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

Complexity Theory of Polynomial-Time Problems Lecture 11: Nondeterministic SETH Karl Bringmann

Complexity Inside P 3SUM n 2 SAT 2 n APSP n 3 APSP equivalent Colinearity n 2 OV n 2 Radius n 3 3SUM-hard Negative Triangle n 3 BMM 𝑜 " LCS n 2 SETH-hard EDIT n 2 Fréchet n 2 SlidingWindowHD 𝑜 " BMM-hard diameter n 2 can we relate SAT / 3SUM / APSP / BMM?

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

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 guesses on an accepting path = proof that we have a YES-instance NP: all problems solvable in polytime by a nondet. Turing machine k-SAT algorithm: guess a satisfying assignment, check correctness 𝑃(𝑜 + 𝑛) „guess a short proof of satisfiability and check it“ 3SUM algorithm: guess 𝑏, 𝑐, 𝑑 ∈ 𝐵 and check 𝑏 + 𝑐 + 𝑑 = 0 𝑃(1)/𝑃(𝑜)

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 guesses on a rejecting path = proof that we have a NO-instance co-NP: all problems solvable in polytime by a co-nondet. Turing machine k-SAT: „guess a short proof of unsatisfiability and check it“ – Is this possible ? classic computational complexity: 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

(Co-)Nondeterministic SETH if NP ≠ co-NP, then k-SAT has no 𝑃(poly(𝑜)) co-nondet. algorithm not even a 𝑃(2 ?)* @ ) co-nondet. algorithm is known! k-SAT has no no 𝑃(2 ?)* @ ) co-nondet. algorithm Nondeterministic SETH: [CGIMPS’16] do not allow randomization! 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

Potential Reduction from SAT to 3SUM an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.: 3SUM reduction k-SAT instance 𝐽 1 size 𝑜 ? fomula 𝜚 … 𝑜 variables, … 𝑛 ≤ 𝑜 % clauses total time instance 𝐽 𝑙 𝑠(𝑜) size 𝑜 % Properties: 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 % ()* ≤ 2 ?)N @ for any 𝜁 > 0 there is a 𝜀 ∈ (0, 𝛿) s.t. ∑ 𝑜 K KL? ()* ≤ 2 ?)*/( @ 𝑜 O% ≤ 2 ?)*/P @ e.g. 𝑙 = 1 and 𝑜 ? = 2 @/( 𝑛 O , then 𝑜 ?

Potential Reduction from SAT to 3SUM an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.: 3SUM reduction k-SAT instance 𝐽 1 size 𝑜 ? fomula 𝜚 … 𝑜 variables, … 𝑛 ≤ 𝑜 % clauses total time instance 𝐽 𝑙 𝑠(𝑜) size 𝑜 % Properties: 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 % ()* ≤ 2 ?)N @ for any 𝜁 > 0 there is a 𝜀 ∈ (0, 𝛿) s.t. ∑ 𝑜 K KL? 𝑃(2 ?)N @ ) algorithm 𝑃(𝑜 ()* ) algorithm ⟸

Potential Reduction from SAT to 3SUM an deterministic algorithm 𝐵 for k-SAT with oracle access to 3SUM s.t.: 3SUM reduction k-SAT instance 𝐽 1 size 𝑜 ? fomula 𝜚 … 𝑜 variables, … 𝑛 ≤ 𝑜 % clauses total time instance 𝐽 𝑙 𝑠(𝑜) size 𝑜 % Properties: 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 % ()* ≤ 2 ?)N @ for any 𝜁 > 0 there is a 𝜀 ∈ (0, 𝛿) s.t. ∑ 𝑜 K KL? nondet. 𝑃(2 ?)N @ ) algorithm and nondet. 𝑃(𝑜 ()* ) algorithm and ⟸ co-nondet. 𝑃(𝑜 ()* ) algorithm 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 reduction k-SAT instance 𝐽 1 size 𝑜 ? fomula 𝜚 … 𝑜 variables, … 𝑛 ≤ 𝑜 % clauses total time instance 𝐽 𝑙 𝑠(𝑜) size 𝑜 % for each instance 𝐽 R : guess whether it is YES- or NO-instance if we guessed YES: guess a proof 𝜌 R that 𝐽 R is a YES-instance if we guessed NO: guess a proof 𝜌 R 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 reduction k-SAT instance 𝐽 1 size 𝑜 ? fomula 𝜚 … 𝑜 variables, … 𝑛 ≤ 𝑜 % clauses total time instance 𝐽 𝑙 𝑠(𝑜) size 𝑜 % nondet. 𝑃(𝑜 ()* ) algorithm and nondet. 𝑃(2 ?)N @ ) algorithm and ⟸ co-nondet. 𝑃(𝑜 ()* ) algorithm co-nondet. 𝑃(2 ?)N @ ) algorithm no nondet. 𝑃(2 ?)N @ ) algorithm or no nondet. 𝑃(𝑜 ()* ) algorithm or ⟹ no co-nondet. 𝑃(𝑜 ()* ) algorithm no co-nondet. 𝑃(2 ?)N @ ) algorithm =NSETH

