Day 7: Camel Cards

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    • 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).