I have a question that is similar, but not identical, to the one answered here.
I would like a function generate all of the k-combinations of elements from
I tried creating an enumerable that can accomplish this task in VB. This is the result:
Public Class CombinationEnumerable(Of T)
Implements IEnumerable(Of List(Of T))
Private m_Enumerator As CombinationEnumerator
Public Sub New(ByVal values As List(Of T), ByVal length As Integer)
m_Enumerator = New CombinationEnumerator(values, length)
End Sub
Public Function GetEnumerator() As System.Collections.Generic.IEnumerator(Of List(Of T)) Implements System.Collections.Generic.IEnumerable(Of List(Of T)).GetEnumerator
Return m_Enumerator
End Function
Private Function GetEnumerator1() As System.Collections.IEnumerator Implements System.Collections.IEnumerable.GetEnumerator
Return m_Enumerator
End Function
Private Class CombinationEnumerator
Implements IEnumerator(Of List(Of T))
Private ReadOnly m_List As List(Of T)
Private ReadOnly m_Length As Integer
''//The positions that form the current combination
Private m_Positions As List(Of Integer)
''//The index in m_Positions that we are currently moving
Private m_CurrentIndex As Integer
Private m_Finished As Boolean
Public Sub New(ByVal list As List(Of T), ByVal length As Integer)
m_List = New List(Of T)(list)
m_Length = length
End Sub
Public ReadOnly Property Current() As List(Of T) Implements System.Collections.Generic.IEnumerator(Of List(Of T)).Current
Get
If m_Finished Then
Return Nothing
End If
Dim combination As New List(Of T)
For Each position In m_Positions
combination.Add(m_List(position))
Next
Return combination
End Get
End Property
Private ReadOnly Property Current1() As Object Implements System.Collections.IEnumerator.Current
Get
Return Me.Current
End Get
End Property
Public Function MoveNext() As Boolean Implements System.Collections.IEnumerator.MoveNext
If m_Positions Is Nothing Then
Reset()
Return True
End If
While m_CurrentIndex > -1 AndAlso (Not IsFree(m_Positions(m_CurrentIndex) + 1)) _
''//Decrement index of the position we're moving
m_CurrentIndex -= 1
End While
If m_CurrentIndex = -1 Then
''//We have finished
m_Finished = True
Return False
End If
''//Increment the position of the last index that we can move
m_Positions(m_CurrentIndex) += 1
''//Add next positions just after it
Dim newPosition As Integer = m_Positions(m_CurrentIndex) + 1
For i As Integer = m_CurrentIndex + 1 To m_Positions.Count - 1
m_Positions(i) = newPosition
newPosition += 1
Next
m_CurrentIndex = m_Positions.Count - 1
Return True
End Function
Public Sub Reset() Implements System.Collections.IEnumerator.Reset
m_Finished = False
m_Positions = New List(Of Integer)
For i As Integer = 0 To m_Length - 1
m_Positions.Add(i)
Next
m_CurrentIndex = m_Length - 1
End Sub
Private Function IsFree(ByVal position As Integer) As Boolean
If position < 0 OrElse position >= m_List.Count Then
Return False
End If
Return Not m_Positions.Contains(position)
End Function
''//Add IDisposable support here
End Class
End Class
...and you can use my code this way:
Dim list As New List(Of Integer)(...)
Dim iterator As New CombinationEnumerable(Of Integer)(list, 3)
For Each combination In iterator
Console.WriteLine(String.Join(", ", combination.Select(Function(el) el.ToString).ToArray))
Next
My code gives combinations of a specified length (3 in my example) though, I just realized that you wish to have combinations of any length (I think), but it's a good start.
My twist, delivering a sorted list, first by length - then by alpha
Imports System.Collections.Generic
Public Class LettersList
Public Function GetList(ByVal aString As String) As List(Of String)
Dim returnList As New List(Of String)
' Start the recursive method
GetListofLetters(aString, returnList)
' Sort the list, first by length, second by alpha
returnList.Sort(New ListSorter)
Return returnList
End Function
Private Sub GetListofLetters(ByVal aString As String, ByVal aList As List(Of String))
' Alphabetize the word, to make letter key
Dim tempString As String = Alphabetize(aString)
' If the key isn't blank and the list doesn't already have the key, add it
If Not (String.IsNullOrEmpty(tempString)) AndAlso Not (aList.Contains(tempString)) Then
aList.Add(tempString)
End If
' Tear off a letter then recursify it
For i As Integer = 0 To tempString.Length - 1
GetListofLetters(tempString.Remove(i, 1), aList)
Next
End Sub
Private Function Alphabetize(ByVal aString As String) As String
' Turn into a CharArray and then sort it
Dim aCharArray As Char() = aString.ToCharArray()
Array.Sort(aCharArray)
Return New String(aCharArray)
End Function
End Class
Public Class ListSorter
Implements IComparer(Of String)
Public Function Compare(ByVal x As String, ByVal y As String) As Integer Implements System.Collections.Generic.IComparer(Of String).Compare
If x.Length = y.Length Then
Return String.Compare(x, y)
Else
Return (x.Length - y.Length)
End If
End Function
End Class
I can offer the following solution - not yet perfect, not fast, and it assumes the input is a set, hence contains no duplicate items. I am going to add some explanation later.
