Graph Theory Practice Questions

The chromatic number of a graph is the minimum number of colors needed to color its vertices so that no two adjacent vertices share the same color.

First, we observe that the vertices {b, c, d} form a triangle, which is a clique of size 3. Therefore, at least 3 colors are required, and the chromatic number $\chi(G) \ge 3$.

Next, we check if the graph can be colored with 3 colors. A possible 3-coloring is:

• Color 1 (e.g., Blue): {a, f}
• Color 2 (e.g., Green): {b, e}
• Color 3 (e.g., White): {c, d}
  Since a valid 3-coloring exists, $\chi(G) \le 3$.
  Combining both conditions, the chromatic number is exactly 3.

The degree of a vertex, y, is the number of neighbors it has. A neighbor is formed by swapping two adjacent numbers in the permutation. For a sequence of 100 numbers, there are $100 - 1 = 99$ adjacent pairs. Each adjacent swap results in a unique neighboring vertex. Therefore, the degree of any vertex is $y = 99$.

The set of all permutations can be generated from any single permutation through a series of adjacent swaps (this is the principle behind bubble sort). This means the graph is connected, so it has only one connected component. Thus, $z = 1$.

The required value is $y + 10z = 99 + 10(1) = 109$.

Statement (I) is true. During a Depth First Search (DFS) on an undirected graph, any edge not part of the DFS tree must be a back edge, connecting a vertex to one of its ancestors. Cross edges, which connect vertices in different subtrees where neither is an ancestor of the other, do not occur in a DFS of an undirected graph. Statement (II) is false. A Breadth First Search (BFS) explores level by level. An edge can connect two vertices at the same depth, for which $|i-j|=0$. This violates the condition $|i-j|=1$.

Breadth-First Search (BFS) explores a graph level by level, starting from a source node. It uses a queue to manage the nodes to be visited. A valid BFS traversal visits all nodes at a certain distance (level) from the source before visiting any nodes at the next level.
Let's trace the option POQNMR starting with P:

1. Start at P. Visit P. Add its neighbors (O, Q) to the queue.
2. Visit nodes at the next level. Dequeue and visit O, then Q.
3. As O and Q are visited, their unvisited neighbors (N, R) are added to the queue.
4. Visit nodes at the next level. Dequeue and visit N, then M (a neighbor of N).
5. Finally, dequeue and visit R.
   The sequence POQNMR is a valid BFS traversal as it explores nodes layer by layer: {P}, then {O, Q}, then {N, R, M}. The order within a layer depends on the order neighbors are added to the queue.

According to the handshaking lemma for an undirected graph, the sum of the degrees of all vertices is equal to twice the number of edges.
$\sum_{i=1}^{n} \text{deg}(v_i) = 2|E|$
Given that each of the $n$ vertices has a degree of at least 3, the sum of degrees is at least $3n$. The number of edges $|E|$ is 25.
Therefore, we have the inequality:
$3n \le 2 \times 25 \implies 3n \le 50$
$n \le \frac{50}{3} \approx 16.66$
Since the number of vertices $n$ must be an integer, the maximum possible value for $n$ is 16.

For any tree with V vertices, the number of edges E is always V-1.
In this case, the tree T has V = 10 vertices, so it must have E = 10 - 1 = 9 edges.
The handshaking lemma states that the sum of the degrees of all vertices in a graph is equal to twice the number of edges.
Sum of degrees = 2 * E
Substituting the number of edges, we get:
Sum of degrees = 2 * 9 = 18.

Breadth-First Search (BFS) visits nodes level by level. A vertex at a distance of 4 from the root is at level 4 (assuming the root is at level 0). To maximize the traversal order n for a vertex at this level, we need to maximize the number of nodes visited before it. This occurs when the binary tree is a complete binary tree up to level 4, and the vertex t is the last node visited at that level.
The total number of nodes in a complete binary tree up to level k is $\sum_{i=0}^{k} 2^i = 2^{k+1}-1$.
For level 4, the total number of nodes is $2^{4+1}-1 = 31$.

This question relates to the Four Color Theorem, a fundamental concept in graph theory. The theorem states that any planar graph-a graph that can be drawn on a plane without any edges crossing-can be vertex-colored with at most four colors. This ensures that no two adjacent vertices share the same color. Therefore, four colors are always sufficient for vertex-coloring any planar graph.

A self-complementary graph with $n$ vertices must have exactly half the number of edges of a complete graph on $n$ vertices. The number of edges in a complete graph is $\frac{n(n-1)}{2}$. Therefore, a self-complementary graph must have $\frac{n(n-1)}{4}$ edges. For this to be an integer, the product $n(n-1)$ must be divisible by 4. This condition is satisfied only when $n$ is a multiple of 4 (i.e., $n \equiv 0 \pmod{4}$) or when $n-1$ is a multiple of 4 (i.e., $n \equiv 1 \pmod{4}$).

For a connected planar graph, Euler's formula states that $v - e + f = 2$, where $v$ is the number of vertices, $e$ is the number of edges, and $f$ is the number of faces.
We are given $v = 10$.

The problem also states that each face is bounded by 3 edges. Since every edge in a planar graph is shared by exactly two faces, we can establish a relationship between the number of edges and faces: $3f = 2e$, which implies $f = \frac{2e}{3}$.

Substituting these values into Euler's formula:
$10 - e + \frac{2e}{3} = 2$
$8 = e - \frac{2e}{3}$
$8 = \frac{e}{3}$
$e = 24$
Thus, the number of edges in G is 24.

A bridge is an edge whose removal increases the number of connected components of a graph. If an edge is part of a cycle, its two endpoints are connected by the edge itself and also by an alternative path formed by the rest of the cycle. Therefore, removing this edge will not disconnect the graph, as the alternative path still exists. Consequently, an edge that is part of a simple cycle cannot be a bridge.

To find the maximum recursion depth for a Depth First Search (DFS), we need to identify the longest possible path in the graph that DFS could take without revisiting nodes. Starting from an arbitrary vertex, DFS explores as far as possible along each branch before backtracking. By tracing a path that sequentially visits unvisited nodes, we can determine the maximum depth. The provided solution shows a path that visits 19 distinct nodes. Each visit represents a recursive call, contributing to the recursion depth. Therefore, the maximum recursion depth is 19.

[CORRECT_OPTION: 19]