On the Complexity of Computing the Justification Status of an Argument

Transcription

1 On the Complexity of Computing the Justifiction Sttus of n Argument dbi Reserch Seminr, Vienn Wolfgng Dvořák Institute of Informtion Systems, Vienn University of Technology Oct 13, 2011 Supported by the Vienn Science nd Technology Fund (WWTF) under grnt ICT Slide 1

2 1. Motivtion Motivtion We dress the problem of: Determining the cceptnce sttus of n rgument in bstrct rgumenttion (Given semntics for computing the extensions). Slide 2

3 1. Motivtion Motivtion We dress the problem of: Determining the cceptnce sttus of n rgument in bstrct rgumenttion (Given semntics for computing the extensions). Trditionl: Skepticl nd/or Credulous Acceptnce. Wu nd Cmind recently proposed new pproch: The Justifiction Sttus of n Argument. Slide 2

4 1. Motivtion Motivtion We dress the problem of: Determining the cceptnce sttus of n rgument in bstrct rgumenttion (Given semntics for computing the extensions). Trditionl: Skepticl nd/or Credulous Acceptnce. Wu nd Cmind recently proposed new pproch: The Justifiction Sttus of n Argument. Their originl pproch is stted in terms of complete semntics. We generlize it to rbitrry semntics Computtionl issues where neglected. We provide n comprehensive complexity nlysis. Slide 2

5 1. Motivtion Outline 1. Motivtion 2. Bckground 3. Justifiction Sttus of n Argument 4. The Complexity of Computing the Justifiction Sttus 5. Conclusion Slide 3

6 2. Bckground Dung s Abstrct Argumenttion Frmeworks Definition An rgumenttion frmework (AF) is pir (A, R) where A is set of rguments R A A is reltion representing the conflicts ( ttcks ) Exmple F=( {,b,c,d,e}, {(,b),(c,b),(c,d),(d,c),(d,e),(e,e)} ) Slide 4

7 2. Bckground Bsic Properties Conflict-Free Sets Given n AF F = (A, R). A set S A is conflict-free in F, if, for ech, b S, (, b) / R. Exmple cf (F ) = { {, c}, Slide 5

8 2. Bckground Bsic Properties Conflict-Free Sets Given n AF F = (A, R). A set S A is conflict-free in F, if, for ech, b S, (, b) / R. Exmple cf (F ) = { {, c}, {, d}, Slide 5

9 2. Bckground Bsic Properties Conflict-Free Sets Given n AF F = (A, R). A set S A is conflict-free in F, if, for ech, b S, (, b) / R. Exmple cf (F ) = { {, c}, {, d}, {b, d}, Slide 5

10 2. Bckground Bsic Properties Conflict-Free Sets Given n AF F = (A, R). A set S A is conflict-free in F, if, for ech, b S, (, b) / R. Exmple cf (F ) = { {, c}, {, d}, {b, d}, {}, {b}, {c}, {d}, } Slide 5

11 2. Bckground Bsic Properties (ctd.) Admissible Sets [Dung, 1995] Given n AF F = (A, R). A set S A is dmissible in F, if S is conflict-free in F ech S is defended by S in F A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple dm(f ) = { {, c}, Slide 6

12 2. Bckground Bsic Properties (ctd.) Admissible Sets [Dung, 1995] Given n AF F = (A, R). A set S A is dmissible in F, if S is conflict-free in F ech S is defended by S in F A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple dm(f ) = { {, c}, {, d}, Slide 6

13 2. Bckground Bsic Properties (ctd.) Admissible Sets [Dung, 1995] Given n AF F = (A, R). A set S A is dmissible in F, if S is conflict-free in F ech S is defended by S in F A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple dm(f ) = { {, c}, {, d}, {b, d}, Slide 6

14 2. Bckground Bsic Properties (ctd.) Admissible Sets [Dung, 1995] Given n AF F = (A, R). A set S A is dmissible in F, if S is conflict-free in F ech S is defended by S in F A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple dm(f ) = { {, c}, {, d}, {b, d}, {}, {b}, {c}, {d}, } Slide 6

