If (a+b) means a OR b, then how does (a+b)* gives a string ab, why is it not just a's OR b's, how does it become a combination ? Kindly solve my doubt if anyone can explain.
Bro (a+b)* denotes all the strings possible of a and b of any length. Eg put 2 in place of *. Therefore (a+b)(a+b) when u expand it aa+ab+ba+bb Hope u got it.
* can be replaced with 0,1,2,3,4....,n ; + represents logical OR ; so when * = 0, it accepts epsilon ; * = 1, it accepts 'a' OR 'b' ; * = 2, it accepts 'aa' OR 'ab' OR 'ba' OR 'bb' ; . . until * = n; therefore (a+b)* is a reg expression which accepts all strings of all length containing a,b
sir plz tell the sol. to the following ques- the set of all strings with atmost one pair of consecutive zero and one pair of consecutive ones. set of all strings not containing 101 as substring.
if we use above expression (b+ab)*(Ɛ+a) would it able to generate baba, bababa..... the same for expression (b+ba)*(Ɛ+a) it would not be able to generate abab, ababab...strings my thought it would not be able.so pls clerify
+Anita Ahuja for "baba" in (b+ab)^*(Ɛ+a) ,(b+ab)^* will generate "bab" and the concatenated part i.e (Ɛ+a) will generate "a". Same can be applied for bababa
hello sir i am looking for answer of this question could you please help me with tat ,Regular Expression that describes the set of Strings over (0,1) where every 0 is always followed by a 1 .....
Hello sir. your lectures are really very nice and informative. I was searching for a DFA which i could not solve if you can help. The problem is to draw a DFA which accepts the strings starting with 01 or ending with 01 but not both. Please help me
+Nishu Bali ..I think you should draw DFA for both starting with 01 and ending with 01 separately and connect both with cross product in this all those state will be final state where you found any one final state of previous two DFA but avoid taking that state which contains the final state of both the DFAs together .. ex..IF you have q1 is a final state of of first previous DFA and q4 is of second then final states of final DFA will be all those states where q1 come or q4 come but not both(q1q4)
+Nishu Bali it is simple..go for 0 along one side from initial and for 1 to another side.. make a complete connection of 01 along 0..& along 1 make self transition then end it with 01
1st problem will accept this string and the solution that he has provided will generate the given string. (b+ab) (b+ab) (b+ab) (b+ab) (b+ab) (b+ab) (epsilon +a). Take ab from 1st and b of 2nd then ab of 3rd, b of 4th, ab of 5th, b of 6th and epsilon of last one. According to one of the identities of regular expression, epsilon.language = language. I'm not sure if youre getting my point but the soltn he has provided is correct.
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You anticipate all my questions when you explain and I try to learn from you. You have a talent. Thank you for your excellent teaching.
I guess it is pretty randomly asking but do anybody know a good website to stream new series online?
TH-cam is such a better teaching medium than the classroom. Ravi, keep going!
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Pkka fail hui hogi...
crystal clear explanation!!!
should b awarded best teacher globally !!
I LOVE THIS TUTORIAL. YOU'VE JUST SAVED MY TOMORROW'S PAPER :')
If (a+b) means a OR b, then how does (a+b)* gives a string ab, why is it not just a's OR b's, how does it become a combination ?
Kindly solve my doubt if anyone can explain.
Bro (a+b)* denotes all the strings possible of a and b of any length.
Eg put 2 in place of *.
Therefore (a+b)(a+b) when u expand it aa+ab+ba+bb
Hope u got it.
Truthfully,the method that you used (start with ends with ) has help me to find the re easily,thanks a lot
Would (b +ab)* represent the string of just b's? Like for example is the string bbbb represented by (b+ab)*?
thank you so much sir ....... all the examples are good... i help me alot :D
(b+ab)*(e+a) does'nt except {abb}. So i guess this RE is wrong.Maybe
It does. Take ab first followed by b from the first bracket (since it's repeating using *) and then take e from the next one to get abb.
Suggestion for rephrasing the first question - no two consecutive a's
(ab+a)*+(ba+b)* is it correct? Please anyone
The 2nd Q is
*RE with no 2 consecutive a's or b's* , right ?
Can a solution be (b* a b+)* . (e + a)?
when this machine produce "bab" language so ther how this regular expression work??
Sir , How is phi and epsilon different in identities of regular expression.
could you explain (a+b)* intuitively!?
* can be replaced with 0,1,2,3,4....,n ;
+ represents logical OR ;
so when * = 0, it accepts epsilon ;
* = 1, it accepts 'a' OR 'b' ;
* = 2, it accepts 'aa' OR 'ab' OR 'ba' OR 'bb' ;
.
.
until * = n;
therefore (a+b)* is a reg expression which accepts all strings of all length containing a,b
Could someone pls explain how is (a+b)*=a*(ba*)*? Eg. for the string abbba. Thanks
a¹(ba*)³=a(ba*)(ba*)(ba*)=a(ba⁰)(ba⁰)(ba)=abbba
at 6:24 (Q : No. of two a's should not come together) Can i write this answer ? ANSWER = (( b^* a b^*))^* + b^*
@13:29 can't we write + instead of or?
what if we write (ab)*a+a(ba)* does it generate two a's together!?
no way!
then it would contain aa,bb
Awsome sir ur way of teaching awsome
😘
Sir I can't find the lecture of DFA NDFA..Mili mechine more machine...state diagram videos...pls reply me those links... thank you sir
Will the answer for question 2 a's should not come together generate aba? How?
(ab*)* + (b*a)* + b* ................is this right?
CHecking my concepts.
