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Check whether right angled triangle is valid or not for large sides

Given three integers a, b and c as triplets. Check if it is possible to make right angled triangle or not. Print Yes if possible, else No. 10-18 <= a, b, c <= 1018

Input: 3 4 5
Output: Yes
Explanation:
Since 3*3 + 4*4 = 5*5
Hence print "Yes"

Input: 8 5 13
Since 8 + 5 < 13 which violates the property of
triangle. Hence print "No"


For a right angled triangle to be valid it must satisfies the following criteria:-

  1. a, b and c should be greater then 0.
  2. Sum of any two sides of triangle must be greater than the third side.
  3. Pythagorean Theorem i.e., a2 + b2 = c2.

First two conditions can be easily checked but for third condition we have to take care of overflow. Since a, b and c can be large so we can’t compare them directly unless we use python or BigInteger library in Java. For languages like C and C++, we have to reduce the expression in fraction form.
 implies a^2 + b^2 = c^2  implies a^2 = c^2 - b^2  implies dfrac{a}{c-b}=dfrac{c+b}{a}
Before comparing the fraction we need convert them in simplified form by dividing the numerator and denominator by gcd of both of them. Now compare both numerator and denominator of both the fractions of LHS and RHS such that if both would become same then it signifies the valid right angled triangle otherwise not.

// C++ program to check validity of triplets
#include <bits/stdc++.h>
using namespace std;
  
// Function to check pythagorean triplets
bool Triplets(long long a, long long b, long long c)
{
    if (a <= 0 || b <= 0 || c <= 0)
        return false;
  
    vector<long long> vec{ a, b, c };
    sort(vec.begin(), vec.end());
  
    // Re-initialize a, b, c in ascending order
    a = vec[0], b = vec[1], c = vec[2];
  
    // Check validation of sides of triangle
    if (a + b <= c)
        return false;
  
    long long p1 = a, p2 = c - b;
  
    // Reduce fraction to simplified form
    long long div = __gcd(p1, p2);
    p1 /= div, p2 /= div;
  
    long long q1 = c + b, q2 = a;
  
    // Reduce fraction to simplified form
    div = __gcd(q1, q2);
    q1 /= div, q2 /= div;
  
    // If fraction are equal return
    // 'true' else 'false'
    return (p1 == q1 && p2 == q2);
}
  
// Function that will return 'Yes' or 'No'
// according to the correction of triplets
string checkTriplet(long long a, long long b, long long c)
{
    if (Triplets(a, b, c))
        return "Yes";
    else
        return "No";
}
  
// Driver code
int main()
{
    long long a = 4, b = 3, c = 5;
    cout << checkTriplet(a, b, c) << endl;
  
    a = 8, b = 13, c = 5;
    cout << checkTriplet(a, b, c) << endl;
  
    a = 1200000000000, b = 1600000000000,
    c = 2000000000000;
    cout << checkTriplet(a, b, c) << endl;
  
    return 0;
}

Output:
Yes
No
Yes

Time complexity: O(log(M)) where M is the Maximum value among a, b and c.
Auxiliary space: O(1)



This article is attributed to GeeksforGeeks.org

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