quiz 1 : queue based search
DESCRIPTION
Quiz 1 : Queue based search. q1a1 : Any complete search algorithm must have exponential (or worse) time complexity. q1a2 : DFS requires more space than BFS. q1a3 : BFS always returns the optimal cost path. q1a4: BFS uses a FIFO queue as fringe. q1a5: UCS uses a LIFO queue as fringe. - PowerPoint PPT PresentationTRANSCRIPT
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Quiz 1 : Queue based search
q1a1: Any complete search algorithm must have exponential (or worse) time complexity.
q1a2: DFS requires more space than BFS. q1a3: BFS always returns the optimal cost path. q1a4: BFS uses a FIFO queue as fringe. q1a5: UCS uses a LIFO queue as fringe. q1a6: BFS has the best time complexity out of UCS,
BFS, DFS, ID.
Yes
No No Yes No No
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CSE511a: Artificial IntelligenceSpring 2013
Lecture 3: A* Search
1/24/2012
Robert Pless – Wash U.
Multiple slides over the course adapted from Kilian Weinberger, originally by Dan Klein (or Stuart Russell or
Andrew Moore)
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Announcements
Projects: Project 0 due tomorrow night.
Project 1 (Search) is out soon, will be due Thursday 2/7.
Find project partner
Try pair programming, not divide-and-conquer
Exercise more and try to eat more fruit.
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Today
A* Search
Heuristic Design
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Recap: Search
Search problem: States (configurations of the world) Successor function: a function from states to
lists of (action, state, cost) triples; drawn as a graph Start state and goal test
Search tree: Nodes: represent plans for reaching states Plans have costs (sum of action costs)
Search Algorithm: Systematically builds a search tree Chooses an ordering of the fringe (unexplored nodes)
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Machinarium
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Machinarium: Search Problem
Youtube
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A*-Search
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General Tree Search
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Uniform Cost Search
Strategy: expand lowest path cost
The good: UCS is complete and optimal!
The bad: Explores options in every
“direction” No information about goal
location Start Goal
…
c 3
c 2
c 1
[demo: countours UCS]
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Best First (Greedy)
Strategy: expand a node that you think is closest to a goal state Heuristic: estimate of
distance to nearest goal for each state
A common case: Best-first takes you
straight to the (wrong) goal
Worst-case: like a badly-guided DFS
…b
…b
[demo: countours greedy]
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Example: Heuristic Function
h(x)
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Combining UCS and Greedy Uniform-cost orders by path cost, or backward cost g(n) Best-first orders by goal proximity, or forward cost h(n)
A* Search orders by the sum: f(n) = g(n) + h(n)
S a d
b
Gh=5
h=6
h=2
1
5
11
2
h=6h=0
c
h=7
3
e h=11
Example: Teg Grenager
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Three Questions
When should A* terminate? Is A* optimal? What heuristics are valid?
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Should we stop when we enqueue a goal?
No: only stop when we dequeue a goal
When should A* terminate?
S
B
A
G
2
3
2
2h = 1
h = 2
h = 0
h = 3
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Is A* Optimal?
A
GS
1
3h = 6
h = 0
5
h = 7
What went wrong? Actual bad goal cost < estimated good goal cost We need estimates to be less than actual costs!
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Admissible Heuristics
A heuristic h is admissible (optimistic) if:
where is the true cost to a nearest goal
Example:
Coming up with admissible heuristics is most of what’s involved in using A* in practice.
15
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Example: Pancake Problem
Cost: Number of pancakes flipped
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Pancake: State Space Object
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Example: Pancake Problem
3
2
4
3
3
2
2
2
4
State space graph with costs as weights
34
3
4
2
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Example: Heuristic FunctionHeuristic: the largest pancake that is still out of place
4
3
0
2
3
3
3
4
4
3
4
4
4
h(x)
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Example: Pancake Problem
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Optimality of A*: Blocking
…Notation:
g(n) = cost to node n
h(n) = estimated cost from n
to the nearest goal (heuristic)
f(n) = g(n) + h(n) =
estimated total cost via n
G*: a lowest cost goal node
G: another goal node
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Optimality of A*: Blocking
Proof: What could go wrong? We’d have to have to pop a
suboptimal goal G off the fringe before G*
This can’t happen: Imagine a suboptimal goal
G is on the queue Some node n which is a
subpath of G* must also be on the fringe (why?)
