Hartmut Klauck

Nanyang Technological University

and Centre for Quantum Technologies

Singapore

Direct Product Problems Suppose we have to compute k instances of some

function f on separate inputs

How do the necessary resources scale with k?

A direct sum theorem holds for a resource in somecomputational model,if computing k instances of any function f takesroughly k times the resources as computing oneinstance of f

The Direct Sum Question Direct Sum Theorems hold for several simple models of

computation, but are unknown for most models

Examples: deterministic and randomized one-waycommunication complexity, decision trees It is not known whether a direct sum theorem holds for the

deterministic two-way communication complexity of all relations

The Direct Product Question Consider computations which may make an error Suppose we can solve a problem with success probability

2/3 We say a Direct Product Theorem (DPT) holds in a

computational model, when for any function the successprobability scales to 2-(k) when computing k instances

We have Weak Direct Product Theorems (WDPT):

The resources allowed to solve k instances are the same as for one instance

Strong Direct Product Theorems (SPDT):Resources allowed to scale with k

Comparing the notions Direct sum and WDPT are incomparable

Direct sum states that k times the resources are neededto compute k instances with prob 2/3

WDPT states that with 1 times the resources the successprobability goes to 2-(k)

SDPT‘s imply both WDPT and Direct Sum

Motivation This is a fundamental question in any model of computation,

namely, „How well can distinct computations be correlated?“

There are many applications of DPT‘s:

Interactive proofs: the Parallel Repetition Theorem withapplications to inapproximability of optimization problems

Cryptography: Yao‘s XOR lemma, bounded storage model

Complexity: Time-Space tradeoffs, computations with advice

A general direct sum theorem in communication complexitywould separate NC1 and NC2

Communication complexity: the role of entanglement, communication-space tradeoffs

Recent Progress SDPT for randomized query complexity [D 11]

SDPT for quantum query complexity [LR 11]

An Obstacle Shaltiel [01] gives a general counterexample Consider the model of decision trees Algorithms are allowed to query their input and the

complexity measure is the number of queries

f(x1,…,xn)=x1_ (x2©© xn)

Inputs are distributed uniformly f(x) requires n queries to be computed with success 5/6 k instances of f can be computed with roughly (k/2)n

queries and success probability close to 1

An Obstacle The counterexample can be realized in most models of computation However, it refers to an average case SDPT In the worst case, randomized algorithms cannot compute f(x) with

less than k(n-2) queries and success probability better than 2-(k)

We can conclude:

It is usually not true, that, for every distribution on the inputs,

success·2/3 when computing f under using resources r

implies

sucess 2-(k) with resources kr/2 on k instances of f on k

However, this might still be true for the hardest distribution for f

The model of communication complexity Players Alice and Bob, inputs x,y

Def: necessary number of communication bits in order to compute a function/relation when each party knows only part of the input

f(x,y)

The model of communication complexity

The deterministic communication complexity of a functionf will be denoted D(f)

The randomized communication complexity with error will be denoted R(f) Randomized protocols are allowed to have a public source of

randomness

The quantum communication complexity with error will be denotedQ(f)

Allow entanglement between Alice and Bob

In this talk we concentrate on randomized and quantumcommunication complexity

k-fold problems Let f:X£Y! {0,1} be a Boolean function

Alice is given x2X, Bob y2Y, and they have to outputf(x,y)

In the for the function f­k they get k inputs xi,yi andhave to come up with a k-tuple of outputs f(xi,yi)

The output is correct if all f(xi,yi) are correctsimultaneously.

What is known? General statements for Direct Sum:

D(f­k)¸ (k D(f)1/2) for all total Boolean f [FKN91]

R(f­k)¸ (k1/2 R(f) ) for all f [BBCR10]

Strongest Result known for Direct Product:[Raz et al. 97] : if communication complexity of f is c, then success probability goes to 2-k/c (in a restricted

model)

Note that the above results fall short of holding forrelations

What can we do then? Consider Restrictions:

Restrict the model (one-way, simultaneous messagepassing …)

Restrict the functions (specific classes, specificfunctions)

Consider DPT‘s for lower bounds on the actualcommunication complexity

Restricted Models Direct Sum:

In one-way communication: D(f­k)=k D(f) for all f [Jain et al.05] show direct sum for randomized communication

Direct Product: In one-way communication:

A weak DPT holds for randomized communication [K04] Strong DPT holds for randomized one-way communication [J11]

Various results also for quantum communication, and the SMP-model

Weaker Statements Restrict the function

[Klauck, Spalek, de Wolf 04] prove a SDPT for the (quantum) two-way cc of the Disjointness problem

