GATE CS 2017 Set 2 Q.23 (CS Q.23): Graph Theory

$G$ is an undirected graph with $n$ vertices and 25 edges such that each vertex of $G$ has degree at least 3. Then the maximum possible value of $n$ is

Your answer

Official answer and solution

Official answer: 16
From: Official GATE 2017 answer key

Solution

Apply the Handshaking Lemma relating the sum of vertex degrees to the number of edges in an undirected graph.

By Euler's Handshaking Lemma for any undirected graph $G = (V, E)$ :

\sum_{v \in V} \deg(v) = 2|E|

Given $|E| = 25$ , the sum of all vertex degrees is:

\sum_{v \in V} \deg(v) = 2 \times 25 = 50

We are given that each vertex has degree at least 3, so $\deg(v) \ge 3$ for all $v \in V$ .
Therefore:

50 = \sum_{v \in V} \deg(v) \ge 3n

n \le \frac{50}{3} \approx 16.67

Since $n$ must be an integer, the maximum possible value is $n = 16$ .

We verify that a graph with $n = 16$ vertices exists: for example, 14 vertices of degree 3 and 2 vertices of degree 4 give sum of degrees $14 \times 3 + 2 \times 4 = 42 + 8 = 50$ . The number of odd-degree vertices is 14 (an even number), so such a graph is valid.
Thus, the maximum possible value of $n$ is 16.

Final answer

Common trap

Rounding 50/3 up to 17, which would require $17 \times 3 = 51 > 50$ degree sum.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 23 of 65, printed as CS Q.23
Section: Computer Science and Information Technology

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