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# Berkeley CS 188 - Section Handout 3 Solutions

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CS 188Fall 2019 Section Handout 3 SolutionsNote: we have provided more material than can be covered in discussion that can be used as practice at home.Q1. CSP: Air Traffic ControlWe have five planes: A, B, C, D, and E and two runways: international and domestic. We would like to schedulea time slot and runway for each aircraft to either land or take off. We have four time slots: {1, 2, 3, 4} for eachrunway, during which we can schedule a landing or take off of a plane. We must find an assignment that meetsthe following constraints:• Plane B has lost an engine and must land in time slot 1.• Plane D can only arrive at the airport to land during or after time slot 3.• Plane A is running low on fuel but can last until at most time slot 2.• Plane D must land before plane C takes off, because some passengers must transfer from D to C.• No two aircrafts can reserve the same time slot for the same runway.(a) Complete the formulation of this problem as a CSP in terms of variables, domains, and constraints (bothunary and binary). Constraints should be expressed implicitly using mathematical or logical notationrather than with words.Variables: A, B, C, D, E for each plane.Domains: a tuple (runway type, time slot) for runway type ∈ {international, domestic} and time slot∈ {1, 2, 3, 4}.Constraints:B = 1D ≥ 3A ≤ 2D < CA 6= B 6= C 6= D 6= E(b) For the following subparts, we add the following two constraints:• Planes A, B, and C cater to international flights and can only use the international runway.• Planes D and E cater to domestic flights and can only use the domestic runway.(i) With the addition of the two constraints above, we completely reformulate the CSP. You are giventhe variables and domains of the new formulation. Complete the constraint graph for this problemgiven the original constraints and the two added ones.1Variables: A, B, C, D, E for each plane.Domains: {1, 2, 3, 4}Explanation of Constraint Graph:We can now encode the runway information intothe identity of the variable, since each runwayhas more than enough time slots for the planesit serves. We represent the non-colliding timeslot constraint as a binary constraint betweenthe planes that use the same runways.Constraint Graph:A BC DE(ii) What are the domains of the variables after enforcing arc-consistency? Begin by enforcing unaryconstraints. (Cross out values that are no longer in the domain.)A 1 2 3 4B 1 2 3 4C 1 2 3 4D 1 2 3 4E 1 2 3 42 Games(a) Consider the zero-sum game tree shown below. Triangles that point up, such as at the top node (root),represent choices for the maximizing player; triangles that point down represent choices for the minimizingplayer. Assuming both players act optimally, fill in the minimax value of each node.4152 7 56 48 310324(b) Which nodes can be pruned from the game tree above through alpha-beta pruning? If no nodes can bepruned, explain why not. Assume the search goes from left to right; when choosing which child to visitfirst, choose the left-most unvisited child.2(c) (optional) Again, consider the same zero-sum game tree, except that now, instead of a minimizing player,we have a chance node that will select one of the three values uniformly at random. Fill in the expectimaxvalue of each node. The game tree is redrawn below for your convenience.8152 7 56 48 31057 8(d) (optional) Which nodes can be pruned from the game tree above through alpha-beta pruning? If no nodescan be pruned, explain why not. No nodes can be pruned. There will always be the possibility that anas-yet-unvisited leaf of the current parent chance node will have a very high value, which increases theoverall average value for that chance node. For example, when we see that leaf 4 has a value of 2, which ismuch less than the value of the left chance node, 7, at this point we cannot make any assumptions abouthow the value of the middle chance node will ultimately be more or less in value than the left chance node.As it turns out, the leaf 5 has a value of 15, which brings the expected value of the middle chance node to8, which is greater than the value of the left chance node. In the case where there is an upper bound to thevalue of a leaf node, there is a possibility of pruning: suppose that an upper bound of +10 applies only tothe children of the rightmost chance node. In this case, after seeing that leaf 7 has a value of 6 and leaf 8has a value of 5, the best possible value that the rightmost chance node can take on is6+5+103= 7, whichis less than 8, the value of the middle chance node. Therefore, it is possible to prune leaf 9 in this case.3 Nonzero-sum Games1. Let’s look at a non-zero-sum version of a game. In this formulation, player A’s utility will be representedas the first of the two leaf numbers, and player B’s utility will be represented as the second of the two leafnumbers. Fill in this non-zero game tree assuming each player is acting optimally.2. Which nodes can be pruned from the game tree above through alpha-beta pruning? If no nodes can be3pruned, explain why not. No nodes can be pruned. Because this game is non-zero-sum, there can exist aleaf node anywhere in the tree that is good for both player A and player

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