The Chomsky Hierarchy(review)

Transcription

1 The Chomsky Hierarchy(review) Languages Machines Other Moels Recursively Enumerable Sets Turing Machines Post System Markov Algorithms, µ-recursive Functions Context-sensitive Languages Context-free Languages Linear-boune Automata Pushown Automata..... Regular Languages Finite State Automata Regular Expression 127

2 Proof of the Proper Containment of the Language Hierarchy Now we will prove the vertical relationship (i.e., the proper containment) of the Chomsky hierarchy. We only prove the lower two levels. (Refer the text or other relate books for proofs of the remaining upper two levels.) It is trivial to prove that every regular language is also a context-free language, an every language accepte by a finite state machine can also be accepte by a pushown automaton. Hence, for the proof it is enough to show a context-free language that is not regular. To o this we first show the following popular, so calle, pumping lemma. Pumping lemma. Let L be a regular language. Then there exists a constant n such that, for every string z L with z n, there exist u,v,w Σ * such that z = uvw with uv n an v 1, such that for all i 0, uv i w L. 128

3 Proof of the lemma. If L is regular, then it is accepte by a FA M = (Q, Σ, δ, q 0, F ). Let Q = n, an consier a string z Σ * of length m n. Let z = a 1 a 2... a m, an for every i, 1 i m, let δ( q 0, a 1 a 2... a i ) = q i. Since n m, i.e., the number of states is less than or equal to the length of the string, there shoul be two integers j an k, 0 j < k n, such that q j = q k. (We can prove this using the pigeonhole principle or proof by inuction.) This implies that path q j... q k is a cycle of the state transition graph. q 0 a q a 2 a q j..... j a q k..... q m j+1 a k a m Let u = a 1 a 2... a j, v = a j+1 a j+2... a k an w = a k+1 a k+2... a m. Obviously, uv n an v 1. Clearly, if a 1 a 2... a m = uvw L (i.e, q m is in F), then we have a 1 a 2... a j (a j+1... a k ) i a k+1... a m = uv i w L, for all i

4 Application of the Pumping Lemma We nee to unerstan clearly what the lemma says before we use it. The lemma can be rewritten as follows. It shows important logical segments to work with. (1) For all regular languages L, (2) there exists a constant n such that (3) for all z L with z n, (4) there exist u, v an w such that (i) z = uvw, (ii) uv n, (iii) v 1, an (iv) for all i 0, uv i w L. 130

5 Application of the Pumping Lemma Application Example. L = {a i b i i 0 } is not regular. Proof (by contraiction). Suppose L is regular. Then it shoul satisfy parts (2) - (4) of the pumping lemma. Let n be the constant (of part (2)) an pick z = a n b n from L. Clearly, z meets the length conition of part (3). Now, we examine if part (4) above is true for z. If not, L oes not satisfies the lemma, which implies that L cannot be a regular language. Consier any u, v an w such that (i) z = a n b n = uvw (ii) uv n (iii) v 1. n n aa abb b uv w Notice that by conitions (i), (ii) an (iii) above, v = a j, 1 j n, i.e., string v contains only a s. We o not know exactly how many a s are in v. It is enough to know that it has only a s. Now, consier string z' = uv 2 w = uvvw. Clearly, z' has more a s than b s, an hence, z' cannot be a member of L. We are in a contraiction. L is not regular. 131

6 Ruminating on the Application of the Pumping Lemma Recall that, to isprove a statement which uses a universal quantifier (for example for all z ), it is enough to fin one counter example. The lemma uses two universal quantifiers in parts (3) an (4)-(iv). For these parts, it is enough to pick one counter example, because our objective is to isprove the statement of the lemma. We picke z = a n b n for part (3), an i = 2 for part (4)-(iv). You may choose others if you like. But for these parts, one counter example is enough. To isprove a statement which uses an existential quantifier (for example there is n.., or there exists u, v, w ), we nee to consier all such cases. Hence, in the proof we use n like a variable. Whatever the value of the constant n is, we are using that value of n. Notice that for u, v an w, we shoul consier all possible u, v an w which satisfy conitions (i), (ii) an (iii). 132

7 Ruminating on the Application of the Pumping Lemma (cont e) The following example shows a typical case of malicious application of the pumping lemma, which we frequently come across while graing tests for this course. Question: Is the language L = {xyx x = abc, y {a,b,c} * } regular? Prove your answer. Answer (wrong): L is not regular. Suppose that it is, an let n be the constant of the pumping lemma. Consier z = abcyabc, where y {a,b,c} * such that y > n. Obviously, z n. Let z = uvw, where u = a, v = b an w = cyabc. Then uv 2 w = abbcyabc is not in L! Contraiction! What is wrong with this proof? Actually L is a regular language. We can easily construct an FA which accepts L. 133

8 A non-regular language which satisfies the pumping lemma We learne that all regular languages satisfy the pumping lemma. We may ask the following questions: Is there a non-regular language which satisfies the pumping lemma? The answer is positive. Here is an example. Let L 1 = { x x {a, b} *, an x has no substring of the form aa or bb}. L 1, which is regular, can also be expresse in terms of a regular expression, for example; (ab) * + b(ab) * + (ab) * a +(ba) *. Note that L 1 has strings of all lengths. Let L 2 = { x x {a, b} * an x is a prime }, an let L = {a,b} * 1 - L 1, the complement of L 1. Consier the language; L = (L 1 L 2 ). We will show that L is not regular an satisfies the pumping lemma. Suppose L is regular, then L - must be regular because is regular. Use the pumping lemma with L 1 L 1 a z L - L by pumping z = uvw up to z' = uv p+1 1 w, where p = z. Let j = v. Then we have z' = uv p+1 w = uvw + v p = p + jp = p(j+1), which is not prime. (Notice that p is a prime number.) This implies that z is not in L - L 1. Hence, L - L 1 is not regular. We are in a contraiction. It follows that L cannot be regular. L 1 134

