TL;DR

Here we are with TASK #1 from the Perl Weekly Challenge #121. Enjoy!

# The challenge

You are given integers 0 <= $m <= 255 and 1 <= $n <= 8.

Write a script to invert $n bit from the end of the binary representation of $m and print the decimal representation of the new binary number.

Example

Input: $m = 12,$n = 3
Output: 8

Binary representation of $m = 00001100 Invert 3rd bit from the end = 00001000 Decimal equivalent of 00001000 = 8 Input$m = 18, $n = 4 Output: 26 Binary representation of$m = 00010010
Invert 4th bit from the end = 00011010
Decimal equivalent of 00011010 = 26


# The questions/assumptions

From the examples, it seems that from the end of the binary representation means that we start from the least significant bit and move towards the most significant bit. This will be our assumption.

It’s interesting that input $n starts from 1, meaning first. As an amateur programmer, I’m used to seeing the first item to get an index of 0, but of course it’s just a matter of knowing what is what. # The solution This bit munching is becoming a trend! To address this challenge we can put to good use an old friend, i.e. the exclusive or (XOR) operator. In a nutshell, it takes two input bits, and tells us if they differ (giving out a 1) or are equal (giving out a 0). This is the so-called truth table for this operator:  A | 0 | 1 | ---+---+---+ 0 | 0 | 1 | B ---+---+---+ 1 | 1 | 0 | ---+---+---+  Now, let’s consider the effect of… • fixing B to 0: we see that the output is the same as A; • fixing B to 1: we see that the output is the inverse of A. This means that we can apply our XOR operator to a whole bunch of bits, making sure that the value we XOR our input with has 0 in the positions of bits that we want to preserve, and 1 where we want to invert to bit. So, for example, if we want to invert the third bit of the input value 12, we can apply XOR with a value that has all 0 except in the third bit, where we put 1:  12 -----> 00001100 XOR 00000100 -------- 8 <----- 00001000  Now we are left with the task of finding the right mask to apply the XOR operator. As we are required to invert one single bit, it will be like the example above, i.e. all 0 except for 1 in one single position. This can be done with the help of another bitwise operator, i.e. the shift left operator (let’s call it SHL). It takes two numbers, the first is the starting value, the second is the number of positions that its bits have to be shifted to the left (appending 0 in the least significant positions). So, for example, if we start from 1 (i.e. 00000001) and shift it 2 times to the left we obtain 4 (i.e. 00000100). If you’re wondering… yes, shifting to the left is the same as multiplying by powers of 2. Now we have all ingredients for our bit inversion recipe: $m XOR (1 SHL ($n - 1))  We have to use $n - 1 because our starting point (i.e. 1, a.k.a. 00000001) already has 1 in the first position, so it needs no shifting.

Let’s get to the code then, Raku first:

#!/usr/bin/env raku
use v6;
subset IntByte of Int where 0 <= * <= 255;
subset BitNum  of Int where 1 <= * <= 8;
sub invert-bit (IntByte:D $m, BitNum:D$n) { $m +^ (1 +< ($n - 1)) }

put "m=12 n=3 -> " ~ invert-bit(12, 3);
put "m=18 n=4 -> " ~ invert-bit(18, 4);


The XOR operator is +^ and the shift operator is +<, so the solution is the direct translation of our generic recipe above.

Perl now, with its own operator XOR (^) and SHL (<<):

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';

sub invert_bit ($m,$n) { $m ^ (1 << --$n) }
say "m=12 n=3 -> " . invert_bit(12, 3);
say "m=18 n=4 -> " . invert_bit(18, 4);


As we have seen a few times in the past, we lose the simplicity to have input parameters validation in a declarative way. That’s life.

On the other hand. though, input parameters in Perl’s (still experimental) signatures are copies of the original values, so we can pre-decrement \$n instead of subtracting 1 and spare a whopping 4 characters!!!

Well… enough is enough, stay safe everyone!