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

2. Communication complexity𝑥 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.

3. 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.

4. 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)

5. 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.

6. 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.

7. 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 .

8. 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 ≠ Fourier sparsity.
Cauchy-Schwartz: 𝑓 1 ≤ 𝑓 0 1/2 for 𝑓: 0,1 𝑛 →±1

9. 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
Thm.*1 𝐷 𝑓∘⊕ ≤ 2 𝑑 2 /2 log 𝑑−2 𝑓 1
Thm.*1 𝐷 𝑓∘⊕ ≤𝑂 𝑑 𝑓
𝑑= deg 2 (𝑓) : degree of 𝑓 as polynomial over 𝔽 2 .
Fact*2. 𝑑≤ log 𝑓
*1. Tsang, Wong, Xie, Zhang, FOCS, 2013.
*2. Bernasconi and Codenotti. IEEE Transactions on Computers, 1999.

10. 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
𝑄 𝐸 𝑓∘⊕ =Ω log 𝑓 *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.

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

12. Goal: compute 𝑓(𝑥+𝑦) where 𝑓: 0,1 𝑛 → ±1 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 𝑓(𝑥+𝑦+𝑡) 𝑡

13. 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 𝑓(𝑥+𝑦+𝑡) 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
(That’s where the factor of 2 𝑑 comes from.)

14. 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 𝑓(𝑥+𝑦+𝑡) 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
(That’s where the factor of 2 𝑑 comes from.)

15. 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 𝑓(𝑥+𝑦+𝑡) 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
(That’s where the factor of 2 𝑑 comes from.)

16. 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 𝑓(𝑥+𝑦+𝑡) 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

17. 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 ?

18. Thanks