Random numbers, but not the same. [GIVEN UP]

[martw]

OK, I got this code:

Dim p1 As Integer = RandomNo.Next(0, 4)
    Dim p2 As Integer = RandomNo.Next(0, 4)
    Dim p3 As Integer = RandomNo.Next(0, 4)
    Dim p4 As Integer = RandomNo.Next(0, 4)
    Dim p5 As Integer = RandomNo.Next(0, 4)

What I want to do, is generate a random number between 0 and 4, yet make none of the choices the same. So what can I do to exclude the numbers from the previous P's as the numbers are generated? This is outside of the Event Procedures so I can't use any Loops to check and corect it. I haven't the slightest clue on how to go about this. Any help appreciated.

[grilkip]

You mean they're declared public?

You could declare them public and assign them on the Load event and put your loop there.

[martw]

Whoops, I just realised I can't assign a random number to a declared integer outside of an event procedure. I think I can fix this now, thanks

[martw]

p1 = RandomNo.Next(0, 4)
        Do
            p2 = RandomNo.Next(0, 4)
        Loop Until p2 <> p1
        Do
            p3 = RandomNo.Next(0, 4)
        Loop Until p3 <> p2 And p1
        Do
            p4 = RandomNo.Next(0, 4)
        Loop Until p4 <> p3 And p2 And p1
        Do
            p5 = RandomNo.Next(0, 4)
        Loop Until p5 <> p4 And p3 And p2 And p1

OK, now I got this code. Will it actually work as intended by assigning a value to each p where the same number is not repeated?

[grilkip]

No because

p2 And p1

evaluates to a Boolean

And you could make something much neater if you would use an array.

[martw]

Damn, I thought so. Do you have any reccomendations?

[grilkip]

Well, I'm still in my learning stage myself, but here's what I came up with:

Const MaxNumGrilkips As Integer = 3
    Dim p(MaxNumGrilkips) As Integer
 
    Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
         Dim RandomNo As New Random
        Dim I As Integer
        Dim Location As Integer
        Dim NewVal As Integer
 
        For I = 0 To MaxNumGrilkips
            Do
                NewVal = RandomNo.Next(0, 5)
            Loop Until Array.IndexOf(p, NewVal) = -1
            p(I) = NewVal
        Next
    End Sub

Note that the upperbound is 5.

[martw]

OK, since the operations are too complicated for me to use, I have decided to give up on this. Thank you to all who helped.

[wild_bill]

I'm still going to put in my two cents. What about declaring an array with all the possible combinations of 1234 (24 possible I think), then use a random number to select which element of the array to use?

[taxes]

Hi,

You will find that virtually all questions have been previously covered in this forum. Have a look at my final post in

http://www.vbforums.com/showthread.php?t=338279

Now, usw the random number generator to fill the array rather than the textboxes and you're there.


BUT Are you seriously using your code to generate 4 random numbers which must be 1,2,3 or 4?????

If so, althought there is a theoretic probability if it requiring only 256 loops, the nature of Random is that you might get a loop that lasts for hours!!! This reasoning applies if you MUST generate ALL the numbers in a given range.

What you need to do then is to set up all your numbers in a 1 dimension array, e.g.

arrNumbers(1) = 1
arrNumbers(2) = 2 etc
(there are quicker ways of doing this; loops for larger ranges, array assignment within the declaration for smaller)

At the start, all the stored numbers will be the same as the index number of the array element.
Then if the random generated is 3 you make a note of the number held in arrNumbers(3) and then loop through the array from the top downwards, pushing all the contents of the elements 4 and greater down 1. Now from element 3 upwards, the number held in the array elements from 3 upwards will NOT be the same as the element index number.
You then reduce the random generator limit by 1

Repeat this until you are left with 1 array element only, which will contain the final number you want.

BTW NEVER give up on a coding problem. If you develop that habit you will greatly limit your progress. You may have to change your approach, but all problems are solvable one way or another - even if it means throwing away your computor

[martw]

Thanks taxes, but I can't afford to waste anymore time on things that I don't understand. It would take me a day to figure out how to use the code you have provided through the link and unfortunately that's all I have left, a day to complete my work. Thank you for your help, I will read through it as soon as I get the chance and maybe I can implement it in some other software some other time. For now, I have settled with a non-random method where everything is hard-coded into the program. That works fine for what I am doing. Oh, and I didn't give up completely, I gave up with that solution and settled for another one. Thanks guys

[Shaggy Hiker]

Whoof, here's a possibility that might be an improvement on taxes code, though I have never proven that this is correct:

Create the 1D array as he said:

array(1)=1
array(2)=2
etc.

Get two random integers: valu and direction.

if array(valu) >0 then
 'you've got your number
 array(valu)=0
else
 'Find a number.
   'ON even values of direction, look up.
   if direction mod 2 >0 then
    do
     valu+=1
     'Need to set valu back to the begninning if it goes over the top of the array.
     if array(valu)>0 then
       array(valu)=0
       exit do
     end if
     loop
   else
    'Do basically the same thing, but start with
    valu-=1
     'Set valu to the top of the array if it has gone below the base of the array.
    'Much left out here, but same as earlier loop
    end if

The basic idea is that you get a random number. If that slot in the array has not been used, use it (set the slot value to 0 or -1). If the slot has been used, look either forwards or backwards in the array for the next unused slot, and use that one.

