All questions of Graph Theory for Engineering Mathematics Exam

The maximum degree of any vertex in a simple graph with n vertices is n - 1.

Explanation:
A simple graph is a graph that does not contain any loops (edges that connect a vertex to itself) or multiple edges (more than one edge between the same pair of vertices). In other words, each edge in a simple graph connects two distinct vertices.

In a simple graph with n vertices, each vertex can be connected to at most n - 1 other vertices. This is because a vertex cannot be connected to itself, so it can have at most n - 1 neighboring vertices.

To understand why the maximum degree of any vertex is n - 1, let's consider the worst-case scenario. In this scenario, each vertex in the graph is connected to every other vertex, creating a complete graph.

Key Points:
- A complete graph is a graph in which every pair of distinct vertices is connected by an edge.
- In a complete graph with n vertices, each vertex is connected to every other vertex.
- Therefore, the degree of each vertex in a complete graph is n - 1.

Example:
Let's consider a simple graph with 5 vertices. The maximum degree of any vertex in this graph is 4 (n - 1).

(2)------(3)
| |
| |
(1)------(4)
|
(5)

In the above example, each vertex is connected to at most 4 other vertices. Therefore, the maximum degree of any vertex is 4.

Conclusion:
In a simple graph with n vertices, the maximum degree of any vertex is n - 1. This is because each vertex can be connected to at most n - 1 other vertices.

Explanation:
In order to determine which properties of the circuits of a graph are correct, let's analyze each option:

Option 1: The minimum number of branches possible in a circuit will be equal to the number of elements in a circuit.
This statement is incorrect. The number of branches in a circuit is always greater than the number of elements. Each element in a circuit contributes at least one branch.

Option 2: There are exactly two paths between any pair of vertices in a circuit.
This statement is incorrect. A circuit is a closed loop, so there is only one path between any pair of vertices. A path refers to a sequence of edges, and in a circuit, there is only one path that connects two vertices.

Option 3: There are at least two branches in a circuit.
This statement is correct. In a circuit, there must be at least two branches to form a closed loop. A branch is a connection between two vertices, and in a circuit, there must be at least two connections to form a closed loop.

Therefore, the correct answer is Option B: 1 and 3 only.

True Statements:

(II) Every wheel graph is Hamiltonian:
A wheel graph is formed by connecting a single central vertex to all the vertices of a cycle graph. In a wheel graph, there is a Hamiltonian cycle that visits every vertex exactly once. The cycle can start from any vertex in the outer cycle and then move to the central vertex before returning to the starting vertex. Therefore, every wheel graph is Hamiltonian.

False Statements:

(I) Every complete graph is Hamiltonian:
A complete graph is a graph in which every pair of distinct vertices is connected by an edge. While it is true that a complete graph with three or more vertices is Hamiltonian, this statement does not hold for complete graphs with one or two vertices. In a complete graph with one vertex, there is no Hamiltonian cycle as there are no other vertices to visit. In a complete graph with two vertices, there is only one edge and no Hamiltonian cycle can be formed. Therefore, not every complete graph is Hamiltonian.

(III) Every complete bipartite graph is Hamiltonian:
A complete bipartite graph is a graph in which the vertices can be partitioned into two sets such that every vertex in one set is connected to every vertex in the other set. In a complete bipartite graph, there is no Hamiltonian cycle. This can be proven by considering the number of vertices and edges in the graph. The number of vertices in a complete bipartite graph is the product of the sizes of the two sets, which can be large. On the other hand, the number of edges is the product of the sizes of the two sets, which can be much smaller. Therefore, the number of edges is significantly less than the number of vertices, making it impossible to form a Hamiltonian cycle. Therefore, not every complete bipartite graph is Hamiltonian.

Conclusion:
The only true statement is that every wheel graph is Hamiltonian. The other two statements are false, as not every complete graph or complete bipartite graph is Hamiltonian.

