Day 17: Chronospatial Computer

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  • @sjmulder
    link
    214 hours ago

    C

    Was looking forward to a VM! Really took my leisure for part 1, writing a disassembler, tracing, all pretty.

    Part 2 reminded me of 2021 day 24 where we also had to reverse an input. It’s been on my mind recently and I was thinking if there would be a way to backfeed an output through a program, yielding a representation like: “the input plus 3498 is a multiple of 40, and divisible by a number that’s 5 mod 8” (considering lossy functions like modulo and integer division).

    Today’s input didn’t lend itself to that, however, but analysing it having the solution ‘click’ was very satisfying.

    Code
    #include "common.h"
    
    #define MEMZ 32
    #define OUTZ 32
    #define BUFZ 64
    
    enum { ADV, BXL, BST, JNZ, BXC, OUT, BDV, CDV };
    
    struct vm {
    	int64_t mem[MEMZ]; int nm, pc;
    	int64_t out[OUTZ]; int no;
    	int64_t a,b,c;
    } vm;
    
    /* returns 0 if halted, 1 otherwise */
    static int
    step(void)
    {
    	int64_t op, ar, ac;	/* operator, lit. arg, combo arg */
    
    	assert(vm.pc >= 0);
    	assert(vm.pc+2 <= MEMZ);
    
    	if (vm.pc >= vm.nm)
    		return 0;
    
    	op = vm.mem[vm.pc++];
    	ar = vm.mem[vm.pc++];
    	ac = ar==4 ? vm.a : ar==5 ? vm.b : ar==6 ? vm.c : ar;
    
    	switch (op) {
    	case ADV: vm.a = vm.a >> ac; break;
    	case BDV: vm.b = vm.a >> ac; break;
    	case CDV: vm.c = vm.a >> ac; break;
    	case BXL: vm.b = vm.b ^ ar; break;
    	case BXC: vm.b = vm.b ^ vm.c; break;
    	case BST: vm.b = ac % 8; break;
    	case JNZ: if (vm.a) vm.pc = ar; break;
    	case OUT: assert(vm.no < OUTZ); vm.out[vm.no++] = ac%8; break;
    	default: assert(!"invalid opcode"); return 0;
    	}
    
    	return 1;
    }
    
    static int64_t
    recur_p2(int64_t a0, int pos)
    {
    	int64_t a, i;
    
    	/*
    	 * The code in the input uses up to 7 low bits of the A register
    	 * to produce a single number, then chops off the low 3 bits and
    	 * continues.
    	 *
    	 * That means bits above the current triplet influence what
    	 * number it generates, but bits below don't. To generate a
    	 * desired sequence then, we append, not prepend,  candidate
    	 * triplets.
    	 *
    	 * We may end up in a situation where, given the prefix found
    	 * for the numbers so far, no triplet yields the desired next
    	 * number. Then we have to backtrack and find another prefix,
    	 * hence the recursion.
    	 */
    
    	if (pos >= vm.nm)
    		return a0 >> 3;
    
    	for (i=0; i<8; i++) {
    		vm.a=a= a0+i;
    		vm.b=vm.c=vm.pc=vm.no= 0;
    
    		while (step() && !vm.no)
    			;
    
    		if (vm.no && vm.out[0] == vm.mem[vm.nm-pos-1])
    			if ((a = recur_p2(a << 3, pos+1)))
    				return a;
    	}
    
    	return 0;
    }
    
    int
    main(int argc, char **argv)
    {
    	char b[BUFZ], *tok, *rest;
    	int i;
    
    	if (argc > 1)
    		DISCARD(freopen(argv[1], "r", stdin));
    
    	fgets(b, BUFZ, stdin); sscanf(b, "Register A: %lld", &vm.a);
    	fgets(b, BUFZ, stdin); sscanf(b, "Register B: %lld", &vm.b);
    	fgets(b, BUFZ, stdin); sscanf(b, "Register C: %lld", &vm.c);
    	fgets(b, BUFZ, stdin);
    
    	assert(vm.b == 0);	/* assumption for part 2 */
    	assert(vm.c == 0);
    
    	rest = fgets(b, sizeof(b), stdin);
    	strsep(&rest, ":");
    
    	while ((tok = strsep(&rest, ","))) {
    		assert(vm.nm < MEMZ);
    		vm.mem[vm.nm++] = atoll(tok);
    	}
    
    	while (step())
    		;
    	for (i=0; i<vm.no; i++)
    		printf(i ? ",%lld" : "17: %lld", vm.out[i]);
    
    	printf(" %lld\n", recur_p2(0, 0));
    	return 0;
    }
    

    https://github.com/sjmulder/aoc/blob/master/2024/c/day17.c