How to resolve the algorithm Count in factors step by step in the Raku programming language

Published on 12 May 2024 09:40 PM
#Raku

How to resolve the algorithm Count in factors step by step in the Raku programming language

Table of Contents

Problem Statement

Write a program which counts up from   1,   displaying each number as the multiplication of its prime factors. For the purpose of this task,   1   (unity)   may be shown as itself.

2   is prime,   so it would be shown as itself.       6   is not prime;   it would be shown as

2 × 3

{\displaystyle 2\times 3}

. 2144   is not prime;   it would be shown as

2 × 2 × 2 × 2 × 2 × 67

{\displaystyle 2\times 2\times 2\times 2\times 2\times 67}

.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Count in factors step by step in the Raku programming language

Source code in the raku programming language

constant @primes = 2, |(3, 5, 7 ... *).grep: *.is-prime;

multi factors(1) { 1 }
multi factors(Int $remainder is copy) {
  gather for @primes -> $factor {

    # if remainder < factor², we're done
    if $factor * $factor > $remainder {
      take $remainder if $remainder > 1;
      last;
    }

    # How many times can we divide by this prime?
    while $remainder %% $factor {
        take $factor;
        last if ($remainder div= $factor) === 1;
    }
  }
}

say "$_: ", factors($_).join(" × ") for 1..*;


sub factor($n is copy) {
    $n == 1 ?? 1 !!
    gather {
	$n /= take 2 while $n %% 2;
	$n /= take 3 while $n %% 3;
	loop (my $p = 5; $p*$p <= $n; $p+=2) {
	    $n /= take $p while $n %% $p;
	}
	take $n unless $n == 1;
    }
}

say "$_ == ", join " \x00d7 ", factor $_ for 1 .. 20;


use Prime::Factor;

say "$_ = {(.&prime-factors || 1).join: ' x ' }" for flat 1 .. 10, 10**20 .. 10**20 + 10;

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