Informed search algorithms

Chapter 4

Material

Chapter 4 Section 1 - 3

Excludes memory-bounded heuristic search

Outline

Best-first search

Greedy best-first search

A^* search

Heuristics

Local search algorithms

Online search problems

Review: Tree search

A search strategy is defined by picking the order of node expansion

Best-first search

Idea: use an evaluation function f(n) for each node

estimate of "desirability"

Expand most desirable unexpanded node

Implementation:

Order the nodes in fringe in decreasing order of desirability

Special cases:

greedy best-first search

A^* search

Romania with step costs in km

Greedy best-first search

Evaluation function f(n) = h(n) (heuristic)

= estimate of cost from n to goal

e.g., h_SLD(n) = straight-line distance from n to Bucharest

Greedy best-first search expands the node that appears to be closest to goal

Greedy best-first search example

Properties of greedy best-first search

Complete? No – can get stuck in loops,
e.g., Iasi à Neamt à Iasi à Neamt à …

Time? O(b^m), but a good heuristic can give
dramatic improvement

Space? O(b^m): keeps all nodes in memory

Optimal? No

A^* search

Idea: avoid expanding paths that are already expensive

Evaluation function f(n) = g(n) + h(n)

g(n) = cost so far to reach n

h(n) = estimated cost from n to goal

f(n) = estimated total cost of path through n to goal

A^* search example

Admissible heuristics

A heuristic h(n) is admissible if for every node n,

h(n) ≤ h^*(n), where h^*(n) is the true cost to reach the goal state from n.

An admissible heuristic never overestimates the cost to reach the goal, i.e., it is optimistic

Example: h_SLD(n) (never overestimates the actual road distance)

Theorem: If h(n) is admissible, A^* using TREE-SEARCH is optimal

Optimality of A^* (proof)

Suppose some suboptimal goal G₂ has been generated and is in the fringe. Let n be an unexpanded node in the fringe such that n is on a shortest path to an optimal goal G.

f(G₂)　 = g(G₂) since h(G₂) = 0

g(G₂) > g(G) since G₂ is suboptimal

f(G)　　 = g(G) since h(G) = 0

f(G₂)　 > f(G) from above

Optimality of A^* (proof)

Suppose some suboptimal goal G₂ has been generated and is in the fringe. Let n be an unexpanded node in the fringe such that n is on a shortest path to an optimal goal G.

f(G₂) > f(G) from above

h(n) ≤ h^*(n) since h is admissible

g(n) + h(n) ≤ g(n) + h^*(n)

f(n) ≤ f(G)

Hence f(G₂) > f(n), and A^* will never select G₂ for expansion

Consistent heuristics

A heuristic is consistent if for every node n, every successor n' of n generated by any action a,

h(n) ≤ c(n,a,n') + h(n')

If h is consistent, we have

f(n') = g(n') + h(n')

　　　　　 = g(n) + c(n,a,n') + h(n')

　　　　　 ≥ g(n) + h(n)

　　　　　 = f(n)

i.e., f(n) is non-decreasing along any path.

Theorem: If h(n) is consistent, A* using GRAPH-SEARCH is optimal

Optimality of A^*

A^* expands nodes in order of increasing f value

Gradually adds "f-contours" of nodes

Contour i has all nodes with f=f_i, where f_i < f_i+1

Properties of A^*

Complete? Yes (unless there are infinitely many nodes with f ≤ f(G) )

Time? Exponential

Space? Keeps all nodes in memory

Optimal? Yes

Admissible heuristics

E.g., for the 8-puzzle:

h₁(n) = number of misplaced tiles

h₂(n) = total Manhattan distance

(i.e., no. of squares from desired location of each tile)

h₁(S) = ?

h₂(S) = ?

Dominance

If h₂(n) ≥ h₁(n) for all n (both admissible)

then h₂ dominates h₁

h₂ is better for search

Typical search costs (average number of nodes expanded):

d=12 IDS = 3,644,035 nodes
A^*(h₁) = 227 nodes
A^*(h₂) = 73 nodes

d=24 IDS = too many nodes
A^*(h₁) = 39,135 nodes
A^*(h₂) = 1,641 nodes

Relaxed problems

A problem with fewer restrictions on the actions is called a relaxed problem

The cost of an optimal solution to a relaxed problem is an admissible heuristic for the original problem

If the rules of the 8-puzzle are relaxed so that a tile can move anywhere, then h₁(n) gives the shortest solution

If the rules are relaxed so that a tile can move to any adjacent square, then h₂(n) gives the shortest solution

