permutation

Question

how to generate the permutation of any set of numbers 11,22,23,45,34 without repetition in vb.net

Ashley Argile · Answer

Here's the VB version as promised. Sorry about the quality of the code but as I mentioned in my last post I'm a C# developer. Feel free to refactor! I've included a Main() so that you can see the output. Imports System.Collections.Generic Module Module1 Sub Main() Dim values() As Integer = {1, 3, 5} Dim permList As New List( Of Integer ()) permList = Permute(values) Dim perm As Integer () For Each perm In permList Dim output As String = String .Empty Dim value As Integer For Each value In perm output += value.ToString Console.WriteLine(output) Next Next Console.ReadLine() End Sub Function Permute( ByVal values() As Integer ) As List( Of Integer ()) Dim permList As New List( Of Integer ()) PermuteWorker(values, 0, values.Length, permList) Return permList End Function Sub PermuteWorker( ByVal values() As Integer , ByVal start As Integer , ByVal n As Integer , ByRef permList As List( Of Integer ())) If start = n - 1 Then Dim thisValues(values.Length) As Integer values.CopyTo(thisValues, 0) permList.Add(thisValues) Else Dim i As Integer For i = start To n - 1 Dim tmp As Integer = values(i) values(i) = values(start) values(start) = tmp PermuteWorker(values, start + 1, n, permList) values(start) = values(i) values(i) = tmp Next End If End Sub End Module Regards Ashley

Ashley Argile · Answer

Here's an algorithm in C#, based on one published by the University of Exeter school of physics . You need to call the Permute function with an array of integers and a list of integer arrays will be returned. permuteWorker recursively does the actual work. I know you wanted it in VB but I'm a C# guy primarily and nobody else has responded. I'll post a VB version this evening (UK time), just thought if you wanted it quick that you might be able to convert it. public List Permute(int[] values) { List permList = new List (); permuteWorker(values, 0, values.Length, ref permList); return permList; } private void permuteWorker(int[] values, int start, int n, ref List permList) { if (start == n - 1) { int[] thisValues = new int[values.Length] ; values.CopyTo(thisValues, 0); permList.Add(thisValues); } else { for (int i = start; i { int tmp = values[i]; values[i] = values[start]; values[start] = tmp; permuteWorker(values, start + 1, n, ref permList); values[start] = values[i]; values[i] = tmp; } } } Regards Ashley

permutation

Answers (2)