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June 21st, 2015, 08:23 AM   #1
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Is it possible to create a circle with all Domino tiles?

Is it possible, according to Domino rules, to create a circle with all Domino tiles from the set, when the symbols:
a) 1, 2, 3, 4, 5, 6
b) 0, 1, 2, 3, 4, 5, 6
are allowed?

What I was trying to do first is brute force it with a computer program, but that kinda got out of hand, I have a working domino set now with no brute forcing yet, but that took me 4 hours.

Is there a way to solve this using graph theory and/or combinatorics? (Since those are the topics we are doing right now)

I'm thinking it's related to the Euler Circle or something similar.
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June 21st, 2015, 08:45 AM   #2
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I would go:
(0,1) (1,2) (2,3) ... (5,6) (6,0) (steps of 1 round the clock)
(0,2) (2,4) (4,6) (6,1) ... (3,5) (5,0) (steps of 2)
(0,3) (3,6) (6,2) ... (1,4) (4,0) (steps of 3)

That's 21 pieces and the seven doubles can be added anywhere appropriate.

This plan works where $n$, the number of different designs is coprime to all of 2, 3, ..., $\tfrac12 (n-1)$. It is this feature that allows each set to return to the zero in exactly $n$ steps. We have $\tfrac12 (n-1)$ sets of $n$ plus the $n$ doubles to get $\tfrac12 n(n+1)$ dominoes in the path.
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June 21st, 2015, 09:26 AM   #3
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Could I do it like this?

I am counting for a) 21 vertices and for b) 28 vertices.
Then for the degree I will subtract one from a) 21, 21-1 = 20 and b) 28-1 = 27.

Euler says the degree of each vertex has to be even in order for there to be a circle.
So the circle should work for a), but nor for b), right?
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June 21st, 2015, 09:40 AM   #4
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You shouldn't subtract 1. The "circle" is possible (and extremely easy) for (b), but not for (a).

For any circle, the numbers occur in pairs, but case (a) has each number occurring 5 times, which is impossible if they occur in pairs.
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June 21st, 2015, 09:51 AM   #5
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Okay, thank you for the answers!
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