Proof: A graph that does not contain any closed walks with odd length is bipartite.

189 Views Asked by At

I came across this proof in the book Mathematics for CSE by E.Lehman and F.T. Leighton(2010 version). They prove that a graph with no walks of odd length is bipartite. Here is the proof: The proof

4 IMPLIES 1 means that the property 4 implies property 1. Here is a photo for reference: The description of the properties of Graph G

So, basically in the proof they use contradiction and go onto say that G' is not bipartite there must be a pair of nodes u1 and u2 whose length of the shortest path to an arbitrary node v in G' is either odd or even. Does this statement not hold true for a bipartite graph? If not how?

And can anyone please simplify the proof for me?

Thanks in advance.