Re: Controlling ('rand') random number upper limit

[Jim Langston <jimlangston@xxxxxxxxxxxxxxxx>]

Chris, Paul,

That being said, Paul's solution can still be screwed, depending
on how the system is adjusting for math.  One out of 32767 times
it will return a value 1 greater than his range.

For a range 1 to 10 (normal return):
32766 / 32767 = 0.999969481 * 10 = 9.99969481.  Truncate returns
9.  +1 = 10.  Top range.
But, that 32767'th time, 
32767 / 32767 = 1 * 10 = 10.  + 1 = 11.  Ooops.

The larger the range returned the less this will happen, but it is
still a possibility.  In my functions, I had always looked for this,
if RangeReturn = TopRange + 1, recursively call the subroutine.

Other options include just throwing that "extra" one into the toprange.
if RangeReturn = TopRange + 1, RangeReturn = TopRange.

With one returning a number from 0 to .9999999999 this isn't a problem.

Regards,

Jim Langston

Me transmitte sursum, Caledoni!

Chris Rehm wrote:
> 
> Paul Jackson wrote:
> 
> Paul, the solution you found is the most accurate one. You can, as been
> pointed out, simply divide the random number generated by the range of
> the number you want and take the remainder. But that is not exactly
> accurate. Let's suppose you wish to generate a number between 1 and
> 100.  327 times out of 328 it will return a number between 0 and 99, add
> one and you have the random number you wanted to generate. But once out
> of 328 tries it will generate a number between 32700 and 32768, not
> 32799. So there is no chance you will generate a number between 69-100
> using this method. So your random generation is slightly skewed.
> 
> The method you have found, using the ratio of the number to it's range
> and applying that to the range of number that you want will give as a
> distribution that is as random as the numbers provided by the generator.
> 
> All that said, it might be that the generator isn't all that flat
> either, and you might not care if you are all that perfect in your
> distribution (I don't know your application). So you just need to decide
> between accuracy and ease of coding.
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