Law of cosines - triples: Difference between revisions
(Added C++ solution) |
m (Minor correction) |
||
| Line 460: | Line 460: | ||
int c2 = a * a + b * b - a * b; |
int c2 = a * a + b * b - a * b; |
||
int c = isqrt(c2); |
int c = isqrt(c2); |
||
if ((c |
if ((c != a || c != b) && c <= max2 && c * c == c2) |
||
++count; |
++count; |
||
} |
} |
||
Revision as of 14:29, 29 March 2020
You are encouraged to solve this task according to the task description, using any language you may know.
The Law of cosines states that for an angle γ, (gamma) of any triangle, if the sides adjacent to the angle are A and B and the side opposite is C; then the lengths of the sides are related by this formula:
A2 + B2 - 2ABcos(γ) = C2
- Specific angles
For an angle of of 90º this becomes the more familiar "Pythagoras equation":
A2 + B2 = C2
For an angle of 60º this becomes the less familiar equation:
A2 + B2 - AB = C2
And finally for an angle of 120º this becomes the equation:
A2 + B2 + AB = C2
- Task
- Find all integer solutions (in order) to the three specific cases, distinguishing between each angle being considered.
- Restrain all sides to the integers 1..13 inclusive.
- Show how many results there are for each of the three angles mentioned above.
- Display results on this page.
Note: Triangles with the same length sides but different order are to be treated as the same.
- Optional Extra credit
- How many 60° integer triples are there for sides in the range 1..10_000 where the sides are not all of the same length.
- Related Task
- See also
- Visualising Pythagoras: ultimate proofs and crazy contortions Mathlogger Video
Ada
<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;</lang>
- Output:
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.
ALGOL 68
<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 #
INT max side = 13; # max triangle side to consider #
INT max square = max side * max side; # max triangle side squared to consider #
[ 1 : max square ]INT root; # table of square roots #
FOR s TO UPB root DO root[ s ] := 0 OD;
FOR s TO max side DO root[ s * s ] := s OD;
INT tcount := 0;
[ 1 : max square ]INT ta, tb, tc, tangle;
# prints solutions for the specified angle #
PROC print triangles = ( INT angle )VOID:
BEGIN
INT scount := 0;
FOR t TO tcount DO IF tangle[ t ] = angle THEN scount +:= 1 FI OD;
print( ( whole( scount, -4 ), " ", whole( angle, -3 ), " degree triangles:", newline ) );
FOR t TO tcount DO
IF tangle[ t ] = angle THEN
print( ( " ", whole( ta[ t ], -3 ), whole( tb[ t ], -3 ), whole( tc[ t ], -3 ), newline ) )
FI
OD
END # print triangles # ;
# stores the triangle with sides a, b, root[ c2 ] and the specified angle, #
# if it is a solution #
PROC try triangle = ( INT a, b, c2, angle )VOID:
IF c2 <= max square THEN
# the third side is small enough #
INT c = root[ c2 ];
IF c /= 0 THEN
# the third side is the square of an integer #
tcount +:= 1;
ta[ tcount ] := a; tb[ tcount ] := b; tc[ tcount ] := root[ c2 ];
tangle[ tcount ] := angle
FI
FI # try triangle # ;
# find all triangles #
FOR a TO max side DO
FOR b FROM a TO max side DO
try triangle( a, b, ( a * a ) + ( b * b ) - ( a * b ), 60 );
try triangle( a, b, ( a * a ) + ( b * b ), 90 );
try triangle( a, b, ( a * a ) + ( b * b ) + ( a * b ), 120 )
OD
OD;
# print the solutions #
print triangles( 60 );
print triangles( 90 );
print triangles( 120 )
END</lang>
- Output:
15 60 degree triangles:
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
3 90 degree triangles:
3 4 5
5 12 13
6 8 10
2 120 degree triangles:
3 5 7
7 8 13
AWK
<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])
}
} </lang>
- Output:
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
C
A brute force algorithm, O(N^3)
<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 use fast computer to obtain the "extra credit" badge. * * Obviously, there are better algorithms. */
- include <stdio.h>
- include <math.h>
- define MAX_SIDE_LENGTH 13
//#define DISPLAY_TRIANGLES 1
