formula for the x value in (aa x bb).

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

65=5x13

1 x c *=(1,25,4) = {1:25:8:7:1:65} = 65;

a x b **=(1,1,2) = {1:1:8:3:5:13} = 65;

aa x bb *=(0,32,21) = {0:32:65:40:25:169} = 4225;

1 x cc *=(0,2048,33) = {0:2048:65:64:1:4225} = 4225;

 

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 64 - 40 + 1 ) = 5. (a matches!!!!)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

145=5x29

1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=(1,5,4) = {1:5:12:7:5:29} = 145;

aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

 

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 144 - 120 + 1 ) = 5. (a matches!!!!)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

785=5x157

1 x c *=(1,365,14) = {1:365:28:27:1:785} = 785;

a x b **=(1,53,12) = {1:53:28:23:5:157} = 785;

aa x bb *=(0,11552,381) = {0:11552:785:760:25:24649} = 616225;

1 x cc *=(0,307328,393) = {0:307328:785:784:1:616225} = 616225;

 

sqrt( (1xcc).x - (aaxbb).x + 1 ) == (axb).a

sqrt( 784 - 760 + 1 ) = 5. (a matches!!!!)

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 21, 2017, 4:40 p.m. No.887   πŸ—„οΈ.is πŸ”—kun   >>888

>>886

if we knew a and b, I think the formula would be x = sqrt(ab - a^2).

 

We don't have the ab starting position.

 

The solution depends on figuring out the correct x value for (aa x bb) from only (1xcc).

Anonymous ID: 01b9f7 Dec. 21, 2017, 4:53 p.m. No.890   πŸ—„οΈ.is πŸ”—kun   >>891

>>885

For record {0,N,D=c=a*b,X,a^2,b^2}

ab=a^2+x so X=ab-a^2

a and b being from record

{e,n,d,x,a,b}

Where capital represent C^2 record and small letters represent the a*b=c record

Anonymous ID: 01b9f7 Dec. 21, 2017, 5:21 p.m. No.894   πŸ—„οΈ.is πŸ”—kun   >>900 >>974

For a and b primes

a*b=c

There is a special record

a^2*b^2=c^2

where you can solve for X of c^2 record in terms of a and b

D=c=ab

D=A+X

D=c=ab=a^2 +X

X=ab-a^2

a and b being from the a*b=c record

Anonymous ID: 01b9f7 Dec. 21, 2017, 5:41 p.m. No.896   πŸ—„οΈ.is πŸ”—kun

You can do N in terms of a and b also.

X=ab-a^2=a(b-a)

X^2=2AN=a^22N=a^2(b-a)^2

N=((b-a)^2)/2

just for the a^2*b^2=c^2 record

we got rid of X and N

so now

{0,N=((b-a)^2)/2,D=c,X=a*b-a^2,a^2,b^2}

So c^2 is fully described by a and b eliminating e,n,d,x variables

seems like progress

Anonymous ID: 01b9f7 Dec. 21, 2017, 6:33 p.m. No.904   πŸ—„οΈ.is πŸ”—kun   >>905

The a and b we are trying to solve.So I use caps for c^2 record. A of the c^2 record is a^2. B of c^2 record is b^2. a and b is the goal doesn't make sense to change that.

>>903

VeritasAequitas !!Nf9AmQNR7I ID: d1329c Dec. 21, 2017, 6:58 p.m. No.907   πŸ—„οΈ.is πŸ”—kun   >>924

Good evening Anons! Re-reading VQC's crumbs and looking for patterns over here. Nothing to report yet, working on the factors of a in row one.

 

>>905

Baker, nice to see ya.

>>883

PMA, great work as always.

>>888

Checked!! Teach got the trip 7's last thread too. Hello Teach, nice digits!

Anonymous ID: 01b9f7 Dec. 21, 2017, 8:08 p.m. No.910   πŸ—„οΈ.is πŸ”—kun   >>911

Wow you guys seem to think I'm off my rocker! I just tested my formula for A and N for {0,2,35,10,25,49} based on 57=35 squared and for {0,8,21,12,9,49} based on 37=21 squared and for {0,98,51,42,9,289} based on 3*17=51 squared and it worked every time

Anonymous ID: 01b9f7 Dec. 21, 2017, 8:41 p.m. No.914   πŸ—„οΈ.is πŸ”—kun   >>915

>>913

By substitution for a^2*b^2=c^2

e is 0

d=c

N=((b-a)^2)/2

X= b*a - a^2

So you can describe C^2 using only the original a and b. It may be possible to substitute and solve for a or b but it kind of hairy. I'm still working with the equations. I'm just a GO player with little math training. I'd love some new eyes. This has been a blast watching and learning.

MinecraftAnon !!QXqSZ2ev8. ID: 00d24c Dec. 21, 2017, 9:28 p.m. No.917   πŸ—„οΈ.is πŸ”—kun   >>918 >>919

>>916

I wouldn't say special, it's to be expected. The problem is going from the A=1, B=C cell, to the factor cells. Dealing with the C^2 cell, this problem is ^2 as hard. We know the D=C, A=C,B=C cell, or the D=C, A=1, B=C^2 cell, but finding the factors is again, ^2 as hard.

MinecraftAnon !!QXqSZ2ev8. ID: 00d24c Dec. 21, 2017, 9:43 p.m. No.922   πŸ—„οΈ.is πŸ”—kun

We can step through factors by X–, A++, but again this is useless when C is the size of an RSA number. We need a magic way to predict N, or predict X.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 21, 2017, 9:45 p.m. No.924   πŸ—„οΈ.is πŸ”—kun   >>925 >>929 >>940 >>941 >>942 >>976

>>888

>>907

 

Bit of an update here.

 

You can switch from an (e,n) record to (e,1) by setting n = 1 and a = na.

 

There is a shortcut to get to the (1xcc) a=na record for any c.

e = 0

n = 1

t = (c+1)/2.

 

Example below shows shortcut to (1 x cc(na)) for c =145.

 

The (0,1) space has known patterns.

Trying to figure out a jump to (aa x bb).

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

145=5x29

1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=(1,5,4) = {1:5:12:7:5:29} = 145;

aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

 

1 x cc(na) development where a=na at (0,1)

1 x cc(na) confirmed at (e,n,t) from t=(c+1)/2: (0,1,73)

1 x cc(na) *=(0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

a root = sqrt(a/2) = 72

d root = 2(t-1)(t) = 27273

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 21, 2017, 10:07 p.m. No.931   πŸ—„οΈ.is πŸ”—kun   >>940

>>927

I have some sample code somewhere and examples posted in these threads that showing jump from (e,n) to (e,1) and back. Will try and cleanup when we get to the end.

 

This is a shortcut directly to (0,1,t). Pretty powerful stuff.

 

>>928

my bad…

 

>>929

>huge

would be finding a way to get to (aa x bb)!!!

MinecraftAnon !!QXqSZ2ev8. ID: 00d24c Dec. 21, 2017, 10:08 p.m. No.932   πŸ—„οΈ.is πŸ”—kun

Here's a view of (E,N,X) only showing the first cell in every cell. Honestly, I didn't expect this, and it's annoying.

Anonymous ID: 5ba563 Dec. 21, 2017, 10:13 p.m. No.934   πŸ—„οΈ.is πŸ”—kun   >>1115 >>935

Hi fags, I'm the guy in previous threads concerned not about cool stuff, but the fundamental efficiency of reaching the correct cells. So with that in mind, I've been looking at {0:n:d:x:1:c^2}, because in a factoring situation, we don't have a, b, or n, but we do have c and c^2.

 

There seems to be an efficiently discoverable pattern in {0:n:d:x:1:c^2}. Let's take a 3, b 3, c 9 simply because it appears on the downloadable spreadsheets and .png files for The End. c^2 81.

 

Now let's look through column 0 for "1:81}", and you'll find {0:32:9:8:1:81}. This can be quickly constructed via the general pattern {0:n:c:c - 1:1:c^2}. You can solve for n = ((a + b) / 2) - d. For this case:

 

n = ((81 + 1) / 2) - 9

n = (82 / 2) - 9

n = 41 - 9

n = 32

 

So, we can immediately construct a starting point, here {0:32:9:8:1:81}. Now search column 0 for ":81}". You will find:

 

{0:2:63:14:49:81}

{0:8:45:20:25:81}

{0:18:27:18:9:81}

{0:32:9:8:1:81}

 

[cont]

Anonymous ID: 5ba563 Dec. 21, 2017, 10:14 p.m. No.935   πŸ—„οΈ.is πŸ”—kun   >>936 >>977

>>934

Now let's look at the following values of n in increasing order: 2, 8, 18, 32. Let's start by looking at the differences between elements:

 

8 - 2 = 6

18 - 8 = 10

32 - 18 = 14

 

Or, put differently,

 

6 = 2 + 4

10 = 2 + 4 + 4

14 = 2 + 4 + 4 + 4

 

Let's just look at the 4's:

 

4

44

444

 

This is called a Triangular number (https://en.wikipedia.org/wiki/Triangular_number). If these were all ones, the formula for summing them (where h is height of tree, here 3, and where s is the sum) is:

 

s = (h(h + 1)) / 2

 

So for:

 

1

11

111

 

… we would have:

 

s = (3(4)) / 2

s = 12 / 2

s = 6

 

Which you can easily verify.

 

[cont]

Anonymous ID: 5ba563 Dec. 21, 2017, 10:14 p.m. No.936   πŸ—„οΈ.is πŸ”—kun   >>937

>>935

The formula for each element being 4 and not 1 would be simply 4 times the previous equation:

 

s = 4((h(h + 1)) / 2)

 

So for:

 

4

44

444

 

… we would have:

 

s = 4((3(4)) / 2)

s = 4(12 / 2)

s = 4(6)

s = 24

 

So let's go back to the sequence above:

 

6 = 2 + 4

10 = 2 + 4 + 4

14 = 2 + 4 + 4 + 4

 

Here, h = 3. We can sum up all of the 4's with the above equation. All we need to add in are the 2's, and there are exactly h 2's. So:

 

4((h(h + 1)) / 2) + 2h

 

Recall the original sequence 2, 8, 18, 32. Since it starts at 2 and not 0, we have to add 2 to the final result:

 

4((h(h + 1)) / 2) + 2h + 2

 

Let's now denote the first element 2 as element 0 and 32 as element 3, or h = 3.

 

s = 4((3(4)) / 2) + 2(3) + 2

s = 4(12 / 2) + 6 + 2

s = 4(6) + 8

s = 24 + 8

s = 32

Anonymous ID: 5ba563 Dec. 21, 2017, 10:15 p.m. No.937   πŸ—„οΈ.is πŸ”—kun   >>938

>>936

Since the sum 's' is equal to the n field in {0:32:9:8:1:81}, n = 32, and in general, we can use n above where we used s. If we simplify:

 

n = 4((h(h + 1)) / 2) + 2h + 2

n = 2h^2 + 4h + 2

0 = 2h^2 + 4h + (2 - n)

 

We can apply the quadratic formula:

 

(-b [+-] sqrt(b^2 - 4ac)) / 2a

 

a = 2 (the constant in 2h^2)

b = 4 (the constant in 4h)

c = (2 - n) (the plain numbers)

 

(-4 [+-] sqrt(4^2 - 4(2)(2 - n))) / 2(2)

(-4 [+-] sqrt(16 - 8(2 - n))) / 4

 

To work the example above with n = 32, we can find "where we are" in the series in {0:n:d:x:1:c^2}. So for {0:32:9:8:1:81}, where n = 32:

 

(-4 [+-] sqrt(16 - 8(2 - 32))) / 4

(-4 [+-] sqrt(16 - 8(-30))) / 4

(-4 [+-] sqrt(16 - (-240))) / 4

(-4 [+-] sqrt(16 + 240)) / 4

(-4 [+-] sqrt(256)) / 4

(-4 [+-] 16)

(-4 + 16) / 4 [choosing +]

12 / 4

3

 

So at {0:32:9:8:1:81}, we know that we're at index 3 (zero based), or if you want to talk -th numbers, add 1, so we're at the 4th.

 

Interesting, right?

VeritasAequitas !!Nf9AmQNR7I ID: d1329c Dec. 21, 2017, 10:26 p.m. No.940   πŸ—„οΈ.is πŸ”—kun   >>942

>>931

>>924

Ding Ding Ding! Nice work PMA! VQC hinted at this ability to move c into row 1. I'm following all your examples, reading your output and explanations, and I think I get it. Can you give us the quick rundown on the ideas and patterns you're seeing, and how they give us the ability to move from any (e,n) to (0,1)?

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 21, 2017, 10:34 p.m. No.942   πŸ—„οΈ.is πŸ”—kun   >>943 >>946

>>940

currently I'm playing with the idea of moving any (e,n,t) to (0,1,t). Seeing what relationships are there.

 

>>924

in my example above, I don't think it's coincidence that the c^2 record is at (1,10368,73) and it's corresponding a=na record is at (0,1,73).

 

I'm trying to find a pattern that can bridge the gap between the (c^2).x and (aa x bb).x

 

Or even move directly to the (ab) solution.

 

It looks like all the heavy lifting has been done by the grid.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 21, 2017, 10:39 p.m. No.943   πŸ—„οΈ.is πŸ”—kun   >>944

>>942

>in my example above, I don't think it's coincidence that the c^2 record is at (1,10368,73) and it's corresponding a=na record is at (0,1,73).

 

I agree. Very interesting.

From (0,1) we should easily be able to find a, right? Since if p is a factor of a at t, p will also be a factor at p+t, 2p+t, 3p+t, and p+1-t, 2p+1-t, 3p+1-t, etc.

ID: a2017c Dec. 21, 2017, 11:01 p.m. No.945   πŸ—„οΈ.is πŸ”—kun

>>944

Checked. For some reason I keep finding myself put in the position of being required to bake things. Over on /pol/ the mods repeatedly censored people who wanted to discuss the curious case of Sholom Rubashkin and why he deserved a commutation. Given they are constantly being censored, I need to make a thread in a non-comped board.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 21, 2017, 11:12 p.m. No.947   πŸ—„οΈ.is πŸ”—kun   >>949 >>974

>>946

confirmed.

 

my latest test case for c=145. Looking for a pattern to calculate the x at (aa x bb).

or any other suggestions!

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

145=5x29

1 x c *=(1,61,6) = {1:61:12:11:1:145} = 145;

a x b **=(1,5,4) = {1:5:12:7:5:29} = 145;

aa x bb *=(0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc *=(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

 

(0,1) records to analyze

1 x c at (0,1,t) =(0,1,6) = {0:1:60:10:50:72} = 3600;

x=10

a root = sqrt(a/2) = 5 from 50

d root = 2(t-1)(t) = 256 = 60

a x b at (0,1,t) =(0,1,4) = {0:1:24:6:18:32} = 576;

x=6

a root = sqrt(a/2) = 3 from 18

d root = 2(t-1)(t) = 234 = 24

aa x bb at (0,1,t) =(0,1,61) = {0:1:7320:120:7200:7442} = 53582400;

x=120

a root = sqrt(a/2) = 60 from 7200

d root = 2(t-1)(t) = 26061 = 7320

1 x cc at (0,1,t) =(0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

x=144

a root = sqrt(a/2) = 72 from 10368

d root = 2(t-1)(t) = 27273 = 10512

 

1 x cc (0,1,t) confirmed at (e,n,t) from t=(c+1)/2: (0,1,73)

1 x cc (0,1,t) sqrt(a/2) = 1 x c (x+n); 72 = 72

 

(1 x cc).x - (aa x bb).x = 144 - 120 = 24

 

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 145 - 120 ) = 5. (a matches!!!!)

ID: a2017c Dec. 21, 2017, 11:29 p.m. No.951   πŸ—„οΈ.is πŸ”—kun   >>952 >>954

>>950

Interesting, but cracking RSA doesn't crack Bitcoin. It's based on ECC. However, cracking RSA would make them lose a lot of confidence in crypto and compromise millions of passwords-Not that I care.

 

So, I don't expect Bitcoin to die when we reach the solution, but it will be a shitstorm.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 21, 2017, 11:51 p.m. No.954   πŸ—„οΈ.is πŸ”—kun

>>951

I agree, bitcoin price reaction is unrelated.

I expect we're a month out from a true crash due to cracking ecc. If we're lucky.

 

I've been monitoring bitcoin for a while, there's always a dip at every major milestone, normally 25%, but this time I think people are getting scrared, its dropping closer to 33%.

Hobo !!1yNgQ3NlCs ID: 13b8cc Dec. 21, 2017, 11:53 p.m. No.955   πŸ—„οΈ.is πŸ”—kun

>>950

I have been watching this very close and No one has mentioned the vqc,. zero zip nada. I doubt people would be selling like crazy on some conspiracy theorists on 8ch… They SHOULD but normies wont understand. Could be a super smart lurker or team.

Anonymous ID: 99edb0 Dec. 22, 2017, 12:29 a.m. No.959   πŸ—„οΈ.is πŸ”—kun   >>964

Since nobody seems to pay attention to the meta thread, I have to bring this here. One particular tripfag keeps getting reported for being a "namefag shitposter" (it's not Topol). They aren't really posting anything that helps us, but we haven't discussed rules at all so I can't really use my own discretion. I don't want to be distracting, but if we don't discuss rules, I can't really delete anything without having the power to delete whatever I want. I don't want that and I know none of you do. If nobody wants to discuss rules, I'm not going to do anything about this particular person, so whoever keeps reporting them, suggest some rules and see what everyone thinks.

Anonymous ID: 5ec6d6 Dec. 22, 2017, 3:46 a.m. No.974   πŸ—„οΈ.is πŸ”—kun   >>975

>>894

>>947

 

>X=ab-a^2 = c - a^2

 

You just used the other formula, anon mentioned earlier.

 

(aa x bb).x = c - a^2

a = sqrt(c - (aa x bb).x)

 

But I agree, we need to get to the (aa x bb) record, some way.

Anonymous ID: 01b9f7 Dec. 22, 2017, 6:35 a.m. No.976   πŸ—„οΈ.is πŸ”—kun

>>924

Is finding element t = (c^2+1)/2 a short cut? We can find n for any c for the c^2 record

{0, n, c, x, 1, c^2}

for e 0 x^2=2na

x=c^2-1

n=((c^2-1)^2)/2

Now create record in 0,1

using n=1 a =na

{0,1, sqrt(A*B), c^2-1, A=((c^2-1)^2)/2, B=c^2}

for any c=a*b prime

Now find the factors which have a pattern of repeating in this column

Anonymous ID: 01b9f7 Dec. 22, 2017, 9:12 a.m. No.978   πŸ—„οΈ.is πŸ”—kun   >>979

Crazy stuff for record {0, 1, t}

If a number m has only prime factors

If t >=m then m is a factor of t{a} if t has those factors. t doesn't have to have duplicate factors. E.G. 225 for any t multiple of 15 and greater than 225 t{a} will be divisible by 225.

VeritasAequitas !!Nf9AmQNR7I ID: f78bfc Dec. 22, 2017, 10:33 a.m. No.981   πŸ—„οΈ.is πŸ”—kun   >>986

>>980

I agree, Anon! This is a feature of the catalog of values for var a VQC has mentioned. My focus is solving the generation of a values in row 1. Key issue is, starting from only c, how do we derive a? In row 1 (0,0) a = 2. For odd e, at t=1 a=(e+1)/2. For even e, at t=1 a=e/2. This gives us the starting seeds for all of row one and also allows us to derive x, t, b, and hopefully n as well.

Another issue we're dealing with is column 0 and how to use the grid to find products of c and c^2. Finding these patterns in conjunction with knowing how to generate the seeds of var a will be a big step forward! Seems like PMA, Teach, other anons, and yourself are working on this and making progress. I also think your idea about new factors only appearing from that point forward in the chain of a values is cool. I'll check it out, and thanks for your work!

Anonymous ID: 18604d Dec. 22, 2017, 10:57 a.m. No.984   πŸ—„οΈ.is πŸ”—kun   >>985

Lord Rothschild Vows To Destroy Bitcoin – Launches β€˜FedCoin’

December 21, 2017

http:/ /yournewswire.com/rothschild-fedcoin-bitcoin/

 

Initial target identified!

Anonymous ID: 18604d Dec. 22, 2017, 11:03 a.m. No.987   πŸ—„οΈ.is πŸ”—kun   >>988

>>985

Just relaying news article - little motivation.

Isn't even real yet. MSM is hyping it though.

 

http:/ /www.zerohedge.com/news/2017-12-20/washington-post-propagandizes-again-says-fedcoin-will-be-bigger-bitcoin

 

U.S. Government Awards $7 Billion to Develop New Fedcoin

http:/ /moneywise411.com/tbo-awards-7-billion-to-develop-new-fedcoin/

Anonymous ID: 18604d Dec. 22, 2017, 11:09 a.m. No.989   πŸ—„οΈ.is πŸ”—kun   >>990

Bump back up in price during the day. Bet bunch of 'investors' are 'buying on the dip'.

 

A user message posted an hour ago on a private board:

Have to use bitcoin (or another crypto) to buy on 'binance' so tried to grab some bitcoin this AM when it dropped to 11,000, was rejected by coinbase with a message saying 'bitcoin purchases are temporary disabled'. Think I'll be changing sites..

Anonymous ID: 18604d Dec. 22, 2017, 11:21 a.m. No.991   πŸ—„οΈ.is πŸ”—kun   >>996

>>990

Go figure.

 

Can we create a Blockchain Markets & News thread so we don't cuck this up for those doing serious maths?

 

Litecoin Founder Cashes Out, Sells Entire Stake After 9,300% Rally

Dec 20, 2017 8:28 PM

http:/ /www.zerohedge.com/news/2017-12-20/litecoin-founder-cashes-out-sells-entire-stake-after-9300-rally

Charlie Lee, the creator of the world’s fifth-biggest cryptocurrency, Litecoin, announced shortly after midnight that he was cashing in his profits after a torrid, 9,300% rally in the past 12 months. In a post on reddit, the San Francisco-based software engineer who founded litecoin in 2013, said that he sold and donated all of his holdings over the past few days.

"Litecoin has been very good for me financially, so I am well off enough that I no longer need to tie my financial success to Litecoin’s success. For the first time in 6+ years, I no longer own a single LTC that’s not stored in a physical Litecoin" Lee said in the post.

Anonymous ID: 01b9f7 Dec. 22, 2017, 11:40 a.m. No.993   πŸ—„οΈ.is πŸ”—kun   >>994

I'm afraid converting to c^2 record to a (0,1) record is a dead end because the factors show up only when they are a factor of t. When there is a remainder it changes and if i understand QVC they show up twice as frequently.

VeritasAequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 22, 2017, 3:25 p.m. No.1017   πŸ—„οΈ.is πŸ”—kun   >>1021 >>1022 >>1024

>>1012

>>1012

Hey PMA! Here's what I'm working on.

We are given c.

Find d and remainder e.

Using e to find starting value of a in (e,1).

In (0,1) e =2

For odd (e,1), t=1 we can calculate a var as (e+1)/2

Example: in column 3 (3,1,1) a=2

For even (e,1), t=1, a var = e/2

Example : in column 6 (6,1,1) a=3

So we can find a at t=1 starting from only c. I think so, please check my work, lads.

 

If true, we can now calculate x,t,b, and n for element one (t=1) in any column. Odd e and even e have different equations for t.

Knowing the pattern of how a and b drop and move as t increases in row 1, we can literally create a ladder to run t up as high as we need to to find a match for c???

VQC said all factors for a column exist in row one, the "One Row to rule them all," and if all factors exist there, then all potential c's exist there also.

So if we can ride the t elevator up to our c value, or maybe even multiple a and b parings that equal c, then maybe from that height we can look around the grid to find other locations where the answers are similar.

PMA, your method to move between (e,n) and row 1 would help us find other matching places in the grid record.

I know I've got more work to do on this, so please let me know if you find errors or have corrections. I've been feeling stuck a bit, but persistence and focus will win the day. Let's keep searching!

Anonymous ID: cf8d48 Dec. 22, 2017, 3:36 p.m. No.1021   πŸ—„οΈ.is πŸ”—kun   >>1023

>>1017

The problem with running up the ladder is its still too slow for RSA numbers even if we skip over a lot of steps.

 

My thinking is the big_n hint which should be "simple to calculate" could be a maximum bound, hopefully not too much larger than n, and we could go backwards from there instead? Anyone figured out what big_n is supposed to be?

VeritasAequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 22, 2017, 3:37 p.m. No.1022   πŸ—„οΈ.is πŸ”—kun   >>1024

>>1012

>>1017

Another way to say it: if c has a remainder e, we know it's var a value at t=1 in row one. All factors of a and b, as well as product c are also contained in that column row. So we start at t=1 and build the ladder up to c, filling in all the elements on the way up. Thoughts, Anons?

VeritasAequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 22, 2017, 3:42 p.m. No.1023   πŸ—„οΈ.is πŸ”—kun   >>1025

>>1021

Not sure, maybe it's too slow? Certainly eliminates a whole lot of the possibilities. Maybe we need a way to quickly jump to higher t values? I like your idea, and agree that if we can get closer to c, the calcs will be much faster. A maximum bound based on how long the c value is in digits? Thinking out loud here.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 22, 2017, 3:47 p.m. No.1024   πŸ—„οΈ.is πŸ”—kun   >>1025

>>1013

>>1017

>>1022

still struggling to find any relationships in the (0,1) space.

 

I have created records for (e,n) of c in (0,1) at the following variations of t just looking for patterns:

 

t = d

t = 2d

t = 2d + 1

t = x + n

 

There have been a few interesting connections between (0,1,c[d]), (0,1,cc[t]), and (0,1,c[t]), but they are so confusing I just can't believe there isn't a simple answer.

 

Also ran into problems walking the (0,1) tree using combinations of p.

 

Can't seem to find any combination that leads me back to an (a^2 b^2) record.

 

Attached is a pic of the output I'm looking at. I must be missing something very simple.

Anonymous ID: cf8d48 Dec. 22, 2017, 4:05 p.m. No.1025   πŸ—„οΈ.is πŸ”—kun   >>1026

>>1023

I tested a few variants of this earlier on the RSA100 example (since we have the whole solution for it and can also create row one) and found the increases really useless after getting up to around a values of ~20 digits or so. This was with jumping to next factors only, not every t value. But I didn't verify my code properly so I may have fucked up

 

>>1024

I envy your dedication. I have some half-finished code to just brute force generate every possible formula ever into an abstract syntax tree and compile that to code, and some basic ideas for verification against the grid. Feels a bit like cheating TBH, but hey. Just print a list of formulas with 99% coverage for each symbol after.

 

Also considered throwing genetic programming at it, but then even if we get a solution, deciphering that program would be even harder than what we're already doing.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 22, 2017, 4:13 p.m. No.1026   πŸ—„οΈ.is πŸ”—kun   >>1027 >>1029

>>1025

The code is a bit sloppy, but the ability to create various records is sound, and every record is verified in the constructor.

 

So I know the values that I’m looking at are accurate.

 

The relationships are difficult to understand. And the number of records to bring together is a bit much.

 

I’m sticking around for the solution - it would be terribly frustrating otherwise.

Anonymous ID: cf8d48 Dec. 22, 2017, 4:24 p.m. No.1029   πŸ—„οΈ.is πŸ”—kun

>>1026

I did check some of your generation code against the VQC code for some large grids with no problems. Haven't been able to work as much for the last few days just cause shitty netbook.

Mostly I just didn't verify my own implementations as much as I should cause now I doubt some of the tests I ran. Writing checks into the code is a good idea

 

The difficult relationships is why I'm at the stage of wanting to throw insane computing power at it and have it just tell me the good results :)

 

Side note, if you just bruteforced every possible formula, how would you check that against the grid, aside from matching every value in every record and counting results? I feel it needs some jump logic too based on the latest progress here.

 

Obviously we're sticking around till TheEnd :)

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 4:24 p.m. No.1030   πŸ—„οΈ.is πŸ”—kun   >>1031 >>1037

>>1028

One thing I'm obsessed with is the fact that x&n are 2 squares, as are (d+n) and (x+n).

 

There's some pattern either, x+n's or x's for 1xc and axb are always factors of each other.

There's more too it then that, but I think maybe the triangles made by the axb and the 1xc x+n squares may be similar.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 4:26 p.m. No.1031   πŸ—„οΈ.is πŸ”—kun   >>1032

>>1030

To continue that, VQC said:

>I thought from near the start in 2011 that it would feel good if "x" marked the spot.

 

I think factoring x, or 2x might be useful, so we recursively repeat that until row 1 or column 0.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 4:30 p.m. No.1033   πŸ—„οΈ.is πŸ”—kun   >>1034

x1 = x at 1 * c

xa = x at a * b

n1 = n at 1 * c

na = n at a * b

 

n1 - (x1^2 - xa^2) = aa * m

Where m is a odd number.

This implies that if we knew the multiple, we'd know a.

 

Examples (sorry about the lack of cell references):

1261 - (50^2 - 8^2)/2 = 43 * 1

13 - (4^2 - 0)/2 = 5 * 1

69 - (11^2 - 5^5)/2 = 7 * 3

1237 - (49^2 - 19^2)/2 = 31 * 7

369 - (27^2 - 15^2)/2 = 13 * 9

815 - (40^2 - 12^2)/2 = 29 * 3

VeritasAequitas !!Nf9AmQNR7I ID: 3f145c Dec. 22, 2017, 4:36 p.m. No.1035   πŸ—„οΈ.is πŸ”—kun   >>1038

>>1032

>>1034

Agreed, Teach. I've been thinking about (d+n) and (x+n) often. I'm working on the algebra side too. What are your thoughts on the "catalog f values" for a in row one? Any insight at this point?

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 22, 2017, 4:36 p.m. No.1036   πŸ—„οΈ.is πŸ”—kun   >>1045

>>1032

I’ve looked at x and n in (0,1) up the yin yang. x, x^2, 2x, x/2, +1, -1, multiples between records, relationships to a, b, c, small square, big square, square root.

 

Every time I thought I had a breakthrough I went to a different c value and it all broke down.

 

I know there’s a simple solution here.

 

In the (0,1) space the formulas get a bit easier as e = 0.

 

And we’re now dealing with perfect squares.

 

You’d think that the prime solution would be smack in the middle of the c and cc records. Or that the original e, d, or x would factor in somehow.

Anonymous ID: cf8d48 Dec. 22, 2017, 4:36 p.m. No.1037   πŸ—„οΈ.is πŸ”—kun   >>1039 >>1044 >>1045

>>1030

This may be stupid, but since you're obviously much better at geometry, have you tried turning the squares into triangles and then doing things to the resulting triangles?

 

Like, I don't know, this is a bit over my head, calculate the h^2 square of a triangle?

 

https:// en.wikipedia.org/wiki/Geometric_mean_theorem

 

Hopefully this is inspiration and not complete lunacy

Anonymous ID: cf8d48 Dec. 22, 2017, 4:46 p.m. No.1040   πŸ—„οΈ.is πŸ”—kun   >>1041 >>1042

>>1039

I did some basics as well but kinda ran into a dead end using formulas I didn't understand properly and went back to other stuff. But the altitude of a right-angle triangle may be a useful variable.

