Timeline for Lower bound for a combinatorial problem ($N$ students taking $n$ exams)
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2019 at 17:23 | vote | accept | Morteza | ||
| Apr 12, 2019 at 10:45 | vote | accept | Morteza | ||
| Apr 12, 2019 at 10:45 | |||||
| Apr 12, 2019 at 10:45 | vote | accept | Morteza | ||
| Apr 12, 2019 at 10:45 | |||||
| Apr 12, 2019 at 10:15 | comment | added | Morteza | Unfortunately, I don't remember the idea for upper bound. I have to think again! | |
| Apr 12, 2019 at 9:12 | comment | added | Ilya Bogdanov | What is the constant at $n\log_2 n$ in your upper bound? Is there still a gap? | |
| Apr 12, 2019 at 1:39 | answer | added | Ilya Bogdanov | timeline score: 3 | |
| Apr 11, 2019 at 21:05 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
attempt at a more informative title
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| Apr 11, 2019 at 20:58 | answer | added | Will Brian | timeline score: 3 | |
| Apr 11, 2019 at 19:19 | comment | added | YCor | Could the title be related to the mathematical contents of the question? right now it seems to be an appeal for closing votes... | |
| Apr 11, 2019 at 19:18 | comment | added | Morteza | Yes an easy way. But can it be better than linear? | |
| S Apr 11, 2019 at 17:48 | history | suggested | user64494 | CC BY-SA 4.0 |
The title style is improved, a \LaTeX typo in the body is corrected.
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| Apr 11, 2019 at 16:35 | review | Suggested edits | |||
| S Apr 11, 2019 at 17:48 | |||||
| Apr 11, 2019 at 15:04 | comment | added | Will Brian | An easy way to get $2n-1$ as a lower bound: Fix $n-1$ students that are the best in every one of the $n$ exams (i.e., these $n-1$ students occupy the top $n-1$ spots for each exam, in some (irrelevant) order). Then let there be $n$ more students, each one in the $n^{\mathrm{th}}$ spot on exactly one exam. The first $n-1$ students will be selected for any ordering of the exams, and one of the other $n$ students will be selected if and only if his/her exam is held last. | |
| Apr 11, 2019 at 14:46 | comment | added | Morteza | Yes, exactly... | |
| Apr 11, 2019 at 14:46 | history | edited | Morteza | CC BY-SA 4.0 |
added 6 characters in body
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| Apr 11, 2019 at 13:57 | comment | added | Andreas Blass | Does "all the student are hopeful" mean that, for every student, there exists an ordering of the exams under which that student will be among the $n$ selected? | |
| Apr 11, 2019 at 13:50 | review | Close votes | |||
| Apr 12, 2019 at 10:40 | |||||
| Apr 11, 2019 at 13:18 | history | asked | Morteza | CC BY-SA 4.0 |