Timeline for Computing the equivalence classes of the symmetric transitive closure of a relation
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 29, 2012 at 21:14 | vote | accept | a06e | ||
| Jun 29, 2012 at 11:14 | answer | added | user1066 | timeline score: 4 | |
| Jun 29, 2012 at 7:36 | history | tweeted | twitter.com/#!/StackMma/status/218608933919666176 | ||
| Jun 29, 2012 at 5:57 | answer | added | Mr.Wizard | timeline score: 6 | |
| Jun 29, 2012 at 5:39 | answer | added | Mr.Wizard | timeline score: 5 | |
| Jun 29, 2012 at 5:38 | history | edited | a06e | CC BY-SA 3.0 |
added 2 characters in body
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| Jun 29, 2012 at 5:34 | answer | added | a06e | timeline score: 2 | |
| Jun 29, 2012 at 5:34 | answer | added | kglr | timeline score: 12 | |
| Jun 29, 2012 at 5:32 | comment | added | a06e | @Jens Thanks for your comment. I see now. I did some drastic edits to the question (prior to seeing your comment). Now it's more to the point of what I want. Your comment now seems out of context, but it does answer the original question. | |
| Jun 29, 2012 at 5:29 | comment | added | Jens |
From {13,10} and {10, 36}, you conclude that 10 and 13 belong together. But 13 also belongs with 1 because {1, 13} exists. However, there is no pair {1, 10} or {10, 1}, so 1 shouldn't be in the same group as 10, which is in the same group as 13 which belongs with 1. So the problem for Gather is that your identity relation is not transitive. As a result, you have to give up some conditions. Maybe you want all gathered groups to be numbers that are at least indirectly connected?
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| Jun 29, 2012 at 5:29 | comment | added | Mr.Wizard | Related: mathematica.stackexchange.com/q/3234/121 | |
| Jun 29, 2012 at 5:25 | history | edited | a06e | CC BY-SA 3.0 |
deleted 414 characters in body; edited title
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| Jun 29, 2012 at 4:56 | history | asked | a06e | CC BY-SA 3.0 |