Here is a solution that tries to minimize the number of calls to rand5() while keeping the implementation simple and efficient; in particular, it does not require arbitrary large integers unlike Adam Rosenfield’s second answer. It exploits the fact that 23/19 = 1.21052... is a good rational approximation to log(7)/log(5) = 1.20906..., thus we can generate 19 random elements of {1,...,7} out of 23 random elements of {1,...,5} by rejection sampling with only a small rejection probability. On average, the algorithm below takes about 1.266 calls to rand5() for each call to rand7(). If the distribution of rand5() is uniform, so is rand7().

uint_fast64_t pool;

int capacity = 0;

void new_batch (void)
{
  uint_fast64_t r;
  int i;

  do {
    r = 0;
    for (i = 0; i < 23; i++)
      r = 5 * r + (rand5() - 1);
  } while (r >= 11398895185373143ULL);  /* 7**19, a bit less than 5**23 */

  pool = r;
  capacity = 19;
}

int rand7 (void)
{
  int r;

  if (capacity == 0)
    new_batch();

  r = pool % 7;
  pool /= 7;
  capacity--;

  return r + 1;
}

该算法将rand5的调用次数减少到理论最小值7/5。通过产生接下来的5个rand7数字来调用它7次。

没有任何随机位的拒绝，也不可能一直等待结果。

#!/usr/bin/env ruby

# random integer from 1 to 5
def rand5
    STDERR.putc '.'
    1 + rand( 5 )
end

@bucket = 0
@bucket_size = 0

# random integer from 1 to 7
def rand7
    if @bucket_size == 0
        @bucket = 7.times.collect{ |d| rand5 * 5**d }.reduce( &:+ )
        @bucket_size = 5
    end

    next_rand7 = @bucket%7 + 1

    @bucket      /= 7
    @bucket_size -= 1

    return next_rand7
end

35.times.each{ putc rand7.to_s }

算法:

7可以用3位的序列表示

使用rand(5)随机地用0或1填充每一位。 例如:调用rand(5)和

如果结果是1或2，则用0填充位 如果结果是4或5，则用1填充位 如果结果是3，则忽略并重新执行(拒绝)

这样，我们可以用0/1随机填充3位，从而得到1-7中的数字。

编辑:这似乎是最简单和最有效的答案，所以这里有一些代码:

public static int random_7() {
    int returnValue = 0;
    while (returnValue == 0) {
        for (int i = 1; i <= 3; i++) {
            returnValue = (returnValue << 1) + random_5_output_2();
        }
    }
    return returnValue;
}

private static int random_5_output_2() {
    while (true) {
        int flip = random_5();

        if (flip < 3) {
            return 0;
        }
        else if (flip > 3) {
            return 1;
        }
    }
}

这里允许作业题吗?

这个函数进行粗略的“以5为基数”的数学运算，生成0到6之间的数字。

function rnd7() {
    do {
        r1 = rnd5() - 1;
        do {
            r2=rnd5() - 1;
        } while (r2 > 1);
        result = r2 * 5 + r1;
    } while (result > 6);
    return result + 1;
}

下面使用随机数发生器在{1,2,3,4,5,6,7}上产生均匀分布，在{1,2,3,4,5}上产生均匀分布。代码很混乱，但逻辑很清晰。

public static int random_7(Random rg) {
    int returnValue = 0;
    while (returnValue == 0) {
        for (int i = 1; i <= 3; i++) {
            returnValue = (returnValue << 1) + SimulateFairCoin(rg);
        }
    }
    return returnValue;
}

private static int SimulateFairCoin(Random rg) {
    while (true) {
        int flipOne = random_5_mod_2(rg);
        int flipTwo = random_5_mod_2(rg);

        if (flipOne == 0 && flipTwo == 1) {
            return 0;
        }
        else if (flipOne == 1 && flipTwo == 0) {
            return 1;
        }
    }
}

private static int random_5_mod_2(Random rg) {
    return random_5(rg) % 2;
}

private static int random_5(Random rg) {
    return rg.Next(5) + 1;
}    

extern int r5();

int r7() {
    return ((r5() & 0x01) << 2 ) | ((r5() & 0x01) << 1 ) | (r5() & 0x01);
}