TheDeveloperBlog.com

Home | Contact Us

C-Sharp | Java | Python | Swift | GO | WPF | Ruby | Scala | F# | JavaScript | SQL | PHP | Angular | HTML

Automata Simplification of CFG

Automata Simplification of CFG with automata tutorial, finite automata, dfa, nfa, regexp, transition diagram in automata, transition table, theory of automata, examples of dfa, minimization of dfa, non deterministic finite automata, etc.

<< Back to AUTOMATA

Simplification of CFG

As we have seen, various languages can efficiently be represented by a context-free grammar. All the grammar are not always optimized that means the grammar may consist of some extra symbols(non-terminal). Having extra symbols, unnecessary increase the length of grammar. Simplification of grammar means reduction of grammar by removing useless symbols. The properties of reduced grammar are given below:

  1. Each variable (i.e. non-terminal) and each terminal of G appears in the derivation of some word in L.
  2. There should not be any production as X → Y where X and Y are non-terminal.
  3. If ε is not in the language L then there need not to be the production X → ε.

Let us study the reduction process in detail./p> Simplification of CFG

Removal of Useless Symbols

A symbol can be useless if it does not appear on the right-hand side of the production rule and does not take part in the derivation of any string. That symbol is known as a useless symbol. Similarly, a variable can be useless if it does not take part in the derivation of any string. That variable is known as a useless variable.

For Example:

T → aaB | abA | aaT
A → aA
B → ab | b
C → ad

In the above example, the variable 'C' will never occur in the derivation of any string, so the production C → ad is useless. So we will eliminate it, and the other productions are written in such a way that variable C can never reach from the starting variable 'T'.

Production A → aA is also useless because there is no way to terminate it. If it never terminates, then it can never produce a string. Hence this production can never take part in any derivation.

To remove this useless production A → aA, we will first find all the variables which will never lead to a terminal string such as variable 'A'. Then we will remove all the productions in which the variable 'B' occurs.

Elimination of ε Production

The productions of type S → ε are called ε productions. These type of productions can only be removed from those grammars that do not generate ε.

Step 1: First find out all nullable non-terminal variable which derives ε.

Step 2: For each production A → a, construct all production A → x, where x is obtained from a by removing one or more non-terminal from step 1.

Step 3: Now combine the result of step 2 with the original production and remove ε productions.

Example:

Remove the production from the following CFG by preserving the meaning of it.

S → XYX
X → 0X | ε
Y → 1Y | ε

Solution:

Now, while removing ε production, we are deleting the rule X → ε and Y → ε. To preserve the meaning of CFG we are actually placing ε at the right-hand side whenever X and Y have appeared.

Let us take

S → XYX

If the first X at right-hand side is ε. Then

S → YX

Similarly if the last X in R.H.S. = ε. Then

S → XY

If Y = ε then

S → XX

If Y and X are ε then,

S → X

If both X are replaced by ε

S → Y

Now,

S → XY | YX | XX | X | Y

Now let us consider

X → 0X

If we place ε at right-hand side for X then,

X → 0
X → 0X | 0

Similarly Y → 1Y | 1

Collectively we can rewrite the CFG with removed ε production as

S → XY | YX | XX | X | Y
X → 0X | 0
Y → 1Y | 1

Removing Unit Productions

The unit productions are the productions in which one non-terminal gives another non-terminal. Use the following steps to remove unit production:

Step 1: To remove X → Y, add production X → a to the grammar rule whenever Y → a occurs in the grammar.

Step 2: Now delete X → Y from the grammar.

Step 3: Repeat step 1 and step 2 until all unit productions are removed.

For example:

S → 0A | 1B | C
A → 0S | 00
B → 1 | A
C → 01

Solution:

S → C is a unit production. But while removing S → C we have to consider what C gives. So, we can add a rule to S.

S → 0A | 1B | 01

Similarly, B → A is also a unit production so we can modify it as

B → 1 | 0S | 00

Thus finally we can write CFG without unit production as

S → 0A | 1B | 01
A → 0S | 00
B → 1 | 0S | 00
C → 01





Related Links:


Related Links

Adjectives Ado Ai Android Angular Antonyms Apache Articles Asp Autocad Automata Aws Azure Basic Binary Bitcoin Blockchain C Cassandra Change Coa Computer Control Cpp Create Creating C-Sharp Cyber Daa Data Dbms Deletion Devops Difference Discrete Es6 Ethical Examples Features Firebase Flutter Fs Git Go Hbase History Hive Hiveql How Html Idioms Insertion Installing Ios Java Joomla Js Kafka Kali Laravel Logical Machine Matlab Matrix Mongodb Mysql One Opencv Oracle Ordering Os Pandas Php Pig Pl Postgresql Powershell Prepositions Program Python React Ruby Scala Selecting Selenium Sentence Seo Sharepoint Software Spellings Spotting Spring Sql Sqlite Sqoop Svn Swift Synonyms Talend Testng Types Uml Unity Vbnet Verbal Webdriver What Wpf