Computing Inconsistency Measurements under Multi-Valued Semantics by - PowerPoint PPT Presentation

Computing Inconsistency Measurements under Multi-Valued Semantics by Partial Max-SAT Solvers Guohui Xiao 1 , 2 Zuoquan Lin 1 Yue Ma 3 Guilin Qi 4 1 Department of Information Science, Peking University, China 2 Knowledge Based Systems Group, Institute of Information Systems, Vienna University of Technology Laboratoire d’Informatique de l’universit´ e Paris-Nord, Universit´ e Paris Nord - CNRS, France 4 School of Computer Science and Engineering, Southeast University, China xiao@kr.tuwien.ac.at, yue.ma@lipn.univ-paris13.fr, gqi@seu.edu.cn, lz@is.pku.edu.cn N U I V G E N R S I K I E T P Y 1 8 8 9 February 8, 2011

Empty Head Outline ➤ Motivation ➤ Inconsistency Measures under Multi-Valued Semantics ➤ Relationship among Different Measurements ➤ Encoding Algorithms ➤ Conclusion and Future Work 1/24

Empty Head Motivation ➤ Consistent KBs serve as useful knowledge resources v.s. inconsistent KBs imply any conclusion (meaningless!) ➤ For handling inconsistent KBs: • paraconsistent reasoning (1960s) • knowledge diagnose and repair (1980s) • Which approach should we take? � inconsistency measurement: a guidance to choose different approaches (2000s) 2/24

Empty Head Motivation ➤ Consistent KBs serve as useful knowledge resources v.s. inconsistent KBs imply any conclusion (meaningless!) ➤ For handling inconsistent KBs: • paraconsistent reasoning (1960s) • knowledge diagnose and repair (1980s) • Which approach should we take? � inconsistency measurement: a guidance to choose different approaches (2000s) ➤ Problem • Relationship among different measurements • Efficient algorithms 2/24

Empty Head Definitions 3/24

Empty Head Definitions ➤ Multi-Valued Semantics • 4-valued, 3-valued, LP m , Quasi-Classical, . . . • I : Var ( K ) → { t , f , Both , None } 3/24

Empty Head Definitions ➤ Multi-Valued Semantics • 4-valued, 3-valued, LP m , Quasi-Classical, . . . • I : Var ( K ) → { t , f , Both , None } ➤ ID of K respect to I under i -semantics ( i = 3 , 4 , LP m , Q ) ID i ( K , I ) = |{ p | p I = B , p ∈ Var ( K ) }| , if I | = i K | Var ( K ) | 3/24

Empty Head Definitions ➤ Multi-Valued Semantics • 4-valued, 3-valued, LP m , Quasi-Classical, . . . • I : Var ( K ) → { t , f , Both , None } ➤ ID of K respect to I under i -semantics ( i = 3 , 4 , LP m , Q ) ID i ( K , I ) = |{ p | p I = B , p ∈ Var ( K ) }| , if I | = i K | Var ( K ) | ➤ ID of K under under i -semantics ( i = 3 , 4 , LP m , Q ) ID i ( K ) = min = K ID i ( K , I ) I | 3/24

Empty Head Inconsistency Degree under 4-valued Semantics ➤ ID 4 ( K , I ) = |{ p | p I = B , p ∈ Var ( K ) }| Truth values: { t , f , B , N } | Var ( K ) | 4-model I : ID 4 ( K ) = min I | = 4 K ID 4 ( K ), K → { t , B } � K = { p , ¬ q , ¬ p ∨ q , r ∨ s } � I 1 : p I 1 = B , q I 1 = f , r I 1 = t , s I 1 = t , I 2 : p I 2 = B , q I 2 = B , r I 2 = t , s I 2 = t I 3 : p I 3 = B , q I 3 = B , r I 3 = t , s I 3 = N � ID 4 ( K , I 1 ) = 1 4 , ID 4 ( K , I 2 ) = 2 4 ID 4 ( K , I 3 ) = 2 4 Figure: FOUR ID 4 ( K ) = 1 4 4/24

