Properties of Context-Free Languages

Transcription

1 Properties of Context-Free Languages Seungjin Choi Department of Computer Science and Engineering Pohang University of Science and Technology 77 Cheongam-ro, Nam-gu, Pohang 37673, Korea 1 / 31

2 Outline Simplification of CFG s: We have already done this. See Lecture 07! Pumping Lemma for CFL s: Similar to the regular case, but slightly more complex. Closure properties: Some but not all, of the closure properties of regular languages carry over to CFL s. Decision properties: Test for membership and emptiness (for instance, CYK algorithm). 2 / 31

3 Before we move on the pumping lemma... Theorem Suppose we have a parse tree according to a CNF, G = (V, T, S, P) and suppose that the yield of the tree is a terminal string w. If the length of the longest path is n, then w 2 n 1. Proof. A simple induction on n. Basis: n = 1. The length of a path in a tree is the number of edges, i.e., one less than the number of nodes. A tree with a maximum path length of 1 consists of only a root and one leaf labeled by a terminal. Thus w = 1. Since 2 n 1 = 2 0 = 1, the basis is proved. Induction: Suppose that the longest path has length n (n > 1). The root of the tree uses a production, which must be of the form A BC. No path in the subtree rooted at B and C can have length greater than n 1, since these path exclude the edge from the root to its child labeled B or C. Thus, by the IH, these two subtrees each have yields of length at most 2 n 2. The yield of entire tree is the concatenation of these two yields, so has length at most 2 n n 2 = 2 n 1. 3 / 31

4 Pumping Lemma for CFL Theorem Let L be a CFL. Then there exists a positive integer n such that if z is any string in L with z n, then we can write z = uvwxy, subject to the following conditions: 1. vwx n. That is, the middle portion is not too long. 2. vx ɛ. Since v and x are the pieces to be pumped, this condition says at least one of the strings we pump must not be empty. 3. i 0, uv i wx i y L. 4 / 31

5 Proof. Start with a CNF, G = (V, T, S, P) such that L(G) = L {ɛ}. Let G have m variables. Choose n = 2 m. Next, suppose that z L is of length at least n. Any parse tree whose longest path is of length m, must have a yield of length 2 m 1 = n 2 or less. Thus, any parse tree with yield z has a path of length at least m + 1. A 0 A 1 A 2. A k a 5 / 31

6 Since k m, there are at least m + 1 occurrences of variables A 0, A 1,..., A k on the path. As there are only m different variables in V, at least two of the last m + 1 variables on the path must be the same. Suppose A i = A j where k m i < j k. S A i = A j A j u v w x y z 6 / 31

7 If A i = A j = A, then we can construct new parse trees. S A A u v w x y 7 / 31

8 We may replace the subtree rooted at A i, which has yield vwx, by the subtree rooted at A j, which has yield w. (i = 0) S A w u y 8 / 31

9 We may replace the subtree rooted at A i by the entire subtree rooted at A i. (i = 2) S A A u v A x y v w x 9 / 31

10 What about the case where vwx n? We picked A i to be close to the bottom of the tree, k i m. Thus the longest path in the subtree rooted at A i is no longer than m + 1. By theorem, the subtree rooted at A i has a yield whose length is no greater than 2 m = n. 10 / 31

11 Examples Prove that the following languages are not context-free: 1. L = {a n b n c n n 0}; 2. L = {ww w {a, b} }. 11 / 31

12 Closure Properties of CFL The family of CFL s is closed under: union concatenation star-closure ( ) and positive closure (+) homomorphisms and inverse homomorphisms The family of CFL s is not closed under: intersection complementation difference 12 / 31

13 Substitutions We introduce an operation substitution (denoted by s( ), in which we replace each symbol in the strings of one language by an entire language. (a generalization of the homomorphism) Consider a mapping s : Σ 2, where Σ and are finite alphabets. Let w Σ where w = a 1 a 2 a n, and define and for L Σ s(a 1 a 2 a n ) = s(a 1 )s(a 2 ) s(a n ), s(l) = w L s(w). 13 / 31

14 Example: Given Σ = {0, 1}, = {a, b}, s(0) = {a n b n n 1}, s(1) = {aa, bb}. Let w = 01. Then s(w) = s(0)s(1) = {a n b n aa n 1} {a n b n+2 n 1}. Let L = {0}. Then s(l) = (s(0)) = {a n1 b n1 a n2 b n2 a n k b n k k 0, n i 1}. 14 / 31

15 Theorem (Substitution theorem) Let L be a CFL over Σ and s( ) a substitution such that s(a) is a CFL, a Σ. Then s(l) is a CFL. Proof. The essential idea is that we may take a CFG for L and replace each terminal a by the start symbol of a CFG for language s(a). The result is a single CFG that generates s(l). We start with grammars for L and for each s(a). G = (V, T, S, P), G a = (V a, T a, S a, P a ), 15 / 31

16 We then construct G for s(l), where ( 1. V = a Σ V a ) V ; G = (V, T, S, P ), 2. T = a Σ T a ; 3. P = a Σ P a plus the productions of P but with each terminal a in their bodies replaced by S a everywhere a occurs. 16 / 31

17 Now we have to show that L(G ) = s(l). (If): Let w s(l). Then x = a 1 a 2 a n in L and x i s(a i ) such that w = x 1 x 2 x n. A derivation tree in G will look like S Sa 1 Sa 2 San x 1 x 2 xn Thus we can generate S a1 S a2 S an x 1 x 2 x n = w. Thus w L(G ). in G and form there we generate 17 / 31

