Timeline for Resolution of multiple edges
Current License: CC BY-SA 4.0
8 events
when toggle format | what | by | license | comment | |
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Mar 18, 2019 at 5:13 | comment | added | Manfred Weis | @FedorPetrov thanks vor the example; I see now, that I habe to rethink my algorithmic solution. Putting aside a maximal set of "correctly" colored balls doesn't solve the problem. | |
Mar 17, 2019 at 22:22 | comment | added | Fedor Petrov | 1) rrb, bbg - rBb, bRg 2)rBb, ggr - rBG, gBr. Capital letter means that the ball can not be used anymore. | |
Mar 17, 2019 at 21:50 | comment | added | Manfred Weis | @FedorPetrov in your question you say that "two girls may exchange a ball"; maybe you could give an example, how a correct exhange for rrb, bbg and ggr would look like when two girl exchange balls until each has rbg balls. | |
Mar 17, 2019 at 21:45 | comment | added | Manfred Weis | @FedorPetrov I have expanded my answer accordingly; I had not read your question careful enough, so I somehow missed that cyclic exchanges along cycles of arbitrary length are permitted. How would the girls agree on such an exchange along a long cycle? | |
Mar 17, 2019 at 21:42 | history | edited | Manfred Weis | CC BY-SA 4.0 |
provided a solution for cyclic exchange
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Mar 17, 2019 at 19:50 | comment | added | Fedor Petrov | The matching in this graph does not always correspond to a family of exchanges, say, a cycle of length 3 is also a matching. Or what do I miss? | |
Mar 17, 2019 at 18:47 | comment | added | Gerhard Paseman | Which would be nice if it could be done (and when k or n is 2 such a matching does exist). Now imagine three girls and three colors, where the arrangement is rrb,bbg,ggr. This can be resolved, but not by the matching you propose. Gerhard "If Only It Were Easy" Paseman, 2019.03.17. | |
Mar 17, 2019 at 17:27 | history | answered | Manfred Weis | CC BY-SA 4.0 |