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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]
ruby
results=[]

1.upto(20) do |a|
1.upto(20) do |b|
c=Math.sqrt(a**2+b**2)
results<<[a, b, c.to_i] if c.to_i==c && !results.index([b, a, c.to_i])
end
end

results=results.sort_by{|r| r[2]}

puts results
def find_pythag( max=20 )
r = []
1.upto max do |n|
n.upto max do |m|
h = Math.sqrt( n**2 + m**2)
r << [n,m,h.to_i] if (h.round - h).zero?
end
end
r.sort_by { |a| a[2] }
end
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;
fantom
triangles := [,]
(1..20).each |Int a|
{
(a..20).each |Int b|
{
c := (a.pow(2) + b.pow(2)).toFloat.sqrt
if (c % c.toInt == 0.0f && !triangles.contains([b,a,c]))
triangles.add([a,b,c.toInt])
}
}
triangles.sort |Int[] x, Int[] y -> Int| { x[2]-y[2] }
echo(triangles)
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).

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.

ruby
135.gcd(30)
# => 15
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;
}
fantom
gcd := |Int a, Int b -> Int| {
pair := [a, b].sort
while (pair.first != 0)
pair.set(1, pair.last % pair.first).swap(0, 1)
return pair.last
}
echo(gcd(12, 8)) // a>b, result == 4
echo(gcd(1029, 1071)) // a<b, result == 21
erlang
-module(gcd).
-export([gcd/2]).

gcd(A, 0) -> A;
gcd(A, B) -> gcd(B, A rem B).