Advent of Code F# – Day 21

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

 

Day 21

See details of the challenge here.

Today’s input looks like this:

swap position 2 with position 7
swap letter f with letter a
swap letter c with letter a
rotate based on position of letter g
rotate based on position of letter f
rotate based on position of letter b
swap position 3 with position 6
swap letter e with letter c
swap letter f with letter h

First, let’s model the different instructions we’ll need to process with union types.

Then, we can use the same Regex active pattern we used in Day 10 to help us parse the input file into the Instruction type.

Now let’s add a few functions to do the actual scrambling:

Most of these are straight forward, but there is one thing I’d like to touch on as it might not be obvious – the first line of the reverse function : let n = n % input.Length.

Suppose the length of the input array is 4 (eg. [| ‘a’, ‘b’, ‘c’, ‘d’ |]) then shifting left or right by 5 spaces is the same as shifting by 1.

To solve part 1 we just need to apply the instructions we parsed earlier on our input.

let part1 = “abcdefgh”.ToCharArray() |> apply instructions

 

Part 2

You scrambled the password correctly, but you discover that you can’t actually
modify the password file on the system. You’ll need to un-scramble one of the
existing passwords by reversing the scrambling process.

What is the un-scrambled version of the scrambled password fbgdceah?

On first thought, it seemed we could reverse the scrambling process by finding the inverse of each of the steps and apply the input in reverse order. For example, the inverse of Reverse, SwapPos and SwapLetter instructions are themselves; the inverse of Rotate(Left, 3) is Rotate(Right, 3) and vice versa; the inverse of Move(1, 3) is Move(3, 1).

But… how do we find the inverse of RotateBasedOn? We will need to know the exact no. of spaces to shift left, there’s no easy way to do it based on the character and the right-shifted result alone. You could, however, go at it from a different angle and try out all possible inputs that would have given us the output that we have – by shifting Left by 1 to input.Length no. of spaces.

That said, given the length of the password we need to unscramble, there are only 40320 permutations of the letters abcdefgh. A brute force approach is feasible and possibly easier in this case, and that’s the approach I decided to go with. For that, I need a function to return all permutations of the letters abcdefgh.

I found a nice implementation of such function on fssnip courtesy of Rick Minerich and to solve part 2 we just need to find the permutation that gives us the scrambled password “fbgdceah”.

 

(ps. if you’re interested in how the first approach would look, I decided to implement it as well just for fun)

 

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1 thought on “Advent of Code F# – Day 21”

  1. Pingback: F# Weekly #52, 2016 – Sergey Tihon's Blog

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