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The chromatic number of the following graph is _____

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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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