如何找到二分查找算法的迭代次数？

Question

如何获取二分查找的迭代次数？

这是我的代码：

int main() 
{
    int target = 11;
    int N = 10;
    std::vector<int> index;
    int num;

    for (int i = 0; i < N; i++) {
        index.push_back(i);
    }

    int counter = 0;
    unsigned int M, L = (unsigned int)-1, R = N;

    M = (R - L) / 2;  // Assume N is not zero

    do {
        int value;
        M = M + L;
        value = index[M];
        if (value < target) L = M; else R = M;
        M = (R - L) / 2;
        counter++;
    } while (M);  // M is the size of the current interval

    std::cout << R << '\n';
    std::cout << "counter: " << counter << '\n';

    system("pause");

    return 0;
}

我想知道 N 的迭代次数。我知道这个算法是如何工作的，但我想要迭代次数用数学表示。

Answer 1

我会通过使用递归二进制搜索函数来递归递增。在二进制检查的每个分支中，只需递增 1 即可递归计算迭代次数。

See live here

#include <iostream>
#include <vector>

std::size_t binarySearch(
    const std::vector<int>& arr,        // pass array as non-modifiyable(const ref)
    std::size_t start, std::size_t end, // start and end indexes of the array
    const int target)                   // target to find
{
    if (arr.size() == 1) return arr[0] == target ? 1 : 0; // edge case

    if (start <= end)
    {
        const std::size_t mid_index = start + ((end - start) / 2);
        return arr[mid_index] == target ? 1 :                         // found the middle element
               arr[mid_index] < target ?
                 binarySearch(arr, mid_index + 1, end, target) + 1:   // target is greater than mid-element
                 binarySearch(arr, start, mid_index - 1, target) + 1; // target is less than mid-element
    }
    return 0;
}

int main()
{
    int target = 11;
    const int N = 10;
    std::vector<int> index;
    index.reserve(N); // reserve some memory
    for (int i = 0; i < N; i++) {
        index.push_back(i);
    }
    std::cout << "counter: " << binarySearch(index, 0, index.size() - 1, target) << std::endl;
    return 0;
}

输出:

counter: 4

Answer 2

数学上可能的最大迭代（假设只有整数类型的情况）是 = ceil( log2 ( initial_r - initial_l ) ) log 的底数是 2，因为每次我们都在潜水我们的范围一半的一半，然后切换到一半。

Answer 3

我也一直在努力思考对数概念化，这就是我试图理解答案的方式

来自 https://en.wikipedia.org/wiki/Binary_search_algorithm

In mathematics, the binary logarithm (log₂n) is the power to which the number 2 must be raised to obtain the value n. That is, for any real number x,
x=log₂n <=equivalent to=> 2^x=n

&

Every binary tree with n leaves has height at least log₂n, with equality when n is a > power of two and the tree is a complete binary tree.

二分搜索有点像沿着二叉搜索树走下去，然后将节点减半，该系列是对数（以 2 为底）系列 _{（个人理解，未引用，可能有误）}

然后从https://www.cct.lsu.edu/~sidhanti/tutorials/data_structures/page305.html

a perfect binary tree of height h has exactly 2^h+1-1 internal nodes. Conversely, the height of a perfect binary tree with n internal nodes is log₂(n+1). If we have a search tree that has the shape of a perfect binary tree, then every unsuccessful search visits exactly h+1 internal nodes, where h=log₂(n+1).

_{（再次以自己的理解跟进...）}
因此，要到达节点 N（根据二叉搜索树的值为 N？），您需要进行 log₂(N+1) 次迭代（遍历树下的多个级别）最坏情况（发现仍然是概率，因此“最坏情况”的措辞）。

在这里干运行（通过构造一个小的BST并手动计数）：https://www.cs.usfca.edu/~galles/visualization/BST.html

_{（当然 phrasing/calculations 中的 review/confirmations/corrections 的答案是开放的，因为我正在尝试整合不同的资源以得出在这种情况下有意义的理论）}

如何找到二分查找算法的迭代次数？

How to find the number of iterations of binary search algorithm?

c++

algorithm

binary-search

c++11