Random Number range extension (such as rand7 () to rand10 ())

Question:
It is known that there is a rand7 () function, returns a random natural number from 1 to 7, and uses this rand7 () to construct rand10 () random 1 ~ 10.
Analysis: To ensure the even distribution of rand10 () on integers 1-10, you can construct a random integer interval with a 1-10 * n uniform distribution (n is any positive integer ). Assume that x is a random integer in the range of 1-10 * n. Then, x % 10 + 1 is an integer evenly distributed between 1 and 10. Because (rand7 ()-1) * 7 + rand7 () can construct a random number evenly distributed between 1 and 49 (for the reason, see the following description), you can set ~ If such a random number is removed, 1-40 is still evenly distributed among 1-40, because each number can be considered as an independent event.
The following explains why (rand7 ()-1) * 7 + rand7 () can create a random number evenly distributed between 1 and 49:
First, rand7 ()-1 gets a discrete integer set {0, 1, 2, 3, 4, 5, 6}, where each integer has a probability of 1/7. Then (rand7 ()-1) * 7 gets A discrete Integer Set A = {1/7, 42}, where each integer has A probability. The probability of each integer in the Set B = {1, 2, 3, 4, 5, 6, 7} obtained by rand7 () is also 1/7. Obviously, the combination of any two elements in A and B can correspond One to One integer between 1 and 49, that is, any number between 1 and 49, A combination of two elements in A and B can be uniquely identified. Since elements A and B can be considered as independent events, according to the probability formula P (AB) = P (A) P (B) of independent events ), the probability of each combination is 1/7*1/7 = 1/49. Therefore, the integers (rand7 ()-1) * 7 + rand7 () are evenly distributed between 1 and 49, and the probability of each number is 1/49.

        x =                x =  * (rand7() - ) +     }(x >       x %  +  }

: Why not use while (x> 40) instead of while (x> 10? The reason is that if you use while (x> 10), there is a 40/49 probability that you need to loop while, and it is very likely that there is an endless loop.

It is known that the random number generator random3 () generates a random number in the range of [1, 3]. Please use random3 () to construct the random5 () function to generate a random number of [1, 5?
Problem Analysis:
How can we construct a larger range of numbers from the range [1-3? At the same time, the probability of occurrence of Numbers meeting the larger range is the same. Two possible operations can be considered: Addition and multiplication.
Consider the following expression:
3*(random3 ()-1) + random3 ();
It can be calculated that the range of the above expression is [1, 9] and the probability of occurrence of the number is the same, that is, 1/9
How can we generate the number of [1, 5] From the number in the range of [1, 9?
You can think of the rejection sampling method, that is, to generate a random number of [1, 9]. If the number range is not within [1, 5], re-sampling

 

        val =                val =  * (random3() - ) +     }(val >       }

 

Abstract This problem further. It is known that the range of the random number generator of random_m () is [1, m]. Calculate random_n () to generate a function in the range of [1, n, m <n & n <= m * m

 

        val =       t;                  val = m * (random_m() - ) +     }(val >      }