/vqc/ - Virtual Quantum Computer

>>1683

Agreed, Baker. No love for famefags. I'm just tripping out that Q is going mainstream. My lady is talking with friends on FB about who Q is, ie Barron, Don Jr, etc. WTF!!! When 60 million eyes get put on Q's posts, the world will redpill itself. AJ is talking openly about Q, chans, etc. Time to FIRE UP THE MEME CANNONS, LADS! I can't believe I'm alive to be seeing this. WHEN we solve this (with Senpai help, of course! Thx Senpai) we are going to help fundamentally change the course of human history. We have green light to go mainstream and fucking blast FB, Twatter, et al. I've been waiting for this moment my whole life, never thought it could really happen. Truth and Justice. Make Earth Great Again!

>>1684

Whoops I thought you were chirping us baker sorry I misinterpreted who you were calling famefags.

>>1684

Yeah, thanks Baker! Can we get some more Victory ponies up in this??

>>1684

I'm not like you. I can't synthesize different areas of maths. I don't know what parabola's and roots are. I haven't had to brush up on them for this.

>>1685

Shame about Baruch. I guess they were right about him. I was in contact with him too.

>>1685

Indeed VA, just waiting for those Gitmo announcements! And for MSM to completely crumble.

Ok fags, here are some test cases and challenges for you:

RSA Test Cases and Challenges

https://pastebin.com/f5Km9K8H

RSA-100 (c) =

1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139

(100 decimal digits, 330 bits)

RSA-100 (a x b) =

37975227936943673922808872755445627854565536638199

×

40094690950920881030683735292761468389214899724061

(2 50 decimal primes)

RSA-100 (c^2) =

231832607105497859450453971149782249091397152352980773038383827802126335737279486733598563928061275517976566803851888280780332815207688528004235456616576727168746464977139224636030180916664018563207

(198 decimal digits)

–

RSA-129 (c) =

114381625757888867669235779976146612010218296721242362562561842935706935245733897830597123563958705058989075147599290026879543541

(129 decimal digits, 426 bits)

RSA-129 (a x b) =

3490529510847650949147849619903898133417764638493387843990820577

×

32769132993266709549961988190834461413177642967992942539798288533

(with a 64 and 65 decimal prime)

RSA-129 (c^2) =

13083156311017746081229837780584604971789185572322940113554957177463495238823718363789678158885976884237362181609103528581150260013472809482392451812847672930017596579807933662233072087386335238374806414971531141586419217198046250532235073881640972514818681

(257 decimal digits)

>>1689

RSA-190 (c) =

1907556405060696491061450432646028861081179759533184460647975622318915025587184175754054976155121593293492260464152630093238509246603207417124726121580858185985938946945490481721756401423481

(190 decimal digits & 629 bits)

RSA-190 (a x b) =

31711952576901527094851712897404759298051473160294503277847619278327936427981256542415724309619

×

60152600204445616415876416855266761832435433594718110725997638280836157040460481625355619404899

(with (2) 95 decimal primes)

RSA-190 (c^2) =

3638771438488087985574274106795079635808387100084863940611975142846520895542297425531641157043983476619993374723043826429566267790036775257657139728945914322148154407595065259459214885711856214268244428701532653022789206595392004837865259854410326840170456159370131034294156696408886672161241549375876666598321636136763746229426218477864974486848489396416573764495046083098157361

(379 decimal digits)

–

RSA-640 (c) = 3107418240490043721350750035888567930037346022842727545720161948823206440518081504556346829671723286782437916272838033415471073108501919548529007337724822783525742386454014691736602477652346609

(193 decimal digits, 640 bits)

RSA-640 (a x b) =

1634733645809253848443133883865090859841783670033092312181110852389333100104508151212118167511579

×

1900871281664822113126851573935413975471896789968515493666638539088027103802104498957191261465571

(with (2) 97 decimal primes)

RSA-640 (c^2) =

9656048121330239196485739047236872733132646152415265572749839063571416858087801139297436736668535657942544532834166699749756556831546825766219543494478915395505001388433032706122580727561545005466558578465273868089134810944324056030967838960018632090507727101247760897728632535169583227345646104139213934877983671194311248207350399320919897813580637534250460303641423529457084273798881

(385 decimal digits)

>>1690

>>1689

Thanks for putting those together. Wikipedia has down syndrome so when I try to copy the RSA numbers it inserts a bunch of newlines.

