Exact (Bichromatic) Maximum Inner Product , - - PowerPoint PPT Presentation

On The Hardness of Approximate and Exact (Bichromatic) Maximum Inner Product

Lijie Chen (Massachusetts Institute of Technology)

𝑁𝑏𝑦 𝑏,𝑐 ∈𝐵×𝐶 𝑏 ⋅ 𝑐

Max-IP and Z-Max-IP

• (Boolean) Max-IP:
• Given sets 𝐵 and 𝐶 of Boolean vectors (each of size n) find 𝑏 in 𝐵 and 𝑐 in 𝐶

with maximum inner product:

• For sets 𝐵 and 𝐶, set 𝑁𝑏𝑦𝐽𝑄 𝐵, 𝐶 ≔

max

𝑏,𝑐 ∈𝐵×𝐶⟨𝑏, 𝑐⟩ .

• Approx. version: find a r-multiplicative approximation to the answer:
• Want an 𝐵𝑀𝐻 s.t. 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ≤ 𝐵𝑀𝐻(𝐵, 𝐶) ≤ 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ⋅ 𝑠.
• Z-Max-IP:
• Two sets of 𝑜 Integer vectors.

Max-IP and Z-Max-IP

• Basic problems, relevant in practice.
• Important theoretical implications as well.
• Approx. Boolean Max-IP:
• [ARW’17]: basis of the recent breakthrough result

in Hardness for Approximation in P, implies hardness for many other problems.

• Z-Max-IP:
• [Wil’18]: Hardness for Z-Max-IP implies hardness

for finding furthest pair in low dimension Euclidean space.

• Bichromatic LCS Closest Pair over permutations,
• Approximate Regular Expression Matching,
• Diameter in Product Metric,
• Approximate Closest Pair in Euclidian Space [Rub’18]

New Hardness for Z-Max-IP (under SETH)

• Z-Max-IP for n vectors of 𝟑𝑷 𝐦𝐩𝐡∗ 𝒐 dimensions

requires 𝑜2−𝑝(1) time under SETH.

• Z-Max-IP for 𝑜 vectors of 𝟑𝑷 𝐦𝐩𝐡∗ 𝒐 in 𝑜1.99 time.
• ⇒Max-IP for 𝑜 vectors of 𝑃(log 𝑜) dim. in 𝑜1.99 time.
• ⇒CNF-SAT for 𝑜 variables and 𝑃(𝑜) clauses in 20.995𝑜 time.
• ⇔SETH is false. [CIP’06]

Z-Max-IP: Given sets 𝐵 and 𝐶 of Integer vectors (each of size n) find 𝑏 in 𝐵 and 𝑐 in 𝐶 with maximum inner product:

• Closer to the upper bound
• [Mat’92] : Z-Max-IP in n2−1/O(d) time.

Lower Bound 𝑜2−o(1) when 𝑒 = 2𝑃 log∗𝑜 [This work] Lower Bound 𝑜2−o(1) when 𝑒 = 𝜕(log2 log 𝑜) Implicit in [Wil’18] Upper Bound 𝑜2−𝜗 when 𝑒 = 𝑃(1) [Mat’92]

?

New Hardness for Z-Max-IP (under SETH)

• 1. New Hardness for Z-Max-IP (under SETH):
• Z-Max-IP for n vectors of 𝟑𝑷 𝐦𝐩𝐡∗ 𝒐 dimensions

requires 𝑜2−𝑝(1) time.

• Separation for Boolean Max-IP / Z-Max-IP:
• Z-Max-IP is much harder than Boolean Max-IP.
• Progress on Open Problem 23 in Dagstuhl workshop
• n Structure and Hardness in P

Z-Max-IP 𝑜2−o(1) when 𝑒 = 2𝑃 log∗𝑜 [This work] Boolean Max-IP 𝑜2−𝜗 when 𝑒 = 𝑑 log 𝑜. [AW15, ACW16] HARD EASY Z-Max-IP: Given sets 𝐵 and 𝐶 of Integer vectors (each of size n) find 𝑏 in 𝐵 and 𝑐 in 𝐶 with maximum inner product:

New Hardness for Z-Max-IP (under SETH)

• 1. New Hardness for Z-Max-IP (under SETH):
• Z-Max-IP for n vectors of 𝟑𝑷 𝐦𝐩𝐡∗ 𝒐 dimensions

require 𝑜2−𝑝(1) time.