int main(void) {
static char description[3][80] = {
"gamma = 90 degrees, a*a + b*b == c*c",
"gamma = 60 degrees, a*a + b*b - a*b == c*c",
"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++)
{
int counter = 0;
for (int a = 1; a <= MAX_SIDE_LENGTH; a++)
for (int b = 1; b <= a; b++)
for (int c = 1; c <= MAX_SIDE_LENGTH; c++)
if (a * a + b * b - coeff[k] * a * b == c * c)
{
counter++;
- ifdef DISPLAY_TRIANGLES
printf(" %d %d %d\n", a, b, c);
- endif
}
printf("%s, number of triangles = %d\n", description[k], counter);
}
return 0;
} </lang>
- Output:
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 = 15 gamma = 120 degrees, a*a + b*b + a*b == c*c, number of triangles = 2
An algorithm with O(N^2) cost
<lang C>/*
* RossetaCode: Law of cosines - triples * * A solutions with O(N^2) cost. */
- include <stdio.h>
- include <math.h>
- define MAX_SIDE_LENGTH 10000
//#define DISPLAY_TRIANGLES
int main(void) {
static char description[3][80] = {
"gamma = 90 degrees, a*a + b*b == c*c",
"gamma = 60 degrees, a*a + b*b - a*b == c*c",
"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);
for (int k = 0; k < 3; k++)
{
int counter = 0;
for (int a = 1; a <= MAX_SIDE_LENGTH; a++)
for (int b = 1; b <= a; b++)
{
int cc = a * a + b * b - coeff[k] * a * b;
int c = (int)(sqrt(cc) + 0.5);
if (c <= MAX_SIDE_LENGTH && c * c == cc)
{
- ifdef DISPLAY_TRIANGLES
printf("%d %d %d\n", a, b, c);
- endif
counter++;
}
}
printf("%s, number of triangles = %d\n", description[k], counter);
}
return 0;
} </lang>
- Output:
MAX SIDE LENGTH = 10000 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 = 28394 gamma = 120 degrees, a*a + b*b + a*b == c*c, number of triangles = 10374
C++
<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 != a || c != b) && 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;
}</lang>
- Output:
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.
Factor
<lang factor>USING: backtrack formatting kernel locals math math.ranges sequences sets sorting ; IN: rosetta-code.law-of-cosines
- triples ( quot -- seq )
[
V{ } clone :> seen
13 [1,b] dup dup [ amb-lazy ] tri@ :> ( a b c )
a sq b sq + a b quot call( x x x -- x ) c sq =
{ b a c } seen member? not and
must-be-true { a b c } dup seen push
] bag-of ;
- show-solutions ( quot angle -- )
[ triples { } like dup length ] dip rot
"%d solutions for %d degrees:\n%u\n\n" printf ;
[ * + ] 120 [ 2drop 0 - ] 90 [ * - ] 60 [ show-solutions ] 2tri@</lang>
- Output:
2 solutions for 120 degrees:
{ { 3 5 7 } { 7 8 13 } }
3 solutions for 90 degrees:
{ { 3 4 5 } { 5 12 13 } { 6 8 10 } }
15 solutions for 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 }
}
FreeBASIC
<lang freebasic>' version 03-03-2019 ' compile with: fbc -s console
- Define max 13
- Define Format(_x) Right(" " + Str(_x), 4)
Dim As UInteger a, b, c, a2, b2, c2, c60 , c90, c120 Dim As String s60, s90, s120
For a = 1 To max
a2 = a * a
For b = a To max
b2 = b * b
' 60 degrees
c2 = a2 + b * b - a * b
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
s60 += Format(a) + Format(b) + Format(c) + Chr(10, 13)
c60 += 1
End If
' 90 degrees
c2 = a2 + b * b
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
s90 += Format(a) + Format(b) + Format(c) + Chr(10, 13)
c90 += 1
End If
' 120 degrees
c2 = a2 + b * b + a * b
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
s120 += Format(a) + Format(b) + Format(c) + Chr(10, 13)
c120 += 1
End If
Next
Next
Print Using "###: 60 degree triangles"; c60
Print s60
Print
Print Using "###: 90 degree triangles"; c90 Print s90 Print
Print Using "###: 120 degree triangles"; c120 Print s120 Print
- Undef max
- Define max 10000
c60 = 0 For a = 1 To max
a2 = a * a
For b = a +1 To max
c2 = a2 + b * (b - a)
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
c60 += 1
End If
Next
Next
Print "For 60 degree triangles in the range [1, 10000]" Print "There are "; c60; " triangles that have different length for a, b and c"
' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