The Hamiltonian path problem is a well-known problem in computer science and graph theory. It involves finding a path in a graph that visits each vertex exactly once. Dynamic programming is a technique that can be used to solve this problem efficiently.

The time complexity of solving the Hamiltonian path problem using dynamic programming depends on the number of vertices in the graph. Let's analyze the given options to understand which one is the correct answer.

a) O(N): This option suggests that the problem can be solved in linear time, where N is the number of vertices in the graph. However, finding a Hamiltonian path is known to be an NP-complete problem, which means there is no known polynomial-time algorithm for it. Therefore, option A is incorrect.

b) O(N log N): This option suggests that the problem can be solved in N log N time. However, this is not possible because the problem is NP-complete. Therefore, option B is incorrect.

c) O(N^2): This option suggests that the problem can be solved in quadratic time, where N is the number of vertices in the graph. This is a possibility because dynamic programming can be used to solve the Hamiltonian path problem in O(N^2 2^N) time. However, we need to analyze the last option to make sure it is not a better choice. Therefore, option C is a possibility but may not be the correct answer.

d) O(N^2 2^N): This option suggests that the problem can be solved in O(N^2 2^N) time, where N is the number of vertices in the graph. This is the correct answer because dynamic programming can be used to solve the Hamiltonian path problem in this time complexity. The 2^N factor is due to the exponential number of subsets of vertices that need to be considered. By using dynamic programming, we can avoid redundant computations and solve the problem efficiently. Therefore, option D is the correct answer.

To summarize, the Hamiltonian path problem can be solved using dynamic programming in O(N^2 2^N) time complexity, where N is the number of vertices in the graph. This is the correct answer to the given question.

A) The diagonal entries of A^2 are not necessarily the degrees of the vertices of the graph. The diagonal entries of A^2 represent the number of length-2 paths between each pair of vertices in the graph. The degrees of the vertices are represented by the diagonal entries of A.

b) If the graph is connected, then all entries of A^(n-1) + A^(n-2) + ... + A + I (where I is the identity matrix) are positive. This is known as the matrix-tree theorem, and it states that the sum of these matrices (up to A^(n-1)) gives the Laplacian matrix of the graph, and the determinant of the Laplacian matrix is equal to the number of spanning trees in the graph. Therefore, if the graph is connected, the determinant of the Laplacian matrix is non-zero, which means none of the entries of A^(n-1) + A^(n-2) + ... + A + I can be zero.

c) If the sum of all the elements of A is at most 2(n - 1), it does not necessarily mean that the graph is acyclic. The sum of all the elements of A represents the total number of edges in the graph. If this sum is at most 2(n - 1), it means that the graph has at most 2(n-1) edges, which is the maximum number of edges in a tree. However, the graph could still have cycles if it is not a tree.

d) If there is at least a 1 in each of A, then it means that each vertex is connected to at least one other vertex in the graph. This does not necessarily imply any specific property of the graph, as it could still have cycles and be a connected graph.

For the complete bipartite graph $K_{m,n}$ to have a Hamilton circuit, we need every vertex to have degree at least 2.

The degree of each vertex in the left part of the graph is $n$, since it is connected to every vertex in the right part. Similarly, the degree of each vertex in the right part is $m$.

Therefore, we need $m \geq 2$ and $n \geq 2$ for every vertex to have degree at least 2.

If $m=1$ or $n=1$, then one of the parts of the graph only has one vertex, and thus it cannot have a Hamilton circuit.

So the answer is: $m \geq 2$ and $n \geq 2$.

Explanation:

A bipartite graph is a graph whose vertices can be divided into two independent sets, such that every edge connects a vertex in one set to a vertex in the other set.

To find the maximum number of edges in a bipartite graph on 12 vertices, we need to consider the case where the two sets of vertices have equal size.

Let the two sets of vertices be A and B, each with 6 vertices. The maximum number of edges that can be formed between vertices in A and B is when every vertex in A is connected to every vertex in B. This gives us a total of 6 * 6 = 36 edges.

Answer: 36