Given any graph $G$, can we find a bipartite subgraph of $G$ with at least $e(G)/2$ edges ($e(G)$ is the number of edges in $G$) by sequentially deleting the edge belonging to the most number of odd cycles? That is, if we take a graph and sequentially delete the edge which belongs to the most odd cycles until we have a bipartite graph, will at least half the edges remain when the graph is bipartite?
The motivation is to find a a simple method for making a graph bipartite making use of the odd cycle-free interpretation of bipartiteness rather than the $2$ color class viewpoint.
Note: I asked this question on M.SE. I expected I was missing some simple argument or counterexample, but no one has given an answer so perhaps it is more tricky:
https://math.stackexchange.com/questions/913366/a-method-of-making-a-graph-bipartite
Edit: clarity
Best Answer
A quick check with Mathematica seems to suggest that a (smallest) counterexample is given by a graph with 6 vertices and 13 edges:
Here the red edges are contained in the most odd cycles (16, 12, 6, 5, 2, 2, 1, 0 respectively) and one possible sequence of edge removals is shown. However, choosing different edges one can end up with a bipartite graph with 7 edges.
Edit: Any sequence of edge removals of the following graph with 8 vertices and 18 edges seems to result in a bipartite graph with at most 8 edges:
Edit: For those interested: here's the Mathematica code to check that the last graph is really a counterexample.
Edit: The first complete graph for which the algorithm fails is the complete graph on 10 vertices. Indeed, it seems that the largest bipartite graph one can obtain from an $n$-vertex ($n>4$) complete graph is the complete bipartite graph on $(n-3)+3$ vertices, which has $3(n-3)$ edges, i.e. smaller than $n(n-1)/4$ for $n\geq 10$.