All Pythagorean Triplets less than equal N in Optimal Time Complexity

All Pythagorean Triplets less than equal N in Optimal Time Complexity

Pythagorean Triplets are set of three ordered positive integers (x,y,z) such that

Pythagorean Triplets are set of three ordered positive integers (x,y,z) such that

                       x^2 + y^2 = z^2    --- (1)
where,
x < y < z <= N

There are several techniques for generating Pythagorean Triplets for example using Gaussian Integers and using Euclid’s formula. They can be used to generate all primitive Pythagorean Triplets, which again can be used to generate all non-primitive Pythagorean triplets by using the fact that the non Primitive Pythagorean triples are a positive integer multiple of primitive Pythagorean Triplets.

Generating Pythagorean triplets simply using two loops by iterating over x and y for all possible x and y, and checking if they satisfy Equation (1), takes time O(N² ). Also using the above methods for generating primitive Pythagorean triplets and from them generating all Pythagorean triples may take time O(N² ).

I used Equation (2) which easily follows from Equation (1) by solving the quadratic equation for x, thus for “ x > 0 ”,

                      x = (b - a) + sqrt{2b(b - a)} --- (2)

where,
a = y - x,
b = z - x and
0 < a < b

To find all Pythagorean Triples, we need to find the all values for a and b such that they are integers and Equation (2) holds. To find values for a and b, first, we will go step by step by finding ranges of each.

From the definition of a, it’s upper and lower limits depends upon b and can be written as

                             1 ≤ a ≤ b − 1

Since b > a and since the lower limit of a is 1 thus lower limit of b is 2. Now we need To find the upper limit of b.

we know 0< z ≤ N, from (b = z − x) _⇒ (_x + b ≤ N), thus from Equation (2) and a = b − 1,

                      (b − a) + sqrt{2b(b − a)} + b ≤ N           
                       ⇒ b + 1 + sqrt{2b} ≤ N
                       ⇒ 0 ≤ b ≤ N − sqrt{2N − 1}
Since first pythagorean Triple is (3,4,5), which makes N ≥ 5

Let,_ v = 2b(b − a), since _x is a positive integer and _1 < a < b, _thus from Equation (2),

                        2b ≤ v ≤ 2b(b − 1) and 
                        sqrt{v} ∈ N        and 
                        v (mod 2b) = 0

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