/vqc/ - Virtual Quantum Computer

Code

C#

BigInteger Square Root —— pastebin.com/rz1SdACZ

Generate Bitmap within original code —— pastebin.com/hMTtJF6E

More on generating a bitmap with the original code —— pastebin.com/JUdtehb4

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

Original VQC code —— pastebin.com/XFtcAcrz

How to run VQC code on Linux —— pastebin.com/6HnN7K5X

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

Traverse the VQC cells in real-time [V2] —— anonfile.com/TeH6q3d8bd/VQCGUI_v2.7z

NodeJS

BigInteger Library and Sqrt —— pastebin.com/y8AXtFFr

Python

College Anon's code (VERY USEFUL) —— pastebin.com/d8xZZnm0

Create the VQC —— pastebin.com/NZkjtnZL

3D VQC —— pastebin.com/vdf8SpYt

3D VQC (v2) —— pastebin.com/wZM5Thzu

Fractal cryptography —— pastebin.com/XuN4U7Dv

Generate cells for a (and more) —— pastebin.com/iAizgLFF

Generate any cell in (0,1) and (0,2) —— pastebin.com/gRTYpdMU

Generate genesis cell —— pastebin.com/GKzcCpMF

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

Calculate variables based on e and t —— pastebin.com/4s6McdbN

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

Solutions (Java)

Binary search for i —— pastebin.com/TAt5bDsR

Calculate factors using -x jumps —— pastebin.com/gKX9GW9r

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 dead) —— archive.fo/eihrQ

RSA General (#5) —— archive.fo/Lr9fP

RSA #6 —— archive.fo/ykKYN

Videos on cryptography —— pastebin.com/9u3hwywe

Chris' posts

(RSA #2 was shoahed)

RSA #3

>>>/cbts/87168

>>>/cbts/87234

>>>/cbts/87300

>>>/cbts/87378

>>>/cbts/87414

RSA #4

>>>/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

RSA #5

>>11

>>12

>>17

>>18

>>19

>>20

>>21

>>23

>>24

>>25

>>26

>>27

>>28

>>29

>>30

>>31

>>32

>>33

>>495

>>699

>>709

>>710

RSA #6

>>1099

>>1380

I'm working on the new VQC crumb map.

rsa2

http://archive.fo/fyzAu

I'm only going to shill this once since it isn't RSA-related, but I know some of you might be interested.

>>2563

>>2567

Its 0+0 right? fuck

Hello PMA! Can you update me on your ideas for how we can use a=1b=C, CC, AA, BB, etc to jump records? Also if anyone else has working formulas or programs for using these to make the jumps between records, jump in too!

On a separate note, what's the latest on mirroring x, lads?

>>2574

Run my program. I'm working on the fact that it's n! speed right now, lol.

Any ideas? I'm going to try looking for other patterns like the x jump. I was completely unaware walking down the tree was as simple as it is.

>>2575

Lol! Thanks for the deep laugh, Topol! That was just what the doctor ordered.

>>2578

If you take any c, and generate the 1*c cell, you can go n down and find a number whose second factorization contains the a you want for c.

145:

start with {1:61:12:11:1:145}

go down n (61) entries in the cell

and you get {1:61:278:133:145:533}

145*533=77285

all you have to do is find out how to get the second factorization of 77285, and you have a:

5*15457 = 772851

145 / 5 = 29

5*29 = 145

Second example: 533

{4:244:23:22:1:533}

getting to this cell is trivial. now go 244 cells down.

{4:244:1043:510:533:2041}

533*2041=1087853

now all you have to do is get the other factorization of 1087853 to factor 533

1087853 = 13 * 83681

533 / 13 = 41

13 * 41 = 533

>>2579

77285, not 772851

>>2580

Based on that I wrote this function. It takes in c and number of steps, then it iterates down n and uses gcd to attempt to find the factors.

def goDownSteps(c, steps=2):
  e, n, d, x, a, b = rowForAB(1, c)
  bb = b
  for i in range(steps):
    ee, nn, dd, xx, aa, bb = rowNegX(n, -(2 * i * n + x), bb)
    g = math.gcd(c, bb)
    if g != 1:
      print(g, c / g)
      break

Unfortunately it's still way too slow.

>>2581

Can you write a function that just gives me the n values down jump? I'm not good with t.

A function that will give me 145*133 when I input 145, because all you have to do is jump n down from 1*145.

>>2582

You mean 145, 533?

That's what rowX already does.

>>> rowForAB(1, 145)
(1, 61, 12, 11, 1, 145)
>>> rowX(61, (2 * 61 + 11), 145)
(1, 61, 278, 133, 145, 533)

The code for rowX is:

def rowX(n, x, a):
  b = a + 2*x + 2*n
  c = a*b
  d = int(math.floor(math.sqrt(c)))
  e = int(c - d*d)
  n = int(((a + b)/2) - d)
  x = int(d - a)
  return (e, n, d, x, a, b)

I've been thinking about what f means and the (-e, n) records and it's quite obvious.

I'm just writing this as I'm thinking loud, but of course (-e, n) records contain the same info, but with d + 1, n - 1 etc.

It's because it's the big square (d + 1)^2.

e is the remainder after removing d^2 from c. So negative e is the "remainder" after removing the upper square (d + 1)^2, or rather the number of empty parts.

n - 1 makes sense because we are now 1 square closer to the n we need to find the factors. Increase of d again makes sense because d represents the square.

Same as x + 1 because n is now (n - 1) so we need to increase x by 1 to maintain the same number (n + x).

I don't see any obvious way to make use of f.

>>2584

So if we extend the idea of f, we can do

e - 3*(2 * d + 3) which should give us n - 2, d + 2, x + 2.

e - 4*(2 * d + 4) which should give us n - 3, d + 3, x + 3 etc..

This is because we are increasing the squares. Note though: This isn't how we can factorize as this is the original fermats factorization.

>>2585

>>2584

We ARE doing Fermat's factorization.

Just a super advanced, instant version.

>>2585

Wait, I think that's wrong. e - 3*(2 * d + 3) should give us n - 3, d + 3, x + 3 since f = e - 1*(2*d + 1).

>>2586

Well yeah, but the e - 3*(2*d + 3), e- 4*(2 * d + 4) etc is the slow, original method.

>>2579

These have more than 2 factorizations, but good idea if you can find how to do it

Somebody tell me how to mirror a cell.

I think that's the solution. Every cell on the positive e side has a counterpart on the negative e side.

>>2590

Found it, I'll work on it tomorrow.

It feels like we're so close. All we need is one last nudge from VQC.

>>2592

I believe the answer is in (e,1).

Look at my picture of the x jump factorization of 145 and compare to (1,1).

You can calculate it, I bet. Brainstorm. I need some sleep.

Wasted a lot of time on (e,n).

I think that the solution is back on (e,1)

Exploring relation between (2c,1) and (2Nc,1)

>>2598

Given the tweet from vqc yesterday Im sure he thinks we're close.

Maybe we found the last pattern, but we continued instead of studying it?

Maybe we should look over the past two days and recheck the patterns discovered?

>>2599

I'm sure there is value in (e,N) but it take massive iteration to find records. I have not found a computational pattern.

Looking at VQC's hints they remain focused on (e,1) on (n,0) and on factor trees

a(t=1) equals c ….

for (2c, 1) and 2c-1,1)!!!!!!!!!!!!!!!!!!

.

The a(2c-1, t) factor tree has twice as many a,b as factors as does a(2c, 1, t) tree.

I'm doing side by side looking at different multiples of e = 2c and (2c-1)

>>2593

Nice work last night on n and na, Anon! I also think (e,1) is the key. How do we USE (e,1) to get what we need in higher values of n? Working to solve that over here, still haven't made the breakthrough. Hopefully we'll find it soon.

>>2592

Hello Isee! Thanks for all your work, anon. We are getting close. Let's keep going!

>>2578

Hello PMA! Thanks for the clear explanation and great output as always. So the left column is used in this example to search for factors of c=65? Cool. Then you can change your search to whatever factors you want. Pulling them out will be the next step?

>>2603

Quiz

for what two values of e

does a(e, 1, 1) = c??

Is this useful in analyzing factor trees?

>>2602

Very impressive!

>>2604

>>2603

Hmmm... Quiz, Ok!

Q1: (1,1) and (1,2).

