Complexity Theory of Polynomial-Time Problems Lecture 7: 3SUM II - PowerPoint PPT Presentation

Complexity Theory of Polynomial-Time Problems Lecture 7: 3SUM II Sebastian Krinninger

Reminder: 3SUM given sets 𝐵, 𝐶, 𝐷 of 𝑜 integers are there 𝑏 ∈ 𝐵, 𝑐 ∈ 𝐶, 𝑑 ∈ 𝐷 such that 𝑏 + 𝑐 + 𝑑 = 0 ? well-known: 𝑃(𝑜 2 ) Conjecture: no 𝑃 𝑜 2−𝜁 algorithm → 3SUM-Hardness Alternative algorithm: 𝑃( 𝐵 ⋅ 𝐶 + |𝐷|) (store negated pairwise sums in hashmap) 2/19 June 16, 2016

Reminder: Hashing Hash function ℎ: 𝑉 → [𝑆] 𝑦 ℎ(𝑦) 2 ⋯ 𝑆 1 Goal: Distribute uniformly, avoid collisions, etc. 3/19 June 16, 2016

Magical hash functions Desired properties for family of hash functions from 𝑉 → 𝑆 (i.e., for every ℎ chosen from family) Uniform difference: Pr ℎ 𝑦 − ℎ(𝑧) = 𝑗 = 1/𝑆 (for any 𝑦, 𝑧 ∈ 𝑉 s.t. 𝑦 ≠ 𝑧 and 𝑗 ∈ 𝑆 ) 𝑦 ∈ 𝑇 ∶ ℎ 𝑦 = 𝑗 ≤ 3𝑜/𝑆 Balanced: (for any set 𝑇 = 𝑦 1 , … , 𝑦 𝑜 ⊆ 𝑉 and any 𝑗 ∈ 𝑆 ) ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑦 + 𝑧 (mod 𝑆) Linear: (for any 𝑦, 𝑧 ∈ 𝑉 ) But: We do not know such a family… 4/19 June 16, 2016

5/19 June 16, 2016

Almost magical hash functions Desired properties for family of hash functions from 𝑉 → 𝑆 (i.e., for every ℎ chosen from family) Uniform difference: Pr ℎ 𝑦 − ℎ(𝑧) = 𝑗 = 1/𝑆 (for any 𝑦, 𝑧 ∈ 𝑉 s.t. 𝑦 ≠ 𝑧 and 𝑗 ∈ 𝑆 ) Almost balanced: Expected number of elements from S hashed to heavy values is 𝑃(𝑆) , where value 𝑗 ∈ 𝑆 is heavy if 𝑦 ∈ 𝑇 ∶ ℎ 𝑦 = 𝑗 > 3𝑜/𝑆 (for any set 𝑇 = 𝑦 1 , … , 𝑦 𝑜 ⊆ 𝑉 and any 𝑗 ∈ 𝑆 ) ℎ 𝑦 + ℎ 𝑧 ∈ ℎ 𝑦 + 𝑧 + 𝑑 ℎ + 0,1 (mod 𝑆) Almost linear: (for any 𝑦, 𝑧 ∈ 𝑉 and some integer 𝑑 ℎ depending only on ℎ ) 6/19 June 16, 2016

Definition of hash function Set 𝑠 = 𝑙𝑛 for some 𝑙 ≥ 𝑉/2 and 𝑉, 𝑆, 𝑠 powers of 2 ℋ 𝑉,𝑆,𝑠 = ℎ 𝑏,𝑐 : 𝑉 → 𝑆 ∣ 𝑏 ∈ 𝑠 odd integer and 𝑐 ∈ 𝑠 ℎ 𝑏,𝑐 𝑦 = 𝑏𝑦 + 𝑐 mod 𝑠 div 𝑠/𝑆 Thm: Family ℋ 𝑉,𝑆,𝑠 is has the uniform difference property, is almost balanced and almost linear with 𝑑 ℎ 𝑏,𝑐 = (𝑐 − 1 mod 𝑠) div 𝑠/𝑆 . (Pairwise independence [Dietzfelbinger ’96] implies uniform difference (easy to check) and almost balanced [Baran et al. ‘08]. Almost linear: easy to check.) Rest of this lecture: ℎ picked randomly from this family 7/19 June 16, 2016

