Transcript more from Thursday (modified from Dan Klein's)

CS 343H: Artificial Intelligence
Week 2b: Informed Search
Quiz : review
 Which are true about DFS? (b is branching
factor, m is depth of search tree.)
1. At any given time during the search, the number of
nodes on the fringe can be no larger than bm.
2. At any given time in the search, the number of
nodes on the fringe can be as large as b^m.
3. The number of nodes expanded throughout the
entire search can be no larger than bm.
4. The number of nodes expanded throughout the
entire search can be as large as b^m.
Quiz : review
 Which are true about BFS? (b is the branching
factor, s is the depth of the shallowest solution)
1. At any given time during the search, the number of
nodes on the fringe can be no larger than bs.
2. At any given time during the search, the number of
nodes on the fringe can be as large as b^s.
3. The number of nodes considered throughout the
entire search can be no larger than bs.
4. The number of nodes considered throughout the
entire search can be as large as b^s.
Quiz: search execution
 Run DFS, BFS, and UCS:
Today
 Informed search
 Heuristics
 Greedy search
 A* search
 Graph search
Recap: Search
 Search problem:
 States (configurations of the world)
 Actions and costs
 Successor function: a function from states to
lists of (state, action, cost) triples (world dynamics)
 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)
 Optimal: finds least-cost plans
Example: Pancake Problem
Cost: Number of pancakes flipped
Example: Pancake Problem
State space graph with costs as weights
4
2
3
2
3
4
3
4
3
2
2
2
4
3
General Tree Search
Action: flip top two
Cost: 2
Action:
four goal:
Pathflip
toall
reach
Cost:
4 flip three
Flip
four,
Total cost: 7
Recall: Uniform Cost Search
 Strategy: expand lowest
path cost
…
c1
c2
c3
 The good: UCS is
complete and optimal!
 The bad:
 Explores options in every
“direction”
 No information about goal
location
Start
Goal
[demo: countours UCS]
Example: Uniform cost search
Another example
Informed search: main idea
Search heuristic
 A heuristic is:
 A function that estimates how close a state is to a goal
 Designed for a particular search problem
10
5
11.2
Example: Heuristic Function
h(x)
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
 How to use the heuristic?
 What about following the “arrow” of the
heuristic?.... Greedy search
Example: Heuristic Function
h(x)
Best First / Greedy Search
 Expand the node that seems closest…
 What can go wrong?
Greedy search
 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
 Worst-case: like a badlyguided DFS
…
b
Quiz: greedy search
 Which solution would greedy search find if
run on this graph?
Enter: A* search
Combining UCS and Greedy
 Uniform-cost orders by path cost, or backward cost g(n)
 Greedy 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
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
Quiz: A* Tree search
Quiz: A* Tree search
 When running A*, the first node expanded is S (like all
search algorithms). After expanding S, two nodes will be
on the fringe: S->A and S->D. Fill in:
1.
g((S->A)) =
2.
h((S->A)) =
3.
f((S->A)) =
4.
g((S->D))=
5.
h((S->D)) =
6.
f((S->D)) =
7.
Which node will be expanded next?
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!
Idea: admissibility
Inadmissible (pessimistic):
break optimality by trapping
good plans on the fringe
Admissible (optimistic):
slows down bad plans but
never outweigh true costs
Admissible Heuristics
 A heuristic h is admissible (optimistic) if:
where
is the true cost to a nearest goal
 Examples:
4
15
 Coming up with admissible heuristics is most of
what’s involved in using A* in practice.
Optimality of A*
Notation:
…
 g(n) = cost to node n
 h(n) = estimated cost from n
A
to the nearest goal (heuristic)
 f(n) = g(n) + h(n) =
estimated total cost via n
 A: a lowest cost goal node
 B: another goal node
Claim: A will exit the fringe before B.
B
Claim: A will exit the fringe before B.
Optimality of A*
 Imagine B is on the fringe.
 Some ancestor n of A must be on
the fringe too (maybe n is A)
 Claim: n will be expanded before B.
1. f(n) <= f(A)
…
A
B
• f(n) = g(n) + h(n) // by definition
• f(n) <= g(A) // by admissibility of h
• g(A) = f(A) // because h=0 at goal
Claim: A will exit the fringe before B.
Optimality of A*
 Imagine B is on the fringe.
 Some ancestor n of A must be on
the fringe too (maybe n is A)
 Claim: n will be expanded before B.
1. f(n) <= f(A)
2. f(A) < f(B)
…
A
• g(A) < g(B) // B is suboptimal
• f(A) < f(B)
// h=0 at goals
B
Claim: A will exit the fringe before B.
Optimality of A*
 Imagine B is on the fringe.
 Some ancestor n of A must be on
the fringe too (maybe n is A)
 Claim: n will be expanded before B.
1. f(n) <= f(A)
2. f(A) < f(B)
3. n will expand before B
…
A
• f(n) <= f(A) < f(B) // from above
• f(n) < f(B)
B
Claim: A will exit the fringe before B.
Optimality of A*
 Imagine B is on the fringe.
 Some ancestor n of A must be on
the fringe too (maybe n is A)
 Claim: n will be expanded before B.
1. f(n) <= f(A)
2. f(A) < f(B)
3. n will expand before B
 All ancestors of A expand before B
 A expands before B
…
A
B
Properties of A*
Uniform-Cost
…
b
A*
…
b
UCS vs A* Contours
 Uniform-cost expands
equally in all directions
 A* expands mainly
toward the goal, but
does hedge its bets to
ensure optimality
Start
Goal
Start
Goal
Recall: greedy
Uniform cost search
A*
Quiz: A* tree search
Quiz
A* applications








