TL;DR

Here we are with TASK #1 from The Weekly Challenge #192. Enjoy!

# The challenge

You are given a positive integer, $n. Write a script to find the binary flip. Example 1 Input:$n = 5
Output: 2

First find the binary equivalent of the given integer, 101.
Then flip the binary digits 0 -> 1 and 1 -> 0 and we get 010.
So Binary 010 => Decimal 2.


Example 2

Input: $n = 4 Output: 3 Decimal 4 = Binary 100 Flip 0 -> 1 and 1 -> 0, we get 011. Binary 011 = Decimal 3  Example 3 Input:$n = 6
Output: 1

Decimal 6 = Binary 110
Flip 0 -> 1 and 1 -> 0, we get 001.
Binary 001 = Decimal 1


# The questions

Well well… not much of a definition, but an operative algorithm in the first example explains it pretty well.

We’re assuming that the positive integer $n fits in a variable, whatever the language and the system. # The solution The result will always be (strictly) lower than the input. Why? The leftmost 1 bit in the input is turned onto a 0, so the resulting number is lower. I know that there must be a clever way of doing it, especially in Raku. I’ll stick to the basics, though. #!/usr/bin/env raku use v6; sub MAIN ($n = 5) { put binary-flip($n) } sub binary-flip (Int$n is copy where * > 0) {
my $mask = 0x01; my$result = 0;
while $n {$result +|= $mask unless$n +& 1;
$n +>= 1;$mask +<= 1;
}
return $result; }  We’re getting bits out from the rightmost part and arranging them in with an always-doubling mask. Just plain bit handling. The translation into Perl if pretty straightforward, with less input checks and slightly different operators: #!/usr/bin/env perl use v5.24; use warnings; use experimental 'signatures'; no warnings 'experimental::signatures'; say binary_flip(shift // 5); sub binary_flip ($n) {
my $mask = 0x01; my$result = 0;
while ($n) {$result |= $mask unless$n & 0x01;
$n >>= 1;$mask <<= 1;
}
return \$result;
}


I guess this is everything, stay safe!