Recursion-
Recursion can be classified into following three types-
- Left Recursion
- Right Recursion
- General Recursion
1. Left Recursion-
- A production of grammar is said to have left recursion if the leftmost variable of its RHS is same as variable of its LHS.
- A grammar containing a production having left recursion is called as Left Recursive Grammar.
Example-
S → Sa / ∈
- Left recursion is considered to be a problematic situation for Top down parsers.
- Therefore, left recursion has to be eliminated from the grammar.
Elimination of Left RecursionLeft recursion is eliminated by converting the grammar into a right recursive grammar. If we have the left-recursive pair of productions- A → Aα / β (Left Recursive Grammar) where β does not begin with an A. Then, we can eliminate left recursion by replacing the pair of productions with- A → βA’ A’ → αA’ / ∈ (Right Recursive Grammar) This right recursive grammar functions same as left recursive grammar. |
2. Right Recursion-
- A production of grammar is said to have right recursion if the rightmost variable of its RHS is same as variable of its LHS.
- A grammar containing a production having right recursion is called as Right Recursive Grammar.
Example-
S → aS / ∈
(Right Recursive Grammar)
- Right recursion does not create any problem for the Top down parsers.
- Therefore, there is no need of eliminating right recursion from the grammar.
3. General Recursion- The recursion which is neither left recursion nor right recursion is called as general recursion.
Example-
S → aSb / ∈
PRACTICE PROBLEMS BASED ON LEFT RECURSION ELIMINATION-
Problem-01: Consider the following grammar and eliminate left recursion-
A → ABd / Aa / a
B → Be / b
Solution- The grammar after eliminating left recursion is-
A → aA’
A’ → BdA’ / aA’ / ∈
B → bB’
B’ → eB’ / ∈
Problem-02: Consider the following grammar and eliminate left recursion-
E → E + E / E x E / a
Solution- The grammar after eliminating left recursion is-
E → aA
A → +EA / xEA / ∈
Problem-03: Consider the following grammar and eliminate left recursion-
E → E + T / T
T → T x F / F
F → id
Solution- The grammar after eliminating left recursion is-
E → TE’
E’ → +TE’ / ∈
T → FT’
T’ → xFT’ / ∈
F → id
Problem-04: Consider the following grammar and eliminate left recursion-
S → (L) / a
L → L , S / S
Solution- The grammar after eliminating left recursion is-
S → (L) / a
L → SL’
L’ → ,SL’ / ∈
Problem-05: Consider the following grammar and eliminate left recursion-
S → S0S1S / 01
Solution- The grammar after eliminating left recursion is-
S → 01A
A → 0S1SA / ∈
Problem-06: Consider the following grammar and eliminate left recursion-
S → A
A → Ad / Ae / aB / ac
B → bBc / f
Solution- The grammar after eliminating left recursion is-
S → A
A → aBA’ / acA’
A’ → dA’ / eA’ / ∈
B → bBc / f
Problem-07: Consider the following grammar and eliminate left recursion-
A → AAα / β
Solution- The grammar after eliminating left recursion is-
A → βA’
A’ → AαA’ / ∈
Problem-08: Consider the following grammar and eliminate left recursion-
A → Ba / Aa / c
B → Bb / Ab / d
Solution- This is a case of indirect left recursion.
Step-01: First let us eliminate left recursion from A → Ba / Aa / c
Eliminating left recursion from here, we get-
A → BaA’ / cA’
A’ → aA’ / ∈
Now, given grammar becomes-
A → BaA’ / cA’
A’ → aA’ / ∈
B → Bb / Ab / d
Step-02: Substituting the productions of A in B → Ab, we get the following grammar-
A → BaA’ / cA’
A’ → aA’ / ∈
B → Bb / BaA’b / cA’b / d
Step-03: Now, eliminating left recursion from the productions of B, we get the following grammar-
A → BaA’ / cA’
A’ → aA’ / ∈
B → cA’bB’ / dB’
B’ → bB’ / aA’bB’ / ∈
This is the final grammar after eliminating left recursion.
Problem-09: Consider the following grammar and eliminate left recursion-
X → XSb / Sa / b
S → Sb / Xa / a
Solution- This is a case of indirect left recursion.
Step-01: First let us eliminate left recursion from X → XSb / Sa / b
Eliminating left recursion from here, we get-
X → SaX’ / bX’
X’ → SbX’ / ∈
Now, given grammar becomes-
X → SaX’ / bX’
X’ → SbX’ / ∈
S → Sb / Xa / a
Step-02: Substituting the productions of X in S → Xa, we get the following grammar-
X → SaX’ / bX’
X’ → SbX’ / ∈
S → Sb / SaX’a / bX’a / a
Step-03: Now, eliminating left recursion from the productions of S, we get the following grammar-
X → SaX’ / bX’
X’ → SbX’ / ∈
S → bX’aS’ / aS’
S’ → bS’ / aX’aS’ / ∈
This is the final grammar after eliminating left recursion.
Problem-10: Consider the following grammar and eliminate left recursion-
S → Aa / b
A → Ac / Sd / ∈
Solution- This is a case of indirect left recursion.
Step-01: First let us eliminate left recursion from S → Aa / b
This is already free from left recursion.
Step-02: Substituting the productions of S in A → Sd, we get the following grammar-
S → Aa / b
A → Ac / Aad / bd / ∈
Step-03: Now, eliminating left recursion from the productions of A, we get the following grammar-
S → Aa / b
A → bdA’ / A’
A’ → cA’ / adA’ / ∈
This is the final grammar after eliminating left recursion.
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