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$r_3=r_1r_2=(a^*b)^*(a+ba)^*bb(a+b)^*$ comes out to be $r_3=r_2=(a+ba)^*bb(a+b)^*$ when i generate the resultant FA and its regex after concatenation i.e. it doesn't include $r_1$
Details:
Consider these two FA
$FA_1$
FA1
$FA_2$
FA2
if $r_1,r_2$ and $r_3$ are regex for $FA_1,FA_2$ and $FA_3$ respectively where
$r_1=(a^*b)^*$
$r_2=(a+ba)^*bb(a+b)^*$
then, mathematically, $r_3=r_1r_2=(a^*b)^*(a+ba)^*bb(a+b)^*$
But $r_3$ turns out different when I actually concatenate $FA_1$ and $FA_2$ and use the new transition table (given below) for $FA_3$ to generate its regex ($r_3$) and the $TG_3$ (given below) and then simplify, $r_3$ becomes $(a+ba)^*bb(a+b)^*$ which is the same as $r_2$ (i.e. $r_1$ not prefixed). and since $r_3=r_1r_2=(a^*b)^*(a+ba)^*bb(a+b)^*$ can't be simplified into $r_2=(a+ba)^*bb(a+b)^*$
I am convinced that the problem is in the concatenation/transition table, especially with the circular references to $x_1$ and $y_1$, which i don't completely get. That also would mean $TG_3$ iswrong.

The Transition Table:
transition table for fa3

$TG_3$
TG3

Now, it'd be very tedious to explain how i created the transition table, it's easier to figure out by looking at it. but key point i can't quite grasp the "$x_1$ and $y_1$ being connected" part
Sidenotes:
- This is the only way to concatenating two FA that i know of. I am on a distant learning platform and it's quite low quality material, so links to proper methods are appreciated. - If you use another method, please keep in mind that the transition table, regex and the TG and all required - The TG looks like an FA to me (barely know the difference) but the site i used to generate it says it's a TG. what's the difference?

Awaisome
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1 Answers1

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You can think of the concatenation of the languages defined by the two automata as the language of the automaton that can be obtained by connecting all final states of the first (in the example, just $x_1$) with the starting state of the second ($y_1$), by an empty transition, and then making them non-final.

A transition table can then be used to convert the NFA just generated into a DFA, which is what you did.

In this case, $r_1r_2$ is indeed equivalent to $r_2$, and this is completely normal. A transition graph represents a transition function, which is part of an automaton. If you minimize $TG_3$, what you get is exactly $FA_2$, which confirms the equivalence.

André Souza Lemos
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  • that was very helpful. i have lots of little misc. questions right now we could discuss in chat. but i posted this question like 48 hours ago and now am in a hurry for these three . 1) how is $r_1r_2 = r_2$. i tried to [simplify] (http://ivanzuzak.info/noam/webapps/regex_simplifier/) it. didn't work. 2) what would be the correct Transition table? what i did in my transition table is whenever a resultant state is $x_1$, i append $y_1$ but also vice versa. (which is where i think i went wrong ). 3) how do i minimize $TG_3$. – Awaisome Dec 04 '18 at 13:08
  • Your "simplifier" has limits. Try to figure out that everything $r_1$ does can be done without it. 2) after reading a symbol, the transition table should contain only the states where it leads to, plus those connected by empty transitions. 3) there are good resources for that, I see that you used JFLAP.
    • – André Souza Lemos Dec 04 '18 at 13:45
  • 1) i am lost here. how do i simplify $r_1r_2$ into $r_2$? 2). i updated the transition table so that when $y_2$ goes to $y_1$ on reading $a$, it doesn't include $x_1$. is it correct now? 3) i am not using JFLAP. i didn't create $FA_1$ and $FA_2$.. i just created the transition table then generated from it $TG_3$ and the $r_3=(a+ba)^bb(a+b)^=r_2$. – Awaisome Dec 04 '18 at 14:20
  • I'm sorry, this is not a tutoring site. If you have other questions, you can ask them as new posts. Important: try to make them, at least potentially, useful to other people, not just a solution to your problems. – André Souza Lemos Dec 04 '18 at 14:24
  • I also strongly encourage you to look for answers to similar questions. – André Souza Lemos Dec 04 '18 at 14:30
  • i tried. i updated the table as you asked. removed the redundant $x_1$ .is the table good now? – Awaisome Dec 04 '18 at 14:35
  • It is. I don't know what the $-$ and $+$ symbols are doing there, though. – André Souza Lemos Dec 04 '18 at 14:44
  • $-$ indicates initial state and $+$ mean final. in this case, those including $y_3$. i don't suppose this is standard nomenclature. as a mentioned in the sidenotes: my college gives us course notes, really substandard ones and video lectures recorded over 10 years ago. no textbook. they're still making us use BGI in computer graphics, you get the point. back the the original and last question : how to simplify $r_1r_2$ into $r_3$. i know, $(a^b^)^* \equiv (a+b)^$ but i only have $(a^b)^*$ to simplify. lost there – Awaisome Dec 04 '18 at 14:54
  • This last question has a complicated answer, that unfortunately doesn't fit here. I have a strong feeling that a very similar question has already been answered. – André Souza Lemos Dec 04 '18 at 15:00
  • is it like really complicated? cuz otherwise, we could go on chat. or i could give my email. cuz that's was the main question, see? and i asked this around 48 hours ago and i only have a couple more. by the way is this the question you're thinking about?. also should $z_1$ the inital state, also be final? (cuz $r_1$ accepts \Lambda as well so i think the concatenated result should as well?) – Awaisome Dec 04 '18 at 15:06
  • Yes, equivalency of regular expressions is complicated, and I gave you the way to skip it using FA conversion. Again, not a tutoring site. Your present question has been answered ("Is the concatenation done right?") Keep looking, or ask other questions. – André Souza Lemos Dec 04 '18 at 15:12
  • that question is not relevant. it simplifies $((aa^)^b)^$ into $((a)^b)^$ where i am trying to eliminate $((a)^b)^*$ – Awaisome Dec 04 '18 at 15:13
  • "Is the concatenation done right?" pointed to the whole problem with the regex. but you said they are equivalent if i could simplify one into the other. and that it would be complicated to explain how. i'll accept.because you said they are equivalent. thanks a lot. – Awaisome Dec 04 '18 at 15:18