Miscellaneous. Closure Properties Decision Properties

Transcription

1 Miscellaneous Closure Properties Decision Properties 1

2 Closure Properties of CFL s CFL s are closed under union, concatenation, and Kleene closure. Also, under reversal, homomorphisms and inverse homomorphisms. But not under intersection or difference. 2

3 Closure of CFL s Under Union Let L and M be CFL s with grammars G and H, respectively. Assume G and H have no variables in common. Names of variables do not affect the language. Let S 1 and S 2 be the start symbols of G and H. 3

4 Closure Under Union (2) Form a new grammar for L M by combining all the symbols and productions of G and H. Then, add a new start symbol S. Add productions S -> S 1 S 2. 4

5 Closure Under Union (3) In the new grammar, all derivations start with S. The first step replaces S by either S 1 or S 2. In the first case, the result must be a string in L(G) = L, and in the second case a string in L(H) = M. 5

6 Closure of CFL s Under Concatenation Let L and M be CFL s with grammars G and H, respectively. Assume G and H have no variables in common. Let S 1 and S 2 be the start symbols of G and H. 6

7 Closure Under Concatenation (2) Form a new grammar for LM by starting with all symbols and productions of G and H. Add a new start symbol S. Add production S -> S 1 S 2. Every derivation from S results in a string in L followed by one in M. 7

8 Closure Under Star Let L have grammar G, with start symbol S 1. Form a new grammar for L* by introducing to G a new start symbol S and the productions S -> S 1 S ε. A rightmost derivation from S generates a sequence of zero or more S 1 s, each of which generates some string in L. 8

9 Closure of CFL s Under Reversal If L is a CFL with grammar G, form a grammar for L R by reversing the right side of every production. Example: Let G have S -> 0S1 01. The reversal of L(G) has grammar S -> 1S

10 Nonclosure Under Intersection Unlike the regular languages, the class of CFL s is not closed under. We know that L 1 = {0 n 1 n 2 n n > 1} is not a CFL (use the pumping lemma). However, L 2 = {0 n 1 n 2 i n > 1, i > 1} is. CFG: S -> AB, A -> 0A1 01, B -> 2B 2. So is L 3 = {0 i 1 n 2 n n > 1, i > 1}. But L 1 = L 2 L 3. 10

11 Nonclosure Under Difference We can prove something more general: Any class of languages that is closed under difference is closed under intersection. Proof: L M = L (L M). Thus, if CFL s were closed under difference, they would be closed under intersection, but they are not. 11

12 Intersection with a Regular Language Intersection of two CFL s need not be context free. But the intersection of a CFL with a regular language is always a CFL. Proof involves running a DFA in parallel with a PDA, and noting that the combination is a PDA. PDA s accept by final state. 12

13 DFA and PDA in Parallel Input DFA PDA Accept if both accept S t a c k Looks like the state of one PDA 13

14 The intersection of a context-free language and a regular language is a context-free language L1 context free L2 regular L1 L 2 context-free 14

15 Formal Construction Let the DFA A have transition function δ A. Let the PDA P have transition function δ P. States of combined PDA are [q,p], where q is a state of A and p a state of P. δ([q,p], a, X) contains ([δ A (q,a),r], α) if δ P (p, a, X) contains (r, α). Note a could be ε, in which case δ A (q,a) = q. 15

16 Formal Construction (2) Accepting states of combined PDA are those [q,p] such that q is an accepting state of A and p is an accepting state of P. Easy induction: ([q 0,p 0 ], w, Z 0 ) * ([q,p], ε, α) if and only if δ A (q 0,w) = q and in P: (p 0, w, Z 0 ) *(p, ε, α). 16

17 Machine M 1 Machine M 2 NPDA for L DFA for 1 L2 context-free regular Construct a new NPDA machine that accepts L1 L 2 M M simulates in parallel M1 and M 2 17

