I wrote a code to output Pascal triangle in a multi-line textbook. The program works fine for inputs between 1 to 12 but gives an overflow error once a value of 13 is inputed.
Is there any modifications I can make to enable the program to accurately give outputs for 13 and higher?
Here is the code I used:
Public Class pascal_triangle
Private Function factorial(ByVal k As Integer) As Integer
If k = 0 Or k = 1 Then
Return 1
Else
Return k * factorial(k - 1)
End If
End Function
Private Sub BtnGen_Click(sender As Object, e As EventArgs) Handles BtnGen.Click
Dim nCr As Integer
Dim i, j, k As Integer
Dim output As String
output = ""
j = Val(TxtColumn.Text)
For k = 0 To j
For i = 0 To k
Dim fact, fact1, fact2 As Integer
fact = factorial(k)
fact1 = factorial(k - i)
fact2 = factorial(i)
nCr = fact / (fact1 * fact2)
TxtOutput.Text += Str(nCr) & output
Next
TxtOutput.Text += vbCrLf
Next
End Sub
End Class
The overflow is because 13! is too big to fit in an integer.
The largest representable integer (32-bit signed) is
2147483647 (0x7FFFFFFF == 01111111 11111111 11111111 11111111b)
so :
12! = 479001600
MaxInt = 2147483647
13! = 6227020800
If you want to use larger numbers than this you need to use a larger number type. The next larger types are Long (64-bit signed, max 9223372036854775807) or, for your purposes, ULong (unsigned 64-bit, since you don't need negative numbers, which is twice that at 18446744073709551615).
This will let you calculate up to20!, which is 2432902008176640000. For numbers larger than this you will need to look into using either BigInteger or other dedicated libraries that allow for holding and calculating arbitrarily large numbers.
Alternatively, you can look into other methods of computing an arbitrary row without using factorials.
Your main problem is that you are using Integer which is too small to hold the factorial of 13. Change your Factorial function to return Long. It would also be a good idea to turn Option Strict On and make nCr a Double.
Private Function factorial(ByVal k As Integer) As Long
If k = 0 Or k = 1 Then
Return 1
Else
Return k * factorial(k - 1)
End If
End Function
Private Sub BtnGen_Click(sender As Object, e As EventArgs) Handles BtnGen.Click
Dim nCr As Double
Dim i, j, k As Integer
Integer.TryParse(TxtColumn.Text, j)
For k = 0 To j
For i = 0 To k
Dim fact, fact1, fact2 As Long
fact = factorial(k)
fact1 = factorial(k - i)
fact2 = factorial(i)
nCr = fact / (fact1 * fact2)
TxtOutput.Text += nCr.ToString & " "
Next
TxtOutput.Text += vbCrLf
Next
End Sub
Related
I am attempting to implement an array using the shell sort algorithm. The program will sort the array and output each element to the Listbox after the button was clicked. However, the first item output is always 0. I have included a piece of my source code and a photo of the form below;
Dim randGen As New Random()
Dim unstArray() As Integer
Dim unstArrayCopy() As Integer
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click 'Generates random number to save in array.
Dim i As Integer = CInt(TextBox1.Text)
ReDim unstArray(i)
ReDim unstArrayCopy(i)
For x = 0 To i
unstArray(x) = randGen.Next(1, 10001)
Next
Array.Copy(unstArray, unstArrayCopy, i)
End Sub
Private Sub ShllSrtBtn_Click(sender As Object, e As EventArgs) Handles shllSrtBtn.Click
shellsort(unstArrayCopy, unstArrayCopy.GetUpperBound(0))
End Sub
Sub shellsort(ByRef shellSort() As Integer, ByVal max As Integer)
Dim stopp%, swap%, limit%, temp%, k%
Dim x As Integer = CInt((max / 2) - 1)
Do While x > 0
stopp = 0
limit = max - x
Do While stopp = 0
swap = 0
For k = 0 To limit
If shellSort(k) > shellSort(k + x) Then
temp = shellSort(k)
shellSort(k) = shellSort(k + x)
shellSort(k + x) = temp
swap = k
End If
Next k
limit = swap - x
If swap = 0 Then stopp = 1
Loop
x = CInt(x / 2)
Loop
For i = 0 To shellSort.GetUpperBound(0)
ListBox1.Items.Add(shellSort(i))
Next i
End Sub
The problem is here:
ReDim unstArray(i)
ReDim unstArrayCopy(i)
In VB, when you initialize an array, you must give it the maximum index you want to use, not the intended array length as in other languages like C#.
