Static Optimality and Dynamic Search Optimality in Lists and Trees

1
Static Optimalityand Dynamic Search
Optimalityin Lists and Trees

• Avrim Blum
• Shuchi Chawla
• Adam Kalai

1/6/2002
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List Update Problem
Query for element 9

• Unordered list
• Access for xi takes i time

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List Update Problem
Query for element 9

• Unordered list
• Access for xi takes i time

• Moving up accessed item is FREE!
• Other reorderings cost 1 per transposition
• Goal minimize total cost
• What should the reordering policy be?

4
Binary Search Tree
7
2
12
9
14
5
4
15

• In-order tree
• Search cost depth of element
• Cost of rotations 1 per rotation
• Replacement policy ?

5
How good is a reordering algorithm?

• Compare against the best offline algorithm
• Dynamic competitive ratio
• Best offline algorithm that cannot change state
  of the list/tree -- static
• Static competitive ratio

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Known results..

• List update
• Dynamic ratio Albers et al95, Teia93 u.b.-
  1.6 l.b.- 1.5
• Trees
• Splay trees Sleator Tarjan 85 static ratio
  3
• No known result on dynamic ratio
• Classic Machine Learning results
• Experts analysis LW94 static ratio (1 ?)
• Computationally inefficient

Recent new result Kalai Vempala01 Efficient
alg for Strong Static Optimality for trees
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Our results..

• List Update
• (1 ?) static ratio efficient variant of
  Experts
• We call this Strong static optimality
• Combining strong static optimality and dynamic
  optimality to get best of both
• Search trees
• Constant Dynamic ratio for trees
• Given free rotations
• ignoring computation time

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Revisiting Experts Algorithm Littlestone
Warmuth 94

• N expert algorithms
• We want to be (1?) wrt the best algorithm
• Weighted Majority algorithm
• Assign weights to each expert by how well it
  performs
• Pick probabilistically according to weights
• Cost incurred lt (1?)m ln(N)/(1-?)
• Applying this to list update
• Each list configuration is an expert
• Too many experts n!
• Can we reduce computation?

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Experts for List Update

• Weight for every static list too much
  computation
• The BIG idea
• Assign weights to every item in list and still
  make an experts style analysis work!

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Candidate algorithm

• Initialize wi for item i
• If ith element accessed, update wi
• Order elements using some rule based on wi

• Need to analyze probability of getting a
  particular static list

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Idea 2 List FactoringBorodin ElYaniv

• Distribute cost of accessing among pairs of
  elements
• Observation
• The relative order of x and y in the list does
  not depend upon accesses to others
• Implication
• Only need to analyze a two element list!

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Algorithm for two element list

• List (x,y)
• Experts (x,y) and (y,x)
• weights wx, wy
• correspond to the respective experts!
• Algorithm
• Initialize wx, wy to rx ry ?R 1..1/?
• If x accessed, wx lt- wx1 else wy lt- wy1
• Always keep the element with higher weight in
  front

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Efficient Experts algorithm for List Update

• Select ri ?R 1..1/? for element i
• Initialize wi lt- ri
• If ith element accessed, wi lt- wi1
• Order elements in decreasing order of weight

• (1?) static competitive
• Kalai Vempala01 give a similar result for trees

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Combining Static Dynamic optimality

• A has strong static optimality, B has dynamic
  optimality
• Combine the two to get the best of both
• Apply Experts again
• Technical difficulties
• Cannot estimate weights running both
  simultaneously defeats our purpose
• Experts maintain state - Huge cost of switching
• Dont switch very often

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How to estimate weights?

