Course Notes on Formal Languages and Compilers

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Proof Let F = {A∈N: A⇒ + ε}. Assume F can be constructed from G.
(Finding a method is an exercise)
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Let G=(N,T,P,S) and G'=(N',T',P',S') where N' = N, T' = T.
S' is obtained from S by:
1) Initializing S' to S.
2) Deleting members of F from elements of S' in all possible ways and
adding the new elements formed to S'.
e.g., if ABC∈S' and A,B,C∈F, then this will result in AB, AC, BC,
A,B,C,ε being added to S'
Similarly P' is obtained from P by:-
1) Initializing P' to P
2) For each rule A→y in P' we examine y and, as before, delete
elements of F from y in all possible ways. Let y 1 ,..,y n be the strings
obtained, then add A→y 1 ,.., A→y n , to P'. For example,
if D→ABC is in P' and A,B,C,∈F
then add the lollowing rules to P':
D→BC, D→AB, D→AC,
D→A, D→B, D→C, D→ε.
3) Delete all rules of the form A→ε from P'.
The previous result allows us to use the context sensitive recognition
method for context free grammars. The next two results will help us in
obtaining improved recognition methods for context free grammars.
Lemma: Removing rules of the form A→B
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Let G be a context free grammar. Then it is possible to construct an
equivalent context free grammar G' having no rules of the form A→B where
A,B∈N' and such that L(G')=L(G).
Proof Let G=(N,T,P,S) and G'=(N',T',P',S') where N'=N, T'=T, and S'=S..
By the previous lemma we may assume there are no rules of the form A→ε.
P' is obtained from P as follows:-
1) P' is initialized to P with rules A→B removed, A,B∈N.
2) If A⇒ + B in G and B→y is in P', add A→y to P'
Repeat (2) till nothing further can be added to P'.
<strong>Notes</strong>:
1) The problem of determining if A⇒ + B in G is an exercise.
2) If the grammar G is like a regular grammar but with extra rules of the
form A→B, the above procedure will work without removing rules of the form
A→ε, and in this case we obtain a regular grammar G' equivalent to G.