>>1689

>>1690

And for some challenges, find a & b:

RSA-240 (c) =

124620366781718784065835044608106590434820374651678805754818788883289666801188210855036039570272508747509864768438458621054865537970253930571891217684318286362846948405301614416430468066875699415246993185704183030512549594371372159029236099

(240 decimal digits & 795 bits)

–

RSA-1024 (c) =

135066410865995223349603216278805969938881475605667027524485143851526510604859533833940287150571909441798207282164471551373680419703964191743046496589274256239341020864383202110372958725762358509643110564073501508187510676594629205563685529475213500852879416377328533906109750544334999811150056977236890927563

(309 decimal digits & 1,024 bits)

RSA-2048 =

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

(617 decimal digits & 2,048 bits)

https://pastebin.com/f5Km9K8H

Godspeed PATHanons.

>>1689

PMA?

>>1693

mr. spice and everything nice.

>>1694

We gotta get to n from c. Wanna hop a Sandworm and cruise on over? ;)

Is anyone in touch with Inner/Outer Heaven?

I have a fresh angle to explore tomorrow, which has been looking me the face this entire time.

>>1696

Do tell, MA. Inner/outer Heaven? What’s your intuition saying?

>>1696

>Is anyone in touch with Inner/Outer Heaven?

>>1697

>Do tell, MA. Inner/outer Heaven? What’s your intuition saying?

It appears VA has outed himself as a newfag?

Ignore your squabbles.

You can watch the entirety of Star Trek: The Next Generation in the Q episodes alone.

VQC isn't "Albino Morpheus".

He's PICARD.

AS A Q!

I'm not even done with the episode, i've been through voyager… DS9 (999) is the ending.

Calling it now.

>>1700

oh this is getting gooooooooood….

>>1701

pffffffffft

As if that wasn't already commented on where I'm seen elsewhere.

Love this.

Oh fuck man I found a very important pattern. I need a minute.

>>1703

Share you faggot. We make progress by banging our heads on this wall together.

>>1704

Do the words "I need a minute" mean nothing to you?

>>1704

FINE!

Linear progression applies, but if you have "the time"…

Tapestry:

Voyager

TNG

DS9

<;3=Q

>>1709

Lowest line

R121C52:R277C130

(121, 52) to (277, 130)

Rise is 156, run is 78, gradient is 2 (6/3)

R133C58:R265C124

(133, 58) to (265, 124)

Rise is 132, run is 66, gradient is 2 (6/3)

R337C160:R457C220

(337, 160) to (457, 220)

Rise is 120, run is 60, gradient is 2 (6/3)

Next line upwards

R169C116:R217C156

(169, 116) to (217, 156)

Rise is 48, run is 40, gradient is 1.2 (6/5)

R253C186:R289C216

(253, 186) to (289, 216)

Rise is 36, run is 30, gradient is 1.2 (6/5)

R289C216:R313C236

(289, 216) to (313, 236)

Rise is 24, run is 20, gradient is 1.2 (6/5)

Next line upwards

R217C204:R229C218

(217, 204) to (229, 218)

Rise is 12, run is 14, gradient is 6/7 (0.857142 repeat)

R229C218:R265C260

(229, 218) to (265, 260)

Rise is 36, run is 42, gradient is 6/7

R265C260:R301C302

(265, 260) to (301, 302)

Rise is 36, run is 42, gradient is 6/7

Next line upwards

R181C210:R205C254

(181, 210) to (205, 254)

Rise is 24, run is 44, gradient is 6/11 (0.545454545454)

Next line up

R181C222:R193C248

(181, 222) to (193, 248)

Rise is 12, run is 26, gradient is 6/13 (0.461538 repeat)

Next line up

R205C290:R217C324

(205, 290) to (217, 324)

Rise is 12, run is 34, gradient is 6/17

>>1601

Thanks, didn't notice until now. Will use those.

Oven is HOT.

>>1709

>>1710

I did the next one too and it was 6/19. The thing you should all be taking away from this is that the denominators are all primes, and from the lowest line to the highest line they increase in order with 6 as the constant numerator.

>>1711

Vibe=Felt

Happening=It's

https://youtu.be/Qk449uj2jgU

>>1713

>>1713

>>1713

>>1713

>>1713

>>1713

Migrate when this thread is over. Good night.

oooh hoo hooo

This feels good, too

https://youtu.be/FJo6ipm7HqY

>>1712

I didn't realize until now but setSize was 12. That means that while those gradient rules still work the numbers are slightly different. I'm currently trying to figure out the mathematical relationship between each line that determines where the next one starts each time.

Tested!

For all records {0, n, c, c-aa, aa, bb}

A Goal record!

There is m such that n=2mm

and …

b-a = 2m

{0, 24200, 669, 660, 9, 49729)

m=110

b=223 a=3 a - b = 2m = 220

{0, 8, 77, 28, 49, 121}

m=2

b=11 a=7 a - b = 2m =4

Same as what I've been doing for constant D values (because for C record we already have correct D, so this could help us navigate through the D's:

pastebin. com/5cd4k2rz

Graph. (+ = from the end, *= extrapolated from pattern).