18 NPDA M M 2 1 DFA a, b c q1 q2 transition a p1 p2 transition NPDA M q1, p 1 a, b c q 2, p 2 transition 18

19 NPDA M M 2 1 DFA λ, b c q1 q2 transition p 1 NPDA M q 1, p 1 λ, b c q 2, p 1 transition 19

20 NPDA M M 2 1 DFA q 0 p 0 initial state initial state NPDA M q 0, p 0 Initial state 20

21 NPDA M M 2 1 DFA q 1 p 1 p 2 final state final states NPDA M q 1, p 1 q 1, p 2 final states 21

22 An Application of Regular Closure Prove that: L n n = { a b : n 100, n 0} is context-free 22

23 We know: n n { a b : n 0} is context-free 23

24 We also know: 100 L 1 = { a b 100 } is regular L1 = {( a + b) } { a b * } is regular 24

25 { a n b n } context-free 100 L1 = {( a + b) } { a b * regular 100 } (regular closure) n n { a b } L 1 context-free n n { a b } L1 = { a n b n : n 100, n 0} = L is context-free 25

26 Another Application of Regular Closure Prove that: L = { w: na = nb = n c } is not context-free 26

27 If L = { w: n = n = nc} a b is context-free Then (regular closure) L { a * b* c*} = { a n b n c n } context-free regular context-free Therefore, L is not context free Impossible!!! 27

28 Closure of CFL s Under String Homomorphism Let L be a CFL with grammar G. Let h be a homomorphism on the terminal symbols of G. Construct a grammar for h(l) by replacing each terminal symbol a by h(a). 28

29 String Homomorphism Map from Σ to * for two alphabets Σ and h: Σ * (assigns a string in to each symbol in Σ) Example: Σ = {0,1} and = {a,b,c} h(0) = abc h(1) = bb The map extends to strings over Σ as well Example: string 010 over Σ h*: Σ* * Example: h*(010) = h(0)h(1)h(0) = abcbbabc 29

30 String Homomorphism h* maps a language over Σ to a language over naturally h*(l) = {h*(w) w L) Theorem: If L Σ* is a CFL, and h: Σ *, then h*(l) is also a CFL, i.e. CFLs are closed under string homomorphism Proof: Consider G = <V, Σ, P, S> generates L. Construct G as follows for h*(l). G = <V Σ,, P, S> where P = P {a h(a) for every a Σ} Note: as many productions as many terminals added Clearly, G generates h*(l). G being a CFG, so is G. 30

31 Example: Closure Under Homomorphism G has productions S -> 0S1 01. h is defined by h(0) = ab, h(1) = ε. h(l(g)) has the grammar with productions S -> abs ab. 31

32 Substitution Let Σ be an alphabet and suppose for every a in Σ, we choose a language L a over alphabet The choice of languages defines a substitution on Σ and we refer to L a as s(a) for each a Σ Example: s(0) = {a n b n n 1}, s(1) = {aa, bb} is a substitution on alphabet Σ ={0, 1}. If w = a 1 a 2 a n is a string in Σ*, then s(w) is the language of all strings x 1 x 2 x n such that string x i s(a i ) for i = 1..n s(w) is a concatenation of the languages s(a 1 )s(a 2 ) s(a n ) s(l) is the union of s(w) for all strings in L 32

33 Substitution Example: s(0) = {a n b n n 1}, s(1) = {aa, bb} is a substitution on alphabet Σ ={0, 1}. s(0) is the set of strings with one or more a s followed by an equal number of b s s(1) is the finite language consisting of the two strings aa and bb Let w = 01. s(w) is the concatenation of the languages s(0)s(1). s(w) consists of all strings of the form a n b n aa and a n b n+2 n 1 Let L = L(0*) i.e. set of all strings of 0 s. s(l) = (s(0))*: all strings of the form a n1 b n1 a n2 b n2 a nk b nk k >= 0 and any sequence of positive integers n1, n2..nk Includes strings such as ε, aabbaaabbb, and abaabbabab 33

34 Substitution Theorem: If a substitution s assigns a CFL to every symbol in the alphabet of a CFL L, then s(l) is a CFL. Proof Let G = (V,, P, S) be grammar for L Let G a = (V a, T a, P a, S a ) be the grammar for each a with V V a =φ G = (V, T, P, S) for s(l) where V = V V a T = union of T a for all a P consists of» All productions in any P a for a» In productions of P, each terminal a is replaced by S a A detailed proof that this construction works Intuition: this replacement allows any string in L a to take the place of any 34 occurrence of a in any string of L.

