org.jgrapht.experimental.permutation

Class CompoundPermutationIter

java.lang.Object

org.jgrapht.experimental.permutation.CompoundPermutationIter

All Implemented Interfaces:

Iterator, ArrayPermutationsIter

public class CompoundPermutationIter
extends Object
implements ArrayPermutationsIter, Iterator

For permutation like this:

1,2 are the same eq.group (numbers) a,b are og the same eq.group (letters) '$' is of its own eq. group (signs) Let the order of the group be (arbitrary): signs,numbers,letters (note that for performance reasons, this arbitrary order is the worst! see Performance section below)

These are the possible compound perm: [$,1,2,a,b,c]

[$,1,2,a,c,b]

[$,1,2,b,a,c]

[$,1,2,b,c,a]

[$,1,2,c,a,b]

[$,1,2,c,b,a]

[$,2,1,a,b,c]

[$,2,1,a,c,b]

[$,2,1,b,a,c]

[$,2,1,b,c,a]

[$,2,1,c,a,b]

[$,2,1,c,b,a]

The overall number is the product of the factorials of each eq. group size; in our example : (1!)x(2!)x(3!)=1x2x6=12. Using the constructor with eq.group sizes and initial order [1,2,3], the result permutations are retrieved as numbers in an array, where [0,1,2,3,4,5] means [$,1,2,a,b,c]:

[0,1,2,3,5,4]

[0,1,2,4,3,5]

etc. etc., till:

[0,2,1,5,4,3] means [$,2,1,c,b,a]

Performance: The implementation tries to advance each time the group zero, if it does not succeed, it tries the next group (1,2 and so on), so: try to put the largest group as the first groups, UNLIKE the example. Performance-wise it is better to do [a,b,c,1,2,$] .The effect is improvement by constant (for example, by 2)

Since:

May 30, 2005

Author:

Assaf