/cbts/ - Calm Before The Storm

Repostan from last thread:

>>49492

>For now, just role with "It's God's Dream".

Which is a reality to us.

But we're Lucid in God's Dream.

I can roll with this. Didn't mean the dream was only shared among humans.

>Yup, just check out my musings on this board, Haramein's there, along with Dan Winter, and a lot of other stuff.

Going to dig through your posts

>Thoughts… are you forming them or simply acknowledging their existence?

My meditation practice has given me the experience that thoughts appear to "me", without any effort on "my" part. What I, the subjective agent, can do is acknowledge or ignore the thoughts that rise up. Our free will gives us power to direct our attention. This is the process of seeing clearly, and what I suspect the third eye and Eye of Horus is referencing.

>What is "time"? What is "not time"? What is Timelessness? It's being outside of the time loop(s). I guess it's the Super Looping Harmony?

My belief of linear time objectively existing has been decreasing steadily, to the point where I'm more leaning towards the Now being the only time there ever was and ever will be. That things happen through cause of effect, combined with observable loops in nature has given us the impression that we are moving forwards in time linearly. I haven't read or thought about this enough to make any certain conclusions, but I believe we create the sense of passing time in our left hemisphere.

The hemi-sync program, if you are familiar with it, involves the lowering of brain frequency, to a low enough level (related to Planck's length) where we start "clicking out" of our subjective time experience.

>The More In+Line with Dream… the more loving you are… the more influence you can manifest

This has been my working assumption as well.

>Now imagine what would happen to 99.+++% of the population ALL this info was sudden dumped into their laps… and on top of that, proven to be REAL.

Agreed, if these kinds of things is what Q has been hinting at. The amount of unfounded assumptions in mainstream science continues to astonish me.

Thanks for the link, looks interesting.

Any information about what role the pineal gland plays, if any? Is type RH- blood relevant in any way?

>>49535

The subjective individuals perception of reality is, for the most part, founded on faulty premises and assumptions. Either the information input is blatantly manipulated (media), or shaped by other agents as to further their own agendas. Most people in this day and age have had their mental models of reality shaped by institutions (such as school), and as such any information moving "through" them is tainted by their perspective. Others are influenced by the sheer scale of the majorities beliefs, that they often doubt themselves over the overarching narrative.

In that sense, life is but a dream - based on a "collective unconcious" - unless you start to see the cracks and break free.

Did I accurately capture your point? I agree with it entirely, and have done so for a number of years now - though you laid it out in a clear and distinct way, much appreciated.

Cheers.

>>49947

>>49941

Would it kill you faggots to do some real tangible work? If yes, make your own thread.

>>49883

RELEVANT CODEPOSTS

>>18284

>>18297

>>29549

>>30692

>>31418

>>32949

>>33791

>>40710

>>45313

>>49406

>>49908

>>49941

Now for the religious side:

God is within each of us, and we are all parts of the infinite whole. As above, so below. Like a hologram. Or a fractal. Or neurons in a cosmic mind. Ancient mystics knew this. They knew that to see beyond, one must look within. The bloodline families suppressed this truth, because it would make us bad slaves. God quietly remained with us, but He bided His time, in accordance with our collective free will- even though satan's cultists hived us to the Beast. But the stench of their sins, and our prayers for freedom, have spread out into the multiverse. This infinitely complex mathematical structure loves us enough to free us from the satanists at our darkest hour, even though we are but a sub-microscopic sub-pixel in its vast fractal consciousness.

The puppetmasters have no idea how cosmically screwed they are. God is about to show His true face to the world. When He does, the tables of the moneychangers will magically flip, and we will also be able to read all the blackmail files, satan stuff, and pizza party invites which the puppetmasters thought was safe because it was hidden behind RSA and elliptic curves. The NSA has been quietly sucking up all data for decades. If they aren't reading it already, they soon will be. Then there will be the death of the old satanic hive mind and (for those who are ready) hiving into God's mind.

Best timeline ever!

We are doing God's work.

In the beginning was the Word, and the Word was with God, and the Word was God. The same was in the beginning with God. All things were made by him; and without him was not any thing made that was made. In him was life; and the life was the light of men. And the light shineth in darkness; and the darkness comprehended it not.

…

For nothing is secret, that shall not be made manifest; neither any thing hid, that shall not be known and come abroad.

DEUS VULT

Related reading:

The Intelligent Universe by James Gardiner

The Holographic Universe by Michael Talbot

A New Kind of Science by Stephen Wolfram

2/2

So how would I use 'n' to factorize RSA100?

>>49985

read the OP kiddo. You can derive d and e from c, if you have an equation for n you can derive x and a and then b, and voila.

>>50003

Then let me rephrase, which variables will contain the primes?

Will 'a' and 'b' contain the prime factors?

>>49729

>;)

>I'm gonna be naughty for a moment.

>If P=NP

>does N=PN? probably not.

>how about

>N=EN?

>if P=¬P

>does ¬=¬e?

>or… !=!e

>Does N complete the jump from .999 to 1?

>What about 0 from .000…1?

>How does it factor in?

>0, "Aleph Null", {.000…1~.999…}, [?e]N, 1)?

>Einstein?

>(ein)"squared"=speed*time???

Further thinking on this, i had it backards

e!n=space*time so...

if n DOES complete the jump from .999... to 1, then that places it on the other side:

(0, !n, Aleph Null, Surreal, .5, Hyperreal Aleph N, e!, 1)

something like that...

So maybe...

e!^n=√({space}{time}) ???

>>49744

>>49774

>>49883

thread theme

>>50009

a and b are the prime factors, kek

>>49947

>>49941

>>49974

Not sure if this is really necessary in this thread. Please refrain from posting this pseudo religious-scientific babble further on.

>>50003

Reading and learning is for fucking faggots dude.

>>50011

Write code || GTFO. Filtered namefag.

>>50013

Write code || GTFO. Filtered namefag.

>>50019

wow such enforcement

many rules

much scary

I'm adding context asshole. You're learning how to decrypt reality.

>>50030

I agree, we're trying to find some fundamental rule here. If only point cloud libraries were actually common and shader programming wasn't the cancer the killed /b/.

We need to take the bmp we have and copy it on 6 axis so we can rotate it. We should see the GR Spiral coming out from all 6 axis'.

>>50011

You literally speak nonsense in all of these posts. Please stop smoking so much weed and post in a less reality based thread.

">Einstein?" Kys

>>49822

>>49834

I believe that x defines index position in the array of values at the specified (x,n) coordinate.

Assuming a = 1, any (c) record can be created with the following c#:

public static TheEndRecord CreateForC( int c ) {

int d = (int)Math.Sqrt(c);

int e = c - ( d * d );

int a = 1;

int b = c;

int x = d - a;

int n = ( (x * x ) + e) / (2 * a);

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

}

Any (f) record can be created with:

public static TheEndRecord CreateForF( TheEndRecord ter ) {

int e = ter.e - ((2 * ter.d) + 1);

int n = ter.n - 1;

int d = ter.d + 1;

int x = ter.x + 1;

int a = ter.a;

int b = ter.b;

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

}

And any combination of a/b can be created by:

public static TheEndRecord CreateForAB( int a, int b ) {

int c = a * b;

int d = (int)Math.Sqrt(c);

int e = c - (d * d);

int x = d - a;

int n = ( ( x * x ) + e) / (2 * a);

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

}

A few days ago, I was trying to compute the genesis / prime record at (e,n) and made an incorrect assumption around the value of x. You need 3 values to compute any key, and the most logical to try and compute was the value of x.

My understanding now, is that the search is on for n.

I'm inclined to think that if an appropriate formula for x can be found, then an n solution is apparent.

With regards to x, I've noticed the following:

For positive values of e

when e = 0, x = 2

when e is odd, x = 1

when e is even, x= 0

For negative values of e, however, it gets interesting:

It appears that the value of x wants to resolve back to 2 with an increasing number the further away from zero.

so for:

e = 0, x = 2

e = -1, x = 3

e = -2, x = 3

e = -3, x = 3

e = -4, x = 4

etc.

Also, as you move from (e,n) to the next valid combination, the index x increments.

