Visual Basic

Numerical accuracy in calculation

10-Oct-09 23:28

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Google[^]

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Re: Export Data to Excel

The Man from U.N.C.L.E.12-Oct-09 4:18

The Man from U.N.C.L.E.

12-Oct-09 4:18

From what?

Though as has already been posted, Google gives thousands of examples.

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www.JacksonSoft.co.uk

A.Najafi10-Oct-09 21:55

A.Najafi

10-Oct-09 21:55

Hi All,
The problem discussed is accuracy of numeric calculation result,in vb. Namely I compare a numerical mathematic calculation, Line-Plan Intersection, in VB.Net and Matlab 2007 the intersection point result specify in below:
Matlab 2007 ---->
Xint=2.000000000000000
Yint=1.333333333333333
Zint=0.666666666666667
But
Visual Studio 2008 ---->
Xint=2.0000000596046448
Yint=1.3333333730697632
Zint=0.66666668653488159
Hand Calculated ----->
Xint=2.000000000000000
Yint=1.333333333333333
Zint=0.666666666666667
why this error occurs?
Thanks.

Andy_L_J10-Oct-09 23:11

10-Oct-09 23:11

What Data Type are you using for the calculation in VB?.

This looks like an issue with floating-point precision.

A.Najafi11-Oct-09 0:13

A.Najafi

11-Oct-09 0:13

Certinly I have been used Double data type for this calculation. The code inserted in below is the same
story that I have been explained.

