In about a week I'll have time to be properly active and not just lurk again, so I was thinking of redoing the introductory guide and creating an explanatory sticky (i.e. "this is the VQC board, we're currently doing x, here is a brief greentexted introduction to the grid, and if you want a more in-depth guide to what we're currently doing and how we got here read the next few posts and the previous threads" kind of thing). If anyone has any suggestions (or if anyone wants to contribute or do it themselves, or if any lurkers want anything in particular explained) then suggest away. Also, should I mention the off-topic Discord in the sticky?

While I've been doing all of this other meaningless bullshit I've had eating away at my time, I've been daydreaming about this somehow turning into a career. I don't mean once RSA's cracked taking those unclaimed buttcoins and being a rich prick, or going public and being an attention whore. I mean the idea of there being an application of this that would involve using it to improve the world, also somehow without the theory having to be kept secret from everyone else, and somehow making us all enough money to live on. That sounds far more meaningful than the corporate black hole that society and the expectations of others is sucking me into, making money for some guy in a suit who eats children's still-beating hearts in his spare time. I hope I don't sound delusional or entitled, because I know this isn't about me in the slightest, but if there is an application of this that would turn into what I'm talking about, VQC, I will survive on canned spaghetti for the rest of my life if that's what it takes.

>>6298

I've been trying to understand this the past couple days since GA's explanation turned out to be for something different. At first instinct, it seems to hint towards there being another cell transform along the path of (e,1), (1,c) etc on the way to (a,b).

>In row 1, the values of a[t] represent na for some c (e.g. RSA 100).

>a[t+n] = nb

>This is true for all c.

We all know this by now, but for example's sake, {2:1:25:6:19:33:4}, c=627, f=-49, a[5] does equal 33*1 and so on through the infinite set.

>For the value of c, at the same t but in cell (-f,1), the value at a[t] = a(n-1) and a[t+n-1] = b(n-1)

>For the value of c

Meaning the same c from whatever example c we used? So 627?

>at the same t but in cell (-f,1)

So you mean at (49,1,4)? That cell is {49:1:42:5:37:49:4}, which has a c of 1813.

>the value at a[t] = a(n-1) and a[t+n-1] = b(n-1)

a[4] in (49,1) =/= 19*0.

>The value of a[t] at -f and e in the first row have the same factor.

a[4] at (2,1) and (49,1) are 19 and 37 respectively. The only factor they share is 1, since they're both primes. That doesn't seem like what you mean. Otherwise you could say every a that exists in the grid has the same factor, and that wouldn't be useful. So obviously it seems like either it's worded weirdly or I'm personally not understanding it.

>>6332

Here's a bitmap of the grid with c and j as the axes, if it helps your understanding at all. In November and December me and another anon who hasn't posted in a while (the baker) were trying to use the gradients in this image with binary search to factorize a and b. It worked almost instantly for c under some certain number of digits, but it was completely useless when it came to RSA numbers (or pretty much all 20+ digit numbers in general, for that matter). Just thought I'd let you know before you waste time in the same way.