Example:

(1, 1 ) = {1:1:2:1:1:5}, x starts at 1.

the next valid group:

(1, 5) = {1:5:4:3:1:17}, the first entry starts at 3.

and so on.

I may, however, be completely crazy. And the solution is somewhere else entirely.

>>49941

>>49974

Concise writeup anon, summarized, combined and expanded on my own thoughts way better than I could. Will be sure to pick up the books you mentioned.

>>50063

You're doing great work, and your input is greatly appreciated. We'll get it.

>>49883

>>50063

oops

>specified (x,n) coordinate.

specified (e,n) coordinate.

>>50093

Hold up…

Metric Ton Second???

Gravity???

>>50073

thanks for the kind words. This is the most interesting problem I've ever seen!

>>50100

I'll check those digits.

Hell yeah it's a cool problem. I don't really care about RSA but there's no denying some fundamental relationship here!

>>50096

Please read a book about physics

it reads "ts" as a metric ton * seconds

>>50017

If you input a number like 147 which has a lot of prime factors, will a be the coprime that doesn't have the largest number, and b is the largest prime factor? That would be the most helpful for RSA

>>50169

It depends on the value of n. When we do find the equation, there will be more than one. Each n value is a separate factorization, because it gives you a and b.

VQC or someone else knowledgeable. Do we need to know assembly lang knowledge in order to figure out this 'rsa crack'?

>>50184

No, just any general purpose programming language, if you actually want to help figure out which equation works for n.

>>50184

I think it's more that you have to know how to assemble language properly. Or in other words… Speak Correctly.

>>50190

> if you actually want to help figure out which equation works for n.

Sounds like it's mostly a mathematics problem. Kek.

>>50157

>>50181

>>50202

Holy frakamolly, shut the fuck up.

Quick clue before I go to bed.

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

>>50228

Sounds good, when's this trial by fire over?

>>50228

That sounds like simple algebra but I have no idea how to put that together

>>50181

>>1

RULE 4

>>50251

It's not simple Algebra, you need The Grid™

>>50228

not quite sure how you can sleep at night with this going on in your head!

>>50277

What I meant was that it sounds like he's pointing us towards a relationship between e and n that we didn't notice. For all possible e values in the first row (that's the (e,1) part) where a = na (I don't know what that means), na - 2d+1 = (n-1)a. That's not the right equation. If I actually knew what the fuck he was talking about I could probably do it, but I'm not 100% sure what he means when he says "element", I don't know what "a at that element equals the na we want" means, and I don't know what "the value at the same element in the first row will be (n-1)a" means.

>>50307

if c = 145.

the a = 1, c = 145 element is c = {1:61:12:11:1:145}.

represented by (e,n) = (1,61).

The genesis/prime record for this is (e,1) = (1,1) = {1:1:2:1:1:5}.

Still trying to process the rest.

>>50319

The genesis records are the first row, right? How do they relate to anything?

>>50324

they're the key!

>>50333

What's that supposed to mean? I'm trying to help here.

>>50333

Nice digits.

I can't believe we're over 50k posts now.

Dam Son.

>>50307

Download and run this: >>18284

(Java Version): https://pastebin.com/M3M1CDRX

It will create a spreadsheet that will help you understand. It's the spreadsheet he's referring to. It's known as the Virtual Quantum Computer.

>>50307

You can use LibreOffice or Microsoft Excel to open it. You will navigate the cells according to {e:n:d:x:a:b}, so when he says "e, 1" it means, the first two letters, where the second one is 1.

>>50343

I'm not new. I'm the one that added the image generator to the Java code. I just don't understand the grid that well.

>>50337

sorry. I don't have much more. My understanding is the (e,1) record is very special. All elements in (e,n) share some properties with it.

Don't believe we know more than that.

>>50349

Well, it'll be a good reference for other people anyway. I don't think there is much to understand in the grid besides the fact that there are a shit ton of patterns and if you know e for a number, then the factorization of that number will be in the e row, and then you use n to find the appropriate column. So, if we had an equation for n we could factor it with equations or by generating cells.

>>50358

That means that: if e for a composite is 1, and if n for that composite is 5, then I'm going to find the complete factorization at (1, 5)

>>50362

yes.

The solution to the c=145 problem is {1:5:12:7:5:29}.

I think VQC just gave us some way to work backwards.

>>50369

If you were desperate you could factor pretty gigantic numbers just by brute forcing n, maybe even above RSA100. But we aren't satisfied with that ;^)

>>50254

Had to.

People were starting to spoof my name.

Thank you, though.

>>50359

>if you know e for a number

You mean for the product of two primes, right? So if you know e for 5*7=35 (which is 10), "the factorization of that number will be in the e row". n is 1 in this case, so are you saying you'd find 5 and 7 in the a and b parts of {e, n, d, x, a, b} at cell (10,1)? I'm I understanding that correctly?

>>50396

So what did VQC mean when he said na?

>>50403

n * a

>>50604

Post the equation that generated the pattern too.

>>50706

f = a - (2 * d + 1)

if f in theend and n in theend[f]:

colors.extend([255, 255, 0])

points.extend([f, n - 1 , d])

else:

colors.extend([255, 165, 0])

points.extend([f, n - 1, d])

>>50751

Awesome! have you tried the golden ratio? No joke, not occult voodoo, it actually works.

Instead of fumbling around with image generation I've been trying to think of everything more closely in context with the RSA algorithm. I've simplified it below for anyone who didn't already know it, and I changed some of the common variable letter names around since some of them are already used in the {e, n, d, x, a, b} thing (a and b are literally the same variable though). Maybe this will help someone to put some of this stuff into closer context or come up with different ideas.

The variables we'll be using are a, b, c, g, l and m. For anyone who's used to d and e being used for the key variables, the public key is l and c and the private key is m and c, because the d and e from endxab aren't the same (unless they magically are I guess).

>a and b

These are two prime numbers. You can pick them at random. They're the same as the ones from endxab.

>c

a * b. Same as the one from the algorithm.

>g

The short explanation is that g = (a-1)(b-1). The longer explanation is that g is the number of numbers <=c that share no common factors greater than 1 with it (also known as phi of c). If c = 8, g would equal 4 (1, 3, 5, 7). With prime numbers, the only number that wouldn't count is the number itself, so phi(prime) = prime-1. Also since phi(a) * phi(b) = phi(a*b), phi(product of primes) = phi(a*b), so g = (a-1)(b-1).

>l

This is a prime number which is relatively prime to g (meaning their highest common factor is 1). It has to be >1 and <g. It can be picked at random as long as it fits that criteria.

>m

This is a number calculated such that (m*1) % g = 1.

The way messages are encrypted involves plugging values into an equation using variables from the public key, and messages are decrypted by plugging values into a similar equation. I've called the ciphertext variable q (since c is used) and the plaintext variable t.

Encrypt: q = t^l % c

Decrypt: t = q^m % c

>>50830

Accurate except for the encryption. It's encrypted by taking it to the power of a Fermat prime, usually 65537 then modulus'ing it to c

>>50830

The decryption variable is also wrong. The decryption exponent is calculated according to phi(n) and phi(n) is impossible to calculate for normal RSA numbers so far.

>>50861

Unless you know the prime factors of c, which is what makes it a private key vs public.

>>50852

>>50861

Whoops. I haven't done this in a year or two so I just got that last part from a website.

>>50852

>>50861

Wait but that is what I said, isn't it? You might be getting confused by the variable names I used, since you mentioned phi(n). n (which I called c) is the product of two primes. phi(prime) = prime-1, and phi is multipliable or whatever the word is that means phi(a) * phi(b) = phi(a*b). That makes the decryption exponent fit the criteria you said. As for the Fermat prime, the thing I read to refresh my memory for that part mentioned it usually being 65537, and I just used their explanation. I haven't read into Fermat primes much so maybe I didn't explain it right.

VQC mentioned this was 'very good'

>>44676

>>47810

The algorithm used to generate that particular image was x % prime = (a + b)/a.

Different primes are different coordinates, so there's multiple routes in the same space on this map. Find an equation that makes a pretty picture, and since the colors of the pretty picture line up with numbers in 2d space all the routes are on top of one another making it look all crazy like. We have to separate the routes through this fractal geometry.

