CS347 FS2003 Final Exam Model Answers

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1 CS347 FS003 Final Exam Model Answers This is a closed-book test. The only item not supplied that you are allowed (and required) to use, is a pen or pencil. Mark all paper you use with your name, the date, and the string cs347fs003 final exam. If you are caught cheating, you will receive a zero grade for this test. The max number of points per question is indicated in square brackets after each question. The sum of the max points is 05, but the exam grade is capped at 00. You have exactly two hours to complete this test. Good luck! Part I - Chapters &. Describe briefly The Turing Test? [6] The Turing Test was designed to provide a satisfactory operational definition of intelligence, based on indistinguishability from human beings; the computer passes the test if a human interrogator, after posing some written questions, cannot tell whether the written responses come from a person or not.. Give one argument for and one argument against The Turing Test being a good test for measuring intelligence [4] For: Giving a closed-form definition of intellingece (i.e., without external references such as comparisons to humans), is an unsolved problem; The Turing Test gives us a workable approximation which in order for a computer to pass will require many advances in AI. Against: It is easy to envision an intelligent entity that could easily be distinguished from humans (e.g., if it can quickly and accurately solve a very complex differential equation it is probably not a human). Therefore, the Turing Test encourages modeling human intelligence, not rational intelligence. 3. Give the PEAS description for the board game Stratego. [5] Performance Measure: Capture the opponent s flag or capture all the opponent s movable pieces Prevent the opponent from taking your flag or all your movable pieces Minimizing the number of moves required to win Minimizing the amount of computational time required to win Minimizing the amount of storage space Environment: 0x0 Grid Game pieces Actuators: Making a move Sensors: Something to detect the current state of the board

2 4. Classify the Statego task environment according to these properties: [5] Fully observable/partially observable Deterministic/stochastic Episodic/sequential Static/dynamic/semi-dynamic Discrete/continuous 5. Explain concisely each of your five choices in the previous question. [5] Partially observable: You have full information about your pieces, but initially no information about your opponents pieces which are only gradually revealed. Deterministic: The next state of the board is completely determined by the move of the agent. Sequential: Later moves will depend on what moves were taken previously. Semi-dynamic: The environment itself does not change with the passage of time, but the agent s performance decreases (the clock is ticking!) Discrete: Stratego has a finite state space. Part II - Chapters 3 & 4 All the questions on this page are about the following state space graph, with A being the start state and T the goal state. The order in which successors are generated is counterclockwise, starting downwards. Example: E generates first T, then G, and finally A. Note the many unidirectional edges in this state space graph! When sorting nodes, whether for path-cost, heuristic value, etc., nodes with equal sorting value are ordered such that the earlier a node is generated, the higher its priority. Nodes already on the open list have higher priority than newly added nodes for equal sorting value. For this exam assume that graph search discards generated nodes if they are already on the closed list, EVEN IF THE ONES ON THE CLOSED LIST HAVE A HIGHER COST! When performing execution traces, let the leftmost column indicate the open list, the next column the closed list (if you are using a closed list), and the rightmost column the node to be evaluated. Two heuristics are defined for this exam: h(node) = minimal number of hops from node to goal (e.g., h(a) = [AET]) h(node) = max number of hops from node to goal without revisiting any state (irrespective of the closed list) (e.g., h(a) = 8 [ABCDGHFET]) with special clause: h(goal) = h(goal) = 0 A 6 E G D B F T 4 3 H C

3 6. What is the diameter of this state space? Explain your answer! [] 3; this is the largest minimal number of hops between any two nodes. 7. Give the execution trace of Iterative-Deepening Depth-First Graph Search (ID-DFGS). [8] depth-limit=0 A - A depth-limit reached and no goal found depth-limit= A - A BE A B E AB E depth-limit reached and no goal found depth-limit= A - A BE A B CFE AB C FE ABC F E ABCF E TG ABCFE T goal found; solution=aet; path-cost(aet)=7 8. Is ID-DFGS complete for this problem? Explain your answer! [] Yes, because it found a solution. 9. Does ID-DFGS find the optimal solution for this problem? Explain your answer! [] No, because it found AET and path-cost(abfet)= 5 < 7 =path-cost(aet). 0. Give the execution trace of A* graph search employing h. [0] A - A B3E7 A B3 F3C6E7 AB F3 E5C6T6 ABF E5 T5C6G7 ABFE T5 goal found; solution=abfet; path-cost(abfet)=5. Give the execution trace of A* graph search employing h. [0] A8 - A8 B9E4 A B9 F0CE4 AB F0 T6CE ABF T6 goal found; solution=abft; path-cost(abft)=6