• New Hardness for ℓ2-Furthest Pair in 𝑆𝑒. (reductions from [Wil18])
• Finding ℓ2-Furthest Pair in 𝑆𝑒 among 𝑜 points for 𝑒 =

2𝑃 log∗ 𝑜 requires 𝑜2−𝑝(1) time.

• Stronger separation between furthest and closest pair.

ℓ2-Furthest Pair 𝑜2−o(1) when 𝑒 = 2𝑃 log∗𝑜 [This work] ℓ2-Closest Pair 2𝑃 𝑒 ⋅ 𝑜 𝑞𝑝𝑚𝑧𝑚𝑝𝑕(𝑜) [BS76, KM95, DHKP97] ℓ2-Furthest Pair 𝑜2−o(1) when 𝑒 = 𝜕(log2 log 𝑜) [Wil’18] EASY HARD Z-Max-IP: Given sets 𝐵 and 𝐶 of Integer vectors (each of size n) find 𝑏 in 𝐵 and 𝑐 in 𝐶 with maximum inner product.

Characterization of Boolean Approx. Max-IP

• 2. Characterization of Approx. Max-IP:
• [ARW’17]: Finding 2(log1−o 1 𝑜) approximation to

Max-IP with 𝑜𝑝 1 dimensions, requires 𝑜2−𝑝(1) 𝑢𝑗𝑛𝑓.

• A more refined question:
• For each vector dimension 𝑒 = 𝑒 𝑜 , what is the

smallest 𝑠 such that Max-IP can be 𝑠-approximated in truly sub-quadratic time?

• Boolean Max-IP:
• For sets 𝐵 and 𝐶 with 𝑜 Boolean

vectors, find 𝑁𝑏𝑦𝐽𝑄 𝐵, 𝐶 ≔ max

𝑏,𝑐 ∈𝐵×𝐶⟨𝑏, 𝑐⟩ .

• Approx. version: find a r-

multiplicative approximation to the answer: 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ≤ 𝐵𝑀𝐻(𝐵, 𝐶) ≤ 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ⋅ 𝑠.

• d = d(n) : vector dimensions
• r = r(n) : approx. ratio

Characterization of Boolean Approx. Max-IP

• 2. Characterization of Approx. Max-IP:
• A more refined question:
• For each vector dimension 𝑒 = 𝑒 𝑜 , what is the

smallest 𝑠 such that Max-IP can be 𝑠-approximated in truly sub-quadratic time?

• We obtain a characterization (under SETH)!
• For all 𝑒 satisfying 𝜕 log 𝑜 < 𝑒 < 𝑜𝑝 1
• Truly sub-quadratic time for 𝑠 =

𝑒 log 𝑜 Ω(1)

EASY!

• Requires 𝑜2−𝑝(1) time for 𝑠 =

𝑒 log 𝑜

• (1)

HARD!

• Boolean Max-IP:
• For sets 𝐵 and 𝐶 with 𝑜 Boolean

vectors, find 𝑁𝑏𝑦𝐽𝑄 𝐵, 𝐶 ≔ max

𝑏,𝑐 ∈𝐵×𝐶⟨𝑏, 𝑐⟩ .

• Approx. version: find a r-

multiplicative approximation to the answer: 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ≤ 𝐵𝑀𝐻(𝐵, 𝐶) ≤ 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ⋅ 𝑠.

• d = d(n) : vector dimensions
• r = r(n) : approx. ratio

Characterization of Boolean Approx. Max-IP

• 2. Characterization of Approx. Max-IP:
• We obtain a characterization!
• 𝑠 =

𝑒 log 𝑜 Ω(1)

𝑜2−𝜗 time. EASY!

• 𝑠 =

𝑒 log 𝑜

• (1)

𝑜2−𝑝 1 time. HARD!

