# RabbitFarm

### 2022-10-30

#### Pairs Divided by Zero

*The examples used here are from the weekly challenge problem statement and demonstrate
the working solution.*

## Part 1

*You are given list of integers @list of size $n and divisor $k. Write a script to
find out count of pairs in the given list that satisfies a set of rules.*

### Solution

```
use v5.36;
use strict;
use warnings;
sub divisible_pairs{
my($numbers, $k) = @_;
my @pairs;
for my $i (0 .. @{$numbers} - 1){
for my $j ($i + 1 .. @{$numbers} - 1){
push @pairs, [$i, $j] if(($numbers->[$i] + $numbers->[$j]) % $k == 0);
}
}
return @pairs;
}
MAIN:{
my @pairs;
@pairs = divisible_pairs([4, 5, 1, 6], 2);
print @pairs . "\n";
@pairs = divisible_pairs([1, 2, 3, 4], 2);
print @pairs . "\n";
@pairs = divisible_pairs([1, 3, 4, 5], 3);
print @pairs . "\n";
@pairs = divisible_pairs([5, 1, 2, 3], 4);
print @pairs . "\n";
@pairs = divisible_pairs([7, 2, 4, 5], 4);
print @pairs . "\n";
}
```

### Sample Run

```
$ perl perl/ch-1.pl
2
2
2
2
1
```

### Notes

The rules, if not clear from the above code are : the pair (i, j) is eligible if and only if

0 <= i < j < len(list)

list[i] + list[j] is divisible by k

While certainly possible to develop a more complicated looking solution using `map`

and
`grep`

I found myself going with nested `for`

loops. The construction of the loop indices
takes care of the first condition and the second is straightforward.

## Part 2

*You are given two positive integers $x and $y. Write a script to find out the number of
operations needed to make both ZERO.*

### Solution

```
use v5.36;
use strict;
use warnings;
sub count_zero{
my($x, $y) = @_;
my $count = 0;
{
my $x_original = $x;
$x = $x - $y if $x >= $y;
$y = $y - $x_original if $y >= $x_original;
$count++;
redo unless $x == 0 && $y == 0;
}
return $count;
}
MAIN:{
say count_zero(5, 4);
say count_zero(4, 6);
say count_zero(2, 5);
say count_zero(3, 1);
say count_zero(7, 4);
}
```

### Sample Run

```
$ perl perl/ch-2.pl
5
3
4
3
5
```

### Notes

The operations are dictated by these rules:

`$x = $x - $y if $x >= $y`

or

`$y = $y - $x if $y >= $x (using the original value of $x)`

This problem seemed somewhat confusingly stated at first. I had to work through the first given example by hand to make sure I really understood what was going on.

After a little analysis I realized this is not as confusing as I first thought. The main
problem I ran into was not properly accounting for the changed value of `$x`

using a
temporary variable `$x_original`

. If you see my
Prolog Solutions for this
problem you can see how Prolog's immutable variables obviate this issue!

## References

posted at: 19:24 by: Adam Russell | path: /perl | permanent link to this entry