Another update on >>6774 and >>7319

>where the product of BigN and c are found in (e,1)

The cell where a[t]=c*BigN in (e,1) appears where x=c-d.

>and where the other value of c is in (e,1)

Another cell in (e,1) where a[t]%c==0 is where x=c+d+1. There are others, but I think it's fairly obvious why this one is probably the significant one (the x formula).

>and what information that gives us

I don't know what information that gives us yet unfortunately.

>the key that is made by column -f with the locations of c in (-f,1) and how to find x for an and a(n-1)

I haven't figured this out yet, but here are some test cases if anyone else spots anything.

—

7*29=203

(7,4,4) = {7:4:14:7:7:29}, f = -22, sqrt(2d) = 5

(7,1,95) = {7,1,18053,189,17864,18244}, f=-36100 (x=c-d)

17864/203=88 (BigN)

(7,1,109) = {7,1,23765,217,23548,23984}, f=-47524 (x=c+d+1)

23548/203=116

(-22,1,95) = {-22,1,17849,188,17661,18039}, e=-22 (x=c-d)

17661/203=87 (BigN-1)

(-22,1,110) = {-22,1,23969,218,23751,24189}, e=-22 (x=c+d+1)

23751/203=117

—

13*43=559

(30,5,6) = {30:5:23:10:13:43}, f = -17, sqrt(2d) = 6

a: (30,1,269) = {30,1,144199,536,143663,144737}, f=-288369 (x=c-d)

143663/559=257 (BigN)

a: (30,1,292) = {30,1,169959,582,169377,170543}, f=-339889 (x=c+d+1)

169377/559=303

a: (-17,1,268) = {-17,1,143639,535,143104,144176}, e=-17 (x=c-d)

143104/559=256 (BigN-1)

a: (-17,1,292) = {-17,1,170519,583,169936,171104}, e=-17 (x=c+d+1)

169936/559=304

>>7341

Also the difference between the x values in these cells is 2d+1, which is also the difference between the e and f values.

>>7321

Nice to see you again Hobo. I don't think it's that simple. This grid is designed specifically to be used to factorize arbitrarily-large numbers. That's the only thing we know it can be used for. You can use it to find a big list of different cells. You can use it to find a bunch of variables related to any number (c) you put into it. But it's not the kind of thing you can apply a smaller problem to. It's definitely a good approach in other contexts, but not this one, unfortunately. If you made a different grid that worked in the same way (with different variables other than endxab), you could maybe create a different VQC to solve a different problem. But that wouldn't help us figure out how to use the location of a[t]%c==0 in (e,1) and (-f,1) to find the x of an and a(n-1) in (e,1) and (-f,1). I think you're right about it not being solved. I'm thinking VQC just saw a bunch of weird graphs from one of the people Topol brought over and thought it looked like an elliptic curve or something. Also, what is this?

>if the problem is never solved

The problem is already solved, in my view. Time is irrelevant. We're all here for a good reason.

>>7341

Okay, I figured out the relevance of the numbers from the other cell in (e,1) where a[t]%c==0. I wasn't sure where 116 mattered, for example, in the relevant cells for 7*29 (23548/203=116). I just realized that 88+116=204, which is c+1. It holds for the other example.

a: (7,1,95) = {7,1,18053,189,17864,18244}, f=-36100 (x=c-d)

17864/203=88 (BigN)

a: (7,1,109) = {7,1,23765,217,23548,23984}, f=-47524 (x=c+d+1)

23548/203=116 (c-BigN+1)

a: (-22,1,95) = {-22,1,17849,188,17661,18039}, e=-22 (x=c-d)

17661/203=87 (BigN-1)

a: (-22,1,110) = {-22,1,23969,218,23751,24189}, e=-22 (x=c+d+1)

23751/203=117 (c-BigN+2)

So in the first relevant cell in (e,1), where x=c-d, a[t]=c*BigN.

In the second relevant cell, where x=c+d+1, a[t]=c(c-BigN+1)

In the equivalent first relevant cell in (-f,1) where x=c-d, a[t]=c*(BigN-1)

In the equivalent second relevant cell in (-f,1) where x=c+d+1, a[t]=c(c-BigN+2).

I still don't know the relevance but it's an answer to VQC's post. So we're somehow meant to use this to find x for an and a(n-1) in (e,1) and (-f,1). For reference,

a[t] = an at (7,1,4) = {7,1,35,7,28,44}, f=-64

x=7

a[t] = a(n-1) at (-22,1,5) = {-22,1,29,8,21,39}, e=-22

x=8