In a previous post, I shared how to build a randomised Tic Tac Toe simulation. The computer plays against itself playing at random positions. In this post, I will share how to teach the computer to play the game strategically.

I love the 1983 classic movie War Games. In this film, a computer plays Tic Tac Toe against itself to learn that it cannot win the game to prevent a nuclear war.

Back in those days, I devoured the wonderful book *Writing Strategy Games on your Atari* by John White which contains an algorithm to play Tic Tac Toe War Games. This is my attempt to relive the eighties using R.

You can find the code on my GitHub page.

## Drawing the Board

A previous post describes the function that draws the Tic Tac Toe board. For completeness, the code is replicated below. The game board is a vector of length nine consisting of either -1 (X), 0 (empty field) or 1 (O). The vector indices correspond with locations on the game board:

1 2 3

4 5 6

7 8 9

draw.board <- function(board) { # Draw the board xo <- c("X", " ", "O") # Symbols par(mar = rep(0,4)) plot.new() plot.window(xlim = c(0,30), ylim = c(0,30)) abline(h = c(10, 20), col="darkgrey", lwd = 4) abline(v = c(10, 20), col="darkgrey", lwd = 4) pieces <- xo[board + 2] text(rep(c(5, 15, 25), 3), c(rep(25, 3), rep(15,3), rep(5, 3)), pieces, cex = 6) # Identify location of any three in a row square <- t(matrix(board, nrow = 3)) hor <- abs(rowSums(square)) if (any(hor == 3)) hor <- (4 - which(hor == 3)) * 10 - 5 else hor <- 0 ver <- abs(colSums(square)) if (any(ver == 3)) ver <- which(ver == 3) * 10 - 5 else ver <- 0 diag1 <- sum(diag(square)) diag2 <- sum(diag(t(apply(square, 2, rev)))) # Draw winning lines if (hor > 0) lines(c(0, 30), rep(hor, 2), lwd=10, col="red") if (ver > 0) lines(rep(ver, 2), c(0, 30), lwd=10, col="red") if (abs(diag1) == 3) lines(c(2, 28), c(28, 2), lwd=10, col="red") if (abs(diag2) == 3) lines(c(2, 28), c(2, 28), lwd=10, col="red") }

## Human Players

This second code snippet lets a human player move by clicking anywhere on the graphic display using the locator function. The click location is converted to a number to denote the position on the board. The entered field is only accepted if it has not yet been used (the *empty* variable contains the available fields).

# Human player enters a move move.human <- function(game) { text(4, 0, "Click on screen to move", col = "grey", cex=.7) empty <- which(game == 0) move <- 0 while (!move %in% empty) { coords <- locator(n = 1) # add lines coords$x <- floor(abs(coords$x) / 10) + 1 coords$y <- floor(abs(coords$y) / 10) + 1 move <- coords$x + 3 * (3 - coords$y) } return (move) }

## Evaluate the Game

This code snippet defines the *eval.game* function which assesses the current board and assigns a score. Zero means no outcome, -6 means that the *X* player has won and +6 implies that the *O* player has won.

# Evaluate board position eval.game <- function(game, player) { # Determine game score square <- t(matrix(game, nrow = 3)) hor <- rowSums(square) ver <- colSums(square) diag1 <- sum(diag(square)) diag2 <- sum(diag(t(apply(square, 2, rev)))) eval <- c(hor, ver, diag1, diag2) # Determine best score minimax <- ifelse(player == -1, "min", "max") best.score <- do.call(minimax, list(eval)) if (abs(best.score) == 3) best.score <- best.score * 2 return (best.score) }

## Computer Moves

The computer uses a modified Minimax Algorithm to determine its next move. This article from the Never Stop Building blog and the video below explain this method in great detail.

The next function determines the computer’s move. I have not used a brute-force minimax algorithm to save running time. I struggled building a fully recursive minimax function. Perhaps somebody can help me with this. This code looks only two steps deep and contains a strategic rule to maximise the score.

The first line stores the value of the players move, the second remainder of the matrix holds the evaluations of all the opponents moves. The code adds a randomised variable, based on the strategic value of a field. The centre has the highest value because it is part of four winning lines. Corners have three winning lines and the rest only two winning lines. This means that the computer will, all things being equal, favour the centre over the corners and favour the other fields least. The randomised variables in the code ensure that the computer does not always pick the same field in a similar situation.

# Determine computer move move.computer <- function(game, player) { empty <- which(game == 0) eval <- matrix(nrow = 10, ncol = 9, data = 0) for (i in empty) { game.tmp <- game game.tmp[i] <- player eval[1, i] <- eval.game(game.tmp, player) empty.tmp <- which(game.tmp ==0) for (j in empty.tmp) { game.tmp1 <- game.tmp game.tmp1[j] <- -player eval[(j + 1), i] <- eval.game(game.tmp1, -player) } } if (!any(abs(eval[1,]) == 6)) { # When winning, play move # Analyse opponent move minimax <- ifelse(player == -1, "max", "min") # Minimax best.opponent <- apply(eval[-1,], 1, minimax) eval[1,] <- eval[1,] * -player * best.opponent } # Add randomisation and strategic values board <- c(3, 2, 3, 2, 4, 2, 3, 2, 3) # Strategic values board <- sapply(board, function(x) runif(1, 0.1 * x, (0.1 * x) + 0.1)) # Randomise eval[1, empty] <- eval[1, empty] + player * board[empty] # Randomise moves # Pick best game minimax <- ifelse(player == -1, "which.min", "which.max") # Minimax move <- do.call(minimax, list(eval[1,])) # Select best move return(move) }

This last code snippet enables computers and humans play each other or themselves. The *players* vector contains the identity of the two players so that a human can play a computer or vice versa. The human player moves by clicking on the screen.

