CS154 Midterm Solutions

Question #1:

(a) Maybe; (b) Maybe (c) Maybe (d) Certain (e) Certain (f) Never (g) Maybe.

An important principle to remember is that a limit on how hard the tail (starting point) of a reduction is does not limit how hard the head is. For example, many people answered ``Certain'' for (a), but saying A is NP-complete does not force B to be NP-complete; B could be harder, even non-RE.

(b) is a Maybe because P=NP is still possible. (f) is Never, because if C is in P, then D is in P. Since P is closed under complementation, ~D is in P as well.

Question #2:

(a) Never (b) Maybe (c) Maybe (d) Never (e) Maybe (f) Certain (g) Certain (h) Certain. The same principle as in Q1 explains most of these answers. For (g), note that the RE languages are closed under union (use a nondeterministic TM to guess one of the languages and test that language). Also, in (h), the RE languages are closed under intersection (test one, and if it is ``yes,'' then test the other).

Question #3:

This is probably the world's simplest polynomial-time reduction, but some people still didn't get the sense of what it means to do a reduction, and lost all credit on this question.

(a) Turn input expression E into the expression NOT(E). Note, incidentally, that this takes O(n) time, not O(1), as several people claimed, since you have to at least read E to copy it.

(b) If there is a truth assignment T that makes NOT(E) false, then T also makes E true. Thus, if NOT(E) is not a tautology, then E is satisfiable. Conversely, if E is satisfiable, the same truth assignment will show NOT(E) is not a tautology.

(c) This part proved rather difficult. First, many people forgot the definition of NP-completeness: L is NP-complete if every problem in P polynomial-time reduces to L. Colloquially, we often say ``L is NP-complete if the statement `if L is in P then P=NP' holds.'' In fact, I noticed that several proofs in the reader use that ``definition,'' even though it is really a consequence of the correct definition and may not be equivalent to the real definition (we don't know). I accepted either definition, but still few people got this part right.

Using the correct definition, the argument is as follows. We know every language L in NP polynomial-time reduces to SAT. Follow that reduction by the polynomial-time reduction from SAT to NT that we gave in (a). Since the cascade of two polynomial-time reductions is polynomial-time, we see that every L polynomial-time reduces to NT.

I also accepted: Suppose NT is in P. Then because of the reduction from (a), SAT is in P. Since SAT in NP complete, every L in NP is in P. Thus, NT-in-P implies P=NP, so NT is NP-complete.

I did not accept arguments about how NT was ``at least as hard as'' SAT, since that argument, while intuitively correct, begs the question; it is essentially what the skeptic in (c) does not believe.

Question #4:

Many people tried to place a coordinate or coordinates in single tape cells, or even in the finite control. You can't do either; the first because the tape alphabet becomes infinite and the latter because the state set becomes infinite. Although we might need only a finite number of tape symbols or states at any finite time, we would need symbols or states beyond limit to keep the simulation going. No finite set of states/symbols allows that.

(a) There are many ways that work: by rows, by columns, or by diagonals, for example. I'll be clever and make (c) easy by using columns. That is, record on the 1D tape each column, up to the infinite sequence of blanks that must appear in the column, followed by a special marker #. List each column, in order, up to the last column that contains a nonblank. Keep the head of the 1D tape at the same cell as the head of the 2D tape.

(b) #0110#110#0011#0B01#BBB...

(c) Make the required symbol change and move one position right. If we meet the #, then shift the tape over one cell and insert a blank there.

Question #5:

(a) The solution must start with 1, because choices 2 and 3 result in an immediate mismatch. But then, the A-string is shorter than the B-string, and there is no pair that allows the length of the A-string to grow faster than that of the B-string.

(b) How about (1, 1) for pair 4. Then 4 is a solution. You were rewarded for this ``outside the box'' thinking. If you created a solution with two index integers, you lost 2 points; if your shortest solution was of length 3 or more, you lost 3 points.