• Example:
• 𝑒 = 𝑑 log 𝑜 , 𝑃(1)-approximation is EASY.
• 𝑒 = log2 𝑜 , 𝑃(log0.1 𝑜)−approximation is EASY.
• 𝑒 = log2 𝑜 , (log𝑝(1) 𝑜)-approximation is HARD.
• Upper Bound via polynomial method.
• Lower Bound follows from [Rub’18].
• Boolean Max-IP:
• For sets 𝐵 and 𝐶 with 𝑜 Boolean

vectors, find 𝑁𝑏𝑦𝐽𝑄 𝐵, 𝐶 ≔ max

𝑏,𝑐 ∈𝐵×𝐶⟨𝑏, 𝑐⟩ .

• Approx. version: find a r-

multiplicative approximation to the answer: 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ≤ 𝐵𝑀𝐻(𝐵, 𝐶) ≤ 𝑁𝑏𝑦𝐽𝑄(𝐵, 𝐶) ⋅ 𝑠.

• d = d(n) : vector dimensions
• r = r(n) : approx. ratio

New Merlin-Arthur Protocol for Set- Disjointness

• 3. A new MA Protocol for Set-Disjointness
• [AW’09]: An O( 𝑜 log 𝑜) MA protocol.
• [Kla’03]: Ω( 𝑜) Lower Bound.
• This work: an O( 𝑜 log 𝑜 log log 𝑜) protocol.
• MA Communication Protocol:
• Alice holds x, Bob holds y, want

to compute F(x,y).

• F(x,y) = 1 ⇒ exists a proof, Pr 𝑏𝑑𝑑 ≥

2 3 .

• F(x,y) = 0 ⇒ for all proofs, Pr 𝑏𝑑𝑑 ≤ 1

3 .

• Complexity = Proof Length +

Communication Ω( 𝑜) [Lower Bound] [Kla’03] O( 𝑜 log 𝑜) [Upper Bound] [AW’09] O( 𝑜 log 𝑜 log log 𝑜) [Upper Bound] [This work]

?

New Connection with Communication Complexity

• 4. New Connection with Communication Complexity
• [ARW’17]:
• 𝑜 log 𝑜 MA protocol for Set-Disjointness
• ⇒ SETH-Hardness for Approx. Boolean Max-IP.
• Open Question from [ARW’17]:
• There is a 𝑜 BQP protocol for Set-Disjointness. Does it also

imply some hardness results?

• [This work]: YES!
• 𝑜 BQP protocol for Set-Disjointness
• ⇒ SETH-Hardness for Approx. {−1,1}-Max-IP

Quantum! {−𝟐, 𝟐}-Max-IP: Given sets 𝐵 and 𝐶 of vectors with {−1,1} entries (each of size 𝑜) find 𝑏 in 𝐵 and 𝑐 in 𝐶 with maximum inner product.

Proof Overview: SETH-Hardness of Z-Max-IP

• Starting Point：SETH implies OV Conjecture.
• Orthogonal Vectors (OV) Problem:
• Given two sets A,B of Boolean vectors, find an orthogonal pair between them.

OV Conjecture: OV with sets of 𝑜 vectors, 𝜕(log 𝑜) dimensions requires 𝑜2−𝑝(1) time.

Our Goal: A “dimensionality” reduction from 𝜕(log 𝑜) dimensional OV to 2𝑃 log∗ 𝑜 dimensional Z-Max-IP

Reduction RoadMap

• First cover a baby version which shows 𝜕 log2 log 𝑜

dimensional Z-Max-IP is hard. (same as [Wil’18])

• Then outline the key ideas to get the 2𝑃 log∗ 𝑜 dimensional

hardness.

• An intermediate problem:
• Z-OV: Given two sets A,B of Integer vectors, find an orthogonal

pair between them. 𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Hard Easy Orthogonal Vectors (OV) Problem: Two sets A,B of Boolean vectors, find an

• rthogonal pair

between them. Z-Max-IP: Two sets of 𝑜 Integer vectors. find a pair between them which maximize their inner product.

