CS347 SP2014 Exam 3 Key

This is a closed-book, closed-notes exam. The only items you are permitted to use are writing implements. Mark each sheet of paper you use with your name and the string “cs347sp2014 exam3”. If you are caught cheating, you will receive a zero grade for this exam. The max number of points per question is indicated in square brackets after each question. The sum of the max points is 45. You have 75 minutes to complete this exam. Good luck!

Multiple Choice Questions - write the letter of your choice on your answer paper 1. The complexity of where b, d, and m have their usual AI meanings and n is the number of distinct stochastic outcomes (e.g., dice rolls) is: [2]

(a) O(bmnm) d d 1 (b) O(b n ) [1 2 ] (c) O(bdnm) [1] (d) O(bndm) [0] (e) none of the above [0]

2. Given the following local search algorithms: Steepest Ascent , Stochastic Hill Climbing, First-Choice Hill Climbing, Random-Restart Hill Climbing, Local , Stochastic Beam Search, . Which of these algorithms are suitable for optimizing real-valued problems? [2] (a) Local Beam Search and Stochastic Beam Search [0] (b) Stochastic Hill Climbing, Stochastic Beam Search, and Simulated Annealing [1] (c) Only Random-Restart Hill Climbing [0] 1 (d) Any of the hill-climbing algorithms [ 2 ] (e) Any of the non-hill-climbing algorithms [1] (f) None of the above (The correct answer is First Choice Hill Climbing and Simulated Annealing, because those are the only ones that limit their branching factors which prevents a real-valued search space from causing an infinite branching factor.)

Regular Questions

3. Alice is given a maximization problem with a search space of size 1012. She decides to tackle it by running Steepest- Ascent Hill Climbing 1000 times, each time with a randomly selected starting point. The lowest value produced by her 1000 runs is 1.3, the highest value is 4.9, and the average is 3.2. The algorithm on average takes 5 steps/loops to converge and return a result. She is satisfied with her search and reports 4.9 as the global optimum.

(a) Is Alice justified in reporting 4.9 as the global optimum? Explain your answer! [4] No, because her runs are very quickly converging to significantly different values, indicating that the search landscape contains many local optima. Steepest-Ascent Hill Climbing is not an appropriate for this type of search landscape, and therefore it is highly unlikely that she has actually found the global optimum. (b) If you were asked to validate Alice’s result with further experiments, which local search algorithms would you use? Explain your answer! [3] For this type of search landscape containing many local optima, a search algorithm with the ability to escape local optima is needed such as stochastic beam search, simulated annealing, or evolutionary algorithms. The final two questions are about the following state space graph. Heuristic h(n) is defined by the values following the node labels in the state space graph. Step cost values follow the action labels. Nodes are expanded counter-clockwise, ending at exactly 9 o’clock; example: A generates first action c, then action d, and finally action e, while B generates first action f and then action g. When multiple actions with equal LRTA*-COST are found, use the one found first.

4. Give the LRTA* trace terminating either when the goal is found or after the 15th call to LRTA*-COST. [30]

current last previous cost world LRTA*-COST min action, state action state estimate knowledge cost S - - H[S]=4 - (S,a,-)=4 a,4 - (S,b,-)=4 a ,4 A a S H[A]=3 R[S,a]=A (S,a,A)=1+3=4 a,4 (S,b,-)=4 a, 4 H[S]=4 (A,c,-)=3 c,3 (A,d,-)=3 c,3 (A,e,-)=3 c ,3 B c A H[B]=2 R[A,c]=B (A,c,B)=1+2=3 c,3 (A,d,-)=3 c,3 (A,e,-)=3 c, 3 H[A]=3 (B,f,-)=2 f,2 (B,g,-)=2 f ,2 E f B H[E]=1 R[B,f]=E (B,f,E)=2+1=3 f,3 (B,g,-)=2 g, 2 H[B]=2 (E,i,-)=1 i,1 LRTA*-COST call limit reached 5. What is the Competitive Ratio (CR) based on the final state of your LRTA* trace? Explain your answer and make sure to list the final state you are using. Note that in the case of call limit termination, if the LRTA*-COST call you terminated on returned an action, then for the purpose of computing the CR, the action is assumed to have been executed. [4]

c(S,a,A)+c(A,c,B)+c(B,f,E) 4 CR = c∗(S,E) = 4 = 1