Why do multiple consecutive unequal conditions not work in vba?
excel, if-statement, vba
Solution
Here is an implementation of the Johnson-Trotter algorithm for enumerating permutations. It is a small modification of one that I wrote once when playing around with brute-force solutions to the Traveling Salesman Problem. Note that it returns a 2-dimensional array, which might consume a lot of memory. It is possible to refactor it so that it is a sub where the permutations are consumed rather than stored. Just replace the part of the code near the bottom (where the current permutation, `perm`, is stored in the array `perms`) by the code that uses the permutation.
Function Permutations(n As Long) As Variant
'implements Johnson-Trotter algorithm for
'listing permutations. Returns results as a variant array
'Thus not feasible for n > 10 or so
Dim perm As Variant, perms As Variant
Dim i As Long, j As Long, k As Long, r As Long, D As Long, m As Long
Dim p_i As Long, p_j As Long
Dim state As Variant
m = Application.WorksheetFunction.Fact(n)
ReDim perm(1 To n)
ReDim perms(1 To m, 1 To n) As Integer
ReDim state(1 To n, 1 To 2) 'state(i,1) = where item i is currently in perm
'state(i,2) = direction of i
k = 1 'will point to current permutation
For i = 1 To n
perm(i) = i
perms(k, i) = i
state(i, 1) = i
state(i, 2) = -1
Next i
state(1, 2) = 0
i = n 'from here on out, i will denote the largest moving
'will be 0 at the end
Do While i > 0
D = state(i, 2)
'swap
p_i = state(i, 1)
p_j = p_i + D
j = perm(p_j)
perm(p_i) = j
state(i, 1) = p_j
perm(p_j) = i
state(j, 1) = p_i
p_i = p_j
If p_i = 1 Or p_i = n Then
state(i, 2) = 0
Else
p_j = p_i + D
If perm(p_j) > i Then state(i, 2) = 0
End If
For j = i + 1 To n
If state(j, 1) < p_i Then
state(j, 2) = 1
Else
state(j, 2) = -1
End If
Next j
'now find i for next pass through loop
If i < n Then
i = n
Else
i = 0
For j = 1 To n
If state(j, 2) <> 0 And j > i Then i = j
Next j
End If
'record perm in perms:
k = k + 1
For r = 1 To n
perms(k, r) = perm(r)
Next r
Loop
Permutations = perms
End Function
Tested like:
Sub test()
Range("A1:G5040").Value = Permutations(7)
Dim A As Variant, i As Long, s As String
A = Permutations(10)
For i = 1 To 10
s = s & " " & A(3628800, i)
Next i
Debug.Print s
End Sub
The first 20 rows of output look like:
Also, `2 1 3 4 5 6 7 8 9 10` is printed in the immediate window. My first version used a vanilla variant away and caused an out-of-memory error with `n = 10`. I tweaked it so that `perms` is redimensioned to contain integers (which consume less memory than variants) and is now able to handle `10`. It takes about 10 seconds on my machine to run the test code.
Problem
I was wondering why the following syntax does not work the way I thought it would in VBA, and what I should do to ensure it does; ``` For a = 1 To 10 For b = 1 To 10 For c = 1 To 10 If a <> b <> c Then MsgBox (a & " " & b & " " & c) End If Next c Next b Next a ``` This is a simplified example, which can still be manually obtained with: ``` if a<>b and b<>c and c<>a then ``` But my actual intended code has 10 such variables multiple times, which makes it unfeasible with 55 unequal conditions, or likely for me to make a typo. I think there is a more efficient way but I have not found it. Ps. My goal is to only have a message box pop up if all the variables are unique. I have obtained my goal, though it can probably be done much more efficient than: ``` For a = 1 To 10 check(a) = True For b = 1 To 10 If check(b) = False Then check(b) = True For c = 1 To 10 If check(c) = False Then check(c) = True For d = 1 To 10 If check(d) = False Then check(d) = True For e = 1 To 10 If check(e) = False Then check(e) = True MsgBox (a & " " & b & " " & c & " " & d & " " & e) End If check(e) = False check(a) = True check(b) = True check(c) = True check(d) = True Next e End If check(d) = False check(a) = True check(b) = True check(c) = True Next d End If check(c) = False check(a) = True check(b) = True Next c End If check(b) = False check(a) = True Next b Next a ```