Random Letters

Random Letters

• Kiranvukkadala

  Reply to Whoisit

  You're creating an array of 16 letters. Do you need to check that the letters are not duplicated in the array

  If you do, how do you do it

  I ask because I want some random alphabets. It's a question of generating a series of numbers with no duplicates. One example is the numbers 1 to 26; another one is numbers 1 to 49 (lottery); and a third is taking a random number of questions from a list.

  I'm stuck on the duplication check. I guess it must be done by looping until you hit a number that isn't already in the array. So far I've come up with several infinite loops.

  Mongkut

  
  

• vomjom

  Hi,

  Is your problem solved If no, can you be more clear on problem definition

  Thank you,
  Bhanu.

  
  
  

• coenzyme77

  Any letter Will any English CAPITAL letter do

  Dim letter As String 'for any CAPITAL letter in English

  Randomize()

  ' Generate random value between 1 and 26

  Dim rndLetter As Integer = CInt(Int((26 * Rnd()) + 1))

  letter = ChrW(rndLetter + 65) 'A to Z ASCII 65 to 90

  --------------------------------------------------------------------------

  I want to randomize the whole English alphabet and ca't remember how to do it.

  Mongkut

  
  

• dragon6158

  Thanks for ideas. That got me started, but, am I right, your code sample is for selecting six lottery numbers out of 49

  I want to rearrange the alphabet and have come up with a possible answer - though it may look a bit long.

  One button and one textbox:

  Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click

  Dim vNumberString As String = "" 'to hold the alphabet

  Dim vNumberArr(25) As Integer 'to hold random 1 to 26

  Dim vNumberTemp As Integer = 0 'to hold each temp random

  Dim i, j, k As Integer 'counters

  For i = 0 To 25 'generate all numbers between 1 and 26

  Randomize()

  For j = 0 To i 'to i or to 25

  'k loop produces a number

  For k = 0 To i 'to i or to 25

  Do Until vNumberTemp <> vNumberArr(k)

  'which it can't be on the first pass

  vNumberTemp = CInt(Int((26 * Rnd()) + 1))

  Loop

  Next k

  'j loop checks if the number is present already

  If vNumberArr(j) = vNumberTemp Then

  Do While vNumberTemp = vNumberArr(j)

  vNumberTemp = CInt(Int((26 * Rnd()) + 1))

  Loop

  End If

  Next j

  'put the number into the array

  vNumberArr(i) = vNumberTemp

  'convert numbers to letters A-Z

  vNumberString = vNumberString & ChrW(vNumberArr(i) + 64) & ","

  Next i

  'display comma-separated random alphabet

  TextBox1.Text = vNumberString

  End Sub

  The next question is how to compare two strings that ought to contain the same elements but in a different order.

  Mongkut

  
  

• dschnare

  This might help:

  Dim str As String = ""

  Dim i, n As Integer

  Dim c As Char

  Dim rand As New Random

  For i = 0 To 9 ' Length of string required

  n = rand.Next(65, 65 + 26) ' A thru Z

  c = System.Convert.ToChar(n)

  str = str + c

  Next

  MessageBox.Show(str)

  It's perhaps not the best way of achieving this, but it works...

  HTH

  
  
  

• John_W_F

  Try this, there is probabaly a better way, but this works.
  
  Public Class Form1
  
  Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
  Dim intNumber As Integer
  Dim arrNumber(0 To 5) As Integer
  Dim x, y As Integer
  '------------------------------------------------
  
  For x = 0 To 5
  Start:
  Randomize()
  intNumber = Int((49 * Rnd()) + 1) ' Random number 1 to 49
  For y = 0 To 5
  ' If number already been selected select another one
  If intNumber = arrNumber(y) Then
  GoTo Start
  End If
  Next y
  arrNumber(x) = intNumber
  Next x
  
  '----------------------------------------------------
  'Display non duplicated Numbers
  
  TextBox1.Text = (arrNumber(0))
  TextBox2.Text = (arrNumber(1))
  TextBox3.Text = (arrNumber(2))
  TextBox4.Text = (arrNumber(3))
  TextBox5.Text = (arrNumber(4))
  TextBox6.Text = (arrNumber(5))
  
  
  End Sub
  
  End Class
  

• severo1980

  Well caught, I was just about to post up a comment on it.

  
  

• RBone

  MISTAKE

  Sorry, the random letter example returns a number in the range 66 to 91; meaning no letter A and and a [ after Z.

  The random number is in the range 1 to 26, so add 64 not 65.

  Another way to do it ought to be

  anyNumber Mod 26 + 65

  Mongkut (after breakfast)

  
  

• viru234

  Sorry about this but the Alert me to replies must have been down as I have tonight had a load of alerts.
  
  Here is what I am using and it works a treat, so thanks to all those contributed to this thread.
  
  For Letter = 0 To 16
  intLetter = CInt(Int((90 - 65 + 1) * Rnd() + 65)) 'GenerateRandom Letters
  arrLetter(Letter) = Chr(intLetter)
  Next
  
  

• TimHiggison

  This is shorter
  
  Public Class Form1
  
  Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
  Dim intLetter As Integer
  Dim arrNumber(0 To 25) As Integer
  Dim Number As Integer
  Dim i, y As Integer
  'Clear(Label)
  Label1.Text = ""
  '------------------------------------------------
  Randomize()
  For Number = 0 To 25
  Start:
  
  intLetter = CInt(Int((90 - 65 + 1) * Rnd() + 65)) 'GenerateRandom Letters
  For y = 0 To 25
  ' If number already been selected select another one
  If intLetter = arrNumber(y) Then
  GoTo Start
  End If
  Next y
  arrNumber(Number) = (intLetter)
  Next Number
  
  '----------------------------------------------------
  'Display non duplicated Letters i a list 1 = w, 2 = r, etc
  For i = 0 To 25
  Label1.Text = Label1.Text & (i + 1) & " = " & Chr(arrNumber(i)) & Chr(13)
  Next
  
  End Sub
  
  End Class
  
  
• Random Letters