complete part 8 (switched to bitmasks).
This commit is contained in:
12
2021/OUTPUT
12
2021/OUTPUT
@@ -91,3 +91,15 @@ aoc-c : res=349769
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aoc-c : res=99540554
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aoc-c : res=99540554
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time: 0:00.00 real, 0.00 user, 0.00 sys
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time: 0:00.00 real, 0.00 user, 0.00 sys
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context-switch: 1+1, page-faults: 0+93
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context-switch: 1+1, page-faults: 0+93
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=========================================
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================= day08 =================
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=========================================
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aoc-c : res=543
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time: 0:00.00 real, 0.00 user, 0.00 sys
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context-switch: 0+1, page-faults: 0+85
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aoc-c : res=994266
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time: 0:00.00 real, 0.00 user, 0.00 sys
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context-switch: 0+1, page-faults: 0+86
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1
2021/day08/EXAMPLE2.txt
Normal file
1
2021/day08/EXAMPLE2.txt
Normal file
@@ -0,0 +1 @@
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acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab | cdfeb fcadb cdfeb cdbaf
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@@ -24,7 +24,7 @@ LDLIB := -l$(LIB)
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export LD_LIBRARY_PATH = $(LIBDIR)
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export LD_LIBRARY_PATH = $(LIBDIR)
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CFLAGS += -std=gnu99
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CFLAGS += -std=gnu99
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#CFLAGS += -O2
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CFLAGS += -O2
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CFLAGS += -g
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CFLAGS += -g
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CFLAGS += -Wall
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CFLAGS += -Wall
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CFLAGS += -Wextra
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CFLAGS += -Wextra
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@@ -71,8 +71,6 @@ Because the digits 1, 4, 7, and 8 each use a unique number of segments, you shou
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In the output values, how many times do digits 1, 4, 7, or 8 appear?
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In the output values, how many times do digits 1, 4, 7, or 8 appear?
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Your puzzle answer was 543.
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Your puzzle answer was 543.
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The first half of this puzzle is complete! It provides one gold star: *
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--- Part Two ---
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--- Part Two ---
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Through a little deduction, you should now be able to determine the remaining digits. Consider again the first example above:
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Through a little deduction, you should now be able to determine the remaining digits. Consider again the first example above:
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@@ -128,3 +126,7 @@ Following this same process for each entry in the second, larger example above,
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Adding all of the output values in this larger example produces 61229.
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Adding all of the output values in this larger example produces 61229.
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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?
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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?
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Your puzzle answer was 994266.
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Both parts of this puzzle are complete! They provide two gold stars: **
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@@ -20,17 +20,25 @@
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#include "bits.h"
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#include "bits.h"
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#include "list.h"
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#include "list.h"
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typedef struct {
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/* bitmask (last 7 bits): ... g f e d c b a
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int len;
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*/
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char *code;
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typedef u64 token;
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} token;
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typedef struct {
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typedef struct {
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token unique[10];
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token unique[10];
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token output[4];
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token output[4];
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} code;
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} code;
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//#ifdef DEBUG
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#ifdef DEBUG
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inline static char *bits_str(u64 c)
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{
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static char str[9];
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for (int i = 7; i >= 0; --i)
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str[7 - i] = c & (1 << i) ? '1': '0';
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return str;
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}
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static void print_code(code *code)
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static void print_code(code *code)
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{
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{
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int i = 0;
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int i = 0;
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@@ -38,14 +46,26 @@ static void print_code(code *code)
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//printf("crabs=%d max=%d\n", ncrabs, crab_max);
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//printf("crabs=%d max=%d\n", ncrabs, crab_max);
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printf("unique: ");
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printf("unique: ");
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for (i = 0; i < 10; ++i)
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for (i = 0; i < 10; ++i)
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printf("[%d]%s ", code->unique[i].len, code->unique[i].code);
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printf("[%d]%s ", popcount64(code->unique[i]), bits_str(code->unique[i]));
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printf("\n");
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printf("\n");
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printf("output: ");
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printf("output: ");
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for (i = 0; i < 4; ++i)
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for (i = 0; i < 4; ++i)
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printf("[%d]%s ", code->output[i].len, code->output[i].code);
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printf("[%d]%s ", popcount64(code->unique[i]), bits_str(code->unique[i]));
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printf("\n");
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printf("\n");
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}
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}
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//#endif
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#endif
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#define BIT(c) (1 << ((c) - 'a'))
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inline static u64 a2bit(char *token)
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{
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u64 res = 0;
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while (*token) {
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res |= BIT(*token);
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token++;
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}
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return res;
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}
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static code *read_code()
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static code *read_code()
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{
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{
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@@ -64,47 +84,127 @@ static code *read_code()
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while (token) {
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while (token) {
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if (*token == '|')
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if (*token == '|')
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break;
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break;
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code.unique[i].code = token;
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code.unique[i] = a2bit(token);
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code.unique[i].len = strlen(token);
