CMPSCI 401: Theory of Computation

Solutions to Final Exam, Spring 2013

David Mix Barrington

9 May 2013

Directions:

• Answer the problems on the exam pages.
• There are eight problems for 125 total points plus 10 extra credit. Actual scale was A = 105, C = 70.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
• The first five questions are statements -- in each case say whether the statement is true or false and give a convincing justification of your answer -- a proof, counterexample, quotation from the book or from lecture, etc. You get five points for the correct boolean answer (so there is no reason not to guess if you don't know) and up to five for the justification.

  Q1: 10 points
  Q2: 10 points
  Q3: 10 points
  Q4: 10 points
  Q5: 10 points
  Q6: 25 points
  Q7: 30+10 points
  Q8: 20 points
 Total: 125+10 points

Question text is in black, solutions in blue.

If C is any class of computers, such as DFA's, CFG's, LBA's TM, strange variant TM's, etc.:

• A_C = {(M, w): M is a computer in C and w ∈ L(M)}
• E_C = {(M): M is a computer in C and L(M) = ∅}
• ALL_C = {(M): M is a computer in C and L(M) = Σ^*}
• ALLSTAR_C = {(M): M is a computer in C and L(M)^* = Σ^*}
• INF_C = {(M): M is a computer in C and L(M) contains infinitely many strings}

A is Turing decidable if A = L(M) for some Turing machine M that always halts.

A function f is Turing computable if there exists a Turing machine M such that for any string w, M when started on w halts with f(w) on its tape.

Recall that if A and B are two languages, A is mapping reducible to B, written A ≤_m B, if there exists a function f: Σ^* → Σ^* such that for any string w, w ∈ A ↔ f(w) ∈ B.

Although we did not prove this in lecture, you may assume both:

• There exists a Turing decidable language that is not in PSPACE.

• There exists a language in PSPACE that is not in L = DPSACE(log n).

Define a weighted graph to be an undirected graph where each vertex has a positive integer weight -- you may assume that each weight is at most n, the number of vertices. The language WC (weighted clique) is the set {(G, S, k): S is a clique in the weighted graph G and the sum of the weights of the vertices in S is at least k}. The language MWC (maximum weighted clique) is the set {(G, k): there exists a clique in the weighted graph G such that the sum of the vertex weights in the clique is at least k}.

A directed graph is strongly connected if for every pair of vertices u and v, there is a (directed) path from u to v. The language STRCONN is the set of all strongly connected directed graphs.

• Question 1 (10): True or false with justification: The language INF_CFG is Turing decidable.

  TRUE. A CFG generates an infinite language if and only if it contains a string w such that p ≤ |w| ≤ 2p, where p is the pumping length of the grammar. If it contains such a string, pumping up shows infinitely many other strings to be in the langauge. And if the language is infinite, there must be a string of length at least 2p, and we can pump that string down until it is in the desired range.

  Since A_CFG is TD, there is a Turing machine that can test all possible strings in that range.

• Question 2 (10): True or false with justification: There exist languages X and Y such that X ≤_m Y is true but X ≤_p Y is false.

  TRUE. Let X be a TD language that is not in P, and let Y = {a} for some letter a in Σ. Then X ≤_m Y because the function can test whether its input w is in X, and output a or aa accordingly. But if X ≤_p Y, X would be in P because Y is in P.

  The fact that there exist TD languages that are not in P is a consequence of the Time Hierarchy Theorem, mentioned but not proved in class.

• Question 3 (10): True or false with justification: Every language in the class L = DSPACE(log n) is context-free.

  FALSE. Our standard non-CFL {aⁿbⁿcⁿ: n ≥ 0} is easily seen to be in L, since a TM can count the number of each letter, keeping the answers in binary on its work tape, and compare the three numbers for equality.

• Question 4 (10): True or false with justification: For any language X, X is Turing decidable if and only if X^* is Turing decidable.

  FALSE. It is true that X^* is TD if X is. But let X be any non-TD language that happens to contain all one-letter strings. Then X^* = Σ^* which is TD.

• Question 5 (10): True or false with justification: The language of the grammar with rules S → Ta, T → bU, T → Ub, U → c, and U → S is regular.

  FALSE. The intersection of this CFL with the regular language b^*ca^* is {bⁿcaⁿ: n ≥ 0}, because every string in the CFL has an equal number of b's and a's. But this intersection is easily shown not regular by Myhill-Nerode: the set {λ, b, bb, bbb,...} is pairwise inequivalent.

• Question 6 (25): These questions use the definition of the languages MWC and WC above.

  • (a, 10) Prove that the language WC is in the class L = DSPACE(log n).

