Special pythagorean triplet

Special pythagorean triplet is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Task

The following problem is taken from Project Euler

Related task

• Pythagorean triples

ALGOL 68

Translation of: Wren

...but doesn't stop on the first solution (thus verifying there is only one). <lang algol68># find the Pythagorian triplet a, b, c where a + b + c = 1000, a < b < c # FOR a TO 1000 OVER 3 DO

   INT a2 = a * a;
   FOR b FROM a + 1 TO 1000 WHILE INT a plus b = a + b;
                                  INT c        = 1000 - a plus b;
                                  c > b
   DO
       IF a2 + b*b = c*c THEN
           print( ( "a = ", whole( a, 0 ), ", b = ", whole( b, 0 ), ", c = ", whole( c, 0 ), newline ) );
           print( ( "a + b + c = ", whole( a plus b + c, 0 ), newline ) );
           print( ( "a * b * c = ", whole( a * b * c, 0 ), newline ) )
       FI
   OD

OD</lang>

Output:

a = 200, b = 375, c = 425
a + b + c = 1000
a * b * c = 31875000

ALGOL W

Translation of: Wren

...but doesn't stop on the first solution (thus verifying there is only one). <lang algolw>% find the Pythagorian triplet a, b, c where a + b + c = 1000 % for a := 1 until 1000 div 3 do begin

   integer a2, b;
   a2 := a * a;
   for b := a + 1 until 1000 do begin
       integer c;
       c := 1000 - ( a + b );
       if c <= b then goto endB;
       if a2 + b*b = c*c then begin
           write( i_w := 1, s_w := 0, "a = ", a, ", b = ", b, ", c = ", c );
           write( i_w := 1, s_w := 0, "a + b + c = ", a + b + c );
           write( i_w := 1, s_w := 0, "a * b * c = ", a * b * c )
       end if_a2_plus_b2_e_c2 ;
       b := b + 1
   end for_b ;

endB: end for_a .</lang>

Output:

a = 200, b = 375, c = 425
a + b + c = 1000
a * b * c = 31875000

Go

Translation of: Wren

<lang go>package main

import (

   "fmt"
   "time"

)

func main() {

   start := time.Now()
   for a := 3; ; a++ {
       for b := a + 1; ; b++ {
           c := 1000 - a - b
           if c <= b {
               break
           }
           if a*a+b*b == c*c {
               fmt.Printf("a = %d, b = %d, c = %d\n", a, b, c)
               fmt.Println("a + b + c =", a+b+c)
               fmt.Println("a * b * c =", a*b*c)
               fmt.Println("\nTook", time.Since(start))
               return
           }
       }
   }

}</lang>

Output:

a = 200, b = 375, c = 425
a + b + c = 1000
a * b * c = 31875000

Took 77.664µs

Julia

julia> [(a, b, c) for a in 1:1000, b in 1:1000, c in 1:1000 if a < b < c && a + b + c == 1000 && a^2 + b^2 == c^2]
1-element Vector{Tuple{Int64, Int64, Int64}}:
 (200, 375, 425)

or, with attention to timing:

julia> @time for a in 1:1000
           for b in a+1:1000
               c = 1000 - a - b; a^2 + b^2 == c^2 && @show a, b, c
           end
       end
(a, b, c) = (200, 375, 425)
  0.001073 seconds (20 allocations: 752 bytes)

Python

Python 3.8.8 (default, Apr 13 2021, 15:08:03)
Type "help", "copyright", "credits" or "license" for more information.
>>> [(a, b, c) for a in range(1, 1000) for b in range(a, 1000) for c in range(1000) if a + b + c == 1000 and a*a + b*b == c*c]
[(200, 375, 425)]

Raku

<lang perl6>hyper for 1..998 -> $a {

   my $a2 = $a²;
   for $a + 1 .. 999 -> $b {
       my $c = 1000 - $a - $b;
       last if $c < $b;
       say "$a² + $b² = $c²\n$a  + $b  + $c = 1000" and exit if $a2 + $b² == $c²
   }

}</lang>

Output:

200² + 375² = 425²
200  + 375  + 425 = 1000

REXX

Some optimizations were done such as pre-computing the squares of all possible A's, B's, and C's.

