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# Approximation Algorithms for Unique Games Luca Trevisan Slides by Avi Eyal.

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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. v3v3 v1v1 v2v2 v4v4

Linear Unique Game The permutation constraints are linear v3v3 v1v1 v2v2 v4v4

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:................ Just create a spanning tree for each connected component! Try for every s ∈ S

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: 1.The Unique Game Conjecture with completeness 1-cε 4 /log 3 n and soundness 1- ε is false. 2.For linear games the conjecture with completeness 1-cε 2 /logn and soundness 1- ε is false.

ε 1-ε ε 1-cε 4 /log 3 n 1-cε 2 /logn General Unique Games Linear Unique Games ε1 General Games (for example Label Cover)

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: clique Constraints that are not satisfied by A

If we dealt with expanders… The chances that a random route would hit a polluted edge is relatively small.

A few definitions 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. From now on we assume that all vertices have the same degree d. A lazy walk on G with matrix M is: M’ = 1/2(I+M) From now on, “walk” = “lazy walk”!

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: For every vector p (over the vertices): pM = the probability distribution after the first step pM k = the probability distribution after k steps. We denote p - the probability distribution vector for the first vertex of the walk.

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 δ close to the stationary distribution.

So G is (t, δ)- mixing if for every distribution p (for a start vertex), |pM t - π| < δ. Select by p Lazy random walk t steps or more Final vertex Vertex selected by π ≈

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.

r selected from π v – the last vertex t-long random walk r selected from π v selected from π The following 2 walks have a δ-close probability:

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.

For a set S ⊆ V, π(S) := Σ u ∈ S π(u) = Σ u ∈ 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) := Σ u ∈ 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 Some more… Let λ(G) be the second largest eigenvalue of M, then ( 1- λ(G)) is the spectral gap of G. Fact: All M ’s eigenvalues are real, and ≤ 1. π has the maximal eigenvalue.

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 1 1 1 1 11 1/3

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). Simple Facts:

The charge of an edge is increased by a factor of at most (1+3h) each time. S C-S d|S | e Increase each edge by From Lemma 4:

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-cε 4 /log 3 n 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(log 3 n/ε 2 ) and δ = 1/n. Apply the “random walk” algorithm on each connected component.

ε/3 constraints removed average error ε/3 A satisfies less than 1-3γ/ε constraints ε/3 A satisfies at least 1-3γ/ε constraints.

Linear Games Definition: Graph G has diameter d if there is a vertex r ∈ V such that every vertex v ∈ V has distance at most d from r. 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 : 1.Let v be a vertex in that component. 2.i=0 3.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++. 4.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.

Example: we want d=3, t~3

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.

“bad” edge Let r get a free variable x. Each node v in the tree is assigned with a function of the form: f v (x) = ax+b If for (u,v) the equation f u,v (f u (x))=f v (x) has no solution, then (u,v) is a “bad” edge and the whole loop cannot be satisfied.

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-cε 2 /logn feasible, then only 2k*cε 2 /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

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