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Search Search plays a key role in many parts of AI. These algorithms provide the conceptual backbone of almost every approach to the systematic exploration of alternatives. There are four classes of search algorithms, which differ along two dimensions: –First, is the difference between uninformed (also known as blind) search and then informed (also known as heuristic) searches. Informed searches have access to task-specific information that can be used to make the search process more efficient. –The other difference is between any solution searches and optimal searches. Optimal searches are looking for the best possible solution while any- path searches will just settle for finding some solution.

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Graphs Graphs are everywhere; E.g., think about road networks or airline routes or computer networks. In all of these cases we might be interested in finding a path through the graph that satisfies some property. It may be that any path will do or we may be interested in a path having the fewest "hops" or a least cost path assuming the hops are not all equivalent.

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Romania graph

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Formulating the problem On holiday in Romania; currently in Arad. Flight leaves tomorrow from Bucharest. Formulate goal: –be in Bucharest Formulate problem: –states: various cities –actions: drive between cities Find solution: –sequence of cities, e.g., Arad, Sibiu, Fagaras, Bucharest

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Another graph example However, graphs can also be much more abstract. A path through such a graph (from a start node to a goal node) is a "plan of action" to achieve some desired goal state from some known starting state. It is this type of graph that is of more general interest in AI.

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Problem solving One general approach to problem solving in AI is to reduce the problem to be solved to one of searching a graph. To use this approach, we must specify what are the states, the actions and the goal test. A state is supposed to be complete, that is, to represent all the relevant aspects of the problem to be solved. We are assuming that the actions are deterministic, that is, we know exactly the state after the action is performed.

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Goal test In general, we need a test for the goal, not just one specific goal state. –So, for example, we might be interested in any city in Germany rather than specifically Frankfurt. Or, when proving a theorem, all we care is about knowing one fact in our current data base of facts. –Any final set of facts that contains the desired fact is a proof.

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Vacuum cleaner?

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Vacuum cleaner

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Formally… A problem is defined by four items: 1.initial state e.g., "at Arad" 2.actions and successor function S: = set of action-state tuples –e.g., S(Arad) = {(goZerind, Zerind), (goTimisoara, Timisoara), (goSilbiu, Silbiu)} 3.goal test, can be –explicit, e.g., x = "at Bucharest" –implicit, e.g., Checkmate(x) 4.path cost (additive) –e.g., sum of distances, or number of actions executed, etc. –c(x,a,y) is the step cost, assumed to be ≥ 0 A solution is a sequence of actions leading from the initial state to a goal state

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Example: The 8-puzzle states? actions? goal test? path cost?

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Example: The 8-puzzle states? locations of tiles actions? move blank left, right, up, down goal test? = goal state (given) path cost? 1 per move

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Romania graph

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Tree search example

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Tree search algorithms Basic idea: –Exploration of state space by generating successors of already-explored states (i.e. expanding states)

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Implementation: states vs. nodes A state is a (representation of) a physical configuration A node is a bookeeping data structure constituting of state, parent node, action, path cost g(x), depth The Expand function creates new nodes, filling in the various fields and using the SuccessorFn of the problem to create the corresponding states.

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The fringe The collection of nodes that have been generated but not yet expanded is called fringe. (outlined in bold) We will implement collection of nodes as queues. The operations on a queue are as follows: Empty?(queue) check to see whether the queue is empty First(queue) returns the first element Remove-First(queue) returns the first element and then removes it Insert(element, queue) inserts an element into the queue and returns the resulting queue InsertAll(elements, queue) inserts a set of elements into the queue and returns the resulting queue

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Implementation public class Problem { Object initialState; SuccessorFunction successorFunction; GoalTest goalTest; StepCostFunction stepCostFunction; HeuristicFunction heuristicFunction; …

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Some pseudocode interpretations Successor-Fn[problem](State[node]) is in fact: problem. Successor-Fn(node.State) or problem.getSuccessorFunction().getSuccessors( node.getState() );

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Implementation: general tree search The queue policy of the fringe embodies the strategy.

