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Law of cosines - triples (view source)

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* [https://youtu.be/p-0SOWbzUYI?t=12m11s Visualising Pythagoras: ultimate proofs and crazy contortions] Mathlogger Video
<br><br>
 
=={{header|11l}}==
{{trans|Python}}
 
<syntaxhighlight lang="11l">-V n = 13
 
F method1(n)
V squares = (0..n).map(x -> x ^ 2)
V sqrset = Set(squares)
Set[(Int, Int, Int)] tri90, tri60, tri120
 
L(a) 1..n
V a2 = squares[a]
L(b) 1..a
V b2 = squares[b]
V c2 = a2 + b2
I c2 C sqrset
tri90.add(tuple_sorted((a, b, Int(sqrt(c2)))))
V ab = a * b
c2 -= ab
I c2 C sqrset
tri60.add(tuple_sorted((a, b, Int(sqrt(c2)))))
c2 += 2 * ab
I c2 C sqrset
tri120.add(tuple_sorted((a, b, Int(sqrt(c2)))))
R [sorted(Array(tri90)),
sorted(Array(tri60)),
sorted(Array(tri120))]
 
print(‘Integer triangular triples for sides 1..#.:’.format(n))
L(angle, triples) zip([90, 60, 120], method1(n))
print(" #3° has #. solutions:\n #.".format(angle, triples.len, triples))
V t60 = method1(10'000)[1]
V notsame = sum(t60.filter((a, b, c) -> a != b | b != c).map((a, b, c) -> 1))
print(‘Extra credit: ’notsame)</syntaxhighlight>
 
{{out}}
<pre>
Integer triangular triples for sides 1..13:
90° has 3 solutions:
[(3, 4, 5), (5, 12, 13), (6, 8, 10)]
60° has 15 solutions:
[(1, 1, 1), (2, 2, 2), (3, 3, 3), (3, 7, 8), (4, 4, 4), (5, 5, 5), (5, 7, 8), (6, 6, 6), (7, 7, 7), (8, 8, 8), (9, 9, 9), (10, 10, 10), (11, 11, 11), (12, 12, 12), (13, 13, 13)]
120° has 2 solutions:
[(3, 5, 7), (7, 8, 13)]
Extra credit: 18394
</pre>
 
=={{header|Action!}}==
<syntaxhighlight lang="action!">PROC Test(INT max,angle,coeff)
BYTE count,a,b,c
 
PrintF("gamma=%B degrees:%E",angle)
count=0
FOR a=1 TO max
DO
FOR b=1 TO a
DO
FOR c=1 TO max
DO
IF a*a+b*b-coeff*a*b=c*c THEN
PrintF("(%B,%B,%B) ",a,b,c)
count==+1
FI
OD
OD
OD
PrintF("%Enumber of triangles is %B%E%E",count)
RETURN
 
PROC Main()
Test(13,90,0)
Test(13,60,1)
Test(13,120,-1)
RETURN</syntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Law_of_cosines_-_triples.png Screenshot from Atari 8-bit computer]
<pre>
gamma=90 degrees:
(4,3,5) (8,6,10) (12,5,13)
number of triangles is 3
 
gamma=60 degrees:
(1,1,1) (2,2,2) (3,3,3) (4,4,4) (5,5,5) (6,6,6) (7,7,7) (8,3,7) (8,5,7) (8,8,8) (9,9,9) (10,10,10) (11,11,11) (12,12,12) (13,13,13)
number of triangles is 15
 
gamma=120 degrees:
(5,3,7) (8,7,13)
number of triangles is 2
</pre>
 
=={{header|Ada}}==
<syntaxhighlight lang="ada">with Ada.Text_IO;
with Ada.Containers.Ordered_Maps;
 
procedure Law_Of_Cosines is
 
type Angle_Kind is (Angle_60, Angle_90, Angle_120);
 
function Is_Triangle (A, B, C : in Positive;
Angle : in Angle_Kind) return Boolean
is
A2 : constant Positive := A**2;
B2 : constant Positive := B**2;
C2 : constant Positive := C**2;
AB : constant Positive := A * B;
begin
case Angle is
when Angle_60 => return A2 + B2 - AB = C2;
when Angle_90 => return A2 + B2 = C2;
when Angle_120 => return A2 + B2 + AB = C2;
end case;
end Is_Triangle;
 
procedure Count_Triangles is
use Ada.Text_IO;
Count : Natural;
begin
for Angle in Angle_Kind loop
Count := 0;
Put_Line (Angle'Image & " triangles");
for A in 1 ..13 loop
for B in 1 .. A loop
for C in 1 .. 13 loop
if Is_Triangle (A, B, C, Angle) then
Put_Line (A'Image & B'Image & C'Image);
Count := Count + 1;
end if;
end loop;
end loop;
end loop;
Put_Line ("There are " & Count'Image & " " & Angle'Image &" triangles");
end loop;
end Count_Triangles;
 
procedure Extra_Credit (Limit : in Natural) is
use Ada.Text_IO;
 
package Square_Maps is new Ada.Containers.Ordered_Maps (Natural, Boolean);
Squares : Square_Maps.Map;
 
Count : Natural := 0;
begin
for C in 1 .. Limit loop
Squares.Insert (C**2, True);
end loop;
 
for A in 1 .. Limit loop
for B in 1 .. A loop
if Squares.Contains (A**2 + B**2 - A * B) then
Count := Count + 1;
end if;
end loop;
end loop;
Put_Line ("There are " & Natural'(Count - Limit)'Image &
" " & Angle_60'Image &" triangles between 1 and " & Limit'Image & ".");
end Extra_Credit;
 
begin
Count_Triangles;
Extra_Credit (Limit => 10_000);
end Law_Of_Cosines;</syntaxhighlight>
 
{{out}}
<pre>ANGLE_60 triangles
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 3 7
8 5 7
8 8 8
9 9 9
10 10 10
11 11 11
12 12 12
13 13 13
There are 15 ANGLE_60 triangles
ANGLE_90 triangles
4 3 5
8 6 10
12 5 13
There are 3 ANGLE_90 triangles
ANGLE_120 triangles
5 3 7
8 7 13
There are 2 ANGLE_120 triangles
There are 18394 ANGLE_60 triangles between 1 and 10000.</pre>
 
=={{header|ALGOL 68}}==
<langsyntaxhighlight lang="algol68">BEGIN
# find all integer sided 90, 60 and 120 degree triangles by finding integer solutions for #
# a^2 + b^2 = c^2, a^2 + b^2 - ab = c^2, a^2 + b^2 + ab = c^2 where a, b, c in 1 .. 13 #
Line 84 ⟶ 275:
print triangles( 90 );
print triangles( 120 )
END</langsyntaxhighlight>
{{out}}
<pre>
Line 110 ⟶ 301:
3 5 7
7 8 13
</pre>
 
=={{header|AWK}}==
<syntaxhighlight lang="awk">
# syntax: GAWK -f LAW_OF_COSINES_-_TRIPLES.AWK
# converted from C
BEGIN {
description[1] = "90 degrees, a*a + b*b = c*c"
description[2] = "60 degrees, a*a + b*b - a*b = c*c"
description[3] = "120 degrees, a*a + b*b + a*b = c*c"
split("0,1,-1",coeff,",")
main(13,1,0)
main(1000,0,1) # 10,000 takes too long
exit(0)
}
function main(max_side_length,show_sides,no_dups, a,b,c,count,k) {
printf("\nmaximum side length: %d\n",max_side_length)
for (k=1; k<=3; k++) {
count = 0
for (a=1; a<=max_side_length; a++) {
for (b=1; b<=a; b++) {
for (c=1; c<=max_side_length; c++) {
if (a*a + b*b - coeff[k] * a*b == c*c) {
if (no_dups && (a == b || b == c)) {
continue
}
count++
if (show_sides) {
printf(" %d %d %d\n",a,b,c)
}
}
}
}
}
printf("%d triangles, %s\n",count,description[k])
}
}
</syntaxhighlight>
{{out}}
<pre>
maximum side length: 13
4 3 5
8 6 10
12 5 13
3 triangles, 90 degrees, a*a + b*b = c*c
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 3 7
8 5 7
8 8 8
9 9 9
10 10 10
11 11 11
12 12 12
13 13 13
15 triangles, 60 degrees, a*a + b*b - a*b = c*c
5 3 7
8 7 13
2 triangles, 120 degrees, a*a + b*b + a*b = c*c
 
maximum side length: 1000
881 triangles, 90 degrees, a*a + b*b = c*c
1260 triangles, 60 degrees, a*a + b*b - a*b = c*c
719 triangles, 120 degrees, a*a + b*b + a*b = c*c
</pre>
 
=={{header|C}}==
=== A brute force algorithm, O(N^23) ===
<syntaxhighlight lang="c">/*
<lang C>/*
* RossetaCode: Law of cosines - triples
*
* An quick and dirty brute force solutions with O(N^3) cost.
* Anyway it is possible set MAX_SIDE_LENGTH equal to 10000 and
* and use fast computer to obtain the "extra credit" badge.
*
* Obviously, there are better algorithms.
Line 137 ⟶ 397:
"gamma = 120 degrees, a*a + b*b + a*b == c*c"
};
static int coeff[3] = { 0, -1, -1 };
 
for (int k = 0; k < 3; k++)
Line 157 ⟶ 417:
return 0;
}
</syntaxhighlight>
</lang>
{{out}}
<pre>
gamma = 90 degrees, a*a + b*b == c*c, number of triangles = 3
gamma = 60 degrees, a*a + b*b - a*b == c*c, number of triangles = 215
gamma = 120 degrees, a*a + b*b + a*b == c*c, number of triangles = 152
</pre>
 
=== An algorithm with O(N^32) cost ===
<syntaxhighlight lang="c">/*
<lang C>/*
* RossetaCode: Law of cosines - triples
*
Line 185 ⟶ 445:
"gamma = 120 degrees, a*a + b*b + a*b == c*c"
};
static int coeff[3] = { 0, -1, -1 };
 
printf("MAX SIDE LENGTH = %d\n\n", MAX_SIDE_LENGTH);
Line 210 ⟶ 470:
return 0;
}
</syntaxhighlight>
</lang>
{{output}}
<pre>
Line 216 ⟶ 476:
 
gamma = 90 degrees, a*a + b*b == c*c, number of triangles = 12471
gamma = 60 degrees, a*a + b*b - a*b == c*c, number of triangles = 1037428394
gamma = 120 degrees, a*a + b*b + a*b == c*c, number of triangles = 2839410374
</pre>
 
