Home / Questions / Intelligent Agents 20pts Circle True or False for each statement True False An agent that ...
Intelligent Agents (20pts)Circle True or False for each statement.1. [True/False] An agent that senses only partial information about the state cannot be rational.2. [True/False] There exist task environments in which no pure reflex agent can behaverationally.3. [True/False] There exists a task environment in which every agent is rational.4. [True/False] The input to an agent program is the same as the input to the agent function.5. [True/False] Every agent function is implementable by some program/machine combination.6. [True/False] Suppose an agent selects its action uniformly at random from the set of possibleactions. There exists a deterministic task environment in which this agent is rational.7. [True/False] It is possible for a given agent to be perfectly rational in two distinct taskenvironments.8. [True/False] Every agent is rational in an unobservable environment.9. [True/False] A perfectly rational pokerplayingagent never loses.10. [True/False] A logical agent can behave rationally even in partially observable environmentConstraint Satisfaction (20pts)Use the given the constraint graph of the 4-queens problem for the board shownto answer the questions below. Each variable represents a queen in itscorresponding column, e.g. Q1 is for the queen assign to column 1. The domain ofeach variable is the number of the row.1. [10pts] Draw the backtracking search tree performing Arc Consistency between eachassignment to solve this 4 queens board, given the position of the first queen Q1 is row 2.one possible tree:? no backtrack searching tree 4? change the meanings of symbols Q1,Q2,Q3,Q4 3? illegal node on the tree 2? some essential node missing from the tree 2? after finding the solution, continue search 2? wrong node position in the tree 1? no conclusion of the solution 2? wrong domain value after Arc Consistency check1per variable? typo 1? only right conclusion of the solution +42. [10pts] Draw the backtracking search tree using forward checking to solve this 4queens board, given the position of Q1 is row 1.? no tree 4? wrong conclusion2? backtracking not involved/Q2=4 branch miss 5? extra work(Q1=2) 1? extra nodes(no forward checking for duplicate and diagonal)3? extra nodes(no forward checking for diagonal)2? not a single tree. 1? start point miss 2?? diagonal constraint totally ignore + 4? Violation assigning a value to a particular variable at each tier.+2Search(20pts)The game of Nim involves two players (A & B) who alternately take matchsticks from piles.On every turn, the player to move can take any number of matchsticks >=1 (so must take atleast one) from one pile, up to and including the total number in the pile. The goal, at least in thisvariant, is to force the opponent to take the last matchstick. In what comes below, we willalways assume the initial state has three piles, with one matchstick in each of the firsttwo piles and two matchsticks in the third pile. Assume for now that there is only one playerwhose goal is to remove all of the matchsticks from the piles using only legal operators.1. [10pts] Draw the state space as a graph, annotating each state with the number ofmatchsticks in each pile, such as 012 for 0 in the first pile, 1 in the second, and 2 in the third.You need not annotate the links with operator names.? When used above form, there are 12 nodes with 24 edges-1 when missed 1~3 edges (-2 when missed 4~6 edges, â€¦)? When used â€˜Treeâ€™ form, there are 46 nodes with 45 edges.-1 when missed 1~5 edges (-2 when missed 6~10 edges, â€¦)? -1 for the bidirected graph2. [5pts] Consider the heuristic h = total number of matchsticks left in the three piles, whichwould yield 3 for the example state above 012. Is this heuristic admissible for A*? Why orwhy not?No, an admissible heuristic never overestimates the actual cost to get to the goalfrom any state. As a counterexample, consider a state in which there are twomatchsticks in one pile and the two other piles are empty (002). In this state, wehave h=2, while the actual cost is 1 because the player can take both matchsticks atthe same time (in one move). Thus, there exist some states in which the heuristicfunction h overestimates the actual cost of reaching to the goal.? +2: Saying No? +3: correct reason? +1: Saying Yes but mentioning â€˜admissible heuristic never overestimates theactual cost to get to the goal from any stateâ€™3. [5pts] Now suppose we include an operator that adds a single matchstick to any one of thepiles. If the heuristic above is used as the state evaluation function, will using a hill climbingsearch technique be