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28

The best you can do is one with an area of 30 (5 x 6): Disproving smaller cases 2 x 2 and 2 x 3 2 x anything 3 x 3 3 x 4 3 x anything 4 x 4 4 x 5 4 x 6 5 x 5 So that's definitely not an elegant mathematical proof, but I think it'll hold up. And since the image at the top of this answer shows that 5x6 can be done, that is the smallest possible ...


24

Here is a proof that 5x6 is the smallest possible rectangle. A rectangle of size $x$ by $y$ has $\frac{xy}{2}$ dominoes and $x+y-2$ potential lines. All of these lines must be blocked by at least one domino which has one square on each side. However, if the line divides the rectangle into two even areas, then one domino blocking it would leave an odd area ...


13

I'll one-up you guys and prove this stronger statement: If $m\geq n$ are the dimensions of a rectangle that admits a nonsplittable domino tiling using more than one domino, then $m \geq 6$ and $n \geq 5$. It's stronger because it also proves the nonexistence of very long yet thin nonsplittable domino tilings, such as $60\times3$ or $999 \times 4$. Proof: ...


11

There was one idea I had that made constructing a solution a lot easier. Here is the solution I came up with.


10

Final Solution(Though its already answered, I wanted to solve it on my own as a part of learning. Due to an ongoing problem with imgur, it took too long for uploading the solution) SUDOKU Solution. Updated Better solution Solution(Getting the idea from Ivo about the marked spots in the puzzle)


10

I count: reasoning:


10

With a friendly nod to @jafe (who came very close), since an hour has gone by I figured it was fair game to post my own independent solution… Didn't want to be seen as sniping or piggy-backing! This is indeed a: As follows (double moves abridged in the diagram - read along each row; moves in square brackets can be carried out in an alternative order): ...


9

Partial answer The first clue indicates Using this, we can convert the dominoes to


9

Yes, it is possible. One configuration is as follows. The 20 horizontally oriented are all of one type; the 12 lowercase dominos are the other type. With these orientations, all upper-left and lower-right corners (in absolute terms) are red, and all lower-left and upper-right corners are blue. Also, there are no four-corners intersections in the ...


9

In a ring, The given dominoes have three 1s, three 2s, four 3s, three 4s, five 5s. So 1, 2, 4 and 5 occur an odd number of times.


8

I worked with @Techidiot's sudoku solution and solved the Dominosa: The key to understand is that unlike regular Dominosa there are no tiles with same digits. and the grey cells indicate cells that shouldn't be included in the solution


8

It is as demonstrated by this image:


8

The generalization of this problem is already known: https://en.wikipedia.org/wiki/De_Bruijn_sequence. A De Bruijn sequence given an alphabet and a word-length is a string containing, as substrings, all words of that word length with letters from that alphabet exactly once. We can apply the existing knowledge on De Bruijn sequences (which can be found on ...


8

Here's an ugly paint drawing of a working pattern.


7

So I think it's Some reasoning.


7

It is So the picture is


7

As noted in the comments to the question, the tiles in the picture form a standard set except for an extra 5/6 and a missing 6/6. If we assume the picture to be correct, Otherwise, making the assumption that we are supposed to arrange a standard set:


7

Here's another solution. I also used Jaap's clever restriction; it's much easier to verify the solution by hand after using that particular limitation. This solution has twelves in the inside corners too.


7

the answer is There are mendatory dominoes with the line :


7

Since the 'donimoes' are limited to moving along their long axis, I believe this can be done in 12 moves as follows: Visually:


7



7

Note that So there are Now just The final answer:


6

Before the question was edited, this was a valid answer: Here's my new answer:


6

This puzzle appears to originate in many places, primarily under the name "Dominosa." I can't seem to find anything called "Dominotion" on the web besides on the UCLA website. (It's not actually a name I've heard before for this puzzle.) One of the oldest reference I can find online comes from Simon Tatham's puzzle collection, for which there is a working ...


6



5

The word is Found by searching a list of ten letter words on this site.


5

For $n = 1$, it's trivial that first player can't win since he can't make any move. Also for $n = 2$, both players can make the moves so the board will be full (by placing two $1 \times 2$ tiles), hence the first player can't win too. But for $n \geq 3$, For more details:


5

The player with the winning strategy is Strategy


5

This seemed to work. There are a couple of spots where there are disconnected pieces during a move, but IIRC that was acceptable.


5

Answer: First of all, Second, Finally:


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