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1. Karger’s min-cut Algorithm Help
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2. QUESTION1: How can Karger's algorithm be used to find
the minimum k-cut in an undirected graph, where a k-
cut is a partition of the vertices into k disjoint sets such
that removing k-1 edges separates the sets?
Answer: Karger's algorithm can be adapted to find the
minimum k-cut in an undirected graph by using a
modified version of the algorithm called Karger’s-
Stein algorithm. This algorithm was proposed by
David Karger’s and Clifford Stein in 1996.
The Karger-Stein algorithm works by recursively
partitioning the graph into smaller and smaller
subgraph

3. until the size of each subgraph is sufficiently small. At
each step, the algorithm randomly selects a small
number of vertices and contracts them into a single
vertex, effectively reducing the size of the graph. The
algorithm then runs Karger's algorithm on the
contracted graph to find a minimum cut. Finally, the
algorithm expands the contracted vertices to reveal
the original vertices in the minimum cut.
To find the minimum k-cut, the Karger-Stein
algorithm runs Karger's algorithm on the original
graph k times, each time identifying a minimum cut.
However, since the minimum cut found may not be a
k-cut, the algorithm recursively partitions the graph
and runs

4. Karger's algorithm on the subgraphs until it
identifies a k-cut.
The Karger-Stein algorithm has a running time of
O(n^2 log k) and can be implemented using either a
recursive or iterative approach. It is a practical and
efficient algorithm for finding minimum k-cuts in
real-world applications, such as in VLSI design and
image segmentation.

5. Question 2: Show that the probability of Karger's algorithm
failing to find the minimum cut in an undirected graph
decreases exponentially with the number of iterations.
Answer: To show that the probability of Karger's
algorithm failing to find the minimum cut in an
undirected graph decreases exponentially with the
number of iterations, we need to use a probabilistic
argument based on the law of total probability.
Suppose that we have an undirected graph G = (V, E)
with a minimum cut of size k, and we run Karger's
algorithm for t iterations. Let P(t) denote the
probability that Karger's algorithm fails to find the

6. minimum cut after t iterations. Then, by the law of
total probability, we have:
(t) = Σi=1,k-1 Pr[min-cut has size i] * Pr[Karger's
algorithm fails to find a cut of size i after t iterations]
Here, Pr[min-cut has size i] denotes the probability
that the minimum cut in the graph has size i. Since
there are C(n, i) ways to choose a cut of size i, and
there are C(n-1, i-1) ways to choose a cut that
separates a particular vertex from the rest of the graph,
we have:
Pr[min-cut has size i] = C(n, i) * C(n-1, i-1) / 2^(n-1)
This expression gives the probability that a randomly
chosen cut has size i.

7. Now, let us consider the second term in the equation
for P(t), which is the probability that Karger's
algorithm fails to find a cut of size i after t iterations.
By the analysis of Karger's algorithm, we know that
the probability of failure after t iterations is at most
1/n^2. Therefore, we have:
Pr[Karger's algorithm fails to find a cut of size i after t
iterations] ≤ (1/n^2)^t
Substituting these expressions into the equation for
P(t), we get:
P(t) ≤ Σi=1,k-1 C(n, i) * C(n-1, i-1) / 2^(n-1) *
(1/n^2)^t
Now, we can simplify the expression by using the fact
that C(n, i) is maximized when i = n/2 (or n/2-1 when

8. P(t) ≤ (n-1) * C(n, n/2) / 2^(n-1) * (1/n^2)^t
Using Stirling's approximation for factorials, we can
show that C(n, n/2) is approximately 2^(nH(1/2)),
where H(x) is the entropy function. Thus, we have:
P(t) ≤ (n-1) * 2^(nH(1/2)-n+1) * (1/n^2)^t
Finally, we can see that the expression on the right-
hand side is a decreasing exponential function of t,
which implies that the probability of failure decreases
exponentially with the number of iterations. Therefore,
as we run Karger's algorithm for more iterations, the
probability of failure becomes smaller and smaller,
and the algorithm becomes more likely to find the
minimum cut.