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  1. Why L={wxw^R| w, x belongs to {a,b}^+ } is a regular language

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    W X W^R And Regular Expression for this language is: a(a + b) + a + b(a + b) + b Don't mix WXW R with WCW R, its X with + that makes language regular. Think by including X that is (a + b)* we can have finite choice for W that is a and b (finite is regular).
    Don't mix WXW R with WCW R, its X with + that makes language regular. Think by including X that is (a + b)* we can have finite choice for W that is a and b (finite is regular). Language WXW R can be say: if start with a ends with a and if start with b end with b. so correspondingly we need two final states. ITs DFA is as given below.
    To prove this, you should formalize the argument that "if y is in WxW, then y is in a (a+b) (a+b)*a + b (a+b) (a+b)*b; and if y is in a (a+b) (a+b)*a + b (a+b) (a+b)*b, then y is in WxW". It doesn't work in the other case since c is a fixed symbol, and can't include all but the characters on the ends.
    Language WXW R can be say: if start with a ends with a and if start with b end with b. so correspondingly we need two final states. ITs DFA is as given below. Does it mean we can not just pick one of x strings? As the example you mention above, If we just select x = “ababab” and w = a^n, then we get wxw^R = (a^n)ababab (a^n).
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