15: 60 degree triangles 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 3: 90 degree triangles 3 4 5 5 12 13 6 8 10 2: 120 degree triangles 3 5 7 7 8 13 For 60 degree triangles in the range [1, 10000] There are 18394 triangles that have different length for a, b and c
Go
<lang go>package main
import "fmt"
type triple struct{ a, b, c int }
var squares13 = make(map[int]int, 13) var squares10000 = make(map[int]int, 10000)
func init() {
for i := 1; i <= 13; i++ {
squares13[i*i] = i
}
for i := 1; i <= 10000; i++ {
squares10000[i*i] = i
}
}
func solve(angle, maxLen int, allowSame bool) []triple {
var solutions []triple
for a := 1; a <= maxLen; a++ {
for b := a; b <= maxLen; b++ {
lhs := a*a + b*b
if angle != 90 {
switch angle {
case 60:
lhs -= a * b
case 120:
lhs += a * b
default:
panic("Angle must be 60, 90 or 120 degrees")
}
}
switch maxLen {
case 13:
if c, ok := squares13[lhs]; ok {
if !allowSame && a == b && b == c {
continue
}
solutions = append(solutions, triple{a, b, c})
}
case 10000:
if c, ok := squares10000[lhs]; ok {
if !allowSame && a == b && b == c {
continue
}
solutions = append(solutions, triple{a, b, c})
}
default:
panic("Maximum length must be either 13 or 10000")
}
}
}
return solutions
}
func main() {
fmt.Print("For sides in the range [1, 13] ")
fmt.Println("where they can all be of the same length:-\n")
angles := []int{90, 60, 120}
var solutions []triple
for _, angle := range angles {
solutions = solve(angle, 13, true)
fmt.Printf(" For an angle of %d degrees", angle)
fmt.Println(" there are", len(solutions), "solutions, namely:")
fmt.Printf(" %v\n", solutions)
fmt.Println()
}
fmt.Print("For sides in the range [1, 10000] ")
fmt.Println("where they cannot ALL be of the same length:-\n")
solutions = solve(60, 10000, false)
fmt.Print(" For an angle of 60 degrees")
fmt.Println(" there are", len(solutions), "solutions.")
}</lang>
- Output:
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.
Haskell
<lang haskell>import qualified Data.Map.Strict as Map import qualified Data.Set as Set import Data.Monoid ((<>))
triangles
:: (Map.Map Int Int -> Int -> Int -> Int -> Int -> Maybe Int) -> Int -> [(Int, Int, Int)]
triangles f n =
let mapRoots = Map.fromList $ ((,) =<< (^ 2)) <$> [1 .. n]
in Set.elems $
foldr
(\(suma2b2, a, b) triSet ->
(case f mapRoots suma2b2 (a * b) a b of
Just c -> Set.insert (a, b, c) triSet
_ -> triSet))
(Set.fromList [])
([1 .. n] >>=
(\a -> (flip (,,) a =<< (a * a +) . (>>= id) (*)) <$> [1 .. a]))
-- TESTS ------------------------------------------------------------------------
f90, f60, f60ne, f120 :: Map.Map Int Int -> Int -> Int -> Int -> Int -> Maybe Int f90 dct x2 ab a b = Map.lookup x2 dct
f60 dct x2 ab a b = Map.lookup (x2 - ab) dct
f120 dct x2 ab a b = Map.lookup (x2 + ab) dct
f60ne dct x2 ab a b
| a == b = Nothing | otherwise = Map.lookup (x2 - ab) dct
main :: IO () main = do
putStrLn
(unlines $
"Triangles of maximum side 13\n" :
zipWith
(\f n ->
let solns = triangles f 13
in show (length solns) <> " solutions for " <> show n <>
" degrees:\n" <>
unlines (show <$> solns))
[f120, f90, f60]
[120, 90, 60])
putStrLn "60 degrees - uneven triangles of maximum side 10000. Total:"
print $ length $ triangles f60ne 10000</lang>
- Output:
Triangles of maximum side 13 2 solutions for 120 degrees: (5,3,7) (8,7,13) 3 solutions for 90 degrees: (4,3,5) (8,6,10) (12,5,13) 15 solutions for 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) 60 degrees - uneven triangles of maximum side 10000. Total: 18394
J
Solution: <lang j>load 'trig stats' RHS=: *: NB. right-hand-side of Cosine Law LHS=: +/@:*:@] - cos@rfd@[ * 2 * */@] NB. Left-hand-side of Cosine Law
solve=: 4 :0
adjsides=. >: 2 combrep y
oppside=. >: i. y
idx=. (RHS oppside) i. x LHS"1 adjsides
adjsides ((#~ idx ~: #) ,. ({~ idx -. #)@]) oppside
)</lang> Example: <lang j> 60 90 120 solve&.> 13 +--------+-------+------+ | 1 1 1|3 4 5|3 5 7| | 2 2 2|5 12 13|7 8 13| | 3 3 3|6 8 10| | | 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| | | +--------+-------+------+
