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Question

L={a$^{2*m}$  b$^{4*n}$  c$^{n}$  d$^{m}$ |m,n>=0}

L={xwxw$^{r}$|x,w $\epsilon$ {0,1}} machine

Comments

I think the regular expression of that language will be like this.

 

$(0+1)^{*} 0(0+1)^{*} 0 + (0+1)^{*} 1 (0+1)^{*} 1$

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@Sayan Bose @kumar.dilip the language is finite right why isn't it Regular?

As it is said w ϵ {0,1}, w can be either 0 or 1 and the reverse of it will also be same as it is unit length string.

So why can't we say its as " L is set of strings with 1st & 3rd positions same and 2 nd and 4th position same" 

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yes you are right and here are only 4 such strings possible

so it is a finite language hence regular