Nondeterministic extensions of the Strong Exponential Time ... · Edit Distance in (~ n2)...

Nondeterministic extensions of the StrongExponential Time Hypothesis andconsequences for non-reducibility

Marco L. Carmosino, Jiawei Gao, Russell Impagliazzo, IvanMikhailin, Ramamohan Paturi, Stefan Schneider

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Fine-Grained Hardness

For many problems progress has stalled:

Edit Distance in Ω(n2) [WagnerFischer74]3-SUM in Ω(n2)CKT-SAT in Ω(2n)3-Points-On-a-Line in Ω(n2)

Proving these lower bounds seems out of reachGoal: Explain bounds from common principle

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For many problems progress has stalled:Edit Distance in Ω(n2) [WagnerFischer74]

3-SUM in Ω(n2)CKT-SAT in Ω(2n)3-Points-On-a-Line in Ω(n2)

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For many problems progress has stalled:Edit Distance in Ω(n2) [WagnerFischer74]3-SUM in Ω(n2)

CKT-SAT in Ω(2n)3-Points-On-a-Line in Ω(n2)

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For many problems progress has stalled:Edit Distance in Ω(n2) [WagnerFischer74]3-SUM in Ω(n2)CKT-SAT in Ω(2n)

3-Points-On-a-Line in Ω(n2)

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For many problems progress has stalled:Edit Distance in Ω(n2) [WagnerFischer74]3-SUM in Ω(n2)CKT-SAT in Ω(2n)3-Points-On-a-Line in Ω(n2)

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Proving these lower bounds seems out of reach

Goal: Explain bounds from common principle

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Conditional Lower Bounds

Conditional lower bounds are an attempt to tackle theproblem

Relate hard problems by reductions that respect resourcesGoal 1: Explain hardness of as many problems as possibleGoal 2: Explain hardness using a common principle

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Conditional lower bounds are an attempt to tackle theproblemRelate hard problems by reductions that respect resources

Goal 1: Explain hardness of as many problems as possibleGoal 2: Explain hardness using a common principle

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Conditional lower bounds are an attempt to tackle theproblemRelate hard problems by reductions that respect resourcesGoal 1: Explain hardness of as many problems as possible

Goal 2: Explain hardness using a common principle

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Conditional lower bounds are an attempt to tackle theproblemRelate hard problems by reductions that respect resourcesGoal 1: Explain hardness of as many problems as possibleGoal 2: Explain hardness using a common principle

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Important Conjectures: 3-Sum

3-SUM: Given n integers a1, . . . ,an ∈ [−M,M] find i , j , ksuch that ai + aj + ak = 0

Simple algorithm: O(n2)

Fastest known algorithms: O(n2/polylog n) [BDP08,GP14]3-SUM conjecture: There is no O(n2−ε) algorithm for3-Sum

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Fastest known algorithms: O(n2/polylog n) [BDP08,GP14]

3-SUM conjecture: There is no O(n2−ε) algorithm for3-Sum

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Important Conjectures: All-Pairs Shortest Path

APSP: Given a weighted directed graph, find the distanceof every pair u, v

Dynamic Programming: O(n3)

Fastest known algorithms: O(n3/2√

log n) [W14]All-pairs shortest path conjecture: There is no O(n3−ε)algorithm for APSP

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APSP: Given a weighted directed graph, find the distanceof every pair u, vDynamic Programming: O(n3)

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log n) [W14]

All-pairs shortest path conjecture: There is no O(n3−ε)algorithm for APSP

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Important Conjectures: Strong Exponential TimeHypothesis

k -SAT: Given a k -CNF, find a satisfying assignment

Brute force: O(2n)

Fastest known algorithm: O(2(1− ck )n) [PPSZ98]

Strong Exponential Time Hypothesis: For every s > 0,there is a k such that k -SAT cannot be solved in time2(1−s)n

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k -SAT: Given a k -CNF, find a satisfying assignmentBrute force: O(2n)

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APSPVector Orthogonality

Edit Distance...

