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Reduce the grammar to Chomsky normal formS→ ABAC | aCAA→ aA | εB→ bB | εC→ dD→ f

Question

Reduce the grammar to Chomsky normal form

S → ABAC | a
A → aA | ε
B → bB | ε
C → d
D → f

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Solution

To reduce the given grammar to Chomsky normal form, we need to follow these steps:

Step 1: Eliminate ε-productions (productions that can derive the empty string ε). In this grammar, we have ε-productions for the variables B and C. To eliminate them, we replace B with ε in all the productions where it appears and remove the ε-production for B. Similarly, we replace C with ε in all the productions where it appears and remove the ε-production for C. After this step, the grammar becomes:

S → ABAC | ABACAA | aA | a | bB | b | dD | f

Step 2: Eliminate unit productions (productions of the form A → B). In this grammar, we have unit productions for the variables S, A, and B. To eliminate them, we replace S with ABAC in the first production, A with aA in the second production, and B with bB in the fifth production. After this step, the grammar becomes:

S → ABAC | ABACAA | aA | a | bB | b | dD | f A → aA | a B → bB | b

Step 3: Eliminate productions with more than two variables on the right-hand side. In this grammar, we have the production S → ABACAA that has more than two variables on the right-hand side. To eliminate it, we introduce a new variable X and rewrite the production as follows:

S → ABACX X → AA

After this step, the grammar becomes:

S → ABACX | aA | a | bB | b | dD | f A → aA | a B → bB | b X → AA

Step 4: Convert all remaining productions to have exactly two variables or terminals on the right-hand side. In this grammar, all the remaining productions already have exactly two variables or terminals on the right-hand side. Therefore, no further modifications are needed.

The final grammar in Chomsky normal form is:

S → ABACX | aA | a | bB | b | dD | f A → aA | a B → bB | b X → AA

This problem has been solved

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