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## Context Free Grammars & Languages

Context free grammars (CFG) and languages (CFL):: CFGs and CFLs are models of computation which define the syntax of hierarchical formal notations as utilized in programming or mark-up languages. Recursion is an essential characteristic which distinguish CFGs and CFLs from FAs and regular languages. Properties, weakness and strengths of CFLs. Equivalence of the CFGs and NPDAs. Non-equivalence of the deterministic and non-deterministic PDAs, Parsing and Context sensitive grammars CSG.Aims of this section:Context free grammars and languages (CFG, CFL)Algol 60 pioneered the CFGs and CFLs to state the syntax of programming languages (Backus-Naur Form).

Example: Arithmetic expression E, term T, factor F, primary P, a-op A = {+, -}, m-op M = {•, /}, exp-op = ^.

E -> T | EAT | AT, T -> F | TMF, F -> P | F^P,

P -> unsigned number |variable| function designator | (E)

[Notice the recursion: E ->* (E)]

Example of Recursive data structures and their traversals:

Binary tree T, leaf L, node N: T -> L | N T T (prefix) or T -> L | T N T (infix) or T -> L | T T N (suffix).

Such definitions can be turned directly to the recursive traversal procedures, example: procedure traverse (p: ptr); begin when p ≠ nil then start visit(p); traverse(p.left); traverse(p.right); end; end;

Df CFG: G = (V, A, P, S)

V: non-terminal symbols, “variables”; A: terminal symbols; S ∈ V: start symbol, “sentence”;

P: set of productions or rewriting rules of the form X -> w, Here X ∈ V, w ∈ (V ∪ A)*

Rewriting the step: for u, v, x, y, y’, z ∈ (V ∪ A)*: u -> v if u = xyz, v = xy’z and y -> y’ ∈ P.

Derivation: “->*” is the transitive, reflexive closure of “->”, that is, u ->* v iff ∃ w0, w1, .. , wk with k ≥ 0 and u = w0, wj-1 -> wj, wk = v.

L(G) context free language produce by G: L(G) = {w ∈ A* | S ->* w}.

Example Symmetric structures: L = {0n 1n | n ≥ 0}, or even palindromes L0 = {w wreversed | w ∈ {0, 1}*}

G(L) = ({S}, {0, 1}, { S -> 0S1, S -> ε}, S); G(L0) = ({S}, {0, 1}, { S -> 0S0, S -> 1S1, S -> ε}, S)

Palindromes (length even or odd): L1 = {w | w = wreversed}. G(L1): add all the rules: S -> 0, S -> 1 to G(L0).

Ex Parenthesis expressions: V = {S}, T = {(, ), [, ] }, P = { S -> ε, S -> (S), S -> [S], S -> SS }

Sample derivation: S -> SS -> SSS ->* ()[S][ ] -> ()[SS][ ] ->* ()[()[ ]][ ]

The rule S -> SS builds this grammar ambiguous. The ambiguity is undesirable in practice, as the syntactic structure is usually employed to convey the semantic information.

Ambiguous structures in natural languages:

‘Time flies like an arrow’ versus. ‘Fruit flies like a banana’.

Bad news: There exist CFLs which are inherently ambiguous, that is each and every grammar for them is ambiguous. Furthermore, the problem of deciding whether the given CFG G is ambiguous or not, is not decidable.

Good news: For practical purposes it is simple to design the unambiguous CFG’s.

Exercise:

i) For Algol 60 grammar G (simple arithmetic expressions) above, describe the aim of the rule E -> AT and show illustrations of its use. Prove or disprove: G is unambiguous.

ii) Build an unambiguous grammar for the language of above parenthesis expressions.

iii) The ambiguity of ‘dangling else’. Some programming languages (example: Pascal) assign to nested if-then[-else] statements an ambiguous structure. This is then left to the semantics of language to disambiguate.

Let E represents Boolean expression, S statement, and let consider the 2 rules:

S -> if E then S, and S -> if E then S else S. Explain the trouble with this grammar and fix it.

iv) Provide a CFG for L = {0

^{i}1^{j }2^{k}| i = j or j = k}. Try to verify: L is inherently ambiguous.Latest technology based Theory of Computation Online Tutoring AssistanceTutors, at the

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