notebook/notes/algorithms/binary-search.md

title	TARGET DECK	FILE TAGS	tags
Binary Search	Obsidian::STEM	algorithm	algorithm

Overview

Property	Value
Best Case	O(1)
Worst Case	O(\lg{n})
Aux. Memory	O(1)

%%ANKI Basic What precondition must the input of binary search satisfy? Back: It must already be sorted. Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

%%ANKI Basic What is the best case running time of binary search? Back: \Omega(1) Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

%%ANKI Basic What input does binary search perform best on? Back: One in which the value being searched for is already in the middle. Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

%%ANKI Basic What is the worst case running time of binary search? Back: O(\lg{n}) Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

%%ANKI Basic What input does binary search perform worst on? Back: One in which the value does not exist. Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

%%ANKI Basic What is the typical output of binary search? Back: The index of the element in the array being searched for, if found. Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

A recursive solution looks as follows:

static int aux(const int needle, const int i, const int j, int *A) {
  if (i > j) {
    return -1;
  }
  int mid = (i + j) / 2;
  if (A[mid] == needle) {
    return mid;
  } else if (A[mid] < needle) {
    return aux(needle, mid + 1, j, A);
  } else {
    return aux(needle, i, mid - 1, A);
  }
}

int binary_search(const int needle, const int n, int A[static n]) {
  return aux(needle, 0, n - 1, A);
}

We can also write this iteratively:

int binary_search(const int needle, const int n, int A[static n]) {
  int i = 0;
  int j = n - 1;
  while (i <= j) {
    int mid = (i + j) / 2;
    if (A[mid] == needle) {
      return mid;
    } else if (A[mid] < needle) {
      i = mid + 1;
    } else {
      j = mid - 1;
    }
  }
  return -1;
}

%%ANKI Basic In binary search, when could using floor for midpoint calculations yield different answers than ceiling? Back: When there exist duplicate members. Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

%%ANKI Basic In binary search, what ensures left pointer L and right pointers R eventually satisfy L > R? Back: The found midpoint is always excluded from the next binary search invocation. Reference: Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).

END%%

References

• Thomas H. Cormen et al., Introduction to Algorithms, 3rd ed (Cambridge, Mass: MIT Press, 2009).