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1 CPS Mathematical Foundations of CS Dr. S. Rodger Section: Properties of Context-Free Languages èhandoutè Section 3.5 Which of the following languages are CFL? æ L=fa n b n c j j 0 énçjg æ L=fa n b j a n b j j né0;j é0g æ L=fa n b j a k b p j n + j ç k + p; n é 0;j é0;k é0;pé0g Pumping Lemma for Regular Language's: Let L be a regular language, Then there is a constant m such that w 2 L, jwj çm,w=xyz such that æ jxyj çm æ jyjç1 æ for all i ç 0, xy i z 2 L Pumping Lemma for CFL's Let L be any inænite CFL. Then there is a constant m depending only on L, such that for every string w in L, with jwj çm,wemay partition w = uvxyz such that: jvxyj çm, èlimit on size of substringè jvyj ç1, èv and y not both emptyè For all i ç 0, uv i xy i z 2L æ Proof: èsketchè There is a CFG G s.t. L=LèGè. Consider the parse tree of a long string in L. For any long string, some nonterminal N must appear twice in the path. 1

a constant m such that w 2 L, jwj çm,w=xyz such that æ jxyj çm æ jyjç1 æ for all i ç 0, xy i z 2 L Pumping Lemma for CFL's Let L be any inænite CFL.

2 . Example: Consider L = fa n b n c n : n ç 1g. Show L is not a CFL. æ Proof: èby contradictionè Assume L is a CFL and apply the pumping lemma. Let m be the constant in the pumping lemma and consider w = a m b m c m. Note jwj çm. Show there is no division of w into uvxyz such that jvyj ç1, jvxyj çm, and uv i xy i z 2L for i =0;1;2;:::. Case 1: Neither v nor y can contain 2 or more distinct symbols. If v contains a's and b's, then uv 2 xy 2 z=2l since there will be b's before a's. Thus, v and y can be only a's, b's, or c's ènot mixedè. Case 2: v = a t1, then y = a t2 or b t3 èjvxyj çmè If y = a t2, then uv 2 xy 2 z = a m+t1+t2 b m c m =2 L since t 1 + t 2 é 0, nèaèénèbè's ènumber of a's is greater than number of b'sè If y = b t3, then uv 2 xy 2 z = a m+t1 b m+t3 c m =2 L since t 1 + t 3 é 0, either nèaèénècè's or nèbèénècè's. Case 3: v = b t1, then y = b t2 or c t3 If y = b t2, then uv 2 xy 2 z = a m b m+t1+t2 c m =2 L since t 1 + t 2 é 0, nèbèénèaè's. If y = c t3, then uv 2 xy 2 z = a m b m+t1 c m+t3 Case 4: v = c t1, then y = c t2 then, uv 2 xy 2 z = a m b m c m+t1+t2 =2 L since t 1 + t 3 é 0, either nèbèénèaè's or nècèénèaè's. =2 L since t 1 + t 2 é 0, nècèénèaè's. Thus, there is no breakdown of w into uvxyz such that jvyj ç1, jvxyj çmand for all i ç 0, uv i xy i z is in L. Contradiction, thus, L is not a CFL. Q.E.D. 2

Case 1: Neither v nor y can contain 2 or more distinct symbols. If v contains a's and b's, then uv 2 xy 2 z=2l since there will be b's before a's.

3 Example Why would we want to recognize a language of the type fa n b n c n : n ç 1g? Example: Consider L = fa n b n c p : péné0g. Show L is not a CFL. æ Proof: Assume L is a CFL and apply the pumping lemma. Let m be the constant in the pumping lemma and consider w = Note jwj çm. Show there is no division of w into uvxyz such that jvyj ç1, jvxyj çm, and uv i xy i z 2L for i =0;1;2;:::. Thus, there is no breakdown of w into uvxyz such that jvyj ç1, jvxyj çmand for all i ç 0, uv i xy i z is in L. Contradiction, thus, L is not a CFL. Q.E.D. 3

Let m be the constant in the pumping lemma and consider w = Note jwj çm.

4 Example: Consider L = fa j b k : k = j 2 g. Show L is not a CFL. æ Proof: Assume L is a CFL and apply the pumping lemma. Let m be the constant in the pumping lemma and consider w = Show there is no division of w into uvxyz such that jvyj ç1, jvxyj çm, and uv i xy i z 2L for i =0;1;2;:::. Case 1: Neither v nor y can contain 2 or more distinct symbols. If v contains a's and b's, then uv 2 xy 2 z=2l since there will be b's before a's. Thus, v and y can be only a's, and b's ènot mixedè. Thus, there is no breakdown of w into uvxyz such that jvyj ç1, jvxyj çmand for all i ç 0, uv i xy i z is in L. Contradiction, thus, L is not a CFL. Q.E.D. Exercise: Prove the following is not a CFL by applying the pumping lemma. èanswer is at the end of this handoutè. Consider L = fa 2n b 2p c n d p : n; p ç 0g. Show L is not a CFL. 4

Case 1: Neither v nor y can contain 2 or more distinct symbols. If v contains a's and b's, then uv 2 xy 2 z=2l since there will be b's before a's. Thus, v and y can be only a's, and b's ènot mixedè.

