How to return the best first level in this F# minimax?

Question

This question is more a semantic-algorithmic-data-structure question than a F# syntactically question. I have a Minimax algorithm. The minimax algorithm should return the best next move, from a start position. To do this, it calculus all next moves, then the next-next-moves until a determined depth or until there is no more moves. It builds a tree like this:

     P  
   /  \  
 a      b  
/ \  
c d

I have the fallowing data struct to handle the tree:

type TreeOfPosition =
    | LeafP   of Position * int
    | BranchP of Position * TreeOfPosition list

In the exemple tree above, P and a are Branchs and b, c and d are Leafs. The code below is my minimax algorithm:

let evaluateTree ( tree : TreeOfPosition, player : int) =
    let rec loop minOrmax node =
        match node with
        | LeafP(position, 0) -> 
            LeafP(position, evaluateLeaf(position))
        | BranchP(position, children)  -> 
            minimax.[minOrmax](List.map (loop (1 - minOrmax)) children)
    loop player tree

This code are returning me a Leaf, for example, c. When I changed the recursion call to

| BranchP(position, children)  -> 
    LeafP(position, 
          getStaticEvalFromNode(minimax.[minOrmax](
                       List.map (loop (1 - minOrmax)) children)))

And this modification makes the static value of a good leaf go up. I need to return the best second level node. Hope somebody can help! Pedro Dusso

EDIT 1

Thanks for all answers guys, they help me a lot. Sorry about didn't specified the things very much. Let's go in parts:

1) I’m matching my LeafP like LeafP(position, 0) because when I create my tree I set the leafs with a default value of 0 as its static value. As I’m going up my static values, eliminating the leaf and making the (before Branches) leafs with (min or max) static values I thought that this way I would prevent to evaluate a ex-Branch leaf (because it would not have the 0 value).

2) My biggest problem was to get the second level (the next move which has to be played) best position back. I solved it this way:

let evaluateTreeHOF ( tree, player : int) =
    let rec loop minOrmax node =
        match node with
        | LeafP(position, 0) -> LeafP(position, evaluateLeaf(position))
        | BranchP(position, children) -> LeafP(position,(children 
                                                         |> List.map (loop (1 - minOrmax)) 
                                                         |> minimax.[minOrmax] 
                                                         |> getStaticEvalFromNode))
    match tree with
    | BranchP(position, children) -> children |> List.map (loop (1 - player)) |> minimax.[player]

Instead of passing the entire tree, I’m passing just the children’s of the start node, and filtering the resulted list (a list of ex-Branches with the static values which went up for be the best for its current level) again. This way I’m getting the node I wanted.

I thought the kvb answers very interesting, but a little complicated to me. The other ones I understudied, but they just give me back the static value – and I could not make them to work for me :(

Thanks a lot for all the answers, all of them inspired me a lot.

Here is my full code: (http://www.inf.ufrgs.br/~pmdusso/works/Functional_Implementation_Minimax_FSharp.htm)

Pedro Dusso

Answer 1

I don't quite understand some aspects of your sample (e.g. why do you match only against leaves with 0s in them?), so I'll make a few changes below. First of all, let's generalize the tree type a bit, so that it can store any types of data in the leaves and branches:

type Tree<'a,'b> = 
| Leaf of 'a 
| Branch of 'b * Tree<'a,'b> list

Let's also use a dedicated player type, rather than using 0 or 1:

type Player = Black | White

Finally, let's generalize the evaluation of the best move a bit, so that the leaf evaluation function is passed in as an argument:

let bestMove evalPos player tree =
  // these replace your minimax function array
  let agg1,agg2,aggBy = 
    match player with
    | Black -> List.min, List.max, List.maxBy
    | White -> List.max, List.min, List.minBy

  // given a tree, this evaluates the score for that tree
  let rec score agg1 agg2 = function
  | Leaf(p) -> evalPos p
  | Branch(_,l) -> agg1 (List.map (score agg2 agg1) l)

  // now we use just need to pick the branch with the highest score
  // (or lowest, depending on the player)
  match tree with
  | Leaf(_) -> failwith "Cannot make any moves from a Leaf!"
  | Branch(_,l) -> aggBy (score agg1 agg2) l 

Answer 2

I think you can use mutually recursive functions:

let maxTree t = 
  match t with
  | child -> xxx
  | subtrees s ->
      s |> Seq.map minTree |> Seq.max

and minTree t = 
  match t with
  | child -> xxx
  | subtrees s ->
      s |> Seq.map maxTree |> Seq.min

Answer 3

The solution to this problem was described in the F#.NET Journal article Games programming: tic-tac-toe (31st December 2009) and uses the following pattern:

type t = Leaf | Branch of t seq

let aux k = function
  | Leaf -> []
  | Branch s -> k s

let rec maxTree t = aux (Seq.map minTree >> Seq.max) t
and minTree t = aux (Seq.map maxTree >> Seq.min) t

See also the playable demo.