# Midterm exam

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COMS W4701x: Artificial Intelligence

MIDTERM EXAM

October 26th, 2006

## DIRECTIONS

This exam is closed book and closed notes. It consists of three parts. Each part is labeled with the amount of time you should expect to spend on it. If you are spending too much time, skip it and go on to the next section, coming back if you have time.

The first part is multiple choice. The second part is short answer and problem solving. The third part is an essay question.
Important: Answer Part I by circling answers on the test sheets and turn in the test itself. Answer Part II and Part III in separate blue books. In other words, you should be handing in the test and at least two blue books.
Part I - Multiple Choice. 20 points total. 15 minutes.
1. Searle defines strong AI as

a. a computer program that produces the strongest possible rational for a decision

b. a computer program that models human cognitive reasoning

c. a computer program that can outperform a human on the Turing test

d. a computer program that can speak Chinese
b
2. Which of the following search algorithms finds the optimal solution?

a. breadth first c. depth first

b. hill climbing d. greedy search
a
3. A search algorithm is complete if it

a. always finds the optimal solution

b. always finds a solution if there is one

c. finds all possible solutions

d. never finds a solution
b

4. Which of the following techniques uses randomness for avoiding getting trapped in local maxima?
a. Best first search

b. Local beam search

c. Simulated annealing

c
5. Which of the following is a problem that occurs in hill climbing?
a. Cliff

b. Ridge

c. Valley
b
6. . Given that

P(a) = 0.2

and

P(b) = 0.6
in which of the following joint probability tables are P(A) and P(B) absolutely independent?

a)
 a ¬a b 0.16 0.42 ¬b 0.04 0.28

b)

 a ¬a b 0.10 0.50 ¬b 0.10 0.30

c)
 a ¬a b 0.12 0.48 ¬b 0.08 0.32

d)
 a ¬a b 0.20 0.40 ¬b 0.00 0.40

c

7. Inference by enumeration

(a) is based on conditional probabilities between atomic events

(b) is based on .circumstantial evidence derived from atomic events

(c) is based on the full joint distribution of atomic events

(d) is based on a list of atomic events alone

c
8. A greedy search uses a heuristic function to

a. expand the node that appears to be closest to the goal

b. expand the node that is closest to the goal

c. expand the node that is the most expensive

d. expand the leftmost node
a
9. What is the difference between the cost function and heuristic function portions of the A* evaluation function?

a. A cost function returns the actual cost from current node to goal, while the heuristic function returns the estimated cost from current node to goal

b. A cost function returns the estimated cost from start node to current node, while the heuristic function returns the estimated cost from current node to goal

c. A cost function returns the estimated cost from current node to goal, while the heuristic function returns the actual cost from start node to current node

d. A cost function returns the actual cost from start node to current node, while the heuristic function returns the estimated cost from current node to goal.
d
10. An utility function is used in game playing to

a. estimate the expected utility of the game from a given position

b. compute the winner of the game based on number of pieces left

c. compute the utility of a terminal leaf in a game tree

d. compute the best move for the successor function
c

Part II. Problem Solving. 60 points. 45 minutes.

1. [20 points] Constraint Satisfaction. Consider the crossword puzzle given below:

Suppose we have the following words in our dictionary: ant, ape, big, bus, car, has, bard, book, buys, hold, lane, year, rank, browns, ginger, symbol, syntax. The goal is to fill the puzzle with words from the dictionary.

(10 pts) Formalize the problem as a constraint satisfaction problem. Be sure to specify the variables, their domains, constraints and give a sketch of how the problem would be solved.
Variables will be positions in the crossword puzzle: (2)
e.g., 4 across, 1 down, etc.
Domain of each variable will be the set of words that can fit in that slot.

So: (3)

1A,4A: ant, ape, big, bus, car, has

1D, 3A: bard, book, buys, hold, lane, year, rank

2D: browns, ginger, symbol, syntax
Constraints: at each place where two words intersect, the letters must be equal. So, the first letter of 1A = 1D. last letter of 1A = 1st letter of 2D, etc. (2)
Sketch of how problem will be solved: (3 – initial state, successor function, backtracking, heuristics)
The initial state will have an empty assignment of values to variable.

