Day 3: Mull It Over
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6 points
Sorry for the delay posting this one, Ategon seemed to have it covered, so I forgot :D I will do better.
5 points
I couldn’t figure it out in haskell, so I went with bash for the first part
Shell
cat example | grep -Eo "mul\([[:digit:]]{1,3},[[:digit:]]{1,3}\)" | cut -d "(" -f 2 | tr -d ")" | tr "," "*" | paste -sd+ | bc
but this wouldn’t rock anymore in the second part, so I had to resort to python for it
Python
import sys
f = "\n".join(sys.stdin.readlines())
f = f.replace("don't()", "\ndon't()\n")
f = f.replace("do()", "\ndo()\n")
import re
enabled = True
muls = []
for line in f.split("\n"):
if line == "don't()":
enabled = False
if line == "do()":
enabled = True
if enabled:
for match in re.finditer(r"mul\((\d{1,3}),(\d{1,3})\)", line):
muls.append(int(match.group(1)) * int(match.group(2)))
pass
pass
print(sum(muls))
2 points
1 point
5 points
*
C
Yay parsers! I’ve gotten quite comfortable writing these with C. Using out pointers arguments for the cursor that are only updated if the match is successful makes for easy bookkeeping.
Code
#include "common.h"
static int
parse_exact(const char **stringp, const char *expect)
{
const char *s = *stringp;
int i;
for (i=0; s[i] && expect[i] && s[i] == expect[i]; i++)
;
if (expect[i])
return 0;
*stringp = &s[i];
return 1;
}
static int
parse_int(const char **stringp, int *outp)
{
char *end;
int val;
val = (int)strtol(*stringp, &end, 10);
if (end == *stringp)
return 0;
*stringp = end;
if (outp) *outp = val;
return 1;
}
static int
parse_mul(const char **stringp, int *ap, int *bp)
{
const char *cur = *stringp;
int a,b;
if (!parse_exact(&cur, "mul(") ||
!parse_int(&cur, &a) ||
!parse_exact(&cur, ",") ||
!parse_int(&cur, &b) ||
!parse_exact(&cur, ")"))
return 0;
*stringp = cur;
if (ap) *ap = a;
if (bp) *bp = b;
return 1;
}
int
main(int argc, char **argv)
{
static char buf[32*1024];
const char *cur;
size_t nr;
int p1=0,p2=0, a,b, dont=0;
if (argc > 1)
DISCARD(freopen(argv[1], "r", stdin));
nr = fread(buf, 1, sizeof(buf), stdin);
assert(!ferror(stdin));
assert(nr != sizeof(buf));
buf[nr] = '\0';
for (cur = buf; *cur; )
if (parse_exact(&cur, "do()"))
dont = 0;
else if (parse_exact(&cur, "don't()"))
dont = 1;
else if (parse_mul(&cur, &a, &b)) {
p1 += a * b;
if (!dont) p2 += a * b;
} else
cur++;
printf("03: %d %d\n", p1, p2);
}
2 points
Got the code a little shorter:
Code
#include "common.h"
static int
parse_mul(const char **stringp, int *ap, int *bp)
{
const char *cur = *stringp, *end;
if (strncmp(cur, "mul(", 4)) { return 0; } cur += 4;
*ap = (int)strtol(cur, (char **)&end, 10);
if (end == cur) { return 0; } cur = end;
if (*cur != ',') { return 0; } cur += 1;
*bp = (int)strtol(cur, (char **)&end, 10);
if (end == cur) { return 0; } cur = end;
if (*cur != ')') { return 0; } cur += 1;
*stringp = cur;
return 1;
}
int
main(int argc, char **argv)
{
static char buf[32*1024];
const char *p;
size_t nr;
int p1=0,p2=0, a,b, dont=0;
if (argc > 1)
DISCARD(freopen(argv[1], "r", stdin));
nr = fread(buf, 1, sizeof(buf), stdin);
assert(!ferror(stdin));
assert(nr != sizeof(buf));
buf[nr] = '\0';
for (p = buf; *p; )
if (parse_mul(&p, &a, &b)) { p1 += a*b; p2 += a*b*!dont; }
else if (!strncmp(p, "do()", 4)) { dont = 0; p += 4; }
else if (!strncmp(p, "don't()", 7)) { dont = 1; p += 7; }
else p++;
printf("03: %d %d\n", p1, p2);
}
5 points
I started poking at doing a proper lexer/parser, but then I thought about how early in AoC it is and how low the chance is that the second part will require proper parsing.
So therefore; hello regex my old friend, I’ve come to talk with you again.
C#
List<string> instructions = new List<string>();
public void Input(IEnumerable<string> lines)
{
foreach (var line in lines)
instructions.AddRange(Regex.Matches(line, @"mul\(\d+,\d+\)|do\(\)|don't\(\)").Select(m => m.Value));
}
public void Part1()
{
var sum = instructions.Select(mul => Regex.Match(mul, @"(\d+),(\d+)").Groups.Values.Skip(1).Select(g => int.Parse(g.Value))).Select(cc => cc.Aggregate(1, (acc, val) => acc * val)).Sum();
Console.WriteLine($"Sum: {sum}");
}
public void Part2()
{
bool enabled = true;
long sum = 0;
foreach(var inst in instructions)
{
if (inst.StartsWith("don't"))
enabled = false;
else if (inst.StartsWith("do"))
enabled = true;
else if (enabled)
sum += Regex.Match(inst, @"(\d+),(\d+)").Groups.Values.Skip(1).Select(g => int.Parse(g.Value)).Aggregate(1, (acc, val) => acc * val);
}
Console.WriteLine($"Sum: {sum}");
}
5 points
*
Nim
From a first glance it was obviously a regex problem.
I’m using tinyre here instead of stdlib re library just because I’m more familiar with it.
import pkg/tinyre
proc solve(input: string): AOCSolution[int, int] =
var allow = true
for match in input.match(reG"mul\(\d+,\d+\)|do\(\)|don't\(\)"):
if match == "do()": allow = true
elif match == "don't()": allow = false
else:
let pair = match[4..^2].split(',')
let mult = pair[0].parseInt * pair[1].parseInt
result.part1 += mult
if allow: result.part2 += mult
4 points
