Transcript Document 7840566

Study Group
Randomized Algorithms
21st June 03
Topics Covered
• Game Tree Evaluation
– its expected run time is better than the worstcase complexity of any deterministic algorithm
– demonstrates a technique to derive a lower
bound on running time of any randomized
algorithm for a problem
• Introduction to Game Theory
– leads to the Minimax Principle
Definition of Game Tree
• a Game Tree Td,k is uniform tree in which
the root and the internal nodes has d
children and every leaf is at distance 2k
from the root
• internal nodes at even distance from the
root are labeled MIN and at odd distance
are labeled MAX
• each leaf is associated with a value
Example of a Game Tree T2,2
MIN
MAX
MAX
MIN
MAX
0
1
MIN
MAX
0
0
MAX
0
1
MIN
MAX
0
MAX
1
0
MIN
MAX
1
0
MAX
0
0
MAX
1
0
1
Observations
• Every root-to-leaf path goes through the
same number of MIN and MAX nodes
(including the root)
• If the depth of the tree is 2k, there are 22k
= 4k leaves
Game Tree Evaluation
• MIN (AND) Node
– returns the lesser of the two children
0
0
0
0
1
MIN
MIN
MIN
MIN
0
0
1
1
0
1
1
Game Tree Evaluation
• MAX (OR) Node
– returns the greater of the two children
0
0
1
1
1
MAX
MAX
MAX
MAX
0
0
1
1
0
1
1
What is the value returned by the root?
AND
OR
OR
AND
OR
0
AND
OR
1
0
OR
0
0
AND
OR
1
0
OR
1
0
AND
OR
1
0
OR
0
0
OR
1
0
1
A Deterministic Algorithm
• Depth-first manner
– always visit the left child before the right child
AND
OR
0
AND
OR
1
0
OR
1
worst case – need to visit ALL 4k leaves
1
OR
0
1
0
a better case – visit 2 leaves is enough
A Randomized Algorithm
• Coin toss
– 0.5 probability choosing the left child and 0.5
probability choosing the right child
AND
OR
0
OR
1
0
1
Expected cost (number of leaves visited)  3
Design Rationale
• Suppose AND node were to return 0
– at least one of the leaves is 0
– if deterministic algorithm is used, your
opponent can always “hide” this 0 and make
your algorithm visit both leaves
– if randomized algorithm is used, you foils your
opponent’s strategy. The expected number of
steps (leaf visits) is 3/2
• Similar for OR node were to return 1
Design Rationale
• Expected cost
– EAND_0 = EOR_1 = 3/2
• What if AND(OR) node were to return 1(0)?
– both children are 1(0), it seems that the
randomized algorithm doesn’t improve much
since we need to visit both children anyway
– however, it benefits the parent level
Analysis of the Randomized Algo.
• Claim:
The expected cost of the randomized
algorithm for evaluating any T2,k game tree
is at most 3k
• Proof by induction:
– consider k = 1
– expected cost  3
Analysis of the Randomized Algo.
• Case I – root evaluated to 0
– at least one of the subtrees
(OR nodes) gives 0
– you have 0.5 probability that
this particular node is checked
first
– E(T) = ½  2 + ½  (3/2 + 2)
= 2.75
return 0
AND
OR
0
OR
0
0
1
Analysis of the Randomized Algo.
• Case II – root evaluated to 1
– both subtrees give 1
– E(T) = 2  3/2
=3
return 1
AND
• Both cases give  3
expected cost, so the claim
is true for k=1
OR
0
OR
1
0
1
Analysis of the Randomized Algo.
• Assume that for all T2,k-1, the expected
cost  3k-1
• First, consider the OR-root tree
either gives 1 or 0
OR
T2,k-1
T2,k-1
Analysis of the Randomized Algo.
• Case I: OR-root gives 1
– at least one subtree gives 1
– 0.5 probability we use it first
– E(T)  ½  3k-1 + ½  2  3k-1
= 3/2  3k-1
either 1 or 0
OR
• Case II: OR-root gives 0
– both subtrees give 0
– E(T)  2  3k-1
T2,k-1
T2,k-1
Analysis of the Randomized Algo.
