day 8 part 1 (C) + init part 2

2021/day08/EXAMPLE.txt

@@ -1 +1,10 @@
-acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab | cdfeb fcadb cdfeb cdbaf
+be cfbegad cbdgef fgaecd cgeb fdcge agebfd fecdb fabcd edb | fdgacbe cefdb cefbgd gcbe
+edbfga begcd cbg gc gcadebf fbgde acbgfd abcde gfcbed gfec | fcgedb cgb dgebacf gc
+fgaebd cg bdaec gdafb agbcfd gdcbef bgcad gfac gcb cdgabef | cg cg fdcagb cbg
+fbegcd cbd adcefb dageb afcb bc aefdc ecdab fgdeca fcdbega | efabcd cedba gadfec cb
+aecbfdg fbg gf bafeg dbefa fcge gcbea fcaegb dgceab fcbdga | gecf egdcabf bgf bfgea
+fgeab ca afcebg bdacfeg cfaedg gcfdb baec bfadeg bafgc acf | gebdcfa ecba ca fadegcb
+dbcfg fgd bdegcaf fgec aegbdf ecdfab fbedc dacgb gdcebf gf | cefg dcbef fcge gbcadfe
+bdfegc cbegaf gecbf dfcage bdacg ed bedf ced adcbefg gebcd | ed bcgafe cdgba cbgef
+egadfb cdbfeg cegd fecab cgb gbdefca cg fgcdab egfdb bfceg | gbdfcae bgc cg cgb
+gcafb gcf dcaebfg ecagb gf abcdeg gaef cafbge fdbac fegbdc | fgae cfgab fg bagce

2021/day08/EXAMPLE2.txt

@@ -1,20 +0,0 @@
-be cfbegad cbdgef fgaecd cgeb fdcge agebfd fecdb fabcd edb |
-fdgacbe cefdb cefbgd gcbe
-edbfga begcd cbg gc gcadebf fbgde acbgfd abcde gfcbed gfec |
-fcgedb cgb dgebacf gc
-fgaebd cg bdaec gdafb agbcfd gdcbef bgcad gfac gcb cdgabef |
-cg cg fdcagb cbg
-fbegcd cbd adcefb dageb afcb bc aefdc ecdab fgdeca fcdbega |
-efabcd cedba gadfec cb
-aecbfdg fbg gf bafeg dbefa fcge gcbea fcaegb dgceab fcbdga |
-gecf egdcabf bgf bfgea
-fgeab ca afcebg bdacfeg cfaedg gcfdb baec bfadeg bafgc acf |
-gebdcfa ecba ca fadegcb
-dbcfg fgd bdegcaf fgec aegbdf ecdfab fbedc dacgb gdcebf gf |
-cefg dcbef fcge gbcadfe
-bdfegc cbegaf gecbf dfcage bdacg ed bedf ced adcbefg gebcd |
-ed bcgafe cdgba cbgef
-egadfb cdbfeg cegd fecab cgb gbdefca cg fgcdab egfdb bfceg |
-gbdfcae bgc cg cgb
-gcafb gcf dcaebfg ecagb gf abcdeg gaef cafbge fdbac fegbdc |
-fgae cfgab fg bagce

2021/day08/Makefile

@@ -24,7 +24,7 @@ LDLIB   := -l$(LIB)
 export LD_LIBRARY_PATH = $(LIBDIR)
 
 CFLAGS += -std=gnu99
-CFLAGS += -O2
+#CFLAGS += -O2
 CFLAGS += -g
 CFLAGS += -Wall
 CFLAGS += -Wextra

2021/day08/README.txt

@@ -69,3 +69,62 @@ fgae cfgab fg bagce
 Because the digits 1, 4, 7, and 8 each use a unique number of segments, you should be able to tell which combinations of signals correspond to those digits. Counting only digits in the output values (the part after | on each line), in the above example, there are 26 instances of digits that use a unique number of segments (highlighted above).
 
