RabbitFarm

The examples used here are from The Weekly Challenge problem statement and demonstrate the working solution.

Part 1

You are given two decimal-coded binary numbers, $a and $b. Write a script to simulate the addition of the given binary numbers.

Solution

use strict;
use warnings;
sub add_binary{
    my($x, $y) = @_;
    my $sum = ""; 
    my @a = reverse(split(//, $x));            
    my @b = reverse(split(//, $y));            
    if(@b > @a){
        my @c = @b;
        @b = @a;
        @a = @c;   
    } 
    my $carry = 0; 
    for(my $d = 0; $d <= @a - 1; $d++){ 
        my $d0 = $a[$d]; 
        my $d1 = $b[$d];
        if($d1){
            $sum = "0$sum", $carry = 0 if $d0 == 1 && $d1 == 1 && $carry == 1;  
            $sum = "1$sum", $carry = 0 if $d0 == 1 && $d1 == 0 && $carry == 0; 
            $sum = "0$sum", $carry = 1 if $d0 == 1 && $d1 == 1 && $carry == 0; 
            $sum = "0$sum", $carry = 1 if $d0 == 0 && $d1 == 1 && $carry == 1; 
            $sum = "0$sum", $carry = 0 if $d0 == 0 && $d1 == 0 && $carry == 0; 
            $sum = "1$sum", $carry = 0 if $d0 == 0 && $d1 == 0 && $carry == 1; 
            $sum = "0$sum", $carry = 1 if $d0 == 1 && $d1 == 0 && $carry == 1; 
            $sum = "1$sum", $carry = 0 if $d0 == 0 && $d1 == 1 && $carry == 0; 
        } 
        else{
            $sum = "0$sum", $carry = 1, next if $d0 == 1 && $carry == 1;  
            $sum = "1$sum", $carry = 0, next if $d0 == 0 && $carry == 1;  
            $sum = "0$sum", $carry = 0, next if $d0 == 0 && $carry == 0;  
            $sum = "1$sum", $carry = 0, next if $d0 == 1 && $carry == 0;  
        }  
    } 
    $sum = "$carry$sum" if $carry == 1;  
    return $sum; 
}

MAIN:{
    print add_binary(11, 1) . "\n"; 
    print add_binary(101, 1) . "\n"; 
    print add_binary(100, 11) . "\n"; 
}

Sample Run

$ perl perl/ch-1.pl
100
110
111

Notes

I have an unusual fondness for Perl's right hand conditional. But that is pretty obvious from the way I wrote this, right?

Part 2

You are given 3 positive integers, $i, $j and $k. Write a script to print the $kth element in the sorted multiplication table of $i and $j.

Solution

use strict;
use warnings;
sub nth_from_table{
    my($i, $j, $k) = @_;
    my @table;
    for my $x (1 .. $i){
        for my $y (1 .. $j){
            push @table, $x * $y; 
        }  
    }  
    return (sort {$a <=> $b} @table)[$k - 1];   
} 

MAIN:{
    print nth_from_table(2, 3, 4) . "\n";  
    print nth_from_table(3, 3, 6) . "\n";  
} 

Sample Run

$ perl perl/ch-2.pl 
3
4

Notes

Full Disclosure: At first I wanted to do this in some convoluted way for fun. After experimenting with, like, nested maps for a few minutes I lost all interest in "fun" and just went with a couple of for loops!

References

Challenge 140