equivalent grammars

Two formal grammars $G_1=({\mathrm{\Sigma }}_1,N_1,P_1,{\sigma }_1)$ and $G_2=({\mathrm{\Sigma }}_2,N_2,P_2,{\sigma }_2)$ are said to be equivalent if they generate the same formal languages:

$$L(G_1)=L(G_2).$$

When two grammars $G_1$ and $G_2$ are equivalent, we write $G_1\cong G_2$.

For example, the language $\{a^n{b}^{2n}\mid n\text{ is a non-negative integer}\}$ over $T=\{a,b,c\}$ can be generated by a grammar $G_1$ with three non-terminals $x,y,\sigma$, and the following productions:

$$P_1=\{\sigma \to \lambda ,\sigma \to x\sigma y,x\to a,y\to b^2\}.$$

Alternatively, it can be generated by a simpler grammar $G_2$ with just the starting symbol:

$$P_2=\{\sigma \to \lambda ,\sigma \to a\sigma b^2\}.$$

This shows that $G_1\cong G_2$.

An alternative way of writing a grammar $G$ is $(T,N,P,\sigma )$, where $T=\mathrm{\Sigma }-N$ is the set of terminals of $G$. Suppose $G_1=(T_1,N_1,P_1,{\sigma }_1)$ and $G_2=(T_2,N_2,P_2,{\sigma }_2)$ are two grammars. Below are some basic properties of equivalence of grammars:

1. Suppose $G_1\cong G_2$. If $T_1\cap T_2=\mathrm{\varnothing }$, then $L(G_1)=\mathrm{\varnothing }$.

2. Let $G=(T,N,P,\sigma )$ be a grammar. Then the grammar $G^{\prime }=(T,N^{\prime },P,\sigma )$ where $N\subseteq N^{\prime }$ is equivalent to $G$.

3. If $(T_1,N_1,P_1,{\sigma }_1)\cong (T_1,N_1,P_2,{\sigma }_2)$, then $(T,N,P_1,{\sigma }_1)\cong (T,N,P_2,{\sigma }_2)$, where $T=T_1\cup T_2$ and $N=N_1\cup N_2$.

4. If we fix an alphabet $\mathrm{\Sigma }$, then $\cong$ is an equivalence relation on grammars over $\mathrm{\Sigma }$.

So far, the properties have all been focused on the underlying alphabets. More interesting properties on equivalent grammars center on productions. Suppose $G=(\mathrm{\Sigma },N,P,\sigma )$ is a grammar.

1. A production is said to be trivial if it has the form $v\to v$. Then the grammar $G^{\prime }$ obtained by adding trivial productions to $P$ is equivalent to $G$.

2. If $u\in L(G)$, then adding the production $\sigma \to u$ to $P$ produces a grammar equivalent to $G$.

3. Call a production a filler if it has the form $\lambda \to v$. Replacing a filler $\lambda \to v$ by productions $a\to va$ and $a\to av$ using all symbols $a\in \mathrm{\Sigma }$ gives us a grammar that is equivalent to $G$.

4. $G$ is equivalent to a grammar $G^{\prime }$ without any fillers. This can be done by replacing each filler using the productions mentioned in the previous section. Finally, if $\lambda \in L(G)$, we add the production $\sigma \to \lambda$ to $P$ if it were not already in $P$.

5. Let $S(P)$ be the set of all symbols that occur in the productions in $P$ (in either antecedents or conseqents). If $a\in S(P)$, then $G^{\prime }$ obtained by replacing every production $u\to v\in P$ with production $u[a/b]\to v[a/b]$ and $b\to a$, with a symbol $X\notin \mathrm{\Sigma }$, is equivalent to $G$. Here, $u[a/b]$ means that we replace each occurrence of $a$ in $u$ with $X$.

6. $G$ is equivalent to a grammar $G^{\prime }$, all of whose productions have antecedents in $N^{+}$ (non-empty words over $N$).

Remark. In fact, if we let

1. $X\to XY$,

2. $XY\to ZT$,

3. $X\to \lambda$,

4. $X\to a$

be four types of productions, then it can be shown that

• any formal grammar is equivalent to a grammar all of whose productions are in any of the four forms above

• any context-sensitive grammar is equivalent to a grammar all of whose productions are in any of forms 1, 2, or 4, and

• and any context-free grammar is equivalent to a grammar all of whose productions are in any of forms 1, 3, or 4.

Title	equivalent grammars
Canonical name	EquivalentGrammars
Date of creation	2013-03-22 17:37:17
Last modified on	2013-03-22 17:37:17
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	6
Author	CWoo (3771)
Entry type	Definition
Classification	msc 91F20
Classification	msc 68Q42
Classification	msc 68Q45
Classification	msc 03D05