Stronger Connections Between Circuit Analysis and Circuit Lower - - PowerPoint PPT Presentation

Stronger Connections Between Circuit Analysis and Circuit Lower Bounds, via PCPs of Proximity

Lijie Chen Ryan Williams

Context: The Algorithmic Method for Proving Circuit Lower Bounds

Proving limitations on non-uniform circuits is extremely hard. Prior approaches (restrictions, polynomial approximations, etc.) face barriers (Relativization, Algebrization, Natural Proofs).

Algorithmic Method

• Non-trivial circuit-analysis algorithm ⇒ Circuit Lower Bounds.
• Breakthroughs where previous approaches failed (NEXP ⊄ ACC0).
• Believed to be possible for strong circuits (even 𝑄/𝑞𝑝𝑚𝑧).

Context: A Frontier of Circuit Complexity, Depth-2 Threshold Circuits

THR gates : 𝑔 𝑦 = 𝑥 ⋅ 𝑦 ≥ 𝑢 𝑥 ∈ 𝑎𝑜, 𝑢 ∈ 𝑎. MAJ gates : when 𝑥𝑗’s and 𝑢 are bounded by poly(n). THR∘THR

THR THR THR THR

We can also define 𝑼𝑰𝑺 ∘ 𝑵𝑩𝑲 𝑵𝑩𝑲 ∘ 𝑼𝑰𝑺 𝑵𝑩𝑲 ∘ 𝑵𝑩𝑲

Context: A Frontier of Circuit Complexity, Depth-2 Threshold Circuits

Exponential Lower Bounds are known for 𝑁𝐵𝐾 ∘ 𝑁𝐵𝐾 [Hajnal-Maass-Pudlák-Szegedy-Turán’93] 𝑁𝐵𝐾 ∘ 𝑈𝐼𝑆 [Nisan’94] 𝑈𝐼𝑆 ∘ 𝑁𝐵𝐾 [Forster-Krause-Lokam-Mubarakzjanov-

Schmitt-Simon’01]

Frontier Open Question: Is NEXP ⊆ 𝑼𝑰𝑺 ∘ 𝑼𝑰𝑺? Potential Approaches in this talk.

NEXP Non-deterministic Exponential Time.

Motivation: Apply the Algorithmic Method to THR of THR?

ℭ-SAT

𝐷 ∃𝑦 ?

∃ x s.t. 𝐷 𝑦 = 1?

ℭ-CAPP

𝐷

Estimate quantity Pr

𝑦∼𝑉𝑜[𝐷 𝑦 = 1],

with additive error 𝜁

𝑦 ∼ 𝑉𝑜 What Circuit-Analysis Tasks?

Non-trivial Circuit- Analysis Algorithms ⇒Circuit Lower Bounds

Derandomization!!

𝜁: constant or inverse polynomial

2𝑜/𝑜𝜕(1) time?

Most previous work on the algorithmic method exploits SAT algorithms.

Problem

SAT of THR of THR is probably very hard. A special case is MAX-𝑙-SAT, for which no non- trivial (2𝑜/𝑜𝜕(1) time) algorithm is known for 𝒍 = 𝝏(log 𝒐) and 𝑞𝑝𝑚𝑧(𝑜) clauses. Considered to be a barrier for the Algorithmic Approach.

THR∘THR

THR THR THR THR

MAX-𝑙-SAT

MAJ 𝑃𝑆𝑙 𝑃𝑆𝑙 𝑃𝑆𝑙

Motivation: Apply the Algorithmic Method to THR of THR?

SAT of THR of THR : probably very hard But derandomization is widely believed to be possible.

From Derandomization (CAPP) ⇒ Circuit Lower Bounds For a circuit class ℭ,

• 2𝑜/𝑜𝜕(1)-time CAPP for (𝐁𝐎𝐄𝐪𝐩𝐦𝐳(𝒐) ∘ 𝐏𝐒𝟒 ∘ 𝕯)

⇒ 𝑂𝐹𝑌𝑄 ⊄ ℭ [Williams’13/14, Santhanam Williams’14, Ben-Sasson Viola’14]

• 2𝑜/𝑜𝜕(1)-time CAPP for (𝑩𝑫𝟏 ∘ 𝕯)

⇒ 𝑂𝐹𝑌𝑄 can’t be

1 2 + 𝑝(1)-approximated by ℭ [R. Chen Oliveira

Santhanam’18]

