Find pythagorean triplets from array

Find the set of Pythagorean triples from the given array
Pythagorean triplets condition: a^2 + b^2 = c^2

Example

Input array: [3,4,6,5,7,8]
Pythagorean triplets: 3, 4, 5
3^2 + 4^2 = 5^2

Method 1

We use a brute force algorithm here :

Algorithm

Time complexity: O(n^3)

Step 1 : We use 3 loops such that we take set of 3 different elements from the array.
a. We run 3 for loops. Such that for every a we take all the values b except itself. For every b we take all the values of a.
b. a, b, c are elements from the array.

Step 2 : we run the Pythagorean condition that is a*a + b*b = c*c, for all the sets of (a, b, c). we print them when it is true.
a. If a^2 + b^2 = c^2, print a, b, c.

Code

[code language=”cpp”]

#include <bits/stdc++.h>
using namespace std;

int main()
{

int arr[] = { 3,8,4,10,6,5,12,13,27}; // array
int N = sizeof(arr)/sizeof(arr[0]);//size of array

int a,b,c;
for(int i = 0; i < N-2; i++)//select an element
{
for(int j=i+1;j <N-1; j++)//select an element in front of the considered element
{
for(int k =i+2; k<N;k++)// this element will be one ahead of the previously selected element in the jus touter loop
{
a = arr[i];
b = arr[j];
c = arr[k];
if(a*a + b*b == c*c) // if the chosen elements satisfy the pythagoras theorem then simply print the three values.
cout << a <<" "<<b<<" "<<c<<endl;

}
}
}
return 0;
}

[/code]

Method 2
Algorithm:

Time complexity: O(n^2)

Step 1 : Sort the given array first using the function sort.
Step 2 : Instead of storing the numbers store the square of each element to directly check the Pythagorean theorem.
Step 3 : Take a as the smallest side, for every a check the elements from the array which satisfy the condition (a = c – b). if they satisfy this condition they form Pythagorean triplet as they satisfy the condition
a2 + b2 = c2
a. for all elements in the array from start, store the first element as “a”.
b. store the last two elements as “b” and “c” respectively.
c. Check the condition “a = c – b”. if true print the sqrts of a, b, c as a set of Pythagorean triplets.
d. If “c – b” is greater than “a”, decrease the variable pointing at the larger element(c) so that we are checking for all “c” is this condition true or not. If “c – b” is less than a decrease the variable pointing at the smaller element so that we are checking for all b is this condition true or not.
e. continue this loop for all a`s
f. If not found any, print no triplets.

code:-

[code language=”cpp”]

#include <bits/stdc++.h>
using namespace std;

int main()
{

int arr[] = { 3,8,4,10,6,5,12,13,27,117,165,19,176,169,44,113,24,145,143,51,149,52,173,181,125}; // array
int N = sizeof(arr)/sizeof(arr[0]);//size of array

int a,b,c;
sort(arr,arr+N); //sort the array
for(int i=0; i < N; i++)
arr[i] = (arr[i] * arr[i]); //store the square of each element to directly check the pythagoras theorem

for(int i=0; i<N; i++)
{
int left = N-2 , right = N-1;
a = arr[i]; // first side of the triangle

while(left > i)
{
b = arr[left];
c = arr[right];

int calculated_side = c – b; //if a*a + b*b = c*c then obviously c*c – b*b = a*a , we utilize this to check the condition
if(calculated_side == a)
{
cout << sqrt(a) << " " << sqrt(b) << " " << sqrt(c) << endl;
left++; right–;
}
else if (calculated_side > a) //if side is larger than expected then decrease the variable pointing at the larger element
right–;
else // if side is smaller than expected then decrease the variable pointing at the smaller element
left–;
}
}

return 0;
}

[/code]