{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Computer Science 188 - Spring 1996 - Russell - Midterm 1

# Computer Science 188 - Spring 1996 - Russell - Midterm 1 -...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CS 188 Spring 1996 Introduction to AI Stuart Russell Midterm You have 1 hour and 20 minutes. The exam is open-book, open-notes. 80 points total (1 point per minute). You will not necessarily nish all questions, so do your best ones rst. Write your answers in blue books. Check you haven't skipped any by accident. Hand them all in. Panic not. 1. (18 pts.) True/False 2. (14 pts.) Search Decide if each of the following is true or false. If you are not sure you may wish to provide some explanation to follow your answer. (a) (3) Breadth- rst is an optimal search algorithm. (b) (3) Truth tables can be used to establish the truth or falsehood of any propositional sentence. (c) (3) Forward chaining is complete for rst-order logic. (d) (3) 9x8y x = y is a satis able sentence. (e) (3) Minimax and alpha-beta can sometimes return di erent results. (f) (3) Simple re ex agents cope well with inaccessible environments. (a) (4) In a map-colouring problem, the aim is to colour a map using a given set of colours so that no two adjacent countries are the same colour. Give a precise formulation of map-colouring as a search problem. (b) (4) Provide a rigorous critique of each step of the following argument, which appeared in a recent submission to the European Conference on AI: \Given two admissible heuristics h1 and h2 where h1(n) h2(n) for all nodes n, it is obvious that A* using h1 will be more e cient than A* using h2 . Now suppose I am given an admissible heuristic h2. If one can nd a constant c such that the heuristic h1 (n) = h2 (n) + c is still admissible, then searching with h1 is better than searching with h2." (c) (6) The following diagram shows a partially expanded search tree. Each arc is labelled with the corresponding step cost, and the leaves are labelled with the h value. Arad 3 Sibiu 3 Arad h = 10 5 6 Oradea h = 8 19 Timisoara h = 5 6 Rimnicu h = 10 5 Zerind h = 13 Fagaras h = 12 3. (8 pts.) Game-playing i. Which leaf will be expanded next by a greedy search? ii. Which leaf will be expanded next by a uniform-cost search? iii. Which leaf will be expanded next by an A* search? Draw the smallest possible game tree on which alpha-beta will prune at least one leaf node. Make sure to label the leaves with values, and circle the leaf (or leaves) that will be pruned. 1 4. (12 pts.) Simple knowledge representation 5. (13 pts.) Logical Inference Translate each of the following English sentences into the language of standard rst-order logic, stating the intended interpretation for any predicate, function or constant you use. (a) (3) \No one at UCB listens to KBLX." (b) (3) \No one is at both UCB and Stanford." (c) (3) \If Joe knows someone at Stanford then Joe knows someone who is not at UCB." (d) (3) \Everyone at Stanford listens to the same radio stations." (a) (3) Give a uni er for each of the following pairs of sentences, if possible, or write \None" if not. i. Between(1 x 2) Between(y F (y) 2) ii. > (x y) < (y x) iii. Related(x Father(x)) Related(Father(y) y) (b) (2) Let S be the sentence 9x 1 + x = 1. Let S be the result of applying Existential Elimination to S . Write down S . (c) (2) Suppose that S and S are now part of our knowledge base, and that we now apply Existential Elimination to S again to obtain S . Write down S . (d) (4) If Existential Elimination is a sound rule, then both S and S follow from S . Yet S and S taken together seem to say there are two numbers x such that 1 + x = 1, which seems to be di erent from the intention of the original sentence. Can you resolve this apparent paradox? (e) (2) Let C1 be the clause :At(Father(x) Stanford) _ At(x Stanford), and let C2 be the clause :At(y Stanford) _ Owns(y BMW ) _ :Happy(y). Write down the result of applying resolution to C1 and C2. 0 0 0 00 00 0 00 0 00 6. (15 pts.) Planning In Stanford's amazing new VRISC machine, there are only three registers, called A, B , and C . Initially, they contain 1, 2, and 3 respectively. There is only one instruction, called Assign(x y), which copies the value contained in register y into register x. The only predicate we need to describe the situation is V alue(x v), which says that register x contains value v. We would like to use a partial-order planning algorithm to construct a plan to switch the contents of A and B . (a) (3) Using the pictorial notation for STRIPS operators, show the Assign operator with its precondition(s) and e ect(s). (b) (3) Leaving about 3 inches of empty space in the middle, draw the initial plan for this problem, again using the standard pictorial notation. (c) (3) Now the planner decides to achieve the goal condition that B should contain a 1. Add an Assign step to your diagram so that this condition is achieved. Make sure to bind variables as appropriate. (d) (3) Now suppose that the planner decides to achieve the precondition of this Assign step by connecting it to the Start step. Update your diagram to re ect this, marking any changes in bindings. (e) (3) Now suppose the planner decides to achieve the other goal condition, namely that A should contain a 2. Explain in words (with changes to your diagram if that makes it easier to explain) the sequence of events that occur when it adds a new Assign step to achieve this, up to the point where the addition of the step with all its rami cations is complete. 2 ...
View Full Document

{[ snackBarMessage ]}