Randomized Communication Complexity for Linear Algebra Problems over Finite Fields

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Algebra Problems over Finite Fields

Xiaoming Sun, Chengu Wang

To cite this version:

Xiaoming Sun, Chengu Wang. Randomized Communication Complexity for Linear Algebra Problems

over Finite Fields. STACS’12 (29th Symposium on Theoretical Aspects of Computer Science), Feb

2012, Paris, France. pp.477-488. �hal-00678209�

Linear Algebra Problems over Finite Fields ^∗

Xiaoming Sun

and Chengu Wang

1 Institute of Computing Technology, Chinese Academy of Sciences Beijing, China

sunxiaoming@ict.ac.cn 2 IIIS, Tsinghua University

Beijing, China wangchengu@gmail.com

Abstract

Finding the singularity of a matrix is a basic problem in linear algebra. Chu and Schnitger [3]

first considered this problem in the communication complexity model, in which Alice holds the first half of the matrix and Bob holds the other half. They proved that the deterministic com- munication complexity is Ω(n²logp) for ann×nmatrix over the finite fieldFp. Then, Clarkson and Woodruff [4] introduced the singularity problem to the streaming model. They proposed a randomized one pass streaming algorithm that uses O(k²logn) space to decide if the rank of a matrix is k, and proved an Ω(k²) lower bound for randomized one-way protocols in the communication complexity model.

We prove that the randomized/quantum communication complexity of the singularity prob- lem overFp is Ω(n²logp), which implies the same space lower bound for randomized streaming algorithms, even for a constant number of passes. The proof uses the framework by Lee and Shraibman [8], but we choose Fourier coefficients as the witness for the dual approximate norm of the communication matrix. Moreover, we use Fourier analysis to show the same random- ized/quantum lower bound when deciding if the determinant of a non-singular matrix is aorb for non-zeroaandb.

1998 ACM Subject Classification F.2.1 Numerical Algorithms and Problems; G.1.3 Numerical Linear Algebra

Keywords and phrases communication complexity, streaming, matrix, singularity, determinant Digital Object Identifier 10.4230/LIPIcs.STACS.2012.477

1 Introduction

Communication complexity, introduced by Yao [16], is a powerful tool to solve a variety of problems in areas as disparate as VLSI design, decision trees, data structures and circuit complexity [7]. It is a game between two parties, Alice and Bob, with unlimited computing power, that want to compute the value of a functionf :X×Y 7→ {0,1}, but Alice only knows x∈Xwhile Bob only knowsy∈Y. The communication complexity is the minimal amount of bits they transfer. We denote the randomized and quantum communication complexity which succeeds with probability at least 1−byR(f) andQ(f) (orQ^∗(f)) respectively, where

∗ This work was supported in part by the National Basic Research Program of China Grant 2011CBA00300, 2011CBA00301, the National Natural Science Foundation of China Grant 61033001, 61061130540, 61073174. The first author was also supported in part by the National Natural Science Foundation of China Grant 61170062 and Tsinghua University Initiative Scientific Research Program 2009THZ02120.

© Xiaoming Sun and Chengu Wang;

licensed under Creative Commons License NC-ND

R(f) is with private coin, Q(f) is without entanglement andQ^∗(f) is with entanglement.

The three functions have the following relationship: Q^∗(f)≤Q(f)≤O(R(f)) [6].

In the streaming model, the input is presented as a sequence and can be examined by the algorithm in only a few passes (typically just one). We are interested in the size of memory the algorithm uses. It can be proved by a reduction that the size of memory times the number of passes can be bounded by the communication complexity if Alice holds the first part of the stream and Bob holds the remaining.

We focus on linear algebra problems, because they are fundamental problems in mathem- atics, and the matrix computation is used everywhere. Furthermore, the singularity problem is the most basic problem, because it can be reduced to many linear algebra problems, e.g. to determine whether linear equations have a solution, to compute the diagonal of the LU decomposition or QR decomposition and to determine whether two subspaces intersect.

