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Inapproximability of Hypergraph Vertex-Cover

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A k-uniform hypergraph H= : V – a set of vertices E - a collection of k-element subsets of V Example: k=3

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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

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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)

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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

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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!

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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:

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Lecture’s outline: Intersecting families and a helpful lemma Reduction from Label Cover to IS Proof of completeness and soundness

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Intersecting Families Denote A family is called pairwise intersecting if for every 2 sets,

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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

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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

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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 -

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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

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(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)

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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

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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

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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

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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.

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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

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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

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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:

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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

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From the Label Cover instance, we construct a weighted 4 -uniform hypergraph G The weight of each vertex: G ’s Vertices: ynyn y2y2 y1y1

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G ’s Hyperedges: iff there’s no and such that y1y1 z y2y2 For each pair of constraints sharing a common z

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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

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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!

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What is the Independent Set’s weight? =Pr choosing a vertex out of this IS, when picked according to A’s weight

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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:

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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’

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What is the size of ? If less than of the blocks have an intersection smaller than, S ’s size can’t be at least

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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 ( )

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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.

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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…

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Let denote the set of all Z variables That participate in some constraint with some. ynyn zmzm y2y2 y1y1 z1z1 z2z2 Y’ Z’

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We associate each with an arbitrary for which there exists y1y1 z y2y2 Y’ Z’ Denote

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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

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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

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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’=

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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

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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?

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The reduction’s complexity: The number of vertices in the hypergraph is at most:

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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

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And for an unweighted Hypergraph? It is proved possible to turn a weighted Hypergraph into an unweighted one, in polynomial time.

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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

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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.

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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

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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

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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

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