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Vertex cover might be hard to approximate within 2 - ε Subhash Khot, Oded Regev Slides by: Ofer Neiman

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Given a graph G(V,E), VC(G) is the minimal set of vertices touching every edge. Vertex Cover Assuming only P≠NP the best known hardness factor is 1.36. 2-approximation is easy.

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Independent set Given a Graph G(V,E), IS(G) is the maximal set of vertices without edges between them. We saw it’s the complement of Vertex Cover.

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Bottom Line With certain assumption, it is NP- hard to distinguish between: IS(G)>½-2ε IS(G)<δ VC is NP-hard to approximate within 2-ε’ VC(G)≤½+ε VC(G)≥1-δ

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Overview Unique games conjecture Unique label cover Strong label cover Graph construction Completeness Soundness

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Unique games conjecture Unique: answer of one prover determines the answer of the other. Prover 2 Prover 1 verifier Acceptance map Probabilistically choose q1 q2 Strategy p 1 :X→R Strategy p 2 :Y→R a1 a2

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Unique Games Conjecture There’s a permutation for every pair of questions: π xy :R→R. verifier accepts iff π xy (p 1 (x))=p 2 (y). NP-hard to decide whether Success probability >1-ς Success probability <γ

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Unique Label Cover An assignment A :XUY→R that satisfies the maximal weight of constraints. x1x1 xnxn x2x2 y1y1 y2y2 ymym Labeled from R π xy constraints Π xy :R→R One to one And onto! w xy Weights w xy ≥0 Complete bi- partite graph

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Notations A constraint π xy is satisfied if π xy )A(x))=A(y). w(x) = w A (x) is the weight of the satisfied x -constraints. is the total weight of satisfied constraints.

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Unique game=unique-LC Unique-LC conjecture: For any ς,γ>0 there is constant |R| such that it is NP-hard to distinguish between: 1.There is an assignment A, such that w A >1-ς. 2.For all assignments A, w A <γ. For a unique-LC with label set R such that

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Why 1-ς ?? The gap of (γ,1) is easy: There’s a linear algorithm determining whether a Unique- LC can have all its tests satisfied. x1x1 xnxn x2x2 y1y1 y2y2 ymym {1,2,…R} π xy x1x1 y1y1 y2y2 x2x2 xnxn ymym

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Is this enough?? New hardness results: Min-2SAT-Deletion. Not-All-Equal predicate on 3 variables. a hardness for vertex cover. Closing the gap requires stronger tools….

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Strong Label Cover x1x1 xnxn x2x2 y1y1 y2y2 ymym Labeled from R π xy constraints Π xy :R→R One to one And onto! No weights! Degree d d left degree bi- partite graph

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Theorem Assuming the unique games conjecture: For any ς,γ>0 there are constants d,|R| such that it is NP- hard to distinguish between: 1.There is an assignment in which at least 1-ς fraction of the X vertices have all of their tests satisfied. 2.No assignment satisfies more than γ fraction of the tests. For a Strong-LC with label set R and left degree d.

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Road map Unique-LC Unique-Lc + Weights adjustements Strong-LC Unique-Lc + Weights rounding

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Lemma 1 For any ς,γ,β>0 there is constant |R| such that it is NP-hard to distinguish between: 1. There is an assignment A such that 1-β fraction of the X vertices have w A (x)>1-ς. 2. For any assignment A at most β fraction of the X vertices have w A (x)>γ. For a Unique-LC with label set R and the property that for all w(x)=1.

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Construction Given (X’,Y’,Π’,W’) with parameters ς’,γ’ to be chosen later, let l be a large constant. x1x1 x2x2 X k(x) xy y

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Proof Fact 2: (l-1)|X’|≤|X|≤l|X’| Fact 1: w(x)=1

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Completeness Assume there’s an assignment A’ for which w’ A ≥1-ς’. Define A(x i )=A’(x), A(y)=A’(y). Objective: 1-β fraction of the X vertices have w A (x)>1-ς. Step 1: Step 2: for at least X vertices

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Soundness Assume by contradiction: There is an assignment A for which β fraction of the X vertices have w A (x)>γ. Objective: build an assignment A’(x)=A(x 1 ) for which w’ A >γ’. w.l.o.g: A(x 1 )=A(x 2 )=…=A(x k(x) ) claim

