asked on 12/8/2004

random number generator and related possibility

Given a function which can generate 5 random numbers: 1,2,3,4,5 with equal possibility. Based on this function, how to write a function which generates only 1,2,3 with equal possibility? what will be the complexity of the function?

Thanks

Programming

Last Comment

ozo

12/8/2004

Mikal613

12/8/2004

Dim MyValue
MyValue = Int((3 * Rnd) + 1) ' Generate random value between 1 and 3.

Mikal613

12/8/2004

without repeating the same number twice

https://www.experts-exchange.com/questions/10036364/Series-of-different-random-numbers.html

kindy

12/8/2004

ASKER

sorry , I don't understand this one, can that generate a number of 3*5+1
=16? How to get 1, 2, 3 with equal possibility?
Thanks

MyValue = Int((3 * Rnd) + 1)

Mikal613

12/8/2004

This will get you random numbers between 1 and 3!!!!

Dim MyValue

MyValue = Int((3 * Rnd) + 1) ' Generate random value between 1 and 3.

kindy

12/8/2004

ASKER

oh, I got you, but I think it was not the solution to my question. I am not asking a function which can generate 1, 2, 3 by its own, what I want is this new function must be based on a given function which can generate 1, 2, 3, 4, 5 randomly with equal possibility, but output 1, 2, 3 with eaqual possiblity

Mikal613

12/8/2004

"but output 1, 2, 3 with eaqual possiblity" Please Explain

kindy

12/8/2004

ASKER

I mean there is already a function which can generate 5 random numbers: 1,2,3,4,5 with equal possibility, we have to use this function to generate only random number 1, 2, 3 with equal possibility

ASKER CERTIFIED SOLUTION

Mike Tomlinson

12/8/2004

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kindy

12/8/2004

ASKER

this seems work, while what is the complexity?

Thanks

kindy

12/8/2004

ASKER

btw, will the loop probably be infinite ?

Mike Tomlinson

12/8/2004

The loop won't be infinite...you have a 3 in 5 chance of getting a number that will cause the loop to drop out. If you get a 4 or 5, you pick another number. It will pick a number between 1 and 3 in the blink of an eye.

As far as complexity goes...I don't know. It has been so long since I did any algorithm analysis...

~IM

ozo

12/8/2004

Another way is to write a decimal fraction in base 5, and stop when the precision is sufficient to confine the range of the number to one of three regions. (This has the advantage that you need not throw away any random information, but can make use of the left over information on the next call)

In either method, consider the probability of stoping on the first step, the probability of stopping on the second step, etc. What will be the average number of steps before the algorithm is complete?