A standard set of double-six dominoes has 28 tiles with 2 numbers on each side from 0 to 6. Tiles can be placed next to each other if the numbers at each end match. Can you place all the 28 tiles such that they form a loop? Note that the end number must also match the front number to complete the loop.
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asked Oct 23, 2019 at 23:56
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1 Answer
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6
Note that
the numbers, $0-6$, are elements of the finite field $\mathbb{F}_7$.
So there are
circles like this: $(0)(d)|(d)(2d)|\cdots|(6d)(0)$ for every $d\in\mathbb{F}_7^\times$.
Now just
connect all the circles, and insert the remaining $7$ tiles of the form $(d)(d)$ anywhere you like.
The final answer:
00 01 11 12 22 23 33 34 44 45 55 56 66 60 02 24 46 61 13 35 50 03 36 62 25 51 14 40
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1$\begingroup$ You don't need the finite field, as any set of cycles will do. It is essentially an Eulerian circuit through a complete graph (with a loop on each vertex), and the method you describe is a general algorithm for any graph that admits a Eulerian circuit. $\endgroup$ Oct 24, 2019 at 4:14
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$\begingroup$ @JaapScherphuis Of course I understand the interpretation of eulerian circuit on the complete graph, but the question is exactly to describe such a circuit, right? And I found it most convenient to interpret it as finite field elements... So I don't quite get your point. $\endgroup$– WhatsUpOct 24, 2019 at 5:22
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$\begingroup$ @JaapScherphuis Well... upon carefully reading the question, it does say "Can you place ...". So perhaps you mean a better answer should be "yes I can, because $7$ is an odd number"? $\endgroup$– WhatsUpOct 24, 2019 at 5:26
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$\begingroup$ I just meant that you can just start laying out a line of dominoes until you can't extend it any further. At that point it will be a circle (yes, because 7 is odd). It just seemed to be so oddly specific to construct the circles the way you did, making it seem as if there was a reason for it, as if only those circles will do. $\endgroup$ Oct 24, 2019 at 5:39