/vqc/ - Virtual Quantum Computer

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

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

>>4155

Can you elaborate? Iterating the tree in the cell?

>>4160

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

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

d_1 = d_0^2 + e

d_2 = d_1^2 + e

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

>>4168

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

BAKER!

FFS!

YOU HAD PUN JOB TO DO!!!

https://www.reddit.com/r/Sax/

NEEEEEEERDS!

WHERE IS YA?!

https://youtu.be/gZvizWDewV0

>>4144

baker - thank you. I put together "test cases" for all the unsolved rsa numbers. This project is mind boggling.

>>4191

Working through some tree ideas. Nothing to share yet.

>>4191

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

Hey all - I'm around too. No progress to report either.

;)

>>4194

>>4192

>>4198

>>4193

Likewise with the lack of ideas.

>>4200

I'll go smoke a toke for those digits!

*can't think of anyone* Damn typo!

>>4200

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

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

>>4204

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

Here's my High School Topol LARP:

>Hey Guise, I'm Topol

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

>I got high during art class in the back room

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

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

>I love truth and I seek it out

>Wanna hang out?

>>4204

https:/ /hooktube.com/watch?v=Ug_v8HNgGf8

>>4206

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

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

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

>>4210

>>4208

Make one for me, Topol!

>High School VA

Anyone else noticed that (x+n)^2=f in (e,1)?

>>4213

Under what conditions?

>>4212

Q said they say a prayer for him every day in the oval office.

I still check daily, but I haven't had any spare time to work on it.

>>4213

Appears to check out.

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

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

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

What does this look like?

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

All odd squares = 1 + 8T

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

Visualise.

2d(n-1)

Why is (n-1) important?

f = 2d + 1 - e

Odd (x+n) - Examples

Let's use some real examples.

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

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

Homework.

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

We will use that one.

>>4140

>>4213

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

>>4219

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

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

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

>>4222

>>4223

>>4225

Afternoon!

Should be some fun times coming up!

>>4210

The legendary artist known as Topolanon.

Always a pleasure!

Hint:

Divide c by four.

The remainder tells you whether (x+n) is odd or even.

0,1 = even

2,3 = odd

Hint (update):

That should be, divide e by 4

>>4230

From the grid, it can be seen that columns follow a pattern mod 4. 0,1,2,3;0,1,2,3;..

(x+n): even,even,odd,odd;..

This pattern in respect of the difference of two squares where c is odd.

>>4221

First, welcome back!

I'm very excited for today!

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

>>4235

When computing the remainder of 4 for e RSA 490 we end up with 3, which means it has an odd (x + n).

>>4235

Thank you!

>>4236

Sounds good.

We'll walk through RSA 100

and then RSA 490

The factor tree is used to factor d and e.

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

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

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

>>4239

Welcome back. And thank you. Will get caught up soon.

>>4239

Good to see you VQC! Will get to work on the new hints.

>>4240

>>4241

Thanks!

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

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

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

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

>>4246

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

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

Maybe VQC can clarify.

>>4239

I think I found a way to generate primes. Unconfirmed but I hope you get your hopes up

>>4250

I think this only applies to (e,1).

>>4251

Teach - I was hoping you’d draw up something like this. Awesome!

>>4252

CA - all ears!!

>>4254

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

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

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

>>4252

That in an of itself would be a great result.

>>4257

Also it looks A LOT like our grid but shifted the other way

>>4252

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

>>4254

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

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

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

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

Example c=145, in base 2 = 10010001

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

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

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

c = 8(2(8(△1) + 1)) + 1

>>4257

Typed out my previous comment before I saw this.

>>4251

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

145 = 12^2 + 1

= 4*6^2 + 1

=4^2*3^2 + 1

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

>>4255

Unless this applies only to a=1, b=c.

Could you double check that PMA?

>>4261

See my latest post :)

But you do have an interesting idea here…

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

I like that thought.

>>4261

= 4*4*3*3 + 1

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

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

= 10*6 + 10*3 + 6*6 + 6*3 + 1

= 60 + 30 + 36 + 18 + 1

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

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

= 18(5+3) + 1

Something like this process for the factoring 145.