15 2. Bckground Semntics (ctd.) Grounded Extension [Dung, 1995] Given n AF (A, R). The unique grounded extension is defined s the smllest set S such tht: ech rgument A which is not ttcked in F belongs to S ech A defended by S in F is contined in S Exmple ground(f ) = { {}} Slide 7

16 2. Bckground Semntics (ctd.) Complete Extension [Dung, 1995] Given n AF (A, R). A set S A is complete in F, if S is dmissible in F ech A defended by S in F is contined in S A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple comp(f ) = { {, c}, Slide 8

17 2. Bckground Semntics (ctd.) Complete Extension [Dung, 1995] Given n AF (A, R). A set S A is complete in F, if S is dmissible in F ech A defended by S in F is contined in S A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple comp(f ) = { {, c}, {, d}, Slide 8

18 2. Bckground Semntics (ctd.) Complete Extension [Dung, 1995] Given n AF (A, R). A set S A is complete in F, if S is dmissible in F ech A defended by S in F is contined in S A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple comp(f ) = { {, c}, {, d}, {}, Slide 8

19 2. Bckground Semntics (ctd.) Complete Extension [Dung, 1995] Given n AF (A, R). A set S A is complete in F, if S is dmissible in F ech A defended by S in F is contined in S A is defended by S in F, if for ech b A with (b, ) R, there exists c S, such tht (c, b) R. Exmple comp(f ) = { {, c}, {, d}, {}, {c}, {d}, } Slide 8

20 2. Bckground Semntics (ctd.) Preferred Extensions [Dung, 1995] Given n AF F = (A, R). A set S A is preferred extension of F, if S is dmissible in F for ech T A dmissible in F, S T Exmple pref (F ) = { {, c}, {, d}, {}, {c}, {d}, } Slide 9

21 2. Bckground Semntics (ctd.) Stble Extensions [Dung, 1995] Given n AF F = (A, R). A set S A is stble extension of F, if S is conflict-free in F for ech A \ S, there exists b S, such tht (b, ) R Exmple stble(f ) = { {, c} Slide 10

22 2. Bckground Semntics (ctd.) Stble Extensions [Dung, 1995] Given n AF F = (A, R). A set S A is stble extension of F, if S is conflict-free in F for ech A \ S, there exists b S, such tht (b, ) R Exmple stble(f ) = { {, c}, {, d}, Slide 10

23 2. Bckground Semntics (ctd.) Stble Extensions [Dung, 1995] Given n AF F = (A, R). A set S A is stble extension of F, if S is conflict-free in F for ech A \ S, there exists b S, such tht (b, ) R Exmple stble(f ) = { {, c}, {, d}, {b, d}, Slide 10

24 2. Bckground Semntics (ctd.) Stble Extensions [Dung, 1995] Given n AF F = (A, R). A set S A is stble extension of F, if S is conflict-free in F for ech A \ S, there exists b S, such tht (b, ) R Exmple stble(f ) = { {, c}, {, d}, {b, d}, {}, {b}, {c}, {d},, } Slide 10

25 3. Justifiction Sttus of n Argument Argumenttion Lbelings Let F = (A, R) be n AF. Definition A lbeling for F is function L : A {in, out, undec}. We denote lbelings by triples (L in, L out, L undec ), with L l ={ A L() = l}. Slide 11

26 3. Justifiction Sttus of n Argument Argumenttion Lbelings Let F = (A, R) be n AF. Definition A lbeling for F is function L : A {in, out, undec}. We denote lbelings by triples (L in, L out, L undec ), with L l ={ A L() = l}. The rnge of set S A is defined s S + R =S {b S : (, b) R}. We define the induced lbeling Ext2Lb F (E) of n extension E A: Ext2Lb F (E) = (E, E + R \ E, A \ E + R ) Slide 11