Easy Part: Z-OV ⇒Z-Max-IP

• Implicit in [Wil’18].
• 𝑏, 𝑐 ∈ 𝑎𝑒. (Squaring trick)
• 𝑏 ⋅ 𝑐 = 0 ⇒ − 𝑏 ⋅ 𝑐 2= 0
• 𝑏 ⋅ 𝑐 ≠ 0 ⇒ − 𝑏 ⋅ 𝑐 2< 0
• To solve Z-OV, it suffices to calculate the maximum

value of − 𝑏 ⋅ 𝑐 2 for 𝑏, 𝑐 ∈ 𝐵 × 𝐶.

• − 𝑏 ⋅ 𝑐 2 = − σ𝑗 𝑏𝑗 ⋅ 𝑐𝑗 2 = − σ𝑗,𝑘 𝑏𝑗𝑏𝑘𝑐𝑗𝑐

𝑘

• ത

𝑏𝑗,𝑘 = 𝑏𝑗 ⋅ 𝑏𝑘, ത 𝑐𝑗,𝑘 = −𝑐𝑗 ⋅ 𝑐

𝑘.

• Maximize ത

𝑏 ⋅ ത 𝑐, Z-Max-IP with 𝑒2 dim.

Z-OV: Two sets A,B of Integer vectors, find an orthogonal pair between them. Z-Max-IP: Given sets 𝐵 and 𝐶

• f Integer vectors

(each of size n) find 𝑏 in 𝐵 and 𝑐 in 𝐶 with maximum inner product.

𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Hard Easy

Hard Part: OV ⇒Z-OV

Z-OV: Two sets A,B of Integer vectors, find an orthogonal pair between them. OV: Two sets A,B of Boolean vectors, find an orthogonal pair between them.

• Want to reduce the dimension:
• E.g. use few integers to represent a long Boolean vector
• Key Idea: Chinese Remainder Theorem (CRT)!
• 𝑢 primes 𝑟1, 𝑟2, ⋯ , 𝑟𝑢.
• 𝑢 remainders 𝑠

1, 𝑠 2, ⋯ , 𝑠 𝑢.

• CRT: exists a unique integer 0 ≤ 𝑦 < ς𝑗 𝑟𝑗, s.t.
• 𝑦 ≡ 𝑠

𝑗 (𝑛𝑝𝑒 𝑟𝑗).

Use a number to represent a vector

𝑦 𝑠

1, 𝑠2, ⋯ , 𝑠𝑢

𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Hard Easy

Chinese Remainder Theorem

𝑑𝑠𝑠 𝑠

1, 𝑠2, ⋯ , 𝑠𝑢 ≔

the unique integer 0 ≤ 𝑦 < ς𝑗 𝑟𝑗, s.t. 𝑦 ≡ 𝑠𝑗 (𝑛𝑝𝑒 𝑟𝑗). crr: Chinese Remainder Representation

• Fix 𝑢 primes 𝑟1, 𝑟2, ⋯ , 𝑟𝑢.
• 𝑏 = 𝑑𝑠𝑠(𝑦1, 𝑦2, ⋯ , 𝑦𝑢), i.e. 𝑏 ≡ 𝑦𝑗(𝑛𝑝𝑒 𝑟𝑗).
• 𝑐 = 𝑑𝑠𝑠 𝑧1, 𝑧2, ⋯ , 𝑧𝑢 , i.e. b ≡ 𝑧𝑗(𝑛𝑝𝑒 𝑟𝑗).

𝑏 ⋅ 𝑐 ≡ 𝑦𝑗 ⋅ 𝑧𝑗 𝑛𝑝𝑒 𝑟𝑗 for 1 ≤ 𝑗 ≤ 𝑢.

Multiplication between Two integers Multiplication between Two vectors simulate

Exactly what we want!

𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Hard Easy

Hard Part: OV ⇒Z-OV

𝑦1,1 𝑦1,2 ⋯ 𝑦1,𝑐 ⋮ ⋱ ⋱ ⋮ 𝑦ℓ,1 𝑦ℓ,2 ⋯ 𝑦ℓ,𝑐 ℓ rows 𝑐 columns

• The reduction:
• d : vector dimension (𝑒 = 𝜕(log 𝑜)).
• ℓ ⋅ 𝑐 = 𝑒
• represent a vector 𝑦 in 0,1 d by a ℓ × 𝑐

table.

