FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

Transcription

1 5-453 FORMAL LANGUAGES, AUTOMATA AN COMPUTABILITY

2 FA NFA EFINITION Regular Language Regular Expression

3 How can we prove that two regular expressions are equivalent? How can we prove that two FAs (or two NFAs) are equivalent? How can we prove that two regular languages are equivalent? (oes this question make sense?)

4 How can we prove that two FAs (or two NFAs) are equivalent?

5 MINIMIZING FAs THURSAY Jan 23

6 IS THIS MINIMAL?

7 IS THIS MINIMAL?

8 THEOREM For every regular language L, there exists a UNIQUE (up to re-labeling of the states) minimal FA M such that L = L(M)

9 THEOREM For every regular language L, there exists a UNIQUE (up to re-labeling of the states) minimal FA M such that L = L(M) Minimal means wrt number of states Given a specification for L, via FA, NFA or regex,this theorem is constructive.

10 NOT TRUE FOR NFAs

11 NOT TRUE FOR RegExp

12 EXTENING δ Given FA M = (Q, Σ, δ, q, F) extend δ to δ ^ : Q Σ* Q as follows: δ(q, ^ ε) = q δ(q, ^ σ) = δ(q, σ) δ(q, ^ ^ σ σ k+ ) = δ( δ(q, σ σ k ), σ k+ ) ^ Note: δ(q, w) F M accepts w String w Σ* distinguishes states p and q iff δ(p, ^ w) F δ(q, ^ w) F

13 EXTENING δ Given FA M = (Q, Σ, δ, q, F) extend δ to δ ^ : Q Σ* Q as follows: δ(q, ^ ε) = q δ(q, ^ σ) = δ(q, σ) δ(q, ^ ^ σ σ k+ ) = δ( δ(q, σ σ k ), σ k+ ) ^ Note: δ(q, w) F M accepts w String w Σ* distinguishes states p and q iff exactly ONE of δ(p ^, w), δ(q, ^ w) is a final state

14 Fix M = (Q, Σ, δ, q, F) and let p, q Q EFINITION: p is distinguishable from q iff there is a w Σ* that distinguishes p and q p is indistinguishable from q iff p is not distinguishable from q iff for all w Σ*, ^ δ(p, w) F δ(q, ^ w) F

15 Fix M = (Q, Σ, δ, q, F) and let p, q Q EFINITION: p is distinguishable from q iff there is a w Σ* that distinguishes p and q p is indistinguishable from q iff p is not distinguishable from q iff for all w Σ*, ^ δ(p, w) F ^ δ(q, w) F

16 q, q q 2 q 3 ε distinguishes accept from non-accept states

17 Fix M = (Q, Σ, δ, q, F) and let p, q, r Q efine relation ~ : p ~ q iff p is indistinguishable from q p ~ q iff p is distinguishable from q / Proposition: ~ is an equivalence relation

18 Fix M = (Q, Σ, δ, q, F) and let p, q, r Q efine relation ~ : p ~ q iff p is indistinguishable from q p ~ q iff p is distinguishable from q / Proposition: ~ is an equivalence relation p ~ p (reflexive) p ~ q q ~ p (symmetric) p ~ q and q ~ r p ~ r (transitive) Proof (of transitivity): for all w, we have: δ(p, ^ w) F δ(q, ^ w) F δ(r, ^ w) F

19 Fix M = (Q, Σ, δ, q, F) and let p, q, r Q so ~ partitions the set of states of M into disjoint equivalence classes Proposition: ~ is an equivalence relation p ~ p (reflexive) p ~ q q ~ p (symmetric) p ~ q and q ~ r p ~ r (transitive) Proof (of transitivity): for all w, we have: δ(p, ^ w) F δ(q, ^ w) F δ(r, ^ w) F

20 Fix M = (Q, Σ, δ, q, F) and let p, q, r Q so ~ partitions the set of states of M into disjoint equivalence classes Proposition: ~ is an equivalence relation [q] = { p p ~ q } Q q q

21

22 Input: FA M Algorithm MINIMIZE Output: FA M MIN such that: M M MIN (that is, L(M) = L(M MIN )) M MIN has no inaccessible states M MIN is irreducible all states of M MIN are pairwise distinguishable

23 Input: FA M Algorithm MINIMIZE Output: FA M MIN such that: M M MIN (that is, L(M) = L(M MIN )) M MIN has no inaccessible states M MIN is irreducible all states of M MIN are pairwise distinguishable Theorem: M MIN is the unique minimum FA equivalent to M

24 NOTE: Theorem not true for NFAs What does this say about Regexs?

25 Intuition: States of M MIN will be blocks of equivalent states of M We ll find these equivalent states with a Table-Filling Algorithm

26 TABLE-FILLING ALGORITHM Input: FA M = (Q, Σ, δ, q, F) Output: () M = { (p,q) p,q Q and p ~ q } / (2) E M = { [q] q Q }

