This code always through Arithmetic overflow exception. Whats wrong ?
Function ChannelSum(ByVal C As System.Drawing.Color) As Integer
Dim temp As Integer : temp = (C.R + C.G + C.B)
Return temp
End Function
...
Dim x, y, R, G, B, a As Integer : Dim tmp As Color
bmp = New Bitmap(picBox.Tag.ToString)
xMax = bmp.Width - 1 : yMax = bmp.Height - 1
For x = 0 To xMax Step 1
For y = 0 To yMax Step 1
tmp = bmp.GetPixel(x, y) : a = ChannelSum(tmp)
Next y
Next x
The loop breaks in the first encounter !
C.R and the others are byte fields and can only hold a value up to 255. Adding byte fields together will result in a number larger than 255. Use CInt() on each color element first.
temp = (CInt(C.R) + CInt(C.G) + CInt(C.B))
Related
i am a bit new to this but I'm trying to create a randomly generated 3d coordinate points with equal spacing, I've tried using for each loop but im confused on how to use in. the purpose is to generate sphere around that point but some sphere are overlapping each other. thanks in advance. the code below is to show how I'm generating the sphere
For i = 0 To noofsp - 1
x = Rnd(1) * maxDist
ws1.Cells(i + 5, 2) = x
y = Rnd(1) * maxDist
ws1.Cells(i + 5, 3) = y
z = Rnd(1) * maxDist
ws1.Cells(i + 5, 4) = z
centers.Add({x, y, z})
Next
You'll need to check the new point against all the other points to make sure that your new point is at a greater distance that the sum of the radii of your new sphere and each sphere you're checking against
You'll need to use pythagoras' theorem to check that the distances and I found the code below from this site. The code on the site is written in c#, but here is the vb.net version.
Public Function Distance3D(x1 As Double, y1 As Double, z1 As Double, x2 As Double, y2 As Double, z2 As Double) As Double
' __________________________________
'd = √ (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2
'
'Our end result
Dim result As Double
'Take x2-x1, then square it
Dim part1 As Double = Math.Pow((x2 - x1), 2)
'Take y2-y1, then square it
Dim part2 As Double = Math.Pow((y2 - y1), 2)
'Take z2-z1, then square it
Dim part3 As Double = Math.Pow((z2 - z1), 2)
'Add both of the parts together
Dim underRadical As Double = part1 + part2 + part3
'Get the square root of the parts
result = Math.Sqrt(underRadical)
'Return our result
Return result
End Function
To generate the spheres, you would need to expand your code to include checking the new point against all the previously generated points. That code is below with comments.
I have assumed the definition of a variable called minDistance to specify how far apart the centre of the spheres should be. I'm also assuming that all the spheres are the same size. The number should be twice the radius of the spheres
Private Sub GenerateSpheres()
Randomize
For i As Integer = 0 To noofsp - 1
Dim distanceOK As Boolean = False
Dim x, y, z As Integer
'keep generating points until one is found that is
'far enough away. When it is, add it to your data
While distanceOK = False
x = Rnd(1) * maxDist
y = Rnd(1) * maxDist
z = Rnd(1) * maxDist
'If no other points have been generated yet, don't bother
'checking your new point
If centers.Count = 0 Then
distanceOK = True
Else
'If other points exist, loop through the list and check distance
For j As Integer = 0 To centers.Count - 1
'if the point is too close to any other, stop checking,
'exit the For Loop and the While Loop will generate a new
'coordinate for checking, and so on
Dim dist As Integer = Distance3D(centers(j)(0), centers(j)(1), centers(j)(2), x, y, z)
If dist <= minDistance Then
distanceOK = False
'exit the For loop and start the next iteration of the While Loop
Continue While
End If
Next
'If all previous points have been checked none are too close
'flag distanceOK as true
distanceOK = True
End If
End While
'ws1.Cells(i + 5, 2) = x
'ws1.Cells(i + 5, 3) = y
'ws1.Cells(i + 5, 4) = z
centers.Add({x, y, z})
Next
End Sub
I am working on a simple program to return all factors of a given input integer. factors of 32 with while loop array Unfortunately I am stuck.
