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    <p>First of all, I would store the relationships as arrays so that you can have duplicate "keys." Key methods: an initial dictionary including every word related to each individual word; a recursive chain expander using map and reduce; filtering chains based on equivalency.</p> <pre><code>Array.prototype.getUnique = function(){ var u = {}, a = []; for(var i = 0, l = this.length; i &lt; l; ++i){ if(u.hasOwnProperty(this[i])) { continue; } a.push(this[i]); u[this[i]] = 1; } return a; } var links = {}; var pairs = [ ["car", "wheel"], ["wheel", "tyre"], ["bed", "sheets"], ["guitar", "strings"], ["guitar", "pickup"], ["rubber", "tyre"], ["truck", "wheel"], ["pickup", "car"] ]; pairs.map(function(pair) { links[pair[0]] = links[pair[0]] || []; links[pair[1]] = links[pair[1]] || []; links[pair[0]].push(pair[1]); links[pair[1]].push(pair[0]); }); var append = function(list) { var related = list.map(function(item) { return links[item]; }).reduce(function(listA, listB) { return listA.concat(listB); }).filter(function(item) { // make sure related only includes new links return list.indexOf(item) == -1 }).getUnique(); return related.length ? append(list.concat(related)) : list.concat(related); }; var branches = []; for( var word in links ) { branches.push(append(links[word].concat(word))); } var compareArrays = function(listA, listB) { if( listA.length != listB.length ) return false; return listA.map(function(element) { if( listB.indexOf(element) == -1 ) return 0; return 1; }).filter(function(el) { return el == 1; }).length == listA.length; }; var _branches = branches; var chains = branches.filter(function(branch1, i) { var isUnique = _branches.filter(function(branch2) { // are they equivalent return compareArrays(branch1, branch2); }).length == 1; delete _branches[i]; return isUnique; }); </code></pre>
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    1. POJavascript group linked key values into arrays
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    1. COthis seems to give me 4 empty arrays and an array containing just rubber http://jsfiddle.net/r4tdN/
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    2. COthanks for the heads up, it's now fully working - at least with the example.
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    3. COSometimes the relationships are quite deep. How do I know how many times to loop through match_chain so that it does not miss these out? http://jsfiddle.net/smghf/
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