asked on Combination Generator
Lets say I have two 3 digit numbers 123 ans 234. So If I break it up there are 6 numbers say 1,2,3,2,3,4
So as per the formula as shown below if I want to generate all 3 digit combinations from the above 6 numbers. I believe the possible number of combination is 11 combinations?
(n+r-1)! / (r!(n-1)!)
n = 6
r = 3
(6+3-1)! / 3!(6-1)!
If that is correct then how do I code the same in VBA? Can a generic Function can be written where I pass an Array of n numbers and it outputs the result in an excel sheet?
Sid
So as per the formula as shown below if I want to generate all 3 digit combinations from the above 6 numbers. I believe the possible number of combination is 11 combinations?
(n+r-1)! / (r!(n-1)!)
n = 6
r = 3
(6+3-1)! / 3!(6-1)!
If that is correct then how do I code the same in VBA? Can a generic Function can be written where I pass an Array of n numbers and it outputs the result in an excel sheet?
Sid
Microsoft ExcelAlgorithmsMath / Science
Last Comment
SiddharthRout
ASKER
Few you missed.
221, 331 etc...
Sid
221, 331 etc...
Sid
221 is the same as 122 if you are doing COMBINATIONS (order doesn't matter).
If order matters, you are doing PERMUTATIONS.
Which is it?
If order matters, you are doing PERMUTATIONS.
Which is it?
ASKER
Then it should be permutations. My bad.
Sid
Sid
ASKER
Now this confuses me. For example we have 1,2,3 then as per the formula
(n+r-1)! / (r!(n-1)!)
5! / (3!*2!) = 120/(6*2) = 10
So we are supposed to get 10 permutations or (combination with repetitions) but when I do it manually I get this
111
123
132
222
213
231
333
312
321
Which is just 9. What am I missing?
Sid
(n+r-1)! / (r!(n-1)!)
5! / (3!*2!) = 120/(6*2) = 10
So we are supposed to get 10 permutations or (combination with repetitions) but when I do it manually I get this
111
123
132
222
213
231
333
312
321
Which is just 9. What am I missing?
Sid
If, in the example above, you are looking for these combinations :
12, 13, 14, 21 ,22, 23, 24, 31, 32, 33, 34, 41, 42, 43
note : you are actually getting 14 combinations (of two digits, when combining to 3 digits you will have many more possibilities)
you could use this setup (paste in a module and run the start macro)
12, 13, 14, 21 ,22, 23, 24, 31, 32, 33, 34, 41, 42, 43
note : you are actually getting 14 combinations (of two digits, when combining to 3 digits you will have many more possibilities)
you could use this setup (paste in a module and run the start macro)
Option Base 1
Sub start()
Dim list() As Integer
ReDim list(2)
list(1) = 123
list(2) = 234
combine list
End Sub
Sub combine(numbers() As Integer)
Dim r As Integer
Dim c As Integer
Dim r2 As Integer
Dim c2 As Integer
Dim pos As Integer
Dim text As String
Dim base() As String
Dim result() As String
Dim candidate As String
'-- determine length & size of numbers
number_count = UBound(numbers)
number_len = 0
For Each Number In numbers
If Len(Trim(Str(Number))) > number_len Then number_len = Len(Trim(Str(Number)))
Next Number
'-- tear numbers apart
ReDim base(number_count, number_len)
For r = 1 To number_count
For c = 1 To number_len
text = Trim(Str(numbers(r)))
base(r, c) = Mid(text, c, 1)
Next c
Next r
'-- start with first result
ReDim result(1)
result(1) = base(1, 1) & base(2, 1)
'-- combine digits
For r = 1 To number_count
For c = 1 To number_len
For r2 = 1 To number_count
For c2 = 1 To number_len
'-- do not allow numbers to combine with themselves
If r <> r2 Then
candidate = base(r, c) & base(r2, c2)
'-- check if candidate is already present
invalid = False
For Each Number In result
invalid = (candidate = Number)
If invalid Then Exit For
Next Number
'-- if candidate is valid, add to results
If Not invalid Then
ReDim Preserve result(UBound(result) + 1)
result(UBound(result)) = candidate
End If
End If
Next c2
Next r2
Next c
Next r
'-- save results to worksheet
ActiveSheet.UsedRange.Delete
For Each Item In result
ActiveSheet.Range("A" & ActiveSheet.UsedRange.Rows.Count + 2) = Item
Next Item
End Sub
ASKER
Thanks akoster.
But I wanted for 3 digits and not two digits.
Sid
But I wanted for 3 digits and not two digits.
