Now to find out how I was ideally supposed to solve it.

Pt2 was easier for me because while at first it took me a bit to land on lifting stuff from Djikstra's algo to solve the challenge maze before the sun turns supernova, as I tend to store the paths for debugging anyway it was trivial to group them by score and count by distinct tiles.

I randomly got it by sorting for the most robots in the bottom left quadrant while looking for robot concentrations, it was number 13. Despite being in the centre of the grid it didn't show up when sorting for most robots in the middle 30% columns of the screen, which is kind of wicked, in the traditional sense.

The first things I tried was looking for horizontal symmetry (find a grid where all the lines have the same number of robots on the left and on the right of the middle axis, there is none, and the tree is about a third to a quarted of the matrix on each side) and looking for grids where the number of robots increased towards the bottom of the image (didn't work, because turns out tree is in the middle of the screen).

I thinks I was on the right track with looking for concentrations of robots, wish I'd thought about ranking the matrices according to the amount of robots lined up without gaps. Don't know about minimizing the safety score, sorting according to that didn't show the tree anywhere near the first tens.

Realizing that the patterns start recycling at ~10.000 iterations simplified things considerably.

In pt2 I had some floating point weirdness when solving for keypress count, I was checking if the key presses where integers (can't press button A five and half times after all) by checking if A = floor(A) and sometimes A would drop to the number below when floored, i.e. it was in reality (A-1).999999999999999999999999999999999999999999999. Whatever, I rounded it away but I did spend a stupid amount of time on it because it didn't happen in the example set.

// F#
let isLegit ((total: int64), (calibration : int64 array)) = 

    let rec search (index : int) (acc: int64) =
        let currentValue = calibration.[index]
        
        [Add; Times; Concat] // operators - remove 'Concat' to solve for 7-1
        |> List.exists (fun op -> // List.exists returns true the first time the lambda returns true, so search stops at first true
                match op with // update accumulator
                | Add -> acc + currentValue
                | Times -> acc * currentValue
                | Concat -> int64 (sprintf "%d%d" acc currentValue)
                |> function // stop search on current accumulator value (state) exceeding total, or being just right
                | state when state > total -> false
                | state when state = total && index < (calibration.Length-1) -> false // this was the problem
                | state when state = total && index = (calibration.Length-1) -> true
                | state -> // stop if index exceeds input length, or continue search
                    if index+1 = calibration.Length
                    then false
                    else search (index+1) state
        )
     
    // start search from second element using the first as current sum
    search 1 calibration.[0]

EDIT: total && index < (calibration.Length-1) -> false -- i.e. stop if you reach the total before using all numbers, well, also stops you from checking the next operator, So, removing it worked.

Rubber ducking innocent people on the internets works, who knew.

So ideally 1|2, 2|3, 3|4 -> 1|2|3|4 -> 1|2, 2|3, 3|4, 1|3, 1|4, 2|4

Except I forgot the last part and just returned the branch elements pairwise in sequence, which is just the original rules, which I noticed accidentally after the fact since I was getting correct results, because apparently it was completely unnecessary and I was just overthinking myself into several corners at the same time.

let comparerFactory (ruleset: (int*int) list) :int -> int -> int = 
    let leftIndex = 
        ruleset 
        |> List.groupBy fst 
        |> List.map (fun (key,grp)-> key, grp |> List.map snd)
        |> Map.ofList

    fun page1 page2 -> 
        match (leftIndex  |> Map.tryFind page1) with
        | Some afterSet when afterSet |> List.contains page2 -> -1
        | _ -> 1

The memoization pattern is for caching an index of rules grouped by the before page, so I can sort according to where each part of the comparison appears. I started out with having a second index where the key was the 'after' page of the rule which I would check if the page didn't appear on the left side of any rule, but it turned out I could just return the opposite case, so again unnecessary.

Would have been pretty useful too if 4-2 turned out to be about finding longer patterns, instead of smaller and in half the directions.

let rec isXmas (row:int) (col:int) (dir:int) (depth:int) (arr: char array2d) : bool =
    let value = arr.[row,col]
    if depth = 3
    then value = 'S'
    else if  [|'X';'M';'A';'S'|].[depth] = value
    then
        let (nextRow, nextCol) =
            match dir with
            | 1 -> row + 1, col - 1
            | 2 -> row + 1, col
            | 3 -> row + 1, col + 1
            | 4 -> row, col - 1
            | 6 -> row, col + 1
            | 7 -> row - 1, col - 1
            | 8 -> row - 1, col
            | 9 -> row - 1, col + 1
            | _ -> failwith $"{dir} isn't a numpad direction." 

        let rowCount = arr |> Array2D.length1
        let colCount = arr |> Array2D.length2
       
        if nextRow >= 0 && nextRow < rowCount && nextCol >= 0 && nextCol < colCount
        then isXmas nextRow nextCol dir (depth+1) arr
        else false
    else false

Then you run this for every appropriately pruned 'X' times every direction and count the trues.