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May 20, 2014 at 10:33 vote accept Joseph O'Rourke
May 20, 2014 at 4:58 history edited Jan Kyncl CC BY-SA 3.0
now considering also separators formed by two noncontractible curves
May 20, 2014 at 3:15 comment added Jan Kyncl You are absolutely right, I missed those disconnected disconnectors; they should be included in the calculations. Fortunately, we have to consider only pairs of boundary curves. Since we are on the torus, at least one of the components of every disconnected subgraph has at most two boundary components (either one contractible curve of a pair of parallel noncontractible curves).
May 19, 2014 at 10:41 comment added Joseph O'Rourke Brilliant idea to use polyomino counts! That leaps the hurdle that was blocking me. However, are there not some non-polyomino disconnectors, e.g., two parallel horizontal rows? Of course their effect would be negligible because they need $2n$ edges to break simultaneously.
May 19, 2014 at 7:07 history edited Jan Kyncl CC BY-SA 3.0
added the second sentence, corrected the upper bound
May 19, 2014 at 6:59 history edited Jan Kyncl CC BY-SA 3.0
added the second sentence
May 19, 2014 at 1:24 history answered Jan Kyncl CC BY-SA 3.0