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    <p>Kim, your problem description sounds very much like a (homework) attempt to describe the simplest solution for enumerating all k-combinations of a set of n elements - without giving the actual solution away too easily. Anyway, see below for my shot. I used Java but the important parts are not different from C.</p> <pre><code>public class Homework { /** * Prints all k-combinations of a set of n elements. Answer to this * question: http://stackoverflow.com/questions/2698551 */ public static void main(String[] args) { Combinations combinations = new Combinations(7, 3); System.out.printf( "Printing all %d %d-combinations of a set with %d elements:\n", combinations.size(), combinations.k, combinations.n); for (int[] c : combinations) System.out.println(Arrays.toString(c)); } /** * Provides an iterator for all k-combinations of a set of n elements. */ static class Combinations implements Iterable&lt;int[]&gt; { public final int n, k; public Combinations(int n, int k) { if (n &lt; 1 || n &lt; k) throw new IllegalArgumentException(); this.n = n; this.k = k; } @Override public Iterator&lt;int[]&gt; iterator() { return new Iterator&lt;int[]&gt;() { private int[] c; @Override public void remove() { throw new UnsupportedOperationException(); } @Override public int[] next() { if (c == null) { c = new int[k]; for (int i = 0; i &lt; k; i++) c[i] = i; } else { int i = c.length - 1; while (i &gt;= 0 &amp;&amp; c[i] == n - k + i) i--; if (i &lt; 0) throw new NoSuchElementException(); c[i]++; for (int j = i + 1; j &lt; c.length; j++) c[j] = c[i] + j - i; } return c.clone(); // remove defensive copy if performance is more important } @Override public boolean hasNext() { return c == null || c[0] &lt; n - k; } }; } /** * Returns number of combinations: n! / (k! * (n - k)!). */ public BigInteger size() { BigInteger s = BigInteger.valueOf(n); for (int i = n - 1; i &gt; n - k; i--) s = s.multiply(BigInteger.valueOf(i)); for (int i = k; i &gt; 1; i--) s = s.divide(BigInteger.valueOf(i)); return s; } } } </code></pre> <p>Here is the output for your example:</p> <pre><code>Printing all 35 3-combinations of a set with 7 elements: [0, 1, 2] [0, 1, 3] [0, 1, 4] [0, 1, 5] [0, 1, 6] [0, 2, 3] [0, 2, 4] [0, 2, 5] [0, 2, 6] [0, 3, 4] [0, 3, 5] [0, 3, 6] [0, 4, 5] [0, 4, 6] [0, 5, 6] [1, 2, 3] [1, 2, 4] [1, 2, 5] [1, 2, 6] [1, 3, 4] [1, 3, 5] [1, 3, 6] [1, 4, 5] [1, 4, 6] [1, 5, 6] [2, 3, 4] [2, 3, 5] [2, 3, 6] [2, 4, 5] [2, 4, 6] [2, 5, 6] [3, 4, 5] [3, 4, 6] [3, 5, 6] [4, 5, 6] </code></pre>
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    1. POMinimal-change algorithm which maximises 'swapping'
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    1. USWerner Lehmann
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    1. COThank you for your answer. Unfortunately you do not appear to have read my question with sufficient care. I want to generate k-out-of-n combinations in a minimal-change order, i.e. each new generation should change exactly one character. There are well-known algorithms with this property, but your solution does not exhibit it: note the transition [0,1,6] [0,2,3], which changes two elements. The specific requirement in my question — to maximise the immediate replacement of new elements — is not addressed at all.
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      1. USKim Bastin
    2. COYou are right, I should have paid more attention to your minimal-change maximal-swapping requirement. I did not know about minimal-change algorithms (a brief web search proved their existance but I did not see any code without ACM membership etc). Mostly replacing the last elements seems to guarantee a high swap rate for those - but other orderings may maximize that. If the number of combinations is manageable I am sure that at least some linear programming approach can optimize the combination order as desired. A direct solution is preferable, of course. May I ask why you need this?
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      1. USWerner Lehmann
 

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