Formally speaking, for ann×nmatrixxwhose entries are in the the finite field Fp, the singularity problem is to decide whetherxis singular overF^p, and the determinant problem is to compute the determinant ofxoverFp for non-singularx. In the streaming model, the inputxcomes row by row sequentially, while in the communication complexity model, Alice holds the firstn/2 rows, and Bob holds the remainingn/2 rows.

These problems have a trivial deterministic algorithm that usesO(n²logp) spaces in one pass or the same number of communications in one-way. Chu and Schnitger [3] proved an Ω(n²logp) communication complexity for deterministic protocols of the singularity problem.

Luo and Tsitsiklis [9] proved that a deterministic protocol must transfer Ω(n²) real numbers for the matrix inversion problem. Clarkson and Woodruff [4] proposed a randomized one pass streaming algorithm that usesO(k²logn) space to decide if the rank of a matrix isk, and proved an Ω(k²) lower bound for randomized one-way protocol in the communication complexity model by reducing from the index function, which implies an Ω(n²) space lower bound in the streaming model with one pass. Deciding the disjointness of twon/2 dimensional subspaces is actually the singularity problem. Miltersen et al. [10] showed a tight lower bound when deciding whether a vector is in a subspace of Fⁿ2 in the one-sided error randomized asymmetric communication complexity model, by using the Richness Lemma.

In the communication model, there is another way to distribute the input: Alice and Bob each holds ann×nmatrixxandy, respectively, and they want to compute the singularity or determinant ofx+y. The two ways are equivalent up to a constant factor, because

det(x+y) = det

x+y 0_n×n y In×n

= det

x −I_n×n y In×n

.

This way is more beautiful and symmetric. When p = 2, f(x+y) = f(x⊕y) is a important block-composed function with good properties and attracted lots of attentions recently [11, 18, 14, 17]. Here, we formally define the problems in this way.

IProblem 1 (Singularity). Alice and Bob hold two n×n matrices x and y over Fp, separately. They want to determine whetherx+y is singular overFp.

IProblem 2(Deta,b). Alice and Bob hold twon×nmatricesxand y overFp, separately.

Fora, b∈Fp\ {0}, we promise thatdet_p(x+y)is eitheraorb, wheredet_p is the determinant overFp. They want to computedetp(x+y).

1.1 Our Results

ITheorem 3. The randomized/quantum communication complexity of Singularity is Ω(n²logp).

We prove it using the duality of the approximate norm [8]. We compute all the Fourier coefficients of the singularity function and use them as a witness in the Duality Theorem.

This result implies the same lower bound when deciding if two subspaces ofFⁿp intersect.

Also, it implies that Ω(k²) communication is required to decide if the rank ofx+y isk, by paddingn²−k²zeros, which improves the determinant result in [4].

ITheorem 4. The randomized/quantum communication complexity of Deta,b isΩ(n²logp), for all non-zeroaandb.

For this problem, we prove a small spectral norm of the matrix representing the problem, by decomposing the matrix onto Fourier basis which has good properties.

All these results imply the same space lower bound for randomized streaming algorithms, even if the algorithm reads the stream a constant number of passes.

1.2 Outline

In Section 2, we define the basic notations. We prove Theorem 3 in Section 3 and Theorem 4 in Section 4. Finally, we discuss the open problem in Section 5.

2 Preliminaries

For primep,Fp is a finite field. For a functionf :F^Np 7→R, the Fourier coefficient of f is fˆ(s) = 1

p^N X

x∈F^N_p

ω^−hs,xif(x),

whereω=e^2πi/p. The inverse transform is f(x) = X

s∈F^Np

ω^hs,xif(s).ˆ

The Kronecker delta, denoted byδi,j, is 1 ifi=j and 0 otherwise.

IFact 5. The number ofn×nmatrices of rankroverFpisp^{r(r−1)/2 n}_r

p

Qn

k=n−r+1(p^k−1).

Especially, the number of non-singularn×nmatrices overFp isQn−1

k=0(pⁿ−p^k).

