TAFL 1 (ECS-403) Unit- III. 3.1 Definition of CFG (Context Free Grammar) and problems. 3.2 Derivation. 3.3 Ambiguity in Grammar

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1 TAFL 1 (ECS-403) Unit- III 3.1 Definition of CFG (Context Free Grammar) and problems 3.2 Derivation 3.3 Ambiguity in Grammar Inherent Ambiguity Ambiguous to Unambiguous CFG 3.4 Simplification of CFGs Removal of useless symbols Elimination of Ԑ production Removal of unit production 3.5 Normal Forms Chomsky s Normal Forms (CNF) Greibach Normal Form (GNF) 3.6 Closure Properties of CFL 3.7 Decision Properties of CGL 3.8 Undecidable Problems 3.9 Applications of CFG 3.10 Pumping Lemma for CFL 3.11 CFG to finite Automata

2 TAFL 2 (ECS-403) 3.1 Definition of CFG (Context Free Grammar): The CFG can be formally defined by G = {V,T,P,S} where V = set of non terminals or variables T = set of terminals P = set of production S = start symbol Problems based on CFG 1. Write CFG for the following languages: a) Having any number of a s over the set {a}. S as S Ԑ b) Having regular expression (0+1)*. S 0S 1S S Ԑ c) Containing strings of at least two a s. S Aa Aa A A aa ba Ԑ d) L = wcw T where w Ԑ (a,b)* S asa S bsb S C e) Which has all the strings which are all palindromes over {a,b}. S asa S bsb S a b Ԑ

3 TAFL 3 (ECS-403) f) Which consists of all the strings having at least one occurrence of 000. S ATA A 0A 1A Ԑ T 000 g) There are no consecutive b s, the string may or may not have consecutive a s. S as ba a b Ԑ A as a Ԑ h) At least one occurrence of double a. S BAB A aa B ab bb Ԑ i) All the string of different first and last symbols over {0,1}. S 0A1 1A0 A 0A 1A Ԑ j) L = a n b 2n where n 1. S asbb abb k) L = a x b y where x y S asb A B A aa a B bb b l) For the regular expression (110+11)*(10)* S AB A 110A 11A Ԑ B 10B Ԑ m) For generating the integers. S GI G + - I DI D D

4 TAFL 4 (ECS-403) n) L = {0 i 1 j 2 k j > i+k}. S ABC A 0A1 Ԑ B 1B 1 C 1C2 Ԑ o) L = {0 i 1 j 2 k i = j}. S AB A 0A1 Ԑ B 2B Ԑ p) L = {0 i 1 j 2 k j k}. S AB A 0A Ԑ B 1B2 C C 2C Ԑ

5 TAFL 5 (ECS-403) 3.2 Derivation: a. Left Most Derivation b. Right Most Derivation Example: Derive the string for leftmost and rightmost derivation. S T00T T 0T 1T Ԑ Left Most Derivation S T00T S T00T 1T00T T 1T 10T00T T 0T 10Ԑ00T T Ԑ 1000T 10001T T 1T T T 1T T T 1T Ԑ T Ԑ Right Most Derivation S T00T S T00T T001T T 1T T0011T T 1T T00111T T 1T T00111Ԑ T Ԑ T T00111 T 1T 10T00111 T 0T 10Ԑ00111 T Ԑ Derivation Tree (Parse Tree): Derivation Tree is a graphical representation for the derivation of the given production rules for a given CFG. Following are properties of any derivation tree: a. The root node is always a node starting symbol. b. The derivation is read from left to right. c. The leaf nodes are always terminal nodes. d. The interior nodes are always the non terminal nodes.

6 TAFL 6 (ECS-403) Example: Construct the derivation tree for the string aabbabba from the CFG given below: S ab ba A a as baa B b bs abb Solution: 3.3 Ambiguity in Grammar: The grammar can be derived in either leftmost or rightmost derivation but if there exists more than one left parse tree or more than one right parse tree than the grammar is said to be ambiguous grammar.

7 TAFL 7 (ECS-403) Example: Example: Remove ambiguity from the following example: E E+E E*E id Solution: E E + T E T T T * F T F F id

8 TAFL 8 (ECS-403) Inherent Ambiguity: A context free grammar is called inherently ambiguous if all the productions rules in the grammar are ambiguous. Example: S XYZ aaybb X aay aa Y baz ba Z bzb bb Ambiguous to Unambiguous CFG: 3.4 Simplification of CFGs: There are three steps for simplification of CFG (reduced grammar): a) Removal of useless symbols. b) Elimination of Ԑ production. c) Removal of unit production Removal of useless symbols: A symbol P is useful if there exists some derivation in the following form: * S αpβ * αpβ w Then p is said to be useful symbol.

