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    <p>It is not that simple, you need to create a utility function that lets you know which is better matching. For example, which will be better: <br>(1) one pair with excellent match, the other is terribly poor. <br>(2) two pairs, with an average match.</p> <p>I suggest using some well proven <a href="http://en.wikipedia.org/wiki/Artificial_intelligence" rel="nofollow">artificial intelligence</a> tools for this problem. <br>first, some definitions: </p> <ul> <li><code>P={all persons}</code> </li> <li><code>S={all possibilities to match all people}</code></li> <li>let <code>u:PxP-&gt;R</code> be a scoring function for a pair: <code>u(p1,p2)=their score as a match</code> [depends on your data of course]</li> <li>let <code>U:S-&gt;R</code> be a scoring function for the whole matching: <code>U(aMatch) = Sigma(u(p1,p2)) for each paired couple</code>.</li> <li>let <code>next:S-&gt;2^S</code> be: <code>next(S)={all possible matchings which are different from S by swapping two people only}</code></li> </ul> <p>Now, with the above definition you can use <a href="http://en.wikipedia.org/wiki/Hill_climbing#Variants" rel="nofollow">steepest ascent hill climbing</a> or a <a href="http://en.wikipedia.org/wiki/Genetic_algorithm" rel="nofollow">genethic algorithm</a>, which are proven Artificial Intelligence tools for getting a good result.</p> <p><strong>Steepest Ascent Hill Climbing</strong> [SAHC] is simple, start with as random matching, and make a small change, until you reach a point where you cannot make a local improvement. This point is a local maximum, and a candidate to be the best solution. restart the process with a different initial state. <br>pseudo code:</p> <pre><code>1. best&lt;- -INFINITY 2. while there is more time 3. choose a random matching 4. NEXT &lt;- next(s) 5. if max{ U(v) | for each v in NEXT} &lt; U(s): //s is a local maximum 5.1. if u(s) &gt; best: best &lt;- u(s) //if s is better then the previous result - store it. 5.2. go to 2. //restart the hill climbing from a different random point. 6. else: 6.1. s &lt;- max { NEXT } 6.2. goto 4. 7. return best //when out of time, return the best solution found so far. </code></pre> <p><strong>Note1:</strong> that this algorithm is <a href="http://en.wikipedia.org/wiki/Anytime_algorithm" rel="nofollow">anytime</a>, which means it will get better results as you give it more time. and if your time->infinity, the algorithm will eventually find the optimal solution. <br><strong>Note2:</strong> the utility function <code>U</code> is just a suggestion, you could also use <code>max{u(p1,p2) | for each pair in the match}</code>, or any other utility function you find fits.</p> <p><strong>EDIT:</strong> <br>Even the simpler problem, which is: two people can simply "match" or "not match" [<code>u(p1,p2)=0</code> or <code>u(p1,p2)=1</code>] is actually the <a href="http://en.wikipedia.org/wiki/Matching_%28graph_theory%29#Maximal_matchings" rel="nofollow">maximal matching problem</a>, which is <a href="http://en.wikipedia.org/wiki/NP-hard" rel="nofollow">NP-Hard</a>, thus there is no known polynomial solution for this problem, and a heuristical solution, such as suggested above should probably be used. <Br>Note that if your graph is <a href="http://en.wikipedia.org/wiki/Bipartite_graph" rel="nofollow">bipartite</a> [i.e. you cannot match male with male/female with female], this problem is solveable in polynomial time. More info: <a href="http://en.wikipedia.org/wiki/Matching_%28graph_theory%29#Maximum_matchings_in_bipartite_graphs" rel="nofollow">matching in bipartite graphs</a></p>
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    1. COMaximum Matching, even with weights in non-bipartite graphs, can actually be solved in polynomial time using for example Edmonds's matching algorithm. "Minimum Maximal Matching" is about finding a maximal matching (that is, one where no edge can be added) that is of minimum size. This doesn't really have anything to do with the problem at hand.
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      1. USFalk Hüffner
 

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