Transcript Heuristic Search - University of Illinois at Urbana

CS 440 / ECE 448
Introduction to Artificial Intelligence
Spring 2010
Lecture #4
Instructor: Eyal Amir
Grad TAs: Wen Pu, Yonatan Bisk
Undergrad TAs: Sam Johnson, Nikhil Johri
When to Use Search Techniques?
• The search space is small, and
– There is no other available techniques, or
– It is not worth the effort to develop a more
efficient technique
• The search space is large, and
– There is no other available techniques, and
– There exist “good” heuristics
Models, |=, math proofs
• Alpha |= Beta
– Alpha, Beta – propositional formulas
– M |= Alpha “M models Alpha” means “Alpha
evaluated to TRUE in model M”
• Math. Proofs: example
– A  B |= A
– (another one later in today’s class)
Search Algorithms
• Blind search – BFS, DFS, ID, uniform cost
– no notion concept of the “right direction”
– can only recognize goal once it’s achieved
• Heuristic search – we have rough idea of how
good various states are, and use this knowledge
to guide our search
Types of heuristic search
• Best First
– A* is a special case
– BFS is a special case
• ID A*
– ID is a special case
• Hill climbing
• Simulated Annealing
A* Example: 8-Puzzle
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
3+3
1+5
2+3
3+4
5+2
0+4
3+2
1+3
2+3
5+0
3+4
1+5
2+4
4+1
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=4
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=4
4
6
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=4
4
5
6
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
5
4
Cutoff=4
4
5
6
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
6
5
4
5
6
6
4
Cutoff=4
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
4
6
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
4
5
6
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
4
5
7
6
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
5
4
5
7
6
6
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
5
4
5
7
6
6
5
ID A*: 8-Puzzle Example
f(N) = g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
5
4
5
7
6
6
5
Hill climbing example
start
2 8 3
1 6 4
7
5
-5
h = -4
-5
2 8 3
1
4 h = -3
7 6 5
-3
h = -3
goal
1 2 3
8
4 h=0
7 6 5
-2
1 2 3
8 4 h = -1
7 6 5
-4
2
3
1 8 4
7 6 5
2 3
1 8 4 h = -2
7 6 5
-4
f(n) = -(number of tiles out of place)
Best-first search
• Idea: use an evaluation function f(n) for
each node n
• Expand unexpanded node n with min f(n)
• Implementation: FRINGE is queue sorted
by decreasing order of desirability
– Greedy search
– A* search
Greedy Search
• h(n) – a ‘heuristic’ function estimating the
distance to the goal
• Greedy Best First: expand argmin_n h(n)
thus,
f(v) = h(v)
Informed Search
• Add domain-specific information to select
the best path along which to continue
searching
• h(n) = estimated cost (or distance) of
minimal cost path from n to a goal state.
• The heuristic function is an estimate,
based on domain-specific information that
is computable from the current state
description, of how close we are to a goal
Robot Navigation
Robot Navigation
f(N) = h(N), with h(N) = Manhattan distance to the goal
8
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Robot Navigation
f(N) = h(N), with h(N) = Manhattan distance to the goal
8
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00 1 2
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What happened???5
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Greedy Search
• f(N) = h(N)  greedy best-first
• Is it complete?
If we eliminate endless loops, yes
• Is it optimal?
More informed search
• Our goal is not to minimize the distance from the
current head of our path to the goal, we want to
minimize the overall length of the path to the
goal!
• Let g(N) be the cost of the best
path found so far between the initial
node and N
• f(N) = g(N) + h(N)
Robot Navigation
f(N) = g(N)+h(N), with h(N) = Manhattan distance to goal
8+3
8 7+4
7 6+3
6+5
6 5+6
5 4+7
4 3+8
3 2+9
2 3+10
3 4
7+2
7
6+1
6
5
5+6
5 4+7
4 3+8
3
3 2+9
2 1+10
1 0+11
0 1
6
5
2
4
7+0
7 6+1
6
8+1
8 7+2
7 6+3
6 5+4
5 4+5
4 3+6
3 2+7
2 3+8
3 4
5
5
6
Can we Prove Anything?
