TL;DR

On with TASK #2 from the Perl Weekly Challenge #085. Enjoy!

# The solution

This challenge is interesting because it can have some twists. For example, we can factorize $N and check the result, but we will have to be careful that$2^4 \cdot 5^2$can be written as$4^2 \cdot 5^2$, so… it’s not immediate. Well, it’s not difficult either. If we get the minimum exponent in the factorization, we can verify that: • it is greater than 1 - because of the constraint that b > 1; • every other exponent is a multiple of it (possibly equal to it). So let’s put this in code:  1 sub power_of_two_integers ($N) {
2    my $factors = factor($N);
3    my ($min, @others) = sort {$a <=> $b} values$factors->%*;
4    return 0 if $min == 1; 5 for my$exponent (@others) {
6       return 0 if $exponent %$min;
7    }
8    return 1;
9 }


The call in line 2 assumes that there is a factor function that returns the factorization of the input number. Looking at line 3, we can see that we expect this function to give back a reference to a hash, whose keys are the prime numbers in the factorization and the associated values are the exponents.

As said, we are interested into the exponents only, so we sort them in increasing numerical order and assign the lowest number to variable $min and collect the rest in @others (line 3). At this point, we first make sure that $min is greater than 1 (line 4), then check that all other exponents in @others are divisible by it (lines 5 and 6).

Now we just have to implement the factorization into prime factors. This is the complicated stuff to do, at lest for big numbers; we’ll just put a simple implementation to work:

 1 sub factor ($N) { 2 my %retval; 3 my @ps = (2, 3); 4 my$k = 1;
5    while ($N > 1) { 6 if (! @ps) { 7 push @ps, 6 *$k - 1, 6 * $k + 1; 8$k++;
9       }
10       my $p = shift @ps; 11 while ($N % $p == 0) { 12$retval{$p}++; 13$N /= $p; 14 } 15 } 16 return \%retval; 17 }  We leverage the fact that, after$2$and$3$, every prime number is of the form$6 \cdot k \pm 1$. So we keep a small array of prime numbers @ps that we pre-load with$2$and$3$(the outliers); afterwards, as this reservoir of candidate primes empties (line 6), we put the next two ones (line 7) and prepare for the next pair (line 8). The astute reader might object that what we extract from @ps in line 10 is not always a prime. No worries: if it’s not a prime, then its factors have already been analyzed and the test in line 11 will fail immediately. So there’s no need to check it’s a prime! The loop in lines 11 through 14 takes care to extract as many $p factors out of $N as possible, counting them as it goes. The count is the same as the exponent for $p in the factorization.

Separating the factorization process and the check process has its advantages (clean code, more readable and maintainable) but it also has its downsides. As an example, we already know that any factor appearing with exponent $1$ would invalidate our test, so in that case it might not be needed to have a full factorization.

As usual, if you’re interested in the whole program…

#!/usr/bin/env perl
use 5.024;
use warnings;
use experimental qw< postderef signatures >;
no warnings qw< experimental::postderef experimental::signatures >;

sub factor ($N) { my %retval; my @ps = (2, 3); my$k = 1;
while ($N > 1) { if (! @ps) { push @ps, 6 *$k - 1, 6 * $k + 1;$k++;
}
my $p = shift @ps; while ($N % $p == 0) {$retval{$p}++;$N /= $p; } } return \%retval; } sub power_of_two_integers ($N) {
my $factors = factor($N);
my ($min, @others) = sort {$a <=> $b} values$factors->%*;
return 0 if $min == 1; for my$exponent (@others) {
return 0 if $exponent %$min;
}
return 1;
}

my $N = shift // 8; say power_of_two_integers($N);


See you soon!

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