At (1,1) we have a,b factors of 1,5 and c=5

At (2,2) we have a,b factors of 1,3 and c=3

Q2: Seems like a repeat of Q1?

Yes, this is very useful for analyzing factor trees. Not sure what you're hinting at. Have you found something new to report, Anon?

>>2604

>>2606

a(e,1,1) = c

can be found at both:

(2c,1,1)

(2c-1,1,1)

>>2603

(e,1,1) = c I'm having a bit more trouble finding.

>>2606

>>2607

Sorry, deleted first post forgot the a.

I was hoping to generate some interest in analyzing factor trees. VQC gave a few hints regarding them.

Deleted post for first time found myself in alternate universe full of deleted posts. Found a VQC post and got excited until I realized it was a deleted post!

>>2608

shit… any here I was thinking all my code was buggy.

So you're confirming that the relevant records are at (2c,1,1) and (2c-1,1,1) correct?

Do you have any additional insight into how they are useful?

>>2610

Knowing how the factor trees work in the VQC lets you factor it.

When you make an xjump from a 1c cell you get a number which has a for c as a factor.

>>2608

Definitely interested over here in factor trees. I've been working for weeks to understand the crumbs about this. Kinda stuck.

>>2610

Can you guys quickly explain your notation? Are we using (e,n,t) as usual? If so, what does a(e,1,1) =c mean? I'm working to follow.

>>2612

my understanding: a(e,1,1) means the a value for the record (e,n,t)

>>2612

>>2613

As if it would kill you to type "a of (e,1,1)".

>>2614

dude… wasn't me. But I like your notation better!

>>2613

Hmm. Ok, that's what I thought. Thanks!

>>2614

No worries over here. Just wanted to make sure I could follow along.

>>2616

Just keep working on it. We're closer than we've ever been. I finally understand the na is n apart crumb.

quick question. if we have multiple records with a values that share factors a, b and c, all with different values.

Then divide the c out, we're left with some combinations of a*b.

With 2 records, with different remainders, can we solve for a?

>>2618

Give me an example. I think this is how you factor it.

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

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

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

>>2618

It could but as t gets bigger more extraneous factors get added.

>>2622

>>2621

Yes, there's somewhat of a wall when you try to factor it by moving around and ignoring the e,1 cell, because you can't "go back." You can go up the branch of the tree and get more numbers who share prime factors of the c you are using, but you can't get the factors. I really think we just have to pay attention to e,1. e,1 contains the factors for every x jump from 145.

>>2622

I understand, I expanded the search to 10*c*n just to pick up more data.

>>2623

yeah, screen shot was testing cc, 2c, 2c-1, and various other (e,1) combinations.

>>2618

Gcd will find the greatest common divider

>>2625

In log(n) time?

>>2625

Given two numbers a, b where a factorized is k, j and b factorized is j, l and k, j and l are primes gcd(a, b) will return j. It's also in On(log n)

>>2621

Any multiple of c will contain at least the same multiple of a and b. Since you still don't know a and b we are no better off.

>>2627

O(log n) stupid auto correct

>>2628

thanks. back to digging.

>>2628

I disagree that we are "no better off." Have you learned nothing from Fermat? And yes, that's a very important rule, that multiples of c will contain a and b.

>>2629

>>2627

O(log(n)) or O(log(n)*n)? If it's in O(logn) we can use it.

>>2620

I thought a lot about this hint. In row one n = 1 so na is just a. There was also a hint about subtract d from a to get n-1. Well at t=1 a = d and n=1 so we get a=d=n=0. If we subtract 2d we get a n = -1.

>>2623

I was looking at some of the records you were generating. Didn't follow all of them but they tested. Do you think you give me a couple (30, 98) records

is it possible that the records we are searching for in (e,1) have a values equal to either a^2 or b^2?

>>2636

scratch that. doesn't apply everywhere.

>>2636

>>2637

Just because it doesn't apply everywhere doesn't mean we can't make it apply everywhere. What did you learn from factoring c^2 instead of c?

That you can make e = 0.

>>2636

theoretically we know where the aa and bb records are. I've looked at odd next to even. I've looked at 2Nc versus 2c records but if you just know c the a and b remain hidden. I think Chris has some magic sauce that will seem easy when someone figures it out.

>>2638

>>2639

ok. here's what I'm looking at, still for c=145.

starting position at (e,1,1) where a = n*c

(1,1,67) = {1:1:8978:133:8845:9113}

factors of a are a,b,c

negative x move from original (1,61,4)

(1,1,-66) = {1:1:8712:-133:8845:8581}

factors of a are also a,b,c

These search boundaries can be created for any (e,1).

Between these t=67 and t=-66, there are 2 instances where a = a^2 (at (1,1,-3) and (1,1,4)), and a = b^2 (at (1,1,-20) and (1,1,21)).

Not sure what to make of these, as I didn't find them in other examples.

Just to give you an idea how little the world understands about factorization, How RSA 155 was solved:

pastebin.com/8BnqvvtT

Ghost in our machine??

For e=2c-1 and factor tree A of (e,1, t) and c=a*b

Using A for value of a in tree because its bigger than c=a*b

a is factor of A when t/a is integer or has remainder 1

b is factor of A when t/b is integer or has remainder 1

c is factor of A when t/c is integer or has remainder 1

here's the weird part when a is a factor of t but b has a remainder of 1 (or vice versa) c is

still a factor so c is a factor more often than a or b

>>2633I love Oolong Tea! Is Teach still working here? Haven't seen him back in a while. Iv been lurking. Will do another chapter and some cleanup tonight if I can think of a gripping topic. There seems to be a major lull in the action here and on the CBTS/Thestorm boards. Loving what you guys are making! Sad that I can't contribute more. Keep up the good work!

>>2648

Yes, they're huge. If e wasn't large enough the number would be vulnerable to primitive versions of Fermat's factorization. You don't really have to worry about how big RSA numbers are. Just solve it for humanly-understandable c's, program it and let the computer do the work.

Thanks for the quick reply, anon! Is this Baker?

>>2648

I'm not missing, computer problems, life problems. Hope to pick it back up this weekend.

>>2653

>>2654

playing along. I'm with you on the up step. not so much on moving down. Are you manipulating something other than x? If so, can you share that formula?

also, your 290 example isn't a valid record.

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

21025 - 20736 = 289.

>>2655

I must have not checked that. I'm using this method. If I want to go down, I pass in n,x,a for 1c

If I want to go up, I pass in n, -(2n+x), c

 
>>> def rowNegX(n, x, a):
...   b = a + 2*(-x) + 2*n
...   c = a*b
...   d = int(math.floor(math.sqrt(c)))
...   e = int(c - d*d)
...   n = int(((a + b)/2) - d)
...   x = int(d - a)
...   return (e, n, d, x, a, b)

>>2657

Also, I think it says it's an invalid record because 290 is not the difference of two squares because it's even, but if you divide it by two it becomes the difference of two squares.

I wonder if that's important.

Passing in -(2n+x) as the formula makes it make the na jump like the crumb says.

>>2657

Thanks.

fyi, an alternate to moving up:

BigInteger newA = ter.b;

BigInteger newD = 2 * (ter.b - ter.d) + ter.d;

TheEndRecord up = TerFactory.CreateForNDA( ter.n, newD, newA );

I got similar exceptions with other even values.

And what logic are you using to resolve a and b?

>>2658

I think it might be.

Evening all.

So as you may have noticed, I haven't posted much lately, because I haven't made any progress.

But I think I just had a mini breakthrough.

I can't explain it better than this at the moment:

We have the grid organized by (e,n). Any number that is e bigger than a perfect square will be listed in column e.

What about f though? For each entry in the grid, what is the f value and how does it change in the column e?

Our semiprime value c has a constant e & f. There are many many entries in column e, but only 2 entries will share the same f.

So lets look at f in (e, 1).

If e is odd, the series of f is:

f = 4, 16, 36, 64

f = 2^2, 4^2, 6^2, 8^2

If e is even, the series of f is:

f = 1, 9, 25, 49, 81

f = 1^2, 3^2, 5^2, 7^2, 9^2

This is super interesting, since we should be able to find our axb record by simply searching for a matching f.

I'm trying to figure out a number of things as next steps:

* How does this pattern change as n grows?

