Which of the following statement(s) is/are correct regarding about the undirected graph? I. Number of odd degree vertices is even. II. Sum of degrees of all vertices is even.
Both I and II
Let's analyze the given statements about undirected graphs. An undirected graph consists of a set of vertices (nodes) and a set of edges connecting pairs of vertices. In an undirected graph, the edges do not have a direction. The degree of a vertex is the number of edges incident to it.
Statement I says that in any undirected graph, the number of vertices with an odd degree is always even. Let's consider the sum of the degrees of all vertices in the graph.
We know from the Handshaking Lemma (discussed in detail in Statement II) that the sum of the degrees of all vertices in an undirected graph is always equal to twice the number of edges. This means the sum of all degrees is always an even number.
Let $V$ be the set of vertices in the graph. We can split the set of vertices into two subsets: $V_{\text{odd}}$ containing vertices with odd degrees, and $V_{\text{even}}$ containing vertices with even degrees.
The sum of degrees can be written as:
$\sum_{v \in V} \text{deg}(v) = \sum_{v \in V_{\text{odd}}} \text{deg}(v) + \sum_{v \in V_{\text{even}}} \text{deg}(v)$
We know that $\sum_{v \in V} \text{deg}(v)$ is even (from Handshaking Lemma).
The sum of degrees of vertices in $V_{\text{even}}$ is a sum of even numbers, which is always an even number.
So, the equation becomes:
Even Number = $\sum_{v \in V_{\text{odd}}} \text{deg}(v) + \text{Even Number}$
For this equation to hold, $\sum_{v \in V_{\text{odd}}} \text{deg}(v)$ must also be an even number (Even - Even = Even).
The sum of degrees of vertices in $V_{\text{odd}}$ is the sum of several odd numbers. The only way the sum of odd numbers can be even is if there is an even count of those odd numbers.
For example:
Therefore, for the sum of degrees of odd-degree vertices ($\sum_{v \in V_{\text{odd}}} \text{deg}(v)$) to be even, the number of terms in the sum, which is the number of odd degree vertices ($|V_{\text{odd}}|), must be even.
Hence, Statement I is correct.
Statement II claims that the sum of the degrees of all vertices in an undirected graph is always an even number. This is a fundamental result in graph theory known as the Handshaking Lemma.
Consider an undirected graph $G = (V, E)$, where $V$ is the set of vertices and $E$ is the set of edges.
The degree of a vertex, $\text{deg}(v)$, is the number of edges incident to vertex $v$.
When we calculate the sum of degrees of all vertices ($\sum_{v \in V} \text{deg}(v)$), we are essentially counting how many times each edge is accounted for.
In an undirected graph, every edge connects exactly two distinct vertices. When we sum the degrees, each edge $(u, v) \in E$ contributes exactly 1 to the degree of vertex $u$ and exactly 1 to the degree of vertex $v$. Therefore, each edge $(u, v)$ is counted exactly twice in the total sum of degrees.
So, the sum of degrees of all vertices is equal to twice the number of edges in the graph. Let $|E|$ be the number of edges.
The Handshaking Lemma states:
$\sum_{v \in V} \text{deg}(v) = 2 \cdot |E|$
Since the sum of degrees is equal to $2 \times (\text{number of edges})$, and $2 \times (\text{any integer})$ is always an even number, the sum of degrees of all vertices in an undirected graph is always even.
Hence, Statement II is correct.
Both Statement I (Number of odd degree vertices is even) and Statement II (Sum of degrees of all vertices is even) are correct properties of undirected graphs, derived from the fundamental Handshaking Lemma.
| Property | Description | Relation to Handshaking Lemma | Status |
|---|---|---|---|
| Sum of Degrees | The total sum of degrees of all vertices in an undirected graph. | $\sum_{v \in V} \text{deg}(v) = 2 \cdot |E|$. Always an even number. | Correct (Statement II) |
| Number of Odd Degree Vertices | The count of vertices that have an odd degree. | Derived from the fact that the sum of degrees is even. Requires an even number of odd degree terms to sum to an even number. | Correct (Statement I) |
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