60 #@(solve -. _3 ]\ 3 # >:@i.@]) 10000 NB. optional extra credit
18394</lang>
Java
<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);
}
} </lang>
- Output:
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
JavaScript
<lang JavaScript>(() => {
'use strict';
// main :: IO ()
const main = () => {
const
f90 = dct => x2 => dct[x2],
f60 = dct => (x2, ab) => dct[x2 - ab],
f120 = dct => (x2, ab) => dct[x2 + ab],
f60unequal = dct => (x2, ab, a, b) =>
(a !== b) ? (
dct[x2 - ab]
) : undefined;
// triangles :: Dict -> (Int -> Int -> Int -> Int -> Maybe Int)
// -> [String]
const triangles = (f, n) => {
const
xs = enumFromTo(1, n),
fr = f(xs.reduce((a, x) => (a[x * x] = x, a), {})),
gc = xs.reduce((a, _) => a, {}),
setSoln = new Set();
return (
xs.forEach(
a => {
const a2 = a * a;
enumFromTo(1, 1 + a).forEach(
b => {
const
suma2b2 = a2 + b * b,
c = fr(suma2b2, a * b, a, b);
if (undefined !== c) {
setSoln.add([a, b, c].sort())
};
}
);
}
),
Array.from(setSoln.keys())
);
};
const
result = 'Triangles of maximum side 13:\n\n' +
unlines(
zipWith(
(s, f) => {
const ks = triangles(f, 13);
return ks.length.toString() + ' solutions for ' + s +
' degrees:\n' + unlines(ks) + '\n';
},
['120', '90', '60'],
[f120, f90, f60]
)
) + '\nUneven triangles of maximum side 10000. Total:\n' +
triangles(f60unequal, 10000).length
return (
//console.log(result),
result
);
};
// GENERIC FUNCTIONS ----------------------------
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) =>
xs.reduce((a, x) => a.concat(f(x)), []);
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
m <= n ? iterateUntil(
x => n <= x,
x => 1 + x,
m
) : [];
// iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]
const iterateUntil = (p, f, x) => {
const vs = [x];
let h = x;
while (!p(h))(h = f(h), vs.push(h));
return vs;
};
// Returns Infinity over objects without finite length // this enables zip and zipWith to choose the shorter // argument when one non-finite like cycle, repeat etc
// length :: [a] -> Int const length = xs => xs.length || Infinity;
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
xs.constructor.constructor.name !== 'GeneratorFunction' ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// Use of `take` and `length` here allows zipping with non-finite lists // i.e. generators like cycle, repeat, iterate.
// Use of `take` and `length` here allows zipping with non-finite lists // i.e. generators like cycle, repeat, iterate.
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) => {
const
lng = Math.min(length(xs), length(ys)),
as = take(lng, xs),
bs = take(lng, ys);
return Array.from({
length: lng
}, (_, i) => f(as[i], bs[i], i));
};
// MAIN --- return main();
})();</lang>
- Output:
Triangles of maximum side 13: 2 solutions for 120 degrees: 3,5,7 13,7,8 3 solutions for 90 degrees: 3,4,5 10,6,8 12,13,5 15 solutions for 60 degrees: 1,1,1 2,2,2 3,3,3 4,4,4 5,5,5 6,6,6 7,7,7 3,7,8 5,7,8 8,8,8 9,9,9 10,10,10 11,11,11 12,12,12 13,13,13 Uneven triangles of maximum side 10000. Total: 18394 [Finished in 3.444s]
Julia
<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))
function filtertriangles(N)
sqd = sqdict(N)
t60 = Vector{Vector{Int}}()
t90 = Vector{Vector{Int}}()
t120 = Vector{Vector{Int}}()
for x in 1:N, y in 1:x
xsq, ysq, xy = (x*x, y*y, x*y)
if haskey(sqd, xsq + ysq - xy)
push!(t60, sort([x, y, sqd[xsq + ysq - xy]]))
elseif haskey(sqd, xsq + ysq)
push!(t90, sort([x, y, sqd[xsq + ysq]]))
elseif haskey(sqd, xsq + ysq + xy)
push!(t120, sort([x, y, sqd[xsq + ysq + xy]]))
end
end
t60, t90, t120
end
tri60, tri90, tri120 = filtertriangles(13) println("Integer triples for 1 <= side length <= 13:\n") println("Angle 60:"); for t in tri60 println(t) end println("Angle 90:"); for t in tri90 println(t) end println("Angle 120:"); for t in tri120 println(t) end println("\nFor sizes N through 10000, there are $(numnotsame(filtertriangles(10000)[1])) 60 degree triples with nonequal sides.")
