I can think of 2 appraches.
[1] Similar to something already suggested. Just adding my 2 cents.
1 to 7 digits can be represented by 3 bits. So going by that,
int rand_1_7() {
b1 = rand_1_5()*(7/5) > (7/2) ? 1 : 0;
b2 = rand_1_5()*(7/5) > (7/2) ? 1 : 0;
b3 = rand_1_5()*(7/5) > (7/2) > 1 : 0;
return (b1*4 + b2*2 + b3*1);
}
[2] What we are trying to do is map numbers 1 to 5 to numbers 1 to 7. Of course
mapping 5 items to 7 items is not straight forward. So lets not map items, but
map transitions. Suppose an imaginary initial state x. Now when we call
rand_1_5(), we have one of the following transition,
1) x -> 1
2) x -> 2
3) x -> 3
4) x -> 4
5) x -> 5
Now, lets call rand_1_5() again, and remember the last transition, viz.
1) x -> 1 -> 1
2) x -> 1 -> 2
....
24) x -> 5 -> 4
25) x -> 5 -> 5
So we have 25 transitions. Divide it by 7, we get 4 as remainder and 3 as
quotient i.e. if we each number 1 to 7, three of these 25 transitions each, we
get 4 unused transitions. Lets take it to next level. We get 25*5 = 125
transitions. Divide by 7, we get 6 unused transitions. Still another level, get
us 125*5 = 625 transition, divide by 7, we get 2 as remainder. So nearest we
get is, 625*5*5 = 15624. Dividing by 7, we get 1 as remainder. So in 15625
transitions, (by calling rand_1_5() 6-times), we can assign 2232
unique-transitions to each on of 1,2,..,7. With just 1 un-assigned transition,
we get 1/15625 part-error, or 0.0064% error.
Comments?
On Sep 7, 10:56 am, ankur aggarwal <[email protected]> wrote:
> Given a random number generator that generates numbers in the range 1 to
> 5, how can u create a random number generator to generate numbers in the
> range 1 to 7. (remember that the generated random numbers should follow a
> uniform distribution in the corresponding range)
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