+1 vote
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Convert the given Finite Automata to Regular Expression.

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r= (bc*a)*.(bc*d) there is no need to write bc*d further whenever we want bc*d only we substitute ∊ in place of (bc*a)*.
$b(c+ab)^{*}d$

Check this R.E.
Can you please mention the steps?

.....

Just look how it is working.It is starting with b and ends with d. In between, that there are two loops c and ab.
@LeenSharma

We know that regular Expressions are not unique (for the same language we can write more than one Regular Expression) but when we convert FA to RE, so in this case also we will get the same RE or can be different.

Does it depends on the order in which we eliminate state??
Is there any sequence that we can follow for state elimination method that gives us a more better evaluated RE.

Like to get b(c+ab)∗d rather than getting (bc*a)*.(bc*d)??

Pardon for handwriting .. Hope this helps :)

answered by Junior (575 points)
Is there any sequence that we can follow for state elimination method that gives us a more better evaluated RE.

Like to get b(c+ab)∗d rather than getting (bc*a)*.(bc*d)??
prefer to eliminate the state with epsilon first.
First of all it is not DFA

In every state for every input alphabet transition is not defined
answered by Loyal (3.2k points)

+1 vote