Complexity Theory of Polynomial-Time Problems Lecture 11: - - PowerPoint PPT Presentation

Complexity Theory of Polynomial-Time Problems

Karl Bringmann

Lecture 11: Nondeterministic SETH

Complexity Inside P

SAT 2n EDIT n2 LCS n2 Fréchet n2 diameter n2 OV n2 Colinearity n2 Negative Triangle n3 Radius n3 3SUM-hard APSP equivalent SETH-hard APSP n3 3SUM n2 BMM 𝑜" SlidingWindowHD 𝑜" BMM-hard

can we relate SAT / 3SUM / APSP / BMM?

Relating Hypotheses

• nly very weak relations are known

e.g. ETH implies that k-SUM has no 𝑜#(%) algorithm today we will see a barrier for tighter connections OPEN: SETH implies that 3SUM has no 𝑃(𝑜()*) algorithm ?

• I. Nondeterministic SETH

Nondeterministic Algorithms

Turing machine: can choose any applicable transition at any point in time RAM: operation guess() fills a cell with an integer YES-instance: at least one accepting path NO-instance: all paths reject NP: all problems solvable in polytime by a nondet. Turing machine k-SAT algorithm: guess a satisfying assignment, check correctness 𝑃(𝑜 + 𝑛) 3SUM algorithm: guess 𝑏, 𝑐, 𝑑 ∈ 𝐵 and check 𝑏 + 𝑐 + 𝑑 = 0 𝑃(1)/𝑃(𝑜) „guess a short proof of satisfiability and check it“ guesses on an accepting path = proof that we have a YES-instance

Co-Nondeterministic Algorithms

Turing machine: can choose any applicable transition at any point in time RAM: operation guess() fills a cell with an integer YES-instance: all paths accept NO-instance: at least one path rejects co-NP: all problems solvable in polytime by a co-nondet. Turing machine if NP ≠ co-NP, then k-SAT has no 𝑷(po poly(𝒐)) co-nondet. algorithm we believe that NP ≠ co-NP, since otherwise the polynomial hierarchy collapses k-SAT: „guess a short proof of unsatisfiability and check it“ – Is this possible ? guesses on a rejecting path = proof that we have a NO-instance classic computational complexity:

(Co-)Nondeterministic SETH

NSETH implies SETH (without randomization) barely anyone believes that NSETH is true but it formalizes a current barrier NSETH can be used to conditionally rule out reductions not even a 𝑃(2 ?)* @) co-nondet. algorithm is known! Nondeterministic SETH: k-SAT has no no 𝑃(2 ?)* @) co-nondet. algorithm do not allow randomization! if NP ≠ co-NP, then k-SAT has no 𝑃(poly(𝑜)) co-nondet. algorithm

[CGIMPS’16]

Potential Reduction from SAT to 3SUM

an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.:

3SUM

total time 𝑠(𝑜) size 𝑜? reduction instance 𝐽1 size 𝑜% instance 𝐽𝑙

… …

for any fomula 𝜚, algorithm 𝐵(𝜚) correctly solves k-SAT on 𝜚 𝐵 runs in time 𝑠(𝑜) = 𝑃(2 ?)E @) for some 𝛿 > 0 for any 𝜁 > 0 there is a 𝜀 ∈ (0, 𝛿) s.t. ∑ 𝑜K

()* % KL?

≤ 2 ?)N @ Properties:

k-SAT

𝑜 variables, 𝑛 ≤ 𝑜% clauses fomula 𝜚 e.g. 𝑙 = 1 and 𝑜? = 2@/(𝑛O, then 𝑜?

()* ≤ 2 ?)*/( @𝑜O% ≤ 2 ?)*/P @

Potential Reduction from SAT to 3SUM

an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.:

3SUM

total time 𝑠(𝑜) size 𝑜? reduction instance 𝐽1 size 𝑜% instance 𝐽𝑙

… …

for any fomula 𝜚, algorithm 𝐵(𝜚) correctly solves k-SAT on 𝜚 𝐵 runs in time 𝑠(𝑜) = 𝑃(2 ?)E @) for some 𝛿 > 0 for any 𝜁 > 0 there is a 𝜀 ∈ (0, 𝛿) s.t. ∑ 𝑜K

()* % KL?

