Satisfiability of Ordering CSPs Above Average Is Fixed-Parameter - - PowerPoint PPT Presentation
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 variables and constraints. Variables 1 ,
Ordering CSP
Given a set of π variables and π constraints.
- Variables π¦1, β¦ , π¦π
- Constraints π1, β¦ , ππ
Find a linear ordering of π¦1, β¦ , π¦π that maximizes the number of satisfied constraints.
π¦5 π¦1 π¦8 π¦3 π¦7 π¦2 π¦4 π¦7
Ordering CSP
Given a set of π variables and π constraints.
- Variables π¦1, β¦ , π¦π
- Constraints π1, β¦ , ππ
Find a linear ordering of π¦1, β¦ , π¦π that maximizes the number of satisfied constraints.
- Each constraints ππ has arity at most π.
- ππ (π¦π1, π¦π2, β¦ , π¦ππ) specifies a list of orderings of
π¦π1, β¦ , π¦ππ.
- ππ is satisfied if the relative ordering of π¦π1 β¦ , π¦ππ is in
the list.
Example 1: Max Acyclic Subgraph
- Given a directed graph π» on π¦1, β¦ , π¦π.
- 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 π» on π¦1, β¦ , π¦π.
- Find a linear ordering of vertices so as to
maximize the number of edges going forward. Each edge (π¦π, π¦π) defines constraint π¦π < π¦π #forward edges = #satisfied constraints The problem is an ordering CSP of arity 2.
Example 2: Betweenness
- Given a set of vertices π¦1, β¦ , π¦π and
- a set of betweenness constraints. Each constraint
is of the form βπ¦π lies between π¦π and π¦πβ π¦π < π¦π < π¦π or π¦π < π¦π < π¦π Find an ordering that maximizes the number of satisfied constraints.
π¦5 π¦1 π¦8 π¦3 π¦7 π¦2 π¦4 π¦7
NP-hardness
Max Acyclic Subgraph
- If all the constraints are satisfiable, the problem
can be easily solved.
- If πππ = 1 β π π, 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
- rdering CSP:
- rder π¦1, β¦ , π¦π randomly.
Max Acyclic Subgraph: each constraint is satisfied with probability Β½. Satisfy π΅ππ» = π/2 constraints in expectation. Betweenness: each constraint is satisfied with probability 1/3. Satisfy π΅ππ» = π/3 constraints in expectation.
Hardness of Approximation (UGC)
[Guruswami, HΓ₯stad, Manokaran, Raghavendra, Charikar]
There is no non-trivial multiplicative approximation algorithm for ordering CSP of any arity π. For every π > 0: No polynomial-time algorithm can find a solution satisfying at least 1 + π π΅ππ» constraints if πππ = 1 β π π.
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 πππ β₯ π΅ππ» + π’ or not.
Fixed Parameter Tractability
Conjecture of Gutin, van Iersel, Mnich, and Yeo. For every π, there a fixed-parameter tractable that decides whether πππ β₯ π΅ππ» + π’ or not. The running time of the algorithm is π
π π’ ππππ§π(π + π)
Fixed Parameter Tractability
[Alon, Gutin, Kim, Szeider, Yeo] Satisfiability above average is fixed-parameter tractable for all βregularβ (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] βit appears technically very difficult to extend results obtained for arities π = 2 and 3 to π > 3β
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 πππ’2 variables.
Then try all possible orderings of these variables in time 2π(π’2 log π’) and find the optimal solution. 2. πππ β₯ π΅ππ» + π’. (in case 1, there is a kernel on πππ’2 variables)
Approach
Consider a random ordering of π¦1, β¦ , π¦π. Let r.v. π be the number of constraints satisfied by the random ordering. E π = π΅ππ». Theorem 1 If the instance (non-trivially) depends
- n at least π variables then Var π β₯ π π .
Corollary If Var π < πβ² π’2 then the instance depends on at most πππ’2 variables. We are in case 1.
Approach
Consider a random ordering of π¦1, β¦ , π¦π. Let r.v. π be the number of constraints satisfied by the random ordering. E π = π΅ππ». Theorem 2 If Var π β₯ π then πππ β₯ π΅ππ» + π π
Var π β₯ π implies that π deviates by at least π from π΅ππ». For arbitrary r.v. π, it doesnβt follow that max π β₯ π΅ππ» + π π .
!
Consider a random ordering of π¦1, β¦ , π¦π. Let r.v. π be the number of constraints satisfied by the random ordering. E π = π΅ππ». Theorem 2 If Var π β₯ π then πππ β₯ π΅ππ» + π π Corollary If Var π β₯ π π’2 then πππ β₯ π΅ππ» + π’. We are in case 2.
Approach
Main Theorems
Theorem 1 If the instance (non-trivially) depends
- n at least π variables then Var π β₯ π π .
Theorem 2 If Var π β₯ π then πππ β₯ π΅ππ» + π π
Use the Fourier analysis: the EfronβStein decomposition. Prove a Bonami-type lemma for the EfronβStein decomposition.
EfronβStein Decomposition
To use the Fourier analysis β want to work with a product space. The product space should be large, but shouldnβt depend on π.
- Assume that each π¦π β [0,1].
- Each assignment (π¦1, β¦ , π¦π) β 0,1 π defines a
linear ordering of π¦1, β¦ , π¦π a.s.
- Random assignment defines a random ordering.
Fourier Analysis on the Boolean Cube
ES decomposition is similar to Fourier decomposition of functions on β1,1 π. For π: β1,1 π β β
π =
πβ{1,β¦,π}
π
πππ
- Function
π
πππ depends only on variables in π.
- Functions
π
πππ are mutually orthogonal.
- Var π = Var[
π
πππ]
- Have only
π
π with π β€ π for CSPs of arity π.
EfronβStein Decomposition
Consider π: [0,1]πβ β. There is a decomposition
π =
πβ{1,β¦,π}
π
π
Such that
- Function π
π depends only on variables in π.
- Functions π
π are mutually orthogonal.
- Var π = Var[π
π]
- Have only π
π with π β€ π for CSPs of arity π.
EfronβStein Decomposition
Consider π = 1.Then π(π¦1) = π
β + π 1(π¦1)
where
- π
β = E π
- π
1 π¦1 = π π¦1 β π β
EfronβStein Decomposition
Consider π = 2. Assume
π = π π¦1 β β(π¦2)
Then
π = πβ + π1 π¦1 ββ + β2 π¦2 = πβ ββ + π1 π¦1 ββ + πβ β2 π¦2 + π1 π¦1 β2 π¦2
Let
- π
β = πβ ββ
- π
{1} = π1(π¦1)ββ
- π
{2} = πβ β2(π¦2)
- π
{1,2} = π1(π¦1)β2(π¦2)
EfronβStein Decomposition
For π > 2. Assume π = π(1) π¦1 β π 2 π¦2 β¦ π π (π¦π) Decompose each π(π) π(π) = πβ
π + ππ π π¦π
Expand the expression for π, get 2π termsβone for each set π. Extend by linearity to all functions π: [0,1]πβ β.
Explicit Formulas
Define π
βπ = E π π¦π with π β π]
Then π
π = πβπ
β1 |πβπ| π
βπ
ES decomposition of π
- π is a sum of indicators π½of elementary events of
the form π¦1 < π¦2 < β― < π¦π.
- Use explicit formulas to compute the ES
decomposition of π½. π½βπ = 1 π΅π ππ π¦π π½{π¦π π‘1 < π¦π π‘2 < β― } where ππ are polynomials with integer coefficients
- f degree at most π.
- By linearity, π½π and ππ are of the same form.
ES decomposition of π
πS = 1 π΅π πβ²π π¦π π½{π¦π π‘1 < π¦π π‘2 < β― } Thus if ππ β 0 Var ππ β₯ πΆπ > 0
Proof of Theorem 1
Theorem 1 If the instance (non-trivially) depends
- n at least π variables then Var π β₯ π π .
Proof Idea:
- Consider the ES decomposition of π
- Each ππ depends on at most k variables
- There are at least π /π non-zero terms.
- For each of them, Var ππ β₯ πΆπ
- Thus Var π = Var [ππ] β₯ π πΆπ/π
Proof of Theorem 2
Theorem 2 If Var π β₯ π then πππ β₯ π΅ππ» + π π Need to show that E π β π΅ππ» 4 < π·πVar π 2 Then by [Alon et al] π > π΅ππ» + π π with positive probability.
Bonami Lemma
Bonami Lemma. Let π be a polynomial of degree π on β1, 1 π E π4 β€ 9πE π2 2 Bonami Lemma for ES. Let π be a function with ES decomposition of degree π. Then E π4 β€ 81ππ·π E π2 2 If πΉ π
π1π π2π π3π π4 β€ π·π π
Var π
ππ 1/2
Proof of Theorem 2
Theorem 2 If Var π β₯ π then πππ β₯ π΅ππ» + π π Letting π = π β π΅ππ», get E π β π΅ππ» 4 < π·π81π Var π 2 Then π > π΅ππ» + π π with positive probability.
Other Results
Result applies to CSPs on domain [0, 1] with constraints defined by linear or bounded degree inequalities with bounded integer coefficients. E.g., constraints of the form
- π¦π is greater than the average of π¦π and π¦π, or
- π¦π is closer to π¦π than to π¦π
Summary
- Proved the conjecture of Gutin, van Iersel, Mnich,
and Yeo
- Proved a Bonami-type Inequality for the Efronβ
Stein decomposition.
- Proved that the satisfiability beyond average is