RL and DCFL

Question

the answer is given that the statement 2 is correct?

But how…

even if we create a DCFL by final state condition like :

q(b,z0| z0)-→ final state ,q(null,a|z0)  ----→   final state [Thats what was mentioned in the video solution]

it will accept the string aab

 

Comments

{${a^{n} b^{n+k}}$ } U {${a^{n+k} b^{n}}$}  => {${a^{m} b^{m}}$} where m=n+k or n+0,

now n>=0 and k>=1 or 3 becomes K>=1 then

possible values are {${a^{n+k} b^{n+k}}$} , {${a^{n+0} b^{n+k}}$},{${a^{n+k} b^{n+0}}$},{${a^{n+0} b^{n+0}}$}

from this string aab is accepted.

any idea?

 

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aab will be accepted when n=1 and k =2

but this aab doesn't belong to the L given in sentence 2:

so why is it true then

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the language is union of two languages bro.

so, it will work. first combine both the languages as above and then execute