* This seems to be related to x+1, more analysis needed for n>1

* How do we use the na cell in (e,1) to locate our axb for n>1?

A little more info. This came out of analyzing pythagorean triplets. They can only be 3 even numbers, or an even and 2 odds. We only care about the second case.

In the c=145 example, the na records in (e,1) both have the even side of the triangle = f (24), and the hypotenuse = a value in (e, 1).

>>2662

But if you are able to find a matching e,f in (e,1) you don't need to analyze (e,n) do you?

>>2664

Correct, this is in the case that you have f = a perfect square, also n = 1.

>>2665

But I mean even for a c with n!=1, if you find the matching f in (e,1) you've already got the factorization. Right?

>>2666

Sorry, maybe i said it wrong. It wont match in (e,1). It will just match elsewhere in e. In the (e,1), all f's are perfect squares.

There are patterns in f too for higher n. I think I'll have an equation soon. Once we have that we can calculate n from f.

>>2667

Can you provide some examples? I haven't tried jumping with f.

>>2668

Take our example c=145.

d=12

e=1

f=-24 (or 24 if you use absolute value)

If we look at (1,1), we get the following elements:

{1:1:2:1:1:5} (f=-4)

{1:1:8:3:5:13} (f=-16)

{1:1:18:5:13:25} (f=-36)

{1:1:32:7:25:41} (f=-64)

{1:1:50:9:41:61} (f=-100)

{1:1:72:11:61:85} (f=-144)

{1:1:98:13:85:113} (f=-196)

{1:1:128:15:113:145} (f=-256)

{1:1:162:17:145:181} (f=-324)

In each cell -f = (x+1)^2. This equation works for any (e,1).

I don't have the full process yet for jumping to the correct n, thats the piece thats missing.

But what I'm testing out right now, is to jump to the n = a record for the largest square under our f.

Then traverse that list.

Its obvious that thats not all there is to it, because c=85 doesn't work.

But if we figure out how f grows with n, we should be able to find the correct n given f.

If you have code to generate the grid, add f to the output and you'll see the patterns i'm talking about.

>>2669

Just use geometry if you want to figure that out.

Reminds me, I need to draw a few things.

Got a lot on my plate right now.

>>2672

This is why it's good that those of us here on this board (as far as I'm aware) don't want to be known for this (we're not famefags, as one you said). I have a feeling since most of the /cbts/ people came from other places on the internet and haven't always necessarily been part of anonymous imageboard culture that they got in over their heads and their egos took over when they became part of a 'super awesome secret club for special people'. That might be why some of them talked to Infowars etc, and it's obviously why so many 50+ year olds are making YouTube videos about it constantly. I have a feeling it's part of why the mods are being dicks. Either they're overwhelmed or they're compromised. I wouldn't want to run a board like that.

Why doesn't Q just make his own board? That way he could lock and sticky all of his own posts and everyone would know it could only be him.

>>2673

Either way, I love Q even more for BTFOing Zion agent Baruch. Lots of new ideas in /thestorm/

Maybe they'll even convince Q to verify himself with GPG..

but oh dear.. One problem.. Kek.

>>2669

When we square c there are 5 possible factorizations. They all have different n's but d=c so f is the same for all 5 records including target record {0,n,c, c-aa, aa, bb}. We use f to move to between n's.

If we look at records with different f like moving down the (e,1) factor tree we have to be careful that we know how to get back to original f.

>>2672

Disinfo necessary.

Loop capital legit crumb!

>>2663

all the e's equal 1?

>>2661

I have a couple of questions regarding your and other transformations I see all over this board.

Some review first.

We start with c Looking to find a and b such that a*b =c

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

VQC suggested squaring c.

New record {0, (cc+1)/2-c, c, c-1, 1, cc}

Are a and b factors of new c?

Questions about your transformations

Is original c a factor of new c's?

Are both original a and b factors of the new c's?

Are we looking to determine a and b?

If a or b or both aren't factors of new c how do we find a and b????????????????

If c isn't factor of new c is transformation legit???????????????

>>2687 (You)

After testing it turns out that

(e = 2c, 1, 1) = {2c, 1, c, 0, c, (c+2)}

(e = 2c-1, 1, 1) = {2c-1, 1, c+1, 1, c, ((c+1)^2+2c-1)/c}

for both new records a of (e,1, 1) = c

So new.c = old.c * new.b

Is old.c a factor of new.c??

Are these transformations legit??

>>2690

That's a (1,1) record but I think you're right it works for (e,1) also. I'll put some work into it. Could be what we're looking for because we know how to get to (e,1).

>>2690

VA - not to be a downer here, but isn’t that just our standard n = (a+b)/2 - d formula??

I haven't made any progress yet, but I'm glad to see you guys are spurring with ideas!

I made this functions for generating the content of (e, n).

def generateN(e, n, rows=10):
  t = getTFromA(e, n)
  initX = 2 * t - 1
  e, nn, d, x, a, b = rowX(n, initX, 1)
  print((e, nn, d, x, a, b))
  for i in range(1, rows):
    e, nn, d, x, a, b = rowX(n, 2*i*n + initX, b)
    print((e, nn, d, x, a, b))

I'm going to see if I can make another similar function.

As you'll see, it only generates the one's that are 'native' to that n. Take for example (1, 5):

>>> generateN(1, 5)
(1, 5, 4, 3, 1, 17.0)
(1, 5, 30, 13, 17.0, 53.0)
(1, 5, 76, 23, 53.0, 109.0)
(1, 5, 142, 33, 109.0, 185.0)
(1, 5, 228, 43, 185.0, 281.0)
(1, 5, 334, 53, 281.0, 397.0)
(1, 5, 460, 63, 397.0, 533.0)
(1, 5, 606, 73, 533.0, 689.0)
(1, 5, 772, 83, 689.0, 865.0)
(1, 5, 958, 93, 865.0, 1061.0)

It doesn't include a=5, b=29. The reason for this is that a=5 is the result of 25 / 5 (or n*a / n yielding a).

This only uses the rowX method for finding the next values in the chain. As a result it will not be a complete (e, n). Nevertheless it's a starting point.

>>2693

I don't think it's actually possible to generate complete (e, n) record. Every (e, n) record should potentially grow infinitely, right?

Every number in an (e, n) is there because there exists a number in (e, 1) that is a*n. If that a is a prime, then it should start a new "chain" in the n records. Take for example (0, 8). It contains 4 chains based on the initial map generation program by VCQ. It's only 4 chains, because we have a limited generation of numbers in the initial map program. If you extend this further then you should see even more chains in (0, 8)

In my function I had a limit of 10 rows, so for a given amount of rows it should be possible to generate the entire (e, n), but since every (e, n) contains these chains, it's not necessarily useful to do so.

>>2694

I hope I'm making sense.

To explain the term chain look at these records:

{0:8:5:4:1:25}
{0:8:12:8:4:36}
{0:8:21:12:9:49}
{0:8:32:16:16:64}
{0:8:45:20:25:81}
{0:8:60:24:36:100}
{0:8:77:28:49:121}
{0:8:96:32:64:144}
{0:8:117:36:81:169}
{0:8:140:40:100:196}
{0:8:165:44:121:225}
{0:8:192:48:144:256}

A "chain" is something that links a row that contains a, b to b, c where a != b, b != c.

In (0, 8) we can see 4 different chains.

# Chain 1

1, 25 -> 25, 81 -> 81, 169

# Chain 2

4, 36 -> 36, 100 -> 100, 196

# Chain 3

9, 49 -> 49, 121 -> 121, 225

#Chain 4

16, 64 -> 64, 144 -> 144, 256

>>2695

This rowX seems to hold a lot of potential.

We can generate every instance where a = c or b = c using it.

