Exam 1
                                      CS 451
                               Wednesday, March 1, 2000

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Problem 1:  

Recall the maze problem from assignment 1.  You are given a maze 
(a list of (x,y) coordinates), and a start and stop coordinate.  
The problem is to discover if you can get from the start to the 
stop coordinate by continually moving up, down, left or right.
one space.  You are only allowed to move to coordinates in the 
maze.  

Consider the sample maze ((1 2)(1 4)(2 1)(2 2)(2 3)(2 4)(2 5)(3 3)
(4 1)(4 2)(4 3)(4 4)(4 5)(5 2)(5 4)), with start coordinate (3 3).

I will ask some questions about search algorithms in this problem.
Always assume that your search algorithm does not allow you to go
back to the parent node.  Additionally, when you are expanding nodes
and you have a choice, assume that the algorithm prefers to expand
the node with the smallest x coordinate.  If the x coordinates are 
the same, it prefers to expand the node with the smallest y coordinate.

A.  Suppose that (5 4) is the stop coordinate.  For each of BFS,
DFS and IDS, state how many nodes will be examined.  (Obviously,
nodes should be counted twice if the algorithm destroys and recreates
the search tree)  Also, what is the largest queue size for each of
the three algorithms?

B.  Same question where (1 2) is the stop coordinate.

C.  Same question where (4 3) is the stop coordinate.



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Problem 2:  

Consider a different kind of maze problem.  This time the maze
is one-dimenasional.  It is represented by a list telling you 
that if you are a given position, then there are two possible
positions you can move to.  Consider the maze:

( (1 (3 5)) (2 (3 6)) (3 (2 4)) (4 (2 3)) (5 (3 8))
  (6 (7 9)) (7 (4 8)) (8 (7 10)) (9 (10 11)) (10 (5 7)) )

This says that if you are at position 1, then you can go to position 3 or 5.
If you are in position 2, then you can go to position 3 or 6, etc.
Starting at position 1, the object is to get to position 11.
Each move has a cost of 1.  So the path cost is the number of edges
on that path.

A.  You will notice that a move can at most move you 4 spaces
closer to the goal.  Give an admissible heuristic based on that fact.

B.  Show the tree that would be constructed by the A* algorithm
using the heuristic in part A.

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Problem 3:  

Consider the following two person game.  There are two piles of 
marbles.  Each person on his turn chooses one pile and takes as
many marbles as he wants out of that pile (but he must take at
least one marble out of that pile).  Suppose that you go first.
The object of the game is not to take the last marble.  When
somebody takes the last marble, the opponent gets as many points
as the number of moves in the game.  For example, if I go first,
then my opponent goes, and then I go again and take the last marble,
then my opponent gets three points because I took the last marble
and there were three moves in the game.

Let (x y) represent the state with x marbles in one pile and y marbles 
in the other.  For simplicity when you draw your tree, think of (x y)
and (y x) as the same state.  

A.  Give the minmax tree for the example when there are originally
two marbles in each pile.

B.  Again consider the example where there are two marbles in each pile.
Give the tree produced by the alpha beta pruning algorithm and
explain why you pruned it wherever you pruned it.  For this problem,
when you have a choice of what to do next you should prefer the state
that leaves the fewest total marbles.  If you have a choice among two
states with the same number of total marbles, choose the one whose largest
pile is the smallest.