Monday, August 8, 2016

Google – Implement Rand6 by Rand2


Google – Implement Rand6 by Rand2
Rand6() return a random number in [0, 6)
Rand2() return a random number in [0, 2)
用Rand2()来实现Rand6()

[Solution]
思路分为两步:
1.第一步要解决如何用rand2()来算其他的,比如rand3, rand4, rand5。
因为我们只能用rand2(),所以无非就是对rand2()进行加减乘除来得到randK. 但是不要天真的以为
rand4() = 2 * rand2(), 因为2 * [0, 2) = [0, 4)。 错!!
请问上面的式子中,2 * rand2()如何得到 3?
正确的rand4()应该等于2 * rand2() + rand2(), 但是这个范围可不是[0, 6),而就是[0, 4).
2.第二步就是要保证等概率。
通过上面的分析,如果现在问要计算rand6()怎么算,可能看着上面现学现卖就会出现这样的回答:
rand6() = 2 * rand4() + rand4() 又错!!
这个方法有两个问题:
(1). 范围实际超过了[0, 6), 变成了[0, 9)
(2). 算一下会发现有两种不同的方法可以得到 4, 比如2 * 2 + 0或者2 * 1 + 2。而只有一种方法可以得到 1,就是2 * 0 + 1。
这样就说明[0, 9)之间每个数并不是等概率的。事实上范围超过[0, 6)并不是大问题,只要保证每个数都是等概率,如果random到678,丢掉继续random就是了,直到random出[0, 6)之间的。
关键是第二个问题,如何得到等概率。
我们可以证明下面的公式
I * randN + randI 得到的范围 [0, N * I) 内所有的数,都是evenly distributed
[0, I), randN = 0
[I, 2I), randN = 1
[2I, 3I), randN = 2

[kI, (k+1)I), randN = k
上面每个区间内的每个数都只有一种方法可以得到。
现在再来看rand6()怎么实现的话,就能得到以下两种方法:
rand6() = 2 * rand4() + rand2()
rand6() = 4 * rand4() + rand4()
不过还有一点需要注意,我们只有rand2()可以用,rand4()需要通过rand2()来计算,而上面第二种方法两个rand4()不是同一个数喔,要通过rand2()计算两次,得到两个不同的rand4()。
class Solution {
  public int rand6() {
    while (true) {
      int rand4 = 2 * rand2() + rand2();
      int result = 2 * rand4 + rand2();
      if (result < 6) {
        return result;
      }
    }
  }
}
 
class Solution2 {
  public int rand6() {
    while (true) {
      int rand4_1 = 2 * rand2() + rand2();
      int rand4_2 = 2 * rand2() + rand2();
      int result = 4 * rand4_1 + rand4_2;
      if (result < 12) {
        return result % 6;
      }
    }
  }
}
Read full article from Google – Implement Rand6 by Rand2

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