.. _RegGrammar:
.. raw:: html
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.. |---| unicode:: U+2014 .. em dash, trimming surrounding whitespace
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.. odsalink:: AV/VisFormalLang/Regular/RILinearGramNFACON.css
.. odsalink:: DataStructures/FLA/FLA.css
.. odsalink:: AV/VisFormalLang/Regular/REtoFACON.css
.. odsalink:: AV/VisFormalLang/Regular/RlRegGramDFACON.css
.. odsalink:: AV/VisFormalLang/Regular/FAtoRegGrammmarCON.css
.. This file is part of the OpenDSA eTextbook project. See
.. http://algoviz.org/OpenDSA for more details.
.. Copyright (c) 2012-2016 by the OpenDSA Project Contributors, and
.. distributed under an MIT open source license.
.. avmetadata::
:author: Susan Rodger, Cliff Shaffer, and Mostafa Mohammed
:requires:
:satisfies: Regular Grammar
:topic: Finite Automata
Regular Grammars
================
Regular Grammars
----------------
Regular grammars are another way to describe regular languages.
Recall that a grammar is made of of terminals, variables, and
production rule.
As the name implies, a **regular** grammar is a special type of
grammar (we will see plenty of grammars later that are not regular).
Which begs the question: What makes a grammar regular?
Before we can properly answer that, we need some definitions.
Suppose we have the following Grammar :math:`G = (V, T, S, P)` where,
.. math::
\begin{array}{lll}
& & \mbox{represented by} \\
V & \mbox{variables (nonterminals)} & A,B,..,Z (that is, capital letters)\\
T & \mbox{terminals} & a,b,..,z,0,1,...9 (lower case letters and numbers)\\
S & \mbox{start symbol} \\
P & \mbox{productions (rules)}\\
\end{array}
:math:`V`, :math:`T`, and :math:`P` are finite sets.
:term:`Linear grammar`:
A grammar is linear if has a single variable
in the RHS of every production rule.
.. math::
\begin{array}{c}
\mbox{All productions are of the form} \\
A \rightarrow xB \\
A \rightarrow Cx \\
A \rightarrow x \\
\mbox{where}\ A,B,C \in V, x \in T^*
\end{array}
In this grammar, each production rule has at most one variable on the
RHS.
(Note: It does not need to be the same terminal :math:`x` in every production!)
:term:`Right-linear grammar`: This is a special case of linear
grammar.
If a grammar is linear and any variable, if it exists, always occurs at
the right end of the RHS, then the grammar is called a Right-linear grammar.
For example, the grammar:
.. math::
\begin{array}{c}
A \rightarrow xB \\
A \rightarrow xC \\
A \rightarrow y \\
\mbox{where}\ A,B,C \in V, x,y \in T^*
\end{array}
Note: :math:`x` or :math:`y` is a string of length 0 or more.
Each production has at most one variable, so the grammar is linear.
The variables (B and C) are at the end of the RHS,
so it is a right linear grammar.
:term:`Left-linear grammar`: This is the same as :term:`Right-linear grammar`,
but the occurance of any variable, if it exists, is at the begining of each
production RHS.
For example,
.. math::
\begin{array}{c}
A \rightarrow Bx \\
A \rightarrow Cy \\
A \rightarrow x \\
\mbox{where}\ A,B,C \in V, x,y \in T^*
\end{array}
Each production has at most one variable, so the grammar is linear.
The variables (B and C) are at the start of the RHS,
so it is a left linear grammar.
OK, so now we have the definitions that we need to define a regular
grammar.
**Definition:**
A :term:`regular grammar` is a right-linear or left-linear grammar.
..
.. .. note::
..
There is a more restrictive definition in which the length of
:math:`x` is :math:`\leq 1`. (Exercise in book.)
.. topic:: Example
.. math::
\begin{eqnarray*}
G &=& (\{S\},\{a,b\},S,P),\\
P &=& \\
&&S \rightarrow abS \\
&&S \rightarrow \lambda \\
&&S \rightarrow Sab \\
\end{eqnarray*}
The language is :math:`(ab)*`.