In this paper, we don’t distinguish vectors (matrices) from discrete unary (bivariate) functions. For example, for vectorv,v(x) means thex-th element ofv, and for matrixA, A(x, y) means the entry atx-th row andy-th column.

For a vectorx, we define the`1-normkxk1=P

i|xi|, and the`_∞-normkxk_∞= maxi|xi|.

For a matrixA, we also define the`₁-normkAk₁=P

i,j|x_i,j|, and the`_∞-normkAk_∞= maxi,j|Ai,j|. Letσ= (σ1,· · ·, σ_rank(A)) be the vector of nonzero singular values ofA. The trace norm ofAiskAk_tr =kσk₁. The spectral norm iskAk=kσk_∞, which is also the square root of the largest eigenvalue of the positive-semidefinite matrix A^†A [5], whereA^† is the conjugate transpose ofA.

3 Lower Bound for Singularity Problem

We first introduce the approximate rank and norm. Then, for XOR composed function g(x⊕y), the trace norm is equal to the`₁norm of the Fourier coefficients. This property still holds for approximate norm and forg(x+y) inFp. After that, we present the Duality Theorem, which converts the definition of the approximate norm from min to max. Finally, we compute Fourier coefficients of the singularity function, and choose it as the witness.

3.1 Approximation Norms

The matrix rank and matrix norm can give a lower bound for deterministic communica- tion complexity. Similarly, the approximate rank and norm can prove a lower bound for randomized/quantum protocols.

Forα≥1 and a sign matrix A, we define the approximate trace norm by kAk^α_tr= min

B:1≤Ai,jB_i,j≤αkBktr, and the approximate rank by

rank^α(A) = min

B:1≤Ai,jB_i,j≤αrank(B).

Forα≥1 and a sign vectorx, we define the approximate Fourier`1-norm by kˆxk^α₁ = min

y:1≤xiy_i≤αkˆyk1.

ITheorem 6. [2] Let A be a sign matrix and 0< < 1/2, and α = 1/(1−2), then Q(A)≥ ¹₂log rank^α(A).

Because the approximate rank can be bounded by the approximate trace norm [8]:

rank^α(A)≥ (kAk^α_tr)² α²·size(A),

the approximate trace norm can give a lower bound for the communication complexity. This result can also be found in [12].

ILemma 7.Letg:F^Np 7→ {−1,1}be a sign function, andf =g◦+^⊗N, i.e. f(x, y) =g(x+y).

Let Abe the sign matrix representing f. Then,kAk^α_tr≥ kˆgk^α₁·p^N.

Thep= 2 case of Lemma 7 can be found in [8, Theorem 85]. It can be easily generalized to Fp with little changes. As a result, we omit this proof.

Combining them together, the randomized/quantum communication complexity can be bounded by thekˆ·k^α₁ norm.

ICorollary 8. For g : F^Np 7→ {−1,1} and f = g◦+^⊗N, Q(f) ≥ logkˆgk^α₁ −2α, where α= 1/(1−2).

3.2 Duality

The definition of the approximate Fourier`₁ norm begins with miny. In such a definition, we have to check every yif we want to prove a lower bound. However, the Duality Theorem converts min_y to max_y. As a result, a particulary, called the witness, is enough to prove a lower bound.

IDefinition 9. For a general normk · kon R^N, the dual norm onR^N, denoted byk · k^∗, is defined by

kxk^∗= max

y∈R^N:kyk≤1hx, yi.

ITheorem 10(Duality Theorem). [8, Theorem 64] For a general norm k · konR^N, kxk^α= max

y:kyk^∗≤1

1 +α

2 hx, yi+1−α 2 kyk1.

3.3 Choosing Fourier Coefficients as Witness

It is difficult to find a useful witness. The first choice that comes to mind is to chooseh=g, which can be used to prove the inner product problem. The discrepancy method is the special case of takingh=µ◦gfor a distributionµ[8]. Here we propose a new choice: taking h= ˆg. Now we calculate ˆg first.