9 Now, we show that L satisfies the pumping lemma. We choose n = 3 as the constant of the lemma. For z L such that z 3, let z = c 1 c 2 c 3 x, where c 1, c 2, c 3 {a, b}, an x Σ *. We will show that there exist u, v, w Σ * such that (i) z = uvw, (ii) uv n, (iii) v 1, (iv) uv i w L, for all i 0. Consier the following three cases epening on c 1, c 2, an c 3, : Case 1: c 1 c 2 c 3. Since c 1, c 2, c 3 {a, b}, we have c 1 = c 3. Pump c 2, i.e., let v = c 2. Clearly, for all i 0, z' = uv i w is in L (actually, is in L 1 except for the case i = 1). (Notice that in this case z can be either in L 1 L 2 or.) Case 2: c 1 = c 2. Then z is in L. Let v = c 3. Then z' = uv i 1 w is in L (actually, in L 1 ), for all i 0. Case 3: c 2 = c 3. Let v = c 1. Then z' = uv i w is in L, for all i 0 as in Case 2. L 1 135

10 The Pumping lemma for context-free languages (uvwxy Theorem) Theorem. For every CFL L whose size is not finite, there is a positive integer p such that, for every z L with z p, there exist u, v, w, x, y Σ * such that z = uvwxy with vwx p an vx 1 such that for all i 0, uv i wx i y L. This lemma can be rewritten as shown below to show the logical context. Notice that in this lemma p has the same role as n has in the pumping lemma for regular languages, an v an x, the pumping sites, can be anywhere in z. Recall that the pumping site v in the regular language pumping lemma can only be locate in the prefix of z of length n. (1) For all context-free language L, (2) there exists a constant p such that (3) for all z L such that z p, (4) there exist u, v, w, x, y, an z such that (i) z = uvwxy, (ii) uwx p, (iii) ux 1, an (iv) for all i 0, uv i wx i y L. 136

11 Proof of the Pumping lemma for context-free languages Proof. (Sketch. See the text for etaile proof.) Consier the following CFG in the Chomsky normal form. S AB A E B F E FG G B g F HG H h The erivation tree for string z = (h) 3 hg() 3 ghgh is given in the following page. Notice the pattern of nonterminal sequence F-G-B-F-G-B-F-G-B which occurs on the longest stem of the parse tree. It has a repeating pattern of F-G-B, which has leaves h to the left an to the right. It is easy to see that we can grow the stem by aing as many repeating patterns of F-G-B as we want or cut the it short by eleting the pattern. The resulting tree is still a parse tree of the grammar. It follows that the grammar has a parse tree for z' = (h) i hg() i ghgh, for all i 0, which implies that z' is also in the language of the grammar. 137

12 138 S B A F E G B F G B F G B F G H h H h H h H h g G B F G H h g F G H h g Z = (h ) 3 h g () 3 h g h g u v w x y S AB A E E FG G B g F HG H h

13 Proof of the Pumping lemma for context-free languages(cont e) In general, consier a parse tree of a CFG in CNF whose height is greater than the number of nonterminal symbols of the grammar. Along the longest stem of the tree, at least one nonterminal symbol, say A, appears more than once. Suppose that this parse tree erives a sting z. Clearly, repeating (or eleting) the pattern from A to A along the stem will result in another parse tree of the grammar which erives string z' = uv i wx i y, i 0, where uvwxy = z, vx 1 an vwx V N, the number of nonterminal symbols. 139

14 Applications of the pumping lemma for CFL's Example: The language L = { a n b n c n n 0 }is not a CFL. Suppose L is a CFL, an consier z = a p b p c p, where p is the constant of the pumping lemma for CFL's. Clearly, z is in L. Let z = uvwxy, where u,v,w,x an y are some substrings of z which satisfy conitions (i) - (iii) above. Since vwx p, substrings v an x together contain at most two ifferent symbols (see the illustration below). Since vx 1, v an x together have at least one symbol. It follows that z' = uv 2 wx 2 y oes not have the same number of a's, b's an c's. Hence, z' is not in L. This contraicts the lemma. Thus language L is not a CFL. (i) z = a p b p c p = uvwxy (ii) vwx p (iii) vx 1. p p p aa abb bcc c u vwx y 140

15 Ruminating on the Application of the Pumping Lemma for Contex-free Languages Notice the ifference between the pumping lemma for regular languages an the lemma for CFL's. With the pumping lemma for regular languages, you have privilege of pumping only on the prefix of length n of the string z that you have chosen. With the pumping lemma for CFL's, you can no longer confine the pumpable site within the prefix of z of length p. Notice that vwx, uner the conition that uvx p, can be positione anywhere because there is no restriction on the length of the strings u an y. Hence, in general, we shoul eal with more cases that result from pumping a string of a CFL than those when we pump a string of a regular language. To eal with this ifficulty, they evelope more complex pumping lemmas for CFL's (for example, see the Ogen's lemma in Introuction to Automata Theory, Languages, an Computation by Hopcroft, Motwani an Ullman). 141