• Map each row into a single number using

Chinese Remainder Theorem

𝑑𝑠𝑠(𝑦1,1, 𝑦1,2, ⋯ , 𝑦1,𝑐) ⋮ 𝑑𝑠𝑠(𝑦ℓ,1, 𝑦ℓ,2, ⋯ , 𝑦ℓ,𝑐) 𝑦 ∈ 0,1 𝑒 𝜒(𝑦) ∈ 𝑎ℓ Want ℓ to be as small as possible 𝑑𝑠𝑠 𝑠

1, 𝑠2, ⋯ , 𝑠𝑢 ≔

the unique integer 0 ≤ 𝑦 < ς𝑗 𝑟𝑗, s.t. 𝑦 ≡ 𝑠𝑗 (𝑛𝑝𝑒 𝑟𝑗). crr: Chinese Remainder Representation

𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Hard Easy

Hard Part: OV ⇒Z-OV

𝑦1,1 𝑦1,2 ⋯ 𝑦1,𝑐 ⋮ ⋱ ⋱ ⋮ 𝑦ℓ,1 𝑦ℓ,2 ⋯ 𝑦ℓ,𝑐 ℓ rows 𝑐 columns

• The reduction:
• d : vector dimension
• ℓ ⋅ 𝑐 = 𝑒
• represent a vector 𝑦 in

0,1 d by a ℓ × 𝑐 table.

• Map each row into a

single number using Chinese Remainder Theorem 𝑑𝑠𝑠(𝑦1,1, 𝑦1,2, ⋯ , 𝑦1,𝑐) ⋮ 𝑑𝑠𝑠(𝑦ℓ,1, 𝑦ℓ,2, ⋯ , 𝑦ℓ,𝑐) 𝑦 ∈ 0,1 𝑒 𝜒(𝑦) ∈ 𝑎ℓ For 𝑗 ∈ [ℓ] and 𝑘 ∈ [𝑐], 𝜒 𝑦 𝑗 ≡ 𝑦𝑗,𝑘 (𝑛𝑝𝑒 𝑟𝑘) 𝜒 𝑦 ⋅ 𝜒 𝑧 𝑛𝑝𝑒 𝑟𝑘 = ෍

𝑗=1 ℓ

𝜒 𝑦 𝑗 ⋅ 𝜒 𝑧 𝑗 (𝑛𝑝𝑒 𝑟𝑘) = ෍

𝑗=1 ℓ

𝑦𝑗,𝑘 ⋅ 𝑧𝑗,𝑘 (𝑛𝑝𝑒 𝑟𝑘) The inner product of 𝑘-th column of 𝑦 and 𝑘-th column of 𝑧. Set all 𝑟𝑘 > ℓ. 𝑦 ⋅ 𝑧 = 0 ⇔ 𝜒 𝑦 ⋅ 𝜒 𝑧 ≡ 0 (𝑛𝑝𝑒 𝑟𝑘) for all 𝑘.

𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Hard Easy

Hard Part: OV ⇒Z-OV

𝜒 𝑦 ⋅ 𝜒 𝑧 𝑛𝑝𝑒 𝑟𝑘 = ෍

𝑗=1 ℓ

𝜒 𝑦 𝑗 ⋅ 𝜒 𝑧 𝑗 (𝑛𝑝𝑒 𝑟𝑘) = ෍

𝑗=1 ℓ

𝑦𝑗,𝑘 ⋅ 𝑧𝑗,𝑘 (𝑛𝑝𝑒 𝑟𝑘) The inner product of 𝑘-th column of 𝑦 and 𝑘-th column of 𝑧. Set all 𝑟𝑘 > ℓ. 𝑦 ⋅ 𝑧 = 0 ⇔ 𝜒 𝑦 ⋅ 𝜒 𝑧 ≡ 0 (𝑛𝑝𝑒 𝑟𝑘) for all 𝑘.