Ruling Out Reductions If NSETH holds and we will 3SUM has a 𝑃(𝑜 ()* ) co-nondeterministic algorithm show this then there is no deterministic reduction from k-SAT to 3SUM either 3SUM has strongly subquadratic algorithms or 3SUM is hard for a different reason than k-SAT or NSETH fails has drawbacks, but this is the only tool for negative results in this area

Co-Nondeterministic Algorithm for 3SUM 3SUM: given set 𝐵 of integers in {−𝑜 Y , … , 𝑜 Y } , are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0 ? V(𝑜 P/( ) 3SUM has a co-nondeterministic algorithm in time 𝑃 Thm: [CGIMPS’16] V hides polylogarithmic factors in n 𝑃 V 𝑔 𝑜 𝑃 = 𝑃(𝑔 𝑜 ⋅ polylog 𝑜) = \ 𝑃 𝑔 𝑜 log O 𝑜 V 𝑔 𝑜 𝑃 O^_

Co-Nondeterministic Algorithm for 3SUM 3SUM: given set 𝐵 of integers in {−𝑜 Y , … , 𝑜 Y } , are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0 ? V(𝑜 P/( ) 3SUM has a co-nondeterministic algorithm in time 𝑃 Thm: [CGIMPS’16] 1) guess prime 𝑞 ≤ 𝑜 P/( log𝑜 V(𝑞) 𝑃 𝑏,𝑐,𝑑 ∈ 𝐵 P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 V(𝑞) 2) compute 𝑢 = 𝑃

Recall: 3SUM for Small Numbers V(𝑉) for numbers in {0, …, 𝑉} 3SUM is in time 𝑃 𝑜 + 𝑃 𝑌 k define polynomial 𝑄 𝑌 ≔ ∑ k∈l has degree at most 𝑉 𝑌 k 𝑌 k 𝑌 k compute 𝑅 𝑌 ≔ 𝑄 𝑌 ⋅ 𝑄 𝑌 ⋅ 𝑄 𝑌 = (∑ )(∑ )(∑ ) k∈l k∈l k∈l ( 𝑌 k ⋅ 𝑌 p ⋅ 𝑌 O = 𝑌 kqpqO ) what is the coefficient of 𝑌 n in 𝑅(𝑌) ? it is the number of (𝒃,𝒄, 𝒅) summing to 𝒖 use efficient polynomial multiplication (via Fast Fourier Transform): V(𝑒) polynomials of degree 𝑒 can be multiplied in time 𝑃

Co-Nondeterministic Algorithm for 3SUM 3SUM: given set 𝐵 of integers in {−𝑜 Y , … , 𝑜 Y } , are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0 ? V(𝑜 P/( ) 3SUM has a co-nondeterministic algorithm in time 𝑃 Thm: [CGIMPS’16] 1) guess prime 𝑞 ≤ 𝑜 P/( log𝑜 𝑏,𝑐,𝑑 ∈ 𝐵 P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 V(𝑞) 2) compute 𝑢 = 𝑃 let 𝐶 ≔ 𝑏 mod 𝑞 𝑏 ∈ 𝐵} (in general 𝐶 is a multi-set!) 𝑏, 𝑐, 𝑑 ∈ 𝐶 P | 𝑏 + 𝑐 + 𝑑 = 0 let 𝑠 _ ≔ universe size 𝑉 = 𝑞 𝑏, 𝑐, 𝑑 ∈ 𝐶 P | 𝑏 + 𝑐 + 𝑑 = 𝑞 let 𝑠 ? ≔ 𝑏, 𝑐, 𝑑 ∈ 𝐶 P | 𝑏 + 𝑐 + 𝑑 = 2𝑞 let 𝑠 ( ≔ then 𝑢 = 𝑠 _ + 𝑠 ? + 𝑠 (

Co-Nondeterministic Algorithm for 3SUM 3SUM: given set 𝐵 of integers in {−𝑜 Y , … , 𝑜 Y } , are there 𝑏, 𝑐, 𝑑 ∈ 𝐵 s.t. 𝑏 + 𝑐 + 𝑑 = 0 ? V(𝑜 P/( ) 3SUM has a co-nondeterministic algorithm in time 𝑃 Thm: [CGIMPS’16] 1) guess prime 𝑞 ≤ 𝑜 P/( log𝑜 𝑏,𝑐,𝑑 ∈ 𝐵 P | 𝑏 + 𝑐 + 𝑑 = 0 mod 𝑞 V(𝑞) 2) compute 𝑢 = 𝑃 3) if 𝑢 > 𝛽 ⋅ 𝑜 P/( log𝑜 : accept (constant 𝛽 to be fixed later) 4) guess distinct 𝑏 ? ,𝑐 ? ,𝑑 ? ,… , 𝑏 n ,𝑐 n , 𝑑 n ∈ 𝐵 P such that 𝑏 K + 𝑐 K + 𝑑 K = 0 mod 𝑞 ∀𝑗 5) check that for all 𝑏 K , 𝑐 K , 𝑑 K we have 𝑏 K + 𝑐 K + 𝑑 K ≠ 0 6) if everything works out: reject (otherwise accept) V(𝑜 P/( ) ✔ time 𝑃 ✔ if we reject then we have a NO-instance YES -instance: all paths accept NO -instance: at least one path rejects ✔