• If the state space is finite and we avoid
repeated states, the search is complete,
but in general is not optimal
• Proof: ?
• If the state space is finite and we do not
avoid repeated states, the search is in
general not complete
Admissible heuristic
• Let h*(N) be the true cost of the optimal
path from N to a goal node
• Heuristic h(N) is admissible if:
0  h(N)  h*(N)
• An admissible heuristic is always
optimistic
A* Search
• Evaluation function:
f(N) = g(N) + h(N)
where:
– g(N) is the cost of the best path found so far to N
– h(N) is an admissible heuristic
• Then, best-first search with this evaluation function
is called A* search
• Important AI algorithm developed by Fikes and Nilsson in
early 70s. Originally used in Shakey robot.
Completeness & Optimality of A*
• Claim 1: If there is a path from the initial
to a goal node, A* (with no removal of
repeated states) terminates by finding the
best path, hence is:
– complete
– optimal
• requirements:
– 0 <   c(N,N’)
- c(N,N’) – cost of going from N to N’
Completeness of A*
• Theorem: If there is a finite path from the
initial state to a goal node, A* will find it.
Proof of Completeness
• Intuition (not math. Proof):
• Let g be the cost of a best path to a goal
node
• No path in search tree can get longer than
g/, before the goal node is expanded
Optimality of A*
• Theorem: If h(n) is admissable, then A* is
optimal (finds an optimal path).
Proof of Optimality
f(G1) = g(G1)
N
G1
G2
Cost of best path
to a goal thru N
f(N) = g(N) + h(N)  g(N) + h*(N)
f(G1)  g(N) + h(N)  g(N) + h*(N)
Heuristic Function
• Function h(N) that estimates the cost of
the cheapest path from node N to goal
node.
• Example: 8-puzzle
5
8
4 2 1
7 3 6
N
1 2 3 h(N) = number of misplaced tiles
=6
4 5 6
7 8
goal
Heuristic Function
• Function h(N) that estimate the cost of the
cheapest path from node N to goal node.
• Example: 8-puzzle
5
8
4 2 1
7 3 6
N
1 2 3 h(N) = sum of the distances of
every tile to its goal position
4 5 6
=2+3+0+1+3+0+3+1
7 8
= 13
goal
8-Puzzle
f(N) = h(N) = number of misplaced tiles
3
5
3
4
3
4
2
4
2
3
3
0
4
5
4
1
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
3+3
1+5
2+3
3+4
5+2
0+4
3+2
1+3
2+3
5+0
3+4
1+5
2+4
4+1
8-Puzzle
f(N) = h(N) =  distances of tiles to goal
6
5
2
5
2
4
3
0
4
6
5
1
8-Puzzle
5
8
1
2
3
6
4
2
1
4
5
7
3
N
6
7
8
goal
• h1(N) = number of misplaced tiles = 6 is admissible
• h2(N) = sum of distances of each tile to goal = 13
is admissible
• h3(N) = (sum of distances of each tile to goal)
+ 3 x (sum of score functions for each tile) = 49
is not admissible
Robot navigation
f(N) = g(N) + h(N), with h(N) = straight-line distance from N to goal
Cost of one horizontal/vertical step = 1
Cost of one diagonal step = 2
Consistent Heuristic
• The admissible heuristic h is consistent
(or satisfies the monotone restriction) if for
every node N and every successor N’ of N:
h(N)  c(N,N’) + h(N’)
N
c(N,N’)
(triangle inequality)
N’
h(N’)
h(N)
8-Puzzle
5
8
1
2
3
6
4
2
1
4
5
7
3
N
6
7
8
goal
• h1(N) = number of misplaced tiles
• h2(N) = sum of distances of each tile to goal
are both consistent
Robot navigation
Cost of one horizontal/vertical step = 1
Cost of one diagonal step = 2
h(N) = straight-line distance to the goal is consistent
Claims
• If h is consistent, then the function f along
any path is non-decreasing:
N
c(N,N’)
f(N) = g(N) + h(N)
f(N’) = g(N) +c(N,N’) + h(N’)
N’
h(N’)
h(N)
Claims
• If h is consistent, then the function f along
any path is non-decreasing:
N
c(N,N’)
f(N) = g(N) + h(N)
f(N’) = g(N) +c(N,N’) + h(N’)
h(N)  c(N,N’) + h(N’)
f(N)  f(N’)
N’
h(N’)
h(N)
Claims
• If h is consistent, then the function f along
any path is non-decreasing:
N
f(N) = g(N) + h(N)
f(N’) = g(N) +c(N,N’) + h(N’)
h(N)  c(N,N’) + h(N’)
f(N)  f(N’)
c(N,N’)
N’
h(N)
h(N’)
