Google interview question: Given a function that generates integers 1 to 5 at random, write a function that generates integers 1 to 7 at random.

We know that a function f can obtain a random number ranging from 1 to 5 with equal probability. How can we obtain a random number ranging from 1 to 7 with equal probability?

Int rand7 ()
{
Int;
While (A = rand5 () * 5 + rand5 ()> 26 );
Return (A-3)/3;
}

AlgorithmThe idea is:
1. Use rand5 () * 5 + rand5 () to generate 6 7 8 9 10 11 ...... 26, 27, 28, 29, 30, and the occurrence rate of each number is equal.
2. Only the first 3*7 numbers are required, so the four numbers are discarded.
3. convert 6 7 8 to 1, 9 10 11 to 2 ,......, 24 25 26 to 7. Formula is (A-3)/3

As long as we can randomly select 1 to n numbers from N numbers, and perform this operation repeatedly until the last number is left.
We can call the given function n times to generate n random numbers between 1 and 5, and select the location of the maximum number to meet the above requirements.

For example
The initial number of 7 numbers [1, 2, 3, 4, 5, 6, 7].
7 random numbers from 1 to 5 [5, 3, 1, 4, 2, 5]
Then we reserve [, 7],
Three random numbers from 1 to 5 [2, 4, 1]
Then we keep the [6]
6 is the random number generated this time.

Compile a random function that generates 0 and 1:
Step1. call the given random function original_rand () to generate a number.
If = 3 goto Step1
If <3 return 0
If> 3 return 1

Compile a random function to generate 1 to 7
Call the Random Functions of 0 and 1 three times to generate 000 or 001 or 010 .......
If ??? Returns if the value is not equal to 0. Otherwise, it is generated again.CodeAs follows:

int rand_01 () {< br> int r = original_rand (); If (r = 3) return rand_01 (); If (r <3) return 0; If (r> 3) return 1; }int rand_17 () {< br> int I = 0; I + = rand_01 (); I + = rand_01 () <1; I + = rand_01 () <2; if (I = 0) return rand_17 (); return I; }