</lang>
- Output:
Integer triples for 1 <= side length <= 13:
Angle 60: [1, 1, 1] [2, 2, 2] [3, 3, 3] [4, 4, 4] [5, 5, 5] [6, 6, 6] [7, 7, 7] [3, 7, 8] [5, 7, 8] [8, 8, 8] [9, 9, 9] [10, 10, 10] [11, 11, 11] [12, 12, 12] [13, 13, 13] Angle 90: [3, 4, 5] [6, 8, 10] [5, 12, 13] Angle 120: [3, 5, 7] [7, 8, 13]
For sizes N through 10000, there are 18394 60 degree triples with nonequal sides.
Kotlin
<lang scala>// Version 1.2.70
val squares13 = mutableMapOf<Int, Int>() val squares10000 = mutableMapOf<Int, Int>()
class Trio(val a: Int, val b: Int, val c: Int) {
override fun toString() = "($a $b $c)"
}
fun init() {
for (i in 1..13) squares13.put(i * i, i) for (i in 1..10000) squares10000.put(i * i, i)
}
fun solve(angle :Int, maxLen: Int, allowSame: Boolean): List<Trio> {
val solutions = mutableListOf<Trio>()
for (a in 1..maxLen) {
inner@ for (b in a..maxLen) {
var lhs = a * a + b * b
if (angle != 90) {
when (angle) {
60 -> lhs -= a * b
120 -> lhs += a * b
else -> throw RuntimeException("Angle must be 60, 90 or 120 degrees")
}
}
when (maxLen) {
13 -> {
val c = squares13[lhs]
if (c != null) {
if (!allowSame && a == b && b == c) continue@inner
solutions.add(Trio(a, b, c))
}
}
10000 -> {
val c = squares10000[lhs]
if (c != null) {
if (!allowSame && a == b && b == c) continue@inner
solutions.add(Trio(a, b, c))
}
}
else -> throw RuntimeException("Maximum length must be either 13 or 10000")
}
}
}
return solutions
}
fun main(args: Array<String>) {
init()
print("For sides in the range [1, 13] ")
println("where they can all be of the same length:-\n")
val angles = intArrayOf(90, 60, 120)
lateinit var solutions: List<Trio>
for (angle in angles) {
solutions = solve(angle, 13, true)
print(" For an angle of ${angle} degrees")
println(" there are ${solutions.size} solutions, namely:")
println(" ${solutions.joinToString(" ", "[", "]")}\n")
}
print("For sides in the range [1, 10000] ")
println("where they cannot ALL be of the same length:-\n")
solutions = solve(60, 10000, false)
print(" For an angle of 60 degrees")
println(" there are ${solutions.size} solutions.")
}</lang>
- Output:
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.
Nim
<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."</lang>
- Output:
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.
