2019 day 9 init from day 7
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Bruno
258
2019/day09/aoc-c.c
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258
2019/day09/aoc-c.c
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/* aoc-c.c: Advent of Code 2019, day 9 parts 1 & 2
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*
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* Copyright (C) 2022 Bruno Raoult ("br")
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* Licensed under the GNU General Public License v3.0 or later.
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* Some rights reserved. See COPYING.
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*
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* You should have received a copy of the GNU General Public License along with this
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* program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
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*
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* SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <unistd.h>
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#include <string.h>
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#include "br.h"
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#include "bits.h"
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#include "debug.h"
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#include "list.h"
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#include "pool.h"
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/* operators codes
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*/
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typedef enum {
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ADD = 1, MUL = 2, /* CALC: add and mult */
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INP = 3, OUT = 4, /* I/O: input and output value */
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JMP_T = 5, JMP_F = 6, /* JUMPS: jump if true / if false */
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SET_LT = 7, SET_EQ = 8, /* COND SETS: set if true/false */
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ADJ_RL = 9, /* ADDRESSING: adjust relative addr */
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HLT = 99 /* HALT */
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} opcode_t;
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/**
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* ops - array of op-codes, mnemo, and number of parameters
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* @op: An integer, the opcode
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* @length: Next instruction offset
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*/
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typedef struct {
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int op;
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u8 length;
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} ops_t;
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typedef struct input {
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int val;
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struct list_head list;
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} input_t;
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#define MAXOPS 1024
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typedef struct {
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int length; /* total program length */
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int cur; /* current position */
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struct list_head input; /* process input queue */
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int mem [MAXOPS]; /* should really be dynamic */
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} program_t;
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static ops_t ops[] = {
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[ADD] = { ADD, 4 }, [MUL] = { MUL, 4 },
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[INP] = { INP, 2 }, [OUT] = { OUT, 2 },
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[JMP_T] = { JMP_T, 3 }, [JMP_F] = { JMP_F, 3 },
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[SET_LT] = { SET_LT, 4 }, [SET_EQ] = { SET_EQ, 4 },
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[HLT] = { HLT, 1 }
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};
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static int _flag_pow10[] = {1, 100, 1000, 10000};
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#define OP(p, n) ((p->mem[n]) % 100)
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#define ISDIRECT(p, n, i) ((((p->mem[n]) / _flag_pow10[i]) % 10))
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#define DIRECT(p, i) ((p)->mem[i])
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#define INDIRECT(p, i) (DIRECT(p, DIRECT(p, i)))
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#define peek(p, n, i) (ISDIRECT(p, n, i)? DIRECT(p, n + i): INDIRECT(p, n + i))
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#define poke(p, n, i, val) do { \
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INDIRECT(p, n + i) = val; } \
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while (0)
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static pool_t *pool_input;
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static __always_inline int prg_add_input(program_t *prg, int in)
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{
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input_t *input = pool_get(pool_input);
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input->val = in;
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list_add_tail(&input->list, &prg->input);
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return in;
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}
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static __always_inline int prg_get_input(program_t *prg, int *out)
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{
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input_t *input = list_first_entry_or_null(&prg->input, input_t, list);
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if (!input)
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return 0;
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*out = input->val;
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list_del(&input->list);
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pool_add(pool_input, input);
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return 1;
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}
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/**
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* permute - get next permutation of an array of integers
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* @len: length of array
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* @array: address of array
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*
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* Algorithm: lexicographic permutations
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* https://en.wikipedia.org/wiki/Permutation#Generation_in_lexicographic_order
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* Before the initial call, the array must be sorted (e.g. 0 2 3 5)
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*
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* Return: 1 if next permutation was found, 0 if no more permutation.