5 Example: Consider L = fw çww : w 2 æ æ g,æ=fa; bg, where çw is the string w with each occurence of a replaced by b and each occurence of b replaced by a. For example, w = baaa, çw = abbb, w çw = baaaabbb. Show L is not a CFL. æ Proof: Assume L is a CFL and apply the pumping lemma. Let m be the constant in the pumping lemma and consider w = Show there is no division of w into uvxyz such that jvyj ç1, jvxyj çm, and uv i xy i z 2L for i =0;1;2;:::. Thus, there is no breakdown of w into uvxyz such that jvyj ç1, jvxyj çmand for all i ç 0, uv i xy i z is in L. Contradiction, thus, L is not a CFL. Q.E.D. 5

Let m be the constant in the pumping lemma and consider w = Show there is no division of w into uvxyz such that jvyj ç1, jvxyj çm, and uv i xy i z 2L

6 Example: Consider L = fa n b p b p a n g. L is a CFL. The pumping lemma should apply! Let m ç 4 be the constant in the pumping lemma. Consider w = a m b m b m a m. We can break w into uvxyz, with: If you apply the pumping lemma to a CFL, then you should ænd a partition of w that works! Chap 8.2 Closure Properties of CFL's Theorem CFL's are closed under union, concatenation, and star-closure. æ Proof: Given2CFGG 1 =èv 1 ;T 1 ;R 1 ;S 1 è and G 2 =èv 2 ;T 2 ;P 2 ;S 2 è í Union: Construct G 3 s.t. LèG 3 è=lèg 1 èëlèg 2 è. G 3 =èv 3 ;T 3 ;R 3 ;S 3 è í Concatenation: Construct G 3 s.t. LèG 3 è=lèg 1 èælèg 2 è. G 3 =èv 3 ;T 3 ;R 3 ;S 3 è 6

2 Closure Properties of CFL's Theorem CFL's are closed under union, concatenation, and star-closure.

7 í Star-Closure Construct G 3 s.t. LèG 3 è=lèg 1 è æ G 3 =èv 3 ;T 3 ;R 3 ;S 3 è QED. Theorem CFL's are NOT closed under intersection and complementation. æ Proof: í Intersection: í Complementation: 7

8 Theorem: CFL's are closed under regular intersection. If L 1 is CFL and L 2 is regular, then L 1 ë L 2 is CFL. æ Proof: èsketchè This proof is similar to the construction proof in which we showed regular languages are closed under intersection. We take a NPDA for L 1 and a DFA for L 2 and construct a NPDA for L 1 ë L 2. M 1 =èq 1 ;æ;,;æ 1 ;q 0 ;z;f 1 èisannpda such that LèM 1 è=l 1. M 2 =èq 2 ;æ;æ 2 ;q 0 0 ;F 2èisaDFA such that LèM 2 è=l 2. Example of replacing arcs ènot a Proof!è: 8

intersection. We take a NPDA for L 1 and a DFA for L 2 and construct a NPDA for L 1 ë L 2.

9 Note this is not a proof, but sketches how we will combine the DFA and NPDA. We must formally deæne æ 3. If then Must show if and only if Must show: w 2 LèM 3 èiæw2lèm 1 è and w 2 LèM 2 è. QED. 9

10 Questions about CFL: 1. Decide if CFL is empty? 2. Decide if CFL is inænite? Example: Consider L = fa 2n b 2m c n d m : n; m ç 0g. Show L is not a CFL. æ Proof: Assume L is a CFL and apply the pumping lemma. Let m be the constant in the pumping lemma and consider w = a 2m b 2m c m d m. Show there is no division of w into uvxyz such that jvyj ç1, jvxyj çm, and uv i xy i z 2L for i =0;1;2;:::. Case 1: Neither v nor y can contain 2 or more distinct symbols. If v contains a's and b's, then uv 2 xy 2 z=2l since there will be b's before a's. Thus, v and y can be only a's, b's, c's, or d's ènot mixedè. Case 2: v = a t1, then y = a t2 or b t3 èjvxyj çmè If y = a t2, then uv 2 xy 2 z = a 2m+t1+t2 b 2m c m d m =2 L since t 1 + t 2 é 0, the number of a's is not twice the number of c's. If y = b t3, then uv 2 xy 2 z = a 2m+t1 b 2m+t3 c m d m =2 L since t 1 + t 3 é 0, either the number of a's èdenoted nèaèè is not twice nècè ornèbè is not twice nèdè. Case 3: v = b t1, then y = b t2 or c t3 If y = b t2, then uv 2 xy 2 z = a 2m b 2m+t1+t2 c m d m =2 L since t 1 + t 2 é 0, nèbèé 2ænèdè. If y = c t3, then uv 2 xy 2 z = a 2m b 2m+t1 c m+t3 d m =2 L since t 1 + t 3 é 0, either nèbèé 2ænèdè or 2ænècèénèaè. Case 4: v = c t1, then y = c t2 or d t3 If y = c t2, then uv 2 xy 2 z = a 2m b 2m c m+t1+t2 d m =2 L since t 1 + t 2 é 0, 2ænècèénèaè. If y = d t3, then uv 2 xy 2 z = a 2m b 2m c m+t1 d m+t3 2ænèdèénèbè. Case 5: v = d t1, then y = d t2 then uv 2 xy 2 z = a 2m b 2m c m d m+t1+t2 =2 L since t 1 + t 3 é 0, either 2ænècèénèaè or =2 L since t 1 + t 2 é 0, 2ænèdèénècè. Thus, there is no breakdown of w into uvxyz such that jvyj ç1, jvxyj çmand for all i ç 0, uv i xy i z is in L. Contradiction, thus, L is not a CFL. Q.E.D. 10

Show there is no division of w into uvxyz such that jvyj ç1, jvxyj çm, and uv i xy i z 2L for i =0;1;2;:::. Case 1: Neither v nor y can contain 2 or more distinct symbols.

Automata and Computability. Solutions to Exercises

Automata and Computability. Solutions to Exercises Automata and Computability Solutions to Exercises Fall 25 Alexis Maciel Department of Computer Science Clarkson University Copyright c 25 Alexis Maciel ii Contents Preface vii Introduction 2 Finite Automata