Empty Head Inconsistency Degree under 3-valued Semantics ➤ ID 3 ( K , I ) = |{ p | p I = B , p ∈ Var ( K ) }| Truth values: { t , f , B } | Var ( K ) | 3-model I : ID 3 ( K ) = min I | = 3 K ID 3 ( K ), K → { t , B } � K = { p , ¬ q , ¬ p ∨ q , r ∨ s } � I 1 : p I 1 = B , q I 1 = f , r I 1 = t , s I 1 = t I 2 : p I 2 = B , q I 2 = B , r I 2 = t , s I 2 = t I 3 : p I 3 = B , q I 3 = B , r I 3 = t , s I 3 = N ———————————————– � ID 3 ( K , I 1 ) = 1 4 , ID 3 ( K , I 2 ) = 2 4 ID 3 ( K , I 3 ) = 2 ——————- 4 Figure: Three ID 3 ( K ) = 1 4 5/24

Empty Head Inconsistency Degree under LPm Semantics ➤ ID LP m ( K , I ) = |{ p | p I = B , p ∈ Var ( K ) }| | Var ( K ) | ID LP m ( K ) = min I | = LPm K ID LP m ( K ), � K = { p , ¬ q , ¬ p ∨ q , r ∨ s } LP m interpretation: � I 1 : p I 1 = B , q I 1 = f , r I 1 = t , s I 1 = t , ➤ 3-valued interpretation I 2 : p I 2 = B , q I 2 = B , r I 2 = t , s I 2 = t ———————————————– ➤ only “most classical” I 3 : p I 3 = B , q I 3 = B , r I 3 = t , s I 3 = N ———————————————– ones are considered � ID LP m ( K , I 1 ) = 1 ID LP m ( K , I 2 ) = 2 4 , ———————— 4 ID LP m ( K , I 3 ) = 2 ———————– 4 ID LP m ( K ) = 1 4 6/24

Empty Head Inconsistency Degree under Quasi-Classical Semantics ➤ ID Q ( K , I ) = |{ p | p I = B , p ∈ Var ( K ) }| | Var ( K ) | ID Q ( K ) = min I | = Q K ID Q ( K ), Quasi-Classical (Q) interpretation: � K = { p , ¬ q , ¬ p ∨ q , r ∨ s } ➤ 4-valued interpretation I 1 : p I 1 = B , q I 1 = f , r I 1 = t , s I 1 = t � ———————————————- ➤ Resolution laws are I 2 : p I 2 = B , q I 2 = B , r I 2 = t , s I 2 = t satisfied I 3 : p I 3 = B , q I 3 = B , r I 3 = t , s I 3 = N I | = Q α ∨ β, I | = Q ¬ β ∨ γ ID Q ( K , I 1 ) = 1 4 , ID Q ( K , I 2 ) = 2 � ——————– 4 ⇒ I | = Q α ∨ γ ID Q ( K , I 3 ) = 2 4 ID Q ( K ) = 2 4 7/24

Empty Head Relationship 8/24

Empty Head Relationship Theorem Given a knowledge base K , then ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K ). 8/24

Empty Head Relationship Theorem Given a knowledge base K , then ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K ). Proof Hints: 8/24

Empty Head Relationship Theorem Given a knowledge base K , then ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K ). Proof Hints: ➤ ID 3 ( K ) ≥ ID 4 ( K ) : Trivial since I | = 3 K ⇒ I | = 4 K 8/24

Empty Head Relationship Theorem Given a knowledge base K , then ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K ). Proof Hints: ➤ ID 3 ( K ) ≥ ID 4 ( K ) : Trivial since I | = 3 K ⇒ I | = 4 K ➤ ID 3 ( K ) ≤ ID 4 ( K ) : Every 4-model can be modified to a 3-model by changing N to t while preserving the inconsistency degree. 8/24

