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there's a proof provided in harary's book but please help me write my own proof.

Proof: (=>)Suppose there's at least one odd cycle, Partitioning the set of vertices V into two distinct sets, we can let V1 be the set of vertices {v1,v3,v5,...} and V2: {v2,v4,v6,...}. Let C* be an arbitrary odd cycle. Since it's an odd cycle then the walk in that cycle would be v1v2v3...v(2n+1)v1 s.t. n is an integer. Therefore since v1 and v(2n+1) belong in the same partition, the graph containing the cycle is not bipartite.

(<=)Conversely, suppose the cycles are all even. Since the length of any cycle in the graph is even, it means there's an even number of edges and vertices for each cycle, then we can create a cycle such that the edges are denoted as v1v2, v2v3, v3v4,...,v(2n)v1. By partitioning any even cycle within the graph by their subscripts such that all even subscripted vertex are separated from the odd subscripted vertex, we have created a bipartite graph.

qed

please point out the errors in the proof. thank you!

TheLast Cipher
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1 Answers1

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You have first dealt with the easy part of the "iff": A bipartite graph cannot have an odd cycle. The correct idea is there, but as a "printed proof" it leaves much to be desired.

Now comes the other part: A (maybe infinite) graph with no odd cycles can be made bipartite. Here I cannot see a satisfying logic line. In your first sentence you talk about "all cycles", which is fine, but already in your second sentence you talk about "that" cycle, which you haven't defined. Note that you cannot assume that there is a single cycle going through all the vertices or edges. But you may assume (with noting this in a preamble) that the given graph is connected.

The second part requires a construction that partitions the vertex set $V$ into two parts. Here is a hint: Call two vertices $v_1$, $v_2\in V$ equivalent if they can be joined with a path of even length.

Christian Blatter
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