• If h is consistent, then whenever A* expands
a node it has already found an optimal path to
the state associated with this node
Avoiding Repeated States in A*
If the heuristic h is consistent, then:
• Let CLOSED be the list of states
associated with expanded nodes
• When a new node N is generated:
– If its state is in CLOSED, then discard N
– If it has the same state as another node in the
fringe, then discard the node with the largest f
Heuristic Accuracy
• h(N) = 0 for all nodes is admissible and
consistent. Hence, breadth-first is a special
case of A* !!!
• Let h1 and h2 be two admissible and consistent
heuristics such that for all nodes N: h1(N) 
h2(N).
• Then, every node expanded by A* using h2 is
also expanded by A* using h1.
• h2 is more informed than h1
Iterative Deepening A* (IDA*)
• Use f(N) = g(N) + h(N) with admissible and
consistent h
• Each iteration is depth-first with cutoff on
the value of f of expanded nodes
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=4
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=4
4
6
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=4
4
5
6
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
5
4
Cutoff=4
4
5
6
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
6
5
4
5
6
6
4
Cutoff=4
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
4
6
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
4
5
6
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
4
5
7
6
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
5
4
5
7
6
6
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
5
4
5
7
6
6
5
f(N)8-Puzzle
= g(N) + h(N)
with h(N) = number of misplaced tiles
4
Cutoff=5
5
4
5
7
6
6
5
About Heuristics
•
•
•
•
•
Heuristics are intended to orient the search along
promising paths
The time spent computing heuristics must be
recovered by a better search
After all, a heuristic function could consist of solving
the problem; then it would perfectly guide the search
Deciding which node to expand is sometimes called
meta-reasoning
Heuristics may not always look like numbers and
may involve large amount of knowledge
What’s the Issue?
• Search is an iterative local procedure
• Good heuristics should provide some
global look-ahead (at low computational
cost)
Another approach…
• for optimization problems
– rather than constructing an optimal solution
from scratch, start with a suboptimal solution
and iteratively improve it
• Local Search Algorithms
– Hill-climbing or Gradient descent
– Potential Fields
– Simulated Annealing
– Genetic Algorithms, others…
Hill-climbing search
• If there exists a successor s for the current state n
such that
– h(s) < h(n)
– h(s) <= h(t) for all the successors t of n,
• then move from n to s. Otherwise, halt at n.
• Looks one step ahead to determine if any successor
is better than the current state; if there is, move to the
best successor.
• Similar to Greedy search in that it uses h, but does
not allow backtracking or jumping to an alternative
path since it doesn’t “remember” where it has been.
• Not complete since the search will terminate at "local
minima," "plateaus," and "ridges."
Hill climbing on a surface of
states
Height Defined
by Evaluation
Function
Hill climbing
• Steepest descent (~ greedy best-first with
no search)  may get stuck into local
minimum
Robot Navigation
Local-minimum problem
f(N) = h(N) = straight distance to the goal
Examples of problems with HC
• applet
Drawbacks of hill climbing
• Problems:
– Local Maxima: peaks that aren’t the highest
point in the space
– Plateaus: the space has a broad flat region
that gives the search algorithm no direction
(random walk)
– Ridges: flat like a plateau, but with dropoffs to
the sides; steps to the North, East, South and
West may go down, but a step to the NW may
go up.
