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# How to make this  looping faster

Posted on 2006-06-21
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Hi! Is there anyway to make this loop faster?

for ctr1 = 1 to NumOfChar
for ctr2 = 1 to NumOfChar
for ctr3 = 1 to NumOfChar
for ctr4 = 1 to NumOfChar
for ctr5 = 1 to NumOfChar

******my statement********

next ctr5
next ctr4
next ctr3
next ctr2
next ctr1
0
Question by:JackOfPH
• 8
• 6
• 5
• +5

LVL 142

Expert Comment

depends alot on what "My Statement" is.
if you are concatenating strings, you might use the stringbuilder class, will be up to 1000% faster
0

LVL 17

Expert Comment

If you are working with strings, with some actions, it is millions% faster to convert them to byte arrays first.

But we need to see or know what your are trying to achieve
0

LVL 10

Expert Comment

hehe, sounds like a comp sci homework question :) (i know that because I've had to answer similar questions
before, and it looks strangely like one of those questions)... if it *is*, you *should* be doing your hmwk yourself...

but technically speaking, yes, to make it loop faster since you have "NumOfChar" as the range each time from 1, the loop
you described is technically equivalent to NumOfChar^5

so
for ctr1 = 1 to (NumOfChar ^ 5)
******my statement********
next ctr1

is technically equivalent to what you wrote, but "faster", because you have less actual "machine" code instructions
being executed each time. It is a very marginal difference (unless you are using a stack each time for the loop, in
which case it is a big difference), but bottomline, the loop above is faster than what you have.

0

LVL 15

Author Comment

to cool12399

This is not a homework... I graduated 2 years ago... I am asking this because I dealing of permutations... in my program... I need to make the program faster.

to others, Angel and inthedark here is the statement...

for ctr1 = 1 to NumOfChar
for ctr2 = 1 to NumOfChar
for ctr3 = 1 to NumOfChar
for ctr4 = 1 to NumOfChar
for ctr5 = 1 to NumOfChar

list1.additem myArray(ctr1) + myArray(ctr2) + myArray(ctr3) + myArray(ctr4) + myArray(ctr5)

next ctr5
next ctr4
next ctr3
next ctr2
next ctr1

0

LVL 15

Author Comment

whats the string builder class? can anyone explain?
0

LVL 15

Author Comment

I am just using vb 6.0? Does string builder comes from vb.net, c#?
0

LVL 26

Expert Comment

>>list1.additem myArray(ctr1) + myArray(ctr2) + myArray(ctr3) + myArray(ctr4) + myArray(ctr5)

If you use listbox, you better change to use listview instead.
0

LVL 85

Expert Comment

It looks like you are basically "counting" using the elements of the "myArray".

What is in "myArray"?

Just to get your creative juices flowing, here is a different approach to a "counting" system:
http://www.experts-exchange.com/Programming/Programming_Languages/Visual_Basic/Q_21857955.html

In the above PAQ, you can specify the characters to use in the counting sequence...
0

LVL 18

Expert Comment

One way of making loops faster is by not using the variable name with the Next clause. So you Next clauses will look like this:

next
next
next
next
next

next ctr5
next ctr4
next ctr3
next ctr2
next ctr1

0

LVL 15

Author Comment

my array holds characters alphbet letters.
0

LVL 18

Expert Comment

What I have mentioned in an optimization tip for any loop and it will work in your case too.
0

LVL 15

Author Comment

Ok trying all your suggestions... will be back in minutes... thanks
0

LVL 10

Expert Comment

hi jack...

if you are trying to enumerate (and create) *all* different possibilities, then there are not really any loop optimizations.
if all you want to do is get the possible #, do something like:
http://www.devx.com/vb2themax/Tip/19017

------->

' number of permutations of N objects in groups of M
'
' Note: requires the FACTORIAL routine

Function Permutations(ByVal Objects As Long, ByVal GroupSize As Long) As Double
Permutations = Factorial(Objects) / Factorial(Objects - GroupSize)
End Function
0

LVL 142

Expert Comment

one of many stringbuilder classes for vb6:
http://www.15seconds.com/howto/pg000929.htm
but as said, this will probably not help here

>my array holds characters alphbet letters.
what kind of combinations? are they in order or what?
what is the max value of NumOfChar?
0

LVL 15

Author Comment

>my array holds characters alphbet letters.
what kind of combinations? are they in order or what?
what is the max value of NumOfChar?

my array holds the characters in the alphabet

for example

myArray(1) holds the value "a" and myArray(2) holds the value "2"
0

LVL 15

Author Comment

Idle mind I tried your comments... and the sample code in the link... but it gave me no speed... is there anyway?
0

LVL 17

Expert Comment

The looping is not the issue here, you could make your code instant. But this will take a bit of time.

