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Oct 2006 Exam Question 7.

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ian.coetzer

Oct 2006 Exam Question 7.
November 02, 2009 03:43PM

Registered: 18 years ago
Posts: 137
Rating: 0

Please see if these Accept, Loop definitions are correct.

~~My Attempt @ Question 7~~

*** Question 7 - revised solution ***

thx

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stevenv

Re: Oct 2006 Exam Question 7.
November 02, 2009 05:05PM

Registered: 18 years ago
Posts: 613
Rating: 0

My solution is ...
Accept(T) = a(aa)*b
Loop(T) = a(aa)*ba

Basically the same except I made "starting with an odd number of a's" into a(aa)*

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ian.coetzer

Re: Oct 2006 Exam Question 7.
November 02, 2009 05:12PM

Registered: 18 years ago
Posts: 137
Rating: 0

Hi, I like your one better, changed the full solution:

*** Question 7 ***

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visnaicker

Re: Oct 2006 Exam Question 7.
November 03, 2009 02:39PM

Registered: 18 years ago
Posts: 518
Rating: 0

loop ( T) = a ( aa ) * ba (a + b)*

as per cos301Y/104 pg5

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stevenv

Re: Oct 2006 Exam Question 7.
November 03, 2009 02:59PM

Registered: 18 years ago
Posts: 613
Rating: 0

Actually thought about adding that last (a+b)* because those words will in fact loop. Thanks for pointing it out.

EDIT: like your sig, very true!!!

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ian.coetzer

Re: Oct 2006 Exam Question 7.
November 03, 2009 03:08PM

Registered: 18 years ago
Posts: 137
Rating: 0

Okay thanks even though (a+b)* will never be reached it could still form part of a valid input string!
Just checked tut 104 ... phew other answers i gave seemed ok, except for Question 5 which still have me bowled out for a duck!
must still check the other chap's solution diagram have not yet tried to trace through it.

*** Question 7 - revised solution ***

cheers

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BurhaanT

Re: Oct 2006 Exam Question 7.
October 31, 2010 12:01PM

Registered: 13 years ago
Posts: 1
Rating: 0

After much battling with trying to understand how to determine what language is accepted by a TM and looking at ian's image.

Would I be wrong in saying that after the b at the end, you could still have some a's and the language would still be accepted by this TM:
So, shouldn't it be:
a(aa)*b(ab +aaab)*

Was wondering why the last set (i.e. (ab + aaab)*) would not be included?