State space 3 chapter 4
1 / 28

State Space 3 Chapter 4 - PowerPoint PPT Presentation

  • Uploaded on

State Space 3 Chapter 4 . Heuristic Search. Backtrack Depth First Breadth First All work if we have well-defined: Goal state Start state State transition rules But could take a long time. Three Algorithms. An informed guess that guides search through a state space

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about 'State Space 3 Chapter 4' - dougal

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
State space 3 chapter 4

State Space 3Chapter 4

Heuristic Search

Three algorithms

  • Backtrack

  • Depth First

  • Breadth First

    All work if we have well-defined:

  • Goal state

  • Start state

  • State transition rules

    But could take a long time

Three Algorithms



Blocks world a stack of blocks

Start Goal





Two rules:

clear(X)  on(X, table)

clear(X) ^ clear(Y)  on(X,Y)

Generate part of the search space BF. We’ll find the answer, but it could take a long time.

Blocks World: A Stack of Blocks

Heuristic 1

Heuristic 1

Hill climbing

  • At every level, generate all children

  • Continue down path with lowest score

    Define three functions:

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


    h(n) is the heuristic estimate for n--guides the search

    g(n) is path length from start to current node—ensures that we choose node closest to root when more than 1 have same h value

Hill Climbing

Problem heuristic is local

Given C and CB




At level n

The f(n) of each structure is the same

f(n) = g(n) = (1+1-1-1) = g(n)

But which is actually better

Problem: heuristic is local

Problem with local heuristics





  • Must be entirely undone

  • Goal Requires 6 moves


  • Goal requires only two moves

Problem with local heuristics

Global heuristic

Takes the entire structure into account

  • Subtract 1 for each block that has correct support structure

  • Add 1 for each block in an incorrect support structure

Global Heuristic

Seems to work

Goal: D




C f(n) = g(n) + (3+2+1+0) =g(n) + 6




CB f(n) = g(n) + (1 + 0 -1+ 0) = g(n)


So the heuristic correctly chose the second structure

Seems to Work

Best first

The Road Not Taken

  • Open contains current fringe of the search

  • Open: priority queue ordered by f(n)

  • Closed: queue of states already visited

  • Nodes could contain backward pointers so that path back to root can be recovered

Best First

State space 3 chapter 4

path best_first(Start)


open = [start], closed = [];

while (!open.isEmpty())


cs = open.serve();

if (cs == goal)

return path;

generate children of cs;

for each child




child is on open; //node has been reached by a shorter path

if (g(child) < g(child) on open)

g(child on open) = g(child);


child is on closed;

if (g(child < g(child on closed))

{ //node has been reached by a shorter path and is more attractive

remove state from closed;






f(child) = g(child) + h(child);//child has been examined yet





closed.enqueue(cs);//all cs’ children have been examined.

open.reorder();//reorder queue because case statement may have affected ordering


return([]); //failure



  • Admissible Search Algorithms

    • Find the optimal path to the goal if a path to the goal exists

  • Breadth-First: Admissible

  • Depth-First: Not admissible


Algorithm a


  • Best First

  • f(n) = g(n) + h(n)

Algorithm A

State space 3 chapter 4

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


  • g*(n) is the cost of the shortest path from start to n

  • h*(n) is the cost of the shortest path from n to goal

    So, f*(n) is the actual cost of the optimal path

    Can we know f*(n)?


The oracle

The Oracle

Consider g n

  • g(n) – actual cost to n

  • g*(n) – shortest path from start to n

    So g(n) >= g*(n)

    When g*(n) = g(n), the search has discovered the optimal path to n

Consider g*(n)

Consider h n

  • Often we can know

    • Ifh(n) is bounded above by h*(n)

    • (This sometimes means finding a function h2(n) such that h(n) <= h2(n) <= h*(n))

  • Meaning: the optimal path is more expensive than the one suggested by the heuristic

  • Turns out this is a good thing (within bounds)

Consider h*(n)

An 8 puzzle intuition

283 123

164 -> 8 4

7 5 765

Invent two heuristics: h1 and h2

h1(n): number of tiles not in goal position

h1(n) = 5 (1,2,6,8,B)

h2(n): number of moves required to move out-of-place tiles to goal

T1 = 1, T2 = 1, T6 = 1, T8 = 2, TB = 1

h2(n) =6

h1(n) <= h2(n) ?

An 8-puzzle Intuition

H1 n is bounded above by h n

h2(n) <= h*(n)

  • Each out-of-place tile has to be moved a certain distance to reach goal

    h1(n) <= h2(n)

  • h2(n) requires moving out-of-place tiles at least as far as h1(n)

    So, h1(n) <= h2(n) <=h*(n)

h1(n) is bounded above by h*(n)

Leads to a definition


If algorithm A uses a heuristic that returns a value h(n) <= h*(n) for all n, then it is called A*

What does this property mean?

Leads to a Definition

State space 3 chapter 4


  • Suppose: h(rst) = 96

  • But we know it is really 1 (i.e. h*(rst) = 1) because we’re the oracle

  • Suppose: h(lst) = 42

  • Left branch looks better than it really is

  • This makes the left branch seem better than it actually is

Claim all a algorithms are admissible

Suppose: path is better than it is

  • h(n) = 0 and so <= h*(n)

    Search will be controlled by g(n)

    If g(n) = 0, search will be random: given enough time, we’ll find an optimal path to the goal

    If g(n) is the actual cost to n, f(n) becomes breadth-first because the sole reason for examining a node is its distance from start.

    We already know that this terminates in an optimal solution

Claim: All A* Algorithms are admissible

State space 3 chapter 4

  • h(n) = h*(n) path is better than it is

    Then the algorithm will go directly to the goal since h*(n) computes the shortest path to the goal

    Therefore, if our algorithm is between these two extremes, our search will always result in an optimal solution

    Call h(n) = 0, h’(n)

    So, for any h such that

    h’(n) <= h(n) <= h*(n) we will always find an optimal solution

    The closer our algorithm is to h’(n), the more extraneous nodes we’ll have to examine along the way


For any two A* heuristics, ha, path is better than it ishb

If ha(n) <= hb(n), hb(n) is more informed.



Comparison of two solutions that discover the optimal path to the goal state:

  • BF: h(n) = 0

  • h(n) = number of tiles out of place

    The better informed solution examines less extraneous information on its path to the goal