7:38 Set difference would contain elements which are only in L1 and not in L2, as E is the only such element won't grammar be L1 - L2 = {a^n b^m/ n,m>= 1} I think sir did intersection but that would still give {E}
U - union n - Intersection E- NULL L1 U L2 = E* ( Explained well) but how come L1 n L2 = PHI ? the example given at 39.40 for which the intersection comes out to be {E} at 7.38 too, sir did similar kind of thing. If anyone could answer, please explain it for better understanding.
may it help you understand it better : let the working set for us is { 1, 2,3,4,5,6} , then let a set A in this domain is {1, 3 } then the A^c will be { 2,4,5,6} so if we do intersection of both A and A^c , it will have nothing common that is {E} , so on the same concept you may relate that to the language , if that make sense , and as we know phi is regular itself , so it align with our purpose of question , . ........... If i am wrong , feel free to correct me , that will be helpful
7:38
Set difference would contain elements which are only in L1 and not in L2, as E is the only such element won't grammar be
L1 - L2 = {a^n b^m/ n,m>= 1}
I think sir did intersection but that would still give {E}
yes you are right.... to be more precise L1-L2={a^n b^m/ n,m>= 0}-{E}
L1 - L2 != {a^n b^m/ n,m>= 1} as strings like aa,bb will not get accepted here
Was about to comment the same!!!
@@sumitdas962 is this is CFL ?
@@NupurDas-vj6cl it is cfl as well as regular.
\Sigma_1 = {a,b} L1 = {a^n b^m | n,m >=0}, \Sigma_2 = {c,d} L2 = {c^n d^n | n >=0}.
L1 - L2 = L1 \cap L2^c
= L1 \cap (L3 := \Sigma_2^* - c^n d^n)
Here, we have nothing in common L1 and L3, so L1 - L2 = L1 - L3 = \phi.
Guys , 40:59 take L1 n,m>=1 . That way, L1 will be regular also and L1 intersection L2 will not be null.
39:40 L1 intersection L2={E} where E=null or epsilon
thnkyu very much sir
Apka kitna rank aaya
Amazing Lecture 🥺❤️😘🔥
Apka kitna rank aaya
41:43
Amazing
nice👏😊👍
At 1:15:00
what about "aaabba"? n=3, m+i=2+0, j=1.
Masta sir 😀😀
Superb analysis of questions.
Thanks sirji, really incredible teaching ways sirji 💓🙏
Apka gate me kitna rank aaya
Thank u sir. I'm new subscribers
U - union n - Intersection E- NULL
L1 U L2 = E* ( Explained well)
but how come L1 n L2 = PHI ? the example given at 39.40 for which the intersection comes out to be {E}
at 7.38 too, sir did similar kind of thing.
If anyone could answer, please explain it for better understanding.
may it help you understand it better : let the working set for us is { 1, 2,3,4,5,6} , then let a set A in this domain is {1, 3 } then the A^c will be { 2,4,5,6} so if we do intersection of both A and A^c , it will have nothing common that is {E} , so on the same concept you may relate that to the language , if that make sense , and as we know phi is regular itself , so it align with our purpose of question , . ........... If i am wrong , feel free to correct me , that will be helpful
7:20 L1-L2= (a^nb^m | n,m>=1) ....? is it right? How L1-L2=empty
24:36 Sir can you please explain how compliment of CFL is CFL and not CSL in this case?
Compliment is any no of as, following any no of b's but n!=m.....since there's one comparison it's a cfl
sir how much TOC syllabus is left after this?
Apka kitna rank aaya
this video loops some content twice