How to design a DFA that accepts the language of pairs of binary words (a,b) with 5a=b?

Question

Let $\begin{bmatrix}0\\ 0\end{bmatrix}$ be a two-column vector with $0$ in the first row and $0$ in the second row.
Let $\Sigma_2 = \left\{ \begin{bmatrix}0\\ 0\end{bmatrix}, \begin{bmatrix}0\\ 1\end{bmatrix}, \begin{bmatrix}1\\ 0\end{bmatrix}, \begin{bmatrix}1\\ 1\end{bmatrix}\right\}$ be the set of all 2-element column vectors of binary digits.

$A =\{\begin{bmatrix}u\\ v\end{bmatrix}\in {\Sigma_2}^*\mid [v]_2 = 5[u]_2\}$

Where $[x]_k$ is the $x$ to the base $k$. So $[x]_2$ is a binary number. $A$ is the language of words where the second row is five times the first row. For example, $\begin{bmatrix}0\\1\end{bmatrix}\begin{bmatrix}0\\ 0\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}$ i.e. $1$ and $5$.

What I've realized so far is that the second row must be the first row shifted twice eg. $001 \to 100$ plus the first row. For example. $[001][101] = 001$ {shift2} $100 + 001$. What I've been trying to do as of late is to design a DFA to recognize a language where the second row is the first row shifted twice and move on from there.

I've been at this for over a day now. Please provide some insight if you can.

Answer 1

Define function $d:{\Sigma_2}^*\to \mathbb Z$, $d(w)=\underline w-5\overline w$, where

• $\overline w$ is the base-$2$ number formed by the digits on the first row of the letters in $w$
• $\underline w$ is the base-$2$ number formed by the digits on the second row of the letters in $w$.

For example, if $w=\begin{bmatrix}0\\1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}$, then $\overline w=(001)_2=1$, $\underline w=(110)_2=6$, $d(w)=6-5\cdot1=1$.

The language in the question $A$ is $\{w\in{\Sigma_2}^*\mid d(w)=0\}$

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How does $d(w)$ change if $w$ is extended by a letter $\sigma=\begin{bmatrix}a\\b\end{bmatrix}\in\Sigma_2$?

Let us check. Let $w'=w\sigma$. By the definition of a base-$2$ number, we have \begin{aligned} \overline{w'} &= 2\times \overline{w} + a\\ \underline{w'} &= 2\times \underline{w} + b\\ \end{aligned} So, $$d(w') = 2\times d(w) + d(\sigma)$$ Since $-5\le d(\sigma)\le 1$, we know

• $d(w)\le -1 \implies d(w')\le -1$
• $d(w)\ge 5 \implies d(w')\ge 5$

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Let us construct a DFA with 7 states $q_{\le-1}, q_0, q_1, q_2, q_3, q_4, q_{\ge5}$, where

• $q_{\le-1}$ for words $w$ such that $d(w)\le-1$.
• $q_{i}$ for words $w$ such that $d(w)=i$, where $0\le i\le 4$.
• $q_{\ge5}$ for words $w$ such that $d(w)=5$.

The transitions $\delta$ is defined by

• $\delta(q_{\le-1}, \cdot)=q_{\le-1}$.
• $\delta(q_{i}, \sigma)=q_{2*i+d(\sigma)}$, for $0\le i\le 4$. Here $q_{2*i+d(\sigma)}$ should be understood as
  • $q_{\le-1}$ if $2*i+d(\sigma)\le -1$.
  • $q_{\ge5}$ if $2*i+d(\sigma)\ge 5$.
• $\delta(q_{\ge5}, \cdot)=q_{\ge5}$.

$q_0$ is the start state and the unique accept state.

By the construction above, it is straightforward to check that the DFA accepts $A$.