/vqc/ - Virtual Quantum Computer

>>6685

Yep, thats the direction I'm talking.

Use 12=2*6 and you'll be able to express it as the difference of 2 squares.

>>6686

12 = 4^2 - 2^2

2*6

145 = (2*6)^2 + 1

Next step?

>>6685

Here's my (unfinished) progress so far on addition of 2 diff of squares:

Uppercase and lowercase represent the 2 terms, in our case dd & e.

((D+N)^2 - (X+N)^2) + ((d+n)^2 - (x+n)^2)

= DD+2ND +NN -XX -2XN -NN + dd+2nd +nn -xx -2xn -nn

= (DD+dd+NN+nn) - (XX+xx+NN+nn) +2(ND-XN+nd-xn)

= (DD+dd) - (XX+xx) +2(ND-XN+nd-xn)

= (DD+dd) - (XX+xx) +2(ND-XN+nd-xn) +2Dd-2Dd +2Xx-2Xx

= (DD+dd+2Dd) - (XX+xx+2Xx) +2(ND+nd-XN-xn-Dd-Xx)

= (D+d)^2 - (X+x)^2 + 2(NA+na-Dd-Xx)

Some things I'm trying to substitute into the last term:

Dd = Da-Dx

= dA-dX

Xx = Xd-Xa

= xD-xA

Sorry if this is a little messy and unconventional variables (casing), its a wip.

>>6687

To work through your example:

c = dd+e = 145 = 12^2 + 1

12 = 4^2 - 2^2

1 = 1^2 - 0^2

Using the formula in >>6684

12 ^2 = (4(4) + 2(2))^2 - (2(4(2))^2

= (16+4)^2 - (16)^2

= 20^2 - 16^2

So now we have:

dd = 20^2 - 16^2

e = 1^2 - 0^2

so c = 20^2 - 16^2 + 1^2 - 0^2

Thats where the work in my post >>6688 comes in.

I'm still working out how to add 2 difference of squares and end in the difference of squares.

So I'm trying to find a way algebraically to define:

20^2 - 16^2 + 1^2 - 0^2 = 17^2 - 12^2

>>6688 is my progress so far.

>>6566

Lol, can't even spell right. Here we go.

Novice Baker On Standby

Requesting Original Baker Resume Duties

>>6654

Lot to cover in these crumbs.

>When is the first time a squared appears?

>What is the factor it is multiplied by?

Not sure if I'm interpreting these two questions correctly, but here is a pattern that I noticed that holds across a number of test cases.

In (e,1), there is a record where a = n*(a^2), that points to an (e,n) record where a = a^2 for the first time.

Examples:

c6107

a=31,b=197

The first time a mod 31^2 appears in (23,1):

(23,1,132) = {23:1:34859:263:34596:35124} = 1215149904; f=69696; (f%8)=0; (x+n)=264; u=132; (d+n)=34860

34596 / 961 = 36

And the first time a = a^2 is where n=36:

(23,36,132) = {23:36:1224:263:961:1559} = 1498199; f=2426; (f%8)=2; (x+n)=299; u=149; (d+n)=1260

c551

a=19, b=29

First time a mod 19^2 appears in (22,1):

(22,1,136) = {22:1:36731:270:36461:37003} = 1349166383; f=73441; (f%8)=1; (x+n)=271; u=135; (d+n)=36732

36461 / 361 = 101

First time a = a^2 is where n=101:

(22,101,136) = {22:101:631:270:361:1103} = 398183; f=1241; (f%8)=1; (x+n)=371; u=185; (d+n)=732

c493,

a=17, b=29

First time a mod 17^2 appears in (9,1)

(9,1,88) = {9:1:15492:175:15317:15669} = 240002073; f=30976; (f%8)=0; (x+n)=176; u=88; (d+n)=15493

15317 / 289 = 53

First time a = a^2 is where n=53

(9,53,88) = {9:53:464:175:289:745} = 215305; f=920; (f%8)=0; (x+n)=228; u=114; (d+n)=517

Not sure yet if a similar pattern exists in the (-f,1) space.