27 3. Justifiction Sttus of n Argument Argumenttion Lbelings Let F = (A, R) be n AF. Definition A lbeling for F is function L : A {in, out, undec}. We denote lbelings by triples (L in, L out, L undec ), with L l ={ A L() = l}. The rnge of set S A is defined s S + R =S {b S : (, b) R}. We define the induced lbeling Ext2Lb F (E) of n extension E A: Ext2Lb F (E) = (E, E + R \ E, A \ E + R ) Definition Let σ be n extension-bsed semntics. The corresponding lbeling-bsed semntics σ L is defined s σ L (F )={Ext2Lb(E) E σ(f )}. Slide 11

28 3. Justifiction Sttus of n Argument Argumenttion Lbelings - Exmple Exmple comp(f ) = {{}, {, c}, {, d}} The complete lbelings re: ({}, {b}, {c, d, e}), ({, c}, {b, d}, {e}), ({, d}, {b, c, e}, {}) Slide 12

29 3. Justifiction Sttus of n Argument Justifiction Sttus of n Argument Definition Let F = (A, R) be n AF nd σ semntic. The justifiction sttus of n A wrt σ is defined s J S σ (F, ) = {L() L σ L (F )}. Slide 13

30 3. Justifiction Sttus of n Argument Justifiction Sttus of n Argument Definition Let F = (A, R) be n AF nd σ semntic. The justifiction sttus of n A wrt σ is defined s J S σ (F, ) = {L() L σ L (F )}. Exmple comp(f ) = {{}, {, c}, {, d}} J S comp (F, ) = {in}, Slide 13

31 3. Justifiction Sttus of n Argument Justifiction Sttus of n Argument Definition Let F = (A, R) be n AF nd σ semntic. The justifiction sttus of n A wrt σ is defined s J S σ (F, ) = {L() L σ L (F )}. Exmple comp(f ) = {{}, {, c}, {, d}} J S comp (F, ) = {in}, J S comp (F, b) = {out}, Slide 13

32 3. Justifiction Sttus of n Argument Justifiction Sttus of n Argument Definition Let F = (A, R) be n AF nd σ semntic. The justifiction sttus of n A wrt σ is defined s J S σ (F, ) = {L() L σ L (F )}. Exmple comp(f ) = {{}, {, c}, {, d}} J S comp (F, ) = {in}, J S comp (F, b) = {out}, J S comp (F, c) = J S comp (F, d) = {in, out, undec} Slide 13

33 3. Justifiction Sttus of n Argument Justifiction Sttus of n Argument Definition Let F = (A, R) be n AF nd σ semntic. The justifiction sttus of n A wrt σ is defined s J S σ (F, ) = {L() L σ L (F )}. Exmple comp(f ) = {{}, {, c}, {, d}} J S comp (F, ) = {in}, J S comp (F, b) = {out}, J S comp (F, c) = J S comp (F, d) = {in, out, undec} J S comp (F, e) = {out, undec} Slide 13

34 3. Justifiction Sttus of n Argument Possible Justifiction Sttuses Ech element of 2 {in,out,undec} is justifiction sttus: {in} {in, undec} {in, out} {undec} {} {in, out, undec} {out, undec} {out} ccept wek ccept borderline wek reject reject Slide 14

35 3. Justifiction Sttus of n Argument Possible Justifiction Sttuses Not ll justifiction sttuses re possible under ech semntics: Slide 15

36 3. Justifiction Sttus of n Argument Possible Justifiction Sttuses Not ll justifiction sttuses re possible under ech semntics: Theorem Let F = (A, R) be n AF nd A. Then we hve tht: J S ground (F, ) {{in}, {out}, {undec}} J S dm (F, ) {{undec}, {in, undec}, {out, undec}, {in, out, undec}} J S comp (F, ) 2 {in,out,undec} \ {, {in, out}} J S stble (F, ) {{in}, {out}, {in, out}, {}} J S pref (F, ) 2 {in,out,undec} \ { } J S semi (F, ) 2 {in,out,undec} \ { } J S stge (F, ) 2 {in,out,undec} \ { } Slide 15