• 2𝑜−𝑜𝜁-time CAPP for (𝐁𝐎𝐄𝐪𝐩𝐦𝐳(𝒐) ∘ 𝐏𝐒𝟒 ∘ 𝕯)

⇒ 𝑂𝑅𝑄 ⊄ ℭ [Murray Williams’18]

• 2𝑜−𝑜𝜁-time CAPP for (𝐁𝐃𝟏 ∘ 𝕯)

⇒ 𝑂𝑅𝑄 can’t be

1 2 + 𝑝(1)-approximated by ℭ [L. Chen’19]

NQP Non-deterministic Quasi-Polynomial

• Time. (𝒐𝒒𝒑𝒎𝒛𝒎𝒑𝒉(𝒐))

Motivation: Apply the Algorithmic Method to THR of THR?

Back to THR of THR

SAT of THR of THR : probably very hard To show 𝑂𝐹𝑌𝑄 ⊄ 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆, we need to derandomize ANDpoly(𝑜) ∘ OR3 ∘ 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆, which could be harder. Our result 1 It suffices to derandomize 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆. Our result 2 Surprisingly, it indeed only suffices to derandomize 𝑈𝐼𝑆 ∘ 𝑁𝐵𝐾 or 𝑁𝐵𝐾 ∘ 𝑁𝐵𝐾!

General Result: A Stronger Connection Between Circuit-Analysis Algorithms and Circuit Lower Bounds

For a circuit class ℭ:

• 𝟑𝒐/𝒐𝝏(𝟐)-time CAPP for ⊕2∘ ℭ, 𝐵𝑂𝐸2 ∘ ℭ, or 𝑃𝑆2 ∘ ℭ

⇒ 𝑂𝐹𝑌𝑄 ⊄ ℭ.

• 𝟑𝒐−𝒐𝜻-time CAPP for ⊕2∘ ℭ, 𝐵𝑂𝐸2 ∘ ℭ, or 𝑃𝑆2 ∘ ℭ

⇒ 𝑂𝑅𝑄 ⊄ ℭ.

Why the constant “2”?

• Short answer: A PCP system needs

to make at least 2 queries.

• Long answer: See the paper☺

Tighter Connections for Algorithms/Lower Bounds for THR of THR

Luckily, the “2” doesn’t matter for 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆 ☺

⊕𝟑∘ 𝑼𝑰𝑺 ∘ 𝑼𝑰𝑺 ⊆ 𝑼𝑰𝑺 ∘ 𝑼𝑰𝑺

2𝑜/𝑜𝜕(1)-time CAPP algorithm for 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆 ⇒ 𝑂𝐹𝑌𝑄 ⊄ 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆. 2𝑜/𝑜𝜕(1)-time CAPP algorithm for 𝑈𝐷𝑒 ⇒ 𝑂𝐹𝑌𝑄 ⊄ 𝑈𝐷𝑒. 𝑼𝑫𝒆: depth-d, poly-size, linear threshold circuits

Let Us Make Our Life Even Easier

THR THR THR THR MAJ MAJ MAJ MAJ

Poly-size 𝑼𝑰𝑺 ∘ 𝑼𝑰𝑺 and 𝑵𝑩𝑲 ∘ 𝑵𝑩𝑲 are equivalent for Non-Trivial (2𝑜/𝑜𝜕(1) time) CAPP Algorithms when 𝜁 = 1/𝑞𝑝𝑚𝑧 𝑜 !

Proved by new structure lemmas for 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆

Let Us Make Our Life Even Easier

THR THR THR THR THR MAJ MAJ MAJ

Poly-size 𝑼𝑰𝑺 ∘ 𝑼𝑰𝑺 and 𝑼𝑰𝑺 ∘ 𝑵𝑩𝑲 are equivalent for Non-Trivial (2𝑜/𝑜𝜕(1) time) CAPP Algorithms for any constant 𝜁 > 0!

Proved by new structure lemmas for 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆

Corollary

If there are

2𝑜/𝑜𝜕(1)-time CAPP for 𝑁𝐵𝐾 ∘ 𝑁𝐵𝐾 with 𝜁 = 1/𝑞𝑝𝑚𝑧(𝑜), or a 2𝑜/𝑜𝜕(1)-time CAPP for 𝑈𝐼𝑆 ∘ 𝑁𝐵𝐾 with constant 𝜁,

then 𝑶𝑭𝒀𝑸 ⊄ 𝑼𝑰𝑺 ∘ 𝑼𝑰𝑺.

