给定一个函数,它产生的是1到5之间的随机整数,写一个函数,它产生的是1到7之间的随机整数。
当前回答
这是我想到的答案,但这些复杂的答案让我认为这是完全错误的/:))
import random
def rand5():
return float(random.randint(0,5))
def rand7():
random_val = rand5()
return float(random.randint((random_val-random_val),7))
print rand7()
其他回答
只需要缩放第一个函数的输出
0) you have a number in range 1-5
1) subtract 1 to make it in range 0-4
2) multiply by (7-1)/(5-1) to make it in range 0-6
3) add 1 to increment the range: Now your result is in between 1-7
上面引用了一些优雅的算法,但这里有一种方法可以接近它,尽管它可能是迂回的。我假设的值是从0开始的。
R2 =给出小于2的随机数生成器(样本空间= {0,1}) R8 =给出小于8的随机数生成器(样本空间= {0,1,2,3,4,5,6,7})
为了从R2生成R8,您将运行R2三次,并将所有3次运行的组合结果作为3位二进制数使用。下面是R2运行三次时的值范围:
0, 0, 0 --> 0 . . 1, 1, 1 --> 7
现在要从R8生成R7,我们只需再次运行R7,如果它返回7:
int R7() {
do {
x = R8();
} while (x > 6)
return x;
}
迂回的解决方案是从R5生成R2(就像我们从R8生成R7一样),然后从R2生成R8,然后从R8生成R7。
首先,我在1点上移动ramdom5() 6次,得到7个随机数。 其次,将7个数相加得到公和。 第三,除法的余数是7。 最后加1,得到从1到7的结果。 这个方法给出了在1到7的范围内获得数字的相等概率,除了1。1的概率略高。
public int random7(){
Random random = new Random();
//function (1 + random.nextInt(5)) is given
int random1_5 = 1 + random.nextInt(5); // 1,2,3,4,5
int random2_6 = 2 + random.nextInt(5); // 2,3,4,5,6
int random3_7 = 3 + random.nextInt(5); // 3,4,5,6,7
int random4_8 = 4 + random.nextInt(5); // 4,5,6,7,8
int random5_9 = 5 + random.nextInt(5); // 5,6,7,8,9
int random6_10 = 6 + random.nextInt(5); //6,7,8,9,10
int random7_11 = 7 + random.nextInt(5); //7,8,9,10,11
//sumOfRandoms is between 28 and 56
int sumOfRandoms = random1_5 + random2_6 + random3_7 +
random4_8 + random5_9 + random6_10 + random7_11;
//result is number between 0 and 6, and
//equals 0 if sumOfRandoms = 28 or 35 or 42 or 49 or 56 , 5 options
//equals 1 if sumOfRandoms = 29 or 36 or 43 or 50, 4 options
//equals 2 if sumOfRandoms = 30 or 37 or 44 or 51, 4 options
//equals 3 if sumOfRandoms = 31 or 38 or 45 or 52, 4 options
//equals 4 if sumOfRandoms = 32 or 39 or 46 or 53, 4 options
//equals 5 if sumOfRandoms = 33 or 40 or 47 or 54, 4 options
//equals 6 if sumOfRandoms = 34 or 41 or 48 or 55, 4 options
//It means that the probabilities of getting numbers between 0 and 6 are almost equal.
int result = sumOfRandoms % 7;
//we should add 1 to move the interval [0,6] to the interval [1,7]
return 1 + result;
}
下面是Adam回答的Python实现。
import random
def rand5():
return random.randint(1, 5)
def rand7():
while True:
r = 5 * (rand5() - 1) + rand5()
#r is now uniformly random between 1 and 25
if (r <= 21):
break
#result is now uniformly random between 1 and 7
return r % 7 + 1
我喜欢把我正在研究的算法扔进Python,这样我就可以摆弄它们,我想我把它贴在这里,希望它对外面的人有用,而不是花很长时间来拼凑。
int ans = 0;
while (ans == 0)
{
for (int i=0; i<3; i++)
{
while ((r = rand5()) == 3){};
ans += (r < 3) >> i
}
}