问题
I'm trying to create a permutation of a multidimensional array in classic asp (vbscript) and I'm seriously stuck. I've tried several functions of my own and also tried copying several php versions over, but I often end up with something that either goes into a buffer overflow / infinite recursion or I get results that are more like a combination than a permutation, if I understand the differences correctly.
Lets say it's for a shirt. The shirt can have colors, sizes, and styles. (The actual system allows for any number of "groups" of options (think color, size, etc) and also any number of options within each group (each particular size, each particular color,etc).
For example:
small med lg xl red blue green white pocket no-pocket
Note that the number of elements in either dimension of the array are unknown beforehand; also, not all second dimensions will have the same number of elements.
I need to iterate through each possible unique option that contains an option from each row. In this particular example, there would be 32 options (because I need to ignore results that have an empty value for any given option, since asp doesn't really handle a jagged array the way I would expect. So: small red pocket small red no-pocket small blue pocket small blue no-pocket etc.
Once I have this part done, I'll need to integrate it with some IDs from the database, but I'm fairly sure I can do that part on my own. It's the recursive function that's killing me.
Anyone able to point me in a good starting place or help me out? Any help is MUCH appreciated!
回答1:
To avoid problems of terminology: I wrote a small program:
Dim aaItems : aaItems = Array( _
Array( "small", "med", "lg", "xl" ) _
, Array( "red", "blue", "green", "white" ) _
, Array( "pocket", "no-pocket" ) _
)
Dim oOdoDemo : Set oOdoDemo = New cOdoDemo.init( aaItems )
oOdoDemo.run 33
and that's its output:
0: small red pocket
1: small red no-pocket
2: small blue pocket
3: small blue no-pocket
4: small green pocket
5: small green no-pocket
6: small white pocket
7: small white no-pocket
8: med red pocket
9: med red no-pocket
10: med blue pocket
11: med blue no-pocket
12: med green pocket
13: med green no-pocket
14: med white pocket
15: med white no-pocket
16: lg red pocket
17: lg red no-pocket
18: lg blue pocket
19: lg blue no-pocket
20: lg green pocket
21: lg green no-pocket
22: lg white pocket
23: lg white no-pocket
24: xl red pocket
25: xl red no-pocket
26: xl blue pocket
27: xl blue no-pocket
28: xl green pocket
29: xl green no-pocket
30: xl white pocket
31: xl white no-pocket
32: small red pocket
If that looks like a seed to a solution of your problem, just say so and I will post the code for the cOdoDemo class.
Code for cOdoDemo:
'' cOdoDemo - Q&D combinations generator (odometer approach)
'
' based on ideas from:
' !! http://www.quickperm.org/index.php
' !! http://www.ghettocode.net/perl/Buzzword_Generator
' !! http://www.dreamincode.net/forums/topic/107837-vb6-combinatorics-lottery-problem/
' !! http://stackoverflow.com/questions/127704/algorithm-to-return-all-combinations-of-k-elements-from-n
Class cOdoDemo
Private m_nPlaces ' # of places/slots/digits/indices
Private m_nPlacesUB ' UBound (for VBScript only)
Private m_aLasts ' last index for each place => carry on
Private m_aDigits ' the digits/indices to spin around
Private m_aaItems ' init: AoA containing the elements to spin
Private m_aWords ' one result: array of combined
Private m_nPos ' current increment position
'' init( aaItems ) - use AoA of 'words' in positions to init the
'' odometer
Public Function init( aaItems )
Set init = Me
m_aaItems = aaItems
m_nPlacesUB = UBound( m_aaItems )
m_nPlaces = m_nPlacesUB + 1
ReDim m_aLasts( m_nPlacesUB )
ReDim m_aDigits( m_nPlacesUB )
ReDim m_aWords( m_nPlacesUB )
Dim nRow
For nRow = 0 To m_nPlacesUB
Dim nCol
For nCol = 0 To UBound( m_aaItems( nRow ) )
m_aaItems( nRow )( nCol ) = m_aaItems( nRow )( nCol )
Next
m_aLasts( nRow ) = nCol - 1
Next
reset
End Function ' init
'' reset() - start afresh: all indices/digit set to 0 (=> first word), next
