random number picker code ??

[Milsnips]

hi there,

i'm after a small function that is kindof like the lottery picker programs,
i pass it some numbers, say (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15) and i want
it to return me 3 random numbers, none being the same.

any help appreciated.
thanks,
Paul.

[Hans Kesting]

> hi there,

i'm after a small function that is kindof like the lottery picker programs, i
pass it some numbers, say (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15) and i want it
to return me 3 random numbers, none being the same.

any help appreciated.
thanks,
Paul.

Start with an array of booleans, index (+1) = lottery number.
Initialize with "false".

Get a random number (see Random class), see if that position is still
"false". If not, get a new (random) number, else set to "true".
Repeat 3 times.

Print all indexes (+1) with a "true" value.
Hans Kesting

[Jan Hyde]

"Milsnips" <mi******@hotmail.com>'s wild thoughts were
released on Thu, 18 May 2006 07:53:35 +0200 bearing the
following fruit:

hi there,

i'm after a small function that is kindof like the lottery picker programs,
i pass it some numbers, say (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15) and i want
it to return me 3 random numbers, none being the same.

any help appreciated.
thanks,
Paul.

I think what you actually want is a shuffling algorithm,
then you just pick the first three numbers.


Jan Hyde (VB MVP)

--
People are like tea bags.
You have to put them in hot water before you know how strong they are. (Bill Rayburn)

[Tasos Vogiatzoglou]

int[] GetNums(int[] nums, int numToReturn)
{
int[] nm = new int[numToReturn];
if (nums.Length <= numToReturn)
{
for (int i=0;i<nums.Length;i++)
{
nm[i] = nums[i];
}
}
else
{
Random rnd = new Random(123);
for (int i=0;i<numToReturn;i++)
{
int idx = rnd.Next() % nums.Length;
nm[i] = nums[idx];
}
}

return nm;
}

Regards,
Tasos

[Hans Kesting]

> int[] GetNums(int[] nums, int numToReturn)

{
int[] nm = new int[numToReturn];
if (nums.Length <= numToReturn)
{
for (int i=0;i<nums.Length;i++)
{
nm[i] = nums[i];
}
}
else
{
Random rnd = new Random(123);
for (int i=0;i<numToReturn;i++)
{
int idx = rnd.Next() % nums.Length;
nm[i] = nums[idx];
}
}

return nm;
}

Regards,
Tasos

But how do you make sure a number is not chosen twice?
(what happens if "idx" is 6 for the second time? It will select
"nums[6]" for a second time)

Hans Kesting

[Tasos Vogiatzoglou]

He can iterrate through the table to check if the number is already
selected.

I'll post in a couple of minutes a revised sample :)

Tasos