AND/OR Algorithm with example

1. Problem Reduction Search-
(AO* Search)
Lecture-21
Hema Kashyap
7 September 2015 1

2. AND/OR Search
• The Depth first search and Breadth first search given earlier for OR trees or
graphs can be easily adopted by AND-OR graph. The main difference lies
in the way termination conditions are determined, since all goals following
an AND nodes must be realized; where as a single goal node following an
OR node will do. So for this purpose we are using AO* algorithm.
• Like A* algorithm here we will use two arrays and one heuristic function.
OPEN: It contains the nodes that has been traversed but yet not been marked
solvable or unsolvable.
CLOSE: It contains the nodes that have already been processed.
h(n):The distance from current node to goal node.

3. The AND/OR Graph Search Problem
• Problem Definition:
– Given: [G , s , T] where
• G: implicitly specified AND/OR graph
• S: start node of the AND/OR graph
• T: set of terminal nodes
• H(n): heuristic function estimating the cost of solving
the sub- problem at n
– To find:
• A minimum cost solution tree

4. Algorithm
Step 1: Place the starting node into OPEN.
Step 2: Compute the most promising solution tree say T0.
Step 3: Select a node n that is both on OPEN and a member of T0.
Remove it from OPEN and place it
Step 4: If n is the terminal goal node then leveled n as solved and
leveled all the ancestors of n as solved. If the starting node is marked
as solved then success and exit.
Step 5: If n is not a solvable node, then mark n as unsolvable. If
starting node is marked as unsolvable, then return failure and exit.
Step 6: Expand n. Find all its successors and find their h (n) value, push
them into OPEN.
Step 7: Return to Step 2.
Step 8: Exit. in CLOSE

5. Example

6. An Example
A(8)

7. An Example
C
DB
A
(8)(1)
(2)
[12]
4 5
5
[13]

8. An Example
C
DB
A
(8)(4)
(2)
[15]
4 5
5
[13]
2

9. An Example
C
DB
A
(3)(4)
G
E
(2)
(1)
(0)
[15]
4 5
5
2
2
4
[8]

10. An Example
C
DB
A
(4)(4)
G
E
(2)
(3)
(0)
[15]
4 5
5
2
2
2
4
[9]
3

11. An Example
C
DB
A
(4)
G
E
(2)
(3)
(0)
[15]
4 5
5
2
2
2
4
Solved
3
Solved
Solved

12. Another Example
Current State = S
f(A) = 3 + 5 = 8
f(B) = 2 + 4 = 6
Current State = B
f(S) = 2 + 8 = 10
f(A) = 4 + 5 = 9
f(C) = 1 + 5 = 6
f(E) = 4 + 2 = 6
Current State = C
f(H) = 2 + 4 = 6
f(B) = 1 + 6 = 7
A B
C
S
(4)
(4) D
(5)
2
(2)
G
(1)
E
(2)H
(0)
3
4
41
(5)
F
2 13
2

13. Current State = H
f(C) = 2 + 7 = 9
Current State = C
f(B) = 1 + 6 = 7
f(H) = 
Current State = B
f(S) = 2 + 8 = 10
f(A) = 4 + 5 = 9
f(E) = 4 + 2 = 6
f(C) = 
Current State = E
f(B) = 4 + 9 = 13
f(D) = 3 + 2 = 5
f(F) = 1 + 1 = 2
A B
C
S
(4)
(4) D
(5)
2
(2)
G
(1)
E
(2)H
(0)
3
4
41
(5)
F
2 13
2
Another Example

14. A B
C
S
(4)
(4) D
(5)
2
(2)
G
(1)
E
(2)H
(0)
3
4
41
(5)
F
2 13
2
Current State = F
f(E) = 1 + 5 = 6
Current State = E
f(D) = 3 + 2 = 5
f(B) = 4 + 9 = 13
f(F) = 
Current State = D
f(G) = 2 + 0 = 2
f(E) = 3 + 13 = 16
Visited Nodes =
S, B, C, H, C, B, E, F, E, D, G
Path = S, B, E, D, G
Another Example

15. Advantages
• It is an optimal algorithm.
• If traverse according to the ordering of nodes.
• It can be used for both OR and AND graph.

16. Disadvantages
• Sometimes for unsolvable nodes, it can’t find
the optimal path.
• Its complexity is than other algorithms.