We define a sign function g:F^n×np 7→ {−1,1}, whereg(x) =−1 if xis full rank overFp

andg(x) = 1 otherwise. Then, we definef(x, y) =g(x+y). In such a definition, f is the function representing theSingularityproblem.

We change the value of gfrom {−1,1} to{0,1}, by definingg01= (1−g)/2. In other words,g₀₁(x) = 1 ifxis full rank, andg₀₁(x) = 0 otherwise.

gc01has a good property that a same rank results in a same value.

ILemma 11. For s, t∈F^n×np , gc₀₁(s) =gc₀₁(t) if rank_p(s) = rank_p(t), where rank_p is the matrix rank over Fp.

Proof. Because rank_p(s) = rank_p(t), there are full rank matrices uandv, such thats=vtu.

Sincev^Tandu^Tare full rank,y=v^Txu^Tis a bijection between matricesxandy.

gc01(s) = P

xω^−tr(sTx)g₀₁(x)

pⁿ² =

P

xω^{−tr(tTvTxuT)}g₀₁(x)

pⁿ² =

P

yω^−tr(tTy)g₀₁(y)

pⁿ² =gc01(t) J ILemma 12. Letr= rankp(s), then

gc₀₁(s) = (−1)^rp^−n(n+1)/2

n−r

Y

k=1

(p^k−1).

Proof. By Lemma 11, we only need to choose one sfor each rank to prove it. For rankr, we chooses= diag(1,· · ·,1,0,· · ·,0), which is a diagonal matrix withr 1’s in the diagonal.

We start from the most simple case of rankp(s) = 0, i.e. sis an all-zero matrix.

pⁿ²·gc₀₁(s) =X

x

ω^−h0,xig₀₁(x) =X

x

g₀₁(x) = # of full rank matrices =

n−1

Y

k=0

(pⁿ−p^k) Then, we consider rank-1 matrixs= diag(1,0,0,· · ·,0), which is a matrix with all zero entries except for the top left one.

For k= 0,1,· · ·, n−1, we denote thek-th row of matrixxbyx(k,·), and the submatrix from thek-th row to the last row byx(k–,·).

pⁿ²·gc₀₁(s) = X

x(0,·)

X

x(1–,·)

ω^−hs,xig₀₁(x) = X

x(0,·)

ω^−x(0,0) X

x(1–,·)

g₀₁(x) P

x(1–,·)g01(x) is the number of the full rank matrices given the first row. It isQn−1

k=1(pⁿ−p^k) if the first row is non-zero, and 0 if the first row is zero. Except for the case that the first row is zero, they are all canceled out becauseP

x(0,0)ω^−x(0,0)= 0. Thus, the remaining is the minus value of the case that the first row is non-zero, i.e. pⁿ²·gc01(s) =−Qn−1

k=1(pⁿ−p^k).

In general, for the rank-rmatrixs= diag(1,· · · ,1,0,· · · ,0) with r1’s in the diagonal, pⁿ²·gc₀₁(s) = X

x(0,·)

ω^−x(0,0) X

x(1,·)

ω^−x(1,1)· · · X

x(r−1,·)

ω−x(r−1,r−1) X

x(r–,·)

g₀₁(x).

P

x(r–,·)g01(x) is the number of full rank matrices given the firstrrows. It isQn−1

k=r(pⁿ−p^k) if the firstrrows is linear independent, and 0 otherwise.

We define ζ as follows: ζ(x, k) = 0 if the first k rows of x are linear dependent and ζ(x, k) = 1 otherwise. Now we have

pⁿ²·gc01(s) = X

x(0,·)

ω^−x(0,0) X

x(1,·)

ω^−x(1,1)· · · X

x(r−1,·)

ω−x(r−1,r−1)ζ(x, r)

n−1

Y

k=r

(pⁿ−p^k) Then we slightly changeζ to ¯ζ: ¯ζ(x, k) = 0 if the firstk−1 rows ofxare linear independent but the first k rows are linear dependent, and ¯ζ(x, k) = 1 otherwise. It is clear that ζ(x, r) =Qr

k=1ζ(x, k), which gives¯ pⁿ²·gc₀₁(s) = X

x(0,·)

ω^−x(0,0)ζ(x,¯ 1) X

x(1,·)

ω^−x(1,1)ζ(x,¯ 2)· · · X

x(r−1,·)

ω−x(r−1,r−1)ζ(x, r)¯

n−1

Y

k=r

(pⁿ−p^k).

