This code doesn't find the correct output
for say n= 1 (although it gives the correct output for say n= 2,3,4..etc.)
if we put n= 1 to find x then the i loop will continue from 1 to 0, hence the first term in x should vanish and leftover should be the second term 5; but it gives 0 ?
Is there any limitation on the input n to run the for loop ?I would appreciate any help.
Function math(n As Integer) As Double
Dim i As Integer
Dim x As Double
For i = 1 To n - 1
x = (n - 1) * 2 + 5
sum = sum + x
Next i
math = sum
End Function
Why not simply:
Function math(n As Integer) As Double
Math = ((n - 1) * 2 + 5) * Abs((n - 1) - (n = 1))
End Function
???
if the answer is correct then Math = (n * 2 + 3) * Abs((n - 1) - (n = 1)) would be easier to understand and make much more sense
In the for loop, if you don't precise the Step, the variable will only increment by 1.
And here, you start at 1 to go to 0, so the loop won't execute, you need to test n to cover both cases :
Function math(n As Integer) As Double
If n < 0 Then Exit Function
Dim i As Integer
Dim x As Double
Dim Summ As Double
Select Case n
Case Is > 1
For i = 1 To n - 1
x = (i - 1) * 2 + 5
Summ = Summ + x
Next i
Case Is = 1
Summ = (n - 1) * 2 + 5
Case Is = 0
Summ = 5
Case Else
MsgBox "This case is not supported", vbInformation + vbOKOnly
Exit Function
End Select
math = Summ
End Function
If n = 1, you end up with For i = 1 To 0 which is incorrect and
should be expressed For i = 1 To 0 STEP -1.
So I suggest you add the STEP BYand make sure it is either 1 to -1 depending on N.
Related
As an example. I want to randomly hand out 100 chocolates to 25 kids. I cannot give any kid more than 10 chocolates.
So here m = 100, n = 25, x = 1 and y = 12.
I have checked these questions.
Dividing a number into m parts uniformly randomly
Dividing a number into random unequal parts
They do give some idea but in these questions x and y are not specified.
So basically,
1) Total No. of Chocolates = 100
2) I can only give minimum 1 and maximum 12 chocolates to each kid
3) Chocolates should be distributed between 25 kids
4) I do not want any distribution (uniform or normal) - it should be purely random. (I am willing to exclude this condition if all else fails.)
Private Function divideUniformlyRandomly(n As Integer, m As Integer) As Integer()
Dim rRandom As New Random
Dim fences As Integer() = New Integer(m - 2) {}
For i As Integer = 0 To m - 3
fences(i) = rRandom.Next(0, n - 1)
Next
[Array].Sort(fences)
Dim result As Integer() = New Integer(m - 1) {}
result(0) = fences(0)
For i As Integer = 1 To m - 3
result(i) = fences(i + 1) - fences(i)
Next
result(m - 1) = n - 1 - fences(m - 2)
Return result
End Function
This does work but I get 0 and 13 as well. I cannot ensure x and y here.
Give each child x chocolate. This will leave you with m - (n * x) to distribute randomly. Keep distributing to children that have less than y chocolates, until there are no more chocolates.
Private Function divideUniformlyRandomly(n As Integer, m As Integer, x As Integer, y As Integer) As Integer()
Dim rRandom As New Random
Dim aResult As Integer() = New Integer(n - 1) {}
Dim i As Integer = 0
Dim remaining As Integer = m
' Every n must have a min of x.
For i = 0 To n - 1
aResult(i) = x
remaining -= x
Next
' distribute the remaining m over the children randomly
While remaining > 0
' pick a child randomly
i = rRandom.Next(0, n)
' if the child has less than y, give them one
If aResult(i) < y Then
aResult(i) += 1
remaining -= 1
End If
End While
' Debug
Dim sum As Integer = 0
For i = 0 To n - 1
Console.WriteLine("{0}: {1}", i, aResult(i))
sum += aResult(i)
Next
Console.WriteLine("Sum: {0}", sum)
divideUniformlyRandomly = aResult
End Function
I am banging my head against the wall for a while now trying different techniques.
