CS 208: Computing Theory Assoc. Prof. Dr. Brahim Hnich Faculty of Computer Sciences Izmir University of Economics

Context Free Languages Context-Free Languages

So far … Methods for describing regular languages Finite Automata Deterministic Non-deterministic Regular Expressions They are all equivalent, and limited Cannot some simple languages like {0 n 1 n | n is positive} Now, we introduce a more powerful method for describing languages Context-free Grammars (CFG)

Are CFGs any useful? Extremely useful! Artificial Intelligence Natural language Processing Programming Languages specification compilation

Example This is a CFG which we call G1 A  0A1 A  B B  #

Example: production rules This is a CFG which we call G1 A  0A1 A  B B  # Each line is a substitution rules or production rules

Example: variables This is a CFG which we call G1 A  0A1 A  B B  # A and B are called variables or non-terminals

Example: variables This is a CFG which we call G1 A  0A1 A  B B  # 0,1, and # are called terminals

Example: variables This is a CFG which we call G1 A  0A1 A  B B  # A is the start variable

Rules We use a CFG to describe a language by generating each string of that language Write down the start variable Pick a variable written down and a production rule that starts with that variable Replace that variable with right-hand side of the production rule Repeat until no variable remain

Derivations This is a CFG which we call G1 A  0A1 A  B B  # Derivations with G1 A  0A1  0B1  0#1 A  0A1  00A11  00B11  00#11 A  0A1  00A11  000A111  000B111  000#111

Parse tree Parse tree for 0#1 in G1 A  0A1  0B1  0#1 A B 0# 1 A

Parse tree Parse tree for 00#11 in G1 A  0A1  00A11  00B11  00#11 A B 0# 1 A A 0 1

Context-free languages All strings generated by a CFG constitute the language of the grammar Example: L(G1)={0 n #1 n | n is positive} Any language generated by a context-free grammar is a context-free language

A useful abbreviation Production rules A  0A1 A  B B  # Can be written as A  0A1 | B B  #

Another example CFG G2 describing a fragment of English   |   |  a | the  boy | girl | flower  touches | likes | sees  with

Another example Examples of strings belonging to L(G2) a boy sees the boy sees a flower a girl with a flower likes the boy with a flower

Another example Derivation of a boy sees   a  a boy  a boy sees

Formal definitions A context-free grammar is a 4-tuple where V is a finite set of variables ∑is a finite set of terminals R is a finite set of rules: each rule is a variable and a finite string of variable and terminals S is the start symbol

Formal definitions If u and v are strings of variable and terminals, and A  w is a rule of the grammar, Then uAv yields uwv, written uAv  uwv We write u  * v if u = v or u  u1  ….  uk  v

Formal definitions The language of grammar G is L(G) = {w | S  * w}

Example Consider G4 = where R is S  (S) | SS | ε What is the language of G4? Examples: (), (()((())), …

Example Consider G4 = where R is S  (S) | SS | ε What is the language of G4? L(G4) is the set of strings of properly nested parenthesis

Example Consider G4 = where R is E  E + T | T T  T X F | F F  (E) | a What is the language of G4? Examples: a+a+a, (a+a) x a

Example Consider G4 = where R is E  E + T | T T  T x F | F F  (E) | a What is the language of G4? E stands for expression, T for Term, and F for Factor: so this grammar describes some arithmetic expressions

Ambiguity Sometimes a grammar can generate the same string in several different ways! This string will have several parse trees This is a very serious problem Think if a C program can have multiple interpretations? If a language has this problem, we say that it is ambiguous

Example Consider G5:  + | x |( ) | a G5 is ambiguous because a+axa has two parse tress!

Example Consider G5:  + | x |( ) | a G5 is ambiguous because a+axa has two parse tress! a+ a xa

Example Consider G5:  + | x |( ) | a G5 is ambiguous because a+axa has two parse tress! a+ a xa a+ a xa

Formal definition: ambiguity A string w is generated ambiguously in CFG G if it has two or more different leftmost derivations! A derivation is leftmost if at every step the variable being replaced is the leftmost one Grammar G is ambiguous if it generates some string ambiguously

Chomsky Normal Form (CNF) Every rule has the form A  BC A  a S  ε Where S is the start symbol, A, B, and C are any variables – except that B and C may not be the start symbol

Theorem Theorem: Any context-free language is generated by a context-free grammar in Chomsky normal form How? Add new start symbol S0 Eliminate all rules of the form A  ε Eliminate all “unit” rules of the form A  B Patch up rules so that grammar denotes the same language Convert remaining rules to proper form

