Effcient quantum protocols for XOR functions Shengyu Zhang The Chinese University of Hong Kong

Communication complexity 𝑥 𝑦 𝐹(𝑥,𝑦) Two parties, Alice and Bob, jointly compute a function 𝑓 on input (𝑥,𝑦). 𝑥 known only to Alice and 𝑦 only to Bob. Communication complexity*1: how many bits are needed to be exchanged? *1. A. Yao. STOC, 1979.

Computation modes Deterministic: Players run determ. protocol. ---𝐷(𝐹) Randomized: Players have access to random bits; small error probability allowed. --- 𝑅(𝐹) Quantum: Players send quantum messages. Share resource? (Superscript.) 𝑄 ∗ (𝐹): share entanglement. 𝑄(𝐹): share nothing Error? (Subscript) 𝑄 𝜖 (𝐹): bounded-error. 𝑄 𝐸 (𝐹): zero-error, fixed length.

Lower bounds Not only interesting on its own, but also important because of numerous applications. to prove lower bounds. Question: How to lower bound communication complexity itself? Communication matrix 𝑀 𝐹 ≝ 𝐹 𝑥,𝑦 𝑥,𝑦 : 𝑦 ↓ 𝑥→ 𝐹(𝑥,𝑦)

Log-rank conjecture Rank lower bound*1 Conj: The rank lower bound is polynomially tight. combinatorial measure ⇔ linear algebra measure. Equivalent to a bunch of other conjectures. related to graph theory*2; nonnegative rank*3, Boolean roots of polynomials*4, quantum sampling complexity*5. ≤ log 𝑂 1 𝑟𝑎𝑛𝑘 𝑀 𝐹 *1. Melhorn, Schmidt. STOC, 1982. *2. Lovász, Saks. FOCS, 1988. *3. Lovász. Book Chapter, 1990. *4. Valiant. Info. Proc. Lett., 2004. *5. Ambainis, Schulman, Ta-Shma, Vazirani, Wigderson, SICOMP 2003.

Log-rank conjecture: quantum version Rank lower bound log 2 𝑟𝑎𝑛𝑘 𝑀 𝐹 ≤𝐷 𝐹 Quantum: rank lower bound *1 log 2 𝑟𝑎𝑛 𝑘 𝜖 𝑀 𝐹 ≤ 𝑄 𝜖 𝐹 ≤ log 𝑂 1 𝑟𝑎𝑛 𝑘 𝜖 𝑀 𝐹 𝑟𝑎𝑛 𝑘 𝜖 𝑀 = min 𝑀 ′ :|𝑀(𝑥,𝑦)− 𝑀 ′ (𝑥,𝑦)|≤𝜖 𝑟𝑎𝑛𝑘( 𝑀 ′ ) ≤ log 𝑂 1 𝑟𝑎𝑛𝑘 𝑀 𝐹 *1. Buhrman, de Wolf. CCC, 2001.

Log-rank conjecture for XOR functions Since Log-rank conjecture appears too hard in its full generality,… let’s try some special class of functions. XOR functions: 𝑓(𝑥⊕𝑦). --- 𝐹=𝑓∘⊕ The linear composition of 𝑥 and 𝑦. Include important functions such as Equality, Hamming Distance, Gap Hamming Distance. Interesting connections to Fourier analysis of functions on 0,1 𝑛 .

Digression: Fourier analysis ∀𝑓: 0,1 𝑛 →ℝ can be written as 𝑓= 𝛼∈ 0,1 𝑛 𝑓 𝛼 𝜒 𝛼 𝜒 𝛼 𝑥 = −1 𝛼⋅𝑥 , and characters are orthogonal 𝑓 𝛼 :𝛼∈ 0,1 𝑛 : Fourier coefficients of 𝑓 Parseval: If 𝑅𝑎𝑛𝑔𝑒 𝑓 = +1,−1 , then 𝛼 𝑓 𝛼 2 =1. Two important measures: 𝑓 1 = 𝛼 𝑓 𝛼 --- Spectral norm. 𝑓 0 = 𝛼: 𝑓 𝛼 ≠0 --- Fourier sparsity. Cauchy-Schwartz: 𝑓 1 ≤ 𝑓 0 1/2 for 𝑓: 0,1 𝑛 →±1