Also, there were multiple shortcuts to limit an otherwise exhausting search;   Once a sum or a square was too big,
the next integer was used. <lang rexx>/*REXX pgm computes/shows integers A, B, and C that solve: A+B+C = 1000; A^2+B^2 = C^2*/ parse arg s hi . /*obtain optional argument from the CL.*/ if s== | s=="," then s= 1000 /*Not specified? Then use the default.*/ if hi== | hi=="," then hi= 1000 /* " " " " " " */

               do j=1  for hi;   @.j= j*j       /*precompute squares up to  HI.        */
               end  /*j*/

     do       a=1   for hi-2;    aa= @.a        /*go hunting for solutions to equation.*/
        do    b=a+1  to hi-1;    ab= a + b      /*calculate sum of two  (A,B)  squares.*/
        if ab>hi          then iterate a        /*Sum of A+B>HI?    Then stop with B's */
        aabb= aa + @.b                          /*compute the sum of:   A^2 + B^2      */
           do c=b+1  to hi                      /*handle all ints that satisfy equation*/
           if @.c > aabb  then iterate b        /*Square> A^2+B^2?  Then stop with C's.*/
           if @.c\==aabb  then iterate          /*Square\=A^2+B^2?  Then keep searching*/
           abc= ab + c                          /*compute the sum of  A  +  B  +  C    */
           if abc ==s     then leave   a        /*Does A+B+C = S?   Then solution found*/
           if abc > s     then iterate b        /*  "    "   > S?   Then stop with C's.*/
           end   /*c*/
        end      /*b*/
     end         /*a*/

say ' a=' a " b=" b ' c=' c exit 0 /*stick a fork in it, we're all done. */</lang>

output   when using the default inputs:

     a= 200      b= 375      c= 425

Ring

<lang ring> load "stdlib.ring" see "working..." + nl

time1 = clock() for a = 1 to 1000

    for b = a+1 to 1000   
        for c = b+1 to 1000
            if (pow(a,2)+pow(b,2)=pow(c,2))
                if a+b+c = 1000
                   see "a = " + a + " b = " + b + " c = " + c + nl
                   exit 3
                ok
            ok
         next
    next

next

time2 = clock() time3 = time2/1000 - time1/1000 see "Elapsed time = " + time3 + " s" + nl see "done..." + nl </lang>

Output:

working...
a = 200 b = 375 c = 425
Elapsed time = 497.61 s
done...

Wren

Very simple approach, only takes 0.013 seconds even in Wren. <lang ecmascript>var a = 3 while (true) {

   var b = a + 1
   while (true) {
       var c = 1000 - a - b
       if (c <= b) break
       if (a*a + b*b == c*c) {
           System.print("a = %(a), b = %(b), c = %(c)")
           System.print("a + b + c = %(a + b + c)")
           System.print("a * b * c = %(a * b * c)")
           return
       }
       b = b + 1
   }
   a = a + 1

}</lang>

Output:

a = 200, b = 375, c = 425
a + b + c = 1000
a * b * c = 31875000

Incidentally, even though we are told there is only one solution, it is almost as quick to verify this by observing that, since a < b < c, the maximum value of a must be such that 3a + 2 = 1000 or max(a) = 332. The following version ran in 0.015 seconds and, of course, produced the same output: <lang ecmascript>for (a in 3..332) {

   var b = a + 1
   while (true) {
       var c = 1000 - a - b
       if (c <= b) break
       if (a*a + b*b == c*c) {
           System.print("a = %(a), b = %(b), c = %(c)")
           System.print("a + b + c = %(a + b + c)")
           System.print("a * b * c = %(a * b * c)")
       }
       b = b + 1
   }

}</lang>

XPL0

<lang XPL0>int N, M, A, B, C; for N:= 1 to sqrt(1000) do

   for M:= N+1 to sqrt(1000) do
       [A:= M*M - N*N; \Euclid's formula
        B:= 2*M*N;
        C:= M*M + N*N;
        if A+B+C = 1000 then
           IntOut(0, A*B*C);
       ]</lang>

Output:

31875000