≤ 2 ?)N @ Properties:

k-SAT

𝑜 variables, 𝑛 ≤ 𝑜% clauses fomula 𝜚

𝑃(𝑜()*) algorithm 𝑃(2 ?)N @) algorithm ⟸

Potential Reduction from SAT to 3SUM

an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.:

3SUM

total time 𝑠(𝑜) size 𝑜? reduction instance 𝐽1 size 𝑜% instance 𝐽𝑙

… …

for any fomula 𝜚, algorithm 𝐵(𝜚) correctly solves k-SAT on 𝜚 𝐵 runs in time 𝑠(𝑜) = 𝑃(2 ?)E @) for some 𝛿 > 0 for any 𝜁 > 0 there is a 𝜀 ∈ (0, 𝛿) s.t. ∑ 𝑜K

()* % KL?

≤ 2 ?)N @ Properties:

k-SAT

𝑜 variables, 𝑛 ≤ 𝑜% clauses fomula 𝜚

• nondet. 𝑃(𝑜()*) algorithm and

co-nondet. 𝑃(𝑜()*) algorithm

• nondet. 𝑃(2 ?)N @) algorithm and

co-nondet. 𝑃(2 ?)N @) algorithm ⟸

Potential Reduction from SAT to 3SUM

an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.:

3SUM

total time 𝑠(𝑜) size 𝑜? reduction instance 𝐽1 size 𝑜% instance 𝐽𝑙

… …

for each instance 𝐽

R: guess whether it is YES- or NO-instance

if we guessed YES: guess a proof 𝜌R that 𝐽

R is a YES-instance

k-SAT

𝑜 variables, 𝑛 ≤ 𝑜% clauses fomula 𝜚 if we guessed NO: guess a proof 𝜌R that 𝐽

R is a NO-instance

if we guessed correctly: 𝜌 = (𝜌?,… , 𝜌%) forms a proof that 𝜚 is satisfiable or unsatisfiable algorithm 𝐵 is the „proof checker“

Potential Reduction from SAT to 3SUM

an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.:

3SUM

total time 𝑠(𝑜) size 𝑜? reduction instance 𝐽1 size 𝑜% instance 𝐽𝑙

… … k-SAT

𝑜 variables, 𝑛 ≤ 𝑜% clauses fomula 𝜚

• nondet. 𝑃(𝑜()*) algorithm and

co-nondet. 𝑃(𝑜()*) algorithm

• nondet. 𝑃(2 ?)N @) algorithm and

co-nondet. 𝑃(2 ?)N @) algorithm ⟸ no nondet. 𝑃(𝑜()*) algorithm or no co-nondet. 𝑃(𝑜()*) algorithm no nondet. 𝑃(2 ?)N @) algorithm or no co-nondet. 𝑃(2 ?)N @) algorithm ⟹ =NSETH

Ruling Out Reductions

either 3SUM has strongly subquadratic algorithms

• r 3SUM is hard for a different reason than k-SAT
• r NSETH fails

then there is no deterministic reduction from k-SAT to 3SUM If NSETH holds and 3SUM has a 𝑃(𝑜()*) co-nondeterministic algorithm

we will show this

has drawbacks, but this is the only tool for negative results in this area

Co-Nondeterministic Algorithm for 3SUM

Thm: 3SUM has a co-nondeterministic algorithm in time 𝑃 V(𝑜P/()

[CGIMPS’16]

3SUM: given set 𝐵 of integers in {−𝑜Y, … , 𝑜Y}, are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0?

𝑃 V hides polylogarithmic factors in n 𝑃 V 𝑔 𝑜 = \ 𝑃 𝑔 𝑜 logO 𝑜

O^_

𝑃 V 𝑔 𝑜 = 𝑃(𝑔 𝑜 ⋅ polylog 𝑜)

Co-Nondeterministic Algorithm for 3SUM

Thm: 3SUM has a co-nondeterministic algorithm in time 𝑃 V(𝑜P/()

[CGIMPS’16]

1) guess prime 𝑞 ≤ 𝑜P/( log𝑜 2) compute 𝑢 = 𝑏,𝑐,𝑑 ∈ 𝐵P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 𝑃 V(𝑞)

3SUM: given set 𝐵 of integers in {−𝑜Y, … , 𝑜Y}, are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0?

