1. State Space 4 Chapter 4
Adversarial Games

2. Two Flavors Games of Perfect Information
Each player knows everything that can be known
Chess, Othello
Games of Imperfect Information
Player’s have partial knowledge
Poker: dispute is settled by revealing the contents of one’s hand
Two Flavors

3. Two Approaches to Perfect Information Games
Use simple heuristics and search many nodes
Use sophisticated heuristics and search few nodes
Cost of calculating the heuristics might outweigh the cost of opening many nodes
The closer h is to h*, the better informed it is. But information can be expensive
Two Approaches to Perfect Information Games

4. A Model: MiniMax on exhaustively searchable graphs
Two Players
min: tries to achieve an outcome of 0
max: tries to achieve an outcome of 1
A Model: MiniMax on exhaustively searchable graphs

5. You are max at node A MAX A C D Min B 1 Max G E F Min K 1 J H I Max OC D
Min B 1
Max G E F
Min K 1 J H I
Max O N 1 1 L M P Q R
Min 1

6. Wins Max would like to end game at F,J,N,Q,L
Min would like to end game at D,H,P,R,M
Propagating Scores: A first pass
Min’s Turn at Node I
Go to M to win
So assign I a 0
Max’s turn at node O
Go to Q to win
So assign 1 to node O
Wins

7. If faced with two labeled choices, you would choose 0 (if min) or 1 (if max)
Assume you’re opponent will play the same way
Conclusion

8. Propagating Scores For each unlabeled node in the tree
If it’s max’s turn, give it the max score of its children
If it’s min’s turn, give it the min score of its children
Now label the tree
Conclusion: Max must choose C at the first move or lose the game
Propagating Scores

9. 7 coins
2 players
Players divide coins into two piles at each move, such that
Piles have an unequal number of coins
No pile is empty
Play ends when a player can no longer move. At that point the other player wins
and his/her score bubbles up
Nim

10. Start of Game
min 7
4,3
5,2
6,1
5,1,1
4,2,1
P. 152 Luger.

11. Problem
For games of any complexity, you can’t search the whole tree
Solution
Look ahead a fixed number of plys (levels)
Evaluate according to some heuristic estimate
Develop a criterion for pruning subtrees that don’t require examination
αβ Pruning

12. Instead of representing wins, numbers represent the relative goodness of nodes.
At any junction
Max chooses highest
Min chooses lowest
Higher means better for max
Lower means better for min
Recast

13. What should Max do? 8 Max g 4 8 j k 9 t n m z 4 8 12 r p q Min 7 3 9
beta n m z
alpha 4 8 12 r p q
Min 7 3 9

14. Situation Max’s turn at node g Left subtree of g has been explored
If max chooses j, min will choose m
So the best max can do by going left is 8. Call this α
Now Examine K and its left subtree n with a value of 4
If max chooses k, the worst min can do is 4.
Why? T may be < 4. If it is min will choose it. If not, min will choose 4
So the worst min can do, if max goes right is 4. Call this β
Situation

15. Question Must max expand the rst(K)? No. min is guaranteed 4.
But if max chose j, min is guaranteed 8
So max is better off by going left
More formally:
Val(k) = min(4, val(t)) <= 4
Val(g) = max(8,val(k))
= max(8, min(4,val(t))
= 8
Question

16. Max Principle If you’re Max:
Search can be stopped below any min node where β <= α of its max ancestor
Max Principle

17. What should Min do?
Min 4 k 9 4 n t d e p q r
Beta 4 3 7 9
alpha 3

18. Situation Min’s turn at node k Left subtree of k has been explored
If min chooses n, max will choose d
So the best min can do by going left is 4. Call this β
Now examine T and its left subtree P with a value of 7
If min chooses T, the worst max can do is 7.
Why?
Q or R may be > 7. If either is Max will choose one of them. If not, max will choose 7.
So the worst max can do, if min goes right is 7. Call this α
Situation

19. Question Should min explore Q and R No
Max is guaranteed 7 if Min chooses T
But if min chooses N, max gets only 4
Val(T) = max(7,val(Q), val(R)) >= 7
Val(k) = min(4,val(T)) = 4
Question

20. Min Principle If you’re min:
Search can be stopped below any max node where α >= β of its min ancestor
Min Principle

21. To Summarize Max’s turn Max principle Min’s turn Min principle
β is min’s guaranteed score (the worst min can do)
α is best max can do
Max principle
Search can be stopped below any min node where
β <= α of its max ancestor
Min’s turn
α is max’s guaranteed score (the worst max can do)
β is best min can do
Min principle
Search can be stopped below any max node where
α >= β of its min ancestor
To Summarize

22. On White Board
A Few Examples

24. AB Prune (Nilsson, p. 205) Similarity between 2 and 2’
Arguments for min/min principles (Slides 15, 19)
AB Prune (Nilsson, p. 205)

25. Need for A-B prune Combinatorial Explosion
Number of nodes produced by a full search of the chess state space: 10120
Not just a big number
Comparable to
Number of molecules in the universe
Number of nanoseconds since the big bang
Need for A-B prune

26. Best Case Performance Call ND the number of terminal nodes
D the depth of the search space
ND the number of terminal nodes
B the branching factor
Best case AB performance:
ND = 2BD/2 – 1 for even D
ND = 2B(D+1)/2 + B(D-1)/2 for odd D
Best Case Performance

27. Suppose B = 5, D = 6 w/out alpha/beta ND= 56 = w/alpha/beta ND= 2* = 249 Approx. 1.6% of the terminal nodes that would have been examined without AB prune
Example

28. Average Performance AB prune reduces branching factor B to B3/4
Suppose B = 5, Bab = 3.34
Suppose D = 6
Then ND = 56 = 15625
NDAB = = 1388 ≈ 8.8% without AB prune
Average Performance

29. Clear relationship between branching factor and the size of the search space
Let T = number of nodes in a full binary tree
T = … + 2D-1 = 2D – 1
Easily proved through induction
Binary Trees

30. T = B0 +B1 + B2 + … + BD = B(BD - 1)/(B-1) + 1 Also provable through induction
Extend to arbitrary B

31. To Sum Up D is the depth of the search space
B is the average number of descendants at each level
Size of search space = B(BD -1)/(B-1) + 1
Grows very fast
As branching factor increases
As depth increases
Combinatorial Explosion: search space grows too fast to be exhaustively searched
But we want to search deeply (large D)
Conclusion: reduce B through AB pruning
To Sum Up