I've got a copy of the textbook if anyone's looking to buy one. I'll have to check the exact edition but I think it's the 2nd ed. Any offers?by stevenv - COS351D
Thanks, 301 was EASY compared to 351 but the latter is passed so it's all good!!!by stevenv - COS301Y
I sneaked through with 59, year mark really helped push it up a little. Considering I found this the most difficult module I've done, very happy with the result.by stevenv - COS351D
Definitely not enough time. Figuring out, drawing and then testing machines takes TIME!!!by stevenv - COS301Y
LOL, I agree: he's been quite a fixture over the past few years! Overall quite a fair paper, as expected but more drawings (hope pencil is acceptable), finished with a few minutes to spare.by stevenv - COS301Y
(1) all words that have an odd number of b's and end in aaa (2) all words of odd length with exactly one a or exactly one b Assume alphabet {a b}by stevenv - COS301Y
(1) S -> bAB A -> B B -> aB | Bb | /\ (2) S -> aX | Yb X -> S | /\ Y -> bY | bby stevenv - COS301Y
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!!!by stevenv - COS301Y
If you encountere a b in the string but none of your states can read that character the machine should crash because nowhere does it say those are valid characters for the tape (should've been given explicitly specified alphabet).by stevenv - COS301Y
Ignore all of that, I wasn't seeing the whole question properly and was still thinking of the ADDER in the textbook.by stevenv - COS301Y
Interesting points, considering that n >= 0 what would the string look like if f(n) = 0 + 0? I see it as containing a singly solitary b and that's all. Beginning at START, the first input character will be b and it will be changed into an a. The machine moves to state 1 where the next character read will be /\ which is left alone and the head is moved left, the machine moves to state 2. Aby stevenv - COS301Y
LOL, was wondering how you posted with superscripts but now I see the toolbar has a button for it, idiot!by stevenv - COS301Y
If we get a question like this, there is no hope for me 'cause only got 16 minutes to figure out and draw itby stevenv - COS301Y
Must be a case of reading between the linesby stevenv - COS301Y
Although probably nearly too late to receive a reply, I would suggest emailing the lecturers as to this question.by stevenv - COS351D
Surely this is the wrong way around? Look at page 574by stevenv - COS301Y
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)*by stevenv - COS301Y
Yip, that's exactly why I stopped wasting my time on it. I think there must be a mistake in the question somewhere otherwise this is the most difficult machine we've had to draw all year and once you've figured out the "trick" it becomes solvable. I'm not wasting another minute on it and rather focusing on understanding everything, hoping the lecturers haven't tby stevenv - COS301Y
I did see that note but even if the steps differ in some way and the proof still makes logical sense and the answer is still the same, how can they deduct the marks?by stevenv - COS301Y
Looks alright, just realised I'm a bit rusty on the whole nullable terminals thingsby stevenv - COS301Y
Are you sure you're not getting mixed up. In chapter 6 of the study guide which relates to chapter 17 of Cohen which is where the union, product and Kleene closures are defined but there are no reformulation of any othe Cohen theorems. In the following chapter of the study guide then refer to reformulation of proofs for emptiness (theorem 42) and membership a.k.a. CYK algorithm (theorem 45).by stevenv - COS301Y
This is my attempt, I'm not so sure I've covered all the bases properly when it comes to vxy being split across two groups of letters. My thinking was that because of the single b after the first group of a's the only sensible way of splitting the word up would be across the third and fourth groups HOWEVER suppose you could still say that splitting it across groups 1 and 2 or 2 andby stevenv - COS301Y
If a state in a PDA has no edge leading from it that can read a character that is on the TAPE (from a READ) or in the STACK (from a POP) then the word is rejected.by stevenv - COS301Y
I don't know about the "reformulated" proofs but I think those proofs in the textbook are pretty clear.by stevenv - COS301Y
I don't think there's really any specific method to find that, I try to imagine different words being run on the TM while isolating what causes those words to be either accepted, rejected or loop. After that its finding the pattern that defines the words that fit into each category i.e. all words that start with a double b end up looping.by stevenv - COS301Y
Quite right, but for 2 marks I wouldn't say too muchby stevenv - COS301Y
I think you can say yes or no and simply state that any language accepted by a 2PDA can be accepted by a (ii) 3PDA and (iii) TM. 2PDA = nPDA = PM = TM (page 491)by stevenv - COS301Y