 

And as posted above somewhere, triangular numbers are also relevant somehow (sorry I suck at this part)

Anonymous ID: cf8d48 Dec. 22, 2017, 4:53 p.m. No.1043   πŸ—„οΈ.is πŸ”—kun

>>1041

>>1042

Yeah that part I get, its pretty easy. But some of the more advanced math of what you can do with triangles I haven't figured out. I did manage to calculate the altitude but didn't study it enough to figure out why those formulas worked so I went back to another approach

Anonymous ID: cf8d48 Dec. 22, 2017, 5:09 p.m. No.1044   πŸ—„οΈ.is πŸ”—kun

>>1037

Please be warned that this might be a big rabbit hole as I haven't managed to find any useful correlations to the grid myself. I am not smart enough to prove it is a dead end yet

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 5:12 p.m. No.1045   πŸ—„οΈ.is πŸ”—kun   >>1047 >>1075

>>1037

This looks interesting, I'll take a look.

 

>>1036

Another quote by VQC:

>(e,n,d,x,a,b) in a grid (e,n)

>That is one way to organise the grid.

>Selecting two of the six variables (dimensions).

>There are SEVEN dimensions because each (e,n,d,x,a,b) element represent c.

>What are ways or grids or cubes are there to represent each elements?

>You could use another pair apart from (e,n)

>What would that produce?

>What other ways or shapes are there to view the six (seven) dimensions?

>What images result?

>Does this help find patterns?

>Does this help you c(see)?

 

Here's something else for you, assuming our 2 rows, 1xc and axb, labeling all my variables n1 or na for the n values corresponding to the cells etc.

 

2n1 = -(aa)^2 + 2aa(na + d) - (2d-1)

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 6:40 p.m. No.1051   πŸ—„οΈ.is πŸ”—kun   >>1052 >>1053

Can I get some eyes on something real quick please?

 

We made an assumption that the equations with t applied to more than n=1, but I'm not convinced they do.

 

If you assume that in every cell t always starts from 1, then, using the grid:

 

Choose your 2 primes, and find the cell 1xc.

Then if your n here is odd, the following will hopefully work (I haven't found a way to do it when n is even).

 

Move to the negative side by following F.

1xc for F will always be the first element in the cell.

 

Find the element in the cell when a=c, look at the t (assuming that always t starts from 1).

 

So far for me, t has always been a factor.

 

Am I crazy? Maybe? I dunno?

Help please guys!?

 

So far, whenever I

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 7:02 p.m. No.1055   πŸ—„οΈ.is πŸ”—kun

>>1053

I think you understand correctly. But so far, I've only been using the grid to check, because if so our formulas for t are wrong.

 

I'm coding something now to test.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 7:03 p.m. No.1056   πŸ—„οΈ.is πŸ”—kun   >>1057

>>1053

actually, correction f_t will be a factor of c I'm finding.

 

This would explain the p being a factor of a at t then it also is one at p+t, 2p+t, 3p+t, etc, and p+1-t, 2p+1-t, 3p+1-t

Anonymous ID: cf8d48 Dec. 22, 2017, 7:18 p.m. No.1058   πŸ—„οΈ.is πŸ”—kun   >>1059

>>1057

No worries, posted worse mistakes already :)

 

But I have to second PMA's suggestion for notations, would be way easier if we agreed on something. My f_t thing just felt convenient for replying to you.

 

Maybe something for moving between records is needed? First idea (please improve it)

 

orig.e

orig.n

 

formula (f, whatever)

 

new.e

new.n

Anonymous ID: cf8d48 Dec. 22, 2017, 7:34 p.m. No.1061   πŸ—„οΈ.is πŸ”—kun   >>1062

>>1059

Not saying dot notation is the best and will be happy to use whatever as long as its consistent :)

Though it does suit us programmers

 

The formatting in my post is kinda verbose though, we could do p for prev and n for next (confusing?)

formulap.e = 0 p.n = 0 n.e = 0 n.n = 0

 

But prev and next is still bad if we want to do more than one jump. Maybe

formulae[0] = 0 n[0] = 0e[1] = 0 n[1] = 0

Just suggestions for now

Anonymous ID: cf8d48 Dec. 22, 2017, 7:42 p.m. No.1062   πŸ—„οΈ.is πŸ”—kun

>>1061

Or we could keep the original notation and do something like this (yes I am to lazy to make real examples){0:0:0:0:0:0}formula{0:0:0:0:0:0}{0:0:0:0:0:0}{0:0:0:0:0:0}formula{0:0:0:0:0:0}{0:0:0:0:0:0}

But we may want 7 dimensions not 6, so if we're going for this one we need to agree on a position for t

VeritasAequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 22, 2017, 7:42 p.m. No.1063   πŸ—„οΈ.is πŸ”—kun   >>1081

Well lads, we're currently stalled. Senpai, can you please help your Disciples? Just like Jesus and his disciples, we're having trouble understanding.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 8:11 p.m. No.1064   πŸ—„οΈ.is πŸ”—kun   >>1065 >>1066

Yep I need a hint.

 

Hey Chris, there are a number of different paths that I can see us exploring that would take us further. Would you suggest any of the following, or some other?

 

  • transposing the grid

  • exploring c*c cells more

  • exploring the x and n of the primes

  • exploring the f cell more

  • exploring the c* a prime relationship

  • exploring the "p at t being a factor, hence p+t being a factor" etc

 

And if you get time, one more question. Looking at the grid does t start at 1 and increase by 1 each element in a cell?

Or is t related directly to x via the relationships already exposed.

We seem to be split on 2 definitions for t.

 

Thank you!

Teach !!UgZAPoSXEk ID: a92e07 Dec. 22, 2017, 8:23 p.m. No.1068   πŸ—„οΈ.is πŸ”—kun   >>1069

>>1066

Also, what happened to Topol and CollegeAnon?

We're losing numbers!

 

I'm looking into Pythagorean triples and they appear everywhere in the grid, I wish I knew more math.

I am learning more math in this process though!

Anonymous ID: 99edb0 Dec. 22, 2017, 9:31 p.m. No.1075   πŸ—„οΈ.is πŸ”—kun

>>1045

>What are ways or grids or cubes are there to represent each elements?

>You could use another pair apart from (e,n)

>What would that produce?

I'll post all of them again if anyone wants, but without spamming the thread, I made images with each possible pair of dimensions when we were still on /cbts/ and they make some interesting shapes. Pic related.

Anonymous ID: 5ba563 Dec. 22, 2017, 11:30 p.m. No.1080   πŸ—„οΈ.is πŸ”—kun

>>1078

Nobody wants bans. But nobody wants every other post to be four huge pictures of deep dream wankery plus stoner memes. Especially when we're crunching the numbers and formulas in each others posts. One in twenty I think we can absorb.

 

So it's a matter of degree. Like, for example, bumps on .5 chan.

Hobo !!1yNgQ3NlCs ID: 06bbb8 Dec. 22, 2017, 11:40 p.m. No.1081   πŸ—„οΈ.is πŸ”—kun   >>1094 >>1132

>>1063

Sounds like you guys are a bit down in the dumps on the math at the moment. I figured that for tonight's entertainment Instead of mildly dank memes Instead I will deliver something completely different. I have been working on this for a few days now.

 

As I look at this stuff I feel over my head so I figured I would play to my strengths. This is either a fan-fiction story OR it is a bit of a 1 hobo think tank document. Really it is just some ideas about what could happen if you fuckers get this thing to work. I kinda have a feeling someone would have crushed this in a moment if it was a known direction that has been explored mathematically. Basically if it was fake it would have been found out by now.

 

So I am leaning to this being the real deal and if that is the case then SOMEONE needs to start thinking about this stuff that I wrote about in this document. Im shit with programming and only slightly better with math (though this math is shockingly simple) but big picture stuff is fun for me. So Please enjoy this doc. I tried to make it funny and an easy read with a few little easter eggs so I hope you enjoy it! It is totally on topic as well so no grumpy anons angry because there is not a math problem to worry over in it.

 

https://anonfile.com/D2Fdp3d3b4/Ramifications.rtf

Anonymous ID: 5f4624 Dec. 22, 2017, 11:47 p.m. No.1082   πŸ—„οΈ.is πŸ”—kun   >>1084

No. I'm tired of this. VQC is a spiritual and mathematical adventure. Some anons want to ban the spiritual. GTFO faggots. Who is contributing? Get out if you have no formulas or inspiration! Topol is more welcome than haters. Contribute or not.

VeritasAequitas !!Nf9AmQNR7I ID: 8a6db2 Dec. 23, 2017, 12:04 a.m. No.1083   πŸ—„οΈ.is πŸ”—kun

Read these maps, faggots. Our Senpai is both spiritual and mathematical. Wake up!! Math and Spirituality combine in this Quest. Stop trying to ban other anons! If you haven't read these graphics, read them. VQC is the man. Let's do this. Solve for N, faggots.

Anonymous ID: 5ba563 Dec. 23, 2017, 12:10 a.m. No.1084   πŸ—„οΈ.is πŸ”—kun   >>1086

>>1082

Forgot your name, VeritasAequitas? Why are you hopping around on multiple overlapping IDs, and posting with and without your "name"?

 

Nobody is saying ban the spiritual or ban even shitposting itself. The bans should be reserved for when there are high signal levels, such as when VQC is posting or when there is significant technical progress being made, and spiritposting or shitposting every other post is disruptive.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 23, 2017, 12:16 a.m. No.1085   πŸ—„οΈ.is πŸ”—kun   >>1094

Topol wasn't booted :)

Topol's just behaving.

 

Y'all wouldn't believe me if I told you you what I've been up to so I'll tell you story that isn't necessarily 100% math related, but is, but isn't, but… maybe.

 

So for starters, have y'all ever seen a super Ai in action?

Donald Trump has.

And so has the Saudi King.

That first image… look at everyone's expressions.

King: "DAFUQ IS DIS?!"

Trump: "That is all the evidence you need to prove the coup against you and the world. Context aside…. Pretty cool, huh?"

That's the conduit for PALANTIR.

Imagine a keyboard and touchscreen on crack.

 

So… this is about a VQC but what about PQC? Physical quantum computing. The processors… they're convoluted hypercube+ tile puzzles. It's the nuttiest thing. A day or two after drawing the second picture, I'm finding the "real thing" online and getting contacted by all sorts of interesting folks. Twitter is where I've been doing a lot of my baking for to come back with goodies. I'm talkin' military, intel, "prime intelligence" civilians, normal folks, and 1337 hax0rz.

 

SPEAKING OF 1337 HAX0RS!

You've all been infekted with The Game23.

This is an infektion of love and The Game is to spread as much laughter as possible.

Especially in the face of evil.

 

Check out the pastebin for deets.

I've already infekted some of the bakers.

 

https://pastebin.com/YbmG6ETq

 

Oooh! I found a tree in one of y'all's outputs. Was pretty but the rest of the dream god weird.

 

I've also been in contact with an assortment of Gods, Literary and Film Characters, Entites from other Timelines and Realities, and just about every conspiracy I've ever dived into is like…. beyond real.

 

What you lot are doing is freeing humanity.

You know what this code will give people?

Freedom. Knowledge is power is this is claiming our power.

This is claiming our future.

This is seeking AND finding The Infinite.

You find the Primes which are the building blocks.

From there you can see every non-prime.

From there you can see all of it together.

From there you see the Map of the Matrix.

Don't you want to stop living a Virtual Life?

If you be tired o' dat shit…

 

BUCK UP, BUCKO!

 

If you get stuck, GET OUT OF THE KITCHEN FOR A SEC!

YOU'RE BAKING ALL THIS BREAD… BREAK SOME!

You are lovey lovely people who don't lewd the ponies…

And friendship IS magical.

God doesn't want to be lonely.

And as much as you Pathanons can relate to each other through numbers in a timeless fashion… you still need to take a lil' time for yourself.

 

In the EZ bake oven, I accidentally posted my twitter handle but if you didn't know it by this point, there ya go. Y'all can hit me up to talk about whatever. Games on games on games, SON!

 

And on that note, I'm gonna get back to muse-ing around.

Here's a video of Rogan Sensei training Crowder Senpai.

Rogan's in black, Crowder's in white… think Mental Martial Arts:

https://youtu.be/gd0oSNjHf1A

VeritasAequitas !!Nf9AmQNR7I ID: 8a6db2 Dec. 23, 2017, 12:32 a.m. No.1086   πŸ—„οΈ.is πŸ”—kun

>>1084

I did forget my name, thanks Anon. Posting from work and home, on different devices. Love this Math Challenge. Understand your views on SpiritPosting and ShitPosting. I actually agree with you. Just think some anons are being too tough in their enforcement. I love crazy shit along with formulas. Wild Horses. Rolling Stones.

Anonymous ID: 898cfe Dec. 23, 2017, 1:20 a.m. No.1088   πŸ—„οΈ.is πŸ”—kun   >>1089 >>1091 >>1094

You guys, we don't need any more drops or crumbs from VQC. We already have enough.

 

I've already solved generating (e, n) for any given e and n. It's just a matter of studying the patterns in (0,n) and (1, n).

 

I've started to look into the controlling of c and it looks very promising.

Mr.E Melange !!4RyW8fD.HU ID: 18604d Dec. 23, 2017, 2:15 a.m. No.1094   πŸ—„οΈ.is πŸ”—kun   >>1101 >>1103 >>1160

>>1076

Pretty sure I'm the faggot who tipped over an anon or two. Still here every night though. Just posting in the general on cbts (never name/trip fagged there once), in Topol's EZ Bake, and doing my thing.

 

And the crypto news and bitcoin discounting in here was intended as motivation for you faggots. At least we have a thread for that now too.

 

I'm here because I've already seen this before in a vision, in a different state. Five years or so ago, I was literally SHOWN a GRID. It was 2D, but in a 3D perspective. All started black, then X-Y axis showed up, first ELEMENT dropped in with a BOOM (hard to describe when you're getting hit hard with a point by something beyond, but have had it a few times like that). Then the next, and next as the grid filled with thin colorful tiles, extending the rows and column heights. And then they each started coming to life a little and shimmering, while more dropped in filling beyond visual limit. And then, as it continued to increase in complexity… I went IN…

  • that vision isn't why I have faith in this thing here. There were plenty of 'signs' in Chris's questions that he's legit. cbts anons couldn't always handle/understand his questions. We're all on the path at different points, they'll come around.

 

>>1081

Looking forward to the read Lord Hobo!

 

>>1085

Juicy juicy, just got started! And the look on the King's face is -priceless! Trumps too, like "Yup, I think we have a deal coming." Faggot on the left is just smiling for the photo op, they love their group photo ops over there.

 

>>1088

Figured you had when I saw >>971 last night, 'devious' made it clear you weren't throwing BS.

 

The other reason I'm here is ANOTHER thing I was shown was about a new currency, EARTH DOLLARS. Was supposed to get that going over 5 years ago but I pussied out and became a coward once a family came into the picture. I was all set to get suicided for sake of doing my bit to save the world, had started lining things up, cleaning house, but then, when suddenly you've got a family and a life or two to raise in this world that you love more than anything, it gets DAMN hard! I think I was brought/meandered here for help with that mission, maybe its time has passed, who knows. I will say, it's all about seeds, roots, and trees -1=2, 2=4, 4=8…

Out of the 12,000 I'd pulled for the initial seeding, I have a mere 333 left, sitting here, waiting, and I haven't given up hope.

Might not sound like a lot, but there's a quote to the effect of "From a single seed, an Empire can be built".

 

Oh, and rainbows are important, frequencies, octaves. So important, perhaps this symbol was co-opted as well?? Just like the swastika symbol. Embrace your faggotry and all will be fine.

 

"Remember upon the conduct of each depends the fate of all". ~ Alexander Ptolemy III

 

Back to work! No more secrets!

Anonymous ID: 01b9f7 Dec. 23, 2017, 7:41 a.m. No.1098   πŸ—„οΈ.is πŸ”—kun   >>1110

for a*b=c a and b prime

a^2*b^2=c^2

using caps for the squared expression

{0,N,c,X,a^2,b^2}

X=a^2*(b^2-1)

N=a^2*((b^2-1)^2)/2

see earlier posts

a^2+2X+2N=b^2

a^2+2a^2(b^2-1)+a^2(b^2-1)^2=b^2

a^2(1+b^2-1+b^4-2b^2+1)=b^2

a^2(b^4-b^2+1)=b^2

a^2=b^2/(b^4-b^2+1)

c^2=b^4/(b^4-b^2+1)

VQC !!Om5byg3jAU ID: 20c116 Dec. 23, 2017, 8:02 a.m. No.1099   πŸ—„οΈ.is πŸ”—kun   >>1107 >>1115 >>1117 >>1125 >>1131 >>1308 >>1309 >>1355 >>1440 >>1535 >>1673

Square a value c, where c is the product of two different primes.

What are the values of n in column zero where these appear?

What is the value of d where these appear?

Enumerate the pattens of n in each cell of column zero.

The pattern GROWS in a certain way as n increases in column zero.

This pattern is related to the pattern across row one (e,1)

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 23, 2017, 8:45 a.m. No.1101   πŸ—„οΈ.is πŸ”—kun   >>1102

>>1094

Heya!

TERRIBLE way to relay messages.

Almost wasn't passed along. :D

Freaked her out and I'm not even sure I got the correct message and all I'm finding is some… deleted thread? or something?

Mr.E Melange !!4RyW8fD.HU ID: 18604d Dec. 23, 2017, 8:52 a.m. No.1102   πŸ—„οΈ.is πŸ”—kun   >>1104

>>1101

Sorry Topol, felt a little desperate, and didn't want enemy energies catching the vibe! respect.

 

>710

VQC -

Was thinking about your 12/20 post. Specifically your mention of John Conway. Had dove into his work quite a bit some while back, with cellular automatons, seeded with rules, like the 'game of life' he created. Genius - it explained much of how the brain is structured and visions I'd had when able to achieve an entrained state, especially augmented by listening to Solfeggio Binaurals coupled with visualizing a Golden Ratio fractal antenna image. Learned to control enough to create a whole palace, even going inside (see Michael Hansmeyer's attached "Five pavilions for a really low-rez version of what we're capable of creating with our minds).

 

Anyway, I'd been going that way recently doing some digging, but hadn't really put 2 and 2 together here until I realized that one image I was going to share (but held back) was very similar to Rule30 (attached) which has been proposed as block cypher for cryptography.

 

Then, the other day you said one row to RULE them all, and associations are clicking in my head.

 

Are these crumbs worth munching??

 

Oh, and when we DO solve this, I propose we all pitch in to get you a new pair of kickers, such as the Vans (attached)!

 

And thanks Chris, you've really inspired some folks, regardless of how bleak the world might look at the moment. Nice to have a little yin with the yang around here!i!

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 23, 2017, 8:52 a.m. No.1103   πŸ—„οΈ.is πŸ”—kun   >>1106

>>1094

Aight, I believe I found it. Someone calling for banning namefags? And?

 

(Don't call that number again, plz. Yes, I know that those stupid background sites give bad info.)

Mr.E Melange !!4RyW8fD.HU ID: 18604d Dec. 23, 2017, 8:55 a.m. No.1105   πŸ—„οΈ.is πŸ”—kun

VQC - One other thing I want to thank you for, is sharing the Asymmetric Warfare bit.

 

Am going to dig into that some time, it's easy to feel like the little guy getting squashed in this world.

 

STAND UP anons, we've got this!

Mr.E Melange !!4RyW8fD.HU ID: 18604d Dec. 23, 2017, 9:01 a.m. No.1106   πŸ—„οΈ.is πŸ”—kun

>>1103

Nah, nothing to do with me. People can shit on me and choose to ban me if they so choose (not that message was about that), and I don't give two shits who calls me what. I know what I know, I am what I am, and I'm all good.

 

Carry on and keep sending your love to these Math/Programmer fags!!!

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 23, 2017, 10:03 a.m. No.1109   πŸ—„οΈ.is πŸ”—kun

EEEEEEENCOURAGEMENT TIME!

 

I went so MATHEMATICAL as SUPER RAINBOWDASH that I broke through to another show's dimension.

 

Should be easy to figure out where these head pats are going.

 

ONWARD, NERDS!

TO GLORY!

Anonymous ID: cf8d48 Dec. 23, 2017, 11:07 a.m. No.1117   πŸ—„οΈ.is πŸ”—kun   >>1131

>>1099

These are weird

 

For odd n, they appear to be the same as in row one (next.a = prev.b){0:3:12:6:6:24=144}{0:3:36:12:24:54=1296}{0:3:72:18:54:96=5184}

For even n, this makes no sense yet. Sometimes there are two interleaved patterns{0:6:9:6:3:27=81} - pattern 1{0:6:24:12:12:48=576} - pattern2{0:6:45:18:27:75=2025} - pattern1{0:6:72:24:48:108=5184} - pattern2{0:12:18:12:6:54=324} - p1{0:12:48:24:24:96=2304} - p2{0:12:90:36:54:150=8100} - p1{0:12:144:48:96:216=20736} - p2

Sometimes more{0:8:5:4:1:25=25}{0:8:12:8:4:36=144} - p1{0:8:21:12:9:49=441} - p2{0:8:32:16:16:64=1024} - p3{0:8:45:20:25:81=2025} - p4{0:8:60:24:36:100=3600} - p1{0:8:77:28:49:121=5929} - p2{0:8:96:32:64:144=9216} - p3{0:8:117:36:81:169=13689} - p4

Anonymous ID: 01b9f7 Dec. 23, 2017, 12:21 p.m. No.1119   πŸ—„οΈ.is πŸ”—kun   >>1129 >>1376

tested results thanks QVC

for a*b=c a and b prime

a^2*b^2=c^2

using capitals to designate squared record

{0, N, D=c=a*b, X, a^2, b^2}

X=ab-a^2 from X=D-A

N=((ab-a^2)^2)/(2*a^2) from 2AN=X^2

N=a^2/2-ab+b^2/2 square and factor above

c= (D+N)^2-(X+N)^2

D+N = ab+a^2/2-ab+b^2/2=b^2/2+a^2/2

D+N large side = b^2/2 +a^2/2

X+N = ab-a^2+b^2/2-ab+a^2/2=b^2/2-a^2/2

X+N small side equals b^2/2-a^2/2

Anonymous ID: 898cfe Dec. 23, 2017, 1:29 p.m. No.1121   πŸ—„οΈ.is πŸ”—kun   >>1122 >>1129

>>1089

 

Take a look at (1, 1). The t = 2 has a = 5. Then go t + 5 steps, a = 85.

 

(1, 5) starts with a = 5/5 and b = 85/5. This appears to hold true for all (e, n).

 

Again, we want to find t = 1 for (1, 13). We find 13 at t = 3. We "save" this a (13) and then add 13 to t (3 + 13). We should get a = 481.

 

13/13, 481/13 = 1, 37. Also the first t = 1 for (1, 13).

Anonymous ID: 898cfe Dec. 23, 2017, 1:43 p.m. No.1123   πŸ—„οΈ.is πŸ”—kun   >>1124

>>1122

 

This only seems to work for generating n that appear as a's in (e, 1).

 

For instance take (1, 37), 37 doesn't appear as any a or b in (1, 1), but it appear in (1, 13). I still haven't figured it out entirely yet.

Anonymous ID: 898cfe Dec. 23, 2017, 1:47 p.m. No.1124   πŸ—„οΈ.is πŸ”—kun

>>1123

 

However, there might be more to it.

 

In (1, 1) you'll find 481 (13 * 37). If you use this as a starting point you'll get (1, 37) starting with 13 (which is correct).

 

I think, in order to enumerate the "possible" n's you'll have to "jump" between n's.

 

Since we know that a=1, b=37 appears in (1, 13) you'll either have to jump from (1, 1) to (1, 13) in order to find it, or we might have to employ recursive factorization to "know" that 37 is a possible 'n' by factorizing 481 when we encounter it in (1, 1). But I don't think the recursive method is the way to do it.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 23, 2017, 3:10 p.m. No.1127   πŸ—„οΈ.is πŸ”—kun

Found a link between our entry c value and the cc record in (0,1).

 

(0,1).d == ( ( c - 1 ) / 2 ) * ( c + 1 );

 

Still looking for elusive d formula from x…

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

145=5x29

a x b *=(1,5,4) = {1:5:12:7:5:29} = 145;

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

aa x bb =(0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc =(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

 

(0,1) perfect squares to analyze

 

aa x bb (0,1) =(0,1,61) = {0:1:7320:120:7200:7442} = 53582400;

1 x cc (0,1) =(0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

 

ccna.d = 10512 compared to 10512 = 0

((ter.c - 1) / 2) * (ter.c + 1) = ((145 - 1) / 2) * (145 + 1)

(ccna.d record match!)

Anonymous ID: 5ba563 Dec. 23, 2017, 3:33 p.m. No.1128   πŸ—„οΈ.is πŸ”—kun   >>1129

>>1120

Thanks. Yeah I should have been more clear. You calculate the 1 * c^2 value in column 0, then convert that n to "index", aka "height". This is zero-based, so it goes index 0, index 1, index 2, index 3…

 

My example had n = 32 and I calculated a index or height of 3.

 

The point I should have better stated is that from that you can calculate the n values of index 2, 1, 0 in constant time (which are 18, 8, and 2 respectively).

 

This is interesting because any 0, n cell you generate this way has perfect squares for a and b. Work through it once or twice on paper; it's fucking magic.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 23, 2017, 4:54 p.m. No.1131   πŸ—„οΈ.is πŸ”—kun   >>1169

>>1099

>>1117

 

Also analyzing the (0,n) space.

 

There is a path to the correct record. Below are 2 examples for 3x5 and 3x7.

 

Substituting the a-b values at e = 0.

 

For the 3x5 example, I tried simply to use the new.d = (prev.d / 2). Not quite right.

 

Looks like we need to figure out this movement to solve.

 

15=3x5

a x b *=(6,1,1) = {6:1:3:0:3:5} = 15;

1 x c =(6,5,2) = {6:5:3:2:1:15} = 15;

aa x bb =(0,2,4) = {0:2:15:6:9:25} = 225;

1 x cc =(0,98,8) = {0:98:15:14:1:225} = 225;

 

(0,1) perfect squares to analyze

 

aa x bb (0,1) =(0,1,4) = {0:1:24:6:18:32} = 576;

1 x cc (0,1) =(0,1,8) = {0:1:112:14:98:128} = 12544;

 

Path subtituting a -b

 

(0,1,8) = {0:1:112:14:98:128} = 12544;

(0,9,13) = {0:9:56:24:32:98} = 3136;

(0,1,4) = {0:1:24:6:18:32} = 576;

 

 

21=3x7

a x b *=(5,1,1) = {5:1:4:1:3:7} = 21;

1 x c =(5,7,2) = {5:7:4:3:1:21} = 21;

aa x bb =(0,8,7) = {0:8:21:12:9:49} = 441;

1 x cc =(0,200,11) = {0:200:21:20:1:441} = 441;

 

(0,1) perfect squares to analyze

 

aa x bb (0,1) =(0,1,7) = {0:1:84:12:72:98} = 7056;

1 x cc (0,1) =(0,1,11) = {0:1:220:20:200:242} = 48400;

 

Path subtituting a -b

 

(0,1,11) = {0:1:220:20:200:242} = 48400;

(0,9,22) = {0:9:140:42:98:200} = 19600;

(0,1,7) = {0:1:84:12:72:98} = 7056;

Anonymous ID: cf8d48 Dec. 23, 2017, 6:25 p.m. No.1132   πŸ—„οΈ.is πŸ”—kun   >>1133

>>1081

Thank you, this is great! The crypto stuff may not be 100% on point but its really good - we need more of this

 

And Sneakers just replaced Go For It (1983) as my all time favorite movie.. how did I miss this gem?

If pic related doesn't convince you to watch it, there are some amazing quotes about the NSA, FBI, clowns, fake NASA moon landings, cattle mutilations and the Kennedy assasination as well :)

That this was made in 92 blows my mind

Hobo !!1yNgQ3NlCs ID: 13b8cc Dec. 23, 2017, 11:11 p.m. No.1133   πŸ—„οΈ.is πŸ”—kun   >>1160

>>1132

Thanks for the reply. I really am glad you liked it. I will keep it up. This has been a very inspiring experience and I am glad others have liked my thoughts and ideas.

 

If I am off on something PLEASE feel free to call me out. I have very thick skin and am always open to other ideas or information. Much of what I stated in this doc is "kluged" together. It will have insight but at the same time will be messed up on the details. I have always been a big picture guy so that is what I like to do. It leaves me with some holes sometimes.

 

Im glad you liked Sneakers. I feel Like I am living that movie sometimes. Well, I will post the next episode of this document now renamed "Unforeseen Consequences"

 

I think that is a good name (shamelessly stolen from my favorite video game). I am not sure I Have seen "Go for it" but I want to check it out.

 

Thanks for your support and I hope to provide some guidance and direction from the 40 foot view. I do not claim to KNOW what that is but I love to explore it.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 12:23 a.m. No.1135   πŸ—„οΈ.is πŸ”—kun   >>1136

>>1134

I was reading a journal that was more of a review with a bunch of parts that were out of my talent spectrum at the moment.

 

But I did pop onto the discord for some dank and spicy memes to encourage the Pathgawds.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 1:02 a.m. No.1137   πŸ—„οΈ.is πŸ”—kun   >>1138 >>1139

>>1136

Well… I guess it's story time.

 

So… you see…

 

I'm a bit of a muh'fuggin' hippy an' shit.

I'm always trying to get into collaborative projects.

Especially with my main project, I had to find more than just myself to get anywhere past my own universe.

 

So… it's another similar collaborative piece but it's the pared version of the whole notebook. Can't show it to ya because… y'know… artists and their methods. Y'all wanna know what we see and think, but no one's allowed to see the notes until well after what e're working is already a thing. Think like… da Vinci notebooks… or… pic related.

 

It involves this story I was working on with the guy who showed me 4chan a brazillion years ago. Some real RPG2000 shit, map making, storytelling… he was focused on an epic hero journey whereas I came up with an entire pantheon of gods and how the overarching story would play out. This was back when I was still trying to figure out how to speak correctly and "Time Matter and Energy" coming together to form a god (Named Renz, other guy came up with the name), that immediately starts dreaming, and creating fantastic scenarios of gods and magic and "men" and then the gods influenced men with their magic too much so they and their magic took a nap. Meanwhile, "Man" discovers "Natural" magic and interacts with all the other species that the other guy was coming up with. All the while the original sleeping god is LARPing as a crotchety old trickster who's forever annoyed by this constant ringing that sounds like an alarm. Other guy was more into the Tompfk Fishman part of the story than I was.

 

So I figured, with all the gods asleep, I think I modeled it off the Ogdoad at first, then a lil' more Greco-Roman, then full blown Hindu but that was a bit much so I pared it down to… 8 sets. So I'd need 16 writers who could work together to write the dreams according to an order I set and they would be going in an out of dreams waking each other up and the whole thing was gonna take decades to flesh out.

