Download presentation

Presentation is loading. Please wait.

Published byEmilia Rundell Modified over 4 years ago

1
Vertex cover might be hard to approximate within 2 - ε Subhash Khot, Oded Regev Slides by: Ofer Neiman

2
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.

3
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.

4
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

5
Overview Unique games conjecture Unique label cover Strong label cover Graph construction Completeness Soundness

6
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

7
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

8
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

9
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.

10
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

11
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

12
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….

13
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

14
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.

15
Road map Unique-LC Unique-Lc + Weights adjustements Strong-LC Unique-Lc + Weights rounding

16
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.

17
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

18
Proof Fact 2: (l-1)|X’|≤|X|≤l|X’| Fact 1: w(x)=1

19
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

20
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

21
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

22
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

23
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

24
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’.

25
The new graph Weightless. d -left regular. |X|=α d |X’|, since:

26
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.

27
Completeness x y1y1 y2y2 ymym yes no yes xixi y1y1 y2y2 ydyd x i connected to Y x only

28
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:

29
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

30
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…

31
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

32
End of part 1 The total fraction of tests satisfied by A is at most γ.

33
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)

34
Construction x1x1 xnxn x2x2 x1x1 x1x1 xnxn x2x2 x1x1 B[x 1 ] B[x 2 ] B[x n ]

35
Construction For each define Fix p=½-ε

36
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

37
Completeness X 0 will be the set of vertices for which all the tests are satisfied. x IS(x) A(x)

38
Let Assume by contradiction: x1x1 y πx1yπx1y x2x2 πx2yπx2y

39
Weight? Weight(IS(x)) = p = ½-ε Weight(IS) = (1-ς)(½-ε) ≥ ½-2ε,For all

40
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

41
Monotone families A family F is called monotone if and implies. F (x) is a monotone family: Reminder:, For all

42
Idea x X X* L[x] Lemma: there is a constant h such that for all for all

43
x1x1 x2x2 X* X L[x 1 ] L[x 2 ] y

44
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)])

45
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

Similar presentations

OK

Graphing Lines Day 0ne. Cover the concepts: Relation Function

Graphing Lines Day 0ne. Cover the concepts: Relation Function

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google