>>> rowBX(0, 0, 145)
(0, 0, 145, 0, 145, 145)
>>> rowBX(1, 0, 145)
(286, 1, 143, 0, 143, 145)
>>> rowBX(2, 0, 145)
(281, 1, 142, 1, 141, 145)
>>> rowBX(3, 0, 145)
(274, 1, 141, 2, 139, 145)
>>> rowBX(4, 0, 145)
(265, 1, 140, 3, 137, 145)
>>> # ...
...
>>> rowBX(72, 0, 145)
(1, 61, 12, 11, 1, 145)

Below is two functions, one for rowX which takes in three parameters (n, x, a) and rowBX which takes in three parameters (n, x, b).

def rowX(n, x, a):
  b = a + 2*x + 2*n
  c = a*b
  d = int(math.floor(math.sqrt(c)))
  e = int(c - d*d)
  n = int(((a + b)/2) - d)
  x = int(d - a)
  return (e, n, d, x, a, b)

def rowBX(n, x, b):
  a = b - 2*x - 2*n
  c = a*b
  d = int(math.floor(math.sqrt(c)))
  e = int(c - d*d)
  n = int(((a + b)/2) - d)
  x = int(d - a)
  return (e, n, d, x, a, b)

The difference between rowX and rowBX is that rowX generates a record for n, x and a while rowBX generates the same, but for b.

If you ignore one of the variables, for example x = 0, and just adjust n it will still generate records and will fill in for n (as x increases).

So ignoring the second (x) parameter and just adjust n you'll see that rowXB(72, 0, 145) is the highest n will go. This is because it will generate the record where a=1, b=145. n = 0 will generate (0, 0, 145, 0, 145, 145). So you can think of n as decreasing the a-variable.

The limit is (b - 1)/2 so for b = 289, the highest n will go is 144. (289 - 1)/2 = 144.

By using rowX you can find every (e, n) where a is equal to whatever value you want.

I haven't fully wrapped my head around it, but if you modify the code to only output e = <some value> you can see VQC's reference to:

> Take the number 71. It appears once in the grid. The number 15, twice. 71 is prime. All prime appear once. 15 appears twice, in the same column. How many times does 105 with three factors appear in one column? How many time for a number with four factors? It's not linear. If you multiple c where it is the product of two primes by say, 105, you're going to have a product, a new c, that appears quite a few times in the same column… This is where we are heading.

>>2695

For (0, n) when n= 2mm for integer m there are multiple chains the rest of (0,n) you can develop from the first record.

These are records which a and b are perfect squares!

>>2690

I tried your formula on some (e,1) records based on various combinations of c and N.

Unfortunately those records have already been solved. It just confirms what we already know. Your formula doesn't get us to the goal record. At least so far.

I don't think this is *the* solution, but is it possible to generate a record for (e, n, x, c)?

that is, use e, n, x and c to generate all possible solutions for c?

VQC said we only need 3 variables to solve for c. It could yield interesting results.

>>2691

>>2692

>>2698

Hey lads, thanks for your replies. So the first part is just the (e,1) factorization we already know. The newer highlighted part is just setting ab to a=1 b=c and then using the basic equation (you weren't being a downer, PMA) to solve for n when ab are set to 1,c.

What I should have taken more time to be clear about is that starting from only C, this is a way to generate two valid cells for C, one at: {1,1,d,x,a,b} which we know using f, and a new one at {1,BigN,d,BigX,1,c}. This gives us an upper bound for n and x. All other factorizations for c should be in lesser n values.

At a conceptual level, I guess I'm trying to take our knowledge in (e,1) and figure out how to expand it. This 1,c bigN may have no value at all, just keeping the mental oven baking over here.

Baker mentioned that he had been able to get the f formula to generate factors for the RSA numbers, although they were in n=1. Baker, any chance you could post some examples of RSA numbers that you've been able to factor in n=1?Thanks!

>>2700

Lmao, ok. I'll post the ones for RSA-240 since this is the lowest that hasn't been factored. I didn't factor the number, just set n = 1. Also, ab != c because there are only 2 factorizations for semiprimes. 1c and ab

I have to rewrite the method because I lost it. Can you repost the formula?

>>2662

Teach, nice to see you! I'm following all your ideas here. Var f seems likely to hold some more goodies for us to discover.

>How does this pattern change as n grows?

>How do we use the na cell in (e,1) to locate our axb for n>1?

>>2651

Hello MA! Hope to see you over the weekend!

>>2701

Sure Baker. Also, thanks for clearing up a question I had. Only two factorizations for semiprime c? 1c and ab. perfect. Here's the formula.

x= floor(SQRT( abs(f)))

then use x to create a and b.

>>2703

>>2702

It assumes n=1 and generates a cell like that.

The c created doesn't equal the original c.

It's like creating a factorization for (e,1).

Might be more useful than I realized if we can find other cells containing same e and f.

>>2704

I don't know if this is useful but I haven't really contributed to this thread yet.

Where n = 1, x = floor(sqrt(abs(f)))

Where n = 1, i - d = 1, which means i = d+1

Where n = 1, x + n = j, which means j = floor(sqrt(abs(f))) + 1

Where n = 1, i scales up linearly with d and j scales up linearly with floor(sqrt(abs(f))).

>>2703

It works for ALL (e,1), but only for (e,1). However, within row 1 it's very powerful. Just start from C, derive d,e,f,x, then calculate a and b. Boom, done. Try it out. grab your grid, pick a random element, multiply a and b to get c, then feed your c var into the calculator. It will solve any c value in row 1, and with basic math. Somehow f is always related to x. Like a "golden ratio" but different. It's pretty incredible.

>>2704

Cool, thanks for looking into it Baker.

>>2705

Thanks for contributing, Anon!

Here are the values if you assume n=1 for rsa240.

It is not the factorization of rsa240.

rsa240c = 124620366781718784065835044608106590434820374651678805754818788883289666801188210855036039570272508747509864768438458621054865537970253930571891217684318286362846948405301614416430468066875699415246993185704183030512549594371372159029236099

if you assume n=1 (it isn't), these are the values you get:

pastebin.com/W40p18aG

I can beat it. But I need to buy a computer and some software. Any recommendations for a windows fag.

>>2712

AMD laptop from eBay with Windows 7 key. Use Windows 7, 10 if you must. You can still get free Windows 10 from a 7 key, but it's a bit of work. Or install and learn linux, you're gonna need it.

>>2713

Thanks MA. Can I have it run both?

>>2714

It's a little more technical, but of course. It's a computer, it'll do whatever you want it to if you try long enough.

>>2711

Thank you Baker!! Appreciate your work. Checked them out carefully in the pastebin link. Is this RSA unsolved?

>>2717

So t+na works within the same cell?

>>2718

The ↑ I was using in my previous was t+na

I changed it to t+na,

and now the ↑ stands for -x jump with parameters

(n, -(2n + x), a)

t+na uses the parameters

(n, -(2n + x), c)

I've been looking at the d for (1, n) records and I have the following:

(1, 5, #1) = 2(5 t^2 - 2t - 1)

(1, 5, #2) = 2(5 t^2 + 2t - 1)

(1, 13, #1) = 2(13 t^2 - 8t - 1)

(1, 13, #2) = 2(13 t^2 + 8t - 1)

(1, 17, #1) = 2(17 t^2 - 4t - 4)

(1, 17, #2) = 2(17 t^2 + 4t - 4)

(1, 25, #1) = 2(25 t^2 - 18 t - 3)

(1, 25, #2) = 2(25 t^2 + 18 t - 3)

Here I use (e, n, #chain).

I was wondering if it's possible to get the t from d only as a point of exploration.

So the t is squared and multiplied by n, but I don't understand the pattern behind the second part.

Although it's interesting to note that the calculation for these d's for t in the different chains are so similar.

I don't think it will be possible to get a way of calculating t of d without knowing n.

I checked a few other (e, n)'s and they appear to follow a similar pattern.

When was the last time Chris posted here?

>>2726

It was in the last thread post >> 2388

Hello lads! Pretty mellow here today. I spent yesterday looking to expand the formulas for f into n=2. Some success, but nothing solid to report. Seems like we're still looking for a way to tie (e,1) and column 0 together.

It would be nice if Senpai could pop in and drop a few tasty crumbs like he promised! VQC, you there? We've made it this far, let's finish this.

In other good news, CodeMonkey got shit sorted out with Q (as Baker mentioned) and there are a bunch of good new Q drops if you haven't seen them. Q is no longer on /cbts/. Might as well read up on Q posts and keep an eye here till VQC pops in. Good updated Q map over there too. New location for Q:

>>>/thestorm/

Saw something in factor trees yesterday that wasn't there. Have to be careful doing math and weed together.

I thought factor trees worked because factors were seeded into the tree and would pile up at their appointed times . It doesn't work like that.

