Since I'm releasing a book on practical lenses and optics later this month I thought it would be fun to do a few of this year's Advent of Code puzzles using as many obscure optics features as possible!

To be clear, the goal is to be obscure, strange and excessive towards the goal of using as many optics as possible in a given solution, even if it's awkward, silly, or just plain overkill. These are NOT idiomatic Haskell solutions, nor are they intended to be. Maybe we'll both learn something along the way. Let's have some fun!

You can find today's puzzle here.

Hey folks! Today's is a nice clean one! The goal is to find all the numbers within a given range which pass a series of predicates! The conditions each number has to match include:

Should be within the range; my range is 307237-769058

Should be six digits long; my range includes only 6 digit numbers, so we're all set here

Two adjacent digits in the number should be the same (e.g. '223456')

The digits should be in monotonically increasing order (e.g. increase or stay the same from left to right)

And that's it!

In normal Haskell we'd make a list of all possibilities, then either chain a series of filter statements or use do-notation with guards to narrow it down. Luckily, folds have filters too!

First things first, since our checks have us analyzing the actual discrete digits we'll convert our Int to a String so we can talk about them as characters:

= ([ 307237 .. 769058 ] :: [ Int ]) main([]) & toListOf (traversed . re _Show) toListOf (traversedre _Show) & print >>> main main [ "307237" , "307238" , "307239" , "307240" , "307241" , "307242" , "307243" , "307244" , "307245" , "307246" , ... ]

_Show is the same prism we've used for parsing in the previous examples, but re flips it in reverse and generates a Getter which calls show ! This is equivalent to to show , but will get you an extra 20 optics points...

Now let's start adding filters! We'll start by checking that the digits are all ascending. I could write some convoluted fold which does this, but the quick and dirty way is simply to sort the digits lexicographically and see if the ordering changed at all:

main :: IO () () = ([ 307237 .. 769058 ] :: [ Int ]) main([]) & toListOf (traversed . re _Show toListOf (traversedre _Show . filtered (\s -> s == sort s) filtered (\ss) ) & print >>> main main [ "333333" , "333334" , "333335" , "333336" , "333337" , "333338" , ... ]

filtered removes any focuses from the fold which don't match the predicate.

We can already see this filters out a ton of possibilities. Not done yet though; we need to ensure there's at least one double consecutive digit. I'll reach for my favourite hammer: lens-regex-pcre :

main :: IO () () = ([ 307237 .. 769058 ] :: [ Int ]) main([]) & toListOf (traversed . re _Show toListOf (traversedre _Show . filtered (\s -> s == sort s) filtered (\ss) . filteredBy (packed . [regex|(\d)\1+|]) filteredBy (packed[regex|(\d)\1+|]) ) >>> main main [ "333333" , "333334" , "333335" , "333336" , "333337" , "333338" , ... ]

Unfortunately we don't really see much difference in the first few options, but trust me, it did something. Let's see how it works:

I'm using filteredBy here instead of filtered , filteredBy is brand new in lens >= 4.18 , so make sure you've got the latest version if you want to try this out. It's like filtered , but takes a Fold instead of a predicate. filteredBy will run the fold on the current element, and will filter out any focuses for which the fold yields no results.

The fold I'm passing in converts the String to a Text using packed , then runs a regex which matches any digit, then requires at least one more of that digit to be next in the string. Since regex only yields matches, if no matches are found the candidate will be filtered out.

That's all the criteria! Now we've got a list of all of them, but all we really need is the count of them, so we'll switch from toListOf to lengthOf :

main :: IO () () = ([ 307237 .. 769058 ] :: [ Int ]) main([]) & lengthOf ( traversed . re _Show lengthOf ( traversedre _Show . filtered (\s -> s == sort s) filtered (\ss) . filteredBy (packed . [regex|(\d)\1+|]) filteredBy (packed[regex|(\d)\1+|]) ) & print >>> main main 889

That's the right answer, not bad!

Part 2

Part 2 only adds one more condition:

The number must have a group of exactly 2 consecutive numbers, e.g. 333 is no good, but 33322 is fine.

Currently we're just checking that it has at least two consecutive numbers, but we'll need to be smarter to check for groups of exactly 2. Luckily, it's not too tricky.

The regex traversal finds ALL non-overlapping matches within a given piece of text, and the + modifier is greedy, so we know that for a given string 33322 our current pattern will find the matches: ["333", "22"] . After that it's easy enough to just check that we have at least one match of length 2!

main :: IO () () = ([ 307237 .. 769058 ] :: [ Int ]) main([]) & lengthOf (traversed . re _Show lengthOf (traversedre _Show . filtered (\s -> s == sort s) filtered (\ss) . filteredBy (packed . [regex|(\d)\1+|] . match . to T.length . only 2 ) filteredBy (packed[regex|(\d)\1+|]matchto T.lengthonly ) & print >>> main main 589

I just get the match text, get its length, then use only 2 to filter down to only lengths of 2. filteredBy will detect whether any of the matches make it through the whole fold and kick out any numbers that don't have a group of exactly 2 consecutive numbers.

That's it for today! Hopefully tomorrow's is just as optical! 🤞

Hopefully you learned something 🤞! If you did, please consider checking out my book: It teaches the principles of using optics in Haskell and other functional programming languages and takes you all the way from an beginner to wizard in all types of optics! You can get it here. Every sale helps me justify more time writing blog posts like this one and helps me to continue writing educational functional programming content. Cheers!