If we have the number we want to find, we can probably prove that you can find it and where it is, the golden ratio is some kind of constant, it's not just a fad or a coincidence it's an actual constant force so let's try to see how we can tie primes to the golden ratio.

At least that's what I'm trying to do, but I don't know how to make pointcloud software. So I have to use another one, but there's not one in my favorite meme lang. Suggestions?

>>50949

I believe that particular image used every 5th prime.

>>50942

Yeah you may be right, there are too many variables required for this stuff. Lol

>>48412

>>48354

>>48424

>>48432

>>48519

Here was another exchange about what this should look like for real. It's some kind of object that has a shape that is a 'prime object' or maybe just a rule that actually works for all primes.

At anyrate, I feel like a monkey smacking stones together to make fire.

What happens if you apply the variables from VQC's code that aren't in the RSA algorithm to the variables in the RSA algorithm that aren't in VQC's code? Are there relationships there? What happens if you graph the extra variables from the RSA algorithm? I'm not saying I know more than any of you but it's useful to think outside the box occasionally.

>>50604

If the light is being cast from the right….

and you look at this at like… meeeeedium size.

You can see a shadow being cast.

>>51069

literally I was in the middle of saving to process when I read the last part.

FIRST OF THE EVENING!

If 2an = xx+e why doesn't n = xx +e / 2a?

values of n for each number up to 20

1—0

2—0

3—1

4—0

5—0, 1

6—0, 1

7—1, 2

8—1, 2

9—1

10—0, 2

11—3

12—1

13—4

14—4

15—1, 5

16—0, 1, 4

17—5

18—0, 1, 5

19—6

20—0, 2, 6

>>51298

it does, maybe you're not adding the parantheses

n = (xx+e) / 2a

https://en.wikipedia.org/wiki/Alcuin%27s_sequence

>>51340

It works for me, but I don't know why we're looking for n when that's the equation we need. That's all.

>>51348

but then we still have to find a

>>51348

and x

if we can find n it gives us those variables.

>>51591

Alcuin's sequence matches up to the first 20 values of n with an accuracy of: YYYYYYYYYNYNYNYNNN

>>51591

That is, if you start off with an extra 0, for c=0 which according to my colleague satisfies the equations.

Doing % golden_ratio seems to really bring some interesting and important things out, lots of variable react positively to it.

http://pointclouds.org/documentation/tutorials/basic_structures.php#basic-structures

This is what we need to be using. Tomorrow I'll rewrite the c# code to c++.

>>52336

What if there is a Java or C# wrapper library? Possible? C++ is pretty low-level for me.

>>52369

keep the autism strong, this is brilliant to watch unfold.

do not give up on this

Anyone know how to actually evaluate that Alcuinn series? It looks confusing to me but the fact that the the terms match up with the first few values of n is a strong indicator that if it were modified somehow it could possibly help solve for n..

So when we do crack RSA, we can break into those Wikileaks insurance files, can't we?

>>52336

https://www.codeproject.com/Articles/839389/Point-Cloud-utility-with-Csharp-and-OpenGL

Boom. You are brilliant!

I don't believe in things without evidence but today I got fortune cookies.

One said I should accept a free gift.

And the other said i would have many friends when I needed them.

I'm tired I'll do this tomorrow if someone doesn't beat me to it.

The golden ratio is the key.

% golden_ratio

= golden_ratio

tough day on this but it'll come together, peace my dudes.

>>52621

Yes but is it even possible. We'll find out.

If we break it he will crumb.

>If we break it he will crumb.

If we break it he will crumb.

>If we break it he will crumb.

If we break it he will crumb.

>If we break it he will crumb.

If we break it he will crumb.

>If we break it he will crumb.

If we break it he will crumb.

>If we break it he will crumb.

If we break it he will crumb.

>If we break it he will crumb.

And I just wanted to write this down.

e * n = z

x y z

Is how I think it should be mapped.

I don't know if anyone has messed around with VQC's algebra from before he posted the code, but I thought I'd try to refresh my high school algebra a bit and I'm wondering if I've either done something wrong or wasn't already aware of the result I got.

Considering the formulae given

>d^2 + e = c

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

>d = a + x

>2a = x^2 + e

for all c where c is the difference of two squares:

d^2 + 2dn - x^2 - 2nx = d^2 + e

2dn - x^2 - 2nx = e

2dn - 2nx = e

2dn - 2nx = 2a

dn - nx = a

d - x = a/n

a + x - x = a/n

a = a/n

n = 1 for all c where c is the difference between two squares.

I didn't use every formula, and obviously they're all necessary, but I've got to get other shit done.

>>53092

all c's are a difference of two squares. I think where you messed up is the 3rd step. Where does -x^2 go?

>>53092

actually, you didn't mess up, I followed your algebra. let me check this out!

>>53092

Or so I thought all c's were a difference of two squares. It seems this equation is true only for some types of c's. Couldn't find any errors in the derivation.

>>50604

Isn't that a nice power function you have there.

I will be posting more today until The End of Monday.

Thank you all for your hard works.

I will be stepping all my input up a gear.

There may be GAPS as to when I can come on and reply but I should be more available.

Thanks again.

Let's go on further ADVENTures!!!

You guys are Awesome.

First we need a way to communicate more easily about certain terms.

If you come up with improved methods, we can use those.

To Begin with, I'll make suggestions, if you're ok with that.

I'll take any feedback.

If you feel like you haven't had your question or query answered, please state it again.

You can even SHOUT if you feel not listened to.

>>53389

Could we communicate on email? Maybe use PGP? Use RSA to break RSA? ;^) (better than Twitter) I see what you're up to!

>>53389

Why does that proof for a = a/n work? Is it only for certain numbers?

>>53384

put your desired file path inside of the quotes and find run somewhere on the IDE

>>53383

VQC. This is really fun and exciting. Thanks man. I love math so much, it's destiny that I found this forum. I have questions for you about the equations and program. The IDEAS come first. Programs later. Pages of notes here.

>>53396

Thx ProgrammerAnon. Feeling dumb. Do I need /file path/ ? I get the math, but learning CSharp now. See the // at top. See the // at string path. Does my path go between // ?

The Grid is something I sometimes call The End (because each element starts with e,n,d,..

It was also a pun because the Key is n.

All whole numbers (integers) c, that give a remainder of 0,1,3 when divided by 4, are the difference of two squares.

Any number that leaves 2 when divided by four can either be divided OR multiplied by 2 to make it the difference of two squares.

Therefore we have a way to include all integers.

For all c, where c is an integer and the difference of two squares:

ab = c, a multiplied by b equals c where a and b are integers

dd+e = c, all integers c can have a square root d, and a remainder e after subtracting the square of d from c, for a perfect square, e is zero

The difference of two squares can be expressed as:

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

The square of (d+n) is the big square.

The square of (x+n) is the little square.

a = d - x

xx + e = 2na (the square of x and the remainder make two copies of a rectangle)

b = a + 2x + 2n

c = aa + 2ax + 2an

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

The original Grid had cartesian coords:

[e] the remainder of c after taking out the biggest square

[n] what is added to the square root (d) to make the big square

[e,n] is the grid

Each cell of the grid is referenced as cartersian coords. e across (columns), n down (rows)

E.g. [1,5] references the first column for integers c where the remainder is 1,

and the cell that contains elements where you add 5 to d to make the big square in the difference of two squares. One across, five down.

Now, if we want to talk about a specific element in a cell in the grid, we need a third identifier. We will call that t : [e,n,t]

t will have values from 1 upwards. A value of t=1 means the first element in the cell. A value of t=2 means the second element in the cell, and so on

[1,1,1] refers to the first element in the first cell in the first row of the first column {1,1,2,1,1,5}

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

The zeroth column refers to the column with all perfect squares. All values of c have an e values of 0.

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?

>>53432

Whoa. Thx KingMathAnonVQC.

>>53399

Pleasure to be working with you.

It's like building a family of fun for math.

I always loved math, though we call it maths here :)

>>53432

set_size = t!

>>53436

MILO VQC wants us to say "Maths" instead of "Math" Good luck with the Islam invasion of your country. We Have GEOTUS DJT over here, my friend. You on the other hand have the solution to RSA, so maybe we won't nuke you. ;)

>>53392

Communicating here is Good.

