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
这相当于Adam Rosenfield的解决方案,但对一些读者来说可能更清楚一些。它假设rand5()是一个函数,返回1到5范围内的统计随机整数。
int rand7()
{
int vals[5][5] = {
{ 1, 2, 3, 4, 5 },
{ 6, 7, 1, 2, 3 },
{ 4, 5, 6, 7, 1 },
{ 2, 3, 4, 5, 6 },
{ 7, 0, 0, 0, 0 }
};
int result = 0;
while (result == 0)
{
int i = rand5();
int j = rand5();
result = vals[i-1][j-1];
}
return result;
}
How does it work? Think of it like this: imagine printing out this double-dimension array on paper, tacking it up to a dart board and randomly throwing darts at it. If you hit a non-zero value, it's a statistically random value between 1 and 7, since there are an equal number of non-zero values to choose from. If you hit a zero, just keep throwing the dart until you hit a non-zero. That's what this code is doing: the i and j indexes randomly select a location on the dart board, and if we don't get a good result, we keep throwing darts.
就像亚当说的,在最坏的情况下,它可以一直运行下去,但从统计上看,最坏的情况永远不会发生。:)
其他回答
就是这样,均匀分布,零rand5调用。
def rand7:
seed += 1
if seed >= 7:
seed = 0
yield seed
需要事先播种。
function rand7() {
while (true) { //lowest base 5 random number > 7 reduces memory
int num = (rand5()-1)*5 + rand5()-1;
if (num < 21) // improves performance
return 1 + num%7;
}
}
Python代码:
from random import randint
def rand7():
while(True):
num = (randint(1, 5)-1)*5 + randint(1, 5)-1
if num < 21:
return 1 + num%7
100000次运行的测试分布:
>>> rnums = []
>>> for _ in range(100000):
rnums.append(rand7())
>>> {n:rnums.count(n) for n in set(rnums)}
{1: 15648, 2: 15741, 3: 15681, 4: 15847, 5: 15642, 6: 15806, 7: 15635}
我想到了一个解决这个问题的有趣方法,想和大家分享一下。
function rand7() {
var returnVal = 4;
for (var n=0; n<3; n++) {
var rand = rand5();
if (rand==1||rand==2){
returnVal+=1;
}
else if (rand==3||rand==4) {
returnVal-=1;
}
}
return returnVal;
}
我构建了一个测试函数,循环rand7() 10,000次,将所有返回值相加,然后除以10,000。如果rand7()工作正常,我们计算的平均值应该是4 -例如,(1+2+3+4+5+6+7 / 7)= 4。在做了多次测试后,平均值确实是4:)
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。
假设rand给予所有位相同的权重,然后用上界进行掩码。
int i = rand(5) ^ (rand(5) & 2);
Rand(5)只能返回:1b, 10b, 11b, 100b, 101b。有时候你只需要考虑设置2位。