Sid
I'm sorry, did not get the right approach.
When combining to 3 digits, I get :
123
122
124
132
133
134
142
143
213
212
214
231
232
233
234
221
223
224
241
243
242
312
313
314
321
322
323
324
331
332
334
341
342
343
412
413
421
423
422
431
432
433
When combining to 3 digits, I get :
123
122
124
132
133
134
142
143
213
212
214
231
232
233
234
221
223
224
241
243
242
312
313
314
321
322
323
324
331
332
334
341
342
343
412
413
421
423
422
431
432
433
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ASKER
Brian: Let me understand it.
akoster: I modified your code to this but I am getting less numbers. I just tried it for 123.
Result I got is
123
132
213
231
312
321
I was expecting this
111
123
132
222
213
231
333
312
321
Sid
akoster: I modified your code to this but I am getting less numbers. I just tried it for 123.
Result I got is
123
132
213
231
312
321
I was expecting this
111
123
132
222
213
231
333
312
321
Sid
Option Base 1
Sub start()
Dim list() As Integer
ReDim list(1)
list(1) = 123
'list(2) = 234
combine list
End Sub
Sub combine(numbers() As Integer)
Dim r As Integer
Dim c As Integer
Dim r2 As Integer
Dim c2 As Integer
Dim pos As Integer
Dim text As String
Dim base() As String
Dim result() As String
Dim candidate As String
'-- determine length & size of numbers
number_count = UBound(numbers)
number_len = 0
For Each Number In numbers
If Len(Trim(Str(Number))) > number_len Then number_len = Len(Trim(Str(Number)))
Next Number
'-- tear numbers apart
ReDim base(1)
For r = 1 To number_count
For c = 1 To number_len
text = Trim(Str(numbers(r)))
ReDim Preserve base(UBound(base) + 1)
base(UBound(base) - 1) = Mid(text, c, 1)
Next c
Next r
ReDim Preserve base(UBound(base) - 1)
ReDim result(1)
For x = 1 To UBound(base)
For y = 1 To UBound(base)
For Z = 1 To UBound(base)
If x <> y And x <> Z And y <> Z Then
candidate = base(x) & base(y) & base(Z)
'-- check if candidate is already present
invalid = False
For Each Number In result
invalid = (candidate = Number)
If invalid Then Exit For
Next Number
'-- if candidate is valid, add to results
If Not invalid Then
ReDim Preserve result(UBound(result) + 1)
result(UBound(result)) = candidate
Debug.Print candidate
End If
End If
Next Z
Next y
Next x
'-- save results to worksheet
ActiveSheet.UsedRange.Delete
j = 1
For Each Item In result
ActiveSheet.Range("A" & j) = Item
j = j + 1
Next Item
End Sub
This code filters out identical numbers, when these are expected as well, comment out lines 34 and 48.
apperently it is getting late, the combine subroutine should be started with
Sub start()
Dim list() As Integer
ReDim list(2)
list(1) = 123
list(2) = 234
combine list
End Sub
Sub start()
Dim list() As Integer
ReDim list(2)
list(1) = 123
list(2) = 234
combine list
End Sub
ASKER
akoster: But that is one example. What if I use this?
Sub start()
Dim list() As Integer
ReDim list(1)
list(1) = 123
combine list
End Sub
Sid
Sub start()
Dim list() As Integer
ReDim list(1)
list(1) = 123
combine list
End Sub
Sid
ASKER
Brian: I checked it. It works but like akoster's code it gives me
123
132
213
231
312
321
I was expecting this
111
123
132
222
213
231
333
312
321
Sid
123
132
213
231
312
321
I was expecting this
111
123
132
222
213
231
333
312
321
Sid
also, note that it relies on input numbers of equal length. In order to combine for example '123' with '23456', update the code to include
and corresponding end if statements
[...]
If x <> y And x <> Z And y <> Z Then
If base(x) <> "" And base(y) <> "" And base(Z) <> "" Then
candidate = base(x) & base(y) & base(Z)
[...]
and corresponding end if statements
ASKER
Not that will never be the case. It will always be 3 digit numbers.
Sid
Sid
Sid,
if you also want results like '111', this is not really a permutation of '123' and '234', because there is only one 1 present...
if you also want results like '111', this is not really a permutation of '123' and '234', because there is only one 1 present...
ASKER
so basically, I need to incorporate an extra loop to get those kind of number?
Sid
Sid
Sid,
You're well advanced with akoster so don't bother with this unless that doesn't work out...
Regards,
Brian.
You're well advanced with akoster so don't bother with this unless that doesn't work out...
