AoC 2021/10 - Syntax scoring


On with Advent of Code puzzle 10 from 2021: playing with parentheses.

Today’s challenge is a bit more abstract than the others, being centered on a matter of programming itself. I guess this might be some recurring these - I mean, matching of parentheses - but it might just be that I did too many programming puzzles around and they are all mixed up in my brain.


We are given a list of parentheses, all of which are wrong in some sense.

Some of them which are the subject of puzzle 1, are just plain wrong. Parentheses have a tradition of liking to be paired with one another in a sort of russian dolls way, so this:


is, usually, a no-no. Well, unless you decide to give some meaning to it, of course.

In this case, anyway, this is considered wrong and we are required to detect all sequences that are wrong in this way in the lot, and calculate a number from them with some weights.

I couldn’t find some evident rule why those weights were chose, but I suspect that they allow anyway to go back from the number to the exact number of illegalities found. At least if there are no more than about 18 illegalities for each type of parenthesis.

This job of finding correspondences, or lack thereof, is usually well addressed by means of a stack data structure, so I didn’t go too far and used it in my solution:

sub part1 ($inputs) {
   state %data-for = ')' => ['(', 3], ']' => ['[', 57],
      '}' => ['{', 1197], '>' => ['<', 25137];
   my $sum = 0;
   my @incomplete;
   for $inputs[0].List -> @seq {
      my @stack;
      for @seq -> $item {
         if %data-for{$item}:exists {
            my $top = @stack.elems ?? @stack.pop !! '';
            if %data-for{$item}[0] ne $top {
               $sum += %data-for{$item}[1];
               next SEQ;
         else {
            @stack.push: $item;

      state %score-for = '(' => 1, '[' => 2, '{' => 3, '<' => 4;
      my $score = 0;
      $score = 5 * $score + %score-for{$_} for @stack.reverse;
      @incomplete.push: $score;
   my $mid = (@incomplete.elems - 1) / 2;
   $inputs[1] = @incomplete.sort[$mid];
   return $sum;

The %data-for at the beginning is useful for figuring out the opening corresponding to each closing, as well as the cost of each illegal closing. Mabye using two hashes would have been clearer, in hindsight.

The algorithm is: every open parenthesis is pushed onto the stack, every close parenthesis is checked against the stack and, if successful, removes one element from it. If the stack is empty… no worries: we assume it’s something that is surely a mismatch (the empty string does not correspond to any closing) and this will ensure a failure.

The second part is implemented here as well. It requires us to estimate how would it cost us to complete an incomplete sequence of otherwise correct parentheses. In this case, the weight is calculated with a base-5 number, where 0 means “good, nothing needed”.

To do this calculation, it suffices to get elements out of the stack in the order the stack gives us (i.e. from last to first) and we can easily calculate our second part score.

Well well well.. a quite boring solution, isn’t it? Yes, it is.

If you want to read a quite illuminating one, though, this Perl solution is brilliant in my opinion. I’ll copy it here, hoping nobody will complain (did I say it’s not mine?):

perl#!/usr/bin/perl -w

use strict;

my %points1 = (')' => 3, ']' => 57, '}' => 1197, '>' => 25137);
my %points2 = ('(' => 1, '[' => 2,  '{' => 3,    '<' => 4);

my $score1 = 0;
my @list2;
while (<>) {
    1 while s/(\(\)|\{\}|\[\]|<>)//;
    if (m/([\]>\}\)])/) {
        $score1 += $points1{$1};
    my $score2 = 0;
    foreach (split //, reverse) {
        $score2 = $score2 * 5 + $points2{$_};
    push @list2, $score2 if $score2;
print $score1, "\n";
@list2 = sort { $::a <=> $::b } @list2;
print $list2[@list2/2], "\n";

The core of part 1 is just this:

1 while s/(\(\)|\{\}|\[\]|<>)//;
if (m/([\]>\}\)])/) {
    $score1 += $points1{$1};

I know it looks like the noise that stains Perl’s reputation for many, but there are a lot of backslashes because we’re dealing with parentheses, most of which have a specific meaning in regular expressions.

The while runs as soon as there are matching consecutive pairs, eliminating them. What we’re left with is only what is illegal or what is incomplete. Genius.

Well, enough for today… stay safe folks!

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