• 𝑊 ≔ all multiples of ς𝑘 𝑟𝑘 between 0 and ℓ ⋅ ς𝑘 𝑟𝑘

2.

• Given an OV instance with sets 𝐵 and 𝐶.
• 𝑦 ⋅ 𝑧 = 0 ⇔ 𝜒 𝑦 ⋅ 𝜒 𝑧 ≡ 0 (𝑛𝑝𝑒 𝑟𝑘) for all 𝑘 ⇔ 𝜒 𝑦 ⋅ 𝜒 𝑧 ∈ 𝑊.
• Let 𝑤 ∈ 𝑊,
• 𝜒 𝑦 ⋅ 𝜒 𝑧 = 𝑤 ⇔ [𝜒 𝑦 , −1] ⋅ 𝜒 𝑧 , 𝑤 = 0.
• 𝐵𝑤 ≔ { 𝜒 𝑦 , −1 : 𝑦 ∈ 𝐵}.
• 𝐶𝑤 ≔

𝜒 𝑧 , 𝑤 : 𝑧 ∈ 𝐶 .

• Therefore,
• There is an orthogonal pair between 𝐵 and 𝐶 ⇔
• There exists 𝑤 s.t. there is an orthogonal pair between 𝐵𝑤 and 𝐶𝑤.
• In summary:
• One 𝑒-dim. OV instance ⇒ |𝑊| instances of ℓ + 1 -dim. Z-OV

𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Hard Easy

Hard Part: OV ⇒Z-OV Informal Analysis

• 𝑊 ≔ all multiplier of ς𝑘 𝑟𝑘 between 0 and

ℓ ⋅ ς𝑘 𝑟𝑘

2.

• 𝐵𝑤 ≔ { 𝜒 𝑦 , −1 : 𝑦 ∈ 𝐵}.
• 𝐶𝑤 ≔

𝜒 𝑧 , 𝑤 : 𝑧 ∈ 𝐶 .

• Therefore,
• There is an orthogonal pair between 𝐵

and 𝐶 ⇔

• There exists 𝑤 s.t. there is an
• rthogonal pair between 𝐵𝑤 and 𝐶𝑤.
• In summary:
• One 𝑒-dim. OV instance ⇒ |𝑊|

instances of ℓ + 1 -dim. Z-OV

• Recall what we have: 𝜕(log 𝑜)-dim. OV requires 𝑜2−𝑝(1) time.
• To preserve hardness, want 𝑊 = 𝑜𝑝 1 .
• 𝑊 = ς𝑘 𝑟𝑘

𝑃 1 = 𝑐O(𝑐).

• Have to set 𝑐 = 𝑝

log 𝑜 log log 𝑜 .

• Therefore, ℓ =

𝜕 log 𝑜 𝑐

= 𝜕(log log 𝑜).

• Q.E.D.

𝜕 log 𝑜 -dim. OV 𝜕 log log 𝑜 -dim. Z-OV 𝜕 log2 log 𝑜 -dim. Z-Max-IP

Easy

2𝑃 log∗ 𝑜 dim. Hardness: Sketch

• What is the bottleneck?
• Not enough small primes!
• 𝑟𝑗

′𝑡 are 𝑐 distinct primes, most of them ≫ 𝑐,

even if we only need them to be > ℓ.

• Idea: Use another CRT to embed small primes

inside big primes.

• Recursive: Then pack even smaller primes

inside small primes, and recurse.

• Pretend we have many small primes, even

though we don’t’.

• Recall what we have: 𝜕(log 𝑜)-dim.

OV requires 𝑜2−𝑝(1) time.

• To preserve hardness, want 𝑊 =

𝑜𝑝 1 .

• 𝑊 = ς𝑘 𝑟𝑘

𝑃 1 = 𝑐O(𝑐).

• Have to set 𝑐 = 𝑝

log 𝑜 log log 𝑜 .

• Therefore, ℓ = 𝜕 log 𝑜

𝑐

= 𝜕(log log 𝑜).

• Q.E.D.