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3 Points on LinePolygon Containment

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Negative TriangleGraph Diameter

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Zero Weight Triangle

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Single-Source FlowTriangle Collection

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Other Problems

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Max FlowMin-Cost Max-Flow

Hitting SetFirst-Order Properties

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Fine-Grained Reductions

Formalized in [VWW10]

Fine-grained reduction from (L1,T1) to (L2,T2)

Turing reduction that respects resources

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Formalized in [VWW10]Fine-grained reduction from (L1,T1) to (L2,T2)

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Formalized in [VWW10]Fine-grained reduction from (L1,T1) to (L2,T2)

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Fine-Grained Reduction

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T2(n1)1−δ

T2(n3)1−δ

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T2(n1)1−δ

T2(n3)1−δ

T1(n)1−ε

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T2(n1)1−δ

T2(n3)1−δ

T1(n)1−ε

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T2(n1)1−δ

T2(n3)1−δ

T1(n)1−ε

∑i T2(ni)

1−δ

≤ T1(n)1−ε

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Proof Idea

Lemma (Basic Idea)Every SETH-hard problem has property X. 3-SUM and APSPdo not have property X.

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Nondeterministic Strong Exponential Time HypothesisFor every s > 0, there is a k such that k -SAT cannot be solvedin co-nondeterministic time 2(1−s)n

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Property X

Det. Nondet. Co-Nondet. X ExampleT T 1−ε T yes CNF-SATT T T 1−ε yes DNF-TAUTT T T yes Exact-Max-SATT T 1−ε T 1−ε no 3-SUM

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Property X

LemmaAssuming NSETH, any problem that is SETH-hard with time Tunder deterministic reductions either

Cannot be solved in nondeterministic time T (1−s)

Cannot be solved in co-nondeterministic time T (1−s)

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Property X

x ∈ L1

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Property X

x ∈ L1

Nondeterministicbits

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Property X

x ∈ L1

L2T2(n3)1−δ

T1(n)1−ε Nondeterministicbits

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Property X

x 6∈ L1

L2T2(n3)1−δ

T1(n)1−ε Nondeterministicbits

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Outline

¬NSETH implies interesting circuit lower bounds

Problems with fast nondeterministic andco-nondeterministic algorithms (¬X )First-order graph properties

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Outline

¬NSETH implies interesting circuit lower boundsProblems with fast nondeterministic andco-nondeterministic algorithms (¬X )

First-order graph properties

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Outline

¬NSETH implies interesting circuit lower boundsProblems with fast nondeterministic andco-nondeterministic algorithms (¬X )First-order graph properties

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¬SETH and ¬NSETH Imply Circuit Lower Bounds

The work of [JVM13] uses a two-part strategy to show¬SETH =⇒ Circuit Lower Bounds:

1 A tight implication from C-CKT-SAT algorithms to C lowerbounds

2 Decomposition of C-circuits into∨

CNF form

“¬NSETH =⇒ Circuit Lower Bounds” is implicit in [JVM13],following from their technical contributions and the proofs of[Williams 2013].

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¬SETH and ¬NSETH are Algorithmic

ETH is false: for every ε > 0,3-SAT is in time 2εn

SETH is false: there is a δ < 1 such that for every k , k -SAT

is in time 2δn

NETH is false: for every ε > 0,3-TAUT is innondeterministic time 2εn

NSETH is false: there is a δ < 1 such that for everyk , k -TAUT is in nondeterministic time 2δn

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is in time 2δn

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is in time 2δn

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is in time 2δn

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Reducing C-CKT-SAT to k -SAT?

For C:

Linear-size circuitsLinear-size series-parallel circuits

There are decompositions: ∀C ∈ C, we have C =∨

CNFk .Execute the faster k -SAT algorithm on each “leaf” of thedecomposition.