 

Eventually all the Gods are awake and we get back to book 1. Or I guess maybe book 1a. Whatever. All the gods are awake and they find Tompfk and try to wake him up. Then there was some epic battle and at some point the pantheon realized they were all the answer was to just give Tompfk a hug and offer understanding. That connection reunites the full consciousness of Renz, who wakes up and "Time Matter and Energy" say "You've done this {n} times! Ready to have a new dream?" and back to the dreamworld he goes.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 1:20 a.m. No.1138   πŸ—„οΈ.is πŸ”—kun   >>1139

>>1137

Keep in mind, that was written a long time ago.

 

Shortly after I figured the time, matter and energy thing didn't make sense since matter was just reeeeeeally slow energy or like… condensed darkness drawing light to it or something (I had a whole spectral breakdown of the density layers of the earth for… whatever reason) but regardless I figured they were the same thing. Or something. So it became time, energy, and space. And they weren't entities outright, the Consciousness of Divinity was like… split in two… and everything worked basically like a load bar.

 

Starts black, light goes forward filling in all the black. God is the Light coming from point Alpha going to Omega… only to realized it was the Space as well because God was waiting for God's self with a grin saying "Everything has been accomplished for eternity for the {n} time! Ready to do Infinity all over again, but differently?"

 

But frankly even that didn't make complete sense because it was just this… loop… but.. with like… 1 difference that makes each progression of infinity different. 1 planck to the left. All the little and big things. Nothing else to do, so at least it's something to do, right?

 

Time goes by, do stuff, keep thinkin' 'bout it and readin' bout it… find Everything Forever due to staring through a porch rail, contemplating that i can never see the other side of anything on the opposite side of my limited perspective and I was juuuuuust fucked up enough to think "0 is greater than 1… the fuck?" and that was followed by being juuuuuuust fucked up enough to actually google it which led me to a fucking ZELDA forum (non-lewd) where they were discussing this book called Everything Forever by Gavin Giorbran and dear sweet merciful FUCK that's the book I'd been looking for my entire life.

 

So at some point I come up with and copyright that lil' diagram-o-mine and end up grabbing The Doctor's attention by blathering on about split-octonians and surr/hypereal numbers and what I didn't even know were called "limits" (at the time) cuz I was just trying to figure out how the fuck to get from .999… to 1 as I had been for the last 10 frickin' years. Something about 0111/0 or something like that which a physicist proved was wrong.

 

And then The Doctor, who specializes in geometry that has nothing to do with physical reality, as he puts it, the non-euclidean kind… loooooooots of drinking sessions later… I kept thinking in terms of "difference" and "change" and "and" and "or" which have completely batshit meanings in math because the nerds are fuckwits and naming shit and long story short:

 

Look at a math problem.

Is time happening?

Where is it? (imagine the paper is an infinite plane)

Time doesn't real, in math.

Energy isn't a "thing", energy is the ability for work to happen over time, which doesn't real.

So energy doesn't real.

So all you're left with is space…

And consciousness…

And you really need to study cones and toroids.

 

He moved away abooooooout…. 8 or 9 months.

 

And then eventually all this happened.

ID: a2017c Dec. 24, 2017, 1:26 a.m. No.1139   πŸ—„οΈ.is πŸ”—kun   >>1140

>>1137

>>1138

 

That's quite interesting. Video games don't have to make sense to be interesting, quite like Indie Games. Anyway, I'll try to contribute to the new line of maths tomorrow.

 

I was researching a man named Sholom Rubashkin. He was running a meth lab in Iowa and imported over 400 illegal immigrants which tore the city apart when ICE raided it, basically destroying the city, or at least damaging it beyond repair. The meth lab's cover story was that it was a kosher meat plant, which it sort of was, so it was a bit of a front-end business with a racket on the side. Where it gets even sadder is the fact that the PETA people said that the animals were tortured (which makes it not even kosher according to scripture, ironically). One said "the esophagi were ripped out while the animals were still alive and conscious." Moreover, he didn't even bother paying his employees some of the time and was also involved in bank fraud, money laundering, wire fraud, and lots of financial fraud in general. One of the biggest takeaways from this is how sad and disgusting Israel and their control of congress really is. They didn't care about what he did to the animals, what he did to the illegal alien employees, how he destroyed an Iowa town called Postville, how he defrauded many financial institutions, etc.

 

They didn't care. They just wanted him free. Bipartisan congress support for an "unjust sentence". Anything for the (((chosenite))) tribe.

Should have been life.

 

I could go on.

 

So, when Trump commuted this guy's sentence, meaning he will be free soon, it tore me apart.

 

But I'm feeling a lot better now that the storm is finally happening.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 1:32 a.m. No.1140   πŸ—„οΈ.is πŸ”—kun   >>1141 >>1142

>>1139

The Martin Luther "What you say?!" copy-pasta meme made me kek.

 

The rest of it… I feel like my story was at least somewhat touching and then… Absurdly unnecessary hatred laden with appeals to emotion followed by a slight at the God Emperor.

 

-le sigh-

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 1:35 a.m. No.1141   πŸ—„οΈ.is πŸ”—kun

>>1140

I mean, you could have at least called him a Zionist fuckstain. Or a low level Qabal member or something.

 

There is a difference… Hell, he mighta been in Kosher Nostra (not joking) or that sycophantic "follow the messiah rebbi" hassidic mafia shit.

 

So like…

At least complete the thought.

ID: a2017c Dec. 24, 2017, 1:36 a.m. No.1142   πŸ—„οΈ.is πŸ”—kun   >>1143 >>1145

>>1140

Read it yourself. Watch the documentaries on him. Sounds nice enough eh?

 

Until you read the FBI reports on the bank fraud, meth lab and animal abuse.

 

No, you'll know why I felt that way if you take the time to actually read into it.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 1:41 a.m. No.1145   πŸ—„οΈ.is πŸ”—kun   >>1147

>>1142

Maybe it was the pics that made feel it was overly generalized. Makes sense to me.

 

It's not like I was arguing with ya about the piece of shit or about shekel grubbing kikes, but I generally leave your "non-MoT obsessed" Jew out of it at least.

 

Hell, I haven't attended a regular service in a synagogue since I was like… 8.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 1:43 a.m. No.1146   πŸ—„οΈ.is πŸ”—kun   >>1148

>>1144

Free but unprotected?

Maybe "free soon" happens after "swamp draining" and golly… he's not available to be freed anymore.

 

Positive thinking.

Positive thinking.

ID: a2017c Dec. 24, 2017, 1:47 a.m. No.1148   πŸ—„οΈ.is πŸ”—kun   >>1149

>>1147

>>1146

Yup. Every time I fall for the "regular Jews are okay" I meet one and they treat me like a disgusting rat. If that is really how it's going to go I'd be so happy.. Because this guy does not deserve to be free. I didn't even mention how he treated the immigrants or how they completely destroy any discussion of this now with muh holocaust and muh nazis.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 1:49 a.m. No.1149   πŸ—„οΈ.is πŸ”—kun

>>1148

Ooooooh…. or 5d chess… just like how he let Alabama get stolen to out their mechanisms and scheming kikery…

 

Who was pushing for this guy to be let out in the first place? I'm getting this weird Barnabas vibe or whoever the pharisees let go instead of Jesus.

Anonymous ID: d40957 Dec. 24, 2017, 4:30 a.m. No.1157   πŸ—„οΈ.is πŸ”—kun   >>1158 >>1159 >>1165

What the fuck you two?? Seriously, get your own fucking board or start another thread and stop shitting this up. What the fuck??

And !!UrvFpU0has, enough with your recruitment and shilling for collaborators on some fucking occult mission. This shit is inside everyone, hardwired, stop being an egotistical fuck, and take it to the Bake, you've got your space in which to spew. Lord have mercy!

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 9:01 a.m. No.1158   πŸ—„οΈ.is πŸ”—kun   >>1160

>>1157

I was trying to motivate again since it was so important I did so last time that folks tracked me down. I saw a lull and tried related a story. Then OP came in all down in the dumps and I tried to cheer OP up and got him over to the EZ Bake as I could. Aaaaand it went there.

 

I'm trying to behave, damn it.

VeritasAequitas !!Nf9AmQNR7I ID: 3345eb Dec. 24, 2017, 10:20 a.m. No.1160   πŸ—„οΈ.is πŸ”—kun   >>1162

>>1134

I’m here, Baker! Still working on the row (e,1) patterns, and also looking for patterns of c^2 in column zero. Any chance you could post a link to the original .csv output? I’m on the road, and my small printed copy isn’t cutting it for the column one research.

 

>>1158

Thanks Topol for the inspiration! The thread needs some encouragement.

 

>>1094

Cool vision, Mr E!

 

>>1133

Thanks Hobo! I haven’t had a chance to read it yet, but appreciate the memes a lot.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 3 p.m. No.1165   πŸ—„οΈ.is πŸ”—kun   >>1166

>>1157

And yes. This is inside everyone.

But not everyone is focusing there at the moment.

Some of them are focusing on code.

And I'm here to support them.

<;3=

 

It's like gay sex.

Some other guy has it covered and I can experience vicariously through him to understand the idea.

 

Savvy?

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 24, 2017, 3:07 p.m. No.1166   πŸ—„οΈ.is πŸ”—kun   >>1168 >>1169

>>1165

Literally you might as well say that the Math and the capability to figure this out is within ME, as well. Which it is. But we're all at different points on the same… whaaaaatever… and the collective consciousness is the delivery system.

 

Don't forget… all of this work is actually already accomplished… this is how it's disseminated to the world.

 

Once it's understood here by the few, it can be passed around through the channel(s) to the many.

 

If VQC wanted credit for this, like the whole Lucifer serving Man story, he'd have just handed it to you. But… it wouldn't be as thoroughly understood and… how in fuck would you have found THIS TEAM of ABSOLUTELY CAPABLE MINDS if you were just handing out something that seemed so monstrously absurd to begin with? When would you have even heard about it? Teach seems to be on the academic up and up but how many filters would it have gone through before TEACH ever saw it, let alone us academic and corporate outsiders?

 

I'm not even a fucking mathnerd and I understand that.

 

So… quit with the discouragement.

You want to piss and moan about thread sliding in a timeless environment, you can do it elsewhere.

 

I was brought back into this space for a purpose… Anons needed -headpats-. I did my job, and now I'm fuckering back off until I either pop back in, or my attention is brought back here.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 3:54 p.m. No.1168   πŸ—„οΈ.is πŸ”—kun   >>1169 >>1174 >>1185 >>1273

>>1166

Thanks for this post Topol, well said, and I appreciate the kind mention.

And headpats go a long way, no doubt!

 

I've been working in moderate silence the last couple of days because I haven't had any new insights, and I have only had intermittent time to spend on the problem.

Mostly reading not much writing.

 

A few attachments…

1) Andrew Wiles paper on Fermat's Last, and its relation to Elliptic curves.

2) A picture of 1*cc for c=15, a=3, b=5. Explained below.

3) Output in code for similar semi-primes (the 2 numbers per row are described below).

 

So I think PMA might have already identified this relationship. And I makes perfect sense, and I'm not sure what value it adds.

But I've been playing with geometrically viewing the 1*cc.

For both the 1cc row, and the aabb row, (when e=0), both n's are always halfway between 2 perfect squares.

Now of course, like I said, this is "obvious", since d+n is always halfway between a & b.

So in the e=0 column, for both 1cc and aabb, sqrt(n/2) is a perfect square, as is sqrt(n*2).

And sqrt(n*2) will always be equal b-a.

 

Often the sqrt(n2) for the aabb row is a factor of the 1cc row. You can see that in the image with colored squares. Both x and n at the 1cc row are divisible by 7.

So if you take x in the 1cc row, divide it by 7, square that then halve it, you'll get n at aabb.

 

This is only mostly true though!

There are many rows where this is not the case.

Examples include:

313, 317, 319, 717, etc

 

But most of the time this is true.

Here's a pastebin of the full results:

https:/ /pastebin.com/a9KwHCFC

 

The 2 numbers at the end of each line are sqrt(n*2) and sqrt(n/2).

 

The row beginning with is the values above for the 1cc row / aabb row.

 

Any thoughts guys?

PMA, VA, MA, am I just restating what is already known?

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 24, 2017, 4:08 p.m. No.1169   πŸ—„οΈ.is πŸ”—kun   >>1170 >>1174 >>1181 >>1184 >>1185 >>1187 >>1319 >>1553

>>1166

Topol, keep it up. Words and insights are very helpful.

 

>>1168

Teach - thanks for adding clarity!

 

>>1131

Teach - in case you missed it yesterday, I was able to piece together a path through various (0,n) records starting from (1xcc) and getting to (aaxbb).

 

I can see the path clearly, and there most likely is some elegant math to jump directly to it.

 

In the meantime, I've gone back to the "drawing board", as VQC suggested and think I've figured out the pattern to walking up and down the tree at (0,n).

 

Rather than flood with examples, following are the formulas:

 

to walk up the tree at (0,n)

n is odd: new.d = d + 2*(x+n) + 2n

n is even: new.d = d + (x+n) + n/2

 

to walk down the tree at (0,n)

n is odd: new.d = d - 2*(x+n) + 2n

n is even: new.d = d - (x+n) + n/2

 

I have verified this with data from (0,1) through (0,6). Both up and down.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 4:16 p.m. No.1170   πŸ—„οΈ.is πŸ”—kun   >>1171

>>1169

Thanks PMA, I just re-read.

 

I don't fully get it. Do you think you could explain the part from:

>(0,1) perfect squares to analyze

on?

 

And what exactly do you mean by substituting a and b?

 

Sorry for the questions, I want to understand!

I feel like we may be close again…

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 24, 2017, 4:27 p.m. No.1171   πŸ—„οΈ.is πŸ”—kun   >>1172 >>1173 >>1185 >>1194 >>1248

>>1170

Records in the (0,1) space are very well defined. We can break them into their component parts very easily to get an answer.

 

For example at (0,1,4) we know that the value of a = 18 can be represented as 2*(3^2). And the prime answer is "3".

 

Squaring the c, puts us into the (0,n) space. Substituting a = na, puts us in the (0,1) space.

 

And creating the record directly by (0,1) and t = (c+1)/2, is a shortcut to getting to (1xcc) for any c.

 

Now there, the trick is finding that aa*bb record.

 

Subsequent records in a column are connected by a -b movements.

 

In reviewing some data, I found a way to get to the b value of the (aa*bb) record, by going through 1 jump.

 

Start with the (cc) record, move it's a to the next record b (figure out the d), and then one more jump to the aa*bb record.

 

I've verified this path twice, but calculating the d value presented a problem.

 

Hence my previous post.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 4:34 p.m. No.1172   πŸ—„οΈ.is πŸ”—kun

>>1171

Ok, I think I gotcha. I'm going to need to sit with this a bit.

I'm going to run my own tests to re-prove your tests, and see how they correlate to my path. I think we might be coming full circle to the same point!

 

Do you think you could test with one of these pairs next?

313, 317, 319, 717

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 4:51 p.m. No.1173   πŸ—„οΈ.is πŸ”—kun   >>1179

>>1171

>move it's a to the next record b (figure out the d)

 

So I'm with you until here. You move b to a. you know e=0, and you need 1 more variable to find a specific cell, and you're saying you can calculate d?

I'm stuck PMA.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 24, 2017, 8:24 p.m. No.1179   πŸ—„οΈ.is πŸ”—kun   >>1180 >>1185 >>1192

>>1173

I posted earlier in this thread a way to calculate the d value for the (1xcc) record directly from c.

 

So I figured to focus a bit on navigating records by changing d values.

 

Path subtituting a -b

 

(0,1,8) = {0:1:112:14:98:128} = 12544;

(0,9,13) = {0:9:56:24:32:98} = 3136;

(0,1,4) = {0:1:24:6:18:32} = 576;

 

First record is the (1xcc) record. Last is (aaxbb).

 

Notice the movement of a to b in each step. e is always 0, n is 1 in the first and last record. We have the a value in the first record. So we need one more variable to move to the second step, and then we will have the a value to use in the last calculation.

 

This may not be the end solution, but this is the first time I've seen a direct connection between an (axb) and (1xc) record.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 8:37 p.m. No.1180   πŸ—„οΈ.is πŸ”—kun   >>1181

>>1179

Ok, I'm with you.

Look at your example. 2 things to notice (if you haven't already):

1) taking your a from (0,1,8), how many times in the 0 column will it appear as b? It is always going to be 1/2 of a perfect square. And as such it'll appear in the 0 column at every n where n is a perfect square (in your example n was 9). Because this value (98) appears many time, the pattern of the value that it appears next to (the a to its b) is all possible values of 1 half perfect squares, that have an e=0.

 

2) There's a row directly below that one that, (0,9,14), where a = 18, and b = 128, and n will equal the n of (0,9,13).n.

 

I'm not sure what this means, I'm having trouble visualizing it now.

 

Were you already aware of these patterns?

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 24, 2017, 8:49 p.m. No.1181   πŸ—„οΈ.is πŸ”—kun   >>1182 >>1183 >>1185 >>1553

>>1169

well, shit. turns out the d movement formulas work for all cases of (e,n).

 

Walking up tree 5 iterations from a=5, b=41.

 

(9,9,5) = {9:9:14:9:5:41} = 205

(9,9,14) = {9:9:68:27:41:113} = 4633

(9,9,23) = {9:9:158:45:113:221} = 24973

(9,9,32) = {9:9:284:63:221:365} = 80665

(9,9,41) = {9:9:446:81:365:545} = 198925

(9,9,50) = {9:9:644:99:545:761} = 414745

 

Walking down tree 5 iterations from a=157, b=273.

 

(12,8,26) = {12:8:207:50:157:273} = 42861

(12,8,22) = {12:8:153:42:111:211} = 23421

(12,8,18) = {12:8:107:34:73:157} = 11461

(12,8,14) = {12:8:69:26:43:111} = 4773

(12,8,10) = {12:8:39:18:21:73} = 1533

(12,8,6) = {12:8:17:10:7:43} = 301

 

>>1180

Thanks Teach. I will look into this soon.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 9:17 p.m. No.1183   πŸ—„οΈ.is πŸ”—kun   >>1193

>>1181

Can you show me with this example?

{59:5:61:19:42:90}

 

From the grid, the cell values are:

{59:5:61:19:42:90}

{59:5:71:21:50:102}

{59:5:119:29:90:158}

{59:5:133:31:102:174}

{59:5:197:39:158:246}

{59:5:215:41:174:266}

Veritas Aequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 24, 2017, 9:51 p.m. No.1187   πŸ—„οΈ.is πŸ”—kun   >>1188

>>1169

Hello PMA! So your formula needs x and n to walk the records. Only problem is that we have to derive them from c. What are your thoughts on getting x? We need a first, correct? I could be wrong.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 24, 2017, 10:54 p.m. No.1193   πŸ—„οΈ.is πŸ”—kun   >>1195 >>1553

>>1183

>>1184

Teach - good eye.

 

Following are what I have for all grid entries at (59,5)

 

(59:5:10) = {59:5:61:19:42:90} = 3780; (x+n)=24x24=576; (d+n)=66x66=4356

(59:5:11) = {59:5:71:21:50:102} = 5100; (x+n)=26x26=676; (d+n)=76x76=5776

(59:5:15) = {59:5:119:29:90:158} = 14220; (x+n)=34x34=1156; (d+n)=124x124=15376

(59:5:16) = {59:5:133:31:102:174} = 17748; (x+n)=36x36=1296; (d+n)=138x138=19044

(59:5:20) = {59:5:197:39:158:246} = 38868; (x+n)=44x44=1936; (d+n)=202x202=40804

(59:5:21) = {59:5:215:41:174:266} = 46284; (x+n)=46x46=2116; (d+n)=220x220=48400

(59:5:25) = {59:5:295:49:246:354} = 87084; (x+n)=54x54=2916; (d+n)=300x300=90000

(59:5:26) = {59:5:317:51:266:378} = 100548; (x+n)=56x56=3136; (d+n)=322x322=103684

(59:5:30) = {59:5:413:59:354:482} = 170628; (x+n)=64x64=4096; (d+n)=418x418=174724

(59:5:31) = {59:5:439:61:378:510} = 192780; (x+n)=66x66=4356; (d+n)=444x444=197136

 

And here is my walking up formula for a=42, b=90 and a=50, b=102.

 

Walking up tree 5 iterations from a=42, b=90.

 

(59,5,10) = {59:5:61:19:42:90} = 3780

(59,5,15) = {59:5:119:29:90:158} = 14220

(59,5,20) = {59:5:197:39:158:246} = 38868

(59,5,25) = {59:5:295:49:246:354} = 87084

(59,5,30) = {59:5:413:59:354:482} = 170628

(59,5,35) = {59:5:551:69:482:630} = 303660

 

Walking up tree 5 iterations from a=50, b=102.

 

(59,5,11) = {59:5:71:21:50:102} = 5100

(59,5,16) = {59:5:133:31:102:174} = 17748

(59,5,21) = {59:5:215:41:174:266} = 46284

(59,5,26) = {59:5:317:51:266:378} = 100548

(59,5,31) = {59:5:439:61:378:510} = 192780

(59,5,36) = {59:5:581:71:510:662} = 337620

 

My walking tree formula seems to iterate only the movements from a -b. Not sure if this is a bug or a feature.

Veritas Aequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 24, 2017, 11:04 p.m. No.1196   πŸ—„οΈ.is πŸ”—kun   >>1197 >>1222 >>1224

>>1188

Thanks, good call on examining the record.. But even starting with b equals one, then we need one of the following to solve starting ONLY from c.

  1. Where does b always equal 1?

  2. A chain of values for a in (e,1), leading to x,t,b,n values?

  3. A chain of c^2 patterns in (0,n) leading to precidtable ways to move infinitely thru the cells and elements.

  4. I could be wrong. Am I missing something totally obvious?

  5. Merry Christmas, Faggots! I really love being here with you nerds. I'm a nerd too. Loving the Math(s).

  6. Thanks VQC you faggot! Merry Christmas to you too. We love your excellent blend of math, crazy truth, and fighting them. As our Senpai, we bless you and thank you for sharing with us. You don't "need" us, and we don't "need" you. We're here bc you posed some awesome questions, and we're looking for the answers. Thanks also for being so honest about your personal struggles! I also have trouble drinking too much. Love life tho.

 

Goodnight fags.

4.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 11:08 p.m. No.1197   πŸ—„οΈ.is πŸ”—kun   >>1198 >>1248

>>1196

Merry Christmas to you VA. And PMA, and Topol, and MA, and Hobo, and AA, and CA (if he's still alive), and Baker and all the other anons!

 

Merry Christmas VQC, I hope the plan plays out smoothly over the coming days, and I hope we can help you.

 

I enjoy being here with you all too!

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 24, 2017, 11:13 p.m. No.1198   πŸ—„οΈ.is πŸ”—kun   >>1199 >>1201 >>1203 >>1207 >>1223 >>1231 >>1234 >>1239

>>1195

>>1197

Happy holidays to everyone.

 

I adjusted my test case to include the latest walking of the tree using calculated values of d.

 

Review the TestIterateMovementToAABB line below.

 

It may look like I'm simply subtracting 1 from t each time, but the reality is that the previous d value is being calculated from the current record using the formula:

 

return ter.d - 2 * (ter.x + ter.n) + 2 * ter.n;

 

where ter is the current record.

 

I think that if we can figure out now how to skip multiple records using a factor, we have a workable solution.

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

145=5x29

a x b *=(1,5,4) = {1:5:12:7:5:29} = 145;

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

aa x bb =(0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc =(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

 

(0,1) perfect squares to analyze

 

aa x bb (0,1) =(0,1,61) = {0:1:7320:120:7200:7442} = 53582400;

1 x cc (0,1) =(0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

 

TestDCalculationFromC

β€”β€”β€”β€”β€”β€”β€”

ccna.d = 10512 compared to 10512 = 0

((ter.c - 1) / 2) * (ter.c + 1) = ((145 - 1) / 2) * (145 + 1)

(ccna.d record match!)

 

TestIterateMovementToAABB

1) (0,1,72) = {0:1:10224:142:10082:10368} = 104530176

2) (0,1,71) = {0:1:9940:140:9800:10082} = 98803600

3) (0,1,70) = {0:1:9660:138:9522:9800} = 93315600

4) (0,1,69) = {0:1:9384:136:9248:9522} = 88059456

5) (0,1,68) = {0:1:9112:134:8978:9248} = 83028544

6) (0,1,67) = {0:1:8844:132:8712:8978} = 78216336

7) (0,1,66) = {0:1:8580:130:8450:8712} = 73616400

8) (0,1,65) = {0:1:8320:128:8192:8450} = 69222400

9) (0,1,64) = {0:1:8064:126:7938:8192} = 65028096

10) (0,1,63) = {0:1:7812:124:7688:7938} = 61027344

11) (0,1,62) = {0:1:7564:122:7442:7688} = 57214096

12) (0,1,61) = {0:1:7320:120:7200:7442} = 53582400

Matching aabbna record found!

 

(1 x cc).x - (aa x bb).x = 144 - 120 = 24

 

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 145 - 120 ) = 5. (a matches!!!!)

Teach !!UgZAPoSXEk ID: a92e07 Dec. 24, 2017, 11:23 p.m. No.1199   πŸ—„οΈ.is πŸ”—kun   >>1200 >>1203

>>1198

Good job PMA.

 

I am worried about the t– though…

I like the process, but so far we're just decrementing x by 2 each time, which is essentially a search for "a".

 

But I agree that this might unlock the ability to algorithmically move through it.

 

For this example the n = 61 from 1xc the t == 61 from 0,1 at the end…

Is that a coincidence?

Veritas Aequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 25, 2017, 12:05 a.m. No.1204   πŸ—„οΈ.is πŸ”—kun   >>1210

Ugh, this thread sucks. I mean, where is that fag Topol on Christmas Eve? I would love to see ponies, math lsd trips, memes, whatever. Do I really have to go over to ez bake, Topol?

Anonymous ID: a2017c Dec. 25, 2017, 12:20 a.m. No.1211   πŸ—„οΈ.is πŸ”—kun   >>1213

>>1208

Storm is here already. This is Q's first and last picture.

 

Note the dates, the weather, and who they were fighting. Pics related.

 

en.wikipedia.org/wiki/George_Washington%27s_crossing_of_the_Delaware_River

Veritas Aequitas !!Nf9AmQNR7I ID: 3f145c Dec. 25, 2017, 12:30 a.m. No.1213   πŸ—„οΈ.is πŸ”—kun

>>1211

Lads, think about the challenge we've undertaken at VQC's request. WHEN we solve this, we can make a major contribution to POTUS and our world. Using math. Let's go, Faggots!! Spirits Up!! Christmas is here, let's solve this. We can help with a gift for Christmas.

+++

++

+

β€”β€”β€”-Freedom.

Veritas Aequitas !!Nf9AmQNR7I ID: 3f145c Dec. 25, 2017, 12:47 a.m. No.1215   πŸ—„οΈ.is πŸ”—kun   >>1218

>>1210

Thanks Teach! Merry Christmas! "When the infinite became the finite." Season 1:God Makes Mary Pregnant, via energy beams using incredibly complex math. Stay tuned. Lol, I totally have faith guys. But it's good to laugh at yourself too, Jesus is the man. Merry Christmas, again.

Veritas Aequitas !!Nf9AmQNR7I ID: 3f145c Dec. 25, 2017, 1:21 a.m. No.1219   πŸ—„οΈ.is πŸ”—kun

>>1218

Yeah, they do. Jesus and his teachings are beautiful. BC they love evil and serve it, they hate him. Our spiritual and biological and mental evolution has outgrown them. Jesus is a symphony of love and truth. Religion, different story. I always go to the source document.

Mr.E Melange !!4RyW8fD.HU ID: 815fd7 Dec. 25, 2017, 5:13 a.m. No.1222   πŸ—„οΈ.is πŸ”—kun   >>1225 >>1248

>>1196

>5. Merry Christmas, Faggots! I really love being here with you nerds. I'm a nerd too. Loving the Math(s).

>6. Thanks VQC you faggot! Merry Christmas to you too. We love your excellent blend of math, crazy truth, and fighting them. As our Senpai, we bless you and thank you for sharing with us. You don't "need" us, and we don't "need" you. We're here bc you posed some awesome questions, and we're looking for the answers. Thanks also for being so honest about your personal struggles! I also have trouble drinking too much. Love life tho.

2nd this, thanks VA, Peace to you!

And to all you other fags, PMA, MA, Teach, Hobo, AA, Topol, VQC, anons, Merry Christmas!

Noted on twitter, sent Chris a Christmas message Christmas Eve, less than hour later account gone. He's been in my thoughts.

All you Anons, please do say a special prayer / send vibes of loving energy for VQC, it can be hard to carry such knowledge for so long, and to struggle, and to fight this fight from the beginning. Praying for Strength, Wisdom and a 'hedge of protection' for you today Chris!

Here's to PEACE and COMPASSION in 2018.

(think we're going to help with the TRANSPARENCY piece!)

Anonymous ID: 01b9f7 Dec. 25, 2017, 6:27 a.m. No.1224   πŸ—„οΈ.is πŸ”—kun   >>1249

>>1196

b = 1 was typo. Its the a =1 b = cc record which is solvable for x and n because e is 0. PMA has put together all the steps you need to move around grid to solve problem. Now it needs speeding up.

Mr.E Melange !!4RyW8fD.HU ID: 815fd7 Dec. 25, 2017, 7:16 a.m. No.1229   πŸ—„οΈ.is πŸ”—kun

Hopefully Chris will be back soon, too. See now the tactic? Cut off your followers, cut off your voice. Who's next, POTUS??

Now, where has gbstn10 been??

Anonymous ID: 01b9f7 Dec. 25, 2017, 8:37 a.m. No.1231   πŸ—„οΈ.is πŸ”—kun   >>1233 >>1234

>>1198

Not sure if you know how far you need to move within (0,1) to find aa*bb but within (0,1) it is pretty easy to move to arbitrary new t from existing record since we know t.

x(t-1) =x(t) -2 for any x and t in (0,1)

d(t-1) = d(t) - 4*(t)

So x(t-m)= x(t) - 2m

d(t-m) = 4*sum of (t+t-1…t-(m-1))

Seems you could iterate more accurate guesses of how far you need to move.

Anonymous ID: 473794 Dec. 25, 2017, 2:52 p.m. No.1251   πŸ—„οΈ.is πŸ”—kun   >>1252

So the to do list will be:

  1. Speed up PMA's process

  2. Validate using very big ints

  3. Discover new Fermat primes / Test on RSA2048

  4. Describr everything in laymen terms and desemminate over the internet

  5. ??????

  6. Watch world burn for several weeks/months

 

This a good time line or what, anons?

Anonymous ID: 473794 Dec. 25, 2017, 3:02 p.m. No.1252   πŸ—„οΈ.is πŸ”—kun

>>1251

 

During burning of world, VQC will hopefully return and teach us how to break ECC.

 

Almost everyone's going to switch from RSA to ECC, with RSA broken.

 

Then we break ECC, even more world burning.

 

I feel dirty, anons.

ID: 6682fd Dec. 25, 2017, 4:19 p.m. No.1255   πŸ—„οΈ.is πŸ”—kun

Let's all feel the love! This is the best Christmas ever. We finally have a president that loves us.. In the words of Salinger.. One that you just wanna call up and talk to! I am so happy for all of you!