35 Example (1) L = {0 n 1 n n 1}, generated by the grammar S 0S1 01, s(0) = {a n b m m n}, generated by the grammar S asb A; A aa ab, s(1) ={ab, abc}, generated by the grammar S aba, A c ε Rename second and third S s to S 0 and S 1 respectively. Rename second A to B. Resulting grammars are: S 0S1 01 S 0 as 0 b A; A aa ab S 1 abb; B c ε 35

36 Example(1) Contd... In the first grammar replace 0 by S 0 and 1 by S 1. The combined grammar: G = ({S, S 0, S 1, A, B}, {a, b}, P, S), where P = {S S 0 SS 1 S 0 S 1, S 0 as 0 b A, A aa ab, S 1 abb, B c ε} 36

37 Application of Substitution Closure under union of CFL s L a and L b Let L a and L b be CFLs. Then L a L b is s(l) where L = {a,b} {a,b} * and s(a) = L a, s(b) = L b Closure under concatenation of CFL s L a and L b Let L a and L b be CFLs. Then L a L b is s(l) where L = {ab} and s(a) = L a, s(b) = L b Work out yourself: Closure under Kleene s star (closure * and positive closure + ) of CFL s L 1 Closure under homomorphism of CFL L i for every a i 37

38 Closure of CFL s Under Inverse Homomorphism If L is a language and h is a homomorphism, then h -1 (L) is the set of strings w such that h(w) is in L. Let L be a CFL and h be a homomorphism. Then h -1 (L) is a CFL. Here, grammars don t help us. But a PDA construction serves nicely. Intuition: Let L = L(P) for some PDA P. Construct PDA P to accept h -1 (L). P simulates P, but keeps, as one component of a twocomponent state a buffer that holds the result of applying h to one input symbol. 38

39 Architecture of P Input: h(0) Buffer State of P Read first remaining symbol in buffer as if it were input to P. Stack of P 39

40 Formal Construction of P States are pairs [q, b], where: 1. q is a state of P. 2. b is a suffix of h(a) for some symbol a. Thus, only a finite number of possible values for b. Stack symbols of P are those of P. Start state of P is [q 0,ε]. 40

41 Construction of P (2) Input symbols of P are the symbols to which h applies. Final states of P are the states [q, ε] such that q is a final state of P. 41

42 Transitions of P 1. δ ([q, ε], a, X) = {([q, h(a)], X)} for any input symbol a of P and any stack symbol X. When the buffer is empty, P can reload it. 2. δ ([q, bw], ε, X) contains ([p, w], α) if δ(q, b, X) contains (p, α), where b is either an input symbol of P or ε. Simulate P from the buffer. 42

43 Proving Correctness of P We need to show that L(P ) = h -1 (L(P)). Key argument: P makes the transition ([q 0, ε], w, Z 0 ) *([q, x], ε, α) if and only if P makes transition (q 0, y, Z 0 ) *(q, ε, α), h(w) = yx, and x is a suffix of the last symbol of w. Proof in both directions is an induction on the number of moves made. 43

44 Equivalence of PDA, CFG Conversion of CFG to PDA Conversion of PDA to CFG 44

45 Overview When we talked about closure properties of regular languages, it was useful to be able to jump between RE and DFA representations. Similarly, CFG s and PDA s are both useful to deal with properties of the CFL s. 45

46 Overview (2) Also, PDA s, being algorithmic, are often easier to use when arguing that a language is a CFL. Example: It is easy to see how a PDA can recognize balanced parentheses; not so easy as a grammar. But all depends on knowing that CFG s and PDA s both define the CFL s. 46

47 Converting a CFG to a PDA Let L = L(G). Construct PDA P such that N(P) = L. P has: One state q. Input symbols = terminals of G. Stack symbols = all symbols of G. Start symbol = start symbol of G. 47

48 Intuition About P Given input w, P will step through a leftmost derivation of w from the start symbol S. Since P can t know what this derivation is, or even what the end of w is, it uses nondeterminism to guess the production to use at each step. 48

49 Intuition (2) At each step, P represents some leftsentential form (step of a leftmost derivation). If the stack of P is α, and P has so far consumed x from its input, then P represents left-sentential form xα. At empty stack, the input consumed is a string in L(G). 49

