Is the regular expression: $(\lambda + a)^$ equivalent to $a^$?

Question

I have a question about convert a regular expression to an automaton. In the middle of the regular expression, appears $(\lambda + a)^*$. What I am wondering is: it is not the same of $ a^* $?

If is so, I could simplify the final automaton. $\lambda$ is the symbol for empty string.

Is $\lambda$ your notation for the empty string? – hmakholm left over Monica Apr 07 '18 at 22:19 — hmakholm left over Monica, Apr 07 '18 at 22:19
yes it is the notation for empty string. – Silva Apr 07 '18 at 22:23 — Silva, Apr 07 '18 at 22:23
Why this is "on hold"? I tagged it incorrectly? – Silva Apr 08 '18 at 15:29 — Silva, Apr 08 '18 at 15:29

score 1 · Accepted Answer · answered Apr 07 '18 at 22:25

Yes, $(\lambda+a)^*$ and $a^*$ describe the same language.

Depending on which context you meet the task in, you may or may not be expected to follow a particular algorithm mindlessly (construct a NFA from the abstract syntax tree of the regex, convert the NFA to a DFA, minimize the DFA).

Otherwise, you can just write down the trivial (one state, one transition) automaton and be done with it.

Is the regular expression: $(\lambda + a)^*$ equivalent to $a^*$?

1 Answers1

Is the regular expression: $(\lambda + a)^$ equivalent to $a^$?