6 ms·
Since it's not provable with pi, then we'd have to do a more circuitous proof of every finite pattern occurring. Inspired by Champernowne's constant, I propose
by pontifier 10mo ago
Since it's not provable with pi, then we'd have to do a more circuitous proof of every finite pattern occurring. Inspired by Champernowne's constant, I propose a Pontifier Pattern that is simple, inefficient, but provably contains every finite pattern.
Starting at the origin, mark off rows of squares. the Nth row would contain NxN^2 squares of size n x n. Each square would be filled in left to right reading order with successive binary numbers with the most significant digit at the top left.
Somewhere in that pattern is the physics simulation of you reading this comment :)
- 7373737373 10mo agominor correction: 2^(NxN) squares per row, right?
- pontifier 10mo agoYeah, what was I thinking?? I really need to slow down sometimes. This should contain every finite pattern, right?
- 7373737373 10mo agoYes, sounds like it! Though I'm thinking that the relative arrangement of patterns would also make a difference. I wonder if such a thing as "all (infinitely many) possible arrangements of all patterns" can exist