=={{header|C++}}==
<syntaxhighlight lang="cpp">#include <cmath>
#include <iostream>
#include <tuple>
#include <vector>
 
using triple = std::tuple<int, int, int>;
 
void print_triple(std::ostream& out, const triple& t) {
out << '(' << std::get<0>(t) << ',' << std::get<1>(t) << ',' << std::get<2>(t) << ')';
}
 
void print_vector(std::ostream& out, const std::vector<triple>& vec) {
if (vec.empty())
return;
auto i = vec.begin();
print_triple(out, *i++);
for (; i != vec.end(); ++i) {
out << ' ';
print_triple(out, *i);
}
out << "\n\n";
}
 
int isqrt(int n) {
return static_cast<int>(std::sqrt(n));
}
 
int main() {
const int min = 1, max = 13;
std::vector<triple> solutions90, solutions60, solutions120;
 
for (int a = min; a <= max; ++a) {
int a2 = a * a;
for (int b = a; b <= max; ++b) {
int b2 = b * b, ab = a * b;
int c2 = a2 + b2;
int c = isqrt(c2);
if (c <= max && c * c == c2)
solutions90.emplace_back(a, b, c);
else {
c2 = a2 + b2 - ab;
c = isqrt(c2);
if (c <= max && c * c == c2)
solutions60.emplace_back(a, b, c);
else {
c2 = a2 + b2 + ab;
c = isqrt(c2);
if (c <= max && c * c == c2)
solutions120.emplace_back(a, b, c);
}
}
}
}
 
std::cout << "There are " << solutions60.size() << " solutions for gamma = 60 degrees:\n";
print_vector(std::cout, solutions60);
 
std::cout << "There are " << solutions90.size() << " solutions for gamma = 90 degrees:\n";
print_vector(std::cout, solutions90);
 
std::cout << "There are " << solutions120.size() << " solutions for gamma = 120 degrees:\n";
print_vector(std::cout, solutions120);
const int max2 = 10000;
int count = 0;
for (int a = min; a <= max2; ++a) {
for (int b = a + 1; b <= max2; ++b) {
int c2 = a * a + b * b - a * b;
int c = isqrt(c2);
if (c <= max2 && c * c == c2)
++count;
}
}
std::cout << "There are " << count << " solutions for gamma = 60 degrees in the range "
<< min << " to " << max2 << " where the sides are not all of the same length.\n";
return 0;
}</syntaxhighlight>
 
{{out}}
<pre>
There are 15 solutions for gamma = 60 degrees:
(1,1,1) (2,2,2) (3,3,3) (3,8,7) (4,4,4) (5,5,5) (5,8,7) (6,6,6) (7,7,7) (8,8,8) (9,9,9) (10,10,10) (11,11,11) (12,12,12) (13,13,13)
 
There are 3 solutions for gamma = 90 degrees:
(3,4,5) (5,12,13) (6,8,10)
 
There are 2 solutions for gamma = 120 degrees:
(3,5,7) (7,8,13)
 
There are 18394 solutions for gamma = 60 degrees in the range 1 to 10000 where the sides are not all of the same length.
</pre>
 
=={{header|C sharp}}==
<syntaxhighlight lang="csharp">using System;
using System.Collections.Generic;
using static System.Linq.Enumerable;
 
public static class LawOfCosinesTriples
{
public static void Main2() {
PrintTriples(60, 13);
PrintTriples(90, 13);
PrintTriples(120, 13);
PrintTriples(60, 10_000, true, false);
}
 
private static void PrintTriples(int degrees, int maxSideLength, bool notAllTheSameLength = false, bool print = true) {
string s = $"{degrees} degree triangles in range 1..{maxSideLength}";
if (notAllTheSameLength) s += " where not all sides are the same";
Console.WriteLine(s);
int count = 0;
var triples = FindTriples(degrees, maxSideLength);
if (notAllTheSameLength) triples = triples.Where(NotAllTheSameLength);
foreach (var triple in triples) {
count++;
if (print) Console.WriteLine(triple);
}
Console.WriteLine($"{count} solutions");
}
 
private static IEnumerable<(int a, int b, int c)> FindTriples(int degrees, int maxSideLength) {
double radians = degrees * Math.PI / 180;
int coefficient = (int)Math.Round(Math.Cos(radians) * -2, MidpointRounding.AwayFromZero);
int maxSideLengthSquared = maxSideLength * maxSideLength;
return
from a in Range(1, maxSideLength)
from b in Range(1, a)
let cc = a * a + b * b + a * b * coefficient
where cc <= maxSideLengthSquared
let c = (int)Math.Sqrt(cc)
where c * c == cc
select (a, b, c);
}
 
private static bool NotAllTheSameLength((int a, int b, int c) triple) => triple.a != triple.b || triple.a != triple.c;
}</syntaxhighlight>
{{out}}
<pre>
60 degree triangles in range 1..13
(1, 1, 1)
(2, 2, 2)
(3, 3, 3)
(4, 4, 4)
(5, 5, 5)
(6, 6, 6)
(7, 7, 7)
(8, 3, 7)
(8, 5, 7)
(8, 8, 8)
(9, 9, 9)
(10, 10, 10)
(11, 11, 11)
(12, 12, 12)
(13, 13, 13)
15 solutions
90 degree triangles in range 1..13
(4, 3, 5)
(8, 6, 10)
(12, 5, 13)
3 solutions
120 degree triangles in range 1..13
(5, 3, 7)
(8, 7, 13)
2 solutions
60 degree triangles in range 1..10000 where not all sides are the same
18394 solutions</pre>
 
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
 
 
<syntaxhighlight lang="Delphi">
 
procedure FindTriples(Memo: TMemo; Max,Angle,Coeff: integer);
var Count,A,B,C: integer;
var S: string;
begin
Memo.Lines.Add(Format('Gamma= %d°',[Angle]));
Count:=0;
S:='';
for A:=1 to Max do
for B:=1 to A do
for C:=1 to Max do
if A*A+B*B-Coeff*A*B=C*C then
begin
Inc(Count);
S:=S+Format('(%d,%d,%d) ',[A,B,C]);
if (Count mod 3)=0 then S:=S+CRLF;
end;
Memo.Lines.Add(Format('Number of triangles = %d',[Count]));
Memo.Lines.Add(S);
end;
 
 
 
procedure LawOfCosines(Memo: TMemo);
begin
FindTriples(Memo,13,90,0);
FindTriples(Memo,13,60,1);
FindTriples(Memo,13,120,-1);
end;
 
 
</syntaxhighlight>
{{out}}
<pre>
Gamma= 90°
Number of triangles = 3
(4,3,5) (8,6,10) (12,5,13)
 
Gamma= 60°
Number of triangles = 15
(1,1,1) (2,2,2) (3,3,3)
(4,4,4) (5,5,5) (6,6,6)
(7,7,7) (8,3,7) (8,5,7)
(8,8,8) (9,9,9) (10,10,10)
(11,11,11) (12,12,12) (13,13,13)
 
Gamma= 120°
Number of triangles = 2
(5,3,7) (8,7,13)
 
Elapsed Time: 9.768 ms.
 
</pre>
 
=={{header|Factor}}==
<langsyntaxhighlight lang="factor">USING: backtrack formatting kernel locals math math.ranges
sequences sets sorting ;
IN: rosetta-code.law-of-cosines
Line 240 ⟶ 728:
[ * + ] 120
[ 2drop 0 - ] 90
[ * - ] 60 [ show-solutions ] 2tri@</langsyntaxhighlight>
{{out}}
<pre>
Line 268 ⟶ 756:
}
</pre>
 
=={{header|Fortran}}==
<syntaxhighlight lang="fortran">MODULE LAW_OF_COSINES
IMPLICIT NONE
 
CONTAINS
 
! Calculate the third side of a triangle using the cosine rule
REAL FUNCTION COSINE_SIDE(SIDE_A, SIDE_B, ANGLE)
INTEGER, INTENT(IN) :: SIDE_A, SIDE_B
REAL(8), INTENT(IN) :: ANGLE
COSINE_SIDE = SIDE_A**2 + SIDE_B**2 - 2*SIDE_A*SIDE_B*COS(ANGLE)
COSINE_SIDE = COSINE_SIDE**0.5
END FUNCTION COSINE_SIDE
 
! Convert an angle in degrees to radians
REAL(8) FUNCTION DEG2RAD(ANGLE)
REAL(8), INTENT(IN) :: ANGLE
REAL(8), PARAMETER :: PI = 4.0D0*DATAN(1.D0)
 
DEG2RAD = ANGLE*(PI/180)
END FUNCTION DEG2RAD
 
! Sort an array of integers
FUNCTION INT_SORTED(ARRAY) RESULT(SORTED)
INTEGER, DIMENSION(:), INTENT(IN) :: ARRAY
INTEGER, DIMENSION(SIZE(ARRAY)) :: SORTED, TEMP
INTEGER :: MAX_VAL, DIVIDE
SORTED = ARRAY
TEMP = ARRAY
DIVIDE = SIZE(ARRAY)
 
DO WHILE (DIVIDE .NE. 1)
MAX_VAL = MAXVAL(SORTED(1:DIVIDE))
TEMP(DIVIDE) = MAX_VAL
TEMP(MAXLOC(SORTED(1:DIVIDE))) = SORTED(DIVIDE)
SORTED = TEMP
DIVIDE = DIVIDE - 1
END DO
END FUNCTION INT_SORTED
 
! Append an integer to the end of an array of integers
SUBROUTINE APPEND(ARRAY, ELEMENT)
INTEGER, DIMENSION(:), ALLOCATABLE, INTENT(INOUT) :: ARRAY
INTEGER, DIMENSION(:), ALLOCATABLE :: TEMP
INTEGER :: ELEMENT
INTEGER :: I, ISIZE
 