Another Application: Inapproximability by Depth-2 Neural Networks

Depth-2 Neural Network

∑ THR THR THR

𝑥1 𝑥2 𝑥3

∑ ReLU ReLU ReLU

𝑥1 𝑥2 𝑥3

Thm For every 𝑙 and constant 𝜀 < 1/2, there is a function 𝑔 ∈ 𝑂𝑄 such that 𝑔 cannot be 𝜀 approximated by Depth-2 Neural Networks of size 𝑜𝑙 Improved [Wil’18], which proved that there is such an 𝑔 ∈ 𝑂𝑄 which cannot be exactly computed by Depth-2 Neural Networks of size 𝑜𝑙.

𝑂 𝑦 ≔ ෍

𝑗

𝑥𝑗 ⋅ 𝑈𝐼𝑆𝑗 𝑦 ∈ ℝ 𝑂 𝑦 ≔ ෍

𝑗

𝑥𝑗 ⋅ 𝑆𝑓𝑀𝑉𝑗 𝑦 ∈ ℝ

Philosophy

Using PCP Algorithmically to Prove Circuit Lower Bounds (Remember: PCPs are algorithms!)

If you want to prove 𝑸 = 𝑶𝑸, then PCPs should make your life much easier (now you only need an algorithm for (

𝟖 𝟗 + 𝜻)-

approximation to 3-SAT!) [Håstad’97]

(Well, I don’t really believe in 𝑄 = 𝑂𝑄.) We only want to derandomize circuits. But PCPs still make our life easier (though in a more indirect way)

Starting Point: Non-deterministic Derandomization Suffices for Circuit Lower Bounds

ℭ-GAP-TAUT (tautology)

𝐷

Distinguish between

• 1. Pr

𝑦 [𝐷 𝑦 = 1] = 1.

(Yes Case)

• 2. Pr

𝑦 [𝐷 𝑦 = 1] ≤ 𝜁.

(No Case)

𝑦 ∼ 𝑉𝑜 [Wil’13] 2𝑜/𝑜𝜕(1) time non-deterministic algorithm for GAP-TAUT

• n poly-size general

circuits with 𝜁 = 1/2 ⇒ 𝑂𝐹𝑌𝑄 ⊄ 𝑄/𝑞𝑝𝑚𝑧.

Non-deterministic Algorithm for GAP-TAUT

Given a general circuit 𝐷, we want a 2𝑜/𝑜𝜕(1) time non-deterministic algo 𝔹, such that:

• 1. If 𝐷 is a tautology, then 𝔹 accepts on some guesses.
• 2. If Pr

𝑦 𝐷 𝑦 = 1 ≤ 1/2, 𝔹 rejects on all guesses.

Proof Overview: Outline

Assume 𝑂𝐹𝑌𝑄 ⊂ ℭ Non-trivial CAPP on OR3 ∘ ℭ with constant 𝜁 2𝑜/𝑜𝜕(1) non- deterministic GAP- TAUT for 𝑄/𝑞𝑝𝑚𝑧 𝑂𝐹𝑌𝑄 ⊄ 𝑄/𝑞𝑝𝑚𝑧 ⇒ 𝑂𝐹𝑌𝑄 ⊄ ℭ Contradiction!

Starting Point [Wil’13]

2𝑜/𝑜𝜕(1) time non-deterministic algorithm for GAP-TAUT

• n poly-size general circuits with 𝜁 = 1/2

⇒ 𝑂𝐹𝑌𝑄 ⊄ 𝑄/𝑞𝑝𝑚𝑧. Key point: make use of this assumption as much as possible!

Think of ℭ as 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆

Goal: Designing the Algorithm under Assumption

Goal

Given an 𝑂𝐵𝑂𝐸 circuit 𝐷, under the two assumptions, design a 2𝑜/𝑜𝜕(1) time non-deterministic algo 𝔹, such that:

• 1. If 𝐷 is a tautology, then 𝔹 accepts on some guesses.
• 2. If Pr

𝑦 𝐷 𝑦 = 1 ≤ 1/2, 𝔹 rejects on all guesses.