• The Bandits approach Burch 00
• Bandit can sample at most one slot machine at a
  time
• Run the (so far) better expert
• Assume good behavior from the other
• - Pessimistic approach
• After a few runs, we have sampled each one
  sufficiently
• Details in the paper

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Binary Search Trees

• Splay trees achieve constant static ratio
• No results on dynamic optimality
• Simplifying the problem
• Allow free rotations
• Allow unlimited computation
• We give a constant dynamic ratio

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Outline of our approach

• Design a probability distribution p over
  accesses
• Low offline cost gt greater probability
• Assume p reflects reality and predict the next
  access from it.
• Construct tree based on conditional probability
  of next access
• Low offline cost gt node closer to root gt low
  online cost

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An observation about offline BSTs

• An access sequence with offline cost k can be
  expressed in 12k bits
• At most 212k sequences of offline cost k.

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An observation about offline BSTs

• An access sequence with offline cost k can be
  expressed in 12k bits
• Express access sequence as rotations performed by
  an offline algorithm
• Start with a fixed tree, make some assumptions
  about offline algorithm gives extra factor of 2
• Express rotations from one tree to another using
  6 bits per rotation
• At most 212k sequences of offline cost k.

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Probability distribution on accesses

• ? Distribution on access sequence a
• p(a) gt 2-13k where k offline cost of a
• Use this to calculate conditional probability of
  next access
• Construct tree such that
• Cost of accessing ? ln(1/p(a))

O(k)
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What next?

• Can we make this algorithm computationally
  feasible?
• True dynamic optimality for BST
• Strong static optimality solved recently KV01
• Lessons to take home
• Experts analysis is a useful tool for data
  structures
• Generic algorithm too slow

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Optimality

• Cost of ALG is at most a constant factor times
  the cost of OPT
• Dynamic optimality
• Static optimality
• Strong static optimality the ratio is (1?)
• Search optimality ignore our rotation cost

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Known results..

• List update
• Dynamic ratio Albers et al95, Teia93 u.b.-
  1.6 l.b.- 1.5
• Trees
• Splay trees Sleator Tarjan 85 static ratio 3

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Back to list update

• How can we hope to achieve strong static
  optimality?
• Soln Use a classic machine learning result!!
• Experts Algorithm

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List Factoring Lemma

• Under a certain condition
• If A performs well on a list of two elements
• it performs well on any arbitrary list
• Condition
• For an arbitrary list, given the same accesses,
• A should order x and y just as in a list with
  only x y

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List Factoring Lemma

• e.g. Move-to-front

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9 and 4 retain the same order LFL applies
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An observation about offline BSTs

• An access sequence with offline cost k can be
  expressed in 12k bits
• At most 212k sequences of offline cost k.

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An observation about offline BSTs

• An access sequence with offline cost k can be
  expressed in 12k bits
• At most 212k sequences of offline cost k.

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Probability distribution on accesses

• ? Distribution on accesses a
• p(a) 2-13k where k offline cost of a
• Use this to calculate probability of next access
• Caveat
• computationally infeasible
• But dynamic search optimal !

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Binary Search Tree
9
7
12
Query for element 9
2
14
15
5
4

• In-order tree
• Search cost depth of element
• Cost of rotations 1 per operation
• Replacement policy ?

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Why is BST harder?

• Decide which nodes should be near the root
• Decide how to move up those nodes
• Not straightforward
• 132 different ways of bringing a node at depth 7
  to the root!!
• We ignore the second issue
• try for Dynamic search optimality

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9
7
7
12
4
12
4
14
9
14
5
2
15
5
2
15
(left,up,right,up)

• Uniquely specifies rotations

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An observation about offline BSTs

• An access sequence with offline cost k can be
  expressed in 12k bits
• Start with a fixed tree
• Express rotations from one tree to another using
  6 bits per rotation
• Assume algorithm first brings accessed element to
  root extra factor of 2
• At most 212k sequences of offline cost k.

34
Outline of our approach

• Design a probability distribution p over
  accesses
• Low offline cost gt greater probability
• Assume p reflects reality and predict the next
  access from it.
• Construct tree based on conditional probability
  of next access
• Low offline cost gt node closer to root gt low
  online cost