(a∪b)* (ab)* (b∪a)* ----> is this generate atleast one substring "ab" ?
yeh yeh yeh yeh ...usher usher :D
would somebody please explain the meaning of the following property of regular expresion: a ≤ b wich is a + b = b
Regular expression in which aab or bba does not occur??
you are a goddamn genius istg
sir plz tell the sol. to the following ques-
the set of all strings with atmost one pair of consecutive zero and one pair of consecutive ones.
set of all strings not containing 101 as substring.
Thanks so much. This really helped clear some things up for my exam
why only use 'a' and 'b'. Would be so much more helpful if you used 'a', 'b', 'c'
For no 2 'a' should come together we write the RE as,
((a+b)* b (a+b)*)*.
Correct me if I am wrong.
Thanks in advance!
(a+b)* makes aa possible so ye
i want to make a regular expression in which start middle and ending word is same and total length is even?
(ab)*(a+null) + (ba)*(b+null)
does this work for no 2a's or 2b's together
if we use above expression (b+ab)*(Ɛ+a) would it able to generate baba, bababa..... the same for expression (b+ba)*(Ɛ+a) it would not be able to generate abab, ababab...strings my thought it would not be able.so pls clerify
+Anita Ahuja for "baba" in (b+ab)^*(Ɛ+a) ,(b+ab)^* will generate "bab" and the concatenated part i.e (Ɛ+a) will generate "a". Same can be applied for bababa
hello sir i am looking for answer of this question could you please help me with tat ,Regular Expression that describes the set of Strings over (0,1) where every 0 is always followed by a 1 .....
+Ankit Wasankar your expression also contain single "0" ,which should not be there i think
Que:-string does not contain two A's together
Ans:-(b*ab*)*+b* will do sir
It will only contain strings which ends with 'b'
Awesome examples! Thanks!
Sir plz upload Turing machine videos
Hands down, Ravi is the best theory of computation tutor. Keep it up!
what if i want to generate "aabbaa" with a*(ba*)* which is equivalent to (a+b)*?
Are you talking about another problem or the one mentioned in the video?
Hello sir. your lectures are really very nice and informative. I was searching for a DFA which i could not solve if you can help. The problem is to draw a DFA which accepts the strings starting with 01 or ending with 01 but not both. Please help me
+Nishu Bali ..I think you should draw DFA for both starting with 01 and ending with 01 separately and connect both with cross product in this all those state will be final state where you found any one final state of previous two DFA but avoid taking that state which contains the final state of both the DFAs together ..
ex..IF you have q1 is a final state of of first previous DFA and q4 is of second then final states of final DFA will be all those states where q1 come or q4 come but not both(q1q4)
+Nishu Bali it is simple..go for 0 along one side from initial and for 1 to another side..
make a complete connection of 01 along 0..& along 1 make self transition then end it with 01
your are a excellent teacher.
if we take common part on left side.....means (ab)*(a+E) (E+b) is this answer is also correct?
+Ashima Mittal the common part is the postfix of second expression so we are bound to take common from right side
no.it is wrong it can accept abb
(b+aa)^* can this be the regular expression for strings containing even no. of a's ??
+Shweta Khara no because here a's will always be consecutive, 'aba' does not satisfy your R.E.
When Ravindrababu killin' it ...
When does (b+ab)* + (b+ab)*a produce b, bb ,bbb as you mentioned.
+Karthik Priyatham (b+ab)^* means any combination of b&ab so b,bb,bbb,bbbbb ..... is there
(a*)*=(ʌ ,a,aa,aaa,aaaa…)
Does the above statement is correct?
Yes , (a*)* is same as a* and with a* you can generate anything
how we will generate bab in this?
what about no 3as should come together
(b*a b+)* + b* will this work ?
abaab will also be accepted, but it not a part of this language
sorry, abaab will not be accepted by this RE. I interpreted it in a wrong way
Isn't (a + b)* = (a*b*)* ??
could (ab)*(a+e)+(ba)*(b+e) be the solution of no 2a 2b
Gate Lectures by Ravindrababu Ravula
for no 2 a's together, can solution be this (@03:37) :
€+a+b*+a
Thanks a lot sir
is this correct for no 2 a's come together-(b*ab*)*+b*
No coz in this we can get two a together
@@priyanaik1088 thanks 😉
thanks a lot sir..you explain so well
nice sir (y)
Babu I like your style
I think (€+RR*)=R^+ and not R*. Is it so?
you cannot generate € with R^+
R^+ = 1 or more
i have more correct ans
Thanks very well that is great
How will the solution ever create abbabbabb?
i don't think so , but did you find a solution to it ?
Actually your string (abbabbabb) isn't recognized by this RE because it contains two consecutive b's.
1st problem will accept this string and the solution that he has provided will generate the given string. (b+ab) (b+ab) (b+ab) (b+ab) (b+ab) (b+ab) (epsilon +a). Take ab from 1st and b of 2nd then ab of 3rd, b of 4th, ab of 5th, b of 6th and epsilon of last one. According to one of the identities of regular expression, epsilon.language = language. I'm not sure if youre getting my point but the soltn he has provided is correct.
(ab*)* + (b*a)* + b* ................is this right?
CHecking my concepts.
The exact answer I was looking for :)
oyaa nan maarai ayye
now all is well
super!!!!
just awesumm...
for my exams
BEST!!!!!!!!!!
Can the answer to the last example be:
*( (b +aba) + (a + bab) )** ?
nope
no
sir upload video on turing machine asap plss before 12 may my exams are near
No 2 a's together - can I write (b* a b+)*?
it wouldn't generate strings ending with 'a'.
this could be the right one:
b* (a b+)* (a + E)