IF (ALLOCATED(ARRAY)) THEN
ISIZE = SIZE(ARRAY)
ALLOCATE(TEMP(ISIZE+1))
 
DO I=1, ISIZE
TEMP(I) = ARRAY(I)
END DO
 
TEMP(ISIZE+1) = ELEMENT
 
DEALLOCATE(ARRAY)
 
CALL MOVE_ALLOC(TEMP, ARRAY)
ELSE
ALLOCATE(ARRAY(1))
ARRAY(1) = ELEMENT
END IF
END SUBROUTINE APPEND
 
! Check if an array of integers contains a subset
LOGICAL FUNCTION CONTAINS_ARR(ARRAY, ELEMENT)
INTEGER, DIMENSION(:), INTENT(IN) :: ARRAY
INTEGER, DIMENSION(:) :: ELEMENT
INTEGER, DIMENSION(SIZE(ELEMENT)) :: TEMP, SORTED_ELEMENT
INTEGER :: I, COUNTER, J
 
COUNTER = 0
 
ELEMENT = INT_SORTED(ELEMENT)
 
DO I=1,SIZE(ARRAY),SIZE(ELEMENT)
TEMP = ARRAY(I:I+SIZE(ELEMENT)-1)
DO J=1,SIZE(ELEMENT)
IF (ELEMENT(J) .EQ. TEMP(J)) THEN
COUNTER = COUNTER + 1
END IF
END DO
 
IF (COUNTER .EQ. SIZE(ELEMENT)) THEN
CONTAINS_ARR = .TRUE.
RETURN
END IF
END DO
 
CONTAINS_ARR = .FALSE.
END FUNCTION CONTAINS_ARR
 
! Count and print cosine triples for the given angle in degrees
INTEGER FUNCTION COSINE_TRIPLES(MIN_NUM, MAX_NUM, ANGLE, PRINT_RESULTS) RESULT(COUNTER)
INTEGER, INTENT(IN) :: MIN_NUM, MAX_NUM
REAL(8), INTENT(IN) :: ANGLE
LOGICAL, INTENT(IN) :: PRINT_RESULTS
INTEGER, DIMENSION(:), ALLOCATABLE :: CANDIDATES
INTEGER, DIMENSION(3) :: CANDIDATE
INTEGER :: A, B
REAL :: C
 
COUNTER = 0
 
DO A = MIN_NUM, MAX_NUM
DO B = MIN_NUM, MAX_NUM
C = COSINE_SIDE(A, B, DEG2RAD(ANGLE))
IF (C .GT. MAX_NUM .OR. MOD(C, 1.) .NE. 0) THEN
CYCLE
END IF
 
CANDIDATE(1) = A
CANDIDATE(2) = B
CANDIDATE(3) = C
IF (.NOT. CONTAINS_ARR(CANDIDATES, CANDIDATE)) THEN
COUNTER = COUNTER + 1
CALL APPEND(CANDIDATES, CANDIDATE(1))
CALL APPEND(CANDIDATES, CANDIDATE(2))
CALL APPEND(CANDIDATES, CANDIDATE(3))
 
IF (PRINT_RESULTS) THEN
WRITE(*,'(A,I0,A,I0,A,I0,A)') " (", CANDIDATE(1), ", ", CANDIDATE(2), ", ", CANDIDATE(3), ")"
END IF
END IF
END DO
END DO
END FUNCTION COSINE_TRIPLES
END MODULE LAW_OF_COSINES
 
! Program prints the cosine triples for the angles 90, 60 and 120 degrees
! by using the cosine rule to find the third side of each candidate and
! checking that this is an integer. Candidates are appended to an array
! after the sides have been sorted into ascending order
! the array is repeatedly checked to ensure there are no duplicates.
PROGRAM LOC
USE LAW_OF_COSINES
 
REAL(8), DIMENSION(3) :: TEST_ANGLES = (/90., 60., 120./)
INTEGER :: I, COUNTER
 
DO I = 1,SIZE(TEST_ANGLES)
WRITE(*, '(F0.0, A)') TEST_ANGLES(I), " degree triangles: "
COUNTER = COSINE_TRIPLES(1, 13, TEST_ANGLES(I), .TRUE.)
WRITE(*,'(A, I0)') "TOTAL: ", COUNTER
WRITE(*,*) NEW_LINE('A')
END DO
 
END PROGRAM LOC</syntaxhighlight>
<pre>
90. degree triangles:
(3, 4, 5)
(5, 12, 13)
(6, 8, 10)
TOTAL: 3
 
 
60. degree triangles:
(1, 1, 1)
(2, 2, 2)
(3, 3, 3)
(3, 7, 8)
(4, 4, 4)
(5, 5, 5)
(6, 6, 6)
(7, 7, 7)
(3, 7, 8)
(8, 8, 8)
(9, 9, 9)
(10, 10, 10)
(11, 11, 11)
(12, 12, 12)
(13, 13, 13)
TOTAL: 15
 
 
120. degree triangles:
(3, 5, 7)
(7, 8, 13)
TOTAL: 2
</pre>
 
=={{header|FreeBASIC}}==
<langsyntaxhighlight lang="freebasic">' version 03-03-2019
' compile with: fbc -s console
 
Line 342 ⟶ 1,018:
Print : Print "hit any key to end program"
Sleep
End</langsyntaxhighlight>
{{out}}
<pre> 15: 60 degree triangles
Line 377 ⟶ 1,053:
 
=={{header|Go}}==
<langsyntaxhighlight lang="go">package main
 
import "fmt"
Line 450 ⟶ 1,126:
fmt.Print(" For an angle of 60 degrees")
fmt.Println(" there are", len(solutions), "solutions.")
}</langsyntaxhighlight>
 
{{out}}
Line 471 ⟶ 1,147:
 
=={{header|Haskell}}==
<langsyntaxhighlight lang="haskell">import qualified Data.Map.Strict as Map
import qualified Data.Set as Set
import Data.Monoid ((<>))
Line 519 ⟶ 1,195:
[120, 90, 60])
putStrLn "60 degrees - uneven triangles of maximum side 10000. Total:"
print $ length $ triangles f60ne 10000</langsyntaxhighlight>
{{Out}}
<pre>Triangles of maximum side 13
Line 555 ⟶ 1,231:
=={{header|J}}==
'''Solution:'''
<langsyntaxhighlight lang="j">load 'trig stats'
RHS=: *: NB. right-hand-side of Cosine Law
LHS=: +/@:*:@] - cos@rfd@[ * 2 * */@] NB. Left-hand-side of Cosine Law
Line 564 ⟶ 1,240:
idx=. (RHS oppside) i. x LHS"1 adjsides
adjsides ((#~ idx ~: #) ,. ({~ idx -. #)@]) oppside
)</langsyntaxhighlight>
'''Example:'''
<langsyntaxhighlight lang="j"> 60 90 120 solve&.> 13
+--------+-------+------+
| 1 1 1|3 4 5|3 5 7|
Line 585 ⟶ 1,261:
+--------+-------+------+
60 #@(solve -. _3 ]\ 3 # >:@i.@]) 10000 NB. optional extra credit
18394</langsyntaxhighlight>
 
=={{header|Java}}==
<syntaxhighlight lang="java">
public class LawOfCosines {
 
public static void main(String[] args) {
generateTriples(13);
generateTriples60(10000);
}
private static void generateTriples(int max) {
for ( int coeff : new int[] {0, -1, 1} ) {
int count = 0;
System.out.printf("Max side length %d, formula: a*a + b*b %s= c*c%n", max, coeff == 0 ? "" : (coeff<0 ? "-" : "+") + " a*b ");
for ( int a = 1 ; a <= max ; a++ ) {
for ( int b = 1 ; b <= a ; b++ ) {
int val = a*a + b*b + coeff*a*b;
int c = (int) (Math.sqrt(val) + .5d);
if ( c > max ) {
break;
}
if ( c*c == val ) {
System.out.printf(" (%d, %d, %d)%n", a, b ,c);
count++;
}
}
}
System.out.printf("%d triangles%n", count);
}
}
 
private static void generateTriples60(int max) {
int count = 0;
System.out.printf("%nExtra Credit.%nMax side length %d, sides different length, formula: a*a + b*b - a*b = c*c%n", max);
for ( int a = 1 ; a <= max ; a++ ) {
for ( int b = 1 ; b < a ; b++ ) {
int val = a*a + b*b - a*b;
int c = (int) (Math.sqrt(val) + .5d);
if ( c*c == val ) {
count++;
}
}
}
System.out.printf("%d triangles%n", count);
}
 
}
</syntaxhighlight>
{{out}}
<pre>
Max side length 13, formula: a*a + b*b = c*c
(4, 3, 5)
(8, 6, 10)
(12, 5, 13)
3 triangles
Max side length 13, formula: a*a + b*b - a*b = c*c
(1, 1, 1)
(2, 2, 2)
(3, 3, 3)
(4, 4, 4)
(5, 5, 5)
(6, 6, 6)
(7, 7, 7)
(8, 3, 7)
(8, 5, 7)
(8, 8, 8)
(9, 9, 9)
(10, 10, 10)
(11, 11, 11)
(12, 12, 12)
(13, 13, 13)
15 triangles
Max side length 13, formula: a*a + b*b + a*b = c*c
(5, 3, 7)
(8, 7, 13)
2 triangles
 
Extra Credit.
Max side length 10000, sides different length, formula: a*a + b*b - a*b = c*c
18394 triangles
</pre>
 
=={{header|JavaScript}}==
<langsyntaxhighlight JavaScriptlang="javascript">(() => {
'use strict';
 
Line 716 ⟶ 1,474:
// MAIN ---
return main();
})();</langsyntaxhighlight>
{{Out}}
<pre>Triangles of maximum side 13:
Line 749 ⟶ 1,507:
18394
[Finished in 3.444s]</pre>
 
=={{header|jq}}==
'''Adapted from [[#Wren|Wren]]'''
{{works with|jq}}
'''Works with gojq, the Go implementation of jq'''
 