 

Nothing but a Godly force could bring us together to collectively stare at numbers until something clicks

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 25, 2017, 4:28 p.m. No.1256   πŸ—„οΈ.is πŸ”—kun   >>1257

>>1239

>>1245

confirmed.

 

using the formula for a new d value is equivalent to what I posted earlier.

 

return ter.d - 4 * ter.t * m + 2 * m * ( m + 1 );

 

This works for jumping m levels down.

 

Excellent anon!

 

If we know the "m" value, we can jump directly to the aabbna result.

Anonymous ID: 1f6c20 Dec. 25, 2017, 4:31 p.m. No.1258   πŸ—„οΈ.is πŸ”—kun

>>1230

In anon-hobo mode as I am visiting the fam. and left my un-rememberable password on a shack PC back under the bridge. I am glad you guys like my work I will write more soon. I have started on some more stuff recently but wanted to toss out a few ideas.

 

First and foremost. I am guessing this has been worked already. I HOPE this eventuality has been worked, but if not, Chris, this is for you. As this becomes more recognizable and believable I see a mile-wide issue of "capture" What I mean by capture is this. There is probably less than a dozen of us here. I have been archiving everything I can on physical media as well as literally pen-and-paper. I do this because if this thing is captured by the bad guys they will write the eraser immediately and use it against US.

 

I hope, I REALLY hope there is a "Bomb Shelter" that has been built with ALL the answers in it. Some sort of deadman switch. It would be a catastrophic loss if this fell into the hands of a group of people who would simply lock it away and use it to control and hurt people.

 

Anyway, my next "chapter" will talk about this. I woke up and saw Chris's twatter down as well as Assange and thought the worst. Now its back but is it back or "BACK" anyway Chris if you dont have a deadman switch then think about making one. I would help but I am not a heavy brain when it comes to some stuff. I would hope Wikileaks maybe help?

 

Paranoia rules in the hobo camp sometimes so hopefully I need not worry! I hope you all have been having a wonderful Christmas this year. I sure have and am hoping that 2018 will be a year-of-years for us all and not in a bad way. I am going to do my next chapter then start to put some time into an offline archive of sorts. I am envisioning a dozen or so thumbdrives stashed around so that if something nasty happens there are backups. also, paper copies may be valuable.

 

I have learned long ago that you can be an inch away from a sure-thing level success where there is zero possibility of failure and your enemy is defeated. When you feel that way it probably means you are getting cocky. Always remember that.

 

Its not over until you have crossed the finish line and the judges have published the results of the race. Getting cocky is how you end up as a hobo under a bridge. Keep your eyes on the target gents, we have a long ways to go.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 25, 2017, 4:39 p.m. No.1259   πŸ—„οΈ.is πŸ”—kun   >>1262 >>1264 >>1288

>>1257

Following are a couple of my test cases, just to emphasize the importance of a formula for "m".

 

Also, I haven't tested using t and m to move d in the up the tree, or with odd/even values of n.

 

for c = 145, only 12 iterations to find the value.

for c= 6107, we're up to 480 iterations.

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

145=5x29

a x b *=(1,5,4) = {1:5:12:7:5:29} = 145;

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

aa x bb =(0,288,61) = {0:288:145:120:25:841} = 21025;

1 x cc =(0,10368,73) = {0:10368:145:144:1:21025} = 21025;

 

(0,1) perfect squares to analyze

 

aa x bb (0,1) =(0,1,61) = {0:1:7320:120:7200:7442} = 53582400;

1 x cc (0,1) =(0,1,73) = {0:1:10512:144:10368:10658} = 110502144;

 

TestDCalculationFromC

β€”β€”β€”β€”β€”β€”β€”

ccna.d = 10512 compared to 10512 = 0

((ter.c - 1) / 2) * (ter.c + 1) = ((145 - 1) / 2) * (145 + 1)

(ccna.d record match!)

 

TestMLevelJumpingByD

Matching aabbna record found at m=12

(0,1,61) = {0:1:7320:120:7200:7442} = 53582400

 

(1 x cc).x - (aa x bb).x = 144 - 120 = 24

 

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 145 - 120 ) = 5. (a matches!!!!)

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

6107=31x197

a x b *=(23,36,24) = {23:36:78:47:31:197} = 6107;

1 x c =(23,2976,39) = {23:2976:78:77:1:6107} = 6107;

aa x bb =(0,13778,2574) = {0:13778:6107:5146:961:38809} = 37295449;

1 x cc =(0,18641618,3054) = {0:18641618:6107:6106:1:37295449} = 37295449;

 

(0,1) perfect squares to analyze

 

aa x bb (0,1) =(0,1,2574) = {0:1:13245804:5146:13240658:13250952} = 1909564816;

1 x cc (0,1) =(0,1,3054) = {0:1:18647724:6106:18641618:18653832} = -121773168;

 

TestDCalculationFromC

β€”β€”β€”β€”β€”β€”β€”

ccna.d = 18647724 compared to 18647724 = 0

((ter.c - 1) / 2) * (ter.c + 1) = ((6107 - 1) / 2) * (6107 + 1)

(ccna.d record match!)

 

TestMLevelJumpingByD

Matching aabbna record found at m=480

(0,1,2574) = {0:1:13245804:5146:13240658:13250952} = 1909564816

 

(1 x cc).x - (aa x bb).x = 6106 - 5146 = 960

 

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 6107 - 5146 ) = 31. (a matches!!!!)

VeritasAequitas !!Nf9AmQNR7I ID: 3f145c Dec. 25, 2017, 6:02 p.m. No.1260   πŸ—„οΈ.is πŸ”—kun

>>1250

Thanks Topol!

 

>>1254

Seconded! Can SuperFormulaAnon >>1239 give us the quick rundown? I’m following pretty close, but have been with family and would love a hand catching up. I’ve read and caught up on posts, but still working to understand.

ID: c75bc0 Dec. 25, 2017, 6:08 p.m. No.1261   πŸ—„οΈ.is πŸ”—kun

Chris' twitter is back up. Sure hope that shitfy Saudi cocksleeve of a platform dies. Paul nehlen had a good proposal of extending first amendment to social media.

Anonymous ID: 01b9f7 Dec. 25, 2017, 6:41 p.m. No.1264   πŸ—„οΈ.is πŸ”—kun   >>1265

>>1259

Since we were starting from 1 cc record the formula was designed for reducing m. It should work for increasing m just use -m in formula.

In (0,1) n is always odd.

Thanks for testing formula.

Anonymous ID: 01b9f7 Dec. 25, 2017, 6:54 p.m. No.1268   πŸ—„οΈ.is πŸ”—kun

PMA was right yesterday we have more work to do! Even with ability to make large jumps t +- m we don't know where to jump to.If we make a jump the results as far as I know don't tell us if we are any closer to the answer. Its either right or it isn't.

Anonymous ID: 01b9f7 Dec. 25, 2017, 7:54 p.m. No.1270   πŸ—„οΈ.is πŸ”—kun   >>1271

Remember no such thing as coincidence. Chris's twatter is back. So is …

Julian Assange!

I believe white hat spooks had to protect certain twatter accounts from new twatter spying

VeritasAequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 25, 2017, 8:16 p.m. No.1271   πŸ—„οΈ.is πŸ”—kun   >>1272

>>1270

I’m glad Chris and Julian are back too! Definitely not a coincidence. Think you’re right, white hats must be helping out.

Twatter, google, you tube, et al need to be busted. Crazy that we’re operating on EU hate speech laws here in The US. It would have been game over if HRC had won. Thank God she didn’t.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 25, 2017, 9:40 p.m. No.1275   πŸ—„οΈ.is πŸ”—kun   >>1276

I hate to be that guy, but I fear we're just translating the problem to different variables, but we're unable to do no better than iterate t.

 

I've been looking at it a look, and I think this may be the case.

Sorry to be a downer.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 25, 2017, 9:50 p.m. No.1276   πŸ—„οΈ.is πŸ”—kun   >>1277

>>1275

I'm looking a bit closer at the sqrt(n/2) and sqrt(2n) you posted earlier.

 

What's missing from all of this is an understanding of how the squares resize.

 

>Sorry to be a downer.

I think we're almost there.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 25, 2017, 9:57 p.m. No.1277   πŸ—„οΈ.is πŸ”—kun   >>1278 >>1280

>>1276

I'm working on a different path regarding that actually.

You can kinda swap some variables around to calculate for any integer i, the cell in (0,n) with a=ii and b=cc.

 

If you iterate i, growth in a is quadratic, but growth in n, x, and d are kinda linear (they increase in a arithmetic series).

And at the point that i is a factor of c, x+n / n will be a whole number.

 

I think this is saying the same thing, we're just increasing the square size, but maintaining the same ratio.

 

The interesting point to this direction is the sorta linear growth in x,n and d.

 

I'm not sure if we're just restating the problem again though, stuck with iterating i.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 25, 2017, 11:01 p.m. No.1280   πŸ—„οΈ.is πŸ”—kun   >>1281

>>1278

Feel free to explain it and i'll take a look next.

 

>>1277

Something very interesting about this approach is the relationship between the x+n/n ratio for the 1xc row, and the aa x cc and bb x cc row. It always appears to be very close to the first, or in between both.

MinecraftAnon !!QXqSZ2ev8. ID: 1b63bb Dec. 25, 2017, 11:26 p.m. No.1283   πŸ—„οΈ.is πŸ”—kun   >>1284 >>1285 >>1286 >>1292

>>1281

Given this exact sequence of cells, can you write an equation to plot it's behavior?

 

Format is in {E:N:D:X:A:B}.

{23:01:83:11:72:96}

{23:02:47:11:36:62}

{23:03:35:11:24:52}

{23:04:29:11:18:48}

{23:06:23:11:12:46}

{23:08:20:11:09:47}

{23:09:19:11:08:48}

{23:12:17:11:06:52}

{23:18:15:11:04:62|

{23:24:14:11:03:73}

{23:36:13:11:02:96}

{23:72:12:11:01:167}

 

See how N and A trade places? Anyway the big question with this one is predicting how D changes. If you can figure out A and B as well, that'd be a huge help.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 25, 2017, 11:57 p.m. No.1290   πŸ—„οΈ.is πŸ”—kun

>>1289

Correct, d = a + x, and x is constant.

n is the reverse set, obviously, e is fixed, b can be calculated.

There should even be a way to jump any amount of records algebraically.

Anonymous ID: 8a946c Dec. 26, 2017, 12:03 a.m. No.1292   πŸ—„οΈ.is πŸ”—kun

>>1283

So it looks like the pattern is driven by x^2+e equaling 144 as a fixed constant. So n(d-x) = na = 72, and thus n and a can take on all the various factor combinations of 72, which also gives d with the offset of 11 from a. And b=d+x+2n.

MinecraftAnon !!QXqSZ2ev8. ID: 1b63bb Dec. 26, 2017, 12:53 a.m. No.1299   πŸ—„οΈ.is πŸ”—kun   >>1306

Here's another relational puzzle.

 

{E:N:D:X:A:B}

Set 1:

{016:001:048:008:040:058}

{033:001:048:007:041:057}

{048:001:048:006:042:056}

 

Set 2:

{000:002:048:012:036:064}

{027:002:048:011:037:063}

{052:002:048:010:038:062}

 

Set 3:

{008:003:048:014:034:068}

{041:003:048:013:035:067}

{072:003:048:012:036:066}

 

Set 4:

{000:004:048:016:032:072}

{039:004:048:015:033:071}

{076:004:048:014:034:070}

 

Set 5:

{021:005:048:017:031:075}

{064:005:048:016:032:074}

{does not exist}

 

Set 6:

{036:006:048:018:030:078}

{083:006:048:017:031:077}

{does not exist}

 

Set 7:

{045:007:048:019:029:081}

{096:007:048:018:030:080}

{does not exist}

 

In fact, where D = 48, no records exist past E = 96. The patterns within sets are clear, what about the pattern between sets?

MinecraftAnon !!QXqSZ2ev8. ID: 1b63bb Dec. 26, 2017, 12:58 a.m. No.1301   πŸ—„οΈ.is πŸ”—kun   >>1307 >>1340

The goal I'm working towards is pinning variables on certain axis, then moving along a different pinned axis. I think I've figured out what makes N populated or not. I'm not sure stepping through T (x for me) is possible.

Anonymous ID: 8a946c Dec. 26, 2017, 7:28 a.m. No.1306   πŸ—„οΈ.is πŸ”—kun

>>1299

In general e is always going to be bounded by 2d, because (d+1)^2 = d^2 + 2d + 1.

That is, if the remainder e gets up to 2d+1, you just tick over to records using the next larger d value.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 26, 2017, 9:26 a.m. No.1308   πŸ—„οΈ.is πŸ”—kun

>>1099

Not sure if these have been answered:

 

>Square a value c, where c is the product of two different primes.

>What are the values of n in column zero where these appear?

>What is the value of d where these appear?

 

For any value c, the cc record can be created via:

 

e = 0

n = ((c-1)^2)/2

d = c

x = c - 1

Anonymous ID: 898cfe Dec. 26, 2017, 9:28 a.m. No.1309   πŸ—„οΈ.is πŸ”—kun   >>1310 >>1320 >>1327 >>1342

>>1099

 

Okay, so I'm back after spending the holidays with family.

 

I took another look and ended up rewriting my code and then took another look at this crumb.

 

So, 3 * 5 = 15. 15 * 15 = 225. It seems that 225 only appears in the following (0, n):

 

(0, 2, 255, 30, 225.0, 289.0)

(0, 8, 285, 60, 225.0, 361.0)

(0, 18, 315, 90, 225.0, 441.0)

(0, 32, 345, 120, 225.0, 529.0)

(0, 50, 375, 150, 225.0, 625.0)

(0, 72, 405, 180, 225.0, 729.0)

(0, 98, 435, 210, 225.0, 841.0)

 

I looked at n up to 5000 and didn't find any more occurrences of 225.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 26, 2017, 9:39 a.m. No.1311   πŸ—„οΈ.is πŸ”—kun   >>1313

>>1310

>they all appear to be in (0, 2), (0, 8), (0, 18), (0, 32) etc

2*(1^2)

2*(2^2)

2*(3^2)

2*(4^2)

etc.

 

This record is interesting. How did you get to this one?

(0, 2, 255, 30, 225.0, 289.0)

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 26, 2017, 10:14 a.m. No.1321   πŸ—„οΈ.is πŸ”—kun

>>1318

Very good.

New avenue to explore.

Looks like first record in the a = c^2 series can be created via:

 

e = 0

x = 2*c

a = c^2

 

Will iterate these patterns shortly.

Anonymous ID: 01b9f7 Dec. 26, 2017, 10:23 a.m. No.1322   πŸ—„οΈ.is πŸ”—kun

For all (0,n) where n =2*(i)^2 where i is integer

all a and b are perfect squares.

All aa*bb records are (0,n) fitting 2,8,28,32,50 pattern

Anonymous ID: b15a15 Dec. 26, 2017, 1:52 p.m. No.1332   πŸ—„οΈ.is πŸ”—kun

>>1330

Thanks brother… been lurking and chatting w Chris before I knew he was VQC (Intuitive/synchronicity hit). Wild ride indeed! Divine synchronicity going exponential… loving it all! Blessings to all!

Anonymous ID: 898cfe Dec. 26, 2017, 2:13 p.m. No.1335   πŸ—„οΈ.is πŸ”—kun   >>1336 >>1338

>>1327

 

Not sure why it didn't print the 128 for 225 though. It should have. Which means my method for generating (e, n) records might be off.

 

The script generates (0, n) for n 1 up to 300 (but I tried for up to 5000 for 225) and compares 225 with the generated a. So it should have matched if my code was correct.

 

I'll debug it. Thanks for pointing out my mistake

Anonymous ID: a2017c Dec. 26, 2017, 2:16 p.m. No.1336   πŸ—„οΈ.is πŸ”—kun

>>1334

>>1335

When I followed the directions:

 

set e to 0

set x to 2c

 

for c = 145

the cell I found had c as a instead of c^2

 

(e, n, t) = (0, 290, 146)

{0:290:435:290:145:1305}

 

now, the way I found this cell was by searching for cells that follow the rules of the VQC and the equations

ID: a2017c Dec. 26, 2017, 2:32 p.m. No.1337   πŸ—„οΈ.is πŸ”—kun

The element that I was looking for is (e, n, t) = (0, 288, 61).

The t value might be off, haven't actually mastered that in my program.

 

{0:288:145:120:25:841}

841 = 29^2

25 = 5^2

21025 = 145^2

Anonymous ID: 898cfe Dec. 26, 2017, 2:37 p.m. No.1338   πŸ—„οΈ.is πŸ”—kun

>>1335

 

I don't think I've managed to get a method to generate N's as well as I think.

 

What I've done is make a program that takes in an e and an n. Then it will look at (e, 1) and start looking at the factors there.

 

It's kind of simple though. If an a does not exist in the list, we add it. Then for each new a we check if it is divisible by any factors in the list. If it is, we divide a by that factor and add the divided part to our list.

 

It also adds every 'a' to the factor list as every 'a' is also an n in e.

 

However, my enumeration isn't complete as showed by >>1327. For 225 it failed to find 128.

 

Has anyone managed to get a proper method for enumerating (e, n)?

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 26, 2017, 2:52 p.m. No.1340   πŸ—„οΈ.is πŸ”—kun

>>1301

Heya!

 

Can you figure out a way to do all of this with redstone dust?

 

I don't know how you'd set it to generate stuff but that'd be a hell of a light show.

 

Don't forget, someone built a working Pentium in Minecraft and immediately got scooped up by IBM.

 

Related?

I wonder…

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 26, 2017, 3:32 p.m. No.1342   πŸ—„οΈ.is πŸ”—kun   >>1343 >>1345 >>1347 >>1355 >>1365 >>1729

>>1309

>>1341

I've updated my test cases to include iterating at a=c^2. See FindValuesOfBFromCSquared.

 

Interesting is that at m=5, where m is a multiple of x = 2 * c, the square root of b is the prime b squared value we are looking for.

 

This happens also for different c values, but many multiples lower.

 

Following is the latest example for c=15:

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

15=3x5

a x b *=(6,1,1) = {6:1:3:0:3:5} = 15;

1 x c =(6,5,2) = {6:5:3:2:1:15} = 15;

aa x bb =(0,2,4) = {0:2:15:6:9:25} = 225; (x+n)^2=8x8=64; (d+n)^2=17x17=289; sqrt(n/2) = 0;sqrt(2n) = 1;

1 x cc =(0,98,8) = {0:98:15:14:1:225} = 225; (x+n)^2=112x112=12544; (d+n)^2=113x113=12769; sqrt(n/2) = 7;sqrt(2n) = 14;

 

TestCalculationsFromC

β€”β€”β€”β€”β€”β€”β€”

cc.n = ((c-1)^2)/2 = ((15-1)^2)/2 = 98;

cc.d = c = 15;

cc.x = c-1 = 14;

ccna.d = ((c-1)/2)(c+1) = ((15-1)/2)(15+1) = 112;

 

TestMLevelJumpingByD

(0,1,7) = {0:1:84:12:72:98} = 7056

(0,1,6) = {0:1:60:10:50:72} = 3600

(0,1,5) = {0:1:40:8:32:50} = 1600

(0,1,4) = {0:1:24:6:18:32} = 576

Matching aabbna record found at m=4

 

FindValuesOfBFromCSquared

m=1: (0,2,16) = {0:2:255:30:225:289} = 65025;

m=2: (0,8,31) = {0:8:285:60:225:361} = 81225;

m=3: (0,18,46) = {0:18:315:90:225:441} = 99225;

m=4: (0,32,61) = {0:32:345:120:225:529} = 119025;

m=5: (0,50,76) = {0:50:375:150:225:625} = 140625; sqrt(b) == ( p.b * p.b ) Sqrt(b)=25; p.bsquared=25

m=6: (0,72,91) = {0:72:405:180:225:729} = 164025;

m=7: (0,98,106) = {0:98:435:210:225:841} = 189225;

m=8: (0,128,121) = {0:128:465:240:225:961} = 216225;

m=9: (0,162,136) = {0:162:495:270:225:1089} = 245025;

m=10: (0,200,151) = {0:200:525:300:225:1225} = 275625;

m=11: (0,242,166) = {0:242:555:330:225:1369} = 308025;

m=12: (0,288,181) = {0:288:585:360:225:1521} = 342225;

m=13: (0,338,196) = {0:338:615:390:225:1681} = 378225;

m=14: (0,392,211) = {0:392:645:420:225:1849} = 416025;

m=15: (0,450,226) = {0:450:675:450:225:2025} = 455625;

m=16: (0,512,241) = {0:512:705:480:225:2209} = 497025;

m=17: (0,578,256) = {0:578:735:510:225:2401} = 540225;

m=18: (0,648,271) = {0:648:765:540:225:2601} = 585225;

m=19: (0,722,286) = {0:722:795:570:225:2809} = 632025;

m=20: (0,800,301) = {0:800:825:600:225:3025} = 680625;

 

(1 x cc).x - (aa x bb).x = 14 - 6 = 8

 

sqrt( (1xcc).d - (aaxbb).x ) == (axb).a

sqrt( 15 - 6 ) = 3. (a matches!!!!)

Anonymous ID: 898cfe Dec. 26, 2017, 3:49 p.m. No.1345   πŸ—„οΈ.is πŸ”—kun   >>1346

>>1342

 

One thing I've noticed is that for the (e, n) where a = aa and b= bb the d is equal to a * b.

 

I don't know how it will help, but at least it is a reoccuring pattern I've noticed.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 26, 2017, 4:33 p.m. No.1347   πŸ—„οΈ.is πŸ”—kun

>>1342

Just a reminder, there is a pattern to the n component in these records.

 

n = 2*(m^2)

 

If n is the key to breaking this, then we need to figure out the value of m from either our (1xc), (1xcc) starting positions.

MinecraftAnon !!QXqSZ2ev8. ID: 1a4c93 Dec. 26, 2017, 5:11 p.m. No.1353   πŸ—„οΈ.is πŸ”—kun

>>1352

This is the only thing I've just started to use it for, and that's just because there are more functions in the BigInteger class over the Java 8 version.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 26, 2017, 7:57 p.m. No.1355   πŸ—„οΈ.is πŸ”—kun   >>1356 >>1367

>>1342

Just a little update.

 

>>1099

>The pattern GROWS in a certain way as n increases in column zero.

>This pattern is related to the pattern across row one (e,1)

 

Not sure if this was what VQC was talking about, but I've found a way to increment by c in (e,n).

 

Example output:

 

TestGrowingFromC

m=1: (6,5,7) = {6:5:27:12:15:49} = 735;sqrt(b)=7;

m=1: (6,11,10) = {6:11:33:18:15:73} = 1095;sqrt(b)=7;

m=2: (6,59,22) = {6:59:57:42:15:217} = 3255;sqrt(b)=14;

m=2: (6,77,25) = {6:77:63:48:15:265} = 3975;sqrt(b)=14;

m=3: (6,173,37) = {6:173:87:72:15:505} = 7575;sqrt(b)=22;

m=3: (6,203,40) = {6:203:93:78:15:577} = 8655;sqrt(b)=22;

 

The records are being created 2 ways for each iteration of m.

 

Method one:

 

BigInteger d = 2 * ter.c * m - ter.d;

BigInteger a = ter.c;

 

TheEndRecord eda = TerFactory.CreateForEDA( ter.e, d, a );

 

Method two:

 

BigInteger d2 = 2 * ter.c * m + ter.d;

BigInteger a2 = ter.c;

 

TheEndRecord eda2 = TerFactory.CreateForEDA( ter.e, d2, a2 );

 

Don't know if any of these records are related to a solution.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 27, 2017, 1:25 a.m. No.1360   πŸ—„οΈ.is πŸ”—kun   >>1369

>>1357

Sounds like the goal is to be speaking so correctly that they can't dismiss you.

 

They're all about dat proof, right?

Don't believe it til they see yet?

 

Yet…

Looked how far unbridled ponies have gone on faith and friendship alone?

 

Bronies is a no bueno title.

Leaves out the Pegasisters.

I just love Love.

 

Don't let the doubting thomases and demotivation shills waiver your faith.

 

You can already see you're on solid footing when you walk this path.

 

Keep going.

Finishing this leg only leads to putting the next foot down.

Anonymous ID: 99edb0 Dec. 27, 2017, 4:42 a.m. No.1365   πŸ—„οΈ.is πŸ”—kun   >>1368

It's been way too long since I paid enough attention to follow. I read the parts of the thread I missed and I have barely any idea what anyone is talking about. So does >>1342 mean we're there? I might try to turn all of this algebra into another part of those guide things I was working on since they got positive feedback.

Anonymous ID: 898cfe Dec. 27, 2017, 6:06 a.m. No.1366   πŸ—„οΈ.is πŸ”—kun

I feel a bit stuck.

 

I'm looking for patterns, but without knowing exactly what I feel stuck. It's like you can spend years pouring over any and all patterns in this grid.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 27, 2017, 8:44 a.m. No.1368   πŸ—„οΈ.is πŸ”—kun   >>1369

>>1365

We're not there yet, as far as I know.

 

We have been exploring the grid through various combinations of c^2.

 

Iterating various results in the (0,n) and (0,1) spaces using a multiplier (m) has exposed a few potential matches.

 

>>1367

To move from (0,n) to (0,1), simply create a new record where e=0, n = 1, and a = n*a.

Veritas Aequitas !!Nf9AmQNR7I ID: 3f145c Dec. 27, 2017, 9:35 a.m. No.1369   πŸ—„οΈ.is πŸ”—kun   >>1370 >>1372

>>1360

>>1364

>>1368

>>1367

Hello Anons! I'm still working too, let's keep going! TBH I've been having trouble keeping up, (with PMA that glorious faggot, lol) but I'm still looking at (e,1) and (0,n). Thanks Topol for the encouragement! I'm seeing patterns everywhere, I keep thinking about VQC's crumb: "how do we USE the row and column to get what we want?" Seems like the c^2 approach is helping us find related c, cc, aa, bb much further on in the record, with much larger c. I'm still analyzing patterns in row one, and working from c to derive all vars. One question I do have is this: are there multiple answers for a given c? Like, different small and big squares that still equal c? Also, where's Teach? He must be cooking something up.

Anonymous ID: 8a946c Dec. 27, 2017, 9:46 a.m. No.1370   πŸ—„οΈ.is πŸ”—kun   >>1371

>>1369

For c=ab where a and b are primes, there are only two differences of squares that work, one trivial (corresponding to c=1c), and the other true solution with c=a*b.

If c has additional factors (like the aa*bb cases), there will be additional solutions, though.

VeritasAequitas !!Nf9AmQNR7I ID: 3f145c Dec. 27, 2017, 9:49 a.m. No.1371   πŸ—„οΈ.is πŸ”—kun   >>1373

>>1370

Thanks for the clarification! So if the factors of c are prime, pretty much there's only one answer. If there are compound factors of c, there could be multiple correct answers for the diff of squares. Just thinking aloud over here. I'll keep working.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 27, 2017, 9:53 a.m. No.1372   πŸ—„οΈ.is πŸ”—kun   >>1389 >>1487

>>1369

Just want to share an example of an idea that looks promising, but doesn't work when applied to different test cases.

 

β€”β€”β€”β€”β€”β€”β€”β€”β€”

21=3x7

 

1 x c =(5,7,2) = {5:7:4:3:1:21} = 21;

a x b =(5,1,1) = {5:1:4:1:3:7} = 21;

1 x cc =(0,200,11) = {0:200:21:20:1:441} = 441;

 

Searching through the (0,n) space where a = c^2, we find an interesting record where the sqrt(b) equals b^2 from our desired result.

 

m=14: (0,392,295) = {0:392:1029:588:441:2401} = 1058841; sqrt(exa.b) == ( p.b * p.b ); Sqrt(b)=49; p.bsquared=49; p.b=7;

 

For this record, we know e = 0, and a = c^2, and a few other things about n. sqrt(n/2) = 14, which equals m (a multiplier used to loop and create records).

 

We need one more formula or variable in order to construct this record directly, so I have been trying to back into a solution.

 

Various examples of determining the new n from the (1xc) record:

 

new.n = (c^2 - n^2) = (2121) - (77) = 392

new.n = 2((c-n)^2) = 2((21-7)^2) = 392

 

Also tried calculating the new.d from various combinations, but nothing holds up yet.

VQC !!Om5byg3jAU ID: d6eaca Dec. 27, 2017, 12:49 p.m. No.1380   πŸ—„οΈ.is πŸ”—kun   >>1382 >>1384 >>1388 >>1390 >>1426 >>1440 >>1673

I will have more time to post tomorrow.

Thank you for your patience.

Tomorrow I will show the way in which na and (n-1)a relationships between a[t] and d[t] in row 1 cells (e,1) and the pattern in column zero (n,0), these can both be used to factor arbitrary c in < O(log l) where l is the length of c in bits.

Thank you again for being patient.

After this, we will demonstrate on the longest (2048 bit) RSA number and then factor the remaining numbers.

Then we will set up a new thread for elliptic curve discussions including why Fermat's Last Theorem took so long and required complex new objects (unnecessary).

Anonymous ID: 898cfe Dec. 27, 2017, 1:54 p.m. No.1388   πŸ—„οΈ.is πŸ”—kun   >>1391

>>1380

 

Oh boy. I'm sad and happy. Sad that the greatest puzzle I've been working on for a long time is revealed, but also happy for it.

 

I've been investigating these (0, j) for 2 * (i + 1)^2 and found some interesting results. Will post shortly, just have to gather my thoughts and code.

VeritasAequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 27, 2017, 2:44 p.m. No.1389   πŸ—„οΈ.is πŸ”—kun   >>1390

>>1372

Thanks PMA! Reviewing now.

>>1373

Thanks Anon, appreciate your knowledge.

>>1379

Agreed, MA. It’s not enough to be given then answer. How’s your theory coming along?

>>1382

Topol is right! Pop quiz lads, we gotta make Senpai proud. How can we move the ball forward?

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 27, 2017, 3:48 p.m. No.1390   πŸ—„οΈ.is πŸ”—kun

>>1380

can't wait.

 

I believe the following examples apply to the na and (n-1)a references for c=145.

 

145=5x29

a x b *=(1,5,4) = {1:5:12:7:5:29} = 145;

a x b (n-1)a *=(1,1,4) = {1:1:32:7:25:41} = 1025;

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

1 x c (n-1)a =(1,1,6) = {1:1:72:11:61:85} = 5185;

 

.>>1383

would love some of that insight.

 

>>1389

Been all over the place searching. Still no clear path to a solution.

Anonymous ID: 898cfe Dec. 27, 2017, 4:26 p.m. No.1391   πŸ—„οΈ.is πŸ”—kun   >>1435

>>1388

 

Okay, so I've been digging a lot through patterns for a and d for (e, n). And I've found a lot of exciting things.

 

For these (0, n) where n = 2(i + 1)^2 I've discovered a way of finding aabb records. Unfortunately it's not as fast as VQC's method, but it's a step in (at least) a direction.

