下面的递归算法是(效率非常低)的方式来计算ñ选择K:
The following recursive algorithm is a (fairly inefficient) way to compute n choose k:
int combinationsOf(int n, int k) { if (k == 0) return 1; if (n == 0) return 0; return combinationsOf(n - 1, k) + combinationsOf(n - 1, k - 1); }
它是基于以下递归洞察:
It is based on the following recursive insight:
其实评价这一函数的函数调用了很多。例如,计算2选2这种方式使这些调用:
Actually evaluating this function takes a LOT of function calls. For example, computing 2 choose 2 this way makes these calls:
combinationsOf(2, 2) | | | +- combinationsOf(1, 2) | | | | | +- combinationsOf(0, 2) | | | +-- combinationsOf(1, 1) | | | | | +- combinationsOf(0, 1) | | | +- combinationsOf(1, 0) +- combinationsOf(2, 1) | | | +- combinationsOf(2, 0) | +- combinationsOf(1, 1) | | | +- combinationsOf(0, 1) | +- combinationsOf(1, 0)有许多的方法来改善该功能的运行时 - 我们可以使用动态编程,使用闭合形式的式NCK =正! /(K(N - !K))等。但是,我很好奇,刚才的如何的低效率这个特殊的算法
There are many ways to improve the runtime of this function - we could use dynamic programming, use the closed-form formula nCk = n! / (k! (n - k)!), etc. However, I am curious just how inefficient this particular algorithm is.
这是什么功能的大O的时间复杂性,n和k的函数?
What is the big-O time complexity of this function, as a function of n and k?
推荐答案让 C(N,K)是计算成本 binom( N,K)以这种方式,以
Let C(n,k) be the cost of computing binom(n,k) in that way, with
C(0,_) = 1 C(_,0) = 1
作为基础的情况下。
as base cases.
复发显然是
C(n,k) = 1 + C(n-1,k-1) + C(n-1,k)如果我们把除了拥有成本1。然后,我们可以很容易地检查
if we take addition to have cost 1. Then, we can easily check that
k C(n,k) = 2 * ∑ binom(n,j) - 1 j=0
感应。
因此,对于 K> = N ,成本 2 ^(N + 1) - 1 , C(N,N-1)= 2 ^(N + 1) - 3 , C(N,1)= 2 * N + 1 , C(N,2)= N *(N + 1)+1 ,(而除此之外,我没有看到整齐的公式)。
So for k >= n, the cost is 2^(n+1) - 1, C(n,n-1) = 2^(n+1)- 3, C(n,1) = 2*n+1, C(n,2) = n*(n+1)+1, (and beyond that, I don't see neat formulae).
我们有明显的上界
C(n,k) < 2^(n+1)
独立 K ,以及 K&LT; N / 2 我们可以粗略估算
independent of k, and for k < n/2 we can coarsely estimate
C(n,k) <= 2*(k+1)*binom(n,k)
这给小得多界小 K ,所以总体
C(n,k) ∈ O(min{(k+1)*binom(n,min(k, n/2)), 2^n})
(需要夹紧 K 的最低,因为 binom(N,K)减小回1为 K&GT; N / 2 )
(need to clamp the k for the minimum, since binom(n,k) decreases back to 1 for k > n/2).
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什么是这个天真code来计算组合的大O的复杂性?
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