Regular expression to NFA-confusing regular expressions-:

[shivajikobardan]

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).
https://lh6.googleusercontent.com/GUZlDA7_UjNWlcNx76e1xHOYhNjYBzbDMnP9qBnMjL8ux_Ntz3SSVIcw-VfMR6jbFlbxYG3nx6ccd2gWObG2hXnmXLE14mKI4PtFf4GtDc_PgenECvC_h_m9Yrbrg99hHZh258F9

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.
https://lh3.googleusercontent.com/naVKgIC_8oj_v6TLAyjAhdOlLEpr4jsUAiYNw59d-2LifF5scOxn7rBw0AwCvmDwtFMZtv-s7DfXZAkt5OTNOksg7PNgzhEhyEXJ1JRmtN4nIELpWrJCIUaKA7uvrJatk6oTbd2k
Source-: https://www.cs.wcupa.edu/rkline/fcs/nfas.html

 

[Evgeny.Makarov]

What will be languages accepted by this NFA?

I believe the regular expression offers the simplest description of this language.

Won’t it be the same as the above one?

The language $(01+010)^*$ contains $01010$, but $(01)^*+(010)^*$ does not.

(for some different question)

Why do you think that this is a different question if the regular expression is the same up to replacement of 0 and 1 by $a$ and $b$?

It was to add epsilon transition, but I don’t understand the need for it.

Well, this is surprising if you understand the meaning of regular expressions. If you learned that the new gaming console would be sold at two stores only and you wanted to buy it, how can it be unclear that the first thing you need to do is to choose one of the stores and go there?