'Coding has over 700 languages', '67% of programming jobs aren’t in the technology industry', 'Coding is behind almost everything that is powered by electricity'
Last Updated: Dec 20, 2023

Simplification of CFG(Context-Free Grammars)

Data structures & algorithms (Beginner to Intermediate)
Free guided path
13 chapters
99+ problems
Earn badges and level up
Simplification of cfg


As we know, various languages can efficiently be represented by context-free Grammar (CFGs). The Grammar is not always optimized, consisting of some extra symbols(non-terminal). Having additional symbols unnecessarily increases the length of Grammar. 

By simplifying Context-Free Grammars, we can remove all the unnecessary symbols while maintaining the converted grammar equivalent to the original Grammar. If two grammars produce the same language, they are equivalent.  

Simplifying Grammars is required to convert them to Normal forms later.In this article we will cover Simplification of CFG(Context-Free Grammars).

Also see, Turing Machine in TOC.

What is the Simplification of CFG?

It is possible in a CFG that the derivation of strings does not require the use of all the production rules and symbols. Moreover, there might be some unit and null productions. The term "simplification of CFGs" refers to the removal of certain productions and symbols.
Context-Free Grammar can be made simpler by removing all the extraneous symbols while yet preserving a converted grammar that is equivalent to the original Grammar.

  1. Each variable (i.e., non-terminal) and terminal of G is used to derive some word in language L.
  2. There should not be any production like X → Y when X and Y are non-terminal.
  3. If ε is not in L, then the production X → ε is unnecessary.

Let’s discuss all the 3 types of reduction processes in detail.

  1. Useless productions 
  2. ε productions
  3. Unit productions
Simplification of CFG

Removal of Useless Symbols

If a symbol does not exist on the right-hand side of the production rule and does not participate in the derivation of any string, that symbol is considered a useless symbol. Similarly, if a variable does not participate in the derivation of any string, it is useless. 


For example, simplify the below-given Grammar G:

S → aaB | abA | aaS  
A → aA  
B → ab | b  
C → ae 


In the above example, first, find the useless productions,

  1. The variable 'C' will never appear in the derivation of any string, so the production C -> ae is useless. Therefore, we will remove it, and the other productions are written so that variable C can never reach from the initial variable 'T.'
  2. Production A -> aA is also useless because there is no way to end it. If it never ends, it can never produce a string. As a result, this production can never participate in any derivation.


The process to remove the above found useless productions:

  1. To eliminate useless productions like A -> aA, we first find all variables that will never lead to a terminal string, such as 'A.' We remove all of the productions in which variable 'B' appears. As a result, the modified grammar:
S → aaB | aaS   
B → ab | b  
C → ae


2. Then we try to find all the variables that can never be reached from the initial variable S, such as 'C.' We remove all of the productions in which variable 'C' appears.

The grammar G is now free of all the useless productions.

S → aaB | aaS  
B → ab | b  

Also read - Arden's theorem

Get the tech career you deserve, faster!
Connect with our expert counsellors to understand how to hack your way to success
User rating 4.7/5
1:1 doubt support
95% placement record
Akash Pal
Senior Software Engineer
326% Hike After Job Bootcamp
Himanshu Gusain
Programmer Analyst
32 LPA After Job Bootcamp
After Job

Elimination of ε Production

Productions of type 'A ->ε' (also known as null productions) is called ε productions. These productions can only be eliminated from grammars that do not generate ε (an empty string). 

Steps to remove ε Productions

Step 1:- First, find all nullable non-terminal variables that derive ε. A variable 'A' is nullable if ε can be derived from 'A.' For any productions of type 'B -> A1A3A3...An', where all 'Ai's are nullable, 'B' is also nullable.

Step 2:- Construct all productions A -> x for each production A -> a, where x is obtained from a by eliminating one or more non-terminals from step 1.

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

Confused? Let’s understand with an example,


Consider the Grammar G:

S -> ABCd                        
A -> BC                                
B -> bB | ε                      
C -> cC | ε  


Find all of the nullable variables. Variables 'B' and 'C' are nullable because they have ε on the RHS of their production. Variable 'A' is also nullable because both variables on the RHS are nullable in A -> BC. Variable 'S' is similarly nullable. As a result, variables 'S,' 'A,' 'B,' and 'C' are all nullable.

Let's make a new grammar. 

  • We begin with the initial production. Include the initial production as it is. Then we generate all available combinations by replacing the nullable variables with ε. 
  • As a result, line (1) is now:
    • 'S -> ABCd | ABd | ACd | BCd | Ad | Bd |Cd | d.' 
  • We apply the same logic to the following lines, but we do not include 'A -> ε.' We eliminate all productions of the type 'V -> ε.' 