Code
Dim x As Integer
x = txtInput.Text
Dim factor As Integer
factor = CInt(txtInput.Text) - 1
Dim i As Integer
i = 1
While factor > 0
Do Until i = x
If factor * i = x Then
ListBox1.Items.Add(factor)
i = i + 1
Else
i = i + 1
End If
Loop
factor = factor - 1
End While
If I have a convex curve, and want to find the minimum point (x,y) using a for or while loop. I am thinking of something like
dim y as double
dim LastY as double = 0
for i = 0 to a large number
y=computefunction(i)
if lasty > y then exit for
next
how can I that minimum point? (x is always > 0 and integer)
Very Close
you just need to
dim y as double
dim smallestY as double = computefunction(0)
for i = 0 to aLargeNumber as integer
y=computefunction(i)
if smallestY > y then smallestY=y
next
'now that the loop has finished, smallestY should contain the lowest value of Y
If this code takes a long time to run, you could quite easily turn it into a multi-threaded loop using parallel.For - for example
dim y as Double
dim smallestY as double = computefunction(0)
Parallel.For(0, aLargeNumber, Sub(i As Integer)
y=computefunction(i)
if smallestY > y then smallestY=y
End Sub)
This would automatically create separate threads for each iteration of the loop.
For a sample function:
y = 0.01 * (x - 50) ^ 2 - 5
or properly written like this:
A minimum is mathematically obvious at x = 50 and y = -5, you can verify with google:
Below VB.NET console application, converted from python, finds a minimum at x=50.0000703584199, y=-4.9999999999505, which is correct for the specified tolerance of 0.0001:
Module Module1
Sub Main()
Dim result As Double = GoldenSectionSearch(AddressOf ComputeFunction, 0, 100)
Dim resultString As String = "x=" & result.ToString + ", y=" & ComputeFunction(result).ToString
Console.WriteLine(resultString) 'prints x=50.0000703584199, y=-4.9999999999505
End Sub
Function GoldenSectionSearch(f As Func(Of Double, Double), xStart As Double, xEnd As Double, Optional tol As Double = 0.0001) As Double
Dim gr As Double = (Math.Sqrt(5) - 1) / 2
Dim c As Double = xEnd - gr * (xEnd - xStart)
Dim d As Double = xStart + gr * (xEnd - xStart)
While Math.Abs(c - d) > tol
Dim fc As Double = f(c)
Dim fd As Double = f(d)
If fc < fd Then
xEnd = d
d = c
c = xEnd - gr * (xEnd - xStart)
Else
xStart = c
c = d
d = xStart + gr * (xEnd - xStart)
End If
End While
Return (xEnd + xStart) / 2
End Function
Function ComputeFunction(x As Double)
Return 0.01 * (x - 50) ^ 2 - 5
End Function
End Module
Side note: your initial attempt to find minimum is assuming a function is discrete, which is very unlikely in real life. What you would get with a simple for loop is a very rough estimate, and a long time to find it, as linear search is least efficient among other methods.
First off, here is what I have so far:
Option Explicit
Dim y As Variant
Dim yforx As Variant
Dim yfork As Variant
Dim ynew As Variant
Dim ymin As Variant
Dim x As Variant
Dim xmin As Variant
Dim k As Variant
Dim kmin As Variant
Dim s As Variant
Dim Z As Variant
Dim Track As Variant
Sub PracticeProgram()
'Selects the right sheet
Sheets("PracticeProgram").Select
'y = k ^ 2 * (x ^ 2 + 2 * x * k - 6) / (x + k) ^ 2
'these are the bounds we are stepping through
Track = 0
x = 1
xmin = 1
k = 1
kmin = 1
y = 100000000
yforx = 100000
yfork = 1000000000
Do
y = 100000000
For x = 0 To 1000 Step 0.1
ynew = kmin ^ 2 * (x ^ 2 + 2 * x * kmin - 6) / (x + kmin) ^ 2
'This checks the new y-value against an absurdly high y-value we know is wrong. if it is less than this y-value, we keep the x-value that corresponds with it.
If ynew < y Then
xmin = x
y = ynew
yforx = y
xmin = Application.Evaluate("=Round(" & xmin & ", 3)")
Else
End If
Next
MsgBox (yforx)
For k = 0 To 1000 Step 0.1
y = k ^ 2 * (xmin ^ 2 + 2 * xmin * k - 6) / (xmin + k) ^ 2
If ynew < y Then
kmin = k
y = ynew
yfork = y
kmin = Application.Evaluate("=Round(" & kmin & ",3)")
Else
End If
Next
MsgBox (yfork)
Loop Until (Abs(yforx - yfork) < 10)
End Sub
This program is supposed to find the values of x and k in order to minimize the value of y. This is a practice for a much more complicated program that will use this same concept. In my actual program y, k, and x will all be greater than zero no matter what, but since it was hard to think of a simple equation whose results would be in the shape of a parabola opening up, I decided to allow negative answers for this practice program.