Dim arrElements As Variant
Dim arrResult As Variant
Dim p As Long
Sub Run_PermutRep()
'http://www.mrexcel.com/forum/showthread.php?t=436932
Dim lRow As Long
arrElements = Array("1", "2", "2", "3", "3", "4")
p = 3
ReDim arrResult(1 To (UBound(arrElements) - LBound(arrElements) + 1) ^ p, 1 To 1)
PermutRep 1, "", lRow
Application.ScreenUpdating = False
Sheets.Add
Range("A1").Resize(UBound(arrResult)) = arrResult
Application.ScreenUpdating = True
End Sub
Sub PermutRep(ByVal lInd As Long, ByVal sresult As String, ByRef lRow As Long)
'http://www.mrexcel.com/forum/showthread.php?t=436932
Dim i As Long
For i = LBound(arrElements) To UBound(arrElements)
If lInd = p Then
lRow = lRow + 1
arrResult(lRow, 1) = sresult & arrElements(i)
Else
PermutRep lInd + 1, sresult & arrElements(i), lRow
End If
Next i
End Sub
Regards,
Brian.
So in effect you are looking for a way to tear apart numbers into digits, and form all possible 3-digit numbers that are based on the available digits ?
that way, the approach could be
that way, the approach could be
Sub combine(numbers() As Integer)
Dim valid1 As Boolean
Dim valid2 As Boolean
Dim valid3 As Boolean
'-- determine length & size of numbers
number_count = UBound(numbers)
number_len = 0
For Each Number In numbers
If Len(Trim(Str(Number))) > number_len Then number_len = Len(Trim(Str(Number)))
Next Number
'-- tear numbers apart
ReDim base(1)
For r = 1 To number_count
For c = 1 To number_len
text = Trim(Str(numbers(r)))
ReDim Preserve base(UBound(base) + 1)
base(UBound(base) - 1) = Mid(text, c, 1)
Next c
Next r
ReDim Preserve base(UBound(base) - 1)
ReDim result(1)
For candidate = 100 To 999
text = Trim(Str(candidate))
valid1 = False
valid2 = False
valid3 = False
For Each Number In base
valid1 = (Mid(text, 1, 1) = Number)
If valid1 Then Exit For
Next Number
For Each Number In base
valid2 = (Mid(text, 2, 1) = Number)
If valid2 Then Exit For
Next Number
For Each Number In base
valid3 = (Mid(text, 3, 1) = Number)
If valid3 Then Exit For
Next Number
If valid1 And valid2 And valid3 Then
ReDim Preserve result(UBound(result) + 1)
result(UBound(result)) = candidate
Debug.Print candidate
End If
Next candidate
'-- save results to worksheet
ActiveSheet.UsedRange.Delete
j = 1
For Each Item In result
ActiveSheet.Range("A" & j) = Item
j = j + 1
Next Item
End Sub
this would lead to
111
112
113
114
121
122
123
124
131
132
133
134
141
142
143
144
211
212
213
214
221
222
223
224
231
232
233
234
241
242
243
244
311
312
313
314
321
322
323
324
331
332
333
334
341
342
343
344
411
412
413
414
421
422
423
424
431
432
433
434
441
442
443
444
111
112
113
114
121
122
123
124
131
132
133
134
141
142
143
144
211
212
213
214
221
222
223
224
231
232
233
234
241
242
243
244
311
312
313
314
321
322
323
324
331
332
333
334
341
342
343
344
411
412
413
414
421
422
423
424
431
432
433
434
441
442
443
444
ASKER
No I don't need the numbers to be repeated like that. I wanted this
111
123
132
222
213
231
333
312
321
for 123
So I have adapted your code to suit my needs. And I am introducing a loop at the end to give me those extra 3 numbers. Thanks very much. I am sharing 50 points with Brian as well. Hope you don't mind :)
Sid
111
123
132
222
213
231
333
312
321
for 123
So I have adapted your code to suit my needs. And I am introducing a loop at the end to give me those extra 3 numbers. Thanks very much. I am sharing 50 points with Brian as well. Hope you don't mind :)
Sid
ASKER
akoster: I found a small glitch. If the number starts with a 0 then it gives me an unexpected output.
I was expecting a 083 (see sheet "Tally Sheet").
Sid
random.xlsm
I was expecting a 083 (see sheet "Tally Sheet").
Sid
random.xlsm
Thanks, Sid!
ASKER
akoster: Also if you see it generates a 2 digit number.
redmondb: You are welcome :)
Sid
redmondb: You are welcome :)
Sid
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