Timeline for What is an explicit bijection in combinatorics?
Current License: CC BY-SA 4.0
6 events
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| Feb 22, 2019 at 22:58 | comment | added | Brendan McKay | If you can't prove that an algorithm implements a bijection, it just means that you can't prove that you have an explicit bijection. I don't think it has anything to do with the definition of an explicit bijection. | |
| Feb 22, 2019 at 13:42 | comment | added | Andrej Bauer | I think this falls under my first counter-example. But more importantly, how do you precisely define "such that an analysis of the algoithm yields its bijectivity"? | |
| Feb 22, 2019 at 9:37 | comment | added | Martin Rubey | I disagree, because there are explicit bijections, which are beautiful, but bijectivity was very hard to prove or only follows because one knows for different reasons that the sets in question have the same cardinality. | |
| Feb 22, 2019 at 9:18 | comment | added | Dima Pasechnik | I don't see how this works with, say, A and B being two isomorphic finite groups - then certainly there is an algorithm giving a bijection, but it would be in general very ugly. An example of this would be arxiv.org/abs/1405.0113 - it's not something one would call explicit... | |
| Feb 22, 2019 at 8:57 | history | edited | Christian Stump | CC BY-SA 4.0 |
added 98 characters in body
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| Feb 22, 2019 at 8:46 | history | answered | Christian Stump | CC BY-SA 4.0 |