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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
CSE511a: Artificial
Intelligence
Spring 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)
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.
Today
A* Search
Heuristic Design
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)
Machinarium
Machinarium: Search Problem
Youtube
A*-Search
General Tree Search
Uniform Cost Search
Strategy: expand lowest
path cost
…
c1
c2
c3
The good: UCS is
complete and optimal!
The bad:
Explores options in every
“direction”
No information about goal
location
Start
Goal
[demo: countours UCS]
Best First (Greedy)
Strategy: expand a node
that you think is closest
to a goal state
…
b
Heuristic: estimate of
distance to nearest goal
for each state
A common case:
Best-first takes you
straight to the (wrong) goal
…
b
Worst-case: like a badlyguided DFS
[demo: countours greedy]
Example: Heuristic Function
h(x)
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)
5
e
h=1
1
S
h=6
c
h=7
1
a
h=5
1
1
3
2
d
h=2
G
h=0
b
h=6
A* Search orders by the sum: f(n) = g(n) + h(n)
Example: Teg Grenager
Three Questions
When should A* terminate?
Is A* optimal?
What heuristics are valid?
When should A* terminate?
Should we stop when we enqueue a goal?
2
A
2
h=2
S
G
h=3
2
B
h=0
3
h=1
No: only stop when we dequeue a goal
Is A* Optimal?
1
A
h=6
3
h=0
S
h=7
G
5
What went wrong?
Actual bad goal cost < estimated good goal cost
We need estimates to be less than actual costs!
Admissible Heuristics
A heuristic h is admissible (optimistic) if:
where
is the true cost to a nearest goal
Example:
15
Coming up with admissible heuristics is most of
what’s involved in using A* in practice.
Example: Pancake Problem
Cost: Number of pancakes flipped
Pancake: State Space Object
Example: Pancake Problem
State space graph with costs as weights
4
2
3
2
3
4
3
4
3
2
2
2
4
3
Example: Heuristic Function
Heuristic: the largest pancake that is still out of place
3
h(x)
4
3
4
3
0
4
4
3
4
4
2
3
Example: Pancake Problem
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
…
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
…
Properties of A*
Uniform-Cost
…
b
A*
…
b
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*]
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
4
15
Inadmissible heuristics are often useful too (why?)
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?
Search State
8 Puzzle I
Heuristic: Number of
tiles misplaced
Why is it admissible?
Average nodes expanded when
optimal path has length…
h(start) = 8
…4 steps …8 steps …12 steps
This is a relaxedproblem heuristic
UCS
112
TILES 13
6,300
39
3.6 x 106
227
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 + … TILES
= 18
MANHATTAN
Average nodes expanded when
optimal path has length…
…4 steps
…8 steps
…12 steps
13
12
39
25
227
73
[demo: eight-puzzle]
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!
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
Other A* Applications
Pathing / routing problems
Resource planning problems
Robot motion planning
Language analysis
Machine translation
Speech recognition
…
Graph Search
Tree Search: Extra Work!
Failure to detect repeated states can cause
exponentially more work. Why?
Graph Search
In BFS, for example, we shouldn’t bother
expanding the circled nodes (why?)
S
e
d
b
c
a
a
e
h
p
q
q
c
a
h
r
p
f
q
G
p
q
r
q
f
c
a
G
Graph Search
Very simple fix: never expand a state twice
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?
Consistency
1
A
h=4
1
h=1
S
h=6
C
1
B
2
h=1
What went wrong?
Taking a step must not reduce f value!
3
G
h=0
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
h=4
A
1
h=0
C
h=1
3
G
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!
Optimality
Tree search:
A* optimal if heuristic is admissible (and nonnegative)
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
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