Timeline for Fixing code for a combinatorics problem
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2023 at 6:17 | vote | accept | matrix42 | ||
| Jun 25, 2023 at 1:50 | answer | added | ubpdqn | timeline score: 4 | |
| Jun 24, 2023 at 23:28 | comment | added | ubpdqn | Your code tests every pairwise intersection of all the 3 element subsets of 8 element set. This is not exactly what you want. Restating your problem, there exists a collection of n different subsets agreeing with condition. You need to find them. A strategy is to partition your 4k set into k 4 element sets and then 3 subset each 4 element set. The 4 element will always satisfy your conditions: {1,2,3}, {1,2,4}, {2,3,4}, {1,3,4}: each intersection length 2. | |
| Jun 24, 2023 at 18:26 | comment | added | ydd |
there are more than n different 3-element subsets that satisify this condition for any integer n>3. In fact (kind of unrelated but interesting), there is a closed form for the number of 3-element subsets satisfying your condition: $\frac{1}{72} (-3 + n) (-2 + n) (-1 + n) n (38 + (-9 +n) n)$
|
|
| S Jun 24, 2023 at 18:08 | history | edited | Domen | CC BY-SA 4.0 |
Improve wording
|
| Jun 24, 2023 at 18:07 | comment | added | JimB | I'm not following. There are 56 different 3-element subsets when $n=8$. Do you mean that one can find a set of 8 of those 56 that satisfy your restriction? Or am I missing something? | |
| Jun 24, 2023 at 16:07 | review | Suggested edits | |||
| S Jun 24, 2023 at 18:08 | |||||
| Jun 24, 2023 at 15:43 | history | asked | matrix42 | CC BY-SA 4.0 |