View Category

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
clojure
(defn pythagorean [a b c] (= (+ (* a a) (* b b)) (* c c)))

(defn intsqrt [cc]
(. (. Math sqrt cc) intValue)
)

(defn triples [maxSize]
(filter not-empty
(for [a (range 1 20) b (range a 20)]
(let [c (intsqrt (+ (* a a) (* b b)))]
(if (pythagorean a b c)
[a b c]
()
)))))

(triples 20)
; -> ([3 4 5] [5 12 13] [6 8 10] [8 15 17] [9 12 15] [12 16 20] [15 20 25])

(defn sortByHypotenuse [triples]
(sort-by #(first (rest (rest %))) triples)
)

(sortByHypotenuse (triples 20))
; -> ([3 4 5] [6 8 10] [5 12 13] [9 12 15] [8 15 17] [12 16 20] [15 20 25])
(doseq [pt (sort-by #(% 2)
(for [a (range 1 21)
b (range a 21)
:let [aa+bb (+ (* a a) (* b b))
c (Math/round (Math/sqrt aa+bb))]
:when (= aa+bb (* c c))]
[a b c]))]
(println pt))
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;
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).
fsharp
let getGoodTri (a,b) =
let h = int(System.Math.Sqrt(float(a*a + b*b)))
if a*a + b*b = h*h then Some(a,b,h)
else None

seq{ for i in 1..20 do yield! seq{for j in i..20 do yield i,j} } |> Seq.choose(getGoodTri) |> Seq.sortBy(fun (_,_,c) -> c);;
ocaml
let is_int v =
v = (snd (modf v))

let sort_by_third tup =
let third (_,_,v) = v in
let cmp a b = compare (third a) (third b) in
List.sort cmp tup

let hypi ia ib =
let hyp a b = sqrt(a**2.0 +. b**2.0) in
hyp (float_of_int ia) (float_of_int ib)

let find_pythag max =
let rec py t = match t with
| (a,_) when a > max -> []
| (a,b) when b > max -> py (a+1,a+1)
| (a,b) ->
let next = (a,b+1) in
let cf = hypi a b in
if (is_int cf) then
( a,b,(int_of_float cf) ) :: (py next)
else
py next
in
sort_by_third ( py (1,1) )

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
clojure
(defn gcd [a b]
(if (zero? b)
a
(recur b (mod b a))))
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;
}
erlang
-module(gcd).
-export([gcd/2]).

gcd(A, 0) -> A;
gcd(A, B) -> gcd(B, A rem B).
fsharp
let rec gcd x y =
if y = 0 then x
else gcd y (x % y)
ocaml
(* tail recursive *)
let rec gcd n m =
if m = 0 then
n
else if n > m then
gcd (n-m) m
else
gcd n (m-n)
;;