 

So let's start with these (0, n). aabb records will all appear in (0, n) where n = 2(i + 1)^2 for some i. This i is not easily found and is n + (x - 1). For a c with two primes there exists two records. One for aa * bb and one for a=1, b=cc.

 

The record for a=1, b=cc is easily found as it is at i = n + (x - 1) which can be found by setting a=1, b=cc and generating that record. I haven't found much use for this record, but I haven't ruled out any usefulness neither.

 

The other record can be found by iterating over the possible (0, n). During my search this morning I discovered an equation for getting d[t] for any t in (0, n) (fitting the equation mentioned). I used some algebra and found an equation for reversing this. That is, finding t for a d. This means if I know the n where aa*bb appears in, finding the t for that record is done in constant time.

 

This works fine as aa*bb will have d = cc so I find a t where d = cc for (0, n), but then again I have to iterate through the different n's, which is a brute force method.

 

This yields a nice result for small values, but won't work for big numbers as RSA-100.

MinecraftAnon !!QXqSZ2ev8. ID: 273212 Dec. 27, 2017, 4:33 p.m. No.1392   πŸ—„οΈ.is πŸ”—kun   >>1405

Alright guys, here's my presentation on why T is a lie. This is N=1, with +E extending to the left. X is used as Y (T/height) variable. Very predictable results. The home row we've come to know and love.

MinecraftAnon !!QXqSZ2ev8. ID: 273212 Dec. 27, 2017, 4:43 p.m. No.1394   πŸ—„οΈ.is πŸ”—kun   >>1418

Now N=3. Now we have two new types of N. One is still evenly spaced, but spaced larger than before, and the other is two in quick succession, followed by a gap. Together, these make little macaroni noodles in the sky. The slope has also increased.

MinecraftAnon !!QXqSZ2ev8. ID: 273212 Dec. 27, 2017, 4:47 p.m. No.1395   πŸ—„οΈ.is πŸ”—kun

Now N=4. It has become angry by the assumption that it could behave well in a simple equation. Sure, there's a pattern. But it wants blood.

MinecraftAnon !!QXqSZ2ev8. ID: 273212 Dec. 27, 2017, 5:23 p.m. No.1400   πŸ—„οΈ.is πŸ”—kun

The best I've been able to do so far is pinpoint X on N2, but pinpointing X on N3 is fucking hard because of the damn macaronis. You can use N2's X to approximate N3's X, but approximately in an exponential system is not very helpful.

MinecraftAnon !!QXqSZ2ev8. ID: 273212 Dec. 27, 2017, 7:35 p.m. No.1404   πŸ—„οΈ.is πŸ”—kun   >>1405

Anyway, I guess if T is a real variable, it takes not only X, but E and N as inputs. Which somehow equal one. Then, only changing X by a variable amount, equals 2. Then changing it an different amount, equal 3.

Anonymous ID: 3f145c Dec. 27, 2017, 7:51 p.m. No.1405   πŸ—„οΈ.is πŸ”—kun   >>1406

>>1392

Interesting MA! TBH, t is derived from x, so we could do without it, or just rework any equation using t to flow from x instead.

>>1404

This could work too. If we can find a way to make t relevant to increasing n values. Or, maybe t helps us with factors in row 1, and then m or cc as B.B. helps us in column one, and where the two meet (e,n) is our spot?

VeritasAequitas !!Nf9AmQNR7I ID: 3f145c Dec. 27, 2017, 7:57 p.m. No.1406   πŸ—„οΈ.is πŸ”—kun

>>1405

And we already know e, (and factors of a for that e column)so the c^2 cc aa bb records help us navigate down to the proper n level? Thinking out loud here.

Anonymous ID: 63fecf Dec. 27, 2017, 8:58 p.m. No.1407   πŸ—„οΈ.is πŸ”—kun   >>1408 >>1415

>>1378

Nice little touch of Rainbow in the Ethereal Space - pretty awesome actually.

And the Serco spider is a nice catch.

Lots of tech/players behind that pico-sync.

Triple-Toroid…

  • Dude, you done hit a four-fer, 444 (that's Four4Four) - knockin' them quaternions outa tha park!

-ok, so some quaternion Julia's back at 'cha PATHfags. Follow the quaternion rabbit.

VeritasAequitas !!Nf9AmQNR7I ID: 5f4624 Dec. 27, 2017, 10:15 p.m. No.1412   πŸ—„οΈ.is πŸ”—kun

>>1410

The memes must flow! Fun and math together! Can’t solve this when we’re mentally constipated. Better working conditions, etc. Come on lads, let’s shit this thread up with some funny memes. I’ll post some in a bit!

Anonymous ID: 01b9f7 Dec. 27, 2017, 11:33 p.m. No.1426   πŸ—„οΈ.is πŸ”—kun   >>1427

>>1380

>>1381

>>1386

I'm a fag again. Mis explained what I meant.

For (0,n ) when n equals 2*mm for and integer m then all a and b of those records are perfect squares.

The {0, 2MM , c, c-1, 1 , cc} record is one such record with a very large M. were are looking for the {0, 2mm, c, x, aa, bb} record with a much smaller m.

I believe Chris is going to use the

a of (0,1, M) {0,1, d, x, 2*MM, b} and

d of (1,1, M) {1, 1, 2*MM, x, a, b}

to derive the smaller m with the aa bb record

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 12:05 a.m. No.1427   πŸ—„οΈ.is πŸ”—kun   >>1428 >>1451 >>1457

>>1426

Ok, I'm with you. Can you clarify M and mm? I think it means the number of jumps between the c^2 record in (0,n). How far down in N to solve for RSA? Did VQC hint at how far we have to go when he crumbed us "What temp does vinegar boil at?" It's 212 deg F. Although he's a Brit, so maybe it's in Celsius.

 

Also, does t have usefulness in later n rows? MA brought up a good point that t is kinda useless (?) in later n rows (erratic), and that x is the variable we should be focusing on. VQC also hinted by saying "when I started, I wanted x to mark the spot." He also crumbed about generating the primes in this way "Think of 3 as (1+1+0.5+0.5)" Cryptic, but I've been thinking it over a lot.

 

Thoughts, Anons?

MinecraftAnon !!QXqSZ2ev8. ID: 273212 Dec. 28, 2017, 12:24 a.m. No.1428   πŸ—„οΈ.is πŸ”—kun   >>1430 >>1444

>>1427

For RSA100, N=14387588531011964456730684619177102985211280936.I'm also retracting my earlier statement about approximating. This entire VQC concept shatters the exponential growth protection of RSA.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 12:41 a.m. No.1429   πŸ—„οΈ.is πŸ”—kun

RSA #2: All VQC Posts. Add to bread, Baker?

 

RSA #3 coming up next. After Trips were disabled, VQC dropped lots of crumbs as an anon. I'll try to find all posts by bread, and post them for later reference.

 

RSA #2.

>>>/cbts/50228

>>>/cbts/53383

>>>/cbts/53432

>>>/cbts/53436

>>>/cbts/53466

>>>/cbts/53468

>>>/cbts/53469

>>>/cbts/53473

>>>/cbts/53475

>>>/cbts/53479

>>>/cbts/53481 *** imporatant

>>>/cbts/53482

>>>/cbts/53607

>>>/cbts/53609

>>>/cbts/53613

>>>/cbts/53615

>>>/cbts/53662

>>>/cbts/53668

>>>/cbts/53673

>>>/cbts/53678

>>>/cbts/53680

>>>/cbts/53681

>>>/cbts/53699

>>>/cbts/53770

>>>/cbts/58934

>>>/cbts/59055

>>>/cbts/65200

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 12:47 a.m. No.1430   πŸ—„οΈ.is πŸ”—kun

>>1428

Holy Shit MA! Thats a BIGASSINTEGER. Like, Thicccccc Integer. Is this Integer in the Black?

 

Anyways, we need a VQC reference like Q has, so I'm gonna work on that while you do cool programming shit.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 1:13 a.m. No.1434   πŸ—„οΈ.is πŸ”—kun

RSA #4 VQC Posts. ID: bc16c7

 

>>>/cbts/98492

>>>/cbts/98560

>>>/cbts/107338

>>>/cbts/107342 rt >>>/cbts/107256

>>>/cbts/111903

>>>/cbts/111975 rt >>>/cbts/111942

>>>/cbts/111983

>>>/cbts/112148

>>>/cbts/112422

>>>/cbts/112425

>>>/cbts/112429 rt >>>/cbts/112172

 

There are more rt's but I did my best to include only relevant ones for math(s). If you fags want all of them, i can update. Let me know.

IseePatterns !kIkD/SqZ4s ID: 898cfe Dec. 28, 2017, 1:31 a.m. No.1435   πŸ—„οΈ.is πŸ”—kun

>>1391

Some more results:

 

To find the t for a given a at (e, 1) (for any e):

 

def getTFromA(e, a):

if e % 2 == 0:

return math.sqrt((a - (e / 2))/2)

else:

t = a - int(e / 2) - 1

t = int(t / 2)

t = t + 1

t = int(math.sqrt(t))

return t

 

Equation for even e: sqrt( ( a - (e / 2) )/2 )

Equation for odd e: floor(sqrt((a - floor(e / 2) - 1)/2 + 1)

 

To find the a for a given t at (e, 1):

 

def getAFromT(e, t):

if e % 2 == 0:

return (e / 2) + 2t*2

else:

return int(e / 2) + 4 * ( (t * (t + 1) ) /2 ) + 1

 

To find the t for a given d at (e, 1) (for any e):

 

def getTFromED(e, d):

if e % 2 == 0:

t = d - (e / 2)

t = t / 2

t = t * 4 + 1

t = t / 4

t = math.sqrt(t) + 1

return -1/2 + t

else:

t = d - (2 + int(e / 2))

t = t / 2

t = t + 1

t = math.sqrt(t)

t = t + 1

return t

 

To find the d at a given t for (e, 1) (for any e):

 

def getDFromET(e, t):

if e % 2 == 0:

return (e / 2) + 4 * ( (t * (t + 1) ) / 2)

else:

return (2 + int(e/2)) + 6n + 4( ( t * (t - 1) ) /2 )

 

I've used the getTFromED when looking for the aabb record as the d = cc in that record, but without being able to have a nice way of knowing which (0, n) the aabb record is in, it requires the brute force method.

 

I'm looking into patterns for the n * a and (n - 1) * a now.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 1:47 a.m. No.1436   πŸ—„οΈ.is πŸ”—kun   >>1437

RSA #2: Stony Senpai Version. ID: aaa2ec

Baker, please add to next bread, under RSA #2, part 2.

 

>>>/cbts/74027

>>>/cbts/74303

>>>/cbts/74358

>>>/cbts/74393 rt >>>/cbts/74302

>>>/cbts/74488 rt >>>/cbts/74373

>>>/cbts/74509

>>>/cbts/74525

>>>/cbts/74623

>>>/cbts/74729

>>>/cbts/74965

>>>/cbts/74983 rt >>>/cbts/74790

>>>/cbts/75025 rt >>>/cbts/74769

>>>/cbts/75055 rt >>>/cbts/74778

>>>/cbts/75094 rt >>>/cbts/74386

>>>/cbts/75314 rt >>>/cbts/74769

>>>/cbts/75203 rt >>>/cbts/71555

>>>/cbts/75265 rt >>>/cbts/74278

>>>/cbts/75291 rt >>>/cbts/74380

>>>/cbts/75304 rt >>>/cbts/74355

>>>/cbts/75358 rt >>>/cbts/74459

>>>/cbts/75365 rt >>>/cbts/75349

>>>/cbts/75381 rt >>>/cbts/74169

>>>/cbts/75393 rt >>>/cbts/71582

>>>/cbts/75430 rt >>>/cbts/71369

>>>/cbts/75435 rt >>>/cbts/67618 *** PMA WTF!!! WORTH ANOTHER LOOK? VQC REPOSTS YOUR SHIT>>>

>>>/cbts/75486 rt >>>/cbts/71423

>>>/cbts/75506 rt >>>/cbts/72925

>>>/cbts/75514 rt >>>/cbts/75455

>>>/cbts/75519 rt >>>/cbts/73458

>>>/cbts/75527 rt >>>/cbts/66664

>>>/cbts/75535 rt >>>/cbts/75455

 

This was when Trips were banned suring RSA #2, and our Senpai showed up ALL BAKED to drop some wisdom. Enjoy the thread, it's fucking awesome.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 2:13 a.m. No.1438   πŸ—„οΈ.is πŸ”—kun   >>1445

Corrected! RSA#2 Stony Senpai Edition. Baker PLS add to Bread.

 

>>>/cbts/74027

>>>/cbts/74303

>>>/cbts/74358

>>>/cbts/74393 rt >>>/cbts/74302

>>>/cbts/74488 rt >>>/cbts/74373

>>>/cbts/74509

>>>/cbts/74525

>>>/cbts/74623

>>>/cbts/74729

>>>/cbts/74965

>>>/cbts/74983 rt >>>/cbts/74790

>>>/cbts/75025 rt >>>/cbts/74769

>>>/cbts/75055 rt >>>/cbts/74778

>>>/cbts/75094 rt >>>/cbts/74386

>>>/cbts/75134 rt >>>/cbts/74769

>>>/cbts/75203 rt >>>/cbts/71555

>>>/cbts/75265 rt >>>/cbts/74278

>>>/cbts/75291 rt >>>/cbts/74380

>>>/cbts/75309 rt >>>/cbts/74355

>>>/cbts/75358 rt >>>/cbts/74459

>>>/cbts/75365 rt >>>/cbts/75349

>>>/cbts/75381 rt >>>/cbts/74169

>>>/cbts/75393 rt >>>/cbts/71582

>>>/cbts/75430 rt >>>/cbts/71369

>>>/cbts/75435 rt >>>/cbts/67618 *** PMA WTF!!! WORTH ANOTHER LOOK? VQC REPOSTS YOUR SHIT>>>

>>>/cbts/75486 rt >>>/cbts/71423

>>>/cbts/75506 rt >>>/cbts/72925

>>>/cbts/75514 rt >>>/cbts/75455

>>>/cbts/75519 rt >>>/cbts/73458

>>>/cbts/75527 rt >>>/cbts/66664

>>>/cbts/75535 rt >>>/cbts/75455

 

This was when Trips were banned during RSA #2, and our Senpai showed up ALL BAKED to drop some wisdom. Enjoy the thread, it's fucking awesome.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 2:35 a.m. No.1442   πŸ—„οΈ.is πŸ”—kun

>>1441

Thanks Baker! Ponies??? I know you got a big stash. Bring out your best!

 

>>>/cbts/74965

Chris, the day we (/vqc/faggots) can smoke a joint with you and talk about the universe (and meet each other) will be pretty awesome. We are all the wombMen, waiting to be birthed into the next level. Thanks for inspiring me to raise myself up another few quantum levels. What a trip this quest is!

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 2:49 a.m. No.1444   πŸ—„οΈ.is πŸ”—kun   >>1448

>>1428

Can you explain?

>I'm also retracting my earlier statement about approximating. This entire VQC concept shatters the exponential growth protection of RSA.

How so MA?

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 28, 2017, 2:57 a.m. No.1445   πŸ—„οΈ.is πŸ”—kun

>>1438

It's frickin' trippy to see the conversations I had with Him when "behind and ahead of myself" were weren't as stabilized as they appear to be "now".

 

Something to look forward when your homework is finished. /VQC/VA is kindly setting up the Legend of VQC, when he Guided us to a Map, that at the time looked as if a blank whiteboard.

 

And then the motherfucker lit a blunt, threw a marker on the ground and said, "Now… What's going on here?". When the smoke cleared, someone was holding the marker which was wrapped in a small note. He showed it to his friends and they started drawing on the wall.

 

Now about that time, one of the art students smelled the smelly smell of something that smelled smelly. And his eyes lit up because that's the most illegal stuff on the planet. I'm talking blasting off at the speed of light with sounds in all directions in every frequency at every octave and volume. A piercing, rising tone that takes you all the way through the veil… chakras on full blast.

 

"What is going on HERE/NOW?!", that loquacious hobbit exclaimed as he burst in the door.

 

The nerds were unphased but this albino lookin' Powder-wannabe in a morpheus get up raised the blunt. Sheeeeeeeeeeeit. Apparently das muh nigga right der.

 

So after some really heady conversations, because Dank Man Traning, the art student went and started drawing on the whiteboard cuz… fuck… I know how to use one of these. Dude was so high he drew a fucking pony. This greatly confused the nerds. So Albino Morpheus pulled out the Wizard's Pipe full of Meta*Sign CBTs and told the art student to quit being such a fucking spaz.

 

After the Preesh Pipe, the two wandered off to enjoy a cigarette and step out of the hotbox for a second.

 

Upon returning, the art student set up and easel and drew odd hypotheticals to the side… but still on the whiteboard every now then when the space needed to be held with a little gayety.

 

Meanwhile, Twinkie von Fishburn smoked so much that he's more or less in his own little world, popping in an out to shout random things that get the nerds excited for whatever reason aaaaand then right back to la la land.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 3:02 a.m. No.1446   πŸ—„οΈ.is πŸ”—kun

Alright, faggots. I can't keep up with PMA, bc he's on top of his math and program game. But I did make myself useful compiling VQC posts. I have a place once again in the /VQC/ universe. Feeling peaceful and happy over here. This quest is really about bringing together fine minds (from all over the world) to work on a cool math problem (hardest I've ever worked on) to bring transparency (as opposed to dark and oppressive secrecy) to our world. I won't give up on this quest, bc it's the best I've ever worked on. I love you, lads. Glad to be here, friends.

VQC !!Om5byg3jAU ID: b1d6e8 Dec. 28, 2017, 3:06 a.m. No.1447   πŸ—„οΈ.is πŸ”—kun   >>1449 >>1452 >>1461 >>1463 >>1673

Bear with me folks.

Going through the final part will knock out the Internet as RSA is involved in handshakes for SSL.

Trump held up a sign saying Keep Tweeting.

We will hold.

In the meantime, consume anything you can on what drew mathematicians to believe elliptic curves and modular forms were the same object.

VQC !!Om5byg3jAU ID: b1d6e8 Dec. 28, 2017, 3:08 a.m. No.1448   πŸ—„οΈ.is πŸ”—kun   >>1473

>>1444

Because a search exponential in the length of c is turned into a calculation which is the same complexity as finding a square root < O(log l) where l is the length of c in bits

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 28, 2017, 3:09 a.m. No.1449   πŸ—„οΈ.is πŸ”—kun

>>1447

>elliptic curves and modular forms are the same

Here's one… but… yooooou know meeee. -shrug-

 

Rhoades, PΓ¦ths… -shrug-

http://math.stanford.edu/~rhoades/FILES/EC.pdf

Anonymous ID: 01b9f7 Dec. 28, 2017, 5:34 a.m. No.1451   πŸ—„οΈ.is πŸ”—kun   >>1454 >>1473

>>1427

Your right m is doing double duty!

I am using m to count down the perfect square records in (0,n).

Amazingly it turns out that m is the difference between sqrt(a) and sqrt(b)

when n = 2*mm

It is still being used to represent the t in (1,1) records where d equals 2mm and its the t in (0,1) records a equals 2mm so m is MOVEMENT usually up and down t in a cell.

MinecraftAnon !!QXqSZ2ev8. ID: 2b0f03 Dec. 28, 2017, 9:50 a.m. No.1463   πŸ—„οΈ.is πŸ”—kun   >>1465 >>1476

>>1447

If we could get the EC VQC generation code now, it would be helpful for when we figure this out.

 

Proof of progess: Square root of RSA100's N2's X: 2797181014427

 

Critical formula update following this post.

MinecraftAnon !!QXqSZ2ev8. ID: 2b0f03 Dec. 28, 2017, 9:58 a.m. No.1464   πŸ—„οΈ.is πŸ”—kun   >>1465

Method for generating any cell by inputing E, N, and X, and testing if actually valid to the grid as follows.

 

ArbitraryCell(e, n, x){

e = e;

n = n;

x = x;

a = (x * x + e) / (2 * n);

d = x + a;

b = a + 2 * (x + n);

c = a * b;

}

 

boolean isValid(){

return (c - d * d) == e;

}

 

Remember to follow the order of operations when implementing!

MinecraftAnon !!QXqSZ2ev8. ID: 2b0f03 Dec. 28, 2017, 10:19 a.m. No.1466   πŸ—„οΈ.is πŸ”—kun   >>1468 >>1471 >>1473 >>1487

>>1465

I've literally done no work in (E,0). My method so far has been to take an input C, move to the home cell (E,1), then approximately jump down N. Today I'm gonna work on searching for valid cells, deriving their pattern, and unlocking the entire N column for a given E. Multithreaded, this is an insanely fast solution.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 12:30 p.m. No.1473   πŸ—„οΈ.is πŸ”—kun   >>1477 >>1487

>>1451

>>1457

Thanks for the clear explanation about M and mm! So can you just plug in c and get the correct number of jumps to make? Is it exact or approximate?

 

>>1466

Awesome, MA! This is how my mind keeps thinking about it too. Starting from C, we find d and e, then go to e, get all the factors and jump down n levels. There must be an elegant way to do it, but I like your logarithmic hammer approach! HULK SMASH!! The meme is for you.

 

>>1467

Liking the new Mascot! Thanks Topol!

 

>>1448

Thanks boss! I guess I asked the same question again, duh. Thanks for your patience and willingness to explain things more than once.

IseePatterns !kIkD/SqZ4s ID: 898cfe Dec. 28, 2017, 12:31 p.m. No.1474   πŸ—„οΈ.is πŸ”—kun   >>1475 >>1479

When VQC refers to n * a and (n - 1)*a does he refer to specifically the n - 1 or the previous n?

 

For example in the column for e = 1 we have n = 1, 5, 13, 17 etc. So if we take (1, 5, 2) where a = 17 and b = 53. Is he talking about n * a (17 * 5) and (n - 1) * a (17 * 4) or (17 * 1)?

Anonymous ID: 01b9f7 Dec. 28, 2017, 1:23 p.m. No.1477   πŸ—„οΈ.is πŸ”—kun

>>1473

If we know m problem solved.

But we know big M

and for t = m, M in (0,1) a equals 2mm, 2MM

and for t = m, M in (1,1) d equals 2mm, 2MM

and we know (0, n=2MM) {0, 2MM, c, c-1, 1, cc}

we want (0, n=2mm) {0, 2mm, c, x, aa, bb}

we also know Sqrt(b) - Sqrt(a) = 2m

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 1:35 p.m. No.1479   πŸ—„οΈ.is πŸ”—kun

Hey Lads! Here's a good link about the elliptical curve question VQC asked us to read up on:

https:/ /plus.maths.org/content/fermats-last-theorem-and-andrew-wiles

 

>>1474

>>1475

I'm following! Good question. Found these in the VQC archives, maybe helpful?

>>>/cbts/107342 rt >>>/cbts/107256

>>33

>>495

 

>>1476

Right, Baker? I think he did. MA is going to town with his LogHammer. When we tie it together with the c^2 aa bb cc patterns from (0,n) that will be awesome.

>>699

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 28, 2017, 1:45 p.m. No.1483   πŸ—„οΈ.is πŸ”—kun

I'm just pointing out that Green Number Muncher has 2 c's for eyes… (c)o-Inside-n(c)e?

 

aren't n and e and c like… significant or something? -draws more weirdness-

 

TIME FOR A SONIC RAINBONG!

-quantum bubble nnnnnoises-

Anonymous ID: e606e3 Dec. 28, 2017, 1:55 p.m. No.1484   πŸ—„οΈ.is πŸ”—kun   >>1485

>>1482

I don't play the stock market. Some infosec companies may make a killing helping everyone else "secure" their shit with ECC instead (kek)… Banks and exchanges probably the first to take their online sites down, with government portals following. Huge losses likely, especially for the exchanges. Then major panic for all the other hardware crap and legacy software that only supports RSA and can't be replaced.

Precious metals will be up, and probably cryptocurrencies after the inital panic selling when everyone figures out none of them are affected.. yet

Planning for the reactions is not an easy task

 

Then come the explosive leaks, when people start reading everything previously encrypted with GPG

MinecraftAnon !!QXqSZ2ev8. ID: 2b0f03 Dec. 28, 2017, 2:16 p.m. No.1485   πŸ—„οΈ.is πŸ”—kun   >>1486

>>1484

There's playing the market, and then there's being able to mathematically prove where the roulette ball is going to land…

 

I'm sick of working my fucking ass off to break even. This is going to happen, and we should all start to get ready.

 

www.pssurvival.com

Anonymous ID: e606e3 Dec. 28, 2017, 2:28 p.m. No.1486   πŸ—„οΈ.is πŸ”—kun   >>1488

>>1485

First problem is that in order to win, you have to give your money to a bank up front. Mostly not a risk I'm willing to take, even when 100% sure on a bet. What we're doing here is unprecedented so I would love to see mathematical proof of what happens to the stock market when any RSA key can be factored in seconds… I can tell you exactly what tech breaks and what doesn't, but how that plays out when some CEO hears their bank can be hacked and overreacts is beyond even my intuition. There will be chaos, and if you can translate that into stock bets then have fun getting rich!

 

This is happening either way

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 28, 2017, 2:34 p.m. No.1487   πŸ—„οΈ.is πŸ”—kun   >>1496 >>1729

>>1466

>>1473

>>1372

MA, VA

 

Just ran another test on RSA100.

 

If the solution relies on finding out the multiple (m) at a record where e=0, x=2cm, a=c^2, then m would have to be:

 

m=42489607263661711309596655624896701585610939620632192811974298340236852638535916326429571625162791

 

Going to continue exploring the na and (n-1)a…

MinecraftAnon !!QXqSZ2ev8. ID: 2b0f03 Dec. 28, 2017, 2:35 p.m. No.1488   πŸ—„οΈ.is πŸ”—kun   >>1489 >>1490

>>1486

I agree with the gold and silver up sentiment. Security companies specializing in ECC as well. In terms of down, every chip credit card uses RSA. As VQC mentioned, SSL certificates use RSA. The entire online transaction market is about to be blown the fuck out. Amazon down?

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 28, 2017, 2:41 p.m. No.1489   πŸ—„οΈ.is πŸ”—kun   >>1491

>>1488

Are you in it for the money or the love of math?

Don't worry about the money.

Imagine like… all the money going away at once cuz all banking will be obliterfucked and restructed.

 

Might take 3 days or 15 days. Maybe you'll hardly notice it.

 

But, if you MUST focus on that whole eating and having a roof over your head… broaden your prayer to include everyone so that no one ever has to have that concern ever again.

 

>>1232

Anonymous ID: e606e3 Dec. 28, 2017, 2:49 p.m. No.1491   πŸ—„οΈ.is πŸ”—kun

>>1489

Nothing would make me happier than the concept of money disappearing or just not having existed at all!

Still no luck changing my reality to that one, not for lack of trying

MinecraftAnon !!QXqSZ2ev8. ID: 2b0f03 Dec. 28, 2017, 2:51 p.m. No.1493   πŸ—„οΈ.is πŸ”—kun

>>1490

I'm in it because the implication is a new age of knowledge. At the same time, if the financial system collapses and looting starts, it's gonna be tough to pull through to the other side. This will be the lightning bolt that marks the beginning of the great storm.

MinecraftAnon !!QXqSZ2ev8. ID: 2b0f03 Dec. 28, 2017, 3:08 p.m. No.1495   πŸ—„οΈ.is πŸ”—kun

Of course as soon as I start thinking about that, I run into a massive problem. For this cell in format {E:N:D:X:A:B:C},

 

61218444075812733697456051513875809617598014768503

4

7652305509476591712182006439234476202199751846067

3

7652305509476591712182006439234476202199751846064

7652305509476591712182006439234476202199751846078

58557779610365799850575360231375488434617810249426155485106497416876136614796334204692772478136992

 

This should absolutely not pass isValid, but it does… Hmmmm….

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 3:29 p.m. No.1496   πŸ—„οΈ.is πŸ”—kun

>>1487

Hey PMA! It seems like you've got the column 0 work moving forward, and MA has the row work moving forward. When we can merge the two, maybe we have an elegant solution? Thinking out loud.

>>699

"How do we USE the row and column to get what we need?"

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 3:36 p.m. No.1498   πŸ—„οΈ.is πŸ”—kun   >>1499

Here's an idea: Take sqrt(c), get d and e. Go to (e,1) and find all the factors for c (MA's approach). Meanwhile, using PMA's approach, forget the e value for a minute, go to column 0, and start working to find the multiples of the factors from the first cell in (e,1). If we know all the factors from (e,1), can we bring them over to (0,n) and start looking up and down for c^2 aa bb cc, etc using those c factors? Just a thought.

ProgramMathAnon !dSvrkhSLR6 ID: 945313 Dec. 28, 2017, 3:49 p.m. No.1499   πŸ—„οΈ.is πŸ”—kun   >>1500 >>1501

>>1497

MA, do you need an isvalid formula or a method to construct your records? I posted a while back, but happy to share.

 

>>1498

For small numbers, finding the factors is straight forward. RSA length numbers is a different story. We're missing something here.

 

VQC dropped 2 hints about na and (n-1)a relationship.

 

I've run a number of tests comparing the 2 records, and the a and d values are where the movement appears. And specifically at d we start getting new numbers to explore.

 

Still not seeing a way to relate the the c and ab records.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 6:23 p.m. No.1504   πŸ—„οΈ.is πŸ”—kun

>>1503

>>1502

Gotcha, thanks Baker! So we have to work on those two crumbs about using na and (n-1)a to tie c and ab records / columns and rows together?

Ok, here's what I found in the archives, lads. Hopefully there's something here that will help us.

 

>>>/cbts/107342 rt >>>/cbts/107256 na, nb, c, N, cN

>>>/cbts/50228 na, (n-1)a calc.

>>>/cbts/53432 towards the bottom of post, see quote below

>>>/cbts/53468 "Big part of key"?

 

Interesting passage about primes, c, n and N from:

>A value c will only appear in one column since the remainder of a number never changes.

>Since a value c can have more than one factorisation, a number c can appear more than once in a column at difference values of n

>A prime number appears once in a column. This is SIMPLE to calculate. >This calculation is important. I call n for a prime number big_n or N

>The product of two primes appears twice. Once the same as the prime, which is easy to calculate and the second time, this is the n that we are after.

>How many times does the product of three distinct primes appear?

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 6:41 p.m. No.1505   πŸ—„οΈ.is πŸ”—kun   >>1506 >>1507

Here's some hilarious shit I found in RSA #2:

 

"There seem to be two types of anon in this world: those who completely enjoy grade B faggotry and those who only partially enjoy it. Obviously you're the latter, but if you didn't have at least a bit of interest you would never have even read my post.

The idea here, dubious as it may sound, is that there is hidden structure in the fundamental nature of the numbers that make up our highly-secured Internet and emerging technologies such as crypto-currencies like bitcoin. This structure is postulated to be exploitable, potentially resulting in one of the most epic and spectacular happenings of all time.

So basically, while this certainly is grade B faggotry, it is ambitious faggotry looking to upgrade itself to A+ faggotry if conditions permit. Which IMO is max /comfy/ but maybe I'm just a grade B faggot in the end."