The successor function will chose a variable to assign a value to and check whether all the constraints are met as it assigns a value. If there is no value which can be assigned without breaking the constraints, the algorithm will backtrack to the last choice point.

When choosing a variable, the successor function will use the MRV to select the variable which has the minimum remaining valueson it. If more than variable have the same # of constraints, the algorithm will choose the variable with the most constraints on remaining variables

(5 pts) Show the resulting constraint network.

Here, you need to show the graph for these constraints.

1A(1) = 1D(1)

1A(3) = 2D(1)

1D(3) = 3A(1)

3A(3) = 2D(3)

4A(1) = 2D (5)

(5 pts) Show how the Minimum Remaining Values heuristic would be used, showing how the first two words would be selected.
The first variable to be selected will be 2D because it has the fewest possible values in its domain.

The value will be chosen as the one which rules out the fewest possible remaining values for other variables. So, the choices are:

browns, ginger, symbol, syntax

Choosing browns, leaves 1 choice for 3A, 0 for 4A, 0 for 1A

Choosing ginger leaves 2 choices for 3A, 0 for 4A, 1 for 1A

Choosing symbol leaves 0 choices for 3A, 0 for 4A, 2 for 1A

Choosing syntax leaves 2 choices for 3A, 2 for 4A, 2 for 1A

So syntax will be chosen.

The second variable to be selected will be either 3A, 1A or 4A by MRV. Given a tie, the one that places the most constraints on remaining variables is either 1A or 3A. So let’s say 1A chosen next (either one is OK). The value will be chosen as the one which rules out the fewest possible remaining values for other variables. So the choices are:
ant, ape, big, bus, car, has
it must be a word that ends in s since 2D begins with s

Choosing bus leaves 3 choices for 1 D

Chososing has leaves 1 choice for 1D
Bus is chosen.
Forward checking would not change the choices for any of these variables at this point in time.
2. [20 points] Game playing and alpha beta pruning

Consider the game of tic-tac-toe (also sometimes called noughts and crosses). We define Xn as the number of rows, columns, or diagonals with exactly n X's and no O's. Similarly, On is the number of rows, columns, or diagonals with exactly n O's and no X's. The utility function assigns +1 to any position with X3 = 1 and -1 to any position with O3 = 1. All other terminal positions have utility 0. For non-terminal positions, we use a linear evaluation function defined as Eval(s)= 3X2(s) + X1(s) - (3O2(s) + O1(s)).

The figure below shows a partial game tree to depth 2.

(10 pts) Given the partial game tree, use the minimax agorithm to choose the best starting move. Mark on the tree the minimax value associated with every node in the tree.

(10 pts) Circle the nodes that would not be evaluated if alpha-beta pruning were applied, assuming the nodes are generated in the optimal order for alpha-beta pruning. (This means that it's ok to (virtually) reorder the leaves and associated internal nodes if they are not in the optimal order.)
For alpha-beta pruning, the best leaf for Max is one of the 1's in the

rightmost branch. So the optimal ordering for alpha-beta pruning

would be to evaluate this branch first (i.e. as if it were the

left-most branch). None of the nodes in this branch can be pruned.

The ordering of the other branches doesn't matter. Within the

1-0-(-1)-1 branch, one of the 1's must be visited, but the remaining 3

leaves will be pruned by alpha-beta pruning. Within the

(-1)-(-1)-0-(-1) branch, (any) one of the leaves must be visited, but

then the rest will be pruned by alph-beta pruning.
Optimal ordering: 2

No pruning of right-most branch: 2

3 branches pruned in left-most: 3

3 branches pruned in middle: 3

Best move: 1

1st levels: 1 each (or 3)

Bottom level: 2 per branch or .5 per evaluation fn.

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