50 Transition Function of P 1. δ(q, a, a) = (q, ε). (Type 1 rules) This step does not change the LSF represented, but moves responsibility for a from the stack to the consumed input. 2. If A -> α is a production of G, then δ(q, ε, A) contains (q, α). (Type 2 rules) Guess a production for A, and represent the next LSF in the derivation. 50

51 Example Grammar G: S asb ab PDA: δ(q, ε, S) contains (q, asb) and (q,ab) nondeterministic move δ(q, a, a) contains (q, ε) δ(q, b, b) contains (q, ε) Simulate it on aaabbb 51

52 Proof That L(P) = L(G) We need to show that (q, wx, S) * (q, x, α) for any x if and only if S =>* lm wα. Part 1: only if is an induction on the number of steps made by P. Basis: 0 steps. Then α = S, w = ε, and S =>* lm S is surely true. 52

53 Induction for Part 1 Consider n moves of P: (q, wx, S) * (q, x, α) and assume the IH for sequences of n-1 moves. There are two cases, depending on whether the last move uses a Type 1 or Type 2 rule. 53

54 Use of a Type 1 Rule The move sequence must be of the form (q, yax, S) * (q, ax, aα) (q, x, α), where ya = w. By the IH applied to the first n-1 steps, S =>* lm yaα. But ya = w, so S =>* lm wα. 54

55 Use of a Type 2 Rule The move sequence must be of the form (q, wx, S) * (q, x, Aβ) (q, x, γβ), where A -> γ is a production and α = γβ. By the IH applied to the first n-1 steps, S =>* lm waβ. Thus, S =>* lm wγβ = wα. 55

56 Proof of Part 2 ( if ) We also must prove that if S =>* lm wα, then (q, wx, S) * (q, x, α) for any x. Induction on number of steps in the leftmost derivation. Ideas are similar; read in text. 56

57 Proof Completion We now have (q, wx, S) * (q, x, α) for any x if and only if S =>* lm wα. In particular, let x = α = ε. Then (q, w, S) * (q, ε, ε) if and only if S =>* lm w. That is, w is in N(P) if and only if w is in L(G). 57

58 From a PDA to a CFG Now, assume L = N(P). We ll construct a CFG G such that L = L(G). Intuition: G will have variables generating exactly the inputs that cause P to have the net effect of popping a stack symbol X while going from state p to state q. P never gets below this X while doing so. 58

59 Variables of G G s variables are of the form [pxq]. This variable generates all and only the strings w such that (p, w, X) *(q, ε, ε). Also a start symbol S we ll talk about later. 59

60 Productions of G Each production for [pxq] comes from a move of P in state p with stack symbol X. Simplest case: δ(p, a, X) contains (q, ε). Then the production is [pxq] -> a. Note a can be an input symbol or ε. Here, [pxq] generates a, because reading a is one way to pop X and go from p to q. 60

61 Productions of G (2) Next simplest case: δ(p, a, X) contains (r, Y) for some state r and symbol Y. G has production [pxq] -> a[ryq]. We can erase X and go from p to q by reading a (entering state r and replacing the X by Y) and then reading some w that gets P from r to q while erasing the Y. Note: [pxq] =>* aw whenever [ryq] =>* w. 61

62 Productions of G (3) Third simplest case: δ(p, a, X) contains (r, YZ) for some state r and symbols Y and Z. Now, P has replaced X by YZ. To have the net effect of erasing X, P must erase Y, going from state r to some state s, and then erase Z, going from s to q. 62

63 Picture of Action of P a w x w x x p r s q X Y Z Z 63

64 Third-Simplest Case Concluded Since we do not know state s, we must generate a family of productions: [pxq] -> a[rys][szq] for all states s. [pxq] =>* awx whenever [rys] =>* w and [szq] =>* x. 64

65 Productions of G: General Case Suppose δ(p, a, X) contains (r, Y 1, Y k ) for some state r and k > 3. Generate family of productions [pxq] -> a[ry 1 s 1 ][s 1 Y 2 s 2 ] [s k-2 Y k-1 s k-1 ][s k-1 Y k q] 65

66 Completion of the Construction We can prove that (q 0, w, Z 0 ) *(p, ε, ε) if and only if [q 0 Z 0 p] =>* w. Proof is in text; it is two easy inductions. But state p can be anything. Thus, add to G another variable S, the start symbol, and add productions S -> [q 0 Z 0 p] for each state p. 66