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Learning Regular Sets Author: Dana Angluin Presented by: M. Andre na Francisco Department of Computer Science Uppsala University February 3, 2014 Minimally Adequate Teachers A Minimally Adequate Teacher (MAT) is an Oracle that must

  1. Learning Regular Sets Author: Dana Angluin Presented by: M. Andre´ ına 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 λ a λ 0 1 a 1 0 States b 1 0 aa 0 1 ab 1 0 5 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 [ w σ ] 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 [ w σ ] Examples q λ Observation Table start λ a Transition Table a a,b λ 0 1 a b a 1 0 q λ q a q a q a b 1 0 q a q λ q a aa 0 1 ab 1 0 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 1 1 8 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 1 0 10 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 1 0 10 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 1 0 10 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 λ a aa λ 0 1 0 a 1 0 * ab 1 0 * b 1 0 * aa 0 1 * aba 0 0 * abb 1 0 * 11 of 28

  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 4.1.1 if not consistent → make it consistent (and complete) 4.1.2 if not closed → close it (and complete it) 4.2 Make a guess! 4.3 If the teacher says no, then 4.3.1 add the counterexample and all its prefixes to RED 4.3.2 update BLUE = RED · Σ 4.3.3 complete the table 12 of 28

  15. An Example Run of L ∗ 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. An Example Run of L ∗ 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. An Example Run of L ∗ We close the table λ λ 1 a 0 b 0 aa * ab * 14 of 28

  18. An Example Run of L ∗ We complete the table λ λ 1 a 0 b 0 aa 1 ab 0 15 of 28

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

  21. An Example Run of L ∗ 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. An Example Run of L ∗ 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. An Example Run of L ∗ 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. An Example Run of L ∗ 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. An Example Run of L ∗ 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 0 0 20 of 28

  26. An Example Run of L ∗ make a guess... q 1 a b a b q 0 q 2 a start b 21 of 28

  27. An Example Run of L ∗ Suppose we get the counterexample ’ abb ’ 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 * * 22 of 28

  28. An Example Run of L ∗ Suppose we get the counterexample ’ abb ’ 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 23 of 28

  29. An Example Run of L ∗ The table is not consistent 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 24 of 28

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

  31. An Example Run of L ∗ We complete the table λ a b λ 1 0 0 a 0 1 0 b 0 0 1 bb 1 0 0 ab 0 0 0 abb 0 1 0 aa 1 0 0 ba 0 0 0 bba 0 1 0 bbb 0 0 1 aba 0 0 1 abba 1 0 0 abbb 0 0 0 26 of 28

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

  33. Questions? 28 of 28

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