Willem Hoek

Solving the Jane Street puzzle of December 2022

Jan 02, 2023

Die Agony

This post describes the approach and solution to the Jane Street Puzzle of December 2022.

Approach

I solved it programmatically using a recursive backtracking algorithm, which works very well for these type of problems. Backtracking is where you incrementally try different paths and abondon a candidate (“backtrack”) as soon as a path is not valid any more.

The provided grid of 36 cells were numbered from 0 to 35 which represent the nodes in the search tree.

 30 31 32 33 34 35
 24 25 26 27 28 29
 18 19 20 21 22 23
 12 13 14 15 16 17
  6  7  8  9 10 11
  0  1  2  3  4  5

The recursive function named solve starts at cell 0 and meanders the grid until it gets to cell 35.

The puzzle stated that we have a “six-sided die, with numbers”. Although with a standard die - the numbers are integers, I decided not to assume this. Hence I used floats in the grid array to store the values. As it turns out – they were all integers.

When you run the program, it should show all the paths evaluated and the solution.

final path =  0 6 7 8 2 1 7 13 19 25 31 32 26 25 24 18 19 20 14 15 9 3 4 5 11 17 16 15 21 27 28 29 35
dice = top:7 bottom:9 up:5 down:-9 left:9 right:-3
unvisited cells = 34 33 30 23 22 12 10
Sum of values in the unvisited cells = 1935

Here is a visual showing the backtracking in action and related Python code to generate the image.

Using OCaml

Since Jane Street is a major user of OCaml, I decided to use OCaml for this puzzle.

The code, listed below, is quite short. Less than half the code solve the actual puzzle. The rest of the code is used to print out the evaluated paths and solution. The evaluated paths from the output was used to generate the pretty backtracking image above – using Python Pillow library.

The good news is – you do not need OCaml installed on your computer to run the code below. There are multiple FREE online OCaml Compilers available on the web. Below is a list of some of them.

Simply copy / paste the source code below into the online OCaml Compiler and execute to see the results.

(*********************************
   Author:   Willem Hoek
   Date:     26 Dec 2022
   Website:  whoek.com
 **********************************)

type move = Up | Down | Left | Right

let roll_dice  (top, bottom, up, down, left, right) d =
  match d with
  | Up    -> down,  up,    top,    bottom, left,   right
  | Down  -> up,    down,  bottom, top,    left,   right
  | Right -> left,  right, up,     down,   bottom, top
  | Left  -> right, left,  up,     down,   top,    bottom

let next_cell cell move =
  match move with
  | Up -> cell + 6
  | Down -> cell - 6
  | Left -> cell - 1
  | Right -> cell + 1

let valid_moves cell  =
  (
    match cell with
    | 0 |  6 | 12 | 18 | 24 | 30 -> [Right]
    | 5 | 11 | 17 | 23 | 29 | 35 -> [Left]
    | _ -> [Right; Left]
  ) @ (
    match cell with
    |  0 |  1 |  2 |  3 |  4 | 5  -> [Up]
    | 30 | 31 | 32 | 33 | 34 | 35 -> [Down]
    | _ -> [Up; Down]
  )

let grid =
  [|
    0.  ;  77.;  32.; 403.; 337.; 452.;
    5.  ;  23.;  -4.; 592.; 445.; 620.;
    -7. ;   2.; 357.; 452.; 317.; 395.;
    186.;  42.; 195.; 704.; 452.; 228.;
    81. ; 123.; 240.; 443.; 353.; 508.;
    57. ;  33.; 132.; 268.; 492.; 732.;
  |]

let print_dice dice =
  let a, b, c, d, e, f = dice in
  Printf.printf "dice = top:%g  bottom:%g  up:%g  down:%g  left:%g  right:%g\n"
    a b c d e f

let print_int_lst p =
  List.iter (Printf.printf "%i ") p;
  Printf.printf "\n"

let unvisited all visited =
  List.filter (fun x -> not @@ List.mem x visited) all

let () = assert (unvisited [1;2;3;4] [1;3] = [2;4])

let sum_values lst  =
  let f acc cell = acc +. grid.(cell) in
  List.fold_left f 0.0 lst

let () = assert (sum_values [6; 35] = 737.0)

let range i = List.init i succ

let () = assert (range 3 = [1; 2; 3])

let print_solution ~new_dice ~cell ~path =
  let all = range 35 in
  let unvisited_cells = unvisited all @@ cell :: path in
  let sum_of_values = sum_values unvisited_cells in
  print_dice  new_dice;
  Printf.printf "unvisited cells = ";
  print_int_lst  unvisited_cells;
  Printf.printf "Sum of values in the unvisited cells = %g\n"
    sum_of_values

let rec solve ~dice ~cell ~n ~path ~score  =
  print_int_lst @@ List.rev @@ cell :: path;
  let new_score = grid.(cell) in
  let new_top = (
      if n = 0.0 then 0.
      else  (new_score -. score) /. n
    ) in
  let top, d2, d3, d4, d5, d6 = dice in
  let new_dice = new_top, d2, d3, d4, d5, d6 in
  let valid = (top = new_top || top = 0.) in
  if cell = 35 && valid then begin
      Printf.printf "Solution found!!!\n";
      print_solution ~new_dice ~cell ~path;
      true                    (* stop looking *)
    end
  else if valid then begin
      let moves = valid_moves cell in
      let check move =
        let new_cell = next_cell cell move in
        solve ~dice:(roll_dice new_dice move)
          ~cell:new_cell
          ~n:(n +. 1.)
          ~path:(cell :: path)
          ~score:new_score
      in
      List.exists check moves
    end
  else false                (* backtrack *)

let main () =
  let dice = 0.0,  0.0,  0.0,  0.0,  0.0,  0.0 in
  let _ = solve ~dice ~cell:0 ~n:0.0 ~path:[] ~score:0.0 in
  Printf.printf "Cheers!\n"

let () = main ()

References and further reading

[1] Jane Street - Puzzle Archive
https://www.janestreet.com/puzzles/archive/

[2] Source code for this post on Github
https://github.com/whoek/janestreet-puzzles

[3] OCaml programming language
https://ocaml.org/

[4] Backtracking Algorithm
https://en.wikipedia.org/wiki/Backtracking

[5] Solving a previous Jane Street puzzle – also using recursive backtracking and Ocaml
https://whoek.com/b/janestreet-puzzle-feb-2021-solved-ocaml-to-the-rescue

[6] OCaml Tips: Implementing a range Function
https://batsov.com/articles/2022/10/31/ocaml-tips-implementing-a-range-function/

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