To save space, we define `squares` as a hash rather than as a JSON array.
<syntaxhighlight lang="jq">def squares(n):
reduce range(1; 1+n) as $i ({}; .[$i*$i|tostring] = $i);
 
# if count, then just count
def solve(angle; maxLen; allowSame; count):
squares(maxLen) as $squares
 
| def qsqrt($n):
$squares[$n|tostring] as $sqrt
| if $sqrt then $sqrt else null end;
 
reduce range(1; maxLen+1) as $a ({};
reduce range($a; maxLen+1) as $b (.;
.lhs = $a*$a + $b*$b
| if angle != 90
then if angle == 60
then .lhs += ( - $a*$b)
elif angle == 120
then .lhs += $a*$b
else "Angle must be 60, 90 or 120 degrees" | error
end
else .
end
| qsqrt(.lhs) as $c
| if $c != null
then if allowSame or $a != $b or $b != $c
then .solutions += if count then 1 else [[$a, $b, $c]] end
else .
end
else .
end
)
)
| .solutions ;
 
def task1($angles):
"For sides in the range [1, 13] where they can all be of the same length:\n",
($angles[]
| . as $angle
| solve($angle; 13; true; false)
| " For an angle of \($angle) degrees, there are \(length) solutions, namely:", .);
 
def task2(degrees; n):
"For sides in the range [1, \(n)] where they cannot ALL be of the same length:",
(solve(degrees; n; false; true)
| " For an angle of \(degrees) degrees, there are \(.) solutions.") ;
 
task1([90, 60, 120]), "", task2(60; 10000)</syntaxhighlight>
{{out}}
<pre>
For sides in the range [1, 13] where they can all be of the same length:
 
For an angle of 90 degrees, there are 3 solutions, namely:
[[3,4,5],[5,12,13],[6,8,10]]
For an angle of 60 degrees, there are 15 solutions, namely:
[[1,1,1],[2,2,2],[3,3,3],[3,8,7],[4,4,4],[5,5,5],[5,8,7],[6,6,6],[7,7,7],[8,8,8],[9,9,9],[10,10,10],[11,11,11],[12,12,12],[13,13,13]]
For an angle of 120 degrees, there are 2 solutions, namely:
[[3,5,7],[7,8,13]]
 
For sides in the range [1, 10000] where they cannot ALL be of the same length:
For an angle of 60 degrees, there are 18394 solutions.
</pre>
 
 
=={{header|Julia}}==
{{trans|zkl}}
<langsyntaxhighlight lang="julia">sqdict(n) = Dict([(x*x, x) for x in 1:n])
numnotsame(arrarr) = sum(map(x -> !all(y -> y == x[1], x), arrarr))
 
Line 779 ⟶ 1,607:
println("Angle 120:"); for t in tri120 println(t) end
println("\nFor sizes N through 10000, there are $(numnotsame(filtertriangles(10000)[1])) 60 degree triples with nonequal sides.")
</langsyntaxhighlight> {{output}} <pre>
 
Integer triples for 1 <= side length <= 13:
Line 812 ⟶ 1,640:
=={{header|Kotlin}}==
{{trans|Go}}
<langsyntaxhighlight lang="scala">// Version 1.2.70
 
val squares13 = mutableMapOf<Int, Int>()
Line 879 ⟶ 1,707:
print(" For an angle of 60 degrees")
println(" there are ${solutions.size} solutions.")
}</langsyntaxhighlight>
 
{{output}}
Line 897 ⟶ 1,725:
 
For an angle of 60 degrees there are 18394 solutions.
</pre>
 
=={{header|Lua}}==
<syntaxhighlight lang="lua">function solve(angle, maxlen, filter)
local squares, roots, solutions = {}, {}, {}
local cos2 = ({[60]=-1,[90]=0,[120]=1})[angle]
for i = 1, maxlen do squares[i], roots[i^2] = i^2, i end
for a = 1, maxlen do
for b = a, maxlen do
local lhs = squares[a] + squares[b] + cos2*a*b
local c = roots[lhs]
if c and (not filter or filter(a,b,c)) then
solutions[#solutions+1] = {a=a,b=b,c=c}
end
end
end
print(angle.."° on 1.."..maxlen.." has "..#solutions.." solutions")
if not filter then
for i,v in ipairs(solutions) do print("",v.a,v.b,v.c) end
end
end
solve(90, 13)
solve(60, 13)
solve(120, 13)
function fexcr(a,b,c) return a~=b or b~=c end
solve(60, 10000, fexcr) -- extra credit
solve(90, 10000, fexcr) -- more extra credit
solve(120, 10000, fexcr) -- even more extra credit</syntaxhighlight>
{{out}}
<pre>90° on 1..13 has 3 solutions
3 4 5
5 12 13
6 8 10
60° on 1..13 has 15 solutions
1 1 1
2 2 2
3 3 3
3 8 7
4 4 4
5 5 5
5 8 7
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
11 11 11
12 12 12
13 13 13
120° on 1..13 has 2 solutions
3 5 7
7 8 13
60° on 1..10000 has 18394 solutions
90° on 1..10000 has 12471 solutions
120° on 1..10000 has 10374 solutions</pre>
 
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<syntaxhighlight lang="mathematica">Solve[{a^2+b^2==c^2,1<=a<=13,1<=b<=13,1<=c<=13,a<=b},{a,b,c},Integers]
Length[%]
Solve[{a^2+b^2-a b==c^2,1<=a<=13,1<=b<=13,1<=c<=13,a<=b},{a,b,c},Integers]
Length[%]
Solve[{a^2+b^2+a b==c^2,1<=a<=13,1<=b<=13,1<=c<=13,a<=b},{a,b,c},Integers]
Length[%]
Solve[{a^2+b^2-a b==c^2,1<=a<=10000,1<=b<=10000,1<=c<=10000,a<=b,a!=b,b!=c,a!=c},{a,b,c},Integers]//Length</syntaxhighlight>
{{out}}
<pre>{{a->3,b->4,c->5},{a->5,b->12,c->13},{a->6,b->8,c->10}}
3
{{a->1,b->1,c->1},{a->2,b->2,c->2},{a->3,b->3,c->3},{a->3,b->8,c->7},{a->4,b->4,c->4},{a->5,b->5,c->5},{a->5,b->8,c->7},{a->6,b->6,c->6},{a->7,b->7,c->7},{a->8,b->8,c->8},{a->9,b->9,c->9},{a->10,b->10,c->10},{a->11,b->11,c->11},{a->12,b->12,c->12},{a->13,b->13,c->13}}
15
{{a->3,b->5,c->7},{a->7,b->8,c->13}}
2
18394</pre>
 
=={{header|Nim}}==
<syntaxhighlight lang="nim">import strformat
import tables
 
# Generate tables at compile time. This eliminates the initialization at
# the expense of a bigger executable.
 
const square13 = block:
var tmp = newSeq[tuple[a, b: int]]()
for i in 1..13:
tmp.add((i * i, i))
toTable(tmp)
const square10000 = block:
var tmp = newSeq[tuple[a, b: int]]()
for i in 1..10000:
tmp.add((i * i, i))
toTable(tmp)
 
proc solve(angle, maxLen: int, allowSame: bool): seq[tuple[a, b, c: int]] =
result = newSeq[tuple[a, b, c: int]]()
for a in 1..maxLen:
for b in a..maxLen:
var lhs = a * a + b * b
if angle != 90:
case angle
of 60:
dec lhs, a * b
of 120:
inc lhs, a * b
else:
raise newException(IOError, "Angle must be 60, 90 or 120 degrees")
case maxLen
of 13:
var c = square13.getOrDefault(lhs)
if (not allowSame and a == b and b == c) or c == 0:
continue
result.add((a, b, c))
of 10000:
var c = square10000.getOrDefault(lhs)
if (not allowSame and a == b and b == c) or c == 0:
continue
result.add((a, b, c))
else:
raise newException(IOError, "Maximum length must be either 13 or 10000")
 
echo "For sides in the range [1, 13] where they can all be the same length:\n"
let angles = [90, 60, 120]
for angle in angles:
var solutions = solve(angle, 13, true)
echo fmt" For an angle of {angle} degrees there are {len(solutions)} solutions, to wit:"
write(stdout, " ")
for i in 0..<len(solutions):
write(stdout, fmt"{solutions[i]:25}")
if i mod 3 == 2:
write(stdout, "\n ")
write(stdout, "\n")
echo "\nFor sides in the range [1, 10000] where they cannot ALL be of the same length:\n"
var solutions = solve(60, 10000, false)
echo fmt" For an angle of 60 degrees there are {len(solutions)} solutions."</syntaxhighlight>
 
{{out}}
<pre>
For sides in the range [1, 13] where they can all be the same length:
 
For an angle of 90 degrees there are 3 solutions, to wit:
(a: 3, b: 4, c: 5) (a: 5, b: 12, c: 13) (a: 6, b: 8, c: 10)
For an angle of 60 degrees there are 15 solutions, to wit:
(a: 1, b: 1, c: 1) (a: 2, b: 2, c: 2) (a: 3, b: 3, c: 3)
(a: 3, b: 8, c: 7) (a: 4, b: 4, c: 4) (a: 5, b: 5, c: 5)
(a: 5, b: 8, c: 7) (a: 6, b: 6, c: 6) (a: 7, b: 7, c: 7)
(a: 8, b: 8, c: 8) (a: 9, b: 9, c: 9) (a: 10, b: 10, c: 10)
(a: 11, b: 11, c: 11) (a: 12, b: 12, c: 12) (a: 13, b: 13, c: 13)
For an angle of 120 degrees there are 2 solutions, to wit:
(a: 3, b: 5, c: 7) (a: 7, b: 8, c: 13)
 
For sides in the range [1, 10000] where they cannot ALL be of the same length:
 
For an angle of 60 degrees there are 18394 solutions.
</pre>
 
=={{header|Perl}}==
{{trans|Perl 6Raku}}
<langsyntaxhighlight lang="perl">use utf8;
binmode STDOUT, "utf8:";
use Sort::Naturally;
Line 934 ⟶ 1,916:
}
 
printf "Non-equilateral n=10000/60°: %d\n", scalar triples(10000,60);</langsyntaxhighlight>
{{out}}
<pre>Integer triangular triples for sides 1..13:
Line 942 ⟶ 1,924:
Non-equilateral n=10000/60°: 18394</pre>
 