The algorithm for generating the factor trees is simple to follow. The initial record for the a b generating factor tree is {2c, 1, c, 0, c, c+2}

generating new records is simple

next.d = d+4t

next.x= x+2

next.a = b or next.d-next.x

next.b = (next.d*next.d + e=2c)/ next.x

Somehow this algorithm that knows nothing about a or b produces records containing a and b as factors without c many times more often than c records.

Somehow the knowledge to produce records with a and b factors exists in the initial record cause it isn't in the algorithm. Ghost in the machine!

>>2731

>>2732

Quantum computer teaching quantum reality

>>2709

Nice!

>>2731

>>2732

Love the number trees, Baker!! Also, I agree that faith in realizing our goal creates that reality. That's why I'm here, I've already seen the end result achieved. This is the persistence part, where winners keep going until they win. Did VQC say that to you on DM?

>>2735

VA - still trying…

I've put a lot of output together, still trying to find any pattern that links to any potential prime match. No such luck yet. Notwithstanding the -x hint earlier. All I've been able to confirm is that the pattern holds for na, nb, nc in negative x and f spaces.

I had a thought earlier that I've been trying to work through in the (e+1) space. Perhaps someone can assist or tell me if I'm barking up another wrong tree…

VQC mentioned that every factor for c exists in (e+1). I originally thought that meant (e+1,1), but I'm leaning now towards (e+1,n). And we also know that the a value at (e,n) moves to the d values at (e+1,n). (I believe this is correct).

For example:

for c=145 (1,61,6), the factor 5 exists as the d value of (2,1,2).

for c=901 (1,421,15), a factor of 17 exists as the d value of (2,33,8).

I thought that if there was some way to link records to e+1, we would have a simpler way to calculate d?

>>2736

ThanksTopol! Are you ready to help FIRE UP SPIRITS?!?

>>2737

PMA, you have worked so hard, man. It'll click and fall into place!! Like the last tumbler on the lock.

Come on, fags, we can do this! Here's your inspirational Literature for this evening. We need to get our levels up to 9999!!

KING HENRY V:

Once more unto the breach, dear friends, once more;

Or close the wall up with our English dead.

In peace there's nothing so becomes a man

As modest stillness and humility:

But when the blast of war blows in our ears,

Then imitate the action of the tiger;

Stiffen the sinews, summon up the blood,

Disguise fair nature with hard-favour'd rage;

Then lend the eye a terrible aspect;

Let pry through the portage of the head

Like the brass cannon; let the brow o'erwhelm it

As fearfully as doth a galled rock

O'erhang and jutty his confounded base,

Swill'd with the wild and wasteful ocean.

Now set the teeth and stretch the nostril wide,

Hold hard the breath and bend up every spirit

To his full height. On, on, you noblest English.

Whose blood is fet from fathers of war-proof!

Fathers that, like so many Alexanders,

Have in these parts from morn till even fought

And sheathed their swords for lack of argument:

Dishonour not your mothers; now attest

That those whom you call'd fathers did beget you.

Be copy now to men of grosser blood,

And teach them how to war. And you, good yeoman,

Whose limbs were made in England, show us here

The mettle of your pasture; let us swear

That you are worth your breeding; which I doubt not;

For there is none of you so mean and base,

That hath not noble lustre in your eyes.

I see you stand like greyhounds in the slips,

Straining upon the start. The game's afoot:

Follow your spirit, and upon this charge

Cry 'God for Harry, England, and Saint George!'

>>2738

+3

https://youtu.be/EOg0d5Wq4i0

new avenue to explore, perhaps.

Records at c and ab both share f and c+f values.

They also share a common record at (e,1, (c+1)/2) which has a different f value.

If you move the 3 records to (e+1, n, t) they each now have different f values. Is there a way to find a difference between these 3 records?

>>2740

Nice PMA! This looks promising, f is part of the key I think. I'll check out the record movement you're describing. Need some sleep, but on it tomorrow!

>>2742

2*(c-1) works for most instances tried.

but not c=54, where o is actually 104

>>2742

I am using this method.

>>> def rowNegX(n, x, a):
...   b = a + 2*(-x) + 2*n
...   c = a*b
...   d = int(math.floor(math.sqrt(c)))
...   e = int(c - d*d)
...   n = int(((a + b)/2) - d)
...   x = int(d - a)
...   return (e, n, d, x, a, b)

Here are the parameters for the method for each element derived from c in the pictures:

→ = 1*c cell

↓ = (n, x, 1)

↓c = (n, x, c)

↓d = (n, x, d)

↓e = (n, x, e)

↓x = (n, x, x)

↓n = (n, x, n)

↓f = (n, x, f)

↑ = (n, -(2n + x), 1)

↑d = (n, -(2n + x), d)

↑e = (n, -(2n + x), e)

↑n = (n, -(2n + x), n)

↑x = (n, -(2n + x), x)

↑f = (n, -(2n + x), f)

t+na = (n, -(2n + x), c)

Wow >>2745

Some excellent work. I need to spend some time to digest it.

Include >>2744 in bread

With no visits from Chris I thought I would try to imitate him. I hope i don't sound like a [D]umb [M]oron

Any factor of any a in (e,1) will be the value of all n at (e,n)

That was fun. Here's another [D]umb [M]oron idea

Patterns. Focus on a[t] and d[t] in (e,1) subtracting d from d[t] gives (n-1)a

If I get any more [D]umb [M]oron ideas I'll be sure to share them!

>>2750

Okay guys. So I've been thinking.

If we do a, b = 5, 29 we have c = 145.

If you "scale" the number by 5 on a, b such that a,b = 5*5, 5 * 25 then c = 25 * 145.

I noticed that when we do that, we can predict the e it will land in. e = 25. Both a and b will contain 5 as a factor => (25, 25, 60, 35, 25, 145).

It doesn't predict the 'n', but shit. We know that we can find a number a % 25 == 0 in (25, 1). Then just use 25 * i + t and 25 * i + 1 - t to search for a, b.

Again, we simply multiply c with a square of our choosing. Then we find the first a in (e, 1) that contains our number (non-squared) as a factor.

This is our prime, we can then use this to factorize c by generating more a's that fit p * i + t, p * i + 1 - t and using gcd.

def solveTest(c, rows=20):
  e, n, d, x, a, b = rowForAB(1, c)
  e, n, dd, xx, a, b = rowForAB(1, x**2 * c)
  li = generateGenesis(e, rows)
  t = 1
  for cell in li:
    e,n,d,xx,a,b = cell
    if a % x == 0:
      break
    t = t + 1
  for i in range(1000):
    p1 = x*x * i + t
    p2 = x*x * i + 1 - t
    pp1 = int(getAFromT(e, p1))
    pp2 = int(getAFromT(e, p2))
    gp1 = gcd(c, pp1)
    gp2 = gcd(c, pp2)
    if gp1 != 1 and gp1 != c:
      print(t,i, pp1, gp1, c, c / gp1)
      break
    if gp2 != 1 and gp2 != c:
      print(t,i, pp2, gp2, c, c / gp2)
      break

Here I use x as an example. It doesn't have to be x, it could even be a chain of primes squared (I think).

I've also been thinking that maybe we can use e again. e will be the k * c, where k is some squared number.

>>2751

> If you "scale" the number by 5 on a, b such that a,b = 5*5, 5 * 25 then c = 25 * 145.

I'm an idiot, here I meant a, b = 5 * 5, 5 * 29

>>2751

This is heavily non-optimized. I don't even think we need to search for the first a that contains our factor.

If we for example use e as our prime (the number we squared) then we could simply use getTFromA(e, e). Then iterate on that to find the number that gcd(c, somenumber) != 1 and gcd(c, somenumber) != c.

I haven't tested that, but I think it should work. We should then be able to skip the first generation of genesis cells and go straight to the prime iteration process.

In my code I added a range(1000), but if we *know* what we are looking for we can just iterate indefinitely until we hit the proper i.

>>2753

> In my code I added a range(1000), but if we *know* what we are looking for we can just iterate indefinitely until we hit the proper i.

Correction, if we *know* that e * i + t, e * i + 1 - t will yield a number where gcd(c, somenumber) == a or b, then we don't need a range-limit.

>>2751

Correction, we can't "predict" e. It appears it doesn't always equal the number we square. But I think we can still use this approach.

>>2755

That is an unacceptable solution.