37 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity - Problems of interest We re interested in the following two problems: The justifiction sttus decision problem JS σ Given: AF F = (A, R), L {in, out, undec} nd rgument A. Question: Does J S σ (F, ) = L hold? The generlized justifiction sttus decision problem GJS σ Given: AF F = (A, R), L, M {in, out, undec} nd rgument A. Question: Does L J S σ (F, ) nd J S σ (F, ) M = hold?. Slide 16

38 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity - Problems of interest We re interested in the following two problems: The justifiction sttus decision problem JS σ Given: AF F = (A, R), L {in, out, undec} nd rgument A. Question: Does J S σ (F, ) = L hold? The generlized justifiction sttus decision problem GJS σ Given: AF F = (A, R), L, M {in, out, undec} nd rgument A. Question: Does L J S σ (F, ) nd J S σ (F, ) M = hold?. Clerly the first problem cn be encoded s instnce of the second one. Slide 16

39 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity - Problems of interest We re interested in the following two problems: The justifiction sttus decision problem JS σ Given: AF F = (A, R), L {in, out, undec} nd rgument A. Question: Does J S σ (F, ) = L hold? The generlized justifiction sttus decision problem GJS σ Given: AF F = (A, R), L, M {in, out, undec} nd rgument A. Question: Does L J S σ (F, ) nd J S σ (F, ) M = hold?. Clerly the first problem cn be encoded s instnce of the second one. To obtin completness for both problems we show membership for GJS σ nd hrdness for JS σ Slide 16

40 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity - Membership Theorem If the problem of verifying σ-extension is in the complexity clss C then the problem GJS σ is in the complexity clss NP C co-np C. Slide 17

41 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity - Membership Theorem If the problem of verifying σ-extension is in the complexity clss C then the problem GJS σ is in the complexity clss NP C co-np C. Proof Ides. We provide NP C lgorithm to decide L J S σ (F, ) For ech l L guess lbeling L l with L l () = l Test whether L l σ(f ) or not, using the C-orcle. Accept if for ech l L, L l σ(f ) nd co-np C lgorithm to decide J S σ (F, ) M =, For ech l M guess lbeling L l with L l () = l Test whether L l σ(f ) or not Accept if there exists n l M such tht L l σ(f ) Slide 17

42 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity - Hrdness Theorem The problems JS comp, GJS comp,js dm, GJS dm re DP-hrd, i.e. NP co-np-hrd. Slide 18

43 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity - Hrdness Theorem The problems JS comp, GJS comp,js dm, GJS dm re DP-hrd, i.e. NP co-np-hrd. Proof Ide. We prove hrdness by reducing the (DP-hrd) SAT-UNSAT problem to JS comp (resp. JS dm ). The reduction builds on slightly modified stndrd trnsltions of both formuls nd dds mutul ttck between them. Slide 18

44 4. The Complexity of Computing the Justifiction Sttus Computtionl Complexity σ ground dm comp stble pref semi stge Cred σ P-c NP-c NP-c NP-c NP-c Σ p 2 -c Σp 2 -c Skept σ P-c trivil P-c co-np/dp-c Π p 2 -c Πp 2 -c Πp 2 -c JS σ P-c DP-c DP-c DP-c P Σp 2 [1] -c DP 2 -c DP 2 -c GJS σ P-c DP-c DP-c DP-c P Σp 2 [1] -c DP 2 -c DP 2 -c Tble: Complexity Results (C-c denotes completeness for clss C) Reltions between the bove complexity clsses: P NP co-np DP Σp 2 Π p 2 P Σp 2 [1] DP 2 Slide 19

45 5. Conclusion Conclusion We generlised the concept of the justifiction sttus of n rgument to rbitrry semntics. Using the Justifiction Sttus in generl increses the complexity. Two sources of complexity: We hve to determine tht some lbels re in the justifiction sttus some lbels re not in the justifiction sttus There re severl problem clsses where these decision problems re esier, e.g. Credulous nd Skepticl Acceptnce. Slide 20