Day 7: Camel Cards

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  • hades@lemm.ee
    ·
    7 months ago

    Python

    Also available on Github with all the support code. Questions and feedback welcome!

    import collections
    
    from .solver import Solver
    
    _FIVE_OF_A_KIND  = 0x100000
    _FOUR_OF_A_KIND  = 0x010000
    _FULL_HOUSE      = 0x001000
    _THREE_OF_A_KIND = 0x000100
    _TWO_PAIR        = 0x000010
    _ONE_PAIR        = 0x000001
    
    _CARD_ORDER            = '23456789TJQKA'
    _CARD_ORDER_WITH_JOKER = 'J23456789TQKA'
    
    def evaluate_hand(hand: str, joker: bool = False) -> int:
      card_counts = collections.defaultdict(int)
      score = 0
      for card in hand:
        card_counts[card] += 1
      joker_count = 0
      if joker:
        joker_count = card_counts['J']
        del card_counts['J']
      counts = sorted(card_counts.values(), reverse=True)
      top_non_joker_count = counts[0] if counts else 0
      if top_non_joker_count + joker_count == 5:
        score |= _FIVE_OF_A_KIND
      elif top_non_joker_count + joker_count == 4:
        score |= _FOUR_OF_A_KIND
      elif top_non_joker_count + joker_count == 3:
        match counts, joker_count:
          case [3, 2], 0:
            score |= _FULL_HOUSE
          case [3, 1, 1], 0:
            score |= _THREE_OF_A_KIND
          case [2, 2], 1:
            score |= _FULL_HOUSE
          case [2, 1, 1], 1:
            score |= _THREE_OF_A_KIND
          case [1, 1, 1], 2:
            score |= _THREE_OF_A_KIND
          case _:
            raise RuntimeError(f'Unexpected card counts: {counts} with {joker_count} jokers')
      elif top_non_joker_count + joker_count == 2:
        match counts, joker_count:
          case [2, 2, 1], 0:
            score |= _TWO_PAIR
          case [2, 1, 1, 1], 0:
            score |= _ONE_PAIR
          case [1, 1, 1, 1], 1:
            score |= _ONE_PAIR
          case _:
            raise RuntimeError(f'Unexpected card counts: {counts} with {joker_count} jokers')
      card_order = _CARD_ORDER_WITH_JOKER if joker else _CARD_ORDER
      for card in hand:
        card_value = card_order.index(card)
        score <<= 4
        score |= card_value
      return score
    
    class Day07(Solver):
    
      def __init__(self):
        super().__init__(7)
        self.hands: list[tuple[str, str]] = []
    
      def presolve(self, input: str):
        lines = input.rstrip().split('\n')
        self.hands = list(map(lambda line: line.split(' '), lines))
    
      def solve_first_star(self):
        hands = self.hands[:]
        hands.sort(key=lambda hand: evaluate_hand(hand[0]))
        total_score = 0
        for rank, [_, bid] in enumerate(hands):
          total_score += (rank + 1) * int(bid)
        return total_score
    
      def solve_second_star(self):
        hands = self.hands[:]
        hands.sort(key=lambda hand: evaluate_hand(hand[0], True))
        total_score = 0
        for rank, [_, bid] in enumerate(hands):
          total_score += (rank + 1) * int(bid)
        return total_score
    
      • hades@lemm.ee
        ·
        7 months ago

        Sure! This generates a number for every hand, so that a better hand gets a higher number. The resulting number will contain 11 hexadecimal digits:

        0x100000 bbbbb
          ^^^^^^ \____ the hand itself
          |||||\_ 1 if "one pair"
          ||||\__ 1 if "two pairs"
          |||\___ 1 if "three of a kind"
          ||\____ 1 if "full house"
          |\_____ 1 if "four of a kind"
          \______ 1 if "five of a kind"
        
        For example:
         AAAAA: 0x100000 bbbbb
         AAAA2: 0x010000 bbbb0
         22233: 0x001000 00011
        

        The hand itself is 5 hexadecimal digits for every card, 0 for "2" to b for "ace".

        This way the higher combination always has a higher number, and hands with the same combination are ordered by the order of the cards in the hand.

        • snowe@programming.dev
          ·
          7 months ago

          That is a really cool solution. Thanks for the explanation! I took a much more... um... naive path lol.

          • hades@lemm.ee
            ·
            7 months ago

            I think you have the same solution, basically, just the details are a bit different. I like how you handled the joker, I didn't realise you could just multiply your best streak of cards to get the best possible combination.

            • snowe@programming.dev
              ·
              7 months ago

              I didn't multiply the streak, I just took the jokers and added them to the highest hand already in the list. Is that not what you did? It looked the same to me.

              • hades@lemm.ee
                ·
                7 months ago

                This is what I meant, but I phrased it poorly :)

                In my solution I reimplement the logic of identifying the hand value, but with the presence of joker (instead of just reusing the same logic).