27 TABLE-FILLING ALGORITHM Input: FA M = (Q, Σ, δ, q, F) Output: () M = { (p,q) p,q Q and p ~ q } / (2) E M = { [q] q Q } IEA: We know how to find those pairs of states that ε distinguishes Use this and recursion to find those pairs distinguishable with longer strings Pairs left over will be indistinguishable

28 TABLE-FILLING ALGORITHM Input: FA M = (Q, Σ, δ, q, F) Output: () M = { (p,q) p,q Q and p / ~ q } (2) E M = { [q] q Q } q q Base Case: p accepts and q rejects p ~ q / q i q n q q q i q n

29 TABLE-FILLING ALGORITHM Input: FA M = (Q, Σ, δ, q, F) Output: () M = { (p,q) p,q Q and p / ~ q } (2) E M = { [q] q Q } q q q i Base Case: p accepts and q rejects p ~ q / Recursion: if there is σ Σ and states p, q satisfying q n q q q i q n δ (p, σ) = p ~ / p ~ / q δ (q, σ) = q Repeat until no more new s

30 q q q 2 q 3 q q q 2 q 3, q q q 2 q 3

31 q q q 2 q 3 q q q 2 q 3, q q q 2 q 3

32 q q q 2 q 3 q q q 2 q 3, q q q 2 q 3

33 q q q 2 q 3 q q q 2 q 3, q q q 2 q 3

34 q q q 2 q 3 q q q 2 q 3, q q q 2 q 3

35 q q q q q 3 q 2 q 2 q 3 q q q 2 q 3

36 q q q q q 3 q 2 q 2 q 3 q q q 2 q 3

37 q q q q q 3 q 2 q 2 q 3 q q q 2 q 3

38 Claim: If p, q are distinguished by Table-Filling algorithm (ie pair labelled by ), then p ~ / q Proof: By induction on the stage of the algorithm Claim: If p, q are not distinguished by Table-Filling algorithm, then p ~ q Proof (by contradiction):

39 Claim: If p, q are distinguished by Table-Filling algorithm (ie pair labelled by ), then p ~ q / Proof: By induction on the stage of the algorithm If (p, q) is marked at the start, then one s in F and one isn t, so ε distinguishes p and q

40 Claim: If p, q are distinguished by Table-Filling algorithm (ie pair labelled by ), then p ~ q / Proof: By induction on the stage of the algorithm If (p, q) is marked at the start, then one s in F and one isn t, so ε distinguishes p and q Suppose (p, q) is marked at stage n+ Then there are states p, q, string w Σ* and σ Σ such that:. (p, q ) are marked p ~ / q (by induction) δ(p, ^ w) F and ^ δ(q, w) F 2. p = δ(p,σ) and q = δ(q,σ) The string σw distinguishes p and q!

41 Claim: If p, q are not distinguished by Table-Filling algorithm, then p ~ q Proof (by contradiction):

42 Claim: If p, q are not distinguished by Table-Filling algorithm, then p ~ q Proof (by contradiction): Suppose the pair (p, q) is not marked by the algorithm, yet p ~ q (a bad pair ) / Suppose (p,q) is a bad pair with the shortest w. δ(p, ^ w) F and ^ δ(q, w) F So, w = σw, where σ Σ (Why is w >?)

43 Claim: If p, q are not distinguished by Table-Filling algorithm, then p ~ q Proof (by contradiction): Suppose the pair (p, q) is not marked by the algorithm, yet p ~ q (a bad pair ) / Suppose (p,q) is a bad pair with the shortest w. δ(p, ^ w) F and ^ δ(q, w) F (Why is w >?) So, w = σw, where σ Σ Let p = δ(p,σ) and q = δ(q,σ) Then (p, q ) cannnot be marked (Why?) But (p, q ) is distinguished by w! So (p, q ) is also a bad pair, but with a SHORTER w! Contradiction!

44 Input: FA M Algorithm MINIMIZE Output: FA M MIN () Remove all inaccessible states from M (2) Apply Table-Filling algorithm to get: E M = { [q] q is an accessible state of M }

45 Input: FA M Algorithm MINIMIZE Output: FA M MIN () Remove all inaccessible states from M (2) Apply Table-Filling algorithm to get: E M = { [q] q is an accessible state of M } efine: M MIN = (Q MIN, Σ, δ MIN, q MIN, F MIN ) Q MIN = E M, q MIN = [q ], F MIN = { [q] q F } δ MIN ( [q], σ ) = [ δ( q, σ ) ] Must show δ MIN is well defined!