>>4263

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

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

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

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

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

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

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

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

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

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

>>4265

The more I think about it the more this just intuitively follows from n(n+1)/2

>>4266

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

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

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

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

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

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

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

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

^proof this is bunk

>>4262

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

>>4259

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

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

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

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

>>4263

>>4268

Hey Lads! I’m following along, got some family stuff to do all day today. Excited to get working on the new hints later.

>>4267

Trying to analyze this more, in combination with the thoughts in this image from my earlier post:

>>3915

>>4268

Don't forget that △0 = 0.

1^2 = 8.△0 + 1

3^2 = 8.△1 + 1

etc

>>4271

Hey hey VA!

>>4246

Look our fave 145: {1:5:12:7:5:29}

145 / 4 = 36 R1

So x + n should be even, and 7 + 5 = 12, so yep, it is. I think this property is only true for RSA numbers, i.e., product of two coprimes

>>4273

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

>>4273

But VQC corrected himself and changed c mod 4 to e mod 4.

>>4275

Ah Jesus fuck this forum is such a pain in the ass to read.

I guess it still holds true for 145, e = 1, 1 / 4 = 0 R1, x + n = 12, which is even

Not all numbers are a difference of squares.

Sorry. So, I think it doesn't seem to hold in some instances because it is for semiprimes.

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

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

12 = x+n, even

5 = n

7 = x

5 + 7 = 12, odd + odd = even

>>4285

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

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

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

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

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

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

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

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

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

>>4285

Also, that equation can be derived from our existing equations.

Its significance however, is still unknown to me.

>>4288

>>4287

>>4286

Forgot my trip. Ok, another question here. You guise have any insight into this yet?

>Visualise.

>2d(n-1)

>Why is (n-1) important?

>>4289

See the 2nd half of >>4251

>>4290

Thanks Teach!

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

>>4287

You just geometrically proved that 8 triangle nums plus 1 make an odd square

>>4293

See >>4251 & >>4272 from much earlier today.

c = 2d + 2n - 1 for a = 1

>>4286

>>4291

Thanks Baker!

>>4293

I agree. Nice job Teach!

>>4295

>>4296

I'm trying to visualize 2d(n-1).

The best way I can so far is 2 rectangles of (n-1) * d.

I'm trying to turn that into an image now.

>>4283

grid patterns can't just be for semiprimes

>>4276

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

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

>>4281

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

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

thanks to PMA for that breakdown here >>4246

e mod 4 : 0, 1, 2, 3

x+n : odd, even, odd, even

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

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

>>4291

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

odd + odd = even

odd + even = odd

even + even = even

odd^2=odd

even^2=even

odd*even=even

odd*odd=odd

even*even=even

odd-even=odd

even-odd=odd

odd-odd=even

even-even=even

>>4300

as a follow up, when c is odd:

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

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

>>4285

This

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

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

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

The process for producing the first algorithm is clearly ancient.

>>4303

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

/// <summary>

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

/// </summary>

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

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

public static BigInteger Triangle(BigInteger base_t)

{

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

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

return BigInteger.Divide(square_side, two);

}

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

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

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

{

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

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

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

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

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

}

The square root method referenced above:

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

public static BigInteger Sqrt(this BigInteger number)

{

BigInteger n = 0, p = 0;

if (number == BigInteger.Zero)

{

return BigInteger.Zero;

}

var high = number >> 1;

var low = BigInteger.Zero;

while (high > low + 1)

{

n = (high + low) >> 1;

p = n * n;

if (number < p)

{

high = n;

}

else if (number > p)

{

low = n;

}

else

{

break;

}

}

return number == p ? n : low;

}

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

Here is the square root of RSA 2048

158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

And the remainder of RSA 2048

149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

And RSA 2048 itself:

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

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

They are a bit shorter!

public static string Rsa100c =

"1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139";

public static string Rsa100a = "37975227936943673922808872755445627854565536638199";

public static string Rsa100b = "40094690950920881030683735292761468389214899724061";

public static string Rsa100d = "39020571855401265512289573339484371018905006900194";

public static string Rsa100e = "61218444075812733697456051513875809617598014768503";

public static string Rsa100f = "16822699634989797327123095165092932420211999031886";

public static string Rsa100n = "14387588531011964456730684619177102985211280936";

public static string Rsa100x = "1045343918457591589480700584038743164339470261995";

public static string Rsa100x_plus_n = "1059731506988603553937431268657920267324681542931";