For any d, we can start off and say there will be an entry at

(e,n) = (-d^2,-d+1)

We can call this our origin cell for D. Now, to navigate along this graph we would need to generate cells from their (e,n,d) values, which I do not think we can do yet. However this pattern is still notable.

Starting at our origin cell (the Zeroth origin cell), (e,n) I THINK we can generate any cell for this pattern:

(e0,n0) -> (e0,n0+u) for any u [I'll go 70% on this step, cuz I can't generate to test, the other ones are pretty certain though]

Then, from our origin cell, if we go to the right one cell we get a second origin cell (origin cell 1). (e0,n0) -> (e0+1,n0) = (e1,n1) From this cell we can do

(e1,n1) -> (e1+2u, n1+u) to navigate down the line as many times as we want

Then, if we go down the line once (e1+n1)->(e1+2,n1+1) and then shift to the right (e1+2+1, n1+1) we get the third origin cell (e2,n2).

From the third origin cell, we can do:

(e2,n2) -> ( e2 + 4u, n2+u ) to generate more cells for any u.

Generally:

The zeroth origin cell for D can be found at:

(-d^2, 1-d).

For the zth origin cell, (e*,n*) we can generate any other entry with the same d by this equation:

(e*,n*) -> (e* + 2zu, n* + u)

Also, from any origin cell we can generate the next origin cell.

( z to z+1), which boils down to going down the line once and then adding 1 to e.

(e*,n*) -> (e* + 2z + 1, n* + 1)

So now we have a method to determine all the entries (e,n) with the same d. Also, this may be an impossible task, but if anyone can generate an entry just given the (e,n,d) values that would be sweet, but I don't think we can.

This may be the biggest thing I've discovered. This path for valid d values is the same for every d. As in, the only difference between entries for certain d values is the zero origin for d. Other than that you use the same exact steps to figure out what other cells have the same d. Knowing that, if we have a valid d, we can say, WITH SUPREME CONFIDENCE, that.

origin cell is at (-d^2, 1-d)

the origin cell for d-1 is at ( - (d-1)^2, 1-(d-1)) = (-dd +2d -1, -d) (same formula)

So if we have a cell we know for certain we can navigate to a valid cell by:

(e,n,d,x,a,b) -> (e+2d-1, n-1, d-1, x?, a?, b?)

to go down a d value.

>>1725

>>1725

The pattern for the origin cells is more consistent if the zero origin is calculated at (-d^2, -d) because that way you can still do the same pattern (go down the line once, then shift e by 1) to get the next origin cell. Before it was that pattern for all origins not equal to 0, now it is consistent.

>>1725

>Also, this may be an impossible task, but if anyone can generate an entry just given the (e,n,d) values that would be sweet, but I don't think we can.

public static TheEndRecord CreateForEND( BigInteger e, BigInteger n, BigInteger d ) {

BigInteger c = d * d + e;

BigInteger x = ( Lib.Sqrt( ( d + n ) * ( d + n ) - c ) ) - n;

BigInteger a = d - x;

BigInteger b = a + 2 * x + 2 * n;

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

}

>>1727

*holy fuck thanks for this code i would have been at a dead end without you*

a bit of output

>((-36, 0, 6, 6, 0, 12, 0), 'a=0')

>((-36, 1, 6, 6, 0, 14, 0), 'a=0')

>((-36, 2, 6, 6, 0, 16, 0), 'a=0')

>((-36, 3, 6, 6, 0, 18, 0), 'a=0')

>((-36, 4, 6, 6, 0, 20, 0), 'a=0')

>((-36, 5, 6, 6, 0, 22, 0), 'a=0')

Okay so I tested this code and and it only gave me values of A where A = 0. I used to have that a=0 meant an invalid cell, but heres the thing: For A=0, you can have ANY value for (e,n,d) that you want and they all could still be valid entries right? Like 0 is a number so why couldn't it be in entry a? Maybe these are all valid entries.

Suppose they are, our output from AB and C fun

C(21) = (5,7,4,3,1,21,21) = start

AB(3,7) = (5,1,4,1,3,7,21) = goal

So for this we have the correct E and D right off the bat. We may be able to iterate through these n values and search things from there, because these are valid cells, but we cannot do END(5,7,4) because I think it takes root of -1. But for this remember we have the correct D. So we can go to the (-d^2) column, which is basically the root for d ((-d^2,-d) would be the seed I guess). From that column, we can expand out and generate any cell for the same (correct) d. We could do this and iterate through the cells from each D origin cell and go down until we either hit or pass e. Then I think we may have our correct cell at that point.

Tada!

