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Cell Probe Complexity - An Invitation

Peter Bro Miltersen

University of Aarhus

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The “Were-you-last?” Game

• m players wait in separate cubicles.• One by one, they are taken to a game room.• When a player leaves the game room, he is asked if he

was the last of the m players to go there and must answer correctly.

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Inside the Game Room

0 1 2

3 4

6 810

9

11

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Inside the Game Room, II

• The player can open at most t boxes

• He can put a pebble in an empty box or remove the pebble from a full box

• He must close a box before opening the next one

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Winning the Game

• The players win if all players answer correctly.

• How small can t be as a function of m to ensure that the players have a winning strategy?

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A Counter

• The players make boxes 0,1,..,t-1 represent the number of players having already been in the room

• The counter can be incremented and read opening at most t boxes

131101

:log mt

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Can we do much better?

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YES! t = 5 log log m.

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Alternative Counter I

111011100000

)2,5,3,1,1 ;5,1(

Type offirst block

Numberof blocks

Size of each block

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Alternative Counter, II

• Each entry in (1,5; 1,1,3,5,2) can be specified using ≈ log log m bits.

• The Counter can be incremented touching only 4 entries ≈ 4 log log m bits.

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Problem: How can a player tell if the value m has been reached?

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Alternative Counter, III

• Test for 0: The vector should start (0,1; …).

• Start the counter at m – Each player decrements the counter.

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Problem: The players can’t make the counter startat m – initially, all boxes must be empty!

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Alternative Counter, IV

• The players would have liked boxes to contain pebbles initially.

• Patch: They follow their protocol, but maintain that boxes are in the opposite state of what they are supposed to.

Thus, the players win!

kiii ,..,, 21

kiii ,..,, 21

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Can we do much better?

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NO! If t = 0.4 log log m, the players lose.

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Decision Assignment Tree

7

5

Open box 7 If it contains a pebble,Remove it, and go here1/00/1

0/0

Strategy for a player:

yes yesno no

1/0

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Sunflower

1S

4S

2S

3S

ji SS is the same for all i,j

Sunflowercenter

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Erdös-Rado Sunflower Lemma

Every large set system contains a large sunflower.

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Why the players lose:

: The boxes appearing in the decision assignment tree of each player

: A big sunflower.

: The other sets

mSSS ,..,, 21

pmjjj SSS

,..,,21

piii SSS ,..,,21

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Why the players lose, II

plkpm iiiijjj SSSSSSS ,..,,,..,,,..,,1121

piii SSS ,..,,21

Sequence:

Center states: 0C 1C 2C 1pC

Pigeon hole principle: for some k, l, .lk CC

pmjjj SSS

,..,,21

New sequence:

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The Game = A Dynamic Problem

Maintain a subset S of {1,..,m} under

Insert(x): Insert x into S .

Full(): Is S={1,..,m}?

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The Room = Cell Probe Model

• Boxes = memory cells.

• Pebble or no pebble = 0 or 1 (word size w=1)

• Opening a box = accessing a memory cell.

• Decision assignment tree for a player = implementation of an operation.

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The Cell Probe Model

• An information theoretic model for solutions to dynamic data structure problems

• Only memory accesses are counted

• Each operation in the solution is assigned a decision assignment tree – the worst case complexity is the depth of the tree.

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Why study cell probe complexity?

• Lower bounds in the cell probe model are valid for a unit cost random access machine with the same word size, independent of the instruction set.

• Fundamental combinatorial complexity measure.

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This invitation to cell probe complexity

• Focuses on worst case time per operation (rather than amortized, and/or tradeoffs)

• Focuses on dynamic problems (rather than static ones)

• Focuses first on w = 1, i.e., on bit probe complexity

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Why Study Bit Probe Complexity?

• Log cost RAM.

• Large lower bounds would be still be large if divided by w.

• Coding theoretic interpretation: Locally decodable source codes.

• All known results have easy proofs.

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Dynamic Graph Problems

Dynamic Graph Connectivity:

Maintain G =({1,..,n},E) under

Insert(e): Insert e into E.

Delete(e): Delete e from E.

Connected(u,v): Are u and v connected?

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DGC, best known boundsUpper bounds:• Worst case time bit

probes per operation [Henzinger and King]• Amortized time bit probes per

operation [Holm, de Lichtenberg, Thorup]Lower bound:• Ω(log n/log log n) bit probes per operation

[Fredman]

) (log 32 nnO

)(log3 nO

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A lower bound for dynamic connectivity

State 0=(0,0,0,…,0)

State 1 State 2 State 3 State 4

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Lower Bound For DGC, II

• State 1,2,..,n all have Hamming weight at most t.

• We can distinguish between the states using a decision tree of depth d=O(t log n).

