Binary Search for Rational Numbers without using floating point arithmetic

A rational is represented as p/qb, for example 2/3. Given a sorted array of rational numbers, how to search an element using Binary Search. Use of floating point arithmetic is not allowed.


Input:  arr[] = {1/5, 2/3, 3/2, 13/2}
        x = 3/2
Output: Found at index 2

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To compare two rational numbers p/q and r/s, we can compare p*s with q*r.

// C program for Binary Search for Rationalnal Numbers
// without using floating point arithmetic
#include <stdio.h>
struct Rational
    int p;
    int q;
// Utility function to compare two Rationalnal numbers
// 'a' and 'b'. It returns
// 0 --> When 'a' and 'b' are same
// 1 --> When 'a' is greater
//-1 --> When 'b' is greate
int compare(struct Rational a, struct Rational b)
    // If a/b == c/d  then  a*d = b*c:
    // method to ignore division
    if (a.p * b.q == a.q * b.p)
        return 0;
    if (a.p * b.q > a.q * b.p)
        return 1;
    return -1;
// Returns index of x in arr[l..r] if it is present, else
// returns -1. It mainly uses Binary Search.
int binarySearch(struct Rational arr[], int l, int r,
                 struct Rational x)
   if (r >= l)
        int mid = l + (r - l)/2;
        // If the element is present at the middle itself
        if (compare(arr[mid], x) == 0)  return mid;
        // If element is smaller than mid, then it can
        // only be present in left subarray
        if (compare(arr[mid], x) > 0)
            return binarySearch(arr, l, mid-1, x);
        // Else the element can only be present in right
        // subarray
        return binarySearch(arr, mid+1, r, x);
   return -1;
// Driver method
int main()
    struct Rational arr[] = {{1, 5}, {2, 3}, {3, 2}, {13, 2}};
    struct Rational x = {3, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("Element found at index %d",
            binarySearch(arr, 0, n-1, x));


Element found at index 2

Thanks to Utkarsh Trivedi for suggesting above solution.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

This article is attributed to GeeksforGeeks.org

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