complete? Why or why not?Yes. Completeness means that we can always find a solution if one exists. The goalhere is to reach 0 matchsticks, so hill climbing must minimize; that is, in each state,we pick the successor whose evaluation functionâ€™s value is smallest. The key pointhere is that every state except the goal (000) has a successor with a smallerevaluation because, in each state, the player can take one or more matchsticks.Therefore, there are no local minima in the search space and hill climbing alwaysreaches the global minimum, which is the goal.? + 2: Saying Yes? + 3: correct reason? +1: Saying No but mentioning â€˜local optimaâ€™Adversarial Search (20pts)Now letâ€™s shift to the full twopersonNim game (players A & B), with the winner being theplayer who does not take the last available matchstick. On every turn, the player to move cantake any number of matchsticks >=1 (so must take at least one) from one pile, up to andincluding the total number in the pile. The goal, at least in this variant, is to force the opponentto take the last matchstick. In what comes below, we will always assume the initial statehas three piles, with one matchstick in each of the first two piles and two matchsticks inthe third pile.1. [10pts] Redraw the state space in twopersongame form, with alternating tiers for the movesby the two players, annotating each state with the number of matchsticks in each pile. You candraw it as a tree or a graph, whichever you find more convenient. You need not annotate thelinks with operator names.? When used above form, there are 12 nodes with 24 edges-1 when wrong or missed 1~3 edges/nodes (-2 when missed 4~6 edges, â€¦)? When used â€˜Treeâ€™ form, there are 46 nodes with 45 edges.-1 when wrong or missed 1~5 edges/nodes (-2 when missed 6~10 edges/nodes,â€¦)2. [5pts] Assuming that terminal states are evaluated with respect to the first player (A), and thata win is worth 1 and a loss 1,annotate the nodes in your tree/graph above with their minimaxvalues.? +3 for minmax root value? +1 all correct 1nodes? +1 all correct 1 nodes3. [5pts] If alphabetais used, and initialized with [1for alpha/loss, 1 for beta/win], and playerB is to move at some node in the tree/graph, and the first successor node explored from thatnode returns a value of 1,does a cutoff (pruning) occur? Why or why not?Yes, we will have a cutoff. The node where B is to move is a Min node and its value iscomputed as the minimum value of its successors. Since alpha-beta is initializedwith [-1, 1], the smallest possible value is -1. When exploring first successor yields-1, so we do not need to continue exploring the other children.? +2: Yes? +3: Correct Answer? +1: For B is is a Min node or the smallest possible value is -1.? +1: For alpha>=beta? -1: For additional false statementProblem Solving (20pts)Suppose you are a farmer with a boat, on one side of a river. You want to deliver three things: awolf, a sheep, and a cabbage to the other side of the river. You can only take one thing withyou on your boat to cross at a time. If you leave the sheep alone with the cabbage, you can kissyour cabbage goodbye since the sheep will eat it. If the sheep and the wolf are left alonetogether, let’s just say youâ€™ll have a happy wolf that eats up the sheep. But the wolf does not eatcabbage. How do you deliver all the three things without losing any of them?Here we are interested in setting up the problem, and will express it in terms of states, whichare snapshots of the world and actions, which transform one state into another.1. [4pts] Write down the definition of a state in this problem. Give the initial and goal states.Hint: you are suggested to use 4 variables.? +2: Definition of a state? +1: Initial state? +1: Goal stateWhether the things or the farmer is on the starting side.The initial state is {F, W, S, C} which denotes the Farmer, the Wolf, the Sheep, and theCabbage are on the starting side.2. [4pts] Write down all the legal states (in which any of the three items is not lost).[Each two states +1]1.{ F, W, S, C}2.{W, C}3.{F, W, C}4.{W}5.{C}6.{F, W, S}7.{F, S, C}8.{S}9.{F, S}10.{}3. [4pts] Define the actions in this problem.Move one thing 3(or nothing) 1to the other side.4. [5pts] The problem can be represented as a graph which contains the states as nodes, andthe state transitions as edges. Draw the graph for this problem (you can omit the actions on theedges between nodes).The simplest graph with states is given by (The actions are omitted since they are obvious):per 2 transitions +1 Each5. [3pts] Provide one sequence of actions that solves this problem with least times of crossing.F, S
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