AMCAT Coding Question 4 (Unsolved)

A Pythagorean triplet is a set of three integers a, b and c such that a2 + b2 = c2. Given a limit, generate all Pythagorean Triples with values smaller than given limit.

Input : limit = 20
Output : 3 4 5
         8 6 10
         5 12 13
         15 8 17
         12 16 20

Simple Solution is to generate these triplets smaller than given limit using three nested loop. For every triplet, check if Pythagorean condition is true, if true, then print the triplet. Time complexity of this solution is O(limit3) where ‘limit’ is given limit.

An Efficient Solution can print all triplets in O(k) time where k is number of triplets printed. The idea is to use square sum relation of Pythagorean triplet, i.e., addition of squares of a and b is equal to square of c, we can write these number in terms of m and n such that,

       a = m2 - n2
       b = 2 * m * n
       c  = m2 + n2
because,
       a2 = m4 + n4 – 2 * m2 * n2
       b2 = 4 * m2 * n2
       c2 = m4 + n4 + 2* m2 * n2

We can see that a2 + b2 = c2, so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets.

Below is C implementation of above idea.

// A C program to generate pythagorean triplets
// smaller than a given limit
#include <stdio.h>
#include <math.h>
 
//  Function to generate pythagorean triplets
//  smaller than limit
void pythagoreanTriplets(int limit)
{
    // triplet:  a^2 + b^2 = c^2
    int a, b, c=0;
 
    //  loop from 2 to max_limitit
    int m = 2;
 
    // Limiting c would limit all a, b and c
    while (c < limit)
    {
        // now loop on j from 1 to i-1
        for (int n = 1; n < m; ++n)
        {
            // Evaluate and print triplets using
            // the relation between a, b and c
            a = m*m - n*n;
            b = 2*m*n;
            c = m*m + n*n;
 
            if (c > limit)
                break;
 
            printf("%d %d %d\n", a, b, c);
        }
        m++;
    }
}
 
// Driver program
int main()
{
    int limit = 20;
    pythagoreanTriplets(limit);
    return 0;
}
3 4 5
8 6 10
5 12 13
15 8 17
12 16 20

Time complexity of this approach is O(k) where k is number of triplets printed for a given limit (We iterate for m and n only and every iteration prints a triplet)