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copied!<p>Don't use regex. A regex matches only 1 element as you say, instead you need to find all the possible "meanings" of your string. Given the fact that each element's lenght is 1-2 chars, I'd go with this algorythm (forgive the pseudocode):</p> <pre><code>string[][] results; void formulas(string[] elements, string formula){ string[] elements2=elements; if(checkSymbol(formula.substring(0,1))){ elements.append(formula.substring(0,1)); if(formula.length-1 ==0){ results.append(elements); } else { formulas(elements,formula.substring(1,formula.length); } } if(checkSymbol(formula.substring(0,2))){ elements2.append(formula.substring(0,2)); if(formula.length-2 ==0){ results.append(elements2); } else { formulas(elements2,formula.substring(2,formula.length); } } } bool checkSymbol(string symbol){ // verifies if symbol is the name of an element } </code></pre> <p>input <code>"hgas"</code> (let's go depth first)</p> <p>first step: <code>checkSymbol(formula.substring(0,1))</code> is true for <code>"h"</code></p> <pre><code>elements2 = [h] </code></pre> <p>recursive call, <code>if(checkSymbol(formula.substring(0,1))) false</code></p> <p>then it tests <code>ga</code> => true</p> <pre><code>elements2 = [h, ga] </code></pre> <p>third recursive call</p> <p>test <code>s</code>, <code>checksymbol</code> returns true, elements is then <code>[h, ga, s]</code>. Length of the substring is 0 so it appends to results the first array: <code>[h, ga, s]</code></p> <p>-- let's go back to the second "branch" of first step</p> <p>The test <code>checkSymbol(formula.substring(0,2)</code> finds that <code>"hg"</code> is an element as well</p> <pre><code>elements2 = [hg] </code></pre> <p>Then we call <code>formulas([hg],"as")</code></p> <p>The test for <code>"a"</code> fails (it is not an element) and the test for <code>"as"</code> works, the length is totally consumed, the result <code>[hg,as]</code> is appended to <code>results[]</code></p> <p>This algorythm should run in O(n^2) time in the worst case, n being the length of the string.</p>

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