Satisfiability of Ordering CSPs Above Average Is Fixed-Parameter - PowerPoint PPT Presentation

Satisfiability of Ordering CSPs Above Average Is Fixed-Parameter Tractable Yury Makarychev, TTIC Konstantin Makarychev, Microsoft Research Yuan Zhou, MIT

Ordering CSP Given a set of 𝑜 variables and 𝑛 constraints. • Variables 𝑦 1 , … , 𝑦 𝑜 • Constraints 𝜌 1 , … , 𝜌 𝑛 Find a linear ordering of 𝑦 1 , … , 𝑦 𝑛 that maximizes the number of satisfied constraints. 𝑦 5 𝑦 7 𝑦 1 𝑦 4 𝑦 8 𝑦 3 𝑦 7 𝑦 2

Ordering CSP Given a set of 𝑜 variables and 𝑛 constraints. • Variables 𝑦 1 , … , 𝑦 𝑜 • Constraints 𝜌 1 , … , 𝜌 𝑛 Find a linear ordering of 𝑦 1 , … , 𝑦 𝑛 that maximizes the number of satisfied constraints. • Each constraints 𝜌 𝑠 has arity at most 𝑙 . • 𝜌 𝑠 (𝑦 𝑗 1 , 𝑦 𝑗 2 , … , 𝑦 𝑗 𝑙 ) specifies a list of orderings of 𝑦 𝑗 1 , … , 𝑦 𝑗 𝑙 . • 𝜌 𝑠 is satisfied if the relative ordering of 𝑦 𝑗 1 … , 𝑦 𝑗 𝑙 is in the list.

Example 1: Max Acyclic Subgraph • Given a directed graph 𝐻 on 𝑦 1 , … , 𝑦 𝑜 . • Find a linear ordering of vertices so as to maximize the number of edges going forward.

Example 1: Max Acyclic Subgraph • Given a directed graph 𝐻 on 𝑦 1 , … , 𝑦 𝑜 . • Find a linear ordering of vertices so as to maximize the number of edges going forward. Each edge (𝑦 𝑗 , 𝑦 𝑘 ) defines constraint 𝑦 𝑗 < 𝑦 𝑘 #forward edges = #satisfied constraints The problem is an ordering CSP of arity 2.

Example 2: Betweenness • Given a set of vertices 𝑦 1 , … , 𝑦 𝑜 and • a set of betweenness constraints. Each constraint is of the form “ 𝑦 𝑗 lies between 𝑦 𝑘 and 𝑦 𝑙 ” 𝑦 𝑘 < 𝑦 𝑗 < 𝑦 𝑙 or 𝑦 𝑙 < 𝑦 𝑗 < 𝑦 𝑘 Find an ordering that maximizes the number of satisfied constraints. 𝑦 5 𝑦 7 𝑦 1 𝑦 4 𝑦 8 𝑦 3 𝑦 7 𝑦 2

NP-hardness Max Acyclic Subgraph • If all the constraints are satisfiable, the problem can be easily solved. • If 𝑃𝑄𝑈 = 1 − 𝜁 𝑛 , the problem is NP-hard. Betweenness • The problem is NP-hard even when all the constraints are satisfiable.

Random Assignment There is a trivial approximation algorithm for ordering CSP: order 𝑦 1 , … , 𝑦 𝑜 randomly. Max Acyclic Subgraph: each constraint is satisfied with probability ½. Satisfy 𝐵𝑊𝐻 = 𝑛/2 constraints in expectation. Betweenness: each constraint is satisfied with probability 1/3. Satisfy 𝐵𝑊𝐻 = 𝑛/3 constraints in expectation.

Hardness of Approximation (UGC) [Guruswami, Håstad, Manokaran, Raghavendra, Charikar] There is no non-trivial multiplicative approximation algorithm for ordering CSP of any arity 𝑙 . For every 𝜁 > 0 : No polynomial-time algorithm can find a solution satisfying at least 1 + 𝜁 𝐵𝑊𝐻 constraints if 𝑃𝑄𝑈 = 1 − 𝜁 𝑛 .

Advantage over Random [GHMRC] No algorithm performs considerably better than random. Can we get some additive advantage over random?

Advantage over Random [GHMRC] No algorithm performs considerably better than random. Can we get some additive advantage over random? Conjecture of Gutin, van Iersel, Mnich, and Yeo. There a fixed-parameter algorithm that decides whether 𝑃𝑄𝑈 ≥ 𝐵𝑊𝐻 + 𝑢 or not.

