3
$\begingroup$

I'm currently attempting to construct a grammar for the language $L = \{0^a1^b2^c3^d | a,b,c,d \in \mathbb{N} \land a+b = c+d\}$

However I'm getting stuck on constructing the rules in a way that the $2$s and $3$s always appear in the correct order.

My current approach is the following ruleset with $E$ being the initial rule. \begin{align*} E &\rightarrow \epsilon &E &\rightarrow AC\\ F &\rightarrow BC &A &\rightarrow 0\\ A &\rightarrow 0E &A &\rightarrow B\\ B &\rightarrow 1 &B &\rightarrow 1F\\ C &\rightarrow 2 &C &\rightarrow 3 \end{align*} However this also incorrectly accepts the word $w = 0032$. How can I make sure no $2$ ever follows a $3$?

$\endgroup$
  • $\begingroup$ try to resolve ambiguity between the two options for C by introducting D $\endgroup$ – Pieter21 Nov 17 '16 at 16:07
  • $\begingroup$ Yeah introducing D is obviously the way to go here, however I don't see a way to ensure the correct order between C and D right now. $\endgroup$ – ntldr Nov 17 '16 at 16:09
  • $\begingroup$ and can you create a Left rule of $0^a1^b$ and a Right rule? $\endgroup$ – Pieter21 Nov 17 '16 at 16:18
  • $\begingroup$ Ah I see. Instead of using many symbols to insert between I just have to find rules for a more broad case. Thanks to you aqd Brian! $\endgroup$ – ntldr Nov 17 '16 at 16:24
1
$\begingroup$

I’d take a different approach altogether:

$$\begin{align*} E&\to X_{03}\mid X_{02}\mid X_{13}\mid X_{12}\mid\epsilon\\ X_{03}&\to 0X_{03}3\mid 0X_{02}3\mid 0X_{13}3\mid 0X_{12}3\mid\epsilon\\ X_{02}&\to 0X_{02}2\mid 0X_{12}2\mid\epsilon\\ X_{13}&\to 1X_{13}3\mid 1X_{12}3\mid\epsilon\\ X_{12}&\to 1X_{12}2\mid\epsilon \end{align*}$$

$\endgroup$
1
$\begingroup$

Context-free languages are equivalent to push-down automata, and in this particular case constructing an automaton is easier:

$$ \begin{array}{c} \mathtt{0}:\mathrm{push}&&\mathtt{1}:\mathrm{push}&&\mathtt{2}:\mathrm{pop}&&\mathtt{3}:\mathrm{pop} \\ \curvearrowleft && \curvearrowleft && \curvearrowleft && \curvearrowleft \\ s_0& \xrightarrow{\epsilon} &s_1& \xrightarrow{\epsilon} &s_2& \xrightarrow{\epsilon} &s_3 \end{array} $$

where $s_0$ is the initial state and $s_3$ is the accepting state (with empty stack). We only use $\mathrm{push}$ and $\mathrm{pop}$ because we don't need any additional information.

To simulate this automaton with a context-free grammar we have to preserve the symmetry between pushes and pops. In other words, each time we produce one of $0$ or $1$ we will need to produce also $2$ or $3$. To keep track which one we can produce, we need enough states to represent all the combinations:

  • production $A$ will represent pair of states $(s_0,s_3)$,
  • production $B$ will represent pair of states $(s_1,s_3)$,
  • production $C$ will represent pair of states $(s_0,s_2)$,
  • production $D$ will represent pair of states $(s_1,s_2)$.

Observe that the path of the automaton determines possible dependencies in the grammar (with a slight oversimplification we could say that in "push" states we go forward and in "pop" states we go backward):

  • from $A$ we could go to $B$, because we can go forward from $s_0$ to $s_1$, but not back;
  • from $A$ we could go to $C$, because we can backward from $s_3$ to $s_2$, but not the other way around;
  • from $B$ we cannot go to $C$, because we cannot go from $s_1$ to $s_0$ (but we can go back from $s_3$ to $s_2$, in particular $B \to D$ is ok);
  • etc.

The completed grammar looks as follows:

\begin{align} S &\to A \\ A &\to \mathtt{0}\ A\ \mathtt{3} \mid B \mid C \mid D \\ B &\to \mathtt{1}\ B\ \mathtt{3} \mid D \\ C &\to \mathtt{0}\ C\ \mathtt{2} \mid D \\ D &\to \mathtt{1}\ D\ \mathtt{2} \mid \epsilon \end{align}

I hope this helps $\ddot\smile$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.