𝑃 V(𝑞)

Recall: 3SUM for Small Numbers

3SUM is in time 𝑃 𝑜 + 𝑃 V(𝑉) for numbers in {0, …, 𝑉} define polynomial 𝑄 𝑌 ≔ ∑ 𝑌k

k∈l

has degree at most 𝑉 compute 𝑅 𝑌 ≔ 𝑄 𝑌 ⋅ 𝑄 𝑌 ⋅ 𝑄 𝑌 = (∑ 𝑌k

k∈l

)(∑ 𝑌k

k∈l

)(∑ 𝑌k

k∈l

) what is the coefficient of 𝑌n in 𝑅(𝑌)? use efficient polynomial multiplication (via Fast Fourier Transform): polynomials of degree 𝑒 can be multiplied in time 𝑃 V(𝑒) (𝑌k ⋅ 𝑌p ⋅ 𝑌O = 𝑌kqpqO) it is the number of (𝒃,𝒄, 𝒅) summing to 𝒖

Co-Nondeterministic Algorithm for 3SUM

Thm: 3SUM has a co-nondeterministic algorithm in time 𝑃 V(𝑜P/()

[CGIMPS’16]

1) guess prime 𝑞 ≤ 𝑜P/( log𝑜 2) compute 𝑢 = 𝑏,𝑐,𝑑 ∈ 𝐵P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 𝑃 V(𝑞)

3SUM: given set 𝐵 of integers in {−𝑜Y, … , 𝑜Y}, are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0?

let 𝐶 ≔ 𝑏 mod 𝑞 𝑏 ∈ 𝐵} (in general 𝐶 is a multi-set!) let 𝑠

_ ≔

𝑏, 𝑐, 𝑑 ∈ 𝐶P | 𝑏 + 𝑐 + 𝑑 = 0 let 𝑠

? ≔

𝑏, 𝑐, 𝑑 ∈ 𝐶P | 𝑏 + 𝑐 + 𝑑 = 𝑞 let 𝑠

( ≔

𝑏, 𝑐, 𝑑 ∈ 𝐶P | 𝑏 + 𝑐 + 𝑑 = 2𝑞 then 𝑢 = 𝑠

_ + 𝑠 ? + 𝑠 (

universe size 𝑉 = 𝑞

Co-Nondeterministic Algorithm for 3SUM

Thm: 3SUM has a co-nondeterministic algorithm in time 𝑃 V(𝑜P/()

[CGIMPS’16]

1) guess prime 𝑞 ≤ 𝑜P/( log𝑜 2) compute 𝑢 = 𝑏,𝑐,𝑑 ∈ 𝐵P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 4) guess distinct 𝑏?,𝑐?,𝑑? ,… , 𝑏n,𝑐n, 𝑑n ∈ 𝐵P such that 𝑏K + 𝑐K + 𝑑K = 0 mod 𝑞 ∀𝑗 3) if 𝑢 > 𝛽 ⋅ 𝑜P/( log𝑜: accept 5) check that for all 𝑏K, 𝑐K, 𝑑K we have 𝑏K + 𝑐K + 𝑑K ≠ 0 6) if everything works out: reject (otherwise accept) 𝑃 V(𝑞) ✔ time 𝑃 V(𝑜P/() YES-instance: all paths accept NO-instance: at least one path rejects ✔ if we reject then we have a NO-instance ✔

3SUM: given set 𝐵 of integers in {−𝑜Y, … , 𝑜Y}, are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0? (constant 𝛽 to be fixed later)

Co-Nondeterministic Algorithm for 3SUM

show that there exists a prime 𝑞 ≤ 𝑜P/( log𝑜 such that 𝑢 = 𝑏, 𝑐,𝑑 ∈ 𝐵P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 < 𝛽 ⋅ 𝑜P/( log𝑜 NO-instance: at least one path rejects: M := # tuples (𝑏, 𝑐, 𝑑, 𝑞) with 𝑏, 𝑐,𝑑 ∈ 𝐵 and prime 𝑞 s.t. 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞

3SUM: given set 𝐵 of integers in {−𝑜Y, … , 𝑜Y}, are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0?