46 Input: FA M Algorithm MINIMIZE Output: FA M MIN () Remove all inaccessible states from M (2) Apply Table-Filling algorithm to get: E M = { [q] q is an accessible state of M } efine: M MIN = (Q MIN, Σ, δ MIN, q MIN, F MIN ) Q MIN = E M, q MIN = [q ], F MIN = { [q] q F } δ MIN ( [q], σ ) = [ δ( q, σ ) ] ^ Claim: δ MIN ( [q], w ) = [ ^ δ( q, w) ], w Σ*

47 Input: FA M Algorithm MINIMIZE Output: FA M MIN () Remove all inaccessible states from M (2) Apply Table-Filling algorithm to get: E M = { [q] q is an accessible state of M } efine: M MIN = (Q MIN, Σ, δ MIN, q MIN, F MIN ) Q MIN = E M, q MIN = [q ], F MIN = { [q] q F } δ MIN ( [q], σ ) = [ δ( q, σ ) ] ^ So: δ MIN ( [q ], w ) = [ δ( ^ q, w) ], w Σ*

48 Input: FA M Algorithm MINIMIZE Output: FA M MIN () Remove all inaccessible states from M (2) Apply Table-Filling algorithm to get: E M = { [q] q is an accessible state of M } efine: M MIN = (Q MIN, Σ, δ MIN, q MIN, F MIN ) Q MIN = E M, q MIN = [q ], F MIN = { [q] q F } δ MIN ( [q], σ ) = [ δ( q, σ ) ] Follows: M MIN M

49 q q q 4,, q 5 q 2 q 3

50 q q q 4, q 5 q 3

51 q q q 4, q q 5 q q 3 q 3 q 4 q 5 q q q 3 q 4 q 5

52 q q q 4, q q 5 q q 3 q 3 q 4 q 5 q q q 3 q 4 q 5

53 q q q 4, q q 5 q q 3 q 3 q 4 q 5 q q q 3 q 4 q 5

54 q q q 4, q q 5 q q 3 q 3 q 4 q 5 q q q 3 q 4 q 5

55 q q q 4, q q 5 q q 3 q 3 q 4 q 5 q q q 3 q 4 q 5

56 q q q 4, q q 5 q q 3 q 3 q 4 q 5 q q q 3 q 4 q 5

57 MINIMIZE q q q 2

58 MINIMIZE q q q 2

59 PROPOSITION. Suppose M M and M has no inaccessible states and is irreducible Then, there exists a - onto correspondence between M MIN and M (preserving transitions) i.e., M MIN and M are Isomorphic

60 PROPOSITION. Suppose M M and M has no inaccessible states and is irreducible Then, there exists a - onto correspondence between M MIN and M (preserving transitions) i.e., M MIN and M are Isomorphic COR: M MIN is unique minimal FA M

61 PROPOSITION. Suppose M M and M has no inaccessible states and is irreducible Then, there exists a - onto correspondence between M MIN and M (preserving transitions) i.e., M MIN and M are Isomorphic COR: M MIN is unique minimal FA M Proof of Prop: We will construct a map recursively Base Case: q MIN q Recursive Step: If p p q σ q σ Then q q

62 PROPOSITION. Suppose M M and M has no inaccessible states and is irreducible Then, there exists a - onto correspondence between M MIN and M (preserving transitions) i.e., M MIN and M are Isomorphic COR: M MIN is unique minimal FA M Proof of Prop: We will construct a map recursively Base Case: q MIN q Recursive Step: If p p and δ(p, σ) = q and δ(p,σ) = q Then q q

63 We need to show: The map is everywhere defined The map is well defined The map is a bijection ( - and onto) The map preserves transitions

64 Base Case: q MIN q Recursive Step: If p p σ σ q q Then q q The map is everywhere defined: That is, for all q M MIN there is a q M such that q q If q M MIN, there is a string w such that δ^ MIN (q MIN,w) = q (WHY?) Let q = δ (q ^,w). q will map to q (by induction)

65 Base Case: q MIN q Recursive Step: If p p The map is well defined q σ q σ Then q q That is, for all q M MIN there is at most one q M such that q q Suppose there exist q and q such that q q and q q We show that q and q are indistinguishable, so it must be that q = q (Why?)

66 Suppose there exist q and q such that q q and q q Suppose q and q are distinguishable M MIN M q MIN u q w Accept q u q w Accept q MIN v Contradiction! w q Reject q v q w Reject

67 The map is - Suppose there are distinct p and q such that p q and q q p and q are distinguishable (why?) M MIN M q MIN u p w Accept q u q w Accept q MIN v q w Reject q v Contradiction! w q Reject

68 Base Case: q MIN q Recursive Step: If p p σ σ q q The map is onto Then q q That is, for all q M there is a q M MIN such that q q If q M, there is w such that δ (q ^,w) = q Let q = δ^ MIN (q MIN,w). q will map to q (why?)

69 Base Case: q MIN q Recursive Step: If p p σ σ q q Then q q The map preserves transitions That is, if δ(p, σ) = q and p p and q q then, δ (p, σ) = q (Why?)

70 How can we prove that two regular expressions are equivalent?

71 Read Chapters 2. & 2.2 for next time