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Sep 14, 2021 at 18:18 comment added Karagounis Z Isn't the definition of strongly base-orderable that there is a bijection such that for every subset X of B1, that both B1 \ X + phi(X) AND B2 \ phi(X) + X are bases? I get this from here: en.wikipedia.org/wiki/Base-orderable_matroid
Jan 20, 2015 at 17:33 comment added Marek Adamczyk Good point. Suppose we are given two bases, and we want to see whether they satisfy the strong exchange property in time polynomial in the rank. Assume we have an oracle that given a set tells us whether it's independent. For weak orderability it is possible in polynomial time, i.e., just construct a graph of possible exchanges and find a perfect matching. My main interest is in understanding whether it's the same situation here as with Hall's theorem for perfect matchings --- we run an algorithm polynomial in the number of vertices, but then we know Hall's condition holds for all subsets.
Jan 20, 2015 at 1:05 comment added Gordon Royle How is the matroid given?
Jan 16, 2015 at 0:42 history edited Marek Adamczyk
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Jan 15, 2015 at 15:51 review First posts
Jan 15, 2015 at 15:54
Jan 15, 2015 at 15:47 history asked Marek Adamczyk CC BY-SA 3.0