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Notes Dijstra’s Algorithm Corrected syllabus
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Tree Search Implementation Strategies
Require data structure to model search tree Tree Node: State (e.g. Sibiu) Parent (e.g. Arad) Action (e.g. GoTo(Sibiu)) Path cost or depth (e.g. 140) Children (e.g. Faragas, Oradea) (optional, helpful in debugging)
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Queue Methods: Empty(queue) Pop(queue) Insert(queue, element)
Returns true if there are no more elements Pop(queue) Remove and return the first element Insert(queue, element) Inserts element into the queue InsertFIFO(queue, element) – inserts at the end InsertLIFO(queue, element) – inserts at the front InsertPriority(queue, element, value) – inserts sorted by value
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informed Search
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Search start goal Uninformed search Informed search
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Informed Search What if we had an evaluation function h(n) that gave us an estimate of the cost associated with getting from n to the goal h(n) is called a heuristic
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Romania with step costs in km
h(n)
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Greedy best-first search
Evaluation function f(n) = h(n) (heuristic) e.g., f(n) = hSLD(n) = straight-line distance from n to Bucharest Greedy best-first search expands the node that is estimated to be closest to goal
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Romania with step costs in km
h(n) f(n)
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Best-First Algorithm
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Performance of greedy best-first search
Complete? Optimal?
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Failure case for best-first search
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Performance of greedy best-first search
Complete? No – can get stuck in loops, e.g., Iasi Neamt Iasi Neamt Optimal? No
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Complexity of greedy best first search
Time? O(bm), but a good heuristic can give dramatic improvement Space? O(bm) -- keeps all nodes in memory
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What can we do better?
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A* search Ideas: Avoid expanding paths that are already expensive
Consider Cost to get here (known) – g(n) Cost to get to goal (estimate from the heuristic) – h(n)
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A * Evaluation functions
Evaluation function f(n) = g(n) + h(n) g(n) = cost so far to reach n h(n) = estimated cost from n to goal f(n) = estimated total cost of path through n to goal start goal n g(n) h(n) f(n)
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n g(n) h(n) f(n)
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A* Heuristics A heuristic h(n) is admissible if for every node n,
h(n) ≤ h*(n), where h*(n) is the true cost to reach the goal state from n. An admissible heuristic never overestimates the cost to reach the goal, i.e., it is optimistic Example: hSLD(n) (never overestimates the actual road distance)
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What happens if heuristic is not admissible?
Will still find solution (complete) But might not find best solution (not optimal)
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Properties of A* (w/ admissible heuristic)
Complete? Yes (unless there are infinitely many nodes with f ≤ f(G) ) Optimal? Yes Time? Exponential, approximately O(bd) in the worst case Space? O(bm) Keeps all nodes in memory
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The heuristic h(x) guides the performance of A*
Let d(x) be the actual distance between S and G h(x) = 0 : A* is equivalent to Uniform-Cost Search h(x) <= d (x) : guarantee to compute the shortest path; the lower the value h(x), the more node A* expands h(x) = d (x) : follow the best path; never expand anything else; difficult to compute h(x) in this way! h(x) > d(x) : not guarantee to compute a best path; but very fast h(x) >> g(x) : h(n) dominates -> A* becomes the best first search
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Admissible heuristics
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Admissible heuristics
E.g., for the 8-puzzle:
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Admissible heuristics
E.g., for the 8-puzzle: h1(n) = number of misplaced tiles h2(n) = summed Manhattan distance for all tiles (i.e., no. of squares from desired location of each tile) h1(S) = ? h2(S) = ?
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Admissible heuristics
E.g., for the 8-puzzle: h1(n) = number of misplaced tiles h2(n) = total Manhattan distance (i.e., no. of squares from desired location of each tile) h1(S) = ? 8 h2(S) = ? = 18 Which is better?
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Dominance If h2(n) ≥ h1(n) for all n (both admissible)
then h2 dominates h1 h2 is better for search What does better mean? All searches we’ve discussed are exponential in time
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Comparison of algorithms (number of nodes expanded)
Iterative deepening A*(teleporting tiles) A* (manhattan distance) 2 10 6 112 13 12 680 20 18 364035 227 73 14 539 113 1.8 * 108 3056 363 24 8.6 * 1010 39135 1641
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Visually
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Where do heuristics come from?
From people Knowledge of the problem From computers By considering a simpler version of the problem Called a relaxation
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Relaxed problems 8-puzzle If the rules of the 8-puzzle are relaxed so that a tile can move anywhere, then h1(n) gives the shortest solution If the rules are relaxed so that a tile can move to any adjacent square, then h2(n) gives the shortest solution Consider the example of straight line distance (Romania navigation) Is that a relaxation?
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Iterative-Deepening A* (IDA*)
Further reduce memory requirements of A* Regular Iterative-Deepening: regulated by depth IDA*: regulated by f(n)=g(n)+h(n)
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Questions?
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