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
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;
}
其他回答
因为1/7是一个以5为底的无限小数,所以没有(完全正确的)解可以在常数时间内运行。一个简单的解决方案是使用拒绝抽样,例如:
int i;
do
{
i = 5 * (rand5() - 1) + rand5(); // i is now uniformly random between 1 and 25
} while(i > 21);
// i is now uniformly random between 1 and 21
return i % 7 + 1; // result is now uniformly random between 1 and 7
这个循环的预期运行时间为25/21 = 1.19次迭代,但是永远循环的概率非常小。
什么是简单的解决方案?(rand5() + rand5()) % 7 + 1 减少内存使用或在较慢的CPU上运行的有效解决方案是什么?是的,这是有效的,因为它只调用rand5()两次,空间复杂度为O(1)
考虑rand5()给出从1到5(包括)的随机数。 (1 + 1) % 7 + 1 = 3 (1 + 2) % 7 + 1 = 4 (1 + 3) % 7 + 1 = 5 (1 + 4) % 7 + 1 = 6 (1 + 5) % 7 + 1 = 7
(2 + 1) % 7 + 1 = 4 (2 + 2) % 7 + 1 = 5 (2 + 3) % 7 + 1 = 6 (2 + 4) % 7 + 1 = 7 (2 + 5) % 7 + 1 = 1 .
(5 + 1) % 7 + 1 = 7 (5 + 2) % 7 + 1 = 1 (5 + 3) % 7 + 1 = 2 (5 + 4) % 7 + 1 = 3 (5 + 5) % 7 + 1 = 4 .
等等
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;
}
def rand5():
return random.randint(1,5) #return random integers from 1 to 5
def rand7():
rand = rand5()+rand5()-1
if rand > 7: #if numbers > 7, call rand7() again
return rand7()
print rand%7 + 1
我想这将是最简单的解决方案,但到处都有人建议5*rand5() + rand5() - 5,如http://www.geeksforgeeks.org/generate-integer-from-1-to-7-with-equal-probability/。 有人能解释一下rand5()+rand5()-1有什么问题吗
下面是一个利用c++ 11特性的答案
#include <functional>
#include <iostream>
#include <ostream>
#include <random>
int main()
{
std::random_device rd;
unsigned long seed = rd();
std::cout << "seed = " << seed << std::endl;
std::mt19937 engine(seed);
std::uniform_int_distribution<> dist(1, 5);
auto rand5 = std::bind(dist, engine);
const int n = 20;
for (int i = 0; i != n; ++i)
{
std::cout << rand5() << " ";
}
std::cout << std::endl;
// Use a lambda expression to define rand7
auto rand7 = [&rand5]()->int
{
for (int result = 0; ; result = 0)
{
// Take advantage of the fact that
// 5**6 = 15625 = 15624 + 1 = 7 * (2232) + 1.
// So we only have to discard one out of every 15625 numbers generated.
// Generate a 6-digit number in base 5
for (int i = 0; i != 6; ++i)
{
result = 5 * result + (rand5() - 1);
}
// result is in the range [0, 15625)
if (result == 15625 - 1)
{
// Discard this number
continue;
}
// We now know that result is in the range [0, 15624), a range that can
// be divided evenly into 7 buckets guaranteeing uniformity
result /= 2232;
return 1 + result;
}
};
for (int i = 0; i != n; ++i)
{
std::cout << rand7() << " ";
}
std::cout << std::endl;
return 0;
}