Private Sub btnCalc_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button6.Click<br />
<br />
        Dim xint, yint, zint As Double<br />
        Dim ob, oc, ratio As Double<br />
        Dim havint, intree As Boolean<br />
        Dim x0, x1, x2, xb As Double<br />
        Dim y0, y1, y2, yb As Double<br />
        Dim z0, z1, z2, zb As Double<br />
        '3 Points of plane formed triangle too.<br />
        x0 = 2<br />
        x1 = 2<br />
        x2 = 2<br />
        y0 = 0<br />
        y1 = 2<br />
        y2 = 2<br />
        z0 = 2<br />
        z1 = 2<br />
        z2 = 0<br />
        'Points of line, other point is [0 0 0]<br />
        xb = 1.5<br />
        yb = 1<br />
        zb = 0.5<br />
        Line_PlaneIntersection(x0, x1, x2, xb, y0, y1, y2, yb, z0, z1, z2, _<br />
                              zb, xint, yint, zint, ob, oc, ratio, havint, intree)<br />
    End Sub<br />
<br />
<br />
    Public Shared Sub Line_PlaneIntersect(ByVal X0 As Double, ByVal X1 As Double, ByVal X2 As Double, _<br />
                                         ByVal Xb As Double,ByVal Y0 As Double, ByVal Y1 As Double, _<br />
                                         ByVal Y2 As Double,ByVal Yb As Double,ByVal Z0 As Double, _ <br />
                                         ByVal Z1 As Double, ByVal Z2 As Double, ByVal Zb As Double, _<br />
                                         ByRef Xint As Double, ByRef Yint As Double, ByRef Zint As _<br />
                                         Double, ByRef OB As Double, ByRef OC As Double, ByRef Ratio As _<br />
                                         Double,ByRef HaveIntersect As Boolean, ByRef InTriangle As _<br />
                                         Boolean) <br />
                                        <br />
                                         <br />
        '***This formulation is based on the assumption that the line starts from origin***<br />
        Dim t, u, v As Single<br />
        Dim A, B, C, D As Double<br />
        Dim A1, A2, A3 As Double<br />
        Dim B1, B2, B3 As Double<br />
        Dim C1, C2, C3 As Double<br />
        Dim D1, D2, D3, D4, D5, D6 As Double<br />
        Dim Det As Double<br />
        Dim Det1, Det2, Det3, Det4, Det5, Det6 As Double<br />
        Det1 = +Xb * ((Y2 - Y0) * (Z1 - Z0))<br />
        Det2 = +Yb * ((X1 - X0) * (Z2 - Z0))<br />
        Det3 = +Zb * ((X2 - X0) * (Y1 - Y0))<br />
        Det4 = -Xb * ((Y1 - Y0) * (Z2 - Z0))<br />
        Det5 = -Yb * ((X2 - X0) * (Z1 - Z0))<br />
        Det6 = -Zb * ((X1 - X0) * (Y2 - Y0))<br />
        Det = Det1 + Det2 + Det3 + Det4 + Det5 + Det6<br />
        If Det <> 0 Then<br />
            HaveIntersect = True<br />
            A1 = X0 * ((Y1 - Y0) * (Z2 - Z0) - (Y2 - Y0) * (Z1 - Z0))<br />
            A2 = Y0 * ((X2 - X0) * (Z1 - Z0) - (X1 - X0) * (Z2 - Z0))<br />
            A3 = Z0 * ((X1 - X0) * (Y2 - Y0) - (X2 - X0) * (Y1 - Y0))<br />
            A = A1 + A2 + A3<br />
            B1 = X0 * ((Y2 - Y0) * Zb - (Z2 - Z0) * Yb)<br />
            B2 = Y0 * ((Z2 - Z0) * Xb - (X2 - X0) * Zb)<br />
            B3 = Z0 * ((X2 - X0) * Yb - (Y2 - Y0) * Xb)<br />
            B = B1 + B2 + B3<br />
            C1 = X0 * ((Z1 - Z0) * Yb - (Y1 - Y0) * Zb)<br />
            C2 = Y0 * ((X1 - X0) * Zb - (Z1 - Z0) * Xb)<br />
            C3 = Z0 * ((Y1 - Y0) * Xb - (X1 - X0) * Yb)<br />
            C = C1 + C2 + C3<br />
            D1 = +(X1 - X0) * (Y2 - Y0) * Zb<br />
            D2 = +(X2 - X0) * (Z1 - Z0) * Yb<br />
            D3 = +(Y1 - Y0) * (Z2 - Z0) * Xb<br />
            D4 = -(X1 - X0) * (Z2 - Z0) * Yb<br />
            D5 = -(X2 - X0) * (Y1 - Y0) * Zb<br />
            D6 = -(Y2 - Y0) * (Z1 - Z0) * Xb<br />
            D = D1 + D2 + D3 + D4 + D5 + D6<br />
            t = A / D<br />
            u = B / D<br />
            v = C / D<br />
            Xint = Xb * t<br />
            Yint = Yb * t<br />
            Zint = Zb * t <br />
            Dim Bool1, Bool2 As Boolean<br />
            Bool1 = u >= 0 And u <= 1<br />
            Bool2 = v >= 0 And v <= 1<br />
            If Bool1 And Bool2 And (u + v) <= 1 Then<br />
                'means that the intersection point lies in triangle wroughted from 3 points of the plane  <br />
                InTriangle = True<br />
                OB = Math.Sqrt(Xb ^ 2 + Yb ^ 2 + Zb ^ 2)<br />
                OC = Math.Sqrt(Xint ^ 2 + Yint ^ 2 + Zint ^ 2)<br />
                Ratio = OB / OC<br />
            End If<br />
        Else<br />
            HaveIntersect = False<br />
            InTriangle = False<br />
        End If<br />
    End Sub

'***************************************************************************************************

%MAtlab syntax

clear<br />
%Data of line's points<br />
xa=0; ya=0; za=0;<br />
xb=1.5; yb=1; zb=1/2;<br />
Ia=[xa<br />
    ya<br />
    za];<br />
Ib=[xb<br />
    yb<br />
    zb];<br />
%Data of plane<br />
x0=2; y0=0; z0=2;<br />
x1=2; y1=2; z1=2;<br />
x2=2; y2=2; z2=0;<br />
A=      [xa-xb x1-x0 x2-x0<br />
         ya-yb y1-y0 y2-y0<br />
         za-zb z1-z0 z2-z0];<br />
B=[xa-x0<br />
   ya-y0<br />
   za-z0];<br />
<br />
C=A^-1*B<br />
<br />
Iint=Ia+(Ib-Ia).*C(1)<br />

after running
Iint =

2.0000
1.3333
0.6667

Andy_L_J11-Oct-09 0:34

11-Oct-09 0:34

I am not Maths Wizard (I was a terrible student), but this excerpt from Wikipaedia on double floating point precision may shed some light...
Why do you need to return so many digits after the decimal point?