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

(Java) (Fixed a dependency)

Works fine for every number I've ever tested.

 public static BigInteger sqrt(BigInteger i) {
        BigInteger zero = BigInteger.ZERO;
        BigInteger one = BigInteger.ONE;
        BigInteger two = BigInteger.valueOf(2);
        BigInteger n = zero;
        BigInteger p = zero;

        if (i.equals(zero)) {
            return  zero;
        } else if (i.equals(one)) {
            return one;
        } else if (i.equals(two)) {
            return one;
        } else if (i.equals(BigInteger.valueOf(3))) {
            return one;
        } else if (i.equals(BigInteger.valueOf(4))) {
            return two;
        } else if (i.equals(BigInteger.valueOf(5))) {
            return two;
        }

        BigInteger high = i.shiftRight(1);
        BigInteger low = zero;

        //high > low + 1
        while (high.compareTo(low.add(one)) == 1) {
            //n = (high + low) >> 1;
            n = (high.add(low)).shiftRight(1);
            p = n.multiply(n);

            int result = i.compareTo(p);
            if (result == -1) {
                high = n;
            } else if (result == 1) {
                low = n;
            } else {
                break;
            }
        }

        if (i.equals(p)) {
            return n;
        } else {
            return low;
        }
    }

>>4312 in reply to >>4306

>>4303

Perfect choice of words. There are no coincedences, only relevant & irrelevant patterns.

The image above shows 8 triangles of equal size plus a unit square.

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

>>4323

Indeed

>>4320

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

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

>>4322

f = 2d - 1 + e

For RSA 100:

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

=

16822699634989797327123095165092932420211999031886

Correction:

f = 2d + 1 - e

>>4326

Absolutely

>>4328

In the diagram of 8 triangles + 1, the +1 is donated by f.

16822699634989797327123095165092932420211999031885 + 1

From the equation:

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

The value f donates another one unit to give:

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

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

f-2 = 16822699634989797327123095165092932420211999031884

f-2 is going to form part of each of the eight triangles.

f divided by 8 = 2102837454373724665890386895636616552526499878985

f mod 8 = 4

>>4331

>>4332

Following 100%

Guys, we have a master here at work. This could redefine how humans interact…

We don't currently know (x+n) or n.

Since f can be divided by 5, we can make a base for each triangle which is made of (f/40), still with 4 left over.

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

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

{

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

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

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

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

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

}

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

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

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

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

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

>>4336

So, we have the 8 bases of our triangles made of f/40. We know that there are four left over. We then know that if the base of f/40 is too small, then the number of 2d to add must have the same left over on the last d to create the final base of each triangle. THIS is key.

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

>>4334

>Since f can be divided by 5, we can make a base for each triangle which is made of (f/40), still with 4 left over.

What if it weren't an easy 5 but some horrible divisibility problem? What then?

>>4338

Good question.

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

The remainder from each of those 8 bases would be added to the four left over from dividing f-2 into 8.

>>4338

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

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

We will add a method to calculate what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d.

That gap will allow us to use more geometry of triangular numbers to establish what multiple of 2d could be added to make our eight triangles.

That geometry with multiples of 2d will give us the solution, if it exists (which in our example we know it does). If doubling the width of the chunks of (f-2) didn't give us a solution, the number c would be prime, the worst case (doubling the width of (f-2) chunks very quickly determines this as the number of times it doubles is logarithmic in the length of (f-2)).

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

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

>>4303

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

>>4326

Well said, Anon.

>>4344

>>4343

Thank You VQC for all the new hints, info, and code! We are all excited to be here working on this challenge. It's exciting when you stop by and drop knowledge bombs on us!

