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Exponential Search

#!/usr/bin/env python3

"""
Pure Python implementation of exponential search algorithm

For more information, see the Wikipedia page:
https://en.wikipedia.org/wiki/Exponential_search

For doctests run the following command:
python3 -m doctest -v exponential_search.py

For manual testing run:
python3 exponential_search.py
"""

from __future__ import annotations


def binary_search_by_recursion(
    sorted_collection: list[int], item: int, left: int = 0, right: int = -1
) -> int:
    """Pure implementation of binary search algorithm in Python using recursion

    Be careful: the collection must be ascending sorted otherwise, the result will be
    unpredictable.

    :param sorted_collection: some ascending sorted collection with comparable items
    :param item: item value to search
    :param left: starting index for the search
    :param right: ending index for the search
    :return: index of the found item or -1 if the item is not found

    Examples:
    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 0, 0, 4)
    0
    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 15, 0, 4)
    4
    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 5, 0, 4)
    1
    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 6, 0, 4)
    -1
    """
    if right < 0:
        right = len(sorted_collection) - 1
    if list(sorted_collection) != sorted(sorted_collection):
        raise ValueError("sorted_collection must be sorted in ascending order")
    if right < left:
        return -1

    midpoint = left + (right - left) // 2

    if sorted_collection[midpoint] == item:
        return midpoint
    elif sorted_collection[midpoint] > item:
        return binary_search_by_recursion(sorted_collection, item, left, midpoint - 1)
    else:
        return binary_search_by_recursion(sorted_collection, item, midpoint + 1, right)


def exponential_search(sorted_collection: list[int], item: int) -> int:
    """
    Pure implementation of an exponential search algorithm in Python.
    For more information, refer to:
    https://en.wikipedia.org/wiki/Exponential_search

    Be careful: the collection must be ascending sorted, otherwise the result will be
    unpredictable.

    :param sorted_collection: some ascending sorted collection with comparable items
    :param item: item value to search
    :return: index of the found item or -1 if the item is not found

    The time complexity of this algorithm is O(log i) where i is the index of the item.

    Examples:
    >>> exponential_search([0, 5, 7, 10, 15], 0)
    0
    >>> exponential_search([0, 5, 7, 10, 15], 15)
    4
    >>> exponential_search([0, 5, 7, 10, 15], 5)
    1
    >>> exponential_search([0, 5, 7, 10, 15], 6)
    -1
    """
    if list(sorted_collection) != sorted(sorted_collection):
        raise ValueError("sorted_collection must be sorted in ascending order")

    if sorted_collection[0] == item:
        return 0

    bound = 1
    while bound < len(sorted_collection) and sorted_collection[bound] < item:
        bound *= 2

    left = bound // 2
    right = min(bound, len(sorted_collection) - 1)
    return binary_search_by_recursion(sorted_collection, item, left, right)


if __name__ == "__main__":
    import doctest

    doctest.testmod()

    # Manual testing
    user_input = input("Enter numbers separated by commas: ").strip()
    collection = sorted(int(item) for item in user_input.split(","))
    target = int(input("Enter a number to search for: "))
    result = exponential_search(sorted_collection=collection, item=target)
    if result == -1:
        print(f"{target} was not found in {collection}.")
    else:
        print(f"{target} was found at index {result} in {collection}.")
About this Algorithm

Prerequisites

Problem Statement

Given a sorted array of n elements, write a function to search for the index of a given element (target)

Approach

  • Search for the range within which the target is included increasing index by powers of 2
  • If this range exists in array apply the Binary Search algorithm over it
  • Else return -1

Example

arr = [1, 2, 3, 4, 5, 6, 7, ... 998, 999, 1_000]

target = 998
index = 0
1. SEARCHING FOR THE RANGE
index = 1, 2, 4, 8, 16, 32, 64, ..., 512, ..., 1_024
after 10 iteration we have the index at 1_024 and outside of the array 
2. BINARY SEARCH
Now we can apply the binary search on the subarray from 512 and 1_000.

Note: we apply the Binary Search from 512 to 1_000 because at i = 2^10 = 1_024 the array is finisced and the target number is less than the latest index of the array ( 1_000 ).

Time Complexity

worst case: O(log *i*) where *i* = index (position) of the target

best case: O(*1*)

Complexity Explanation

  • The complexity of the first part of the algorithm is O( log i ) because if i is the position of the target in the array, after doubling the search index ⌈log(i)⌉ times, the algorithm will be at a search index that is greater than or equal to i. We can write 2^⌈log(i)⌉ >= i
  • The complexity of the second part of the algorithm also is O ( log i ) because that is a simple Binary Search. The Binary Search complexity ( as explained here ) is O( n ) where n is the length of the array. In the Exponential Search, the length of the array on which the algorithm is applied is 2^i - 2^(i-1), put into words it means '( the length of the array from start to i ) - ( the part of array skipped until the previous iteration )'. Is simple verify that 2^i - 2^(i-1) = 2^(i-1)

After this detailed explanation we can say that the the complexity of the Exponential Search is:

O(log i) + O(log i) = 2O(log i) = O(log i)

Binary Search vs Exponential Search

Let's take a look at this comparison with a less theoretical example. Imagine we have an array with1_000_000 elements and we want to search an element that is in the 4th position. It's easy to see that:

  • The Binary Search start from the middle of the array and arrive to the 4th position after many iterations
  • The Exponential Search arrive at the 4th index after only 2 iterations