,MAT3707 Assignment 3 (COMPLETE ANSWERS)
2024 (156556) - DUE 30 August 2024 ; 100%
TRUSTED Complete, trusted solutions and
explanations.
QUESTION 1 Consider the following two graphs: a b d c e f G1 u
v w y x z G2 Are G1 and G2 isomorphic? A. Yes, because they
both have the same degree sequence. B. Yes, because function
g with g(a) = u, g(b) = v, g(c) = w, g(d) = x, g(e) = y, and g(f) = z is
an isomorphism. C. Yes, because function h with h(a) = z, h(b) =
w, h(c) = x, h(d) = y, h(e) = u, and h(f) = v is an isomorphism. D.
Yes, because they both have the same number of vertices, the
same number of edges and are both connected. E. No, they are
not isomorphic.
The correct answer is E. No, they are not isomorphic.
To determine if two graphs G1G_1G1 and G2G_2G2 are
isomorphic, it's not enough for them to have the same degree
sequence, the same number of vertices and edges, or to be
connected. You must also ensure there is a one-to-one
correspondence between the vertices of G1G_1G1 and
G2G_2G2 that preserves the adjacency (i.e., the structure of the
edges).
While the options B and C provide specific functions mapping
vertices of G1G_1G1 to G2G_2G2, these mappings need to
preserve adjacency. Without seeing the specific structures of
the graphs and verifying the adjacencies under these mappings,
,you can't definitively state that the functions provided are valid
isomorphisms.
Thus, without more specific adjacency information or a
verification process, the safest conclusion is that they are not
isomorphic.
QUESTION 2 Which of the following is a necessary condition for
a graph to be bipartite? A. The graph has no odd cycles. B. The
graph has even cycles. C. The graph has an Euler cycle. D. The
graph is planar. E. The graph has a Hamiltonian circuit.
Necessary Condition for a Graph to be Bipartite
To determine which of the provided options is a necessary
condition for a graph to be bipartite, we first need to
understand what it means for a graph to be bipartite. A
bipartite graph is one in which the set of vertices can be divided
into two distinct sets such that no two graph vertices within the
same set are adjacent.
Step 1: Analyze Each Option
A. The graph has no odd cycles.
This statement is indeed true and is a fundamental property of
, bipartite graphs. A key theorem in graph theory states that a
graph is bipartite if and only if it does not contain any odd-
length cycles. If there exists an odd cycle, it would not be
possible to color the graph using two colors without having two
adjacent vertices share the same color, which contradicts the
definition of a bipartite graph.
B. The graph has even cycles.
While even cycles can exist in bipartite graphs, their presence
alone does not guarantee that a graph is bipartite. A non-
bipartite graph can also contain even cycles; thus, this condition
is not necessary.
C. The graph has an Euler cycle.
An Euler cycle exists in a connected graph if every vertex has an
even degree. However, this property does not imply that the
graph is bipartite or vice versa; therefore, this condition is also
not necessary.
D. The graph is planar.
Planarity refers to whether a graph can be drawn on a plane
without edges crossing each other. While all trees (which are
bipartite) are planar, not all planar graphs are bipartite (for
example, K5 and K3,3 are planar but not bipartite). Thus,
planarity is not a necessary condition for being bipartite.
E. The graph has a Hamiltonian circuit.
A Hamiltonian circuit involves visiting each vertex exactly once
before returning to the starting vertex and does not have any
direct implications regarding whether or not the graph is
2024 (156556) - DUE 30 August 2024 ; 100%
TRUSTED Complete, trusted solutions and
explanations.
QUESTION 1 Consider the following two graphs: a b d c e f G1 u
v w y x z G2 Are G1 and G2 isomorphic? A. Yes, because they
both have the same degree sequence. B. Yes, because function
g with g(a) = u, g(b) = v, g(c) = w, g(d) = x, g(e) = y, and g(f) = z is
an isomorphism. C. Yes, because function h with h(a) = z, h(b) =
w, h(c) = x, h(d) = y, h(e) = u, and h(f) = v is an isomorphism. D.
Yes, because they both have the same number of vertices, the
same number of edges and are both connected. E. No, they are
not isomorphic.
The correct answer is E. No, they are not isomorphic.
To determine if two graphs G1G_1G1 and G2G_2G2 are
isomorphic, it's not enough for them to have the same degree
sequence, the same number of vertices and edges, or to be
connected. You must also ensure there is a one-to-one
correspondence between the vertices of G1G_1G1 and
G2G_2G2 that preserves the adjacency (i.e., the structure of the
edges).
While the options B and C provide specific functions mapping
vertices of G1G_1G1 to G2G_2G2, these mappings need to
preserve adjacency. Without seeing the specific structures of
the graphs and verifying the adjacencies under these mappings,
,you can't definitively state that the functions provided are valid
isomorphisms.
Thus, without more specific adjacency information or a
verification process, the safest conclusion is that they are not
isomorphic.
QUESTION 2 Which of the following is a necessary condition for
a graph to be bipartite? A. The graph has no odd cycles. B. The
graph has even cycles. C. The graph has an Euler cycle. D. The
graph is planar. E. The graph has a Hamiltonian circuit.
Necessary Condition for a Graph to be Bipartite
To determine which of the provided options is a necessary
condition for a graph to be bipartite, we first need to
understand what it means for a graph to be bipartite. A
bipartite graph is one in which the set of vertices can be divided
into two distinct sets such that no two graph vertices within the
same set are adjacent.
Step 1: Analyze Each Option
A. The graph has no odd cycles.
This statement is indeed true and is a fundamental property of
, bipartite graphs. A key theorem in graph theory states that a
graph is bipartite if and only if it does not contain any odd-
length cycles. If there exists an odd cycle, it would not be
possible to color the graph using two colors without having two
adjacent vertices share the same color, which contradicts the
definition of a bipartite graph.
B. The graph has even cycles.
While even cycles can exist in bipartite graphs, their presence
alone does not guarantee that a graph is bipartite. A non-
bipartite graph can also contain even cycles; thus, this condition
is not necessary.
C. The graph has an Euler cycle.
An Euler cycle exists in a connected graph if every vertex has an
even degree. However, this property does not imply that the
graph is bipartite or vice versa; therefore, this condition is also
not necessary.
D. The graph is planar.
Planarity refers to whether a graph can be drawn on a plane
without edges crossing each other. While all trees (which are
bipartite) are planar, not all planar graphs are bipartite (for
example, K5 and K3,3 are planar but not bipartite). Thus,
planarity is not a necessary condition for being bipartite.
E. The graph has a Hamiltonian circuit.
A Hamiltonian circuit involves visiting each vertex exactly once
before returning to the starting vertex and does not have any
direct implications regarding whether or not the graph is
