Shengyu Zhang The Chinese University of Hong Kong
- Slides: 23
Shengyu Zhang The Chinese University of Hong Kong
Algorithms Circuit lb Streaming Algorithms Info. theory crypto Quantum Computing Communication Complexity VLSI Data Structures games … … Question: What’s the largest gap between classical and quantum communication complexities?
Communication complexity • [Yao 79] Two parties, Alice and Bob, jointly compute a function f(x, y) with x known only to Alice and y only to Bob. x y Alice Bob f(x, y) • Communication complexity: how many bits are needed to be exchanged?
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 Alice Bob f(x, y) • One-way: Alice sends a message to Bob. --- D 1(f), R 1(f), Q 1(f)
About one-way model • Power: As efficient as the best two-way protocol. – 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.
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: R 1(EQ) = 2∙Q 1(EQ), – Best bound: R 1(f) = exp(Q 1(f)). • Conjecture: R 1(f) = poly(Q 1(f)), – or even R 1(f) = O(Q 1(f)).
Approaches • Approach 1: Directly simulate a quantum protocol by classical one. – [Aaronson] R 1(f) = O(m∙Q 1(f)). • Approach 2: L(f) ≤ Q 1(f) ≤ R 1(f) ≤ poly(L(f)). – [Nayak 99; Jain, Z. ’ 09] R 1(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 Q 1(f).
Main result • There are three lower bound techniques known for Q 1(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 x {0, 1}n Index(x, i) = xi Alice ρx i [n] Bob • [Nayak’ 99] Q 1(Index) = Ω(n). – ρx contains Ω(1) info of x 1, since i may be 1. – Regardless of x 1, ρx contains Ω(1) info of x 2. – And so on.
Nayak’s info. theo. argument x {0, 1}n Index(x, i) = xi Alice ρx i [n] Bob • ρ = ∑x px∙ρx • S(ρ) = S(½ρ0+½ρ1) // ρb = 2 ∑x: x_1=b px∙ρx ≥ I(X 1, M 1) + ½S(ρ0)+½S(ρ1) // Holevo bound. M 1: Bob’s conclusion about X 1 ≥ 1 – H(ε) + ½S(ρ0)+½S(ρ1) // Fano’s Inequ. ≥ … ≥ n(1 – H(ε)).
Partition tree x {0, 1}n Index(x, i) = xi Alice • ρ = ∑x px∙ρx • ρb = 2 ∑x: x 1=b px∙ρx • ρb 1 b 2 = 4 ∑x: x 1=b 1, x 2=b 2 px∙ρx ρx i [n] Bob ρ00 ρ0 ρ ρ01 ρ10 ρ1 ρ11 000 001 010 011 100 101 110 111
Partition tree x {0, 1}n Index(x, i) = xi Alice ρx • 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) i [n] Bob 0 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]. H(δ) • 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: max. T, p, ε* log(1/ε*)∙∑vp(v)[H(Xv)-H(ε*)]+ • Question: How to calculate this?
Picture clear • max. T, 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 – D 1: (x, y) ← μ. – D 2: y ← μ, x 1, x 2 ← μy. Then Q 1(f) = Ω(log ∥D 2 -D 12∥ 1 -1)
Separation • [Thm] Take a random graph G(N, p) with ω(log 4 N/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(p. N)). • Q*(f) ≥ Q(f) = Ω(log(1/disc(f))). • disc(f) is related to σ2(D-1/2 AD-1/2), which can be bounded by O(1/√p. N) 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. • p. N → p 2 N → p 3 N → … → 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(log 4 N/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(p. N)). • [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(log 4 N/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.
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