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
这个解决方案受到了Rob McAfee的启发。 然而,它不需要循环,结果是一个均匀分布:
// Returns 1-5
var rnd5 = function(){
return parseInt(Math.random() * 5, 10) + 1;
}
// Helper
var lastEdge = 0;
// Returns 1-7
var rnd7 = function () {
var map = [
[ 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 ]
];
var result = map[rnd5() - 1][rnd5() - 1];
if (result > 0) {
return result;
}
lastEdge++;
if (lastEdge > 7 ) {
lastEdge = 1;
}
return lastEdge;
};
// Test the a uniform distribution
results = {}; for(i=0; i < 700000;i++) { var rand = rnd7(); results[rand] = results[rand] ? results[rand] + 1 : 1;}
console.log(results)
结果:[1:99560,2:99932,3:100355,4:100262,5:99603,6:100062,7:100226]
js小提琴
其他回答
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;
}
Here is a solution that tries to minimize the number of calls to rand5() while keeping the implementation simple and efficient; in particular, it does not require arbitrary large integers unlike Adam Rosenfield’s second answer. It exploits the fact that 23/19 = 1.21052... is a good rational approximation to log(7)/log(5) = 1.20906..., thus we can generate 19 random elements of {1,...,7} out of 23 random elements of {1,...,5} by rejection sampling with only a small rejection probability. On average, the algorithm below takes about 1.266 calls to rand5() for each call to rand7(). If the distribution of rand5() is uniform, so is rand7().
uint_fast64_t pool;
int capacity = 0;
void new_batch (void)
{
uint_fast64_t r;
int i;
do {
r = 0;
for (i = 0; i < 23; i++)
r = 5 * r + (rand5() - 1);
} while (r >= 11398895185373143ULL); /* 7**19, a bit less than 5**23 */
pool = r;
capacity = 19;
}
int rand7 (void)
{
int r;
if (capacity == 0)
new_batch();
r = pool % 7;
pool /= 7;
capacity--;
return r + 1;
}
面对这么复杂的答案,我觉得自己很蠢。
为什么不能:
int random1_to_7()
{
return (random1_to_5() * 7) / 5;
}
?
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:)
这类似于@RobMcAfee,除了我使用魔术数字而不是2维数组。
int rand7() {
int m = 1203068;
int r = (m >> (rand5() - 1) * 5 + rand5() - 1) & 7;
return (r > 0) ? r : rand7();
}