Here we are with TASK #1 from the Perl Weekly Challenge #118. Enjoy!

The challenge

You are given a positive integer $N.

Write a script to find out if the binary representation of the given integer is Palindrome. Print 1 if it is otherwise 0.


Input: $N = 5
Output: 1 as binary representation of 5 is 101 which is Palindrome.

Input: $N = 4
Output: 0 as binary representation of 4 is 100 which is NOT Palindrome.

The questions

One question is about the exact representation of the integer as a decimal number. OK, usually any leading 0 is ignored, but in base-2 this means ruling out all even numbers in a single stroke! Whatever, the example seems explicit to this regard.

The solution

Raku first, in the spirit of taking the challenge in the new language and try to learn something:

#!/usr/bin/env raku
use v6;
sub binary-palindrome (Int:D $N where * > 0 --> Bool) {
   return False if $N %% 2;
   my ($M, $n) = (0, $N);
   ($M, $n) = (($M +< 1) +| ($n +& 1), $n +> 1) while $n > 0;
   return so $M == $N;
sub MAIN (*@args is copy) {
   @args = 1 .. 31;
   put $_, ' -> ', binary-palindrome($_) ?? 1 !! 0 for @args;

The approach chosen is to stick to bitwise operations instead of turning the integers into a string representation and then check for palindromeness.

So what we’re doing here is to reverse the input number $N and check whether the result (which is $M, by the way) is the same as $N.

One tricky part here was using the correct operators. Coming from Perl, binary operations are | and & and the bitshifts are << and >>.

The story in Raku is different though, so we have respectively +|, +&, +<, and +>. The error messages were great to this regard.

Porting the solution to Perl is easy:

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

sub binary_palindrome ($N) {
   die "invalid $N (positive integers are OK)\n"
      if $N !~ m{\A [1-9]\d* \z}mxs;
   return unless $N % 2;
   my ($M, $n) = (0, $N);
   ($M, $n) = (($M << 1) | ($n & 1), $n >> 1) while $n > 0;
   return $M == $N;

my @args = @ARGV ? @ARGV : 1 .. 31;
say $_, ' -> ', binary_palindrome($_) ? 1 : 0 for @args;

Well, that’s on the brink of plagiarism!

I decided to put a validation for the input (which I usually avoid) just to make the two solutions more aligned (Raku signatures allow setting validations so it’s almost a sin to not put them).

Then, of course, there’s no is-divisible-by operator %% in Perl, so we have to resort to inverting the result of the rest-by operator %:

# Raku: is $N divisible by 2?
return False if $N %% 2;

# Perl: does integer division of $N by 2 have no rest?
return unless $N % 2;

No big deal, but it’s good to be able and be precise in Raku.

Well.. enough for today, have fun and stay safe!