=={{header|Perl 6Phix}}==
Using a simple flat sequence of 100 million elements (well within the desktop language limits, but beyond JavaScript) proved significantly faster than a dictionary (5x or so).
In each routine, <tt>race</tt> is used to allow concurrent operations, requiring the use of the atomic increment operator, <tt>⚛++</tt>, to safely update <tt>@triples</tt>, which must be declared fixed-sized, as an auto-resizing array is not thread-safe. At exit, default values in <tt>@triples</tt> are filtered out with the test <code>!eqv Any</code>.
<!--<syntaxhighlight lang="phix">(phixonline)-->
<lang perl6>multi triples (60, $n) {
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
my %sq = (1..$n).map: { .² => $_ };
<span style="color: #000080;font-style:italic;">--constant lim = iff(platform()=JS?13:10000)</span>
my atomicint $i = 0;
<span style="color: #008080;">constant</span> <span style="color: #000000;">lim</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">10000</span>
my @triples[2*$n];
<span style="color: #000080;font-style:italic;">--sequence squares = repeat(0,lim*lim)</span>
(1..^$n).race(:8degree).map: -> $a {
<span style="color: #004080;">sequence</span> <span style="color: #000000;">squares</span> <span style="color: #0000FF;">=</span> <span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">platform</span><span style="color: #0000FF;">()=</span><span style="color: #004600;">JS</span><span style="color: #0000FF;">?{}:</span><span style="color: #7060A8;">repeat</span><span style="color: #0000FF;">(</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">lim</span><span style="color: #0000FF;">*</span><span style="color: #000000;">lim</span><span style="color: #0000FF;">))</span>
for $a^..$n -> $b {
<span style="color: #008080;">for</span> <span style="color: #000000;">c</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">lim</span> <span style="color: #008080;">do</span>
my $cos = $a * $a + $b * $b - $a * $b;
<span style="color: #000000;">squares</span><span style="color: #0000FF;">[</span><span style="color: #000000;">c</span><span style="color: #0000FF;">*</span><span style="color: #000000;">c</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">c</span>
@triples[$i⚛++] = $a, %sq{$cos}, $b if %sq{$cos}:exists;
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
}
}
<span style="color: #008080;">function</span> <span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">angle</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">maxlen</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">bool</span> <span style="color: #000000;">samelen</span><span style="color: #0000FF;">=</span><span style="color: #004600;">true</span><span style="color: #0000FF;">)</span>
@triples.grep: so *;
<span style="color: #004080;">sequence</span> <span style="color: #000000;">res</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{}</span>
}
<span style="color: #008080;">for</span> <span style="color: #000000;">a</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">maxlen</span> <span style="color: #008080;">do</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">a2</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">a</span><span style="color: #0000FF;">*</span><span style="color: #000000;">a</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">b</span><span style="color: #0000FF;">=</span><span style="color: #000000;">a</span> <span style="color: #008080;">to</span> <span style="color: #000000;">maxlen</span> <span style="color: #008080;">do</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">c2</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">a2</span><span style="color: #0000FF;">+</span><span style="color: #000000;">b</span><span style="color: #0000FF;">*</span><span style="color: #000000;">b</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">angle</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">90</span> <span style="color: #008080;">then</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">angle</span><span style="color: #0000FF;">=</span><span style="color: #000000;">60</span> <span style="color: #008080;">then</span> <span style="color: #000000;">c2</span> <span style="color: #0000FF;">-=</span> <span style="color: #000000;">a</span><span style="color: #0000FF;">*</span><span style="color: #000000;">b</span>
<span style="color: #008080;">elsif</span> <span style="color: #000000;">angle</span><span style="color: #0000FF;">=</span><span style="color: #000000;">120</span> <span style="color: #008080;">then</span> <span style="color: #000000;">c2</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">a</span><span style="color: #0000FF;">*</span><span style="color: #000000;">b</span>
<span style="color: #008080;">else</span> <span style="color: #7060A8;">crash</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"angle must be 60/90/120"</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">c</span> <span style="color: #0000FF;">=</span> <span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">c2</span><span style="color: #0000FF;">></span><span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">squares</span><span style="color: #0000FF;">)?</span><span style="color: #000000;">0</span><span style="color: #0000FF;">:</span><span style="color: #000000;">squares</span><span style="color: #0000FF;">[</span><span style="color: #000000;">c2</span><span style="color: #0000FF;">])</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">c</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">0</span> <span style="color: #008080;">and</span> <span style="color: #000000;">c</span><span style="color: #0000FF;"><=</span><span style="color: #000000;">maxlen</span> <span style="color: #008080;">then</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">samelen</span> <span style="color: #008080;">or</span> <span style="color: #000000;">a</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">b</span> <span style="color: #008080;">or</span> <span style="color: #000000;">b</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">c</span> <span style="color: #008080;">then</span>
<span style="color: #000000;">res</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">append</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">a</span><span style="color: #0000FF;">,</span><span style="color: #000000;">b</span><span style="color: #0000FF;">,</span><span style="color: #000000;">c</span><span style="color: #0000FF;">})</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">res</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">procedure</span> <span style="color: #000000;">show</span><span style="color: #0000FF;">(</span><span style="color: #004080;">string</span> <span style="color: #000000;">fmt</span><span style="color: #0000FF;">,</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">res</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">bool</span> <span style="color: #000000;">full</span><span style="color: #0000FF;">=</span><span style="color: #004600;">true</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">fmt</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">),</span><span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">full</span><span style="color: #0000FF;">?</span><span style="color: #7060A8;">sprint</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">):</span><span style="color: #008000;">""</span><span style="color: #0000FF;">)})</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">procedure</span>
<span style="color: #7060A8;">puts</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"Integer triangular triples for sides 1..13:\n"</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">show</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"Angle 60 has %2d solutions: %s\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span> <span style="color: #000000;">60</span><span style="color: #0000FF;">,</span><span style="color: #000000;">13</span><span style="color: #0000FF;">))</span>
<span style="color: #000000;">show</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"Angle 90 has %2d solutions: %s\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span> <span style="color: #000000;">90</span><span style="color: #0000FF;">,</span><span style="color: #000000;">13</span><span style="color: #0000FF;">))</span>
<span style="color: #000000;">show</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"Angle 120 has %2d solutions: %s\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span><span style="color: #000000;">120</span><span style="color: #0000FF;">,</span><span style="color: #000000;">13</span><span style="color: #0000FF;">))</span>
<span style="color: #000080;font-style:italic;">--if platform()!=JS then</span>
<span style="color: #000000;">show</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"Non-equilateral angle 60 triangles for sides 1..10000: %d%s\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span><span style="color: #000000;">60</span><span style="color: #0000FF;">,</span><span style="color: #000000;">10000</span><span style="color: #0000FF;">,</span><span style="color: #004600;">false</span><span style="color: #0000FF;">),</span><span style="color: #004600;">false</span><span style="color: #0000FF;">)</span>
<span style="color: #000080;font-style:italic;">--end if</span>
<!--</syntaxhighlight>-->
{{out}}
<pre style="font-size: 11px">
Integer triangular triples for sides 1..13:
Angle 60 has 15 solutions: {{1,1,1},{2,2,2},{3,3,3},{3,8,7},{4,4,4},{5,5,5},{5,8,7},{6,6,6},{7,7,7},{8,8,8},{9,9,9},{10,10,10},{11,11,11},{12,12,12},{13,13,13}}
Angle 90 has 3 solutions: {{3,4,5},{5,12,13},{6,8,10}}
Angle 120 has 2 solutions: {{3,5,7},{7,8,13}}
Non-equilateral angle 60 triangles for sides 1..10000: 18394
</pre>
As noted I had to resort to a little trickery to get that last line to run in JavaScript, should a future version impose stricter bounds checking that will stop working.
 
=={{header|Prolog}}==
multi triples (90, $n) {
{{works with|SWI Prolog}}
my %sq = (1..$n).map: { .² => $_ };
<syntaxhighlight lang="prolog">find_solutions(Limit, Solutions):-
my atomicint $i = 0;
find_solutions(Limit, Solutions, Limit, []).
my @triples[2*$n];
(1..^$n).race(:8degree).map: -> $a {
for $a^..$n -> $b {
my $cos = $a * $a + $b * $b;
@triples[$i⚛++] = $a, $b, %sq{$cos} and last if %sq{$cos}:exists;
}
}
@triples.grep: so *;
}
 
find_solutions(_, S, 0, S):-
multi triples (120, $n) {
!.
my %sq = (1..$n).map: { .² => $_ };
find_solutions(Limit, Solutions, A, S):-
my atomicint $i = 0;
find_solutions1(Limit, A, A, S1, S),
my @triples[2*$n];
A_next is A - 1,
(1..^$n).race(:8degree).map: -> $a {
find_solutions(Limit, Solutions, A_next, S1).
for $a^..$n -> $b {
my $cos = $a * $a + $b * $b + $a * $b;
@triples[$i⚛++] = $a, $b, %sq{$cos} and last if %sq{$cos}:exists;
}
}
@triples.grep: so *;
}
 
find_solutions1(Limit, _, B, Triples, Triples):-
use Sort::Naturally;
B > Limit,
!.
find_solutions1(Limit, A, B, [Triple|Triples], T):-
is_solution(Limit, A, B, Triple),
!,
B_next is B + 1,
find_solutions1(Limit, A, B_next, Triples, T).
find_solutions1(Limit, A, B, Triples, T):-
B_next is B + 1,
find_solutions1(Limit, A, B_next, Triples, T).
 
is_solution(Limit, A, B, t(Angle, A, B, C)):-
my $n = 13;
X is A * A + B * B,
say "Integer triangular triples for sides 1..$n:";
Y is A * B,
for 120, 90, 60 -> $angle {
(
my @itt = triples($angle, $n);
if $angle == 60 {Angle push= @itt90, "$_C is round(sqrt(X)), $_X $_"is forC 1..$n* }C
;
printf "Angle %3d° has %2d solutions: %s\n", $angle, +@itt, @itt.sort(*.&naturally).join(', ');
Angle = 120, C2 is X + Y, C is round(sqrt(C2)), C2 is C * C
}
;
Angle = 60, C2 is X - Y, C is round(sqrt(C2)), C2 is C * C
),
C =< Limit,
!.
 