Log-rank Conj. For XOR functions Interesting connections to Fourier analysis: 1. 𝑟𝑎𝑛𝑘 𝑀 𝑓∘⊕ = 𝑓 0 . Log-rank Conj: 𝐷 𝑓∘⊕ ≤ log 2 𝑂(1) 𝑓 0 . Thm.*1 𝐷 𝑓∘⊕ ≤ 2 𝑑 2 /2 log 𝑑−2 𝑓 1 Thm.*1 𝐷 𝑓∘⊕ ≤𝑂 𝑑 𝑓 1 . 𝑑= deg 2 (𝑓) : degree of 𝑓 as polynomial over 𝔽 2 . Fact*2. 𝑑≤ log 2 𝑓 0 . *1. Tsang, Wong, Xie, Zhang, FOCS, 2013. *2. Bernasconi and Codenotti. IEEE Transactions on Computers, 1999.

Quantum This talk: A simpler case 𝑄 𝐸 𝑓∘⊕ =O 2 𝑑 log 𝑓 0 . 𝑓 1,𝜖 = min 𝑔: 𝑓−𝑔 ∞ ≤𝜖 𝑔 1 𝑟𝑎𝑛 𝑘 𝜖 𝑀 = min 𝑀 ′ :|𝑀(𝑥,𝑦)− 𝑀 ′ (𝑥,𝑦)|≤𝜖 𝑟𝑎𝑛𝑘( 𝑀 ′ ) This paper: 𝑄 𝑓∘⊕ ≤ 𝑂 2 𝑑 log 𝑓 1,𝜖 , where 𝑑= deg 2 (𝑓) . Recall classical: 𝐷 𝑓∘⊕ ≤ 2 𝑑 2 /2 log 𝑑−2 𝑓 1 Confirms quantum Log-rank Conjecture for low-degree XOR functions. This talk: A simpler case 𝑄 𝐸 𝑓∘⊕ =O 2 𝑑 log 𝑓 0 . 𝑄 𝐸 𝑓∘⊕ =Ω log 𝑓 0 . *2 Ω log 𝑟𝑎𝑛 𝑘 𝜖 𝑀 𝑓∘⊕ ≥ *1. Lee and Shraibman. Foundations and Trends in Theoretical Computer Science, 2009. *2. Buhrman and de Wolf. CCC, 2001.

About quantum protocol Much simpler. log 𝑓 0 comes very naturally. Inherently quantum. Not from quantizing any classical protocol.

Goal: compute 𝑓(𝑥+𝑦) where 𝑓: 0,1 𝑛 → ±1 0 + 1 2 ⊗ 0 + 1 2 ⊗ 𝜓 = 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝐸 𝛼 𝜓 Add phase 𝜒 𝛼 (𝑦) 𝜓′ = 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝜒 𝛼 (𝑦) 𝐸 𝛼 𝜓′ 𝐸 𝛼 → 𝛼 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥+𝑦 𝛼 Fourier: 𝛼 → 𝑡 𝜒 𝛼 𝑡 |𝑡〉 𝑡 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥+𝑦 𝜒 𝛼 (𝑡) 𝑡 = 𝑡 𝑓 𝑥+𝑦+𝑡 𝑡 𝑡 0 +𝑓(𝑥+𝑦+𝑡) 1 2 𝑡

Goal: compute 𝑓(𝑥+𝑦) where 𝑓: 0,1 𝑛 → ±1 𝜓 =|+〉 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝐸 𝛼 𝜓 𝜓 =|+〉 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝐸 𝛼 𝜓 Add phase 𝜒 𝛼 (𝑦) 𝜓′ Decoding + Fourier One more issue: Only Alice knows 𝑡! Bob doesn’t. It’s unaffordable to send 𝑡. Obs: 𝑠𝑢𝑝𝑝 Δ 𝑡 𝑓 ⊆𝐴+𝐴,∀𝑡. 𝐴=𝑠𝑢𝑝𝑝 𝑓 𝑡 0 +𝑓(𝑥+𝑦+𝑡) 1 2 𝑡 Measure{|+〉,|−〉} Measure 𝑡 A random 𝑡∈ 0,1 𝑛 and 𝑓(𝑥+𝑦+𝑡). Recall our target: 𝑓(𝑥+𝑦). What’s the difference? The derivative: Δ 𝑡 𝑓 𝑧 =𝑓 𝑧+𝑡 𝑓(𝑧). Good: deg 2 ( Δ 𝑡 𝑓) ≤ deg 2 (𝑓) −1. Bad: Δ 𝑡 𝑓 0 ≤ 𝑓 0 2 . (That’s where the factor of 2 𝑑 comes from.)

