 # Notes Dijstra’s Algorithm Corrected syllabus.

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Notes Dijstra’s Algorithm Corrected syllabus

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)

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

informed Search

Search start goal Uninformed search Informed search

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

Romania with step costs in km
h(n)

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

Romania with step costs in km
h(n) f(n)

Best-First Algorithm

Performance of greedy best-first search
Complete? Optimal?

Failure case for best-first search

Performance of greedy best-first search
Complete? No – can get stuck in loops, e.g., Iasi  Neamt  Iasi  Neamt Optimal? No

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

What can we do better?

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)

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)

n g(n) h(n) f(n)

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)

What happens if heuristic is not admissible?
Will still find solution (complete) But might not find best solution (not optimal)

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

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

E.g., for the 8-puzzle:

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) = ?

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?

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

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

Visually

Where do heuristics come from?
From people Knowledge of the problem From computers By considering a simpler version of the problem Called a relaxation

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?

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)

Questions?

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