day 9.dib

#!meta

{"kernelInfo":{"defaultKernelName":"fsharp","items":[{"aliases":[],"languageName":"fsharp","name":"fsharp"}]}}

#!markdown

# --- Day 9: Mirage Maintenance ---
You ride the camel through the sandstorm and stop where the ghost's maps told you to stop. The sandstorm subsequently subsides, somehow seeing you standing at an oasis!

The camel goes to get some water and you stretch your neck. As you look up, you discover what must be yet another giant floating island, this one made of metal! That must be where the parts to fix the sand machines come from.

There's even a hang glider partially buried in the sand here; once the sun rises and heats up the sand, you might be able to use the glider and the hot air to get all the way up to the metal island!

While you wait for the sun to rise, you admire the oasis hidden here in the middle of Desert Island. It must have a delicate ecosystem; you might as well take some ecological readings while you wait. Maybe you can report any environmental instabilities you find to someone so the oasis can be around for the next sandstorm-worn traveler.

You pull out your handy Oasis And Sand Instability Sensor and analyze your surroundings. The OASIS produces a report of many values and how they are changing over time (your puzzle input). Each line in the report contains the history of a single value. For example:

```
0 3 6 9 12 15
1 3 6 10 15 21
10 13 16 21 30 45
```

To best protect the oasis, your environmental report should include a prediction of the next value in each history. To do this, start by making a new sequence from the difference at each step of your history. If that sequence is not all zeroes, repeat this process, using the sequence you just generated as the input sequence. Once all of the values in your latest sequence are zeroes, you can extrapolate what the next value of the original history should be.

In the above dataset, the first history is 0 3 6 9 12 15. Because the values increase by 3 each step, the first sequence of differences that you generate will be 3 3 3 3 3. Note that this sequence has one fewer value than the input sequence because at each step it considers two numbers from the input. Since these values aren't all zero, repeat the process: the values differ by 0 at each step, so the next sequence is 0 0 0 0. This means you have enough information to extrapolate the history! Visually, these sequences can be arranged like this:

```
0   3   6   9  12  15
  3   3   3   3   3
    0   0   0   0
```

To extrapolate, start by adding a new zero to the end of your list of zeroes; because the zeroes represent differences between the two values above them, this also means there is now a placeholder in every sequence above it:

```
0   3   6   9  12  15   B
  3   3   3   3   3   A
    0   0   0   0   0
```

You can then start filling in placeholders from the bottom up. A needs to be the result of increasing 3 (the value to its left) by 0 (the value below it); this means A must be 3:

```
0   3   6   9  12  15   B
  3   3   3   3   3   3
    0   0   0   0   0
```

Finally, you can fill in B, which needs to be the result of increasing 15 (the value to its left) by 3 (the value below it), or 18:

```
0   3   6   9  12  15  18
  3   3   3   3   3   3
    0   0   0   0   0
```

So, the next value of the first history is 18.

Finding all-zero differences for the second history requires an additional sequence:

```
1   3   6  10  15  21
  2   3   4   5   6
    1   1   1   1
      0   0   0
```

Then, following the same process as before, work out the next value in each sequence from the bottom up:

```
1   3   6  10  15  21  28
  2   3   4   5   6   7
    1   1   1   1   1
      0   0   0   0
```

So, the next value of the second history is 28.

The third history requires even more sequences, but its next value can be found the same way:

```
10  13  16  21  30  45  68
   3   3   5   9  15  23
     0   2   4   6   8
       2   2   2   2
         0   0   0
```

So, the next value of the third history is 68.

If you find the next value for each history in this example and add them together, you get 114.

Analyze your OASIS report and extrapolate the next value for each history. What is the sum of these extrapolated values?

#!fsharp

type PolynomialSequence = int list

let rec parsePolynomialSequence (result: int list) (nums: int array) =
    if nums |> Array.forall ((=) 0)
    then result |> List.rev
    else 
        let next =
            Seq.pairwise nums
            |> Seq.map (fun (x, y) -> y - x)
            |> Array.ofSeq

        parsePolynomialSequence (nums[0]::result) next

let parseLine (input: string) =
    input.Split(' ', StringSplitOptions.RemoveEmptyEntries)
    |> Array.map int
    |> parsePolynomialSequence []

let sampleInput =
    [|
        "0 3 6 9 12 15"
        "1 3 6 10 15 21"
        "10 13 16 21 30 45"
    |]
    |> Array.map parseLine

sampleInput

#!fsharp

let rec fnDoesntWork (c: PolynomialSequence) (i: int) =
    match c, i with
    | _, -1 -> 0
    | [], _ -> 0
    | cn::tail, _ -> i * (fnDoesntWork tail (i-1)) + cn

sampleInput
|> Array.map (fun input -> Seq.map (fnDoesntWork input) { 0..5 })

#!markdown

Need to think through this more appropriately.