Because of that, your code creates an array of length i+1, but you only loop from 0 to i when filling the array. So the last element at index i will always be zero.
You should set the initializer in these lines to i-1.
VB Array Reference
hello im trying to do this calculation : [365!/((365^x)((365-x)!))]
the problem is when i do it it doesn't give me the decimals just the integer it give me 0 or 1 because the answer is 0
Public Class Form1
Private Function fact(ByVal n As Integer) As Numerics.BigInteger
Dim Z As New Numerics.BigInteger(1)
For i As Integer = 1 To n
Z = Z * i
Next
Return Z
End Function
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim min As Integer
Dim max As Integer
Dim ranum As Integer
Dim ind() As Integer
Dim ran As New Random
Dim F365 As New Numerics.BigInteger(0)
F365 = Numerics.BigInteger.Parse("25104128675558732292929443748812027705165520269876079766872595193901106138220937419666018009000254169376172314360982328660708071123369979853445367910653872383599704355532740937678091491429440864316046925074510134847025546014098005907965541041195496105311886173373435145517193282760847755882291690213539123479186274701519396808504940722607033001246328398800550487427999876690416973437861078185344667966871511049653888130136836199010529180056125844549488648617682915826347564148990984138067809999604687488146734837340699359838791124995957584538873616661533093253551256845056046388738129702951381151861413688922986510005440943943014699244112555755279140760492764253740250410391056421979003289600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000")
min = Integer.Parse(Tmin.Text)
max = Integer.Parse(Tmax.Text)
ranum = Integer.Parse(TRan.Text)
ReDim ind(ranum)
For x As Integer = 1 To ranum
ind(x) = ran.Next(min, max + 1)
Answer.Items.Add(ind(x))
Next
Dim P(ranum) As Numerics.BigInteger
Dim facts(ranum) As Numerics.BigInteger
For x = 1 To ranum
P(x) = 365 ^ (ind(x))
facts(x) = fact(365 - ind(x))
Next
Dim phenB(ranum) As Numerics.BigInteger
Dim phen(ranum) As Double
For x = 1 To ranum
phenB(x) = (P(x) * facts(x))
phen(x) = F365 / phenB(x)
tx.Text = phen(x) (here is the aswer)
Next
End Sub
End Class
The BigInteger class does not have a function to give a non-integer result for division. However, it does have BigInteger.Log, so, using these logarithmic identities:
ln(a⋅b) = ln(a) + ln(b)
ln(a/b) = ln(a) - ln(b)
ln(a^b) = b⋅ln(a)
we can perform the calculation like this:
Function SomeCalc(n As Integer) As Double
Dim lnF365 = BigInteger.Log(fact(365))
Dim lnPower = n * Math.Log(365)
Dim lnOtherFact = BigInteger.Log(fact(365 - n))
Return Math.Exp(lnF365 - lnPower - lnOtherFact)
End Function
where fact() is a pre-calculated array:
Option Strict On
Option Infer On
' ... other code ...
Dim fact(365) As BigInteger
' ... other code ...
Private Sub CalcFacts()
Dim z = BigInteger.One
For i = 1 To 365
z *= i
fact(i) = z
Next
End Sub
You could even have an array of pre-calculated logs of the factorials, instead of an array of the factorials. It depends on if you're using them elsewhere, and if there is any need for it to go a tiny tiny bit faster:
Function SomeCalc(n As Integer) As Double
Dim lnF365 = lnFact(365)
Dim lnPower = n * Math.Log(365)
Dim lnOtherFact = lnFact(365 - n)
Return Math.Exp(lnF365 - lnPower - lnOtherFact)
End Function
and
Dim lnFact(365) As Double
' ...