We're not just talking about RSA, we'll be moving onto other encryption methods too.

We should at some point include a separate set of threads for the Mandelbrot Set, which is the most beautiful object in mathematics to me.

It can be used to make physical things.

It can be used as a perfect resonator.

It can be used on images and video to 'wind' back time or change the spatial angle.

It can be used to expose photoshopped images, videos and expose FAKE NEWS.

>>53394

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

>>53435

Thank You

>>53468

Is my code wrong?!

I was sure it was right :(

>>53437

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.

>>53452

Math or maths, it doesn't matter. I have generally used math

>>53463

Yes.

This is about you more than me.

I had a spiritual void.

Filling that led me here.

Money is not very important to me.

Teamwork is.

You are.

>>53467

In the first row, n=1, look left and right of any cell by moving 2n along. Do you see? The cells in row one where n=1 have a relationship with the cells 2n to the right and 2n to the left.

>>53472

Don't worry.

Which code are you referring to?

>>53481

I see the pattern for left/right shifts, but moving in the positive n direction while holding e constant is what I am stuck on

>>53482

i = d + n;

j = x + n;

c - (j + j)² - (d + d)² = (c + n)²

√(c + n)² - c = n

No?

Find n from c.

>>53482

I did send you a DM on Twitter.

>>53493

If you are taking the square root and squaring a value isn't that pointless? Also I couldn't find the RSA 100 example on the grid you showed.

Not sure if these equations are useful but they work

(c-a)/2 = j

j+a = i

n = i - d

^how to easily get n from a and b

i = d+n

j = x+n

>>53521

I wrote the code for this a while ago but I thought it was pointless. All you have to do is add and subtract if I'm not mistaken.

>>53531

kek, I'll rewrite it then

>>53452

VQC=MILO is very delightful. Gay, Brainy, Hot as Shit, Etc. And I'm a happily married man. I'd suck Milo / VQC any time.God we need some humor in this thread!! ;)

>>53551

Yes, very good goyim.

ayyyy. Just trying to make you laugh before we get back to work.

>>53572

Do you like Quantum computers? I was making a non-mathematical explanation of Shor. Seems helpful for this.

>>53482

VQC what do you think of Shor? Does it seem like banging one's head on a wall given that the only thing you need to factorize is simple algebra?

You can actually run Shor on a classical computer, easily. But it might take a million years to find the answer..

>>53493

Keep in mind i and j are used to create the patterns.

For RSA you are only given c to find n

>>53498

Roger roger

>>53551

Humour is good.

I'm a fan of Milo.

I'm married.

>>53595

The virtual quantum computer in the grid representation is equivalent to what the quantum fast fourier transform is trying to achieve.

The clever bit is using constructable infinite sets in a simple pattern which requires a finite small number of variables to control and utilise.

>>53615

If so, is the period of the Euler sequence involved in the VQC somehow?

>>53633

You can come very close the prime factorization of N using the 1*N factorization of N. Use my equations

>>53637

I think you should change it back first to get what the original grid was like. gbn's equation was interesting but wasn't helpful or didn't see what connection it made.

>>53638

Which equations? Could you post a quick example?

>>53644

Specifically (c-a)/2 = j

j+a=i

i and j are very interesting. In this particular case you are solving for just the 1*c factorization but if you compare the n, x, d and e values of the "useless" factorization they are very close, almost like a common thread is missing.

the n value for the 1*c factorization is simply d-1, try comparing it to the real factorisation (like 1*145 and 5*29, but if you want to see more patterns try 11121 as c.)

>>53658

Also compare the i and j values of the real factorization versus the 1c factorization. Important crumb.

>>53633

I like it.

Cell may be better to have as a grid cell.

Elements of that cell are each {..} for each c

Sound Good?

>>53637

Question 1.

At most O(log t), where log t is the natural logarithm of the length of c in bits

In other words, doubling the length of a key would only take twice as long to compute as opposed to a factorial amount of time, i.e. no effort.

>>53637

Question 2:

x squared plus e equals 2an

for odd e in the first row, x = (t+1)/2

for even e in the first row x = 2(t-1)

for the first element in the cell, t=1, second t=2, etc

Fun fact: the values of a in the first cell where e=1 contains ALL factors for odd numbers that are the sum of two squares.

I believe Fermat would have gone nuts with the Grid. Or maybe this is how he worked…

>>53637

Question 3,4,5 answered earlier and above

Questions give answers

>>53639

Thank you

>>53674

Enjoy!

Fun to share!

It is good to eat, drink, sleep and rest though :)

>>53674

Goy, you need to stop thinking about this. It's too dangerous. We wouldn't want you becoming hard to control now..

When doing a Maths problem, gotta read all the instructions. What are these variables and equations straight from QVC , anons?

>>18322

>(last step to find N according to him)

>"O(log t) where t = length of C in bits?" What? on the bitmap output??

>>18247

>"2an=xx+e"

>so WTF N=(xx+e)/2a ??

>this equation appears nowhere in the program or Maths up to this point. Maybe i'm a dumbass, but I don't think so.

>>18322 "t is an index number that allows us to specify an element in a grid location."

>Uh, so {e,n,t} ? VQC gives examples, verify. T is the variable that allows us to transit thru infinity? How appropriate.

>int i, int j, int f, all omitted from maths but included in program. To be fair, i and j are starting points to calculate a,b,and c. But F? F ME!! No answer yet. F My Life!

>also want to know from a CoderAnon what this means: "bool odd=c%" Yeah, boolean something or other. Help me PLS.

>>53683

It's algebra. You're given a total of 9 variables: e, n, d, x, a, b, f, i, j

Each equation is a way to arrange these variables to help you find the variable of another. The end goal is to have enough equations to go straight from c, a very very large number straight to the prime factors of said number.

>>53683

He didn't include i and j and f at first but they are also quite important. Just try these equations, it will surprise you just how flawlessly they come together. It's a bridge and n=??? Is the last piece required to complete the bridge.

>>53683

O(…)

This is Big Oh notation, not an equation.

Big Oh notation describes the complexity class of a problem.

Currently integere factorisation is big oh sub factorial t, we are talking about making it big oh log t

>>53468

So row 1 where all n are 1 is fundamentally important BECAUSE n == 1 and thus n "falls out" and allows a straightforward relationship with a to be leveraged.

>>53432

>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

Is big_n function as an upper bound for n (or other bound) or is it more of a link to a chain/tree/decision tree that can be resolved in O(log) time?

>>53686

Thanks anon, i know it's algebra. After reviewing all the OP, diagrams, threads, etc, I found these unanswered Q's and equations. Still working to understand, haven't seen anyone post these yet, submitting them for discussion and analysis by Clever Autists such as yourself. Variables i, j, and f are in the program. F has me puzzled at the moment.

>>22370

Most curious is Chris' statement: N= "O(log t) where t is the length of C in bits"

So what, we use e^2+n^2=^2 to find the length of the hypotenuse? (distance from 0 to prime on the bitmap grid using e and n as coordinates, solving for the hypotenuse?) and then use Log functions to scale it up or down towards infinity?

>>53704

Just… quantum intuition… but I'm seeing patterns… does this help at all?

>>50764

>>53712

Sorry to spam three replies so quick, but is vertical traversal of t++ a fundamental O(log) situatiom? Is the goal to hit the right cell and ride it down (i.e., ride t++ up)?

>>53714

I have some errors in my algebra! Questions remain, link to >>22370 is correct.

>>53699

Thx VQC. Can you please show it in algebraic notation? Got calculus, but not big O skills. So big O denotes a very complex equation? The key is understanding primes and their multiples, along with all possible combinations, hence the tree of numbers. Feeling small over here, but willing to learn. I'm up so late right now, work will take more focus tomorrow. However my mind is SO AWAKE I can't go to sleep if I tried to. Maths are Excellent.

>>53699

So the N KEYSTONE (lol, Hello Q! and Clowns too!) allows us to unlock a specific set of primes and their derivatives or multiples along an infinite line using N like a magnifying lens to zoom in on one spot in the pattern? Almonds Activated.