When there is one layer, it is a linear equation
`y = mx + c`

When there are two layers, it is a quadratic equation
The bottom layer is linear, still
`y = mx + c`. However, there is one fewer number in the list.

The second layer from the bottom is quadratic. In the sample
```
1   3   6   10  15 21
  2   3   4   5   6
    1   1   1   1
```

The top layer is
```
= sum i=0..4 (2 + i) + 1
= (2 + 0) + (2 + 1) + (2 + 2) + (2 + 3) + (2 + 4) + 1
```

Let's go with the summation approach for now, or even a sequence generator. I'm sure there's a mathematical function that would generate this, but not sure how to construct it at this stage.

#!fsharp

let rec fnSeq' (op: int->int->int) (input: PolynomialSequence) (state: int array) =
    seq {
        yield state[0]

        let intermediate =
            if state.Length = input.Length
            then state
            else Array.append state [| op input[state.Length] 0 |]

        let folder i state = (op i state, op i state)
        let (next, _) =
            (intermediate, 0)
            ||> Array.mapFoldBack folder

        yield! fnSeq' op input next
    }

let forwardFnSeq (input: PolynomialSequence) = fnSeq' (+) input [| input[0] |]

sampleInput
|> Array.map (forwardFnSeq >> Seq.take 7)

#!fsharp

sampleInput
|> Seq.map (forwardFnSeq >> Seq.item 6)
|> Seq.sum

#!markdown

That works. Let's see if it will work for the challenge input

#!fsharp

let challengeInputLines =
    File.ReadAllLinesAsync "day 9 input.txt"
    |> Async.AwaitTask
    |> Async.RunSynchronously

let challengeInput =
    challengeInputLines
    |> Array.map parseLine

challengeInputLines
|> Array.map (fun x -> x.Split(' ', StringSplitOptions.RemoveEmptyEntries))
|> Array.groupBy (fun x -> x.Length)

#!markdown

Cool, every input sequence is 21 entries long. We should be able to solve the challenge by grabbing the 22nd (index=21) entry for each sequence and summing them.

#!fsharp

challengeInput
|> Seq.map (forwardFnSeq >> Seq.item 21)
|> Seq.sum

#!markdown

## --- Part Two ---
Of course, it would be nice to have even more history included in your report. Surely it's safe to just extrapolate backwards as well, right?

For each history, repeat the process of finding differences until the sequence of differences is entirely zero. Then, rather than adding a zero to the end and filling in the next values of each previous sequence, you should instead add a zero to the beginning of your sequence of zeroes, then fill in new first values for each previous sequence.

In particular, here is what the third example history looks like when extrapolating back in time:

```
5  10  13  16  21  30  45
  5   3   3   5   9  15
   -2   0   2   4   6
      2   2   2   2
        0   0   0
```

Adding the new values on the left side of each sequence from bottom to top eventually reveals the new left-most history value: 5.

Doing this for the remaining example data above results in previous values of -3 for the first history and 0 for the second history. Adding all three new values together produces 2.

Analyze your OASIS report again, this time extrapolating the previous value for each history. What is the sum of these extrapolated values?# 

#!fsharp

let revFnSeq (input: PolynomialSequence) = fnSeq' (-) input (Array.ofList input)

sampleInput[2]
|> revFnSeq
|> Seq.take 2

#!fsharp

sampleInput
|> Array.map (revFnSeq >> Seq.item 1)
|> Array.sum

#!fsharp

challengeInput
|> Array.map (revFnSeq >> Seq.item 1)
|> Array.sum

#!markdown

I want to have a go at making the function cleaner. Let's try where a sequence depends on other sequences

#!fsharp

let rec fnSeq'' (op: int->int->int) (skip: int) (input: PolynomialSequence) =
    match input with
    | [] -> seq {
            while true do
                yield 0
        }
    | head::tail ->
        fnSeq'' op skip tail
        |> Seq.scan (fun i next -> op i next) head
        |> Seq.skip skip

let forwardFnSeq' = fnSeq'' (+) 0
let revFnSeq' = fnSeq'' (-) 1

sampleInput
|> Array.map (forwardFnSeq' >> Seq.take 7)

#!fsharp

challengeInput
|> Seq.map (forwardFnSeq >> Seq.item 21)
|> Seq.sum

#!fsharp

sampleInput
|> Array.map (revFnSeq' >> Seq.take 3)

#!fsharp

challengeInput
|> Array.map (revFnSeq' >> Seq.head)
|> Array.sum