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Advent of Code 2024-Day 25: Code Chronicle

from advent import parse, partition, cross_product

pairs = [ (schematic[0][0] == '#', [ len(partition(line, lambda c: c == line[0])[0])
                                     for line in schematic ])
          for schematic in parse(25, lambda s: list(zip(*str.split(s, '\n'))), sep='\n\n') ]
locks = [ lengths for is_lock, lengths in pairs if is_lock     ]
keys  = [ lengths for is_lock, lengths in pairs if not is_lock ]
pins  = 5
total = sum(1 for lock, key in cross_product(locks, keys) if all([ lock[pin] <= key[pin]
                                                                   for pin in range(pins) ]))

assert total == 2618

Parsing

We're given a list of schematics - either for locks or keys. For each schematic, we'll do the following:

Transpose the matrix (using zip(*...)), so the pins are horizontal instead of vertical:

lambda s: list(zip(*str.split(s, '\n')))

Compute the length of the inital list of either . characters or # characters:

[ len(partition(line, lambda c: c == line[0])[0]) for line in schematic ]

Finally, form a 2-tuple of (<boolean>, <len>) where the <boolean> indicates the schematic is a lock, and the <len> is the length of the # characters for locks or . characters for keys:

(schematic[0][0] == '#', [ len(partition(line, lambda c: c == line[0])[0]) for line in schematic ])

Solution

For today, parsing the input is more than half the battle! Once we have our data structures in place via the above, we simply form a cartesian product of locks and keys, and filter the pairs by whether the key fits in the lock:

total = sum(1 for lock, key in cross_product(locks, keys) if all([ lock[pin] <= key[pin]
                                                                   for pin in range(pins) ]))

Conclusion

This was a nice easy puzzle to wrap up Advent of Code 2024. I really enjoyed this years contest - it was a fun way to become more proficient with Python and some of its ecosystem - particularly the networkx library.