# Fun with Perl 6 using Roman Numerals

I recently took up solving some of the Perl weekly challenges, the most recent being the Roman Numeral encoder/decoder. In this post I’m going to go through how I put my solution together and how it works.

**Defining The Numerals**

To start off with, I declared a few constants. The first two being the fundamental set of characters for our numerals:

`constant @letters = ｢IVXLCDM｣.comb;`

constant @overlines = "\c[combining overline]", "\c[combining double overline]"; # These represent x 1000 and x 1000000

constant @letter-pairs = reverse # … the following code

These constants are then used to give us a list of all of our numerals:

`@letters[0], |( @letters, |@overlines.map( @letters X~ * ) ).map({`

gather {

for .rotor(3 => -1) -> @group {

for 1, 2 {

take @group[0, $_].join;

take @group[$_];

}

}

}

}).flat;

We start of by taking our `I`

from the list. Following that, we create a list of three groups of letters:

`@letters, |@overlines.map( @letters X~ * )`

# (I V X L C D M)(I̅ V̅ X̅ L̅ C̅ D̅ M̅)(I̿ V̿ X̿ L̿ C̿ D̿ M̿)

We split each of these into overlapping groups of three using `rotor`

, and then use the elements from each of those lists to `take`

our numerals:

`.rotor(3 => -1)`

# ((I V X) (X L C) (C D M))

# .[0,1], .[1], .[0,2], .[2] on each group gives us IV, V, IX, X etc

Now we want to assign all of these numerals their appropriate values. We do that using the following sequence:

`1, |( * X* 4, 5, 9, 10 ) … ∞`

What this sequence does is take the 1 and gives us that multiplied by 4, 5, 9 and 10. With the 10 now on the end of the list, the sequence is repeated, giving us 40, 50, 90 and 100. This continues ad infinitum for as many numbers as we need.

Taking our list of numerals and zipping ( `z=>`

) them with the numbers gives us the list of pairs we’ll be using:

`:I(1), :IV(4), :V(5), :IX(9), :X(10), :XL(40) …`

# Converting Numbers to Roman Numerals

`given %(@letter-pairs.Map.antipairs) -> %letter-map {`

return [~] gather {

for $number.flip.comb.pairs.reverse {

given 10 ** .key -> $key {

when .value == 4 | 9 {

take %letter-map{ $key * .value };

}

take %letter-map{ $key * 5 } if .value ≥ 5;

take %letter-map{ $key } x .value % 5;

}

}

};

}

We start off with taking the list of pairs we defined earlier, and turning it into a hash with the keys being the Arabic number each Roman numeral represents. We then take the input (being `$number`

here) and transform it with:

`12345.flip.comb.pairs.reverse`

# 4 => 1, 3 => 2, 2 => 3, 1 => 4, 0 => 5

This gives us a place value for each digit we’re going to use, `10 ** .key`

giving us 10000, 1000, 100, etc.

We’re then going to `take`

each Roman numeral based on what we got:

`when .value == 4 | 9 {`

take %letter-map{ $key * .value };

}

take %letter-map{ $key * 5 } if .value ≥ 5;

take %letter-map{ $key } x .value % 5;

}

With the `when`

block, if the condition matches, control is returned to the parent block and any subsequent code is skipped. This would happen with our `1 => 4`

In `$key`

we would have `10`

. We look in the hash for the character assigned to `10 × 4`

, which is `40 => "XL"`

, and take that to be used in our eventual string. If the value is not 4 or 9, then we instead take the character representing our 5s, if we need it, and then a string of 1s, between 0 or 3 depending on what our value is.

12345 gives us X̅MMCCCXLV.

# Converting Roman Numerals to Numbers

`return [+] gather {`

my $str = $roman-string.uc;

for @letter-pairs -> $pair {

if $str ~~ / ^ ( $($pair.key) )+ / {

take ($pair.value xx $0).Slip;

$str.=substr($0.join.chars);

}

}

}

Our letter pairs are sorted from highest to lowest, as that’s typically how they’re given in Roman numeral format. Starting with our largest numeral, we take our string and attempt to match the start of it with a regex. The match is then stored in `$0`

.

If we did get a match, we take the value of that numeral, multiplied by how many we found e.g. if the numeral has XXX we take 10 three times. We then take the number of characters in that match and cut them off our string for our next iteration.

While there is some error checking (we throw an error if the string isn’t empty by the end), the routine is not very strict, and will still give a result if you enter a series of characters like XXXXX. It will however take the resulting value, convert it back to Roman to compare, and ask if you meant L 😉