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Outline

The following problems have fast nondeterministic andco-nondeterministic algorithms:

Max-Flow (O(m))Min-Cost Max Flow (O(m))3-SUM (O(n3/2))All-Pairs Shortest Path (O(n2.69))

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Outline

Max-Flow (O(m))

Min-Cost Max Flow (O(m))3-SUM (O(n3/2))All-Pairs Shortest Path (O(n2.69))

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Outline

Max-Flow (O(m))Min-Cost Max Flow (O(m))

3-SUM (O(n3/2))All-Pairs Shortest Path (O(n2.69))

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Outline

Max-Flow (O(m))Min-Cost Max Flow (O(m))3-SUM (O(n3/2))

All-Pairs Shortest Path (O(n2.69))

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Outline

Max-Flow (O(m))Min-Cost Max Flow (O(m))3-SUM (O(n3/2))All-Pairs Shortest Path (O(n2.69))

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Maximum Flow Problem

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1Licensed under CC BY-SA 3.0 via Commons -https://commons.wikimedia.org/wiki/File:Max flow.svg#/media/File:Max flow.svg

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Is there a flow of value at least k

Fastest known algorithm: O(mn) [Orlin13]O(m) approximation algorithms for undirected case[KLOS14]Can we prove a Ω(mn) lower bound under SETH?

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Is there a flow of value at least kFastest known algorithm: O(mn) [Orlin13]

O(m) approximation algorithms for undirected case[KLOS14]Can we prove a Ω(mn) lower bound under SETH?

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Is there a flow of value at least kFastest known algorithm: O(mn) [Orlin13]O(m) approximation algorithms for undirected case[KLOS14]

Can we prove a Ω(mn) lower bound under SETH?

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Is there a flow of value at least kFastest known algorithm: O(mn) [Orlin13]O(m) approximation algorithms for undirected case[KLOS14]Can we prove a Ω(mn) lower bound under SETH?

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There is a O(m) nondeterministic algorithm

There is a O(m) co-nondeterministic algorithm(Min-cut/Max-flow theorem)Max-Flow is not SETH-hard at time O(m1+ε) underdeterministic reductions (assuming NSETH)Disproving this statement implies new lower bounds forlinear size circuits

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There is a O(m) nondeterministic algorithmThere is a O(m) co-nondeterministic algorithm(Min-cut/Max-flow theorem)

Max-Flow is not SETH-hard at time O(m1+ε) underdeterministic reductions (assuming NSETH)Disproving this statement implies new lower bounds forlinear size circuits

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There is a O(m) nondeterministic algorithmThere is a O(m) co-nondeterministic algorithm(Min-cut/Max-flow theorem)Max-Flow is not SETH-hard at time O(m1+ε) underdeterministic reductions (assuming NSETH)

Disproving this statement implies new lower bounds forlinear size circuits

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There is a O(m) nondeterministic algorithmThere is a O(m) co-nondeterministic algorithm(Min-cut/Max-flow theorem)Max-Flow is not SETH-hard at time O(m1+ε) underdeterministic reductions (assuming NSETH)Disproving this statement implies new lower bounds forlinear size circuits

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Min-Cost Maximum Flow Problem

Edges have a capacity and a cost per unit flow

Is there a flow of value more than k or of value k and costat most c?Fastest algorithm: O(m2) [Orlin88]Special cases: Max-Flow, Min-Cost Perfect BipartiteMatching

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Edges have a capacity and a cost per unit flowIs there a flow of value more than k or of value k and costat most c?

