Timeline for Distance measure on weighted directed graphs
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2010 at 20:26 | comment | added | Mitch | As to design criteria or motivation, I think that would warrant a separately titled question. | |
| Feb 3, 2010 at 20:24 | comment | added | Mitch | That min doesn't work but max does is not obvious to me (and I can't seem to think through it). Is there a short counterexample? | |
| Feb 3, 2010 at 18:10 | answer | added | Greg Kuperberg | timeline score: 6 | |
| Feb 3, 2010 at 17:24 | comment | added | Tom Leinster | Mitch, I meant max (and agree with domotorp). And yes, I also had it in mind that you could use + to symmetrize, rather than max. It sounds like you have some kind of design criteria in mind for how this distance should behave. Maybe you could add something to the question explaining them? (There's an "edit" button.) | |
| Feb 3, 2010 at 16:23 | comment | added | domotorp | Min does not satisfy the triangle inequality, only max does. | |
| Feb 3, 2010 at 16:07 | comment | added | Mitch | Yes...I had considered d(x,y) = (D(x,y)+D(y,x))/2 but that just didn't feel right. Did you mean 'min'? If so that seems much more satisfying for edge distance. but...I don't think it'll work for paths. | |
| Feb 3, 2010 at 15:58 | comment | added | Tom Leinster | Could you not do the same thing as with undirected graphs, which gives a non-symmetric metric D, and then symmetrize, i.e. define a metric d by d(x,y) = max{D(x,y), D(y,x)}? | |
| Feb 3, 2010 at 15:53 | history | asked | Mitch | CC BY-SA 2.5 |