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OEIS sequence A280984 (based on this Math Stack Exchange question) describes the

minimum number of dominoes on an $n \times n$ chessboard to prevent placement of another domino.

The sequence begins:

0, 2, 3, 6, 9, 12, 17, 22, 27, 34, 41, 48, 57, 66, 75

Fifteen terms are known, and a few folks have conjectured that $$ A280984(n) = \left\lceil \frac {n^2}3 \right\rceil \text{ for } n > 1. $$

Andrey Zabolotskiy has a proof that (1) this is a lower bound, and (2) this conjecture holds when $n$ is a multiple of $3$.

In order to prove the conjecture it is sufficient to give a tiling strategy for the $(3n+1)\times(3n + 1)$ and $(3n+2)\times(3n+2)$ boards.

Example of $3n \times 3n$ boards.

To achieve this bound for the $3n \times 3n$ boards you can just do the naive placement, but for the other boards, it's not clear which placement strategy to use.

(Notice that the second and third examples are not optimal: the last two boards have $35$ tiles and $42$ tiles respectively, but there exist $34$ and $41$ tile placements.)

3n by 3n (3n+1) by (3n+1) (3n+2) by (3n+2)

Question

Is there simple strategy for the $(3n+1) \times (3n+1)$ and/or the $(3n+2) \times (3n+2)$ boards that attains the lower bound? Or is the conjecture false?

Peter Kagey
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  • What is conjectured for the $m\times n$ board? – bof Apr 18 '19 at 05:42
  • I don’t know of any standing conjectures, but a perfectly naive guess is $\lceil nm/3 \rceil$ once $n$ and $m$ are large. – Peter Kagey Apr 18 '19 at 19:19
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    I stumbled on this article: https://www.sciencedirect.com/science/article/pii/0012365X88900283 It has a conjecture, and since it is old (1988) you may find newer results by seeing what cited this article etc. Apparently this type of arrangement is called a clumsy packing. – Herman Tulleken May 21 '19 at 08:52
  • TWIMC: Peter has added a reference to the "Clumsy packing of dominoes" paper into the OEIS sequence entry, and I have removed the conjecture mentioned in the question because it is now obvious that it does not hold. – colt_browning May 23 '19 at 20:46
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    Here is a thesis that considers the more general problem: https://pdfs.semanticscholar.org/b876/f89c89f766507642a38da86ca6b65181e8d1.pdf (I haven't read it yet but judging from the images it looks very interesting). – Herman Tulleken May 27 '19 at 07:17
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    Walter Trump has just added the terms 19-33 of the sequence (with $a(19)=122=\lceil 19^2/3\rceil+1$) with some examples of optimal solutions and announced an effective algorithm for finding the optimal solutions. – colt_browning Jun 14 '20 at 20:30

1 Answers1

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Too long for a comment, but here's a solution for $n=16$ that attains the lower bound of $\lceil 16^2/3\rceil=86$: \begin{matrix} 0 &22 &22 &0 &26 &26 &0 &42 &42 &0 &58 &58 &0 &74 &74 &0 \\ 1 &0 &23 &23 &0 &33 &33 &0 &49 &49 &0 &65 &65 &0 &81 &81 \\ 1 &2 &0 &3 &27 &27 &0 &43 &43 &0 &59 &59 &0 &75 &75 &0 \\ 0 &2 &4 &3 &0 &34 &34 &0 &50 &50 &0 &66 &66 &0 &82 &82 \\ 5 &0 &4 &0 &28 &28 &0 &44 &44 &0 &60 &60 &0 &76 &76 &0 \\ 5 &6 &24 &24 &0 &35 &35 &0 &51 &51 &0 &67 &67 &0 &83 &83 \\ 0 &6 &0 &7 &29 &29 &0 &45 &45 &0 &61 &61 &0 &77 &77 &0 \\ 8 &0 &9 &7 &0 &36 &36 &0 &52 &52 &0 &68 &68 &0 &84 &84 \\ 8 &10 &9 &0 &30 &30 &0 &46 &46 &0 &62 &62 &0 &78 &78 &0 \\ 0 &10 &0 &11 &0 &37 &37 &0 &53 &53 &0 &69 &69 &0 &85 &85 \\ 12 &0 &13 &11 &14 &0 &40 &40 &0 &56 &56 &0 &72 &72 &15 &0 \\ 12 &16 &13 &0 &14 &38 &38 &0 &54 &54 &0 &70 &70 &0 &15 &17 \\ 0 &16 &0 &25 &25 &0 &41 &41 &0 &57 &57 &0 &73 &73 &0 &17 \\ 18 &0 &19 &0 &31 &31 &0 &47 &47 &0 &63 &63 &0 &79 &79 &0 \\ 18 &20 &19 &21 &0 &39 &39 &0 &55 &55 &0 &71 &71 &0 &86 &86 \\ 0 &20 &0 &21 &32 &32 &0 &48 &48 &0 &64 &64 &0 &80 &80 &0 \end{matrix}

$n=17$ with $\lceil 17^2/3\rceil=97$ dominoes: \begin{matrix} 0 &89 &89 &0 &1 &0 &2 &0 &93 &93 &0 &3 &0 &4 &0 &5 &0 \\ 88 &88 &0 &6 &1 &7 &2 &92 &92 &0 &8 &3 &9 &4 &10 &5 &11 \\ 12 &0 &13 &6 &0 &7 &0 &14 &0 &15 &8 &0 &9 &0 &10 &0 &11 \\ 12 &16 &13 &0 &17 &0 &18 &14 &19 &15 &0 &20 &0 &21 &0 &22 &0 \\ 0 &16 &0 &23 &17 &24 &18 &0 &19 &0 &25 &20 &26 &21 &27 &22 &28 \\ 29 &0 &30 &23 &0 &24 &0 &31 &0 &32 &25 &0 &26 &0 &27 &0 &28 \\ 29 &33 &30 &0 &34 &0 &35 &31 &36 &32 &0 &37 &0 &38 &0 &39 &0 \\ 0 &33 &0 &40 &34 &41 &35 &0 &36 &0 &42 &37 &43 &38 &44 &39 &45 \\ 46 &0 &47 &40 &0 &41 &0 &48 &0 &49 &42 &0 &43 &0 &44 &0 &45 \\ 46 &50 &47 &0 &51 &0 &52 &48 &53 &49 &0 &54 &0 &55 &0 &56 &0 \\ 0 &50 &0 &57 &51 &58 &52 &0 &53 &0 &59 &54 &60 &55 &61 &56 &62 \\ 63 &0 &64 &57 &0 &58 &0 &65 &0 &66 &59 &0 &60 &0 &61 &0 &62 \\ 63 &67 &64 &0 &68 &0 &69 &65 &70 &66 &0 &71 &0 &72 &96 &96 &0 \\ 0 &67 &0 &73 &68 &74 &69 &0 &70 &0 &75 &71 &76 &72 &0 &97 &97 \\ 77 &0 &78 &73 &0 &74 &0 &79 &0 &80 &75 &0 &76 &0 &81 &0 &82 \\ 77 &83 &78 &0 &91 &91 &84 &79 &85 &80 &0 &95 &95 &86 &81 &87 &82 \\ 0 &83 &0 &90 &90 &0 &84 &0 &85 &0 &94 &94 &0 &86 &0 &87 &0 \end{matrix}

You can think of this problem as computing the independent domination number of a graph with a node for each of the $2n(n-1)$ dominoes and an edge for each pair of dominoes that share a cell.

RobPratt
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