each 𝑏 + 𝑐 + 𝑑 is in −3𝑜Y , …, 3𝑜Y ∖ {0}, so it has at most log(3𝑜Y) prime factors thus 𝑁 ≤ 𝑜P log 3𝑜Y ≤ 3𝑡 ⋅ 𝑜P log 𝑜 by prime number theorem: there are at least 𝑜P/(/𝛾 primes 𝑞 ≤ 𝑜P/( log 𝑜 thus there is a prime 𝑞 contained in at most 𝑁/(𝑜P/(/𝛾) tuples (𝑏, 𝑐, 𝑑, 𝑞) thus there is a prime 𝑞 with 𝑢 ≤ 𝑁/(𝑜P/(/𝛾) ≤ 3𝑡 ⋅ 𝛾 ⋅ 𝑜P/( log(𝑜) set 𝛽 ≔ 3𝑡 ⋅ 𝛾

Co-Nondeterministic Algorithm for 3SUM

Thm: 3SUM has a co-nondeterministic algorithm in time 𝑃 V(𝑜P/()

[CGIMPS’16]

𝑃 V(𝑞) ✔ time 𝑃 V(𝑜P/() YES-instance: all paths accept NO-instance: at least one path rejects ✔ if we reject then we have a NO-instance ✔

3SUM: given set 𝐵 of integers in {−𝑜Y, … , 𝑜Y}, are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0? (constant 𝛽 to be fixed later)

✔ 1) guess prime 𝑞 ≤ 𝑜P/( log𝑜 2) compute 𝑢 = 𝑏,𝑐,𝑑 ∈ 𝐵P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 4) guess distinct 𝑏?,𝑐?,𝑑? ,… , 𝑏n,𝑐n, 𝑑n ∈ 𝐵P such that 𝑏K + 𝑐K + 𝑑K = 0 mod 𝑞 ∀𝑗 3) if 𝑢 > 𝛽 ⋅ 𝑜P/( log𝑜: accept 5) check that for all 𝑏K, 𝑐K, 𝑑K we have 𝑏K + 𝑐K + 𝑑K ≠ 0 6) if everything works out: reject (otherwise accept)

• I. Randomized Nondeterministic SETH

Randomized Nondeterministic SETH

Nondeterministic SETH: k-SAT has no no 𝑃(2 ?)* @) co-nondet. algorithm what if we allow randomization? Thm: k-SAT has a randomized co-nondeterministic 𝑃(2@/(poly(𝑜)) algorithm with error probability 2)€(@) Thm: OV has a randomized co-nondeterministic 𝑃(𝑜 poly(𝑒,log 𝑜)) algorithm with error probability 𝑜)€(?)

⟸

[Williams’16]

then the hypothesis is wrong!

Tools: Basics on Polynomials

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time univariate polynomials 𝑄 𝑌 = ∑ 𝑏K𝑌K

@ KL_

, 𝑅 𝑌 = ∑ 𝑐K𝑌K

ƒ KL_

, 𝑛 ≤ 𝑜 multiplication 𝑄 𝑌 ⋅ 𝑅(𝑌): 𝑃 V 𝑜 (by FFT, without proof) division with remainder: 𝑃 V 𝑜 (without proof) 𝑄 𝑌 = 𝑇 𝑌 ⋅ 𝑅 𝑌 + 𝑆(𝑌), where 𝑆(𝑌) has degree < 𝑛 we write 𝑆 𝑌 = 𝑄 𝑌 mod 𝑅 𝑌 evaluate 𝑄(𝑌) at a given point 𝑦: 𝑃(𝑜) Horner‘s method: 𝑄 𝑦 = 𝑏_ + 𝑦 ⋅ 𝑏? + 𝑦 ⋅ 𝑏( + 𝑦 ⋅ …

Tools: Multipoint Evaluation on Polynomials

multipoint evaluation: 𝑃 V 𝑜 1) let 𝑀 𝑌 ≔ (𝑌 − 𝑦?)⋯(𝑌 − 𝑦@/() and 𝑆 𝑌 ≔ (𝑌 − 𝑦@/(q?)⋯ (𝑌 − 𝑦@) evaluate 𝑄(𝑌) at given points 𝑌 = 𝑦?,… , 𝑦@ 2) let 𝑄

