Welcome! Log In Create A New Profile


Assignment 2 - Question 3

Posted by InSaNiE 
Announcements Last Post
Announcement SoC Curricula 09/30/2017 01:08PM
Announcement Demarcation or scoping of examinations and assessment 02/13/2017 07:59AM
Announcement School of Computing Short Learning Programmes 11/24/2014 08:37AM
Announcement Unisa contact information 07/28/2011 01:28PM
Assignment 2 - Question 3
April 24, 2007 10:12PM
Just want some clarification on how everyone interpreted the question - does it mean it will only accept words that ONLY contain an even number of 'ab' substring, but no other letters (eg: abab, abababab) or can the word contain other letters as well, but as long as has even number of 'ab' substrings (eg: abaab, abababbab)?

Thanks in advance.
Re: Assignment 2 - Question 3
April 26, 2007 01:03PM
It means that the FA will accept all words that contain an even number of occurrences of the ab-substring. Thus, your second interpretation is correct.
Re: Assignment 2 - Question 3
May 12, 2007 03:13PM
Through a spanner in the works and ensure that the FA (or machine) does not leave it's Final state should one or more a's or one or more b's be receieved after the last even numbered 'ab' substring, but msut also leave the Final state should another 'ab' substring (count now odd) be received.

In some sick kind of way, I actually enjoyed this section. Which was a little surprising after the Recursion + Induction section.
Re: Assignment 2 - Question 3
May 12, 2007 03:14PM
Sorry - Maybe "throw a spanner in the works" might be a little easier.
Re: Assignment 2 - Question 3
November 04, 2007 05:46PM
Ok I must be losing it... looking at the solutions for exam purposes does not make sense to me because the first ± state on the left has a 'b' loop and 'b' on its own is NOT 'ab' so this word IS accepted and should not be?!

What am I missing?

Re: Assignment 2 - Question 3
November 05, 2007 09:46AM
This one confuses me too.

From the solution, it seems that 0 is considered an even number. By that I mean a and b and ba are accepted words.

So, if the word has no occurance of the substring ab or an even number of occurances of the substring ab, it is accepted.

Does this sound right?

These questions worry me because there is more than one interpretation of the question...
Re: Assignment 2 - Question 3
November 05, 2007 06:25PM
The machine will accept any word with any number of a's and any number of b's, but if the word contains the substring ab then it will only accept it if there are an even number of them. It's not saying it will exclusively accept only ab's and only an even number of them....and yes it will accept the null word too given by the combination start and final state ie +-.

aaaaaaaaaaaaaaabaaaaaaaaaaababbbbbbbbbbbb is a word in this language...there are 2 substrings ab.


are also words in this language

BTW 0 is an even number, but that is irrelevant in this case.

Re: Assignment 2 - Question 3
November 08, 2007 07:31PM
Thanks Clive - that makes sense.

Sorry, only registered users may post in this forum.

Click here to login