SinceP

x(k,·)ω^−x(k,k)= 0, we have X

x(k,·)

ω^−x(k,k)ζ(x, k¯ + 1) = X

x(k,·): ¯ζ(x,k+1)=1

ω^−x(k,k)=− X

x(k,·): ¯ζ(x,k+1)=0

ω^−x(k,k).

x(k,·) goes over the linear combinations of the firstkrows. At the same time,x(k, k) goes over the linear combinations ofx(0, k), x(1, k),· · · , x(k−1, k). Ifx(0, k), x(1, k),· · · , x(k−1, k) are not all zero,x(k, k) is balanced, so−P

ω^−x(k,k)= 0. Ifx(0, k), x(1, k),· · ·, x(k−1, k) are all zero,x(k, k) = 0, so−Pω^−x(k,k)=−p^k.

Consequently, we have

pⁿ²·gc01(s) = (−1)(−p)(−p²)· · ·(−p^r−1)

n−1

Y

k=r

(pⁿ−p^k) = (−1)^rp^−n(n+1)/2

n−r

Y

k=1

(p^k−1).

J ILemma 13. kˆg(s)k1<1 + 6·p⁻ⁿQn

k=1(p^k−1) Proof.

kˆg(s)k1 ≤ 1 + 2kgc₀₁(s)k1

= 1 + 2

n

X

r=0

X

s:rank_p(s)=r

|gc01(s)|

= 1 + 2

n

X

r=0

p^r(r−1)/2 n

r

p n

Y

k=n−r+1

(p^k−1)·p^−n(n+1)/2

n−r

Y

k=1

(p^k−1)

= 1 + 2p^−n(n+1)/2

n

Y

k=1

(p^k−1)

n

X

r=0

p^r(r−1)/2 n

r

p

= 1 + 2p^−n(n+1)/2

n

Y

k=1

(p^k−1)

n−1

Y

k=0

(1 +p^k)

= 1 + 2p⁻ⁿ

n

Y

k=1

(p^k−1)

n−1

Y

k=0

1 +p^k p^k

< 1 + 2p⁻ⁿ

n

Y

k=1

(p^k−1)·3

J

Now we can prove that the approximate Fourier`1-norm ofg is large.

ILemma 14.

kˆgk^3/2₁ =p^Ω(n2)

Proof. In the proof of Lemma 7, we know kˆhk^∗₁=pⁿ²kˆhk∞. We rewrite Theorem 10:

kˆgk^α₁ = max

h:pⁿ²kˆhk∞≤1

1 +α

2 hg, hi+1−α 2 khk1.

We choose h= (−1)ⁿ⁺¹gc₀₁. So, ˆh= (−1)ⁿ⁺¹g₀₁/pⁿ², andkˆhk∞= 1/pⁿ². hg01, hi = hg01,(−1)ⁿ⁺¹gc01i

= (−1)ⁿ⁺¹X

s

g₀₁(s)gc₀₁(s)

= (−1)ⁿ⁺¹ X

s:rank_p(s)=n

gc01(s)

= (−1)ⁿ⁺¹

n−1

Y

k=0

(pⁿ−p^k)·(−1)ⁿp^−n(n+1)/2

= −p⁻ⁿ

n

Y

k=1

(p^k−1)

hg, hi=−2hg₀₁, hi+X

x

h(x) =−2· −p⁻ⁿ

n

Y

k=1

(p^k−1) + 0 = 2p⁻ⁿ

n

Y

k=1

(p^k−1)

kˆgk^3/2₁ ≥ 1 + 3/2

2 hg, hi+1−3/2 2 khk₁

≥ 5 4 ·2p⁻ⁿ

n

Y

k=1

(p^k−1)−1 4 6p⁻ⁿ

n

Y

k=1

(p^k−1) + 1

!

