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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.
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
General Tree Search
Action: flip top two
Cost: 2
Action: flip all four
Cost: 4
Path to reach goal:Flip four, flip three
Total cost: 7
Recall: 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]
Search heuristic
A heuristic is: A function that estimates how close a state is to a goal Designed for a particular search problem
10
511.2
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)
Greedy search
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
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)
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
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
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?
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!
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:
Coming up with admissible heuristics is most of what’s involved in using A* in practice.
4 15
Optimality of A*
…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
A: a lowest cost goal node
B: another goal nodeClaim: A will exit the fringe before B.
A
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
Claim: A will exit the fringe before 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
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
B
Claim: A will exit the fringe before B.
• g(A) < g(B) // B is suboptimal
• f(A) < f(B) // h=0 at goals
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
B
Claim: A will exit the fringe before B.
• f(n) <= f(A) < f(B) // from above
• f(n) < f(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
Claim: A will exit the fringe before 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
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
Inadmissible heuristics are often useful too (why?)
15366
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?
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
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
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
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
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
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?
S
A
B
C
G
1
1
1
23
h=2
h=1
h=4h=1
h=0
S (0+2)
A (1+4) B (1+1)
C (2+1)
G (5+0)
C (3+1)
G (6+0)
State space graph Search tree
Consistency of Heuristics
Admissibility: heuristic cost <=
actual cost to goal
h(A) <= actual cost from A to G
3
A
C
G
h=4 1
Consistency of Heuristics
Stronger than admissibility
Definition:
heuristic cost <= actual cost per arc
h(A) - h(C) <= cost(A to C)
Consequences:
The f value along a path never decreases
A* graph search is optimal
A
Ch=4
h=1
1
h=2
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, 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
…
f 3
f 2
f 1
There’s a problem with this argument. What are we assuming is true?
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? YES:
What can we require to prevent these inversions?
Consistency: Real cost must always exceed reduction in heuristic Like admissibility, but better!
A
B
G
3h = 0
h = 10
g = 10
h = 8