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Approximation Algorithms for Unique Games. Luca Trevisan Slides by Avi Eyal. What’s in the Lecture:. Definition of “Unique Game” The Unique Game Conjecture General Unique Game Limitation Linear Unique Game Limitation. General Unique Game.
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Approximation Algorithms for Unique Games Luca Trevisan Slides by Avi Eyal
What’s in the Lecture: • Definition of “Unique Game” • The Unique Game Conjecture • General Unique Game Limitation • Linear Unique Game Limitation
General Unique Game A trio (G=(V,E), {f(e)}, S) where S is a set of possible values to V, and f(e) is a permutation between the vertices of e. v1 v4 v3 v2
Linear Unique Game The permutation constraints are linear v1 v4 v3 v2
The Unique Game Conjecture Given a (G(V,E), {f(e)}, S), it is NP hard to decide between: Completeness 1-γ Soundness γ
What about a (γ,1) gap? That would be easy, since a value of a node forces the values of all it’s neighbors: . . . . . . . . Try for every s∈S Just create a spanning tree for each connected component!
What if we new that the graph has a value of at least (1-γ)? From now on we shall denote A as the assignment that satisfies (1-γ) of the constraints.
Theorems: • The Unique Game Conjecture with completeness 1-cε4/log3n and soundness 1- ε is false. • For linear games the conjecture with completeness 1-cε2/logn and soundness 1- ε is false.
General Games (for example Label Cover) ε 1 General Unique Games 1-cε4/log3n ε 1-ε Linear Unique Games 1-cε2/logn ε 1-ε
Algorithm for General Games We would like to use the spanning tree algorithm we’ve seen before. . . . . . . . . Just pick a random vertex and get going!
But what if our graph looks like this: Constraints that are not satisfied by A clique clique
If we dealt with expanders… The chances that a random route would hit a polluted edge is relatively small.
A few definitions From now on we assume that all vertices have the same degree d. A random walk matrix: M(u,v) = 1/d if (u,v)∈E, else M(u,v)=0 A random walk: If the next step starts from vertex u, then each of u’s edges have the same probability to be next. A lazy walk on G with matrix M is: M’ = 1/2(I+M) From now on, “walk” = “lazy walk”!
We denote p - the probability distribution vector for the first vertex of the walk. For every vector p (over the vertices): pM = the probability distribution after the first step pMk = the probability distribution after k steps. π(u) = d/2|E|(=1/n) is the stationary distribution for M. πM = π The statistical distance between 2 distributions p,q is: Alternative definition:
Fact: For every probability distribution and for every random walk matrix we have: Mixing Time(t, δ) - after t steps of a random walk starting on an arbitrary vertex, we come δclose to the stationary distribution.
So G is (t, δ)-mixing if for every distribution p (for a start vertex), |pMt - π| < δ. Final vertex Lazy random walk t steps or more Vertex selected by π Select by p ≈
Working on Expanders • Choose a random node r (by π!) and guess A(r) (A is the best assignment). • Pick random walks (of length t) in the graph to all nodes. Assign values to the nodes consistent with A(r) and with the path.
Lemma 1: Given a 1-γ value unique game such that G is (t, δ)-mixing, the above algorithm finds an assignment that satisfies on average at least 1-2(tγ+ δ) of the constraints.
Lemma 2: The probability that v gets a value different than A(v) is at most tγ+δ.
The following 2 walks have a δ-close probability: r selected from π r selected from π t-long random walk v – the last vertex v selected from π
Every edge (in every step) has the same 1/|E| probability! Up to γ of the edges are spoiled. → The chance to meet a spoiled edge ≤ tγ. → The chance to meet a spoiled edge when v is chosen from π≤ tγ + δ. → At the end of the algorithm each edge has a 2(tγ + δ) probability to be spoiled (tγ + δ from each of it’s vertices).
The Idea Decompose the graph into connected components with good (t, δ). Run the “random walk algorithm” on each connected component.