Perl
<lang perl>use utf8; binmode STDOUT, "utf8:"; use Sort::Naturally;
sub triples {
my($n,$angle) = @_;
my(@triples,%sq);
$sq{$_**2}=$_ for 1..$n;
for $a (1..$n-1) {
for $b ($a+1..$n) {
my $ab = $a*$a + $b*$b;
my $cos = $angle == 60 ? $ab - $a * $b :
$angle == 120 ? $ab + $a * $b :
$ab;
if ($angle == 60) {
push @triples, "$a $sq{$cos} $b" if exists $sq{$cos};
} else {
push @triples, "$a $b $sq{$cos}" if exists $sq{$cos};
}
}
}
@triples;
}
$n = 13; print "Integer triangular triples for sides 1..$n:\n"; for my $angle (120, 90, 60) {
my @itt = triples($n,$angle);
if ($angle == 60) { push @itt, "$_ $_ $_" for 1..$n }
printf "Angle %3d° has %2d solutions: %s\n", $angle, scalar @itt,
join ', ', nsort @itt;
}
printf "Non-equilateral n=10000/60°: %d\n", scalar triples(10000,60);</lang>
- Output:
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, 6 8 10, 5 12 13 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 Non-equilateral n=10000/60°: 18394
Phix
Using a simple flat sequence of 100 million elements (well within the language limits) proved significantly faster than a dictionary (5x or so). <lang Phix>sequence squares = repeat(0,10000*10000) for c=1 to 10000 do
squares[c*c] = c
end for
function solve(integer angle, maxlen, bool samelen=true)
sequence res = {}
for a=1 to maxlen do
integer a2 = a*a
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
procedure show(string fmt,sequence res, bool full=true)
printf(1,fmt,{length(res),iff(full?sprint(res):"")})
end procedure
puts(1,"Integer triangular triples for sides 1..13:\n") show("Angle 60 has %2d solutions: %s\n",solve( 60,13)) show("Angle 90 has %2d solutions: %s\n",solve( 90,13)) show("Angle 120 has %2d solutions: %s\n",solve(120,13)) show("Non-equilateral angle 60 triangles for sides 1..10000: %d%s\n",solve(60,10000,false),false)</lang>
- Output:
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
Python
Sets
<lang python>N = 13
def method1(N=N):
squares = [x**2 for x in range(0, N+1)]
sqrset = set(squares)
tri90, tri60, tri120 = (set() for _ in range(3))
for a in range(1, N+1):
a2 = squares[a]
for b in range(1, a + 1):
b2 = squares[b]
c2 = a2 + b2
if c2 in sqrset:
tri90.add(tuple(sorted((a, b, int(c2**0.5)))))
ab = a * b
c2 -= ab
if c2 in sqrset:
tri60.add(tuple(sorted((a, b, int(c2**0.5)))))
c2 += 2 * ab
if c2 in sqrset:
tri120.add(tuple(sorted((a, b, int(c2**0.5)))))
return sorted(tri90), sorted(tri60), sorted(tri120)
- %%
if __name__ == '__main__':
print(f'Integer triangular triples for sides 1..{N}:')
for angle, triples in zip([90, 60, 120], method1(N)):
print(f' {angle:3}° has {len(triples)} solutions:\n {triples}')
_, t60, _ = method1(10_000)
notsame = sum(1 for a, b, c in t60 if a != b or b != c)
print('Extra credit:', notsame)</lang>
- Output:
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
Dictionaries
A variant Python draft based on dictionaries. (Test functions are passed as parameters to the main function.) <lang python>from itertools import (starmap)
def f90(dct):
return lambda x2, ab, a, b: dct.get(x2, None)
def f60(dct):
return lambda x2, ab, a, b: dct.get(x2 - ab, None)
def f120(dct):
return lambda x2, ab, a, b: dct.get(x2 + ab, None)
def f60unequal(dct):
return lambda x2, ab, a, b: (
dct.get(x2 - ab, None) if a != b else None
)
- triangles :: Dict -> (Int -> Int -> Int -> Int -> Maybe Int)
- -> [String]
def triangles(f, n):
upto = enumFromTo(1)
xs = upto(n)
dctSquares = dict(zip(xs, [x**2 for x in xs]))
dctRoots = {v: k for k, v in dctSquares.items()}
fr = f(dctRoots)
dct = {}
for a in xs:
a2 = dctSquares[a]
for b in upto(a):
suma2b2 = a2 + dctSquares[b]
c = fr(suma2b2, a * b, a, b)
if (c is not None):
dct[str(sorted([a, b, c]))] = 1
return list(dct.keys())
def main():
print(
'Triangles of maximum side 13\n\n' +
unlines(
zipWith(
lambda f, n: (
lambda ks=triangles(f, 13): (
str(len(ks)) + ' solutions for ' +
str(n) + ' degrees:\n' +