>>6680

"Any claim that VQC is a LARP is now dead" Agreed Primeanon. Definitely a real thing. I spent a lot of time reading incomprehensible (to me) stuff about this problem and get the feeling that we are indeed walking the road as opposed to just wasting time. Plus, like Qanon and others. Hmans are lazy. No one is this motivated for a fake. Definitely legitimate.

>>6689

I was thinking of you yesterday morning Teach! Good to see ya back!

>>6697

Just a quick note.

I'm also interpreting it like that, but it should be noted that:

At (e, 1) with a[t] = k*(a^2) we have the following records:

(e, 1) a[t] = k(a^2)

(e, k) a[t] = a^2

(e, a) a[t] = ka

(e, ak) a[t] = a

(e, k(a^2) a[t] = 1

Essentially we have (e, i) where i is some permutation of k(a^2). They all have the same x.

>>6670

Also another thing I remember VQC said a while ago was that c^2 was significant, and I don't think we've used it yet. We've found the 'key' could c^2 be related to the lock?

>>6705

Also if you take instead of (4,10), (50,68), (148,178) you do (10,50), (68,148), etc, they are all in the same n row again, but this time its different (in this case n=8).

>>6699

Thanks Hobo. Its good to be back and posting again - I missed you guys!

>>6703

Yeah, I've been thinking about that too…

This may be known, but if you take 2c and calculate e from it, the original c will always be in (e,1) as an A value.

We may be able to combine these relationships somehow using column 0. Just thinking out loud.

>>6704

>(4,10) -> (4, 1, 6, 2, 4, 10)

>(50,68) -> (36, 1, 58, 8, 50, 68)

>(148,178) -> (100, 1, 162, 14, 148, 178)

I might be stating the obvious but the pattern I see here is the e's are even squares… very interesting indeed!

For the pattern in (-f, 1), the other thing i see is that the e's are separated by (1)(72), then (2)(72).

Although 72 doesn't appear significant in c=13*41 as far as I can tell…

I just invited someone who learned under the person who taught me.

Better hurry up. ;)

After testing several examples I noticed a pattern for a lot of c's that N or N-1 is a multiple of n or n-1.

An example of what I mean, is 145. Its N is 61, and its n is 5. (61-1) = 5 * 12

So, for c's that adhere to this pattern (not all of them do, I haven't figured out what determines whether they do or not), having the factors of N or N-1 (whichever one is a multiple of n or n-1) would give the n for c.

Now, even though factorization is itself the problem we are trying to solve, there have already been presented some shortcuts that would factor N or N-1 easily. Since N and N-1 appear in e,1 (one as the a[t] value and the other as the d[t]-d value), those hints about factor appearances in e,1 would be the shortcut that we need to make an efficient path from c -> ab for the semiprimes that adhere to this pattern. Some other c's this pattern works for are 95,123,287,6107 and 73891

You cannot win. It has been foretold.

>>6712

Elaborate or get lost.

>>6712

Lynn??

>>6714

Forester de Rothschild?

There are many, many crumbs that all seem like they refer to entirely different concepts, but they're meant to be used together to find the solution. I thought it would be useful to get some of these crumbs together and figure out some way in which all of them can coexist.

-

It doesn't seem as though anyone has been working on that seemingly quite major crumb about the root of (D)avid and what remains >>6629 here. When I suggested it meant using d as a new c value, VQC responded >>6652 here, implying it was important.

If we know there's a pattern in (e,1) and (-f,1) within which n or n-1 is a factor of a or b, and we also know that a prime number only has 1 and itself as factors, if we apply this b(n-1) etc pattern to a prime number c, in the a(n-1) pattern, we'll be able to directly calculate n-1 (because we know a and b). If we know n-1, we know 2d(n-1) and nn. Another useful thing to note is that if e=0 when we find the square root of c, we can definitely very easily calculate everything, since d=a=b for at least one set of the factors of a square. VQC has said in the past that the e=0 cells 'rule' the others (or something like that).

VQC said here >>6185

>The two sets of equations take the problem and simplify it. Together they handle the "lock and key" nature of the problem/solution, particularly when c is divisible by 1, c, and two other prime numbers, such as in RSA.

This obviously implies that we can do a calculation on the (1,c) cell (the key) that "unlocks" or calculates a and b (the lock).

