Intelligent CS 5 ? HW 11 (1 problem !) M/T sections W/Th sections due Sunday, 11/14 at midnight due Monday, 11/15 at midnight Recitation for HW11 -- Friday 11/12, 8:00 am 2nd midterm exam -- this Friday, 11/12 Take-home, 2.0 hours, closed-book exam. Practice problems are online… exemption: > 95% HW Exam will be available this Friday; it’s due Sunday evening by 5:00 pm. Chess is the Drosophila of artificial intelligence. The computer that defeated Garry Kasparov (top link) Today’s Lab: M-Z - Alexander Kronrod Games: computers vs. humans…

Two-player games early programs ~ 1960’s Strategic thinking was considered essential for intelligence Computer Chess has a long history: Ranking beginner amateur world ranked world champion MacHack ( 1100 ) ~ 1967 MIT Deep Thought ~ 1989 Carnegie Mellon Slate ( 2070 ) ~ 1970’s Northwestern Deep Blue ~ 1996 IBM Deep Blue rematch ~ 1997 IBM

Computers’ strategy… early programs ~ 1960’s Strategic thinking was considered essential for intelligence Computer Chess has a long history: Ranking beginner amateur world ranked world champion MacHack ( 1100 ) ~ 1967 MIT Deep Thought ~ 1989 Carnegie Mellon Slate ( 2070 ) ~ 1970’s Northwestern Deep Blue ~ 1996 IBM Deep Blue rematch ~ 1997 IBM 100’s of moves/sec 10,000’s of moves/sec 100,000,000 moves/sec 200,000,000 moves/sec how far ahead is this?

Games’ Branching Factors early programs ~ 1960’s Branching Factor Estimates for different two-player games Tic-tac-toe 4 Connect Four 7 Checkers 10 Othello 30 Chess 40 Go 300 On average, there are about 40 possible moves that a chess player can make from any board configuration… Ranking beginner amateur world ranked world champion MacHack ( 1100 ) ~ 1967 MIT Deep Thought ~ 1989 Carnegie Mellon Slate ( 2070 ) ~ 1970’s Northwestern Deep Blue ~ 1996 IBM Deep Blue rematch ~ 1997 IBM 0 Ply 1 Ply 2 Ply game tree for C4

Games’ Branching Factors early programs ~ 1960’s Branching Factor Estimates for different two-player games Tic-tac-toe 4 Connect Four 7 Checkers 10 Othello 30 Chess 40 Go 300 On average, there are about 40 possible moves that a chess player can make from any board configuration… Ranking beginner amateur world ranked world champion MacHack ( 1100 ) ~ 1967 MIT Deep Thought ~ 1989 Carnegie Mellon Slate ( 2070 ) ~ 1970’s Northwestern Deep Blue ~ 1996 IBM Deep Blue rematch ~ 1997 IBM 0 Ply 1 Ply 2 Ply

Games’ Branching Factors early programs ~ 1960’s Branching Factor Estimates for different two-player games Tic-tac-toe 4 Connect Four 7 Checkers 10 Othello 30 Chess 40 Go 300 On average, there are about 40 possible moves that a chess player can make from any board configuration… Ranking beginner amateur world ranked world champion MacHack ( 1100 ) ~ 1967 MIT Deep Thought ~ 1989 Carnegie Mellon Slate ( 2070 ) ~ 1970’s Northwestern Deep Blue ~ 1996 IBM Deep Blue rematch ~ 1997 IBM “solved” games computer-dominated human-dominated 0 Ply 1 Ply 2 Ply

Winning: Details public boolean winsFor(char ch) { for (int r=0 ; r<this.nrows-3 ; ++r) { for (int c=0 ; c<this.ncols-3 ; ++c) { if (this.data[r+0][c+0] == ch && this.data[r+1][c+1] == ch && this.data[r+2][c+2] == ch && this.data[r+3][c+3] == ch) { return true; } … same idea for vert., horiz., other diag. … return false; } which diagonals? which board piece? which curly braces?

public boolean winsFor(char ch) { for (int r=0 ; r<this.nRows-3 ; ++r) for (int c=0 ; c<this.nCols-3 ; ++c) if (this.data[r+0][c+0] == ch && this.data[r+1][c+1] == ch && this.data[r+2][c+2] == ch && this.data[r+3][c+3] == ch) return true; … same idea for vert., horiz., other diag. … return false; } Winning: Details (compact version)

Objects hide details! Class: Board Object: b b.winsFor(‘X’) b.addMove(3,‘X’) ‘X’ ‘O’ capabilities of b so that important things aren’t lost in the shuffle… b.removeMove(3) b.isOver() (the last 3 are new for this week) b.clear()