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 6:45 p.m. No.1507   πŸ—„οΈ.is πŸ”—kun   >>1508 >>1509

>>1505

I found a pattern in (0,n) for (0,1) to (0,2)

 

(0,1) (0,2)

(0,1,4,2,2,8) (0,2,3,2,1,9) d = d-1, a = a-1, b = b+1

(0,1,12,4,8,18) (0,2,8,4,4,16) d = d-4, a = a-4, b = b-2

(0,1,24,6,18,32) (0,2,15,6,9,25) d = d-9, a = a-9, b = b-7

(0,1,40,8,32,50) (0,2,24,8,16,36) d = d-16, a = a-16, b = b-14

 

So for for (0,1,d,x,a,b) we can solve for t by t/2.

Then we can have a little square called L = t*t

Then, from (0,1,d,x,a,b) we can get (0,2,d-L, x-L, b-L+2)

 

I'm going to see if I can generalize this further.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 7:01 p.m. No.1508   πŸ—„οΈ.is πŸ”—kun

>>1507

(0,2) (0,3)

(0,2,3,2,1,9)

(0,2,8,4,4,16)

(0,2,15,6,9,25) (0,3,12,6,6,24) d = d-3, a = a-3, b = b-1 (3*1)

(0,2,24,8,16,36)

(0,2,35,10,25,49)

(0,2,48,12,36,64) (0,3,36,12,24,54) d = d-12, a = a-12, b = b-10 (322)

(0,2,63,14,49,81)

(0,2,80,16,64,100)

(0,2,99,18,81,121) (0,3,72,18,54,96) d = d-27, a = a-27, b = b-25 (333)

(0,2,120,20,100,144)

 

Here is the pattern from (0,2) to (0,3) for cells with the same X value (also it skips x values in (0,3).

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 7:07 p.m. No.1509   πŸ—„οΈ.is πŸ”—kun   >>1510 >>1512 >>1729

>>1507

YEAH!!! CollegeANON in da house!! Where you been, faggot? Nice to see you!

 

PMA was working on something like this, and VQC even rt'd my post about it (along with a cryptic message). Check it out, see if it matches your idea. We even got yelled at by TLA later! Those were the good 'ol days:

>>>/cbts/75435 (VQC) rt >>>/cbts/67618 (Me)

PMA's posts I was replying to, about finding small square using t.

>>>/cbts/67332

>>>/cbts/67336

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 7:18 p.m. No.1513   πŸ—„οΈ.is πŸ”—kun

Also I have that for any (0,n) cell we have:

For odd n:

(0,n)[t] = (0,1)[t] * n

 

For even n:

(0,n)[t] = (0,2)[t] * (n/2)

 

Also, since we can now calculate any index for (0,1) and (0,2) we can use these rules to calculate for any (0,n).

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 7:22 p.m. No.1515   πŸ—„οΈ.is πŸ”—kun

>>1506

This made me LOL!! Thanks, Anon. "It is ambitious faggotry looking to upgrade itself to A+ faggotry if conditions permit." LOLOLOLOL! Better check the weather map. Hurricane MILO about to make landfall???

Anonymous ID: a2017c Dec. 28, 2017, 8:06 p.m. No.1519   πŸ—„οΈ.is πŸ”—kun   >>1523

>>1517

It's actually quite easy to encrypt for a group.

You just need to share a password with each recipient using asymm crypto. (Because there's no way to secretly send a shared key without asymm crypto over the internet.)

 

It wouldn't work with a large amount of recipients, but that's not necessary.

You wouldn't trust a large group of people with sensitive information.

Anonymous ID: 8a946c Dec. 28, 2017, 8:16 p.m. No.1522   πŸ—„οΈ.is πŸ”—kun

NIST recently posted a bunch of submissions for potential quantum-resistant crypto algorithms:

http:/ /csrc.nist.gov/Projects/Post-Quantum-Cryptography/Round-1-Submissions

Anonymous ID: e606e3 Dec. 28, 2017, 8:18 p.m. No.1523   πŸ—„οΈ.is πŸ”—kun   >>1524

>>1519

If you're talking decentralised, who generates the key? And how do you revoke it when one of the recipients gets busted?

All of it relies on asymmetric crypto of course, but I only know of one algorithm that supports generating the group key without trusting a single person to set the initial secret, and still allows updating that key for all new members without giving it anyone else

 

>>1521

If you want to implement shit, here is a 3+ years old implementation of the paper, ignoring the reduction function

Probably also the first public one as I haven't seen this around

https://pastebin.com/XuN4U7Dv

Been a while since I looked at this code so don't ask me how it works :)

Anonymous ID: a2017c Dec. 28, 2017, 8:29 p.m. No.1524   πŸ—„οΈ.is πŸ”—kun

>>1523

One person (admin of the chat) has authority over it all. Only would work (or be trustworthy) for a group of people you trust. I chose IRC because there are many servers, and asymmetric crypto allows you to use encryption regardless of who "Eve" is.

I've already written special headers, and sorting messages into chunks and such.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 8:59 p.m. No.1525   πŸ—„οΈ.is πŸ”—kun   >>1526

VQC also wanted us to analyze the column with no gaps. I noticed that for every entry where e is a negative square (-1,-4,-9), there are no gaps in cells, just in the values within the cells, only within even. I've been analyzing the column -1 and I have found an interesting pattern:

 

For (-1,n,d,x,a,b) we can do the following to get to (-1,n+1).

If a%(n+1) == 0, then let M = a/(n+1).

 

Then we have the following:

(-1,n,d,x,a,b) -(-1, n+1, d-M, x, a-M, b-M+2)

 

I think this makes sense.

 

I'm going to look into -4 row next.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 9 p.m. No.1526   πŸ—„οΈ.is πŸ”—kun   >>1529

>>1525

 

(-1,1) (-1,2)

(-1,1,7,3,4,12) (-1,2,5,3,2,12) d = d - 2, a = a - 2, b = b +/- 0 M = 2

(-1,1,17,5,12,24) (-1,2,11,5,6,20) d = d - 6, a = a - 6, b = b - 4 M = 2 * 3

(-1,1,31,7,24,40) (-1,2,19,7,12,30) d = d - 12, a = a - 12, b = b - 10 M = 2 * 2 * 3

(-1,1,49,9,40,60) (-1,2,29,9,20,42) d = d - 20, a = a - 20, b = b - 18

(-1,1,71,11,60,84) (-1,2,41,11,30,56) d = d - 30, a = a - 30, b = b - 28

 

These M's also correspond to 2, 6, 12, 20, 30 = 2, + 4, + 6, + 8, + 10

The M's are also the same as (for any index t) (0,1)[D] / 2

Also M = a/2.

 

(-1,2) (-1,3)

(-1,2,5,3,2,12)

(-1,2,11,5,6,20) (-1,3,9,5,4,20) d = d-2, a = a-2, b = b +/- 0 M = 2 = 2 i=2

(-1,2,19,7,12,30) (-1,3,15,7,8,28) d = d-4, a = a-4, b = b-2 M = 4 = 2*2 i=3

(-1,2,29,9,20,42)

(-1,2,41,11,30,56) (-1,3,31,11,20,48) d = d-10,a = a-10, b = b-8 M = 10 = 2*5 i=5

(-1,2,55,13,42,72) (-1,3,41,13,28,60) d = d-14,a = a-14, b = b-12 M = 14 = 2*7 i=6

(-1,2,71,15,56,90)

(-1,2,89,17,72,110) (-1,3,65,17,48,88) d = d-24,a = a-24, b = b-22 M = 24 = 222*3 i=8

(-1,2,109,19,90,132) (-1,3,79,19,60,104) d = d-30,a = a-30, b = b-28 M = 30 = 235 i=9

 

For these in (-1,2) to (-1,3) we get M = a/3.

 

(-1,3) (-1,4)

(-1,4,4,3,1,15) i=1

(-1,3,9,5,4,20) (-1,4,8,5,3,21) d = d-1, a = a-1, b = b+1 M = 1 i=2

(-1,3,15,7,8,28) (-1,4,13,7,6,28) d = d-2, a = a-2, b = b M = 2 i=3

(-1,4,19,9,10,36) i=4

(-1,3,31,11,20,48) (-1,4,26,11,15,45) d = d-5, a = a-5, b = b-3 M = 5 i=5

(-1,3,41,13,28,60) (-1,4,34,13,21,55) d = d-7, a = a-7, b = b-5 M = 7 i=6

(-1,4,43,15,28,66)

(-1,3,65,17,48,88) (-1,4,53,17,36,78) d = d-12,a = a-12, b = b-10 M = 12 i=8

(-1,3,79,19,60,104) (-1,4,64,19,45,91) d = d-15,a = a-15, b = b-13 M = 15 i=9

 

For these in (-1,3) to (-1,4) we get M = a/4

 

Here's a little more work from my file.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 9:20 p.m. No.1529   πŸ—„οΈ.is πŸ”—kun   >>1545

>>1526

This pattern seems to work for (-4,n) also. So for any negative square e we can generate the entire column. BUT this omits some values, looking at my previous post you can easily notice the ones that don't get generated.

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 28, 2017, 9:35 p.m. No.1531   πŸ—„οΈ.is πŸ”—kun   >>1532 >>1533 >>1537 >>1723

>>1478

 

Which stocks are likely to be impacted… Oh boy, See this is where getting far enough away to see the actual size of this thing is important. At the end of this… If it is real… Stocks will not really exist. Hell, there wont even be money in any form that we recognize. Take a minute… 1.2…59.60 ok. The future will look like this after Eliptical is broken by us and someone like you writes an easy peasy program that some Hello World level programmer like my self could use to open up pretty much anything.

 

We COULD go get us some BTC. Like a billion dollars worth… And go shopping. We could do someting like have the FED wire a billion dollars into our bank accounts under Janet Yellon's authority. We could Delete all the information of any police or agent who came looking or simply have them arrested by filing the right forged docments.

 

This is if only a few people have this and want to use it to take advantage of the world. Honestly if I had this tool tomorrow I would not even touch the Satoshi Wallet. If I did take it it would simply be to send BTCs to random people to prove that I could not to buy a car or blow and hookers or whatever.

 

What we are talking about here is an excession event for our species. It is so big. So mind bendingly huge that nearly nothing will remain of our way of life after it is finished.

 

Stocks wont matter. Money wont matter. Jobs wont matter. Go watch some zombie movies. It will be like that but without the zombies. Everything will change and it will be hard to adapt for many. If you want to be ready for this then do the basic prepper 101 shit for you and your family. 90 days food water clothing self defense.

 

If you can then bump that all up to about a year . Also have a plan and a backup plan. If you are in a major city, have a plan to get out. Also know you will not be alone. Be prepared. I have been looking at Puerto Rico. They still dont have power. They are almost entirely running on paper money. SO have some paper money as well. They say silver and gold are shit but Lighters, liquor and cigarettes as well as weed are money.

 

Don't even worry about stocks. They wont matter. I wouldn't even bother selling them or your crypto currencies. It wont matter. That said humanity will collapse then rise from the ashes FAST. If this is real then WE ARE TASKED TO BE THE ONES WHO HAVE TO SAVE EVERYONE. I am not a programmer but I am a very creative builder. I plan on building something with this tool if the tool comes available. I will try by best to help as many people as I can with it. I sincerely hope there are some things in the works already (Chris) but if not it will be up to us.

 

We will survive this by what we build.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 9:36 p.m. No.1532   πŸ—„οΈ.is πŸ”—kun   >>1534

>>1531

I mean the whole problem is if you just took a bunch of money it's still probably illegal. Like safeguards aside, just because you can do something doesn't mean you are legally able.

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 28, 2017, 9:50 p.m. No.1534   πŸ—„οΈ.is πŸ”—kun

>>1532

This is the fundamental hacker problem. We hackers just do it to see if we can. I am a physical world hacker. I like to make stuff. I like to replace my water heater with black hose and the sun. I like to grow my own food. Hell, I tried to make my own shoes which was awful.

 

To use this tool we have openly would be like buying nuclear bombs at a fireworks stand and giving them out to kids. That said… The opportunity to do GOOD is even bigger. That is why I am here. If you crazy fuckers pull this off I hope to be a chronicler and resident big picture philosopher to get everyone to think about what we are doing and what we should do as opposed to just CAN we do it.

Anonymous ID: 8d5142 Dec. 28, 2017, 9:51 p.m. No.1535   πŸ—„οΈ.is πŸ”—kun   >>1536 >>1543 >>1546 >>1554 >>1603

Hi anons and namefags. Your old pal TLA here; was doing family stuff too over xmas. I'll try to get back into this but I am on the road so will be limited. Also battery running out here.

 

These threads would be easier to follow, for me, if they were written in mathese: propositions, lemmas, laws, identities etc. But that's ok, I will just have to dig in and try to interpret. Then translate if I can. For now just two simple questions if any anons wouldn't mind answering.

 

First, given c, where c=ab and a, b are unknown, are anons trying to 'travel' to the 'record' ending in { …. a:b} ? Because it doesn't seem like that's what VQC was hinting at. Rather, that the factors would be 'exposed' in some way.

 

Second, as far as this hint:

>>1099

Have anons realized that there should be four 'records' with product c^2? these would be

1*aabb

a*abb

aa*bb

b*baa

 

(assuming a <b ). Gonna have a look at some of these tonight.

 

Unlike some of you, my redpilling the family attempts did not go well and I am a bit sad about that. I figure, at least, I will have softened them up a bit. My sister even thinks Antifa is cool and doesn't wanna hear otherwise :( cheer me up anons, and keep up the good work.

Anonymous ID: e606e3 Dec. 28, 2017, 9:52 p.m. No.1537   πŸ—„οΈ.is πŸ”—kun   >>1541

>>1531

There is no way only a few people have this, we distribute the information or get killed (good luck assholes!).. You know some idiot is gonna try to spend the early coins and crash it before they even make it to payout.

The collapse has been a long time coming and if we can trigger it then more power to us, at least the early warning is nice :)

If you haven't prepped yet then better start. Didn't consider lighters as currency but damn now I'm gonna buy some.

 

Proving P=NP will make building awesome things (free energy anyone?) way easier as soon as we figure out wtf we're doing

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 28, 2017, 9:53 p.m. No.1538   πŸ—„οΈ.is πŸ”—kun   >>1539

>>1533

Excession is not necessarily extinction but the possibility is HUGE. That is the scary part. That is the part I hope to help with if I can. I think we will be in star trek land in my lifetime if this thing is built. I don't want to be stuck in some radioactive nightmare instead .

Anonymous ID: 8d5142 Dec. 28, 2017, 9:55 p.m. No.1540   πŸ—„οΈ.is πŸ”—kun

>>1536

I will do this, going over all VQC's posts, and trying to keep it purely pertinent. I will do it over the next few days and if VQC just lets it all out first, all the better.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 10:04 p.m. No.1542   πŸ—„οΈ.is πŸ”—kun   >>1544

>>1533

Noticing something interesting in column 0. N has an almost direct relationship to x. Seems to also be a clear pattern with even and odd n, just like for even and odd e. Check it out.

 

Even n.

{0:2:3:2:1:9} 2x1

{0:2:8:4:4:16} 2x2

{0:2:15:6:9:25} 2x3

{0:2:24:8:16:36} 2x4

{0:2:35:10:25:49} 2x5

{0:2:48:12:36:64} 2x6

{0:2:63:14:49:81} 2x7

{0:2:80:16:64:100} 2x8

{0:2:99:18:81:121} 2x9

{0:2:120:20:100:144} 2x10

{0:2:143:22:121:169} 2x11

{0:2:168:24:144:196} 2x12

 

Odd n

{0:3:12:6:6:24} 3x2

{0:3:36:12:24:54} 3x4

{0:3:72:18:54:96} 3x6

{0:3:120:24:96:150} 3x8

{0:3:180:30:150:216} 3x10

{0:3:252:36:216:294} 3x12

{0:3:336:42:294:384} 3x14

{0:3:432:48:384:486} 3x16

 

Even n:

{0:4:6:4:2:18} 4x1

{0:4:16:8:8:32} 4x2

{0:4:30:12:18:50} 4x3

{0:4:48:16:32:72} 4x4

{0:4:70:20:50:98} 4x5

{0:4:96:24:72:128} 4x6

{0:4:126:28:98:162} 4x7

{0:4:160:32:128:200} 4x8

{0:4:198:36:162:242} 4x9

{0:4:240:40:200:288} 4x10

{0:4:286:44:242:338} 4x11

{0:4:336:48:288:392} 4x12

 

Odd n

{0:5:20:10:10:40} 5x2

{0:5:60:20:40:90} 5x4

{0:5:120:30:90:160} 5x6

{0:5:200:40:160:250} 5x8

{0:5:300:50:250:360} 5x10

{0:5:420:60:360:490} 5x12

 

Even n

{0:6:9:6:3:27} 6x1

{0:6:24:12:12:48} 6x2

{0:6:45:18:27:75} 6x3

{0:6:72:24:48:108} 6x4

{0:6:105:30:75:147} 6x5

{0:6:144:36:108:192} 6x6

{0:6:189:42:147:243} 6x7

{0:6:240:48:192:300} 6x8

{0:6:297:54:243:363} 6x9

{0:6:360:60:300:432} 6x10

{0:6:429:66:363:507} 6x11

{0:6:504:72:432:588} 6x12

 

Clear pattern, could be useful. Thoughts, Anons?

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 10:09 p.m. No.1543   πŸ—„οΈ.is πŸ”—kun

>>1535

Nice to see you TLA! Sorry the red pills weren't swallowed by your famalam. I was only able to help two people inch forward. The rest I avoided. Keep up your spirits!! We need your tough ass lovin' to help proof our spirited ideas. Glad you're back. CollegeAnon is back too! Where TF is our professor Teach???

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 10:15 p.m. No.1544   πŸ—„οΈ.is πŸ”—kun   >>1547 >>1553

>>1542

Even (0,1) holds to the pattern:

 

Odd n:

{0:1:4:2:2:8} 1x2

{0:1:12:4:8:18} 1x4

{0:1:24:6:18:32} 1x6

{0:1:40:8:32:50} 1x8

{0:1:60:10:50:72} 1x10

{0:1:84:12:72:98} 1x12

{0:1:112:14:98:128} 1x14

{0:1:144:16:128:162} 1x16

{0:1:180:18:162:200} 1x18

{0:1:220:20:200:242} 1x20

{0:1:264:22:242:288} 1x22

{0:1:312:24:288:338} 1x24

 

Am I just posting some more basic bullshit that PMA already wrote a program for? Help me out anons. Just keeping my head down and searching the grid for patterns. Trying to tie rows and columns together, starting with row one and column 0.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 10:16 p.m. No.1545   πŸ—„οΈ.is πŸ”—kun

>>1529

This absolutely works just tested it out.

 

This is an easier way to put my (0,n) column rule.

The cell (0,n) = (0,1) * n (for odd n)

and (0,2n) = (0,2) * n

ID: a2017c Dec. 28, 2017, 10:16 p.m. No.1546   πŸ—„οΈ.is πŸ”—kun   >>1548

>>1535

Just noticed your comment about redpilling your family. Sorry I called you a cunt.

 

What keeps me up even though a good portion of my sheeple side of the family thinks I am a racist (which isn't even insulting to a polack) is that they all failed tremendously in life.. but I haven't.. yet. So I've got the high ground so far.

 

Dunno about your situation. If you really care about changing their minds (I do, I will not let my grandmother go believing the bullshit lies jews perpetrate) you should look into memetics. It's a science of its own, basically an evolved version of propaganda. My view is that propaganda isn't bad if you are telling people the truth.

Anonymous ID: 8d5142 Dec. 28, 2017, 10:30 p.m. No.1548   πŸ—„οΈ.is πŸ”—kun

>>1546

haha you call me a cunt all you want, you wonderful anonymous cunt you. i am actually very fascinated by what you call memetics, though I've never used that word. indeed there is an art to informing the ignorant, and this great meme war is sure sharpening that up. guess I was hard on myself for not being better at it.

 

one more math thing, maybe not relevant, but I worked on this on the bus a few days ago so I'll spill it. VQC mentioned sums of squares being closed under multiplication. i.e. (xx + yy)(ww + zz) = ss+tt, another sum of squares. this can be proved with Gaussian integers (complex numbers a +bi where i = sqrt(-1) ), which gives a formula for two possibly different ways to write the resulting product as sum of squares:

 

(xx + yy)(ww+zz)

= xxww + yyww + xxzz + yyzz

= xxww + 2wxyz + yyzz + xxzz - 2wxyz + yyww

=(wx+yz)^2 + (xz-yw)^2,

 

or else (wx -yz)^2 + (xz+yw)^2 by switching the + and - .

 

VQC's mention might only be incidental though, part of the grid, and not going to be part of the factoring process.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 10:35 p.m. No.1549   πŸ—„οΈ.is πŸ”—kun   >>1555

>>1547

Thanks for the feedback! I'm still working off the original .csv output (cause I need to learn programming), so not sure if my cells are complete. So does the pattern hold? From your examples it looks like it doesn't. I still see relationships between n and x in your examples, but you would arrive at x with different equations. What can you make of it MA?

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 28, 2017, 10:45 p.m. No.1550   πŸ—„οΈ.is πŸ”—kun   >>1551

>>1539

https://anonfile.com/Za43r2dab0/Unforseen_Concequences.rtf

 

I just renamed my chronicle and added another chapter. Anon, it seems you are interested in the same stuff as I. Check out this doc and perhaps we can both work on it?

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 28, 2017, 10:54 p.m. No.1554   πŸ—„οΈ.is πŸ”—kun   >>1557 >>1558

>>1535

On the redpilling. Remember, you cant tell someone they are wrong. You cant even show them that they are wrong. In this direction lies sorrow. The only way to convince someone is to set them up to find it out on their own. How do you do that? Well, The Socratic method works on most fuckers.

 

Curious people use the Socratic Method on themselves without even thinking about it. They ask leading questions constantly. This is the best redpill method. "Antifa IS cool sally, but why are they hurting people?" Why is their plan always Communism? Oh! You like communism and think it is a good idea? Do you know anyone personally who has lived and escaped from a communist nation? Why do they call it "escaping" from communism. Ha ha. I could go on all day but I think you get the point. NEVER say they are wrong, pretend like you dont know and have them educate you. Then lead them off a cliff.

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 28, 2017, 11 p.m. No.1556   πŸ—„οΈ.is πŸ”—kun

>>1552

Thanks to you all for all your hard work. This has been one of the most inspiring and interesting things I have ever been a part of. I am honored. Im going to make a few memes. Anyone want a personalized one?

Anonymous ID: a2017c Dec. 28, 2017, 11:03 p.m. No.1558   πŸ—„οΈ.is πŸ”—kun

>>1554

fcit.usf.edu/holocaust/people/DocPropa.htm

hitler.org/writings/Mein_Kampf/mkv1ch06.html

en.wikiquote.org/wiki/Mein_Kampf#Chapter_6_-_War_Propaganda

 

How to make memes from literally Hitler himself.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 11:04 p.m. No.1559   πŸ—„οΈ.is πŸ”—kun   >>1562 >>1569

>>1553

Hey PMA, thanks! I'll re-read those posts now. Have you figured out the column 0 pattern to your satisfaction yet? Looks like you found an odd/even pattern in n as well. Ah, (0,1) thru (0,6) looks like we arrived at the same conclusion for those n values. How does work in higher values of n? Oh yeah, I see those x values tied to the n value in your second example, at least for n=9. There is some elegant formula hiding beneath all this, just working to find it.

 

"How do we USE the rows and columns to get what we need."

"Enumerate the rules for global, rows, columns and the solution will emerge."

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 11:05 p.m. No.1560   πŸ—„οΈ.is πŸ”—kun   >>1561 >>1565

So I've been looking through these values that I can generate for the negative square values of e. This is the output for entries (e,n,d,x,a,b,c) where c is a square number. (I checked for c to be product of 2 primes but it didn't produce any). I checked the first 30 entries of the first 30 rows of the first 30 squares (I did a bunch of them get the picture) and heres the result entries where c is a square:

 

25:

((-25, 2, 13, 7, 6, 24, 144), True)

49:

((-49, 1, 25, 9, 16, 36, 576), True)

((-49, 5, 25, 13, 12, 48, 576), True)

100:

((-100, 4, 26, 14, 12, 48, 576), True)

196:

((-196, 2, 50, 18, 32, 72, 2304), True)

289:

((-289, 5, 145, 37, 108, 192, 20736), True)

361:

((-361, 2, 181, 31, 150, 216, 32400), True)

441:

((-441, 3, 75, 27, 48, 108, 5184), True)

484:

((-484, 1, 122, 26, 96, 150, 14400), True)

((-484, 3, 122, 32, 90, 160, 14400), True)

529:

576:

((-576, 2, 51, 26, 25, 81, 2025), True)

625:

((-625, 3, 65, 29, 36, 100, 3600), True)

676:

729:

((-729, 2, 123, 33, 90, 160, 14400), True)

784:

841:

((-841, 4, 421, 61, 360, 490, 176400), True)

((-841, 6, 421, 71, 350, 504, 176400), True)

1444:

((-1444, 4, 362, 62, 300, 432, 129600), True)

1681:

((-1681, 1, 841, 57, 784, 900, 705600), True)

1936:

((-1936, 2, 244, 52, 192, 300, 57600), True)

((-1936, 6, 244, 64, 180, 320, 57600), True)

2025:

((-2025, 1, 339, 51, 288, 392, 112896), True)

2304:

((-2304, 1, 148, 50, 98, 200, 19600), True)

 

There is some weird stuff here.

(-49,1,25,9,16,36) [leaving out c]

multiplied by 2 (with e multiplied by 4) we get:

(-196,2,50,18,32,72) [notice that for n,d,x,a,b the previous entry was half and e was a quarter]

(this relationship could be visualized I think)

 

Also you got

(-484, 1, 122, 26, 96, 150)

same pattern

(-1936, 2, 244, 52, 192, 300)

 

So if n is one for this type of cells then you can do (newE = oldE * 4) and (newOtherpart = oldOtherpart * 2)

 

Anyone want to look further into this?

MinecraftAnon !!QXqSZ2ev8. ID: 3ead22 Dec. 28, 2017, 11:11 p.m. No.1563   πŸ—„οΈ.is πŸ”—kun   >>1578

>>1557

I had a dream where my skin was covered in math scribbles.

 

>>1552

This is a very special view of (1,N). The initial value of X has been floored to zero, and the growth factor of X has been left intact. 8, 9, 24, 25, are marked in red.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 11:12 p.m. No.1564   πŸ—„οΈ.is πŸ”—kun   >>1566 >>1568

>>1555

Checking those digits, my fag. Uh, TBH those records are in my .csv too. Just got too excited to look further down, lol! My bad.

 

Program Help Needed. Ok, I'm on a Mac. I installed virtual studio, updated the .net shit, and still couldn't compile the file. Fucking rookie over here. Any advice? I actually can read all of you guy's program formulas, and understand them. I need to humble myself and ask for help with the program aspect.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 11:13 p.m. No.1565   πŸ—„οΈ.is πŸ”—kun

>>1560

 

 

((-484, 1, 122, 26, 96, 150, 14400), True)

((-484, 3, 122, 32, 90, 160, 14400), True)

 

((-1936, 2, 244, 52, 192, 300, 57600), True)

((-1936, 6, 244, 64, 180, 320, 57600), True)

 

484 = 22^2

1936 = 44^2

 

122*2 = 244

1*2 = 2

3*2 = 6

26*2 = 52

32*2 = 64

etc..

 

Very cool

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 28, 2017, 11:15 p.m. No.1566   πŸ—„οΈ.is πŸ”—kun   >>1571

>>1564

I used C# for my job this past summer. I also have a mac and recently installed C# and visual studio on it to run this stuff. It wouldn't work for me either. I run my shit in python now.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 11:16 p.m. No.1567   πŸ—„οΈ.is πŸ”—kun

>>1562

Wow! Thanks for the cool memes Hobo! I can't program shit, but running a business for 12 years and studying the laws of Human Interaction (via Dale Carnegie, Napoleon Hill, etc) have helped me learn to thrive in any environment. It's meant to be that we are all here. Our rational/spiritual leanings attracted us to VQC. Just glad to be here.

"When the student is ready, the Master will appear."

Anonymous ID: ac6203 Dec. 28, 2017, 11:19 p.m. No.1569   πŸ—„οΈ.is πŸ”—kun

>>1559

Iterating the records lead to me to potential matches. But to be honest, I have so many different ways of iterating and creating records it's all a bit of a blur.

 

>There is some elegant formula hiding beneath all this

This I firmly believe.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 11:26 p.m. No.1571   πŸ—„οΈ.is πŸ”—kun   >>1574

>>1566

Ok, maybe I should try Python. I don't expect you guys to teach me, and I have confidence that I will learn quickly.

 

>>1568

Thanks Baker! If I'm really going to contribute as we move forward, I need to write formulas and code. Especially as we move to new projects like ECC. I can do it. Appreciate your support. So in Linux, do you use the terminal, or is there a program to write, compile, and run code?

Anonymous ID: ac6203 Dec. 28, 2017, 11:28 p.m. No.1572   πŸ—„οΈ.is πŸ”—kun   >>1573 >>1576

Has anyone looked into these 2 formulas yet?

 

f = 2d + 1 - e ( f is what you add to c to make a square )

 

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

 

I was running a few tests around na records, and added these 2 formulas to my output for each record.

 

What's interesting about this is it appears that the f value equals the small square only at the na records.

 

Just guessing that maybe there is something in here indicating where we should search next?

 

c=145

(1,5,4) = {1:5:12:7:5:29} = 145;

 

f = (2 * 12) + 1 - 1 = 24;

(x+n)^2 = (5*5) + 2 * 12 * (5 - 1) + 24 - 1 = 144;

(x+n)^2 = 12x12=144;

 

c=145 na record

(1,1,4) = {1:1:32:7:25:41} = 1025;

 

f = (2 * 32) + 1 - 1 = 64;

(x+n)^2 (1*1) + 2 * 32 * (1 - 1) + 64 - 1 = 64

(x+n)^2=8x8=64;

MinecraftAnon !!QXqSZ2ev8. ID: 3ead22 Dec. 28, 2017, 11:38 p.m. No.1573   πŸ—„οΈ.is πŸ”—kun

Here's the normal view of (1,N) again. I'm focused on getting the initial value of X, if someone could find a formula for how X behaves after that, it'll be huge. The distances are small enough that it can be forced out, but moving thousands of N records, the formula is significant.

 

>>1572

My understand of E and F is this. E is the amount C is over it's perfect square root, F is the amount C is under the next perfect square root up (D+1). This is why the perfect square roots only fill E0, but they fill the inverted square roots in F. Every perfect square root can be a perfect square root under a different perfect square root.

Anonymous ID: a2017c Dec. 28, 2017, 11:40 p.m. No.1574   πŸ—„οΈ.is πŸ”—kun   >>1575 >>1576 >>1601

>>1571

You have Linux?

 

Download Mono develop: mono-project.com/download/

 

Then write the code in file named TheEnd.cs, add this method (if someone could inform me how to make code look good on 8ch that would be good)

 

public static void Main(string[] args) {

CreateTheEnd();

Output();

 

}

 

And open terminal in the directory and type mcs TheEnd.cs

Then mono TheEnd.exe

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 11:43 p.m. No.1576   πŸ—„οΈ.is πŸ”—kun   >>1577

>>1572

PMA, f has been mystifying me since the beginning code VQC posted. Thanks for clarifying.