Basically, it should bounce back and forth between the equations finding the ideal values for x and k until finally it has a minimal answer for y using ideal answers for both x and k. I'm not sure what the actual answer is, so I'm letting it stop within a range of 10. If it works, I'll make it smaller, but I don't want the program going for forever, just in case.
MY PROBLEM: I keep getting overflow errors! I'm trying to round the values for xmin and kmin to three figures after the decimal, but it doesn't seem to be helping. Am I using them wrong? Can someone help me get this program working?
You're doing a division by zero. xmin = 0, k = 0, (xmin + k) ^ 2 = 0. (I'm not sure why it isn't reporting division by zero.)
A suggestion: use the Locals pane to see the value of local variables. You can also use the Watch pane to see the value of expressions you want to monitor.
During some iterative optimization, the following VBA code for the computation of the bivariate normal CDF sometimes throws an Overflow error on the line with z = hx * hy * c inside the while loop of the upper function.
I debugged the code and the overflow occurs when the numbers being multiplied result in a number bigger than what a double can hold.
Can you show me how to handle the problem by ignoring the iterations of the loop with such high values - I guess that's the only feasible solution (?). I tried myself with a On Error Goto nextiteration line before the multiplication and placing the nextiteration jump point before the Wend, but the error persists.
Function tetrachoric(x As Double, y As Double, rho As Double) As Double
Const FACCURACY As Double = 0.0000000000001
Const MinStopK As Integer = 20
Dim k As Integer
Dim c As Double
Dim z As Double
Dim s As Double
Dim hx As Double
Dim hx1 As Double
Dim hx2 As Double
Dim hy As Double
Dim hy1 As Double
Dim hy2 As Double
Dim CheckPass As Integer
hx = 1
hy = 1
hx1 = 0
hy1 = 0
k = 0
c = rho
z = c
s = z
CheckPass = 0
While CheckPass < MinStopK
k = k + 1
hx2 = hx1
hy2 = hy1
hx1 = hx
hy1 = hy
hx = x * hx1 - (k - 1) * hx2
hy = y * hy1 - (k - 1) * hy2
c = c * rho / (k + 1)
z = hx * hy * c
s = s + z
If Abs(z / s) < FACCURACY Then
CheckPass = CheckPass + 1
Else
CheckPass = 0
End If
Wend
tetrachoric = s
End Function
Public Function bivnor(x As Double, y As Double, rho As Double) As Double
'
' bivnor function
' Calculates bivariat normal CDF F(x,y,rho) for a pair of standard normal
' random variables with correlation RHO
'
If rho = 0 Then
bivnor = Application.WorksheetFunction.NormSDist(x) * _
Application.WorksheetFunction.NormSDist(y)
Else
bivnor = Application.WorksheetFunction.NormSDist(x) * _
Application.WorksheetFunction.NormSDist(y) + _
Application.WorksheetFunction.NormDist(x, 0, 1, False) * _
Application.WorksheetFunction.NormDist(y, 0, 1, False) * _
tetrachoric(x, y, rho)
End If
End Function
Source: Available for download at http://michael.marginalq.com/
you're hitting on the limits of the computer architecture. Many complex algorithms can't be implemented 1:1 with their mathematical representation because of performance reasons and/or erroneous behavior when overflowing. There's an exceptionally good blog about these issues - John D. Cook.
Please take a look here for a better implementation.
You can also try binding an external library, that gives you arbitrary precision number handling, of course implemented using very expensive (in terms of CPU time) software algorithms. More can be found here.
Updated code using On Error Resume Next instead of On Error Goto:
While CheckPass < MinStopK
k = k + 1
hx2 = hx1
hy2 = hy1
hx1 = hx
hy1 = hy
hx = x * hx1 - (k - 1) * hx2
hy = y * hy1 - (k - 1) * hy2
c = c * rho / (k + 1)
On Error Resume Next
z = hx * hy * c
If Err.Number = 0 Then
s = s + z
If Abs(z / s) < FACCURACY Then
CheckPass = CheckPass + 1
Else
CheckPass = 0
End If
Else
Err.Clear
End If
Wend
http://www.codeproject.com/KB/recipes/float_point.aspx treats how to "Use Logarithms to Avoid Overflow and Underflow", which is a simple but quite effective way of working around overflow problems. In fact, it's so simple yet logical, why haven't we thought of that solution ourselves? ;)