ECS 170- Introduction to Artificial Intelligence; Homework Assignment #2 - Solutions

Winter 2003

Rao Vemuri

(70 points) – A third problem, discussed in the class, has been added at the end           

Assigned: 21 January 2003

Due: 28 January 2003


(1) (25 points, 5 pts for each part) Consider the game of Tic-tac-toe. The game involves playing on a board with 9 squares arranged in a 3 x 3 grid. Two players take turns; one player puts a token (a cross) and the other puts another token (a circle) in one of the available empty squares of the game board. The play continues until one of the players gets all three of his/her tokens in a row, column or along a diagonal.

 

(a)    Make an estimate of the number of states of this game. Do not just give a number. Show how you got it.

At the first level the branching factor = 9. Next level it is 8, and so on. So, one way of looking at this problem is to say that there are 9! states.

 

(b)    Show the whole game tree starting from an empty board down to a depth 2 (i.e., one X and one O on the board, taking symmetry into account (you should have 3 positions at level 1 and 12 at level 2)

Let us define the canonical positions as 1, 2, 3, 4, 5, 6, 7, 8, 9 going from the top left square to the bottom right square as a raster.

At the first level, there are is a state with an x in one corner (square 1) one state with an x in the middle (square 5), and a state with an x in the middle of a side (square 2). All others are symmetrical.

At the second level, there are 5 children to the node with an x in the corner,

These children get a “o” in positions 2, 3, 5, 6, 9.

five children to the node with an x in the middle of the side and

These children get a “o” in positions 3, 4, 5, 7, 8.

only two children  for the node with a x in the center.

These children get a “o” in positions 1, 2.

© Now it is an easy matter to calculate the evaluation function for these using the given formula.

 

(d)    Mark on your tree the backed-up values for the positions at levels 1 and 0, using the minimax algorithm and then use them to choose the best move.

(e)   Circle the nodes that at level 2 that would not be evaluated if alpha-beta pruning were applied, assuming that the nodes are generated in the optimal order for alpha-beta pruning.

 

(2) (25 points, 8/8/9 points for the parts) Let us continue with the tic-tac-toe problem. Now we are interested in looking at the programming aspects of this problem. One of the first issues we have to tackle is the representation issue. Toward this goal let us make a few attempts.

(a) First attempt: TTT1: Let us define a data structure as shown below.


Data structures:

·                       | 1 | 2 | 3 |
·                       | 4 | 5 | 6 |
·                       | 7 | 8 | 9 |

Each element contains the values


Algorithm:


Comments:

Advantages:

Disadvantages:


(b) Second attempt: TTT2: Let us define a data structure as shown below.


Data Structures:

·                       value 2 === blank
·                       value 3 === X
·                       value 5 === O
·                why?...
·                       turn = 1 first move
·                       turn = 9 last move
  

Algorithm:

Three procedures.

·                               if player p cannot win in the next move
·                                Return 0
·                              else
·                                Returns the number of square that
·                                 constitutes a winning move
  

We first call posswin(us), if we can win we make the winning move. If we cannot win we call posswin(opponent) and block opponents move if he/she can win.
posswin checks each row, column, and diagonal

That is, Scan row/column/diag to find blank space to move to.

X starts!!


Algorithm in more detail:

·                               if(board[5] is blank) 
·                                 go(5) 
·                               else 
·                                 go(1)
       
·                               if(board[9] is blank)
·                                 go(9)
·                               else
·                                 go(3)
         
·                               if(posswin(X) != 0 )
·                                 go(posswin(X))    /*block opponent*/
·                               else
·                                 go(make2)
         

Comments:

Disadvantages:

Advantages:

 

 


(c)    Third attempt: TTT3: Suggest an alternative formulation and an alternate algorithm (just state the algorithm in a few steps, each step in informal English) and its advantages and disadvantages.

Solution

Data Structures:


Algorithm:

Deciding which move is best:

For each move

 


 

(3) (20 points, you will lose one point for each error) In  the class, I drew a tree on the board and demonstrated how to do the alpha-beta pruning. Complete that procedure for the rest of the tree and show the best move sequence for the maximizing player.