January 28, 2016
Nearly one year ago, I made a program called jing-logic. A tic-tac-toe game written in miniKanren. I remembered when a friend looked at the source codes, listening to me talking about the AI in it that you can play against with, “So you’ve already got an AI in this?” She asked, disbelieving that the AI part was so short that it was nearly nothing, even though the game was also simple.
I always regard that as a fun experiment, and in this post, I will introduce the game, the AI, as well as the source code.
The game was written in miniKanren, a logic programming language that is simple and fun to play with.
A logic program has two parts, some logic expressions, and a logic solver. Those logic expressions set up constraints about some logic variables. The logic solver is usually provided, and resolves those logic expressions to find possible assignments to those logic variables that satisfy those constraints.
In a “pure” miniKanren program, only several primitive constraints are provided.
Using those primitive constraints, together with Lisp’s
expression, you can create many complex constraints that represents anything
that is computable (which means miniKanren is Turing-complete1). A special
run is provided to trigger the logic solver to compute the results. We
run’s parameters, about what logic variables we want to query.
The most fundamental operator is
(run* (q) (== q 1))
The program prints out
((1)) representing that it can figure out one value that
makes the program succeeds, that is
q equals to
Not all logic variables are for querying, so we have
fresh, that creates an
unexposed logic variable.
(run* (q) (fresh (p) (== p q) (== p 1)))
The final constraint is
conde that represents logical disjunction (OR).
(run* (q) (conde ((== q 1)) ((== q 2))))
The above program would return
((1) (2)). The logic solver can find two
results that satisfy the constraints, that is
As you may see, there’s
run. That’s nearly
everything about miniKanren.
A tic-tac-toe game contains 9 positions in a 3x3 board. Two players (let’s call
o) put piece on the board in turn. The first player that connects
three pieces in a line no matter horizontally, vertically, or diagonally, wins.
So first let’s define those two players:
(define (playero p) (conde ((== p 'o)) ((== p 'x))))
A value in a board is either occupied by a player, or is nothing:
(define (valueo x) (conde ((playero x)) ((nullo x))))
As for a board, there are 3 columns and 3 rows, so we use
represent those indexes.
(define (indexo x) (conde ((== x 1)) ((== x 2)) ((== x 3))))
Therefore we can define different positions in a board:
(define (positiono row column value board) (fresh (p11 p12 p13 p21 p22 p23 p31 p32 p33) (boardo p11 p12 p13 p21 p22 p23 p31 p32 p33 board) (conde ((== row 1) (== column 1) (== value p11)) ((== row 1) (== column 2) (== value p12)) ((== row 1) (== column 3) (== value p13)) ((== row 2) (== column 1) (== value p21)) ((== row 2) (== column 2) (== value p22)) ((== row 2) (== column 3) (== value p23)) ((== row 3) (== column 1) (== value p31)) ((== row 3) (== column 2) (== value p32)) ((== row 3) (== column 3) (== value p33)))))
By definition, horizontal positions are at the same row:
(define (horizontalo x y z board) (fresh (row) (indexo row) (positiono row 1 x board) (positiono row 2 y board) (positiono row 3 z board)))
And vertical positions are at the same column:
(define (verticalo x y z board) (fresh (column) (indexo column) (positiono 1 column x board) (positiono 2 column y board) (positiono 3 column z board)))
For diagonal positions, they are either, in order, ‘
3’ or ‘
(define (diagonalo x y z board) (fresh (column-x column-y column-z) (conde ((== 1 column-x) (== 2 column-y) (== 3 column-z)) ((== 3 column-x) (== 2 column-y) (== 1 column-z)) (positiono 1 column-x x board) (positiono 2 column-y y board) (positiono 3 column-z z board)))
Now as we have all the necessary definition here, we can know the meaning of
winner – a player that occupies three pieces either horizontally, vertically,
(define (winnero player board) (playero player) (conde ((horizontalo player player player board)) ((verticalo player player player board)) ((diagonalo player player player board))))
Now if you execute
(run* (board) (winnero 'o board)), the solver will output
all possible board configurations that
o is a winner.
The game is simple, as a result, there’s a perfect strategy that a player can follow to never lose the game.
- Win: If the player has two in a row, they can place a third to get three in a row.
- Block: If the opponent has two in a row, the player must play the third themselves to block the opponent.
- Fork: Create an opportunity where the player has two threats to win (two non-blocked lines of 2).
- Blocking an opponent’s fork (Option 1): The player should create two in a row to force the opponent into defending, as long as it doesn’t result in them creating a fork. For example, if “X” has a corner, “O” has the center, and “X” has the opposite corner as well, “O” must not play a corner in order to win. (Playing a corner in this scenario creates a fork for “X” to win.)
- Blocking an opponent’s fork (Option 2): If there is a configuration where the opponent can fork, the player should block that fork.
- Center: A player marks the center. (If it is the first move of the game, playing on a corner gives “O” more opportunities to make a mistake and may therefore be the better choice; however, it makes no difference between perfect players.)
- Opposite corner: If the opponent is in the corner, the player plays the opposite corner.
- Empty corner: The player plays in a corner square.
- Empty side: The player plays in a middle square on any of the 4 sides.
So we can implement that:
(define (strategy-wino player board move) (fresh (row column new-board) (moveo row column player move) (surpose-boardo board move new-board) (winnero player new-board))) (define (strategy-blocko player board move) (fresh (opponent row column imaginary-opponent-move) (opponento player opponent) (strategy-wino opponent board imaginary-opponent-move) (moveo row column opponent imaginary-opponent-move) (moveo row column player move))) (define (strategy-randomo player board move) (fresh (row column next-board) (moveo row column player move) (surpose-boardo board move next-board)))
And create an AI player:
(define (ai-playo player current-board next-board) (conda ((full-boardo current-board) (== current-board next-board)) ((fresh (move) (surpose-boardo current-board move next-board) (conda ((strategy-wino player current-board move)) ((strategy-blocko player current-board move)) ((strategy-randomo player current-board move)))))))
We don’t consider optimization here. Using those primitive constraints, everything is computable, but probably not fast. ↩