Read the pictures I sent and try taking the difference between two ↑d's or two ↓d's

It's predictable based on a. I'm working on whether you can derive it from c and work backwards.

Always multiple of 2.

>>2748

I think I know what this one means.

Look at records where e=n

For simplicity sake I am going to refer to o as the difference between a ↓ element and a ↑ element, and I'm going to refer to the difference between the ↑ element of the a*b element and the ↑ element of the 1*c element as p

Two↓ elements also works here.

If you don't recognize this notation just look at the pictures I sent.

This difference, p, may be directly related to a and maybe can be derived from c.

I don't know the linear formula, but most of the values of p for a can be found when you put a into the formulas for all the elements.

For example, you get 8*22 when you type in 8, and p is always 22 when a=8

For example, you get 4*10 when you type in 4, and p is always 10 when a=4

Full list of relationships:

a = 1, p = 0

a = 2, p = 4

a = 3, p = 6

a = 4, p = 10

a = 5, p = 12

a = 6, p = 16

a = 7, p = 18

a = 8, p = 22

a = 9, p = 24

a = 10, p = 28

>>2758

It might be congruent to this sequence.

oeis.org/A006093

Try entering d for any c into the element deriver that I sent the formulas for. I noticed the value of p is always there, but I haven't confirmed this. If it is, then you'd be able to factor the number by finding what term p is of the sequence above. (E.g 16 is the 6th term of that sequence if you ignore 1).

>>2757

interesting. thanks.

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

(1,61,56) = {1:61:212:111:101:445}

(1,61,67) = {1:61:278:133:145:533}

The following formula works for generating these next records for (e,n).

a=a+2*(x-n) will generate (1,61,56) - this one can also come from (e,n,t+n-x)

a=a+2*(x+n) will generate (1,61,67) - this one can also come from (e,n,t+n)

My previous move by d formulas stepped over (1,61,67).

>>2761

The way to generate the 101*445 element was the puzzle piece I needed.

>>2762

not sure if this is of any value, but I found a few more records for (1,61) at -t.

(1,61,-66) = {1:61:12:-133:145:1} = 145; f=24; (x+n)=-72; f+c=169; -x: 11

(1,61,-55) = {1:61:-10:-111:101:1} = 101; f=-20; (x+n)=-50; f+c=81; -x: -11

(1,61,-5) = {1:61:-10:-11:1:101} = 101; f=-20; (x+n)=50; f+c=81; -x: -111

(1,61,6) = {1:61:12:11:1:145} = 145; f=24; (x+n)=72; f+c=169; -x: -133

(1,61,56) = {1:61:212:111:101:445} = 44945; f=424; (x+n)=172; f+c=45369; -x: -233

(1,61,67) = {1:61:278:133:145:533} = 77285; f=556; (x+n)=194; f+c=77841; -x: -255

>>2763

Please take a look at my method.

If you can get p from c you just know what term of the sequence it is to find a.

Has anyone figured out how O(log n) factors into the grid navigation? I noticed you're mostly all working on calculations and things. Surely if we knew how the idea of this being done in O(log n) time makes sense it would help immensely.

>>2765

log(n) means it's instant.

>>2764

if you are referring to the rowNegX method, yes, I'm familiar with what you're doing. I have some sample code that follows along.

I'm not quite understanding what you mean with p. How is it different to the x or t values that we already know, or how we can navigate between records by factors?

>>2767

It's all there. Value of p seems to be close to d.

a is equal to the number of the sequence I posted that p is, minus one.

>>2766

What are you talking about? O(1) means it's instant. Also what does that have to do with anything? VQC said it would be in O(log n) time. I'm not aware of many ways in which an algorithm can work in O(log n) time. Wouldn't that be a constructive thing to think about? It might more clearly show what we should be looking for in these relationships.

>>2769

You have a good point, but you can't draw many conclusions from it being log(n), because that n can be different things.

O notation is just meant to analyze how fast something scales with input.

>>2770

How many different O(log n) algorithms can you list? What n is is irrelevant to what I'm trying to say. Even if we don't apply any existing O(log n) algorithms to the grid, what we do with the grid is meant to fit the criteria of an O(log n) algorithm according to Chris. Does that mean there has to be some variable or relationship between variables that allows us to halve n's search pool every time? Does that mean doing something weird with powers? If we know what criteria are necessary for something to be O(log n), we'll know more about what kind of mathematical relationship we're looking for, so that whenever we find an interesting relationship, we can figure out if we're going in the right direction.

>>2769

There is a linear relationship between log n and the number of digits or bits of n. (Using base 2 or e or 10 would just give a different scale ). So a log n algorithm would have solving routine that might:

-narrow down (magnify) search area by 10 times per step, so step 1 tells you first digit, step two tells you second digit, and so on

-narrow down by 2 times per step, e.g a binary tree. If there is a Tree of Knowledge, and the first step is whether e is even or odd, and the following steps iterate some similar decision, then this would be log n

-being able to decide first bit, second bit, third bit in constant time each

One point about what VQC said, either the VQC works in log n time (really we mean log c, since it is c which is given in our problem), and VQC misspoke by saying log n where n is the size of the digits of the input. If VQC meant that literally and wasn't being sloppy then since the number of digits is already log of the number, the VQC's speed is log (log c).

>>2764

Baker these r legit but maybe [D]ont [M]ention

Up to you.

>>2748

>>2749

>>2775

Oops. C=145

Hey guise, I'm here working away too. Nothing to report yet tonight. Checking out all your N and -x patterns.

>>2775

Excellent! I couldn't figure out what that meant.

>>2749

>>2775

Why did you select (1, 29)?

>>2782

Nevermind re-read.

Btw great work! This looks very interesting

>>2775

8700 - 160 = 8540

8540 - 1464 = 7076

>>2784

yes.

I also thought that the d difference between (1,29,67) -> (1,5,67) equals (1,61,6) n * 2d or n*f. But that formula doesn't apply to different test cases.

So for these records, if we can figure out the a or d formula…

>>2785

Yeah, this shit feels soooo god damn close. If only we can get to one of those records, we got it.

>>2786

I've also looked into a formula for b. Haven't found a combination of x+n, x-n, multiple of n, etc. that will work.

Interesting thing about these records is the shared e and x. and the 2an is all the same - as you would expect.

Pity we don't yet understand moving between n's other than to 1.

So the (n-1)*a takes you to the n=1 value and is a shortcut for the "f transition" (whatever that means).

What's the formula to take you to another n value other than 1?!???

Guess that's the Integer Factorization secret…

>>2786

important formula I believe:

n = (xx + e) / (f - 2x)

>>2788

Based on (4, 34, 161, 76, 85, 305)

n = (xx + e) / ((f + a) - 2x)

>>2789

Damnit, clicked reply too fast:

n = (xx + e) / ((2 * (d + a)) - 2x)

>>2790

Doesn't work for c=145, so might be a fluke, but then again there might be a different equation for odd e

>>2791

my equation worked for c=145. I'm trying to get a formula for n in terms of known variables. e, x, d movement we have some formulas for already.

>>2792

I know, but for (4, 34, 161, 76, 85, 305) I calculated the following rows based on the factors of 34 * 85:

(4, 1, 2966, 76, 2890, 3044)

(4, 2, 1521.0, 76, 1445.0, 1601.0)

(4, 5, 654.0, 76, 578.0, 740.0)

(4, 10, 365.0, 76, 289.0, 461.0)

(4, 17, 246.0, 76, 170.0, 356.0)

and

(76 ** 2 + 4) / (2 * (161 + 85) - 2*76) = 17.

I was thinking maybe that's the key.

>>2791

>>2793

works for c=65 also.

more testing required for e other than 1.

n = (xx + e) / (f - 2x)

n = (xx + e) / (2d + 1 - e - 2x)

if this holds, we have a formula for n in terms of known x and e variables. And now we need to understand better the d jump.

>>2793

(76 ** 2 + 4) / (2 * (161 + 85) - 2*76) =17.0

(76 ** 2 + 4) / (2 * (161 + 85 + (85 + 34)) - 2*76) = 10.0

(76 ** 2 + 4) / (2 * (161 + 85 + (85 + 34) + (34 + 85 + 85 + 85)) - 2*76) = 5

>>2795

Adding more 34 + 85.. terms doesn't get to the (4, 2) record though.

