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Re: [tlaplus] Recursion without StackOverflowError



Interesting. Gravitated to the RECURSIVE approach because it looked familiair, but it felt a bit bad that I wasn't expressing it with memoization. But maybe that TLC already uses memoization internally for functions?

After reading the reference and surrounding sections, I think that I now also have a better understanding of CHOOSE. Maybe the time has come to read it completely, instead of diving in head first and messing around.

Tried to update the model with your approach, and ended up with this:
GoodBoard[B_0 \in Board] ==
    IF Win(B_0) \/ Draw(B_0) THEN TRUE
    ELSE \A B_1 \in {[B_0 EXCEPT ![counter] = "O"]: counter \in PossibleMoves(B_0)}:
        IF Lose(B_1) THEN FALSE
        ELSE \E B_2 \in {[B_1 EXCEPT ![y] = "X"]: y \in PossibleMoves(B_1)}:
            GoodBoard[B_2]

Unfortunately this fails to terminate within 10sec. Maybe this is because the recursion doesn't visit all boards due the "short circuiting" on Win, Draw or Lose. 
However I'm still a bit puzzled as to why TLC hangs, I feel like it should be able detect that it "exhausted" the recursion but is still missing part of the mapping. Or that it is not making any progress, and error out?

Op maandag 22 juli 2024 om 23:19:02 UTC+2 schreef Hillel Wayne:

Generally when doing these kinds of recursive things, I find it easier to generate all possible values as a value, instead of as a computation. There are only 3^9 (19683) possible tic-tac-toe boards. This means you can do things like

GoodBoard[B \in Board] ==

  IF Win(B) \/ Draw(B)
            THEN TRUE

            ELSE \A B_1 \in {[B_0 EXCEPT ![counter] = "O"]: counter \in PossibleMoves(B_0)]}:
                    \E y \in PossibleMoves(B_1): GoodBoard(B_1)

I think this will make TLC generate the whole function as one value and avoid a stackoverflow error. See Specifying Systems 6.4 for the trick of using "Function Definitions" to make recursive functions.

H

On 7/19/2024 12:29 PM, Isaac Dynamo wrote:
Hi,

Looked around online for a TLA+ community and this looks like the best place for a beginner to ask some questions. 

I'm trying to model tic-tac-toe with the end goal to show that the starting player doesn't have to lose.

I started with a specification that has all possible game states, and did some sanity checks on it. This all worked as expected.

Next I wanted to show than by restricting the moves of the first player to "good moves" that the system doesn't deadlock and Lose(board) is unreachable. 

I wrote a recursive function GoodMove() that tries to ensure that after every counter move there is still a good move to make. Resulting in a Win or a Draw.

However with this addition TLC crashes with an java.lang.StackOverflowError. I understand that recursion can result in a stack overflow, but the recursion in this case should be bounded. Each iteration sets a tile, and at one point all tiles are set and this must be a Win or a Draw. So I don't understand why this results in a stack overflow.

Can somebody explain what is going on? And what techniques can be used to fix this?

I get the feeling that recursion in TLA+ works differently compared to regular programming languages and that I'm missing something.

Thanks,

------------------------------- MODULE tictactoe -------------------------------
EXTENDS Integers

N == 2
Players == {"X", "O"}
Tiles == (0..N) \X (0..N)
Token == {" ", "X", "O"}
Board == [Tiles -> Token]

VARIABLES board, turn
vars == <<board, turn>>

NextPlayer == [X |-> "O", O |-> "X"]

ThreeInARow(B, T) ==
    \/ \E x \in 0..N: \A y \in 0..N: (B[<<x, y>>] = T \/ B[<<y, x>>] = T)
    \/ \A x \in 0..N: B[<<x, x>>] = T
    \/ \A x \in 0..N: B[<<x, N-x>>] = T

Win(B) == ThreeInARow(B, "X")
Lose(B) == ThreeInARow(B, "O")
Draw(B) ==  ~Win(B) /\ ~Lose(B) /\ \A t \in Tiles: B[t] /= " "
Done == Draw(board) \/ Win(board) \/ Lose(board)

PossibleMoves(B) == {x \in Tiles: B[x] = " "}

RECURSIVE GoodMove(_, _)
GoodMove(B, M) ==
    LET B_0 == [B EXCEPT ![M] = "X"] IN
        IF Win(B) \/ Draw(B)
            THEN TRUE
            ELSE \A counter \in PossibleMoves(B_0):  
                LET B_1 == [B_0 EXCEPT ![counter] = "O"] IN
                    \E y \in PossibleMoves(B_1): GoodMove(B_1, y)

Move ==
    /\ turn' = NextPlayer[turn]
    /\ \E x \in PossibleMoves(board):
        /\ board' = [board EXCEPT ![x] = turn]
        /\ turn = "X" => GoodMove(board, x) \* Comment out this line for all possible games and no stack overflow

Init ==
    /\ board = [t \in Tiles |-> " "]
    /\ turn = "X"

Next ==
    \/ ~Done /\ Move
    \/ Done /\ UNCHANGED vars

   
Fairness == WF_vars(Move)

Spec == Init /\ [][Next]_vars /\ Fairness

TypeInvariant ==
    /\ turn \in Players
    /\ board \in Board

_OneOutcomeInvariant_ ==
    /\ Win(board) => (~Lose(board) /\ ~Draw(board))
    /\ Lose(board) => (~Win(board) /\ ~Draw(board))
    /\ Draw(board) => (~Lose(board) /\ ~Win(board))

TerminationInvariant == <>[] Done

CanWinContradiction  == []~Win(board)
CanLoseContradiction == []~Lose(board)
CanDrawContradiction == []~Draw(board)

=============================================================================




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