Writing GNF grammar for a CFL

Question

Hello I would like to ask you this question.

I was supposed to compute (manually) a grammar in Greibach Normal Form, that generates the language

L = {a i b j c k | i + j = 2k and k >= 1}

I really have no idea. Can someone please help me?

Thanks in advance
Chriss

Answer 1

Context Free Grammar CFG for your language L a = {a i b j c k | i + j = 2k and k >= 1} L a = {a i b j c k | i + j = 2k and k >= 1} .

Below is answer with language L = {a i b j c k | i + j = k and k >= 1} L = {a i b j c k | i + j = k and k >= 1} .

CFG:

S --> aAc | bBc |
A --> aAc | B   |  ^
B --> bBc | ^ 

What is GNF?

One important form of CFG is Greibach Normal Form GNF:

   A --> aα   
   Where α ∈ V* (any number of variables including zero)

Note: nul ^ can't be a symbol on RHS of any production accept start symbol S with constrict that if S --> ^ is a production in grammar then S can't be appear on RHS of any other production in grammar.

Any CFG can be written in GNF form.

How to convert CFG in GNF?

Note in CFG I written above have nul productions A --> ^ and B --> ^ and a Unit production A --> B . Unit productions and nul productions are not allowed in GNF form. Although other productions can easily written in GNF form by introduction in GNF productions in grammar eg S --> aAc can be rewrite as S --> aAC and C --> c .

So below I am rewriting equivalent CFG for language and removing nul and unit productions called simplified CFG.

Simplified CFG:

S --> aAc | bBc | ac | bc
A --> aAc | bBc 
B --> bBc | bc

Now this grammar can easily converted into GNF form by introducing new GNF production C --> c and replace c by C in other production rules.

GFN for language L:

S --> aAC | bBC | aC | bC
A --> aAC | bBC 
B --> bBC | bC
C --> c

^{By mistake I written a wrong grammar I will update answer for language L _a}

Edit

L a = {a i b j c k | i + j = 2k and k >= 1} L a = {a i b j c k | i + j = 2k and k >= 1} .

CFG for L _a :

S --> aaAc | bbBc | abBc
A --> aaAc | B    | abBc |  ^
B --> bbBc | ^ 

Simplified CFG:

S --> aaAc | bbBc | abBc | aac | bbc | abc 
A --> aaAc | bbBc | abBc | aac | abc
B --> bbBc | bbc 

GFN for language L _a :

Add three new production rules: X --> a , Y --> b and Z --> c .

Change programmer and replace terminal by variables:

S --> aXAZ | bYBZ | aYBZ | aAZ | bYZ | aYZ  
A --> aXAZ | bYBZ | aYBZ | aXZ | aYZ
B --> bYBZ | bYZ
X --> a
Y --> b
Z --> c

Answer 2

About Below answer CFG of La = {ai bj ck | i + j = k and k >= 1} La = {ai bj ck | i + j = k and k >= 1} is wrong

Your Simplified CFG:

S --> aAc | bBc | ac | bc . A --> aAc | bBc . B --> bBc | bc .

Because the above language L = {ai bj ck | i + j = k and k >= 1} L = {ai bj ck | i + j = k and k >= 1} produce the language aabccc but your Simplified CFG not produce it .

Right CFG is

Simplified CFG

S --> aAc | bBc | ac | bc A --> aAc | bBc | bc | ac B --> bBc | bc

Correct it, Second language has the same problem.

Thanks!