What I see is that we put blank on the first 0 so we have 2^n-1 zeroes left over... (For the case of 2^n) and numbers of the form 2^n - 1 can be expressed as the sum of consecutive powers of 2(check it's binary form) so while alternately crossing zeroes every time we are removing zeroes in powers of 2 (for eg in 7 , we cross 4 0s, then 2 0s and finally 1 0.
Dear Ma'am, It is really good and works well. But I found a bit mistake in solution of this TM, that while transition is at state q1 and next symbol on tape is x where it will move to ??? I tried this at JFLAP, which doesn't accept the given language and even crashes the above said state. Please give its solution. I want a sloution for the L = { 0^2^2n+1 , n >= 0}. It's paid task. kindly contact back asap (within an hour). Quick response will be appreciated.
What I see is that we put blank on the first 0 so we have 2^n-1 zeroes left over... (For the case of 2^n) and numbers of the form 2^n - 1 can be expressed as the sum of consecutive powers of 2(check it's binary form) so while alternately crossing zeroes every time we are removing zeroes in powers of 2 (for eg in 7 , we cross 4 0s, then 2 0s and finally 1 0.
I want to know why this method works. What's the logic behind this?
Thank you so much. Your explanation is clear.
You killed the ant😭😭😭 10:42..
Thank you so much! This really helps a lot. Keep up the good work.
You are doing a great job sister now I understand the subject thank you sister
mam if n>1 then whats logic behind this ?
Thank you so much
mam for any one question in TM can u do all the three mam viz state diag, TT, ID Description ?, state diagram is vunderstandable rest two mam.
Thank you!
LOL
Thank u .......
thanks mam it helped me a lot
Where n is a square?? How to do that?
thanks
Thankyou maam
such a sweet voice ma'am
Dear Ma'am,
It is really good and works well.
But I found a bit mistake in solution of this TM, that while transition is at state q1 and next symbol on tape is x where it will move to ???
I tried this at JFLAP, which doesn't accept the given language and even crashes the above said state.
Please give its solution.
I want a sloution for the L = { 0^2^2n+1 , n >= 0}. It's paid task.
kindly contact back asap (within an hour). Quick response will be appreciated.
So....did you ever get a response?
length 16
which si 2^4
you require 4 iterations to cancel all zeroes
reject if you skip a zero and cannot cancel anymore
Hafiz tayyeb squad like maro
Thok diya Jani
Thank you so much