However, the (4, 2) record, when divided by 2 is equal to 85 * 34 (2890)

Banged through a bunch of e=n records.

Discovery!

Guess z in c terms?

a of (z, z, 1) = c if c semi prime

>>2787

Hello PMA! This is exactly what I've been working on. How to get out of row 1 starting only from c.

>>2789

Where'd you derive this? I'll check it out, could be very useful.

>>2797

Interesting, Anon! PMA, remember "Use the z on y at x" ??

>>2787

I think the secret of getting out of row one could be multiples of f. An expandable search pattern for a simultaneous lock on x,a,b, and n that works outward from d. A very restricted search algorithm using f as its base. Thinking out loud here.

>>2798

>PMA, remember "Use the z on y at x" ??

VA - I still have no idea what this means…

>>2800

Lol, PMA! Neither do I. However, it was in relation to your idea about solving for the small square using t. We have been one variable away from success for weeks!

>>2794

Elaborate. You evaluate that formula with x from 1c to get n from ab?

>>2802

I've been trying to get further along with the (n-1)*a hint.

I'm starting with a record for c at (e,n,t). I move up to a record at (e,n,t+n). This record contains all factors of a,b,c.

It is also similar to a record at (e,1,t+n) in that e and x are the same.

Then if you find records at (e,a t+n) and (e,b,t+n) where a and b are the values we are searching for, you'll notice that they also share e and x values. And 2an are the same value for all of these records.

We know how to take our (e,n,t) for any c and translate it to (e,1,t). That's the (n-1)*a transformation.

So I've been researching relationships between these records to try and learn how to transform the n value to a or b instead of 1.

In analyzing those differences, and searching for a formula to define the change in a or d, I stumbled upon a way to represent that change in terms of known variables for e, x, and d.

Those are the formulas posted at >>2794.

Unfortunately, they only appear to work where e=1.

>>2803

What is the point of that formula? It gives you n for 1c

>>2804

>>2803

Wait.. it gave me 2*5 for 10..hmm

but 1*145 for 145

Looks like it can be improved

>>2804

no. it gives you any n from any (e,1) and it may provide a way to get to n for a,b in terms of e, x and a change in d values.

>>2806

It gave me the factorization of 10 which has n of 0.

>>2807

I may not be explaining myself. That formula doesn't solve for n yet.

It is a step I am using to break down variables into smaller pieces to try and analyze the jump to a valid d that will solve for n.

>>2808

But you might find it interesting that I was unable to find the factorization of 10 in the VQC output and when I factored 10 using that formula it give me a non-existent element

{1:2:3:1:2:5}

>>2809

sure. I'll look into it.

There are a lot of things I missed in my -x jump brute force factorization program. Try calculating change between each b it prints out. It also sometimes gives a instead of n. I think there is something valuable there. I just noticed it but I have to go.

>>2814

That's some good shit. Send the spreadsheet!

Add

 System.out.println(v); 

after line 46 in my x jump code to see what it actually does

>>2787

What do you mean "move between n's other than 1"?

While right now we are a bit limited in the moving regard, but we can, for example, move from (1, 61) to (123, 61).

>>> rowX(61, 2 * 61 + 11, 145)

(1, 61, 278, 133, 145, 533)

>>> createForENX(1 + 2 * 61, 61, 133)

(123, 61, 279.0, 133, 146.0, 534.0)

>>> createForENX(123 + 2 * 61, 61, 133)

(245, 61, 280.0, 133, 147.0, 535.0)

>>> createForENX(1 - 2 * 61, 61, 133)

(-121, 61, 277.0, 133, 144.0, 532.0)

>>> createForENX(-121 - 2 * 61, 61, 133)

(-243, 61, 276.0, 133, 143.0, 531.0)

>>2818

It even keeps the logic behind xx + e = 2na

>>> rowForAB(1, 13 * 19)
(22, 109, 15, 14, 1, 247)
>>> rowX(22, 2 * 22 + 14, 247)
(40, 10, 317, 70, 247, 407)
>>> createForENX(40 - 2*10, 10, 70)
(20, 10, 316.0, 70, 246.0, 406.0)
>>> createForENX(20 - 2*10, 10, 70)
(0, 10, 315.0, 70, 245.0, 405.0)
>>> 245 * 10
2450
>>> getTFromA(0, 2450)
35.0
>>> generateNthRowFromGenesis(0, 35)
(0, 1, 2380, 68, 2312, 2450)
>>> generateNthRowFromGenesis(0, 36)
(0, 1, 2520, 70, 2450, 2592)
>>> createForENX(0 + 2 * 1, 1, 70)
(2, 1, 2521.0, 70, 2451.0, 2593.0)
>>> createForENX(2 + 2 * 1, 1, 70)
(4, 1, 2522.0, 70, 2452.0, 2594.0)
>>> createForENX(38 + 2 * 1, 1, 70)
(40, 1, 2540.0, 70, 2470.0, 2612.0)
>>> (70**2 + 40)/2
2470.0

>>2819

Oh boy I messed that code up. I was trying to look at the next row for 247, but I used e instead of n and it jumped from e=22, to e=40. I don't trust the result from the code above. Need to verify

Wasn't there an equation for moving around (e, n)'s?

(e + 2n, n) or (e - 2n, n)?

>>2812

I can't tell you why, but I did some minor algebra on it:

b - 1 = 4x

b = 4x + 1

b = a + 2x + 2n

4x + 1 = a + 2x + 2n

4x - 2x = a + 2n - 1

2x = a + 2n - 1

x = (a + 2n - 1)/2

x = a/2 + n - 1/2

For n:

n = -a/2 + x + 1/2

For a:

a = 2x - 2n + 1

Appears to check out for multiple records:

c = 145

c = 1313

c = 3905

I don't know if these equations will be helpful, but at least here they are.

>>2821

yes. for moving right or left.

>>2822

thanks.

>>2823

I played some more with the numbers

We know x = a/2 + n - 1/2 for some c's (at least the ones I've been making by taking two primes and multiplying them).

We also know x*x + e = 2*a*n. I ran this through wolframalpha:

So by substituting x*x we get:

(a/2 + n - 1/2)^2 + 1 = 2 * a * n (for (1, n))

This yields:

n = 1/2 (a - 2*sqrt(a - 1) + 1)

Integer solutions:

a = 4 m^2 + 8 m + 5

n = 2 m^2 + 2 m + 1

m element Z

>>2824

So for c = 145,

a = 4 * 5^2 + 8 * 5^2 + 5 = 145

n = 2 * 5^2 + 2 * 5^2 + 1 = 61

(1, 1, 5) = {1:1:50:9:41:61}

c = 901

a = 4 * 14^2 + 8 * 14 + 5 = 901

n = 2 * 14^2 + 2 * 14 + 1 = 421

Again, not sure if this is of any actual use.

>>2824

Look at the attached image in >>2815

The mirrored (e,n) at x and -x follow a pattern where the difference in x is always 2x, and the difference in b is always 4x.

The equations make this obvious. Just restating that there are 2 possible ways to get to a solution, and 2 hints about movement in b.

>>2816

Thanks Baker! Here's the spreadsheet, anons. Couple of quick notes:

1. The formula for x works for all (e,1) but not yet for higher values of n. Replace the x value if you're inspecting an element that has n>1. It will auto fill the visual and re-calc all values.

2. The third box on the right is actually just the top left box (a * (x+n)) moved down to more easily see a*b = c in rectangular form. This is where b = a+2x+2n comes from. I left both boxes for accuracy, but you only use one.

3. Hope you lads find it helpful for visualizing the problem at hand!

https://anonfile.com/H4Qawed4be/DiffOfSquares.xls

https://anonfile.com/J1Qcw5dab4/DiffOfSquares.xlsx

>>2827

>"Use the z on y at x"

this hint was from VQC, correct?

One thing I've been thinking of.

How would these patterns look if we look at numbers we know are primes?

For example, take a,b = 1,17. We know 17 is a prime. So patterns should diverge from when the case it's not.

My point is, have we discovered a better primality test than the ones that exists so far? If so, then it lends creedence to the truthability of the whole VQC.

Another thing I'm curious about, VQC said he was going to show us a lot of things. This grid is just one virtual quantum computer. He never said anything else about the second one which we are supposed to use to show us the language of math.

>>2828

Nope.

>>2831

I honestly thought it was an AI.

