CFLs and Regular Languages. CFLs and Regular Languages. CFLs and Regular Languages. Will show that all Regular Languages are CFLs. Union.

Transcription

1 We can show that every RL is also a CFL Since a regular grammar is certainly context free. We can also show by only using Regular Expressions and Context Free Grammars That is what we will do in this half. Note: Much of this lecture is not in the text! Will show that all Regular Languages are CFLs - If L 1 and L 2 are CFLs then - L 1 L 2 is a CFL - L 1 L 2 is a CFL - L * 1 is a CFL - With the above shown, showing every Regular Language is also a CFL can be shown using a basic inductive proof. Formally, Let L 1 and L 2 be CFLs. Then there exists CFGs: G 1 = (V 1, T, S 1, P 1 ) G 2 = (V 2, T, S 2, P 2 ) such that L(G 1 ) = L 1 and L(G 2 ) = L 2 Assume that V 1 V 2 = We will define: G u = (V u, T, S u, P u ) such that L(G u ) = L 1 L 2 G c = (V c, T, S c, P c ) such that L(G c ) = L 1 L 2 G k = (V k, T, S k, P k ) such that L(G c ) = L 1 * Union Basic Idea Define the new CFG so that we can either start with the start variable of G 1 and follow the production rules of G 1 or start with the start variable of G 2 and follow the production rules of G 2 Union Formally G u = (V u, T, S u, P u ) V u = V 1 V 2 {S u } S u = S u P u = P 1 P 2 {S u S 1 S 2 } The first case will derive a string in L 1 The second case will derive a string in L 2 1

2 Concatenation General Idea Define the new CFG so that We force a derivation staring from the start variable of G 1 using the rules of G 1 After that We force a derivation staring from the start variable of G 2 using the rules of G 2 Concatenation Formally G c = (V c, T, S c, P c ) V c = V 1 V 2 {S c } S c = S c P u = P 1 P 2 {S c S 1 S 2 } Kleene Star General Idea Define the new CFG so that We can repeatedly concatenate derivations of strings in L 1 Since L * contains λ, we must be careful to assure that there are productions in our new CFG such that λ can be derived from the start variable Kleene star Formally G k = (V k, T, S k, P k ) V k = V 1 {S k } S k = S k P k = P 1 {S k S 1 S k λ } Now we can complete the proof Use an inductive proof Regular Expression Recursive definition of regular languages / expression over Σ : 1. is a regular language and its regular expression is 2. {λ} is a regular language and λ is its regular expression 3. For each a Σ, {a} is a regular language and its regular expression is a 2

3 Regular Expression 4. If L 1 and L 2 are regular languages with regular expressions r 1 and r 2 then -- L 1 L 2 is a regular language with regular expression (r 1 + r 2 ) -- L 1 L 2 is a regular language with regular expression (r 1 r 2 ) -- L 1 * is a regular language with regular expression (r 1* ) Only languages obtainable by using rules 1-4 are regular languages. RE -> CFG Base cases 1. can be expressed as a CFG with no productions 2. {λ} can be expressed by a CFG with the single production S λ 3. For each a Σ, {a} can be expressed by a CFG with the single production S a RE -> CFG Assume R 1 and R 2 are regular expressions that describe languages L 1 and L 2. Then, by the induction hypothesis, L 1 and L 2 are CFLs and as such there are CFGs that describe L 1 and L 2 Create CFGs that describe the the languages: L 1 L 2 Context Free Languages Regular Languages Finite Languages L 1 L 2 L 1 * Which we just did We are done! What have we learned? CFLs are closed under union, concatenation, and Kleene Star Every Regular Language is also a CFL We now have an algorithm, given a Regular Expression, to construct a CGF that describes the same language Find a CFG for the L = ( ) * (01) * ( ) can be described by the CFG with productions: A ( ) * can be described by the CFG with productions: B AB λ A

4 Find a CFG for the L = ( ) * (01) * (01) can be described by the CFG with productions: D 01 (01) * can be described by the CFG with productions: C DC λ D 01 Find a CFG for the L = ( ) * (01) * Putting it all together ( ) * (01) * can be described by the CFG with productions: S BC B AB λ A C DC λ D 01 Questions? You can use proof of closure properties in building CFLs: : Find a CFL for L = {0 i 1 j 0 k j > i + k} Number of 1s is greater than the combined number of 0s This language can be expressed as L = {0 i 1 i 1 m 1 k 0 k m > 0} : Find a CFL for L = {0 i 1 j 0 k j > i + k} This language can be expressed as L = {0 i 1 i 1 m 1 k 0 k m > 0} This is concatenation of 3 languages L 1 L 2 L 3 where L 1 = {0 i 1 i i 0} L 2 = {1 m m > 0} L 3 = {1 k 0 k k 0} CFG for L 1 = {0 i 1 i i 0} A 0A1 λ CFG for L 2 = {1 m m > 0} B 1B 1 CFG for L 3 = {1 k 0 k k 0} C 1C0 λ CFG for L S ABC A 0A1 λ B 1B 1 C 1C0 λ Formally G = (V, T, S, P) where V = {S, A, B, C} Σ = {0, 1} P = {S ABC A 0A1 λ B 1B 1 C 1C0 λ} 4

5 Questions? Practical uses for grammars How a compiler works Source file lexer Stream parser Parse of tokens codegen Tree Object code Theory Hall of Fame The Bell Labs Gang A real practical example Ken Thompson Regular expressions in UNIX / grep / vi Eric E. Schmidt Mike Lesk lex Stephen C Johnson yacc Grammars for programming languages <> <for-> <if-> <> { <> <> } ε <if-> if ( <expr> ) then <> <for-> for ( <expr>; <expr>; <expr>) <> A real practical example Grammars for programming languages Keywords and punctuation are terminals Program constructs are variables Production rules define the syntax of the language Dangling else <> if (<expr>) <> if (<expr>) <> else <> <some_other_> This is really the second step in building a compiler! To which if does the else belong? 5

6 else else In this derivation, the else belongs to the 1 st if A way to fix this <> <matched> <unmatched> <matched> if (<expr>) <matched> else <matched> <other> <unmatched> if (<expr>) <matched> if (<expr>) <unmatched> if (<expr>) <matched> else <unmatched> <matched> represents if statements with matching else <unmatched> represents if statements with at least 1 unmatched if Note what productions are not defined: <> if (<expr>) <unmatched> else <matched> <> if (<expr>) <unmatched> else <unmatched> <unmatched> can not come between if and else matched if ( expr ) unmatched else unmatched if ( expr ) matched NOT ALLOWED if ( expr ) s1 else s1 In this derivation, the else belongs to the 1 st if 6

7 Summary All Regular Languages are CFLs Use regular language operations in constructing CFGs CFGs in compiler design 7