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

Question

How I can get the number of iterations of binary search?

This is my code:

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;
}

I want to know the number of iterations depending on N . I know how this algorithm works but I want the number of iterations represented mathematically.

Answer 1

I would go for a recursive increment of by using a recursive binary search function. In each branch of the binary checks, just increment by one and that will count the iterations recursively.

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;
}

Output :

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 ) ) 对数的基数是 2 因为每次我们通过取一个中间值并切换到一半来将我们的范围减半。

Answer 3

I have been trying to wrap my head around the logarithmic conceptualization too, and this is how I try to understand the answer

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

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

&

Every binary tree with n leaves has height at least log ₂ n, with equality when n is a > power of two and the tree is a complete binary tree.

Binary search is kind of like walking down a binary search tree, and halving the nodes, and that series is a logarithmic (base 2) series _{(my own understanding, no citation, can be erroneous)}

And then from 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 log ₂ (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=log ₂ (n+1).

_{(again following up with my own understanding...)}
So, to reach a node N (with value N as per binary search tree?), you would make log ₂ (N+1) iterations (traverse that many levels down the tree) in worst case (finding is still a probability, hence "worst case" phrasing).

Dry run here (by constructing a small BST and counting manually): https://www.cs.usfca.edu/~galles/visualization/BST.html

_{(answer open for review/confirmations/corrections in phrasing/calculations too ofcourse as I am trying to consolidate different resources to reach a theory that makes sense in this context)}