unlines(ks) + '\n'
)
)()
)([f120, f90, f60])
([120, 90, 60])
) + '\n\n' +
'60 degrees - uneven triangles of maximum side 10000. Total:\n' +
str(len(triangles(f60unequal, 10000)))
)
- GENERIC --------------------------------------------------------------
- enumFromTo :: Int -> Int -> [Int]
def enumFromTo(m):
return lambda n: list(range(m, 1 + n))
- unlines :: [String] -> String
def unlines(xs):
return '\n'.join(xs)
- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
def zipWith(f):
return lambda xs: lambda ys: (
list(starmap(f, zip(xs, ys)))
)
if __name__ == '__main__':
main()</lang>
- Output:
Triangles of maximum side 13 2 solutions for 120 degrees: [3, 5, 7] [7, 8, 13] 3 solutions for 90 degrees: [3, 4, 5] [6, 8, 10] [5, 12, 13] 15 solutions for 60 degrees: [1, 1, 1] [2, 2, 2] [3, 3, 3] [4, 4, 4] [5, 5, 5] [6, 6, 6] [7, 7, 7] [3, 7, 8] [5, 7, 8] [8, 8, 8] [9, 9, 9] [10, 10, 10] [11, 11, 11] [12, 12, 12] [13, 13, 13] 60 degrees - uneven triangles of maximum side 10000. Total: 18394
Raku
(formerly Perl 6)
In each routine, race is used to allow concurrent operations, requiring the use of the atomic increment operator, ⚛++, to safely update @triples, which must be declared fixed-sized, as an auto-resizing array is not thread-safe. At exit, default values in @triples are filtered out with the test !eqv Any.
<lang perl6>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);</lang>
- Output:
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
REXX
using some optimization
Instead of coding a general purpose subroutine (or function) to solve all of the
task's requirements, it was decided to
write three very similar do loops (triple nested) to provide the
answers for the three requirements.
Three arguments (from the command line) can be specified which indicates the
maximum length of the triangle sides
(the default is 13, as per the task's requirement) for each of the
three types of angles (60º, 90º, and 120º) for
the triangles. If the maximum length of the triangle's number of
sides is positive, it indicates that the triangle sides are
displayed, as well as a total number of triangles found.
If the maximum length of the triangle sides is negative, only
the number of triangles are
displayed (using the
absolute value of the negative number).
<lang rexx>/*REXX pgm finds integer sided triangles that satisfy Law of cosines for 60º, 90º, 120º.*/
parse arg s1 s2 s3 . /*obtain optional arguments from the CL*/
if s1== | s1=="," then s1= 13 /*Not specified? Then use the default.*/
if s2== | s2=="," then s2= 13 /* " " " " " " */
if s3== | s3=="," then s3= 13 /* " " " " " " */
w= max( length(s1), length(s2), length(s3) ) /*W is used to align the side lengths.*/
if s1>0 then do; call head 120 /*────120º: a² + b² + ab ≡ c² */
do a=1 for s1; aa = a*a
do b=a+1 to s1; x= aa + b*b + a*b
do c=b+1 to s1 until c*c>x
if x==c*c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s1
end
if s2>0 then do; call head 90 /*────90º: a² + b² ≡ c² */
do a=1 for s2; aa = a*a
do b=a+1 to s2; x= aa + b*b
do c=b+1 to s2 until c*c>x
if x==c*c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s2
end
if s3>0 then do; call head 60 /*────60º: a² + b² ─ ab ≡ c² */
do a=1 for s3; aa = a*a
do b=a to s3; x= aa + b*b - a*b
do c=a to s3 until c*c>x
if x==c*c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s3
end
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ foot: say right(# ' solutions found for' angle "(sides up to" arg(1)')', 65); say; return head: #= 0; parse arg deg; angle= ' 'deg"º "; say center(angle, 65, '═'); return show: #= # + 1; say ' ('right(a, w)"," right(b, w)"," right(c, w)')'; return</lang>
- output when using the default number of sides for the input: 13
═════════════════════════════ 120º ══════════════════════════════
( 3, 5, 7)
( 7, 8, 13)
2 solutions found for 120º (sides up to 13)
══════════════════════════════ 90º ══════════════════════════════