>>1728

Iterating to find the solution isn't going to work.

See a couple of my previous posts:

>>1342

>>1487

>>1509

VA - thanks for reposting this. Am thinking this through a bit more.

We've been one variable or formula away from a solution for some time now.

The "Use Y on C at X" from >>>/cbts/75435 needs more attention.

I think we know what C and X are. And we know how to create a small square from t. What is Y? y axis? is that n?

CA got me thinking a bit more about the END formula.

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

This can be solved a couple of ways. And given that we know c, d, and e for our target aa*bb and a*b records, perhaps if we can identify just the small square movement, we can solve for n.

For example:

Solving for x given n, d and c.

x = ( Lib.Sqrt( ( d + n ) * ( d + n ) - c ) ) - n;

Solving for n given e, d, and (x+n):

c = d*d + e;

n = Lib.Sqrt( c - (x+n)^2 ) - d

Maybe all we have to do is figure out the "small square" movement?

>>1729

When you're talking about the small square and the big square, do you mean X+N and D+N?

>>1730

yes. I'm focusing specifically on the small square (x+n). It's relationship to t is simple. And it remains constant when you change e at (e,1,t).

>>1731

forgot trip.

Also, if anyone can help explain "Use Y on C at X".

Hey guys, its been a while between posts, but I've been following along with the progress.

I've been focused on my own theories mostly, and none have been fruitful.

>>1731

Hey PMA, the size of the small square has been where my attention has been lately.

I've also been spending some time going through reading for ECC.

I want to point out something that I've realized too…

n = (x^2 + e) / 2(d - x)

This is just a simple rewrite of x^2+e=2na.

But what I like about this representation, is that d and e are constants and so we have a relationship between n and x for any c.

I'm trying to use this equation to figure out the possible sizes of the small square.

Anyway, how is everyone?

Also welcome back CollegeAnon! How were your exams?

>>1731

>>1732

I like the idea. How do we deal with erratic t vars as n increases? Maybe we need a rewrite of t equations for larger n, or maybe the t vars are accurate in later n are accurate because the d,a,x vars create t? Basically, does t represent small square values as n increases and t begins to jump upward in patters rather than linear growth?

>>1733

>I've also been spending some time going through reading for ECC.

I tried to read some of those documents… realized I'm not that smart.

>n = (x^2 + e) / 2(d - x)

I like the concept. After all, this seems to be an exercise in taking a really small square to it's perfect middle size.

>>1734

I believe this get's solved in the (e,1) space where there aren't any gaps in t. Remember, looking for one more variable or formula. t is just an indexer on values of x. Don't think we want to start changing it's definition based on n.

>>1733

All is good, not posted much either since I'm little out of ideas and none of my attempts have gone anywhere. Still here banging my head against the grid with you!

Pretty sure I solved it in a dream already but not sure how. Doh.

>>1735

I was having a really hard time with understanding what VQC meant when he said:

>At the correct element in the grid at (e,1) where the value of a at that element equals the na we want, if you subtract 2d+1 from na then you are in the negative half of the grid in terms of e and the value at the same element in the first row will be (n-1)a

I believe the "na" that we want refers to the na in the 2na = x^2 + e.

This is why I think the equation is important.

I'm trying to work out what the (n-1)a cell means now.

>>1733

Teach! Nice to see you!

>>1736

Nice to see you too, Anon!

>>1735

Well, if we’re able to solve for the small square using the t equations in row one that’s great! How do we tie that together with the column 0 knowledge? Or is it needed? If we know c, d and the small square, are we back to solving for n with the quadratic we derived? Column 0 still needs to get tied in? Thoughts, Anons?

>>1737

Teach - likewise. I'm not sure I fully understand, but here's where my head is by way of an example:

c=145

1 x c => (1,61,6) = {1:61:12:11:1:145} = 145; (x+n)^2=72x72=5184; (d+n)^2=73x73=5329;

1 x c na => (1,1,6) = {1:1:72:11:61:85} = 5185; (x+n)^2=12x12=144; (d+n)^2=73x73=5329;

In the first record, calculate (n-1)*a = (61-1)*1 = 60.

If you look at the differences in the n, d, and a values between (1,61,6) and (1,1,6) you will notice:

n = 1

d = d + 60

a = a + 60

So essentially the amount you take off of n gets added into the d and a.

This "transform" enables us to move from any (e,n) into the (e,1) space, and explains what's happening to d and a along the way.

I have played with this attempting to take off different values from n. (n-2), 2(n-1), etc. Haven't had much luck there.

>>1733

Ah, nice Teach! There’s that clear relationship between n and x we’ve been looking for. Nice rewrite on the formula! We’ve been banging our heads on that for a while, I think you just made it crystal clear.