Course Notes on Formal Languages and Compilers

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So clearly L 0 ij can be defined by a regular expression for all i, j.
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Inductive step: Let us suppose that all the sets L k ij can be defined by a
regular expression.
What about the sets L (k+1) ij
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In fact L (k+1) ij = L k ij | (L k i (k+1) (L k (k+1) (k+1))* L k (k+1) j)
which is in fact is a regular expression built from the regular expressions for
L k ij.
Hence by induction, L k ij can be defined by a regular expression for
k = 0,1,2...m.
Now recall that A e is the only non terminal for which A e →ε and that
S ={A 1 , A 2 , .....A s }
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So L(G) = {t∈T*: A j ⇒*t for 1≤ j ≤s}
= {t∈T*: A j ⇒*A e t⇒t for 1≤ j ≤s}
(This is because the last step must use A e →ε.)
= L m e1 | L m e2 ... | L m es
which is a regular expression.
Q.E.D.
Exercise: It is highly desirable not to convert the grammar to backward
deterministic form in the above, as the set construction can create large
grammars. Explain how you would rewrite the previous proof without having
to assume that the grammar is backward deterministic.
Hint: Assume that that the non terminals are numbered A 1 , A 2 , ..., A m in
such a way that S = {A 1 , ..., A s } and {A: A→ε} = {A e , A e+1 , ..., A f }. Why is this
always possible Now how do we proceed
Example: Consider the grammar G=( {A 1 , A 2 }, {d}, P, {A 1 } ) where P consists
of
A 1 →A 1 d A 1 →A 2 d A 2 →ε
G is backward deterministic and the non terminals are numbered as
required by the theorem. Of course m = 2, e = 2, s=1.