Quotient of a formal language

In mathematics and computer science, the right quotient (or simply quotient) of a language $L_1$ with respect to language $L_2$ is the language consisting of strings $w$ such that $wx$ is in $L_1$ for some string $x$ in {{nowrap| $L_2$ ,}} where $L_1$ and $L_2$ are defined on the same alphabet {{nowrap| $\Sigma$ .}} Formally:{{cite book |last=Pin |first=J-É. |author-link=Jean-Éric Pin |translator-last=Howie |translator-first=A. |date=1986 |title=Varieties of Formal Languages |location=New York |publisher=Plenum Press |isbn=0306422948 |pages=14}}{{cite book |last1=Linz |first1=Peter |last2=Rodger |first2=Susan H. |author-link2=Susan H. Rodger |name-list-style=amp |date=2023 |title=An Introduction to Formal Languages and Automata |edition=Seventh |location=Burlington, MA |publisher=Jones & Bartlett Learning |isbn=978-1284231601 |pages=112–117}} ([https://books.google.com/books?id=hsxDiWvVdBcC&dq=right+quotient+automata&pg=PA104 Fifth ed.] at Google Books)

$L_1 / L_2 = \{w \in \Sigma^* \mid wL_2 \cap L_1 \neq \varnothing\} = \{w \in \Sigma^* \mid \exists x \in L_2 \ \colon\ wx \in L_1\}$

where $\Sigma^*$ is the Kleene star on $\Sigma$ , $wL_2$ is the language formed by concatenating $w$ with each element of {{nowrap| $L_2$ ,}} and $\varnothing$ is the empty set.

In other words, for all strings in $L_1$ that have a suffix in $L_2$ , the suffix (right part of the string) is removed.

Similarly, the left quotient of $L_1$ with respect to $L_2$ is the language consisting of strings $w$ such that $xw$ is in $L_1$ for some string $x$ in $L_2$ . Formally:

$L_2 \backslash L_1 = \{w \in \Sigma^* \mid L_2w \cap L_1 \neq \varnothing\} = \{w \in \Sigma^* \mid \exists x\in L_2 \ \colon\ xw \in L_1\}$

In other words, for all strings in $L_1$ that have a prefix in $L_2$ , the prefix (left part of the string) is removed.

Note that the operands of $\backslash$ are in reverse order, so that $L_2$ preceeds $L_1$ .

The right and left quotients of $L_1$ with respect to $L_2$ may also be written as $L_1 L_2^{-1}$ and $L_2^{-1} L_1$ respectively.{{cite book |last=Simovici |first=Dan A. |date=2024 |title=Introduction to the Theory of Formal Languages |location=Singapore |publisher=World Scientific |isbn=978-9811294013 |doi=10.1142/13862 |pages=11–12}}

Example

Consider

$L_1 = \{ a^n b^n c^n \mid n \ge 0 \}$

and

$L_2 = \{ b^i c^j \mid i,j \ge 0 \}.$

If an element of $L_1$ is split into two parts, then the right part will be in $L_2$ if and only if the split occurs somewhere after the final {{nowrap| $a$ .}} Assuming this is the case, if the split occurs before the first $c$ then $0 \le i \le n$ and {{nowrap| $j=n$ }}, otherwise $i=0$ and {{nowrap| $0 \le j \le n$ .}} For instance:

$aa \mid bbcc \ \ (n=2, i=j=2)$

$aaab \mid bbccc \ \ (n=3, i=2, j=3)$

$aabbcc \mid \epsilon \ \ (n=2, i=j=0)$

where $\epsilon$ is the empty string.