>>53740

Put simply, big O describes how long something can be expected to take. Let's say you have n integers, random and unsorted, say 5, -2, 10, 1, 400, -12000. So here n == 6. Lets say you wanted to calculate the biggest. You would have to crawl through the whole list, all 6 (or all n) to do so. Therefore your algorithm is O(n) which means roughly "you have to look at all n once" orhere, you have to look at all 6.

What if your list was ordered, though. Same list but 400, 10, 5, 1, -2,-12000. You know how the list is ordered and you can exploit it. In this case you have O(1) or "constant time" because you know the biggest is first, so no matter how many n you have, this operation will never take longer or shorter, be itn == 6 or n == 600,000,000. Whereas if you're at O(n) for 6 life is good but if n == 600,000,000 it might not be, have fun examining all 600,000,000.

Does this make sense?

>>53759

>being this mad

t. someone who thinks people getting interested in math is bad

>>53765

Methinks this namefag doth protest too much

>>53765

Kek

>>53770

Yes!! Made VQC Kek! Fuck Math(s), I can rest in peace now.

>>53767

Our Senpai Hath Kek'd. Maths+ Keks = MK (Math Knerds)

>>53773

Please do, trip related

>>53786

How the fuck did you do that? Even when you're making fun of me i'm impressed.Fucking MathAnons are the best.

>>53792

You're on an open source image board, of course the secure tripcode code is known.

>what is OpenCL

>>53801

Shit. How does Q keep his/her trip private then, ParanoidAussie?

How does one code a Ghost into the Machine?

What about an Angel?

>>53762

>getting interested in math

As someone who held no interest beyond math exists in math. I can safely and with confidence say that this thread did not increase my interest in math. All it did was reinforce in my mind that this thread is filled with some grade B faggotry and some difficult looking math.

Can someone please attempt to explain in simple words what this thread attempts to convey? I've read too many elaborate posts that just further cement confusion.

>>53832

We're figuring out the code to The Matrix so we can seize control of it as our birthright, and then spread the information, selflessly, to everyone.

We Serve Man.

We're figuring out "N" and "P=NP".

Not only that, it will be proofed.

And it will be literally drawn out for you.

Freely available… handed out to everyone like quantum bread… where no matter how much you take from it, there is always more.

When you add or subtract something to/from Infinity, what do you have?

Still Infinity.

>>53832

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 thr 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.

>>53838

Free lunch?

I'll believe it when I see it.

>>53806

Because it still takes an umpteenth amount of computer power to calculate an exact tripcode.

>>53847

Oops, forgot to stop namefagging, sorry.

You guys should do the real math and figure out the importance of 3, 6, and 9.

The answer is beautiful.

Good to see you've all been focusing more on the math and less on making interesting plots since I left.

>>53224

Like I said with the answer, it's where c is the difference between two squares. Making that the criteria of a plot might be helpful to us.

>>53463

Is there any reason why nobody has done anything with this? This looks like our breakthrough. I want to mess with it but I also want to go to sleep.

>>53452

>Islamic invasion

People say "maths" in Australasia too you know.

>>53466

>>53389

>if you feel like you haven't had your question or query answered

It's pretty irrelevant but you seem to know about a lot of crazy shit: do you know if New Ageism was a plot by the deep state? I've always felt like it might have been an intentional maneuvering away of attention from more evidence-based forms of esotericism since New Thought and Golden Dawn were bringing disciplinary/learning-based occultism more attention in the early 1900s (regardless of their flaws) and suddenly New Age comes along and seems to encourage its believers to never actually look into things and just blindly believe what they hear about. I'm enjoying getting closer to that 40,000ft you and Q keep mentioning.

>using the Mandelbrot set to do all of that stuff

Oh man that sounds so fucking cool.

>>53849

Codes have been cracked before. I recall a tripfag from /sp/ getting outed by an autist who bruteforced using his gpus to number crunch.

>>53389

>Communicate more easily about certain terms.

Please let that way be code. I'll switch to any lang you want c# is fine. Java is fine.

>>54204

Shot in the dark I was showering fuck it:

Supposed to take log(N) time if N is number. So the time is proportional to the amount of bits. If it's proportional, is there a function related to the time of bits?

Maybe if we use the BIT MAP and use the BITS of the number to guide us into the MAP we can solve this problem.

If anyone uses this to figure it out I want my bitcoins 1J7GaAdXHMryKJ1hdkQvwY6kXqJoho6YfN

>>22370

Will be.

Final piece for calculating arbitrary n.

O(log t) where t is the length of c in bits.

so if c = 145 it's length in bits is 8.

>>54256

idk, I'll make some more headway with the c# code.

>>53681

Everytime our threads get dubs you have to drink water.

>>53633

Look at those fine digits and that fine math. Good work my dude.

>>53637

Math anon, tell me what you need and I will write it. You break the math down, and I will put it into a program.

>>50381

You could just not be a namefag…

So lets say I find the formula for n, how would you find the primes?

Wouldnt you need to generate a huuuuuuge table?

>>53755

Yes, good explanation. Thanks Anon!

>>54569

You probably don't need to calculate the whole thing, just a specific part of it. Maybe specifically the length of the spiral which is 'n' for O(log c) which is 'O(log a * b)'

>>54610

Beer is acceptable probably. I spent a few minutes contemplating actually drinking water.

>>54613

Mathanon, tell me what you need code wise.

>>54658

Are you saying:

moving 2n to the right, the new a = a + 1, and the new b = b + 1

moving -2n to the left, the new a = a - 1 and the new b = b - 1

>>54697

Yes, and same for d. And x = x

It relates to t in some way too which I haven't figured out yet. But the pattern holds for same rows!

>>54714

Similar pattern for the negative (f) record.

(f)e = e - (2*d + 1)

(f)n = n - 1

(f)d = d + 1

(f)x = x + 1

VQC clue from last night was a relationship between na

>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

Didn't quite understand that until >>53704

(2d + 1) looks to be a pattern for movements in any direction with +1 or -1 changes to the other components.

Does the real value of PI helps in any way?

Its 3.14269680527

* Formula = 4 / SQRT(1.62) ****

Real value of PI

https://www.youtube.com/watch?v=LnfykAdoEL0&t

We got it from the ancient masters

https://www.youtube.com/watch?v=ULGnc9bO4BU

>>54652

Thanks Anon, basic help needed! So I have Visual Studio Code with the Csharp extension installed. VQC's code is copied properly. Pretty sure I got the file path correct set it as /users/desktop. From this point what are the last few steps I need to complete to get the output? Do I have to create an executable file?

oops

>>53633

not >>53704

>>53922

Yoooou…. want us to learn Vortex Math?

If yes…

Something like this?

http://vbm.isteaching.com/

>>54803

You have to write a main function I'm working on it now. gimme a few minutes.

Then you compile it and it should produce an executable and a .csv file. Open the csv file in excel and you should see the grid.

https://pastebin.com/bPHikDaw

That should get you the .csv file. I'm working on making a point cloud now.

So yeah, create an executable and run it, then open the file it produces.

>>54831

I'm really not an expert. Danny Wilton, the guy who has "figured out" the real value of PI, really did it in an amazing way. By studying ancient structures. I cant help myself but think that his work is very very important for math/spiritual guys.

>>54946

cool video. thanks.

>>54972

Preciatecha

If anyone is having trouble seeing Vortex Math at Play, this might be a little more clear:

https://youtu.be/N05u54WLhL0

Once you can see it, here's a more advanced and refined demonstration that includes what I was calling The Crab and the Tesla Coil

https://youtu.be/cOyjBXOz5d4

>>54972

Here's another way of looking at it:

https://youtu.be/iS1xkieyRWQ

Here's another way of seeing Vortex Math using a BUDDHIST MONK'S SPADE

https://youtu.be/wREW99kJnt4

Is this link shilling or not?

https://www.theproblemsite.com/vortex/

>>55146

If you understand what you're looking at, you'll understand why I linked them. I have so many others but those 3 props are significantly connected to what's going on here.

Also, if you like the link you posted, here's a door to pop your head through.

https://vortex369math.wordpress.com/2013/02/28/randy-powell/

VQC, I've been trying (maybe) the naive approach by applying Fermats factorization to the problem of finding 'n', however I have a gut feeling telling me I'm not supposed to brute force it.

