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    <p>Working answer: <a href="http://jsfiddle.net/y3h9L/" rel="nofollow">http://jsfiddle.net/y3h9L/</a></p> <p>OK, so if I understand the requirement A is a one-dimensional array that is assumed to have an even number of elements in x,y pairs.</p> <pre><code>A = [1,2, 1,3, 2,3, 9,10, 9,11, 10,11] // output should be [ [1,2,3], [9,10,11] ] // but if you add an extra pair that links the two halves, say add 2,11 A2 = [1,2, 1,3, 2,3, 9,10, 9,11, 10,11, 2,11] // then all are related so output should be [ [1,2,3,9,10,11] ] </code></pre> <p>I've made no effort to pretty-up or optimise the following code, but it works:</p> <pre><code>// single dimensional array of x,y pairs var A = [1,2, 1,3, 2,3, 9,10, 9,11, 10,11]; // create a working copy of A so that we can remove elements // and still keep the original A intact. var workingCopy = A.slice(0, A.length), matchedPairs = [], currentMatches, finalCombinations = [], x, y, i, j, tempArray; while (workingCopy.length &gt; 0) { currentMatches = []; currentMatches.push([workingCopy.shift(),workingCopy.shift()]); workingCopyLoop: for (x=0,y=1; x &lt; workingCopy.length;) { for (i=0; i &lt; currentMatches.length; i++){ if (workingCopy[x] === currentMatches[i][0] || workingCopy[y] === currentMatches[i][1]) { currentMatches.push([workingCopy.shift(),workingCopy.shift()]); // go back to the beginning of workingCopyLoop x=0; y=1; continue workingCopyLoop; } } x += 2; y += 2; } matchedPairs.push(currentMatches); } for (i=0; i&lt;matchedPairs.length; i++){ tempArray = []; for (j=0; j&lt;matchedPairs[i].length; j++) { // I assume you have a new enough version of JS that you have Array.indexOf() if (-1 === tempArray.indexOf(matchedPairs[i][j][0])) tempArray.push(matchedPairs[i][j][0]); if (-1 === tempArray.indexOf(matchedPairs[i][j][1])) tempArray.push(matchedPairs[i][j][1]); } finalCombinations.push(tempArray); } for (i=0; i&lt;finalCombinations.length; i++) console.log(finalCombinations[i]); // console.log shows that finalCombinations = [ [1,2,3], [9,10,11] ] </code></pre> <p>If it's not obvious how this works, follow it through with a debugger and/or pencil and paper.</p>
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    1. COThis worked so well! I've learned several things while studying this code. Thank you for the assistance.
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    2. COYou're welcome. Since you said you were studying my code note that I did some things with the for loop labeled "workingCopyLoop" which I do not recommend for _most_ for loops, but which happened to work out nicely for the way I happened to tackle your problem: (1) I modified the array being iterated over within the loop (shift() removes an array's first item and returns that item), though it worked fine because each time I did it (2) inside the loop I reset the loop counter to 0 to start the loop again, which is also why...
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    3. CO(3) I omitted the "final expression" in the initial for statement, i.e., the bit just before the closing parenthesis that normally increments the loop counter (since I was fiddling the counters inside the loop). So I guess what I'm saying is that it probably should've been a while loop, but I think (trying to remember back to yesterday) it started out as a more conventional for loop then I realised it wasn't quite doing what I wanted and by the time I had it working I couldn't be bothered changing it to a while loop.
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