Thus, the left part will be either $a^n b^{n-i}$ or {{nowrap| $a^n b^n c^{n-j}$ }} {{nowrap| $(0 \le i,j \le n$ ),}} and $L_1 / L_2$ can be written as:

$\{\ a^p b^q c^r \ \mid\ (p \ge q \text{ and } r = 0) \ \ \text{or} \ \ p = q \ge r \ \ ; \ \ p, q, r \ge 0 \ \}.$

Basic properties

If $L, L_1, L_2$ are languages over the same alphabet then:

$(L_1 \cup L_2) / L \ = \ L_1 / L \ \cup \ L_2 / L$

$(L_1 \cup L_2) \backslash L \ = \ L_1 \backslash L \ \cup \ L_2 \backslash L$

$(L_1 \cap L_2) / L \ \subseteq \ L_1 / L \ \cap \ L_2 / L$

$(L_1 \cap L_2) \backslash L \ \subseteq \ L_1 \backslash L \ \cap \ L_2 \backslash L$

$L \backslash (L_1 \cup L_2) \ = \ L \backslash L_1 \ \cup \ L \backslash L_2$

$L \backslash (L_1 \cap L_2) \ \subseteq \ L \backslash L_1 \ \cap \ L \backslash L_2$

$L_1 / L - L_2 / L \ \subseteq \ (L_1 - L_2) / L$

$L \backslash L_1 - L \backslash L_2 \ \subseteq \ L \backslash (L_1 - L_2)$

=Example proof=

As an example, the third property is proved as follows:

If {{nowrap| $w \in (L_1 \cap L_2) / L$ ,}} there exists $x \in L$ such that {{no wrap| $wx \in L_1 \cap L_2$ .}} Since then $wx \in L_1$ and {{nowrap| $wx \in L_2$ ,}} it must be that {{nowrap| $w \in L_1 / L \cap L_2 / L$ .}} Conversely, let $w \in L_1 / L$ and {{nowrap| $w \in L_2 / L$ ,}} then there exists $x_1, x_2 \in L$ such that $wx_1 \in L_1$ and $wx_2 \in L_2$ (given {{nowrap| $w$ ,}} if $L_1 \neq L_2$ then $x_1,x_2$ may differ). Now $wx_1 \in L_1 \cap \ L_2$ and $wx_2 \in L_1 \cap \ L_2$ only if {{nowrap| $x_1 = x_2$ ,}} hence {{nowrap| $(L_1 \cap L_2) / L \subseteq L_1 / L \cap L_2 / L$ .}}

For instance, let $L_1 = \{aab, bbb\},$ $L_2 = \{abb, bbb\},$ {{nowrap| $L = \{ab, bb\}$ .}}

Then {{no wrap| $L_1 \cap L_2 = \{bbb\}$ }}, hence {{nowrap| $(L_1 \cap L_2) / L = \{b\}$ .}}

Also $L_1 / L = \{a, b\}$ and {{nowrap| $L_2 / L = \{a, b\}$ }}, hence {{nowrap| $L_1 / L \cap L_2 / L = \{a, b\}$ .}}

Relationship between right and left quotients

The right and left quotients of languages $L_1$ and $L_2$ are related through the language reversals $L_1^R$ and $L_2^R$ by the equalities:

$L_1 / L_2 = (L_2^R \backslash L_1^R)^R$

$L_2 \backslash L_1 = (L_1^R / L_2^R)^R$

=Proof=

To prove the first equality, let {{nowrap| $w \in L_1 / L_2$ .}} Then there exists $x \in L_2$ such that {{nowrap| $wx \in L_1$ .}} Therefore, there must exist $y \in L_2^R$ such that {{nowrap| $yw^R \in L_1^R$ ,}} hence by definition {{nowrap| $w^R \in L_2^R \backslash L_1^R$ .}} It follows that {{nowrap| $w \in (L_2^R \backslash L_1^R)^R$ ,}} and so {{nowrap| $L_1 / L_2 = (L_2^R \backslash L_1^R)^R$ .}}

The second equality is proved in a similar manner.

Other properties

Some common closure properties of the quotient operation include:

The quotient of a regular language with any other language is regular.
The quotient of a context free language with a regular language is context free.
The quotient of two context free languages can be any recursively enumerable language.
The quotient of two recursively enumerable languages is recursively enumerable.

These closure properties hold for both left and right quotients.

References

Category:Formal languages