• Remedy:
– Introduce randomness
• Random restart.
• Some problem spaces are great for hill climbing
What’s the Issue?
• Search is an iterative local procedure
• Good heuristics should provide some
global look-ahead (at low computational
cost)
Hill climbing example
start
2 8 3
1 6 4
7
5
-5
h = -4
-5
2 8 3
1
4 h = -3
7 6 5
-3
h = -3
goal
1 2 3
8
4 h=0
7 6 5
-2
1 2 3
8 4 h = -1
7 6 5
-4
2
3
1 8 4
7 6 5
2 3
1 8 4 h = -2
7 6 5
-4
f(n) = -(number of tiles out of place)
Example of a local maximum
start
1 2 5
7 4
8 6 3
1 2 5
7 4
8 6 3
1 2 5
7 4 -4
8 6 3
-3
-4
1 2 5
7 4 -4
8 6 3
goal
1 2 5
7 4 0
8 6 3
Potential Fields
• Idea: modify the heuristic function
• Goal is gravity well, drawing the robot toward it
• Obstacles have repelling fields, pushing the
robot away from them
• This causes robot to “slide” around obstacles
• Potential field defined as sum of attractor field
which get higher as you get closer to the goal
and the indivual obstacle repelling field (often
fixed radius that increases exponentially closer
to the obstacle)
Does it always work?
• No.
• But, it often works very well in practice
• Advantage #1: can search a very large
search space without maintaining fringe of
possiblities
• Scales well to high dimensions, where no
other methods work
• Example: motion planning
• Advantage #2: local method. Can be done
online
Example: RoboSoccer
All robots have same field: attracted to the ball
• Repulsive potential to other players
• Kicking field: attractive potential to the ball and
local repulsive potential if clase to the ball, but
not facing the direction of the opponent’s goal.
Result is tangent, player goes around the ball.
• Single team: kicking field + repulsive field to
avoid hitting other players + player position fields
(paraboilic if outside your area of the field, 0
inside). Player nearest to the ball has the
largest attractive coefficient, avoids all players
crowding the ball.
Simulated annealing
• Simulated annealing (SA) exploits an analogy between
the way in which a metal cools and freezes into a
minimum-energy crystalline structure (the annealing
process) and the search for a minimum [or maximum] in a
more general system.
• SA can avoid becoming trapped at local minima.
• SA uses a random search that accepts changes that
increase objective function f, as well as some that
decrease it.
• SA uses a control parameter T, which by analogy with the
original application is known as the system
“temperature.”
• T starts out high and gradually decreases toward 0.
Simulated annealing (cont.)
• A “bad” move from A to B is accepted with a
probability
e
(f(B)-f(A)/T)
• The higher the temperature, the more likely it
is that a bad move can be made.
• As T tends to zero, this probability tends to
zero, and SA becomes more like hill climbing
• If T is lowered slowly enough, SA is complete
and admissible.
The simulated annealing algorithm
Summary: Local Search Algorithms
• Steepest descent (~ greedy best-first with
no search)  may get stuck into local
minimum
• Better Heuristics: Potential Fields
• Simulated annealing
• Genetic algorithms
When to Use Search Techniques?
• The search space is small, and
– There is no other available techniques, or
– It is not worth the effort to develop a more
efficient technique
• The search space is large, and
– There is no other available techniques, and
– There exist “good” heuristics
Summary
•
•
•
•
•
•
•
Heuristic function
Best-first search
Admissible heuristic and A*
A* is complete and optimal
Consistent heuristic and repeated states
Heuristic accuracy
IDA*
Modified Search Algorithm
1.
2.
INSERT(initial-node,FRINGE)
Repeat:
If FRINGE is empty then return failure
n  REMOVE(FRINGE)
s  STATE(n)
If GOAL?(s) then return path or goal state
For every state s’ in SUCCESSORS(s)
 Create a node n’
 INSERT(n’,FRINGE)