it is regular language.you can check by giving the regular expression for it.it is same as a$(a+b)^{+}$a +b($(a+b)^{+}$b,a($(a+b)^{+}$b,b$(a+b)^{+}$a..

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+1 vote

0

it is regular language.you can check by giving the regular expression for it.it is same as a$(a+b)^{+}$a +b($(a+b)^{+}$b,a($(a+b)^{+}$b,b$(a+b)^{+}$a..

0

**this is not regular.**

** wxw. **Let W=01 X= 100

**wxw= 01 100 01.**Since there is no mention about the lentgh of X,we try to match elements as many as we can to make it regular.In worst case , even if you go till last but one, LHS has 0 and RHS has 1. W cant take two different values 0,1.

If it had been **WXW^R** IT is regular.

+1 vote

It is regular.

Lets see the strings in L

= { a, b, aa, ab, ba, bb, aaa, ..... } (When w is ϵ, wxw generates all these strings and hence we don't need to consider any other case for w)

= Σ* - {ϵ}

As we have **no restriction on the length of x**, we can leave 1st and last symbol as it is and consider the rest part as x.

eg : w=bab x=bb

**wxw **= babbbbab; we may assume x = abbbba so we can read it as : bxb which is finite.

0

.W=01 and X=001 and last two symbol match with 01.but what happen when string is like 0001111 then i think it can't satisfy the condition of WXW.so it is not regular language.

0

it is wxw not wxwr 0r wrxw...

and x is not (a+b)*... i dont think we will have a regular language here.!

and x is not (a+b)*... i dont think we will have a regular language here.!

0

@ arvin ,you are correct it's not a regular language.for ex take string=000111,0011,10,1100,111000 etc.it does not satisfy the condition of WXW.

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