𝑂𝐵𝑂𝐸 𝑦, 𝑧 ≔ 𝑂𝑃𝑈(𝐵𝑂𝐸(𝑦, 𝑧)) It is universal

Assume 𝑂𝐹𝑌𝑄 ⊂ ℭ Non-trivial CAPP on OR3 ∘ ℭ with constant 𝜁 2𝑜/𝑜𝜕(1) non-deterministic GAP-TAUT

• n 𝑄/𝑞𝑝𝑚𝑧

Think of ℭ as 𝑈𝐼𝑆 ∘ 𝑈𝐼𝑆

Review: Approach of [Wil’14] Guess-and-Verify-Equivalence

𝑂𝐹𝑌𝑄 ⊂ ℭ implies 𝑄/𝑞𝑝𝑚𝑧 collapses to ℭ. That is, under assumption, the given general circuit 𝐷 has an equivalent 𝕯 circuit 𝑬. If we can find 𝑬, then we can derandomize 𝐸 instead, where we have algorithms!

Problem: How to find 𝑬?

Allowed to use non-determinism so one can guess 𝐸. But still have to verify 𝐸 is equivalent to 𝐷, which seems HARD.

Solution

Well, just guess more

circuits!

Review: Approach of [Wil’14] Guess-and-Verify-Equivalence

Suppose 𝐷 has 𝑛 gates, let 𝐷1, 𝐷2, ⋯ , 𝐷𝑛 be the corresponding sub-circuits.

• 1. 𝐷𝑛 is the output gate.
• 2. 𝐷1, ⋯ , 𝐷𝑜 are inputs.

𝑂𝐹𝑌𝑄 ⊂ ℭ implies 𝑄/𝑞𝑝𝑚𝑧 collapses to ℭ. We guess ℭ circuits 𝐸1, 𝐸2, ⋯ , 𝐸𝑛, hoping that 𝐸𝑗 ≡ 𝐷𝑗.

We wish to check 𝐸𝑛 ≡ 𝐷 ≡ 𝐷𝑛. To do this, for each 𝑗 ∈ {𝑜 + 1, 𝑜 + 2, ⋯ , 𝑛}, suppose gate-𝑗 has inputs from gate-𝑗1 and gate-𝑗2. We verify 𝑶𝑩𝑶𝑬 𝑬𝒋𝟐 𝒚 , 𝑬𝒋𝟑 𝒚 ≡ 𝑬𝒋 𝒚 . Then run CAPP on 𝐸𝑛. Problem

Checking 𝑶𝑩𝑶𝑬 𝑬𝒋𝟐 𝒚 , 𝑬𝒋𝟑 𝒚 = 𝑬𝒋 𝒚 for all 𝒚 requires solving SAT for 𝑩𝑶𝑬𝟒 ∘ 𝕯.

A Local-checkable Proof System View

Problem: the previous approach requires solving SAT for 𝐵𝑂𝐸3 ∘ ℭ.

Let 𝜌 𝑦 ≔ 𝐸𝑜+1 𝑦 , 𝐸𝑜+2 𝑦 , ⋯ , 𝐸𝑛 𝑦 . This is a Claimed Proof for 𝐷𝑛 𝑦 = 1 by giving values at all gates. Intuitively, it is supposed to be the computation history of 𝑫 on input 𝒚.

Local checks on 𝒚 ∘ 𝝆(𝒚)

• For each 𝑗 ∈ {𝑜 + 1, 𝑜 + 2, ⋯ , 𝑛},

𝑂𝐵𝑂𝐸 𝐸𝑗1 𝑦 , 𝐸𝑗2 𝑦 = 𝐸𝑗(𝑦).

• 𝐸𝑛 𝑦 = 1.

What is so good about this proof 𝜌(𝑦)?