Hashing down the universe Lem: If 3SUM on universe of size 𝑃(𝑜 3 ) solvable in exp. time 𝑃(𝑜 2−𝜗 ) , then 3SUM on arbitrary universe solvable in expect. time 𝑃(𝑜 2−𝜗 ) . Follows from [Baran et al. ‘08] Algorithm: Repeat until output: Pick hash function ℎ: 1 … 𝑉 → 1 … 6𝑜 3 at random • 𝐵 ′ = 𝑏 ∈ 𝐵 , 𝐶 ′ = {ℎ 𝑐 ∣ 𝑐 ∈ 𝐶} , 𝐷 ′ = ℎ 𝑏 ℎ 𝑑 + 𝑑 ℎ 𝑑 ∈ 𝐷 • 𝐵′ ′ = 𝑏 ∈ 𝐵 , 𝐶′ ′ = {ℎ 𝑐 ∣ 𝑐 ∈ 𝐶} , 𝐷′ ′ = ℎ 𝑏 ℎ 𝑑 + 𝑑 ℎ + 1 𝑑 ∈ 𝐷 • Solve two 3SUM instances 𝐵 ′ , 𝐶 ′ , 𝐷 ′ and 𝐵′ ′ , 𝐶′ ′ , 𝐷′ ′ • • If algorithm reports no 3SUM witness: output ‘no 3SUM’ Consider first reported 3SUM witness 𝑦 ′ , 𝑧 ′ , 𝑨 ′ for 𝐵 ′ , 𝐶 ′ , 𝐷 ′ : • If ℎ −1 𝑦′ , ℎ −1 𝑧′ , ℎ −1 𝑨 ′ − 𝑑 ℎ contains witness 𝑦, 𝑧, 𝑨 : output • 𝑦, 𝑧, 𝑨 Consider first reported 3SUM witness 𝑦′ ′ , 𝑧 ′′ , 𝑨 ′′ for 𝐵 ′′ , 𝐶 ′′ , 𝐷 ′′ : • If ℎ −1 𝑦′′ , ℎ −1 𝑧′′ , ℎ −1 𝑨 ′′ − 𝑑 ℎ − 1 contains witness 𝑦, 𝑧, 𝑨 : • output 𝑦, 𝑧, 𝑨 No false negatives: If 𝑦 + 𝑧 = 𝑨 , then ℎ 𝑦 + ℎ 𝑧 ∈ ℎ 𝑨 + 𝑑 ℎ + 0,1 8/19 June 16, 2016

Running Time We need to bound: • Number of iterations 𝑃(1) Number of candidate witnesses 𝑃(1) • Then: number of calls to 3SUM algorithm: 𝑃(1) Number of iterations: Triple 𝑦, 𝑧, 𝑨 gives false positive if 𝑦 + 𝑧 ≠ 𝑨 and one of ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑨 + 𝑑 ℎ or ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑨 + 𝑑 ℎ + 1 Linearity: ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑦 + 𝑧 + 𝑑 ℎ or ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑦 + 𝑧 + 𝑑 ℎ + 1 Thus, probability that fixed 𝑦, 𝑧, 𝑨 (with 𝑦 + 𝑧 ≠ 𝑨 ) gives false positive is: 3 1 Pr ℎ 𝑦 + 𝑧 − ℎ 𝑨 ∈ {−1,0,1} ≤ 6𝑜 3 = (uniform difference) 2𝑜 3 Overall probability of false positive : ≤ 𝑜 3 ⋅ 2𝑜 3 = 1 1 2 In expectation: 2 iterations until no false positive (waiting time bound) (If no false positive, then algorithm certainly stops) 9/19 June 16, 2016

Running Time We need to bound: • Number of iterations 𝑃(1) Number of candidate witnesses 𝑃(1) • Then: number of calls to 3SUM algorithm: 𝑃(1) Number of candidate witnesses: Fix 3SUM witness 𝑦 ′ , 𝑧 ′ , 𝑨′ of instance (𝐵 ′ , 𝐶 ′ , 𝐷 ′ ) Let 𝑦 ∗ ∈ ℎ −1 𝑦′ 1 For every 𝑦 ≠ 𝑦 ∗ : Pr ℎ 𝑦 = ℎ 𝑦 ∗ = (uniform difference) 6𝑜 3 𝐹 ℎ −1 𝑦 ′ 𝑜 ≤ 1 + 4𝑜 3 ≤ 2 Similarly: 𝐹 ℎ −1 𝑧 ′ ≤ 2 , 𝐹 ℎ −1 𝑨 ′ ≤ 2 𝐹 ℎ −1 𝑦 ′ ∪ ℎ −1 𝑧 ′ ∪ ℎ −1 𝑨 ′ ≤ 𝑃(1) (linearity of expectation) In expectation, algorithm manually checks constant number of candidate witnesses per iteration 10/19 June 16, 2016

Convolution 3SUM Given array 𝐵 1 … 𝑜 of integers are there 𝑗, 𝑘 such that 𝐵 𝑗 + 𝐵 𝑘 = 𝐵[𝑗 + 𝑘] ? 𝑦 + 𝑧 𝑦 𝑧 𝑗 𝑘 𝑗 + 𝑘 trivial algorithm: 𝑃(𝑜 2 ) Thm: There is no 𝑃(𝑜 2−𝜗 ) algorithm for Convolution 3SUM unless the 3SUM Conjecture fails. [ Pătrașcu 2010] Stepping stone towards hardness of other “structured” problems 11/19 June 16, 2016

