Which of the following statements is true for every planar graph on $n$ vertices?

- The graph is connected
- The graph is Eulerian
- The graph has a vertex-cover of size at most $\frac{3n}{4}$
- The graph has an independent set of size at least $\frac{n}{3}$

### 11 Comments

@Pratik, I also think the same. It should be minimum vertex cover and largest independent set.

Because if we consider 2 theorems , then answer will be valid in case of **minimum** vertex cover not vertex cover.

1) size of largest independent set + size of minimum vertex cover = no. of vertices

i.e. $\alpha(G) + \beta (G) = n$

2) $\alpha(G) \geq \frac{n}{\chi (G)}$

Since , for planar graphs , $\chi (G) \leqslant 4$ . So, $\frac{n}{\alpha (G)} \leqslant 4$. Therefore, $\frac{n}{4} \leqslant \alpha (G)$.

Now, Since, $\frac{n}{4} \leqslant \alpha (G)$. So, $\;\frac{n}{4} \leqslant (n-\beta(G))$. Therefore, $\beta(G) \leqslant \frac{3n}{4}$.

So, both things should be correct in case of** minimum** vertex cover and **largest** independent set not only vertex cover and independent set.

@Nandkishor3939 that's not vertex cover whatever you've written. One possible vertex cover for your example is $A = \{a, c,e,g\}$ where $|A| \leq 7$

## 6 Answers

Independent Set $\geq$ $\large \left \lceil \frac{n}{k} \right \rceil$ where,

$n$ is the number of vertices and $k$ is the chromatic number

For any planar graph, $k \geq 3$ and $k \leq 4$, therefore, the Independent set is at least $\lceil n/4 \rceil$.

Hence (**D**) is false.

Now we know that size of Vertex Cover + Independent Set Number $=n$.

If Independent set number $\geq \lceil n/4 \rceil$ then,

Vertex cover $\leq n - (n/4)$

Vertex Cover $\leq 3n/4$

Vertex Cover is at most $3n/4.$ So, **(C)** is the correct answer.

### 6 Comments

For the vertex cover, we're missing the "there exists" part. THERE EXISTS a vertex cover of size at most 3n/4. Although, we can take a vertex cover with all the vertices as well, so there is also a vertex cover of size n. Hence these are also are not contradictory.

A) **Incorrect**. Consider the following disconnected planar graph:

B) **Incorrect**. A graph is Eulearian if all vertices have even degree but a planar graph can have vertices with odd degree .

C) **Correct**. Any planar graph can be vertex colored with maximum 4 colors (4-color theorem) so that no two adjacent vertices have same color.

So consider a planar graph with n vertices having chromatic number 4. So atmost we have 4 set of vertices of each color having size n/4 (evenly distributed). If we remove a set, we still will have all the vertices which is atleast one of the endpoint of all the edges i.e. we have a vertex cover having size n-(n/4) = 3n/4.

D) **Incorrect**. Consider K4 graph. It has independent set size 1 which is less than 4/3.

Clearly option (B) is wrong. Take $K_{1,3}$ which is non Euler because not all vertices have even degree [rather no vertices have even degree] but it can be drawn as a plane graph.

Also option (D) is wrong. Take $K_4$. For any complete graph, there is only one possible size of independent set of vertices which is $1$, Because each pair of vertices are adjacent. But option (D) says that we should have an independent set of size at least $\frac{4}{3} = 1.33$ or at least $2$.

So (C) is the correct option. Let’s see why.

$\alpha(G) : \text{Independence Number of G}$

$\beta(G) : \text{Minimum vertex cover of G}$

$n(G): \text{Number of vertices of G}$

$\chi(G): \text{Chromatic Number of G}$

We know,

$$\alpha(G)+\beta(G)=n(G) \tag 1$$

We also know from $\text{Four color theorem}$,

For a planar graph, $$\chi(G) \leq 4 \tag 2$$

Now we also know that,

$$\frac{n(G)}{\alpha(G)}\leq \chi(G)\tag 3$$

From $(2)$ and $(3)$ we have,

$$\frac{n(G)}{\alpha(G)}\leq \chi(G) \leq 4 \implies \frac{n(G)}{\alpha(G)} \leq 4 \implies \alpha(G)\geq \frac{n(G)}{4} \tag 4$$

[The above result in $(4)$ simply states that we have maximum independent set of size at least $\frac{n}{4}$. Since we have a maximum independent set of size at least $\frac{n}{4}$ we can say that we have “an independent set” (i.e. “$\exists$ an independent set”) of size at least $\frac{n}{4}$. But this does not prove option (D) wrong!! There might be an independent set of size at least $\frac{n}{3}$ even if $\alpha(G)\geq\frac{n}{4}$. Who knows? So the counter-example was necessary]

Adding $\beta(G)$ to both sides of $(4)$, we have,

$$\alpha(G)+\beta(G)\geq \frac{n(G)}{4} + \beta(G)$$

But from $(1)$, we have

$$n(G)\geq \frac{n(G)}{4} + \beta(G) \implies \beta(G)\leq \frac{3n}{4}$$

Since we have a minimum vertex cover of size at most $\frac{3n}{4}$, we indeed have “a vertex cover” of size at most $\frac{3n}{4}$.