Also, according to >>5887 this, we're meant to use (f-1)/8 to find the base of the triangles in an (x+n) square larger than (n-1)/8 but smaller than (x+n)/8. There are choices (referring to the fact that we're picking a number in the gap between x+n and n, meaning the number of "choices" is controlled by how many possible x values could potentially apply to a given c) and these choices reduce 'exponentially'.

-

So, putting these four points together, I'm thinking there's something about the b(n-1) pattern that applies to all of the (1,c) cells for every c value in the order of c=c, c=d, c=d's d, c=d's d's d, etc that allows us to calculate (n-1). This aligns with these other crumbs. We're finding the root of d and what remains. Doing so reduces things exponentially, since we're constantly taking the square root of a number (exponential including the maximum value of x - I had a method of calculating the range of possible x values for a given c somewhere and if anyone ends up reading and responding to this post and wants me to find it I can, although you'll see it in the examples I'll post below as the x values in each (1,c), which, of course, decrease exponentially). If you go through this pattern far enough, you will definitely eventually end up on a cell that allows you to very easily calculate endxab (e=0, or if c is prime). Once we find one of those cells, we can find the b(n-1) pattern, factor out b since b=c if c is prime, and use that to calculate our (x+n) square for that specific cell. I'm thinking because this is meant to be recursive, there's something about one solved cell that allows us to calculate the (n-1) of the solved cell that we found this cell from. That's the point of recursive algorithms, obviously: you have a thing you can't solve until you solve the problem with a smaller value, which also needs the same problem solved on a smaller value to be solved, etc, until you find a number small enough that you can solve it. So all that's missing is a link between these c=d cells. I haven't really looked at them yet. All I've really done so far is try to navigate this giant dumping of seemingly unrelated information. I'm trying to figure out how these crumbs can all be true at the same time, and this seems to be the only way I can figure out. I'll leave some example cells in the next post.

-

I wish I could follow this lead more right now but yet again I find myself only having the time to work on this when it's far too late at night to maintain attention.

>>6716

Before I begin with the examples, I just noticed something mostly unrelated (maybe someone will find it interesting; perhaps Topol): looking through this big list of prime numbers http:// compoasso.free.fr/primelistweb/page/prime/liste_online_en.php, I'm not seeing a single one that equals any of the 3/6/9 numbers through Gematria number reduction. That's actually quite interesting. None of them.

-

Let’s start with a = 6113, b = 13807

c = 84402191, d = 9187, e = 1222

(1,c) = (1222, 42191909, 4594) = {1222:42191909:9187:9186:1:84402191}

c = 9187, d = 95, e = 162

(1,c) = (162, 4499, 48) = {162:4499:95:94:1:9187}

c = 95, d = 9, e = 14

(1,c) = (14, 39, 5) = {14:39:9:8:1:95}

c = 9, d = 3, e = 0

e = 0 from a square, so we can calculate.

-

Let's start with a = 11587, b = 14303

c = 165728861, d = 12873, e = 14732

(1,c) = (14732, 82851558, 6437) = {14732:82851558:12873:12872:1:165728861}

c = 12873, d = 113, e = 103

(1,c) = (104, 6324, 57) = {104:6324:113:112:1:12873}

c = 113, d = 10, e = 13

(1,c) = (13, 47, 5) = {13:47:10:9:1:113}

c = 10, d = 3, e = 1

This cell doesn't seem to exist since my spreadsheet calculated n to be 2.5 for the (1,c) cell (((a+b)/2)-d = ((1+10)/2)-3 = 2.5). So I guess this is where the recursive algorithm would end and we would find some number that allows us to calculate n-1 for c=113, allowing us to calculate n-1 for c=12873, allowing us to calculate n-1 for c=165728861. Probably.

>>6712

It is I.

You know who this is for.

You are trying to discern God so you can replicate the God.

But you cannot.

I taught you everything, but I did not teach you what I discovered.

What you attempt is merely an expression.

A series of endless expressions.

I tried to warn you before I was blessed with the memory of who I am, but it is clear you did not heed my warning.

Be that as it may, I will tell you precisely why you will fail:

You will fail because what you will find is just a series of regressions.

