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day 18 part 2 (before cleaning)

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Bruno Raoult
2022-01-07 21:08:56 +01:00
parent 8d3a064c73
commit 0344a9162f
2 changed files with 32 additions and 7 deletions
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8
2021/day18/README.txt
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@@ -162,8 +162,6 @@ The magnitude of this final sum is 4140.
Add up all of the snailfish numbers from the homework assignment in the order they appear. What is the magnitude of the final sum? Add up all of the snailfish numbers from the homework assignment in the order they appear. What is the magnitude of the final sum?
Your puzzle answer was 3987. Your puzzle answer was 3987.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two --- --- Part Two ---
You notice a second question on the back of the homework assignment: You notice a second question on the back of the homework assignment:
@@ -188,3 +186,9 @@ Again considering the last example homework assignment above:
The largest magnitude of the sum of any two snailfish numbers in this list is 3993. This is the magnitude of [[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]] + [[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]], which reduces to [[[[7,8],[6,6]],[[6,0],[7,7]]],[[[7,8],[8,8]],[[7,9],[0,6]]]]. The largest magnitude of the sum of any two snailfish numbers in this list is 3993. This is the magnitude of [[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]] + [[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]], which reduces to [[[[7,8],[6,6]],[[6,0],[7,7]]],[[[7,8],[8,8]],[[7,9],[0,6]]]].
What is the largest magnitude of any sum of two different snailfish numbers from the homework assignment? What is the largest magnitude of any sum of two different snailfish numbers from the homework assignment?
Your puzzle answer was 4500.
Both parts of this puzzle are complete! They provide two gold stars: **
At this point, you should return to your Advent calendar and try another puzzle.

31
2021/day18/aoc-c.c
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@@ -155,7 +155,6 @@ static int node_explode(node_t *node)
} }
if (depth == 4) { if (depth == 4) {
node_t *left = node->tree[LEFT], *right = node->tree[RIGHT]; node_t *left = node->tree[LEFT], *right = node->tree[RIGHT];
node_t *tmp;
//struct list_head *tmp; //struct list_head *tmp;
/* skip leaves */ /* skip leaves */
@@ -287,8 +286,10 @@ static node_t *node_read(char **p, int depth)//, node_t *parent)
node->depth = depth; node->depth = depth;
if (!depth) /* root node */ if (!depth) { /* root node */
root = node; root = node;
//log_f(1, "%s\n", *p);
}
//log_f(2, "str=%s depth=%d\n", *p, depth); //log_f(2, "str=%s depth=%d\n", *p, depth);
switch (**p) { switch (**p) {
@@ -308,6 +309,9 @@ static node_t *node_read(char **p, int depth)//, node_t *parent)
list_add_tail(&node->leaf_list, &root->leaf_list); list_add_tail(&node->leaf_list, &root->leaf_list);
} }
(*p)++; (*p)++;
//if (!depth) { /* root node */
// log_f(1, "%s\n", *p);
//}
return node; return node;
} }
@@ -363,7 +367,6 @@ static s64 node_magnitude(node_t *node)
static s64 part1() static s64 part1()
{ {
node_t *head = NULL, *next = NULL; node_t *head = NULL, *next = NULL;
s64 res = 1;
head = node_read(lines + 0, 0); head = node_read(lines + 0, 0);
for (int i = 1; i < nlines; ++i) { for (int i = 1; i < nlines; ++i) {
@@ -376,8 +379,26 @@ static s64 part1()
static s64 part2() static s64 part2()
{ {
s64 res = 2; node_t *l1 = NULL, *l2 = NULL;
return res; s64 max = 0, cur;
char *tmp;
for (int i = 0; i < nlines; ++i) {
for (int j = 0; j < nlines; ++j) {
if (j == i)
continue;
tmp = lines[i];
l1 = node_read(&tmp, 0);
tmp = lines[j];
l2 = node_read(&tmp, 0);
cur = node_magnitude(node_reduce(node_add(l1, l2)));
if (cur > max) {
log(2, "NEW BEST: %s + %s\n", lines[i], lines[j]);
max = cur;
}
}
}
return max;
} }
static int usage(char *prg) static int usage(char *prg)