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    <p>It seems you are looking for a fast algorithm to calculate all k-combinations. The following Delphi code is a direct translation of the C code found here: <a href="http://compprog.wordpress.com/2007/10/17/generating-combinations-1/" rel="noreferrer">Generating Combinations</a>. I even fixed a bug in that code!</p> <pre><code>program kCombinations; {$APPTYPE CONSOLE} // Prints out a combination like {1, 2} procedure printc(const comb: array of Integer; k: Integer); var i: Integer; begin Write('{'); for i := 0 to k-1 do begin Write(comb[i]+1); if i&lt;k-1 then Write(','); end; Writeln('}'); end; (* Generates the next combination of n elements as k after comb comb =&gt; the previous combination ( use (0, 1, 2, ..., k) for first) k =&gt; the size of the subsets to generate n =&gt; the size of the original set Returns: True if a valid combination was found, False otherwise *) function next_comb(var comb: array of Integer; k, n: Integer): Boolean; var i: Integer; begin i := k - 1; inc(comb[i]); while (i&gt;0) and (comb[i]&gt;=n-k+1+i) do begin dec(i); inc(comb[i]); end; if comb[0]&gt;n-k then// Combination (n-k, n-k+1, ..., n) reached begin // No more combinations can be generated Result := False; exit; end; // comb now looks like (..., x, n, n, n, ..., n). // Turn it into (..., x, x + 1, x + 2, ...) for i := i+1 to k-1 do comb[i] := comb[i-1]+1; Result := True; end; procedure Main; const n = 4;// The size of the set; for {1, 2, 3, 4} it's 4 k = 2;// The size of the subsets; for {1, 2}, {1, 3}, ... it's 2 var i: Integer; comb: array of Integer; begin SetLength(comb, k);// comb[i] is the index of the i-th element in the combination //Setup comb for the initial combination for i := 0 to k-1 do comb[i] := i; // Print the first combination printc(comb, k); // Generate and print all the other combinations while next_comb(comb, k, n) do printc(comb, k); end; begin Main; Readln; end. </code></pre> <p><strong>Output</strong></p> <pre><code>{1,2} {1,3} {1,4} {2,3} {2,4} {3,4} </code></pre>
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    1. POcombination without repetition of N elements without use for..to..do
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    1. USDavid Heffernan
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    1. POcombination without repetition of N elements without use for..to..do
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    1. VO
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    1. COThanks, this solution is more fast. Very very good, thanks again :)
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      1. USMarcello Impastato
    2. COIt evens outputs in the order you are expecting. Nice algorithm.
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      1. USDavid Heffernan
 

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