Problem 18: Find the maximum total from top to bottom of the triangle below.

Wow.  The algorithm just took a little bit of pacing to realize, but I’ve gone through several refinements and variations of the implementation.

First our triangle as a string for parsing.

let tstr = "75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23"

Our first solution is imperative.

let problem18c = 
    //parse
    let tarr = Array2D.create 15 16 0 //make first column filled with zeros so no out of bounds checks required
    tstr.Split('\n') |> Array.iteri (fun i row -> row.Split(' ') |> Array.iteri (fun j cell -> (tarr.[i,j+1] <- (cell |> int))))

    //calc
    for i in 1..14 do //start with second row
        for j in 1..i+1 do //shift orientation to right
            tarr.[i,j] <- (max tarr.[i-1,j-1] tarr.[i-1,j]) + tarr.[i,j]

    //find largest        
    let mutable largest = 0        
    for j in 0..14 do 
        largest <- max largest tarr.[14,j] 
           
    largest

Our second solution uses a nicer recursive functional algorithm for computing the answer, but uses (nearly) the same parsing algorithm and array.  It would be nicer to work with a tree structure, but string.Split is just too easy.

let problem18d = 
    let tarr = Array2D.create 15 15 0
    tstr.Split('\n') |> Array.iteri (fun i row -> row.Split(' ') |> Array.iteri (fun j cell -> (tarr.[i,j] <- (cell |> int))))
    
    let rec find (row,col) = 
        if row = 14 then tarr.[row,col]
        else (max (find (row+1,col)) (find (row+1,col+1))) + tarr.[row,col]
    find (0,0)