Outstanding anon.
Semi-related, who would have believed you could find a way to calculate a chosen decimal digit of PI without calculating the ones before it.
Have you heard of that method? Spigot. It involves one of the most surprising shortcuts in history.
https://en.wikipedia.org/wiki/Spigot_algorithm
https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula
Some beautiful moments in history are behind us.
Many are in front of us!
Brilliant PMA.
The way you are all thinking is great to see.
The recognition of patterns and the visualizations… brilliant.
I have to admit, you're all moving faster, way faster, than I did.
I remember in 2013 staring at the grid (The End) for hours. Hours!
I knew it meant something, I'd tried so many different variables as the axis and went back and fourth, desperate for more insight. Often insight only came when I left it alone or seemingly from nowhere.
Plugged into bitmaps like you have here, hundreds and hundreds of bitmaps. And tables.
Anons are working so well together.
To clear this up slightly, the base of a triangle WITHIN the triangle.
The final steps involve finding a base using (f-1)/8 for each of the eight triangles (some a bit bigger than others by a unit in length) which is larger than (n-1)/8 BUT smaller than (x+n)/8.
The choices reduce 'exponentially' and with less complexity than find the root of C, so the last steps do not push the overall complexity Big Oh above Big Oh for finding the root.
Using the previously supplied C# method before, you construct the first guess of (x+n)/8 which is too small but larger than (n-1)/8.
So, the biggest hint so far.
For odd (x+n) and odd (n-1) to start with.
(n-1)/8, (f-1)/8, (x+n)/8
You are lining these triangular bases up by picking (f-1)/8 that is between (n-1)/8 and (x+n)/8.
The properties of 2d and f limit the factors that will lock these values together. For two prime factors there is one way and for more than two prime factors there are more and more ways depending on the number of factors.
If you can find a way to visualise this like you have above, it will become clearer very quickly.
Appreciate the irony behind the post.
Maybe subtlety will reduced the risk of bans.
I think safety is key.
Switch your router off and on again for qresearch?
Please don't get banned for me anon!
I'll be here more often now.
Thanks again for your support.
And to everyone here.
Lots of adventures ahead!
I'm going to create a C# app to help with the graphics.
Two panes side by side.
On the left will be a bitmap displaying the x+n square.
On the right will be the grid (The End)
Hovering the mouse on a pixel of the grid with entries for that e,n will animate through the x+n squares in that set.
We'll build the program in stages and the code will be shared here.
This is great.
Problem statement (re: building an app).
Why do we gather requirements and thoroughly look at the problem statement?
As my physics teacher said, by fully stating the problem statement, the answer presents itself. Or rather, the solution.
If I was to build an app.
On the left, the picture (like you've been building above), would be the square (x+n)(x+n).
On the right, would be the entries of the grid (the end) to select a cell.
To TEST it, we need to state the problem we are solving.
What test cases would be use?
Say we selected an (x+n) square and to test we wanted to keep the square the same size…
If I pick a grid cell at position t within that cell, how would I pick another cell with a (x+n) square the same size?
Would I pick a cell to the right (increasing e) that is 2n cells to the right?
Would that give me a cell with a value of (x+n) which is the same?
Then 2n from that?
What would be the difference in f?