rand25() =5*(rand5()-1) + rand5()

rand7() { 
   while(true) {
       int r = rand25();
       if (r < 21) return r%3;         
   }
}

为什么这样做:循环永远运行的概率是0。

这里是我的一般实现，在给定一个范围为[0,B-1]的均匀发生器的情况下，生成范围为[0,N-1]的均匀。

public class RandomUnif {

    public static final int BASE_NUMBER = 5;

    private static Random rand = new Random();

    /** given generator, returns uniform integer in the range 0.. BASE_NUMBER-1
    public static int randomBASE() {
        return rand.nextInt(BASE_NUMBER);
    }

    /** returns uniform integer in the range 0..n-1 using randomBASE() */
    public static int randomUnif(int n) {
        int rand, factor;
        if( n <= 1 ) return 0;
        else if( n == BASE_NUMBER ) return randomBASE();
        if( n < BASE_NUMBER ) {
            factor = BASE_NUMBER / n;
            do
                rand = randomBASE() / factor;
            while(rand >= n);
            return rand;
        } else {
            factor = (n - 1) / BASE_NUMBER + 1;
            do {
                rand = factor * randomBASE() + randomUnif(factor);
            } while(rand >= n);
            return rand;
        }
    }
}

不是特别高效，但一般和紧凑。对基生成器的均值调用:

 n  calls
 2  1.250 
 3  1.644 
 4  1.252 
 5  1.000 
 6  3.763 
 7  3.185 
 8  2.821 
 9  2.495 
10  2.250 
11  3.646 
12  3.316 
13  3.060 
14  2.853 
15  2.650 
16  2.814 
17  2.644 
18  2.502 
19  2.361 
20  2.248 
21  2.382 
22  2.277 
23  2.175 
24  2.082 
25  2.000 
26  5.472 
27  5.280 
28  5.119 
29  4.899 

int getOneToSeven(){
    int added = 0;
    for(int i = 1; i<=7; i++){
        added += getOneToFive();
    }
    return (added)%7+1;
}

只要没有剩下7种可能性，就再画一个随机数，将可能性数乘以5。在Perl中:

$num = 0;
$possibilities = 1;

sub rand7
{
  while( $possibilities < 7 )
  {
    $num = $num * 5 + int(rand(5));
    $possibilities *= 5;
  }
  my $result = $num % 7;
  $num = int( $num / 7 );
  $possibilities /= 7;
  return $result;
}

为什么不除以5再乘以7，然后四舍五入呢?(当然，你必须使用浮点数no.)

它比其他解决方案更简单、更可靠(真的吗?)例如，在Python中:

def ranndomNo7():
    import random
    rand5 = random.randint(4)    # Produces range: [0, 4]
    rand7 = int(rand5 / 5 * 7)   # /5, *7, +0.5 and floor()
    return rand7

这不是很容易吗?

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;
}