Context Free Language Properties

Transcription

1 Context Free Language Properties Knowing that the context free languages are exactly those sets accepted by nondeterministic pushdown automata provides us a bit of information about them. We know that their membership problem is solvable, but the Σ * problem (whether all strings over Σ are in the language) and the cofiniteness problems are unsolvable. Our discussion of general language properties brought forth even more. It revealed that this family was closed under the operations of union, concatenation, and Kleene star. In this section we shall look at what has been set aside in our treatment thus far. ll of the remaining decision problems and closure properties. The following lemma shall prove of great value in these endeavors. It is the context free version of the pumping lemma we examined for regular sets. Lemma (Pumping). For every context free language L there is an integer n such that for any string z in L whose length is at least n: a) there exist u, v, w, x, y such that z = uvwxy, b) vx ε, c) the length of vwx 2n, and d) for all i, uv i wx i y L. Proof. Let G = (N,T,P,) be a Chomsky normal form grammar for L with k nonterminal symbols. Let z L(G) be at least 2 k characters long. Let's also set n = 2 k and claim that this is the n we are looking for. Consider the derivation tree for this string z. ince G is in Chomsky Normal form the entire tree except for the edges leading to the leaves is binary. We know that the shortest binary tree with 2 k leaves has height k. Thus the derivation tree for the string z must contain paths of more than k nonterminal symbols since z is longer than 2 k. Let us now select the longest path in the derivation tree. Watch closely. s we stated above, there are more than k nonterminal symbols on this longest path. Thus a nonterminal symbol must be repeated along it. (There are only k of these - remember?) Let us start at the leaf on this path and go up the tree. We now find the first nonterminal symbol which appears twice and call it. We also mark the two occurrences of this symbol which are closest to the bottom of the tree. This is illustrated in figure 1.

2 Context Free Language Properties 2 u v w x y Figure 1 - Derivation tree for z L(G) In figure 1 we also depicted the assignment of u, v, w, x, and y such that: generates uvwxy = z, generates vwx, and generates w. This is all we need. Part (a) of the theorem is obviously true since z = uvwxy. ince the nonterminals on the path from to do generate terminal symbols, both u and v cannot be empty. (In fact, if B, then v still contains some terminals since B will generate a terminal string.) Thus part (b) is correct also. Part (c) depends upon the fact that we selected the longest path and set to be the nonterminal which repeated closest to the bottom of the derivation tree. Thus the path from the top in figure 1 to the leaves cannot contain more than k+1 nonterminals (including the two 's). binary derivation tree of this height can produce at most 2 k+1 = 2n terminals. ll that remains is to verify part (d) of the theorem. Here are some more pictures. In figure 2 we collapse the path from to.

3 Context Free Language Properties 3 w u y Figure 2 - Derivation Tree for uwy This generates uwy = uv 0 wx 0 y. ince it is a valid derivation by our grammar G, uwy must be a member of the language L. The next illustration (figure 3) reveals what takes place when we repeat the path from to one more time. u v x y v w x Figure 3 - Derivation Tree for uvvwxxy The string uvvwxxy = uv 2 wx 2 y is generated this time. This must also be a member of the language L. In a like manner all strings of the form uv i wx i y can be generated by the grammar G for L.

4 Context Free Language Properties 4 Just as the pumping lemma for regular sets was used to demonstrate that there are some nonregular sets, we shall use the context free language pumping lemma to show that not all languages are context free. Theorem 1. The set of strings of the form 0 n 1 n 0 n is not context free. Proof. We use almost the same argument that we invoked to prove that strings of the form a n b n were not regular. We assume that strings of the form 0 n 1 n 0 n are context free. Then we apply the pumping lemma and state that there is some integer m such that for any string at least m in length of the form 0 n 1 n 0 n satisfies the conditions of the pumping lemma. We shall go a bit overboard and let this string be 0 m 1 m 0 m. We know the pumping lemma assures us that: a) There exist u, v, w, x, and y such that uvwxy =0 m 1 m 0 m, b) vx ε, and c) for all i, uv i wx i y is of the form 0 n 1 n 0 n. We now observe that either v and x each contain only one kind of symbol (0 or 1), or at least one of them contains some zeros and some ones. In either case the string uv 2 wx 2 y cannot be of the form 0 n 1 n 0 n, because either the runs of zeros and ones will not be of the same length, or there will be more than three sequences. ince the set of strings of the form 0 n 1 n 0 n is a context sensitive language we may now state a corollary to our last theorem. Corollary. The context free languages are a strict subclass of the context sensitive languages. Other results which follow quickly from the pumping lemma and the fact that 0 n 1 n 0 n is not context free are two nonclosure properties. The second (complement) is just an application of DeMorgan's Laws. Theorem 2. intersection. The family of context free languages is not closed under Proof. Here are two context free grammars. B B B 0B B 0B B 0 B 0