Steps to convert any grammar into CNF Step1 Add a new start symbol S0 Add the rule S0  S

Steps to convert any grammar into CNF Step2: Repeat  Remove some rule of the form A  ε where A is not the start symbol  Then, for each occurrence of A on the right-hand side of a rule, we add a new rule with that occurrence deleted E.g., if R  uAvAu where u and v are strings of variables and terminals We add rules: R  uvAu, R  uAvu, and R  uvu  For R  A add R  ε, except if R  ε has already been removed Until all ε-rules not involving the start symbol have been removed

Steps to convert any grammar into CNF Step3: eliminate unit rules Repeat  Remove some rule of the form A  B  For each B  u, add A  u, except if A  u has already been removed Until all unit rules have been removed

Steps to convert any grammar into CNF Step4: convert remaining rules Replace each rule A  u 1 u 2 …u k, where k >2 and each u i is a terminal or a variable with the rules A  u 1 A 1 A 1  u 2 A 2 A 2  u 3 A 3 …. A k-2  u k-1 u k If k=2, we replace any terminal u i in the preceding rules with the new variable U i and add the rule U i  u i

Example Start with S  ASA | aB A  B | S B  b | ε

Example Step 1: add new start symbol and new rule S0  S S  ASA | aB A  B | S B  b | ε

Example Step 2: remove ε- rule B  ε S0  S S  ASA | aB | a A  B | S | ε B  b

Example Step 2: remove ε- rule A  ε S0  S S  ASA | aB | a | SA | AS | S A  B | S B  b

Example Step 3: remove unit rule S  S S0  S S  ASA | aB | a | SA | AS | S A  B | S B  b

Example Step 3: remove unit rule S0  S S0  S | ASA | aB | a | SA | AS S  ASA | aB | a | SA | AS A  B | S B  b

Example Step 3: remove unit rule A  B S0  ASA | aB | a | SA | AS S  ASA | aB | a | SA | AS A  B | S | b B  b

Example Step 3: remove unit rule A  S S0  ASA | aB | a | SA | AS S  ASA | aB | a | SA | AS A  S | b | ASA | aB | a | SA | AS B  b

Example Step 3: remove unit rule A  S S0  ASA | aB | a | SA | AS S  ASA | aB | a | SA | AS A  b | ASA | aB | a | SA | AS B  b

Example Step 4: convert remaining rules S0  AA1|UB| a| SA | AS S  AA1|UB | a | SA | AS A  b | AA1 | UB | a | SA | AS B  b U  a A1  SA

Pushdown Automata

Pushdown automata Pushdown automat (PDA) are like nondeterministic finite automat but have an extra component called a stack Can push symbols onto the stack Can pop them (read them back) later Stack is potentially unbounded

State control aaba x y z stack input

Formal Definition A pushdown automaton is a 6-tuple (Q,∑,S, ξ,q0,F), where Q is a finite set of states ∑ is a finite set of symbols called the alphabet S is the stack alphabet ξ : Q x ∑ ε x S ε  P(Q x S ε ) is the transition function q0 Є Q is the start state F ⊆ Q is the set of accept states or final states

Conventions Question: when is the stack empty? Start by pushing a $ onto the stack If you see it again, stack is empty Question: when is input string empty Doesn’t matter Accepting states accept only if inputs exhausted

Notation Transition a,b  c means Read a from the input Pop b from stack Push c onto stack Meaning of ε transition If a = ε, don’t read input If b= ε, don’t pop any symbol If c= ε, don’t push any symbols

Example Recall 0 n 1 n which is not regular Consider the following PDA Read input symbols For each 0, push it on the stack As soon as a 1 is seen, pop a 0 for each 1 read Accept if stack is empty when last symbol read Reject if stack non-empty, or if input symbol exist, or if 0 read after a 1, etc…

Example ε,ε$ε,ε$ 0, ε  0 1,0  ε ε,$  ε {0 n 1 n | n is positive}

Example ε,ε$ε,ε$ {a i b j c k | i=j or i=k} a, ε  a ε,ε εε,ε ε ε,ε εε,ε ε ε,ε εε,ε ε ε, $  ε b, ε  ε c, a  ε b, a  ε ε,$  ε c,ε  ε

Theorem Theorem: A language is context-free if and only some pushdown automaton accepts it Proof: we will skip it! (Those interested may read the book) Corollary: Every regular language is a context- free language Regular languages Context-free languages

Conclusions Context-free grammars definition ambiguity Chomsky normal form Pushdown automata definition Next: Part C; Computability Theory