Operations on Regular Languages
Вставка
- Опубліковано 16 жов 2024
- TOC: Operations on Regular Languages in Theory of Computation.
Topics Discussed:
1. Union operation on regular languages.
2. Concatenation operation on regular languages.
3. Star operation on regular languages.
4. Theorems on Union and Concatenation.
Full Course on TOC: goo.gl/f4CmJw
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#TheoryOfComputation #TOCByNeso #RegularLanguages #AutomataTheory
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Theorem: The class of regular languages is closed under union and intersection. That
is, if L1 and L2 are regular languages, the so are L1 ∪ L2 and L1 ∩ L2.
Proof Lets start with the union. For simplicity, let us assume that L1 and L2
are languages over the same alphabet Σ.
Since L1 is regular, there exists a DFA
M1 = (Q1, Σ, δ1, q1, F1) which recognizes L1.
Similarly, there exists a DFA M2 =
(Q2, Σ, δ2, q2, F2) which recognizes L2.
To prove that L1∪L2 is regular, we will construct a DFA M∪ which recognizes L1∪L2 =
{w|w ∈ L1 or w ∈ L2}.
The idea: M∪ = (Q, Σ, δ, q0, F) simulates a parallel execution of M1 and M2. M∪ is
defined as follows:
- Q = Q1 × Q2;
- Σ is the same;
- δ((q1, q2), a) = (δ(q1, a), δ(q2, a));
- q0 = (q1, q2);
- F = {(r1, r2) | r1 ∈ F1 or r2 ∈ F2}
To prove correctness we need to show that w ∈ L1 ∪ L2 if and only if M∪ accepts w.
This, in turn, follows from the fact that (r0, r1, . . . , rn) is a computation of M1 on w, and
(t0, t1, . . . , tn) is a computation of M2 on w, if and only if ((r0, t0),(r1, t1), . . . ,(rn, tn))
is a computation of M∪ on w.
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@@deepakchand6360 Respect and good wishes
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I finished this lecture just now. Please upload those two theorem proofs in Another video, it will be helpful for us.
For proof that Regular Languages are closed under Union Operation see Theorem 1.25 on page 45 of Sipser's textbook. For proof that Regular Languages are closed under Concatenation Operation see Theorem 1.26 of Sipser's textbook on page 47.
Sipser no make sense.
Sir..your videos are great.These helped me a lot.Will it be possible for you to upload videos on C programming,data structures and algorithm?
Hello. I think u messed up the Kleene star operation. It is not just a concatenation of all elements of A. It contains strings of 0 lengths, 1, 2, and so on. So I guess they should even be separated by commas. I suggest it would be better to use union notation. A mathematical definition would suffix for this, if L = {a,b}, L* = {@, a, b, aa, ab, ba, bb, aaa, ...} not just a concatenation of symbols from the alphabet that L is defined. {bbabababbabaabababa} for example is just an element of L*
Thankyou !!
Thank you for the content, as always. Very easy to follow and understand
Thank you ,sir . Very easily understandable . But can you please provide the proofs of the theorems also ?
Thank you so much sir
Welcome beta
please upload a video on proofs of the theorms at last
for union purpose dont you think we should mention and like x belongs to a and x belongs to b. please correct me if i am wrong.
Hello, question not specific to current subject but I must ask you:
Will you record data structures and algorithms tutorial series in the future?
Wow, nice! I would like to see this series because you are great.
Please bring up the algorithm part as quick as you can. It would be very kind of you.
Awesome! Where can we find these videos?
for proof of the following theorems I will use the fact that regular languages are those that can be described by regular expression, let's say we have 2 regular languages A and B, and their regex be p and q, then concatenation of those languages can be described by pq (first part will describe string from A and second part string from B), as for union we can use | operator for regular expressions and describe language by (p | q) since union contains strings that are present in both languages.
Id like proofs of those 2 theorems please. Thank you for this video as well sir!!!
Should the AUB and A•B included the Epsilon ?
Such that
AUB ={ε,pq ,r,t,uv}
A•B={pqt,pquv,rt,ruv, ε,pq(=pqε) ,r,u,t}
No
For anyone wondering about this in the future, and to be more helpful than just "No", only A* can contain epsilon because in the definition it says there are k elements of A where k can be any positive number including zero. k being zero is how you get epsilon aka no elements at all.
sir give me the proof of this
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Sir please make a video on theorems.🙏🏻
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Sir please record videos on Computer Networks of CS branch
Sir please make a video of proving the
all theorems of regular languages
dear sir kindly explain the theorem of class of regular language is closed under the "kalena" star operation
kleene*
please upload videos of turing machine,recursive and recursively enumerable language
Best Tuts Ever........!!!!
hi sir,
i want
proof of theorems
Sure sir can you make a video for Proof of Theorem1 and Theorem2
Please make a video on proof of the theorems
Helps me alot....tnk u sir ji
This is so well explained ❤️❤️❤️❤️❤️❤️❤️
hii
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Amazing! thank you
concatenation: cartesian prod, order of elements in pair immaterial
hi sir nice TOC series...plz also upload profs of these theorems...?
thanks in advance...
Union, Concatenation & star.
Thanks you sir
Sir please provide proof for those 2 theorems, imp for exam
The star one is also called kleen closure .
sir, could you please explain proof of two theorems
We want proof...... Can you please explain
Really appreciate if you could do the proof too
Can you explain to me this example, please?
A*={a*,b*,a*,b*}
sir can I use this answer for closure properties of regular sets.
what is the significance of putting k => 0 for A*?
nice concept
Thankyou sir
please please make vedios on data structures
Love from muzaffer garh Pakistan
dear sir are you also teaching any other subjects for cse students
Yes 👍
How to visualise practically the concept of regular languages
THANK YOU..
Sir, can you provide the proof of Regular language expression?
sir i can find one ambiguity over here.When you are declaring definition of star,i.e. A*={x1x2x3...xk:k>=0 and all xi belongs to A}.So when k=0,what would be the concatenating string?and if you are saying it is a null string then again ambiguity arise because as you said all xi,including x0 should belong to A,but that is not in the actual case.Please make it clear.
Kishan Kanabar Hey... Are you not aware of the fact that all sets have a null element i.e. null element belongs to each and every set....
sir i want regular language's many types of problem with solution for better understand.thank you.
Sir pls give the proof for theorems
Sir I need proves of two theorem of regular languages. If you send me then I will be thankful to you.
pls clarify our doubts
If possible please give the proof of theorems.
Sir plz..show some prove. Of this theorem
plz.. explain proofs of those theorems
If L={ab, c}...what will be L^3....??
Can i please get a proof of the two theoram?
Sir please explain Kleen's theorem...
Sir , give us the prove of this 2 theorem ... please
In A* prq also be there
sir will pqpqr is also in A* ??
Yes
Video was awesome but You should provide proof to complete topic
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Set of regular languages over a given alphabet is not closed under
a) union b) complementation
c) intersection d) none of the above
what about (A+B)*
Sir please give the proof of the theorem.
Please prove the theorem 1& 2 sir
can we have the proofs please?
what is regular language
Thank you :)
What is it mean every one commenting like this
Sir please do proo
5:35/5:57 : 🦭🦭
❤❤❤
Proof of the therom sir pls
Proof for Theorem 1,Theorem 2 plz
thermos of regular languages expain sir
Explain hota hai 😜
sir proof please on union operation
my teacher plays your videos in lectures
sir proofs please
5:52 your device is fine
sir please upload proof of this video
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Proof please
Want the proof
Want proof
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Theorm : Stating the obvious facts with fancy english .
And fancy maths.
prooof