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Finding Search Heuristics Henry Kautz

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if State[node] is not in closed OR g[node] < g[LookUp(State[node],closed)] then A* Graph Search for Any Admissible Heuristic

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Inconsistent but Admissible

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Memory Efficient Versions of A* Iterative Deepening A* (IDA*) –Just like IDS, except that: Stopping condition: g(n)+h(n)=f(n) > Cutoff Increment: Min{ f(n) | n was cutoff last iteration} –Optimal –Memory linear in cheapest path –Will re-expand nodes frequently

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Memory-Bounded A* Throw nodes out of Closed and/or Fringe when memory gets low –Dropping nodes from Closed can make extra work, but doesn’t hurt optimality Dropping nodes from Fringe: –Choose worst node n from Fringe –Set f(Parent(n)) = Max { f(Parent(n)), Min{ f(n’) | n=Parent(n’) } –If Parent is not in Fringe, put it in Fringe Many variations – MA*, SMA*, SMAG*, …

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Properties of Heuristics Let h1 and h2 be admissible heuristics if for all s, h1(s) h2(s), then –h1 dominates h2 –h1 is better than h2 h3(s) = max(h1(s), h2(s)) is admissible –h3 dominates h1 and h2

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Exercise: Rubik’s Cube State: position and orientation of each cubie –Corner cubie: 8 positions, 3 orientations –Edge cubie: 8 positions, 2 orientations –Center cubie: 1 position – fixed Move: Turn one face –Center cubits never move! Devise an admissible heuristic

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Heuristics # cubies out of place / 8 –Must divide up 8 so is admissible –Too optimistic / weak Better: Korf 1997 – Solves Rubik’s cube optimally

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Automatically Deriving Heuristics for STRIPS planning Goal: quickly compute upper bound on distance from S to a state containing Goals Consider: |Goals – S|/m where m is the maximum number of goals added by any one action Admissible? Accurate? Yes! No, ignores preconditions!

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Another Attempt Create relaxed planning problem: Solve –Initial = S –Same Goals –Eliminate negative effects from all operators Use length of shortest solution to relaxed problem as h –Admissible? –Accurate? –Easy to compute? Yes! No, hard as set covering! Pretty good

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Plan Graph Heuristic Idea: Quickly estimate the minimum length of solution of relaxed problem Method: 1.For each literal, create a dummy “persistence” action with that literal as precondition and effect 2.Compute a Plan Graph: S0: initial literals A0: actions whose preconditions are in S0 S1: add effects of actions in A0 A1: actions whose preconditions are in S1 …

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Plan Graph Heuristic Method (continued): 3. Propagate mutual exclusions (mutexes): Two non-persistence actions at the same level are mutex A persistence action is mutex with an action that whose effect negates it Two literals A and B are mutex at level i if all the actions that add A are mutex with those that add B 4. Estimated length = lowest level containing all goals with no mutexes between them

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Example Plan Graph

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Is Planning Solved? A* with plan graph heuristic works well for domains with relatively few bad interactions between actions For “harder” domains: –Solve non-optimally using non-admissible heuristics –Solve optimally using other approaches – such as satisfiability testing More later in the course!

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