For now I've made a simple loop that works well enough to generate 'n' for smaller values. It goes from d + 1 and to some upper limit I hardcoded.

I'm guessing I this is the wrong approach and I'm rather supposed to find the relationship between 'n' and the blocks.

>>54927

Thanks for your help, Anon! I'll compile and run it when I get home from work.

>>55192

If you know how to see him outside of here, you can bake in a different oven.

That's what I've been doing.

There's a recipe.

Hey, i did a python version of the basic implementation because I suck in CSharp.

>>https://pastebin.com/NZkjtnZL

It took me a little time. I lurked a little and I saw you are going deep on math and I stopped following.

Is there something interesting to implement / focus on right now ?

>>55915

Very good. Does it produce the same spreadsheet as the C# code?

>>55969

I think so far, it really looks like the first images I saw in the first thread but I did not implement complex image generation yet

>>56065

Thanks for the summary, I'll catch up on that !

>>56466

Checked!

Same!

DRINK WATER!

>>56466

me too. going around in circles.

>>50220

i can almost see myself looking back in there

>>54256

>>50220

If you do this you pattern and follow f for 0's and a,b for 1's you eventually get two arrays (for just the first entry in the graph, maybe the rest will help?) but one entry is a negative value and one is a positive value (as in, the entire array is just filled with that one value)

Hope that helped

>>56741

didn't finish.

I was thinking these two values (pos and neg) could be new entries for an (a,b) and maybe you could generate something with that. I still don't understand the generating thing someone mentioned earlier about how all cells generate something.

>>53673

>for odd e in the first row, x = (t+1)/2

>for even e in the first row x = 2(t-1)

looks like we can figure out t for any x in (e,1).

I think that when you're done, the image produced should look something like this… flashed over and over again in the video:

https://youtu.be/hJa1Vl1j1qo

>>49883

According to the graphic, there is a relationship between n, d, a, and b. Not sure if this has been previously noticed.

>n + d = value between a and b

based on that, I've made an assumption:

( a + b ) / 2 = n + d

therefore:

n = (a + b) / 2 - d

I've iterated all the values in the grid and found that this holds true, and further appears that a + b are always even.

Can someone verify please?

>>58040

I'll verify. One question, is it

(a + b) / 2 - d

or

(a + b) / (2 - d) ?

be clear with your parantheses.

>>58050

It's ((a+b)/2) - d = n

>>58040

So how do you go from that to the variables in a public key?

>>58059

Leeeeeeet me check that out first. That's a pretty big statement you're making.

>>58059

not sure. Seems like we were struggling to find a way to calculate n.

Don't think we've seen this formula before.

And yes:

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

>>58071

yep. makes me a bit nervous.

>>58081

Lol, I must have not been thinking there for a second because you still need a and b to solve that equation.

>>58076

The whole point of this is to get n from only c. You can't use a and b because you don't know those.

>>58096

We can already get an n from c assuming a = 1.

Determining the correct n for the correct prime record is the problem.

This relationship may assist in determining the genesis records or otherwise navigating the grid.

>>58143

You make a good point. Did my

j = (c-a)/2

help you with that? I hope it did. Your equation is true. I think I might be able to do something with it.

>>58157

everything helps!

This is a really tough problem.

>>58513

You are doing something fantastically beautiful.

>>58564

Looks interesting, where's the rest of the code?

>>58592

>>58592

I'm going to perfect it and repost

>>58600

0: (*,*)

1: ([-3],*)

2: ([-2],[1])

3: ([-1, -1],*)

4: ([-5],[0, 0])

5: ([-4, -4],[1])

6: ([-3, -3],[2])

7: ([-2, -2, -2],*)

8: ([-1],[4, 4, 4])

9: ([-7, -7],[0, 0])

10: ([-6, -6],[1, 1])

11: ([-5, -5, -5],[2])

12: ([-4, -4],[3, 3])

13: ([-3, -3, -3],[4])

14: ([-2, -2, -2],[5])

15: ([-1, -1, -1, -1],*)

16: ([-9],[0, 0, 0, 0])

17: ([-8, -8],[1, 1, 1])

18: ([-7, -7],[2, 2, 2])

19: ([-6, -6, -6],[3, 3])

20: ([-5, -5],[4, 4, 4])

21: ([-4, -4, -4],[5, 5])

22: ([-3, -3, -3],[6, 6])

23: ([-2, -2, -2, -2],[7])

24: ([-1, -1],[8, 8, 8])

25: ([-11, -11, -11],[0, 0])

26: ([-10, -10, -10],[1, 1])

27: ([-9, -9, -9, -9],[2])

28: ([-8, -8, -8],[3, 3])

29: ([-7, -7, -7, -7],[4])

30: ([-6, -6, -6, -6],[5])

31: ([-5, -5, -5, -5, -5],*)

32: ([-4],[7, 7, 7, 7, 7])

33: ([-3, -3],[8, 8, 8, 8])

34: ([-2, -2],[9, 9, 9, 9])

35: ([-1, -1, -1],[10, 10, 10])

36: ([-13, -13],[0, 0, 0, 0])

37: ([-12, -12, -12],[1, 1, 1])

38: ([-11, -11, -11],[2, 2, 2])

39: ([-10, -10, -10, -10],[3, 3])

40: ([-9, -9],[4, 4, 4, 4])

41: ([-8, -8, -8],[5, 5, 5])

42: ([-7, -7, -7],[6, 6, 6])

43: ([-6, -6, -6, -6],[7, 7])

44: ([-5, -5, -5],[8, 8, 8])

45: ([-4, -4, -4, -4],[9, 9])

46: ([-3, -3, -3, -3],[10, 10])

47: ([-2, -2, -2, -2, -2],[11])

48: ([-1, -1],[12, 12, 12, 12])

49: ([-15, -15, -15],[0, 0, 0])

50: ([-14, -14, -14],[1, 1, 1])

51: ([-13, -13, -13, -13],[2, 2])

52: ([-12, -12, -12],[3, 3, 3])

53: ([-11, -11, -11, -11],[4, 4])

54: ([-10, -10, -10, -10],[5, 5])

55: ([-9, -9, -9, -9, -9],[6])

def fromCgetN(c):

#(e,n,d,x,a,b)

cellC = createForC(c)

factors1 = []

factors2 = []

while(c>0):

if(c & 1):

cell = createForF(cellC)

factors1.append(cell[0])

else:

cell = createForAB(cellC[4], cellC[5])

factors2.append(cell[0])

c = c >> 1

return (factors1, factors2)

Sum the entries in the arrays and put it back into the function (if negative do absolute value), eventually you get a factor. Also this is python. Sorry I thought nobody was online so I was going to post it tomorrow when more people were on

>>58618

def createForC(c):

d = int(math.sqrt(c))

e = c - (d*d)

a = 1

b = c

x = d-a

n = ((x*x)+e/(2*a))

return (e,n,d,x,a,b)

def createForF(rec):

e = rec[0] - ((2*rec[2])+1)

n = rec[1] - 1

d = rec[2] + 1

x = rec[3] + 1

a = rec[4]

b = rec[5]

return (e,n,d,x,a,b)

def createForAB(a,b):

c = a*b

if(c<0):

d = int(math.sqrt(-c))

else:

d = int(math.sqrt(c))

e = c - (d*d)

x = d-a

n = ((x*x)+e)/(2*a)

return (e,n,d,x,a,b)

>>58635

>>58618

Like keep putting the sum of the factors back into the function and eventually you'll get a factor

Y'all are doing the Lord's work.

Damn low wage workers stealing everyone's jobs.

Peace and Protection to all anons working diligently for Truth.

May the presence of God surround you and may your mind be granted +10 Sharpness!

love seeing the progress that gets made every time i sign on.

Peace be with you.

Cras es Noster

Deus Vult

Let me make a correction to my equations.

x+n = j

ONLY when c is not the difference of squares or a perfect square. Other than that it seems objectively right ;^)

Which means?!

That if you have (x+n) you have one of the addends and subtrahends of the factors.

