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1 Query Learning of Derived ω-tree Languages in Polynomial Time Dana Angluin, Tımos Antonopoulos & Dana 1

2 Query Learning a set of possile concepts Question Answer tc Learner Teacher Tries to identify a target concept 2

3 Query Learning w. MQ & EQ Regular Languages MQ = memership queries EQ = equivalence queries. Master Is w in L? title Yes / No Is [H] same as L? L w 4 w 3 w 5 w 1 w 2 Learner H s.t. [H] =L Yes / No, c.e: w Teacher 3

4 Requirements from Learner Work well regardless of the given counterexample Complexity is measured wrt. l Length of the maximum counterexample l Size of a minimal description for L

5 Learning Regular Languages In 1987 Angluin showed that a regular language can e learned in polynomial time using MQ and EQ. And proposed the learning algorithm L* for that. Regular languages cannot e learned in polynomial time using just MQ or just EQ [Angluin 90]

6 Learning Regular Languages Many improvements to the original L* algorithm since Many applications Master to L* in particular title in analysis, verification and synthesis of programs, protocols and hardware. See recent Model Learning CACM article y Frits Vaandrager

7 Query Learning of Richer Formalisms Finite Words L* Finite Words AL* ω-words L ω DFA Words over Infinite alphaets Λ* AFA Weighted Words W* ω-aut 7 Sym. Aut DWA

8 Our Interest a Master title T ω ω-tree Aut 8

9 ω-words and ω-trees ω-word Infinite length word over a finite alphaet ω-tree Infinite depth tree, with a finite fixed ranching degree, over a finite alphaet aaaaaaaaa a a a a What was accomplished in learning regular ω-languages? a a 9

10 Hierarchy of Regular ω-languages Learnale [Farzan et al. 08] [Angluin & Fisman 14] Difficulties: - No Myhill-Nerode Thm - No minimization Alg Master D-Parity title ω-reg Langs D-Buchi D-coBuchi Poly-Learnale [Maler & Pnueli 95] DW-Buchi DW-Parity DW-coBuchi Weak ω-reg Langs 10 Expressive Power

11 Our Focus Derived ω-tree Languages: ω-tree languages consisting of trees all of whose paths satisfy a certain property on ω-words GF a GF a GF a Let L e an ω-language. Trees d (L) denotes the set of all d-ary ω-trees where all paths are in L GF a GF a Hmm Can this e reduced to learning corr. ω-word langs? 11

12 Main Result Click THM: to edit Learning of Derived(D-Buchi) langs w MQ & EQ can e polynomially reduced to learning D-Buchi langs w MQ & EQ 12

13 Corollary Derived(D-Parity) Poly- Reducile Derived(D-Buchi) Derived(D-coBuchi) Poly- Learnale Derived(DW-Parity) Derived(DW-Buchi) Derived(DW-coBuchi) 13

14 Perspective NPT Master NBT title DBT Poly- Reducile Derived(D-Buchi) 14 Poly- Learnale Expressive Power

15 We need some definitions 15

16 ω-word Acceptor (DBW) Deterministic Buchi Word Acceptor (DBW) look like DFA ut the Click acceptance criterion to edit requires that an accepting state is visited infinitely often a a 2 L = a* ω L = (a) ω 16

17 !sman D.!sman Asking MQ reg.. langs Is aa in L? Learner ord w ω reg. gs. lan Teacher Is aaaaaaaaaaaaaa Learner Teacher tree ω. reg gs. lan Learner 17 Is Teacher

18 Representing Examples Ultimately periodic words are words of the form u v v v v v v v... How can we represent infinite words or infinite trees in a finite manner? denoted uv ω Master represented title (u, v) [McNaughton 66] THM Two regular ω-word languages are equivalent iff they agree on the set of ultimately periodic words 18

19 Representing Examples Regular ω-trees are ω-trees that have a finite numer of non-isomorphic complete infinite sutrees How can we represent infinite words or infinite trees in a finite manner? [Rain 72] THM: Two regular ω-tree languages are equivalent iff they agree on the set of regular ω-trees 19

20 Representing Examples Regular ω-trees are ω-trees that have a finite numer of non-isomorphic complete infinite sutrees a a a [Rain 72] THM: Two regular ω-tree languages are equivalent iff they agree on the set of regular ω-trees 20

21 Representing Examples Regular ω-trees can e represented using ω-tree automaton a a Complete DFAs over the alphaet of tree directions, accompanied y an output function laeling transitions a L/a A t R/ 21 L/ R/ Σ in = {L, R} Σ out = {a, } t a a

22 Derived Langs. ω-word lang its derived ω-tree lang L = a* ω aaaa... Trees 2 (L) a aaaaaaaaaaa... ω-word acceptor a 1 2 ω-tree acceptor??? 22

23 ω-tree acceptors ω-word acceptor ω-tree acceptor Runs on a ω-word Runs on a ω-tree a a a... q 0 q 1 q 2 q 1 q 1 q 8 q 0 q 2 q 0 a q 2 a a q 0 q 1 q 4 δ : Q Σ Q δ : Q Σ 2 Q 2 23

24 M => M T Given an ω-word aut M can we uild an ω-tree aut M T s.t. M T accepts Trees d ([M])? 24

25 M for L => M T for Trees(L) 25 Given an ω-word aut M for L a 1 We can construct an ω-tree aut. M T for Trees 2 (L) y mimicking M in each direction independently L = a* ω 2 From State T = Trees 2 (L) Upon Reading To States q1 (a,a) (q1, q1) q1 (a,) (q1, q2) q1 (,a) (q2, q1) q1 (,) (q2, q2) q2 (a,a) (q2, q2) q2 (a,) (q2, q ) q2 (,a) (q, q2) q2 (,) (q, q )

