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Heap's algorithm with Go Lang for possibile permutations of N objects.
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// Suppose we have a permutation containing N different elements. | |
// Heap found a systematic method for choosing at each step a pair of elements to switch, | |
// in order to produce every possible permutation of these elements exactly once. | |
// Let us describe Heap's method in a recursive way. First we set a counter i to 0. | |
// Now we perform the following steps repeatedly until i is equal to N. | |
// We use the algorithm to generate the (N − 1)! permutations of the first N − 1 elements, | |
// adjoining the last element to each of these. This generates all of the permutations that | |
// end with the last element. Then if N is odd, we switch the first element and the last one, | |
// while if N is even we can switch the i th element and the last one (there is no difference between | |
// N even and odd in the first iteration). We add one to the counter i and repeat. In each iteration, | |
// the algorithm will produce all of the permutations that end with the element that has just been | |
// moved to the "last" position. The following pseudocode outputs all permutations of a data array of length N. | |
// Reference https://en.wikipedia.org/wiki/Heap%27s_algorithm | |
package main | |
import "fmt" | |
func HeapPermutation(a []int, size int) { | |
if size == 1 { | |
fmt.Println(a) | |
} | |
for i := 0; i < size; i++ { | |
HeapPermutation(a, size-1) | |
if size%2 == 1 { | |
a[0], a[size-1] = a[size-1], a[0] | |
} else { | |
a[i], a[size-1] = a[size-1], a[i] | |
} | |
} | |
} | |
func main() { | |
a := []int{1, 2, 3, 4} | |
HeapPermutation(a, len(a)) | |
} |
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https://en.wikipedia.org/wiki/Heap%27s_algorithm#Frequent_mis-implementations