Study Group Randomized Algorithms

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Study GroupRandomized Algorithms

• 21st June 03

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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

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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

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Example of a Game Tree T2,2
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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

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Game Tree Evaluation

• MIN (AND) Node
• returns the lesser of the two children

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Game Tree Evaluation

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

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What is the value returned by the root?
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A Deterministic Algorithm

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

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A Randomized Algorithm

• Coin toss
• 0.5 probability choosing the left child and 0.5
  probability choosing the right child

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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
  opponents strategy. The expected number of
  steps (leaf visits) is 3/2
• Similar for OR node were to return 1

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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 doesnt improve much since
  we need to visit both children anyway
• however, it benefits the parent level

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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

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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

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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 k1

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Analysis of the Randomized Algo.

• Assume that for all T2,k-1, the expected cost ?
  3k-1
• First, consider the OR-root tree

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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

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Analysis of the Randomized Algo.

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

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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

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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

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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 Yaos technique ? the Minimax
Theorem ? Game Theory Basics
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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

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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

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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

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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 Cs
  strategy is
• Optimal strategy for R is an strategy i that
  maximize minjMij.

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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
  Rs 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

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Inequality for All Payoff Matrix

• Minimax Inequality
• maximinjMij ? minjmaxiMij
• Proof

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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

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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)

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Mixed Optimal Strategy

• Expected Payoff
• Epayoff 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

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von Neumanns 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