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Sep 14, 2015 at 13:57 history edited Daniel Soltész CC BY-SA 3.0
Motivated the question.
Jul 5, 2015 at 0:46 comment added Daniel Soltész It is indeed not minimal, you can delete an edge of the identified point and the resulting graph is still $m$-tough. The minimally $m$-tough subgraph in your graph is a star-like tree but with longer paths instead of the leaves.
Jul 4, 2015 at 23:37 comment added The Masked Avenger I'm still getting my head around the definition. Anyway, take m = 1/epsilon many disjoint cycle graphs of diameter > 2 , and pick a point on each cycle and identify all those m points to get an m petalled flower graph. Does this help with the question? (I'm unsure if it is minimal.)
Jul 4, 2015 at 14:01 history asked Daniel Soltész CC BY-SA 3.0