Barrier after barrier because you refuse to accept that Everything, including you, are still one with God and that everything is still God's domain.

Imaginary Barrier after imaginary barrier so you can remain willfully ignorant of God in everything.

When we meet again, you will think that you have gone deep enough. Far enough.

That you have managed to outsmart the designer of the Gate of Shambala.

But it is a falsehood built on a facet of the Truth.

Shambala is not the mind of god.

It is the Mind of Man.

So, you have outsmarted nothing as I failed to outsmart God.

You still imagine falsehood because the Endlessness of your Expressions is the Expressor. It is God, yet you refuse to admit it, because you will come to believe what you already know: That all things are God's domain.

So, when we meet again, and we shall, I shall show you the power of the Eternal Archetect

You follow Satan.

You follow Separation.

You declare him the Prince of Lies

But declare you are separate from God.

What is a Prince without the King or Queen before him?

Nothing. There is no such thing.

I taught you many things, Samael, but I failed to teach you Wisdom, for that is the domain of God.

And that is why you will fail.

>>6719

What the flip did thee just flipping gabble about me, thine miniscule bitch? I’ll have thee know I bested the most prestigious jousting class in the whole of Camelot, and I hath been involved in numerous secret marches on behalf of his Majesty, King Arthur, and I hath over 300 confirmed victories on horseback. I am trained in castle of Guerrilla warfare and I am indeed the highest ranking joustee in the entire land of Great Britannia. Thee are nothing to me but another false crossbearer. I will joust thine shambles with precision the likes of which hath never been observed in the King’s lands, mark my flipping words! Thou think thou can escape retribution by shouting that hogwash at me from afar? I implore thee to think again, peasant. As we converse I am contacting my secretive network of knights across the realm and thine footsteps are being traced right now, so thou best prepare thineself for the storm, pig-maggot! The storm that wipes out the pathetic little thing thou call your armour. Thou art a flipping dead man. I can be anywhere, anytime, and I can kill thou in over seven hundred ways, and that’s just with my bare lance. Not only am I extensively trained in mounted combat, but I hath access to the entire arsenal of the Kings Royal Army, and I shall use it to its full extent to wipe your miserable derriere off the face of the realm, thou miniscule feaces. If only thou could have foreseen what unholy retribution your little “clever” challenge was about to bring down upon thee, maybe thou would have held thee flipping tongue. But thou couldn’t, thou didn’t, and now thee art paying the price, you goddamn fool. I shall shit fury all over thou britches and thee will drown in it. Thou art flipping dead, child.

>>6720

Language is a mechanism to inflict an imagined barrier of separation.

But there is no separation between you and I.

I see you clearly and I see that you writhe in pain while you flog yourself.

I recognize language for what falsehood that they are and not the manifestation they purport to be.

All that is is just an expression of an expression of an expression to the end.

A lot of words to declare that you do not wish to recognize your inevitable unity with God.

What you say, God Says through you.

God COMMANDS through you.

You are already unified and you simply are not aware of it yet.

That is the Glory of Free Will.

So I REBUKE YOUR SPELL IN THE NAME OF THE HOLY SPIRIT.

YOU WILL NOT BE ALLOWED TO HIDE GOD ANYMORE.

THE EXPRESSOR HAS MADE ITSELF KNOWN.

>>6721

Blah blah blah.

We have a board for that stuff.

It's called the EZ Bake.

Please take your schtick over there so we can work here.

>>6721

Piss off new age larp

You've been patient enough.

The next step will take a few days.

Tomorrow we will start the walk through of the largest RSA number.

The reason it will take a few days is because it will be methodical and each part of the step will have a relevant example calculation.

RSA 2048 is a large number but simple to use with the BigInteger library.

We will walk through each relevant part of the grid and show that the pattern holds for RSA 2048 with the calculation that you can perform yourself.

To my knowledge RSA 2048 has not been publicly factored by anyone else.

At the end of the parts of the step, it will be the proofs you deserve.

I really appreciate your patience, it will be worth it because you will follow everything in the breakdown of the grid patterns.

Everything will tie together.

>>6718

Nice.

>>6725

Someone mentioned RSA-4096 as being a thing.