Fastest algorithm: O(m2) [Orlin88]Special cases: Max-Flow, Min-Cost Perfect BipartiteMatching

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Edges have a capacity and a cost per unit flowIs there a flow of value more than k or of value k and costat most c?Fastest algorithm: O(m2) [Orlin88]

Special cases: Max-Flow, Min-Cost Perfect BipartiteMatching

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Edges have a capacity and a cost per unit flowIs there a flow of value more than k or of value k and costat most c?Fastest algorithm: O(m2) [Orlin88]Special cases: Max-Flow, Min-Cost Perfect BipartiteMatching

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Simple O(m) nondeterministic algorithm

O(m) co-nondeterministic algorithm based on Klein’s cyclecanceling algorithm:

There is a flow of the same value and smaller cost if andonly if there is a negative cost cycle in the residual graphCo-Nondeterministic O(m) algorithm for negative cycles

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Simple O(m) nondeterministic algorithmO(m) co-nondeterministic algorithm based on Klein’s cyclecanceling algorithm:

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There is a flow of the same value and smaller cost if andonly if there is a negative cost cycle in the residual graph

Co-Nondeterministic O(m) algorithm for negative cycles

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Certifying Negative Cycles

Co-nondeterministic algorithm follows from propertiesBellman-Ford algorithm

Potential: p : V → R such that for all(u, v) ∈ Ep(u) + l(u, v) ≥ p(v)

There is a negative weight cycle if and only if there is nopotential

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Co-nondeterministic algorithm follows from propertiesBellman-Ford algorithmPotential: p : V → R such that for all(u, v) ∈ Ep(u) + l(u, v) ≥ p(v)

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Co-nondeterministic algorithm follows from propertiesBellman-Ford algorithmPotential: p : V → R such that for all(u, v) ∈ Ep(u) + l(u, v) ≥ p(v)

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Nondeterministically guess the min-cost max flow

Certify that it is a max flow by guessing a cutCertify that it is minimum cost by guessing a potentialTime complexity: O(m)

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Nondeterministically guess the min-cost max flowCertify that it is a max flow by guessing a cut

Certify that it is minimum cost by guessing a potentialTime complexity: O(m)

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Nondeterministically guess the min-cost max flowCertify that it is a max flow by guessing a cutCertify that it is minimum cost by guessing a potential

Time complexity: O(m)

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Nondeterministically guess the min-cost max flowCertify that it is a max flow by guessing a cutCertify that it is minimum cost by guessing a potentialTime complexity: O(m)

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Min-Cost Max-Flow is not SETH-hard at time O(m1+ε)under deterministic reductions (assuming NSETH)

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Min-Cost Max-Flow is not SETH-hard at time O(m1+ε)under deterministic reductions (assuming NSETH)Disproving this statement implies new lower bounds forlinear size circuits

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Co-Nondeterministic List and Count Algorithm

Embed original problem into one that is much easier, butmay have some false positive solutions

Nondeterministically list all false positivesCheck that all of them are indeed false positivesCount number of solutions and check that given list iscomplete

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Embed original problem into one that is much easier, butmay have some false positive solutionsNondeterministically list all false positives

Check that all of them are indeed false positivesCount number of solutions and check that given list iscomplete

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Embed original problem into one that is much easier, butmay have some false positive solutionsNondeterministically list all false positivesCheck that all of them are indeed false positives

Count number of solutions and check that given list iscomplete

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Embed original problem into one that is much easier, butmay have some false positive solutionsNondeterministically list all false positivesCheck that all of them are indeed false positivesCount number of solutions and check that given list iscomplete

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3-Sum problem

3-Sum: Given n integers a1, . . . ,an ∈ [−M,M] find i , j , ksuch that ai + aj + ak = 0

M = poly(n)

FFT based counting algorithm: O(M)

Fast nondeterministic algorithm

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3-Sum problem

3-Sum: Given n integers a1, . . . ,an ∈ [−M,M] find i , j , ksuch that ai + aj + ak = 0M = poly(n)

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3-Sum problem

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3-Sum problem

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3-Sum problem

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List and Count Algorithm for 3-SUM

Nondeterministically pick a prime p such that there are tsolutions modulo p

Nondeterministically guess t triples (iq, jq, kq).Check that∀q ≤ t : a[iq] + a[jq] + a[kq] = 0(mod p),buta[iq] + a[jq] + a[kq] 6= 0.Count number of solutions of 3-SUM(mod p) and checkthat it is equal to t .