Local search algorithms

In many optimization problems, the path to the goal is irrelevant; the goal state itself is the solution

State space = set of "complete" configurations

Find configuration satisfying constraints, e.g., n-queens

In such cases, we can use local search algorithms

keep a single "current" state, try to improve it

Example: n-queens

Put n queens on an n × n board with no two queens on the same row, column, or diagonal

Hill-climbing search

"Like climbing Everest in thick fog with amnesia"

Hill-climbing search

Problem: depending on initial state, can get stuck in local maxima

Hill-climbing search: 8-queens problem

h = number of pairs of queens that are attacking each other, either directly or indirectly

h = 17 for the above state

Hill-climbing search: 8-queens problem

Simulated annealing search

Idea: escape local maxima by allowing some "bad" moves but gradually decrease their frequency

Properties of simulated annealing search

One can prove: If T decreases slowly enough, then simulated annealing search will find a global optimum with probability approaching 1

Widely used in VLSI layout, airline scheduling, etc

Local Beam Search

Why keep just one best state?

Can be used with randomization too

Genetic algorithms

A successor state is generated by combining two parent states

Start with k randomly generated states (population)

A state is represented as a string over a finite alphabet (often a string of 0s and 1s)

Evaluation function (fitness function). Higher values for better states.

Produce the next generation of states by selection, crossover, and mutation

Genetic algorithms

Fitness function: number of non-attacking pairs of queens (min = 0, max = 8 × 7/2 = 28)

24/(24+23+20+11) = 31%

23/(24+23+20+11) = 29% etc.

Genetic algorithms

Online search and exploration

Many problems are offline

Do search for action and then perform action

Online search interleave search & execution

Necessary for exploration problems

New observations only possible after acting

Competitive Ratio

Actual cost of path

Best possible cost

(if agent knew space in advance)

30/20 = 1.5

For cost, lower is better

Exploration problems

Exploration problems: agent physically in some part of the state space.

e.g. solving a maze using an agent with local wall sensors

Sensible to expand states easily accessible to agent (i.e. local states)

Local search algorithms apply (e.g.,
hill-climbing)

Hex!

Assignment

Build a game player

Restricted by time per move (specifics TBA)

Interact with the game driver through command line

Each turn, we will run your program, providing a a grid as input.

Your output will be just the coordinates of your stone placement.

Homework #1 - Hex

Note: we haven’t yet covered all of the methods to solve this problem

This week: start thinking about it, discuss among yourselves (remember the G.I. Rule!)

What heuristics are good to use?

What type of search makes sense to use?


	Chapter 4 Section 1 - 3
	Excludes memory-bounded heuristic search


	Best-first search
	Greedy best-first search
	A^* search

	Heuristics

	Local search algorithms

	Online search problems







	A search strategy is defined by picking the order of node expansion


	Idea: use an evaluation function f(n) for each node
		estimate of "desirability"
		Expand most desirable unexpanded node

	Implementation:
	Order the nodes in fringe in decreasing order of desirability

	Special cases:
		greedy best-first search
		A^* search


	Evaluation function f(n) = h(n) (heuristic)
	= estimate of cost from n to goal
	e.g., h_SLD(n) = straight-line distance from n to Bucharest
	Greedy best-first search expands the node that appears to be closest to goal


	Complete? No – can get stuck in loops, e.g., Iasi à Neamt à Iasi à Neamt à …
	Time? O(b^m), but a good heuristic can give dramatic improvement
	Space? O(b^m): keeps all nodes in memory
	Optimal? No


	Idea: avoid expanding paths that are already expensive
	Evaluation function f(n) = g(n) + h(n)
	g(n) = cost so far to reach n
	h(n) = estimated cost from n to goal
	f(n) = estimated total cost of path through n to goal


	A heuristic h(n) is admissible if for every node n,
	h(n) ≤ h^(n), where h^(n) is the true cost to reach the goal state from n.
	An admissible heuristic never overestimates the cost to reach the goal, i.e., it is optimistic
	Example: h_SLD(n) (never overestimates the actual road distance)
	Theorem: If h(n) is admissible, A^* using TREE-SEARCH is optimal


	Suppose some suboptimal goal G₂ has been generated and is in the fringe. Let n be an unexpanded node in the fringe such that n is on a shortest path to an optimal goal G.





	f(G₂)　 = g(G₂) since h(G₂) = 0
	g(G₂) > g(G) since G₂ is suboptimal
	f(G)　　 = g(G) since h(G) = 0
	f(G₂)　 > f(G) from above


