I am trying to implement a 2D Perlin Noise in VB.Net. I've spent the whole day searching for sources that explain the subject and one of the most notable was this article by Hugo Elias
Most of the implementation went well. On the exception of a very important part that did not seem to work in VB.Net, causing overflows.
function Noise1(integer x, integer y)
n = x + y * 57
n = (n<<13) ^ n;
return ( 1.0 - ( (n * (n * n * 15731 + 789221) + 1376312589) & 7fffffff) / 1073741824.0);
end function
In VB.net I translated it to
Private Function Noise(tX As Integer, tY As Integer) As Double
'Return a double between -1 & 1 according to a fixed random seed
Dim n As Integer = tX + tY * 57
n = (n << 13) Xor n
Return (1.0 - ((n * (n * n * 15731 + 789221) + BaseSeed) And &H7FFFFFFF) / 1073741824.0)
End Function
Which cause overflows.
Since the idea seem to be to simply generate a fractional number between -1 and 1. I've made this little function which create a Integer Number based on the coordinates and BaseSeed. BaseSeed(999999) being the base for every noise I'll create in this particular part of my game.
Private Function Noise(tX As Integer, tY As Integer) As Double
Dim tSeed As Integer
tSeed = WrapInteger(789221, BaseSeed, (tX * 1087) + (tY * 2749))
RandomGenerator = New Random(tSeed)
Return (RandomGenerator.Next(-10000, 10001) / 10000)
End Function
WrapInteger simply makes sure that the number will always be in the range of an integer, to avoid overflow errors.
Public Function WrapInteger(ByVal Lenght As Integer, ByVal Position As Integer, ByVal Movement As Integer) As Integer
Lenght += 1
Return ((Position + Movement) + (Lenght * ((Abs(Movement) \ Lenght) + 1))) Mod Lenght
End Function
When I fire it up with a Persistence of 0.25, 6 Octaves and a starting frequency of 2. this is what I get. This is a 128x128 pixel bitmap that I scaled.
Result
Anyone have an idea of why it would be so linear? When I look at this picture I have the feeling that it's not far from the truth, as if it only worked in 1D. All suposition.
Below you will find my entire PerlinNoise Class. I think the rest of it is just fine, but I added it for reference purpose. Beside, I haven't been able to find a single VB implementation of Perlin Noise on the internet. So I guess if I can fix this one, it might help others. There seem to be alot of question about Perlin noise malfunction on StackOverflow
Public Class cdPerlinNoise
Private RandomGenerator As Random
Private BaseSeed As Integer
Private Persistence As Double
Private Frequency As Integer
Private Octaves As Integer
Public Sub New(tSeed As Integer, tPersistence As Double, tOctaves As Integer, tFrequency As Integer)
Frequency = tFrequency
BaseSeed = tSeed
Persistence = tPersistence
Octaves = tOctaves
End Sub
Private Function Noise(tX As Integer, tY As Integer) As Double
Dim tSeed As Integer
tSeed = WrapInteger(789221, BaseSeed, (tX * 1087) + (tY * 2749))
RandomGenerator = New Random(tSeed)
Return (RandomGenerator.Next(-10000, 10001) / 10000)
End Function
Private Function SmoothNoise(tX As Integer, tY As Integer) As Double
Dim Corners As Double = (Noise(tX - 1, tY - 1) + Noise(tX + 1, tY - 1) + Noise(tX - 1, tY + 1) + Noise(tX + 1, tY + 1)) / 16
Dim Sides As Double = (Noise(tX - 1, tY) + Noise(tX + 1, tY) + Noise(tX, tY - 1) + Noise(tX, tY + 1)) / 8
Return (Noise(tX, tY) / 4) + Corners + Sides
End Function