Addition and subtraction

A simple method to add floating-point numbers is to first represent them with the same exponent. In the example below, the second number is shifted right by three digits, and we then proceed with the usual addition method:

123456.7 = 1.234567 * 10^5
  101.7654 = 1.017654 * 10^2 = 0.001017654 * 10^5
  
  Hence:
  123456.7 + 101.7654 = (1.234567 * 10^5) + (1.017654 * 10^2)
                      = (1.234567 * 10^5) + (0.001017654 * 10^5)
                      = (1.234567 + 0.001017654) * 10^5
                      =  1.235584654 * 10^5

In detail:

e=5;  s=1.234567     (123456.7)
+ e=2;  s=1.017654     (101.7654)

  e=5;  s=1.234567
+ e=5;  s=0.001017654  (after shifting)
--------------------
  e=5;  s=1.235584654  (true sum: 123558.4654)

This is the true result, the exact sum of the operands. It will be rounded to seven digits and then normalized if necessary. The final result is

e=5;  s=1.235585    (final sum: 123558.5)

Note that the low 3 digits of the second operand (654) are essentially lost. This is round-off error. In extreme cases, the sum of two non-zero numbers may be equal to one of them:

 e=5;  s=1.234567
+ e=-3; s=9.876543

  e=5;  s=1.234567
+ e=5;  s=0.00000009876543 (after shifting)
----------------------
  e=5;  s=1.23456709876543 (true sum)
  e=5;  s=1.234567         (after rounding/normalization)

Another problem of loss of significance occurs when two close numbers are subtracted. In the following example e = 5; s = 1.234571 and e = 5; s = 1.234567 are representations of the rationals 123457.1467 and 123456.659.

e=5;  s=1.234571
- e=5;  s=1.234567
----------------
  e=5;  s=0.000004
  e=-1; s=4.000000 (after rounding/normalization)

The best representation of this difference is e = −1; s = 4.877000, which differs more than 20% from e = −1; s = 4.000000. In extreme cases, the final result may be zero even though an exact calculation may be several million. This cancellation illustrates the danger in assuming that all of the digits of a computed result are meaningful. Dealing with the consequences of these errors is a topic in numerical analysis; see also Accuracy problems.

A.Najafi11-Oct-09 1:42

A.Najafi

11-Oct-09 1:42

I find answer! Because I had been defined three of variable as Single.
However this variables were small but when I change those to Double this problem solved.
Dim t, u, v As Single -------->Dim t, u, v As Double
Sorry

Dave Kreskowiak11-Oct-09 6:57

Dave Kreskowiak

11-Oct-09 6:57

Using the Double datatype does not "solve" the problem. It mearly masks it. It still has the same problem as the Single type. Google and read up on "IEEE 754" and "IEEE 854" to understand how number representation works in computers and why you're getting the results you are.

A guide to posting questions on CodeProject[^]

Dave Kreskowiak
Microsoft MVP
Visual Developer - Visual Basic
2006, 2007, 2008
But no longer in 2009...

inserting input into matlab and run m-file from vb

Luc Pattyn11-Oct-09 8:13

Luc Pattyn

11-Oct-09 8:13

Hmm. In this case I think single/double is the root cause; the OP complained about the numbers his code produced (which displayed some 16 digits of which obviously only 7 or so were correct). He wasn't really worried about the very last digit (yet).

Smile | :)

Luc Pattyn

I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages

Local announcement (Antwerp region): Lange Wapper? Neen!

waiting 4 some110-Oct-09 21:17

waiting 4 some1

10-Oct-09 21:17

i want to insert some values "from visual basic into one m-file in matlab" then run another m-file from visual basic that will get benefit from the first m-file with the inserted values. i really would apprecitae any help. this is what i wrote so far in visual basic:

Dim Matlab As Object
Dim Result As String
Dim MReal (0,3) As Double
Dim MImag() As Double

Set Matlab = createObject("Matlab.Application")

' here where i saved the two m-files
Result = Matlab.Excute("C:\Porgram Files\MATLAB71\work")

'this is the second m-file which i want to run
Result = Matlab.Excute("solvingnle")

Call Matlab.GetFullMatrix("X", "base", MReal, MImag)

'here where i will show the result of my m-file, each result of the matrix seperate
Text1.Text = MReal(0, 0)
Text2.Text = MReal(0, 1)
Text3.Text = MReal(0, 2)
Text4.Text = MReal(0, 3)
End Sub

How to add headings to a dataset

Dhruva Hein10-Oct-09 9:27

Dhruva Hein

10-Oct-09 9:27

I'm building a dataset by querying several different Oracle tables with the same query.

