Learning Regular Sets

Transcription

1 Learning Regular Sets Author: Dana Angluin Presented by: M. Andreína Francisco Department of Computer Science Uppsala University February 3, 2014

2 Minimally Adequate Teachers A Minimally Adequate Teacher (MAT) is an Oracle that must answer correctly two types of questions: Membership queries the answer must be yes or no Strong equivalence queries the answer is yes or any counterexample 2 of 28

3 Definitions Prefix-closed Every prefix of every member is also a member, that is uv S u S i.e. the set {λ, a, ab} Suffix-closed Every suffix of every member is also a member, that is uv S v S i.e. the set {λ, a, ba} 3 of 28

4 Observation Table An observation table is a triple STA, EXP, OT where: STA = RED BLUE RED Σ is a finite set of states BLUE = RED Σ \ RED is the set of successor states of RED that are not RED EXP Σ is the experiment set. OT : STA EXP {0, 1, } is a function such that: 1 if ue L OT[u][e] = 0 if ue / L otherwise (not known) 4 of 28

5 Observation Table Examples Experiment set States λ a λ 0 1 a 1 0 b 1 0 aa 0 1 ab of 28

6 How to build an automaton from an OT Q {q r : r RED} F A {q we : we RED OT[w][e] = 1} F R {q we : we RED OT[w][e] = 0 q w Q σ Σ δ(q w, σ) q u : u RED OT[u] = OT[wσ] 6 of 28

7 How to build an automaton from an OT F A {q we : we RED OT[w][e] = 1} q w Q σ Σ δ(q w, σ) q u : u RED OT[u] = OT[wσ] Examples Observation Table start q λ λ a λ 0 1 a 1 0 b 1 0 aa 0 1 ab 1 0 Transition Table a b q λ q a q a q a q λ q a a q a b a,b 7 of 28

8 Closed Table A table is closed if for every row u BLUE there is a row v RED such that row(u) = row(v) Examples The following table is not closed because of row ab λ a λ 0 1 a 1 0 b 1 0 aa 0 1 ab of 28

9 How do we close a table? Let s be the row of BLUE that is not in RED Move s to RED a Σ, add sa to BLUE Examples The following table is now closed but it is not complete. λ a λ 0 1 a 1 0 ab 1 1 b 1 0 aa 0 1 aba * * abb * * 9 of 28

10 Consistent Table A table is consistent if every pair of equivalent rows in RED remains equivalent after appending any symbol. Examples The following table is inconsistent λ a λ 0 1 a 1 0 ab 1 0 b 1 0 aa 0 1 aba 0 0 abb of 28

11 Consistent Table A table is consistent if every pair of equivalent rows in RED remains equivalent after appending any symbol. Examples The following table is inconsistent λ a λ 0 1 a 1 0 ab 1 0 b 1 0 aa 0 1 aba 0 0 abb of 28

12 Consistent Table A table is consistent if every pair of equivalent rows in RED remains equivalent after appending any symbol. Examples The following table is inconsistent λ a λ 0 1 a 1 0 ab 1 0 b 1 0 aa 0 1 aba 0 0 abb of 28

13 How do we make a table consistent? Let x Σ be the differentiating string such that OT[s 1 ] = OT[s 2 ] but OT[s 1 x] OT[s 2 x]. Let e be the experiment for which OT[s 1 x][e] OT[s 2 x][e]. Adding the experiment xe will differentiate OT[s 1 ] and OT[s 2 ]. Examples 11 of 28 λ a aa λ a 1 0 * ab 1 0 * b 1 0 * aa 0 1 * aba 0 0 * abb 1 0 *

14 The Learner L 1. Initialize RED = EXP = {λ} 2. Initialize BLUE = RED Σ 3. Complete the table 4. while Teacher says NO 4.1 while table is not closed or not consistent if not consistent make it consistent (and complete) if not closed close it (and complete it) 4.2 Make a guess! 4.3 If the teacher says no, then add the counterexample and all its prefixes to RED update BLUE = RED Σ complete the table 12 of 28

15 Suppose the unknown regular set is the set of all strings over {a, b} with an even number of a s and an even number of b s Initial table λ λ 1 a 0 b 0 13 of 28

16 Suppose the unknown regular set is the set of all strings over {a, b} with an even number of a s and an even number of b s Initial table λ λ 1 a 0 b 0 This table is not closed 13 of 28

17 We close the table λ λ 1 a 0 b 0 aa * ab * 14 of 28

18 We complete the table λ λ 1 a 0 b 0 aa 1 ab 0 15 of 28

19 We complete the table λ λ 1 a 0 b 0 aa 1 ab 0 This table is closed, consistent and complete, so we can make a guess 15 of 28

20 The guess is... start q λ a a,b q a (note that it is the same automaton as in the construction example, but the observation tables are different) b 16 of 28

21 Assume we get the counter example bb. 1. add the counterexample and all its prefixes to RED 2. update BLUE = RED Σ λ λ 1 a 0 b 0 bb * aa 1 ab 0 ba * bba * bbb * 17 of 28

22 Assume we get the counter example bb. 1. add the counterexample and all its prefixes to RED 2. update BLUE = RED Σ λ λ 1 a 0 b 0 bb 1 aa 1 ab 0 ba 0 bba 0 bbb 0 18 of 28

23 Assume we get the counter example bb. 1. add the counterexample and all its prefixes to RED 2. update BLUE = RED Σ λ λ 1 a 0 b 0 bb 1 aa 1 ab 0 ba 0 bba 0 bbb 0 18 of 28

24 Add the experiment a λ a λ 1 0 a 0 1 b 0 0 bb 1 0 aa 1 * ab 0 * ba 0 * bba 0 * bbb 0 * 19 of 28

25 Add the experiment a λ a λ 1 0 a 0 1 b 0 0 bb 1 0 aa 1 0 ab 0 0 ba 0 0 bba 0 1 bbb of 28

26 make a guess... a q 1 b a start q 0 b q 2 a b 21 of 28

27 Suppose we get the counterexample abb 22 of 28 λ a λ 1 0 a 0 1 b 0 0 bb 1 0 ab 0 0 abb * * aa 1 0 ba 0 0 bba 0 1 bbb 0 0 aba * * abba * * abbb * *

28 Suppose we get the counterexample abb 23 of 28 λ a λ 1 0 a 0 1 b 0 0 bb 1 0 ab 0 0 abb 0 1 aa 1 0 ba 0 0 bba 0 1 bbb 0 0 aba 0 0 abba 1 0 abbb 0 0

29 The table is not consistent 24 of 28 λ a λ 1 0 a 0 1 b 0 0 bb 1 0 ab 0 0 abb 0 1 aa 1 0 ba 0 0 bba 0 1 bbb 0 0 aba 0 0 abba 1 0 abbb 0 0

30 The table is not consistent 25 of 28 λ a b λ a b bb ab abb aa 1 0 * ba 0 0 * bba 0 1 * bbb 0 0 * aba 0 0 * abba 1 0 * abbb 0 0 *

31 We complete the table 26 of 28 λ a b λ a b bb ab abb aa ba bba bbb aba abba abbb 0 0 0

32 We make a guess... and the Teacher says YES! :-) δ a b q 0 q 1 q 2 q 1 q 0 q 3 q 2 q 3 q 0 q 3 q 2 q 1 start q 0 a b a b q 1 q 2 b a b a q 3 27 of 28

33 Questions? 28 of 28