Private Function InterpolateCosine(tA As Double, tB As Double, tX As Double) As Double
Dim f As Double = (1 - Cos(tX * 3.1415927)) * 0.5
Return tA * (1 - f) + tB * f
End Function
Private Function Interpolate2D(tX As Double, tY As Double) As Double
Dim WholeX As Integer = CInt(Fix(tX))
Dim RemainsX As Double = tX - WholeX
Dim WholeY As Integer = CInt(Fix(tY))
Dim RemainsY As Double = tY - WholeY
Dim v1 As Double = SmoothNoise(WholeX, WholeY)
Dim v2 As Double = SmoothNoise(WholeX + 1, WholeY)
Dim v3 As Double = SmoothNoise(WholeX, WholeY + 1)
Dim v4 As Double = SmoothNoise(WholeX + 1, WholeY + 1)
Dim i1 As Double = InterpolateCosine(v1, v2, RemainsX)
Dim i2 As Double = InterpolateCosine(v3, v4, RemainsX)
Return InterpolateCosine(i1, i2, RemainsY)
End Function
Public Function PerlinValue(tX As Double, tY As Double) As Double
Dim Total As Double = 0
Dim Frequency As Double
Dim Amplitude As Double
For i = 0 To Octaves - 1
Frequency = Frequency ^ i
Amplitude = Persistence ^ i
Total = Total + (Interpolate2D(tX * Frequency, tY * Frequency) * Amplitude)
Next
Return Total
End Function
Public Function ScaleNoise(ByVal tX As Double, ByVal tY As Double, ByVal OutputLow As Double, ByVal OutputHigh As Double) As Double
Dim Range1 As Double
Dim Range2 As Double
Dim Result As Double
Range1 = 1 - -1
Range2 = OutputHigh - OutputLow
'(B*C - A*D)/R1 + n1*(R2/R1)
Result = (((1 * OutputLow) - (-1 * OutputHigh)) / Range1) + ((PerlinValue(tX, tY) * (Range2 / Range1)))
If Result < OutputLow Then
Return OutputLow
ElseIf Result > OutputHigh Then
Return OutputHigh
Else
Return Result
End If
End Function
End Class
Related
Hey I have no idea why I get an error "run time error 13 type mismatch". Thats my code and the place where I get an error:
EDIT: That is my code:
Function payoff(S_T, K, CallPut As String)
If CallPut = "call" Then
omega = 1
Else: omega = -1
End If
payoff = WorksheetFunction.Max(omega * (S_T - K), 0)
End Function
Function BS_trajektoria(S_0 As Double, T As Double, r As Double, q As Double, sigma As Double, N As Long) As Double()
Randomize
Dim S() As Double
Dim delta_t As Double
Dim i As Long
ReDim S(N)
S(0) = S_0
delta_t = T / N
For i = 1 To N
S(i) = S(i - 1) * Exp((r - q - 0.5 * sigma ^ 2) * delta_t + sigma * delta_t ^ 0.5 * Application.NormSInv(Rnd))
Next i
BS_trajektoria = S
End Function
Function barrier_MC(S_0 As Double, K As Double, T As Double, r As Double, q As Double, sigma As Double, _
B As Double, N As Long, num_of_sim As Long, CallPut As String, BarType As String) As Double
Randomize
Dim max_value As Double
Dim suma_wyplat As Double
Dim wyplata As Double
Dim i As Long
Dim S() As Double
suma_wyplat = 0
If (BarType = "DO" Or BarType = "DI") And B > S_0 Then
MsgBox "Too high barrier!"
Exit Function
ElseIf (BarType = "UO" Or BarType = "UI") And B < S_0 Then
MsgBox "Too low barrier!"
Exit Function
End If
With WorksheetFunction
For i = 1 To num_of_sim
S = BS_trajektoria(S_0, T, r, q, sigma, N)
max_value = .Max(S)
If max_value >= B Then
wyplata = 0
Else
wyplata = payoff(S(N), K, CallPut)
End If
suma_wyplat = suma_wyplat + wyplata
Next i
End With
barrier_MC = Exp(-r * T) * suma_wyplat / num_of_sim
End Function
Sub test3()
MsgBox barrier_MC(100, 100, 1, 0.05, 0.02, 0.2, 120, 1000, 1000000, "call", "UO")
End Sub
Anyone know where is the problem? For smaller value of N and num_of_sim everything works fine, the problem is when I use bigger values for these variables.
