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Official PYQ · 2014

GATE CS 2014 Set 3 Q52 — Graphs: Connectivity, Matching, Colouring

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Let δ\delta denote the minimum degree of a vertex in a graph. For all planar graphs on nn vertices with δ≥3\delta \geq 3, which one of the following is TRUE?

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Answer: A

To find the correct statement, we can use the Handshaking Lemma and Euler's formula for planar graphs.

Step 1: Relate Edges and Vertices using Minimum Degree

Let nn be the number of vertices, ee be the number of edges, and ff be the number of faces in a planar embedding of the graph.

By the Handshaking Lemma, the sum of the degrees of all vertices is equal to twice the number of edges:

∑v∈Vdeg(v)=2e\sum_{v \in V} \text{deg}(v) = 2e

Since the minimum degree of any vertex is δ≥3\delta \geq 3, we have deg(v)≥3\text{deg}(v) \geq 3 for all v∈Vv \in V. Therefore:

2e=∑v∈Vdeg(v)≥3n  ⟹  e≥3n22e = \sum_{v \in V} \text{deg}(v) \geq 3n \implies e \geq \frac{3n}{2}

Step 2: Apply Euler's Formula

For any connected planar graph, Euler's formula states:

n−e+f=2  ⟹  f=e−n+2n - e + f = 2 \implies f = e - n + 2

Substituting the inequality e≥3n2e \geq \frac{3n}{2} into Euler's formula:

f≥3n2−n+2f \geq \frac{3n}{2} - n + 2 f≥n2+2f \geq \frac{n}{2} + 2

Note: If the graph is disconnected with kk components, Euler's formula is n−e+f=1+kn - e + f = 1 + k. Since k≥1k \geq 1, we get:

f=e−n+1+k≥3n2−n+1+k=n2+1+k≥n2+2f = e - n + 1 + k \geq \frac{3n}{2} - n + 1 + k = \frac{n}{2} + 1 + k \geq \frac{n}{2} + 2

Thus, in any planar embedding, the number of faces is at least n2+2\frac{n}{2} + 2.

Conclusion

  • Option A is TRUE.
  • Options B and C are false because ff cannot be strictly less than n2+2\frac{n}{2} + 2.
  • Option D is false because the number of faces must be at least n2+2\frac{n}{2} + 2, which is generally larger than nδ+1\frac{n}{\delta+1}.

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