CS
Lecture-03-04-Uninformed_Search

# Not queuing up a node that has a state which we have

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- not queuing-up a node that has a state which we have already visited: may yield suboptimal solution - simply avoiding to go back to our parent: looks promising, but we have not proven that it works Solution? do not enqueue a node if its state matches the state of any of its parents (assuming path costs>0). Indeed, if path costs > 0, it will always cost us more to consider a node with that state again than it had already cost us the first time. Is that enough?? CS561 - Lectures 3-4 - Macskassy - Fall 2010

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Example G From: CS561 - Lectures 3-4 - Macskassy - Fall 2010
Breadth-First Search Solution From: CS561 - Lectures 3-4 - Macskassy - Fall 2010

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Uniform-Cost Search Solution From: CS561 - Lectures 3-4 - Macskassy - Fall 2010
Queueing in Uniform-Cost Search In the previous example, it is wasteful (but not incorrect) to queue-up three nodes with G state, if our goal if to find the least-cost solution: Although they represent different paths, we know for sure that the one with smallest path cost (9 in the example) will yield a solution with smaller total path cost than the others. So we can refine the queueing function by: - queue-up node if 1) its state does not match the state of any parent and 2) path cost smaller than path cost of any unexpanded node with same state in the queue (and in this case, replace old node with same state by our new node) Is that it?? CS561 - Lectures 3-4 - Macskassy - Fall 2010

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Example G 100 5 D 5 E 5 F 5 S A C 1 5 B 1 1 # State Depth Cost Parent 1 S 0 0 - CS561 - Lectures 3-4 - Macskassy - Fall 2010
Example G 100 5 D 5 E 5 F 5 S A C 1 5 B 1 1 # State Depth Cost Parent 1 S 0 0 - 2 A 1 1 1 3 C 1 5 1 Insert expanded nodes Such as to keep open queue sorted Black = open queue Grey = closed queue CS561 - Lectures 3-4 - Macskassy - Fall 2010

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Example G 100 5 D 5 E 5 F 5 S A C 1 5 B 1 1 # State Depth Cost Parent 1 S 0 0 - 2 A 1 1 1 4 B 2 2 2 3 C 1 5 1 Node 2 has 2 successors: one with state B and one with state S. We have node #1 in closed with state S; but its path cost 0 is smaller than the path cost obtained by expanding from A to S. So we do not queue-up the successor of node 2 that has state S. CS561 - Lectures 3-4 - Macskassy - Fall 2010
Example G 100 5 D 5 E 5 F 5 S A C 1 5 B 1 1 # State Depth Cost Parent 1 S 0 0 - 2 A 1 1 1 4 B 2 2 2 5 C 3 3 4 6 G 3 102 4 Node 4 has a successor with state C and Cost smaller than node #3 in open that Also had state C; so we update open To reflect the shortest path. CS561 - Lectures 3-4 - Macskassy - Fall 2010

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Example G 100 5 D 5 E 5 F 5 S A C 1 5 B 1 1 # State Depth Cost Parent 1 S 0 0 - 2 A 1 1 1 4 B 2 2 2 5 C 3 3 4 7 D 4 8 5 6 G 3 102 4 CS561 - Lectures 3-4 - Macskassy - Fall 2010
Example G 100 5 D 5 E 5 F 5 S A C 1 5 B 1 1 # State Depth Cost Parent 1 S 0 0 - 2 A 1 1 1 4 B 2 2 2 5 C 3 3 4 7 D 4 8 5 8 E 5 13 7 6 G 3 102 4 CS561 - Lectures 3-4 - Macskassy - Fall 2010

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Example G 100 5 D 5 E 5 F 5 S A C 1 5 B 1 1 # State Depth Cost Parent 1 S 0 0 - 2 A 1 1 1 4 B 2 2 2 5 C 3 3 4 7 D 4 8 5 8 E 5 13 7 9 F 6 18 8 6 G 3 102 4 CS561 - Lectures 3-4 - Macskassy - Fall 2010
Example G 100 5 D 5

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• Fall '09
• Artificial Intelligence, Depth-first search, Search algorithms, Search algorithm, Graph algorithms, Uniform-Cost Search

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