Problem #6: q_0 0110 |- 1q_1 110 |- q_2 1010 |- 1q_0 010 |- 11q_1 10 |- 1q_2 100 |- 11q_0 00 |- 111q_1 0

Question #7:

The answer is simple: the reduction is not (known to be) polynomial-time. There were no other answers that I accepted.

Question #8:

(a) (1) contained in (2). (2) generates all strings of 0's and 1's, while (1) doesn't generate strings like 11, or any odd-length string.

(b) (1) contained in (2). These almost look the same. (2) is the set of strings of the form 0^n1^m such that 0 <= m <= n. However, the PDA of (1) cannot reach state p without reading at least one 0 and one 1, so it accepts the set of strings of the form 0^n1^m such that 1 <= m <= n.

(c) The languages are equal.

(d) (1) contains (2). (1) is all strings of 0's and 1's with two consecutive 1's. (2) misses some of these strings, e.g., something ending with 110101.

(e) Equal. Both define the language of regular expression (0(0+1)*0)*.

Question #9:

(a) REG. With 5 states, we can keep track of the difference in the number of 0's and 1's, as long as that difference is -2, -1, -, 1, or 2. If the difference ever goes outside that range, we can go to a dead state.

(b) CF-CF. Here is how a PDA works. Let m be the number of 0's seen so far and n the number of 1's. If m > 2n, we'll have m-2n X's on the stack, and if m < 2n we'll have 2n-m Y's on the stack. You should be able to figure out how to adjust X's or Y's as 0's and 1's come in. This strategy lets you design PDA's for both the language and its complement; you just need to decide differently when to accept what you've seen so far.

(c) This language is not context-free.

(d) CF-CF. The trick is to realize that 1's and 2's can each be regarded as 1's. Then the problem becomes like (b), only easier because we can match one 0 against one 1, instead of two 1's.

(e) Imagine a DFA that has states of the form [i,j,k], where each of i, j, and k are in the range 0 to 100. Use one component to count each of the three input symbols, but top out at 100; i.e., if i=100, and another 0 comes in (assuming i counts 0's), leave i at 100. [100,100,100] is the only accepting state.

Question #10:

a	b	a	a	a
There was a small flaw in that B is useless --- the result of a last minute change in the grammar. However, I accepted the answer above, as well as the answer in which B doesn't appear.

Question #11:

A surprising number of people tried to do this with a short string z like 00. You need to pick something that is bigger than the pumping-lemma constant, as I was forced to remind you at the exam because someone pointed out a mistake in the reader. I do not have much sympathy for someone who didn't get the point (even if you were taking the exam elsewhere or at another time), because the condition was stated correctly in class, in the notes, and in the ``adversary'' rules on p. 232. In fact, Example 4.3 on p. 113 is a proof of nonregularity of the same language. You can get the answer to this question by mimicking the proof there.

Specifically, for (a) pick z to be 0^p for the the smallest prime p that is at least n+2. For (b), pick i = p-m, where m = |vx|. The answer to (c) is in Example 4.3, where we show that uv^{p-m}wx^{p-m}y has a length that is the product of two factors neither of which is 1.

Question #12:

This problem was intended to test your mettle, and indeed it did. Only one person got full credit, but a number of people got most of the ideas. The biggest mistake was not reading the definitions (a common error that transcends automata theory and is actually seen in the real world). Thus, many people thought ``nice'' meant ``if L is regular then M is regular.'' Actually, it's the other way round.

(a) Not nice. The union with 0* can obscure a hard problem and make f(L) be regular even when L isn't. For instance, use L = {0^p | p is a prime}, which we know is not regular.

(b) Nice. The homomorphism h(0)=1 and h(1)=0 turns f(L) into L. It also turns L into f(L), but that is not the point you needed to make to prove this function is nice.

(c) Nice. L = (f(L)/1)/1, using the operation covered in HW2, problem 6.

(d) Not nice. Similarly to (a), we can hide the difficulty by throwing away the first symbol. A commonly chosen example is L = {a0^n | a = 1 and n is prime or a = 0 and n is not prime}. Then f(L) = 0*, which is regular, but it is easy to show L itself is not regular.