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

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

edited | 4k views
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Answer is (A) The graph is connected
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Yeah it should be C. I came to the same conclusion, following the same reasoning
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Planar graph dont need to be connected.
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This is not out of syllabus.
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not out of syllabus. 2016 question came from here
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Shouldn't option C be like:

The graph has a minimum vertex-cover of size at most 3n/4.

Because each graph has a vertex-cover of size n, as set of all vertices is also a vertex cover.

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@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.

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exactly correct explaination

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Is it out of syllabus now?
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the option C said that the vertex cover size cannot be greater than 3n/4

but when I took n=8 in worst case the vertexcover size came out to be 7

which is > 3*8/4.

What's wrong here?

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

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 $k \geq 4$, therefore, the Independent set is at least ceil$(n/4)$.
Hence (D) is false.

Now we know that Vertex cover + Independent Set Number $=n$.
If Independent set $\geq$ ceil$(n/4)$ 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.

by 1 flag:
✌ Edit necessary (Aalok8523 “Sir, please review the 3rd line of your answer. Correction :- for a planer graph chromatic number can be almost 4.”)

edited
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I can't find any source for the formula for independent set$\geq \left \lceil n/k \right \rceil$. Please cite your source.
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@habedo, we can use our Intuition here.IF we can color a graph using $k$ colors then it means every set of $k$ vertices are dependent (or adjacent ) to each other so  total size of independent set  will be $n/k$
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I didn't get your sentence. Can't every planar be colored with at most $4$ color only? why it is at least 3 or 4?
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@Mk Utkarsh did you edit 1st line in the answer , if yes did you change 2 in place of k

Planar graph not  necesarily be connected. So option (A) is false. Options (B) and (D) can be proven false with the example of K4 . Now only one option (C)  left  .

So, answer should be (C). But I don't know how to prove it.

by
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How is option D proved false with example of K4? K4 has an independent set of size 1 which is what you get by doing n/3 => 4/3 = 1.
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For vertex cover of K4 size will be 2 maximal two edges would be sufficient to cover all vertices here and according to formula it would be less than or equal to (3*4)/4 equals to 3 bt how size of vertex cover would be 3 not getting it 2 becoz in that case it wouldn't be maximal set
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We can prove C using every planar graph is 4 colourable.
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A wheel with 7 vertices is counter example for option D.

The independent set can consist of centre vertex which alone forms an independent set, whose size is 1. (given size is atleast n/3, here 7/3 = 2).So it is wrong.

Is this right?
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you can also take any example of wheel graph with vertices greater 6 u should take 6

all have independence set 1

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.