= p⁻ⁿ

n

Y

k=1

(p^k−1)−1 4

≥ p⁻ⁿ

n

Y

k=1

p^k−1−1 4

= p^n(n−3)/2−1 4

J ITheorem 15(Theorem 3 Restated). The randomized/quantum communication complexity of the Singularityproblem isΩ(n²logp).

Proof. By Corollary 8,Q_1/6(f)≥logkˆgk^3/2₁ −3 = Ω(n²logp). R_1/6(f) = Ω(Q_1/6(f)). J

4 Lower Bound for Determinant of Non-singular Matrix

We use a small spectral norm of the matrix representing the Deta,b problem to prove the communication complexity lower bound. To prove the lower bound of spectral norm, we decompose the matrix onto Fourier basis, because the Fourier basis has good properties and a small spectral norm.

4.1 Spectral Norm Method

In the previous section, a large trace norm implies a large communication complexity lower bound. However, here we use a small spectral norm to prove a large communication complexity lower bound.

The spectral norm method is based on the discrepancy method, which can derive the com- munication complexity lower bound by giving an upper bound for a value called discrepancy defined below.

I Definition 16 (Discrepancy). Let f : F^Np ×F^Np 7→ {0,1} be a function, S×T be a rectangle, and µ be a probability distribution on F^Np ×F^Np. Denote discµ(S×T, f) =

|P

(x,y)∈S×Tµ(x, y)(−1)^f(x,y)|, and discµ(f) = max_S,T⊆FN

p discµ(S×T, f).

The discrepancy is widely used in proving communication complexity lower bound [1, 15, 7], with many applications. It was also used to prove the quantum lower bound [6, 13], and could be phrased in the following theorem.

ITheorem 17. [6] For any function f and any distributionµ, we have Q^∗(f) = Ω

log 1−2 disc_µ(f)

.

Furthermore, the discrepancy can be bounded by the spectral norm.

ITheorem 18. [7, Example 3.29] Letf :F^Np ×F^Np 7→ {0,1,⊥}be a partial Boolean function.

We define the corresponding partial sign matrixF by its entries

F(x, y) =











1 if f(x, y) = 0,

−1 if f(x, y) = 1, 0 otherwise.

For the uniform distributionµon the defined inputs off, we havedisc_µ(f)≤p^N· kFk/kFk₁.

4.2 Fourier Basis Matrix

For the determinant problem, it is difficult to computekFk directly. We will decomposeF into Fourier basis: F =P

kλkHk. The spectral norm of the Fourier basisHk is easier to compute. At last, we will use the triangle inequality to boundkFk.

IDefinition 19(Discrete logarithm). F^∗p=Fp\ {0} is a cyclic multiplicative group, in which 2 is a primitive element (the generator for the multiplicative group). For a∈F^∗p, we say k= log₂aif 2^k =a.

Letη=e^2πi/(p−1). Fork= 0,1,· · ·, p−2 anda∈Fp, we define a family of τ as below.

τ_k(a) =

(0 ifa= 0, η^klog2a otherwise.

ILemma 20. Fora, b∈Fp, 1. τ_k(a)τ_k(b) =τ_k(ab);

2. Pp−1

a=0τk(a) = 0, if k6= 0;

3. τk(a)τk(a)^∗τk(a) =τk(a).

Proof. 1. If a = 0 or b = 0, both sides are 0. If a 6= 0 and b 6= 0, τk(a)τk(b) = η^klog2aη^klog2b=η^klog2(ab)=τk(ab).

2.

p−1

X

a=0

τk(a) =

p−1

X

a=1

τk(a) =

p−1

X

a=1

η^klog2a =

p−2

X

l=0

η^kl = 0

3. If a= 0, both sides are 0. Ifa6= 0, τ_k(a)^∗τ_k(a) =|τk(a)|²= 1. J Then, we define a family of hk. Forx∈F^n×np , we define hk(x) =τk(detp(x)).

