Timeline for When does a `distinguished matching' exist?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2012 at 9:42 | vote | accept | Nick Gill | ||
| Nov 2, 2012 at 21:33 | answer | added | Nick Gill | timeline score: 1 | |
| Oct 24, 2012 at 12:40 | comment | added | Nick Gill | Emil, true enough. Patricia, sure, condition (3) is equivalent to something like "restricting the graph to DX and DY induces a bijective function between DX and DY". (This lies behind the applications I mention later.) | |
| Oct 24, 2012 at 12:36 | comment | added | Patricia Hersh | This is looking more and more like the definition of bijective function, phrased in graph theory language. | |
| Oct 24, 2012 at 12:26 | comment | added | Emil Jeřábek | Well, if you want to minimize the conditions, then in (3), the uniqueness of $y$ is redundant (it follows from (2)), and the existence of $x$ is also redundant (it follows from (1)). | |
| Oct 24, 2012 at 12:14 | history | edited | Nick Gill | CC BY-SA 3.0 |
corrected the third condition. again :-(
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| Oct 24, 2012 at 12:11 | comment | added | Nick Gill | Ben, condition (1) requires the friendship groups cover $Y$ so in general one cannot take $DX$ and $DY$ to both be empty. | |
| Oct 24, 2012 at 12:09 | comment | added | Nick Gill | heck, i've not made a very good job of this - sorry! i'm going to have another go in a moment at writing down all of the required conditions. my aim was to have as few as possible, so that the definition did not become hideous, but i have overdone it... | |
| Oct 24, 2012 at 12:03 | comment | added | Ben Barber | Here's a restatement of the three conditions. Every $y \in DY$ has a friend in $DX$. The friendship groups are disjoint. The friendship groups cover $DX$. Which of these conditions do you want to hold, and what constraints do you have on $DX$, $DY$? If you only want the existence of some $DX$, $DY$ then you can take them both empty, but this doesn't seem to have any application to your group theory problem. | |
| Oct 24, 2012 at 10:32 | comment | added | Emil Jeřábek | Patricia’s example satisfies the adjusted definition as well. | |
| Oct 24, 2012 at 9:03 | history | edited | Nick Gill | CC BY-SA 3.0 |
deleted 53 characters in body
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| Oct 24, 2012 at 8:49 | comment | added | Nick Gill | Patricia and Johan, thank you for your comments which are absolutely correct. I have adjusted the definition and I hope everything now makes sense. | |
| Oct 24, 2012 at 8:48 | history | edited | Nick Gill | CC BY-SA 3.0 |
Corrected definition in response to comments.
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| Oct 23, 2012 at 13:46 | comment | added | Johan Wästlund | Is there a typo or am I missing something? The first condition doesn't involve DY, and the second one is always satisfied when DY is empty or a singleton, so they can't possibly imply that |DX| = |DY|. | |
| Oct 23, 2012 at 13:41 | comment | added | Patricia Hersh | How do you rule out the possibility that $DY = \{ y_1 \} $ and $DX = \{ x_1,x_2,x_3 \} $ with edges $(x_i,y_1)$ for $i=1,2,3$? I guess I must be missing something? Thanks! | |
| Oct 23, 2012 at 9:06 | history | asked | Nick Gill | CC BY-SA 3.0 |