Day 14: Parabolic Reflector Dish
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FAQ
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C
Chose not to do transposing/flipping or fancy indexing so it’s rather verbose, but it’s also clear and (I think) fast. I also tried to limit the number of search steps by keeping two cursors in the current row/col, rather than shooting a ray every time.
Part 2 immediately reminded me of that Tetris puzzle from day 22 last year so I knew how to find and apply the solution. State hashes are stored in an array and (inefficiently) scanned until a loop is found.
One direction of the shift function:
/* * Walk two cursors i and j through each column x. The i cursor * looks for the first . where an O can go. The j cursor looks * ahead for O's. When j finds a # we start again beyond it. */ for (x=0; x<w; x++) for (i=0, j=1; i<h && j<h; ) if (j <= i) j = i+1; else if (g[j][x] == '#') i = j+1; else if (g[j][x] != 'O') j++; else if (g[i][x] != '.') i++; else { g[i++][x] = 'O'; g[j++][x] = '.'; vis14_emit(); }
The main loop:
for (nleft = 1*1000*1000*1000; nleft; nleft--) { shift_all(); if (!period) { assert(nhist < (int)LEN(hist)); hist[nhist++] = hash_grid(); for (i=0; i<nhist-1; i++) if (hist[i] == hist[nhist-1]) { period = nhist-1 - i; nleft = nleft % period; break; } } }
Full solution: https://github.com/sjmulder/aoc/blob/master/2023/c/day14.c