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CS880: Approximations Algorithms Scribe: Chi Man Liu Lecturer: Shuchi Chawla Topic: Inapproximability: Vertex Cover and Set Cover Date: 5/3/
In the previous lecture we introduced probabilistically checkable proofs (PCPs) and saw how they could be used to obtain a tight inapproximability result for MAX-3SAT. We also introduced Has- tad’s 3-bit PCP, a very useful tool for proving inapproximability results. In this lecture we apply Hastad’s 3-bit PCP to show that approximating Vertex Cover within any constant factor less than 7/6 is NP-hard. We also show that Set Cover cannot be approximated within a factor of β · log n for some constant β, unless NP ⊆ DTIME(nlog log^ n).
29.1 Vertex Cover
Last time we introduced Hastad’s 3-bit PCP and out of it we obtained the following result.
Theorem 29.1.1 For all constants > 0 there is a reduction from 3SAT (or any NP-complete problem) to MAX-3LIN that maps satisfiable formulas to MAX-3LIN instances where a 1 − fraction of the constraints can be satisfied simultaneously, and maps unsatisfiable formulas to MAX-3LIN instances where no assignment satisfies more than a 1 /2 + fraction of the constraints.
We used Theorem 29.1.1 to show that it is NP-hard to approximate MAX-3SAT within any constant factor greater than 7/8. In the following, we show that approximating Vertex Cover within any constant factor less than 7/6 is NP-hard using a similar technique.
Theorem 29.1.2 For all constants > 0 , Vertex Cover is NP-hard to approximate within a factor of 7 / 6 − .
Proof: Our goal is to construct a gap-preserving reduction from MAX-3LIN to Vertex Cover. Given a MAX-3LIN instance with m constraints, we want to construct a graph G. For each linear constraint xp ⊕ xq ⊕ xr = b, we add to G a clique with 4 vertices, where each vertex represents a setting of (xp, xq , xr) satisfying the constraint. Then, for any two vertices, if their corresponding settings of variables conflict with each other, we add an edge between them. We now show that this polynomial-time reduction is gap-preserving. Suppose that the MAX-3LIN instance has an assignment satisfying at least a 1 − fraction of the constraints. For each satisfied constraint, pick the vertex in G that corresponds to its setting of variables in this assignment. These vertices form an independent set since they all come from different cliques, and all settings of variables agree. This independent set has size at least (1 − )m. There are 4m vertices in G, hence G has a vertex cover of size at most (3 + )m. Now suppose that no assignment satisfies more than a 1/2 + fraction of constraints. We claim that any independent set of G has at most (1/2 + )m vertices. From this it follows that G has minimum vertex cover of size at least (7/ 2 − )m, and we have achieved a gap of 7/ 6 − . It still remains to prove the claim. Let S be an independent set of G with k vertices. The vertices in S correspond to distinct linear constraints as they must lie in different cliques. Moreover, these vertices represent settings of variables that have no conflict. Thus, we can augment the settings to a total assignment that satisfies those k constraints. We finish the
argument by setting k = (1/2 + )m.
29.2 Set Cover
In this section, we show that Set Cover is unlikely to have a sublogarithmic approximation. For- mally, we have the following theorem.
Theorem 29.2.1 For some constant β, approximating Set Cover within a factor of β · log n is NP-hard unless NP ⊆ DTIME(nlog log^ n).
To prove this theorem, we consider the 2-Prover 1-Round system for 3SAT in the previous lecture. Given a 3CNF with m clauses, the verifier selects a clause and one of the variables in that clause uniformly at random, queries that clause and that variable to provers 1 and 2 respectively, and accepts if and only if the two responses agree on the value of the chosen variable. This system uses O(log m) random bits, has completeness 1 and soundness 1 − ′^ for some cosntant ′^ > 0. We take k parallel repetitions to reduce the soundness to αk^ for some constant α < 1, while increasing the amount of randomness to O(k log m). We will determine the value of k later. Now, a query to prover 1 is a k-tuple of clauses, while a query to prover 2 is a k-tuple of variables. The numbers of different queries to provers 1 and 2 are mk^ and nk^ respectively.
We introduce the Label Cover problem in the next section. It will soon be clear that finding optimal provers (ones that maximize the accepting probability) in 2P1R systems can be reduced to solving Label Cover instances.
29.2.1 Label Cover
Consider a bipartite graph G = (Q 1 , Q 2 ; E) with |Q 1 | = mk^ and |Q 2 | = nk. Each vertex in Q 1 (resp. Q 2 ) represents a possible query to prover 1 (resp. prover 2). There is an edge between u ∈ Q 1 and v ∈ Q 2 if and only if the verifier can ask prover 1 query u and prover 2 query v simultaneously for some sequence of random bits. To each vertex w ∈ Q 1 ∪ Q 2 we associate a label set Lw containing all correct answers to the query w. For every edge e = (u, v), let Re be the relation containing all pairs (a, b) (a ∈ Lu, b ∈ Lv) such that (a, b) is an accepting answer pair to query pair (u, v). Our goal is to pick an answer from each label set so that the probability of acceptance is maximized. In other words, we want to assign to each vertex w a label aw ∈ Lw such that the following quantity is maximized: |{e | e = (u, v) ∈ E, (au, av) ∈ Re}| |E|
We can generalize the above problem in the following way. Let L be a set of labels. For each vertex w, the label set Lw is a nonempty subset of L. For each edge e, the relation Re is a subset of L × L. This generalized problem is known as Label Cover.
