Day 13: Point of Incidence

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  • @sjmulder
    link
    311 months ago

    C

    Implementing part 1 with a bunch of for loops made me wonder about elegant NumPy solutions but then part 2 came along and it was a perfect fit! Just change a flag to a counter and remove the if-match-early-exit.

    https://github.com/sjmulder/aoc/blob/master/2023/c/day13.c

    int main()
    {
    	static char g[32][32];
    	int p1=0,p2=0, w,h, x,y,i, nmis;
    	
    	while (!feof(stdin)) {
    		for (h=0; ; h++) {
    			assert(h < (int)LEN(*g));
    			if (!fgets(g[h], LEN(*g), stdin)) break;
    			if (!g[h][0] || g[h][0]=='\n') break;
    		}
    
    		assert(h>0); w = strlen(g[0])-1;
    		assert(w>0);
    
    		for (x=1; x<w; x++) {
    			nmis = 0;
    			for (i=0; i<x && x+i<w; i++)
    			for (y=0; y<h; y++)
    				nmis += g[y][x-i-1] != g[y][x+i];
    			if (nmis==0) p1 += x; else
    			if (nmis==1) p2 += x;
    		}
    
    		for (y=1; y<h; y++) {
    			nmis = 0;
    			for (i=0; i<y && y+i<h; i++)
    			for (x=0; x<w; x++)
    				nmis += g[y-i-1][x] != g[y+i][x];
    			if (nmis==0) p1 += y*100; else
    			if (nmis==1) p2 += y*100;
    		}
    	}
    
    	printf("13: %d %d\n", p1, p2);
    	return 0;
    }