[Gavinsky 06] gives an SDPT for the randomized cc of a class ofrelations

Weaker statements DPT‘s for lower bounds

[Shaltiel01] SDPT for the discrepancy bound

later generalized to disc for all distributions [Lee and Shraibman]

[Beame et al. 05] prove a SDPT for the rectangle bound under product

distributions (for functions f)

Open whether the rectangle bound is polynomially close to R(f)

[Jain, Klauck, Nayak 08] generalize this to all relations

[Klauck 04] shows a weak DPT for the rectangle bound under all

distributions (for functions) [Harsha et al. 07] prove (roughly):

R (f­k)¸ k ¢ max D(f), where ranges over product distributions

[Sherstov 11] proves SDPT for smooth discrepancy

Weaker Statements Further work combines the restrictions, and e.g. shows

SDPT for one-way cc of the Index function [Ben-Aroyaet al. 08]

„If the function is important enough we still care!“

The Disjointness Problem The Disjointness Problem: f(x1,…,xn;y1,…,yn)=_i=1,...,n(xi^yi)

[KSW04] show Q1-exp (-k)(f­k)¸ (k n1/2)

The quantum cc of f is (n1/2) Same result bound is shown for R(f) by [Beame et al.

05] The rectangle bound under product distributions is(n1/2)

[K04] shows R1- exp (-k)(f­k)¸ (n)

The Disjointness Problem The Disjointness Problem (or rather its complement):

f(x1,…,xn;y1,…,yn)=_i=1,...,n(xi^yi)

Most important single problem in communicationcomplexity

„NP-complete“ in the world of cc

Lower bounds have applications to other models, e.g. monotone circuit depth for matching

The Disjointness Problem First result by [Babai et al. 86] : R(f)=(n1/2)

[KS87] showed that R(f)=(n)

Simpler proof by [Razborov 90]

Quantum upper bound is O(n1/2) [AA03], and is tight[Razborov02]

DPT‘s and Disjointness

[KSW04] show Q1-exp (-k)(f­k)¸ (k n1/2)

The quantum cc of f is (n1/2)

Same result bound is shown for R(f) by [Beame et al. 05] The rectangle bound for f under product distributions is(n1/2)

[K04] shows R1-exp (-k)(f­k)¸ (n)

Main Result We resolve the complexity of this problem by showing

R1-exp (-k)(f­k)¸ (kn)

Matches the trivial upper bound

Applications Communication-Space Tradeoffs:

Alice and Bob are now space bounded machines Can we get simultaneous bounds? Generalizes Time-Space tradeoffs.

Boolean matrix product: Alice gets matrix A Bob gets matrix B The want to compute the Boolean matrix product C

Quantum case resolved in [KSW04 ]: CS2=£(n5)

Randomized case open problem of [Beame et al.94] Resolve here to the optimal CS=(n3)

Applications Multiparty communication:

Alice knows strings x,z

Bob knows y,z

Charlie knows x,y

They want to compute 3-party Disjointness

Recent lower bound: (n1/4) [LS09]

Previous one-way bound: (n1/2) [VW07]

Charlie drops dead after one message: (n1/3) [BPSW06]

New bound: (n1/2) in the Charlie drop dead model

Applications Let f be a symmetric function and denote by df the

minimum i such that f changes value between inputswith i ones and with i+1 ones

Then is is easy to see that R(f(xÆy)) = £(n-df)

within log factor

Our result implies that all functions f(xÆ y) withsymm. f have an SDPT

Lower bound from DISJ SDPT

Proof Part 1: Massaging theProblem We are interested in the best success probability achievable

with communication ²kn for f­k(x1,y1,…, xk,yk).

f(x,y)=_i=1,...,n(x(i)Æ y(i))

Instead we prove a lower bound for the following problem:

Alice received an N-bit string x, Bob an N-bit string y, theyoutput k indices i1,…,ik such that xij=yij=1 for all j, where N=kn.

If there aren‘t k such indices, reject

W.l.o.g. they never err, but might reject unnecessarily

Success probability ¾

Proof Part 1: Massaging theProblem Proof is by 2 simple inductions

k-fold search problem can be reduced to f­k by binaryseach

k out of N search problem reduces to k-fold search bypermuting N=kn inputs randomly and hoping thatmany blocks get a good index

Same start of the proof as in [KSW04]

Proof Part 2: The LP The Problem: Alice received an N-bit string x, Bob an N-bit string y, they

output k indices i1,…,ik such that xij=yij=1 for all j, where N=kn.

