Shengyu Zhang The Chinese University of Hong Kong

On the power of lower bound techniques for 1-way quantum communication complexity Shengyu Zhang The Chinese University of Hong Kong

Algorithms Circuit lb Streaming Algorithms Info. theory Quantum Computing Communication Complexity crypto VLSI Data Structures games … … Question: What’s the largest gap between classical and quantum communication complexities?

Communication complexity • [Yao79] Two parties, Alice and Bob, jointly compute a function f(x,y) with x known only to Alice and y only to Bob. • Communication complexity: how many bits are needed to be exchanged? x y Alice Bob f(x,y) f(x,y)

Various protocols • Deterministic: D(f) • Randomized: R(f) • A bounded error probability is allowed. • Private or public coins? Differ by ±O(log n). • Quantum: Q(f) • A bounded error probability is allowed. • Assumption: No shared Entanglement. (Does it help? Open.)

Communication complexity: one-way model x y • One-way: Alice sends a message to Bob. --- D1(f), R1(f), Q1(f) Alice Bob f(x,y)

About one-way model • Power: • Efficient protocols for specific functions such as Equality, Hamming Distance, and in general, all symmetric XOR functions. • Applications: • Lower bound for space complexity of streaming algorithms. • Lower bound? Can be quite hard, especially for quantum. As efficient as the best two-way protocol.

Question • Question: What’s the largest gap between classical and quantum communication complexities? • Partial functions, relations: exponential. • Total functions, two-way: • Largest gap: Q(Disj) = Θ(√n), R(Disj) = Θ(n). • Best bound: R(f) = exp(Q(f)). • Conjecture: R(f) = poly(Q(f)).

Question • Question: What’s the largest gap between classical and quantum communication complexities? • Partial functions, relations: exponential. • Total functions, one-way: • Largest gap: R1(EQ) = 2∙Q1(EQ), • Best bound: R1(f) = exp(Q1(f)). • Conjecture: R1(f) = poly(Q1(f)), • or even R1(f) = O(Q1(f)).

Approaches • Approach 1: Directly simulate a quantum protocol by classical one. • [Aaronson] R1(f) = O(m∙Q1(f)). • Approach 2: L(f) ≤Q1(f) ≤R1(f) ≤ poly(L(f)). • [Nayak99; Jain, Z.’09] R1(f) = O(Iμ∙VC(f)), where Iμ is the mutual info of any hard distribution μ. • Note: For the approach 2 to be possibly succeed, the quantum lower bound L(f) has to be polynomially tight for Q1(f).

Main result • There are three lower bound techniques known for Q1(f). • Nayak’99: Partition Tree • Aaronson’05: Trace Distance • The two-way complexity Q(f) • [Thm] All of these lower bounds can be arbitrarily weak. • Actually, random functions have Q(f) = Ω(n), but the first two lower bounds only give O(1).

Next • Closer look at the Partition Tree bound. • Compare Q and Partition Tree (PT) and Trace Distance (TD) bounds.

Nayak’s info. theo. argument Index(x,i) = xi • [Nayak’99] Q1(Index) = Ω(n). • ρx contains Ω(1) info of x1, since i may be 1. • Regardless of x1, ρx contains Ω(1) info of x2. • And so on. x{0,1}n i[n] Alice Bob ρx

Nayak’s info. theo. argument Index(x,i) = xi • ρ = ∑x px∙ρx • S(ρ) = S(½ρ0+½ρ1) // ρb = 2 ∑x:x_1=b px∙ρx ≥ I(X1,M1) + ½S(ρ0)+½S(ρ1) // Holevo bound. M1: Bob’s conclusion about X1 ≥ 1 – H(ε) + ½S(ρ0)+½S(ρ1)// Fano’s Inequ. ≥ … ≥ n(1 – H(ε)). x{0,1}n i[n] Alice Bob ρx

Partition tree Index(x,i) = xi • ρ = ∑x px∙ρx • ρb = 2 ∑x:x1=b px∙ρx • ρb1b2 = 4 ∑x:x1=b1,x2=b2 px∙ρx x{0,1}n i[n] Alice Bob ρx 000 001 010 011 100 101 110 111 ρ00 ρ0 ρ01 ρ ρ10 ρ1 ρ11

Partition tree Index(x,i) = xi • ρ = ∑x px∙ρx • In general: • Distri. p on {0,1}n • Partition tree for {0,1}n • Gain H(δ)-H(ε) at v • v is partitioned by (δ,1-δ) x{0,1}n i[n] Alice Bob ρx 0 0 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1

Issue • [Fano’s inequality] I(X;Y) ≥ H(X) – H(ε). • X,Y over {0,1}. • ε = Pr[X ≠ Y]. • What if H(δ) < H(ε)? • Idea 1: use success amplification to decrease ε to ε*. • Idea 2: give up those vertices v with small H(X). • Bound: maxT,p,ε* log(1/ε*)∙∑vp(v)[H(Xv)-H(ε*)]+ • Question: How to calculate this? H(δ)

Picture clear • maxT,p,ε* log(1/ε*)∙∑vp(v)[H(Xv)-H(ε*)]+ • Very complicated. Compare to Index where the tree is completely binary and each H(δv) = 1 (i.e. δv=1/2). • [Thm] the maximization is achieved by a complete binary tree with δv=1/2 everywhere.

Two interesting comparisons • Comparison to decision tree: • Decision tree complexity: make the longest path short • Here: make the shortest path long. • Comparison to VC-dim lower bound: [Thm] The value is exactly the extensive equivalence query complexity. • A measure in learning theory. • Strengthen the VC-dim lower bound by Nayak.

Trace distance bound • [Aaronson’05] • μ is a distri on 1-inputs • D1: (x, y) ←μ. • D2: y ←μ, x1, x2←μy. Then Q1(f) = Ω(log ∥D2-D12∥1-1)

Separation • [Thm] Take a random graph G(N,p) with ω(log4N/N) ≤ p ≤ 1-Ω(1). Its adjacency matrix, as a bi-variate function f, has the following w.p. 1-o(1) Q(f) = Ω(log(pN)). • Q*(f) ≥Q(f) = Ω(log(1/disc(f))). • disc(f) is related to σ2(D-1/2AD-1/2), which can be bounded by O(1/√pN) for a random graph.

[Thm] For p = N-Ω(1), PT(f) = O(1) w.h.p. • By our characterization, it’s enough to consider complete binary tree. • For p = N-Ω(1), each layer of tree shrinks the #1’s by a factor of p. • pN → p2N →p3N →… →0: Only O(1) steps. • [Thm] For p = o(N-6/7), TD(f) = O(1) w.h.p. • Quite technical, omitted here.

Putting together • [Thm] Take a random graph G(N,p) with ω(log4N/N) ≤ p ≤ 1-Ω(1). Its adjacency matrix, as a bi-variate function f, has the following w.p. 1-o(1) Q(f) = Ω(log(pN)). • [Thm] For p = o(N-6/7), TD(f) = O(1) w.h.p. • [Thm] For p = N-Ω(1), PT(f) = O(1) w.h.p. • Taking p between ω(log4N/N) and o(N-6/7) gives the separation.

Discussions • Negative results on the tightness of known quantum lower bound methods. • Calls for new method. • Somehow combine the advantages of these methods? • Hope the paper shed some light on this by identifying their weakness.

Shengyu Zhang The Chinese University of Hong Kong