Regular expression to NFA -- confusing regular expressions-:

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shivajikobardan
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Homework Statement
Regular expression to NFA -- confusing regular expressions-:
Relevant Equations
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R=(01+010)*
For it I made the below nfa which i believe seems correct. Plus the tutorials that I am following also make sure it’s correct. Q0 is initial state(forgot to mention in figure).
nmXLE14mKI4PtFf4GtDc_PgenECvC_h_m9Yrbrg99hHZh258F9.jpg


R=(01)*+(010)*

But idk how to convert this to NFA
What will be languages accepted by this NFA? Won’t it be the same as the above one?

(for some different question)
I got small hint about this. It was to add epsilon transition, but I don’t understand the need for it.
sg7PNgzhEhyEXJ1JRmtN4nIELpWrJCIUaKA7uvrJatk6oTbd2k.png

Source-: https://www.cs.wcupa.edu/rkline/fcs/nfas.html
 
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shivajikobardan said:
But idk how to convert this to NFA
I don't know is not good enough: make a start and show your efforts.

shivajikobardan said:
What will be languages accepted by this NFA? Won’t it be the same as the above one?
Test both expressions with the string 01010.

shivajikobardan said:
(for some different question)
I got small hint about this. It was to add epsilon transition, but I don’t understand the need for it.
The material you linked includes exactly the same pair of expressions as you have, just with different symbols. The need for the epsilon transition is to deal with the fact that (01+010)* (Example 2 (ab∪aba)*) is different from (01)*+(010)* (Example 3 (ab)*∪(aba)*. Work through that material and the examples until you understand it.
 
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