ELIMINATING LEFT RECURSIVENESS

Abbreviation. “cfg” stands for “context free grammar” Definition. A cfg is left recursive if it contains a production of the form A  A α i.e. a production in which the lhs occurs as the head of the rhs.

Theorem 1. Eliminating left recursivenes. For any cfg G we can construct an equivalent cfg G’ such that G’ is not left-recursive

PROOF. For each nonterminal A that occurs as the lhs of a left-recursive production of G, do the following:

Let the left-recursive productions in which A occurs as lhs be A  A  1 ………. A  A  r and the remaining productions in which A occurs as lhs be A   1 …………. A   s

Let K A denote a symbol which does not already occur in the grammar. Replace the above productions by: A   1 K A |... |  s K A K A  ε |  1 K A |... |  r K A Clearly the grammar G' produced is equivalent to G.

EXAMPLE. S  R a | A a | a R  a b A  A  R | A T | b T  T b | a A non-left recursive grammar equiv. to the above is: S  R a | A a | a R  a b K A  ε | R K A | T K A A  b K A T  a K T K T  ε | b K T Clearly the grammar G’ obtained is equivalent to G, and has the required two properties

Definition. A grammar is indirectly left recursive if, for some production with left hand side (say) A, it is possible to derive a string whose head is A. Example. A  Db D  Ecc E  A t

Theorem 2. Eliminating indirect left recursiveness. For any cfg G we can construct an equivalent cfg G’ such that G’ is not indirectly left-recursive

Note. The theorem 1 assertion is a subset of that of theorem 2. Theorem 2 follows directly from the following lemma.

LEMMA. For any cfg G, and any ordering of the nonterminals of G, denoted here as A 1,...,A m, we can construct an equivalent cfg G’ such that: If A i  A j  is any production of G’, then i < j

PROOF. The lemma is true for i  1, since we can eliminate left recursion in productions (if any) in which A 1 occurs as lhs. Assume that the lemma is true for all i  t where t<m, and let the grammar so formed from G be denoted as G t.

Consider any production of G t with A t+1 as LHS and A j as the head of its rhs, where j < t+1, e.g. A t+1  A j . By the inductive assumption, all productions of the form A j   have as their head either a terminal or A k for some k > j.

So if we substitute for A j in A t+1  A j  all the rhs’s with A j as the LHS, then we will get productions of the form: A t+1  a rhs with a terminal as head, or A t+1  a rhs with A k as its head where k > j

By iterating the above process, we will end up with a grammar G t ’ in which all productions with A t+1 as lhs either have a terminal as the head of the rhs or a nonterminal A k for some k  t+1. The lemma follows by induction

Productions of this kind in which k=t+1 are left recursive, and can be eliminated to produce a grammar which we will call G t+1.

EXAMPLE GRAMMARS G 1 S  A A | 0 A  S S | 1 G 2 A 1  A 2 A 3 A 2  A 3 A 1 | b A 3  A 1 A 1 | a