The Alpha-Beta Procedure
There are two rules for terminating search:
 Search can be stopped below any MIN node having
a beta value less than or equal to the alpha value of
any of its MAX ancestors.
 Search can be stopped below any MAX node
having an alpha value greater than or equal to the
beta value of any of its MIN ancestors.
The Alpha-Beta Procedure
Example:
max
min
max
min
The Alpha-Beta Procedure
Example:
max
min
max
=4
4
min
The Alpha-Beta Procedure
Example:
max
min
max
=4
4 5
min
The Alpha-Beta Procedure
Example:
max
min
=3
=3
4 5 3
max
min
The Alpha-Beta Procedure
Example:
max
min
=3
=3
max
=1
4 5 3 1
min
The Alpha-Beta Procedure
Example:
max
=3
min
=3
=3
max
=1
=8
4 5 3 1
8
min
The Alpha-Beta Procedure
Example:
max
=3
min
=3
=3
max
=1
4 5 3 1
=6
8 6
min
The Alpha-Beta Procedure
Example:
max
=3
=3
=3
min
=6
=1
4 5 3 1
=6
8 6 7
max
min
The Alpha-Beta Procedure
=3
Example:
=3
=3
=3
min
=6
=1
4 5 3 1
max
=6
8 6 7
max
min
The Alpha-Beta Procedure
=3
Example:
max
=3
=3
=3
min
=6
=1
4 5 3 1
=6
8 6 7
max
=2
2
min
The Alpha-Beta Procedure
=3
Example:
max
=3
=3
=3
min
=6
=1
4 5 3 1
=6
8 6 7
=3
=2
2
max
min
The Alpha-Beta Procedure
=3
Example:
max
=3
=3
=3
min
=6
=1
4 5 3 1
=6
8 6 7
=3
max
=2
=5
2
5
min
The Alpha-Beta Procedure
=3
Example:
max
=3
=3
=3
min
=6
=1
4 5 3 1
=6
8 6 7
=3
=2
2
max
=4
5 4
min
The Alpha-Beta Procedure
=3
Example:
max
=3
=3
=3
=4
=6
=1
4 5 3 1
=6
8 6 7
=4
=2
2
min
max
=4
5 4 4
min
The Alpha-Beta Procedure
=3
Example:
max
=3
=3
=3
=4
=6
=1
4 5 3 1
=6
8 6 7
min
=4
=2
2
max
=4
=6
5 4 4
6
min
The Alpha-Beta Procedure
=3
Example:
max
=3
=3
=3
=4
=6
=1
4 5 3 1
=6
8 6 7
min
=4
=2
2
max
=4
=6
5 4 4
6 7
min
The Alpha-Beta Procedure
=4
Example:
max
=3
=3
=3
=4
=6
=1
4 5 3 1
=6
8 6 7
min
=4
=2
2
=6
=4
=6
5 4 4
6 7 7
max
min
The Alpha-Beta Procedure
=4
Example:
Done!
=3
=3
=3
=4
=6
=1
4 5 3 1
=6
8 6 7
min
=4
=2
2
=6
=4
max
=6
5 4 4
6 7 7
max
min
Real Time A*
 Considers the cost (> 0) for switching from one branch to
another in the search
 Example: path finding in real life
11
4
1
F
D
B
f(B) = 1 + 1 = 2
f(D) = 1 + 4 = 5
f(B) = 1 + 5 = 6
f(A) = 1 + 6 = 7
f(B) = 1 + 5 = 6
f(D) = 1 + 4 = 5
f(F)=1+11= 12 (D)
A
2
7
16
C
E
G
(A)
(B)
(A)
(C)
(A)
(B)
f(C) =
f(A) =
f(C) =
f(E) =
f(C) =
f(A) =
f(B) =
1+2=3
1+3=4
1+2=3
1+7=8
1+8=9
1 + 9 = 10
1 + 10 = 11