However, cannot mix left/right rules!
So this is not a regular grammar.
.. topic:: Example
.. math::
\begin{eqnarray*}
G &=& (\{S\},\{a,b\},S,P),\\
P &=& \\
&&S \rightarrow aB | bS | \lambda \\
&&B \rightarrow aS | bB \\
\end{eqnarray*}
This is a right linear grammar representing the language
:math:`L = \{ \mbox{strings with an even number of a's}\}, \Sigma = \{a,b\}`
Our Next Step
~~~~~~~~~~~~~
| What we have already done:
| Definition: DFA represents regular language
| Theorem: NFA :math:`\Longleftrightarrow` DFA
| Theorem: RE :math:`\Longleftrightarrow` NFA
| What we will do next:
| Theorem: DFA :math:`\Longleftrightarrow` regular grammar
NFA from Regular Grammar
~~~~~~~~~~~~~~~~~~~~~~~~
**Theorem:** L is a regular language if and only if there exists a regular
grammar :math:`G` such that :math:`L = L(G)`.
| (Doing here for RR grammar)
| (:math:`\Longleftarrow`) Given a regular grammar G,
Construct NFA M such that :math:`L(G)=L(M)`
| Make a state for each non-terminal.
| Make a transition on each terminal in that production rule.
| Make it final if there is a production without non-terminals.
| For rules with multiple terminals, need intermediate states.
.. topic:: Example
| :math:`S \rightarrow aB | bS | \lambda`
| :math:`B \rightarrow aS | bB`
|
| This is a right linear grammar representing the language
| :math:`L = \{` strings with an even number of a's :math:`\}, \Sigma = \{a,b\}`
.. inlineav:: RILinearGramNFACON dgm
:align: center
:output: show
What about a rule like :math:`S \rightarrow abB`?
Make two states (S to intermediate state on a, then intermediate
state to B on b).
Or :math:`S \rightarrow ab`?
Make two states (S to intermediate state on a, then intermediate
state to an accepting state on B.
.. inlineav:: REtoFACON ss
:points: 0.0
:required: False
:threshold: 1.0
:id: 188699
:output: show
Right-linear Regular Grammar from DFA
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
**Theorem:** L is a regular language if and only if there exists a
regular grammar G such that :math:`L = L(G)`.
(:math:`\Longrightarrow`) Given a DFA :math:`M`,
construct regular grammar :math:`G` such that :math:`L(G)=L(M)`
| The process is pretty much the same as when we made an NFA from
RRG:
| Each DFA state gets a non-terminal.
| Each transition gets a production rule.
Construct the Regular Grammar for the NFA
.. inlineav:: RlRegGramDFACON dgm
:align: center
:output: show
| :math:`G = (\{S,B\},\{a,b\},S,P)`,
| :math:`P =`
| :math:`Q0 \rightarrow a Q1`
| :math:`Q1 \rightarrow a Q0 | b Q1 | \lambda`
.. inlineav:: FAtoRegGrammmarCON ss
:points: 0.0
:required: False
:threshold: 1.0
:id: 188702
:output: show
Something to Think About
~~~~~~~~~~~~~~~~~~~~~~~~
.. topic:: Example
:math:`L = \{a^nb^n \mid n>0\}`
Is language :math:`L` regular?
Can you draw a DFA, regular expression, or Regular grammar for this
language?
Consider this grammar:
:math:`S \rightarrow aSb \mid ab`
Nice and easy... but this grammar is not regular!
We will come back to this question later.
.. odsascript:: AV/VisFormalLang/Regular/RILinearGramNFACON.js
.. odsascript:: lib/underscore.js
.. odsascript:: DataStructures/FLA/Discretizer.js
.. odsascript:: DataStructures/FLA/FA.js
.. odsascript:: AV/VisFormalLang/Regular/REtoFACON.js
.. odsascript:: AV/VisFormalLang/Regular/RlRegGramDFACON.js
.. odsascript:: AV/VisFormalLang/Regular/FAtoRegGrammmarCON.js