Presentation on theme: "Inapproximability of Hypergraph Vertex-Cover. A k-uniform hypergraph H= : V – a set of vertices E - a collection of k-element subsets of V Example: k=3."— Presentation transcript:
A k-uniform hypergraph H= : V – a set of vertices E - a collection of k-element subsets of V Example: k=3
The Vertex-Cover problem (Ek-Vertex-Cover) : Given a k-uniform hypergraph, find the smallest subset of vertices that intersects every hyperedge. The matching decision problem is NP-Complete
We will show that the problem is hard to approximate within a factor of (The actual known bound is ) We will show this by using a reduction to the Independent Set Problem Both results by Dinur, Guruswami, Khot, and Regev (2004)
Independent Set : Find the maximal set of vertices which doesn’t contain an edge. This problem is VC’s complement: is the minimal VC is the maximal IS
Proof: – a vertex cover. Suppose that isn’t an independent set. Then there is an edge e whose vertices are all in. So e isn’t represented in C – a contradiction! So all of e ‘s vertices are in S – a contradiction! - an independent set. Suppose that isn’t a vertex cover. Then there is an edge e which isn’t represented in So, if we have a maximal Independent Set, its complement is a minimal Vertex Cover!
Given a 4 -uniform hypergraph G, for every, It is NP-hard to distinguish between the following: VC is hard to approximate within We will see:
Lecture’s outline: Intersecting families and a helpful lemma Reduction from Label Cover to IS Proof of completeness and soundness
Intersecting Families Denote A family is called pairwise intersecting if for every 2 sets,
What is the maximal size of a pairwise intersecting family? Of every pair we can only choose one set Is this bound tight? i All sets such that For odd n All sets such that
For a bias parameter and a set, the weight of a set is The weight of a family is When clear from the context, we will omit the R
What is the weight of the family ? i It’s the probability that i appears in a given set: p What is the weight of the family ? Depends on p If - nearly 0 If - nearly 1 If -
3. For any, all sets such that A family is called pairwise t-intersecting if for every 2 sets, Examples: 1. All sets such that [t] 2. All sets such that t+2
(i,j)- shift on a family: replacing j with i in all sets such that and Left-shifted family: a family which is invariant with respect to ( i,j )-shift for any By iterating the ( i,j )-shift for all we eventually get a left-shifted family S(F)
Lemma 1: Left-shifting a family maintains its size and the size of the sets In the family. If F is a pairwise t -intersecting family, then so is S(F). Lemma 2: Let be a left-shifted pairwise t -intersecting family. Then, for every, there exists a with Proof
Lemma 3: For arbitrary, let. Then, there exists s.t. for any pairw ise t -intersecting Family,. Meaning, for any, a family of non-negligible weight ( ) can’t be pairwise t -intersecting for a large enough t
Correctness of Lemma3: For every, there exists a with the probability that such a j exists for a random set chosen according to. if F is pairwise t -intersecting then so is S(F), and
Look at For some Since, the prob. of a set having a big intersection with a big section ( [ t+2j ] ) is very small. So any pairwise t –intersecting family can’t be heavy for a large enough t.
Label Cover Constraints - Functions from to Variable Set Y Over Variable Set Z Over The graph is bi-regular Goal : find a labeling for X and Y that satisfies the maximal fraction of constraints a constraint is satisfied if ynyn zmzm y2y2 y1y1 z1z1 z2z2
Theorem (PCP theorem + Raz’s parallel Repetition Theorem): There exists a universal constant such that for every (large) constant R it is NP-hard to distinguish between the following two cases: No: no labeling can satisfy more than of the constraints. Yes: There is a labeling such that every constraint is satisfied by : for every
We will use the PCP theorem to prove a factor hardness for E4-Vertex-Cover, or in other words: Given a 4 -uniform hypergraph G, for every, it is NP-hard to distinguish between the following cases:
We start with a Label Cover instance as described - A local set of constraints over variables whose respective ranges are Parameters: (from Lemma 3) ( - the constant from the PCP theorem) The Reduction
From the Label Cover instance, we construct a weighted 4 -uniform hypergraph G The weight of each vertex: G ’s Vertices: ynyn y2y2 y1y1
G ’s Hyperedges: iff there’s no and such that y1y1 z y2y2 For each pair of constraints sharing a common z
Completeness: if there’s a labeling satisfying all constraints then Proof: Assume a satisfying labeling The following set is an independent set of G : We need to show : there’s no edge which consists solely of vertices of the above form y y, a a aiai ajaj akak
y1,y1, a aiai ajaj akak Suppose that an edge like that exists: y1,y1, a ajaj alal y2,y2, b bibi bjbj y2,y2, b bkbk y1y1 a y2y2 b But we have: By the definition of the edges, this is a contradiction!
