AA !LF1mmWigHQ ID: 774bd8 July 1, 2019, 3:32 a.m. No.9420   🗄️.is 🔗kun

Based on the posts from the person at the end of the last thread saying they would tell us how to use the grid, and what they said about there being a link between the elements in (e,1)/(f,1) where d is between d[t], the solution cells for dd in (0,n), and something about (0,e) and (0,f). I've put some code together to analyze. It doesn't always work, so I'll probably need to go over it before putting it here, but here are two examples of it working. Hopefully I got the negative n stuff right for the (0,f) cells. Does anyone see any links anywhere? Or maybe if that person comes back and continues, you could possibly clarify?

AA !LF1mmWigHQ ID: 774bd8 July 10, 2019, 4:16 p.m. No.9453   🗄️.is 🔗kun   >>9454

>>9447

I'm quite skeptical of this being the way forward. From my testing, it scales almost identically to just looking for i[t]=i in d between d[t] recursing through na.

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 12:51 a.m. No.9885   🗄️.is 🔗kun   >>9887

>>9884

2na, but anyway. I'm not really seeing any unifying patterns in this i/j difference stuff. It seems like usually the numbers decrease but not always. If i and j are estimates, they're never going to be correct, since c isn't always going to be in (0,n), and even then if you do this with a square you're going to be looking at different n values. So are you saying whatever we're meant to find to calculate the distance between our estimate i and estimate j has something to do with xx+0=2na? Because that's very cryptic.

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 12:55 a.m. No.9888   🗄️.is 🔗kun

>>9887

Run the same recursive algorithm on each individual c estimate? Use i^2-xx or j^2-xx? I don't know what you're trying to say.

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 1:02 a.m. No.9892   🗄️.is 🔗kun   >>9894

>>9891

>Root finding square root function

Are you saying there's a different "root" that isn't a square root? I know you're all saying that you're trying to be more helpful than Chris but this is starting to get Chris-tier, at least in the lucidity of the clues.

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 1:17 a.m. No.9896   🗄️.is 🔗kun   >>9897

>>9894

>>9895

>root of a function

>zero as a noun related to c

>e is basically n somehow

>n and d+n are somehow related to the root of david element

Like I'm not trying to be an asshole here but can you see it from my perspective? I have no idea what to do with literally anything you just said.

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 1:39 a.m. No.9903   🗄️.is 🔗kun   >>9905

>>9897

Surely when you say my "binary search code" you aren't talking about this? It wasn't based on anything to do with the grid at all. Plus I'd be surprised if anyone remembered that all that much. What are you referring to?

AA !LF1mmWigHQ ID: 774bd8 July 25, 2019, 12:26 p.m. No.9908   🗄️.is 🔗kun

>>9905

Okay if you even remember me making that code then you'd probably also remember that it didn't work very well. I'll have to remind myself of how it actually worked. But my point back then was that we were looking for a log(n) algorithm, and binary search is a log(n) algorithm. Is there anything specific to my code that was actually on the right track, or is it just the fact that it's log(n)?