Study GroupRandomized Algorithms

21st June 03

Topics Covered

• Game Tree Evaluation– its expected run time is better than the worst-

case 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

MINMINMIN

MAXMAX

MIN

MAXMAXMAX MAX MAXMAXMAX MAX

0 1 0 0 0 1 0 1 0 1 0 0 0 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

MIN

0 0

0

MIN

0 1

0

MIN

1 0

0

MIN

1 1

1

Game Tree Evaluation

• MAX (OR) Node– returns the greater of the two children

MAX

0 0

0

MAX

0 1

1

MAX

1 0

1

MAX

1 1

1

What is the value returned by the root?

AND

ANDANDAND

OROR

AND

OROROR OR OROROR OR

0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1

A Deterministic Algorithm

• Depth-first manner– always visit the left child before the right child

AND

OR OR

0 1 0 1

AND

OR OR

1 0 1 0

worst case – need to visit ALL 4k leaves 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 OR

0 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

AND

OR OR

0 0 0 1

return 0

Analysis of the Randomized Algo.

• Case II – root evaluated to 1– both subtrees give 1– E(T) = 2 3/2

= 3

• Both cases give 3 expected cost, so the claim is true for k=1

AND

OR OR

0 1 0 1

return 1

Analysis of the Randomized Algo.

• Assume that for all T2,k-1, the expected cost 3k-1

• First, consider the OR-root tree

OR

T2,k-1 T2,k-1

either gives 1 or 0

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

• Case II: OR-root gives 0– both subtrees give 0– E(T) 2 3k-1

OR

T2,k-1 T2,k-1

either 1 or 0

Analysis of the Randomized Algo.

• Now, consider the AND-root game tree, T2,k

OR

T2,k-1 T2,k-1

AND

OR

T2,k-1 T2,k-1

OR

T2,k-1 T2,k-1

T2,k

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

• Case II: AND-root gives 1– both subtrees give 1

– E(T2,k) 2 3/2 3k-1

= 3k

either 1 or 0

AND

OR

T2,k-1 T2,k-1

OR

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 zer

o, i.e., the amount of money is not increased or decreased among them

• Every two-person zero-sum game can be represented by a nm 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 unkno

wn

• Naturally, the goal of the row (column) player is to maximize (minimize) the payoff– If R chooses strategy i, then she is guarantee

d 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 InequalitymaximinjMij minjmaxiMij

• Proof:i

j

Vr

Vc

z

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 saddle-point 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