Category: Perl Weekly Challenges

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Yes, it’s been a long while! Many a curveball has been thrown at my face. This week’s tasks are pretty quick and easy, so let’s dive right in.

Count Even Digits Number

The first task is simple. Given a list of integers, return the count of integers that have an even number of digits. So, for example, (10, 1, 111, 24, 1000) has 3 values with an even number of digits.

I was in a regex-y mood, so here’s what I came up with (Perl):

sub even_re { ()= "@_" =~ /\b(\d\d)+\b/g }

The pattern is simple: match two digits (\d\d) one or more times (+), separated by a word boundary (\b). That is matched against "@_", which interpolates to the list, separated by $".

Since we use the /g (global) match, we get all matches back. The ()= pseudo-operator turns that into the count of matches. If that were being assigned to a variable, you’d see it written like this instead:

$result =()= $var =~ /pattern/g

Raku

In Raku, I decided to do the slightly more sane thing and grep the list rather than interpolating first:

sub even_re { +@_.grep(/^(\d\d)+$/) }

Without the +, we would return the list of matches. The + forces it to a scalar, giving us the count of matches instead.

Sum of Values

Task 2 has us take in a list of integers and a number $k, and add up the integers whose index in the array contain $k ones. So, for example, given $k = 1 and @ints = (2, 5, 9, 11, 3), we have to look at the binary representation of the array indicies:

The bold rows (values: 5, 9, 3) have a base-2 index containing only one 1, so the result is 17. Why anyone would want to do this, I couldn’t tell you.

Still feeling a slight bit of regex fever, I based my second task solution around a very simple /1/g regex for counting the 1s in the binary representation of a number. Here’s the full solution:

use List::Util qw< sum >;

sub sum_idx_bit_set {
    my $k = pop;
    sum map { $_[$_] } grep { $k == ( ()= sprintf('%b',$_) =~ /1/g ) } 0..$#_
}

The whole thing is a grep over all of the indices for any where the binary representation contains $k ones. Those indices are fed into a simple map to get the list element, and the whole thing is sum()med up.

Raku

The Raku solution is conceptually similar, but there are a few language differences:

sub sum_idx_bit_set($k, @n) {
    @n[ @n.keys.grep({ (TR:d/10/1/ with .base(2)) == $k }) ].sum
}

I decided to use TR// here, which I could have also done with Perl. Converting to binary is easier in Raku thanks to the base() method. The grep({...}) over the indices of @n sets us up to take a slice of @n and sum() it up.

This solution feels a bit less Raku-ish to me, although my Raku is very rusty at the moment, not having looked at Raku code for nearly two years. Time to get hacking, eh!

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

This week’s tasks include a simple number theory calculation, and a language feature. Here’s a look at task 1.

Odd Abundant Numbers

Abundant numbers are numbers where the sum of the proper divisors is greater than the number. The first odd abundant number is 945. 945’s proper divisors are 1, 3, 5, 7, 9, 15, 21, 27, 35, 45, 63, 105, 135, 189, and 315. (Recall that we exclude 945 itself as a proper divisor.)

The sum of those divisors is 975, so therefore 945 is an abundant number.

Equivalently, a number is abundant if the sum of all divisors is greater than twice the number. We’ll see both variations in the solutions below.

I’m going to take you through a few different approaches. Since I don’t like repeating myself, let’s get some foreshadowing for task #2 going, by using a first class function:

sub n_odd_abundant(&$) {
    my ($is_abundant, $N) = @_;

    my @r;
    for (my $n = 3; $N; $n += 2) {
        if ($is_abundant->($n)) {
            push @r, $n;
            $N--
        }
    }

    @r;
}

The above code takes in a code ref ($is_abundant) and a limit ($N). We loop over all odd numbers, pushing any that pass the $is_abundant check to our result.

Although the above version suited my purposes better, it’s also possible to do this with an iterator, to avoid having to store the intermediate result:

sub odd_abundant_iterator(&) {
    my $is_abundant = shift;
    my $n = 1;
    
    sub {
        do { $n += 2 } until $is_abundant->($n);

        $n;
    }
}

Now that we have a framework for gathering abundant numbers, let’s try it a few different ways.