‰ 𝑌 ≔ 𝑄 𝑌 mod 𝑀(𝑌)

and 𝑄

Š 𝑌 ≔ 𝑄 𝑌 mod 𝑆(𝑌)

3) recursively compute 𝑄

‰(𝑦?),…, 𝑄 ‰(𝑦@/() and 𝑄 Š(𝑦@/(q?),…, 𝑄 Š(𝑦@)

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time univariate polynomial 𝑄 𝑌 = ∑ 𝑏K𝑌K

@ KL_

𝑄 𝑌 = 𝑇 𝑌 ⋅ 𝑀 𝑌 + 𝑄

‰(𝑌)

𝑄 𝑦K = 𝑇 𝑦K ⋅ 𝑀 𝑦K + 𝑄

‰ 𝑦K

= 0 for 𝑗 ≤ 𝑜/2 polynomial division: 𝑈 𝑜 = 2𝑈 𝑜/2 + 𝑃 V 𝑜 = 𝑃 V 𝑜 = 𝑄

‰ 𝑦K

Tools: Multipoint Evaluation on Polynomials

computing 𝑀 𝑌 ≔ (𝑌 − 𝑦?)⋯(𝑌 − 𝑦@/(): fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time univariate polynomial 𝑄 𝑌 = ∑ 𝑏K𝑌K

@ KL_

(𝑌 − 𝑦K) 𝑄

K,R 𝑌 : = • 𝑌 − 𝑦% R %LK

computes canonical polynomials 𝑄Y⋅(Žq?, Yq? (Ž 𝑌 defined by straight-forward binary tree log𝑜 in layer 𝑗: 𝑜/2K multiplications of polynomials of degree 2K total time 𝑃 V 𝑜

Tools: Polynomial Interpolation

polynomial interpolation: 𝑃 V 𝑜 given pairs 𝑦?,𝑧? ,… , (𝑦@, 𝑧

@) find a polynomial 𝑄(𝑌) with 𝑄 𝑦K = 𝑧K for all 𝑗

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time univariate polynomial 𝑄 𝑌 = ∑ 𝑏K𝑌K

@ KL_

Lagrange‘s formula: 𝑄 𝑌 = • 𝑧K ⋅ • 𝑌 − 𝑦R 𝑦K − 𝑦R

R‘K K

Caveat: „division by 𝑦“ in ℤ‚ means multiplication with the inverse 𝑦)? extended Euclidean algorithm: computes 𝑡,𝑢 with 𝑡 ⋅ 𝑦 + 𝑢 ⋅ 𝑞 = gcd 𝑦, 𝑞 = 1 modulo 𝑞: 𝑡 ⋅ 𝑦 = 1 so 𝑡 = 𝑦)? is the inverse of 𝑦

Tools: Polynomial Interpolation

1st goal: compute 𝑄 𝑌 = • 𝑧K

“ ⋅ • 𝑌 − 𝑦R R‘K K

• 𝑧K

“ ⋅ • 𝑌 − 𝑦R R‘K K

=

• 𝑧K

“ ⋅

• 𝑌 − 𝑦R

?”R”@/( R‘K ?”K”@/(

⋅ 𝑆 𝑌 recursion: +

• 𝑧K

“ ⋅

• 𝑌 − 𝑦R

@/(•R”@ R‘K @/(•K”@

⋅ 𝑀 𝑌 (for 𝑧K

“:= 𝑧K ⋅ ∏ ? —˜)—™ R‘K

) let 𝑀 𝑌 ≔ (𝑌 − 𝑦?)⋯(𝑌 − 𝑦@/() and 𝑆 𝑌 ≔ (𝑌 − 𝑦@/(q?)⋯ (𝑌 − 𝑦@)

Tools: Polynomial Interpolation

2nd goal: compute factors 𝑡K = • 1 𝑦K − 𝑦R

R‘K

𝑅“ 𝑌 = •• 𝑌 − 𝑦R

R‘K K

= • 𝑗 ⋅ 𝑏K𝑌K)?

