Author: Ryan Thompson

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 challenges are very quick, so I’m putting them in one blog post. The title is a bit of a play on words, given the tail recursion in part 2.

Full solutions are available at GitHub.

Task 1 › Bitwise Sum

For the first task, we’re asked to first bitwise AND (&) each unique pair of numbers, and then compute the sum of those operations. So, given an input of (a, b, c), we would compute:

a & b + a & c + b & c

To formalize this a bit, in preparation for writing the code, consider a pair (n_i, n_j) where i and j are the indices in the input array. Unique pairs could be interpreted to mean j ≥ i. However, the provided example clearly implies that self pairs are to be avoided, so we’ll include pairs where j > i. So for an array of length k (n₁ … n_k), this looks like:

\({ \displaystyle \sum \limits_{i=1}^{k} } \sum_{j=i+1}^{k} n_i\ \&\ n_j
\)

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.)

Personal note: I’ve had to take a break from participating in the PWC, but I’m back for this week, at least. Hopefully I’ll be able to contribute more again.

The first task this week is a sort of simple phone number validation, based on provided templates. Numbers must match the following, where n is any decimal digit:

+nn  nnnnnnnnnn
(nn) nnnnnnnnnn
nnnn nnnnnnnnnn

Based on the provided sample output, it seems clear that leading and trailing whitespace are ignored. Internal whitespace is also compressed, as the first provided template has two spaces after +nn, yet the phone number +44 1148820341 is supposed to match.

Let’s try two different methods of matching, with Perl and Raku.

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.)

The second task this week is simple: given a (simplified) comma-separated-value (CSV) file, transpose its rows and columns. For example:

\(
\begin{bmatrix}
A & B & C & D \\
1 & 2 & 3 & 4 \\
w & x & y & z
\end{bmatrix}^T
\Rightarrow
\begin{bmatrix}
A & 1 & w \\
B & 2 & x \\
C & 3 & y \\
D & 4 & z
\end{bmatrix}
\)

The challenge task does not actually refer to the input as CSV, so I’m using that term loosely, with simplified parsing to match the input specification. If more compliant parsing is needed, one could use the usual Text::CSV module in Perl, or ports in Raku.

There are a couple of ways to transpose files, which I’ll explore in Raku and Perl.

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 a simple tree traversal:

You are given a binary tree and a sum, write a script to find if the tree has a path such that adding up all the values along the path equals the given sum. Only complete paths (from root to leaf node) may be considered for a sum.

For both my Perl and Raku versions, I’m going super-lean with the implementation, using only array references. The “node,” which recursively defines an entire (sub)tree, looks like this:

• Element 0: Node’s value
• Elements 1..N: References to child nodes

Thus, the (Raku) syntax my @tree = [10, [18, [5], [2]], [8, [16, [18]], [9]]] describes a tree that looks like this:

              10
            /    \
           18     8
          / \    / \
         5   2  16  9
               /
              18

If we look for a path sum of 30, there is precisely one path with that sum: 10 → 18 → 2.

It’s worth noting that, although the task is limited to binary trees, my implementation will handle m-ary trees. Forcing it to handle only binary trees would actually be slightly more difficult, and a lot less useful.

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 is to implement the Diff-K algorithm, as explained by Mohammad:

You are given an array @N of positive integers (sorted) and another non negative integer k. Write a script to find if there exists 2 indices i and j such that A[i] – A[j] = k and i != j.

This one is pretty easy. We can boil down the solution into two operations for each element ($_) of @N:

• First, filter @N for elements where $k+$_ exists in @N.
• For the remaining elements, return an array containing the indexes of $k+$_ and $_.

To make this easier and more efficient, we’ll create an %idx_of hash to store the index of each element in @N. This can be created in linear time, and gives us O(1) lookups for both operations, above.

I really like how easy it is to create a reverse hash like this in Raku:

my %idx_of = @N.antipairs;

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 was one of my devising:

It is thought that the following sequence will always reach 1:

For example, if we start at 23, we get the following sequence:

23 → 70 → 35 → 106 → 53 → 160 → 80 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

Write a function that finds the Collatz sequence for any positive integer. Notice how the sequence itself may go far above the original starting number.

Extra Credit

Have your script calculate the sequence length for all starting numbers up to 1000000 (1e6), and output the starting number and sequence length for the longest 20 sequences.

I’ve always liked the Collatz conjecture. It is simple enough for schoolchildren to play with, yet the math to prove the conjecture is still beyond our greatest mathematicians.

Here is how I solved this task.