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Prove that if $L$ is regular over $\Sigma=\{0,1,2\}$ then the following language over $\{0,1,2,\$\}$ is also regular: $$ G=\{\$x\$|\exists w\in L: x\text{ is derived from }w\text{ by substituting } 01 \text{ with }\$\$ \} $$ For example, if $10112\in L$ then $\$1\$\$12\$\in G$.

I think this can be solved using closure properties of regular languages.

1) Let $H=\$L\$$. $H$ is also regular because it's concatenation.

2) Let $h:\Sigma\to \Sigma^*$ be defined as follows: $$ h(\$)=h(0)=h(1)=\$ $$ Then: $$ G=h^{-1}(H)\cap \$(\Sigma^*\$\$\Sigma^*)^+\$ $$ which is regular because $h^{-1}$, intersection and regular expressions are closed in regular languages.

I wonder if my proof using closure is correct or automaton should be built in this case? In addition if I managed to think of a regular expression to describe $G$ then this alone would've proved that $G$ is regular?

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Unfortunately, your argument breaks down at step 2). For example, let $L=\{01\}$. Then
$G =\{\$\$\$\$\}.$
$H=\{\$01\$\}$.
$h^{-1}(H)=\emptyset$ since every word in the range of $h$ contains neither 0 or 1.
$h^{-1}(H)\cap \$(\Sigma^*\$\$\Sigma^*)^+\$=\emptyset.$
However, $G$ is not empty.

If I managed to think of a regular expression to describe G then would this alone have proved that G is regular?

Of course. However, it looks like it is not immediate to figure out a regular expression for $G$ even if we have been given the regular expression for $L$ and DFA for $L$.

Here is a way to show $G$ is regular by DFA.

Let the DFA for $L$ be $(\Sigma,Q, q_0,\delta_L, F)$.

Define (an incomplete) DFA $D$ with alphabet $\{0,1,2,\$\}$, states $Q\times \{s_0, s_1, o\}$, initial state $(q_0, s_0)$, accepting state $F\times\{s_0, o\}$, transition function $\delta_D$ such that
$\delta_D((q, s_0), 0)= (\delta_L(q, 0), o) $
$\delta_D((q, s_0), 1)= (\delta_L(q, 1), s_0) $
$\delta_D((q, s_0), 2)= (\delta_L(q, 2), s_0) $
$\delta_D((q, s_0), \$)= (\delta_L(q, 0), s_1) $

$\delta_D((q, s_1), \$)= (\delta_L(q, 1), s_0) $

$\delta_d((q, o), 0)= \delta_L(q, 0), o) $
$\delta_d((q, o), 2)= \delta_L(q, 2), s_0) $
$\delta_d((q, o), \$)= \delta_L(q, 1), s_1) $

Here is how we can understand the states

  • state $(q,o)$ corresponds to the states that are reached by words that end with $0$.
  • state $(q,s_1)$ corresponds to the states that are reached by words that end with odd number of $\$$'s.
  • state $(q,s_0)$ corresponds to the states that are reached by words that end with $1$ or $2$ or even number of $\$$'s.

We can check that $G=\$L(D)\$$.

Exercise. Prove that if $L$ is regular over $\Sigma=\{0,1\}$ then the following language over $\{0,1,2\}$ is also regular. $$ G=\{x\mid \exists w\in L: x\text{ is derived from }w\text{ by substituting } 00 \text{ and } 11 \text{ with } 2 \text{ from left to right} \}$$ For example, if $w=1000111110$, then $x=1202210$.

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  • $\begingroup$ Can you please explain why do you have states based on what letter reached the end of the words? For example, I'd define the transition functions as follows: $$\delta_D((q,2),0)=(\delta(q,2),0)\\\delta_D((q,\$),0)=(\delta(q,0),1)\\\delta_D((q,\$),1)=(\delta(q,1),0)$$.In my example we start out with state $0$. When we read the first $\$$ we flip the state to $1$. When we read the second $\$$ we flip the state back to $0$. When we read $2$ nothing changes so we stay in state $0$ as well. In the end we can add $\$$'s from the beginning and end of each word. $\endgroup$
    – Yos
    Jan 28 '19 at 20:38
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    $\begingroup$ If we know the last letters of the words that reaches a state $s$, then we can specified what will be the next state if the word is extended with another letter. Note that we must ensure words containing 01 and words with isolated $\$$ should not be accepted. $\endgroup$
    – John L.
    Jan 28 '19 at 20:47
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    $\begingroup$ Exactly. That is why incomplete DFA is popular since there is no need to write any transition that goes to a dead state. In fact, we do not have to specify that dead state. $\endgroup$
    – John L.
    Jan 28 '19 at 20:58
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    $\begingroup$ I understand the transitions now, thanks! I still don't why you concatenate dollar sign here: $\delta_D((q, s_1), \$)= (\delta_L(q, 1), s_0)\$$? $\endgroup$
    – Yos
    Jan 28 '19 at 21:05
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    $\begingroup$ Updated with a few corrections of typos. $\endgroup$
    – John L.
    Jan 28 '19 at 21:13

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