Permutations by swapping: Difference between revisions
Added Kotlin
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"There are 32 permutations of 5 items."</lang>
=={{header|Kotlin}}==
This is based on the recursive Java code found at http://introcs.cs.princeton.edu/java/23recursion/JohnsonTrotter.java.html
<lang scala>// version 1.1.2
fun johnsonTrotter(n: Int): Pair<List<IntArray>, List<Int>> {
val p = IntArray(n) { it } // permutation
val q = IntArray(n) { it } // inverse permutation
val d = IntArray(n) { -1 } // direction = 1 or -1
var sign = 1
val perms = mutableListOf<IntArray>()
val signs = mutableListOf<Int>()
fun permute(k: Int) {
if (k >= n) {
perms.add(p.copyOf())
signs.add(sign)
sign *= -1
return
}
permute(k + 1)
for (i in 0 until k) {
val z = p[q[k] + d[k]]
p[q[k]] = z
p[q[k] + d[k]] = k
q[z] = q[k]
q[k] += d[k]
permute(k + 1)
}
d[k] *= -1
}
permute(0)
return perms to signs
}
fun printPermsAndSigns(perms: List<IntArray>, signs: List<Int>) {
for ((i, perm) in perms.withIndex()) {
println("${perm.contentToString()} -> sign = ${signs[i]}")
}
}
fun main(args: Array<String>) {
val (perms, signs) = johnsonTrotter(3)
printPermsAndSigns(perms, signs)
println()
val (perms2, signs2) = johnsonTrotter(4)
printPermsAndSigns(perms2, signs2)
}</lang>
{{out}}
<pre>
[0, 1, 2] -> sign = 1
[0, 2, 1] -> sign = -1
[2, 0, 1] -> sign = 1
[2, 1, 0] -> sign = -1
[1, 2, 0] -> sign = 1
[1, 0, 2] -> sign = -1
[0, 1, 2, 3] -> sign = 1
[0, 1, 3, 2] -> sign = -1
[0, 3, 1, 2] -> sign = 1
[3, 0, 1, 2] -> sign = -1
[3, 0, 2, 1] -> sign = 1
[0, 3, 2, 1] -> sign = -1
[0, 2, 3, 1] -> sign = 1
[0, 2, 1, 3] -> sign = -1
[2, 0, 1, 3] -> sign = 1
[2, 0, 3, 1] -> sign = -1
[2, 3, 0, 1] -> sign = 1
[3, 2, 0, 1] -> sign = -1
[3, 2, 1, 0] -> sign = 1
[2, 3, 1, 0] -> sign = -1
[2, 1, 3, 0] -> sign = 1
[2, 1, 0, 3] -> sign = -1
[1, 2, 0, 3] -> sign = 1
[1, 2, 3, 0] -> sign = -1
[1, 3, 2, 0] -> sign = 1
[3, 1, 2, 0] -> sign = -1
[3, 1, 0, 2] -> sign = 1
[1, 3, 0, 2] -> sign = -1
[1, 0, 3, 2] -> sign = 1
[1, 0, 2, 3] -> sign = -1
</pre>
=={{header|Lua}}==
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