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Nondeterministically pick a prime p such that there are tsolutions modulo pNondeterministically guess t triples (iq, jq, kq).

Check that∀q ≤ t : a[iq] + a[jq] + a[kq] = 0(mod p),buta[iq] + a[jq] + a[kq] 6= 0.Count number of solutions of 3-SUM(mod p) and checkthat it is equal to t .

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Nondeterministically pick a prime p such that there are tsolutions modulo pNondeterministically guess t triples (iq, jq, kq).Check that∀q ≤ t : a[iq] + a[jq] + a[kq] = 0(mod p),buta[iq] + a[jq] + a[kq] 6= 0.

Count number of solutions of 3-SUM(mod p) and checkthat it is equal to t .

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Nondeterministically pick a prime p such that there are tsolutions modulo pNondeterministically guess t triples (iq, jq, kq).Check that∀q ≤ t : a[iq] + a[jq] + a[kq] = 0(mod p),buta[iq] + a[jq] + a[kq] 6= 0.Count number of solutions of 3-SUM(mod p) and checkthat it is equal to t .

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Analysis

Nondeterministically listing all the false positive can bedone in linear time: O(t)

Counting all the false positive can be done by FFT-basedalgorithm in time O(p)

One can always pick t ,p = O(n3/2)The running time is O(n3/2)

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Analysis

Nondeterministically listing all the false positive can bedone in linear time: O(t)Counting all the false positive can be done by FFT-basedalgorithm in time O(p)

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Analysis

Nondeterministically listing all the false positive can bedone in linear time: O(t)Counting all the false positive can be done by FFT-basedalgorithm in time O(p)

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3-SUM modulo p

Consider the first n3/2 primes, ≤ O(n3/2)

ai + aj + ak has at most log(3M) = O(log n) prime factors

On average, a prime p has O(

n3 log(n)n3/2

)= O(n3/2) false

positivesThere is a prime p ≤ O(n3/2) such that t ≤ O(n3/2)solutions

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3-SUM modulo p

n3 log(n)n3/2

)= O(n3/2) false

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3-SUM modulo p

n3 log(n)n3/2

)= O(n3/2) false

positives

There is a prime p ≤ O(n3/2) such that t ≤ O(n3/2)solutions

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3-SUM modulo p

n3 log(n)n3/2

)= O(n3/2) false

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3-SUM is not SETH-hard at time O(n3/2+ε) underdeterministic reductions (assuming NSETH)

Disproving this statement implies new lower bounds forlinear size series-parallel circuits

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3-SUM is not SETH-hard at time O(n3/2+ε) underdeterministic reductions (assuming NSETH)Disproving this statement implies new lower bounds forlinear size series-parallel circuits

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Better co-nondeterministic for 3-SUM will give nontrivialco-nondeterministic for SUBSET SUM. This will prove thatSUBSET SUM is not SETH -hard at time O(2n/2) underdeterministic reductions (assuming NSETH).

Nice coincidence: decision tree complexity of 3-SUM isalso O(n3/2).

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Better co-nondeterministic for 3-SUM will give nontrivialco-nondeterministic for SUBSET SUM. This will prove thatSUBSET SUM is not SETH -hard at time O(2n/2) underdeterministic reductions (assuming NSETH).

Nice coincidence: decision tree complexity of 3-SUM isalso O(n3/2).

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All-Pairs Shortest Path Problem

APSP: Given a weighted directed graph, find the distanceof every pair u, v

Dynamic Programming: O(n3) [FloydWarshall62]We give a co-nondeterministic algorithm for Zero-WeightTriangle

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APSP: Given a weighted directed graph, find the distanceof every pair u, vDynamic Programming: O(n3) [FloydWarshall62]

We give a co-nondeterministic algorithm for Zero-WeightTriangle

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APSP: Given a weighted directed graph, find the distanceof every pair u, vDynamic Programming: O(n3) [FloydWarshall62]We give a co-nondeterministic algorithm for Zero-WeightTriangle

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Zero-Weight Triangle

Edit Distance...