'' increment at utmost right
Public Sub reset()
For m_nPos = 0 To m_nPlacesUB
m_aDigits( m_nPos ) = 0
Next
m_nPos = m_nPlacesUB
End Sub ' reset
'' tick() - increment the current position and deal with carry
Public Sub tick()
m_aDigits( m_nPos ) = m_aDigits( m_nPos ) + 1
If m_aDigits( m_nPos ) > m_aLasts( m_nPos ) Then ' carry to left
For m_nPos = m_nPos - 1 To 0 Step -1
m_aDigits( m_nPos ) = m_aDigits( m_nPos ) + 1
If m_aDigits( m_nPos ) <= m_aLasts( m_nPos ) Then ' carry done
Exit For
End If
Next
For m_nPos = m_nPos + 1 To m_nPlacesUB ' zero to right
m_aDigits( m_nPos ) = 0
Next
m_nPos = m_nPlacesUB ' next increment at utmost right
End If
End Sub ' tick
'' map() - build result array by getting the 'words' for the
'' indices in the current 'digits'
Private Sub map()
Dim nIdx
For nIdx = 0 To m_nPlacesUB
m_aWords( nIdx ) = m_aaItems( nIdx )( m_aDigits( nIdx ) )
Next
End Sub ' map
'' run( nMax ) - reset the odometer, tick/increment it nMax times and
'' display the mapped/translated result
Public Sub run( nMax )
reset
Dim oPad : Set oPad = New cPad.initWW( Len( CStr( nMax ) ) + 1, "L" )
Dim nCnt
For nCnt = 0 To nMax - 1
map
WScript.Echo oPad.pad( nCnt ) & ":", Join( m_aWords )
tick
Next
End Sub ' run
End Class ' cOdoDemo
Some hints/remarks: Think of an odometer that genererates all combinations for 6 (7?) places/digits in numerical order. Now imagine an odometer that lets you specify a sequence/ordered set of 'digits'/words/items for each place/slot. This specification is done by aaItems.
This is the code for cPad, used in .run():
''= cPad - Q&D padding
Class cPad
Private m_nW
Private m_sW
Private m_sS
Private m_nW1
Public Function initWW( nW, sW )
m_nW = nW
m_nW1 = m_nW + 1
m_sW = UCase( sW )
m_sS = Space( nW )
Set initWW = Me
End Function
Public Function initWWC( nW, sW, sC )
Set initWWC = initWW( nW, sW )
m_sS = String( nW, sC )
End Function
Public Function pad( vX )
Dim sX : sX = CStr( vX )
Dim nL : nL = Len( sX )
If nL > m_nW Then
Err.Raise 4711, "cPad::pad()", "too long: " & nL & " > " & m_nW
End If
Select Case m_sW
Case "L"
pad = Right( m_sS & sX, m_nW )
Case "R"
pad = Left( sX & m_sS, m_nW )
Case "C"
pad = Mid( m_sS & sX & m_sS, m_nW1 - ((m_nW1 - nL) \ 2), m_nW )
Case Else
Err.Raise 4711, "cPad::pad() Unknown m_sW: '" & m_sW & "'"
End Select
End Function
End Class ' cPad
Sorry for the missing documentation. I'll try to answer all your question.
回答2:
Generic solution in 20 lines!
Function Permute(parameters)
Dim results, parameter, count, i, j, k, modulus
count = 1
For Each parameter In parameters
count = count * (UBound(parameter) + 1)
Next
results = Array()
Redim results(count - 1)
For i = 0 To count - 1
j = i
For Each parameter In parameters
modulus = UBound(parameter) + 1
k = j Mod modulus
If Len(results(i)) > 0 Then _
results(i) = results(i) & vbTab
results(i) = results(i) & parameter(k)
j = j \ modulus
Next
Next
Permute = results
End Function
回答3:
If you only have to worry about those four fixed categories, just use nested for loops.
If the number of categories may change, a recursive solution is easy to define:
permute(index, permutation[1..n], sources[1..n])
1. if index > n then print(permutation)
2. else then
3 for i = 1 to sources[index].length do
4. permutation[index] = sources[index][i]
5. permute(index+1, permutation, sources)
Call with index=0 and permutation empty for best results (sources is an array of arrays containing your categories).
Example:
index = 1
sources = [[blue, red, green], [small, medium, large], [wool, cotton, NULL], [shirt, NULL, NULL]].
permutation = [NULL, NULL, NULL, NULL]
permute(index, permutation, sources)
note: n = 4 because that's how many categories there are
index > n is false, so...
compute length of sources[1]:
sources[1][1] isn't NULL, so...
sources[1][2] isn't NULL, so...
sources[1][3] isn't NULL, so...
sources[1].length = 3
let i = 1... then permutation[1] = sources[1][1] = blue
permute(2, permutation, sources)
etc.
来源：https://stackoverflow.com/questions/6764744/permutation-of-jagged-array