5 Context Free Language Properties 5 It is evident that the languages generated by these grammars are the sets of strings of the form 0 n 1 n 0 and 0 1 n 0 n for values of n 1. The intersection of these languages is merely our old friend 0 n 1 n 0 n. o, the context free languages cannot be closed under intersection. Theorem 3. The family of context free languages is not closed under complement. To close out our examination of context free language properties we turn to decision problems. Our study of pushdown machines lead to the unsolvability of equivalence, cofiniteness, and whether the language contained all strings. Now we yet again bring out the pumping lemma to show that emptiness is solvable. Theorem 4. The emptiness problem is solvable for context free languages. Proof. Given a Chomsky Normal Form grammar, we claim that it will generate a string of length less than 2 k (where as in the pumping lemma, k is the number of nonterminals) if it generates any strings at all. The justification for this claim is that if the shortest string in the language is longer than 2 k symbols, we can apply the pumping lemma and shorten it. Thus a check of all strings up to 2 k in length forms an inefficient algorithm for emptiness. In order to show that finiteness is also solvable for the family of context free languages we could merely trot out the pumping lemma and look at strings longer than 2 k in length. (In fact, prove this as an interesting exercise!) Instead, we shall carry out some transformations on context free grammars which make them nice enough to almost solve finiteness for us. Definition. useless nonterminal symbol is one which generates no strings of terminals. Theorem 5. Every context free language can be generated by a grammar which contains no useless nonterminals. Proof. First we detect the useless symbols and then discard them. To find out if a symbol is useless, just make it the starting symbol and check for emptiness. Easy as that. Now we toss out all productions containing useless nonterminals and claim that the grammar generates the same language. (Note that if a nonterminal never generates a terminal string, then productions containing it will not lead to terminal strings either.)

6 Context Free Language Properties 6 Definition. n unreachable nonterminal symbol is one which cannot be generated from the starting symbol. Theorem 6. Every context free language can be generated by a grammar which contains no unreachable nonterminal. The proof of this last theorem is left as an exercise. (The diagram defined below is of great use in this though.) For now we note that we have removed two kinds of nasty nonterminals from our grammars. We need one other bit of business before unveiling our finiteness algorithm. transition diagram for context free grammars. This is just a graph with nonterminals as vertices and directed edges going from one nonterminal to another if they are on different sides of the same production. (The direction is from left to right.) In other words, an edge in the graph means that one nonterminal generates another. Figure 5 illustrates this for a grammar which generates our constant companion: strings of the form a n b n. B C C B a B b C B Figure 5 - Grammar and Transition Diagram Theorem 7. The finiteness problem is solvable for context free languages. Proof. We begin by making some observations. If a language is infinite then it has strings of arbitrary length. Long strings have high derivation trees which have repeating nonterminals on paths through the tree. nd, repeating nonterminals signify infinite languages. o, we must detect repeating nonterminals. First, we produce a grammar for the language which contains no useless or unreachable nonterminals. Next, we draw the transition diagram for the grammar. t this point we maintain that if there is a cycle in the transition diagram then a nonterminal can be repeated in a derivation. nd, if there is a derivation with a repeating nonterminal, then there must be a cycle in the diagram because the nonterminal eventually generates itself. Thus, detecting cycles in the transition diagram reveals whether or not a grammar generates an infinite language.