Hartmut KlauckCentre for Quantum TechnologiesNanyang Technological University

Singapore

An introduction to lower bound methods in

communication complexity

Two players Alice and Bob want to cooperatively compute a function f(x,y)

Alice knows x, Bob knows yHow much communication is needed?

The communication model

f(x,y)

Equality: EQ(x,y)=1 iff x=yDisjointness: DISJ(x,y)=1 iff x and y are

disjoint setsInner Product mod 2: IP(x,y)=1 iff i xiÆyi is

odd

Representation as matrices:M(x,y) contains entries f(x,y)

Examples, the communication matrix

The set of inputs that share the same message sequence form a combinatorial rectangle in the communication matrix

Proof sketch: Alice’s first message depends on her input only,

partitions the rowsThen Bobs message partitions the columns etc.

The messages partition the communication matrix into combinatorial rectangles

A key insight

For any integer matrix M the partition number is the minimum number of monochromatic rectangles needed to partition M

Clearly: D(M)¸ log part(M)The covering number is the minimum

number of rectangles needed to cover the entries of M with monochromatic rectangles

This corresponds to nondeterministic protocols

The partition number/coverings

D(EQ)=n+1Consider the inputs x,xNo two inputs x,x and y,y can be in the same

combinatorial rectangleOtherwise x,y is also in the same rectangle!

Hence we need at least 2n 1-rectangles to cover the 1-inputs of EQ

On the other hand n rectangles are enough to cover the 0-inputs of EQ

Example

It is known that log part(f) cannot be too far from D(f) for Boolean f

But part(f) is not always easy to determineWe will consider different relaxations of

part(f) that are easier to calculate

Lower bound methods

We can relax the partition requirement for Boolean functions

If M can be partitioned into k 1-rectangles, then M can be written as the sum of k rank 1 matricesIe. rank(M) · k

Examples:rank(EQ)=2n, rank(DISJ)=2n-1, rank(IP)=2n-1

Relaxation 1: the rank bound

Let M be a Boolean matrixWe know that D(M)¸ log rank(M)Conjecture: D(M)· poly(log rank(M))Best known upper bound is rank(M)There is a polynomial gap

D(f)=(n), log rank(f)=n0.61

Conjecture: quadratic gap

The log rank conjecture

Observation: 1-rectangles are rank one matrices with nonnegative entries only

prank(M) is the minimum k such that M can be written as the sum of k rank 1 matrices with nonnegative entries

Clearly D(M)¸ log prank(M) for all Boolean MNow we get a bound that is polynomially tight:

D(M)· O(log prank(M)¢ log prank(J-M)) for all Boolean M

J is the all ones matrix

The nonnegative rank

log prank(M)· poly(log rank(M)) for all Bool. M

Every Bool. m£ n rank r matrix M has a monochromatic submatrix of size mn/2polylog(r)

Every Bool. m£ n rank r matrix M has a submatrix of size mn/2polylog(r) that has rank <.99r

The rank conjecture has also been related to some open problems in additive combinatorics

Different formulations of the conjecture

Computing prank is NP-hardAlways gives polynomially tight bounds, but

hard to show

Problems with prank

In general it is hard to estimate the partition number (number of rectangles in a partition of the comm. matrix)

Idea: write as an integer program, relax into a lin. program, estimate via the dualThe dual is a max. problem!

Relaxation 2: the rectangle bound

Consider the set R of all 1-monochromatic rectangles in M

Every R2R gets a weight wR2 {0,1}

Minimize wR

such that for all x,y with f(x,y)=1: R:x,y2R wR=1(implicit: for all x,y with f(x,y)=0: R: x,y2R

wR=0)

The optimum is the partition number

The 0-1-program

R: set of all 1-monochromatic rectangles in MEvery R2R gets a nonnegative real weight wR

Minimize wR

such that for all x,y with f(x,y)=1: R: x,y2R wR=1(implicit: for all x,y with f(x,y)=0: R: x,y2R wR=0)

The optimum of this LP lower bounds the partition number

The linear program

A variant of the LP lower bounds the (one-sided) nondeterministic CC:Minimize wR

such that for all x,y with f(x,y)=1: R: x,y2R wR¸ 1

Denote the optimal value B(M)Use max of B(M), B(J-M) to show lower bounds on det.

CCThen: D(M)· O(log B(M)¢ log B(J-M)) + log2 n

Bounds for D(M) obtainied this way are never worse than quadratically smaller than D(M)

Comments

In the dual there is one real variable for every input x,y

Max Áx,y

such that for all 1-chromatic R: x,y2R Áx,y· 1

In other words, put weights on inputs to “balance” the weights on each 1-chromatic rectangle

For the dual of the nondeterministic LP bound the variables must have nonnegative values

The dual

Max Áx,y

such that for all 1-chromatic R: x,y2R Áx,y· 1Suppose that all Áx,y ¸ 0After rescaling the Áx,y can be regarded as a

probability distributionThe scaling factor is the max size of matrices

that are 1-monochromatic under the distribution

The dual

Equality:Weights Áx,x = 1The only 1-monochromatic rectangles contain exactly

one input x,x Thus B(EQ)¸ 2n

Inner Product modulo 2:Consider f(x,y)=1-IP(x,y)1-chromatic rectangles A£B satisfy that A ? BHence dim(A)+dim(B)· n ) |A|¢|B|· 2n All 1-inputs get weight ½n