What is the Independent Set’s weight? =Pr choosing a vertex out of this IS, when picked according to A’s weight
Soundness: if no assignment satisfies more than of the constraints in the Label Cover instance then The same as: if then there exists an assignment that satisfies more than a fraction of the constraints in the Label Cover instance. We need to prove:
Proof: Let be an independent set of size at least. Look at, the set of all variables y for which the weight of in is at least.S ynyn y2y2 y1y1 Y’
What is the size of ? If less than of the blocks have an intersection smaller than, S ’s size can’t be at least
For each, define. y y y y FyFy So, we have: for each, We saw (L3), that for, a family of weight can’t be pairwise t -intersecting for our choice of t ( )
Therefore, there must exist s.t. We shall call the core of assignments for y. Intuitively, for any y for which V[y] has a large intersection with S, we will match B(y) as potential values. We will translate the cores into an assignment satisfying More than a fraction of the constraints.
For every and such that are constraints, we must have y1,y1, A1A1 y1,y1, y2,y2, y2,y2, A2A2 C1C1 C2C2 An edge in the IS – a contradiction! Otherwise…
Let denote the set of all Z variables That participate in some constraint with some. ynyn zmzm y2y2 y1y1 z1z1 z2z2 Y’ Z’
We associate each with an arbitrary for which there exists y1y1 z y2y2 Y’ Z’ Denote
We have matched each variable a set of values B(y) We have matched each variable a set of values B(z) We will now use these values to define a labeling Which will satisfy a sufficient number of constraints
We define a random assignment : For each we independently select a random value from B(y) For each we independently select a random value from B(z) For the rest of the variables we assign any arbitrary value ynyn zmzm y2y2 y1y1 z1z1 z2z2 B(y1)B(y1) B(yn)B(yn) B(z2)B(z2) RZRZ RZRZ RyRy
We now show that for every, each constraint is satisfied by with probability We saw: so there is at least one value s.t. Look at for some Assume that z is associated with some y z y’y’ or y z y’=
For every, and so So there is at least prob. of having y z B(y)B(y) aiai ajaj akak B(z)B(z) blbl b
Variables from Y’ participate in of the constraints And we are done! There exists some assignment that meets the expectation. satisfies of the constraints We chose: so the expected number of local constraints satisfied by is We saw: How many constraints are satisfied?
The reduction’s complexity: The number of vertices in the hypergraph is at most:
And for any k? We prove for any even k as follows: Pairwise t-intersecting K/2-wise t-intersecting And build a k-uniform hypergraph. To prove for any odd k, we add a new vertex to each edge in a (k-1)-uniform hypergraph. Obviously, that doesn’t change the size of the vertex cover
And for an unweighted Hypergraph? It is proved possible to turn a weighted Hypergraph into an unweighted one, in polynomial time.
Lemma 2: Let be a left-shifted s -wise t -intersecting family. Then, for every, there exists a with Proof: For two sets A, B s.t. |A|=|B|=l, We say that if for all
We first prove: For, if then also Proof by induction: for i=0,…,l, let and therefore We now show that implies that : If then, and the claim holds. Otherwise, and hence F is left-shifted and => Since, we are done.
Assume, by contradiction, that there exists such that for all, Let, and A has at least i “holes” in Therefore, A has at least one “hole” for each s elements from t to n Therefore, Since F is s -wise t -intersecting, the size of each set must be greater than t, and therefore
For, define the set by Since are all integers, And so, or in other words, So, and therefore Notice that: We saw: Since, we have
We saw: For, if then also A contradiction! But F is s -wise t -intersecting Therefore, since, for all, We get: Lemma 2 is proved: For every, there exists a with