NFA-E to DFA conversion. (which is the correct solution?)

Question

1. Which solution is correct? (or both wrong!)
2. Does every 'DFA equivalent' of any NFA has same starting state? if not, please give any smallest example.

Comments

(i)The above Epsilon-NFA represents language L = 0*1*...So, I think 1st solution  is correct and 2nd solution does not accept string "1" which is accepted by the original non deterministic machine by null move(e-move)...

(ii) Yes , start state will be the same in both NFA and its equivalent DFA but if we reverse  the DFA than we toggle start and final states then we may get more than 1 start state and new machine model can be  non-deterministic ..In this case , we have to connect those start state from one start state by null moves...

Answer 1

1. Solution 1 is correct and You could either do it directly by practice Or Could use NFA to DFA conversion Algorithm to see that. . Solution 2 is wrong and to check that You can use "Null string" as a counter example.

2.  Does every 'DFA equivalent' of any NFA has same starting state??

No, Not Necessarily. In case of Epsilon Move Free NFA (NFA without Epsilon/Null Move), You could say that 'DFA equivalent' of any "Epsilon free NFA (NFA without Null moves)" has same starting state. 

But in case of Epsilon-NFA (NFA with Null Moves), Starting state of Equivalent DFA is the Epsilon Closure of the Starting state of that NFA. So, Starting state May or May not be exactly same (But It would definitely contain the Starting state of that NFA).

For example, You could use NFA in your diagram. In the given NFA (It is Epsilon-NFA), Starting state is A but in the Equivalent DFA (Solution 1), Starting state is AB. 

Answer 2

The first one seems to be quite correct.

Can you please explain the logic of second solution.

Comments

Bro , in the given link ,they relabeled the node...how is it possible that same machine start with 2 different state ? we can relabel the node...In above , I explained the reverse procedure , after reversing , finally we get only one start state...

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Sir, now I think what the problem is, what I am referring to as 'diffrent start state' must be same as 'relabeling the start State' !

Thanks , conusion cleared I think. :)

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@ashish bro ,please don't call me sir .. here I have written what I understood..may be I am wrong..others will explain better...