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UH COSC 4368 - COSC 6368: Artificial Intelligence

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G1CDG2Review 10/17/2017COSC 6368: Artificial Intelligence1) Best first Search and A* [10]a) Best-First-Search (using function h only) [3]b) A* (using f=g+h)[4]3. Comparision of Seach AlgorithmsReview 10/17/2017COSC 6368: Artificial Intelligence1) Best first Search and A* [10]Consider the search space below, where S is the start node and G1 and G2 satisfy the goaltest. Arcs are labeled with the cost of traversing them and the estimated cost to a goal (the h function itself) is reported inside nodes.For each of the following search strategies, indicate which goal state is reached (if any) and list, in order, all the states popped off of the OPEN list. When all else is equal, nodes should be removed from OPEN in alphabetical order.a) Best-First-Search (using function h only) [3]Goal state reached: G2 [1]States popped off OPEN: S, E, G2 [2]b) A* (using f=g+h)[4]Goal state reached: G2 [1]States popped off OPEN: S, A, B, G1 [3]d) Assume you have 2 admissible heuristics h1(s) and h2(s) are given for a given seach problem. Is h3(s)=min(h1(s),h2(s)) also admissible? Would you prefer using h2 or using h3 in conjuction with A*? Give reasons for your anwers[4]. 27121529 B 5A 90AAG1 0 E8 C8D4G2 0S23841451Yes, h3 is admissible. If h1 and h2 always underestimate the “true” cost thenthe lesser of the two will certainly underestimate the true cost as well; therefore, h3 is admissible.I will prefer h2, because h2 is always greater equal than h3 and therefore it provides a closer approximation of the true cost. As a matter of fact, h2 dominates h3, which translates into equal or better efficiency of the search, as discussed on the bottom of page 106 of our textbook.a) Assume you apply randomized hill climbing to a minimization involving a continuous, differentiable function that has 3 minima. Will it always find the optimal solution? Give reasons for your answer! [3]2No, HC might climb down the wrong minimum depending on the chosen starting pointb) What is the “main” difference between simulated annealing and randomized hill climbing? [2]3… SA does allow downward steps…d) Assume you apply a version of depth first search that checks for repeated states on the current path1, but which does not use a depth bound to the 8-puzzle. Will it always find a solution if a solution exists (assuming that there are enough computational resources)? [6]1 This version backtracks, if a loop in the current path is encountered.4Yes!Because the search-tree for the 8-puzzle is finite and because the algorithm checks for loops in the current path the algorithm will sooner or later stop moving forward, backtrack, and find the solution eventually.e) When does A* terminate? Be precise! 52.Reinforcement Learning]Consider the following World called ABC is given:Give the Bellman equation for states 1 and 4 of the ABC world! [3]U(1)= 5 + *U(4) [1]U(4)= 3 + *max (U(2)*0.3+ U(3)*0.1+U(5)*0.6, U(1)*0.4+U(5)*0.6) [2]No partial credit! b) Now we apply temporal difference learning, assuming the agent starts in state 2 and applies the operator sequence w-y(ending up in state 2)-w; assume the initial utilities are0; what are the new utilities; also assume =0.5 and =0.5)?6General Update Formula: U(s):=(1-)*U(s)+ *(R(s)+ *U(s’))U(2)=0+ 0.5(-2+0))=-1U(3)=0+0.5*(3+0.5*-1)=1.25U(2)=0.5*-1 + 0.5*(-2+0.5*1,25)=-0.5+0.5*-1.375=-1.1875Remark; if  would have been 1, U(2) would be greater than 1!c) Assume you use temporal difference learning in conjunction with a random policy which choses actions randomly assuming a uniform distribution. Do you believe that the estimations obtained are a good measurement of the “goodness” of states, that tell an intelligent agent (assume the agent is smart!!) what states he/she should/should not visit? Give reasons for your answer! [3]7Not always; as we assume an intelligent agent will take actions that lead to good states and avoids bad states, the agent that uses the random policy might not recognize that a state is a good state if both good and bad states are successors of this state; for example, S2: R=+100S1:R=-1 S3: R=-100Due to the agent’s policy the agent will fail to realize the S1 is a good state, as the agent’saverage reward for visiting the successor states of S1 is 0; an intelligent agent would almost always go from S1 to S2, making S1 a high utility state with respect to TD-learning. d) What role does the learning rate  play in temporal difference learning; how does running temporal difference learning with low values of  differ from running it with high values of ? [2]It determines how quickly our current beliefs/estimations are updated based on new evidence. e) Assume you run temporal difference learning with high values of —what are the implications of doing that? [2]If  is high the agent will more focus on its long term wellbeing, and will shy away from taking actions—although they lead to immediate rewards—that will lead to the medium and long term suffering of the agent. 83. Comparision of Seach AlgorithmsCompare Traditional Hill Climbing/Randomized Hill Climbing and Best-first Search! What are the main differences between the two approaches?9n be the size of the search space, the number of nodes in the search tree b the branching factor, the number of successors m is the length of the longest path in the search tree Randomized HCBest First SearchThe way they searchExplore a single pathCan explore multiple paths in generalOnly moves forward, cannot move backwardJumps between states; can explore multiple paths in parallel!Storage O(1) / O(m)2O(n)Runtime O(m) but might stop prematurely, fastO(n) in the worst case, not fastFinding solutions in finite search spacesmight terminate prematurely; might go into the wrong direction and get stuckYesFind global optimumno no, terminates permaturelyParameter SelectionChoosing neighbor hood size andsampling ratefor RHC is Straight Forward2 Only if it is necessary to return the solution path, as in the 8-puzzle problem!10challengingIncorperatingHeuristics Needs good state evaluation functionNeed good state evaluation function Other RHC is a probabilistic algorithm (usually) returns different solutions in different


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