• Pick 𝑥 primes 𝑞1, 𝑞2, ⋯ , 𝑞𝑥, such that ς𝑘 𝑞𝑘

2 ⋅ ℓ < 𝑐.

Ԧ 𝑦1,1 Ԧ 𝑦1,2 ⋯ Ԧ 𝑦1,𝑐/𝑥 ⋮ ⋱ ⋱ ⋮ Ԧ 𝑦ℓ,1 Ԧ 𝑦ℓ,2 ⋯ Ԧ 𝑦ℓ,𝑐/𝑥 ℓ rows 𝑑𝑠𝑠(𝑑𝑠𝑠

𝑛( Ԧ

𝑦1,1), 𝑑𝑠𝑠

𝑛( Ԧ

𝑦1,2), ⋯ , 𝑑𝑠𝑠

𝑛( Ԧ

𝑦1,𝑐/𝑥)) ⋮ 𝑑𝑠𝑠(𝑑𝑠𝑠

𝑛( Ԧ

𝑦ℓ,1), 𝑑𝑠𝑠

𝑛( Ԧ

𝑦ℓ,2), ⋯ , 𝑑𝑠𝑠

𝑛( Ԧ

𝑦ℓ,𝑐/𝑥)) 𝑦 ∈ 0,1 𝑒 𝜒(𝑦) ∈ 𝑎ℓ 𝑐/𝑥 columns

• d : vector dimension (𝑒 = 𝜕(log 𝑜))
• ℓ ⋅ 𝑐 = 𝑒 (want ℓ as small as possible)
• represent a vector 𝑦 in 0,1 d by a ℓ × (𝑐/𝑥) table, each

entry is a vector in 0,1 𝑥.

inner 𝑑𝑠𝑠

𝑛: CRT w.r.t small primes 𝑞𝑗 ′𝑡

• uter 𝑑𝑠𝑠: CRT w.r.t. big primes 𝑟𝑗’s, 𝑟𝑗 > 𝑐.

One Step of Recursion

Ԧ 𝑦1,1 Ԧ 𝑦1,2 ⋯ Ԧ 𝑦1,𝑐/𝑥 ⋮ ⋱ ⋱ ⋮ Ԧ 𝑦ℓ,1 Ԧ 𝑦ℓ,2 ⋯ Ԧ 𝑦ℓ,𝑐/𝑥 ℓ rows 𝑑𝑠𝑠(𝑑𝑠𝑠

𝑛( Ԧ

𝑦1,1), 𝑑𝑠𝑠

𝑛( Ԧ

𝑦1,2), ⋯ , 𝑑𝑠𝑠

𝑛( Ԧ

𝑦1,𝑐/𝑥)) ⋮ 𝑑𝑠𝑠(𝑑𝑠𝑠

𝑛( Ԧ

𝑦ℓ,1), 𝑑𝑠𝑠

𝑛( Ԧ

𝑦ℓ,2), ⋯ , 𝑑𝑠𝑠

𝑛( Ԧ

𝑦ℓ,𝑐/𝑥)) 𝑦 ∈ 0,1 𝑒 𝜒(𝑦) ∈ 𝑎ℓ 𝑐/𝑥 columns inner 𝑑𝑠𝑠

𝑛: CRT w.r.t small primes 𝑞𝑗 ′𝑡

• uter crr: CRT w.r.t. big primes 𝑟𝑗’s, 𝑟𝑗 > 𝑐.