Objects hide details! class CS5App { public static void main(String[] args) { H.pl("Hi! Welcome to Connect 4..."); H.pl("How many rows/columns ? (4-15)"); int R = H.ni(); int C = H.ni(); Board b = new Board(R,C); char player = 'X'; while (true) { b.print(); int c = H.ni(); // gets next move b.addMove(c,player); if (b.winsFor(player)) break; if (player == 'X') player = '0'; else player = 'X'; } // end of while } Hw10Hw11

Where we’re headed… Player playerForX Details (data and methods) Player playerForO Details (data and methods) Ask what kind of players should play for X and O Create two objects of class Player with appropriate inputs Ask each of these objects to findScores for X and O and then breaktie s. See who wins! demo… what details are needed?

char checker Picture of a Player object Player int lookahead Player playerForX Player(char ch, int lk, int tbk) int tiebreakType void printScores(double[] s) double evaluate(Board b) double[] plyHuman(Board b) double[] ply0(Board b) int breaktie(double[] s) double[] findScores(Board b) Imagine if Board weren’t a Class… ! double[] ply1,2,3,4,N(Board b) methods

Player code private char checker; private int lookahead; private int tiebreakType; public Player(char ch, int la, int tbk) // constructor { } public char getChecker() // accessor “getter” method { } public char me() // short for getChecker() { } public char opp() // returns the opponent’s checker { } class Player {

Where we’re headed… Player playerForX Details (data and methods) Player playerForO Details (data and methods) Ask what kind of players should play for X and O Create two objects of class Player with appropriate inputs Ask each of these objects to findScores for X and O and then breaktie s. See who wins! demo… what details are needed?

Hw10 class CS5App { public static void main(String[] args) { H.pl("Hi! Welcome to Connect 4..."); int R = H.ni(); int C = H.ni(); Board b = new Board(R,C); char player = 'X'; while (true) { b.print(); int uc = H.ni(); // user’s column b.addMove(uc,player); if (b.winsFor(player)) break; if (player == 'X') player = '0'; else player = 'X'; } // end of while } Hw11

Choosing a move 1) Find scores at appropriate lookahead… 2) Print the scores. 3) Break ties to determine the next move. ply0 : 0 ply of lookahead ply1 : 1 ply of lookahead ply2,3,4 : 2,3,4 ply of lookahead plyHuman : ask the user printScores : prints the scores to each column breaktie : chooses ONE maximum score findScores chooses one of these methods to run

ply0 class Player { // returns scores with no lookahead // public double[] ply0(Board b) { char checker int lookahead int tiebreakType this b

ply1 class Player { // returns scores with 1ply lookahead // public double[] ply1(Board b) { char checker int lookahead int tiebreakType this b Which column should have the best score? Lookahead 1 move Evaluate the results for each column Later, we’ll choose the best column to move…

Evaluating a board Assigns a score to any Board b Score for X Score for O Score for X Score for O Score for X Score for O for a win -1.0 for an invalid move 0.0 for a loss 50.0 for a “tie” not possible in evaluate

evaluate class Player { // returns the appropriate score for b // remember: all of Player’s methods are available public double evaluate(Board b) { for a win -1.0 for an invalid move 0.0 for a loss 50.0 for a “tie” Improvements? Write tournamentEvaluate for Ex. Cr.!

b 0-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 1-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 2-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 3-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 “Quiz” It is X’s move.. Compute the score that X would find for each column for each of these lookaheads: no moves at all! X moves O moves X moves O moves X moves

class Player { private int tiebreakType; private int lookahead; private char checker; public int breaktie(double[] s) { double maxScore = getMax(s); /* assume getMax is already written */ if (this.tiebreakType == 2) /* random tie breaker is tiebreakType == 2 */ { } Write breaktie to return a randomly chosen best score (max score) from an array of scores named s.

‘X’ ‘O’ new‘X’ b Looking ahead 1 ply… (1) For each possible move (2) Add the column’s move (3) Evaluate the boards (4) Choose one of the best ‘X’ to move

‘X’ ‘O’ new‘X’ Col 6 Col 5 Col 4 Col 3 Col 2 Col 1 Col 0 b Looking ahead 1 ply… ‘X’ to move (1) For each possible move (2) Add the column’s move

‘X’ ‘O’ new‘X’ Col 6 Col 5 Col 4 Col 3 Col 2 Col 1 Col 0 b Looking ahead 1 ply… (1) For each possible move (2) Add the column’s move (3) Evaluate the boards ‘X’ to move NONE

