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Given an undirected graph G=(V,E), the MAX-CUT problem is to partition the vertex set V into two disjoint sets U and W so that the number of edges crossing the cut from U to W is maximized. (An edge crosses the cut if it has one endpoint in U and one in W.)

A simple (1/2)-approximation algorithm for this problem is:

  1. Choose a random subset U of V, let W=V-U.

Then each edge has probability 1/2 of crossing the cut, so the expected size of the cut is |E|/2. Since OPT ≤ |E|, this is a (1/2)-approximation algorithm.

This was the best algorithm known until Goemans and Williamson gave a better performance guarantee for an algorithm based on SemidefiniteProgramming .

Given an undirected graph G=(V,E) with n vertices, consider the following optimization problem:


Embed the vertices of G on the surface of the unit sphere in ℜn,
so as to maximize u,v [1-y(u).y(v)]/2,
where y(v) denotes the point where vertex v is embedded.

Note: [1-y(u).y(v)]/2 is the Euclidean distance between u and v, squared, over 4. So the cost is the sum, over the edges, of the distances between the endpoints squared, over 4.


  1. Let G = path of length 3.
  2. Let G = the triangle.
  3. Let G = 5-cycle.

A feasible solution to EMBED is an embedding (assigning each vertex to a point on the sphere).
The cost of the solution is u,v [1-y(u).y(v)]/2.

claim 1: This optimization problem is a relaxation of max-cut.
proof sketch:

Prove that, for each cut in G, there is a feasible solution whose cost equals the size of the cut.

claim 2: This optimization problem can be solved in polynomial time using SDP.
proof sketch: Here is the corresponding SDP:

maximize u,v [1-Xuv]/2 subject to
Xuu = 1 for each vertex u
X is positive semidefinite.

Each feasible solution to EMBED gives a feasible solution to the SDP: X_{uv} = y(u).y(v).
Conversely, since X is positive semidefinite if and only if it can be expressed as X = YT Y,
each feasible solution to the SDP gives a feasible solution to the SDP.
The costs of the feasible solutions correspond.

The algorithm:
  1. Given the graph G with n vertices, compute y*∈ℜn - the optimal embedding onto the unit sphere.
  2. Choose a random hyperplane in ℜn through the origin.
  3. Let U be the vertices v such that y(v) is on one side of the hyperplane, let W be the vertices on the other side.

thm: This is a .878-approximation algorithm for MAX-CUT.
Proof sketch:

Consider a single edge (u,v) in the graph, with vertices embedded at y(u) and y(v) respectively.
What is the probability that (u,v) crosses the cut? We claim this is at least .878*[1-y(u).y(v)]/2.
This proves the result because then the expected number of edges cut is at least
.878 uv [1-y(u).y(v)]/2, which is at least .878 OPT, because EMBED is a relaxation of MAX-CUT.
Let θ denote the angle between y(u) and y(v). That is, cos(theta) = y(u).y(v).
Then the probability that y(u) and y(v) lie on opposite sides of the cut is 2θ/2π. (Draw a picture.)
So, we need to show 2θ/2π ≥ .878 [1-y(u).y(v)]/2 = .878 [ 1-cos(θ)/2]..
We leave this as an exercise.

  1. Chapter 26 of Approximation Algorithms by Vijay Vazirani
  2. [M. Goemans and D. Williamson. A 0.878 approximation algorithm for MAX-2SAT and MAXCUT. In Proc. 26th ACM Symp. on Theory of Computing, pages 422--431, 1994.]

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Last edited December 6, 2004 3:38 pm by NealYoung (diff)