Timeline for Ten tetrominoes inside an 8x8 grid
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2022 at 15:13 | comment | added | Jaap Scherphuis | @JLee I have now posted it | |
| Jun 15, 2022 at 14:08 | comment | added | JLee | @JaapScherphuis my brain hurts. Please post the 3rd solution so i can kick myself | |
| Jun 14, 2022 at 6:31 | vote | accept | Dmitry Kamenetsky | ||
| Jun 14, 2022 at 11:02 | |||||
| Jun 13, 2022 at 20:29 | comment | added | JLee | Let us continue this discussion in chat. | |
| Jun 13, 2022 at 20:20 | comment | added | RobPratt | @JLee To see that $4$ is a lower bound for the independent domination number, note that no two of the following four tetrominoes can be blocked by a single tetromino: $$\begin{matrix} 1 &. &. &. &. &2 &2 &2 \\ 1 &1 &. &. &. &. &2 &. \\ 1 &. &. &. &. &. &. &. \\ . &. &. &. &. &. &. &. \\ . &. &. &. &. &. &. &. \\ . &. &. &. &. &. &. &3 \\ . &4 &. &. &. &. &3 &3 \\ 4 &4 &4 &. &. &. &. &3 \\ \end{matrix} $$ | |
| Jun 13, 2022 at 18:23 | comment | added | RobPratt | @Jlee It's fundamentally the same problem, just on a different graph. For $n\in\{1,\dots,8\}$, the optimal values are $0, 0, 1, 1, 1, 3, 4, 4$. | |
| Jun 13, 2022 at 15:11 | comment | added | RobPratt | @JLee You are talking about the independent domination number. You might be interested in my answer to this closely related problem. | |
| Jun 13, 2022 at 10:32 | comment | added | Jaap Scherphuis | I have confirmed by computer that there are two other solutions (ignoring rotation/reflection). If no one finds these, I'll post them in a day or two. | |
| Jun 13, 2022 at 7:47 | history | edited | Bass | CC BY-SA 4.0 |
You can have my picture, since you beat me to the answer :-)
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| Jun 13, 2022 at 7:46 | comment | added | Dmitry Kamenetsky | Well done! Can you find any other solutions? | |
| Jun 13, 2022 at 7:35 | history | answered | franck vivien | CC BY-SA 4.0 |