 In the output values, how many times do digits 1, 4, 7, or 8 appear?
+
+Your puzzle answer was 543.
+
+The first half of this puzzle is complete! It provides one gold star: *
+--- Part Two ---
+
+Through a little deduction, you should now be able to determine the remaining digits. Consider again the first example above:
+
+acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
+cdfeb fcadb cdfeb cdbaf
+
+After some careful analysis, the mapping between signal wires and segments only make sense in the following configuration:
+
+ dddd
+e    a
+e    a
+ ffff
+g    b
+g    b
+ cccc
+
+So, the unique signal patterns would correspond to the following digits:
+
+    acedgfb: 8
+    cdfbe: 5
+    gcdfa: 2
+    fbcad: 3
+    dab: 7
+    cefabd: 9
+    cdfgeb: 6
+    eafb: 4
+    cagedb: 0
+    ab: 1
+
+Then, the four digits of the output value can be decoded:
+
+    cdfeb: 5
+    fcadb: 3
+    cdfeb: 5
+    cdbaf: 3
+
+Therefore, the output value for this entry is 5353.
+
+Following this same process for each entry in the second, larger example above, the output value of each entry can be determined:
+
+    fdgacbe cefdb cefbgd gcbe: 8394
+    fcgedb cgb dgebacf gc: 9781
+    cg cg fdcagb cbg: 1197
+    efabcd cedba gadfec cb: 9361
+    gecf egdcabf bgf bfgea: 4873
+    gebdcfa ecba ca fadegcb: 8418
+    cefg dcbef fcge gbcadfe: 4548
+    ed bcgafe cdgba cbgef: 1625
+    gbdfcae bgc cg cgb: 8717
+    fgae cfgab fg bagce: 4315
+
+Adding all of the output values in this larger example produces 61229.
+
+For each entry, determine all of the wire/segment connections and decode the four-digit output values. What do you get if you add up all of the output values?

2021/day08/aoc-c.c

@@ -0,0 +1,140 @@
+/* aoc-c: Advent2021 game, day 6 parts 1 & 2
+ *
+ * Copyright (C) 2021 Bruno Raoult ("br")
+ * Licensed under the GNU General Public License v3.0 or later.
+ * Some rights reserved. See COPYING.
+ *
+ * You should have received a copy of the GNU General Public License along with this
+ * program. If not, see <https://www.gnu.org/licenses/gpl-3.0-standalone.html>.
+ *
+ * SPDX-License-Identifier: GPL-3.0-or-later <https://spdx.org/licenses/GPL-3.0-or-later.html>
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <unistd.h>
+#include <string.h>
+#include <malloc.h>
+
+#include "debug.h"
+#include "bits.h"
+#include "list.h"
+
+typedef struct {
+    int len;
+    char *code;
+} token;
+
+typedef struct {
+    token unique[10];
+    token output[4];
+} code;
+
+//#ifdef DEBUG
+static void print_code(code *code)
+{
+    int i = 0;
+
+    //printf("crabs=%d max=%d\n", ncrabs, crab_max);
+    printf("unique: ");
+    for (i = 0; i < 10; ++i)
+        printf("[%d]%s ", code->unique[i].len, code->unique[i].code);
+    printf("\n");
+    printf("output: ");
+    for (i = 0; i < 4; ++i)
+        printf("[%d]%s ", code->output[i].len, code->output[i].code);
+    printf("\n");
+}
+//#endif
+
+static code *read_code()
+{
+    int i = 0;
+    static char *buf = NULL;
+    char *token;
+    size_t alloc = 0;
+    static code code;
+
+    if (getline(&buf, &alloc, stdin) < 0)
+        return NULL;
+
+    /* read unique segment data
+     */
+    token = strtok(buf, " \n");
+    while (token) {
+        if (*token == '|')
+            break;
+        code.unique[i].code = token;
+        code.unique[i].len = strlen(token);
+        i++;
+        token = strtok(NULL, " \n");
+    }
+    //printf("cont = %c\n", *token);
+    //print_code(&code);
+    i = 0;
+    while ((token = strtok(NULL, " \n"))) {
+        //printf("output %d = [%s]\n", i, token);
+        code.output[i].code = token;
+        code.output[i].len = strlen(token);
+        i++;
+    }
+    if (i != 4)
+        printf("output = %d\n", i);
+
+    free(buf);
+    return &code;
+}
+
+static u64 doit(int part)
+{
+    code *code;
+    int res = 0;
+
+    while ((code = read_code())) {
+        if (part == 1) {
+            for (int i = 0; i < 4; ++i) {
+                int len = code->output[i].len;
+                /* digits: 1           4           7           8 */
+                if (len == 2 || len == 4 || len == 3 || len == 7) {
+                    printf("%d : %s\n", len, code->output[i].code);
+                    res++;
+                }
+            }
+        }
+    }
+    return res;
+}
+
+static int usage(char *prg)
+{
+    fprintf(stderr, "Usage: %s [-d debug_level] [-p part]\n", prg);
+    return 1;
+}
+
+int main(int ac, char **av)
+{
+    int opt;
+    u32 exercise = 1;
+    u64 res;
+
+    while ((opt = getopt(ac, av, "d:p:")) != -1) {
+        switch (opt) {
+            case 'd':
+                debug_level_set(atoi(optarg));
+                break;
+            case 'p':                             /* 1 or 2 */
+                exercise = atoi(optarg);
+                if (exercise < 1 || exercise > 2)
+                    return usage(*av);
+                break;
+            default:
+                return usage(*av);
+        }
+    }
+    if (optind < ac)
+        return usage(*av);
+
+    res = doit(exercise);
+    printf ("%s : res=%lu\n", *av, res);
+    exit (0);
+}