26 Main Result Click THM: to edit Learning of Derived(D-Buchi) langs w MQ & EQ can e polynomially reduced to learning D-Buchi langs w MQ & EQ 26

27 The reduction framework 27

28 A Trees A M T accepts a tree t iff M accepts all of t s paths

29 A Trees A M T accepts a tree t iff M accepts all of t s paths

30 EQ answered no t Master Trees d (L) [M T ] title t is a counterexample pos. ce t Trees d (L) t [M T ] neg. ce t Trees d (L) t [M T ] t We can give w as a ce to A Not clear how to extract a ce for A 30

31 A Trees A

32 A Trees A

33 Dealing with neg. counterexample 33

34 Dealing w. Negative CE neg. ce Click t Trees d (L) to edit t [M T ] t If only we could have asked suset queries aout L and get a ce.. We would have asked paths(t) L and the ce returned is what we are looking for USQ = Unrestricted suset query returns whether H L and a ce if not RSQ = Restricted suset query returns whether H L and does not return a ce 34

35 Dealing w. Negative CE We show that USQ can e simulated y RSQ neg. ce Click t Trees d (L) to edit t [M T ] t USQ = Unrestricted suset query returns whether H L and a ce if not That is, we can extract the ce ourselves RSQ = Restricted suset query returns whether H L and does not return a ce 35

36 Dealing w. Negative CE We show that USQ can e simulated y RSQ neg. ce Click t Trees d (L) to edit t [M T ] t That is, we can extract the ce ourselves But we don t have RSQ either We show that the RSQ we need can e simulated y MQ 36

37 RSQ => USQ counterexample length = 11 Reg Langs NFW/DFW M USQ for L RSQ for L 37 Idea: first estalish the length of the ce Yes No, ce Is M ε L? Is M Σ L? Is M Σ 2 L? Is M Σ 3 L? Is M Σ 11 L?

38 RSQ => USQ counterexample a a a a a length = 11 Reg Langs NFW/DFW M USQ for L RSQ for L Then work out letter y letter Yes No, ce Is M aσ 10 L? Is M aaσ 9 L? Is M aσ 9 L? 38

39 RSQ => USQ counterexample a a a a a length = 11 Reg Langs NFW/DFW M USQ for L RSQ for L 39 Note: the length must e estalished first Yes No, ce Take L = Σ* \ a* ε is a prefix of a ce a is a prefix of a ce aa is a prefix of a ce aaa is a prefix of a ce

40 RSQ => USQ counterexample prefix ω period ω-reg Langs DBW/NBW M USQ for L RSQ for L Need to estalish the prefix and the period Yes No, ce Once we estalish the period we can estalish the prefix as efore 40 But how do we estalish the period?

41 Estalishing the Period ω q 0 prefix q period Lq 0,q (Lq,q) ω \ L q Suppose we know there is a period of the form (aσ k ) ω The length of the smallest period can still e more than k+1 E.g. L = Σ ω \ (aaccadd) ω 41

42 Estalishing the Period ω q 0 prefix q period Lq 0,q (Lq,q) ω \ L q So what do we do? We can estalish pieces of the period, one y one 42

43 Estalishing the Period prefix ω period q 0 q Lq 0,q (Lq,q) ω \ L q 43

44 Estalishing the Period prefix ω period q 0 Lq 0,q q (Lq,q Lq,q) ω q \ L 44

45 Estalishing the Period prefix ω period q 0 q q Lq 0,q (Lq,q Lq,q) ω \ L q 45

46 Estalishing the Period ω q 0 prefix a a period v 1 q q Lq 0,q (Lq,q Lq,q) ω \ L q 46

47 Estalishing the Period ω q 0 prefix a a period v 1 q q Lq 0,q (Lq,q Lq,q) ω \ L Lq,q q 47

48 Estalishing the Period ω q 0 prefix a a period Master c c title v 1 q q q Lq 0,q (Lq,q Lq,q) ω \ L Lq,q v 2 q 48

49 Estalishing the Period n = #states in DBT for L prefix ω period q 0 v 1 v q q 2 v 3 v n q q q q Lq 0,q ((Lq,q) n Lq,q) ω \ L q Lq 0,q (v1 v2 v3 vn Lq,q) ω \ L eventually for some i,j < n: Lq 0,q v1 v2 (vi vi+1 vj) ω \ L 49

50 Proof Let w Lq 0,q (v1 v2 vn Lq,q) ω Then w is of the form \ L w = u (v1 v2 vn y1) (v1 v2 vn y2) (v1 v2 vn y3) w = u ( y1) ( y2) ( y3) ( y4) ( y5) ( y6) ( y7) ( y8) ( y9) q 0 q q q q q q q q q q Eventually visits only non-accepting states r 1 r 2 r 3 r n r i = r j for some i, j 50

51 Proof w = u (v1 v2 vn y1) (v1 v2 vn y2) (v1 v2 vn y3) w = u ( y1) ( y2) ( y3) ( y4) ( y5) ( y6) ( y7) ( y8) ( y9) q 0 q q q q q q q q q q r 1 r i r j r n = r i u' v' u' (v') ω Lq 0,q (Lq,q) ω \ L 51

52 A Trees A

53 Summary We have shown how to reduce learning Derived(D-Buchi) Click to langs edit w. MQ and EQ to learning D-Buchi langs w. MQ and EQ Together with [MP95] s result we otain polylearnaility of Derived(DW-Parity) langs. Desired: a poly-learnailty result for more than DW-Parity 53

54 Thank you for your attention! Comments / Questions? 54