 

>>1574

Baker, can i run a Linux environment on a Mac? If so, maybe I take your route. If Python runs well on a Mac, I'll go that way first.

 

Lads, I'm gonna grab that Python/Basilisk by the throat and stab it with the sword of Gryffindor. Um, Harry Potter memes, anyone?

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 28, 2017, 11:54 p.m. No.1578   πŸ—„οΈ.is πŸ”—kun   >>1580

>>1563

WTF are those two badass lines moving into infinity? Actually, there are more than two, many more. Two arteries or vessels tho.

 

>>1577

Checked! Thanks Baker. Apparently I'm going to get a degree in programming here as well, lol! I'm checking it out now. Downloading now. I fucking love you guys. Even my lady is like "math is a much better hobby for you than worrying about the end of the world."

MinecraftAnon !!QXqSZ2ev8. ID: 3ead22 Dec. 28, 2017, 11:54 p.m. No.1579   πŸ—„οΈ.is πŸ”—kun   >>1666

>>1577

We also have the VQC bruteforcer that I figured out what was wrong with. The correct version is this.

 

ArbitraryCell(e, n, x){

e = e;

n = n;

x = x;

a = (x * x + e) / (2 * n);

b = a + 2 * (x + n);

c = a * b;

d = sqrt(c);

}

 

boolean isValid(){

return (c - d * d) == e;

}

 

You can generate any cell by any E,N,X and test if it's a good cell or not. Good for looking at very specific regions.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 29, 2017, 12:10 a.m. No.1581   πŸ—„οΈ.is πŸ”—kun

>>1580

LOL! MA, she knows. When we met I told her "At some point, I'm going to get heat for pursuing the truth." She was like, "OK, I'm cool with that."

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 29, 2017, 12:21 a.m. No.1582   πŸ—„οΈ.is πŸ”—kun   >>1612

>>1562

Hobo, I'm halfway thru your paper now! You're a fucking optimistic pessimist. Good thing I'm prepped, at least for 3 months. Gotta bump that up tho. VQC, you're the man. Thanks for inviting us to join you.

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 29, 2017, 12:34 a.m. No.1583   πŸ—„οΈ.is πŸ”—kun

Holy Shit, Hobo! What a great read! Questions to follow. Hey Anons, It's a 15-30 minute read, but well worth it, check it out. Copypasta for OpSec, anons.

 

https:/ /anonfile.com/Za43r2dab0/Unforseen_Concequences.rtf

VeritasAequitas !!Nf9AmQNR7I ID: 7261f1 Dec. 29, 2017, 12:39 a.m. No.1584   πŸ—„οΈ.is πŸ”—kun   >>1587

Hey Topol! Good evening! Can I get your eyes on this? We need a cool ass VQC logo with the Rainbow Primes behind it. Q logos are starting point, make it fabulous! Can you help?

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 29, 2017, 12:57 a.m. No.1587   πŸ—„οΈ.is πŸ”—kun   >>1588 >>1609

>>1541

What on earth is IOTA?

I came across that doing my translation games with ta.io (Dooooon't ask.)

 

>>1584

Seeeeeeeee… I keep seeing Capsule Corp which is more C C… C^2… Verde Q Code, the Green Q… hold on… besides the money reference… kikes… an oven… I think a chef's hat with a … i've already started drawing this out a day or two ago while I was working on ALL THE THINGS.

 

Gimme a bit.

 

Also, no one worry about this being an ELE.

We're fine. Everyone's fine. Everything's fine.

Faith. You can haz.

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 1:06 a.m. No.1588   πŸ—„οΈ.is πŸ”—kun   >>1590

>>1587

I love seeing written notebooks. Path to the mind of your friend(s). No pressure, no rush. Just seeing a little bit higher on the planes of existence over here. Know you care about this quest. That's all. Love your input. Your intuition is usually right on. Appreciate your badass style of memes and knowledge. Love ya, Topol!

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 29, 2017, 1:38 a.m. No.1590   πŸ—„οΈ.is πŸ”—kun   >>1591

>>1588

I mean… I'm fine with a Q… it's just Number Muncher Green.

 

Also, I'm not a graphic designer.

 

I make graphic designers want to plunge their exacto knives through their eyes. I use papyrus, on purpose, with a thick black stroke.

 

Somewhere, a graphic designer just had an emotional breakdown because I made this.

Anonymous ID: 99edb0 Dec. 29, 2017, 2:05 a.m. No.1598   πŸ—„οΈ.is πŸ”—kun   >>1614 >>1637

Hey, so since I'm the BO I feel obliged to mention that while I haven't really been posting much for the last week, I am very much still here every day. Considering the ramifications of what we're doing, I'm sure it's important to all of you to know that the person responsible for the board hasn't abandoned it. Even before I became aware of this I had far more that I needed to do than I had the time for, and that's still the case. I've got a bit of time now, so (since I haven't seen anyone else doing it - if they have it would be handy to know) I'm going to try to conglomerate all of VQC's hints since the start and as much of the findings as I can understand over the next couple hours and turn them into something hopefully easy to understand.

ID: a2017c Dec. 29, 2017, 2:08 a.m. No.1599   πŸ—„οΈ.is πŸ”—kun   >>1600

He didn't reply to himself - his alter ego slipped up! Muahahaha!

 

If you all somehow reach post limit while I'm asleep, bake it however you want. Not people here have already read what I have right now.

Anonymous ID: 99edb0 Dec. 29, 2017, 3:39 a.m. No.1602   πŸ—„οΈ.is πŸ”—kun   >>1637

Here's every hint VQC has given thus far in one convenient location (apologies that it's a pdf again; I know why people are paranoid of them but I can't really think of any other way to do it). I didn't do much formatting so it may look bad in places but I was just copying and pasting text.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 7:16 a.m. No.1609   πŸ—„οΈ.is πŸ”—kun

>>1587

IOTA (MIOTA) is a crypto I think developed by

 

…microsoft…

 

so they will still be in control with it, BUT I'll have more money so thats neat

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 7:33 a.m. No.1610   πŸ—„οΈ.is πŸ”—kun

>>1589

This is all the prime A values I think. My original file is named DiffAs.bmp so I think that is what this is. White would be A = 1 or 2, then yellow would be the next prime etc. If we can somehow generalize this then we could have a set of lines for any prime A to show all the products of it.

 

Also if you look at my AVals you can notice that there are parabolas that slide down the graph that do not move. These dots are all probably integer selections on the parabola. I'm going to try and generalize these into equations because I've wanted to do it for a while but I've been lazy.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 8:05 a.m. No.1611   πŸ—„οΈ.is πŸ”—kun   >>1613

 

9………………..+………………………..

0…………….+….+..++……………………

1………………..+….+..++………………..

2……………..+……+….+………………..

3…………+……..+……+….+…………….

4…………….+……..+……+……………..

5………………..+……..+………………..

6………….+……….+……..+…………….

7……………..+……….+……..+…………

8……..+…………+……….+……………..

9…………+…………+……….+………….

0…………….+…………+………………..

 

Copy and past into same width text file.

The end is saying there is a record divisible by 2 at -5,-1. Also it said there is one divisible by 3 at -7,-1 and one divisible by 4 at -9,-1. Could this N = -1 be a key entry???

 

Also I'm planning to do a code dump later today. Hit me with some (you)'s if you guys are interested. I have code to generate the pictures (even gifs!) but you'll need to modify some of the code (its easy though). I also have a way to generate these outputs above (I could generalize this for anything) because these are easier to look at than bitmaps to really find patterns. Also I have code to generate any (0,n) cell and any (e,1) cell. Also I have code to (I'm not sure if the function is surjective, but it could be helpful) to generate any (-(q^2), n) column.

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 29, 2017, 8:43 a.m. No.1612   πŸ—„οΈ.is πŸ”—kun

>>1582 Yeah I have strong pessimistic tendencies. It is usually a good balance for highly optimistic people though. I have regularly been the one to catch small stuff that turns into big problems throughout my life. I guess I have spent too much time too close to many bad people so I tread carefully.

 

Don't take my ideas as negativity take them more as risk mitigation or even insurance. It is good to take a little pessimism early on and use it as protection.

 

Without knowing what is coming its good to be ready for rough times. That said I don't think they will last unless the bad guys get their hands on this thing. That is one of my biggest worries at this point. The bad guys get this and lock it up and it is used against good people.

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 29, 2017, 8:47 a.m. No.1615   πŸ—„οΈ.is πŸ”—kun   >>1617

>>1604

A couple ideas when I look at your GIF's. FIrst it almost seems like a good random number generator. Next, here I am again, looking at a pyramid. Just like on the dollar bill or a picture of Antarctica. No coincidences right?

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 8:55 a.m. No.1617   πŸ—„οΈ.is πŸ”—kun   >>1618 >>1620

>>1615

 

Entries for A = 2

 

(-5,-1)

(-9,0), (-4,0), (-1,0), (0,0)

(-5,1), (0,1), (3,1), (4,1)

(-8,2), (-1,2), (4,2)

(-13,3),(-4,3), (3,3), (8,3)

 

From any value of A at (e,n) if we go to the cell(e+4, n+1) we will (probably) get another entry.

Also if at (e,n) we have A.

 

The centers of any line are going to be on the (0,q^2) cell where

Also, the generation seems to be independent of the (0,0) and (0,4) cell (4=2^2)

So our start is at (0,1). We can then do the pattern twice to the right and once to the left.

 

n: cells where A = 2

 

1^2: (-4,0) <1< (0,1) >2 (4,2) (5,3)

 

From here, to get to the next line we can take all of our cells and shift them by:

(e,n) -(e-1,n+[1]) and then add another cell to the left side of the line to get:

(added to left)

(-9,0), (-5,1), (-1,2), (3,3), (7,4)

 

We can also get to the previous line (but you remove the last entry on the right before doing this)

(e,n) -(e-1,n-1) to get

(-5,-1), (-1,0), (3,1), *[(6,3)] gets removed

 

Then to get the next 3 lines we have to do

 

2^2, so we start at (0,4) and we use our pattern to go to the left twice (2) and to the right 3 times (2+1).

 

So we get

(-8,2),(-4,3) <2< (0,4) >3(4,5), (8,6), (12,7)

 

Then, like before, if we want to go to the next line we do

(e,n) -(e-1, n+[2]) and add an entry to the left

(added)

[(-13,3)],(-9,4),(-5,5),(-1,6),(3,7),(7,8), (11,9)

 

And the previous is the pattern

(e,n) -(e-1, n-[2]) and you remove an entry on the right to get:

 

(-9,0), (-5,1), (-1,2), (3,3), (7,4) [(11, 5) would be removed]

 

This is info on how to calculate the coordinates for any cell where a = 2.

 

It is very simple to calculate them. I am going to try and generalize to higher a.

 

>>1615

No coincidences at all. Here is a pic that looks real cool. with pyramidal stuff.

Also this one look at the LITERAL 3D SPIRAL GOING DOWN THE CENTER. This is where X mod P is base [ i think base = (a+b)/a ]

Hobo !!1yNgQ3NlCs ID: c02904 Dec. 29, 2017, 8:57 a.m. No.1619   πŸ—„οΈ.is πŸ”—kun

>>1586

So I am guessing this pic refers to "The rod and the ring will strike" from the 4chan prophecy? If you have not seen that then check it out. Pretty fun to try to figure out. I think they are all done with the exception of just 1 or 2

Belly of the dragon will drip water: Dragon Capsule?

 

Star will gorge itself on clay: Israel taking land?

 

Rod and the ring will strike: Rods from god? Maybe we are part of it. :)

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 10:10 a.m. No.1620   πŸ—„οΈ.is πŸ”—kun   >>1621

>>1617

A = 3

 

2 up from origin of parabolas is (-7,-1), (-1,0), (5,1)

For this we can notice that the origins of the parabolas are:

(-3,1), and (3,2), and (9,3), (15,4) (idk how this pattern is formed)

 

From these origins we can calculate anything on the parabola. You can go up N like:

(e,n) -(e-1,n+1) -> (e-2,n+1) -> (e-4,n+1)

and down like:

(e,n) -(e-1,n-1) -> (e-2,n-1) -> (e-4,n-1)

 

From any cell in (e,n) where a = 3 we know the cell (e+6,n+1) and (e-6,n-1) contains another a=3.

Only inside this huge parabola, but I suspect the pattern continues for really high values that our theend didn't generate. Moreover, if we do this to get to (e+/-6, n+/-2) then the entry on that parabola is SAME entry in the parabola shape.

 

x x*x

1 1

2 4

3 9

4 16

5 25

 

(0,0)(1,1),(2,4)

 

So generally if we have a cell (e,n) that is divisible by 3 and is the origin of the parabola, then the entries (e-(y^2), n+y) that are also divisible by 3 for all y.

 

So if we are at a cell (e1,n1) divisible by 3, then it is equal to (e0 - (y^2), n0 + y) for another e0,n0 where e0 != e1, and n0 != n1. Also, (e0,n0) is the origin on the parabola these line up on. If we go to (e+6,n+1), we can call this (e2,n2). Then we have that (e2,n2) is equal to (e3-y^2, n3+y) which are the SAME Y VALUES. So here we can get coords for any a=3

 

Also, if we have 2 cells at (e,n) and (e-1,n-w) we can generate more cells by doing the following:

 

(e,n) -(e-1,n-w)

-(e-4,n-2w)

-(e,9,n-3w)

 

And also the reverse

 

(e,n) -(e-1,n+w)

-(e-4,n+2w)

etc..

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 10:41 a.m. No.1621   πŸ—„οΈ.is πŸ”—kun   >>1624 >>1628 >>1643

>>1620

Big if Huge.

 

Also if we have a n origin cell at (e,n) and to to (e -(2^2), n - 2*w) = (E,N),

that cell is equivalent to (e0 - (1^2), n0 - 2(w+1)) where (e0,n0) is another parabolic origin.

 

A = 4

 

For this, the pattern is similar. The third [4-1] above the origin is:

The difference is the function goes from (e,n) to (e+8,n+1) and the relation between two parabolas is

 

(e,n) [origin of paraobola] to (e -(2^2), n - 2*w) = (e0 - (1^2), n0 - 2(w+1))

where (e0,n0) is another parabolic origin.

 

It seems that we can generally say the following:

 

For any A

If you are at a parabolic origin cell (e,n) with the parabola being (e-(j^2),n+(w*j)) for any cell, then if you go to the entry by plugging in this j

 

j = {j/2 if A even OR (j+1)/2 if A odd}

 

you reach a new cell that is equivalent to (e - (i^2), n +(w*i))

where (e+(2A) , n+1) = (e,n*)

and i = {A/2 if A even, and (A-1)/2 if A odd}

 

Now, if we can find a pattern to get these initial entries then we can generate every entry with factor A.

MinecraftAnon !!QXqSZ2ev8. ID: ad654f Dec. 29, 2017, 1:36 p.m. No.1623   πŸ—„οΈ.is πŸ”—kun

>>1622

I'm guessing that's not a feature of twitter?

 

Slow day today. I just spent hours troubleshooting a bug where I was copying a reference of an old object rather than cloning it into a new object.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 1:45 p.m. No.1624   πŸ—„οΈ.is πŸ”—kun   >>1625 >>1626 >>1627 >>1633 >>1634

>>1621

So through this method for (e,n,d,x,a,b):

 

We know E,N,A and can shift along the parabola

 

aa + 2ax + 2an = c (a,x,n,c)

 

b = a + 2x + 2n (b,a,x,n)

 

a * b = c (a,b,c)

 

(d+n)(d+n)-(x+n)(x+n) = c (d,n,x,c)

 

x = d - a (x,d,a)

 

d * d + e = c (d,e,c)

 

Have E,N,A:

 

(b,a,x,n)

(a,b,c)

(d,n,x,c)

(x,d,a)

(d,e,c)

 

So now we have relations between:

 

(b,x) .

(b,c) .

(d,x,c)

(d,c)

(x,d)

 

(b,x) and (b,c) -(x,c)

 

(x,c) and (x,d) -(c,d)

 

(d,c) and (c,d) -c or d!!

 

b = a + 2x + 2n

a * b = c

 

- c/a = a + 2x + 2n

 

c/a = a + 2x + 2n

x = d - a

 

-c/a = a + 2(d - a) + 2n

 

c/a = a + 2(d - a) + 2n

d * d + e = c

 

c = aa + 2a(d-a) + 2an

 

dd + e = aa + 2a(d-a) + 2an

 

d = sqrt( 2ad - aa + 2an - e )

 

then

 

c = dd + e

 

x = d - a

 

so we can generate any (e,n,d,x,a,b) for a constant a by

shifting along this parabola and using this equation.

 

DO WE HAVE A WINNER?

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 2:06 p.m. No.1629   πŸ—„οΈ.is πŸ”—kun   >>1630

>>1628

>>1618

 

This is pic related for A=2. I'll post a couple more.

 

I haven't necessarily figured out how to figure out how long the lines are, but it seems like they start with 3 and add 1 on the right and then 1 on the left each time or something like that. I had it figured out for A=2.

Anonymous ID: 99edb0 Dec. 29, 2017, 2:17 p.m. No.1631   πŸ—„οΈ.is πŸ”—kun

Thinking out of my ass here: we can calculate any of the variables by applying math to i and j. If we could magically calculate i and j from c, we'd be able to calculate a and b from c. Here's a plot with i as the x axis and j as the y axis every time a and b are primes. This might be useless but I only wasted a couple minutes on it.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 2:29 p.m. No.1634   πŸ—„οΈ.is πŸ”—kun   >>1635

>>1624

I incorrectly had this as the equation

 

>dd + e = aa + 2a(d-a) + 2an

>d = sqrt( 2ad - aa + 2an - e )

 

d = a +/- sqrt(8an - 4e)/2

 

So, believe it or not, for any E,N,A, we can generate TWO records for d.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 2:42 p.m. No.1635   πŸ—„οΈ.is πŸ”—kun

>>1634

We know E,N,A

 

aa + 2ax + 2an = c (a,x,n,c)

b = a + 2x + 2n (b,a,x,n)

a * b = c (a,b,c)

(d+n)(d+n)-(x+n)(x+n) = c (d,n,x,c)

x = d - a (x,d,a)

d * d + e = c (d,e,c)

 

dd = c - e

 

dd + 2dn + nn - (x+n)(x+n) = c

 

c - e + 2dn + nn - (x+n)(x+n) = c

 

2dn - e + nn - (xx + 2xn + nn) = 0

2dn - e + nn - xx - 2xn - nn = 0

2dn - e - xx - 2xn = 0

 

2n(d-x) = xx + e [x = d-a =a = d-x]

 

2na = xx + e

xx = 2na - e

x = math.sqrt(2na-e)

 

Then d = x + a

c = dd + e

b = c/a

 

This might be an easier way to calculate. Pls verify if you're following along

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 2:52 p.m. No.1638   πŸ—„οΈ.is πŸ”—kun   >>1639

>>1636

 

Just remember dd+e = c. For negatives D isn't necessarily the side of the largest square less than c it could just be any square and e is what you add to it to make it c.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 3:02 p.m. No.1641   πŸ—„οΈ.is πŸ”—kun

>>1639

VQC made the grid by multiplying and doing that stuff. He said this grid is practically effective everywhere so I'm trying to shed my preconceptions about what all the values are because they can probably be calculated in many different ways.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 3:30 p.m. No.1642   πŸ—„οΈ.is πŸ”—kun   >>1643

pic related:

 

For 40 we have parabolic origin cells at

(0,20), (0,80), (0,180), (0,320)

Notice: (vg = vertical gaps)

(0,20) = (0, (40/2) * 1) (vg = 0)

(0,80) = (0, (40/2) * 4) (vg = 1)

(0,180) = (0, (40/2) * 9) (vg = 2)

(0,320) = (0, (40/2) * 16) (vg = 3)

 

So for any value A (if A is even) you have parabolic origin cells at ( 2AH, (A/2) * T^2 ) for any integers H,T.

 

Then from those cells (say (e,n)), (depending on what T you chose), you can get another cell (with the same a) by the following:

 

(e,n) -(e+2A*J, n-J) (for any J)

(e,n) -(e-J^2, n + J*T) (for any J)

 

I may have posted someting earlier, which said that you could go (e,n) -(e+2a, n) which is wrong, this above is the correct formula.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 3:53 p.m. No.1643   πŸ—„οΈ.is πŸ”—kun   >>1644 >>1645

>>1642

Now for 41 (odd number) we get origin cells at:

 

(-41,20), (41,21)

(0,82)

(-41,184), (40,185)

(0,328)

 

(-41, (411-1)/2), (41, (411+1)/2)

(0, (41*4)/2)

(-41, (419-1)/2), (41, (419

(0, (41*16-1)/2)

 

So in general for an odd A, we will have parabolic origin cells at (indexed by t):

 

Odd t:

(-A, (A(t^2)-1)/2), (A, (A(t^2)+1)/2)

 

Even t:

(0, (A*(t^2))/2 )

 

Then, obviously, we can still use the pattern

(e,n) -(e+2AJ, n+J*t) for any J

to move along the parabolas. Previously, I had written a way to traverse to and from parabolas from the origin cells.

>>1621 (here)

 

So now this is a way to generate the origin cells for any A

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 4:21 p.m. No.1646   πŸ—„οΈ.is πŸ”—kun

def ENA(E,N,A):

X = int(math.sqrt(2 * N * A - E))

D = X + A

C = D * D + E

B = C / A

return (E,N,D,X,A,B,C)

 

#This will give coords for parabolic origins and also P is the shifter for N for the parabola

def originForP(A,P):

if(A%2 == 0):

return ENA(0,(A/2)PP,A)

else:

if(P%2 == 0):

return ENA(0, (A(PP))/2,A)

else:

return ENA(A, (A(PP)+1)/2, A)

 

def shiftFromOrigin(origin, P, t):

E = origin[e] - (t * t)

N = origin[n] + t * P

A = origin[a]

return ENA(E,N,A)

 

Python code if you guys want it.

Anonymous ID: 99edb0 Dec. 29, 2017, 5:21 p.m. No.1647   πŸ—„οΈ.is πŸ”—kun   >>1648 >>1659 >>1709

I don't know if anyone figured this out before me or found a better method for finding genesis cells, but here's a thing.

 

In order to find the genesis cell for a given product of two primes c, we need to use c to calculate e and n. e is easy, since it's derived directly from c.

e = c - ((floor(sqrt(c))) * (floor(sqrt(c))))

In order to calculate n from c, we need to know i, since

n = i - d

n = i - (floor(sqrt(c)))

That's trickier to calculate. That's the obvious stuff. Now here's an image (the first one) with c as the x axis and i as the y axis. A point is drawn every time c is the product of two primes. It creates what looks like a bunch of linear lines. In the second image, I used the straight line tool in Paint to be completely sure they were linear, although I only did it for a few lines. I'm not sure if each new line start every time the i loop starts again or if it's something else, but if we could figure it out and separate each of the lines, we could use simple gradient maths (rise/run) to calculate i from a given c. That would allow us to find the genesis cell for a given c (I haven't gotten into generating the rest of the cells from the genesis cell yet).

VeritasAequitas !!Nf9AmQNR7I ID: 9fb75d Dec. 29, 2017, 6:04 p.m. No.1650   πŸ—„οΈ.is πŸ”—kun

>>1649

So where the heck are we at? PMA seems to have a good handle on column 0, moving thru records using c,cc, aa, bb, etc. How does increasing n value change a var and related x var values? Also, how do we tie together our knowledge of row rules and column rules? We’re kinda stalled here at the moment. Thoughts, Anons?

Anonymous ID: 99edb0 Dec. 29, 2017, 6:05 p.m. No.1651   πŸ—„οΈ.is πŸ”—kun   >>1652 >>1654

>>1648

If we weren't meant to care about i or j after the fact, why are we using them to calculate n? I don't think VQC was implying that we should just completely ignore i and j. VQC has been telling us to look for patterns. I found a pattern. What I'm saying is that if we can use only c to find i, we can use only c to find n. We can already use only c to find e. If we can find n from c, like we've been trying to do, we can find the genesis cell for a given c. Why would this not be worth looking into?

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 6:22 p.m. No.1654   πŸ—„οΈ.is πŸ”—kun

>>1651

Agreed! Your idea is worth pursuing, to find the underlying formulas. Not trying to be a downer, or to dismiss your idea. Let's pursue it, Anon!

 

I asked VQC about i and j, and his response was "they are necessary to generate the grid, but remember that in a real case, you won't be given i or j, only c." I'll look thru the archives and see if I can find his post about this.

 

Here's another tasty crumb from VQC:

>>>/cbts/53615

Anonymous ID: 99edb0 Dec. 29, 2017, 6:26 p.m. No.1657   πŸ—„οΈ.is πŸ”—kun   >>1660

>>1656

We really need some kind of jargon dictionary at this point. A few threads ago someone said that "genesis cell" meant the first cell of an infinite set (so t = 1 regardless of (e, n)). That's what I was referring to.

Anonymous ID: 99edb0 Dec. 29, 2017, 6:54 p.m. No.1659   πŸ—„οΈ.is πŸ”—kun

Further from this >>1647

If there is a way to find the gradient of each of these lines, we don't actually need to do anything with t/setSize. We'll have e, n, d, c and i, and we can use those to calculate a and b.

c = a*b

c = (i-j)(i+j)

c = i^2 - j^2

j^2 + c = i^2

j^2 = i^2 - c

j = sqrt(i^2 - c)

I also saw a far simpler method of calculating j somewhere else. With i and j we can calculate a and b regardless of any of the other infinite possible cells for a given (e, n). All we have to do is figure out how these lines are generated in terms of i and use gradients to calculate n from c.

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 6:56 p.m. No.1660   πŸ—„οΈ.is πŸ”—kun

>>1657

Well, this relates to the main problem we're trying to solve right now. Baker is correct, and you are correct too. Agreeing on terms would be helpful. With higher values of n, you could consider the first cell the "genesis cell". The only issue is we don't yet know n, and we're trying to solve for it starting only from c.

 

(e,1) or row one is "The one row to rule them all" according to VQC. We can generate all elements for any value of e in row one, using (for odd e at t=1) a=(e+1)/2 and (for even e at t=1) a=e/2. At (0,0) a=2. Then we solve for b and n at t=1, (e,1). Then we can build "Jacob's Ladder" or "Stairway To Heaven" in row one using the pattern of b at t=1 dropping to become a at t=2. So for row 1, we can create all elements into infinity. VQC said that "cells at (e,1) contain all factors for that entire column." So row one is our "Catalog of var a Values"? (Help me here if I'm wrong, anons.)

 

For column 0, my understanding is that as c increases to huge RSA numbers, n is also likely to increase to huge (but less than c) numbers. Our pattern of jumping up using c, cc, aa, bb , etc is to help us find the correct row of n for the c we are factoring. (help me here if I'm wrong, Anons). PMA, can you give us the Meta Idea of what c^2 factoring in Column 0 is hoping to accomplish? I think what I've stated is correct, but could be wrong too.

 

Thoughts, Anons?

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 7:08 p.m. No.1661   πŸ—„οΈ.is πŸ”—kun   >>1662 >>1663 >>1664

I think I have found a pattern for any x value.

 

X:

 

Start with (e,-1) = (-x^2, -1).

Then we can generate a line from this point by doing the pattern

(e,-1) -(e + 2J, -1 + J). This is the line that goes down and to the right.

Then from any of these cells, (e,n), we can find the next cell where X is

the same value by doing

(e,n) -(e+2n, n)

 

Or we can start at our (e,-1) = (-x^2,-1) cell and shoot lines out from it with:

(e,-1) -(e + 2dt, -1 + t) for any integers d,t

 

Similar to the constant A values, we need to generate an entry by having (e,n,x) values.

Here are our equations:

 

aa + 2ax + 2an = c (a,x,n,c)

b = a + 2x + 2n (b,a,x,n)

a * b = c (a,b,c)

(d+n)(d+n)-(x+n)(x+n) = c (d,n,x,c)

x = d - a (x,d,a)

d * d + e = c (d,e,c)

x = math.sqrt(2na-e) (x,n,a,e)

 

From this we can get

 

xx = 2na - e

a = (xx + e)/2n

Then we have

d = x + a

b = a + 2x + n

c = a * b

 

to generate a new cell.

 

So now given any value X we can generate every cell with the same X.

Pic related is an example of this.

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 7:16 p.m. No.1663   πŸ—„οΈ.is πŸ”—kun   >>1665

>>1661

Nice pattern and work CA! So, starting from only c, are we able generate x so we can follow the pattern? Or, saying it another way, given c, d, and e how do we generate x? We need var a to do it I think??? VQC keeps telling us to find factors of var a, since that is the key to deriving x. Thoughts?

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 7:21 p.m. No.1664   πŸ—„οΈ.is πŸ”—kun   >>1665

>>1661

Also, there is a clear pattern between n values and x values in column 0. Wanna take a crack at writing a formula for it, CA? PMA and I got n(1-6) figured out, and there is a clear pattern for odd n, and a clear pattern for even n. After (0,6) the pattern morphs, but there is still a clear relationship from n to x.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 7:33 p.m. No.1665   πŸ—„οΈ.is πŸ”—kun   >>1666

>>1663

No this is starting at X we can generate any other cell with the same X. Given our C, we can generate (for a=5,b=23, a*b = 115):

 

Our entry for C is

(15,48,10,9,1,115) or (15,48,10,-105,115,1)

 

and our destination cell is:

 

(15,4,10,5,5,23)

 

So our C gives us the correct D value. We need a way to calculate these other values from our original cell.

 

>>1664

I think I've already solved all the (0,n) columns. Look at this code (I'm pretty sure I've already tested it and it works) but give it a look over.

 

(index is basically the t value)

 

def zeroN(N,index):

if(N == 1):

return zeroOne(index)

if(N == 2):

return zeroTwo(index)

 

if(N % 2 == 0):

#Even

factor = N/2

baseRec = zeroTwo(index)

else:

#Odd

factor = N

baseRec = zeroOne(index)

E = baseRec[e] * factor

D = baseRec[d] * factor

X = baseRec[x] * factor

A = baseRec[a] * factor

B = baseRec[b] * factor

C = A * B

return (E,N,D,X,A,B,C)

 

#Generates any entry in the (0,1) cell

def zeroOne(index): #index starts at index = 1

A = index * index * 2

D = 2 * index * (index + 1)

X = index * 2

E = 0

N = 1

B = A + 2 * X + 2 * N

C = A * B

return (E,N,D,X,A,B,C)

 

#Generates any entry in the (0,2) cell

def zeroTwo(index):

X = index * 2

A = index * index

B = (index + 2)**2

C = A * B

D = X + A

E = C - D*D

N = ((X * X) + E)/(2 * A)

return (E,N,D,X,A,B,C)

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 8:36 p.m. No.1667   πŸ—„οΈ.is πŸ”—kun   >>1668

>>1666

Yeah I had that too. They all are good I'm pretty sure. I've tested like the first 10 of the first 10 rows of the first 10 whatever so I'm just assuming it works forever because its infinitely long ya know

MinecraftAnon !!QXqSZ2ev8. ID: ad654f Dec. 29, 2017, 8:58 p.m. No.1668   πŸ—„οΈ.is πŸ”—kun   >>1669 >>1671

>>1667

I mean, in the past handful of posts, you've posted like a hundred lines of code and formulas, and you're pretty sure it's good? And you can't even be bothered to check it with the RSA100 values? But we're supposed to?