Empty Head Relationship Theorem Given a knowledge base K , then ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K ). Proof Hints: ➤ ID 3 ( K ) ≥ ID 4 ( K ) : Trivial since I | = 3 K ⇒ I | = 4 K ➤ ID 3 ( K ) ≤ ID 4 ( K ) : Every 4-model can be modified to a 3-model by changing N to t while preserving the inconsistency degree. ➤ ID 3 ( K ) ≥ ID LP m ( K ) : Assume that ID 3 ( K ) < ID LP m ( K ). Then we can find a contradiction. 8/24

Empty Head Relationship Theorem Given a knowledge base K , then ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K ). Proof Hints: ➤ ID 3 ( K ) ≥ ID 4 ( K ) : Trivial since I | = 3 K ⇒ I | = 4 K ➤ ID 3 ( K ) ≤ ID 4 ( K ) : Every 4-model can be modified to a 3-model by changing N to t while preserving the inconsistency degree. ➤ ID 3 ( K ) ≥ ID LP m ( K ) : Assume that ID 3 ( K ) < ID LP m ( K ). Then we can find a contradiction. ➤ ID 3 ( K ) ≤ ID LP m ( K ) : Trivial since I | = LP m K ⇒ I | = 3 ( K ). 8/24

Empty Head Relationship Theorem Given a knowledge base K , then ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K ). Proof Hints: ➤ ID 3 ( K ) ≥ ID 4 ( K ) : Trivial since I | = 3 K ⇒ I | = 4 K ➤ ID 3 ( K ) ≤ ID 4 ( K ) : Every 4-model can be modified to a 3-model by changing N to t while preserving the inconsistency degree. ➤ ID 3 ( K ) ≥ ID LP m ( K ) : Assume that ID 3 ( K ) < ID LP m ( K ). Then we can find a contradiction. ➤ ID 3 ( K ) ≤ ID LP m ( K ) : Trivial since I | = LP m K ⇒ I | = 3 ( K ). ➤ ID 4 ( K ) ≤ ID Q ( K ) : Trivial since I | = Q K ⇒ I | = 4 ( K ). 8/24

Empty Head Partial Max-SAT Problem 9/24

Empty Head Partial Max-SAT Problem ➤ Partial Max-SAT Problem: • Optimized Version of SAT problem • P = ( H , S ) • H : hard clauses, all must be satisfied • S : soft clauses, should be satisfied as many as possible • ˆ I = arg max I |{ γ | γ ∈ S , I | = γ, I | = H }| . 9/24

Empty Head Partial Max-SAT Problem ➤ Partial Max-SAT Problem: • Optimized Version of SAT problem • P = ( H , S ) • H : hard clauses, all must be satisfied • S : soft clauses, should be satisfied as many as possible • ˆ I = arg max I |{ γ | γ ∈ S , I | = γ, I | = H }| . ➤ Solvers: • SAT4j MaxSAT, MSUnCore, Clone, MiniMaxSAT, . . . 9/24

Empty Head Partial Max-SAT Problem ➤ Partial Max-SAT Problem: • Optimized Version of SAT problem • P = ( H , S ) • H : hard clauses, all must be satisfied • S : soft clauses, should be satisfied as many as possible • ˆ I = arg max I |{ γ | γ ∈ S , I | = γ, I | = H }| . ➤ Solvers: • SAT4j MaxSAT, MSUnCore, Clone, MiniMaxSAT, . . . ➤ Max-SAT Competition • http://www.maxsat.udl.cat/09/ • http://www.maxsat.udl.cat/10/ 9/24

Empty Head Computing Inconsistency Degrees ➤ only KB as a set of clauses (CNF) considered ➤ consider ID 4 and ID Q (Since ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K )) 10/24

Empty Head Computing Inconsistency Degrees ➤ only KB as a set of clauses (CNF) considered ➤ consider ID 4 and ID Q (Since ID 3 ( K ) = ID 4 ( K ) = ID LP m ( K ) ≤ ID Q ( K )) Road Map 1 Multi-valued semantics ⇒ 2-valued semantics. 2 Represent ID i by 2-valued semantics. 3 ID i ⇒ partial Max-SAT problem. 10/24