- 54 Views
- Uploaded on
- Presentation posted in: General

Classic AI Search Problems

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

Classic AI Search Problems

- Sliding tile puzzles
- 8 Puzzle (3 by 3 variation)
- Small number of 8!/2, about 1.8 *105 states

- 15 Puzzle (4 by 4 variation)
- Large number of 16!/2, about 1.0 *1013 states

- 24 Puzzle (5 by 5 variation)
- Huge number of 25!/2, about 7.8 *1025 states

- 8 Puzzle (3 by 3 variation)
- Rubik’s Cube (and variants)
- 3 by 3 by 3 4.3 * 1019 states

- Navigation (Map searching)

2

1

3

4

7

6

5

8

Classic AI Search Problems

- Invented by Sam Loyd in 1878
- 16!/2, about1013 states
- Average number of 53 moves to solve
- Known diameter (maximum length of optimal path) of 87
- Branching factor of 2.13

3*3*3 Rubik’s Cube

- Invented by Rubik in 1974
- 4.3 * 1019 states
- Average number of 18 moves to solve
- Conjectured diameter of 20
- Branching factor of 13.35

Navigation

start

end

Arad

Zerind

Sibiu

Timisoara

Oradea

Fagaras

Arad

Rimnicu Vilcea

Sibiu

Bucharest

function general-search(problem, QUEUEING-FUNCTION)

nodes = MAKE-QUEUE(MAKE-NODE(problem.INITIAL-STATE))

loop do

if EMPTY(nodes) then return "failure"

node = REMOVE-FRONT(nodes)

if problem.GOAL-TEST(node.STATE) succeeds then return node

nodes = QUEUEING-FUNCTION(nodes, EXPAND(node, problem.OPERATORS))

end

A nice fact about this search algorithm is that we can use a single algorithm to do many kinds of search. The only difference is in how the nodes are placed in the queue.

Completeness

solution will be found, if it exists

Time complexity

number of nodes expanded

Space complexity

number of nodes in memory

Optimality

least cost solution will be found

Breadth first

Uniform-cost

Depth-first

Depth-limited

Iterative deepening

Bidirectional

QUEUING-FN:- successors added to end of queue (FIFO)

Arad

Zerind

Sibiu

Timisoara

Oradea

Fagaras

Arad

Rimnicu Vilcea

Arad

Arad

Oradea

Lugoj

Complete ?

Yes if branching factor (b) finite

Time ?

1 + b + b2 + b3 +…+ bd = O(bd), so exponential

Space ?

O(bd), all nodes are in memory

Optimal ?

Yes (if cost = 1 per step), not in general

Assuming b = 10, 1 node per ms and 100 bytes per node

QUEUING-FN:- insert in order of increasing path cost

Arad

75

118

140

Zerind

Sibiu

Timisoara

140

151

99

80

Oradea

Fagaras

Arad

Rimnicu Vilcea

118

75

71

111

Arad

Arad

Lugoj

Oradea

Complete ?

Yes if step cost >= epsilon

Time ?

Number of nodes with cost <= cost of optimal solution

Space ?

Number of nodes with cost <= cost of optimal solution

Optimal ?- Yes

QUEUING-FN:- insert successors at front of queue (LIFO)

Arad

Zerind

Zerind

Sibiu

Sibiu

Timisoara

Timisoara

Arad

Oradea

Complete ?

No:- fails in infinite- depth spaces, spaces with loops

complete in finite spaces

Time ?

O(bm), bad if m is larger than d

Space ?

O(bm), linear in space

Optimal ?:- No

Choose a limit to depth first strategy

e.g 19 for the cities

Works well if we know what the depth of the solution is

Otherwise use Iterative deepening search (IDS)

Complete ?

Yes if limit, l >= depth of solution, d

Time ?

O(bl)

Space ?

O(bl)

Optimal ?

No

function ITERATIVE-DEEPENING-SEARCH():

for depth = 0 to infinity do

if DEPTH-LIMITED-SEARCH(depth) succeeds

then return its result

end

return failure

Complete ?

Yes

Time ?

(d + 1)b0 + db1 + (d - 1)b2 + .. + bd = O(bd)

Space ?