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*
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*/
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static int permute_next(int len, int *array)
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{
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int k, l;
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/* 1. Find the largest index k such that a[k] < a[k + 1] */
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for (k = len - 2; k >= 0 && array[k] >= array[k + 1]; k--)
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;
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/* No more permutations */
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if (k < 0)
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return 0;
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/* 2. Find the largest index l greater than k such that a[k] < a[l] */
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for (l = len - 1; array[l] <= array[k]; l--)
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;
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/* 3. Swap the value of a[k] with that of a[l] */
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swap(array[k], array[l]);
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/* 4. Reverse sequence from a[k + 1] up to the final element */
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for (l = len - 1, k++; k < l; k++, l--)
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swap(array[k], array[l]);
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return 1;
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}
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static int run(program_t *p, int *end)
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{
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int out = -1;
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while (1) {
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int op = OP(p, p->cur), cur = p->cur, input;
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if (!(ops[op].op)) {
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fprintf(stderr, "PANIC: illegal instruction %d at %d.\n", op, p->cur);
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return -1;
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}
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switch (op) {
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case ADD:
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poke(p, p->cur, 3, peek(p, p->cur, 1) + peek(p, p->cur, 2));
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break;
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case MUL:
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poke(p, p->cur, 3, peek(p, p->cur, 1) * peek(p, p->cur, 2));
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break;
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case INP:
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if (prg_get_input(p, &input))
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poke(p, p->cur, 1, input);
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else
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/* we need an input which is not yet avalaible, so we need
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* to put the program in "waiting mode": We stop it (and
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* return output value) without setting end flag.
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*/
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goto sleep;
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break;
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case OUT:
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out = peek(p, p->cur, 1);
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break;
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case JMP_T:
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if (peek(p, p->cur, 1))
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p->cur = peek(p, p->cur, 2);
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break;
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case JMP_F:
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if (!peek(p, p->cur, 1))
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p->cur = peek(p, p->cur, 2);
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break;
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case SET_LT:
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poke(p, p->cur, 3, peek(p, p->cur, 1) < peek(p, p->cur, 2) ? 1: 0);
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break;
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case SET_EQ:
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poke(p, p->cur, 3, peek(p, p->cur, 1) == peek(p, p->cur, 2) ? 1: 0);
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break;
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case HLT:
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*end = 1;
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sleep:
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return out;
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}
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if (p->cur == cur)
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p->cur += ops[op].length;
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}
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}
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static void parse(program_t *prog)
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{
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while (scanf("%d%*c", &prog->mem[prog->length++]) > 0)
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;
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}
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static int usage(char *prg)
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{
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fprintf(stderr, "Usage: %s [-d debug_level] [-p part] [-i input]\n", prg);
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return 1;
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}
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int main(int ac, char **av)
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{
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int phase1[] = {0, 1, 2, 3, 4}, phase2[] = {5, 6, 7, 8, 9}, *phase;
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int opt, max = 0, part = 1;
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program_t p = { 0 }, prg[5];
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while ((opt = getopt(ac, av, "d:p:o:")) != -1) {
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switch (opt) {
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case 'd':
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debug_level_set(atoi(optarg));
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break;
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case 'o':
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for (ulong i = 0; i < strlen(optarg); ++i)
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phase1[i] = optarg[i] - '0';
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break;
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case 'p': /* 1 or 2 */
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part = atoi(optarg);
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if (part < 1 || part > 2)
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return usage(*av);
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break;
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default:
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return usage(*av);
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}
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}
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pool_input = pool_create("input", 128, sizeof(input_t));
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if (optind < ac)
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return usage(*av);
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phase = part == 1? phase1: phase2;
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parse(&p);
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do {
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int out = 0, end = 0;
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/* reset programs initial state, and add phase to their input
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*/
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for (unsigned i = 0; i < ARRAY_SIZE(prg); ++i) {
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prg[i] = p;
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INIT_LIST_HEAD(&prg[i].input);
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prg_add_input(&prg[i], phase[i]);
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}
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/* run the 5 processes in order (0, 1, 2, 3, 4, 0, 1, etc...),
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* until end flag is set by the process 4 (HLT instruction)
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*/
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while (!end) {
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for (int i = 0; i < 5; ++i) {
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/* add last process output in current process input queue
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*/
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prg_add_input(&prg[i], out);
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out = run(&prg[i], &end);
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}
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}
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max = max(max, out);
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} while (permute_next(5, phase));
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printf("%s : res=%d\n", *av, max);
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pool_destroy(pool_input);
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exit(0);
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}
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