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28 votes

The chromatic number of the following graph is _____

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@Aks9639 bhai no sir.. and yes, you are right. :)

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It simply can’t be 2. The chromatic number has to be 3
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5 Answers

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66 votes
Best answer

Here, Independent sets, $S_1 = \{a,d\}, S_2 = \{b,e\}, S_3 = \{c,f\}$

Therefore, vertices of $S_1$ has no connection between each other [∵ $a$ & $d$ are not connected by an edge]

$ S_1 ⇒  $put ${\color{red}{RED}}$


Vertices of $S_2$ has no connection between each other [∵ $b$ & $e$ are not connected by an edge]

$ S_2 ⇒  $put ${\color{GREEN}{GREEN}}$


Vertices of $S_3$ has no connection between each other [∵ $c$ & $f$ are not connected by an edge]

$ S_3 ⇒  $put ${\color{BLUE}{BLUE}}$


∴ These graph has chromatic number as $3$.

Explanation: Why solving by independent sets?

Independent set means a set containing vertices & each & every vertex of this set is independent to each other i.e. if there are $3$ vertices in an independent set, then each vertex of this set does not connected to other vertex of this set by an edge.

Suppose, there are two independent sets $(S_1$ & $S_2)$. Then any vertex of $S_1$ has connection(share an edge) to any vertex of set $S_2$.

∵ $S_1,S_2, S_3$ are different independent sets & they share some connection between them, we put different colours to different sets.

$S_1$ share connections to $S_2$ & $S_3$

∴ Put  ${\color{RED}{RED}}$ to $S_1$, not $S_2$ & $S_3$

Similarly, $S_2$ share connections to $S_1$ & $S_3$

∴ Put  ${\color{GREEN}{GREEN}}$ to $S_2$, not $S_1$ & $S_3$

$S_3$ share connections to $S_1$ & $S_2$

∴ Put  ${\color{BLUE}{BLUE}}$ to $S_3$, not $S_1$ & $S_2$

The advantage of following this method is when we have a complicated graph then we do not need to continuously see whether any vertex is adjacent to each other when we colour any vertex.

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4 Comments

In @Ayush Upadhyaya comment,

The maximal independent sets of this graph are

{a,d}, {a,f}, {b,e}, {c,f}, {d,e}.

How we area getting 3 terms left from it ??

{a,d} , {b,e} , {c,f}

 Anyone help!!

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 we have to keep the minimum sets covering all the vertices in the graph!

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@Ayush Upadhyaya Sir, your answer is of a great help in understanding Chromatic Partitioning. I feel this question must have a tag: “Chromatic partitioning” as it is difficult to find such an easy and clear explanation on this topic.

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13 votes
13 votes

$3$ is answer. 

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1 comment

nice!yes

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6 votes
6 votes

Answer is 3

2 votes
2 votes

Ans is 3