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i++;
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i++;
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token = strtok(NULL, " \n");
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token = strtok(NULL, " \n");
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}
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}
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//printf("cont = %c\n", *token);
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//print_code(&code);
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i = 0;
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i = 0;
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while ((token = strtok(NULL, " \n"))) {
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while ((token = strtok(NULL, " \n"))) {
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//printf("output %d = [%s]\n", i, token);
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code.output[i] = a2bit(token);
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code.output[i].code = token;
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code.output[i].len = strlen(token);
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i++;
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i++;
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}
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}
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if (i != 4)
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printf("output = %d\n", i);
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free(buf);
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free(buf);
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return &code;
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return &code;
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}
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}
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static u64 doit(int part)
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static u64 part1()
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{
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{
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code *code;
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code *code;
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int res = 0;
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int res = 0;
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while ((code = read_code())) {
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while ((code = read_code())) {
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if (part == 1) {
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for (int i = 0; i < 4; ++i) {
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for (int i = 0; i < 4; ++i) {
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int len = popcount64(code->output[i]);
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int len = code->output[i].len;
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/* digits: 1 4 7 8 */
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/* digits: 1 4 7 8 */
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if (len == 2 || len == 4 || len == 3 || len == 7)
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if (len == 2 || len == 4 || len == 3 || len == 7) {
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res++;
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printf("%d : %s\n", len, code->output[i].code);
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res++;
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}
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}
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}
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}
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}
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}
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return res;
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return res;
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}
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}
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static u64 part2()
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{
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code *code;
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u64 bits;
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int res = 0, tmp;
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while ((code = read_code())) {
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u64 digits[10] = { 0 };
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/* find digits 1, 4, 7, 8 */
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for (int i = 0; i < 10; ++i) {
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bits = code->unique[i];
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switch (popcount64(bits)) {
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case 2:
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digits[1] = bits;
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code->unique[i] = 0;
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break;
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case 3:
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digits[7] = bits;
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code->unique[i] = 0;
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break;
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case 4:
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digits[4] = bits;
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code->unique[i] = 0;
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break;
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case 7:
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digits[8] = bits;
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code->unique[i] = 0;
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break;
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}
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}
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/* find digits 3 & 6: 2 and 1 bits in common with 1 */
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for (int i = 0; i < 10; ++i) {
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bits = code->unique[i];
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if (popcount64(bits) == 5) {
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if (popcount64(bits & digits[1]) == 2) {
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digits[3] = bits;
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code->unique[i] = 0;
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}
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}
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if (popcount64(bits) == 6) {
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if (popcount64(bits & digits[1]) == 1) {
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digits[6] = bits;
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code->unique[i] = 0;
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}
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}
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}
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/* find digits 9 and 0: 9 has 5 bits in common with 5 */
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for (int i = 0; i < 10; ++i) {
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bits = code->unique[i];
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if (popcount64(bits) == 6) {
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if (popcount64(bits & digits[3]) == 5)
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digits[9] = bits;
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else
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digits[0] = bits;
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code->unique[i] = 0;
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}
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}
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/* find digits 2 and 5: 2 has 4 bits in common with 9 */
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for (int i = 0; i < 10; ++i) {
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bits = code->unique[i];
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if (popcount64(bits) == 5) {
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if (popcount64(bits & digits[9]) == 4)
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digits[2] = bits;
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else
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digits[5] = bits;
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code->unique[i] = 0;
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}
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}
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tmp = 0;
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for (int i = 0; i < 4; ++i) {
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for (int j = 0; j < 10; ++j) {
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if (code->output[i] == digits[j]) {
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tmp = tmp * 10 + j;
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break;
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}
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}
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}
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res += tmp;
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}
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return res;
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}
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static u64 doit(int part)
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{
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return part == 1? part1(): part2();
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}
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static int usage(char *prg)
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static int usage(char *prg)
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{
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{
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fprintf(stderr, "Usage: %s [-d debug_level] [-p part]\n", prg);
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fprintf(stderr, "Usage: %s [-d debug_level] [-p part]\n", prg);
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