Private Sub CalcLnFacts()
Dim z = BigInteger.One
For i As Integer = 1 To largestNum
z *= i
lnFact(i) = BigInteger.Log(z)
Next
End Sub
That number 365 should be a named variable - I had no idea what a sensible name for it would be.
I need help with one simple task:
Input an integer number n and output the sum: 1 + 2^2 + 3^2 + ... + n^2. Use input validation for n to be positive.
My code does not work and till now is:
Sub Main()
Dim inputNumber As Integer
Console.WriteLine("Please enter a positive number.")
inputNumber = Console.ReadLine()
If inputNumber <= 0 Then
Console.WriteLine("Please use only positive numbers > 0 !")
End If
Dim sum As Integer
Dim i As Integer = 1
For i = 1 To i <= inputNumber
sum = sum + (i * i)
i = i + 1
Next
Console.WriteLine(sum)
Console.ReadLine()
End Sub
Try these changes:
Dim inputNumber as Long ' not Integer. Also change sum, i.
...
inputNumber = CLng(Console.ReadLine) ' make it a number, not a string
...
Dim sum as Long ' yum
dim i as Long ' don't assign it here
for i = 1 to inputNumber ' don't use "<=" in a for loop
...
' i = i+1 ' Don't increment i within the loop, since the loop does that for you.
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim Input As Integer
Input = TextBox1.Text
Dim i As Integer
i = 0
Dim x As Integer
x = 0
Dim y As Integer
y = 0
For a = 1 To Input
y = Math.Pow(a, (Input - i))
x = Math.Pow(a, (Input - i)) + x
i = i + 1
Next
Label1.Text = x
End Sub
With reference to this link
Calculate CRC32 of an String or Byte Array
I modified the code in order to calculate the CRC16 instead of CRC32, however I am getting wrong result, can some one point me where is the mistake?
Private Sub Main()
Crc16.ComputeChecksum(Encoding.UTF8.GetBytes("Some string"))
End Sub
Public Class CRC16
Shared table As UShort()
Shared Sub New()
Dim poly As UShort = &HA001US 'calculates CRC-16 using A001 polynomial (modbus)
table = New UShort(255) {}
Dim temp As UShort = 0
For i As UShort = 0 To table.Length - 1
temp = i
For j As Integer = 8 To 1 Step -1
If (temp And 1) = 1 Then
temp = CUShort((temp >> 1) Xor poly)
Else
temp >>= 1
End If
Next
table(i) = temp
Next
End Sub
Public Shared Function ComputeChecksum(ByVal bytes As Byte()) As UShort
Dim crc As UShort = &H0US ' The calculation start with 0x00
For i As Integer = 0 To bytes.Length - 1
Dim index As Byte = CByte(((crc) And &HFF) Xor bytes(i))
crc = CUShort((crc >> 8) Xor table(index))
Next
Return Not crc
End Function
End Class
Try this, it's working VB6 code for Instrument control. (sCommand is a temp string which contains all Bytes, Result is added to sCommand, Modbus is using LSB first, TextToString and StringToAscii are functions to convert a readable string "FF EE" into ASCII and back, thus they are not of interest here.):
Private Sub cmdCRC16_Click()
Dim sCommand As String
Dim x As Long
Dim y As Long
Dim lCRC As Long
sCommand = TextToString(txtASCII)
'Initial value
lCRC = 65535 '(&HFFFF results in Integer -1)
For x = 1 To Len(sCommand)
lCRC = lCRC Xor Asc(Mid(sCommand, x, 1))
For y = 1 To 8
If (lCRC Mod 2) > 0 Then
lCRC = (lCRC And 65534) / 2
lCRC = lCRC Xor 40961 '(&HA001 results in whatever negative integer)
Else
lCRC = (lCRC And 65534) / 2
End If
Next y
Next x
'Add CRC with LSB first
sCommand = sCommand + Chr(lCRC And 255)
sCommand = sCommand + Chr((lCRC And 65280) / 256)
txtASCII = StringToASCII(sCommand)
End Sub
I just came accross the same issue. Simple solution is to omit negation at the end, so just change your "Return Not crc" to "Return crc" and you be fine.