• Now, consider the AND-root game tree,
T2,k
OR
T2,k-1
T2,k
AND
T2,k-1
OR
T2,k-1
OR
T2,k-1
T2,k-1
T2,k-1
Analysis of the Randomized Algo.
• Case I: AND-root gives 0
– at least one subtree gives 0
– 0.5 probability we use it first
– E(T2,k)  ½  2  3k-1
+ ½  (3/2  3k-1 + 2  3k-1)
= 2.75  3k-1  3k
either 1 or 0
AND
• Case II: AND-root gives 1
– both subtrees give 1
– E(T2,k)  2  3/2  3k-1
= 3k
OR
T2,k-1
OR
T2,k-1
T2,k-1
T2,k-1
Analysis of the Randomized Algo.
• Proved the claim:
The expected cost of the randomized
algorithm for evaluating any T2,k game tree
is at most 3k
• A tree has n = 4k leaves, then k = log4n.
Substitute log4n for k in the expected cost,
then the cost  3log4n. By xlogab = blogax, the
cost  nlog43 =n0.793
Question
Our randomized algorithm for the game
tree evaluation of any uniform binary tree
with n leaves is n0.793. Can we establish
that no randomized algorithm can have a
lower expected running time?
YES! Using Yao’s technique  the
Minimax Theorem  Game Theory Basics
Introduction to Game Theory
• Consider the stone-paper-scissors game
between 2 players
– loser pays $1 to the winner
– payoff matrix M : Mij denotes the payoff by the
Column player to the Row player
Scissors
Paper
Stone
Scissors
0
1
-1
Paper
-1
0
1
Stone
1
-1
0
Two-person Zero-sum Game
• Zero-sum game
– the net amount won by C and R is exactly
zero, i.e., the amount of money is not
increased or decreased among them
• Every two-person zero-sum game can be
represented by a nm payoff matrix
Pure Strategy v.s. Mixed Strategy
• Pure (Deterministic) strategy
– always uses the same strategy or a
deterministic pattern, e.g., R always chooses
‘stone’ while C always chooses ‘paper’
• Mixed (Randomized) strategy
– the strategy chosen by a player is randomized,
i.e., a probability distribution among all
possible strategies
Pure Optimal Strategy
• Zero-information game
– the strategy chosen by the opponent is
unknown
• Naturally, the goal of the row (column)
player is to maximize (minimize) the payoff
– If R chooses strategy i, then she is
guaranteed a payoff of minjMij, regardless of
what C’s strategy is
– Optimal strategy for R is an strategy i that
maximize minjMij.
Pure Optimal Strategy
• Similarly, the optimal strategy of C is
– If C chooses strategy j, then he is guaranteed
a loss of no more than maxiMij, regardless of
what R’s strategy is
– Optimal strategy for C is an strategy j that
minimize maxiMij.
• Let Vr = maximinjMij and Vc = minjmaxiMij
be the lower bound of payoff R can get
and the upper bound of loss C can ensure
respectively
Inequality for All Payoff Matrix
• Minimax Inequality
maximinjMij  minjmaxiMij
j
• Proof:
i
Vr
z
Vc
Clearly, Vr  z  Vc
Saddle-point
• If Vr = Vc, we say the game has a solution
and the value of the game is V = Vr = Vc
• the solution is also known as the saddlepoint of the game
• If no saddle-point exists, it means there is
no clear-cut pure optimal strategy for any
player
Mixed Strategy Game
• The Row player picks a vector p = (p1, …,
pn), which is a probability distribution on
the rows of M. (i.e., pi is the probability
that R will choose strategy i)
• Similarly, the Column player picks a vector
q = (q1, …, qm), i.e., qj is the probability
that R will choose strategy j)
Mixed Optimal Strategy
• Expected Payoff
– E[payoff] = pT M q
– R aims to maximize it while C aims to minimizes it
• As before, let Vr = maxpminq pT M q be the lower
bound of the expected payoff R can get using a
strategy p
• Let VC = minqmaxp pT M q be the upper bound of
the expected payoff C need to pay using a
strategy q
von Neumann’s Mininmax Theorem
• For any two person, zero-sum game
specified by a matrix M
maxpminqptMq = minqmaxppTMq
• the optimal strategy for R will yield the
same payoff as the optimal strategy for C!
• if either player uses his optimal strategy,
the opponent cannot improve the payoff