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ZWT: Given a weighted graph with all weights ∈ [−M,M],find a triangle of total weight equal to 0.

Trivial algorithm : O(n3)

ZWT modulo p in time O(pnω)

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ZWT: Given a weighted graph with all weights ∈ [−M,M],find a triangle of total weight equal to 0.Trivial algorithm : O(n3)

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ZWT: Given a weighted graph with all weights ∈ [−M,M],find a triangle of total weight equal to 0.Trivial algorithm : O(n3)

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List and Count Algorithm for ZWT

Nondeterministically pick t and a prime p

Nondeterministically guess t triples (xi , yi , zi) of verticesCheck that∀i ≤ t : W [i , j] + W [j , k ] + W [j , i] = 0(mod p),butW [i , j] + W [j , k ] + W [j , i] 6= 0.Count number of solutions of ZWT (mod p) and check thatit is equal to t .

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Nondeterministically pick t and a prime pNondeterministically guess t triples (xi , yi , zi) of vertices

Check that∀i ≤ t : W [i , j] + W [j , k ] + W [j , i] = 0(mod p),butW [i , j] + W [j , k ] + W [j , i] 6= 0.Count number of solutions of ZWT (mod p) and check thatit is equal to t .

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Nondeterministically pick t and a prime pNondeterministically guess t triples (xi , yi , zi) of verticesCheck that∀i ≤ t : W [i , j] + W [j , k ] + W [j , i] = 0(mod p),butW [i , j] + W [j , k ] + W [j , i] 6= 0.

Count number of solutions of ZWT (mod p) and check thatit is equal to t .

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Nondeterministically pick t and a prime pNondeterministically guess t triples (xi , yi , zi) of verticesCheck that∀i ≤ t : W [i , j] + W [j , k ] + W [j , i] = 0(mod p),butW [i , j] + W [j , k ] + W [j , i] 6= 0.Count number of solutions of ZWT (mod p) and check thatit is equal to t .

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Counting Solutions

Define matrix A with A[i , j] = xW [i,j] mod p

Compute A3 in time O(pnω)

For all entries A3[i , i] sum the coefficients of x0, xp, x2p

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Counting Solutions

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Counting Solutions

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Analysis

Same argument as for 3-SUM gives t ≤ O(n3

Nondeterministically listing all the false positive can bedone in linear time: O(t)Counting takes time O(pnω)

Pick p = n0.31

The running time is O(n2.69)

It immediately yields O(n2.9) for APSP [VW09]Nondeterministic reduction yields O(n2.69) for APSP

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Analysis

Nondeterministically listing all the false positive can bedone in linear time: O(t)

Counting takes time O(pnω)

Pick p = n0.31

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Analysis

Pick p = n0.31

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Analysis

Pick p = n0.31

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Analysis

Pick p = n0.31

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Analysis

Pick p = n0.31

It immediately yields O(n2.9) for APSP [VW09]

Nondeterministic reduction yields O(n2.69) for APSP

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Analysis

Pick p = n0.31

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All-Pairs Shortest Path

All-Pairs Shortest Path is not SETH-hard at time O(n2.69+ε)under deterministic reductions (assuming NSETH)

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All-Pairs Shortest Path

All-Pairs Shortest Path is not SETH-hard at time O(n2.69+ε)under deterministic reductions (assuming NSETH)Disproving this statement implies new lower bounds forlinear size circuits

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Quantifier Structures of Hard Problems

Many SETH-hard problems have similar quantifierstructures:

Orthogonal Vectors (O(n2))(∃v1)(∃v2)(∀i) [(v1[i] = 0) ∨ (v2[i] = 0)]Graph k -Dominating Set (O(nk ))(∃v1) . . . (∃vk )(∀vk+1)

[∨ki=1 E(vi , vk+1)