Pathing / routing problems
Video games
Resource planning problems
Robot motion planning
Language analysis
Machine translation
Speech recognition
…
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
366
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?
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
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
Quiz 1
 Consider the problem of Pacman eating all dots in a
maze. Every movement action has a cost of 1. Select all
heuristics that are admissible, if any.





the total number of dots left
the distance to the closest dot
the distance to the furthest dot
the distance to the closest dot plus distance to the furthest dot
none of the above
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!
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
 Idea: never expand a state twice
 How to implement:




Tree search + set of expanded states (“closed set”)
Expand the search tree node-by-node, but…
Before expanding a node, check to make sure its state is new
If not new, skip it
 Important: store the closed set as a set, not a list
 Can graph search wreck completeness? Why/why not?
 How about optimality?
Warning: 3e book has a more complex, but also correct, variant
A* Graph Search Gone Wrong?
State space graph
Search tree
S (0+2)
A
1
1
h=4
S
h=1
h=2 1
2
B
C
3
h=1
G
h=0
A (1+4)
B (1+1)
C (2+1)
C (3+1)
G (5+0)
G (6+0)
Consistency of Heuristics
 Admissibility: heuristic cost <=
A
h=4
actual cost to goal
1
C
3
G
 h(A) <= actual cost from A to G
Consistency of Heuristics
 Stronger than admissibility
A
h=4
h=2
 Definition:
1
 heuristic cost <= actual cost per arc
C
h=1
 h(A) - h(C) <= cost(A to C)
 Consequences:
 The f value along a path never
decreases
 A* graph search is optimal
Optimality
 Tree search:
 A* is 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, most natural admissible heuristics tend to be
consistent, especially if from relaxed problems
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
Summary: A*
 A* uses both backward costs and
(estimates of) forward costs
 A* is optimal with admissible / consistent
heuristics
 Heuristic design is key: often use relaxed
problems
Optimality of A* Graph Search
 Sketch: Consider what A* does with a consistent
heuristic:
 Fact 1: In tree search, A* expands nodes in
increasing total f value (f-contours)
 Fact 2: For every state s, nodes that reach s optimally
are expanded before nodes that reach s suboptimally.
 Result: A* graph search is optimal.
Optimality of A* Graph Search
Proof:
 New possible problem: some n on path to
G* isn’t in queue when we need it,
because some worse n’ for the same state
dequeued and expanded first (disaster!)
 Take the highest such n in tree
 Let p be the ancestor of n that was on the
queue when n’ was popped
 f(p) < f(n) because of consistency
 f(n) < f(n’) because n’ is suboptimal
 p would have been expanded before n’
 Contradiction!
Graph Search
 Very simple fix: never expand a state twice
 Can this wreck completeness? Optimality?
Graph Search, Reconsidered
 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?
Optimality of A* Graph Search
 Consider what A* does:
 Expands nodes in increasing total f value (f-contours)
 Proof idea: optimal goals have lower f value, so get
expanded first
We’re making a stronger
assumption than in the last
proof… What?
Optimality of A* Graph Search
 Consider what A* does:
 Expands nodes in increasing total f value (f-contours)
Reminder: f(n) = g(n) + h(n) = cost to n + heuristic
 Proof idea: the optimal goal(s) have the lowest f
value, so it must get expanded first
…
There’s a problem with
this argument. What are
we assuming is true?
f1
f2
f3
Course Scheduling
 From the university’s perspective:
 Set of courses {c1, c2, … cn}
 Set of room / times {r1, r2, … rn}
 Each pairing (ck, rm) has a cost wkm
 What’s the best assignment of courses to rooms?
 States: list of pairings
 Actions: add a legal pairing
 Costs: cost of the new pairing
 Admissible heuristics?
Consistency
 Wait, how do we know parents have better f-vales than
their successors?
 Couldn’t we pop some node n, and find its child n’ to
have lower f value?
h=0 h=8
 YES:
B
3
g = 10
G
A
h = 10
 What can we require to prevent these inversions?
 Consistency:
 Real cost must always exceed reduction in heuristic
 Like admissibility, but better!