write_triples(Angle, Solutions):-
my ($angle, $count) = 60, 10_000;
find_triples(Angle, Solutions, List, 0, Count),
say "\nExtra credit:";
writef('There are %w solutions for gamma = %w:\n', [Count, Angle]),
say "$angle° integer triples in the range 1..$count where the sides are not all the same length: ", +triples($angle, $count);</lang>
write_triples1(List),
{{out}}
nl.
<pre>Integer triangular triples for sides 1..13:
Angle 120° has 2 solutions: 3 5 7, 7 8 13
Angle 90° has 3 solutions: 3 4 5, 5 12 13, 6 8 10
Angle 60° has 15 solutions: 1 1 1, 2 2 2, 3 3 3, 3 7 8, 4 4 4, 5 5 5, 5 7 8, 6 6 6, 7 7 7, 8 8 8, 9 9 9, 10 10 10, 11 11 11, 12 12 12, 13 13 13
 
find_triples(_, [], [], Count, Count):-
Extra credit:
!.
60° integer triples in the range 1..10000 where the sides are not all the same length: 18394</pre>
find_triples(Angle, [Triple|Triples], [Triple|Result], C, Count):-
Triple = t(Angle, _, _, _),
!,
C1 is C + 1,
find_triples(Angle, Triples, Result, C1, Count).
find_triples(Angle, [_|Triples], Result, C, Count):-
find_triples(Angle, Triples, Result, C, Count).
 
write_triples1([]):-!.
=={{header|Phix}}==
write_triples1([t(_, A, B, C)]):-
Using a simple flat sequence of 100 million elements (well within the language limits) proved significantly faster than a dictionary (5x or so).
writef('(%w,%w,%w)\n', [A, B, C]),
<lang Phix>sequence squares = repeat(0,10000*10000)
for c=1 to 10000 do!.
write_triples1([t(_, A, B, C)|Triples]):-
squares[c*c] = c
writef('(%w,%w,%w) ', [A, B, C]),
end for
write_triples1(Triples).
 
main:-
function solve(integer angle, maxlen, bool samelen=true)
find_solutions(13, Solutions),
sequence res = {}
write_triples(60, Solutions),
for a=1 to maxlen do
write_triples(90, Solutions),
integer a2 = a*a
write_triples(120, Solutions).</syntaxhighlight>
for b=a to maxlen do
integer c2 = a2+b*b
if angle!=90 then
if angle=60 then c2 -= a*b
elsif angle=120 then c2 += a*b
else crash("angle must be 60/90/120")
end if
end if
integer c = iff(c2>length(squares)?0:squares[c2])
if c!=0 and c<=maxlen then
if samelen or a!=b or b!=c then
res = append(res,{a,b,c})
end if
end if
end for
end for
return res
end function
 
{{out}}
procedure show(string fmt,sequence res, bool full=true)
<pre>
printf(1,fmt,{length(res),iff(full?sprint(res):"")})
There are 15 solutions for gamma = 60:
end procedure
(1,1,1) (2,2,2) (3,3,3) (3,8,7) (4,4,4) (5,5,5) (5,8,7) (6,6,6) (7,7,7) (8,8,8) (9,9,9) (10,10,10) (11,11,11) (12,12,12) (13,13,13)
 
There are 3 solutions for gamma = 90:
puts(1,"Integer triangular triples for sides 1..13:\n")
(3,4,5) (5,12,13) (6,8,10)
show("Angle 60 has %2d solutions: %s\n",solve( 60,13))
 
show("Angle 90 has %2d solutions: %s\n",solve( 90,13))
show("AngleThere 120are has %2d2 solutions: %s\n",solve(for gamma = 120,13)):
(3,5,7) (7,8,13)
show("Non-equilateral angle 60 triangles for sides 1..10000: %d%s\n",solve(60,10000,false),false)</lang>
{{out}}
<pre style="font-size: 11px">
Integer triangular triples for sides 1..13:
Angle 60 has 15 solutions: {{1,1,1},{2,2,2},{3,3,3},{3,8,7},{4,4,4},{5,5,5},{5,8,7},{6,6,6},{7,7,7},{8,8,8},{9,9,9},{10,10,10},{11,11,11},{12,12,12},{13,13,13}}
Angle 90 has 3 solutions: {{3,4,5},{5,12,13},{6,8,10}}
Angle 120 has 2 solutions: {{3,5,7},{7,8,13}}
Non-equilateral angle 60 triangles for sides 1..10000: 18394
</pre>
 
=={{header|Python}}==
===Sets===
<langsyntaxhighlight lang="python">N = 13
 
def method1(N=N):
Line 1,083 ⟶ 2,090:
_, t60, _ = method1(10_000)
notsame = sum(1 for a, b, c in t60 if a != b or b != c)
print('Extra credit:', notsame)</langsyntaxhighlight>
 
{{out}}
Line 1,098 ⟶ 2,105:
A variant Python draft based on dictionaries.
(Test functions are passed as parameters to the main function.)
<langsyntaxhighlight lang="python">from itertools import (starmap)
 
 
Line 1,178 ⟶ 2,185:
 
if __name__ == '__main__':
main()</langsyntaxhighlight>
{{Out}}
<pre>Triangles of maximum side 13
Line 1,211 ⟶ 2,218:
60 degrees - uneven triangles of maximum side 10000. Total:
18394</pre>
 
=={{header|Quackery}}==
 
<syntaxhighlight lang="Quackery"> [ dup 1
[ 2dup > while
+ 1 >>
2dup / again ]
drop nip ] is sqrt ( n --> n )
 
[ dup * ] is squared ( n --> n )
 
[ dup sqrt squared = ] is square ( n --> b )
[ say "90 degrees" cr
0 temp put
13 times
[ i^ 1+ dup times
[ i^ 1+ squared
over squared +
dup square
over sqrt 14 < and
iff
[ i^ 1+ echo sp
over echo sp
sqrt echo cr
1 temp tally ]
else drop ]
drop ]
temp take echo
say " solutions" cr cr ] is 90deg ( --> )
 
[ say "60 degrees" cr
0 temp put
13 times
[ i^ 1+ dup times
[ i^ 1+
2dup * dip
[ squared
over squared + ]
- dup square
over sqrt 14 < and
iff
[ i^ 1+ echo sp
over echo sp
sqrt echo cr
1 temp tally ]
else drop ]
drop ]
temp take echo
say " solutions" cr cr ] is 60deg ( --> )
[ say "120 degrees" cr
0 temp put
13 times
[ i^ 1+ dup times
[ i^ 1+
2dup * dip
[ squared
over squared + ]
+ dup square
over sqrt 14 < and
iff
[ i^ 1+ echo sp
over echo sp
sqrt echo cr
1 temp tally ]
else drop ]
drop ]
temp take echo
say " solutions" cr cr ] is 120deg ( --> )
 
90deg 60deg 120deg</syntaxhighlight>
 
{{out}}
 
<pre>90 degrees
3 4 5
6 8 10
5 12 13
3 solutions found
 
60 degrees
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
3 8 7
5 8 7
8 8 8
9 9 9
10 10 10
11 11 11
12 12 12
13 13 13
15 solutions
 
120 degrees
3 5 7
7 8 13
2 solutions
</pre>
 
=={{header|R}}==
This looks a bit messy, but it really pays off when you see how nicely the output prints.
<syntaxhighlight lang="rsplus">inputs <- cbind(combn(1:13, 2), rbind(seq_len(13), seq_len(13)))
inputs <- cbind(A = inputs[1, ], B = inputs[2, ])[sort.list(inputs[1, ]),]
Pythagoras <- inputs[, "A"]^2 + inputs[, "B"]^2
AtimesB <- inputs[, "A"] * inputs[, "B"]
CValues <- sqrt(cbind("C (90º)" = Pythagoras,
"C (60º)" = Pythagoras - AtimesB,
"C (120º)" = Pythagoras + AtimesB))
CValues[!t(apply(CValues, MARGIN = 1, function(x) x %in% 1:13))] <- NA
output <- cbind(inputs, CValues)[!apply(CValues, MARGIN = 1, function(x) all(is.na(x))),]
rownames(output) <- paste0("Solution ", seq_len(nrow(output)), ":")
print(output, na.print = "")
cat("There are",
sum(!is.na(output[, 3])), "solutions in the 90º case,",
sum(!is.na(output[, 4])), "solutions in the 60º case, and",
sum(!is.na(output[, 5])), "solutions in the 120º case.")</syntaxhighlight>
{{out}}
<pre> A B C (90º) C (60º) C (120º)
Solution 1: 1 1 1
Solution 2: 2 2 2
Solution 3: 3 4 5
Solution 4: 3 5 7
Solution 5: 3 8 7
Solution 6: 3 3 3
Solution 7: 4 4 4
Solution 8: 5 8 7
Solution 9: 5 12 13
Solution 10: 5 5 5
Solution 11: 6 8 10
Solution 12: 6 6 6
Solution 13: 7 8 13
Solution 14: 7 7 7
Solution 15: 8 8 8
Solution 16: 9 9 9
Solution 17: 10 10 10
Solution 18: 11 11 11
Solution 19: 12 12 12
Solution 20: 13 13 13
There are 3 solutions in the 90º case, 15 solutions in the 60º case, and 2 solutions in the 120º case.</pre>
 