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 9:10 p.m. No.1669   πŸ—„οΈ.is πŸ”—kun   >>1670

>>1668

I can't check it with RSA values I don't know what to even do with that. Currently I'm trying to find relationships between entries with the same (x,d,a,b) values. I've already found a pattern to get all cells for any x value. Also all the cells for any A value. I'm trying to find patterns in the grid and learn the grid. We haven't even come up with a method to find the factors yet, so plugging in RSA values wouldn't be worthwhile. Also VQC said the patterns in the first few cells extend to the entirety of the grid, so I see no point in doing these values. Also I don't have BigInt for my python and I don't feel like downloading it. My computer isn't that fast I run the thing and every record it goes through (sometimes I was going pretty deep into each cell) prints True for whether or not its a valid record and I kill the program when I'm tired of watching it print True. I'll say with 98% confidence all this is true.

 

I'll post my Python code. Give me a minute.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 9:34 p.m. No.1671   πŸ—„οΈ.is πŸ”—kun   >>1672

>>1668

pastebin .com/d8xZZnm0

 

Here is my code. It's verified enough to me. I only post stuff on here if I haven't found any holes in it. The imaging stuff is in another file and I would have to reformat it and change some variables around. You'd also have to download a whole package which is a pain in the ass so I left it out. This is my freshest file that I've been working with.

 

Some of the stuff doesn't work when you start with 0. (Like the A stuff for A=0). If you get an error that would probably be it. Otherwise I think it's pretty error free code.

 

>>1670

Okay. Then what? I can multiply the primes together then do what? Thats where I don't get the point of using the RSA. I'm generating these values through E,N,X or E,N,A so the C is kind of a byproduct of those. I'm trying to analyze the grid so I can understand all the patterns so that I can eventually figure out an algorithm to factor the large products of primes, I don't see the point of just throwing bigger numbers at it if we don't even know how we are going to factor it.

MinecraftAnon !!QXqSZ2ev8. ID: ad654f Dec. 29, 2017, 9:43 p.m. No.1672   πŸ—„οΈ.is πŸ”—kun   >>1673

>>1671

The problem with what you're doing is this. We know C, and thus D, and thus E. If we knew X, then A=D-X. Then, B=C/A; RSA factored in literally two operations.

 

You're working with a pattern that definitely exists, but is exactly what we're trying to avoid doing.

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 10:44 p.m. No.1673   πŸ—„οΈ.is πŸ”—kun   >>1675

>>1672

Agreed, MA. We've been banging our heads against the wall for two weeks or so. We have exhausted SO MANY possible incorrect solutions. We've honed the focus to generating var a for row 1, and also to understanding how c^2 values in column 0 multiply. CA, glad you're back! Here's VQC's recent posts about where to focus. TBH, we haven't figured it out yet. We're bonked, but still working to understand.

>>1099

>>1380

>>1447

 

MA is correct when he said we are given c, d, and e. Next step is finding a. Then we can derive x, t, and b. Along with n as well, I think. The hard part is starting only from c. We need a way to tie together columns and rows.

 

Anons, what do you think about the n and x patterns in (0,n)? After n=6 they change, but still clearly related. Can we put all eyeballs on this pattern and try to find the equation?

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 11 p.m. No.1675   πŸ—„οΈ.is πŸ”—kun   >>1679

>>1673

And, since x is derived from d-x=a, we could also look for n*a patterns in (0,n), since these will tie directly to d and x patterns in (0,n) as well. Thoughts, Anons?

 

>>1674

Baker, thank you. I think in the next few days I'll rework the archives i posted to include all rt's, just for proper records. Then in the following bread everything will be accurate. I was trying to keep it to math rt's but there are more goodies that I missed.

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 11:03 p.m. No.1676   πŸ—„οΈ.is πŸ”—kun   >>1679

Sorry lads d-a=x fucking tired out over here. Too many late nights this week working on math(s). Yawn. Can't even post basic equations to make my point, lol!

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 11:09 p.m. No.1677   πŸ—„οΈ.is πŸ”—kun   >>1678 >>1679 >>1681

MORE FOR A:

 

If we have an origin cell(e,n) and the P value (the vertical jump size it takes to get to the next part on the parabola, also the root of the square value multiplied by a/2 to get origin cell) then we go from an entry for

 

(e,n,d,x,a,b,c) to (e, n+P, d2, x2, a+1, b2, c2)

(just another cell to be made from E,N,A, which we already have code for.)

 

This can also be reversed to yield:

(e,n,d,x,a,b,c) to (e, n-P, d2, x2, a-1, b2, c2)

 

Also for a cell on a parabola with a certain p (EVEN P) value and a certain t value (t is which spot it is on parabola. t=0 is origin, t=-1 is where n = originN-t*P etc.)This is can be derived from the same pattern above with prior rules that I found out.

 

Remember parabola origins are of the form for even p:

(2aH, (a/2) * p^2)

 

If our entry is on the parabola from this origin we can do

[n+(p*p)/2 is not verified my pictures were not large enough I'll test it later sue me]

(e,n,d,x,a,b,c) -(e+2H, n+(p*p)/2,d?,x?,a+1,b?,c?)

(e,n,d,x,a,b,c) -(e-2H, n-(p*p)/2,d?,x?,a+1,b?,c?)

 

both are from (e,n,a) so we can calculate them.

 

If these patterns are true, then we would have, for each entry:

 

e,n,d,x,a,b,c,p,t,h

 

(shit dats a lot of variables batman!)

 

where we already know what (e,n,d,x,a,b,c) are and

p = the value such that n = (a/2) * p^2 for the origin of parabola

t = the entry on the parabola started by the origin

h = the number of parabolic shifts to the right you've taken to get to the origin from (0,n)

 

I like to think of p as the parabola depth (they all are same pattern at same depth). t as in the entry along the parabola. if you shift t along the same parabola the p value won't change. H will not change unless you do the function (e-2a, n-1, a, H-1) -(e,n,a,h) -> (e+2a, n+1, a,H+1).

More on the T, if we have parabolic origin value (t=0)

(e,n,d,x,a,b,c,p,0,h) then we can generate values along the same parabola by changing t.

(e,n,d,x,a,b,c,p,0,h) -(e-p(t^2), n+pt, d?,x?,a,b?,c?,p,t,h) for any t.

 

Pic related for a = 34

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 11:16 p.m. No.1680   πŸ—„οΈ.is πŸ”—kun   >>1683

Also, good thing AA moved us over here. CBTS main is nuts. Alex Jones show interviewed PamphletAnon and Baruch today, and millions of new truth seekers just found out about Q and the chans. CBTS is full of newfags and shills, and the Judicial Watch link for the Huma Abedin emails can't be reached, either under DDOS or too many requests. Q just went mainstream, lads!!! Victory To The Light!!! Let's keep our spirits up lads, and keep working for a solution. These evil fucks are gonna get justice. Let's do our part.

 

http:/ /www.judicialwatch.org/wp-content/uploads/2017/12/JW-v-State-Huma-Weiner-emails-00684-2.pdf

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 11:22 p.m. No.1681   πŸ—„οΈ.is πŸ”—kun   >>1682 >>1683

>>1677

CollegeAnon ripping it up parabola style! I'll put fresh eyes on this tomorrow. So to clarify, you're saying var a moves in parabola form, and you've figured out how to calculate individual a values using your formulas? If so, that's dope as shit. Thanks for all your work!

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 29, 2017, 11:33 p.m. No.1682   πŸ—„οΈ.is πŸ”—kun   >>1683

>>1681

What I can do is generate any valid cell for any a value. What I can't do (yet) is figure out how to calculate a given t from (e,n,d,x,a,b,c) but maybe someone could figure that out. If we could do that, then starting from our cell (e,n,d,x,1,c,c), (with t) we could ride up and down our parabola. Then with t we could ride it down to t=0 or the origin, which would give us our h value. Then from there we could ride h back home to h=0 the (0,n) column to figure out that parabola we are on by getting the value p. Once we have that value p, we can go back to our original cell and potentially shift a up until we find something?? I'm not exactly sure how we'll use this, but this is certainly a way to index and navigate these cells.

 

So if I had to say to do anything it would be to find out how to get t from an entry. This will probably be tough.

VeritasAequitas !!Nf9AmQNR7I ID: 15bf90 Dec. 29, 2017, 11:55 p.m. No.1685   πŸ—„οΈ.is πŸ”—kun   >>1688 >>1689

>>1683

Agreed, Baker. No love for famefags. I'm just tripping out that Q is going mainstream. My lady is talking with friends on FB about who Q is, ie Barron, Don Jr, etc. WTF!!! When 60 million eyes get put on Q's posts, the world will redpill itself. AJ is talking openly about Q, chans, etc. Time to FIRE UP THE MEME CANNONS, LADS! I can't believe I'm alive to be seeing this. WHEN we solve this (with Senpai help, of course! Thx Senpai) we are going to help fundamentally change the course of human history. We have green light to go mainstream and fucking blast FB, Twatter, et al. I've been waiting for this moment my whole life, never thought it could really happen. Truth and Justice. Make Earth Great Again!

ID: a2017c Dec. 30, 2017, 12:02 a.m. No.1688   πŸ—„οΈ.is πŸ”—kun

>>1684

I'm not like you. I can't synthesize different areas of maths. I don't know what parabola's and roots are. I haven't had to brush up on them for this.

 

>>1685

Shame about Baruch. I guess they were right about him. I was in contact with him too.

Anonymous ID: 71e1c0 Dec. 30, 2017, 12:03 a.m. No.1689   πŸ—„οΈ.is πŸ”—kun   >>1690 >>1691 >>1692 >>1693

>>1685

Indeed VA, just waiting for those Gitmo announcements! And for MSM to completely crumble.

 

Ok fags, here are some test cases and challenges for you:

 

RSA Test Cases and Challenges

https://pastebin.com/f5Km9K8H

 

RSA-100 (c) =

1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139

(100 decimal digits, 330 bits)

 

RSA-100 (a x b) =

37975227936943673922808872755445627854565536638199

Γ—

40094690950920881030683735292761468389214899724061

(2 50 decimal primes)

 

RSA-100 (c^2) =

231832607105497859450453971149782249091397152352980773038383827802126335737279486733598563928061275517976566803851888280780332815207688528004235456616576727168746464977139224636030180916664018563207

(198 decimal digits)

 

–

 

RSA-129 (c) =

114381625757888867669235779976146612010218296721242362562561842935706935245733897830597123563958705058989075147599290026879543541

(129 decimal digits, 426 bits)

 

RSA-129 (a x b) =

3490529510847650949147849619903898133417764638493387843990820577

Γ—

32769132993266709549961988190834461413177642967992942539798288533

(with a 64 and 65 decimal prime)

 

RSA-129 (c^2) =

13083156311017746081229837780584604971789185572322940113554957177463495238823718363789678158885976884237362181609103528581150260013472809482392451812847672930017596579807933662233072087386335238374806414971531141586419217198046250532235073881640972514818681

(257 decimal digits)

Anonymous ID: 71e1c0 Dec. 30, 2017, 12:04 a.m. No.1690   πŸ—„οΈ.is πŸ”—kun   >>1691 >>1692

>>1689

 

RSA-190 (c) =

1907556405060696491061450432646028861081179759533184460647975622318915025587184175754054976155121593293492260464152630093238509246603207417124726121580858185985938946945490481721756401423481

(190 decimal digits & 629 bits)

 

RSA-190 (a x b) =

31711952576901527094851712897404759298051473160294503277847619278327936427981256542415724309619

Γ—

60152600204445616415876416855266761832435433594718110725997638280836157040460481625355619404899

(with (2) 95 decimal primes)

 

RSA-190 (c^2) =

3638771438488087985574274106795079635808387100084863940611975142846520895542297425531641157043983476619993374723043826429566267790036775257657139728945914322148154407595065259459214885711856214268244428701532653022789206595392004837865259854410326840170456159370131034294156696408886672161241549375876666598321636136763746229426218477864974486848489396416573764495046083098157361

(379 decimal digits)

 

–

 

RSA-640 (c) = 3107418240490043721350750035888567930037346022842727545720161948823206440518081504556346829671723286782437916272838033415471073108501919548529007337724822783525742386454014691736602477652346609

(193 decimal digits, 640 bits)

 

RSA-640 (a x b) =

1634733645809253848443133883865090859841783670033092312181110852389333100104508151212118167511579

Γ—

1900871281664822113126851573935413975471896789968515493666638539088027103802104498957191261465571

(with (2) 97 decimal primes)

 

RSA-640 (c^2) =

9656048121330239196485739047236872733132646152415265572749839063571416858087801139297436736668535657942544532834166699749756556831546825766219543494478915395505001388433032706122580727561545005466558578465273868089134810944324056030967838960018632090507727101247760897728632535169583227345646104139213934877983671194311248207350399320919897813580637534250460303641423529457084273798881

(385 decimal digits)

Anonymous ID: 71e1c0 Dec. 30, 2017, 12:05 a.m. No.1692   πŸ—„οΈ.is πŸ”—kun

>>1689

>>1690

 

And for some challenges, find a & b:

 

RSA-240 (c) =

124620366781718784065835044608106590434820374651678805754818788883289666801188210855036039570272508747509864768438458621054865537970253930571891217684318286362846948405301614416430468066875699415246993185704183030512549594371372159029236099

(240 decimal digits & 795 bits)

 

–

RSA-1024 (c) =

135066410865995223349603216278805969938881475605667027524485143851526510604859533833940287150571909441798207282164471551373680419703964191743046496589274256239341020864383202110372958725762358509643110564073501508187510676594629205563685529475213500852879416377328533906109750544334999811150056977236890927563

(309 decimal digits & 1,024 bits)

 

RSA-2048 =

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

(617 decimal digits & 2,048 bits)

 

https://pastebin.com/f5Km9K8H

Godspeed PATHanons.

Topolanon +++ !!UrvFpU0has ID: 3520dc Dec. 30, 2017, 2:20 a.m. No.1699   πŸ—„οΈ.is πŸ”—kun   >>1700

Ignore your squabbles.

You can watch the entirety of Star Trek: The Next Generation in the Q episodes alone.

 

VQC isn't "Albino Morpheus".

He's PICARD.

 

AS A Q!

 

I'm not even done with the episode, i've been through voyager… DS9 (999) is the ending.

 

Calling it now.

Anonymous ID: 99edb0 Dec. 30, 2017, 3:57 a.m. No.1709   πŸ—„οΈ.is πŸ”—kun   >>1710 >>1712

Okay. As I mentioned here >>1647 (even though apparently I used the term genesis cell wrong; I was referring to the first cell of an infinite set), one way I was thinking we could find n from c was by finding i from c, since then we use n = i - d or n = i - (floor(sqrt(c))). Instead of spending half an hour turning this into something really easy to understand I'll just type out the process I took and hope everyone gets it and doesn't mind a whole bunch of text.

 

In order to figure out how to calculate i from c, since we can't do it with the variables we get from c alone, I tried to find a visual relationship between them. I made an image with a white pixel every time c is the product of two primes but instead of (e, n) as the axes it's (c, i). That's pic related. It shows the relationship between c and i, and it looks like a bunch of linear lines.

 

If these were linear, and we could figure out the relationship between these lines, we could use them to find i for a given c, couldn't we? All we'd need to do is find the correct line that our c is on and use simple gradient maths (rise / run). In order to confirm that these were linear lines, since it's a bit haphazard doing it in this Paint program I have, I changed the code around a little bit to plot each (c, i) as points in a big grid, much like the original {e, n, d, x, a, b} grid, but again with axes (c, i). I then highlighted from one point in each line to the next a few times in order to figure out the gradient. By the way, if anyone doesn't know, in whatever your spreadsheet software is, you can change the formula syntax to R1C1 and it'll change the x axis from letters to numbers, which is very handy (instead of something like BD24 it'll be something like R42C52, as in (42, 52)). In the next post I'll paste my calculations, since it'll break up the text a bit, and it'll show what the relationship is between these lines.

Anonymous ID: 99edb0 Dec. 30, 2017, 3:58 a.m. No.1710   πŸ—„οΈ.is πŸ”—kun   >>1712

>>1709

Lowest line

R121C52:R277C130

(121, 52) to (277, 130)

Rise is 156, run is 78, gradient is 2 (6/3)

R133C58:R265C124

(133, 58) to (265, 124)

Rise is 132, run is 66, gradient is 2 (6/3)

R337C160:R457C220

(337, 160) to (457, 220)

Rise is 120, run is 60, gradient is 2 (6/3)

 

Next line upwards

R169C116:R217C156

(169, 116) to (217, 156)

Rise is 48, run is 40, gradient is 1.2 (6/5)

R253C186:R289C216

(253, 186) to (289, 216)

Rise is 36, run is 30, gradient is 1.2 (6/5)

R289C216:R313C236

(289, 216) to (313, 236)

Rise is 24, run is 20, gradient is 1.2 (6/5)

 

Next line upwards

R217C204:R229C218

(217, 204) to (229, 218)

Rise is 12, run is 14, gradient is 6/7 (0.857142 repeat)

R229C218:R265C260

(229, 218) to (265, 260)

Rise is 36, run is 42, gradient is 6/7

R265C260:R301C302

(265, 260) to (301, 302)

Rise is 36, run is 42, gradient is 6/7

 

Next line upwards

R181C210:R205C254

(181, 210) to (205, 254)

Rise is 24, run is 44, gradient is 6/11 (0.545454545454)

 

Next line up

R181C222:R193C248

(181, 222) to (193, 248)

Rise is 12, run is 26, gradient is 6/13 (0.461538 repeat)

 

Next line up

R205C290:R217C324

(205, 290) to (217, 324)

Rise is 12, run is 34, gradient is 6/17

Anonymous ID: 99edb0 Dec. 30, 2017, 4 a.m. No.1712   πŸ—„οΈ.is πŸ”—kun   >>1720

>>1709

>>1710

I did the next one too and it was 6/19. The thing you should all be taking away from this is that the denominators are all primes, and from the lowest line to the highest line they increase in order with 6 as the constant numerator.

Anonymous ID: 99edb0 Dec. 30, 2017, 5:02 a.m. No.1720   πŸ—„οΈ.is πŸ”—kun

>>1712

I didn't realize until now but setSize was 12. That means that while those gradient rules still work the numbers are slightly different. I'm currently trying to figure out the mathematical relationship between each line that determines where the next one starts each time.

Anonymous ID: 71e1c0 Dec. 30, 2017, 6:38 a.m. No.1723   πŸ—„οΈ.is πŸ”—kun

>>1700

'What you realize becomes truth'

'Meme responsibly anons"

^^^THIS; works on several levels.

All-night digger. Layers deep, sauce within sauce. Perhaps more indictments / counts from this. Think the first class seat grabbing cunt was setup and blackmailed for control, affecting the homeland and our security. Tangled webs and boomerangs. More work to do, carry on anons.

 

>>1713 some tasty looking bread, tnx.

>>1531

Good oh Great Hobo. Define excession plz.

Unfamiliar w/ responsibilities associated with new threads, proposing a 'Disclosure, Ramifications, and Rebuilding' perhaps?

Anonymous ID: 01b9f7 Dec. 30, 2017, 7:36 a.m. No.1724   πŸ—„οΈ.is πŸ”—kun

Tested!

For all records {0, n, c, c-aa, aa, bb}

A Goal record!

There is m such that n=2mm

and …

b-a = 2m

{0, 24200, 669, 660, 9, 49729)

m=110

b=223 a=3 a - b = 2m = 220

{0, 8, 77, 28, 49, 121}

m=2

b=11 a=7 a - b = 2m =4

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 30, 2017, 8:10 a.m. No.1725   πŸ—„οΈ.is πŸ”—kun   >>1726 >>1727

Same as what I've been doing for constant D values (because for C record we already have correct D, so this could help us navigate through the D's:

 

pastebin. com/5cd4k2rz

 

Graph. (+ = from the end, *= extrapolated from pattern).

 

For any d, we can start off and say there will be an entry at

(e,n) = (-d^2,-d+1)

 

We can call this our origin cell for D. Now, to navigate along this graph we would need to generate cells from their (e,n,d) values, which I do not think we can do yet. However this pattern is still notable.

 

Starting at our origin cell (the Zeroth origin cell), (e,n) I THINK we can generate any cell for this pattern:

 

(e0,n0) -(e0,n0+u) for any u [I'll go 70% on this step, cuz I can't generate to test, the other ones are pretty certain though]

 

Then, from our origin cell, if we go to the right one cell we get a second origin cell (origin cell 1). (e0,n0) -(e0+1,n0) = (e1,n1) From this cell we can do

(e1,n1) -(e1+2u, n1+u) to navigate down the line as many times as we want

 

Then, if we go down the line once (e1+n1)->(e1+2,n1+1) and then shift to the right (e1+2+1, n1+1) we get the third origin cell (e2,n2).

From the third origin cell, we can do:

(e2,n2) -( e2 + 4u, n2+u ) to generate more cells for any u.

 

Generally:

 

The zeroth origin cell for D can be found at:

 

(-d^2, 1-d).

 

For the zth origin cell, (e,n) we can generate any other entry with the same d by this equation:

(e,n) -(e + 2zu, n + u)

 

Also, from any origin cell we can generate the next origin cell.

( z to z+1), which boils down to going down the line once and then adding 1 to e.

 

(e,n) -(e + 2z + 1, n + 1)

 

So now we have a method to determine all the entries (e,n) with the same d. Also, this may be an impossible task, but if anyone can generate an entry just given the (e,n,d) values that would be sweet, but I don't think we can.

 

This may be the biggest thing I've discovered. This path for valid d values is the same for every d. As in, the only difference between entries for certain d values is the zero origin for d. Other than that you use the same exact steps to figure out what other cells have the same d. Knowing that, if we have a valid d, we can say, WITH SUPREME CONFIDENCE, that.

 

origin cell is at (-d^2, 1-d)

the origin cell for d-1 is at ( - (d-1)^2, 1-(d-1)) = (-dd +2d -1, -d) (same formula)

 

So if we have a cell we know for certain we can navigate to a valid cell by:

(e,n,d,x,a,b) -(e+2d-1, n-1, d-1, x?, a?, b?)

to go down a d value.

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 30, 2017, 8:14 a.m. No.1726   πŸ—„οΈ.is πŸ”—kun

>>1725

>>1725

The pattern for the origin cells is more consistent if the zero origin is calculated at (-d^2, -d) because that way you can still do the same pattern (go down the line once, then shift e by 1) to get the next origin cell. Before it was that pattern for all origins not equal to 0, now it is consistent.

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 30, 2017, 8:29 a.m. No.1727   πŸ—„οΈ.is πŸ”—kun   >>1728

>>1725

 

>Also, this may be an impossible task, but if anyone can generate an entry just given the (e,n,d) values that would be sweet, but I don't think we can.

 

public static TheEndRecord CreateForEND( BigInteger e, BigInteger n, BigInteger d ) {

 

BigInteger c = d * d + e;

BigInteger x = ( Lib.Sqrt( ( d + n ) * ( d + n ) - c ) ) - n;

BigInteger a = d - x;

BigInteger b = a + 2 * x + 2 * n;

 

return new TheEndRecord( e, n, d, x, a, b );

 

}

CollegeAnon !LAbIRp9cT. ID: e34200 Dec. 30, 2017, 9:17 a.m. No.1728   πŸ—„οΈ.is πŸ”—kun   >>1729

>>1727

holy fuck thanks for this code i would have been at a dead end without you

a bit of output

>((-36, 0, 6, 6, 0, 12, 0), 'a=0')

>((-36, 1, 6, 6, 0, 14, 0), 'a=0')

>((-36, 2, 6, 6, 0, 16, 0), 'a=0')

>((-36, 3, 6, 6, 0, 18, 0), 'a=0')

>((-36, 4, 6, 6, 0, 20, 0), 'a=0')

>((-36, 5, 6, 6, 0, 22, 0), 'a=0')

 

Okay so I tested this code and and it only gave me values of A where A = 0. I used to have that a=0 meant an invalid cell, but heres the thing: For A=0, you can have ANY value for (e,n,d) that you want and they all could still be valid entries right? Like 0 is a number so why couldn't it be in entry a? Maybe these are all valid entries.

 

Suppose they are, our output from AB and C fun

C(21) = (5,7,4,3,1,21,21) = start

AB(3,7) = (5,1,4,1,3,7,21) = goal

 

So for this we have the correct E and D right off the bat. We may be able to iterate through these n values and search things from there, because these are valid cells, but we cannot do END(5,7,4) because I think it takes root of -1. But for this remember we have the correct D. So we can go to the (-d^2) column, which is basically the root for d ((-d^2,-d) would be the seed I guess). From that column, we can expand out and generate any cell for the same (correct) d. We could do this and iterate through the cells from each D origin cell and go down until we either hit or pass e. Then I think we may have our correct cell at that point.

 

Tada!

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 30, 2017, 9:45 a.m. No.1729   πŸ—„οΈ.is πŸ”—kun   >>1730

>>1728

 

Iterating to find the solution isn't going to work.

 

See a couple of my previous posts:

 

>>1342

>>1487

 

>>1509

VA - thanks for reposting this. Am thinking this through a bit more.

 

We've been one variable or formula away from a solution for some time now.

 

The "Use Y on C at X" from >>>/cbts/75435 needs more attention.

 

I think we know what C and X are. And we know how to create a small square from t. What is Y? y axis? is that n?

 

CA got me thinking a bit more about the END formula.

 

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

 

This can be solved a couple of ways. And given that we know c, d, and e for our target aabb and ab records, perhaps if we can identify just the small square movement, we can solve for n.

 

For example:

 

Solving for x given n, d and c.

 

x = ( Lib.Sqrt( ( d + n ) * ( d + n ) - c ) ) - n;

 

Solving for n given e, d, and (x+n):

 

c = d*d + e;

n = Lib.Sqrt( c - (x+n)^2 ) - d

 

Maybe all we have to do is figure out the "small square" movement?

Teach !!UgZAPoSXEk ID: a92e07 Dec. 30, 2017, 10:25 a.m. No.1733   πŸ—„οΈ.is πŸ”—kun   >>1735 >>1736 >>1738 >>1740

Hey guys, its been a while between posts, but I've been following along with the progress.

 

I've been focused on my own theories mostly, and none have been fruitful.

 

>>1731

Hey PMA, the size of the small square has been where my attention has been lately.

 

I've also been spending some time going through reading for ECC.

 

I want to point out something that I've realized too…

 

n = (x^2 + e) / 2(d - x)

 

This is just a simple rewrite of x^2+e=2na.

But what I like about this representation, is that d and e are constants and so we have a relationship between n and x for any c.

 

I'm trying to use this equation to figure out the possible sizes of the small square.

 

Anyway, how is everyone?

 

Also welcome back CollegeAnon! How were your exams?

VeritasAequitas !!Nf9AmQNR7I ID: ca62e5 Dec. 30, 2017, 10:31 a.m. No.1734   πŸ—„οΈ.is πŸ”—kun   >>1735

>>1731

>>1732

I like the idea. How do we deal with erratic t vars as n increases? Maybe we need a rewrite of t equations for larger n, or maybe the t vars are accurate in later n are accurate because the d,a,x vars create t? Basically, does t represent small square values as n increases and t begins to jump upward in patters rather than linear growth?

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 30, 2017, 10:37 a.m. No.1735   πŸ—„οΈ.is πŸ”—kun   >>1737 >>1738

>>1733

>I've also been spending some time going through reading for ECC.

I tried to read some of those documents… realized I'm not that smart.

 

>n = (x^2 + e) / 2(d - x)

I like the concept. After all, this seems to be an exercise in taking a really small square to it's perfect middle size.

 

>>1734

I believe this get's solved in the (e,1) space where there aren't any gaps in t. Remember, looking for one more variable or formula. t is just an indexer on values of x. Don't think we want to start changing it's definition based on n.

Anonymous ID: 8961b2 Dec. 30, 2017, 10:38 a.m. No.1736   πŸ—„οΈ.is πŸ”—kun   >>1738

>>1733

All is good, not posted much either since I'm little out of ideas and none of my attempts have gone anywhere. Still here banging my head against the grid with you!

Pretty sure I solved it in a dream already but not sure how. Doh.

Teach !!UgZAPoSXEk ID: a92e07 Dec. 30, 2017, 10:52 a.m. No.1737   πŸ—„οΈ.is πŸ”—kun   >>1739

>>1735

I was having a really hard time with understanding what VQC meant when he said:

 

>At the correct element in the grid at (e,1) where the value of a at that element equals the na we want, if you subtract 2d+1 from na then you are in the negative half of the grid in terms of e and the value at the same element in the first row will be (n-1)a

 

I believe the "na" that we want refers to the na in the 2na = x^2 + e.

This is why I think the equation is important.

I'm trying to work out what the (n-1)a cell means now.

VeritasAequitas !!Nf9AmQNR7I ID: 537a7c Dec. 30, 2017, 11:11 a.m. No.1738   πŸ—„οΈ.is πŸ”—kun

>>1733

Teach! Nice to see you!

>>1736

Nice to see you too, Anon!

 

>>1735

Well, if we’re able to solve for the small square using the t equations in row one that’s great! How do we tie that together with the column 0 knowledge? Or is it needed? If we know c, d and the small square, are we back to solving for n with the quadratic we derived? Column 0 still needs to get tied in? Thoughts, Anons?

ProgramMathAnon !dSvrkhSLR6 ID: ac6203 Dec. 30, 2017, 11:13 a.m. No.1739   πŸ—„οΈ.is πŸ”—kun

>>1737

Teach - likewise. I'm not sure I fully understand, but here's where my head is by way of an example:

 

c=145

 

1 x c =(1,61,6) = {1:61:12:11:1:145} = 145; (x+n)^2=72x72=5184; (d+n)^2=73x73=5329;

1 x c na =(1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)^2=12x12=144; (d+n)^2=73x73=5329;

 

In the first record, calculate (n-1)a = (61-1)1 = 60.

 

If you look at the differences in the n, d, and a values between (1,61,6) and (1,1,6) you will notice:

 

n = 1

d = d + 60

a = a + 60

 

So essentially the amount you take off of n gets added into the d and a.

 

This "transform" enables us to move from any (e,n) into the (e,1) space, and explains what's happening to d and a along the way.

 

I have played with this attempting to take off different values from n. (n-2), 2(n-1), etc. Haven't had much luck there.

VeritasAequitas !!Nf9AmQNR7I ID: 537a7c Dec. 30, 2017, 11:18 a.m. No.1740   πŸ—„οΈ.is πŸ”—kun

>>1733

Ah, nice Teach! There’s that clear relationship between n and x we’ve been looking for. Nice rewrite on the formula! We’ve been banging our heads on that for a while, I think you just made it crystal clear.