Advent of Code F# – Day 6

Yan Cui

I help clients go faster for less using serverless technologies.

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ps. look out for all my other solutions for Advent of Code challenges here.

 

Day 6

See details of the challenge here.

The input for today’s challenge looks like this:

cmezkqgn
nmzrgcft
ydpndcps
zjihhows
kvptxsrx
ubbvugwq

Since the only difference between the 2 parts of this challenge is how the characters in a column are sorted by count, so we can capture the common flow to solving both parts in a solve function below.

Let’s walk through what’s happening here step by step.

  1. first we collect all the characters of each column into a seq<char[]>
  2. then for each char[] we
    1. group by the chars by itself
    2. map the resuling seq<char * seq<char>> into a sequence of char and count, ie seq<char * int>
    3. use the supplied sortBy function to sort this sequence by count
    4. take the first (most common/least common depending on the sortBy function)
    5. return the char
  3. stich the chars together into a good old fashioned string

 

So now, solving part 1 is a simple one-liner:

solve Seq.sortByDescending

 

Part 2

Of course, that would be the message – if you hadn’t agreed to use a modified
repetition code instead.

In this modified code, the sender instead transmits what looks like random data,
but for each character, the character they actually want to send is slightly
less likely than the others. Even after signal-jamming noise, you can look at
the letter distributions in each column and choose the least common letter to
reconstruct the original message.

In the above example, the least common character in the first column is a; in
the second, d, and so on. Repeating this process for the remaining characters
produces the original message, advent.

Given the recording in your puzzle input and this new decoding methodology, what
is the original message that Santa is trying to send?

Because of the work we had done earlier, solving part 2 is also a one-liner:

solve Seq.sortBy

 

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