n will be popped before G
…
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Properties of A*
…b
…b
Uniform-Cost A*
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UCS vs A* Contours
Uniform-cost expanded in all directions
A* expands mainly toward the goal, but does hedge its bets to ensure optimality
Start Goal
Start Goal
[demo: countours UCS / A*]
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Creating Admissible Heuristics
Most of the work in solving hard search problems optimally is in coming up with admissible heuristics
Often, admissible heuristics are solutions to relaxed problems, where new actions are available
Inadmissible heuristics are often useful too (why?)
415
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Example: 8 Puzzle
What are the states? How many states? What are the actions? What states can I reach from the start state? What should the costs be?
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Search State
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8 Puzzle I
Heuristic: Number of tiles misplaced
Why is it admissible?
h(start) =
This is a relaxed-problem heuristic
8 Average nodes expanded when optimal path has length…
…4 steps …8 steps …12 steps
UCS 112 6,300 3.6 x 106
TILES 13 39 227
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8 Puzzle II
What if we had an easier 8-puzzle where any tile could slide any direction at any time, ignoring other tiles?
Total Manhattan distance
Why admissible?
h(start) =
3 + 1 + 2 + …
= 18
Average nodes expanded when optimal path has length…
…4 steps …8 steps …12 steps
TILES 13 39 227
MANHATTAN 12 25 73[demo: eight-puzzle]
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8 Puzzle III
How about using the actual cost as a heuristic? Would it be admissible? Would we save on nodes expanded? What’s wrong with it?
With A*: a trade-off between quality of estimate and work per node!
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Trivial Heuristics, Dominance
Dominance: ha ≥ hc if
Heuristics form a semi-lattice: Max of admissible heuristics is admissible
Trivial heuristics Bottom of lattice is the zero heuristic (what
does this give us?) Top of lattice is the exact heuristic
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Other A* Applications
Pathing / routing problems Resource planning problems Robot motion planning Language analysis Machine translation Speech recognition …
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Graph Search
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Tree Search: Extra Work!
Failure to detect repeated states can cause exponentially more work. Why?
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Graph Search
In BFS, for example, we shouldn’t bother expanding the circled nodes (why?)
S
a
b
d p
a
c
e
p
h
f
r
q
q c G
a
qe
p
h
f
r
q
q c G
a
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Graph Search
Very simple fix: never expand a state twice
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Graph Search
Idea: never expand a state twice
How to implement: Tree search + list of expanded states (closed list) Expand the search tree node-by-node, but… Before expanding a node, check to make sure its state is new
Python trick: store the closed list as a set, not a list
Can graph search wreck completeness? Why/why not?
How about optimality?
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Consistency
A
CS
1
1h = 4
h = 1h = 6
What went wrong? Taking a step must not reduce f value!
B
h = 0
Gh = 11 2 3
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Consistency
Stronger than admissability Definition:
C(A→C)+h(C) h(A)≧ C(A→C) h(A)-h(C)≧
Consequence: The f value on a path never
decreases A* search is optimal
A
G
1
h = 4
h = 0
Ch = 1 3
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Optimality of A* Graph Search
Proof: New possible problem: nodes on path to G*
that would have been in queue aren’t, because some worse n’ for the same state as some n was dequeued and expanded first (disaster!)
Take the highest such n in tree Let p be the ancestor which was on the
queue when n’ was expanded Assume f(p) f(n) ≦ (consistency!) f(n) < f(n’) because n’ is suboptimal p would have been expanded before n’ So n would have been expanded before n’,
too Contradiction!
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Optimality
Tree search: A* optimal if heuristic is admissible (and non-negative) UCS is a special case (h = 0)
Graph search: A* optimal if heuristic is consistent UCS optimal (h = 0 is consistent)
Consistency implies admissibility
In general, natural admissible heuristics tend to be consistent
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Summary: A*
A* uses both backward costs and (estimates of) forward costs
A* is optimal with admissible heuristics
Heuristic design is key: often use relaxed problems