>>4332

>f mod 8 = 4

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

>>4342

>>4343

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

>>4348

Checks out here too, always good to have your screenshots to debug code against! Heres a python example for those wanting to play along

from gmpy2 import mpz, isqrt, isqrt_rem, t_div_2exp

def tri(bs):
  return t_div_2exp(bs ** 2 + bs, 1)

def tri_n_odd(bs, c, d):
  return isqrt(tri(bs) * 8 + 1 + c) - d

c = mpz(1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139)
d, e = isqrt_rem(c)

x = mpz(1045343918457591589480700584038743164339470261995)
n = mpz(14387588531011964456730684619177102985211280936)

print(n)
print(tri_n_odd((x + n) // 2, c, d))

>>4351

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

The fixed point decimal module is annoying for what we're doing but it'll give you arbitrary precision square roots when needed. Its also much slower and uglier as the precision has to be specified beforehand, and using the length of c is probably overkill since we're rounding back to integers but whatever.

import decimal

def sqrt(n):
  return int(decimal.Decimal(n).sqrt().quantize(1))

def tri(bs):
  return (bs ** 2 + bs) // 2

def tri_n_odd(bs, c, d):
  return sqrt(tri(bs) * 8 + 1 + c) - d

c = 1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139

ctx = decimal.getcontext()
ctx.rounding = decimal.ROUND_FLOOR
ctx.prec = len(str(c))

d = sqrt(c)
e = c - (d ** 2)

x = 1045343918457591589480700584038743164339470261995
n = 14387588531011964456730684619177102985211280936

print(n)
print(tri_n_odd((x + n) // 2, c, d))

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

>>4346

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

>>4347

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

>>4355

You just missed it! We're on #421 now.

>>4346

Right? Also, there can never be enough programmers!

>>4355

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

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

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

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

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

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

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

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

>>4357

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

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

>>4357

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

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

>>4358

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

>>4359

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

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

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

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

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

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

>>4353

4,10,20…

>>4362

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

>>4362

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

>>4359

>>4360

>>4364

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

>>4360

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

>>4367

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

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

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

>>4371

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

>>4366

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

>>4372

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

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

>>4373

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

>>4370

You sir, have received a gift from your time here. An education is someone no one can ever take away. Very few things are like that.

>>4374

Again, I'd like to point out that I'm about as anonymous as the Sun. I'll happily voiceover whatever we need to do.

https:// youtu.be/GV95vo4CPRY

>>4376

pretty sweet swordplay!

>>4376

Cool as shit.. so the fire poi video earlier is you? Awesome

>>4377

'Preciatcha

>>4378

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

>>4379

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

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

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

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

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

>He threatened to fail my project.

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

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

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

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

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

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

>Wanna join or help? All ideas welcome!

>>4382

Great stuff. Since your programming skills hacked my tripcode already I'm gonna go ahead and post a new one… mm'kay

>>4401

I've always put it:

MEGA

Make Everything Great Again

Hey guys!

Thanks VQC for the guidance!

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

>>4376

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

>>4382

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

>>4407

This is cool CA! Multidimensional Triangles!

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

>>4337

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

a = 3

b = 29

c = 87

d = 9

e = 6

f = 13

(f-2)/8 = 1

(f-2)%8 = 3

x = 6

n = 7

xpn = 12

half_xpn = 6

test_correct_base = 7

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

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

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

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

>>4334

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

Comments on my thinking guys?

>>4402

Great video Topol! I'll check out more of your vids!

>>4412

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

>>4412

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

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

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

Chapter and verse: >>4339

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

Ponder……. OK, choice may be arbitrary.

Ponder………. Now read: >>4340

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

Hmm

>The objective is to find a base larger than n and smaller than x+n at this stage.

So 5 is a "base". The objective I think is analogous to Newton's method; you need an initial guess to get the process started. The process will transform our initial guess to be n < HERE < x + n

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

Does this make any sense?

>>4426

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

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

>>4360

Name is changeable - open to suggestions. Sorry for ruining it ;)

>>4423

>>4426

Ok, I think I'm getting it. Also, no need for disclaimers here. Thank you for your insights, they are very helpful.

>>4360

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

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

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

>>4429

How would your code identify you? Is there anything other than filepaths that you'd have to delete? Unless you're worried about the mannerisms in your code.

>>4430

>How would your code identify you?

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

Look what happened to this guy!

>>4430

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

I could host the git server on:

1) my home server - linked to my ip

2) my amazon server - linked to my cc

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

Any suggestions?

