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The edges of Kun are painted red and ble in such way that each red edge 1S in at most one monochro matic red triangle_ Frove there is subgrapgh K; withont red trian...

Question

The edges of Kun are painted red and ble in such way that each red edge 1S in at most one monochro matic red triangle_ Frove there is subgrapgh K; withont red triangles_

The edges of Kun are painted red and ble in such way that each red edge 1S in at most one monochro matic red triangle_ Frove there is subgrapgh K; withont red triangles_



Answers

In Exercises 21–25 determine whether the graph is bipartite. You may find it useful to apply Theorem 4 and answer the question by determining whether it is possible to assign either red or blue to each vertex so that no two adjacent vertices are assigned the same color.

In this question we asked, Is this grab by partite? And the answer is yes. To see this, we just have to find a way to assign these witnesses into two groups. Right? Such that in the same group, the witness is in the same group cannot. How many ages And we can do this by grouping Yancey. You can see that they have no, it just connecting each other. And the rest will be in the same partite. And you can see there's no witnesses. So there's no edges between b e or the so So these gonna be our part I and see to get the end mead e e together. And that is it. Thank you.

Were given a graph and were asked to determine if the graph is bi partite. So to do this will use theorem before from the book. So I have here a graph courtesies A, B, C, D, E and F. And to answer the question of whether it's apartheid, those that color without loss of generality, the color of vortex a read. Then we see that I did color every adjacent edge to a referee, juicing for text to a blue, any color. Be blue and you just color deep blue and into color e blue. And this doesn't make sense, because B is not Jason to D India's Maj. Jason T. E. He is not adjacent to E, and now, since B is blue, it follows. That seems to be read and F needs to be read. However, this is a problem. We can't have that both sea and death or red. And so by theorem four. It follows that this graph is not bi partite

In this question, we are asked to determine if this graph by apartheid and so we want to fly away. No, we want to find out if we can separate these branches is into two groups suspect in the same group. There's no it just connecting them. And it is, in fact, by Pat I we can go maybe d e e together. You can see that there's no just between them and the rest that the two what is left will be in the same partite again. F and C has no it just connecting them. So he's gonna be out to partite. So we have a baby big and gift and see, and that is it.

Were given a craft were asked to determine if the graph is by party. So suppose that loss of generality that we color Vertex a read and it follows that we need to color overseas me and F blue since they're adjacent to a This is for text. B is blue. It follows that need to color courtesies. Do you and e Red is already read since these air adjacent to be And we already have a contradiction because we have that is adjacent to D But both he and your red near her follows that this graph is not bi partite like here um four from the book.


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