4 . Is for this problem h admissible? Explain your answer. [] Yes, because the step cost is or larger and h predicts that the step cost is, therefore it never overestimates the path cost to the goal. 3. Is for this problem h consistent? Explain your answer. [] Yes, because n, n : h(n) step-cost(n, n ) + h(n ) with n being a successor node of node n. 4. Is A* graph search employing h optimal for this problem? Explain your answer. [] Yes, because A* graph is optimal for consistent heuristics. (Note: answering this with because it found the optimal path would require proving that it found the optimal path!) 5. Is for this problem h admissible? Explain your answer. [] No, proof by counterexample: h(e) = 8 > = path-cost(et ) (it overestimates) 6. Is for this problem h consistent? Explain your answer. [] No, it is not admissible and therefore not consistent. 7. Is A* graph search employing h optimal for this problem? Explain your answer. [] No, because it found ABFT and path-cost(abft)= 6 > 5 =path-cost(abfet). 8. Explain briefly how the local search technique Simulated Annealing works (this is a general question which has nothing to do with the above state space graph). [4] Simulated Annealing search is inspired by the process of annealing in metallurgy, with the aim to combine the benefits of hill-climbing search (efficiency) and random search (completeness). It works by selecting randomly from the successor moves, always accepting the selected move if it resulted in an improvement, and otherwise accepting it with a probability less than. The probability decreases exponentially with the amount by which the evaluation is worsened; it also decreases as the temperature goes down.

5 Part III - Chapter 6 All the questions on this page are about the following adversarial search tree. State evaluation heuristic values for the max player are provided in the form of numbers following the letter labels of the states (e.g., A3 indicates that the heuristic value of state A is 3). The order in which successors are generated is from left to right. Example: A generates first B, then C, and finally D. The non-quiescent states are circled in bold; assume that all moves from these states are a threat or counter a threat (which implies that they need to be explored). A3 B3 C4 D4 E F3 G H5 I J3 K3 L3 M8 N6 O P Q R7 S 9. What is the advantage of adding a move-ordering heuristic to a minimax algorithm with alpha-beta pruning? [] It on average speeds up the algorithm by reducing the number of nodes that need to be examined. This is achieved by increasing the chance that good moves are examined early on, which results in increased alpha-beta pruning. 0. Give the execution trace of Depth Limited Minimax employing Alpha-Beta pruning with depth-limit= and root-node=a and Quiescence Search with QSdepth-limit= (QSABDLM(A,,,, )). [9] Define: DLM = QSABDLM, MinV = QSABMinV, MaxV = QSABMaxV call open eval value α, β best action,value DLM(A,,,, ) BCD B MinV(B,,,, )= (SSS), B, CD C MinV(C,,,, )=, B, D D MinV(D,,,, )=6 (SSS & QS) 6, D,6 AD MinV(C,,,, ) FGH F MaxV(F,0,,, )=3, 3 F,3 GH G MaxV(G,0,,, 3)= (prune), 3 G,. What is the Principal Variant (PV) found by QSABDLM(A,,,, )? [] ADI

6 . Give the execution trace of Iterative Deepening Minimax employing Alpha-Beta pruning with depthlimit= and root-node=a and History Table Move Ordering Heuristic and Quiescence Search with QSdepth-limit= (QSHTABIDM(A,,,, )). [0] Define: DLM = QSHTABDLM, MinV = QSHTABMinV, MaxV = QSHTABMaxV call open eval value α, β best action,value Iteration DLM(A,,,, ) B0C0D0 B MinV(B,0,,, )=3 3, B,3 C0D0 C MinV(C,0,,3, )= 3, B,3 D0 D MinV(D,0,,3, )=4 4, D,4 [AD:] AD MinV(C,0,,3, ) F0G0H0 F MaxV(F,0,0,3, )=3 3, 3 F,3 G0H0 G MaxV(G,0,0,3, 3)= (prune) 3, 3 G, [CG:] Iteration DLM(A,,,, ) DB0C0 D MinV(D,,,, )=6 (SSS & QS) 6, D,6 [DI:,IN:] B0C0 B MinV(B,,,6, )= (SSS) 6, D,6 [BE:] C0 C MinV(C,,,6, )= 6, D,6 [AD:] AD MinV(C,,,6, ) GF0H0 G MaxV(G,0,,6, )= (prune) 6, G, [CG:] 3. What is the PV found by QSHTABIDM(A,,,, )? [] ADI 4. From ALL the information available to you in the adversarial search tree, did having the Quiescence Search in the above traces you executed improve the search results? [] No, because the Quiescence Search caused AD to be returned as best move instead of AB, while AB is the better move if you look at all the information provided: DLM(A,4)= with PV=ABEJO.

7 The final three questions are about the following adversarial chance tree. MAX CHANCE MIN A a a 3 a B C D Calculate the EXPECTIMINIMAX values for nodes B, C and D in the above adversarial chance tree. Show your calculations! [3] EXPECTIMINIMAX(B) =.5 * * = =.5 EXPECTIMINIMAX(C) =.3 * * +. * 7 = = 3. EXPECTIMINIMAX(D) =.8 * +. * 8 = =.4 6. Which action will MAX choose, a, a, or a 3? Explain your answer! [] MAX will choose action a because it has the highest EXPECTIMINIMAX value. 7. If the utility values given for MIN were multiplied with a positive constant c, which action would MAX then choose? [] MAX would still choose action a because multiplying with a positive constant is a positive linear transformation and such transformations do not change decisions made on the basis of EXPECTIMINIMAX values.