Some more… For a set S⊆V, π(S) := Σu∈Sπ(u) = Σu∈S d/2|E| = |S|d/2|E| For a cut (S, V-S), Q(S, V-S) is half the fraction of edges that cross the cut. Q(S, V-S) := Σu∈S, v∉S π(u)M(u,v) = |e(S, V-S)|/2|E| For a cut with π(S) ≤ ½, the conductance of the cut: h(S):= Q(S, V-S) / π(S) = |e(S, V-S)|/|S|d Fact: All M’s eigenvalues are real, and ≤ 1. π has the maximal eigenvalue. Let λ(G) be the second largest eigenvalue of M, then (1- λ(G)) is the spectral gap of G.
Relations: Lemma 3 (fact): Every graph G with λ(G)<1 is (k, δ)-mixing with Lemma 4 (fact): It is possible to find in polynomial time a cut of conductance
Lemma 5: Given G=(V,E) and 0<ε≤1/2, it is possible to find in polynomial time a subset E’⊆E, with at least(1- ε)|E| edges, such that in the graph G’=(V,E’) every connected component C gives:
Algorithm for Lemma 5: Set: For each connected component C, if λ(C)≤λ we are done. Otherwise, use Lemma 4 to find a cut of conductance ≤ h, and remove the edges of that cut from E’.
Charges • Each edge gets a charge 1 to begin with. • When we remove the edges of (S, C-S), we distribute the charges of the deleted edges among the edges of S. 1 1 1 1/3 1 1 1/3 1 1 1 1 1 1/3 1 1
Simple Facts: • The sum of the edges charges in E’ is |E| throughout the algorithm. • In every connected component, all edges have the same charge. • The charge of an edge is increased at most log|E| times throughout the algorithm. (because π(S)≤1/2).
C-S d|S| e The charge of an edge is increased by a factor of at most (1+3h) each time. S From Lemma 4: Increase each edge by
Conclusion At the end, each edge has weight at most Which means that |E’|≥(1-ε)|E|.
Back to Theorem 1 The Unique Game Conjecture with completeness 1-cε4/log3n and soundness 1- ε is false.
Remove ε/3 of the edges to get connected components in which : • 1-λ(G) ≥ O((ε/logn)2) • (t, δ)-mixing with t = O(log3n/ε2) and δ = 1/n. • Apply the “random walk” algorithm on each connected component.
ε/3 constraints removed A satisfies at least 1-3γ/ε constraints. average error ε/3 ε/3 A satisfies less than 1-3γ/ε constraints
Linear Games Definition: Graph G has diameter d if there is a vertex r∈Vsuch thatevery vertexv∈Vhas distance at mostdfromr. Lemma: There is a polynomial time algorithm that on input G=(V,E) and t>1, returns a subset E’⊆E and |E’|≥|E|/t, such that every connected component in the graph G’=(E’,V) has diameter at most (1+log|E|)/(log t).
The algorithm E’ = E While there is a connected component with diameter > d: • Let v be a vertex in that component. • i=0 • While the number of edges in the cut B(v,i), V-V(v,i) > t-1 times the number of edges in B(v,i), i++. • Remove from E’ the edges in the cut. We increase i only if the number of edges is increased by a factor of t. Therefore the radius is always ≤ 1 + log|E|/log t.
Lemma: Given a Linear Unique Game with diameter ≤ t, and given r ∈ V, root of a spanning tree of depth ≤ t, it is possible to either find a satisfactory assignment, or to find 4t+2 edges that cannot be all satisfied at one time.
Let r get a free variable x. Each node v in the tree is assigned with a function of the form: fv(x) = ax+b If for (u,v) the equation fu,v(fu(x))=fv(x) has no solution, then (u,v) is a “bad” edge and the whole loop cannot be satisfied. “bad” edge
Algorithm for Theorem 2 Given G and ε>0, delete up to ε|E|/2 edges to get connected components each with diameter k ≤ O(log|E|/ ε). Find a spanning tree of depth ≤ k. Remove every unsatisfiable loop found. If the graph is 1-cε2/logn feasible, then only 2k*cε2/logn ~ ε/2 of the edges will be deleted. In total we have deleted only ε of the edges.
d-to-d Game If (u,v) are neighbors then every value of u can match d values in v 3-coloring of a graph is a 2-to-2 game