>>58857

I'm a macfag

Can someone generate a map using:

z = a + bω

please?

>>58861

IF it isn't a perfect square or the difference of two squares

>>58857

Hash it with sha256 or sha512 and search the thread for the hash.

>>58862

>macfag

The code has been ported to several other languages that work flawlessly on a Mac. Why aren't you using one of those?

>>58618

>>58620

Translated to C#

private static int[][] FromCgetN(int c)

{

//#(e,n,d,x,a,b)

int[] cellC = CreateForC(c);

List<int> factors1 = new List<int>();

List<int> factors2 = new List<int>();

while (c > 0)

{

if (c % 2 != 0)

{

int[] cell = CreateForF(cellC);

factors1.Add(cell[0]);

}

else

{

int[] cell = CreateForAB(cellC[4], cellC[5]);

factors2.Add(cell[0]);

}

c = c >> 1;

}

return new int[][] { factors1.ToArray(), factors2.ToArray() };

}

private static int[] CreateForC(int c)

{

int d = (int)Math.Sqrt(c);

int e = c - (d * d);

int a = 1;

int b = c;

int x = d - a;

int n = ((x * x) + e / (2 * a));

return new int[] { e, n, d, x, a, b };

}

private static int[] CreateForF(int[] rec)

{

int e = rec[0] - ((2 * rec[2]) + 1);

int n = rec[1] - 1;

int d = rec[2] + 1;

int x = rec[3] + 1;

int a = rec[4];

int b = rec[5];

return new int[] { e, n, d, x, a, b };

}

private static int[] CreateForAB(int a, int b)

{

int d;

int c = a * b;

if (c < 0)

d = (int)Math.Sqrt(-c);

else

d = (int)Math.Sqrt(c);

int e = c - (d * d);

int x = d - a;

int n = ((x * x) + e) / (2 * a);

return new int[] { e, n, d, x, a, b };

}

Also what specific variable do I change where to get the negative side to plot on the image? I've been ignoring it until now since it isn't completely necessary.

>>58873

Because I haven't asked how to do it.

How do I do that think (You) say?

couldn't get rustup (rustle up, jimmies) to do anything.

>>55192

Let the grid do the work for you.

What about negative values of x for cell (1,1) and then cells 2 to the right?

Enumerate the patterns.

Once you state everything you know about the problem, the solution presents itself.

Create a map of the grid's properties.

The key to the map is the solution.

Sound familiar?

Enumerate the grid properties.

The solution will present itself.

i think I found something, useful?

for all c's that aren't perfect squares

i = d+n

j = x+n

:D

It's not ICQ, it's IQC

Intuitive Quantum Comprehension

So… shift your perspective a little.

Imagine:

Verified Quantum Comprehension

We're working on the key that unlocks all doors.

Without this key…

You're a dead man.

So what needs to be found is the key to trigger the switch.

Via the bitmaps…

You'll know It when you see It.

What is an I what is a V

Where do they meet?

Get the point.

>>58944

Can you elaborate?

Yes and no.

No because that's past the point of no return and we're trying to earn the point we're getting to. No need to go there.

Yes….

What is an I?

What is a V?

How did they meet?

What point are you getting?

That's as far in as I can stick it.

Also, maybe, let me see if it lets me post, if not, i'll create new image that explains it.

It involves Vortex Based Math, Phi, and the Perfect Squirt Gun

3175397 prime search result:

If we transform the parabola y = (x + 1)2 + 2 by shifting 7 units to the right and 5 units down, what is the vertex of the resulting parabola?

The answer would be (6,-3)

ooh la la 3397157

search that leads here:

http://effbot.org/imagingbook/concepts.htm

I solved the puzzle!

n=2(OwO)

>>58635

Very nice!!!

Three images.

This one.

Even values of e only

Odd values of e only

For real though

What if instead of trying to solve for n we tried to solve for x+n or d+n? They're equivalent to i and j for most principals and if you knew either one you could find n.

Possibly related… can anyone explain what I'm looking at?

http://www.mysql-password.com/database/40879

also,

I

F

PIL

N

Z

Posted this in the wrong thread

Specifically this one… what am I looking at?

408,780,073

*29497527F69434447AF237B24FA36EBB613ACCE7

r3sumd

>>58878

Bump

>>58923

There's Java and Python as well. If you can't figure out how to make Java or Python work, there's no way you'll figure out any of this other stuff.

>>58944

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

I want to try an summarize a bit:

We want to find n, which represents the difference between d and the larger square block used in Fermats Factorization.

Once we know e, we can calculate (e, 1) and we know that n exists at (e, n). We just need to figure out how to find the correct n.

We have have to test different potential 'n', which again is the same method used in Fermats Factorization. However, we are supposed to have an algorithm that can test and find the correct n within O(log).

This should mean that there is a relationship between one (or more) values within (e, 1) to allow us to continue our search.

Am I missing something, or have I misunderstood something?

>>58618

>>58620

>>58698

Good Job my dude.

>>58923

Re: Rustup. I intended on releasing a point cloud visualization thing yesterday but it turns out shader programming is the cancer that killed /b/. I'll get it worked out today.

>>53759

Why is there a rothschild shitposting in here?

Friend of yours VQC?

>>58635

>>58635

How did you color this?

>>59658

Create a new bitmap and color it black:

Bitmap bmp = new Bitmap(x_max * 2, y_max);

using (Graphics g = Graphics.FromImage(bmp))

{

g.FillRectangle(Brushes.Black, 0, 0, bmp.Width, bmp.Height);

}

Then inside the main loop of VQC's Output function in TheEnd.cs code:

if (theend.ContainsKey(x) && theend[x].ContainsKey(y) && theend[x][y].Count > z)

{

tw.Write(theend[x][y][z] + ",");

String[] values = theend[x][y][z].Split(':');

int xx = Int32.Parse(values[3]);

int a = Int32.Parse(values[4]);

int b = Int32.Parse(values[5].Split('}')[0]);

bmp.SetPixel(x + x_max, y, Color.FromArgb(255, xx * 255 / i_max, a * 255 / (i_max * 2), b * 255 / (i_max * 4)));

//bmp.SetPixel(x + x_max, y, Color.White);

}

>>59658

>>59747

Then save it:

bmp.Save("output.png");

>>58934

Quaternions?

%4?

What should I do with these?

I know if m%4==2 then its not a diff of 2 squares but otherwise it is. What do I do with this? Also someone in an earlier thread mentioned quaternions so I'm stuck on those

For negative x we must have:

x = d-a < 0

d-a < 0

d < a

(let d = a-t)

(a-t)(a-t) + e = ab

aa - 2ta + tt + e = ab

aa - (2t+b)a + (tt+e) = 0

(solve for a)

[-(2t+b) +- sqrt((2t+b)^2 - 4(1)(tt+e))]/2

= [-2t+b +- sqrt( 4t^2 + 4tb + b^2 - 4t^2+e)]/2

= [-2t+b +- sqrt(b^2 + 4tb + e)]/2

if(e=4t)

= [-2tb + b +- (b + 2t)]/2

= [-2tb + b + b + 2t]/2 OR [-2tb + b - b - 2t]/2

= [2b + 2t - 2tb]/2 OR [-2tb -2t]/2

= (b+t-tb) OR -t(b-1)

= b + t(1-b) or t(1-b)

(solve for t)

(a-t)(a-t) + e = ab

aa - 2ta + tt + e = ab

tt - 2at + (e+aa-ab) = 0

(quad)

t=(2a +- sqrt[ 4a^2 - 4(1)(e+aa-ab)])/2

t = (2a +- sqrt[ 4aa - 4e - 4aa + 4ab]/2

t = (2a +- sqrt[4ab-4e])/2

t = a +- sqrt[ab-e]

So for this we have:

a = b + t(1-b) or t(1-b)

t = a+- sqrt[ab-e]

(if its in cell (1,1) then we have e = 1) so

t = a +- sqrt[ab-1]

(this next part might not even be valid)

If we want to solve for t we can try pluggin in

t = a +- sqrt[ab-1]

t = (b+t(1-b) +- sqrt[ (b+t(1-b))b-1)]

t = (b + t - tb +- sqrt( (b+t-tb)b - 1))

t = b + t - tb +- sqrt( bb + tb - tbb - 1)

t - b - t + tb = +- sqrt (bb+tb-tbb-1)

b(t-1) = +- sqrt(bb+tb-tbb-1)

b(tt - 2t + 1) = bb + tb - tbb - 1

btt - tbt + b = bb + tb - tbb - 1

btt - btt + b - bb - tb + tbb + 1 = 0

b - bb - tb + tbb + 1 = 0

t(bb-b) = bb - b + 1

t = (bb-b+1)/(bb-b)

Closer?

