AA !dTGY7OMD/g ID: 38e508 July 23, 2018, 1:35 p.m. No.7021   🗄️.is 🔗kun   >>7022

>>7020

It might be. I've "found" things that turned out to be super obvious or explainable with algebra and made myself look like an idiot in the past.

Anonymous ID: 38e508 Aug. 5, 2018, 10:14 p.m. No.7127   🗄️.is 🔗kun   >>7128

Some transparency from VQC would go a very long way right about now… hint hint

 

I didn't think it would be super useful to post what I'd been working on since none of it went anywhere, but I guess I might as well. We're meant to calculate the "value" at (e,1) where x=f-1. One specific value. There are several values, so who knows not only which value we're looking for but also why we're looking for it. So I thought I'd just generate some of these cells and see if any values stand out. e doesn't change, so it wouldn't be that. n will always be 1 at this cell, obviously, so it wouldn't be that. We know x is equal to the original f, so it wouldn't be that, since we wouldn't need to go to a new cell to figure that out. t is directly related to x, so it wouldn't be that either. So it's going to be something to do with d, a, b, c or f. I wouldn't think it would be f, since if we continually do this f becomes exponential, and that's the opposite of what we want (logarithmic). I can't see anything else about d, a, b or c that has any relevance to literally anything else, though. I have found that in quite a few test cases, c'/a is a whole number. It isn't in every case, but it has been in most of my test cases. I don't think that's useful but I was blindly looking or something.

 

I was also thinking about diagonals again, while looking at this image >>6952 here. Maybe whatever diagonals we're looking for for whatever obscure reason we're looking for them have a gradient of 1/2, rather than being a direct 1:1 relationship between the change in e and the change in d. There are diagonals in this image but they're all across two down one.