How to make a DFA given a regular grammar ?

Question

A-->aB/bA/b , B-->aC/bB , C-->aA/bC/a

what is the approach ?

Answer 1

here A is initial and final state of the dfa

Comments

 A->b, C->a are equivalent to A->bF, C->aF, F-> ^, as per definition F is final state becoz of null production.

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@srestha

your dfa is wrong
here it accepts $\epsilon$ which is not the case for this grammar