But for a quick fix try this.....it will make the code run way much faster

1) Qucik solution

Before the loop

FreezeSet Me
for ctr1 = 1 to NumOfChar
For ctrl2
etc.....
Next
Next

' After all loops are complete
FreezeClear

See next post for an instant solution......

' change private to public if you want to put code in a module.

Private Declare Function LockWindowUpdate Lib "user32" (ByVal hwndLock As Long) As Long

Private Sub FreezeSet(FormToFreeze As Object)
' lock as form so that windows will not update display
' to increase speed or make whole screen appear as if by magic
'Example:  FreezSet Me
LockWindowUpdate FormToFreeze.Hwnd
End Sub
Private Sub FreezeClear()
' release a locked screen
' Example: FreezeClear
LockWindowUpdate 0&
End Sub
0

LVL 142

Expert Comment

I tried some combinations, but with NumOfChar=9, it takes already 2 seconds.
what value of NumOfChar do you use?
0

LVL 142

Expert Comment

inthedark, fyi, the LockWindowUpdate did not change anything in my case...
0

LVL 15

Author Comment

No, change...
0

LVL 17

Expert Comment

Here is a handy permutations class I created for generating permutations for either alpha or numeric values.
It loops like a recordset, for simplicity and will return either a key value or an array of numeric values.

See examples in the declarations...

----------Class: zPermutations.cls
Option Explicit

' Class: zPermutations
' Author: Nick Young nyoung@vipintersoft.com
' Copyright (c) 2003 Nick Young
' You may use & distribute freely but provided that the author and copyright are acknowledged.

' WARNING: Read The Notes for the item property

'EXAMPLES OF USAGE:

'' example 1 using numeric keys
'Dim Px As zPermutations
'Set Px = New zPermutations
'Px.Elements = 3
'Px.Numeric = True
'Px.MoveFirst
'Do While Not Px.EOF
'    Debug.Print Px.CurrentKey
'    Px.MoveNext
'Loop
'Stop
'
'' example 2 using aplha keys
'Set Px = New zPermutations
'Px.Elements = 3
'Px.Numeric = False
'Px.MoveFirst
'Do While Not Px.EOF
'    Debug.Print Px.CurrentKey
'    Px.MoveNext
'Loop
'Stop
'
'' Example 3 using array of values
'Set Px = New zPermutations
'Px.Elements = 3
'Px.Numeric = True
'Px.ArrayBase = 0 ' set
'
'Px.MoveFirst
'
'Dim CurrentKey() As Long
'Dim lc As Long
'
'Do While Not Px.EOF
'    For lc = LBound(CurrentKey) To UBound(CurrentKey)
'        Debug.Print CStr(CurrentKey(lc)); " ";
'    Next lc
'    Debug.Print
'    Px.MoveNext
'Loop
'Stop

Dim mlElements As Long

Dim B() As Boolean
Dim Sequence() As Long

Dim mbComplete As Boolean
Dim mlCurrent As Long

Public ElementsFound As Long

Public Numeric As Boolean

Dim mlKeyCount
Dim mvKeys() As Variant

Dim mlArrayBase As Long

' adds a key value for the next element

mlKeyCount = mlKeyCount + 1
ReDim Preserve mvKeys(mlKeyCount - 1)
mvKeys(mlKeyCount - 1) = Key

End Function

Public Property Let ArrayBase(plNewValue As Long)
mlArrayBase = plNewValue
End Property

Public Function Char(Item) As String
' converts a binary value into a letter
' so zero is A, 1 is B etc.
Char = Chr(Item + 65)
End Function

Public Function CurrentKey() As String

' display the current key value
' this may key may be meaningless if numeric keys above 9 are used.

If Numeric Then
Dim sKey As String
Dim lc As Long

sKey = Space(mlElements + 1)

For lc = 0 To mlElements
Mid(sKey, lc + 1, 1) = Chr(48 + mvKeys(Sequence(lc)))
Next lc
CurrentKey = sKey
Else

sKey = ""

For lc = 0 To mlElements
sKey = sKey & mvKeys(Sequence(lc))
Next lc
CurrentKey = sKey
End If

End Function

' display the current key value
' this may key may be meaningless if numeric keys above 9 are used.
Static bDone As Boolean
If Not bDone Then
bDone = True
ReDim rLongArray(mlArrayBase To mlArrayBase + mlElements)
End If

Dim sKey As String
Dim lc As Long

sKey = ""

For lc = 0 To mlElements
rLongArray(lc + mlArrayBase) = mvKeys(Sequence(lc))
Next lc

End Sub
Public Function EOF() As Boolean

' returns true if all possibles been found?

