問題描述

我有兩種方法可以在 [0..n-1] 范圍內(nèi)生成 m 個不同的隨機數(shù)

I have two methods of generating m distinct random numbers in the range [0..n-1]

方法一:

//C++-ish pseudocode
int result[m];
for(i = 0; i < m; ++i)
{
   int r;
   do
   {
      r = rand()%n;
   }while(r is found in result array at indices from 0 to i)
   result[i] = r;   
}

方法二:

//C++-ish pseudocode
int arr[n];
for(int i = 0; i < n; ++i)
    arr[i] = i;
random_shuffle(arr, arr+n);
result = first m elements in arr;

當(dāng) n 遠大于 m 時，第一種方法更有效，否則第二種方法更有效.但是更大"不是一個嚴格的概念，是嗎?:)

The first method is more efficient when n is much larger than m, whereas the second is more efficient otherwise. But "much larger" isn't that strict a notion, is it? :)

問題: 我應(yīng)該使用 n 和 m 的什么公式來確定方法 1 還是方法 2 更有效?(就運行時間的數(shù)學(xué)期望而言)

Question: What formula of n and m should I use to determine whether method1 or method2 will be more efficient? (in terms of mathematical expectation of the running time)

推薦答案

純數(shù)學(xué):
讓我們計算兩種情況下 rand() 函數(shù)調(diào)用的數(shù)量并比較結(jié)果:

Pure mathematics:
Let's calculate the quantity of rand() function calls in both cases and compare the results:

案例 1:當(dāng)您已經(jīng)選擇了 k 個數(shù)字時，讓我們看看在步驟 i = k 上調(diào)用的數(shù)學(xué)期望.通過一次 rand() 調(diào)用獲得一個數(shù)字的概率等于 p = (n-k)/n.我們需要知道此類調(diào)用數(shù)量的數(shù)學(xué)期望，這導(dǎo)致獲得我們還沒有的數(shù)字.

Case 1: let's see the mathematical expectation of calls on step i = k, when you already have k numbers chosen. The probability to get a number with one rand() call is equal to p = (n-k)/n. We need to know the mathematical expectation of such calls quantity which leads to obtaining a number we don't have yet.

使用1 調(diào)用獲得它的概率是p.使用 2 調(diào)用 - q * p，其中 q = 1 - p.在一般情況下，在 n 次調(diào)用之后準確獲得它的概率是 (q^(n-1))*p.因此，數(shù)學(xué)期望為
Sum[ n * q^(n-1) * p ], n = 1 -->INF.這個總和等于 1/p(由 wolfram alpha 證明).

The probability to get it using 1 call is p. Using 2 calls - q * p, where q = 1 - p. In general case, the probability to get it exactly after n calls is (q^(n-1))*p. Thus, the mathematical expectation is
Sum[ n * q^(n-1) * p ], n = 1 --> INF. This sum is equal to 1/p (proved by wolfram alpha).

因此，在步驟 i = k 上，您將執(zhí)行 1/p = n/(nk) 調(diào)用 rand()功能.

So, on the step i = k you will perform 1/p = n/(n-k) calls of the rand() function.

現(xiàn)在讓我們總結(jié)一下:

Sum[ n/(n - k) ], k = 0 -->m - 1 = n * T - 方法 1 中的 rand 調(diào)用次數(shù).
這里 T = Sum[ 1/(n - k) ], k = 0 -->m - 1

Sum[ n/(n - k) ], k = 0 --> m - 1 = n * T - the number of rand calls in method 1.
Here T = Sum[ 1/(n - k) ], k = 0 --> m - 1

情況 2:

這里 rand() 在 random_shuffle 內(nèi)被調(diào)用 n - 1 次(在大多數(shù)實現(xiàn)中).

Here rand() is called inside random_shuffle n - 1 times (in most implementations).

現(xiàn)在，要選擇方法，我們必須比較這兩個值: n * T ?n - 1.
因此，要選擇合適的方法，請按上述方法計算 T.如果 T <(n - 1)/n 最好使用第一種方法.否則使用第二種方法.

Now, to choose the method, we have to compare these two values: n * T ? n - 1.
So, to choose the appropriate method, calculate T as described above. If T < (n - 1)/n it's better to use the first method. Use the second method otherwise.

這篇關(guān)于在 [0..n-1] 范圍內(nèi)生成 m 個不同的隨機數(shù)的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！