=={{header|Raku}}==
(formerly Perl 6)
In each routine, <tt>race</tt> is used to allow concurrent operations, requiring the use of the atomic increment operator, <tt>⚛++</tt>, to safely update <tt>@triples</tt>, which must be declared fixed-sized, as an auto-resizing array is not thread-safe. At exit, default values in <tt>@triples</tt> are filtered out with the test <code>!eqv Any</code>.
<syntaxhighlight lang="raku" line>multi triples (60, $n) {
my %sq = (1..$n).map: { .² => $_ };
my atomicint $i = 0;
my @triples[2*$n];
(1..^$n).race(:8degree).map: -> $a {
for $a^..$n -> $b {
my $cos = $a * $a + $b * $b - $a * $b;
@triples[$i⚛++] = $a, %sq{$cos}, $b if %sq{$cos}:exists;
}
}
@triples.grep: so *;
}
 
multi triples (90, $n) {
my %sq = (1..$n).map: { .² => $_ };
my atomicint $i = 0;
my @triples[2*$n];
(1..^$n).race(:8degree).map: -> $a {
for $a^..$n -> $b {
my $cos = $a * $a + $b * $b;
@triples[$i⚛++] = $a, $b, %sq{$cos} and last if %sq{$cos}:exists;
}
}
@triples.grep: so *;
}
 
multi triples (120, $n) {
my %sq = (1..$n).map: { .² => $_ };
my atomicint $i = 0;
my @triples[2*$n];
(1..^$n).race(:8degree).map: -> $a {
for $a^..$n -> $b {
my $cos = $a * $a + $b * $b + $a * $b;
@triples[$i⚛++] = $a, $b, %sq{$cos} and last if %sq{$cos}:exists;
}
}
@triples.grep: so *;
}
 
use Sort::Naturally;
 
my $n = 13;
say "Integer triangular triples for sides 1..$n:";
for 120, 90, 60 -> $angle {
my @itt = triples($angle, $n);
if $angle == 60 { push @itt, "$_ $_ $_" for 1..$n }
printf "Angle %3d° has %2d solutions: %s\n", $angle, +@itt, @itt.sort(&naturally).join(', ');
}
 
my ($angle, $count) = 60, 10_000;
say "\nExtra credit:";
say "$angle° integer triples in the range 1..$count where the sides are not all the same length: ", +triples($angle, $count);</syntaxhighlight>
{{out}}
<pre>Integer triangular triples for sides 1..13:
Angle 120° has 2 solutions: 3 5 7, 7 8 13
Angle 90° has 3 solutions: 3 4 5, 5 12 13, 6 8 10
Angle 60° has 15 solutions: 1 1 1, 2 2 2, 3 3 3, 3 7 8, 4 4 4, 5 5 5, 5 7 8, 6 6 6, 7 7 7, 8 8 8, 9 9 9, 10 10 10, 11 11 11, 12 12 12, 13 13 13
 
Extra credit:
60° integer triples in the range 1..10000 where the sides are not all the same length: 18394</pre>
 
=={{header|REXX}}==
===using(Using some optimization=== and memoization.)
 
Instead of coding a general purpose subroutine (or function) to solve all of the
task's requirements, &nbsp; it was decided to
Line 1,231 ⟶ 2,448:
displayed &nbsp; (using the
<br>absolute value of the negative number).
<langsyntaxhighlight lang="rexx">/*REXX pgm finds integer sided triangles that satisfy Law of cosines for 60º, 90º, 120º.*/
parse arg s1os1 s2os2 s3os3 os4 . /*obtain optional arguments from the CL*/
if s1os1=='' | s1os1=="," then s1os1= 13; s1=abs(os1) /*Not specified? Then use the default.*/
if s2os2=='' | s2os2=="," then s2os2= 13; s2=abs(os2) /* " " " " " " */
if s3os3=='' | s3os3=="," then s3os3= 13; s3=abs(os3) /* " " " " " " */
if os4=='' | os4=="," then os4= -0; s4=abs(os4) /* " " " " " " */
w= max( length(s1), length(s2), length(s3) ) /*W is used to align the side lengths.*/
@.= /*@: array holds squares, max of sides*/
 
if s1>0 then do; call head 120 do j=1 for max(s1, s2, s3, s4); @.j= j /*────120º: j a² + b² + ab ≡ c²/*use memoization.*/
end do a=1 for s1; aa = a/*j*a/
if s1>0 then call s1 /*handle the triangle docase for b=a+1 to s1; x= aa + b*b +120º. a*b/
if s2>0 then call s2 /*handle the triangle case for do c=b+1 to90º. s1 until c*c>x/
if s3>0 then call s3 if x==c/*chandle the thentriangle do;case for call show; iterate60º. b; end*/
if s4>0 then call s4 /*handle the case for unique sides. end /*c*/
exit 0 end /*bstick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
end /*a*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
call foot s1
dAng: w= length(s); ang= ' 'd"º " uq' '; ss= s * s; @sol= " solutions found for"; return
end
foot: say right(commas(#) @sol ang "(sides up to" commas(arg(1) +0)')', 65); say; return
 
head: #= 0; parse arg d,uq,s; @= ','; call dAng; say center(ang, 65, '═'); return
if s2>0 then do; call head 90 /*────90º: a² + b² ≡ c² */
show: #=#+1; arg p; if p>0 then say ' ('right(a,w)@ right(b,w)@ right(c,w)")"; return
do a=1 for s2; aa = a*a
/*──────────────────────────────────────────────────────────────────────────────────────*/
do b=a+1 to s2; x= aa + b*b
s1: call head 120,,s1 /*────────── 120º: a² + b² do c=b+1 ab to s2≡ c² until c*c>x/
do a=1 for s1; if xap1==c*c a then do; call show; iterate b; + end1
do b=ap1 for s1-ap1+1; x= @.a + @.b + a*b; if x>ss then enditerate /*c*/a
do c=b+1 for s1-b+1 until end /*b*/@.c>x
if x==@.c then do; call show os1; iterate b; end /*a*/
end call foot s2/*c*/
end /*b*/
end /*a*/
 
if s3>0 then do; call head 60 foot s1; /*────60º: a² + b² ─ ab ≡ c² */return
/*──────────────────────────────────────────────────────────────────────────────────────*/
do a=1 for s3; aa = a*a
s2: call head 90,,s2 /*────────── 90º: do a² + b=a² to s3; x=≡ aac² + b*b - a*b/
do a=1 for s2; ap1= do c=a + to s3 until c*c>x1
do b=ap1 for s2-ap1+1; x= @.a + @.b; if x==c*c>ss then do; call show; iterate b; enda
do c=b+1 for s2-b+2 until end /*@.c*/>x
if x==@.c then do; call show os2; iterate b; end /*b*/
end /*ac*/
end call foot s3/*b*/
end /*a*/
exit call foot s2; /*stick a fork in it, we're all done. */return
/*──────────────────────────────────────────────────────────────────────────────────────*/
s3: call head 60,,s3 /*────────── 60º: a² + b² ─ ab ≡ c² */
do a=1 for s3; s3ma= s3 - a + 1
do b=a for s3ma; x= @.a + @.b - a*b; if x>ss then iterate a
do c=a for s3ma until @.c>x
if x==@.c then do; call show os3; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s2; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
s4: call head 60, 'unique', os4 /*────────── 60º: a² + b² ─ ab ≡ c² */
foot: say right(# ' solutions found for' angle "(sides up to" arg(1)')', 65); say; return
head: #= 0; parse arg deg; angle do a=1 ' 'deg"º " for s4; say center(angle, 65, '═') ap1= a + 1; s4map1= s4 - ap1 + return1
show: #= # + 1; say ' ('right(a, w)"," do right(b,=ap1 w)"," for s4map1; right(c, w)')' x= @.a + @.b - a*b; if return</langx>ss then iterate a
do c=ap1 for s4map1 until @.c>x
if x==@.c then do; call show os4; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s4; return</syntaxhighlight>
{{out|output|text=&nbsp; when using the default number of sides for the input: &nbsp; &nbsp; <tt> 13 </tt>}}
<pre>
Line 1,306 ⟶ 2,539:
15 solutions found for 60º (sides up to 13)
</pre>
{{out|output|text= &nbsp; when using the inputs of: &nbsp; &nbsp; <tt> 0 &nbsp; 0 &nbsp; 0 &nbsp; -10000 </tt>}}
 
Note that the first three computations are bypassed because of the three zero ('''0''') numbers, &nbsp; the negative ten thousand indicates to find all the triangles with sides <u>up</u> to 10,000, &nbsp; but not list the triangles, it just reports the &nbsp; ''number'' &nbsp; of solutions found.
===using memoization===
<pre>
<lang rexx>/*REXX pgm finds integer sided triangles that satisfy Law of cosines for 60º, 90º, 120º.*/
══════════════════════════ 60º unique ══════════════════════════
parse arg s1 s2 s3 s4 . /*obtain optional arguments from the CL*/
18,394 solutions found for 60º unique (sides up to 10,000)
if s1=='' | s1=="," then s1= 13 /*Not specified? Then use the default.*/
</pre>
if s2=='' | s2=="," then s2= 13 /* " " " " " " */
if s3=='' | s3=="," then s3= 13 /* " " " " " " */
if s4=='' | s4=="," then s4= -10000 /* " " " " " " */
parse value s1 s2 s3 s4 with os1 os2 os3 os4 . /*obtain the original values for sides.*/
s1=abs(s1); s2=abs(s2); s3=abs(s3); s4=abs(s4) /*use absolute values for the # sides. */
@.=
do j=1 for max(s1, s2, s3, s4); @.j = j*j
end /*j*/ /*build memoization array for squaring.*/
 
=={{header|RPL}}==
if s1>0 then do; call head 120,,os1 /*────120º: a² + b² + ab ≡ c² */
{{works with|HP|28}}
do a=1 for s1
≪ → formula max
do b=a+1 to s1; x= @.a + @.b + a*b
≪ { }
if x>z then iterate a
1 max '''FOR''' a a max '''FOR''' b
do c=b+1 to s1 until @.c>x
formula EVAL √ RND
if x==@.c then do; call show; iterate b; end
'''IF''' DUP max ≤ OVER FP NOT AND end /*c*/
'''THEN''' a b ROT 3 →LIST 1 →LIST + end /*b*/
'''ELSE''' DROP '''END'''
end /*a*/
'''NEXT call foot s1NEXT'''
≫ ≫ ‘<span style="color:blue">TASK</span>’ STO
end
 