I want to output the dataset to an Excel file with a separate heading for the results from each Oracle table.

To do this, so far I've added a 'Heading" column to the datatable, but I can't figure out how to fill that column to identify the datarows that came from each Oracle table.

If anyone can shed some light on this it would be appreciated.

For i = 0 To 3

Try

CN = New OracleConnection("Data Source=xe;User Id=" & Username(i) & ";Password=" & Password(i))

'create the dataadapter object

Dim adp As New OracleDataAdapter(SQL(i), CN)

adp.Fill(ds)

If i = 0 Then

ds.Tables(0).Columns.Add("Headings")

End If

Henry Minute10-Oct-09 11:18

Henry Minute

10-Oct-09 11:18

I don't think that you need to add a 'Headings' column.

A DataSet has a Tables collection, each DataTable in that collection has a TableName property. so that is accessible by iterating over the Tables collection.

Each DataTable has a Columns collection which is a collection of DataColumns, each of which has a ColumnName property.

So iterate over the Tables collection and for each DataTable returned iterate over its Columns collection. For each DataColumn, get its ColumnName.

Henry Minute

Do not read medical books! You could die of a misprint. - Mark Twain
Girl: (staring) "Why do you need an icy cucumber?"
“I want to report a fraud. The government is lying to us all.”

Dhruva Hein10-Oct-09 12:49

Dhruva Hein

10-Oct-09 12:49

Hi. Thanks for your input.

I'm iterating through a list of users of an Oracle database and running the same query on each. I create a new Oracle dataadapter for each user.

I would have thought that VB would fill a separate table with each user's query results. For some reason it's putting them all in the same table: Table(0).

It would be much easier to handle things if they did go into separate tables, but I can't see how to do this. Do you have any suggestions?

Thanks again.

Henry Minute10-Oct-09 13:38

Henry Minute

10-Oct-09 13:38

Without seeing your code, it is difficult to be certain, and I am also not familiar with Oracle from a .Net point of view.

But if you are iterating a user list, it should be possible to create a new Table on each loop and have the query results added to that Table which then gets added to the DataSet.

Open File with my application

Dave Kreskowiak10-Oct-09 16:15

Dave Kreskowiak

10-Oct-09 16:15

Creating a new table for each user in your list would be exceedingly tedious to manage and also adds a lot of weight to your data in memory. DataTables are not light-weight objects. So, if you have thousands of users, you're going to consuming vast amounts of memory, needlessly.

From your limited description, it sounds like you have a serious design flaw in your data model.

A guide to posting questions on CodeProject[^]

Dave Kreskowiak
Microsoft MVP
Visual Developer - Visual Basic
2006, 2007, 2008
But no longer in 2009...

Anubhava Dimri10-Oct-09 0:24

Anubhava Dimri

10-Oct-09 0:24

Hello to All,

I am creating file with .ttg extenction. And i want to open those files with my application(Default) by using programming.

So how can i do that?

Thanks for help

If you can think then I Can.

Re: Open File with my application

Steven J Jowett10-Oct-09 0:41

Steven J Jowett

10-Oct-09 0:41

look here[^]

Steve Jowett
-------------------------
Real programmers don't comment their code. If it was hard to write, it should be hard to read.

Need Help

Purish Dwivedi9-Oct-09 22:53

Purish Dwivedi

9-Oct-09 22:53

I am a beginner in VB.
I am creating a Window Application. I want to authenticate it with username and password, on first screen.
How can it be done?
Thank You.

Re: Need Help

Andy_L_J9-Oct-09 23:03