A Local-checkable Proof System View

Let 𝜌 𝑦 ≔ 𝐸𝑜+1 𝑦 , 𝐸𝑜+2 𝑦 , ⋯ , 𝐸𝑛 𝑦 . A Claimed Proof for 𝐷𝑛 𝑦 = 1 by giving values at all gates. One can get ℓ = 𝑃 𝑛 = 𝑞𝑝𝑚𝑧(𝑜) functions 𝐺

1, 𝐺2, ⋯ , 𝐺ℓ on

𝑦 ∘ 𝜌(𝑦), such that

• Each 𝐺𝑗 is an 𝑷𝑺 of 3 bits (or their negations) from 𝑦 ∘ 𝜌(𝑦).
• If 𝐷 𝑦 = 1, on the correct guesses 𝐸𝑜+1, ⋯ , 𝐸𝑛, all 𝐺𝑗’s are

satisfied by 𝑦 ∘ 𝜌 𝑦 . (Completeness)

• If 𝐷 𝑦 = 0, for all possible 𝜌 𝑦 , at least one 𝐺𝑗 is not

satisfied by 𝑦 ∘ 𝜌 𝑦 . (Soundness)

Guess circuits 𝐸𝑜+1, , ⋯ , 𝐸𝑛, let 𝜌 𝑦 ≔ 𝐸𝑜+1 𝑦 , 𝐸𝑜+2 𝑦 , ⋯ , 𝐸𝑛 𝑦 . Estimate 𝔽𝑗∈[ℓ]𝔽𝑦[𝐺𝑗(𝑦 ∘ 𝜌(𝑦))]. (𝐺𝑗 𝑦 ∘ 𝜌 𝑦 ∈ 𝑃𝑆3 ∘ ℭ.) (ℓ:number of 𝐺𝑗’s)

• If 𝐷 is a tautology. Then on the correct guess,

𝔽𝑗∈[ℓ]𝔽𝑦 𝐺𝑗 𝑦 ∘ 𝜌 𝑦 = 1.

• If Pr

𝑦 𝐷 𝑦 = 1 ≤ 1/2, then on all guesses,

𝔽𝑗∈[ℓ]𝔽𝑦 𝐺𝑗 𝑦 ∘ 𝜌 𝑦 ≤ 1 −

1 2ℓ.

An Attempt

• To distinguish the above two cases, we need a CAPP algo with

error

1 2ℓ = 1 𝑞𝑝𝑚𝑧(𝑜).

• But we only assume a CAPP algo with constant error!

What Went Wrong?

One can get ℓ = 𝑃(𝑛) functions 𝐺

1, 𝐺 2, ⋯ , 𝐺 ℓ on 𝑦 ∘ 𝜌(𝑦), such that

• Each 𝐺

𝑗 is an 𝑃𝑆 of 3 bits (or their negations) from 𝑦 ∘ 𝜌(𝑦).

• If 𝐷 𝑦 = 1, on the correct guess 𝐸𝑜+1, ⋯ , 𝐸𝑛, all 𝐺

𝑗’s are satisfied by 𝑦 ∘ 𝜌 𝑦 .

(Completeness is 1)

• If 𝐷 𝑦 = 0, for all possible 𝜌 𝑦 , at least one 𝐺

𝑗 is not satisfied by 𝑦 ∘ 𝜌 𝑦 .

(Soundness is 𝟐 − 𝟐/ℓ)

This is an extremely ``bad’’ PCP! Why not just use the PCP theorem?

If there is a verifier who picks a random 𝑗 ∈ ℓ , and checks whether 𝐺

𝑗 𝑦 ∘ 𝜌 𝑦

= 1. She detects an error only with probability 𝟐/ℓ when 𝐷 𝑦 = 0.

Proof System Vie iew 𝜌(𝑦) : a claimed proof of 𝐷 𝑦 = 1 𝐺𝑗 : local check of the verifier

Issues When Applying PCPs Directly

Recall that in the end we want to estimate 𝔽𝑗∈[ℓ]𝔽𝑦 𝐺𝑗 𝑦 ∘ 𝜌 𝑦 . Key properties being used in previous attempt:

These local checks 𝑮𝒋 (verifier’s queries positions) do not depend on the input 𝒚!

Use PCPs of Proximity!

Like PCPs but both input and proof are given as oracles. PCPs V 𝑦 (input) 𝜌(𝑦) (proof)

Unlimited access 3 queries

Now, 𝐺

𝑗(𝑦 ∘ 𝜌(𝑦)) can depend on many bits of 𝑦.

PCPs of Proximity V 𝑦 (input) 𝜌(𝑦) (proof)

3 queries in total

𝐺𝑗 𝑦 ∘ 𝜌 𝑦 ∈ 𝑃𝑆3 ∘ ℭ

Issues When Applying PCP Directly

Therefore, we want a proof system for verifying 𝐷 𝑦 = 1, such that given the random bits, verifier 𝑊 queries both input 𝒚 and proof 𝝆(𝒚).