The loop keeps running until the board is full or a winner has been identified. A previous Tic Tac Toe post explains the *draw.board* function.

# Main game engine tic.tac.toe <- function(player1 = "human", player2 = "computer") { game <- rep(0, 9) # Empty board winner <- FALSE # Define winner player <- 1 # First player players <- c(player1, player2) draw.board(game) while (0 %in% game & !winner) { # Keep playing until win or full board if (players[(player + 3) %% 3] == "human") # Human player move <- move.human(game) else # Computer player move <- move.computer(game, player) game[move] <- player # Change board draw.board(game) winner <- max(eval.game(game, 1), abs(eval.game(game, -1))) == 6 # Winner, winner, chicken dinner? player <- -player # Change player } }

You can play the computer by running all functions and then entering *tic.tac.toe()*.

I am pretty certain this simplified minimax algorithm is unbeatable—why don’t you try to win and let me know when you do.

## Tic Tac Toe War Games

Now that this problem is solved, I can finally recreate the epic scene from the WarGames movie. The Tic Tac Toe War Games code uses the functions explained above and the animation package. Unfortunately, there are not many opportunities to create sound in R.

# WAR GAMES TIC TAC TOE source("Tic Tac Toe/Tic Tac Toe.R") # Draw the game board draw.board.wargames <- function(game) { xo <- c("X", " ", "O") # Symbols par(mar = rep(1,4), bg = "#050811") plot.new() plot.window(xlim = c(0,30), ylim = c(0,30)) abline(h = c(10, 20), col = "#588fca", lwd = 20) abline(v = c(10, 20), col = "#588fca", lwd = 20) text(rep(c(5, 15, 25), 3), c(rep(25, 3), rep(15,3), rep(5, 3)), xo[game + 2], cex = 20, col = "#588fca") text(1,0,"r.prevos.net", col = "#588fca", cex=2) # Identify location of any three in a row square <- t(matrix(game, nrow = 3)) hor <- abs(rowSums(square)) if (any(hor == 3)) hor <- (4 - which(hor == 3)) * 10 - 5 else hor <- 0 ver <- abs(colSums(square)) if (any(ver == 3)) ver <- which(ver == 3) * 10 - 5 else ver <- 0 diag1 <- sum(diag(square)) diag2 <- sum(diag(t(apply(square, 2, rev)))) # Draw winning lines if (all(hor > 0)) for (i in hor) lines(c(0, 30), rep(i, 2), lwd = 10, col="#588fca") if (all(ver > 0)) for (i in ver) lines(rep(i, 2), c(0, 30), lwd = 10, col="#588fca") if (abs(diag1) == 3) lines(c(2, 28), c(28, 2), lwd = 10, col = "#588fca") if (abs(diag2) == 3) lines(c(2, 28), c(2, 28), lwd = 10, col = "#588fca") } library(animation) player <- -1 games <- 100 saveGIF ({ for (i in 1:games) { game <- rep(0, 9) # Empty board winner <- 0 # Define winner #draw.board.wargames(game) while (0 %in% game & !winner) { # Keep playing until win or full board empty <- which(game == 0) move <- move.computer(game, player) game[move] <- player if (i <= 12) draw.board.wargames(game) winner <- max(eval.game(game, 1), abs(eval.game(game, -1))) == 6 player <- -player } if (i > 12) draw.board.wargames(game) } }, interval = c(unlist(lapply(seq(1, 0,-.2), function (x) rep(x, 9))), rep(0,9*94)), movie.name = "wargames.gif", ani.width = 1024, ani.height = 1024)

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There is a way to beat the algorithm. Given I have the first move, place it on one of the corners. The algorithm will chose the center since it has the highers score on “board”. Then chose the opposite corner to the one I chose on my first move. The algorithm will chose one of the two available corners since they have higher scores on “board” after the center. Place my third move on the last available corner. I have won the game by this point since I have two potential winning moves and the other player can only block one.

I win because the algorithm only looks two steps ahead and it needs to look three to figure out my strategy. Given that this might be the only way to fool the algorithm, we could solve it with brute force (i.e. feeding it an if statement that changes the values on “board” whenever I have placed my first two moves on opposite corners) or by teaching it to look three steps ahead (this is considerably harder). I am currently working on implementing three steps; however, there are bugs in the algorithm and at the moment I am too tired to find them.

I loved the article by the way. It taught me a lot. I hope to be able to learn from cool projects of yours in the future.

Bummer, I thought I had fixed that. I let the computer play itself lots but this did not come up.

I struggled writing a proper minimax algorithm – one that reviews all moves. If somebody can help me with this that would be great.

Back to the drawing board. Thanks for finding the bug.

I started writing code that generates and evaluates all possible moves:

Next step will be to work out how to implement MiniMax to find the best move to make it truly unbeatable.