'a^2+b^2' 13 <span style="color:blue">TASK</span>
if s2>0 then do; call head 90,, os2 /*────90º: a² + b² ≡ c² */
'a^2+b^2-a*b' 13 <span style="color:blue">TASK</span>
do a=1 for s2
'a^2+b^2+a*b' 13 <span style="color:blue">TASK</span>
do b=a+1 to s2; x= @.a + @.b
{{out}}
if x>z then iterate a
<pre>
do c=b+1 to s2 until @.c>x
3: { { 4 3 5 } { 8 6 10 } { 12 5 13 } }
if x==@.c then do; call show; iterate b; end
2: { { 1 1 1 } { 2 2 2 } { 3 3 3 } { 4 4 4 } { 5 5 5 } { 6 6 6 } { 7 7 7 } { 8 3 7 } end{ 8 5 /*c*/7 } { 8 8 8 } { 9 9 9 } { 10 10 10 } { 11 11 11 } { 12 12 12 } { 13 13 13 } }
1: { { 5 3 7 } { 8 7 13 } }
end /*b*/
</pre>
end /*a*/
call foot s2
end
 
=={{header|Ruby}}==
if s3>0 then do; call head 60,, os3 /*────60º: a² + b² ─ ab ≡ c² */
<syntaxhighlight lang="ruby">grouped = (1..13).to_a.repeated_permutation(3).group_by do |a,b,c|
do a=1 for s3
sumaabb, ab = a*a + b*b, a*b
do b=a to s3; x= @.a + @.b - a*b
case c*c
if x>z then iterate a
when sumaabb then 90
do c=a to s3 until @.c>x
when sumaabb - ab then 60
if x==@.c then do; call show; iterate b; end
when sumaabb + ab then 120
end /*c*/
end
end /*b*/
end
end /*a*/
call foot s3
end
 
grouped.delete(nil)
if s4>0 then do; call head 60, 'unique', os4 /*────60º: a² + b² ─ ab ≡ c² */
res = grouped.transform_values{|v| v.map(&:sort).uniq }
do a=1 for s4
do b=a to s4; x= @.a + @.b - a*b
if x>z then iterate a
do c=a to s4 until @.c>x
if x==@.c then do; if a==b&a==c then iterate b
call show; iterate b
end
end /*c*/
end /*b*/
end /*a*/
call foot s4
end
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
foot: say right(# ' solutions found for' ang "(sides up to" arg(1)')', 65); say; return
head: #=0; arg d,,s;z=s*s;w=length(s); ang=' 'd"º " arg(2); say center(ang,65,'═'); return
show: #= # + 1; if s>0 then say ' ('right(a,w)"," right(b,w)"," right(c,w)')'; return</lang>
{{out|output|text= &nbsp; when using the inputs of: &nbsp; &nbsp; <tt> 0 &nbsp; 0 &nbsp; 0 &nbsp; -10000 </tt>}}
 
res.each do |k,v|
Note that the first three computations are bypassed because of the three zero ('''0''') numbers, &nbsp; the negative ten thousand indicates to find all the triangles with sides up to 10,000, &nbsp; but not list the triangles, it just reports the &nbsp; ''number'' &nbsp; of triangles found.
puts "For an angle of #{k} there are #{v.size} solutions:"
puts v.inspect, "\n"
end
</syntaxhighlight>
{{out}}
<pre>For an angle of 60 there are 15 solutions:
[[1, 1, 1], [2, 2, 2], [3, 3, 3], [3, 7, 8], [4, 4, 4], [5, 5, 5], [5, 7, 8], [6, 6, 6], [7, 7, 7], [8, 8, 8], [9, 9, 9], [10, 10, 10], [11, 11, 11], [12, 12, 12], [13, 13, 13]]
 
For an angle of 90 there are 3 solutions:
[[3, 4, 5], [5, 12, 13], [6, 8, 10]]
 
For an angle of 120 there are 2 solutions:
[[3, 5, 7], [7, 8, 13]]
</pre>
Extra credit:
<syntaxhighlight lang="ruby">n = 10_000
ar = (1..n).to_a
squares = {}
ar.each{|i| squares[i*i] = true }
count = ar.combination(2).count{|a,b| squares.key?(a*a + b*b - a*b)}
 
puts "There are #{count} 60° triangles with unequal sides of max size #{n}."
</syntaxhighlight>
{{out}}
<pre>There are 18394 60° triangles with unequal sides of max size 10000.
 
</pre>
 
=={{header|Wren}}==
{{trans|Go}}
<syntaxhighlight lang="wren">var squares13 = {}
var squares10000 = {}
 
var initMaps = Fn.new {
for (i in 1..13) squares13[i*i] = i
for (i in 1..10000) squares10000[i*i] = i
}
 
var solve = Fn.new { |angle, maxLen, allowSame|
var solutions = []
for (a in 1..maxLen) {
for (b in a..maxLen) {
var lhs = a*a + b*b
if (angle != 90) {
if (angle == 60) {
lhs = lhs - a*b
} else if (angle == 120) {
lhs = lhs + a*b
} else {
Fiber.abort("Angle must be 60, 90 or 120 degrees")
}
}
if (maxLen == 13) {
var c = squares13[lhs]
if (c != null) {
if (allowSame || a != b || b != c) {
solutions.add([a, b, c])
}
}
} else if (maxLen == 10000) {
var c = squares10000[lhs]
if (c != null) {
if (allowSame || a != b || b != c) {
solutions.add([a, b, c])
}
}
} else {
Fiber.abort("Maximum length must be either 13 or 10000")
}
}
}
return solutions
}
 
initMaps.call()
System.write("For sides in the range [1, 13] ")
System.print("where they can all be of the same length:-\n")
var angles = [90, 60, 120]
var solutions = []
for (angle in angles) {
solutions = solve.call(angle, 13, true)
System.write(" For an angle of %(angle) degrees")
System.print(" there are %(solutions.count) solutions, namely:")
System.print(" %(solutions)")
System.print()
}
System.write("For sides in the range [1, 10000] ")
System.print("where they cannot ALL be of the same length:-\n")
solutions = solve.call(60, 10000, false)
System.write(" For an angle of 60 degrees")
System.print(" there are %(solutions.count) solutions.")</syntaxhighlight>
 
{{out}}
<pre>
For sides in the range [1, 13] where they can all be of the same length:-
 
For an angle of 90 degrees there are 3 solutions, namely:
[[3, 4, 5], [5, 12, 13], [6, 8, 10]]
 
For an angle of 60 degrees there are 15 solutions, namely:
[[1, 1, 1], [2, 2, 2], [3, 3, 3], [3, 8, 7], [4, 4, 4], [5, 5, 5], [5, 8, 7], [6, 6, 6], [7, 7, 7], [8, 8, 8], [9, 9, 9], [10, 10, 10], [11, 11, 11], [12, 12, 12], [13, 13, 13]]
 
For an angle of 120 degrees there are 2 solutions, namely:
[[3, 5, 7], [7, 8, 13]]
 
For sides in the range [1, 10000] where they cannot ALL be of the same length:-
 
For an angle of 60 degrees there are 18394 solutions.
</pre>
 
=={{header|XPL0}}==
<syntaxhighlight lang="xpl0">proc LawCos(Eqn);
int Eqn;
int Cnt, A, B, C;
 
proc Show;
[Cnt:= Cnt+1;
IntOut(0, A); ChOut(0, ^ );
IntOut(0, B); ChOut(0, ^ );
IntOut(0, C); CrLf(0);
];
 
[Cnt:= 0;
for A:= 1 to 13 do
for B:= 1 to A do
for C:= 1 to 13 do
case Eqn of
1: if A*A + B*B = C*C then Show;
2: if A*A + B*B - A*B = C*C then Show;
3: if A*A + B*B + A*B = C*C then Show
other [];
IntOut(0, Cnt); Text(0, " results^m^j");
];
 
proc ExtraCredit;
int Cnt, A, B, C, C2;
[Cnt:= 0;
for A:= 1 to 10_000 do
for B:= 1 to A-1 do
[C2:= A*A + B*B - A*B;
C:= sqrt(C2);
if C*C = C2 then Cnt:= Cnt+1;
];
Text(0, "Extra credit: ");
IntOut(0, Cnt);
CrLf(0);
];
 
int Case;
[for Case:= 1 to 3 do LawCos(Case);
ExtraCredit;
]</syntaxhighlight>
 
{{out}}
<pre>
4 3 5
══════════════════════════ 60º unique═══════════════════════════
8 6 10
18394 solutions found for 60º unique (sides up to 10000)
12 5 13
3 results
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 3 7
8 5 7
8 8 8
9 9 9
10 10 10
11 11 11
12 12 12
13 13 13
15 results
5 3 7
8 7 13
2 results
Extra credit: 18394
</pre>
 
=={{header|zkl}}==
<langsyntaxhighlight lang="zkl">fcn tritri(N=13){
sqrset:=[0..N].pump(Dictionary().add.fp1(True),fcn(n){ n*n });
tri90, tri60, tri120 := List(),List(),List();
Line 1,403 ⟶ 2,785:
tri.sort(fcn(t1,t2){ t1[0]<t2[0] })
.apply("concat",",").apply("(%s)".fmt).concat(",")
}</langsyntaxhighlight>
<langsyntaxhighlight lang="zkl">N:=13;
println("Integer triangular triples for sides 1..%d:".fmt(N));
foreach angle, triples in (T(60,90,120).zip(tritri(N))){
println(" %3d\U00B0; has %d solutions:\n %s"
.fmt(angle,triples.len(),triToStr(triples)));
}</langsyntaxhighlight>
{{out}}
<pre>
Line 1,421 ⟶ 2,803:
</pre>
Extra credit:
<langsyntaxhighlight lang="zkl">fcn tri60(N){ // special case 60*
sqrset:=[1..N].pump(Dictionary().add.fp1(True),fcn(n){ n*n });
n60:=0;
Line 1,429 ⟶ 2,811:
}
n60
}</langsyntaxhighlight>
<langsyntaxhighlight lang="zkl">N:=10_000;
println(("60\U00b0; triangle where side lengths are unique,\n"
" side lengths 1..%,d, there are %,d solutions.").fmt(N,tri60(N)));</langsyntaxhighlight>
{{out}}
<pre>
PureFox
9,482

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