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Let $h: \{ a,b,c,d \}^* \rightarrow \{a,b\}^* $ be a homomorphism such, that $h(a) = aa, h(b) = ab, h(c) = ba, h(d) = b $ . Determine: $h^{-1}((bab)^*ba^*b).$

I have trying do it by 4 hours. I don't have idea. I observed, what is obvious, that: $h^{-1}(bab)^*= (db)^*$

Please help me!

user180834
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    I don't know how to solve this either, but I do notice that $h^{-1}(bab)^\ne (db)^$; I think $h^{-1}(bab)^$ maybe is $(db+cd)^$. Observe that $h(cd) = bab$ also. – MJD Mar 22 '15 at 18:44

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As MJD points out in a comment, we have $$h^{-1}(bab)^*=(db+cd)^*$$ which can be proven noting that two consecutive $b$'s must not be "generated" by the same element of $\{a,b,c,d\}^*$, since none of the images of those elements has consecutive $b$'s. Thus, we can factor, for instance $$h^{-1}(babbab)=h^{-1}(bab)h^{-1}(bab)$$ and then we just note that $h^{-1}(bab)=(db+cd)$ and apply the above factoring inductively. We can similarly factor $$h^{-1}((bab)^*ba^*b)=h^{-1}(bab)^*h^{-1}(ba^*b)$$ $$h^{-1}((bab)^*ba^*b)=(db+cd)^*h^{-1}(ba^*b)$$ so we just need to figure out that last term; I don't see any easy way to do this beyond splitting into cases; if $X$ is a string such that $h(X)$ matches $ba^*b$, then it's clear that $X$ starts either with $c$ or $d$, followed by a string $Y$ with $h(Y)$ matching $a^*b$ - this gives us: $$h^{-1}((bab)^*ba^*b)=(db+cd)^*(c+d)h^{-1}(a^*b)$$ where the last term of possible strings $Y$ clearly ends with either $b$ or $d$ and can have as many $a$'s preceding it as desired. This yields $$h^{-1}((bab)^*ba^*b)=(db+cd)^*(c+d)a^*(b+d)$$

Milo Brandt
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  • The part I was worried about is concluding that $h^{-1}(PQ) = h^{-1}(P)h^{-1}(Q)$; you did that with $P=(bab)^$ and $Q=ba^b$. How do you know that you can deal with them separately like that? – MJD Mar 22 '15 at 18:58
  • @MJD Because $P$ (if not empty) ends in $b$, and $Q$ starts with $b$ and $h$, having no letter $\ell$ in the alphabet with $h(\ell)=XbbY$, must have that $h^{-1}(XbbY)=h^{-1}(Xb)h^{-1}(bY)$. – Milo Brandt Mar 22 '15 at 19:04
  • Got it, thanks! – MJD Mar 22 '15 at 19:05