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Search strategies A search strategy is defined by picking the order of node expansion Strategies are evaluated along the following dimensions: –completeness: does it always find a solution if one exists? –time complexity: number of nodes generated –space complexity: maximum number of nodes in memory –optimality: does it always find a least-cost solution? Time and space complexity are measured in terms of –b: maximum branching factor of the search tree –d: depth of the least-cost solution –m: maximum depth of the state space

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Uninformed search strategies Uninformed do not use information relevant to the specific problem. Breadth-first search Uniform-cost search Depth-first search Depth-limited search Iterative deepening search

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Breadth-first search TreeSearch(problem, FIFO-QUEUE()) results in a breadth-first search. The FIFO queue puts all newly generated successors at the end of the queue, which means that shallow nodes are expanded before deeper nodes. –I.e. Pick from the fringe to expand the shallowest unexpanded node

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Breadth-first search Expand shallowest unexpanded node Implementation: –fringe is a FIFO queue, i.e., new successors go at end

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Breadth-first search Expand shallowest unexpanded node Implementation: –fringe is a FIFO queue, i.e., new successors go at end

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Breadth-first search Expand shallowest unexpanded node Implementation: –fringe is a FIFO queue, i.e., new successors go at end

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Properties of breadth-first search Complete? –Yes (if b is finite) Time? –1+b+b 2 +b 3 +… +b d + b(b d -1) = O(b d+1 ) Space? –O(b d+1 ) (keeps every node in memory) Optimal? –Yes (if cost is a non-decreasing function of depth, e.g. when we have 1 cost per step)

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DepthNodesTimeMemory sec1 Megabyte 4111,10011 sec106 Megabyte min10 Gigabyte hours1 Terabyte days101 Terabyte years10 Petabyte ,5231 Exabyte Suppose b=10, 10,000 nodes/sec, 1000 bytes/node

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Uniform-cost search Expand least-cost unexpanded node. The algorithm expands nodes in order of increasing path cost. Therefore, the first goal node selected for expansion is the optimal solution. Implementation: –fringe = queue ordered by path cost (priority queue) Equivalent to breadth-first if step costs all equal Complete? Yes, if step cost ≥ ε (I.e. not zero) Time? number of nodes with g ≤ cost of optimal solution, O(b C*/ ε ) where C * is the cost of the optimal solution Space? Number of nodes with g ≤ cost of optimal solution, O(b C*/ ε ) Optimal? Yes – nodes expanded in increasing order of g(n)

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Try it here

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Depth-first search Expand deepest unexpanded node Implementation: –fringe = LIFO queue, i.e., put successors at front

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Properties of depth-first search Complete? No: fails in infinite-depth spaces, spaces with loops –Modify to avoid repeated states along path complete in finite spaces Time? O(b m ): terrible if m is much larger than d – but if solutions are dense, may be much faster than breadth-first Space? O(bm), i.e., linear space! Optimal? No

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DepthLimitedSearch (int limit) { stackADT fringe; insert root into the fringe do { if (Empty(fringe)) return NULL; /* Failure */ nodePT = Pop(fringe); if (GoalTest(nodePT->state)) return nodePT; /* Expand node and insert all the successors */ if (nodePT->depth < limit) { insert into the fringe Expand(nodePT) } while (1); } Depth-limited search

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Iterative deepening search IterativeDeepeningSearch () { for (int depth=0; ; depth++) { node=DepthLimitedtSearch(depth); if ( node != NULL ) return node; } }

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Iterative deepening search l =0

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Iterative deepening search l =1

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Iterative deepening search l =2

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Iterative deepening search l =3

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Iterative deepening search Number of nodes generated in a depth-limited search to depth d with branching factor b: N DLS = b 0 + b 1 + b 2 + … + b d-2 + b d-1 + b d Number of nodes generated in an iterative deepening search to depth d with branching factor b: N IDS = (d+1)b 0 + d b 1 + (d-1)b 2 + … + 3b d-2 +2b d-1 + 1b d For b = 10, d = 5, –N DLS = , , ,000 = 111,111 –N IDS = , , ,000 = 123,456 Overhead = (123, ,111)/111,111 = 11%

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Properties of iterative deepening search Complete? Yes Time? (d+1)b 0 + d b 1 + (d-1)b 2 + … + b d = O(b d ) Space? O(bd) Optimal? Yes, if step cost = 1

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Summary of algorithms

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Repeated states Failure to detect repeated states can turn a linear problem into an exponential one!

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Graph search

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Class problem You have three jugs, measuring 12 gallons, 8 gallons, and 3 gallons, and a water faucet. You can fill the jugs up, or empty them out from one another or onto the ground. You need to measure out exactly one gallon.

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