Brute force

The first way you might think to try is simply to brute force your way through every divisor of every number. This is O(n) for each number, and takes over a second to find the first 20 numbers:

sub n_abundant_naive {
    n_odd_abundant {
        my $n = shift;
        $n < sum grep { $n % $_ == 0 } 1..$n/2;
    } $_[0];
}

Using sqrt

Stopping at \(\sqrt{n} \) perhaps unsurprisingly brings the asymptotic time down to O(\(\sqrt{n} \)). Since divisors come in pairs, we can simply calculate the other divisor and avoid looping through most of the numbers:

sub n_abundant_sqrt {
    n_odd_abundant {
        my $n    = shift;
        my $sqrt = sqrt($n);

        my $sum  = sum map { $_,  $n / $_ } 
                      grep { $n % $_ == 0 } 1..$sqrt;

        $sum -= $sqrt if $sqrt == int $sqrt;
        
        2*$n < $sum;
    } $_[0];
}

It might not seem like a huge change, but the above code runs about 27 times faster than the naïve version, when asked to find the first 20 odd abundant numbers.

Using Math::Prime::Util divisor_sum

Just for fun, our old friend, Math::Prime::Util has a function that seems perfect for our needs: divisor_sum. It does what it says on the tin: it calculates the sum of the divisors of whatever number we give it.

sub n_abundant_mpu {
    n_odd_abundant {
        my $n = shift;
        my $sum = divisor_sum($n);

        2*$n < $sum;
    } $_[0];
}

This one is another 12 times faster than the sqrt solution, and 356 times faster than the naïve method. Great, right? Well, not so fast. Under the hood, the divisor_sum function is still finding all divisors for every number, so we’re still at \(O(n \sqrt{n}) \) time. It’s only faster because of the very tightly optimized C code under the hood.

Sieve

We can still do quite a bit better by realizing that since we’re looking for a whole bunch of abundant numbers, we’re repeating the same calculations over and over again, for every multiple of a number we’ve seen already. So as long as we’re willing to tweak the requirement slightly to find all abundant numbers below a given limit (although we’ll see how we can still accommodate the old calling syntax if we really want to), we can do much better:

sub n_abundant_sieve {
    my $lim = shift;
    my @r;

    my @div_sum; # Sum of divisors for each number
    for my $n (1..$lim) {
        $div_sum[$n*$_] += $n for 1..$lim/$n+1;
        push @r, $n if $n % 2 and 2*$n <= $div_sum[$n];
    }

    @r;
}

When called as n_odd_abundant_sieve(10000), this returns the first 23 odd abundant numbers, about twice as fast as the sqrt version returns 20. That’s because this algorithm runs in \(O(n \log{n}) \) time, which is strictly less than \(O(n \sqrt{n}) \) time for all \(n > 0 \). It grows significantly slower.

Sneaking up on the limit

In number theory circles, finding all numbers below a limit is usually just fine, but if we wanted to be a stickler for the parameters of the challenge and return only the first 20, we could simply call n_odd_abundant_sieve multiple times, doubling the limit every time, until we have at least 20 results. I wouldn’t bother, though.

You might be wondering, “how does this perform compared to the MPU version?”

For small values (the first 23 numbers in the sequence), the MPU version is faster by about 200%. However, when given more realistic limits, our better algorithm pulls ahead. It breaks even at about 100 numbers on my machine. By the time we ask for 150 numbers, the sieve is already 60% faster.

Benchmarks are an incredibly useful—and essential—tool when you need to write high performance code. Knowing that one algorithm is faster than another doesn’t always translate to real code. When dealing with implementation details, those hidden constants can really make a difference. What’s also really important, however, is knowing your requirements. If we’re only ever going to need to find less than 100 abundant primes, the MPU version might be a worthy choice. However, if we want to push it beyond that, then the better algorithm wins.

Of course, translating the sieve algorithm to tight C code would blow both of them out of the water at any value of n, since the sieve version grows strictly slower than the MPU version.

Raku

Here’s a quick port to Raku. If you’re only concerned with outputting the results, there’s no point in storing them first:

sub MAIN(Int $lim = 10000) {
    my @div_sum; # Sum of divisors for each number

    for 1...$lim -> $n {
        @div_sum[$n*$_] += $n for 1..$lim/$n+1;
        $n.say if $n % 2 and 2*$n <= @div_sum[$n];
    }

}

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #2 this week is no doubt about first class functions, but gets more specific, asking us to have a go at function composition:

Create sub compose($f, $g) which takes in two parameters $f and $g as subroutine refs and returns subroutine ref i.e. compose($f, $g)->($x) = $f->($g->($x))

Task #2

Before we get too far ahead of ourselves, let’s briefly review what these terms mean.