Goal: compute 𝑓(𝑥+𝑦) where 𝑓: 0,1 𝑛 → ±1 0,1 𝑛 0,1 𝑛 𝜓 =|+〉 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝐸 𝛼 𝜓 Add phase 𝜒 𝛼 (𝑦) 𝐴+𝐴 ≝ 𝑎+𝑏:𝑎,𝑏∈𝐴 𝜓′ Decoding + Fourier One more issue: Only Alice knows 𝑡! Bob doesn’t. It’s unaffordable to send 𝑡. Obs: 𝑠𝑢𝑝𝑝 Δ 𝑡 𝑓 ⊆𝐴+𝐴,∀𝑡. 𝐴=𝑠𝑢𝑝𝑝 𝑓 𝑡 0 +𝑓(𝑥+𝑦+𝑡) 1 2 𝑡 𝐴 Measure{|+〉,|−〉} Measure 𝑡 A random 𝑡∈ 0,1 𝑛 and 𝑓(𝑥+𝑦+𝑡). Recall our target: 𝑓(𝑥+𝑦). What’s the difference? The derivative: Δ 𝑡 𝑓 𝑧 =𝑓 𝑧+𝑡 𝑓(𝑧). Good: deg 2 ( Δ 𝑡 𝑓) ≤ deg 2 (𝑓) −1. Bad: Δ 𝑡 𝑓 0 ≤ 𝑓 0 2 . (That’s where the factor of 2 𝑑 comes from.)

Goal: compute 𝑓(𝑥+𝑦) where 𝑓: 0,1 𝑛 → ±1 𝜓 =|+〉 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝐸 𝛼 𝜓 𝜓 =|+〉 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝐸 𝛼 𝜓 Add phase 𝜒 𝛼 (𝑦) 𝜓′ Decoding + Fourier One more issue: Only Alice knows 𝑡! Bob doesn’t. It’s unaffordable to send 𝑡. Obs: 𝑠𝑢𝑝𝑝 Δ 𝑡 𝑓 ⊆𝐴+𝐴,∀𝑡. 𝐴=𝑠𝑢𝑝𝑝 𝑓 Thus in round 2, Alice and Bob can just encode the entire 𝐴+𝐴. 𝑡 0 +𝑓(𝑥+𝑦+𝑡) 1 2 𝑡 Measure{|+〉,|−〉} Measure 𝑡 A random 𝑡∈ 0,1 𝑛 and 𝑓(𝑥+𝑦+𝑡). Recall our target: 𝑓(𝑥+𝑦). What’s the difference? The derivative: Δ 𝑡 𝑓 𝑧 =𝑓 𝑧+𝑡 𝑓(𝑧). Good: deg 2 ( Δ 𝑡 𝑓) ≤ deg 2 (𝑓) −1. Bad: Δ 𝑡 𝑓 0 ≤ 𝑓 0 2 . (That’s where the factor of 2 𝑑 comes from.)

At last, deg 2 𝑓 𝑑 =0, a constant function. Goal: compute 𝑓(𝑥+𝑦) where 𝑓: 0,1 𝑛 → ±1 𝜓 =|+〉 𝛼∈𝐴 𝑓 𝛼 𝜒 𝛼 𝑥 𝐸 𝛼 𝜓 Add phase 𝜒 𝛼 (𝑦) 𝜓′ Decoding + Fourier 𝑡 0 +𝑓(𝑥+𝑦+𝑡) 1 2 𝑡 Measure{|+〉,|−〉} Measure 𝑡 A random 𝑡∈ 0,1 𝑛 and 𝑓(𝑥+𝑦+𝑡). Compute 𝑓 2 ≝ Δ 𝑡 𝑓 𝑧 =𝑓 𝑧+𝑡 𝑓(𝑧). At last, deg 2 𝑓 𝑑 =0, a constant function. Cost: log 𝐴 + log 2𝐴 + log 4𝐴 +…+ log 2 𝑑−1 𝐴 ≤ 2 𝑑 log |𝐴| . Used trivial bound: 𝑘𝐴 ≤ 𝐴 𝑘

Open problems Get rid of the factor 2 𝑑 ! What can we say about additive structure of supp 𝑓 for Boolean functions 𝑓? Say, 𝐴+𝐴 ≤ 𝐴 1.5 ?

Thanks