There are various variants of CRC-16, where "CRC-16" normally refers to the IBM variant, also called "ARC". It uses an XorOut value of zero. See Catalogue of parametrised CRC algorithms with 16 bits.
I recently wrote this program in Visul Basic 13.
it searchs for the nth catalan number but after 48 even Decimal type is too short.
Is there any other way to represent them? like in the form of A*10^n?
Public Class Try_Catalan_Number
'Catalan numbers form a sequence of natural numbers that occur in various counting problems,
'often involving recursively defined objects.
Inherits Base_Number
Public Overrides Sub Test()
Dim Return_Catalan_Value As Decimal
If Function_Catalan(Return_Catalan_Value) = False Then
Return_To_Form_Boolean = False
Else
Return_To_Form_Boolean = True
End If
Return_To_Form_Value = Function_Catalan(Return_Catalan_Value)
End Sub
Private Function Function_Catalan(Return_Catalan_Value As Decimal) As Decimal
'We return a Decimal function because catalan numbers can be very big and decimal is the biggest type.
Dim Binomial_Cofficients As Decimal
Dim Result As Decimal
Dim Number_Of_Loops As Integer
Dim tmpNumber As Object
Dim K As Decimal
Dim N As Decimal
If (Number > 48) Then
Return False
Exit Function
End If
'48 is the largest catalan number position which can be displayed...any position above 48 is too big.
tmpNumber = Number - 1
N = 2 * tmpNumber
K = tmpNumber
Result = 1
For Number_Of_Loops = 1 To K
Result = Result * (N - (K - Number_Of_Loops))
Result = Result / Number_Of_Loops
Next Number_Of_Loops
Binomial_Cofficients = Result
tmpNumber = Number - 1
tmpNumber = ((1 / (1 + tmpNumber)) * Binomial_Cofficients)
Return_Catalan_Value = tmpNumber
Return Return_Catalan_Value
End Function
End Class
[I assume by "Visul Basic 13" you mean the VB which is associated with Visual Studio 2013, i.e. VB version 12.0.]
You can use System.Numerics.BigInteger (you'll have to add a reference to System.Numerics):
Imports System.Numerics
Module Module1
Friend Function Factorial(n As Integer) As BigInteger
If n < 2 Then Return 1
If n = 2 Then Return 2
Dim f As BigInteger = BigInteger.Parse("2")
For i = 3 To n
f *= i
Next
Return f
End Function
Friend Function CatalanNumber(n As Integer) As BigInteger
Return Factorial(2 * n) / (Factorial(n + 1) * Factorial(n))
End Function
Sub Main()
For i = 0 To 550
Console.WriteLine(CatalanNumber(i).ToString())
Next
Console.ReadLine()
End Sub
End Module
I did not test to see the maximum Catalan number it can calculate, and I have no inclination to verify the results beyond those shown on the Wikipedia page.
Optimisations are left as an exercise for the reader ;)
Edit: FWIW, I can get it to run a bit faster by using
Function CatalanNumber(n As Integer) As BigInteger
Dim nFactorial = Factorial(n)
Dim twonFactorial = nFactorial
For i = (n + 1) To 2 * n
twonFactorial = BigInteger.Multiply(twonFactorial, i)
Next
Return twonFactorial / (BigInteger.Pow(nFactorial, 2) * (n + 1))
End Function
The speed increase varies from roughly 50% (n=50) to 20% (n=5000). If you're only using the function a few times for fairly small n, there may be little point worrying about it.
Edit2 Re-writing your function a bit to make it easier to read and removing the off-by-one error, we get:
Private Function Function_Catalan(a As Integer) As BigInteger
If a = 0 Then Return 1
Dim binomialCofficient As BigInteger = BigInteger.One
Dim n As Integer = 2 * a
Dim k As Integer = a - 1
For i As Integer = 1 To k
binomialCofficient = binomialCofficient * (n - (k - i)) / i
Next i
Return binomialCofficient / a
End Function
to get this format you could use:
String.Format("{0:E4}", InputNumber)