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Algorithms
Geometry

Find all Pythagorean triangles with length or height less than or equal to 20

Pythagorean triangles are right angle triangles whose sides comply with the following equation:

a * a + b * b = c * c

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. Find all such triangles where a, b and c are non-zero integers with a and b less than or equal to 20. Sort your results by the size of the hypotenuse. The expected answer is:

[3, 4, 5]
[6, 8, 10]
[5, 12, 13]
[9, 12, 15]
[8, 15, 17]
[12, 16, 20]
[15, 20, 25]
scala
val res = for (
x <- 1 to 20 ;
y <- x to 20 ;
z = Math.sqrt(x*x + y*y) ;
if (z.toInt == z) )
yield (x, y, z.toInt)

res.toList.sortWith { (t1, t2) =>
t1._3 < t2._3
} foreach (println(_))
(for(x <- 1 to 20;
y<- x to 20;
z<- 1 to 30;
if(z*z == x*x + y*y)) yield(x, y, z)
).sortWith(_._3 < _._3) foreach println
( for (
a <- 1 to 20 ;
b <- a to 20 ;
c = math.sqrt( a*a + b*b )
if c.toInt == c
) yield ( a, b, c.toInt )
).sortBy {_._3} foreach println
erlang
find_all_pythagorean_triangles(L) ->
lists:sort(fun({_, _, H1}, {_, _, H2}) -> H1 =< H2 end,
[ { X, Y, Z } ||
X <- lists:seq(1,L),
Y <- lists:seq(1,L),
Z <- lists:seq(1,2*L),
X*X + Y*Y =:= Z*Z,
Y > X,
Z > Y
]).

main(_) ->
List = find_all_pythagorean_triangles(20).
cpp
vector<solution> solutions;

for (int a = 1; a <= 20; ++a)
for (int b = a + 1; b <= 20; ++b)
{
int c_squared = a*a + b*b;
int c = b + 1;
while (c * c < c_squared)
++c;
if (c * c == c_squared)
solutions.push_back(make_tuple(a, b, c));
}

sort(begin(solutions), end(solutions),
[](const solution& s1, const solution& s2) { return get<2>(s1) < get<2>(s2); });

for (const auto &s: solutions)
cout << '[' << get<0>(s) << ", " << get<1>(s) << ", " << get<2>(s) << ']' << endl;
Arithmetic

Greatest Common Divisor

Find the largest positive integer that divides two given numbers without a remainder. For example, the GCD of 8 and 12 is 4.

scala
def gcd(x: Int, y: Int): Int =
if (b == 0) x
else gcd(b, x % y)
erlang
-module(gcd).
-export([gcd/2]).

gcd(A, 0) -> A;
gcd(A, B) -> gcd(B, A rem B).
csharp
public static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
cpp
#include <iostream>
#include <cstdlib>
#include <algorithm>

using namespace std;

int gcd_recursive(int i, int j) {
if (min(i, j) == 0)
return max(i, j);
else
return gcd_recursive(min(i, j), abs(i - j));
}

int gcd_recursive2(int x, int y) {
if (y == 0)
return x;
else
return gcd_recursive2(y, (x % y));
}

int gcd_iterative(int i, int j) {
while (min(i, j) != 0) {
i = min(i, j);
j = abs(i - j);
}
return max(i, j);
}

int main() {
std::cout << gcd_recursive(8, 12) << std::endl;
std::cout << gcd_recursive2(8, 12) << std::endl;
std::cout << gcd_iterative(8, 12) << std::endl;
return 0;
}