First Class Functions

A language that supports first class functions simply allows you to pass functions around like any other variable. Passing anonymous functions (also known as lambda functions) around is usually included in this definition as well. Perl makes this easy:

my $add2 = sub { $_[0] + 2 }; # Returns a sub that adds 2 to its argument
$sub->(5); # Returns 7

That example may not be the most compelling, but for some motivation, look no farther than Perl’s map or grep builtins. When you call something like map { $_ * $_ } 1..10 to get the first ten square numbers, that block { $_ * $_ } is an anonymous subroutine.

First class functions are incredibly useful, and deserve more discussion than I can cram into this blog post, so perhaps I’ll do them justice with a longer dedicated post in the future.

Function Composition

Function composition is a distinct concept in mathematics. In computer science, it depends on first class functions, but is otherwise not related. Function composition, often denoted with the ∘ operator, takes two functions f and g and produces a new function:

\(h = g \circ f \text{ such that } h(x) = g(f(x)) \)

The reason it’s usually written as g ∘ f is that in plain English, g follows f, because we are feeding the output of f into g. Of course, f and g are just symbols, so they can be swapped to match the task description with no issues:

\(h = f \circ g \text{ such that } h(x) = f(g(x)) \)

Perl

Now that we’ve gotten all of the pesky definitions out of the way, the code is … well, there’s hardly any code at all, really. Here’s a function that generates the composition h = f ∘ g:

sub comp {
    my ($f, $g) = @_;

    sub { $f->($g->(@_)) }
}

Here’s an example usage that calculates the sum of squares of a list of numbers:

use List::Util qw< sum0 >;

my $squares = sub { map { $_ * $_ } @_ };
my $h = comp( \&sum0, $squares );

say "The sum of squares for 1..10 = " . $h->(1..10); # 385

I chose to use List::Util‘s sum0 function so I could demonstrate how to pass in a reference to a named function. The $squares function shows how to use a variable. I could have also done this as an anonymous function:

my $h = comp( \&sum0, sub { $_ * $_ } @_ } );

Raku

Raku’s first class function support is very good. In fact, the language was designed around higher order features like this, so there are some built-in helpers we can use, such as the composition operator. That’s right, we can use ∘ or o right in our code to do the function composition. I could stick this into a comp function like I did with the Perl example, but that seems less expressive to me.

my &sum    = sub {    [+] @_ };
my &square = sub { @_ »*« @_ };

my &h = &sum ∘ □

say &h(1..10); # 385

First class functions open up endless possibilities in your code.

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #1 this week asks us to find the first 13 Perrin primes. “What’s a Perrin prime,” I can’t hear you asking? To answer that, we first have to look at the Perrin sequence, as described in OEIS A001608. It’s easy to generate:

Starting with [3, 0, 2], each new term is determined by adding the 2^nd and 3^rd last terms. So, the 4^th number is 3 + 0 = 3, giving us [3, 0, 2, 3]. The 5^th number is 0 + 2 = 2, and so on.

Perrin primes are simply the elements of the Perrin sequence that also happen to be prime.

Normally (and per the example output in the task) we are to find the unique Perrin primes, in order. So, we’ll just seed the first prime in our @r results, and then rely on the fact that the sequence is strictly increasing after the first five terms.

Building the sequence is very simple from there:

This post is part of a series on Mohammad Anwar’s excellent Weekly Challenge, where hackers submit solutions in Perl, Raku, or any other language, to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)

Task #2 this week asks us to calculate a so called Home prime. Home primes are found by factoring a number and concatenating the prime factors (including powers, so 20 = 5×2×2), and repeating this until the result is a prime number.

The given example, HP(10) can be found via the following steps: HP(10) = HP(25) = HP(55) = HP(511) = HP(773), and we stop, since 773 is a prime number.

This is a natural problem for recursion:

sub home_prime_recursive {
    my @fac = factor($_[0]);

    @fac == 1 ? $_[0] : home_prime(join '', @fac);
}

I like this solution for its expressiveness, but it’s about 20% slower than the following iterative version: