Download presentation

Presentation is loading. Please wait.

Published bySavannah Lawrence Modified over 3 years ago

1
The Connectivity of Boolean Satisfiability: Computational & Structural Dichotomies Parikshit GopalanGeorgia Tech Phokion KolaitisIBM Almaden Elitza ManevaUC Berkeley Christos PapadimitriouUC Berkeley

2
Boolean Satisfiability Problems 3-SAT (x 1 Ç x 2 Ç x 3 ) Æ (:x 2 Ç : x 3 Ç x 4 ) 2-SAT (x 1 Ç x 2 ) Æ (x 3 Ç x 4 ) 3-LIN-SAT (x 1 + x 2 + x 3 = 1) Æ (x 2 + x 4 + x 5 = 0) NotAllEqual-SAT NAE(x 1, x 2, x 3 ) Æ NAE(x 2, x 3, x 4 ) NP-complete P P

3
Boolean Satisfiability Problems Q: What makes some SAT problems hard? A: Classification of easy and hard SAT problems: Schaefers Dichotomy Theorem [1978] Easy Classes in P: 1.2-SAT 2.LIN-SAT 3.Horn-SAT 4.Dual-Horn-SAT Everything else is NP-complete.

4
Boolean Satisfiability Problems What is a Boolean SAT problem? Boolean relation : R1R1 R2R2 00 0110 11 Given a set of Boolean Relations S = {R 1, …, R k }. CNF(S) consists of formulae: x 1,…,x n ) = R 1 (x 1, x 16 ) Æ R 2 (x 4, x 5, x 32 ) … Æ R 1 (x n-1, x n )

5
Boolean Satisfiability Problems Example S = {R}. R 000 111 Gives NAE-3SAT.

6
Schaefers Dichotomy Theorem R is bijunctive if it is a 2-SAT formula. R(x 1,x 2 ) : : x 1 Ç : x 2

7
Schaefers Dichotomy Theorem R is bijunctive if it is a 2-SAT formula. R(x 1, x 2, x 3 ) : (x 1 Ç x 2 ) Æ (x 2 Ç x 3 ) Æ (x 1 Ç x 3 ) S = {R 1,…,R k } is bijunctive if all R i s are bijunctive. CNF(S) is easy, run any 2-SAT algorithm.

8
Schaefers Dichotomy Theorem R is linear if it is a system of linear equations mod 2. S = {R 1,…,R k } is linear if all R i s are linear. CNF(S) is easy, run Gaussian elimination. R(x 1,x 2 ) : x 1 © x 2 = 1 mod 2

9
Schaefers Dichotomy Theorem The Schaefer sets of relations: 1. Bijunctive 2. Linear 3. Horn 4. Dual Horn Satisfiability is in P. For non-Schaefer sets, CNF(S) is NP-complete.

10
Dichotomy Theorems SatisfiabilitySchaefer CountingCreignou-Hermann Approximability Khanna-Sudan-Trevisan- Williamson, Creignou- Khanna-Sudan 3-valued SAT problemsBulatov Minimal satisfiabilityKirousis-Kolaitis Inverse satisfiabilityKavvadias-Sideri Propositional abductionCreignou-Zanuttini

11
Structure of the Solution Space Given (x 1,…,x n ) 2 CNF(S), its solutions induce a subgraph G( ) of the hypercube. For which relations S does G( ) have nice structure ? What is nice structure ? Why is this interesting ?

12
Random formula with m clauses. Phase transition in solution space structure as m increases. Seems to control performance of heuristics. Number of components for k-SAT : 2-SAT: a single component 3-SAT to 7-SAT: not known 8-SAT above : exponential number of components [Achlioptas, Ricci-Tersenghi `06] [Mezard, Mora, Zecchina `05] Solution Space Structure for Random Instances

13
Structure of the Solution Space What is nice structure? Structural Properties : What kinds of graphs G( ) are possible? Number of Components, Diameter, … Computational Properties : Is G( ) connected ? Are two solutions in the same component ? All give the same answer...

14
A New Dichotomy Connectivity is PSPACE- complete. Connectivty is in co-NP. st-Connectivity is PSPACE- complete. st-Connectivity is in P. Diameter can be exponential in n. Diameter is linear in number of variables n. G( ) can be arbitrary.G( ) has nice structure. Non-Tight SetsTight Sets

15
A New Dichotomy Three kinds of Tight sets of relations: 1.OR-free 2.NAND-free 3.Componentwise Bijunctive Each has a nice structural property. Inherited by all CNF formulas. Results in small diameter, connectivity algorithms…

16
OR-free Relations Cannot set variables in R(x 1,…,x k ) to constants and get x 1 Ç x 2. G(R) does not contain :

17
OR-free Relations Cannot set variables in R(x 1,…,x k ) to constants and get x 1 Ç x 2. Example of an OR-free relation :

18
OR-free Relations Cannot set variables in R(x 1,…,x k ) to constants and get x 1 Ç x 2. Example of relation which contains OR: NAE(x 1, x 2, 0) = x 1 Ç x 2

19
OR-free Relations Cannot set variables in R(x 1,…,x k ) to constants and get x 1 Ç x 2. G(R) does not contain : Structural Property : Every component has a unique minimum. If S is OR-free, every CNF(S) formula inherits this property.

20
OR-free Relations Structural Property : Every component has a unique minimum. Diameter of a Component, st-Connectivity:

21
NAND-free Relations Cannot set variables in R(x 1,…,x k ) to constants and get : x 1 Ç : x 2. G(R) does not contain : Structural Property : Every component has a unique maximum.

22
C up = x i Ç x j C down = : x i Ç : x j Each component is bijunctive. (2-SAT formula). Examples : All bijunctive relations, Componentwise Bijunctive Relations C itself is not bijunctive. C:

23
Each component is bijunctive. (2-SAT formula). Componentwise Bijunctive Relations Structural Property : Within each component, graph distance equals Hamming distance.

24
Each component is bijunctive. (2-SAT formula). Componentwise Bijunctive Relations Structural Property : Within each component, graph distance equals Hamming distance. Not a characterization. NAE has this property, it is not bijunctive.

25
Each component is bijunctive. (2-SAT formula). Componentwise Bijunctive Relations Structural Property : Within each component, graph distance equals Hamming distance. Inherited by CNF formulas. Characterization Results in small diameter, algorithms for st- Connectivity …

26
Relation to Schaefer Classes Dual - Horn Horn OR-freeNAND-free Comp. Bijunctive Linear Bijunctive

27
A New Dichotomy Three kinds of Tight sets of relations: 1.OR-free 2.NAND-free 3.Componentwise Bijunctive Each has a nice structural property. Inherited by all CNF formulas. If S is not Tight, G( ) can be arbitrary.

28
Schaefers Expressibility Theorem Theorem : If S is not Schaefer, every relation R is expressible as a CNF(S) formula. Expressibility : There is a formula 2 CNF(S) s.t. R(x 1,…, x k ) = 9 y 1, …, y t s.t. (x 1,…,x k,y 1,…, y t ) Taking R to be 3-SAT clauses; gives the hard part of Schaefers Dichotomy. Need a notion of expressibility that is faithful to the solution space structure.

29
Faithful Expressibility R(x 1,…,x k ) y 1,…,y 3,x 1,…,x k ) 1.Projecting onto x gives R. 2.Witness space is connected. 3.Matching witness property.

30
Faithful Expressibility Theorem Faithful Expressibility : Preserves diameter up to polynomial factors. Reduces Connectivity and st-Connectivity. Theorem : If S is not tight, every relation R is faithfully expressible as a CNF(S) formula. For 3-SAT formulas Diameter : exponential. Connectivty, st-Connectivity : PSPACE-complete.

31
A New Dichotomy Connectivity is PSPACE- complete. Connectivty is in co-NP. st-Connectivity is PSPACE- complete. st-Connectivity is in P. Diameter can be exponential in n. Diameter is linear in number of variables n. G( ) can be arbitrary.G( ) has nice structure. Non-Tight SetsTight Sets

32
Open Questions 2. Trichotomy for Connectivity ? Known to be in coNP for Tight sets of relations. Conjecture : In P for Schaefer sets; coNP- complete otherwise. 1. Are Tight but non-Schaefer sets easy on average?

33
Faithful Expressibility Theorem Theorem : If S is not tight, every relation R is faithfully expressible as a CNF(S) formula. Proof in 4 steps. Step 0: Express 2-SAT clauses. Some relation contains OR, NAND. Other 2-SAT clauses by resolution.

34
Faithful Expressibility Theorem Theorem : If S is not tight, every relation R is faithfully expressible as a CNF(S) formula. Proof in 4 steps. Step 1: Express T where some distance expands. Some R 2 S is not componentwise bijunctive. Æ (x 1 Ç x 3 ) =

35
Faithful Expressibility Theorem Theorem : If S is not tight, every relation R is faithfully expressible as a CNF(S) formula. Proof in 4 steps. Step 2 : Express a path of length 4 between vertices at distance 2.

36
Faithful Expressibility Theorem Theorem : If S is not tight, every relation R is faithfully expressible as a CNF(S) formula. Proof in 4 steps. Step 3: Express all 3-SAT clauses from such paths. [Demaine-Hearne]

37
Faithful Expressibility Theorem Theorem : If S is not tight, every relation R is faithfully expressible as a CNF(S) formula. Proof in 4 steps. Step 4: Express all relations R from 3-SAT clauses.

38
A New Dichotomy st-Connectivity is PSPACE- hard. st-Connectivity is in P. Connectivity is PSPACE- hard. Connectivty is in NP. Diameter can be exponential in n. Diameter is linear in number of variables n. G( ) can be arbitrary.G( ) has nice structure. Non-Tight SetsTight Sets

39
Open Questions 2. Trichotomy for Connectivity ? Known to be in coNP for Tight sets of relations. Conjecture : In P for Schaefer sets; coNP- complete otherwise. 1. Are Tight but non-Schaefer sets easy on average?

Similar presentations

Presentation is loading. Please wait....

OK

The Theory of NP-Completeness

The Theory of NP-Completeness

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on public speaking tips Download ppt on coal and petroleum for class 8 Ppt on polynomials for class 8 Ppt on political parties and electoral process in ghana Ppt on cross-sectional study meaning Ppt on obesity prevention policy A ppt on satellite communication English 8 unit 13 read ppt on iphone Ppt on real estate sector in india Ppt on leverages resources