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    <p>If a User never "wants" a type he "has" for counting the way it can be done combinatorics is enough, no algorithm needed. If you want to "show" the ways the swapping takes place, then :</p> <p>For visualizing the problem, arrange the types in the following way:</p> <pre><code> TYPE A ........................... TYPE N | | | | | | | | | USER 2 | | | | USER 1 | | | | USER 1 | +..........+ +----------+ HAS WANTS +----------+ | USER 3 | | USER 4 | | USER 5 | | | </code></pre> <p>So in our example User 1 has two Type A cards, user 2 has one, and users 3,4 and 5 wants one type A card each. </p> <p>Let's think at this moment of just Type A cards. </p> <p>You need to form pairs for exchanging cards such as the sequential one:</p> <pre><code> {{usr 1,usr3},{usr1, usr 4}, {usr2,usr5}} </code></pre> <p>If every user will be satisfied, there will be same owners than askers. </p> <p>Now, for forming pairs you know you have to pick one element of each set without reposition. As there may be repetitions (users wants or has more than one card of a type), you have to account for that also. This is for counting the ways each type should be "processed". And combinatorics is enough. </p> <p>For forming the pairs you may: </p> <p>1) Construct all different permutations of each set: </p> <pre><code> {2, 1, 1} {1, 2, 1} {1, 1, 2} # == 3!/2! {3, 4, 5} {5, 3, 4} {4, 5, 3} {4, 3, 5} {5, 4, 3} {3, 5, 4} # == 3! </code></pre> <p>2) For each pair of permutations you have a different result set 3 x 6 = 18 in our example. ....</p> <p>Do the same for all card types ...</p> <p>And finally you have N possible result sets for each card type. For getting ALL possible ways to exchange all the cards, you have to combine all possible ways for each type ... </p>
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    1. POFinding all possible ways to swap cards
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    1. USDr. belisarius
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