There are ¸ 22n-1 1-inputs Hence B(f)¸ 2n/2

Some examples

Disjointness:1-inputs satisfy i xiÆyi = 0There are 3n 1-inputs1-chrom. rectangles A£B have A?B Hence still |A|¢ |B|· 2n

Give weights 1/2n

We get the bound B(DISJ)¸ 3n/2n

Some examples

Recall the primal program:

Minimize wR

such that for all x,y with f(x,y)=1: R: x,y2R wR=1

We don’t allow x,y to be “covered too much”In the dual this means we can use negative weights Áx,y

Makes it easier to satisfy constraintsHave not used that in our examples

The partition constraint

Promise-Nondisjointness: f(x,y)=0 if i xiÆ yi =0

f(x,y)=1 if i xi Æ yi =1

otherwise f is undefinedThe LP for f with the constraint R: x,y2R wR¸ 1

has optimum · nUse the following LP:

Minimize wR

such that for all x,y with f(x,y)=1: R:x,y2R wR = 1for all x,y with f(x,y)=0: R: x,y2R wR = 0for all other x,y: R: x,y2R wR · 1

z

Another example

We have to exhibit a solution to the dualWe should put positive weights on inputs with

intersections size 1Negative weights on inputs with larger

intersection sizeChoose 2

Weight 0 elsewhere

Example cont.

The following fact is useful [Razborov 92]Let ¹k denote the uniform distribution on x,y with

|xÅ y|=k and |x|=|y|=n/4Then for all large enough R=A£ B¹ x1(R)¸ (1-²)¹0(R)

This mean all large rectangles are corruptedSimilarly ¹2(R)¸ (1-²) 2 ¹1(R) for all 1-chrom. RHence putting 1-² the weight on x,y with intersection 2 as

on x,y with intersection 1 is enough to satisfy the constraints for all large 1-chrom. rectangles

Constraints are also true for small rectangles if weights are not too large

Total weight is 2 (n)

Example cont.

Readily generalizes to bounded error protocols

Can be generalized to deal with quantum protocols

Proof consist of exhibiting a dual solution

For randomized: Use contraints ¸ 1-² and · 1 in the primal program

Advantages of the rectangle bound

Primal:Min wR

For all x,y with f(x,y)=1: R:x,y2R wR¸ 1-²For all x,y with f(x,y)=0: R: RwR · ²

This is the rectangle/corruption bound

Example

Razborov showed that there is a distribution on inputs such that for all large R the fraction of 0-inputs is ² time the fraction of 1-inputs

This corresponds to a solution of the dual

Example DISJ

Previously we bounded the size of monochromatic rectangles

In the bounded error scenario we want to bound the size of almost monochromatic rectangles

Often this is much harder!What about the bias?Often easier to bound

But usually not good enough

Relaxation 3: Discrepancy

Fix a distribution ¹ on inputsThen the discrepancy of a rectangle is

|¹(RÅ 1)-¹(RÅ 0)|

Disc(f)=min log of the aboveQuite easy to show that Disc (f) is a lower

bound on D(f)Even for randomized, quantum, and (weakly)

unbounded error

Discrepancy

Disc(IP)=(n)All rectangles are almost balanced

Disc(DISJ)=O(log n)Nondeterministic complexity is small, hence

large monochromatic rectangles exist

The method fails to capture either randomized or quantum communication complexity

Examples

Re-visit the result about inner productIndeed it is hard to compute IP even with very

small errorError 1/2-½²n

BUTmany functions are close to IP even if their discrepancy is small

“Generalized” or “Smooth” discrepancy

What to do?

Take the function Maj(x,y)=1 iff i xiÆ yi¸ n/2Easy to see that Disc(Maj)=O(log n)Nevertheless Maj is close enough to IP to

inherit the lower bound (n) for quantum protocols

An Idea

Every rectangle R gets a real weight wR

Min wR

such thatfor all x,y with f(x,y)=1: R contains x,y wR2 [1-²,1]for all x,y with f(x,y)=0: R contains x,y wR2[0,²]

Dual: Put weights on inputs such that all large rectangles are almost balanced

Difference to the rectangle bound: two-sided balance condition

Another LP bound

Relaxing the rank bound furtherWhy?Motivated by quantum communicationNorm based methods allow to deal with

entanglement in quantum protocolsWe will arrive at (almost) the same quantity

as above (the LP bound) (via Grothendieck’s inequality)

Relaxation 4

D(f)¸ log rank(f)¸ ||A||2/mnThere is another relaxation of the rank: °2=max u,v ||M ± u¢v|| tr

rank(M)¸ °2 (M)2

Then define °2® as the minimum °2 of any M that is ®-close to M

This method subsumes all previous methods for lower bounding quantum cc

Norm based ideas