None of them are working well.
I have two strings.
I need to compare them and get an exact percentage of match,
ie. "four score and seven years ago" TO "for scor and sevn yeres ago"
Well, I first started by comparing every word to every word, tracking every hit, and percentage = count \ numOfWords. Nope, didn't take into account misspelled words.
("four" <> "for" even though it is close)
Then I started by trying to compare every char in each char, incrementing the string char if not a match (to count for misspellings). But, I would get false hits because the first string could have every char in the second but not in the exact order of the second. ("stuff avail" <> "stu vail" (but it would come back as such, low percentage, but a hit. 9 \ 11 = 81%))
SO, I then tried comparing PAIRS of chars in each string. If string1[i] = string2[k] AND string1[i+1] = string2[k+1], increment the count, and increment the "k" when it doesn't match (to track mispellings. "for" and "four" should come back with a 75% hit.) That doesn't seem to work either. It is getting closer, but even with an exact match it is only returns 94%. And then it really gets screwed up when something is really misspelled. (Code at the bottom)
Any ideas or directions to go?
Code
count = 0
j = 0
k = 0
While j < strTempName.Length - 2 And k < strTempFile.Length - 2
' To ignore non letters or digits '
If Not strTempName(j).IsLetter(strTempName(j)) Then
j += 1
End If
' To ignore non letters or digits '
If Not strTempFile(k).IsLetter(strTempFile(k)) Then
k += 1
End If
' compare pair of chars '
While (strTempName(j) <> strTempFile(k) And _
strTempName(j + 1) <> strTempFile(k + 1) And _
k < strTempFile.Length - 2)
k += 1
End While
count += 1
j += 1
k += 1
End While
perc = count / (strTempName.Length - 1)
Edit: I have been doing some research and I think I initially found the code from here and translated it to vbnet years ago. It uses the Levenshtein string matching algorithm.
Here is the code I use for that, hope it helps:
Sub Main()
Dim string1 As String = "four score and seven years ago"
Dim string2 As String = "for scor and sevn yeres ago"
Dim similarity As Single =
GetSimilarity(string1, string2)
' RESULT : 0.8
End Sub
Public Function GetSimilarity(string1 As String, string2 As String) As Single
Dim dis As Single = ComputeDistance(string1, string2)
Dim maxLen As Single = string1.Length
If maxLen < string2.Length Then
maxLen = string2.Length
End If
If maxLen = 0.0F Then
Return 1.0F
Else
Return 1.0F - dis / maxLen
End If
End Function
Private Function ComputeDistance(s As String, t As String) As Integer
Dim n As Integer = s.Length
Dim m As Integer = t.Length
Dim distance As Integer(,) = New Integer(n, m) {}
' matrix
Dim cost As Integer = 0
If n = 0 Then
Return m
End If
If m = 0 Then
Return n
End If
'init1
Dim i As Integer = 0
While i <= n
distance(i, 0) = System.Math.Max(System.Threading.Interlocked.Increment(i), i - 1)
End While
Dim j As Integer = 0
While j <= m
distance(0, j) = System.Math.Max(System.Threading.Interlocked.Increment(j), j - 1)
End While
'find min distance
For i = 1 To n
For j = 1 To m
cost = (If(t.Substring(j - 1, 1) = s.Substring(i - 1, 1), 0, 1))
distance(i, j) = Math.Min(distance(i - 1, j) + 1, Math.Min(distance(i, j - 1) + 1, distance(i - 1, j - 1) + cost))
Next
Next
Return distance(n, m)
End Function
Did not work for me unless one (or both) of following are done:
1) use option compare statement "Option Compare Text" before any Import declarations and before Class definition (i.e. the very, very first line)
2) convert both strings to lowercase using .tolower
Xavier's code must be correct to:
While i <= n
distance(i, 0) = System.Math.Min(System.Threading.Interlocked.Increment(i), i - 1)
End While
Dim j As Integer = 0
While j <= m
distance(0, j) = System.Math.Min(System.Threading.Interlocked.Increment(j), j - 1)
End While
i created a code in VBA to calculate the amount of combinations for stirling numbers of second kind. But in following example only half of the values are correct.
