Day 7: Camel Cards

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  • hades
    1·
    3 months ago

    Python

    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
        3·
        1 year 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.

        • SteveDinn@lemmy.ca
          2·
          1 year ago

          Wow, this is exactly what I did, but in C#. That’s cool.

              public class Hand
    {
        public string Cards;
        public int Rank;
        public int Bid;
    }

    public static HandType IdentifyHandType(string hand)
    {
        var cardCounts = hand
            .Aggregate(
                new Dictionary(),
                (counts, card) => 
                {
                    counts[card] = counts.TryGetValue(card, out var count) ? (count + 1) : 1;
                    return counts;
                })
            .OrderByDescending(kvp => kvp.Value);

        using (var cardCount = cardCounts.GetEnumerator())
        {
            cardCount.MoveNext();
            switch (cardCount.Current.Value)
            {
                case 5: return HandType.FiveOfAKind;
                case 4: return HandType.FourOfAKind;
                case 3: { cardCount.MoveNext(); return (cardCount.Current.Value == 2) ? HandType.FullHouse : HandType.ThreeOfAKind; }
                case 2: { cardCount.MoveNext(); return (cardCount.Current.Value == 2) ? HandType.TwoPairs : HandType.OnePair; }
            }
        }

        return HandType.HighCard;
    }

    public static Hand SetHandRank(Hand hand, Dictionary cardValues)
    {
        int rank = 0;
        int offset = 0;

        var cardValueHand = hand.Cards;
        for (int i = cardValueHand.Length - 1; i >= 0; i--)
        {
            var card = cardValueHand[i];
            var cardValue = cardValues[card];
            var offsetCardValue = cardValue << offset;
            rank |= offsetCardValue;
            offset += 4; // To store values up to 13 we need 4 bits.
        }

        // Put the hand type at the high end because it is the most
        // important factor in the rank.
        var handType = (int)IdentifyHandType(hand.Cards);
        var offsetHandType = handType << offset;
        rank |= offsetHandType;

        hand.Rank = rank;
        return hand;
    }
        • snowe@programming.dev
          2·
          1 year ago

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

          • hades
            2·
            1 year 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
              1·
              1 year 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
                1·
                1 year 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).