	Suppose some suboptimal goal G₂ has been generated and is in the fringe. Let n be an unexpanded node in the fringe such that n is on a shortest path to an optimal goal G.





	f(G₂) > f(G) from above
	h(n) ≤ h^*(n) since h is admissible
	g(n) + h(n) ≤ g(n) + h^*(n)
	f(n) ≤ f(G)
	Hence f(G₂) > f(n), and A^* will never select G₂ for expansion


	A heuristic is consistent if for every node n, every successor n' of n generated by any action a,

	h(n) ≤ c(n,a,n') + h(n')

	If h is consistent, we have
	f(n') = g(n') + h(n')
	= g(n) + c(n,a,n') + h(n')
	≥ g(n) + h(n)
	= f(n)

	i.e., f(n) is non-decreasing along any path.

	Theorem: If h(n) is consistent, A* using GRAPH-SEARCH is optimal


	A^* expands nodes in order of increasing f value

	Gradually adds "f-contours" of nodes
	Contour i has all nodes with f=f_i, where f_i < f_i+1


	Complete? Yes (unless there are infinitely many nodes with f ≤ f(G) )
	Time? Exponential
	Space? Keeps all nodes in memory
	Optimal? Yes


	E.g., for the 8-puzzle:
	h₁(n) = number of misplaced tiles
	h₂(n) = total Manhattan distance
	(i.e., no. of squares from desired location of each tile)




	h₁(S) = ?
	h₂(S) = ?


	If h₂(n) ≥ h₁(n) for all n (both admissible)
	then h₂ dominates h₁
	h₂ is better for search

	Typical search costs (average number of nodes expanded):

	d=12 IDS = 3,644,035 nodes A^(h₁) = 227 nodes A^(h₂) = 73 nodes
	d=24 IDS = too many nodes A^(h₁) = 39,135 nodes A^(h₂) = 1,641 nodes


	A problem with fewer restrictions on the actions is called a relaxed problem
	The cost of an optimal solution to a relaxed problem is an admissible heuristic for the original problem
	If the rules of the 8-puzzle are relaxed so that a tile can move anywhere, then h₁(n) gives the shortest solution
	If the rules are relaxed so that a tile can move to any adjacent square, then h₂(n) gives the shortest solution


	In many optimization problems, the path to the goal is irrelevant; the goal state itself is the solution

	State space = set of "complete" configurations
	Find configuration satisfying constraints, e.g., n-queens

	In such cases, we can use local search algorithms
	keep a single "current" state, try to improve it


	Put n queens on an n × n board with no two queens on the same row, column, or diagonal


	Problem: depending on initial state, can get stuck in local maxima



	h = number of pairs of queens that are attacking each other, either directly or indirectly
	h = 17 for the above state


	Idea: escape local maxima by allowing some "bad" moves but gradually decrease their frequency


	One can prove: If T decreases slowly enough, then simulated annealing search will find a global optimum with probability approaching 1

	Widely used in VLSI layout, airline scheduling, etc


	Why keep just one best state?






	Can be used with randomization too


	A successor state is generated by combining two parent states

	Start with k randomly generated states (population)

	A state is represented as a string over a finite alphabet (often a string of 0s and 1s)

	Evaluation function (fitness function). Higher values for better states.

	Produce the next generation of states by selection, crossover, and mutation







	Fitness function: number of non-attacking pairs of queens (min = 0, max = 8 × 7/2 = 28)
	24/(24+23+20+11) = 31%
	23/(24+23+20+11) = 29% etc.


	Many problems are offline
		Do search for action and then perform action

	Online search interleave search & execution
		Necessary for exploration problems
		New observations only possible after acting


	Actual cost of path
	Best possible cost
	(if agent knew space in advance)


	30/20 = 1.5

	For cost, lower is better


Exploration problems: agent physically in some part of the state space.
	e.g. solving a maze using an agent with local wall sensors
	Sensible to expand states easily accessible to agent (i.e. local states)
		Local search algorithms apply (e.g., hill-climbing)


	Build a game player
	Restricted by time per move (specifics TBA)

	Interact with the game driver through command line
		Each turn, we will run your program, providing a a grid as input.
		Your output will be just the coordinates of your stone placement.



	Note: we haven’t yet covered all of the methods to solve this problem

	This week: start thinking about it, discuss among yourselves (remember the G.I. Rule!)
		What heuristics are good to use?
		What type of search makes sense to use?