>>53473

I can generate any element in a cell from just one along with their t, but only for the first row (n = 1)

Is there a trick I have missed for n > 1 ?

$ ./t.py {17:1:400:27:373:429}

4 {17:1:40:7:33:49}

5 {17:1:58:9:49:69}

6 {17:1:80:11:69:93}

7 {17:1:106:13:93:121}

8 {17:1:136:15:121:153}

9 {17:1:170:17:153:189}

10 {17:1:208:19:189:229}

11 {17:1:250:21:229:273}

12 {17:1:296:23:273:321}

13 {17:1:346:25:321:373}

14 {17:1:400:27:373:429}

15 {17:1:458:29:429:489}

16 {17:1:520:31:489:553}

17 {17:1:586:33:553:621}

18 {17:1:656:35:621:693}

19 {17:1:730:37:693:769}

20 {17:1:808:39:769:849}

21 {17:1:890:41:849:933}

22 {17:1:976:43:933:1021}

23 {17:1:1066:45:1021:1113}

24 {17:1:1160:47:1113:1209}

$

>>61184

Program.cs

using System;

namespace IntegerFactorisation

{

class Program

{

static void Main(string[] args)

{

TheEnd.CreateTheEnd();

TheEnd.Output();

}

}

}

Point clouds will be the death of me.

Seriously VQC, you already have some of this code if you're not LARPing, hook us up with what we need to render this shit properly I'm not working for weeks for free just to render this thing. You've put up no code beyond the original and you said you would TWICE.

I'm not here for RSA.

I'm not here for BTC.

I'm here for whatever the lesson is supposed to be.

>>59832

t = (bb-b-1)/(bb-b)

Jesus the last line was wrong (2nd to last line I had I didn't flip the 1)

Also if we put in t(1-b) instead of

t = a +- sqrt[ab-1]

t = (b+t(1-b) +- sqrt[ (b+t(1-b))b-1)]

t = (b + t - tb +- sqrt( (b+t-tb)b - 1))

t = b + t - tb +- sqrt( bb + tb - tbb - 1)

t - b - t + tb = +- sqrt (bb+tb-tbb-1)

b(t-1) = +- sqrt(bb+tb-tbb-1)

b(tt - 2t + 1) = bb + tb - tbb - 1

btt - tbt + b = bb + tb - tbb - 1

btt - btt + b - bb - tb + tbb + 1 = 0

b - bb - tb + tbb + 1 = 0

t(bb-b) = bb - b + 1

t = (bb-b+1)/(bb-b)

maybe solve for b,t in Z to find this?

t = a +- sqrt[ab-1]

t = a +- sqrt[a(t(1-b)) - 1]

t = a +- sqrt[a(t-tb)-1]

t = a +- sqrt[at-atb-1]

t-a = +- sqrt[at-atb-1]

(t-a)^2 = at-atb-1

tt-2ta+aa=at-atb-1

tt-3ta+atb+aa+1=0

tt+t(ab-3a) + aa+1 = 0

t = (1/2)[-(ab-3a) +- sqrt( (ab-3a)(ab-3a) - 4(1)(aa+1) )]

(1/2)[a(3-b) +- sqrt( aabb -6aab + 9aa - 4aa + 4 )]

(1/2)[a(3-b) +- sqrt(aabb - 6aab + 5aa + 4)]

aa(bb-6b+5) + 4

aa(b(b-5)+5)+4

t = (1/2) * [ a * (3-b) +/- sqrt(a*a*b*(b-1)

>>59747

Python script updated with the new image generation equation

https://pastebin.com/DdmwSsZx

>>61483

Have patience faggot. Without this guy you would still trust (((RSA)))

>>61483

And if you think this doesn't apply to you then you better stop using

—ANY SECURE WEBSITE

—BANKS

—CREDIT CARDS OR DEBIT

—EMAIL

—ANY MESSAGING SERVICE

Basically anything important.

>>61956

Crypto doesn't have to be fundamentally broken to be untrusted. Your 'random number generators' have been fucking you over plenty for decades.

>>61981

It's already that way bro, fundamental flaws don't need to exist for crypto to be effectively subverted. Look all those RSA hw generators that just 'oopsie' have a backdoor in them. HTTPS is a fucking joke, and subverted. Trusting 'trust' is not security my dude.

From an engineering perspective, if it can be done you can bet your ass somebody is doing it. Weak RNG, poisoned numbers, NIST/NSA collusion, bad implementations, timing analysis, other sidechannels, rubber hose, crypto's already done son. Your rig was designed to calculate, not be a secure extension of yourself.

pointless argument but don't think I don't already know.

Then you hear about 'wire tapping' in the news regarding the POTUS "wiretap" and nobody dares to mention it was probably 'The Hammer' which isn't a wiretap at all but an active program that exploits all manner of side channels, hardware exploits, and Julian Assange knows what else. To say it it a 'wiretap' is like comparing a match to the sun. Totally disingenuous!

Whatever. I'll quit bitching and keep making my point cloud.

Just noted that in some image you can see some pattern in 3D perspective

>>53432

Did anyone make progress on the big_n hint? If its simple then I'm apparently retarded

>>59055

Hi VQC! Been studying the chart. Even and odd values of variable e tie into the stable but different increases in variable x, which could be used to help generate index variable t for each element. Can you please confirm and maybe drop some crumbs? Also, what does "t= length of c in bits" mean?

>>22371

Seems like being able to create variable t is the key to creating ability to traverse any part of data set. Am I on the right track?

>>62510

In the original VQC, t = 12. It is represented by set size. Notice how there is more than one value for each e,n? It's because the t variable makes it calculate more than one. There are infinite entries in each e,n that has elements. He is stating that t = the length of c in bits.

>>58878

pls halp

>>62510

>>53673

>for odd e in the first row, x = (t+1)/2

t = x / 2 + 1

>>62607

Add half the width of your image to both e and f when drawing (x + 1024 in image coordinates if its 2048 wide)

>>62510

> Also, what does "t= length of c in bits" mean?

if t = 12, it's binary represenation will be (1100) base 2, so in this case length of t in bits will be 4 (eg you need four bits to code its internal representation)

>>62655

replace 't' with 'c', for c=12, t=4

>>62908

Wow, beautiful! Damn!

Legit question. Doesn't a free energy device totally defeat energy trade? Why would Donald Trump put so much time into energy trade and energy independence if there's a free energy device?

>>63004

I would imagine, if true, that Trump would have to tread carefully with this one. Imagine how the energy industry and corporations would take to this sort of news? They would be put out of business and the stock market would take a hit.

>>63031

You're right. I'm just trying to decide if I should continue on with this because it's worthwhile or if it's a math black hole and I'm too stupid to know it because my math is weak. It'd be nice to have some kind of proof, really anything, that any part of what VQC claims is possible. It seems like our resident newfriends are getting lost in the Q stuff which I know most of us here could help with but aren't because we're tired of talking to a wall and this is more of building the future instead of healing the past which most of us would probably enjoy considering PG is awful.

VQC, I'm stuck between helping my President and my countrymen or taking them to the next level by working on this. If this is real I know I'd rather be doing this, if it's fake, I'm not doing all I can by helping meme some shit. What do my dude? Can you offer any proof what you claim is true?

>>63082

Easy to get burnt out with this stuff, evident by the lack of activity here compared to earlier. If the math is just a distraction its the most elaborate one I've ever seen, you cannot create something like this by accident or just by trying to fake it

I'm also not that interested in the RSA applications but it is a great, irrefutable way to prove the equations are real and get more minds working on it

I'm here just for the knowledge, if it pans out or not is irrelevant. That said, the hints are getting tedious and we need some tangible progress