]Problems with other quantifier structures:

k -Clique (Solvable in time O(nωk/3))(∃v1) . . . (∃vk )

[∧i 6=j E(vi , vj )

]Hitting Set (not known to be SETH-hard)(∃H)(∀S)(∃x) [(u ∈ H) ∧ (u ∈ S)]

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Orthogonal Vectors (O(n2))(∃v1)(∃v2)(∀i) [(v1[i] = 0) ∨ (v2[i] = 0)]

Graph k -Dominating Set (O(nk ))(∃v1) . . . (∃vk )(∀vk+1)

[∨ki=1 E(vi , vk+1)

[∧i 6=j E(vi , vj )

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[∨ki=1 E(vi , vk+1)

Problems with other quantifier structures:

[∧i 6=j E(vi , vj )

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[∨ki=1 E(vi , vk+1)

[∧i 6=j E(vi , vj )

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[∨ki=1 E(vi , vk+1)

[∧i 6=j E(vi , vj )

Hitting Set (not known to be SETH-hard)(∃H)(∀S)(∃x) [(u ∈ H) ∧ (u ∈ S)]

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[∨ki=1 E(vi , vk+1)

[∧i 6=j E(vi , vj )

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First-order formula ϕ with k quantifiers

ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Each Qi ∈ ∃, ∀.

Model checking problem on graphs

Input: Sparse graph G, given by its edge list of size m.Output: Whether G |= ϕ.Can be done in O(mk−1)

Our results:

All SETH-hard problems have

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

If NSETH holds, there is no reduction from this quantifierstructure to other quantifier structures.

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Each Qi ∈ ∃, ∀.Model checking problem on graphs

Our results:

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Input: Sparse graph G, given by its edge list of size m.

Output: Whether G |= ϕ.Can be done in O(mk−1)

Our results:

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Input: Sparse graph G, given by its edge list of size m.Output: Whether G |= ϕ.

Can be done in O(mk−1)

Our results:

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Our results:

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Our results:

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Our results:All SETH-hard problems have

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Our results:All SETH-hard problems have

Q1 = Q2 = · · · = Qk−1 = ∃,Qk = ∀

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ϕ = (∃x1)(Q2x2) . . . (Qkxk )ψ

Quantifierstructure

Result Hardness

∃ . . . ∃∀ If solvable in O(mk−1−ε) co-nondeterministic time, thenNSETH is false.

SETH-hard

All k quanti-fiers are ∃’s

solvable in O(mk−1.5) time Easy

More thanone ∀’s

faster co-nondeterministic algo-rithms

Not SETH-hard underNSETH

Exactly one∀, but not atQk

faster co-nondeterministic algo-rithms

Not SETH-hard underNSETH

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Example: Hitting Set

Decide if ∃S ∀T ∃x (x ∈ S ∧ x ∈ T )

Exactly one ∀ quantifier, but not at Qk .O(m) nondeterministic algorithm

Guess S, enumerate T , guess x .

O(m) co-nondeterministic algorithm

Decide if ∀S ∃T ∀x (x /∈ S ∨ x /∈ T ) nondeterministicallyFor each S, guess T , for each (x ∈ S), check if (x ∈ T )If none element x in S is also in T , accept.Otherwise, reject.

Hitting Set ≤FGR Orthogonal Vectors [AVWW16]Orthogonal Vectors 6≤FGR Hitting Set, under NSETH.

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Exactly one ∀ quantifier, but not at Qk .

O(m) nondeterministic algorithm

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Guess S, enumerate T , guess x .O(m) co-nondeterministic algorithm

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Decide if ∀S ∃T ∀x (x /∈ S ∨ x /∈ T ) nondeterministically

For each S, guess T , for each (x ∈ S), check if (x ∈ T )If none element x in S is also in T , accept.Otherwise, reject.

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Decide if ∀S ∃T ∀x (x /∈ S ∨ x /∈ T ) nondeterministicallyFor each S, guess T , for each (x ∈ S), check if (x ∈ T )

If none element x in S is also in T , accept.Otherwise, reject.

