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I have drawn the DFA for language L1 containing strings starting with 01 and language L2 containing strings ending in 11. For the final DFA, I have concatenated both DFA's. The finals DFA does not accept 011. Kindly help!

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The reason why your DFA doesn't work is that not all strings that start with 01 and end with 11 can be written as a concatenation of a string $x$ that start with 01 and a string $y$ that ends with 11, as you discovered.

This can happen because $x$ and $y$ might intersect. However the only way for this to happen is for $x$ and $y$ to share a single "$1$". I.e., the only problematic string is $011$. To fix this you can just union your DFA with one that recognizes $011$.

Rephrasing your problem using a regular expression, you cannot obtain a regular expression for you language by simply concatenating $01(0\mid1)^*$ with $(0\mid1)^*11$ which would yield $01(0\mid1)^*(0\mid1)^*11=01(0\mid1)^*11$. However, the following regular expression is correct:

$$ ( 01(0 \mid 1)^*11 ) \mid (011) $$

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  • $\begingroup$ Thank you! That is very helpful! $\endgroup$ Jun 26 '21 at 4:27

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