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What is the chromatic number of an $n$ vertex simple connected graph which does not contain any odd length cycle? Assume $n > 2$.

1. 2
2. 3
3. n-1
4. n
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n-1
What is minimum number for EDGE coloring? (I'm getting 3 as I find Edge-coloing instead of vertex during test, so dumb!)

Lemma 1.- G is bipartite, if and only if it does not contain any cycle of odd length.

Proof. Suppose G has an odd cycle. Then obviously it cannot be bipartite, because no odd cycle is 2-colorable. Conversely, suppose G has no odd cycle. Then we can color the vertices greedily by 2 colors, always choosing a different color for a neighbor of some vertex which has been colored already. Any additional edges are consistent with our coloring, otherwise they would close a cycle of odd length with the edges we considered already. The easiest extremal question is about the maximum possible number of edges in a bipartite graph on n vertices. 1 [email protected] http://math.mit.edu/~fox/MAT307-lecture07.pdf

Bipartite Graph: A graph which is 2-colorable is called bipartite. We have already seen several bipartite graphs, including paths, cycles with even length, and the graph of the cube (but not any other regular polyhedra)

3. Bipartite graphs: By definition, every bipartite graph with at least one edge has chromatic number 2. (otherwise 1 if graph is null graph )

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What if I have a graph like below ? These would be simple graphs without any odd length, right ? So should we not have (n - 1) as the answer ?

Yes in this case you are geting (n-1) as a chromatic number, but this is subset.
No, in this case you always getting 2 as a chromatic number . The middle node of  graph is 1 and its surroundings are 2 ,2, 2... same color.
Star graphs always have chromatic no 2

Registered user 31  for given above graphs chromatic no is 2 not (n-1), one colour for middle node and another colour for all the remaining nodes.

2 draw some random graph and you will realise that 2 is the chromatic number
as every complete graph is simple

using k4 no chromatic = n

im wrong?

@

yes we can assume 2 regular graph for no odd length cycle graph

What if I have a graph like below ? These would be simple graphs without any odd length, right ? So should we not have (n - 1) as the answer ?

Registered user 31  for given above graphs chromatic no is 2 not (n-1), one colour for middle node and another colour for all the remaining nodes.

No odd length cycle means no 3,5,7,... Length cycle should be there. So it means we can color this with less than 3 colors. Becz a presence of 3 length cycle will atlst need 3 colors to be colored. So here 2 color will work..

What if I have a graph like below ? These would be simple graphs without any odd length, right ? So should we not have (n - 1) as the answer ?

well, for these graphs, chromatic no. will be 2 only, because all the degree one vertices are not adjacent to any other vertex, so they can take the same color, and center can take another.

But I have a concern with wheel graph like the one below. Its chromatic no. will be 3. then why is the ans 2?

Wheel graph contains odd length cycle.
ohh, two outer vertices connecting with inner (central) vertex forms odd length cycle.
A big miss. :P
Thanks.
Question is about not having odd length cycle. You have a cycle of 3 here
Consider this Graph as composition of even length(0, 2, 4 etc) cycles. And each even length cycle could be colored using two colors without creating any conflict. Process is as following -->

(1) Choose any vertices give color X.

(2) Give color Y to its neighbors.

Now this Y can not create conflict with X otherwise ood length cycle will appear. We can repeat this alternate coloring process until all vertices are not colored.

Means all the vertices which are odd no of  edges  away from First vertex will get Y color and remaining will get X color. During this process at any point  if problem comes it means an odd length cycle is present in our graph which is failing our assumption.
–1 vote
it's even-length cycle graph i.e C2, C4,C6......all need max 2 colour so chromatic number X(G)=2
No its nt ... Question has said that  simple connected graph which does not contain any odd length cycle ...It does nt imply that it has even length cycle ...