@ KL_

compute the derivative of can compute 𝑅“ 𝑌 in time 𝑃 V 𝑜 𝑅 𝑌 ≔ • 𝑌 − 𝑦K

@ KL?

= • 𝑏K𝑌K

@ KL_

then we have 𝑅“ 𝑦% = • • 𝑦% − 𝑦R

R‘K K

= • 𝑦% − 𝑦R

R‘%

= 𝑡%

)?

compute all 𝑅“ 𝑦% for 𝑙 = 1, … ,𝑜 in time 𝑃 V 𝑜 by multipoint evaluation

Tools: Polynomial Interpolation

polynomial interpolation: 𝑃 V 𝑜 given pairs 𝑦?,𝑧? ,… , (𝑦@, 𝑧

@) find a polynomial 𝑄(𝑌) with 𝑄 𝑦K = 𝑧K for all 𝑗

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time univariate polynomial 𝑄 𝑌 = ∑ 𝑏K𝑌K

@ KL_

Lagrange‘s formula: 𝑄 𝑌 = • 𝑧K ⋅ • 𝑌 − 𝑦R 𝑦K − 𝑦R

R‘K K

can be computed in time 𝑃 V 𝑜 !

Tools: Arithmetic Circuits

arithmetic circuits are a (succinct) representation of (multivariate) polynomials fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time

× + × 42 𝑌? 𝑌( + × 𝑌P 7

input gates labeled with variables 𝑌?,…, 𝑌%

• univariate if 𝑙 = 1

each other gate is a “+” or “×” (unbounded fanin)

• r a “−” (fanin 1)
• r a constant (fanin 0)

fanout is unbounded

• ne output gate

given an input 𝑦?,… ,𝑦% ∈ ℤ‚ (no cyclic dependencies) the output 𝐷(𝑦?,… ,𝑦%) is the number computed

42 + 𝑌?𝑌( 𝑌?𝑌( −𝑌( 𝑌P + 𝑌P ⋅ 7

−

by the output gate (in ℤ‚) = 7 = 1 = 1 = −2 = 7 = 8 = −1 = −2 = 40 = 42 = 640

Tools: Arithmetic Circuits

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time

× + × 2 𝑌 + 3

1 + 2𝑌 2𝑌 3 − 𝑌

− 1

−4𝑌P + 10𝑌( + 6𝑌 =

𝑌 × 6 10 −4

representation as circuit is not unique

+ × + ×

circuit = unstructured, succinct polynomial = structured, verbose

Tools: Arithmetic Circuits

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time

2

working modulo 𝑞 is necessary

• ver ℤ numbers can get very large:

× × × × × × = 2(

Tools: Arithmetic Circuits

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time

× + × 42 𝑌? 𝑌( + × 𝑌P 7

42 + 𝑌?𝑌( 𝑌?𝑌( −𝑌( 𝑌P + 𝑌P ⋅ 7

−

size 𝑡 = number if wires 𝑙 inputs degree(input gate) = 1 degree(constant gate) = 0 degree(“−” gate) = degree of child degree(“+” gate) = maximum of degrees of children degree(“×” gate) = sum of degrees of children degree of circuit = degree(output gate) degree 𝑒 = „largest degree of any monomial assuming no cancelations“ 1 1 1 1 1 1 2 2 6

Tools: Evaluation of Circuits

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time size 𝑡 = number if wires 𝑙 inputs

× + × 2 𝑌 + 3

1 + 2𝑌 2𝑌 3 − 𝑌

− 1

evaluating a circuit at given input: 𝑃(s) degree 𝑒

Tools: Identity Testing

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time size 𝑡 = number if wires 𝑙 inputs given two circuits 𝐷?,𝐷(, do they represent the same polynomial over ℤ‚? degree 𝑒 assume that 𝐷?,𝐷( are univariate and assume 𝑞 ≥ 2𝑒 this has a randomized 𝑃 V(𝑡)-time algorithm with error probability 1 − 𝑡)€(?) this problem is called polynomial identity testing 1) for log𝑡 rounds: 2) pick random 𝑦 ∈ ℤ‚ 3) if 𝐷?(𝑦) ≠ 𝐷((𝑦): return „not identical“ 4) return „identical“ 𝐷? 𝑌 − 𝐷((𝑌) is a poly- nomial 𝑅(𝑌) of degree 𝑒 if 𝐷?(𝑌) ≠ 𝐷((𝑌) then 𝑅(𝑌) is not the 0-polynomial and thus has at most 𝑒 roots with probability ≥ 1/2 we pick a non-root 𝑦 Schwarz-Zippel-Lemma:

Tools: Evaluation of Circuits

fix field ℤ‚ and assume that field operations can be performed in 𝑃(1) time size 𝑡 = number if wires 𝑙 inputs

× + × 2 𝑌 + 3

1 + 2𝑌 2𝑌 3 − 𝑌

− 1

evaluating a circuit at given input: 𝑃(s) no near-linear time algorithm known converting a univariate circuit to a polynomial: 𝑃 V(s ⋅ d) degree 𝑒 (write each gate as a degree polynomial) multipoint evaluation (at 𝑜 points): trivial algorithm: 𝑃(𝑜 ⋅ s) conversion + multipoint evaluation for polynomials = multipoint evaluation for univariate circuits in 𝑃 V(s ⋅ d + 𝑜 + 𝑒) (for multivariate: degree 𝑒 polynomial has up to 𝑒 + 𝑙 + 1 𝑙 monomials ☹)

OV as Multipoint Evaluation on Circuits

circuit 𝐷(𝑏, 𝑐) for testing orthogonality of 𝑏, 𝑐: Given sets 𝐵,𝐶 ⊆ 0,1 ¤ of size 𝑜 Decide whether there are 𝑏 ∈ 𝐵, 𝑐 ∈ 𝐶 such that 𝑏 ⊥ 𝑐 OV: 𝐷(𝑏, 𝑐) = •(1 − 𝑏K𝑐K)

¤ KL?

× + × 1 𝑏? 𝑐? − 1 … − × 𝑏¤ 𝑐¤ −

OV as Multipoint Evaluation on Circuits

circuit 𝐷(𝑏) for testing orthogonality of 𝑏 with any 𝑐 ∈ 𝐶: Given sets 𝐵,𝐶 ⊆ 0,1 ¤ of size 𝑜 Decide whether there are 𝑏 ∈ 𝐵, 𝑐 ∈ 𝐶 such that 𝑏 ⊥ 𝑐 OV: 𝐷(𝑏) = • 𝐷(𝑏, 𝑐)

p∈¦

= • •(1 − 𝑏K𝑐K)

¤ KL? p∈¦

+ …

for each 𝑐 ∈ 𝐶

𝐷(𝑏, 𝑐) …

• 𝑒 inputs (for the coordinates of 𝑏)
• size 𝑃(𝑜𝑒)
• degree ≤ 2𝑒
• for any 𝑏 ∈ {0,1}¤: 0 ≤ 𝐷(𝑏) ≤ 𝑜
• pick prime 𝑞 ≥ 𝑜 and work modulo 𝑞, i.e., over field ℤ‚

(Co-)Nondet. Multipoint Evaluation on Circuits

Given circuit 𝐷 on inputs 𝑌?,… ,𝑌% with size 𝑡 and degree 𝑒 over ℤ‚ can assume 𝑞 ≥ 2𝑜𝑒, 𝑞 ≤ 𝑜§(?) Given inputs 𝑨?,… ,𝑨@ ∈ ℤ‚

%

want to evaluate 𝐷 on each 𝑨R = (𝑨R 1 , …, 𝑨R 𝑙 ) 1) compute polynomials 𝑆? 𝑌 , …, 𝑆%(𝑌) such that 𝑆K 𝑘 = 𝑨R[𝑗] by polynomial interpolation 𝑃 V(𝑙𝑜) new goal: evaluate univariate circuit 𝐷“ 𝑌 = 𝐷(𝑆? 𝑌 , …, 𝑆%(𝑌)) on 𝑌 = 1, … ,𝑜 2) guess a polynomial 𝑅 𝑌 of degree at most 𝑒𝑜 3) check that 𝑅 𝑌 = 𝐷(𝑆? 𝑌 , … ,𝑆%(𝑌)) by “polynomial identity testing” 𝑃(𝑒𝑜) 𝑃 V(𝑡 + 𝑙𝑜 + 𝑒𝑜) 4) multipoint evaluate 𝑅 𝑌 on 𝑌 = 1, …, 𝑜 and return these values 𝑃 V(𝑒𝑜) 𝐷’ has size ≤ 𝑃(𝑡 + 𝑙𝑜) and depth ≤ 𝑒𝑜

Co-Nondet. Algorithm for OV

5) ACCEPT if ∑ 𝑅(𝑘)

@ RL?

