Question

• karger's min cut algorithm in the class has probability at least 2/n2 of returning a min-cut. how many times do you have to rerun this algorithm to get the probability of an error to be at most 1/n?

Answer 1

The Karger's algorithm relates to graph theory where G=(V,E)  is an undirected graph with |E| edges and |V| vertices.  The objective is to find the minimum number of cuts in edges in order to separate G into two disjoint graphs.  The algorithm is randomized and will, in some cases, give the minimum number of cuts.  The more number of trials, the higher probability that the minimum number of cuts will be obtained.

The Karger's algorithm will succeed in finding the minimum cut if every edge contraction does not involve any of the edge set C of the minimum cut.

The probability of success, i.e. obtaining the minimum cut, can be shown to be ≥ 2/(n(n-1))=1/C(n,2),  which roughly equals 2/n^2 given in the question.Given: EACH randomized trial using the Karger's algorithm has a success rate of P(success,1) ≥ 2/n^2.

This means that the probability of failure is P(F,1) ≤ (1-2/n^2) for each single trial.

We need to estimate the number of trials, t, such that the probability that all t trials fail is less than 1/n.

Using the multiplication rule in probability theory, this can be expressed as
P(F,t)= (1-2/n^2)^t < 1/n 

We will use a tool derived from calculus that 
Lim (1-1/x)^x as x->infinity = 1/e, and
(1-1/x)^x < 1/e   for x finite.  

Setting t=(1/2)n^2 trials, we have
P(F,n^2) = (1-2/n^2)^((1/2)n^2) < 1/e

Finally, if we set t=(1/2)n^2*log(n), [log(n) is log_e(n)]

P(F,(1/2)n^2*log(n))
= (P(F,(1/2)n^2))^log(n) 
< (1/e)^log(n)
= 1/(e^log(n))
= 1/n

Therefore, the minimum number of trials, t, such that P(F,t)< 1/n is t=(1/2)(n^2)*log(n)    [note: log(n) is natural log]