Timeline for Transversals and almost transversals of a finite family of sets
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 9, 2020 at 19:10 | vote | accept | Adam Přenosil | ||
| Oct 9, 2020 at 19:00 | comment | added | Adam Přenosil | Antoine's understanding is correct: you have exactly one almost transversal for each of the sets $S_i$, while $n$ is the bound of the size of each $S_i$. In other words, the original index $l$, rather than $n$, was indeed correct. | |
| Oct 9, 2020 at 18:59 | history | edited | Adam Přenosil | CC BY-SA 4.0 |
The index l was in fact correct. n is the bound on the size of each S_i.
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| Oct 9, 2020 at 17:42 | comment | added | Michael Engelhardt | I think we should stop editing this question and wait for the OP to clarify. It is quite possible that the size of a family of $i$-transversals doesn't have to be tied to $l$, but that the OP then ultimately wants to consider such a family that happens to be of size $l$. | |
| Oct 9, 2020 at 16:02 | review | Suggested edits | |||
| Oct 9, 2020 at 17:30 | |||||
| Oct 9, 2020 at 15:59 | answer | added | Antoine Labelle | timeline score: 3 | |
| Oct 9, 2020 at 14:38 | comment | added | Antoine Labelle | @AlonYariv If I understand well I think that the index really should be $l$ and not $n$, $n$ is the maximum size of each $S_i$ | |
| S Oct 9, 2020 at 14:34 | history | suggested | Alon Yariv | CC BY-SA 4.0 |
I changed the index in the question, otherwise $n$ is meaningless. and fixed spelling mistakes and some grammatical errors
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| Oct 9, 2020 at 9:48 | review | Suggested edits | |||
| S Oct 9, 2020 at 14:34 | |||||
| Oct 9, 2020 at 6:01 | history | asked | Adam Přenosil | CC BY-SA 4.0 |