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CS 385 Fall 2006 Chapter 3. Structures and Strategies for State Space Search. Where are we?. Predicate calculus: a way to describe objects and relationships Inference rules: a way to infer new knowledge, defining a state space that is searched to find a solution to a problem Strategy
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CS 385 Fall 2006Chapter 3 Structures and Strategies for State Space Search
Where are we? Predicate calculus: • a way to describe objects and relationships Inference rules: • a way to infer new knowledge, defining a state space that is searched to find a solution to a problem Strategy • generate all possible elements of the state space and see if your answer is there • works for tic-tac-toe • doesn't work for chess Goal: intelligent ways to search a state space Tool: state space graphs.
Famous example: Konigsburg bridge problem Is there a walk that crosses each bridge exactly once? Can we represent this in a better way?
Graph of the Königsberg bridge system: Predicate calculus representation: connect(i1,i2, b1) ....
Euler's Proof A node is odd or even depending on the number of arcs leading into it Odd degree nodes can only be at the beginning of the path A walk must contain 0 or 2 odd nodes (why?) Is there a path? Does predicate calculus suffice for this argument? No, no notion of odd/even We need graph theory to do more with this.
What are N, A, S, and GD for tic-tac-toe? N could be all possible configurations of 0s and 1s or just reachable ones. A: allowable moves between boards S: empty board, GD winning boards
The 8 Puzzle N = 3 x 3 configurations of tiles 1-8 and 1 blank Start state: Goal state: Arcs: move the blank up/down/left/right
Figure 3.6:State space of the 8-puzzle generated by “move blank” operations. Is this like tic-tac toe? No, you might go around in circles
Traveling Salesperson N= cities A = paths between cities with weights (mileage, time, cost) S = home GD = home Path: visit each exactly once Goal: minimize something
Figure 3.8:Search of the traveling salesperson problem. Each arc is marked with the total weight of all paths from the start node (A) to its endpoint.
Traveling Salesperson How many potential paths? Can we solve this for 50 states? Of great interest to people who like algorithms because it gets large so fast. Exhaustive search out of the question
Figure 3.9:The traveling salesperson problem with the nearest neighbor path in bold. Note that this path (A, E, D, B, C, A), at a cost of 550, is not the shortest. The comparatively high cost of arc (C, A) defeated the heuristic.
Data- versus Goal-Driven for Finding a Route from Aurora to LA Where you start. E.g. routes from Aurora to LA Data driven (forward chaining) From Aurora one can get to x, y, z... From x one can get to .. From y one can get to ... Keep checking to see if LA ends up in one of the destinations Goal driven (backward chaining) One can get to LA from x, y, z... One can get to x from .. One can get to y from Keep checking to see if Aurora ends up in one of the sources
Better examples Lineage: am I related to Reverend Thomas Carter? Which is better, data or goal driven? Could I prune extraneous paths? Is this the same as "Am I related to Thomas Jefferson"?
Medical Diagnosis "Do I have strep throat" vs. "What disease do I have" strep symptoms vs. lots of symptoms take a culture vs. run a spectrum of tests Which is data-driven/forward chaining? Which is goal-driven/backward chaining?
Backtracking A technique for systematically trying all paths through a state space. Begin at start and pursue a path until goal or dead end If dead end, backtrack to most recent node with unexamined siblings E.g. is 6 in this tree? 1 Possible trials: 1 2 4 5 3 6 2 3 1 3 7 6 1 3 6 4 5 6 7 How do we pick the "best" path?