If you declare a new Double variable called rand and modify the main loop so that it looks like:
For i = 1 To N
rand = Rnd
S(i) = S(i - 1) * Exp((r - q - 0.5 * sigma ^ 2) * delta_t + sigma * delta_t ^ 0.5 * Application.NormSInv(rand))
Next i
you will see that the problem always happens when rand = 0. Why it throws that particular error is a bit of a mystery, but it is what it is. As a fix, what you could do is to keep the code as modified above with the following twist:
For i = 1 To N
rand = Rnd
If rand = 0 Then rand = 0.0000001
S(i) = S(i - 1) * Exp((r - q - 0.5 * sigma ^ 2) * delta_t + sigma * delta_t ^ 0.5 * Application.NormSInv(rand))
Next i
Then the code will run without error. It is still somewhat slow, but optimizing it (if possible) would be for a different question.
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.
I am trying a fairly simple function to calculate an exponentially weighted moving average volatility in Excel VBA, following. However, I think there is some error in my function that I can't pinpoint, because I don't get the correct solution.
Function EWMA(numbers As Range, Lambda As Single) As Double
Dim mean As Double
Dim x As Double
Dim c As Range
Dim n As Integer
mean = WorksheetFunction.Average(numbers)
n = WorksheetFunction.Count(numbers)
For Each c In numbers
x = x + (Lambda ^ (n - c.Count)) * ((c.Value - mean) ^ 2)
Next c
EWMA = (1 - Lambda) * x
End Function
The values I am using and the target volatility (calculated using a spreadsheet EWMA) are here.
What am I doing wrong?
Update: using Ron Rosenfeld's solution:
Option Explicit
Function EWMA(Zero As Range, Lambda As Double) As Double
Dim vZero As Variant
Dim SumWtdRtn As Double
Dim I As Long
Dim vPrices As Variant
Dim LogRtn As Double, RtnSQ As Double, WT As Double, WtdRtn As Double
vZero = Zero
For I = 2 To UBound(vZero, 1)
vPrices = 1 / ((1 + vZero(I, 1)) ^ (3 / 12))
LogRtn = Log(vPrices(I - 1, 1) / vPrices(I, 1))
RtnSQ = LogRtn ^ 2
WT = (1 - Lambda) * Lambda ^ (I - 2)
WtdRtn = WT * RtnSQ
SumWtdRtn = SumWtdRtn + WtdRtn
Next I
EWMA = SumWtdRtn ^ (1 / 2)
End Function
Here's a little bit different way of doing it as a VBA function. The inputs are an array of prices, which is assumed to be in descending order as you show, and Lambda. Hopefully, the names of the variables will let you see the logic:
Option Explicit
Function EWMA2(Prices As Range, Lambda As Double) As Double
Dim vPrices As Variant
Dim dSumWtdRtn As Double
Dim I As Long
Dim dLogRtn As Double, dRtnSQ As Double, dWT As Double, dWtdRtn As Double
vPrices = Prices
For I = 2 To UBound(vPrices, 1)
dLogRtn = Log(vPrices(I - 1, 1) / vPrices(I, 1))
dRtnSQ = dLogRtn ^ 2
dWT = (1 - Lambda) * Lambda ^ (I - 2)
dWtdRtn = dWT * dRtnSQ
dSumWtdRtn = dSumWtdRtn + dWtdRtn
Next I
EWMA2 = dSumWtdRtn ^ (1 / 2)
End Function
With your data, it gives the same results as the spreadsheet calculation (within the limits of precision of the data types)
EDIT
If you want to input the 3M CAD Zero Rates as the range input, and not the pricing, then you could modify the above to compute the two relevant prices from the returns data. In this case, it would be:
