HEADLINES

Today is the first Monday of the June 2020 and you guessed it correctly. It is time to announce the Champion of the Month. With great pleasure, we announce the Champion of last month, Shahed Nooshmand, currently ranked #042 with score of 62. He joined the Team PWC in the Week #053 and has never missed a week since. As of today, he has contributed Perl (1), Raku (20) and Blog (10).

Perl Weekly Challenge - 053: Blog (1), Raku (2)
Perl Weekly Challenge - 054: Blog (1), Raku (2), Perl (1)
Perl Weekly Challenge - 055: Blog (1), Raku (2)
Perl Weekly Challenge - 056: Blog (1), Raku (2)
Perl Weekly Challenge - 057: Blog (1), Raku (2)
Perl Weekly Challenge - 058: Blog (1), Raku (2)
Perl Weekly Challenge - 059: Blog (1), Raku (2)
Perl Weekly Challenge - 060: Blog (1), Raku (2)
Perl Weekly Challenge - 061: Blog (1), Raku (2)
Perl Weekly Challenge - 062: Blog (1), Raku (2)

The Week #062 would get the honour of least (50) contributions week since the launch. Prior to that, it was the Week #006 when we had 52 contributions by the Team PWC. I have a feeling that the “N Queens” task was too big to fit in one-week schedule. Our aim is always to have a task that can be solved in no more than 2 hours as this is what I expect Team PWC members to spend on each task. If it breaches the deadline of 2 hours then it is not fit for the weekly challenge. We will keep in mind next time. I blame myself for this as I also couldn’t solve the task as I didn’t know how 3D Chess Game works. To be honest with you, I still don’t know how it works. I am still looking for better visualisation of 3D Chess Game.

I was pleasantly surprised when Javier Luque shared codepen for the 3D N Queens task. It was easier to follow but I haven’t had chance to go through yet. I will re-visit sometimes next week when I have free moment. Well, Javier didn’t stop there, he next shared codepen for the 2D N Queens version of the same. Please do check, if you still fancy to peak into the finer details. Thanks Javier for making an extra effort for this and sharing with us.

Last but not the least, I would like to thank each and every member for their support and encouragement.

RECAP

Quick recap of the “Perl Weekly Challenge - 062” by Mohammad S Anwar.

PERL REVIEW

If you missed any past reviews then please checkout the collection.

RAKU REVIEW

If you missed any past reviews then please checkout the collection.

CHART

Please take a look at the charts showing interesting data.

I would like to thank every member of the team for their valuable suggestions. Please do share your experience with us.

NEW MEMBERS

Please find out How to contribute?, if you have any doubts.

Please give it a try to an excellent tool EZPWC created by respected member Saif Ahmed of Team PWC.

GUESTS

Richard Park shared solution to Task #1 in APL.

Sangeet Kar shared solutions to Task #1 and Task #2 in Python.

Please find out past solutions by respected guests. Please do share your creative solutions in other languages.

TASK #1 › Last Word

Submitted by: Mohammad S Anwar

Reviewed by: Ryan Thompson

Define sub last_word($string, $regexp) that returns the last word matching $regexp found in the given string, or undef if the string does not contain a word matching $regexp.

For this challenge, a “word” is defined as any character sequence consisting of non-whitespace characters (\S) only. That means punctuation and other symbols are part of the word.

The $regexp is a regular expression. Take care that the regexp can only match individual words! See the Examples for one way this can break if you are not careful.

Examples

last_word('  hello world',                qr/[ea]l/);      # 'hello'
last_word("Don't match too much, Chet!",  qr/ch.t/i);      # 'Chet!'
last_word("spaces in regexp won't match", qr/in re/);      #  undef
last_word( join(' ', 1..1e6),             qr/^(3.*?){3}/); # '399933'

TASK #2 › Rotate String

Submitted by: Mohammad S Anwar

Reviewed by: Ryan Thompson

Given a word made up of an arbitrary number of x and y characters, that word can be rotated as follows: For the ith rotation (starting at i = 1), i % length(word) characters are moved from the front of the string to the end. Thus, for the string xyxx, the initial (i = 1) % 4 = 1 character (x) is moved to the end, forming yxxx. On the second rotation, (i = 2) % 4 = 2 characters (yx) are moved to the end, forming xxyx, and so on. See below for a complete example.

Your task is to write a function that takes a string of xs and ys and returns the minimum non-zero number of rotations required to obtain the original string. You may show the individual rotations if you wish, but that is not required.

Example

Input: $word = 'xyxx';

• Rotation 1: you get yxxx by moving x to the end.
• Rotation 2: you get xxyx by moving yx to the end.
• Rotation 3: you get xxxy by moving xxy to the end.
• Rotation 4: you get xxxy by moving nothing as 4 % length(xyxx) == 0.
• Rotation 5: you get xxyx by moving x to the end.
• Rotation 6: you get yxxx by moving xx to the end.
• Rotation 7: you get xyxx by moving yxx to the end which is same as the given word.

Output: 7

Last date to submit the solution 23:59 (UK Time) Sunday 7th June 2020.