using System;
using System.Linq;
using System.Collections.Generic;
class Program
{
static void Main()
{
Int32 n = 5;
Int32 k = 3;
Boolean[] falseTrue = new[] { false, true };
Boolean[] pattern = Enumerable.Range(0, n).Select(i => i < k).ToArray();
Int32[] items = Enumerable.Range(1, n).ToArray();
do
{
Int32[] combination = items.Where((e, i) => pattern[i]).ToArray();
String[] stringItems = combination.Select(e => e.ToString()).ToArray();
Console.WriteLine(String.Join(" ", stringItems));
var right = pattern.SkipWhile(f => !f).SkipWhile(f => f).Skip(1);
var left = pattern.Take(n - right.Count() - 1).Reverse().Skip(1);
pattern = left.Concat(falseTrue).Concat(right).ToArray();
}
while (pattern.Count(f => f) == k);
Console.ReadLine();
}
}
It generates a sequence of boolean patterns that determine if an element belongs to the current combination starting with k
times true (1) at the very left and the rest all false (0).
n = 5 k = 3 11100 11010 10110 01110 11001 10101 01101 10011 01011 00100
The next pattern is generated as follows. Assume the current pattern is the following.
00011110000110.....
Scan from left to right and skip all zeros (false).
000|11110000110....
Scan further over the first block of ones (true).
0001111|0000110....
Move all the skipped ones besides the rightmost one back to the very left.
1110001|0000110...
And finally move the rightmost skipped one a single position to the right.
1110000|1000110...
It's not clear to me in what form you want your VB code to return the combinations it generates, but for simplicity let's assume a list of lists. VB does allow recursion, and a recursive solution is simplest. Doing combinations rather than permutations can be obtained easily, by simply respecting the ordering of the input list.
So, the combinations of K items out of a list L that's N items long are:
In pseudocode (using for example .size to give a list's length, [] as an empty list, .append to add an item to a list, .head to get a list's first item, .tail to get the list of "all but the first" items of L):
function combinations(K, L):
if K > L.size: return []
else if K == L.size:
result = []
result.append L
return result
else:
result = []
for each sublist in combinations(K-1, L.tail):
subresult = []
subresult.append L.head
for each item in sublist:
subresult.append item
result.append subresult
for each sublist in combinations(K, L.tail):
result.append sublist
return result
This pseudocode can be made more concise if you assume more flexible list-manipulation syntax. For example, in Python ("executable pseudocode") using "slicing" and "list comprehension" syntax:
def combinations(K, L):
if K > len(L): return []
elif K == len(L): return [L]
else: return [L[:1] + s for s in combinations(K-1, L[1:])
] + combinations(K, L[1:])
Whether you need to verbosely construct lists by repeated .append, or can concisely construct them by list comprehension notation, is a syntax detail (as is the choice of head and tail vs list slicing notation to get the first item of the list vs the rest): the pseudocode is intended to express exactly the same idea (which is also the same idea expressed in English in the previous numbered list). You can implement the idea in any language that is capable of recursion (and, of course, some minimal list operations!-).
Code in C# that produces list of combinations as arrays of k elements:
public static class ListExtensions
{
public static IEnumerable<T[]> Combinations<T>(this IEnumerable<T> elements, int k)
{
List<T[]> result = new List<T[]>();
if (k == 0)
{
// single combination: empty set
result.Add(new T[0]);
}
else
{
int current = 1;
foreach (T element in elements)
{
// combine each element with (k - 1)-combinations of subsequent elements
result.AddRange(elements
.Skip(current++)
.Combinations(k - 1)
.Select(combination => (new T[] { element }).Concat(combination).ToArray())
);
}
}
return result;
}
}
Collection initializer syntax used here is available in VB 2010 (source).