>>4426

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

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

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

>>4432

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

>>4433

I'm talking my whole project, hundreds of files. All run different tests and produce different outputs.

>>4427

Revelation 5:5

>>4434

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

>>4432

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

Looks that way to me.

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

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

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

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

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

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

>We will add a method to calculate what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d.

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

>>4436

For posterity:

>Revelation 5:5

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

>>4440

Fucking retards didn't know

>He is able to open the scroll and its seven seals

ITS EIGHT TRIANGLE NUMBERS YOU STUPID FUCKS

>>4437

The second I have something small thats worth sharing I will. Otherwise, I'll upload my work to dropbox or something - thanks.

>>4439

Thanks, I'm going to head to bed to think about this more, and hopefully dream of triangles.

Hey Lads! I'm here working on the new VQC crumbs. Anyone else here yet?

>>4454

>>4456

Howdy Topol!

>>4459

>>4460

It's true, that's why I don't use a trip. We are all Topolanon. So are you, VA.

>>4457

VA - good job!

Just reading and rereading crumbs, considering where I can be helpful next. So much to process.

>>4463

>>4462

>>4461

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

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

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

*drops mic*

>>4465

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

>how many have you had to weed out so far

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

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

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

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

>>4464

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

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

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

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

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

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

>>4467

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

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

>>4475

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

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

-brohoof-

Semiprimes are bae.

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

>>4479

"Topol… why in fuck would you even be paying attention to that?"

Don't worry about the details.

>>4478

Yeah, I remember the bit about pi as well. That and the grid structure really got me thinking.

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

>>4334

>>4335

>>4336

>>4337

>>4339

>>4342

>>4343

>>4344

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

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

>>4473

I agree Topol! We will show the Path to all who want to learn it.

>>4482

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

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

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

>>4346

>>4350

>>4412

Thanks!

>>4358

We will get to the sono experiments.

>>4407

Coincidence? ;)

>>4426

Base in that context referred to the base of a triangle

>>4467

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

>>4482

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

>>4485

Thank you, that's great.

>>4407

>>4489

Should I try and extend this into the negative?

>>4492

if you need help with the image, just think about Heisenberg.

>>4494

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

Totally … not…

-sips just made coffee cuz morning shift apparently-

>>4444

>I hereby pronounce this day the day of 4

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

>>4498

Nice work Isee! Very clear graphics and ideas. I'm working to understand over here too. Thanks for walking through an actual grid example!

>>4498

Great walkthrough.

n0 = 1

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

+1 from the origin 8Tu + 1

total: 6

Is this the n we are looking for?

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

>>4500

>fit GDC

gcd…

>>4500

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

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

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

>>4502

I'm writing some test cases now. I believe this only works for odd (x+n) and even n.

>>4503

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

>>4504

I use spreadsheets and just colour the squares.

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

>>4500

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

Here you will have the following:

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

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

c = 1159

e = 3

d = 34

f = 66

f2 = 64

f2 % 8 = 0

f2/8 = 8

n0 = 4

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

>>4497

Funny that you'd say that. They say there have been 4 great awakenings.

>>4504

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

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

Anyone here yet? Seems like we all have free time beginning around this point.

>>4514

I'm lurking but I'm a bit busy for the next hour or two. How's it going?

>>4515

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

>>4504

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

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

>>4506

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

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

For n0=4 in your example:

n0*n0 + n0 + 1 = x+n

4*4 + 4 + 1 = 21.

The 2d value would be too large.

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

>>4343

>>4344

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

>We will add a method to calculate what would be missing from the triangles if we could only fill them with n0 squared and multiples of 2d.

>That gap will allow us to use more geometry of triangular numbers to establish what multiple of 2d could be added to make our eight triangles.

>That geometry with multiples of 2d will give us the solution, if it exists (which in our example we know it does). If doubling the width of the chunks of (f-2) didn't give us a solution, the number c would be prime, the worst case (doubling the width of (f-2) chunks very quickly determines this as the number of times it doubles is logarithmic in the length of (f-2)).

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

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

>>4518

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

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

>>4517

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

>>4518

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

>>4519

>Recursively grow until we find a match?

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

n0 is the triangle seed for the c var tree!

>>4522

The factor tree has to play into this somehow.