EOF = mbComplete
End Function
Public Property Get Item(Element) As Variant

' ** WARNING: READ THESE NOTES:
' make this the default proeprty
' Select Tools, Provedure attributes, select the Item propery,
' then Advanced, change the procedure ID to Default,
' then click APPLY

' this will return the key value for an element
' or if no keys were specified it will return
' the numeric value of the sequence

If mlKeyCount > 0 Then
Item = mvKeys(Sequence(Element))
Else
Item = Sequence(Element)
End If

End Property

Public Property Get Value(Element) As Long

' this will return the numeric value for an element

Value = Sequence(Element)

End Property

Public Sub MoveFirst()

' move to the start of the sequence

ReDim B(mlElements)
ReDim Sequence(mlElements)

Dim lc As Long

' SET UP LOWEST POSSIBLE VALUE
For lc = 0 To mlElements
Sequence(lc) = lc
Next

' setup starting point for next move
mlCurrent = mlElements
ElementsFound = 1
mbComplete = False

End Sub

Public Sub MoveNext()

' setup key values for the next item

Dim bInvalid As Boolean
Dim lc As Long
Dim lVal As Long

' keep looping until the next valid permutation is found

' example 42345 is invalid so add 1
' butthe thing that makes this routine fast
' is that it knows that the 2nd 4 is repeated and must be changed next

Do

If mbComplete Then Exit Sub

' now see if the current sequence is a valid permutation

' clear an incidator so see if a value has been found
For lc = 0 To mlElements
B(lc) = False
Next

bInvalid = False

For lc = 0 To mlElements
lVal = Sequence(lc)
If B(lVal) Then
bInvalid = True
Exit For
Else
B(lVal) = True
End If
Next

If Not bInvalid Then
' make sure next advance will work on the last change
mlCurrent = mlElements
Exit Do
End If

Loop

ElementsFound = ElementsFound + 1

End Sub

' to see what this functions does is imagine you
' are working in base 10 and you have the number
' 12349
' add 1 to this number and you get 12350
' this function does the same but in the specified base

' ABC
' ACA
' ACB

Do
' add 1 to the last returned sequence
Sequence(mlCurrent) = Sequence(mlCurrent) + 1
If Sequence(mlCurrent) <= mlElements Then
mlCurrent = mlElements
' Form1.List1.AddItem "N: " + CurrentKey
Exit Sub
End If
Sequence(mlCurrent) = 0
mlCurrent = mlCurrent - 1
If mlCurrent < 0 Then
mbComplete = True
Exit Sub
End If
Loop

End Sub
Public Property Get Elements() As Long

' returns the number of elements

Elements = mlElements + 1

End Property

Public Property Let Elements(NewValue As Long)

' Sets the number of elements to be used

' example value passed as 5
' is elements 0 to 4 so the number 4 is stored

' but don't allow less than 2 elements
If NewValue < 2 Then
mlElements = 1
Else
mlElements = NewValue - 1
End If

ReDim mvKeys(mlElements)
Dim lc As Long
For lc = 0 To mlElements
mvKeys(lc) = Chr(65 + lc)
Next lc
mlKeyCount = 0

MoveFirst

End Property

Public Property Get Permutations() As Double

' calculate the possibilities

Dim P As Double
Dim lc As Double

P = 1
For lc = mlElements + 1 To 2 Step -1
P = P * lc
Next

Permutations = P

End Property

Private Sub Class_Initialize()
mlArrayBase = 0
End Sub

0

LVL 17

Expert Comment

Here is a method to display as you were trying but not using a list box (which is slow) but using a virtual window VP

The results display instantly.

1) New project
2) Add picture box called VP

When you run the app press page/down and page/up to scroll through data.

