How to resolve the algorithm Pythagorean triples step by step in the Factor programming language

Published on 12 May 2024 09:40 PM
#Factor

How to resolve the algorithm Pythagorean triples step by step in the Factor programming language

Table of Contents

Problem Statement

A Pythagorean triple is defined as three positive integers

( a , b , c )

{\displaystyle (a,b,c)}

where

a < b < c

{\displaystyle a<b<c}

, and

a

2

b

2

=

c

2

.

{\displaystyle a^{2}+b^{2}=c^{2}.}

They are called primitive triples if

a , b , c

{\displaystyle a,b,c}

are co-prime, that is, if their pairwise greatest common divisors

g c d

( a , b )

g c d

( a , c )

g c d

( b , c )

1

{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}

. Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (

g c d

( a , b )

1

{\displaystyle {\rm {gcd}}(a,b)=1}

).   Each triple forms the length of the sides of a right triangle, whose perimeter is

P

a + b + c

{\displaystyle P=a+b+c}

.

The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.

Deal with large values.   Can your program handle a maximum perimeter of 1,000,000?   What about 10,000,000?   100,000,000? Note: the extra credit is not for you to demonstrate how fast your language is compared to others;   you need a proper algorithm to solve them in a timely manner.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Pythagorean triples step by step in the Factor programming language

Source code in the factor programming language

USING: accessors arrays formatting kernel literals math
math.functions math.matrices math.ranges sequences ;
IN: rosettacode.pyth

CONSTANT: T1 { 
  {  1  2  2 }
  { -2 -1 -2 }
  {  2  2  3 }
}
CONSTANT: T2 {
  {  1  2  2 }
  {  2  1  2 }
  {  2  2  3 }
}
CONSTANT: T3 {
  { -1 -2 -2 }
  {  2  1  2 }
  {  2  2  3 }
}

CONSTANT: base { 3 4 5 }

TUPLE: triplets-count primitives total ;
: <0-triplets-count> ( -- a ) 0 0 \ triplets-count boa ;
: next-triplet ( triplet T -- triplet' ) [ 1array ] [ m. ] bi* first ;
: candidates-triplets ( seed -- candidates )
  ${ T1 T2 T3 } [ next-triplet ] with map ;
: add-triplets ( current-triples limit triplet -- stop )
  sum 2dup > [
   /i [ + ] curry change-total
   [ 1 + ] change-primitives drop t 
  ] [ 3drop f ] if ;
: all-triplets ( current-triples limit seed -- triplets )
  3dup add-triplets [ 
    candidates-triplets [ all-triplets ] with swapd reduce
  ] [ 2drop ] if ;
: count-triplets ( limit -- count )
  <0-triplets-count> swap base all-triplets ;
: pprint-triplet-count ( limit count -- )
  [ total>> ] [ primitives>> ] bi 
  "Up to %d: %d triples, %d primitives.\n" printf ;
: pyth ( -- )
  8 [1,b] [ 10^ dup count-triplets pprint-triplet-count ] each ;

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