    Given (G, S, k), we need to verify that S is a clique and that the sum of the weights of the vertices in S is at least k. Since each weight is at most n, the sum of weights is at most n² and we can be sure that we can add up the weights on our worktape. To verify that S is a clique, we need merely loop through all pairs of distinct vertices {x, y} in S and verify that there is an edge between x and y. This takes two loop variables of O(log n) bits each.

  • (b, 15) Prove that the language MWC is NP-complete.

    MWC is in NP because any clique S with the proper weight, so that (G, S, k) is in WC, is a certificate for (G, k) being in MWC. Our verifier tests membership in WC, which is in P because we showed it to be in L above and L ⊆ P. To prove NP-completeness we reduce from CLIQUE, which was shown to be NP-complete in the text. Given an input (G, k), our reduction function will output (G', k), where G' is the weighted graph made from G by giving each vertex a weight of 1. Then (G, k) is in CLIQUE if and only if a clique of size at least k exists in G, which is true if and only if a clique of weight at least k exists in G'.

• Question 7 (30+10): These questions all involve the language STRCONN of strongly connected directed graphs, defined above.
  • (a, 10): Prove that STRCONN is in the class P.

    On input graph G, loop over all vertices x. For each x, carry out a depth-first search of G starting from x and verify that every vertex is reachable from x. If every vertex is reachable from every other, accept G, otherwise reject it. This is polynomial time because we know that DFS takes O(e) = O(n²) operations on a graph with n vertices, and each basic operation takes only polynomial time on a Turing machine.

  • (b, 10): Prove that STRCONN is in the class DSPACE(log² n).

    For every vertex x and for every vertex y ≠ x, test whether there is a path from x to y using the Savitch's Theorem argument (middle-first search). This means testing each predicate PATH(s, t, 2^k) by looping through all possible vertices u and testing whether both PATH(s, u, 2^k-1) and PATH(u, t, 2k-1) are true for any u. This recursion takes O(log² n) space as the depth of recursion is O(log n) and only O(log n) bits are needed to record each intermediate vertex.

  • (c, 10): Prove that STRCONN is in the class NL.

    We know that PATH(x, y) is in the class NL for any vertices x and y. Our nondeterministic log-space Turimg machine must be able to accept a directed graph G if and only if it is strongly connected. What it does is to loop over all pairs of vertices (x, y), guess a path out of x for each, and accept if and only if each of these guessed paths reaches its respective y. If the graph is strongly connected, it is possible for the machine to guess correctly. If it is not, for some choice of x and y there will be no path from x to y, and thus whatever path the machine guesses will fail and the machine will reject G no matter how it guesses.

  • (d, 10XC): Prove that STRCONN is complete for the class NL under ≤_L reductions.

    We have shown that STRCONN is in NL -- it remains to show that PATH ≤_L STRCONN. Let (G, s, t) be an input to PATH, so that G is a directed graph and s and t are vertices. Let G' be a directed graph with the same vertices as G as follows. G' includes all the edges of G, but in addition for every vertex z other than s and t, it has an edge from z to s and an edge from t to z. I claim that G has a path from s to t if and only if G' is strongly connected.

    Clearly if there is a path from s to t in G, then for any vertices x and y in G', there is a path from x to s to t to y in G' and so G' is strongly connected. Now suppose that G' is strongly connected. There must be a path from s to t in G'. But if this path exists, there must also be such a path that does not use any of the new edges and is thus also a path from s to t in G. This is because the shortest path from s to t in G' cannot use any new edges -- if it uses an edge from z to s, then the portion of the path following that edge is a shorter path from s to t, and if it uses an edge from t to z, the portion of the path before that edge is a shorter path from s to t.

• Question 8 (20): These questions both use the definition of the language ALLSTAR_C from above.
  • (a, 10): Prove that the language ALLSTAR_DFA is Turing decidable.

    I claim that M is in ALLSTAR_DFA if and only if L(M) contains all one-letter strings, that is, if and only if Σ ⊆ L(M). If it does, Σ^* ⊆ L(M)^* by the regular language identity X ⊆ Y → X^* ⊆ Y^*. If it does not, some single-letter string a is not in L(M), and that string cannot be in L(M)^* either because there is no way to make up a as a concatenation of other strings.

  • (b, 10): Prove that the language ALLSTAR_TM is not Turing decidable.

    We reduce the undecidable language A_TM to ALLSTAR_TM. Let M be any TM and w be any string over M's input alphabet. We define f(M, w) to be a Turing machine N that ignores its input and simulates M on w. If (M, w) ∈ A_TM, then L(N) = Σ^* so clearly L(N)^* = Σ^* and N ∈ ALLSTAR_TM. If (M, w) ∉ A_TM, then L(N) = ∅ and L(N)^* ≠ Σ^*, so N ∉ ALLSTAR_TM.

Last modified 1 May 2014