ILemma 21. Forx, y∈F^n×np , 1. hk(x)hk(y) =hk(xy);

2. hk(x)hk(x)^∗hk(x) =hk(x).

Proof.

1. hk(x)hk(y) =τk(detp(x))τk(detp(y)) =τk(detp(x)detp(y)) =τk(detp(xy)) =hk(xy);

2. h_k(x)h_k(x)^∗h_k(x) =τ_k(detp(x))τ_k(detp(x))^∗τ_k(detp(x)) =τ_k(detp(x)) =h_k(x). J ILemma 22. Fort∈F^n×np ,hˆk(t) = 0 if the first row oft is all-zero andk6= 0.

Proof.

hˆ_k(t) =X

x

h(x)ω^hx,ti= X

x(1–,·)

ωhx(1–,·),t(1–,·)i X

x(0,·)

h(x)

Ifx(1–,·) is fixed andx(0,·) goes overFⁿp, det_p(x) is balanced on all non-zero values. Thus, P

x(0,·)h(x) = 0. J

ILemma 23. Forw, t∈F^n×np , ifdet(w)6= 0, thenhˆ_k(wt) =h_k(w)^∗hˆ_k(t).

Proof.

hˆk(wt) = hk(w)^∗hk(w) ˆhk(wt)

= hk(w)^∗hk(w) 1 pⁿ²

X

x

hk(x)ω^−hx,wti

= h_k(w)^∗h_k(w^T) 1 pⁿ²

X

x

h_k(x)ω^−hwTx,ti

= h_k(w)^∗ 1 pⁿ²

X

x

h_k(w^Tx)ω^−hwTx,ti

= h_k(w)^∗hˆ_k(t)

J ILemma 24. Fort∈F^n×np , if det(t) = 0andk6= 0, thenhˆk(t) = 0.

Proof. Becauset is singular, we can find an invertible matrixw such that the first row of wt is all zero. By Lemma 22 and Lemma 23, ˆh_k(t) = ˆh_k(wt)/h_k(w)^∗= 0/h_k(w)^∗ = 0. J ILemma 25. Fork6= 0and t∈F^n×np ,hˆk(t) = ˆhk(I)hk(t)^∗, where I is the identity matrix of size n×n.

Proof. If det(t) = 0, ˆhk(t) = 0 = ˆhk(I)·0 = ˆhk(I)hk(t)^∗.

If det(t)6= 0, ˆhk(t) = ˆhk(t·I) =hk(t)^∗hˆk(I). J ILemma 26. Fork6= 0,hˆk(I)^∗hˆk(I) =p⁻ⁿ².

Proof.

hhˆk,hˆki = X

x

hˆk(x)^∗hˆk(x)

= X

x

hˆk(I)hk(x)∗

hˆk(I)hk(x)

= hˆk(I)^∗hˆk(I)X

x

hk(x)^∗hk(x)

= hˆk(I)^∗hˆk(I)hhk, hki

= hˆk(I)^∗hˆk(I)·pⁿ²hhˆk,hˆki

Therefore, ˆhk(I)^∗hˆk(I) =p⁻ⁿ². J

At last, we define a family of matrixHk. Forx, y∈Fp^n×n, we defineHk(x, y) =hk(x+y).

ILemma 27. Fork6= 0,HkH_k^†Hk=pⁿ²·Hk. Proof. Forw, x, y, z, r, s, t∈F^n×np ,

(H_kH_k^†H_k)(w, z)

= X

x

X

y

Hk(w, x)Hk(y, x)^∗Hk(y, z)

= X

x

X

y

hk(w+x)hk(x+y)^∗hk(y+z)

= X

x

X

y

X

r

hˆk(r)ω^hr,w+xi

! X

s

hˆk(s)ω^hs,x+yi

!∗

X

t

hˆk(t)ω^ht,y+zi

!