18 (Only if): Let w L(G ). Then the parse tree for w must again look like S Sa 1 Sa 2 San x 1 x 2 xn Now delete the dangling subtrees. Then you have yield S a1 S a2 S an, where a 1 a 2 a n L(G). Now w is also equal to s(a 1 a 2 a n ) which is in s(l). 18 / 31

19 Theorem The CFL s are closed under: (1) union; (2) concatenation; (3) star-closure ( ) and positive closure (+); (4) homomorphism. Proof. The proof is done using the substitution theorem. 1. Let L 1 and L 2 be CFL s. Let L = {1, 2} and s(1) = L 1, s(2) = L 2. Then s(l) = L 1 L Choose L = {12} and s as before. Then s(l) = L 1 L 2, which is CFL. 3. Suppose L 1 is CFL. Let L = {1}, s(1) = L 1. Then L 1 = s(l). Similar proof for Let L 1 be a CFL over Σ and h a homomorphism on Σ. Then define s by a {h(a)}. Then h(l) = s(l). 19 / 31

20 Theorem If L is a CFL, so is L R. Proof. Suppose L is generated by G = (V, T, S, P). Construct G R = (V, T, S, P R ) where P R = {A α R A α P}. Easy induction on the lengths of derivations in G and G R to show that L(G R ) = L R. 20 / 31

21 The CFL s are not closed under intersection. A simple example is as follows. Consider two CFL s L 1 and L 2 given by L 1 = {0 n 1 n 2 i n 1, i 1}, L 2 = {0 i 1 n 2 n n 1, i 1}. However, L = L 1 L 2 = {0 n 1 n 2 n n 1} is not CFL. (prove it) See Example 7.26 in page 291! 21 / 31

22 Theorem If L is a CFL and R is a RL, then L R is a CFL. Proof. Let L be accepted by PDA M P = (Q P, Σ, Γ, δ P, q P, Z 0, F P ) by final state and let R be accepted by DFA Construct a PDA for L R as shown in the RHS. M D = (Q D, Σ, δ D, q D, F D ). 22 / 31

23 Formally, define M P = (Q P Q D, Σ, Γ, δ, (q P, q D ), Z 0, F P F D ), where δ((q, p), a, X ) = {((r, δd(p, a)), γ) (r, γ) δ P (q, a, X )}. One can easily prove by an induction, both for M P and for M P that (q P, w, Z 0 ) (q, ɛ, γ) in M P, if and only if ((q P, a D ), w, Z 0 ) ((q, δ (p D, w)), ɛ, γ) in M P. 23 / 31

24 Theorem Let L, L 1, L 2 be CFL s and R be regular. Then 1. L R is a CFL; 2. L is not necessarily CFL; 3. L 1 L 2 is not necessarily CFL. Theorem Let L be a CFL and h a homomorphism. Then h 1 (L) is a CFL. 24 / 31

25 Testing Membership: CYK Algorithm Testing membership: Decide membership of a string w in a CFL L. A naive way to test membership, takes time which is exponential in w. We discuss an efficient technique based on the idea of dynamic programming. CYK algorithm: Named after J. Cocke, D. Younger, and T. Kasami. 25 / 31

26 Dynamic Programming vs Divide-and-Conquer Divide-and-conquer Divide-and-conquer works best when all sub-problems are independent. Divide-and conquer is best suited for the case when no overlapping subproblems are encountered. Dynamic programming Dynamic programming is a method for solving complex problems by breaking them down into simpler sub-problems. It is applicable to problems exhibiting the properties of overlapping sub-problems which are only slightly smaller and optimal sub-structure. Dynamic programming is needed when sub-problems are dependent. Dynamic programming solves the sub-problem only once and then stores it in table. Dynamic programming solves the sub-problems bottom up. The problem cant be solved until we find all solutions of sub-problems. The solution comes up when the whole problem appears.. 26 / 31

27 CYK Algorithm Starts with a CNF grammar G = (V, T, S, P). Input: w = a 1 a 2 a n. Construct a triangular table where X ij contains all variables A such that A a i a i+1 a j. 27 / 31

28 To fill the table we work row-by-row, upwards. The first row is computed in the basis, the subsequent ones in the induction. Basis: X ii = {A A a i G}. Induction: Compute X ij which is in row j i + 1. Any derivation A a i a i+1 a j must start with some step A BC. Then B a i a i+1 a k for some k < j and C a k+1 a k+2 a j. We find variables B and C and integer k such that: 1. i k < j. 2. B is in X ik. 3. C is in X k+1 j. 4. A BC is a production of G. 28 / 31

29 Example: See Ex 7.34 in page 306. A CNF grammar G has the following productions: S AB BC, A BA a, B CC b, C AB a. 29 / 31

30 Consider which variables have a production body a or b, leading to X 11 = X 44 = {B} and X 22 = X 33 = X 55 = {A, C}. Compute X 12 : In order for a variable to generate ba it must have a body whose first variable is in X 11 = {B} and whose second variable is in X 22 = {A, C}, leading to variables containing a body BA or BC. Such variables are {S,A}. Compute X 24 : Consider all bodies in X 22 X 34 X 23 X 44. The set of strings is {A, C}{S, C} {B}{B} = {AS, AC, CS, CC, BB}. Only CC is a body and its head is B. Thus, X 24 = {B} 30 / 31

31 Theorem The run time of the CYK algorithm is O(n 3 ). A string w is in L(G) if and only if S is in X 1n. 31 / 31