Timeline for Recovering set of $k$-subsets without specific element $t$ by modifying subsets with element $t$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 28, 2020 at 15:25 | comment | added | timudk | Great, thank you. | |
| Oct 28, 2020 at 14:32 | comment | added | Tony Huynh | Multiset: en.wikipedia.org/wiki/…. | |
| Oct 28, 2020 at 14:00 | comment | added | timudk | One last question: what mathematical "object" should I use for the $|\mathcal{R}|$ copies of $\mathcal{T}$. My issues is that, for example, $\{\{1, 2\}, \{1, 3\}, \{1, 2\}, \{1, 3\}\} = \{\{1, 2\}, \{1, 3\}\}$ so it seems I cannot express those copies as sets? | |
| Oct 28, 2020 at 3:54 | comment | added | Tony Huynh | You're welcome. The answer is yes to your new question. Just use the same proof. This time the bipartite graph $G$ has $|\mathcal R||\mathcal T|$ vertices on each side, and every vertex has degree $k |\mathcal R|=(n-k) |\mathcal T|$. Thus, $G$ has a perfect matching. | |
| Oct 28, 2020 at 2:35 | comment | added | timudk | Thank you so much for the answer and for fixing the formulation of the problem. I have a follow-up question, however, I am not entirely sure how to formulate it. As you said, the theorem just holds if $k \geq n - k$. What if I had $|\mathcal{R}|$ copies of $\mathcal{T}$ (call $\mathcal{T}_R$) and $|\mathcal{T}|$ copies of $\mathcal{R}$ (call $\mathcal{R}_T$). Is there a result and a proper way to connect each element of $\mathcal{T}_R$ with an element of $\mathcal{R}_T$. | |
| Oct 28, 2020 at 2:10 | vote | accept | timudk | ||
| Oct 28, 2020 at 1:12 | history | edited | Tony Huynh | CC BY-SA 4.0 |
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| Oct 28, 2020 at 0:21 | history | edited | Tony Huynh | CC BY-SA 4.0 |
added 9 characters in body
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| Oct 28, 2020 at 0:16 | history | answered | Tony Huynh | CC BY-SA 4.0 |