= X

r

X

s

X

t

hˆk(r) ˆhk(s)^∗hˆk(t)ω^hr,wi X

x

ω^hr−s,xi

! X

y

ω^h−s+t,yi

! ω^ht,zi

= X

r

X

s

X

t

hˆk(r) ˆhk(s)^∗hˆk(t)ω^hr,wi

pⁿ²δr,s pⁿ²δs,t

ω^ht,zi

= p²ⁿ²X

r

hˆk(r) ˆhk(r)^∗hˆk(r)ω^hr,wiω^hr,zi

= p²ⁿ²X

r

( ˆh_k(I)h_k(r)^∗)( ˆh_k(I)^∗h_k(r))( ˆh_k(I)h_k(r)^∗)ω^hr,w+zi

= p²ⁿ²X

r

hˆk(I) ˆhk(I)^∗hˆk(I)hk(r)^∗hk(r)hk(r)^∗ω^hr,w+zi

= p²ⁿ²X

r

p⁻ⁿ²hˆk(I)hk(r)^∗ω^hr,w+zi

= pⁿ²X

r

hˆ_k(r)ω^hr,w+zi

= pⁿ²hk(w+z)

= pⁿ²H_k(w, z)

J ILemma 28. Fork6= 0,

kHkk=p^n2/2.

Proof. We denote the largest eigenvalue of a semi-definite matrixAby max eval(A). Thus, kAk=p

max eval(A^†A).

kH_kH_k^†H_kk = q

max eval((H_kH_k^†H_k)^†·H_kH_k^†H_k)

= q

max eval((H_k^†Hk)³)

= q

max eval(H_k^†H_k) 3

= kHkk³

Comparing to Lemma 27, we have kHkk³=pⁿ²kHkk. Therefore,kHkk=p^n2/2. J

4.3 Lower Bound for Det

ForDeta,b,f(x, y) is 0 if detp(x+y) =a, 1 if detp(x+y) =band undefined for other cases.

We define the corresponding partial sign matrix F as below.

F(x, y) =











1 if detp(x+y) =a,

−1 if detp(x+y) =b, 0 otherwise.

ILemma 29. kFk₁= Ω(p²ⁿ²⁻¹).

Proof. In the uniform distribution, Pr_x∈F^n×n

p [detp(x)6= 0] =Qn

k=1(1−p^−k)> φ(p⁻¹)>0, whereφis the Euler function. This means that the density of the non-singular matrix is greater than a constant. Furthermore, the determinant is balanced on all non-zero values. Thus, Pr_x∈

F^n×np [detp(x) =aor b]> _p−1² φ(p⁻¹). Finally,kFk1≥_p−1² φ(p⁻¹)·p²ⁿ²= Ω(p²ⁿ²⁻¹). J ILemma 30. kFk ≤2p^n2/2.

Proof. It is easy to check that we can decomposeF to the Fourier basis matrices:

F =1 p·

p−2

X

k=1

η^−klog2a−η^−klog2b H_k

. Then, we use the triangle inequality of matrix norm to bound the norm ofF:

kFk ≤ 1 p·

p−2

X

k=1

η^−klog2a−η^−klog2b

kH_kk ≤ 1

p·(p−2)·2·p^n2/2<2p^n2/2.

J ITheorem 31(Theorem 4 Restated). The randomized/quantum communication complexity of Det^a,b isΩ(n²logp).

Proof.

discµ(f)≤pⁿ² kFk kFk1

≤pⁿ²· 2p^n2/2

Ω(p²ⁿ²⁻¹) =O(p^−n2/2+1) R_1/3(f) = Ω(Q^∗_1/3(f)), Q^∗_1/3(f) = Ω

log 1

discµ(f)

= Ω(n²logp).

J

5 Open Problems

One open problem is to distinguish between detp(x) = 0 and detp(x) =a. We guess it has an Ω(n²logp) lower bound even for quantum protocols. The proof could be similar to Section 3.

The other one is to compute the (i, j)-th element of the inverse of matrixx. We conjecture the quantum communication Ω(n²logp) as well. Actually, this problem is as hard as solving linear equations.

We discuss all these problems overFp, but we think they are still hard over integers.

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