The result should be 1,7,6,1 if n is equal to 4. (Wikipedia stirling numbers
I get 1,7,6.5,4.16
Sub stirlingerzahlen()
Dim n As Integer
Dim sum As Double
Dim subsum As Double
Dim k As Long
Dim j As Long
n = 4
For k = 1 To n Step 1
For j = 0 To k Step 1
subsum = 1 / Application.WorksheetFunction.Fact(k) * (-1) ^ (k - j) * Application.WorksheetFunction.Fact(k) / Application.WorksheetFunction.Fact(j) * j ^ n
sum = sum + subsum
Next
Sheets("Tabelle2").Cells(k, 1) = sum
sum = 0
Next
End Sub
Can someone find the mistake?
There is another version of the formula which seems to be easier to implement:
http://home.mathematik.uni-freiburg.de/junker/ss10/DAS-SS10.pdf
(Page 13)
And here the updated code:
Sub stirlingerzahlen()
Dim n As Integer
Dim sum As Double
Dim subsum As Double
Dim k As Long
Dim j As Long
n = 4
For k = 1 To n Step 1
For j = 0 To k
subsum = (((-1) ^ (k - j)) * ((j ^ n) / (Application.WorksheetFunction.Fact(j) * Application.WorksheetFunction.Fact(k - j))))
sum = sum + subsum
Next
Sheets("Tabelle2").Cells(k, 1) = sum
sum = 0
Next
End Sub
I have a function that only call the spline function when something happens..in this case when a division is less than zero..the inputs for the function is the same that for the spline function(called CUBIC), the spline was tested and works well when I call it direct! someone can help me?...follows a party of the code
Function NDF6(T As Variant, dias As Variant, taxas As Variant)
If T <= dias(1) Then
NDF6 = taxas(1)
Exit Function
End If
If T >= dias(tam) Then
NDF6 = taxas(tam)
Exit Function
End If
For i = 1 To tam
If T <= dias(i) Then
If taxas(i) / taxas(i - 1) < 0 Then
Call CUBIC(T, dias, taxas)
Else
i0 = ((taxas(i - 1) * dias(i - 1)) / 360) + 1
i1 = ((taxas(i - 1) * dias(i - 1)) / 360) + 1
irel = i1 / i0
i2 = irel ^ ((T - dias(i - 1)) / (dias(i) - dias(i - 1)))
i2rel = i2 * i0
i2real = i2rel - 1
NDF6 = i2real * (360 / T)
End If
Public Function CUBIC(x As Variant, input_column As Variant, output_column As Variant)
The function returns a zero value when I call the cubic function. The inputs are a cell with a value with a value equivalent a day and two arrays(DUONOFF and ONOFF) equivalent a days and rates, I call the function like:
NDF6(512,DUONOFF,ONOFF)
follows the CUBIC function
Public Function CUBIC(x As Variant, input_column As Variant, output_column As Variant)
'Purpose: Given a data set consisting of a list of x values
' and y values, this function will smoothly interpolate
' a resulting output (y) value from a given input (x) value
' This counts how many points are in "input" and "output" set of data
Dim input_count As Integer
Dim output_count As Integer
input_count = input_column.Rows.Count
output_count = output_column.Rows.Count
Next check to be sure that "input" # points = "output" # points
If input_count <> output_count Then
CUBIC = "Something's messed up! The number of indeces number of output_columnues don't match!"
