Question text is in black, my answers in blue.
4.28 - I'm stumped on this question and it is tying my head in knots. First, is A supposed to contain all decidable machine encodings or just some arbitrary subset of them?
An arbitrary subset, because all it says is that every Mi is a decider, not that every decider is an Mi.
Second, an enumerator for A (assuming its just some set of decider encodings) would just print out all of the encodings in A. How does considering this get us any closer to proving that there is a decidable language that is automatically not in A? It seems the only thing we get from the enumerator is that the complement of A would then be decidable.
I don't understand why you say that -- deciding the complement of A would be essentially the same thing as deciding A.
I'm tempted to say that the machine that decides the complement of A cannot be in A, but I don't think that's the case. I think it could be in there. However, my brain bluescreens on me everytime I try to consider it. Can you provide another hint? I take it we're trying to arrive at a contradiction of some sort?
You want to imitate the diagonalization argument on pages 180-1, or the argument I gave in lecture that LBOS is not TR. That argument said that if LBOS cannot be equal to any L(Mi), because Mi itself is in LBOS iff it is not in L(Mi).
5.17 - I've reduced the PCP problem over a unary alphabet to the problem of deciding whether any linear combination of the lengths of the tops of the dominos is equal to any linear combination of the lengths of the bottoms of the dominos. However, I'm struggling to find a simple way to prove that this new problem is decidable. Is this the right approach? Any suggestions?
Yes, that's the right approach. You might try to construct examples that are solvable or not solvable and see what the difference between them is. There are also things you might recall from number theory about what numbers could possibly be linear combinations of other numbers...
5.20 - Can we safely say that being able to construct a TM that decides any subset of sigma star is a contradiction? If so, what is the best example in the book to point to of this?
No. There are subsets of Sigma* that are decidable -- perhaps I'm not getting your conjecture. Remember that so far all our undecidable sets have been proved to be undecidable by reduction. If you could find a language X such that X was a subset of 1* (all the strings in X consist only of 1's) and ATM ≤m X, you win.
5.35 - It strikes me that the answer to this is no, as we can't even decide if an arbitrary TM will halt; let alone change the first k characters of the tape. However, I'm open to suggestions on how to approach the proof. Should we try to us Rice's theorem?
Rice's Theorem doesn't apply because this language is not a property of Turing machines -- you could have two machines M1 and M2 such that L(M1) = L(M2) but on some w, M1 modifies that portion of the tape and M2 doesn't. If you are right that this language X is undecidable, the way to prove it would be to show that ATM ≤m X.
Last modified 4 April 2008