Adversary Games

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Adversary Games
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Two Flavors

• Perfect Information
• everything that can be known is known
• Chess, Othello
• Imperfect Information
• Players have each have partial knowledge
• Poker dispute is settled by revealing the
  contents of ones hand

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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

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MiniMax on exhaustively searchable graphs

• Two Players
• min tries to achieve an outcome of 0
• max tries to achieve an outcome of 1

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You are max at node A

• Expand the entire search space

MAX
A
0
C
D
Min
B
1
Max
G
E
F
Min
K
1
J
0
H
I
Max
O
N
1
1
L
M
0
0
R
P
Q
0
Min
1
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Wins

• Max F,J,N,Q,L
• Min D,H,P,R,M
• Propagating Scores A first pass
• Mins Turn a Node I
• Go to M to win
• So assign I a 0
• Maxs turn at node O
• Go to Q to win
• So assign 1 to node O

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Conclusion

• If faced with two labeled choices, you would
  choose 0 (if min) or 1 (if max)
• Assume youre opponent will play the same way

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Propagating Scores

• For each unlabeled node in the tree
• If its maxs turn, give it the max score of its
  children
• If its mins turn, give I the min score of its
  children
• Now label the tree
• Conclusion Max must choose C at the first move
  or lose the game

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Nim

• 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

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Start of Game
min
7
4,3
5,2
6,1
5,1,1
4,2,1
Complete the game to see that Max wins only if
min makes a mistake at 6,1 or 5,2
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a-ß Pruning

• Problem
• For games of any complexity, you cant 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
  dont require examination

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Recast

• 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

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Example Maxs Turn
8
Max
g
4
8
j
k
9
t
n
beta
m
z
Alpha
4
8
12
r
p
q
Min
7
3
9
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Situation

• Maxs 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 a
• Now Examine K and its left subtree n with a value
  of 4
• If max chooses k, the worst min can do is 4.
• Because, T may be lt 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 ß

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Question

• Should max expand the right subtree of k.
• No. Because is guaranteed 4. But if max chose
  j, min is only guaranteed 8.
• Val(k) min(4, val(t)) lt 4
• Val(g) max(8,val(k))
• max(8, min(4,val(t))
• 8

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Leads to Max Principle

• Search can be stopped below any min node where ß
  lt a of its max ancestor

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Example Mins Turn
4
min
k
9
4
n
t
d
e
4
p
q
r
Min
Beta
4
3
7
9
3
alpha
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Situation

• Mins 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 ß
• Now examine T and its left subtree P with a value
  of 7
• If min chooses T, the worst max can do is 7.
• Because, Q or R may be gt 7. If it is Max will
  choose it. If not, min will choose 7.
• So the worst max can do, if min goes right is 7.
  Call this a

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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)) gt 7
• Val(k) min(4,val(T)) 4

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Leads to min principle

• Search can be stopped below any max node where a
  gt ß of its min ancestor

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To Summarize

• Maxs turn
• ß is mins guaranteed score (the worst min can
  do)
• a is best max can do
• Max principle
• Search can be stopped below any min node where
• ß lt a of its max ancestor
• Mins turn
• a is maxs guaranteed score (the worst max can
  do)
• ß is best min can do
• Min principle
• Search can be stopped below any max node where
• a gt ß of its min ancestor

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Combinatorial Explosion

• Key Idea is the branching factor
• Sum of Subsets Problem
• 2
• Traveling Salesperson
• (N 1)/2
• 8 puzzle
• 2.67
• 15 puzzle
• Approximately 4
• Let B average branching factor
• Let T total nodes
• Let D depth of search
• Then
• T B(BD 1)/(B 1) 1

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Sum of Subsets

• Given a set, S, of positive integers, find all
  subsets whose sum is m.
• E.G.
• S 7,11,13,24
• m 31
• Solutions
• S-1 7,11,13
• S-2 7,24

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Problem Representation

• Solution is a sequence of 1s and 0s, indicating
  that elements of S have been chosen or not.
• Rep of S-1 (1,1,1,0)
• Rep of S-2 (1,0,0,1)
• State space is a tree where left turn indicates a
  1 and right turn indicates a 0

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Partial Space

S-1
Three left turns and a right turn to get to S-1
is the sequence (1,1,1,0) Clearly the branching
factor is 2
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TSP

• Can also be represented as a state space search.
• Suppose 4 cities
• 4 Choices at level 0
• 3 Choices at level 1
• 2 Choices at level 2
• 1 Choice at level 3
• T 4321 4P4 4!/(4-4)! 4!

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Branching Factor

• B_F (sum of choices at each level)/of levels
• (4321)/4 2.5
• Clearly this increases with the size of the tour
• (123n)/n (n(n1)/2)/n (n1)/2

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Relationship between branching factor and nodes
in the tree

• Whenever the branching factor gt 2, we have an
  exponentially complex problem
• Let T number of nodes in a full binary tree
• T 20 21 22 2d-1 2d 1
• Easily proved through induction

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Replace 2 by branching factor, B

• T B0 B1 B2 BL
• B(BL -1)/(B-1) 1
• Where L is d-1, d being the depth of the tree
• Proof
• Basis T B(B0 1)/(B-1) 1 1
• Inductive hypothesis
• T B0 B1 B2 BL B(BL -1)/(B-1) 1

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• Show that this is true at level L1
• That is,
• Show
• B0 B1 B2 BL1 B(BL1 -1)/(B-1) 1
• B0 B1 B2 BL BL1 B(BL1 -1)/(B-1)
  BL1 1
• (B(BL -1) BL1(B-1))/(B-1) 1
• (B( (BL -1) BL(B-1))/(B-1) 1
• (B(BL -1 BL1 BL )/(B-1) 1
• B(BL1 -1)/(B-1) 1
• Which is what we were trying to prove

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Two Concepts

• B average number of descendents that emerge
  from any state in the space
• Total nodes B(BL -1)/(B-1) 1
• Where L is the deepest level (or, the depth of
  the search)

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Another Problem 8 Puzzle

• A B C
• 1 2 3 1 2 1 2
• 8 4 3 4 5 3 4 5
• 7 6 5 6 7 8 6 7 8
• A 4 moves for blank 1 position 4
• B 2 moves for blank 4 positions 8
• C 3 moves for blank 4 positions 12
• B (4 8 12) /(1 4 4) 2.67
• Does the branching factor of larger (15, 24)
  puzzles approach 4 as the puzzles get larger?

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