Randomized protocols:

For each value of the public coin, a randomized protocolhas to partition the matrix indexed with {0,1}N £ {0,1}N

into rectangles

Rectangles are labelled accepting or rejecting (labelledwith outputs)

Proof Part 2: The primal LP For every rectangle R=A£B with Aµ{0,1}N and

Bµ{0,1}N there is a variable wR

minX

R

wR s.t. (1)

for all x; y with jx \ yj < k :X

R:x;y2RwR = 0 (2)

for all x; y with jx \ yj = k :X

R:x;y2RwR ¸ ¾ (3)

for all x; y with jx \ yj > k :X

R:x;y2RwR · 1 (4)

wR ¸ 0 (5)

Proof Part 2: The primal LP It easy to see that any randomized protocol P leads to

an assignment of weightswR such that

R:x,y 2R wR is the acceptance probabilty of P on x,y

wR is the probability of an accepting rectangle in the

protocol

When a protocol makes an output i1,…,ik it also

accepts (and has a unique witness for this fact)

On inputs with intersection <k the protocol rejectswith certainty

Proof Part 2: The primal LP

Reject if intersection <k Accept intersection k Still a probability otherwise

minX

R

wR s.t. (1)

for all x; y with jx \ yj < k :X

R:x;y2RwR = 0 (2)

for all x; y with jx \ yj = k :X

R:x;y2RwR ¸ ¾ (3)

for all x; y with jx \ yj > k :X

R:x;y2RwR · 1 (4)

wR ¸ 0 (5)

Proof Part 2: The dual LP

Restrict to rectangles with a unique witness

2 types of weights: positive and negative

maxX

x;y

¾Áx;y + Ãx;y s.t. (1)

Áx;y ¸ 0 (2)

Ãx;y · 0 (3)

if jx \ yj 6= k then Áx;y = 0 (4)

for all R 2 Rv :X

x;y2RÁx;y + Ãx;y · 1 (5)

The Solution For the solution we first define probability

distributions on the inputs

Tr,n is the set of input pairs x,y of length n which

intersect on r positions

Distribution ¹r,n is uniform on a „typical“ subset of Tr,n

We will put positive weights on ¹k,n, and negative weights on ¹2k,n

The solution

2¯ n will dominate the objective function

negative weights are smaller by 2® k

Feasibility It is straightforward to compute the cost function

But is the solution feasible?

For this we have to show that for all rectangles the sumof the weights falls below 1

The following Lemma suffices (somewhatsurprisingly)

The Intersection Sampling Lemma Lemma

Let R be a rectangle with ¹0,n(R)¸ 2-² n

Then ¹k,n(R)¸ ¹0,n(R)/2k+1

Razborov‘s Main Lemma in his Disjointness lower bound is essentiallythe same statement for k=1

Induction proof using a slight generalization of Razborov‘s in everystep.

Intuitively this is much weaker than the SDPT statement: Every large rectangle must have an exponentially small amount of

intersection-k-inputs

However, this can be „lifted“ to the k versus 2k setting, and nowintersection-2k-inputs have much more weights than intersection-k-inputs, leading to feasibility of the solution

We get ¹2k,n(R)¸ ¹k,n(R) ¢ (2k/k1/2)

² The positive weight inputs are in Tk;n.

Their weight is de¯ned as Áx;y = 2¯n¹k;n(x; y).

² The negative weight inputs are in T2k;n.

Their weight is Ãx;y = ¡2¯n2¡®k¹2k;n(x; y).

² For all other inputs x; y : Áx;y = Ãx;y = 0.

The solution

The Quantum Case In the quantum SDPT for Disjointness we start with

the same reduction. Then apply a powerful result byRazborov that translates to the domain of polynomialsrepresenting k-fold search problems

Analyze the polynomial degree

Recent progress: Sherstov (2011) shows SDPT forapproximate polynomial degree.

Conclusions We show a strong direct product theorem for

Disjointness

This has nice applications to communication-spacetradeoffs, and to Multiparty communication

The underlying proof method is new, it uses thepartition constraints not used in previous rectanglemethods

Recent Developments Sherstov (2011) proves a strong direct product theorem

for the smooth discrepancy method, which subsumes[KSW04]

Jain (2011) proves a strong direct product theorem for a new lower bound method crent that also applies toDISJ (but it is not clear how powerful the method is in general and the proof is more involved)

Open: SDPT for the rectangle bound (or even directsum) or for srec