Tag: 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 has us build an LRU cache, which is a cache of fixed capacity, whereby when it is full but a new item needs to be added, the item that was least-recently accessed is evicted from the cache.

For example, if a cache of capacity = 3 contains [a b c] (where a was accessed most recently), elements are evicted off the right side of the list. Adding element d, the list would contain [d a b], for example.

Of course, [d a b] are just the keys of the items in the cache. We’ll take an arbitrary scalar (which could itself be a reference) as a value.

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 as follows:

Write a script to accept a positive number as command line argument and print the smallest multiple of the given number consists of digits 0 and 1.

Brute Force

It’s easy enough to brute force this problem, by checking every multiple in order:

$_ += $_[0] while /[^10]/;

$_ now contains the answer. Unfortunately, some numbers—particularly multiples of 9—require extremely large multiples. For example, 99 × 1122334455667789 = 111111111111111111 takes several minutes to discover via this method.

We can do better, though.

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 described as follows:

There are 50 people standing in a circle in position 1 to 50. The person standing at position 1 has a sword. He kills the next person i.e. standing at position 2 and pass on the sword to the immediate next i.e. person standing at position 3. Now the person at position 3 does the same and it goes on until only one survives.

Write a script to find out the survivor.

This is a well known theoretical exercise in computer science, by the name of the Josepheus problem, based on the story of Flavius Josephus, a Jewish historian in the first century.

Algorithm

To me, the most natural way to solve this problem is with a circular linked list. As a very quick review, a linked list is a list of items where each item also contains a pointer, reference, index, whatever to the next item in the list. A single element is usually drawn like this:

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 (and I quote): print all Palindrome Dates between 2000 and 2999. The format of date is mmddyyyy. For example, the first one was on October 2, 2001 as it is represented as 10022001.

It’s pretty easy to avoid using any sort of date library with a couple of key observations about the problem domain. First, we’ll split it up into month (mm), day (dd), century (cc), and 2-digit year (yy). Thus our “baseline” date format is mm dd cc yy.

If we use R[xx] as shorthand for “string reverse of xx“, we can rewrite the date in a couple of different ways:

  mm    dd    cc    yy     # Original
R[yy] R[cc]   cc    yy     # Start with year
  mm    dd  R[dd] R[mm]    # Start with mm/dd
R[yy]   dd  R[dd]   yy     # Start with yy and dd

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 has us print the first 20 “gapful numbers,” as described by OEIS sequence A108343. Gapful numbers are numbers greater than 99 that are evenly divisible by their first and last digits combined. For example, 132 is a gapful number because 132 ÷ 12 = 11.

This is certainly the easier of the two tasks, as both Perl and Raku have convenient ways to index and concatenate string fragments.

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.)

Challenge #2 this week asks for “a script that dumps its own source code”. This is almost a quine, although Mohammad did not name it as such. Specifically, we are given the constraint that perl ch-2.pl | diff - ch-2.pl must return nothing.

What’s a Quine?

A quine, otherwise known as a self-replicating program, is a program that accepts no input, and produces a copy of its source code as its only output.

This is a stronger definition than what Mohammad has asked for this week: specifically, Mohammad did not add any restrictions on input. So, programs that simply read their own source file would be acceptable this week. Since I felt that was too easy, I have decided to go with the stronger definition of an actual quine, not allowing any input. Still, I’ll include a few solutions to show the various options:

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.)

Happy new year! We are on Week 41, and this is Challenge #2.

The Leonardo Numbers (A001595) are a simple recursively defined sequence:

The sequence starts: 1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, …

You’ll note this sequence is very similar to the well-known Fibonacci sequence, which differs only in that the Fibonacci sequence does not have the + 1 term, and starts at F(0) = 0.