how to calculate the number of looping?

[vbwire]

i use vb 6.0


[CODE=vb]
Option Explicit

Dim Error As Double
Dim x As Integer
Dim y As Double
Dim z As Integer
Dim eps As Double
Dim N As Integer

Private Sub cmdCalculate_Cl ick()

'user input

If IsNumeric(txtN) And IsNumeric(txtEp s) And IsNumeric(txtX) Then
N = txtN.Text
eps = txtEps.Text
x = txtX.Text
Else
MsgBox "all input must be numeric", vbExclamation, "numeric test"
txtN.SetFocus
End If

'calculate

Do
y = (x) - (x - (N ^ (1 / 2)))

Error = y - x
If Error < 0 Then
Error = Error * -1
End If

z = z + 1

Loop Until Error < eps

'display the result
lblDisplay.Capt ion = "root : " & y & " iteration : " & z



End Sub

Private Sub cmdExit_Click()
Unload Me
End Sub
[/CODE]

the problems..

the 'eps' is the decimal places.in my program above it need user to insert the number of decimal places like this '0.001'.. but actually i just want the user put the value as '3'...

the iteration value must be in integer data type as the result of number of looping it takes to complete the loop.. the number of looping (iteration) didn't match its actually value..

[Stewart Ross]

Hi vbwire. It would be helpful to know what the loop calculation is for. When I test it the calculated error value remains constant and the loop does not exit.

I wonder if your reference to root N (N ^ (1/2)) in the loop is correct? N is not changing after it is entered by the user, and I guess there should a use of the loop counter z somewhere to change the value of N if the iterative calculation is intended to reduce the error value below the loop threshold. This is just a guess, though.

To refer to the number of decimal places for the loop threshold as an integer value you could define a variable NDecimals as an integer, get the value of NDecimals from the user instead of the double eps, then replace your loop exit test with
[CODE=vb]Loop Until Error < 10^(-NDecimals)[/Code]
Another change is to use the Abs (absolute value) function to replace
[CODE=vb]Error = y - x
If Error < 0 Then
Error = Error * -1
End If[/CODE]
with
[CODE=vb]Error = abs(y-x)[/CODE]
Regards

Stewart

[Killer42]

As far as I could see, z should correctly return the number of iterations.

As for the calculation, I haven't looked into it in detail. But if you have any Integer data types in there, VB may convert interim results to Integer, thus losing precision. Try using all Double data types in there. And that includes for the literal values. If you say "1#" instead of just "1", you will force VB to use double data type for the value. In any case, why use (1/2), when you could just use the actual value, 0.5?

Oh, one other thing. Between lines 20 and 21, I believe you should Exit Sub.

[vbwire]

Hi vbwire. It would be helpful to know what the loop calculation is for. When I test it the calculated error value remains constant and the loop does not exit.

the loop calculate the value for y until the different between y and x is smaller than 'eps' ..

z is the number of looping it takes to complete the loop..

the value of x will take the value of y for the next looping.. for example in the first looping, x=2, then get the value of y.. then it will loop,use the value of y we get to calculate the new y. its continue give a new y until the different of the new y and y before it smaller than 'eps'

[MikeTheBike]

the loop calculate the value for y until the different between y and x is smaller than 'eps' ..

z is the number of looping it takes to complete the loop..

the value of x will take the value of y for the next looping.. for example in the first looping, x=2, then get the value of y.. then it will loop,use the value of y we get to calculate the new y. its continue give a new y until the different of the new y and y before it smaller than 'eps'

Hi
I started to look at this yesterday, but work got in the way (such a pain!).

Apart from the use of Abs() and changing the loop exit condition to

Loop Until Error < 0.1 ^ eps

there does not seem to be any iteration in the loop, ie. nothing changes from one loop to the next! So I've moded the code as follows:-

[code=vb]
Option Explicit

Dim Error As Double
'Dim x As integer
Dim x As Double
Dim y As Double
Dim z As Integer
Dim eps As Double
Dim N As Integer

Private Sub cmdCalculate_Cl ick()

Dim txtX As Integer
'Dim txtY As Double
Dim txtZ As Integer
Dim txtN As Integer
Dim txtEps As Integer


'user input
txtX = 3
txtZ = 5
txtN = 1000
txtEps = 3

z = 0
If IsNumeric(txtN) And IsNumeric(txtEp s) And IsNumeric(txtX) Then
N = txtN
eps = txtEps
x = txtX
Else
MsgBox "all input must be numeric", vbExclamation, "numeric test"
'txtN.SetFocus
Exit Sub
End If

'calculate

Do
y = (x) - (x - (N ^ (1 / 2)))

Error = Abs(y - x)
x = y
z = z + 1

Loop Until Error < 0.1 ^ eps

'display the result
'lblDisplay.Cap tion = "root : " & y & " iteration : " & z
MsgBox "root : " & y & " iteration : " & z

End Sub[/code]

ie have added x=y in the loop (as you indicted in the last post), but this always calculated the squre root of N in 2 iterations !!??

I still don't know what you are trying to do ?

Is this homework ?

ps. as this is not written in a form module I've hard coded values of txtN etc.



MTB

[vbwire]

thank to both of you..

this is the program after i corrected it..

[CODE=vb]
Option Explicit

Dim Error As Double
Dim x As Double
Dim y As Double
Dim z As Integer
Dim eps As Double
Dim a As Integer
Dim N As Integer

Private Sub cmdCalculate_Cl ick()

'user input

If IsNumeric(txtN) And IsNumeric(txtEp s) And IsNumeric(txtX) Then
N = txtN.Text
a = txtEps.Text
x = txtX.Text
Else
MsgBox "all input must be numeric", vbExclamation, "numeric test"
txtN.SetFocus
End If

'calculate

eps = (10 ^ (-a))

Do
y = x - (x - (N ^ (1 / 3)))

Error = Abs(y - x)

x = y

z = z + 1

Loop Until Error < eps

'display the result
lblDisplay.Capt ion = "root : " & y & " iteration : " & z

End Sub

Private Sub cmdExit_Click()
Unload Me
End Sub

[/CODE]

please look at my code for displaying the result.. how to make it in two line but i want it in same label.

also.. i want to display the result for root in the decimal palaces chosen by user,'eps'..

[vbwire]

ie have added x=y in the loop (as you indicted in the last post), but this always calculated the squre root of N in 2 iterations !!??

I still don't know what you are trying to do ?

Is this homework ?

ps. as this is not written in a form module I've hard coded values of txtN etc.



MTB

thank you mike.. =)

i just realise my mistake in the calculation..

[CODE=vb]y = (x) - (x - (N ^ (1 / 2)))
[/CODE]
its actually..

[CODE=vb]y = x - (x - (N ^ (1 / 3)))
[/CODE]

and check my previous post.. i still have something for you..haha..