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PO
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14360328
data
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0
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2013-01-16T14:11:24.473
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2013-01-17T06:37:45.170
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2013-01-17T06:37:45.170
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1796579
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1796579
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14212030
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2
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text
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Here is an algorithm in Java that enumerates the possible sequences by mapping an integer to the sequence. <pre><code>public class Main { private int[] counts = { 3, 2, 1 }; // 3 Symbols A, 2 Symbols B, 1 Symbol C private int n = sum(counts); public static void main(String[] args) { new Main().enumerate(); } private void enumerate() { int s = size(counts); for (int i = 0; i < s; ++i) { String p = perm(i); System.out.printf("%4d -> %s\n", i, p); } } // calculates the total number of symbols still to be placed private int sum(int[] counts) { int n = 0; for (int i = 0; i < counts.length; i++) { n += counts[i]; } return n; } // calculates the number of different sequences with the symbol configuration in counts private int size(int[] counts) { int res = 1; int num = 0; for (int pos = 0; pos < counts.length; pos++) { for (int den = 1; den <= counts[pos]; den++) { res *= ++num; res /= den; } } return res; } // maps the sequence number to a sequence private String perm(int num) { int[] counts = this.counts.clone(); StringBuilder sb = new StringBuilder(n); for (int i = 0; i < n; ++i) { int p = 0; for (;;) { while (counts[p] == 0) { p++; } counts[p]--; int c = size(counts); if (c > num) { sb.append((char) ('A' + p)); break; } counts[p]++; num -= c; p++; } } return sb.toString(); } } </code></pre> The mapping used by the algorithm is as follows. I use the example given in the question (3 x A, 2 x B, 1 x C) to illustrate it. There are 60 (=6!/3!/2!/1!) possible sequences in total, 30 (=5!/2!/2!/1!) of them have an <code>A</code> at the first place, 20 (=5!/3!/1!/1!) have a <code>B</code> at the first place, and 10 (=5!/3!/2!/0!) have a <code>C</code> at the first place. The numbers 0..29 are mapped to all sequences starting with an <code>A</code>, 30..49 are mapped to the sequences starting with <code>B</code>, and 50..59 are mapped to the sequences starting with <code>C</code>. The same process is repeated for the next place in the sequence, for example if we take the sequences starting with <code>B</code> we have now to map numbers 0 (=30-30) .. 19 (=49-30) to the sequences with configuration (3 x A, 1 x B, 1 x C)
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1. POnumber to unique permutation mapping of a sequence containing duplicates
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1. USHenry
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