Copyright (c) 2017 Bart Massey
After thinking about it for a while, here's what I came up with as a series of semi-plausible automated compiler optimizations that would untangle the Day 23 inner loop.
Let's start with the raw loop. I wrote it in Go, because
reasons. The f flag is set to false iff b is
composite. This is what I call "trial multiplication": try
| # Once you know the Cartesian (x, y) coordinate of spiral position i in Part 1, | |
| # you can compute Part 2 really easily using just a map from coordinates to int. | |
| # map[(0,0)] = 1, and then you go around the spiral. At each new coordinate, you look | |
| # at its nine neighbors with a pair of for loops. Most of them (including the new position) | |
| # won't be in there. | |
| # Here's a Python implementation to show what I mean. | |
| # Essentially a Fibonacci-style dynamic | |
| # programming calculation but with 2-3 neighbors. |
| 29 W 29 W | |
| K.... K.... | |
| ..... ..... | |
| ..... ..... | |
| ..... ..... | |
| ..... ..... | |
| ....k ....k | |
| 29 B 29 B | |
| ..... ..... |
| -- Copyright © 2016 Bart Massey | |
| -- Chart Histogram demo | |
| import Control.Lens | |
| import Data.Default.Class | |
| import Graphics.Rendering.Chart | |
| import Graphics.Rendering.Chart.Backend.Cairo | |
| pdfRender :: String -> Renderable a -> IO () |
| M[16],X=16,W,k;main(){T(system("stty cbreak") | |
| );puts(W&1?"WIN":"LOSE");}K[]={2,3,1};s(f,d,i | |
| ,j,l,P){for(i=4;i--;)for(j=k=l=0;k<4;)j<4?P=M | |
| [w(d,i,j++)],W|=P>>11,l*P&&(f?M[w(d,i,k)]=l<< | |
| (l==P):0,k++),l=l?P?l-P?P:0:l:P:(f?M[w(d,i,k) | |
| ]=l:0,++k,W|=2*!l,l=0);}w(d,i,j){return d?w(d | |
| -1,j,3-i):4*i+j;}T(i){for(i=X+rand()%X;M[i%X] | |
| *i;i--);i?M[i%X]=2<<rand()%2:0;for(W=i=0;i<4; | |
| )s(0,i++);for(i=X,puts("\e[2J\e[H");i--;i%4|| | |
| puts(""))printf(M[i]?"%4d|":" |",M[i]);W-2 |