GoTo out
End If
ReDim xin(input_count) As Single
ReDim yin(input_count) As Single
Dim c As Integer
For c = 1 To input_count
xin(c) = input_column(c)
yin(c) = output_column(c)
Next c
values are populated
Dim N As Integer 'n=input_count
Dim i, k As Integer 'these are loop counting integers
Dim p, qn, sig, un As Single
ReDim u(input_count - 1) As Single
ReDim yt(input_count) As Single 'these are the 2nd deriv values
N = input_count
yt(1) = 0
u(1) = 0
For i = 2 To N - 1
sig = (xin(i) - xin(i - 1)) / (xin(i + 1) - xin(i - 1))
p = sig * yt(i - 1) + 2
yt(i) = (sig - 1) / p
u(i) = (yin(i + 1) - yin(i)) / (xin(i + 1) - xin(i)) - (yin(i) - yin(i - 1)) / (xin(i) - xin(i - _1))
u(i) = (6 * u(i) / (xin(i + 1) - xin(i - 1)) - sig * u(i - 1)) / p
Next i
qn = 0
un = 0
yt(N) = (un - qn * u(N - 1)) / (qn * yt(N - 1) + 1)
For k = N - 1 To 1 Step -1
yt(k) = yt(k) * yt(k + 1) + u(k)
Next k
now eval spline at one point
Dim klo, khi As Integer
Dim h, b, a As Single
first find correct interval
klo = 1
khi = N
Do
k = khi - klo
If xin(k) > x Then
khi = k
Else
klo = k
End If
k = khi - klo
Loop While k > 1
h = xin(khi) - xin(klo)
a = (xin(khi) - x) / h
b = (x - xin(klo)) / h
y = a * yin(klo) + b * yin(khi) + ((a ^ 3 - a) * yt(klo) + (b ^ 3 - b) * yt(khi)) * (h ^ 2) _/ 6
CUBIC = y
out:
End Function
First off, here's my code:
Sub SimulatePortfolio()
Dim lambda As Double
Dim num As Integer
Dim cycles As Long
Column = 12
q = 1.5
lambda = 0.05
cycles = 100000
Dim data(1 To 100000, 1 To 10) As Integer
Dim values(1 To 10) As Double
For i = 1 To 10
values(i) = 0
Next i
temp = lambda
For i = 1 To cycles
lambda = temp
num = 10
t = 0
Dim temps(1 To 10) As Integer
For k = 1 To 10
temps(k) = 1000
Next k
Do While (t < 10 And num > 0)
t = t + tsim(lambda, num)
For j = 1 To 10
If (j > t) Then
temps(j) = temps(j) - 50
End If
Next j
num = num - 1
If (num <= 0) Then
Exit Do
End If
lambda = lambda * q
Loop
For l = 1 To 10
values(l) = values(l) + temps(l)
data(i, l) = temps(l)
Next l
Next i
For i = 1 To 10
Cells(i + 1, Column) = values(i) / cycles
'Problem occurs on this line:
Cells(i + 1, Column + 1).Value = Application.WorksheetFunction.Var(Application.WorksheetFunction.Index(data, i, 0))
Next i
End Sub
Function tsim(lambda As Double, num As Integer) As Double
Dim v As Double
Dim min As Double
Randomize
min = (-1 / lambda) * Log(Rnd)
For i = 1 To (num - 1)
Randomize
v = (-1 / lambda) * Log(Rnd)
If (min > v) Then
min = v
End If
Next i
tsim = min
End Function
When I set the value for cycles to 10000, it runs fine without a hitch. When I go to 100000 cycles, it gets an Error 13 at the indicated line of code.
Having been aware that Application.Tranpose is limited to 65536 rows with variants (throwing the same error) I tested the same issue with Index
It appears that Application.WorksheetFunction.Index also has a limit of 65536 rows when working with variants - but standard ranges are fine
So you will need to either need to dump data to a range and work on the range with Index, or work with two arrays
Sub Test()
Dim Y
Dim Z
'works in xl07/10
Debug.Print Application.WorksheetFunction.Index(Range("A1:A100000"), 1, 1)
Y = Range("A1:A65536")
`works
Debug.Print Application.WorksheetFunction.Index(Y, 1, 1)
'fails in xl07/10
Z = Range("A1:A65537")
Debug.Print Application.WorksheetFunction.Index(Z, 1, 1)
End Sub