The 2P1R system for 3SAT discussed above has completeness 1 and soundness αk. Hence, G either has an assignment satisfying all edges, or has no assignment satisfying more than an αk^ fraction of the edges. Note that |L| = 2O(k)^ and G has O(nk) vertices, where n is the number of literals in the formula. Moreover, we can assume that G satisfies certain properties that we will make use of in the reduction to Set Cover. This is formalized in the following lemma.
since the sets are constructed by picking elements independently and uniformly at random. This rough argument suggests that covering all elements in U would require Θ(log |U |) sets.
We can now use I as a building block to transform our Label Cover instance G to a Set Cover instance I˜. For each edge e = (u, v), we construct an instance Ie = (Ue, Ce) equivalent to I with |Ue| = nk. We take t = |Lv| = 2k, so that each label b ∈ Lv gets mapped to a unique set Se,b ∈ Ce. Then, Ce = {Se,b | b ∈ Lv} ∪ { S¯e,b | b ∈ Lv} has size 2k+1. Next, we merge all the “edge” instances to get a large instance I˜ = ( U ,˜ C˜) for the whole graph. Let U˜ =
e∈E Ue.^ For each^ v^ ∈^ Q^2 and b ∈ Lv, let S˜v,b =
e incident on v
Se,b.
For each u ∈ Q 1 and a ∈ Lu, let
S˜u,a =
e incident on u
S^ ¯e,Re(a),
where Re(a) denotes the unique b ∈ Lv satisfying (a, b) ∈ Re. Let C˜ be the collection of all these sets. We have | U˜ | = nO(k)^ and | C|˜ = nO(k). Suppose that uncoordinated solutions to ”edge” instance Ie have size at least ` = Θ(log |Ue|) = Θ(k · log n). We will see that our reduction works if we pick k = O(log log n). Then, intuitively, if we had a “good” approximation algorithm for Set Cover, we could use it to solve our Label Cover problem (and thus 3SAT(5)) exactly in nO(log log^ n) time. Hence, Theorem 29.2.1 follows. In the following, we give a formal proof for the correctness of our reduction.
Suppose that G has an assignment satisfying all edges. By picking the sets in C˜ corresponding to this assignment, we are able to cover U˜ with only |Q 1 | + |Q 2 | = O(nk) sets. It remains to show that if G does not have any good solution, covering U˜ would require Ω(`nk) sets.
Claim 29.2.3 If every assignment to G satisfies at most an αk^ fraction of edges, then every set cover of U˜ has size at least Ω(`nk).
Proof: We start with a minimum set cover C∗. Let G 1 (resp. G 2 ) be the set of vertices in Q 1 (resp. Q 2 ) that have less than /2 sets in C∗. Let B 1 = Q 1 \G 1 and B 2 = Q 2 \G 2. Let e = (u, v) be an edge such that u ∈ G 1 and v ∈ G 2. We pick an assignment to G in the following way. For every u ∈ Q 1 , let Au denote the set of labels a in Lu such that S˜u,a ∈ C∗. Pick an answer uniformly at random from Au. Likewise, for every v ∈ Q 2 , pick an answer uniformly at random from Av. Consider some edge e = (u, v) with u ∈ G 1 and v ∈ G 2. Then, |Au| + |Av| <, so C∗ induces a coordinated solution to Ie, and so Au and Av contain a pair of “corresponding” answers, i.e. there exist a ∈ Au such that Re(a) ∈ Av. Thus, the probability of this assignment satisfying e is at least (^) /^12 · (^)/^12 = 4/`^2. The expected number of edges satisfied by the assignment is at least
#e(G 1 , G 2 ) · 4 /^2 , where #e(G 1 , G 2 ) = |E ∩ (G 1 × G 2 )|. Using the fact that the expectation never exceeds the maximum achievable value, we have #e(G 1 , G 2 ) · 4 /^2 ≤ αk^ · |E|. Therefore,
#e(G 1 , G 2 ) |E| ≤ αk^ ·
`^2
Since ^2 = O(k^2 log^2 n) and α < 1, it suffices to pick k = O(log log n) to make αk^2 / 4 < 1 /2. Recall that G is a d-regular bipartite graph for some d depending on k. If |B 1 | < |Q 1 |/4 and |B 2 | < |Q 2 |/4,
the number of edges with at least one endpoint in B 1 ∪ B 2 would be less than d(|Q 1 | + |Q 2 |)/4, implying that #e(G 1 , G 2 )/|E| > 1 /2 since |E| = d(|Q 1 | + |Q 2 |)/2. Hence, either |B 1 | ≥ |Q 1 |/4 or |B 2 | ≥ |Q 2 |/4. Since |Q 1 |, |Q 2 | > nk, the number of sets in C∗^ is at least 2 · n k 4 = Θ(n k).
Let N = O(nlog log^ n) denote the size of the Set Cover instance. If Set Cover could be approximated to within a factor of β · log N for every β > 0, we could easily tell whether our Label Cover instance has an assignment satisfying all edges in time poly(N ). By Lemma 29.2.2, NP ⊆ DTIME(N ). This proves Theorem 29.2.1.