Reduction from 3SUM Given set 𝑌 ⊆ [𝑉] of integers are there 𝑦, 𝑧, 𝑨 ∈ 𝑌 such that 𝑦 + 𝑧 = 𝑨 ? Preprocessing: Check if there is a solution 2𝑦 = 𝑨 𝑃 𝑜 log 𝑜 Pick random hash function ℎ: 𝑉 → 𝑆 (almost linear, etc.) For this proof: assume ℎ is almost balanced and linear (magically…) 3𝑜/𝑆 ⋮ 1 2 ⋯ 1 𝑆 In expectation: 𝑃 𝑆 elements in buckets with load > 3𝑜/𝑆 (almost bal.) For each such 𝑦 : check for 3SUM triple involving 𝑦 𝑃 𝑆𝑜 (in exp.) 12/19 June 16, 2016

Convolution 3SUM instance 3𝑜 Number elements in each bucket from 0 to 𝑆 − 1 Iterate over all triples 𝑗, 𝑘, 𝑙 ∈ 3𝑜/𝑆 𝑗 𝑘 𝑙 𝑢 ⋯ 𝑆 1 ⋯ ⋯ ⋯ For every bucket 𝑢 : Put 𝑗 -th element to 𝐵[8𝑢 + 1] • Put 𝑘 -th element to 𝐵[8𝑢 + 3] • • Put 𝑙 -th element to 𝐵[8𝑢 + 4] Set all other array entries to ∞ (sufficiently large number) 3 3𝑜 instances of Convolution 3SUM 𝑆 13/19 June 16, 2016

Correctness Assume 𝑦 + 𝑧 = 𝑨 Then ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑨 (mod 𝑆) (linearity) If 𝑦 = 𝑧 , triple found in preprocessing If 𝑦 , 𝑧 , or 𝑨 hashed to heavy bucket: triple found in second step Either ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑨 or ℎ 𝑦 + ℎ 𝑧 = ℎ 𝑨 + 𝑆 Duplicate array for Convolution 3SUM instance 𝐵 8ℎ 𝑦 + 1 + 𝐵 8ℎ 𝑧 + 3 = 𝐵 8ℎ 𝑨 + 4 or 𝐵 8ℎ 𝑦 + 1 + 𝐵 8ℎ 𝑧 + 3 = 𝐵 8(ℎ 𝑨 + 𝑆) + 4 Thus, no false negatives. Also no false positives: 𝐵 𝑗 + 𝐵 𝑘 = 𝐵 𝑗 + 𝑘 only if 𝑗 = 8𝑢 1 + 1 and 𝑘 = 8𝑢 2 + 3 Observation: ( 𝑦 + 𝑧 = 𝑨 mod 8 has unique solution over 1,3,4 and 𝐵 𝑗 ≠ 𝐵[𝑘] )

Running Time Assumption: Convolution 3SUM in time 𝑃(𝑜 2−𝜗 ) 3 𝑜 𝑜 2−𝜗 Total expected running time: 𝑃 𝑜 log 𝑜 + 𝑜𝑆 + 𝑆 Set 𝑆 = 𝑜 1−𝜗/4 Total time: 𝑃 𝑜 2−𝜗/4 Contradicts 3SUM Conjecture

Set Disjointness Problem 1. Preprocess subsets 𝒝, ℬ ⊆ 𝑉 over universe 𝑉 2. Answer queries : Given 𝐵 ∈ 𝒝, 𝐶 ∈ ℬ , is 𝐵 ∩ 𝐶 ≠ ∅ ? Repeated queries (Static) data structure Queries not known in advance Goal: Lower bound on preprocessing and query time Offline Set Disjointness: 𝑟 queries known in advance (part of input)

Reduction to 3SUM [Kopelowitz et al] Thm: Let 𝑔 𝑜 be such that 3SUM requires expected time Ω 𝑜 2 /𝑔(𝑜) . For any constant 0 ≤ 𝛿 < 1 , let ALG be an algorithm for offline Set Disjointness where 𝒝 = ℬ = Θ(𝑜 log 𝑜) , 𝑉 = Θ 𝑜 2−2𝛿 , each set in 𝒝 ∪ ℬ has at most 𝑃(𝑜 1−𝛿 ) elements from 𝑉 , and 𝑟 = Θ 𝑜 1+𝛿 log 𝑜 . Then ALG requires expected time Ω 𝑜 2 /𝑔(𝑜) . Assuming the 3SUM conjecture, for any 0 < 𝛿 < 1 , any data Cor: structure for Set Disjointness has 1+𝛿 2 2−𝛿−𝑝(1) 𝑢 𝑞 + 𝑂 2−𝛿 𝑢 𝑟 = Ω 𝑂 where 𝑂 is the sum of the set sizes, 𝑢 𝑞 is the preprocessing time, and 𝑢 𝑟 is the time per query. (From Thm: 𝑂 = Θ 𝑜 2−𝛿 log 𝑜 ) Example: Data structures with constant query time Make 𝛿 tend to 1 , need 𝑢 𝑞 = Ω 𝑂 2−𝑝(1) Evidence that trivial preprocessing algorithm is optimal (for constant query)