>>2831

thanks. I've reread. Still searching…

>>2831

>use the y on c at x

Are you sure that's meant to mean the variable x? We don't have a variable y, so x and y might refer to e and n, meaning "use the n on c at e".

>>2834

>"Use the Y on C at X."

this is the correct quote. it also mentioned the boiling point of vinegar. 100.6c or 213f. Mentioned 1,1,2,3. Some other stuff I didn't understand.

and this gem also:

112115 is also 5510101020

which I think means {1:1:2:1:1:5} is also {5:5:10:10:10:20}?

>>2835

>{5:5:10:10:10:20}?

doesn't look like a valid record. no clue.

>>2834

Not sure, Anon! It could mean exactly what you think, that's definitely one possibility. I thought it could be that as well.

>>2835

Lol, PMA there are some gems in there for sure. At least it's a fun read. There's a bunch of stuff I don't understand either.

Interesting records to investigate:

For c = a*b prime

(2*c-1, 2*c-1, 3*c-1, 2*c-1, c, b)

(2*c-1, 1, c+1, 1, c, b)

(2*c, 1, c, 0, c, b)

(0, 2*c, 3*c, 2*c, c, b)

For all t (0, 2*c, t) a and b are multiples of c

>>2838

What formula are you using to calculate t?

>>2840

t comes from the c record at (e,n,t)

>>2841

Yes, what formula? And what is EDB, EDA?

>>2838

>>2842

Create the c record normally. Using the values from there to create another record via ent:

e = c.e

n = c.n

t = (1-(c.n+c.t))

This will give you a -x value.

Alternatively, create a record via ent with t equal to n+t, and just use the x value as negative.

>EDB, EDA

just some helper functions to create records.

public static TheEndRecord CreateForEDA( BigInteger e, BigInteger d, BigInteger a ) {

BigInteger c = d * d + e;

BigInteger b = c / a;

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

BigInteger x = d - a;

return new TheEndRecord( e, n, d, x, a, b );

}

public static TheEndRecord CreateForEDB( BigInteger e, BigInteger d, BigInteger b ) {

BigInteger c = d * d + e;

BigInteger a = c / b;

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

BigInteger x = d - a;

return new TheEndRecord( e, n, d, x, a, b );

}

>>2843

excellent!

Thanks PMA

>>2844

you're welcome!

By the way, we can save a calculation at the beginning (if this ends up being a correct approach), by creating our entry record with a,b values of c,1.

This gives us the same e, n, d values, but with the -x we are looking for.

example:

145=5x29

created from a=1, b=c

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

created from a=c, b=1

-c = (1,61,-66) = {1:61:12:-133:145:1} = 145;

>>2845

Wow. It was that simple.

What's the row for a, b = 2, 7?

>>2847

14 mod 4 = 2. So ignore it or divide it by 2.

>>2848

There are similar entries in the -n space:

(-4,0,2) = {-4:0:4:2:2:6} = 12

(-9,0,2) = {-9:0:5:3:2:8} = 16

I don't see one for 2, 7 though.

>>2849

>There are similar entries in the -n space:

-e space.

The formula to go to matching -x records depends on e:

even values of e

(e,n,t) -> (e,n, (2-t))

odd values of e

(e,n,t) -> (e,n, (1-t))

Still working. Rereading crumbs.

>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.

The simple to calculate comment was with regards to the entry point at a=1,b=c.

Anyone made any further progress?

>>2853

No, it wasn't. According to what he wrote, "only appears in one column" was simple to calculate.

This would be a very important discovery, because it would mean being able to instantly identify a prime.

What's in a name?

ChrisrootODavid?

[D]oes it [M]ean anything?

Root of d = {0, 2*d, 3*d, 2*d, d, 9*d}. All a and b of ( 0, 2*d, t) are multiple of d for all t.

Yes and that pattern can be used elsewhere.

>>2853

>>2856

Hey Baker and PMA! I’ve got some good ideas on how to find all the primes. We should work on solving that big N crumb. Even had some verification about my ideas from Stony Senpai back in RSA #2. I’ll post it up later today, away from my notes at the moment.

>>2858

Good idea Anon! Worth exploring for sure. Have you found any patterns in roots of d?

>>2859

>>2858

I thought that record might be root of d.

He verified it!

Plenty of patterns but no tie back to our goal yet.

>>2853

Sorry, I'm still at this project, just real life that is catching up now that the new year is upon us.

I'm not going to give up this easily. I don't have anything new, I've been working for the past few days and after work I have other commitments, but I still read over crumbs and think every day.

So, we can factorize d?

If we can factorize d with (0, 2d), can we factorize c^2, with (0, 2c)?

>>2864

>>2839

Unfortunately it isn't factorizing. Use the pattern for c, d, e, n or whatever. For example e. For any t … e is a factor of every a and b of (0, 2*e, t) using the pattern.

>>2835

What's up Topol!! Nice to see you. Thanks for the inspiration as always!

>>2862

>I'm not going to give up this easily. I don't have anything new, I've been working for the past few days and after work I have other commitments, but I still read over crumbs and think every day.

Thanks IseePatterns! I think we're all feeling pretty much the same. Let's keep going. I'm seriously falling asleep thinking about the grid at night, and going over crumbs every day too. The solution will present itself if we keep adding to and writing down the rules like VQC asked us to.

>>2865

So, instead of going down the tree, you're going up the tree, eh? Factorizing numbers with e instead of finding the factorization of e?

>>2867

Yes its moving up the factor tree. There were a number of hints in that regard in terms of where factors would turn up. For example (2*c-1, 1, t) a and b turn up as factors almost twice as often as c but no way to find them without iteration. But it doesn't help move down from N.

I don't say this to be a demoralization shill but I've really lost enthusiasm for this lately. Chris hasn't even posted for two weeks. I know it's real, and I haven't contributed since I gave up looking for a linear relationship to use in binary search so I have no idea if any progress has really been made. I just wish Chris hadn't kept saying "oh I'm going to tell you how it works now, oh wait I'll tell you in a week, oh wait I'll tell you in a month" etc.

>>2870

and c for these records is d^2.

>>2870

for all t of (0, 10, t)

>>2872

Here are valid (0,10,t) records from t=-10 to t=50.

They only exist 5 t apart.

(0,10,-9) = {0:10:0:-20:20:0} = 0; f=1; (x+n)=-10; f+c=1; -x: 0; (x-n)=-30

(0,10,-4) = {0:10:-5:-10:5:5} = 25; f=-9; (x+n)=0; f+c=16; -x: -10; (x-n)=-20

(0,10,1) = {0:10:0:0:0:20} = 0; f=1; (x+n)=10; f+c=1; -x: -20; (x-n)=-10

(0,10,6) = {0:10:15:10:5:45} = 225; f=31; (x+n)=20; f+c=256; -x: -30; (x-n)=0

(0,10,11) = {0:10:40:20:20:80} = 1600; f=81; (x+n)=30; f+c=1681; -x: -40; (x-n)=10

(0,10,16) = {0:10:75:30:45:125} = 5625; f=151; (x+n)=40; f+c=5776; -x: -50; (x-n)=20

(0,10,21) = {0:10:120:40:80:180} = 14400; f=241; (x+n)=50; f+c=14641; -x: -60; (x-n)=30

(0,10,26) = {0:10:175:50:125:245} = 30625; f=351; (x+n)=60; f+c=30976; -x: -70; (x-n)=40

(0,10,31) = {0:10:240:60:180:320} = 57600; f=481; (x+n)=70; f+c=58081; -x: -80; (x-n)=50

(0,10,36) = {0:10:315:70:245:405} = 99225; f=631; (x+n)=80; f+c=99856; -x: -90; (x-n)=60

(0,10,41) = {0:10:400:80:320:500} = 160000; f=801; (x+n)=90; f+c=160801; -x: -100; (x-n)=70

(0,10,46) = {0:10:495:90:405:605} = 245025; f=991; (x+n)=100; f+c=246016; -x: -110; (x-n)=80

Maybe this means that (0,2d,t) records only exist where t = d apart.

>>2874

not quite correct. Examples where d = 12 at (0,24). Valid records are 6 apart.

(0,24,-23) = {0:24:0:-48:48:0} = 0; f=1; (x+n)=-24; f+c=1; -x: 0; (x-n)=-72