Option Explicit
Function EWMAV(Zeros As Range, Lambda As Double) As Double
Dim vZeros() As Variant
Dim dPrice1 As Double, dPrice2 As Double
Dim dSumWtdRtn As Double
Dim I As Long
Dim dLogRtn As Double, dRtnSQ As Double, dWT As Double, dWtdRtn As Double
vZeros = Zeros
For I = 2 To UBound(vZeros, 1)
dPrice1 = 1 / ((1 + vZeros(I - 1, 1)) ^ (3 / 12))
dPrice2 = 1 / ((1 + vZeros(I, 1)) ^ (3 / 12))
dLogRtn = Log(dPrice1 / dPrice2)
dRtnSQ = dLogRtn ^ 2
dWT = (1 - Lambda) * Lambda ^ (I - 2)
dWtdRtn = dWT * dRtnSQ
dSumWtdRtn = dSumWtdRtn + dWtdRtn
Next I
EWMAV = dSumWtdRtn ^ (1 / 2)
End Function
I am performing linear regression using this data in VB.Net
1411478155,71.9700012207031
1411478150,72.9700012207031
1411478145,73.9700012207031
1411478140,74.9700012207031
1411478135,76.9700012207031
1411478130,78.9700012207031
1411478125,80.9700012207031
1411478120,81.9700012207031
1411478115,82.9700012207031
1411478110,84.9700012207031
1411478105,85.9700012207031
1411478100,88.9700012207031
The formula that I am using is this,
where x = UTC Seconds, y = Values
In the denominator, I am getting a zero value because both expressions in the denominator equal to a value of 2.8688695263517E+20.
I defined my series as,
Dim xs(12) As [Double]
Dim ys(12) As [Double]
I am not sure if the square brackets matter.
For now, I am not able to get results due to zero denominator. What data type should I use?
I expect more rows of data in future.
Edit:
Given below is the sub
`
Public Sub GetLinearRegressionParams(ByVal xs() As Double, ByVal ys() As Double, ByRef a As Double, ByRef b As Double)
Dim sumX As Double = 0
Dim sumY As Double = 0
Dim sumXY As Double = 0
Dim sumX2 As Double = 0
Dim n As Integer
n = 0
For index = 0 To xs.Length - 1
If xs(index) = Nothing Then
Else
sumX = sumX + xs(index)
sumY = sumY + ys(index)
sumXY = sumXY + xs(index) * ys(index)
sumX2 = sumX2 + xs(index) * xs(index)
n = n + 1
End If
Next
a = (sumY * sumX2 - sumX * sumXY) / (n * sumX2 - sumX * sumX)
b = (n * sumXY - sumX * sumY) / (n * sumX2 - sumX * sumX)
End Sub
`
Is there anybody know how to make normal distributed random number in vb.net?
thank you
From this forum post :
Usage:
GaussNumDist(Mean, Standard Deviation, Sample Size)
Code example below, which will populate GaussNumArray() with the sample of numbers, whose distribution will have the mean and standard deviation specified:
Imports System.Math
Module Module1
Friend GaussNumArray() As Double
Friend intICell As Long
Friend Function GaussNumDist(ByVal Mean As Double, ByVal StdDev As Double, ByVal SampleSize As Integer)
intICell = 1 'Loop variable
ReDim GaussNumArray(SampleSize)
Do While (intICell < (SampleSize + 1))
Call NumDist(Mean, StdDev)
Application.DoEvents()
Loop
End Function
Sub NumDist(ByVal meanin As Double, ByVal sdin As Double)
'---------------------------------------------------------------------------------
'Converts uniform random numbers over the region 0 to 1 into Gaussian distributed
'random numbers using Box-Muller algorithm.