Function backtrack algorithm (general, no examination order specified) typo in book, p should be≠ CS: current state SL: states in current path NSL states awaiting evaluation DE: dead ends
Observations No order is specified for adding nodes to NSL (opportunity for intelligence) SL gives us the path to the current solution (hence to the goal at the end) When C is the current state, F is not added to NSL (because it is in DE)
How used on a maze? • ifgoal return success • else • try north • try east • try south • try west • backtrack • Track: • where you are • where you can go • from each state • where you came from • visited states
Function breadth_first search algorithm X is CS in backtrack, closed = DE + SL
breadth_first_search on Figure 3.13 open closed [A] [ ] [B C D] [A] [C D E F] [B A] [D E F G H] [C B A] [E F G H I J] [D C B A] [F G H I J K L] [E D C B A] [G H I J K L M] [F E D C B A] [H I J K L M N] [G F E D B C A] [I J K L M N O P] [F T L S K E B A] [J K L M N O P Q] [M F T L S K E B A] [K L M N O P Q R] [J M F T L S K E B A]
Observations Cleaner But no path to start state Solution: associate the parent with each node. E.g. [B, C, D] → [(B,A), (C,A), (D,A)] [C, D, E, F] → [(C,A), (D,A), (B,A), (F, B)] Is the first algorithm ever better?
depth_first_search on Figure 3.13 Put new nodes at the beginning of the open list. open closed [A] [ ] [B C D] [A] [E F C D] [B A] [K L F C D] [E B A] [S L F C D] [K E B A] [L F C D] [S K E B A] [T F C D] [L S K E B A] [F C D] [T L S K E B A] [M C D] [F T L S K E B A] [C D] [M F T L S K E B A] [G H D] [C M F T L S K E B A]
Comparison BFS finds the shortest path DFS gets quickly into a deep search space good if you know the solution is "far away" wrong path: inefficient DFS with iterative deepening use a depth bound retreat at the bound no luck: increase the bound Later: use knowledge about the problem to order nodes on the open list
Using State Space to Represent Predicate Calculus node: state of the problem arc: inference search: to decide if an assertion is implied by others q → p p r → p v → q q r u s → r t → r v t s s → u s, t Determining truth: path from boxed nodes to proposition Data-driven: start with boxed; goal-driven: start with goal. DFS or BFS?
BFS open closed [t s] [] [s r] [t] [r u] [s t] [u p] [r s t] [p] [u r s t] p q r u v t s Is p true?
DFS open closed [t s] [] [r s] [t] [p u] [r t] p q r u v t s Is p true?
And/Or Graphs in the graph above t s →r: r t s To express t s →r, connect incoming arcs r t s
a b c a b → d a c → e b d → f f→ g a e → h Graph? Example 3.3.2
a b c a b → d a c → e b d → f f→ g a e → h Example 3.3.2
Goal-directed: h: try to prove a and e a is true e is true if c and a a is true c is true ← e is true ← h is true Data-directed: a, b, c are true a and b → d a and c → e a and e → h Search for h:
Symbolic Integration MACSYMA Symbolic algebra, including integration http://integrals.wolfram.com/index.jsp How do you think this works? Decomposition using and/or graphs ∫f + g decomposes to ∫f and ∫g More complicated expressions decompose into possible transformations Graph is generated on the fly Goal-directed How is it searched? BFS? DFS?
Figure 3.26: And/or graph for the financial advisor
Five rules for a simple subset of English grammar (rewrite rules): Does 1 look like AND? Do 2 and 3 this look like OR? Construct the graph
Figure 3.27:And/or graph for the grammar of Example 3.3.6. How do we use this? A sentence is well-formed if it consists of terminal symbols and there is a series of substitutions that reduce it to the sentence symbol
Figure 3.28: Parse tree for “The dog bites the man.” Note that this is a subtree Figure 3.26.
Parse "the dog bites the man" 7: art ↔ the gives art dog bites the man 7: art ↔ the gives art dog bites art man 8: n ↔ man gives art dog bites art n* 3: np ↔ art n gives art dog bites np** 9: n ↔ dog gives art n bites np 3: np↔ art n gives np bites np 11: v ↔ bites gives np v bites 5: vp ↔ v np gives np vp 1: sentence ↔ np vp ... * Why isn't dog rewritten first? ** Why didn't 11 fire instead: v ↔ bites gives art n v the man
Parse "dog the man" 7: art ↔ the gives dog art man 8: n ↔ man gives dog art man 9: n ↔ dog gives n art n There are no rules that rewrite this How does the C++ compiler work? Can you write a grammer for C++? program ↔ declarations body ...
