CHOMSKY NORMAL FORM (CNF) & CONVERSION OF CFG TO CNF IN AUTOMATA THEORY || CFG TO CNF || TOC
ฝัง
- เผยแพร่เมื่อ 11 ต.ค. 2024
- CHOMSKY NORMAL FORM (CNF)
For any non-empty context-free language L, there is a grammar G, such that
L(G) = L and each rule in G is of the form
1. A → a where a ∈ Σ, or
2. A → BC where neither B nor C is the start symbol, or
3. S → where S is the start symbol
STEPS TO BE FOLLOWED IN CONVERSION OF CFG TO CNF
Step 1: Eliminate start symbol from the RHS. If the start symbol S is at the right-hand side of any production, create a new production as.
S1 → S
Step 2: In the grammar, remove the null, unit and useless productions. (Simplification of CFG)
Step 3: Eliminate terminals from the RHS of the production if they exist with other non-terminals or terminals. For example, production A → aB can be decomposed as:
A → XB
X → a
Step 4: Eliminate RHS with more than two non-terminals. For example, A → BCD can be decomposed as:
A → YD
Y → BC
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Best explanation in youtube, understood very easily, thank you sir
One of the best explanation... Thanks sir🙏
In 17:28 it should be A-->B|S|epsilon right? because B is changed the 2nd time only
best explanation sir lot of thanks to you sir😍🤩🤩🤩🤩
Excellent explanation tq sir
Make me to understand it easily.thnks of u sir
Your videos are so good bro.... I want to talk you could you help me....
Best explanation sir
Best Explain in Hardwork
Thank you sir🙏
Thank you very much sir :))
Thank you sir. ❤
"Eliminate unit production" In step 3 you have done mistake.
How
best Explanation ever
Crazy explanation, thank you so much sir:)
THANK YOU SIR.
Thank you ❤sir
You missed S in eliminating unit production.
Really helped ❤️
Thank you sir 🙏🏻......
👌👌 explanation
Bhaiii dimaak kaa bharosaa ho gaya is conversion me🥲🥲
Thankyou so much sir
thank u siir
Super
Try to correct your mistake
s u did a mistake in 3rd step
Sir explain perfectly sir not understand
S-->aS/AB/epsilon
A-->epsilon
B--->epsilon
D--->b
can anyone plz say the answer for this
Step1 - removal of useless symbols
The D production is removed since it do not take place in any step of derivation of a string .
S-->aS | AB | ep
A-->ep
B-->ep
Step 2 - ep productions elimination
We've A-->ep....substitute in rule S-->AB...since A is ep...we get S-->B.
The obtained productions are
(since A-->ep is removed)
S--> aS | AB | B | ep
B-->ep
Now, we've B-->ep
substituting the B in S rules
S-->AB
-->A ep
-->A
Obtained productions are
(B-->ep is eliminated)
S--> aS | AB | B | A | ep
Now we've S-->ep
substituting it in S-->aS
we get S-->a
and S-->ep is eliminated
Obtained productions are
S--> aS | AB | B | A | a | ep
Step 3 ..elimination of unit productions
We've got S-->B & S-->A unit productions.
the values of A and B should be replaced with the rules ...since we don't have any rules or productions left with A and B.....the above mentioned 2 rules S-->A and S-->B can be simply eliminated...
The final productions obtained are S--> aS | AB | a .
@@laavanyasri4730can you help me
@@laavanyasri4730
As sir said in first step we have to right S1-->S because S is present on both RHS and LHS
What if we remove useless production after unit production
S-> aS|AB|a
Here then AB would also be a useless production
Thank you🙏 sir