≥ 1 1) compute polynomials 𝑆? 𝑌 , …, 𝑆%(𝑌) such that 𝑆K 𝑘 = 𝑨R[𝑗] by polynomial interpolation 𝑃 V(𝑙𝑜) new goal: evaluate univariate circuit 𝐷“ 𝑌 = 𝐷(𝑆? 𝑌 , …, 𝑆%(𝑌)) on 𝑌 = 1, … ,𝑜 2) guess a polynomial 𝑅 𝑌 of degree at most 𝑒𝑜 3) check that 𝑅 𝑌 = 𝐷(𝑆? 𝑌 , … ,𝑆%(𝑌)) by “polynomial identity testing”, if not: ACCEPT 𝑃(𝑒𝑜) 𝑃 V(𝑡 + 𝑙𝑜 + 𝑒𝑜) 4) multipoint evaluate 𝑅 𝑌 on 𝑌 = 1, …, 𝑜 and return these values 𝑃 V(𝑒𝑜) 𝐷’ has size ≤ 𝑃(𝑡 + 𝑙𝑜) and depth ≤ 𝑒𝑜 use circuit 𝐷(𝑏) for testing orthogonality of 𝑏 with any 𝑐 ∈ 𝐶 ⊆ {0,1}¤ evaluate 𝐷(𝑏) at each 𝑏 ∈ 𝐵 𝑒 inputs, size 𝑃(𝑒𝑜), degree ≤ 2𝑒 𝑃 V(𝑒𝑜) YES-instance: all paths accept NO-instance: at least one path rejects ...with high probability

Co-Nondet. Algorithm for OV

5) ACCEPT if ∑ 𝑅(𝑘)

@ RL?

≥ 1 1) compute polynomials 𝑆? 𝑌 , …, 𝑆%(𝑌) such that 𝑆K 𝑘 = 𝑨R[𝑗] by polynomial interpolation 𝑃 V(𝑙𝑜) new goal: evaluate univariate circuit 𝐷“ 𝑌 = 𝐷(𝑆? 𝑌 , …, 𝑆%(𝑌)) on 𝑌 = 1, … ,𝑜 2) guess a polynomial 𝑅 𝑌 of degree at most 𝑒𝑜 3) check that 𝑅 𝑌 = 𝐷(𝑆? 𝑌 , … ,𝑆%(𝑌)) by “polynomial identity testing”, if not: ACCEPT 𝑃(𝑒𝑜) 𝑃 V(𝑡 + 𝑙𝑜 + 𝑒𝑜) 4) multipoint evaluate 𝑅 𝑌 on 𝑌 = 1, …, 𝑜 and return these values 𝑃 V(𝑒𝑜) 𝐷’ has size ≤ 𝑃(𝑡 + 𝑙𝑜) and depth ≤ 𝑒𝑜 𝑃 V(𝑒𝑜) NO-instance: If we correctly guess 𝑅 𝑌 then the identity test works with prob. 1 and we correctly report non-existence of an orthogonal pair = REJECT YES-instance: If we correctly guess 𝑅 𝑌 then we ACCEPT If we wrongly guess 𝑅 𝑌 then identity test fails with prob. 1 − 𝑜)€(?) for any guess: we ACCEPT with probability 1 − 𝑜)€(?)

Conclusion

No randomization allowed Nondeterministic SETH: k-SAT has no no 𝑃(2 ?)* @) co-nondet. algorithm If it holds, then there is no deterministic reduction from SETH to 3SUM, This is the only tool for ruling out reductions! „Randomized Nondeterministic SETH“: is wrong! We have seen a 𝑃 V(𝑒𝑜) co-nondeterministic algorithm for OV uses many tools for computing with polynomials and arithmetic circuits since 3SUM has a 𝑃 V(𝑜P/() co-nondeterministic algorithm