‘X’ ‘O’ new‘X’ Col 6 Col 5 Col 4 Col 3 Col 2 Col 1 Col 0 b Looking ahead 1 ply… (1) For each possible move (2) Add the column’s move (3) Evaluate the boards ‘X’ to move NONE

‘X’ ‘O’ new‘X’ Col 6 Col 5 Col 4 Col 3 Col 2 Col 1 Col 0 b Looking ahead 1 ply… (1) For each possible move (2) Add the column’s move (3) Evaluate the boards (4) Choose one of the best ‘X’ to move NONE

ply1 public double[] ply1(Board b) {

Two-player games have been a key focus of AI as long as computers have been around… Strategic thinking = intelligence ? Humans and computers have different relative strengths in these games:

Two-player games have been a key focus of AI as long as computers have been around… Strategic thinking = intelligence ? computers good at looking ahead in the game to find winning combinations of moves Humans and computers have different relative strengths in these games: this week…

Two-player games have been a key focus of AI as long as computers have been around… Strategic thinking = intelligence ? humanscomputers good at evaulating the strength of a board for a player good at looking ahead in the game to find winning combinations of moves Humans and computers have different relative strengths in these games: this week… (extra credit)

How humans play games… - experts could reconstruct these perfectly - novice players did far worse… An experiment (by deGroot) was performed in which chess positions were shown to novice and expert players…

How humans play games… - experts could reconstruct these perfectly - novice players did far worse… Random chess positions (not legal ones) were then shown to the two groups - experts and novices did just as badly at reconstructing them! An experiment (by deGroot) was performed in which chess positions were shown to novice and expert players…

Looking further ahead … 0 ply: 2 ply:3 ply: random (but legal) choice of move ! (1) player will win (2) player will avoid losing (3) player will set up a win by forcing the opponent to avoid losing 1 ply: X ’s move

ply2 public double[] ply2(Board b) { depends on ply1 !

Extra Credit: the plyN method ! Lab this week You’ll need to write (and use) Last Names Problem 1: A Connect Four Player … Extra Credit: tournamentEvaluate & a C4 round-robin 2, 4, 6, and 8-ply lookahead for O will all produce different scores! Player(char ch, int lk, int tbk) char getChecker() char me() char opp() void printScores() double evaluate(Board b) double[] plyHuman(Board b) double[] ply0(Board b) int breaktie(double[] s) int go(Board b) (and the others listed on the HW) M-Z double[] findScores(Board b) O to move

b 0-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 1-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 2-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 3-ply scores for X: col 0col 1col 2col 3col 4col 5col 6 “Quiz” It is X’s move.. Compute the score that X would find for each column for each of these lookaheads: no moves at all! X moves O moves X moves O moves X moves

class Player { private int tiebreakType; private int lookahead; private char checker; public int breaktie(double[] s) { double maxScore = getMax(s); /* assume getMax is already written */ if (this.tiebreakType == 2) /* random tie breaker is tiebreakType == 2 */ { } Write breaktie to return a randomly chosen best score (max score) from an array of scores named s.

‘X’ ‘O’ new‘X’ Col 6 Col 5 Col 4 Col 3 Col 2 Col 1 Col 0 b Looking ahead 1 ply… (1) For each possible move (2) Add the column’s move (3) Evaluate the boards (4) Choose one of the best ‘X’ to move NONE

Winning -- details public boolean winsFor(char ox) { for (int r=0 ; r<this.nRows-3 ; ++r) { for (int c=0 ; c<this.nCols-3 ; ++c) { if (this.data[r+0][c+0] == ox && this.data[r+1][c+1] == ox && this.data[r+2][c+2] == ox && this.data[r+3][c+3] == ox) { return true; } … same idea for vert., horiz., SW-NE diag. … return false; } | | | | | |X| | | | | | | |O|X|O| | | | | |O|X|X| |O|O| |X|O|O|X|X|O|X| finds this diagonal: complete HW10PR2 solutions at

static static methods belong to a class, not an object H.pl(“I’m a static method”); // lots double av = averageArray(stocks); // HW 7 int syl = numSyllables(word); // HW 6 double d = Math.sqrt(343.0); If the static method is in another class, the class name is needed!

opp() and ?: class Player { private char checker; // data member public char opp() // returns opponent’s checker { } ? : is shorthand for if … else …, but only for deciding between values if else

addMove Class: Board Object: b b.addMove(3,‘X’) ‘X’ ‘O’ changes b by adding checker ‘X’ into row 3 new‘X’ b before b after

Adding a move without changing b ! b before b after Board nextb = b.newAddMove(3,‘X’); a new Board with the move added