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    <p>While the problem looks quite simple it could be related to the NP-complete <a href="http://en.wikipedia.org/wiki/Set_cover_problem" rel="nofollow noreferrer">Set Cover Problem</a>. So it could be possible that there is no efficent way to detect valid combinations, hence no efficent way to count them.</p> <p><strong>UPDATE</strong></p> <p>I thought about the list items beeing pairs because it seems to make the problem harder to attack - you have to check two properties for one item. So I looked for a way to reduce the pair to a scalar item and found a way.</p> <p>Map the set of the <code>n</code> symbols to the set of the first <code>n</code> primes - I will call this function <code>M</code>. In the case of the symbols <code>0</code> to <code>9</code> we obtain the following mapping and <code>M(4) = 11</code> for example.</p> <pre><code>{0, 1, 2, 3, 4, 5, 6, 7, 8, 9} =&gt; {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} </code></pre> <p>Now we can map a pair <code>(n, m)</code> using the mapping X to the product of the mappings of <code>n</code> and <code>m</code>. This will turn the pair <code>(2, 5)</code> into <code>X((2, 5)) = M(2) * M(5) = 5 * 13 = 65</code>.</p> <pre><code>X((n, m)) = M(n) * M(m) </code></pre> <p>Why all this? If we have two pairs <code>(a, b)</code> and <code>(c, d)</code> from two lists, map them using the mapping <code>X</code> to <code>x</code> and <code>y</code> and multiply them, we obtain the product <code>x * y = M(a) * M(b) * M(c) * M(d)</code> - a product of four primes. We can extend the product by more factors by selecting a pair from each list and obtain a product of <code>2w</code> primes if we have <code>w</code> lists. The final question is what does this product tell us about the pairs we selected and multiplied? If the selected pairs form a valid selection, we never choose one symbol twice, hence the product contains no prime twice and is <a href="http://en.wikipedia.org/wiki/Square-free_integer" rel="nofollow noreferrer">square free</a>. If the selection is invalid the product contains at least one prime twice and is not square free. And here a final example.</p> <pre><code>X((2, 5)) = 5 * 13 = 65 X((3, 6)) = 7 * 17 = 119 X((3, 4)) = 7 * 11 = 77 </code></pre> <p>Selecting <code>25</code> and <code>36</code> yields <code>65 * 119 = 7735 = 5 * 7 * 13 * 17</code> and is square free, hence valid. Selecting <code>36</code> and <code>34</code> yields <code>119 * 77 = 9163 = 7² * 11 * 17</code> and is not square free, hence not valid.</p> <p>Also note how nicely this preserves the symmetrie - X((m, n)) = X((n, m)) - and prohibites symmetric pairs because <code>X((m, m)) = M(m) * M(m)</code> is not square free.</p> <p>I don't know if this will be any help, but now you know it and can think about it...^^</p> <hr> <p>This is the first part of an reduction of a 3-SAT problem to this problem. The 3-SET problem is the following.</p> <pre><code>(!A | B | C) &amp; (B | !C | !D) &amp; (A | !B) </code></pre> <p>And here is the reduction as far as I got.</p> <ul> <li>m-n represents a pair</li> <li>a line reprresents a list</li> <li>an asterisk represents an abitrary unique symbol </li> </ul> <hr> <pre><code>A1-A1' !A1-!A1' =&gt; Select A true or false B1-B1' !B1-!B1' =&gt; Select B true or false C1-C1' !C1-!C1' =&gt; Select C true or false D1-D1' !D1-!D1' =&gt; Select D true or false A1-* !B1-* !C1-* =&gt; !A | B | C A2-!A1' !A2-A1' =&gt; Create a copy of A B2-!B1' !B2-B1' =&gt; Create a copy of B C2-!C1' !C2-C1' =&gt; Create a copy of C D2-!D1' !D2-D1' =&gt; Create a copy of D !B2-* C2-* D2-* =&gt; B | !C | !D (How to perform a second copy of the four variables???) !A3-* B3-* </code></pre> <p>If I (or somebody else) can complete this reduction and show how to do it in the general case, this will proof the problem NP-complete. I am just stuck with copying the variables a second time.</p>
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    1. POCounting combinations of pairs of items from multiple lists without repetition
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    1. USDaniel Brückner
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    1. COWow, thanks for the update. I'll spend some time experimenting with this idea :)
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