random number?

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Mar 9th, 2003, 03:27 PM
oyad

random number?

please i want to generate random numbers between 1 and 52,i was using someting like this:
randomize
x=int((p)*rnd)
but wasn't getting the desired results
the rnd is not in this format:- 0.rrr ie i just need the three digits after the decimal.
i was seeing something like :-0.rrrrrrrrrrrrrrr and some in negative exponent.i need numbers between 1 and 52 shuffled randomly

VB Code:

Private Sub Command1_Click()
    Randomize
    Debug.Print Int(Format(52 * Rnd + 1))
End Sub

VB Code:

Dim nTry As Integer
    Dim nAddCount As Integer
    Dim bFound As Integer
    Dim nDummy As Integer
 
    Dim MyCollection As New Collection
    
    Randomize
    
    Do Until nAddCount = 52
        nTry = Int((52) * Rnd + 1)
        On Error Resume Next
        nDummy = MyCollection.Item(CStr(nTry))
        bFound = (Err = 0)
        If bFound Then
            Err.Clear
        Else
            MyCollection.Add nTry, CStr(nTry)
            nAddCount = nAddCount + 1
        End If
    Loop

When you want to use the values remember that collections start at 1 and not 0.

Mar 9th, 2003, 03:59 PM
peet

:confused: what did I miss ?? :D
Mar 9th, 2003, 04:02 PM
manavo11

Peet, it should be : 51 * rnd - 1 . If you want a random number between two other specific ones (hypothetically a,b) then you will get it from (b - a) * RND + a.
Mar 9th, 2003, 04:06 PM
MartinLiss

Quote:

Originally posted by peet
:confused: what did I miss ?? :D

I assumed he wanted to shuffle a deck of cards so I gave him a solution that would produce the numbers 1 to 52 without repetition.
Mar 9th, 2003, 04:08 PM
seaweed
Quote:

Originally posted by manavo11
Peet, it should be : 51 * rnd - 1 . If you want a random number between two other specific ones (hypothetically a,b) then you will get it from (b - a) * RND + a.

Actually, Peet was correct. From VB Help:
Quote:
Rnd Function Example
This example uses the Rnd function to generate a random integer value from 1 to 6.

VB Code:

Dim MyValue
MyValue = Int((6 * Rnd) + 1) ' Generate random value between 1 and 6.
Mar 9th, 2003, 04:10 PM
peet
Quote:

Originally posted by manavo11
Peet, it should be : 51 * rnd - 1 . If you want a random number between two other specific ones (hypothetically a,b) then you will get it from (b - a) * RND + a.

no,
VB Code:
```
Debug.Print Int(Format(52 * Rnd + 1))
```
is correct , I was reffering to Marty's coll code :)
Mar 9th, 2003, 04:11 PM
peet

Quote:

Originally posted by MartinLiss
I assumed he wanted to shuffle a deck of cards so I gave him a solution that would produce the numbers 1 to 52 without repetition.

I see :)
Mar 9th, 2003, 04:12 PM
peet
uhh... loose the Format btw :o
VB Code:
```
Debug.Print Int(52 * Rnd + 1)
```
Mar 9th, 2003, 04:15 PM
manavo11

Peet, if I am not mistaken RND generates a random number between 0 and 1. So in your case, if you get the minimum value you get 1 as a result. But if you get the maximum, you get 53 as a result. Right?

no, using my statement will never result in 53.
rnd will never return 1

VB Code:

Private Sub Command1_Click()
    Dim i As Integer
    Randomize
    i = Int(52 * Rnd + 1)
    While i <> 53
        Debug.Print i
        i = Int(52 * Rnd + 1)
    Wend
End Sub

the loop will never end.

Mar 9th, 2003, 04:33 PM
manavo11

Well, if RND never returns the value 1 then it is my mistake.
Mar 9th, 2003, 05:45 PM
MartinLiss

From Help

Quote:

The Rnd function returns a value less than 1 but greater than or equal to zero.
Mar 9th, 2003, 05:55 PM
manavo11

Thanks Martin. I never noticed that.
Mar 10th, 2003, 02:38 AM
Spooner
MSDN also says this for the general solution:

Quote:

Int((upperbound - lowerbound + 1) * Rnd + lowerbound)

So for numbers between 1 and 52 (inclusive I assume) upperbound is 52, lowerbound is 1, so you get
VB Code:
```
'Int((52-1+1)*Rnd + 1)
 
Int (52 * Rnd + 1)
```
as Peet said.
Mar 10th, 2003, 07:43 AM
manavo11

OK I got it! I just thought that RND returned the value 1.

Quote:

Originally posted by MartinLiss

VB Code:

Dim nTry As Integer
Dim nAddCount As Integer
Dim bFound As Integer
Dim nDummy As Integer

Dim MyCollection As New Collection

Randomize

Do Until nAddCount = 52
nTry = Int((52) * Rnd + 1)
On Error Resume Next
nDummy = MyCollection.Item(CStr(nTry))
bFound = (Err = 0)
If bFound Then
Err.Clear
Else
MyCollection.Add nTry, CStr(nTry)
nAddCount = nAddCount + 1
End If
Loop

When you want to use the values remember that collections start at 1 and not 0.

That's a bit over the top Marty ;) Here's a shortened version of the same code:

VB Code:

Dim nTry As Integer
Dim MyCollection As Collection
  
  Set MyCollection = New Collection
 
  Randomize
 
  On Error Resume Next
  Do Until MyCollection.Count = 52
      nTry = Int((52) * Rnd + 1)
      MyCollection.Add nTry, CStr(nTry)
  Loop
  Err.Clear

Mar 10th, 2003, 11:45 AM
MartinLiss

Thanks, that is an improvement and both changes you made (using the collection's count as the limit and eliminating the unnecessary error checking) are obvious if you think about it.
Mar 10th, 2003, 11:59 AM
si_the_geek

Quote:

Originally posted by MartinLiss
Thanks, that is an improvement and both changes you made (using the collection's count as the limit and eliminating the unnecessary error checking) are obvious if you think about it.

No problem, I'll use it myself at some point :) I get used to seeing code around here that is far too long, as I'm sure you know some VB beginners write code that does over 5 times as much as needed.

It's nice to be able to 'correct' somebody who knows what they are doing - even tho it's just because they didn't optimise it fully ;)