Code a function rand7() when we're given rand5() .

[L0s3r]

Hi Guys.

Random no.'s always bother me.So,I want to start a discussion over their uniform distribution.

Given : rand5() which generates randomly distributed no. from 1-5.
To Code: rand7() which generates randomly distributed no. from 1-5.

You have to give code using rand5() as well as without using it.

Thanks,

[vijayan121]

If we call rand5() seven times and add up the results, we would get an integer in the inclusive interval [7,35].
Divide this number by 5.0 and we would get a real number in the inclusive interval [1.0,7.0].

Take care of the round-off correctly (how?) and you have rand7().

[StuXYZ]

First off given your question statement,

To Code: rand7() which generates randomly distributed no. from 1-5

you can do this: ;)

int rand7() {  return rand5(); }

Ok I guess that isn't your question, you actually want to return rand7 where the result is 1-7.

First thing to establish is if rand5() returns an integer value or a floating point number in the rand 1-5. If it is integer you can try this:

Quicker in the long run is to use a reduction algorithm. E.g create the number

//
while(a>20) {
  a=(rand5()-1)*5+(rand5()-1);
}
return 1+a/7;

The issue is is it quicker than adding 7 numbers (and seven calls to rand5()). It can be slower, but most of the time it is quicker.

If on the other hand you have 1.0->5.0 you can actually use the same algorithms, but in that case there is the much quicker approach of just dividing the result, however, this approach looses low bit accuracy.