In the MAXIMUM CUT problem we are given an undirected graph$\mathit{G}\mathbf{=}\left(V,E\right)$ with a weight$\mathbf{w}\left(e\right)$ on each edge, and we wish to separate the vertices into two sets S and V − S so that the total weight of the edges between the two sets is as large as possible. For each$\mathit{S}\mathbf{\subseteq }\mathit{V}$ define$\mathbf{w}\left(S\right)$ to be the sum of all$w\left(e\right)$ over all edges$\left\{u,v\right\}$ such that $\left|S\cap \left\{u,v\right\}\right|\mathbf{=}\mathbf{1}\mathbf{.}$ Obviously, MAX CUT is about maximizing w(S) over all subsets of .

Consider the following local search algorithm for MAX CUT:

start with any$\mathbf{S}\mathbf{\subseteq }\mathbf{V}$

while there is a subset$\mathbf{S}\mathbf{\text{'}}\mathbf{\subseteq }\mathbf{V}$ such that

||S 0 | − |S|| = 1 and$\mathbf{w}\left(S\text{'}\right)\mathbf{>}\mathbf{w}\left(S\right)$ do:

set $\mathbf{S}\mathbf{=}\mathbf{S}\mathbf{\text{'}}$

1. Show that this is an approximation algorithm for MAX CUT with ratio$\mathbf{2}.$

(b) But is it a polynomial-time algorithm?

In the multiway cut problem, we are given a graph$G=\left(V,E\right)$ with k special vertices $s1,s2,...,sk\in V.$Our goal is to find the smallest set of edges which when removed from the graph disconnect the graph into at least k components where each is in a different component. When $k=2$,

this is exactly the min s-t cut problem, but if$k\ge 3$ the problem becomes $NP-hard$. Consider the following algorithm.

Let Fi be the set of edges in the minimum cut with si one one side and all other special vertices on the other side. And output is $F$, the union of all . Also note that this is a multiway cut because removing from$G$ isolates in its own component.

An approximation algorithm for MAX CUT with ratio two is defined as already given in the question In the MAXIMUM CUT problem we are given an undirected graph with a weight$w\left(e\right)$ on each edge, and we wish to separate the vertices into two sets S and V − S so that the total weight of the edges between the two sets is as large as possible. For each$S\in V$ define$w\left(S\right)$ to be the sum of all$w\left(e\right)$ over all edges such that $\left|S\in \left\{u,v\right\}\right|=1.$ Obviously, MAX CUT is about maximizing w(S) over all subsets .

Consider the following local search algorithm for MAX CUT:

while there is ||S 0 | − |S|| = 1 and$\left|S\in \left\{u,v\right\}\right|=1.$

In particular, it appears that they are solving instances of weighted MAX-CUT where the input graph is the complete graph on two thousand vertices with edge weights assigned from $\{-1,1\}$independently and uniformly at random.

$\left|S\in \left\{u,v\right\}\right|=1.$$\left|S\in \left\{u,v\right\}\right|=1.$=$\left|S\in \left\{u,v\right\}\right|=2.$

Anapproximation algorithm for MAX CUT with ratio two is proved.

there are (this is a polynomial algorithms). We can bound this through some very simple observations.

1. Initially $\left|\text{E}\left(\text{S,S¯}\right)\right|\ge \text{0,}$

2. In every iteration,$\left|\text{E}\left(\text{S,S¯}\right)\right|$ increases by at least one, and

3.$\left|\text{E}\left(\text{S,S¯}\right)\right|$ is always at most $m\ll n.$

These facts in the graph$G=\left(V,E\right)$ clearly imply that the number of iterations is at most $m\ll n.$, and thus the total running time is polynomial. To prove the next theorem about the approximation ratio, we will do something which is very common when analyzing local search algorithms. We will not reason directly about the solution which the algorithm outputs,$\left(S,T\right)$ but will instead reason about an arbitrary local optimum. Since the algorithm outputs a local optimum.

This algorithm will take polynomial time for its operation.

And hence, Yes it is a polynomial-time algorithm.