TL;DR

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

# The challenge

You are given a positive integer $N. Write a script to find out if the binary representation of the given integer is Palindrome. Print 1 if it is otherwise 0. Example Input:$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!

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