One Step of Recursion

For 𝑗 ∈ [ℓ] and 𝑘 ∈ [𝑐/𝑥], 𝜒 𝑦 𝑗 ≡ 𝑑𝑠𝑠

𝑛( Ԧ

𝑦𝑗,𝑘) (𝑛𝑝𝑒 𝑟𝑘) 𝜒 𝑦 ⋅ 𝜒 𝑧 𝑛𝑝𝑒 𝑟𝑘 = ෍

𝑗=1 ℓ

𝜒 𝑦 𝑗 ⋅ 𝜒 𝑧 𝑗 (𝑛𝑝𝑒 𝑟𝑘) = ෍

𝑗=1 ℓ

𝑑𝑠𝑠

𝑛( Ԧ

𝑦𝑗,𝑘) ⋅ 𝑑𝑠𝑠

𝑛( Ԧ

𝑧𝑗,𝑘) (𝑛𝑝𝑒 𝑟𝑘) Get to know: ෍

𝑗=1 ℓ

𝑑𝑠𝑠

𝑛( Ԧ

𝑦𝑗,𝑘) ⋅ 𝑑𝑠𝑠

𝑛( Ԧ

𝑧𝑗,𝑘) and whether for all 𝑗, Ԧ 𝑦𝑗,𝑘 ⋅ Ԧ 𝑧𝑗,𝑘 = 0. ς𝑘 𝑞𝑘

2 ⋅ ℓ < 𝑐 ⇒ σ𝑗=1 ℓ

𝑑𝑠𝑠

𝑛( Ԧ

𝑦𝑗,𝑘) ⋅ 𝑑𝑠𝑠

𝑛( Ԧ

𝑧𝑗,𝑘) < 𝑟𝑘.

One Step of Recursion: Informal Analysis

𝜒 𝑦 ⋅ 𝜒 𝑧 𝑛𝑝𝑒 𝑟𝑘 = ෍

𝑗=1 ℓ

𝜒 𝑦 𝑗 ⋅ 𝜒 𝑧 𝑗 (𝑛𝑝𝑒 𝑟𝑘) = ෍

𝑗=1 ℓ

𝑑𝑠𝑠

𝑛( Ԧ

𝑦𝑗,𝑘) ⋅ 𝑑𝑠𝑠

𝑛( Ԧ

𝑧𝑗,𝑘) (𝑛𝑝𝑒 𝑟𝑘) Get to know: ෍

𝑗=1 ℓ

𝑑𝑠𝑠

𝑛( Ԧ

𝑦𝑗,𝑘) ⋅ 𝑑𝑠𝑠

𝑛( Ԧ

𝑧𝑗,𝑘) and whether for all 𝑗, Ԧ 𝑦𝑗,𝑘 ⋅ Ԧ 𝑧𝑗,𝑘 = 0.

• Set ς𝑘 𝑞𝑘

2 ⋅ ℓ = 𝑥𝑃(𝑥) = 𝑐

⇒ w = Θ(log 𝑐 / log log 𝑐).

• 𝑊 = ς𝑘 𝑟𝑘

𝑃 1 = 𝑐O(𝑐/𝑥) = log 𝑐 𝑃(𝑐).

• Want 𝑊 = 𝑜𝑝(1), set 𝑐 = 𝑝(log 𝑜/ log log log 𝑜).
• Therefore, ℓ = 𝜕 log 𝑜

𝑐

= 𝜕(log log log 𝑜).

• Improvement!
• A recursive construction leads

to the final 2𝑃(log∗ 𝑜) dim. hardness.

ς𝑘 𝑞𝑘

2 ⋅ ℓ < 𝑐 ⇒ σ𝑗=1 ℓ

𝑑𝑠𝑠

𝑛( Ԧ

𝑦𝑗,𝑘) ⋅ 𝑑𝑠𝑠

𝑛( Ԧ

𝑧𝑗,𝑘) < 𝑟𝑘.

Open Questions

• Construct an O( 𝑜)-bit MA protocol for Set-Disjointness.
• Show that Z-Max-IP for any 𝜕 1 dimensions requires 𝑜2−𝑝(1) time

under some plausible hypothesis.

• Implies same hardness for 𝜕(1) dimensions ℓ2-Furthest Pair
• NP⋅UPP communication protocol: a potential approach
• NP⋅UPP: a relaxation of MA, where Arthur's error can be arbitrary close to 0.5.
• Our results can be interpreted as a sub-linear proof length, O(log∗ 𝑜)

communication NP⋅UPP protocol for Set-Disjointness.

• O(log∗ 𝑜) to 𝛽 𝑜 ⇒ Z-Max-IP for 2𝛽(𝑜) dim. requires 𝑜2−𝑝(1) under SETH.

Thanks

Any Questions?