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POMahjong winning hand algorithm
 

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    <p>I'm looking for an algorithm that will determine if the current mahjong hand is a winning one. If you are not familiar with the game, here's the basic idea (simplified):</p> <blockquote> <ul> <li>There are three suits of tiles, each containing tiles ranked 1-9. There are also special tiles, seven types of them. Each tile exists in four copies, so there are 36 tiles of each suit and 28 special tiles.</li> <li>A hand has 14 tiles in it.</li> <li>A <em>chow</em> is a set of three tiles of a single rank in sequence.</li> <li>A <em>pong</em> is a set of three identical tiles.</li> <li>A <em>kong</em> is a set of four identical tiles.</li> <li>A <em>pair</em> is a set of two identical tiles.</li> <li>A winning hand is one in which the tiles form any number of chows, pongs, and/or kongs and one pair.</li> </ul> </blockquote> <p>The hand is sorted by suit and then by rank. My idea is something like this:</p> <ol> <li>Mark all tiles as unvisited and nonwinning.</li> <li>Visit first unvisited tile.</li> <li>Check subsequent tiles until a chow, pong, or a kong is encountered, or until there is no possibility of it. If a combination is completed, mark all participating tiles as visited and winning</li> <li>If all tiles have been visited, check if all of them are winning. If not all tiles have been visited, go to 2.</li> </ol> <p>The problem is that once a tile is a part of a combination is cannot be a member of any other combination, one that might make the hand a winning one.</p> <p>Any ideas for a working algorithm?</p>
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    1. COpong and kong are the same thing? You've said they're both "a set of three identical tiles"
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    2. COFixed now; thanks.
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    3. COIf a winning hand must contain a pair you can narrow the search space considerably by selecting each possible pair and seeing if the rest makes chow/pong/kong (as opposed to grouping the hand and seeing if the groups happen to have a pair).
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      1. USBen Jackson
 

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