Lec-32: Closure properties of regular languages in TOC

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The video explains the properties of regular languages and their closure under various operations, including complementation and intersection. It also covers the use of finite automata in regular language processing. [00:00] Regular language properties are important in TOC and competitive exams - Regular languages have properties like union, concatenation, complementation, intersection, difference, reversal, homomorphism, and quotient - Teasers and closed captioning are examples of regular language properties - Regular languages can be finite or infinite and are used in computer science and linguistics [03:07] Regular languages have a closure property and can be represented by finite automata. - Regular languages have a closed operation that accepts the last word - Regular languages are closed under finite union and concatenation - Finite automata can be used to represent regular languages [06:15] Regular languages and finite automata design concepts are discussed in detail - Regular languages can be represented by finite automata - Designing a DFA involves selecting a start state and final state - Complementation of regular languages involves inverting the final and non-final states - The intersection and union of regular languages are also closed under regular languages

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Thanks Sir ... VERY WELL EXPLAINED 😊

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[@yash1152]

8:08 DFA का complement कैसे करके हैं का विवरण

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[@revivingfaith17]

Property 4. The difference of two regular set is regular. Proof − Let us take two regular expressions − RE1 = a (a*) and RE2 = (aa)* So, L1 = {a, aa, aaa, aaaa, ....} (Strings of all possible lengths excluding Null) L2 = { ε, aa, aaaa, aaaaaa,.......} (Strings of even length including Null) L1 - L2 = {a, aaa, aaaaa, aaaaaaa, ....} (Strings of all odd lengths excluding Null) RE (L1 - L2 ) = a (aa)* which is a regular expression. Hence, proved Pls explain this sir

[@spiral546]

on subtracting L1 and L2, we get -ε, but you did not write it?

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Thanku so much sir ji

[@arnabseal9457]

Complementation is defined by difference Intersection is defined by complementation Difference is defined by intersection. Then it is a loop right?

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Sir are you going to cover compiler design in the TOC tutorial also ? Please reply sir

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[@yosh6458]

thanks sir

[@juanjosemorenogiraldo2066]

I wish there was a video like this in full english :(

[@revivingfaith17]

Are you clear all concepts but not difine how to proof this theorem , I don't know ho to proof difference theorem if any body knows pls described for me . Lots of thanks 😊👍

[@simransingh6137]

Thank you sir

[@divyar9150]

If L1=a^n and L2= a^n. Then whether a^n.b^n is regular?

[@AbhishekTiwari-xt1kt]

Yes but a^n is not regular

[@thepriestofvaranasi]

@@AbhishekTiwari-xt1kt It is regular.

[@Patrick-hb9dbj]

#Conceptual_doubt if L1 =lamda, L2={ab} L3={aabb} L4={aaabbb} L5={aaaabbbb} and so on.. then L1 U L2 U L3 U ........ will give l={a^n b^n ; n>=0} which is basically non regular language. then how can we say that union is closed under union?? kindly help!

[@treexs2353]

We will get a^nb^m. So it will be regular

[@rishisahu193]

7:33 there should be loop of 'a' not Epsilon!!

[@mdgulmoajjamfaizy5379]

Given two regular languages L1, L2, prove by construction that : i) L1^R n L2^R is regular, ii) L1* U L2 ^R is regular. Plice sir solution send me

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[@shyamsharma1995]

Sir subset ka liya bhi not closed hai.

[@_glock_69_]

Sir bas ek topic missing hai and that is arden's theorem

[@lalitkumar8445]

Sir in 3rd property i.e closure of a language is regular. U took L= {a} then then L*= a*= €,a,aa,aaa..... Now after apply * operation on L......L* all values not belongs to L so how it's closure? Pls explain how it works here🙏

[@Patrick-hb9dbj]

Sir, when we say Regular languages are closed under union then it implies that union of 2 or more Regular languages is always regular. but union of infinite regular languages might not be regular. #Conceptual_doubt if L1 =lamda, L2={ab} L3={aabb} L4={aaabbb} L5={aaaabbbb} and so on.. then L1 U L2 U L3 U ........ will give l={a^n b^n ; n>=0} which is basically non regular language. then how can we say that union is closed under union?? kindly help!

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link to/name of next video

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[@aabid3375]

Regular Languages are not closed under infinite union...

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Yes he have said it

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Sir subset bhi closed nhi hai

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thanks sir

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L=L* equal hota hai ke nahi

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