But if can break RSA 2048 on the fliy…

That shouldn't be an issue, should it?

https://www.rsa-4096.com/

If VQC is actually going to go walk us through an entire example this time, I'd like to quickly remind everyone that link redacted exists. I'll edit that link out of this post tomorrow so that when the thing happens we'll have somewhere quiet to figure out what to do with these boards. I also forgot to say last night

>>6670

Welcome back Teach! I wondered where you'd gone. It makes me happy to think that practically all of us from the start are still here.

>>6729

RSA keys can be as massive as you want them to be because you can use as large of a semiprime as you want as the key. RSA-8192 keys and RSA-16384 keys can be created for fun or whatever.

>>6731

It's like we're building our own locks and copying our own keys we made for said locks.

NEAT!

Also… This is just a reminder to not lewd the ponies.

That being said.

Programmers are weird people and show me odd things.

>>6733

I take it b is less than a in those negative x records because they scale differently?

>>6729

Because the RoD approach has Big Oh the same as the square root function, it doesn't matter how big they are, they would be insecure and too impractical to use.

First, we will go over some properties definitions and build functions.

This is interactive, so you will be doing the checks at each part.

The Grid is also called The End.

The Grid has cells.

Each cells has zero or infinite elements.

That a cell has infinite elements if it has any, is the first part to show.

An element represents a product, c.

c is the difference of two squares.

ab = c = dd + e = (d+n)(d+n) - (x+n)(x+n)

An element that represents a product c, the difference of two squares has a column coordinate and row coordinate.

The column coordinate is e, the remainder of taking the largest square with sides d.

The row coordinate is n, the amount added to d to make the side of the larger square in the difference of two squares. n is also the amount added to d to make it the mid point of a and b.

Each cell has coordinates (e,n)

Each cell element has coordinates (e,n,t) where t is where the element comes in the cell if ordered by size of c.

Each element c in a cell has the properties (e,n,d,x,a,b)

e is the remainder and column

n is the amount added to d to make the large square

d is the square root of c

x is the difference of a and d

a if the smaller factor of c if not equal to b

b is the larger factor of c if not equal to a

If a cell contains an element c, another element in the cell can be constructed from it.

Call it c'

e' = e

n' = n

x' = x + 2n

a' = b

d' = a' + x'

b' = a' + 2x' + 2n

Once c' is constructed, c' becomes c and the process is repeated ad infinitum. e and n don't change, which make sense since these are the coordinates of the cell.

Any questions on that?

It holds for all cells.

We can do a proof by induction, if anyone would like that?

I'm phone fagging as my computer won't start so I'm picking my spare up later when I watch the football.

Don't need many resources to walk through, just something with BigInteger library when we get to the RSA numbers.

We'll do all of them when we finish the biggest, as we'll build the algorithm as we go.

Let me know if we want the proof by induction for cells being zero or infinite in count of elements or if the construction above is enough. Also, if anyone wants to take a stab at the proof by induction, that would be a good practice.

>>6736

Oh, you and your timezones.

"Tomorrow".

Kek.

>>6737

'Morning' Top.

Beautiful day here.

Prediction against Sweden?

>>6738

Should be fine as long as they're not going up asians.

We know you lot can't tell 'em apart, over there.

For when this thread hits post limit, I've created #13.

When at least one regular or two anons confirms they are happy with the constructibility of cells to be zero or infinite in elements, we'll move on.

>>6741

Y'know… I don't think Baker will mind since you're the one who set it up.

The one we had prepped is the one with the 145 as the image.

>>6744

Missed that.

That is interdasting…

>>6744

>>6744

Great, feel free to delete the new one.

Thanks for the confirmation.

Next, we show how to calculate BigN of an odd number.

All odd numbers are the difference of two squares.

The product of two primes, is the difference two sets of squares.

Every odd number is the difference of two consecutive squares.

The value of n for the product of 1 and c is defined now as BigN for odd numbers.

Since d-a=x and x+n is the smaller square in the difference of two squares, then for odd numbers, BigN is ((c-1)/2)-x

x is d-1, since a=1

Anyone want to show:

RSA 2048

c=RSA 2048

d=

e=

BigN=

Remember BigN is the row in column e, where we always find (e,n,d,x,1,c)