The new grammar is now –

S -> ABCd | ABd | ACd | BCd | Ad  |  Bd  |Cd | d
A -> BC | B | C
B -> bB | b
C -> cC | c

Removing Unit Productions

Unit productions are those in which one non-terminal produces another non-terminal. Unit productions are productions of the type 'A -> B.'

To eliminate unit production, take the following steps:

  1. Step 1:- To remove A -> B, add the production A -> a whenever Y -> a appears in the Grammar.
  2. Step 2:- Remove A -> B from the grammar.
  3. Step 3:- Repeat steps 1 and 2 until all unit productions have been removed.


S → 0A | 1B | C  
A → 0S | 000  
B → 11 | A  
C → 01  


S -> C denotes a unit production. However, before we remove S -> C, we must consider what C provides. As a result, we can add a rule to S.

S → 0A | 1B | 01  


Similarly, B -> A is a unit production; we can change it as

B → 11 | 0S | 000  


As a result, we can finally write CFG without unit production as

S → 0A | 1B | 01   
A → 0S | 000  
B → 11 | 0S | 000   
C → 01  

Read About - Context-Free Grammar and Language

Benefits of Simplifying CFGs

  • Improved Readability: Simplified CFGs are easier to understand and analyze, making it more manageable for developers to work with complex grammars.
  • Enhanced Debugging: Fewer rules and symbols reduce the chances of errors in the grammar and make debugging and troubleshooting more efficient.
  • Better Performance: Simplified CFGs often lead to faster parsing and processing of input, resulting in improved performance for language processing tasks.

Disadvantages to simplification of CFGs

  • Loss of Expressiveness: Over-simplification may lead to a loss of expressive power in the grammar, limiting its ability to describe complex languages accurately.
  • Reduced Precision: Simplifying CFGs may result in less precise parsing, potentially allowing ambiguous or unintended interpretations of input.
  • Increased Complexity: Overly simplified grammars might require more complex semantic analysis to capture the intended meaning of sentences.

Tips to minimize the disadvantages of simplifying CFGs

  • Careful Refinement: Instead of excessive simplification, refine the grammar thoughtfully to strike a balance between simplicity and expressiveness.
  • Thorough Testing: Extensively test the simplified CFG to ensure it accurately handles a variety of inputs and edge cases.
  • Documentation: Provide comprehensive documentation for the grammar to clarify any potential ambiguities or limitations resulting from simplification.

Frequently Asked Questions

What is the formula of CFG?

A four-tuple G = (N,Σ, S, P) is a context-free grammar (CFG) G. Where N is a finite set. Σ is a finite set. P defines the set of production rules. And S is the start non-terminal. 

Why simplify context-free grammar?

In a CFG, it is feasible that not all of the production rules and symbols must be used to derive strings. Because not all grammar is streamlined, some grammar may contain superfluous symbols (non-terminal). Adding extra, pointless symbols lengthens the grammar.

What is simplification of context-free grammar?

Simplification of context-free grammar involves reducing its complexity by removing redundancy, ambiguity, and unnecessary rules while preserving its ability to define the intended language.

What are the steps in simplification of CFG?

1. Remove unreachable symbols.
2. Eliminate epsilon productions.
3. Remove unit productions.
4. Eliminate left recursion.
5. Factor and merge common productions.


In this article, we learned about simplifying the Simplification of CFG(Context-Free Grammars). We learned that in simplified Context-Free Grammar, we remove all the unnecessary symbols from the Grammar, maintaining the meaning the same as that of the original Grammar.

We learned that we could simplify Context-Free Grammar in three ways.

Recommended Readings:

You can use Coding Ninjas Studio to learn about various new concepts in Computer Science. Also check out the Interview guide for Product Based Companies as well as some of the Popular interview problems from top tech companies like Amazon, Adobe, Google, Uber, Microsoft, etc.

Refer to our guided pathways on Code studio to learn more about DSACompetitive ProgrammingJavaScriptSystem Design, etc. Enroll in our courses, and use the accessible sample exams and questions as a guide. For placement preparations, look at the interview experiences and interview package.

Previous article
Difference between Ambiguous and Unambiguous Grammar
Next article
Chomsky Normal Forms(CNF)
Guided path
Data structures & algorithms (Beginner to Intermediate)
13 chapters
109+ Problems
Earn badges and level up
Live masterclass