写一个 C++给定 rand1() 生成 0 到 5 之间的随机分布的程序，随机返回 0 或 1

发布于 2024-11-30 15:26:18 字数 579 浏览 1 评论 0原文

函数

int rand1();

给定一个以相同概率返回 0 或 1 的，实现一个

int rand5();

以相等概率返回 0,1,2,3,4,5 的函数。

！！！扭转！！！在将其标记为重复之前先阅读...

可以调用 rand1() 的次数是固定的。您可以将其设置为 10、20 或 100，但不能设置任意数量的 rand1() 调用。即 rand1() 调用次数有上限。此外，您还必须保证 rand5() 始终以相同的概率返回 o 到 5。代码偏向于少数额外的 0 和 1 是不可接受的。

如果您认为不可能编写这样的函数，那么您可以让我们都知道，为什么它不可能。

编辑：这就是我所拥有的，我认为这还不够

int rand5()
{
bitset<3> b;
b[0] = rand1();
b[1] = rand1();
b[2] = rand1();
int i = b;
if(b >= 6)
 return rand5();
return i;
}

原文

Given a function

int rand1();

which return 0 or 1, with equal probability,
implement a function

int rand5();

which returns 0,1,2,3,4,5 with equal probability.

!!! Twist !!! Read before marking it as duplicate...

Number of times you can call rand1() is fixed. You may decide it to be 10 or 20 or 100 for that matter, but NOT any number of rand1() calls.
i.e. there is a upper limit on on number of rand1() calls.
Also you have to guarantee that rand5() should always return o to 5, with equal probability. It is not acceptable that the code is skewed towards, few extra 0 and 1.

If you think it is NOT POSSIBLE to write such function, then you can let us all know, as to why it is not possible.

EDIT :
this is what i have, which I think is not sufficient

int rand5()
{
bitset<3> b;
b[0] = rand1();
b[1] = rand1();
b[2] = rand1();
int i = b;
if(b >= 6)
 return rand5();
return i;
}

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绝不放开 2024-12-07 15:26:18

不可能。你不能将 2^n 整分为 6。

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柠檬色的秋千 2024-12-07 15:26:18

这就是人们在从 [0, 1) 范围内的随机浮点数生成 [0, N) 范围内的随机整数时所做的事情（无论他们是否知道）：

// assumed sizeof(unsigned int) == sizeof(float) == 32/CHAR_BIT
// assumed 'float' is IEEE single precision with the mantissa in the
// low bits
unsigned int randN(unsigned int high)
{
    union { unsigned int i; float f; } u;

    u.i = 0;
    for (int j = 0; j < 24; j++)
        u.i = u.i*2 + rand1();

    // u.f will be in the range [0, 0.5)
    // so multiply by twice the desired range
    return (unsigned int)floor(u.f * (high * 2));
}

我怀疑这不会产生完全均匀分布，但对于大多数用途来说已经足够了。

This is what people are doing (whether they know it or not) when they generate a random integer in the range [0, N) from a random floating-point number in the range [0, 1):

// assumed sizeof(unsigned int) == sizeof(float) == 32/CHAR_BIT
// assumed 'float' is IEEE single precision with the mantissa in the
// low bits
unsigned int randN(unsigned int high)
{
    union { unsigned int i; float f; } u;

    u.i = 0;
    for (int j = 0; j < 24; j++)
        u.i = u.i*2 + rand1();

    // u.f will be in the range [0, 0.5)
    // so multiply by twice the desired range
    return (unsigned int)floor(u.f * (high * 2));
}

I suspect this does not produce a perfectly uniform distribution, but it's good enough for most purposes.

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