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Decide if ∀S ∃T ∀x (x /∈ S ∨ x /∈ T ) nondeterministicallyFor each S, guess T , for each (x ∈ S), check if (x ∈ T )If none element x in S is also in T , accept.

Otherwise, reject.

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Hitting Set ≤FGR Orthogonal Vectors [AVWW16]

Orthogonal Vectors 6≤FGR Hitting Set, under NSETH.

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Randomized Reductions

x ∈ L1

Nondeterministicbits

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The argument does not extend to randomized reductions

Argument only gives a fast Merlin-Arthur algorithm for SATMASETH?There is a O(2n/2) MA algorithm for CNF-SAT [Williams]Zero-error reductions are ok

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The argument does not extend to randomized reductionsArgument only gives a fast Merlin-Arthur algorithm for SAT

MASETH?There is a O(2n/2) MA algorithm for CNF-SAT [Williams]Zero-error reductions are ok

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The argument does not extend to randomized reductionsArgument only gives a fast Merlin-Arthur algorithm for SATMASETH?

There is a O(2n/2) MA algorithm for CNF-SAT [Williams]Zero-error reductions are ok

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The argument does not extend to randomized reductionsArgument only gives a fast Merlin-Arthur algorithm for SATMASETH?There is a O(2n/2) MA algorithm for CNF-SAT [Williams]

Zero-error reductions are ok

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The argument does not extend to randomized reductionsArgument only gives a fast Merlin-Arthur algorithm for SATMASETH?There is a O(2n/2) MA algorithm for CNF-SAT [Williams]Zero-error reductions are ok

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Idea: Consider a stronger hypothesis that rules outrandomized reductions

Non-Uniform Nondeterministic Strong Exponential TimeHypothesisFor every s > 0, there is a k such that k -SAT does not have2(1−s)n size nondeterministic circuits

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Idea: Consider a stronger hypothesis that rules outrandomized reductions

Non-Uniform Nondeterministic Strong Exponential TimeHypothesisFor every s > 0, there is a k such that k -SAT does not have2(1−s)n size nondeterministic circuits

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Summary

Introduced Nondeterministic Strong Exponential TimeHypothesis

If NSETH is false, then new circuit lower bounds followIf NSETH is true, then non-reducibility results followIf there is a deterministic fine-grained reduction from vectororthogonality to hitting set, then we have new lowerbounds for linear size circuits

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Summary

Introduced Nondeterministic Strong Exponential TimeHypothesisIf NSETH is false, then new circuit lower bounds follow

If NSETH is true, then non-reducibility results followIf there is a deterministic fine-grained reduction from vectororthogonality to hitting set, then we have new lowerbounds for linear size circuits

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Summary

Introduced Nondeterministic Strong Exponential TimeHypothesisIf NSETH is false, then new circuit lower bounds followIf NSETH is true, then non-reducibility results follow

If there is a deterministic fine-grained reduction from vectororthogonality to hitting set, then we have new lowerbounds for linear size circuits

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Summary

Introduced Nondeterministic Strong Exponential TimeHypothesisIf NSETH is false, then new circuit lower bounds followIf NSETH is true, then non-reducibility results followIf there is a deterministic fine-grained reduction from vectororthogonality to hitting set, then we have new lowerbounds for linear size circuits

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Open Questions

Deal with randomized reductions

Find exponential time problems not hard under SETH

Adapt the framework to dynamic problemsConsequences of NETH

Every APSP-hard problem has property X, CNFSAT and3-SUM do not

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Open Questions

Deal with randomized reductionsFind exponential time problems not hard under SETH

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Open Questions

Adapt the framework to dynamic problems

Consequences of NETH

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Open Questions

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Open Questions

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Thank You!

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Nondeterministic extensions of the Strong Exponential Time ... · Edit Distance in (~ n2)...

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