If you are truly starting your search from a random spot in the array, then the next empty slot should be equally random. However, I have a feeling that this is not quite true. With a large array (say 100 items), if five consecutive slots were used, then the odds that the next 'random' number would be the next number above that slot becomes 6/100 when it should be 1/100. By looking either up or down, you cut this non-randomness in half, but the time taken to find a number is reduced.

This may not be an issue. In your case, with values 1,2,3,4, The first number has a 1 in 4 chance of being 1. However, if it isn't 1, then the second number has a 1 in 3 chance of being 1. The final number is NEVER random in this case, so perhaps any scheme to create a complete set of exclusive random numbers will not be random.

[taxes]

Hi martw

Understood. Best of luck.



Hi Shaggy Hiker

With repect, I don't think what you have there covers the problem. What you are suggestiing simply records the fact that a particular number has been already chosen. It does not stop the process from producing previously delivered numbers. Just for interest the following works.

TextBox1.Text = ""
        Dim arrNumber() As Integer = New Integer() {1, 2, 3, 4}
        Dim icount As Integer
        Dim irand As Integer
        Dim r As Random = New Random
        Dim iTop As Integer
        Do
            iTop = arrNumber.GetUpperBound(0)
            irand = r.Next(0, iTop)
 
            TextBox1.Text &= arrNumber(irand).ToString & "; "
            For icount = irand To arrNumber.GetUpperBound(0) - 1
                arrNumber(icount) = arrNumber(icount + 1)
 
            Next
            ReDim Preserve arrNumber(iTop - 1)
            If arrNumber.GetUpperBound(0) = 0 Then Exit Do
        Loop
        TextBox1.Text &= arrNumber(0).ToString

[Shaggy Hiker]

Well, we don't seem to be communicating very well, because I am using the technique I suggested for a much larger array (up to thousands) where each number can only be used once.

In this case, however, I think Wild Bill had the best suggestion. It won't scale well, but if that is not an issue, it will be faster and more random than either of our suggestions.

[taxes]

Hi Shaggy Hiker,

Sorry, I'm not making myself clear.

Both your and my sugestions will take up to any number required, but, unless I have misread your code 'cos I have not actually tried it, every time a new random number is produced your code will simply reject it. My code prevents the random process from producing a previously produced number, without affecting the random nature of the selection. I.E. It will take only 4 random loops to produce a complete result

I think you will find that by ensuring that only the minimum number of Random numbers have to be produced, my code will be faster than WildBill's, but it depends on whether or not a 24 case Select Case is faster than 4 loops through Random.

But this is only academic as martw is going another path.

[Shaggy Hiker]

I assumed that what Wild Bill meant was that you create an array of 24 integers (all possible combinations of the four digits used once each). Then a single random call would select one of the 24, and that would be the order of the random numbers. This would be VERY fast, because it would consist of only 1 call to random, and a look-up table. In fact, you could expand on that to re-generate the 1-4 array very quickly.

My code would always provide one new random number in each run. I wrote out more for martw's benefit, which may have clouded the issue. The basic idea is that you create a list of the possible numbers. Then use Random to get a random spot in the list. If the spot has been used, you look either up the list or down the list, and take the next unused number. Thus for each call to random there are these steps:

1)If the number is new, use it. One if statement is all that is used to determine this.

2)If the number is not new, then one if determines whether to look up or down, then there is a loop that will end within n iterations (where n is the size of the array). The loop body contains a single if.

It is academic at this point, but for something with serious speed requirements, Wild Bill's solution has the edge because the bulk of the effort can be pre-computed. As for our solutions....I guess we'd have to test. My doubt is that the numbers returned from my code is truly random.

[taxes]

HI,

Right. I follow what you are saying. That is quicker than I thought but as you say is not truly random. It limits some choices to 1 either side of another randum number.

Wildbill's code may (we won't know unless we try it) be faster but would be very cumbersome for large numbers. I might give them both a try just to see. But not tonight (as Napoleon said!!!)

[Shaggy Hiker]

His code becomes unwieldly above about 4 numbers.

1:1
2:2
3:6
4:24
5:96

rule Nx = x * (N(x-1)) For all x>1. Assuming 1:1

You sure wouldn't do 5 or more without a program creating the combinations for you.

I was thinking about my solution more. It is fast, but randomness degrades MAY degrade severly. However, if you were to map chosen and non chosen numbers as light and dark zones in a bar, it is likely that you would see some really interesting patterns. I would predict that at first several points of light would appear, and each point would slowly expand. It may be that one point would reach some size, and appear to rapidly gobble the whole region as the increased probability of landing in that zone caused it to grow at a faster rate than surrounding smaller zones. This could potentially create a feedback loop that accelerated growth.

Ok, enough rambling. The technique is quick, and adequately random for my application, but it clearly is not truly random.

[wild_bill]

True, you certainly wouldn't want more than 4 unique numbers. What about setting up a loop to act as kind of an odemeter to populate the array?