9 TAFL 9 (ECS-403) Example: S aa a Bb cc A ab B a Aa C ccd D ddd Solution: S aa a Bb A ab B a Aa Elimination of Ԑ production Example: S XYX X 0X Ԑ Y 1Y Ԑ Solution: S XYX XY YX Y X X 0X 0 Y 1Y Removal of unit production. Example: S 0A 1B C A 0S 00 B 1 A C 01 Solution S 0A 1B 01 A 0S 00 B 1 0S 00 C 01 Exercise: Simplify the following CFGs S A 0C1 A B C Ԑ CD Solution S 01 10

10 TAFL 10 (ECS-403) 3.5 Normal Forms: Chomsky s Normal Forms (CNF): The CNF can be defined in the following form: Non terminal Non Terminal. Non Terminal Non terminal Terminal Example: Convert the following CFG into CNF: 1. S aaaas S aaaa 2. S asa bsb a b Solution: S C a C 1 C 1 C a C 2 C 2 C a C 3 C 3 C a C 4 S C 5 C 5 C 5 C a C a Ca a Solution: S C a C 1 C 1 SC a Ca a S C b C 2 C 2 SC b C b b S a S b Greibach Normal Form (GNF): The GNF can be defined in the following form: Non terminal One Terminal. Any number of non-terminals Non terminal One terminal

11 TAFL 11 (ECS-403) To convert given CFG into GNF we can use two lemmas based on which it is easy to convert given CFG to GNF: A. Lemma 1: G = (V, T, P, S) be a given CFG and if there is a production A Ba and B β 1 β 2. β n Then we can convert A rule to GNF as A β 1 a β 2 a. β n a B. Lemma 2: G = (V, T, P, S) be a given CFG and if there is a production A Aa 1 Aa 2. Aa n β 1 β 2. β n Such that β i do not start with A then equivalent grammar in GNF can be: A β 1 β 2. β n A β 1 Z β 2 Z. β n Z Z a 1 a 2. a n Z a 1 Z a 2 Z. a n Z Example: Convert the following grammar to GNF: Example 1. S absb S aa Solution: S absb S aa B b A a Example 2. S ABA A aa Ԑ B bb Ԑ

12 TAFL 12 (ECS-403) Remove Ԑ in the CFG: S ABA AB BA AA AA A B A aa a B bb b Remove unit production: S ABA AB BA AA AA aa a bb b A aa a B bb b In GNF we can write S aaba aab aba ab aaa aa a bba ba bb b A aa a B bb b Example 3. S AA 0 A SS 1 Step 1: (for A) A SS 1 Put the valve of S in left most A AAS 0S 1 From lemma 2: a 1 = AS, β 1 = 0S, β 2 = 1

13 TAFL 13 (ECS-403) A 0S 1 A 0SZ 1Z Z AS Z ASZ Step 2: S AA 0 Put the above value of A in left most: S 0SA 1A 0SZA 1ZA 0 Put the value of A for Z: Z 0SS 1S 0SZS 1ZS Z 0SSZ 1SZ 0SZSZ 1ZSZ The final solution is S 0SA 1A 0SZA 1ZA 0 A 0S 1 0SZ 1Z Z 0SS 1S 0SZS 1ZS 0SSZ 1SZ 0SZSZ 1ZSZ Exercise: convert the following CFG into GNF S AB A BS b B SA a

14 TAFL 14 (ECS-403) 3.6 Closure Properties of CFL: a) The context free languages are closed under union. b) The context free languages are closed concatenation. c) The context free languages are closed under kleen closure. d) The context free languages are not closed under intersection. e) The context free languages are not closed under complement. 3.7 Decision Properties of CGL: a) Emptiness: The given context free grammar cannot generate any string at all. b) Finiteness: The given context free grammar generates a finite language. c) Membership: whether given string belongs to given grammar. 3.8 Undecidable Problems: There are no algorithms to answer these questions. Hence these problems are known as Undecidable Problems: a) Whether or not two different context free languages define the same language? b) Whether given CFL is ambiguous or not? c) Whether complement of given CFL is context free language? d) Whether the intersection of two context free languages is context free? 3.9 Applications of CFG: When any high level program like C or PASCAL is compiled, the compiler checks the syntax of every programming statement by constructing syntax tree, And for building the syntax tree, it is necessary to write context free grammar for each statement in the program.

15 TAFL 15 (ECS-403) There are various Parsing Techniques: Types of Parser Top Down Parser Bottom Up Parser Backtracking Predictive Parser Shift Reduce LR Parser Parser SLR Parser LALR Parser LR Parser 3.10 Pumping Lemma for CFL: Let L be any context free language, then there is a constant n, which depends only upon L, such that there exist a string w Ԑ L and w n where w = pqrst such that a) qs 1 b) prs n and c) For all i 0 pq i rs i t is in L. Example: Proof whether the given language L = {SS T S Ԑ (a,b)*} is context free or not?

16 TAFL 16 (ECS-403) 3.11 CFG to Finite Automata: S 0A 1B 0 1 A 0S 1B 1 B 0A 1S