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Lemma 2 We can change the unique-LC from lemma 1 a little… For a large integer α=O(Y) and some integer i. w xy

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Proof of Strong-LC So far we have (X’,Y’,Π’,W’) with parameters ς’,β’,γ’>0 to be chosen later, and a constant |R| such that: w’(x)=1. w xy = i / α

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Construction x1x1 xnxn x2x2 y1y1 y2y2 ymym x1x1 Complete bi- partite graph π’ xy w’ xy d -left regular π’ xy X,(y 1,y 2,…,y d ) x1x1 x2x2 X k(x) y1y1 y2y2 ydyd

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The new graph x x x x x y1y1 y2y2 ymym x x x x x x x x x x d -Left regulat For each we create a large (constant) set of vertices. Y=Y’.

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The new graph Weightless. d -left regular. |X|=α d |X’|, since:

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Completeness Assume A’ is an assignment such that 1-β’ fraction of the X’ vertices have w’ A (x)>1-ς’. For each denote Y x to be the set of vertices such that π’ xy is satisfied. We define A(x i )=A’(x) if x i was created from x. A(y)=A’(y). Objective: an assignment A such that for 1-ς of X vertices all the test are satisfied.

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Completeness x y1y1 y2y2 ymym yes no yes xixi y1y1 y2y2 ydyd x i connected to Y x only

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For each having w’ A (x)>1-ς’, there are at least vertices connected only to Y x Counting… How many good vertices are there??? The total number is:

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A’(y)=A(y). A’(x)=i, where is the label maximizing w A’ (x). Soundness Assume no assignment A’ to the unique-LC has more than β’ fraction of the X’ vertices with w A’ (x)>γ’. Assume we have A : an assignment to the Strong-LC, we will build A’ as follows: Objective: show that no assignment satisfy more than γ

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Observations For 1-β’ fraction of the X’ vertices, after fixing labels for Y’ any assignment will satisfy at most γ’ test weight. For any such x, let A vertex created from a tuple in Z x will have at most 1 satisfied test. There are many such vertices…

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Counting 1-β’ of the x in X’ create at least good vertices (one successful test at most). β’ fraction of |X’|: d satisfied edges. fraction 1 satisfied edge fraction d sat edges X

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End of part 1 The total fraction of tests satisfied by A is at most γ.

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Part 2: The Graph (X,Y,Ē,Π) x1x1 xnxn x2x2 y1y1 y2y2 ymym Degree d There is A such that 1-ς of the X vertices have all tests satisfied For all A at most γ tests satisfied ς,γ>0 chosen later w(IS(G)) > ½ -2ε w(IS(G)) < δ G(V,E)

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Construction x1x1 xnxn x2x2 x1x1 x1x1 xnxn x2x2 x1x1 B[x 1 ] B[x 2 ] B[x n ]

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Construction For each define Fix p=½-ε

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Construction x1x1 y πx1yπx1y x2x2 πx2yπx2y x2x2 x1x1 F G B[x 1 ] B[x 2 ] Edge exists between all pairs for whom π x 1 y )F)∩π x 2 y (G)=Ø

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Completeness X 0 will be the set of vertices for which all the tests are satisfied. x IS(x) A(x)

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Let Assume by contradiction: x1x1 y πx1yπx1y x2x2 πx2yπx2y

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Weight? Weight(IS(x)) = p = ½-ε Weight(IS) = (1-ς)(½-ε) ≥ ½-2ε,For all

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Soundness For all A at most γ tests satisfied w(IS(G))<δ Assume by contradiction that there is independent set I of weight δ. Definitions

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Monotone families A family F is called monotone if and implies. F (x) is a monotone family: Reminder:, For all

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Idea x X X* L[x] Lemma: there is a constant h such that for all for all

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x1x1 x2x2 X* X L[x 1 ] L[x 2 ] y

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Assuming lemma Let be the set a vertices having a test with some. For any choose that there is a test π x(y)y x1x1 y1y1 y2y2 x2x2 xnxn ymym X* Y* L[y]=π x(y)y (L[x(y)])

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End of proof Claim For all we have A(x), A(y) are randomly chosen from L[x], L[y] With probability at least we have π xy (A(x))=A(y) The expected fraction of satisfied tests: Choosing yields contradiction

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