Also set the VP.appearance to flat and the scalemode to pixels

-----------------------sample.frm
Option Explicit

Dim mlTextHeight As Long
Dim mlLinesInWindow As Long
Dim mlTopLine As Long
Dim mlTotalLines As Long

Dim MyArray(1 To 5) As String

Private Type udtPagePoint
Positions(1 To 5) As Long
End Type

Dim Pages() As udtPagePoint
Dim Current As udtPagePoint

Dim mlLineHeight As Long
Dim mlLinesInVP As Long
Dim mlPage As Long
Dim mlMaxPages As Long

Dim mbEOF As Boolean

' get next number

Dim lc As Long
lc = UBound(MyArray)
Do

Current.Positions(lc) = Current.Positions(lc) + 1
If Current.Positions(lc) <= UBound(MyArray) Then
mbEOF = False
Exit Do
End If
Current.Positions(lc) = 1
lc = lc - 1
If lc < 1 Then
mbEOF = True
Exit Do
End If
Loop

End Sub

Sub DisplayWindow()
If mlPage > UBound(Pages) Then
ReDim Preserve Pages(UBound(Pages) + 100)
End If
If mlPage > mlMaxPages Then
mlMaxPages = mlPage
Pages(mlPage) = Current
mbEOF = False
Else
If mlPage < 0 Then
mlPage = 0
MoveFirst
End If
Current = Pages(mlPage)
mbEOF = False
End If

Dim lc As Long
VP.Cls

Dim lcc As Long

For lc = 0 To mlLinesInVP
VP.CurrentX = 0
VP.CurrentY = lc * mlLineHeight
For lcc = 1 To UBound(Current.Positions)
VP.Print MyArray(Current.Positions(lcc));
Next
If mbEOF Then
Exit Sub
End If
Next

End Sub

Sub MoveFirst()
Dim lc As Long
For lc = 1 To UBound(MyArray)
Current.Positions(lc) = 1
Next
End Sub

Sub WindowSetup()

mlLineHeight = VP.TextHeight("V") * 1.05
mlLinesInVP = VP.ScaleHeight / mlLineHeight

End Sub

Private Sub Form_Activate()
VP.SetFocus
End Sub

MyArray(1) = "A"
MyArray(2) = "B"
MyArray(3) = "C"
MyArray(4) = "D"
MyArray(5) = "E"
mlPage = 0

mlMaxPages = -1

ReDim Pages(100)

WindowSetup

mlPage = 0
MoveFirst
DisplayWindow

End Sub

Private Sub Form_Resize()

If Me.WindowState = vbMinimized Then Exit Sub

VP.Move 0, 0, Me.ScaleWidth, Me.ScaleHeight

End Sub

Private Sub VP_KeyDown(KeyCode As Integer, Shift As Integer)
Select Case KeyCode
Case Is = vbKeyPageUp
mlPage = mlPage - 1
mbEOF = False
DisplayWindow
Case Is = vbKeyPageDown
mlPage = mlPage + 1
DisplayWindow
End Select
End Sub

0

LVL 17

Expert Comment

Me.Show
DoEvents

Before

DisplayWindow
0

LVL 85

Expert Comment

My code will NOT increase your speed at all.

It was simply intended to show you a different way of generating permutations.  Your current method is limited because you are hard coding a series of For...Next loops which cannot be changed programmatically.

The method I demonstrate in my link can generate any length(s) of permutations given the "character set".  You would simply initialize a string to the starting sequence and then repeatedly pass it in to the function inside a while loop until the desired string length is exceeded.

So my method is more flexible, but definitely not any faster...
0

LVL 10

Accepted Solution

cool12399 earned 500 total points
Ahhhhhhhhhhhhhh... Ok.

Now that you clarified your code, I know how to make it faster. You don't want to make the "loop" faster per se.

list1.visible=false

Then have list1.visible = true

You'll notice a huge difference in speed. So basically:

====================================

list1.visible=false
for ctr1 = 1 to NumOfChar
for ctr2 = 1 to NumOfChar
for ctr3 = 1 to NumOfChar
for ctr4 = 1 to NumOfChar
for ctr5 = 1 to NumOfChar
list1.additem myArray(ctr1) + myArray(ctr2) + myArray(ctr3) + myArray(ctr4) + myArray(ctr5)
next ctr5
next ctr4
next ctr3
next ctr2
next ctr1
list1.visible=true

0

LVL 10

Expert Comment

Anyways, there ya go. You'll notice a huge difference. :D
0

LVL 9

Expert Comment

The listbox never even disappears but about 210% speed increase...
0

LVL 10

Expert Comment

yes, once you put the listbox in the code, I knew exactly what you were talking about, because
I've had exactly the same thing! :) if you want it to 'disappear' temporarily (which might even
be a *little* faster), stick a 'doEvents' after you do the 'visible' statement, i.e.,

listbox1.visible=false
doevents
...
etc
...
listbox1.visible=true
doevents

good luck!
0

LVL 26

Expert Comment

If you use listview you will have 500% faster

8->
0

LVL 17

Expert Comment

Virtual window would be million% faster as it is instant.
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