Day 7: Camel Cards

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  • 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.

    • 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).

    • 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;
    }