• 1. If 𝐷 𝑦 = 1, ∃ 𝜌(𝑦), such that 𝑊𝑦∘𝜌(𝑦) always accept.
• 2. If 𝐷 𝑦 = 0, ∀ 𝜌(𝑦), 𝑊𝑦∘𝜌(𝑦)rejects w.h.p.

Counter-example?

Suppose 𝐷 computes the parity. Parity changes if we flip a random bit of 𝑦. The verifier can’t distinguish unless she queried that bit.

Solution

Give 𝑊 access to an error correcting code of 𝑦!

Combing PCP of Proximity and ECCs

PCP of Proximity

Verifier 𝑊 is given both the input (𝒚) and the proof 𝝆(𝒚) as oracles and makes 3 queries.

• 𝑊𝑦∘𝜌(𝑦) accepts w.p. 1, when 𝐷 𝑦 = 1;
• 𝑊𝑦∘𝜌(𝑦) accepts w.p. 𝜀 < 1, when 𝑦 makes 𝐷 robustly output 𝟏

(𝑫 is zero in a small hamming ball around 𝑦). (like property testing)

How it avoids the parity counter example? No inputs can make parity robustly output 𝟏! V

𝑦 (input) 𝜌(𝑦) (proof)

3 queries in total

PCP of Proximity with ECCs

Verifier 𝑊 is given both the encoded input (𝐹𝐷𝐷(𝑦)) and the proof 𝜌(𝑦) as oracles and makes 3 queries.

• 𝑊𝐹𝐷𝐷(𝑦)∘𝜌(𝑦) accepts w.p. 1, when 𝐷 𝑦 = 1;
• 𝑊𝐹𝐷𝐷(𝑦)∘𝜌(𝑦) accepts w.p. 𝜺 < 𝟐, when 𝐷 𝑦 = 0.

Use 𝑸𝑫𝑸 of Proximity for verifying 𝑭 𝒛 ≔ 𝑫 𝑬𝑭𝑫 𝒛 = 𝟐, 𝐹𝐷𝐷(𝑦) makes 𝐹(⋅) robustly output 𝟏 when 𝐷 𝑦 = 0! DEC(corrupted 𝐹𝐷𝐷(𝑦)) is still 𝑦 ☺

Final Algorithm

Estimate 𝔽𝑗∈[ℓ]𝔽𝑦[𝐺𝑗(𝐹𝐷𝐷(𝑦) ∘ 𝜌(𝑦))]. (𝐺𝑗 𝑦 ∘ 𝜌 𝑦 ∈ 𝑃𝑆3 ∘ ℭ.).

• If 𝐷 is a tautology. Then on the correct guesses,

𝔽𝑗∈[ℓ]𝔽𝑦 𝐺𝑗 𝑦 ∘ 𝜌 𝑦 = 1.

• If Pr

𝑦 𝐷 𝑦 = 1 ≤ 1/2, then on all guesses,

𝔽𝑗∈[ℓ]𝔽𝑦 𝐺𝑗 𝑦 ∘ 𝜌 𝑦 ≤ 1 − 𝜀/2. Guess circuits 𝐸𝑜+1, , ⋯ , 𝐸𝑛, let 𝜌 𝑦 ≔ 𝐸𝑜+1 𝑦 , 𝐸𝑜+2 𝑦 , ⋯ , 𝐸𝑛 𝑦 . Fix 𝐹𝐷𝐷 to be 𝔾2-linear. That is, 𝐹𝐷𝐷 𝑦 𝑗 is a parity on a subset of bits in 𝑦. Suppose there is uniform parity circuit in ℭ for now (this assumption can be avoided) Now constant error CAPP algo for 𝑃𝑆3 ∘ ℭ suffices!

Future Work

NEW Building on the PCPP based approach, [Alman Chen’19] give a construction of Razborov- rigid matrices in 𝑄𝑂𝑄. Can we find non-trivial CAPP algorithms for 𝑼𝑰𝑺 ∘ 𝑵𝑩𝑲 or 𝑵𝑩𝑲 ∘ 𝑵𝑩𝑲 to prove circuit lower bounds for 𝑼𝑰𝑺 ∘ 𝑼𝑰𝑺? Recall: we know exponential lower bounds for these two models! Can we ``mine’’ some algorithms from these proofs?

Thank You