Fixed Parameter Tractability Conjecture of Gutin, van Iersel, Mnich, and Yeo. For every 𝑙 , there a fixed-parameter tractable that decides whether 𝑃𝑄𝑈 ≥ 𝐵𝑊𝐻 + 𝑢 or not. The running time of the algorithm is 𝑔 𝑙 𝑢 𝑞𝑝𝑚𝑧 𝑙 (𝑛 + 𝑜)

Fixed Parameter Tractability [Alon, Gutin, Kim, Szeider, Yeo] Satisfiability above average is fixed- parameter tractable for all “regular” (non-ordering) CSPs Conjecture was proved for: Gutin, Kim, Szeider, Yeo Max Acyclic Subgraph Gutin, Kim, Mnich, Yeo Betweenness Gutin, van Iersel, Mnich, Yeo Ordering CSPs of arity 3 [GIMY] “it appears technically very difficult to extend results obtained for arities 𝑠 = 2 and 3 to 𝑠 > 3 ”

Our Results Prove the conjecture of Gutin et al. Prove that the satisfiability above average is fixed- parameter tractable for a large class of CSPs, which includes ordering CSPs.

Approach Follow the high-level approach of Alon, Gutin, Kim, Szeider, and Yeo. Prove that there are two possibilities: 1. The instance depends on at most 𝑑 𝑙 𝑢 2 variables. Then try all possible orderings of these variables in time 2 𝑃(𝑢 2 log 𝑢) and find the optimal solution. 𝑃𝑄𝑈 ≥ 𝐵𝑊𝐻 + 𝑢. 2. (in case 1, there is a kernel on 𝑑 𝑙 𝑢 2 variables)

Approach Consider a random ordering of 𝑦 1 , … , 𝑦 𝑜 . Let r.v. 𝑎 be the number of constraints satisfied by the random ordering. E 𝑎 = 𝐵𝑊𝐻. Theorem 1 If the instance (non-trivially) depends on at least 𝑠 variables then Var 𝑎 ≥ 𝑏 𝑠 . Corollary If Var 𝑎 < 𝑏′ 𝑢 2 then the instance depends on at most 𝑑 𝑙 𝑢 2 variables. We are in case 1.

Approach Consider a random ordering of 𝑦 1 , … , 𝑦 𝑜 . Let r.v. 𝑎 be the number of constraints satisfied by the random ordering. E 𝑎 = 𝐵𝑊𝐻. Theorem 2 If Var 𝑎 ≥ 𝑠 then 𝑃𝑄𝑈 ≥ 𝐵𝑊𝐻 + 𝑐 𝑠 Var 𝑎 ≥ 𝑠 implies that 𝑎 deviates by at least 𝑠 from ! 𝐵𝑊𝐻 . For arbitrary r.v. 𝑎 , it doesn’t follow that max 𝑎 ≥ 𝐵𝑊𝐻 + 𝑐 𝑠 .

Approach Consider a random ordering of 𝑦 1 , … , 𝑦 𝑜 . Let r.v. 𝑎 be the number of constraints satisfied by the random ordering. E 𝑎 = 𝐵𝑊𝐻. Theorem 2 If Var 𝑎 ≥ 𝑠 then 𝑃𝑄𝑈 ≥ 𝐵𝑊𝐻 + 𝑐 𝑠 Corollary If Var 𝑎 ≥ 𝑏 𝑢 2 then 𝑃𝑄𝑈 ≥ 𝐵𝑊𝐻 + 𝑢 . We are in case 2.

Main Theorems Theorem 1 If the instance (non-trivially) depends on at least 𝑠 variables then Var 𝑎 ≥ 𝑏 𝑠 . Theorem 2 If Var 𝑎 ≥ 𝑠 then 𝑃𝑄𝑈 ≥ 𝐵𝑊𝐻 + 𝑐 𝑠 Use the Fourier analysis: the Efron — Stein decomposition. Prove a Bonami-type lemma for the Efron — Stein decomposition.

Efron — Stein Decomposition To use the Fourier analysis — want to work with a product space. The product space should be large, but shouldn’t depend on 𝑜 . • Assume that each 𝑦 𝑗 ∈ [0,1] . • Each assignment (𝑦 1 , … , 𝑦 𝑜 ) ∈ 0,1 𝑜 defines a linear ordering of 𝑦 1 , … , 𝑦 𝑜 a.s. • Random assignment defines a random ordering.

Fourier Analysis on the Boolean Cube ES decomposition is similar to Fourier decomposition of functions on −1,1 𝑜 . For 𝑔: −1,1 𝑜 → ℝ 𝑔 = 𝑔 𝑇 𝜓 𝑇 𝑇⊂{1,…,𝑜} • Function 𝑔 𝑇 𝜓 𝑇 depends only on variables in 𝑇 . • Functions 𝑔 𝑇 𝜓 𝑇 are mutually orthogonal. • Var 𝑔 = Var[ 𝑔 𝑇 𝜓 𝑇 ] • Have only 𝑔 𝑇 with 𝑇 ≤ 𝑙 for CSPs of arity 𝑙 .

Efron — Stein Decomposition Consider 𝑔: [0,1] 𝑜 → ℝ . There is a decomposition 𝑔 = 𝑔 𝑇 𝑇⊂{1,…,𝑜} Such that • Function 𝑔 𝑇 depends only on variables in 𝑇 . • Functions 𝑔 𝑇 are mutually orthogonal. • Var 𝑔 = Var[𝑔 𝑇 ] • Have only 𝑔 𝑇 with 𝑇 ≤ 𝑙 for CSPs of arity 𝑙 .

Efron — Stein Decomposition Consider 𝑜 = 1 .Then 𝑔(𝑦 1 ) = 𝑔 ∅ + 𝑔 1 (𝑦 1 ) where • 𝑔 ∅ = E 𝑔 • 𝑔 1 𝑦 1 = 𝑔 𝑦 1 − 𝑔 ∅

Efron — Stein Decomposition Consider 𝑜 = 2 . Assume 𝑔 = 𝑕 𝑦 1 ⋅ ℎ(𝑦 2 ) Then 𝑔 = 𝑕 ∅ + 𝑕 1 𝑦 1 ℎ ∅ + ℎ 2 𝑦 2 = 𝑕 ∅ ℎ ∅ + 𝑕 1 𝑦 1 ℎ ∅ + 𝑕 ∅ ℎ 2 𝑦 2 + 𝑕 1 𝑦 1 ℎ 2 𝑦 2 Let • 𝑔 ∅ = 𝑕 ∅ ℎ ∅ • 𝑔 {1} = 𝑕 1 (𝑦 1 )ℎ ∅ • 𝑔 {2} = 𝑕 ∅ ℎ 2 (𝑦 2 ) • 𝑔 {1,2} = 𝑕 1 (𝑦 1 )ℎ 2 (𝑦 2 )

Efron — Stein Decomposition For 𝑜 > 2 . Assume 𝑔 = 𝑕 (1) 𝑦 1 ⋅ 𝑕 2 𝑦 2 … 𝑕 𝑜 (𝑦 𝑜 ) Decompose each 𝑕 (𝑗) 𝑗 + 𝑕 𝑗 𝑗 𝑦 𝑗 𝑕 (𝑗) = 𝑕 ∅ Expand the expression for 𝑔 , get 2 𝑜 terms — one for each set 𝑇 . Extend by linearity to all functions 𝑔: [0,1] 𝑜 → ℝ .

Explicit Formulas Define 𝑔 ⊂𝑈 = E 𝑔 𝑦 𝑗 with 𝑗 ∈ 𝑈] Then −1 |𝑇∖𝑈| 𝑔 𝑔 𝑇 = ⊂𝑈 𝑈⊆𝑇

ES decomposition of 𝑎 • 𝑎 is a sum of indicators 𝐽 of elementary events of the form 𝑦 1 < 𝑦 2 < ⋯ < 𝑦 𝑙 . • Use explicit formulas to compute the ES decomposition of 𝐽 . 𝐽 ⊂𝑇 = 1 𝑞 𝜌 𝑦 𝑇 𝐽{𝑦 𝜌 𝑡 1 < 𝑦 𝜌 𝑡 2 < ⋯ } 𝐵 𝑙 where 𝑞 𝜌 are polynomials with integer coefficients of degree at most 𝑙 . • By linearity, 𝐽 𝑇 and 𝑎 𝑇 are of the same form.

ES decomposition of 𝑎 𝑎 S = 1 𝑞′ 𝜌 𝑦 𝑇 𝐽{𝑦 𝜌 𝑡 1 < 𝑦 𝜌 𝑡 2 < ⋯ } 𝐵 𝑙 Thus if 𝑎 𝑇 ≠ 0 Var 𝑎 𝑇 ≥ 𝐶 𝑙 > 0

Proof of Theorem 1 Theorem 1 If the instance (non-trivially) depends on at least 𝑠 variables then Var 𝑎 ≥ 𝑏 𝑠 . Proof Idea: • Consider the ES decomposition of 𝑎 • Each 𝑎 𝑇 depends on at most k variables • There are at least 𝑠/𝑙 non-zero terms. • For each of them, Var 𝑎 𝑇 ≥ 𝐶 𝑙 • Thus Var 𝑎 = Var [𝑎 𝑇 ] ≥ 𝑠𝐶 𝑙 /𝑙