'Adapted from Numerical Recipes in C
'---------------------------------------------------------------------------------
'Defining variables
Dim dblR1 As Double
Dim dblR2 As Double
Dim mean As Double
Dim var As Double
Dim circ As Double
Dim trans As Double
Dim dblY1 As Double
Dim dblY2 As Double
Dim Pi As Double
Pi = 4 * Atan(1)
'Get two random numbers
dblR1 = (2 * UniformRandomNumber()) - 1
dblR2 = (2 * UniformRandomNumber()) - 1
circ = (dblR1 ^ 2) + (dblR2 ^ 2) 'Radius of circle
If circ >= 1 Then 'If outside unit circle, then reject number
Call NumDist(meanin, sdin)
Exit Sub
End If
'Transform to Gaussian
trans = Sqrt(-2 * Log(circ) / circ)
dblY1 = (trans * dblR1 * sdin) + meanin
dblY2 = (trans * dblR2 * sdin) + meanin
GaussNumArray(intICell) = dblY1 'First number
'Increase intICell for next random number
intICell = (intICell + 1)
GaussNumArray(intICell) = dblY2 'Second number
'Increase intICell again ready for next call of ConvertNumberDistribution
intICell = (intICell + 1)
End Sub
Friend Function UniformRandomNumber() As Double
'-----------------------------------------------------------------------------------
'Outputs random numbers with a period of > 2x10^18 in the range 0 to 1 (exclusive)
'Implements a L'Ecuyer generator with Bays-Durham shuffle
'Adapted from Numerical Recipes in C
'-----------------------------------------------------------------------------------
'Defining constants
Const IM1 As Double = 2147483563
Const IM2 As Double = 2147483399
Const AM As Double = (1.0# / IM1)
Const IMM1 As Double = (IM1 - 1.0#)
Const IA1 As Double = 40014
Const IA2 As Double = 40692
Const IQ1 As Double = 53668
Const IQ2 As Double = 52774
Const IR1 As Double = 12211
Const IR2 As Double = 3791
Const NTAB As Double = 32
Const NDIV As Double = (1.0# + IM1 / NTAB)
Const ESP As Double = 0.00000012
Const RNMX As Double = (1.0# - ESP)
Dim iCell As Integer
Dim idum As Double
Dim j As Integer
Dim k As Long
Dim temp As Double
Static idum2 As Long
Static iy As Long
Static iv(NTAB) As Long
idum2 = 123456789
iy = 0
'Seed value required is a negative integer (idum)
Randomize()
idum = (-Rnd() * 1000)
'For loop to generate a sequence of random numbers based on idum
For iCell = 1 To 10
'Initialize generator
If (idum <= 0) Then
'Prevent idum = 0
If (-(idum) < 1) Then
idum = 1
Else
idum = -(idum)
End If
idum2 = idum
For j = (NTAB + 7) To 0
k = ((idum) / IQ1)
idum = ((IA1 * (idum - (k * IQ1))) - (k * IR1))
If (idum < 0) Then
idum = (idum + IM1)
End If
If (j < NTAB) Then
iv(j) = idum
End If
Next j
iy = iv(0)
End If
'Start here when not initializing
k = (idum / IQ1)
idum = ((IA1 * (idum - (k * IQ1))) - (k * IR1))
If (idum < 0) Then
idum = (idum + IM1)
End If
k = (idum2 / IQ2)
idum2 = ((IA2 * (idum2 - (k * IQ2))) - (k * IR2))
If (idum2 < 0) Then
idum2 = idum2 + IM2
End If
j = (iy / NDIV)
iy = (iv(j) - idum2)
iv(j) = idum
If (iy < 1) Then
iy = (iy + IMM1)
End If
temp = AM * iy
If (temp <= RNMX) Then
'Return the value of the random number
UniformRandomNumber = temp
End If
Next iCell
End Function
End Module
You can use following line
Dim x1 as Double = MathNet.Numerics.Distributions.Normal.Sample(MEAN, STDEV)
Math.Net Numeric package can be installed using following NuGet command
Install-Package MathNet.Numerics -Version 4.9.0
You can found more information on NuGet site