A(iv) Assuming that the graph G = (V, E) is represented in Adjacency List format, justify in detail the fact GreedvIS can be implemented in O(n² + m) worst-case running time, where n = |V|, m = |E|. note: this will require you to take real care in how the adjustment of Adj is done in line 10. The key is to only update/delete what is really necessary for the Algorithm, rather than being concerned with an accurate representation of the residual graph. A(v) Consider an unweighted bipartite graph G = (LUR, E), with L= = {₁,...,U[n/3]}{W₁, W2, W3, W₁}, and R = {₁,...,Vn}. The edge set E=E₁ UE2 consists of the following edges: E₁ = {(ui, v₁): i = 1, ..., [n/3]} U {(ui, Vi+[n/3]): i = 1,..., [n/3]} U {(ui, Vi+2[n/3]): i = 1,..., [n/3]} {(wi, vj) i = 1,. ‚ . . .‚ 4, j = 1, . ,n} E2 = Essentially, the E2 edges form a complete bipartite graph between the 4 special vertices {W₁, W2, W3, W4 and the set R, while the edge set E₁ is a subgraph where each of the vertices {u₁,..., un/3]} have 3 adjacent edges and all of the vertices in R have 1 adjacent edge. Work out the independent set I which will be constructed by GreedylS, justifying your answer with respect to the degrees of the different vertices as the algorithm proceeds. Show that this set will be approximately 1/3 of the size of the true maximum independent set for this graph? note: You are welcome to assume n is a multiple of 3. A(vi) How would you alter the graph construction of (v) to get progressively worse "approximation factors" for the result returned by GreedyIS?
A(iv) Assuming that the graph G = (V, E) is represented in Adjacency List format, justify in detail the fact GreedvIS can be implemented in O(n² + m) worst-case running time, where n = |V|, m = |E|. note: this will require you to take real care in how the adjustment of Adj is done in line 10. The key is to only update/delete what is really necessary for the Algorithm, rather than being concerned with an accurate representation of the residual graph. A(v) Consider an unweighted bipartite graph G = (LUR, E), with L= = {₁,...,U[n/3]}{W₁, W2, W3, W₁}, and R = {₁,...,Vn}. The edge set E=E₁ UE2 consists of the following edges: E₁ = {(ui, v₁): i = 1, ..., [n/3]} U {(ui, Vi+[n/3]): i = 1,..., [n/3]} U {(ui, Vi+2[n/3]): i = 1,..., [n/3]} {(wi, vj) i = 1,. ‚ . . .‚ 4, j = 1, . ,n} E2 = Essentially, the E2 edges form a complete bipartite graph between the 4 special vertices {W₁, W2, W3, W4 and the set R, while the edge set E₁ is a subgraph where each of the vertices {u₁,..., un/3]} have 3 adjacent edges and all of the vertices in R have 1 adjacent edge. Work out the independent set I which will be constructed by GreedylS, justifying your answer with respect to the degrees of the different vertices as the algorithm proceeds. Show that this set will be approximately 1/3 of the size of the true maximum independent set for this graph? note: You are welcome to assume n is a multiple of 3. A(vi) How would you alter the graph construction of (v) to get progressively worse "approximation factors" for the result returned by GreedyIS?
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![A(iv) Assuming that the graph G = (V, E) is represented in Adjacency List format, justify in detail the
fact GreedvIS can be implemented in O(n² + m) worst-case running time, where n = |V], m = |E|.
note: this will require you to take real care in how the adjustment of Adj is done in line 10.
The key is to only update/delete what is really necessary for the Algorithm, rather than being
concerned with an accurate representation of the residual graph.
A(v) Consider an unweighted bipartite graph G = (LUR, E), with L = :{u₁,...,U[n/3]}U{w₁, W2, W3, W₁},
and R = {₁,..., Un}. The edge set E = E₁ UE2 consists of the following edges:
E₁
=
{(ui, v₁): i = 1,
, [n/3]} U
{(ui, Vi+[n/3]): i = 1,..., [n/3]} U
{(Ui, Vi+2[n/3]): i = 1, . . . ,
1,..., [n/3]}
{(wi, vj) i = 1, ..., 4, j = 1, ..., n}
:
E2 =
Essentially, the E2 edges form a complete bipartite graph between the 4 special vertices {w₁, W2, W3, W4}
and the set R, while the edge set E₁ is a subgraph where each of the vertices {u₁,...,U[n/3]} have
3 adjacent edges and all of the vertices in R have 1 adjacent edge.
Work out the independent set I which will be constructed by GreedyIS, justifying your answer with
respect to the degrees of the different vertices as the algorithm proceeds. Show that this set will be
approximately 1/3 of the size of the true maximum independent set for this graph?
note: You are welcome to assume n is a multiple of 3.
A(vi) How would you alter the graph construction of (v) to get progressively worse "approximation
factors" for the result returned by GreedyIS?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe45ffb40-fb97-4856-9f50-b44beb4208a0%2F96beb792-9eb1-4e56-a960-699580735e91%2Fpxn5zh_processed.png&w=3840&q=75)
Transcribed Image Text:A(iv) Assuming that the graph G = (V, E) is represented in Adjacency List format, justify in detail the
fact GreedvIS can be implemented in O(n² + m) worst-case running time, where n = |V], m = |E|.
note: this will require you to take real care in how the adjustment of Adj is done in line 10.
The key is to only update/delete what is really necessary for the Algorithm, rather than being
concerned with an accurate representation of the residual graph.
A(v) Consider an unweighted bipartite graph G = (LUR, E), with L = :{u₁,...,U[n/3]}U{w₁, W2, W3, W₁},
and R = {₁,..., Un}. The edge set E = E₁ UE2 consists of the following edges:
E₁
=
{(ui, v₁): i = 1,
, [n/3]} U
{(ui, Vi+[n/3]): i = 1,..., [n/3]} U
{(Ui, Vi+2[n/3]): i = 1, . . . ,
1,..., [n/3]}
{(wi, vj) i = 1, ..., 4, j = 1, ..., n}
:
E2 =
Essentially, the E2 edges form a complete bipartite graph between the 4 special vertices {w₁, W2, W3, W4}
and the set R, while the edge set E₁ is a subgraph where each of the vertices {u₁,...,U[n/3]} have
3 adjacent edges and all of the vertices in R have 1 adjacent edge.
Work out the independent set I which will be constructed by GreedyIS, justifying your answer with
respect to the degrees of the different vertices as the algorithm proceeds. Show that this set will be
approximately 1/3 of the size of the true maximum independent set for this graph?
note: You are welcome to assume n is a multiple of 3.
A(vi) How would you alter the graph construction of (v) to get progressively worse "approximation
factors" for the result returned by GreedyIS?
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