( 3, 4, 5)
( 5, 12, 13)
( 6, 8, 10)
3 solutions found for 90º (sides up to 13)
══════════════════════════════ 60º ══════════════════════════════
( 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)
15 solutions found for 60º (sides up to 13)
using memoization
<lang rexx>/*REXX pgm finds integer sided triangles that satisfy Law of cosines for 60º, 90º, 120º.*/ parse arg s1 s2 s3 s4 . /*obtain optional arguments from the CL*/ if s1== | s1=="," then s1= 13 /*Not specified? Then use the default.*/ 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.*/
if s1>0 then do; call head 120,,os1 /*────120º: a² + b² + ab ≡ c² */
do a=1 for s1
do b=a+1 to s1; x= @.a + @.b + a*b
if x>z then iterate a
do c=b+1 to s1 until @.c>x
if x==@.c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s1
end
if s2>0 then do; call head 90,, os2 /*────90º: a² + b² ≡ c² */
do a=1 for s2
do b=a+1 to s2; x= @.a + @.b
if x>z then iterate a
do c=b+1 to s2 until @.c>x
if x==@.c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s2
end
if s3>0 then do; call head 60,, os3 /*────60º: a² + b² ─ ab ≡ c² */
do a=1 for s3
do b=a to s3; x= @.a + @.b - a*b
if x>z then iterate a
do c=a to s3 until @.c>x
if x==@.c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s3
end
if s4>0 then do; call head 60, 'unique', os4 /*────60º: a² + b² ─ ab ≡ c² */
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>
- output when using the inputs of: 0 0 0 -10000
Note that the first three computations are bypassed because of the three zero (0) numbers, the negative ten thousand indicates to find all the triangles with sides up to 10,000, but not list the triangles, it just reports the number of triangles found.
══════════════════════════ 60º unique═══════════════════════════
18394 solutions found for 60º unique (sides up to 10000)
Ruby
<lang ruby>grouped = (1..13).to_a.repeated_permutation(3).group_by do |a,b,c|
sumaabb, ab = a*a + b*b, a*b case c*c when sumaabb then 90 when sumaabb - ab then 60 when sumaabb + ab then 120 end
end
grouped.delete(nil) res = grouped.transform_values{|v| v.map(&:sort).uniq }
res.each do |k,v|
puts "For an angle of #{k} there are #{v.size} solutions:"
puts v.inspect, "\n"
end </lang>
- Output:
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]]
Extra credit: <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}." </lang>
- Output:
There are 18394 60° triangles with unequal sides of max size 10000.
zkl
<lang zkl>fcn tritri(N=13){
sqrset:=[0..N].pump(Dictionary().add.fp1(True),fcn(n){ n*n });
tri90, tri60, tri120 := List(),List(),List();
foreach a,b in ([1..N],[1..a]){
aa,bb := a*a,b*b;
ab,c := a*b, aa + bb - ab; // 60*
if(sqrset.holds(c)){ tri60.append(abc(a,b,c)); continue; }
c=aa + bb; // 90*
if(sqrset.holds(c)){ tri90.append(abc(a,b,c)); continue; }
c=aa + bb + ab; // 120*
if(sqrset.holds(c)) tri120.append(abc(a,b,c));
}
List(tri60,tri90,tri120)
} fcn abc(a,b,c){ List(a,b).sort().append(c.toFloat().sqrt().toInt()) } fcn triToStr(tri){ // ((c,d,e),(a,b,c))-->"(a,b,c),(c,d,e)"
tri.sort(fcn(t1,t2){ t1[0]<t2[0] })
.apply("concat",",").apply("(%s)".fmt).concat(",")
}</lang> <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)));
}</lang>
- Output:
Integer triangular triples for sides 1..13:
60° has 15 solutions:
(1,1,1),(2,2,2),(3,8,7),(3,3,3),(4,4,4),(5,8,7),(5,5,5),(6,6,6),(7,7,7),(8,8,8),(9,9,9),(10,10,10),(11,11,11),(12,12,12),(13,13,13)
90° has 3 solutions:
(3,4,5),(5,12,13),(6,8,10)
120° has 2 solutions:
(3,5,7),(7,8,13)
Extra credit: <lang zkl>fcn tri60(N){ // special case 60*
sqrset:=[1..N].pump(Dictionary().add.fp1(True),fcn(n){ n*n });
n60:=0;
foreach a,b in ([1..N],[1..a]){
c:=a*a + b*b - a*b;
if(sqrset.holds(c) and a!=b!=c) n60+=1;
}
n60
}</lang> <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)));</lang>
- Output:
60° triangle where side lengths are unique, side lengths 1..10,000, there are 18,394 solutions.