Take $\Sigma = \set{(0, 0), (0, 1), (1, 0), (1, 1)}$.

Write the function $V(x)$ to denote the value of a string $x \in \set{0,1}^*$ when taken as a binary number with most significant digit first.

If $x = (a_1, b_1) (a_2, b_2) \dots (a_n, b_n)$ let

$$t(x) = V(a_1 a_2 \dots a_n) \quad b(x) = V(b_1 b_2 \dots b_n)$$

Then consider the language $L = \setbuilder{x \in \Sigma^*}{b(x) = 3 \cdot t(x)}$, that is, the binary value of the concatenation of the right-hand numbers is three times that of the left hand numbers.

We build a DFA that accepts $L^R$, that is, strings satisfying the same numerical properties as above, but starting with the least significant digit.

First, suppose $k = k_n k_{n-1} \dots k_1 k_0$, i.e., a binary string with MSB first. Then,

$$\begin{align*} 3V(k) &= 3(k_n 2^n + k_{n-1} 2^{n-1} + \dots + k_1 2^1 + k_0) \\ &= (2+1)(k_n 2^n + k_{n-1} 2^{n-1} + \dots + k_1 2^1 + k_0) \\ &= (k_n 2^{n+1} + k_{n-1} 2^{n} + \dots + k_0 2^1 + 0) + (k_n 2^n + k_{n-1} 2^{n-1} + \dots + k_1 2^1 + k_0) \end{align*}$$

So $3V(k)$ is $k$ added with a shifted copy of itself. So, a naive way to express the condition that $V(b) = 3V(a)$ is:

$$a_0 = b_0 \tand a_i + a_{i-1} = b_i \text{ for all } i$$

Which is almost correct, but we need to account for carry.

Let $x = (a_1, b_1) (a_2, b_2) \dots$ be an input to our DFA. Our machine will emulate a full adder. That is, upon reading $(a_i, b_i)$, it will check that $a_{i-1} + a_i = b_i$ (mod 2), accounting for carry bit, and remember if a carry bit was introduced or used up in that addition. This means we need to remember value of $a_{i-1}$ and whether or not there is currently a carry bit, which will require four states. If ever $a_{i-1} + a_i \neq b_i$, the machine goes into a dead state.

Obviously, states with carry bits cannot be accepting, because the presence of a carry bit means that the addition is not 'complete' without writing down another 1. In addition, if $a_{i-1} = 1$, the current value of $3V(a)$ would certainly require one more digit, so this state cannot be accepting either. The only accepting state is $a_{i-1} = 0$ with no carry bit.

So let $Q = \set{0, 1, 0^c, 1^c, D}$. The number denotes the value of $a_{i-1}$ and superscript $c$ denotes the presence of a carry bit. $D$ is the dead state. Then the following is a DFA $M$ that accepts $L^R$

Then $L^R$ is regular and $L$ is also regular (by 1.31).