Problem Statement

A Pythagorean triple is defined as three positive integers

( a , b , c )

{\displaystyle (a,b,c)}

where

a < b < c

{\displaystyle a<b<c}

, and

a

2

b

2

=

c

2

.

{\displaystyle a^{2}+b^{2}=c^{2}.}

They are called primitive triples if

a , b , c

{\displaystyle a,b,c}

are co-prime, that is, if their pairwise greatest common divisors

g c d

( a , b )

g c d

( a , c )

g c d

( b , c )

1

{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}

. Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (

g c d

( a , b )

1

{\displaystyle {\rm {gcd}}(a,b)=1}

).   Each triple forms the length of the sides of a right triangle, whose perimeter is

P

a + b + c

{\displaystyle P=a+b+c}

.

The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.

Deal with large values.   Can your program handle a maximum perimeter of 1,000,000?   What about 10,000,000?   100,000,000? Note: the extra credit is not for you to demonstrate how fast your language is compared to others;   you need a proper algorithm to solve them in a timely manner.

Let's start with the solution:

var total = 0
var prim = 0
var maxPeri = 100

func newTri(s0:Int, _ s1:Int, _ s2: Int) -> () {
    
    let p = s0 + s1 + s2
    if p <= maxPeri {
        prim += 1
        total += maxPeri / p
        newTri( s0 + 2*(-s1+s2), 2*( s0+s2) - s1, 2*( s0-s1+s2) + s2)
        newTri( s0 + 2*( s1+s2), 2*( s0+s2) + s1, 2*( s0+s1+s2) + s2)
        newTri(-s0 + 2*( s1+s2), 2*(-s0+s2) + s1, 2*(-s0+s1+s2) + s2)
    }
}

while maxPeri <= 100_000_000 {
    prim = 0
    total = 0
    newTri(3, 4, 5)
    print("Up to \(maxPeri) : \(total) triples \( prim) primitives.")
    maxPeri *= 10
}