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POGenerating all n-letter permutations
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  1. POGenerating all n-letter permutations
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    copied!<p>I am attempting to calculate all the possible 3 letter <em>permutations</em>, using the 26 letters (Which amounts to only 26*25*24=15,600). The order of the letters matters, and I don't want repeating letters. (I wanted the permutations to be generated in lexicographical order, but that isn't necessary)</p> <p>So far I attempted to nest <code>for</code> loops, but I ended up iterating through every <em>combination</em> possible. So there are repeating letters, which I do not want, and the <code>for</code> loops can become difficult to manage if I want more than 3 letters.</p> <p>I can flip through the letters until I get a letter that has not been used, but it isn't in lexicographical order and it is <em>much</em> slower than using <code>next_permutation</code> (I cannot use this <code>std</code> method because I'm left calculating all of the subsets of the 26 letters). </p> <p>Is there a more efficient way to do this? To put in perspective of the inefficiency, <code>next_permutation</code> iterates through the first 6 digits instantaneously. However, it takes several seconds to get all the three letter permutations using this method, and <code>next_permutation</code> still quickly becomes inefficient with the 2^n subsets I must calculate.</p> <p>Here is what I have for the nested <code>for</code> loops:</p> <pre><code>char key[] = {'a','b','c','d','e','f','g','h','i','j','k', 'l','m','n','o','p','r','s','t','u','v','w','x','y','z'}; bool used[25]; ZeroMemory( used, sizeof(bool)*25 ); for( int i = 0; i &lt; 25; i++ ) { while( used[i] == true ) i++; if( i &gt;= 25 ) break; used[i] = true; for( int j = 0; j &lt; 25; j++ ) { while( used[j] == true ) j++; if( j &gt;= 25 ) break; used[j] = true; for( int k = 0; k &lt; 25; k++ ) { while( used[k] == true ) k++; if( k &gt;= 25 ) break; used[k] = true; cout &lt;&lt; key[i] &lt;&lt; key[j] &lt;&lt; key[k] &lt;&lt; endl; used[k] = false; } used[j] = false; } used[i] = false; } </code></pre>
 

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