Problem:
Let "output" denote the cut output by Karger's min cut algorithm on a given connected graph with $ n $ vertices, and let $ p=\frac{1}{\left(\begin{array}{c}n \\ 2\end{array}\right)} $. Which of the following statements are true?
[Choices elided]
Solution:
We know that the probability of finding any fixed minimum cut is lower bounded $ p $, so these two options is automatically true.
For every graph G with n nodes, there exists a min cut (A,B) such that
$ Pr[out=(A,B)] \ge p $
For every graph G with n nodes and every min cut (A,B) of G,
$ Pr[out=(A,B)] \ge p $
What is non-trivial is that whether or not there exists a graph and a min-cut such that the probability of finding it is at most $ p $. This is possible, for example, a graph with only two nodes and one edge, the probability of finding the min cut is 1 and that is also the same value for $ p $. Therefore this is also a correct option.
There exists a graph G with n nodes and a min cut (A,B) of G,
$ Pr[out=(A,B)] \le p $
This is a tricky question and I screwed this one the first time.
Let "output" denote the cut output by Karger's min cut algorithm on a given connected graph with $ n $ vertices, and let $ p=\frac{1}{\left(\begin{array}{c}n \\ 2\end{array}\right)} $. Which of the following statements are true?
[Choices elided]
Solution:
We know that the probability of finding any fixed minimum cut is lower bounded $ p $, so these two options is automatically true.
For every graph G with n nodes, there exists a min cut (A,B) such that
$ Pr[out=(A,B)] \ge p $
For every graph G with n nodes and every min cut (A,B) of G,
$ Pr[out=(A,B)] \ge p $
What is non-trivial is that whether or not there exists a graph and a min-cut such that the probability of finding it is at most $ p $. This is possible, for example, a graph with only two nodes and one edge, the probability of finding the min cut is 1 and that is also the same value for $ p $. Therefore this is also a correct option.
There exists a graph G with n nodes and a min cut (A,B) of G,
$ Pr[out=(A,B)] \le p $
This is a tricky question and I screwed this one the first time.
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