O(bd)

Optimal ?

Yes if step cost = 1

Various uninformed search strategies

Iterative deepening is linear in space

not much more time than others

Use Bi-directional Iterative deepening were possible

Suppose that you happen to know that the optimal solution goes thru Rimnicy Vilcea…

Suppose that you happen to know that the optimal solution goes thru Rimnicy Vilcea…

Rimnicy Vilcea

- Uses evaluation function f = g+ h
- g is a cost function
- Total cost incurred so far from initial state
- Used by uniform cost search

- h is an admissible heuristic
- Guess of the remaining cost to goal state
- Used by greedy search
- Never overestimating makes h admissible

- g is a cost function

Our Heuristic

QUEUING-FN:- insert in order of f(n) = g(n) + h(n)

Zerind

Sibiu

Timisoara

Arad

g(Zerind) = 75

g(Timisoara) = 118

g(Sibiu) = 140

g(Timisoara) = 329

h(Zerind) = 374

h(Sibiu) = 253

f(Sibui) = …

f(Zerind) = 75 + 374

- Optimal and complete
- Admissibility guarantees optimality of A*
- Becomes uniform cost search if h= 0

- Reduces time bound from O(b d ) to O(b d - e)
- b is asymptotic branching factor of tree
- d is average value of depth of search
- e is expected value of the heuristic h

- Same as BFS and uniform cost. But an iterative deepening version is possible … IDA*

- Solves problem of A* memory usage
- Reduces usage from O(b d ) to O(bd )
- Many more problems now possible

- Easier to implement than A*
- Don’t need to store previously visited nodes

- AI Search problem transformed
- Now problem of developing admissible heuristic
- Like The Price is Right, the closer a heuristic comes without going over, the better it is
- Heuristics with just slightly higher expected values can result in significant performance gains

- Suppose you have two admissible heuristics…
- But h1(n) > h2(n)
- You may as well forget h2(n)

- Suppose you have two admissible heuristics…
- Sometimes h1(n) > h2(n) and sometimes h1(n) < h2(n)
- We can now define a better heuristic, h3
- h3(n) = max( h1(n) , h2(n) )

- Suppose you have two admissible heuristics…
- h1(n) is h(n) = 0 (same as uniform cost)
- h2(n) is misplaced tiles
- h3(n) is Manhattan distance

- The study of games is called game theory
- A branch of economics

- We’ll consider special kinds of games
- Deterministic
- Two-player
- Zero-sum
- Perfect information

- A zero-sum game means that the utility values at the end of the game total to 0
- e.g. +1 for winning, -1 for losing, 0 for tie

- Some kinds of games
- Chess, checkers, tic-tac-toe, etc.

- Initial state
- Initial board position, player to move

- Operators
- Returns list of (move, state) pairs, one per legal move

- Terminal test
- Determines when the game is over

- Utility function
- Numeric value for states
- E.g. Chess +1, -1, 0

- Each level labeled with player to move
- Max if player wants to maximize utility
- Min if player wants to minimize utility

- Each level represents a ply
- Half a turn

- MAX wants to maximize utility, but knows MIN is trying to prevent that
- MAX wants a strategy for maximizing utility assuming MIN will do best to minimize MAX’s utility

- Consider minimaxvalue of each node
- Utility of node assuming players play optimally

- Calculate minimax value of each node recursively
- Depth-first exploration of tree

- Game tree (aka minimax tree)

Max node

Min node

Min

2

7

5

5

6

7

Max

5

4

2

5

10

4

6

Utility

- Time Complexity?
- O(bm)

- Space Complexity?
- O(bm) or O(m)

- Is this practical?
- Chess, b=35, m=100 (50 moves per player)
- 3510010154 nodes to visit

- Improvement on minimax algorithm
- Effectively cut exponent in half

- Prune or cut out large parts of the tree
- Basic idea
- Once you know that a subtree is worse than another option, don’t waste time figuring out exactly how much worse

3

2

3

0

3

30

a pruned

32

a pruned

3

5

2

0

5

3

53

b pruned

3

5

1

2

2

0

2

1

2

3

5

0