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Lower Bound For DGC, III

7

5

10

0

State 3

1

10

The sequence 001 is uniquefor State 3 and contains at most t zeros

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Lower bound for DGC, IV

n ≤ # ways to arrange at most t ones in a sequence of length at most d

t = Ω(log n/log log n)

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Dynamic Circuit Value

Maintain Boolean Circuit C under

Insert(g,h): Insert wire from output of gate g to input of gate h

Switch(g): Switch gate g between AND,OR

Evaluate(): Return value of circuit

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Dynamic Circuit Value

Best known algorithm for DCV: Reevaluate from scratch.

Best known lower bound in bit probe model:

Ω(log n) bits must be touched in some operation.

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Dynamic Language Membership Problems

Given Boolean language L. Maintain Boolean string x of length n under

Change(i,a): Set ith bit of x to a.

Member(): Is x a member of L?

Dynamic graph connectivity

Dynamic UGAP

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Completeness of DCV

Dynamic circuit value needs Ω(log n) bit probes per operation

The dynamic language membership problem for some language in P needs Ω(log n) bit probes per operation.

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Element distinctness

Element distinctness language: ED = binary strings x consisting of n blocks, each of length 2 log n, all different.

Dynamic ED needs Ω(log n) bit probes per operation.

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Lower bound for dynamic ED

By fixing all blocks, except the first, we create a solution to a subproblem.

#decision assignment tree systems of depth t ≥

#subproblems

Dynamic ED needs Ω(log n) bit probes per operation.

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A sad situation

• We don’t know any lower bounds for DCV better than Ω(log n) bit probes.

• Thus we don’t know any lower bound for any dynamic language membership problem in P better than Ω(log n) bit probes.

• We suspect many problems in P needs Ω(n) bit probes.

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The transdichotomous model

Problem:• Problem over universe of size m.• Problem instance of size (i.e., #elements of

universe) n.

Cell probe solution:• Word size w = log m. • Time bound should be a function of n only.

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Dynamic Search Problems

Maintain subset S of {1,..,m} under

Insert(x): Insert x into S.

Delete(x): Delete x from S.

and queries like

Member(x): Is x in S ?

Predecessor(x): What is max{y| y≤x, y in S} ?

HammingNeighbor(x): Return Hamming neighbor of x in S.

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Transdichotomous upper bounds

Dynamic member and predecessor :• AVL trees: O(log n) cell probes per

operation.• Andersson/Thorup’99:

cell probes per operation.Dynamic Hamming neighbor:• Trivial bound: O(n) cell probes per

operation.

)loglog/log( nnO

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Dynamic algebraic problems

Dynamic polynomial multiplication:

Maintain two polynomials f,g of degree at most n over GF(m) under

Change-f(i,a): Let ith coefficient of f be a.

Change-g(i,a): Let ith coefficient of g be a.

Query(j): What is the jth coefficient of fg?

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Transdichotomous upper bound

Dynamic polynomial multiplication:

• cell probes per operation [Reif and Tate]

)(log nnO

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Transdichotomous lower bounds

Lower bound of Ω(f(n))

No transdichotomous upper bound of o(f(n)).

There are c, , so that any solution with parameters uses at least cell probes.

),...,(),,(),,( 332211 mnmnmn),( ii mn

)( incf

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Transdichotomous lower bounds, II

• We are free to choose the size of the universe as large as we want.

• This makes large lower bounds possible using current techniques.

• It also makes the lower bounds somewhat less interesting.

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Transdichotomous lower bounds, III

• Dynamic predecessor: cell probes per operation is necessary [Beame and Fich]

• Dynamic Hamming neighbor: cell probes per operation is necessary [Barkol and Rabani]

• Dynamic polynomial multiplication: cell probes per operation is necessary [Frandsen,Hansen,Miltersen]

• Dynamic graph connectivity: cell probes per operation is necessary [Fredman and Saks]

)loglog/log( nn

)1(n

)( n

)loglog/(log nn

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Two main techniques

• Communication complexity method [Ajtai]

• Time stamp method [Fredman and Saks]

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Communication Complexity Technique

Given dynamic problem, construct two-party communication problem:

Alice gets state reachable within d operations from the initial state.

Bob gets a query operation.Upper bound for dynamic problem implies upper

bound for communication problemHence, lower bound for communication problem

implies lower bound for dynamic problem.

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Transferring Upper Bounds

• Alice constructs data structure corresponding to her state.

• Alice sends Bob a perfect hash function of the memory cells that have changed since initial state.

• Bob simulates the query operation by asking Alice for values of memory cells: He sends her a hashed address, Alice sends back an address with this hash value that were changed and a new value.

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Challenges for cell probe complexity, I

Show a lower bound for a dynamic language membership problem of the form

)(log nt

)loglog/(log nnt

(w = 1)

(w = log n)

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Static problem

}1,0{: QDf

Set of possibledata

Set of questionsto data

Answers to questions

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Challenges for cell probe complexity, II

}1,0{}1,0{}1,0{: nmf

probesbit )( Time bits )( Space ntmOs

Show a lower bound for a static problem

of the form :