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 Post subject: A domino puzzlePosted: Wed Dec 12, 2012 12:20 pm

Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
Those of you who have been following my posts may notice that yes, I am obsessed with dominoes. This time, here's an actual puzzle/problem.

If you remove the doubles from a double-six set, you are left with 21 pieces. These 21 pieces can be arranged into sets of 3 which match in a loop, such as 0-1, 1-2, 2-0.

The problem is, either find the next double-less set which can be divided into three-piece loops, or prove that no other exists.

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 Post subject: Re: A domino puzzlePosted: Wed Dec 12, 2012 1:18 pm

Joined: Thu Jul 23, 2009 5:06 pm
Location: Berkeley, CA, USA
I don't quite understand the goal. The double-less double six set can be arranged. Although you didn't give the complete partition, but I can see it can be done. This is not your question.

Are you asking, like, whether the double-less double-n set can be arranged for another n!=6? For example, whether the double-nine set works?

Do you require this set is commercially available or any n? At least n=2 works. Double less double-two set has only three ties, 0-1, 0-2 and 1-2.

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 Post subject: Re: A domino puzzlePosted: Wed Dec 12, 2012 1:26 pm

Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
By "next set" I meant any n > 6 is fine.

Obviously only sets with 3k pieces will work, though.

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 Post subject: Re: A domino puzzlePosted: Sat Dec 22, 2012 4:26 am

Joined: Sun Nov 23, 2008 2:18 am
I have not worked out a full solution yet, but I thought I would share my thoughts so far.

First of all, I am interpreting a "Double-N set minus the doubles" of Dominoes as follows:
An individual domino can be represented as a pair of intergers x and y.-
-The set of Dominoes under consideration consists of all pairs of integers x and y such that: 0 <= x < y <= n for n an integer.

It is fairly easy to determine that the nth such set contains a number of Dominoes equal to the nth triangular number.

Since, as Jared said, the desired parition requires a set that is a multiple of three, this can only work for n such that the nth triangular number is a multiple three.

Taking the triangular numbers mod 3 gives the pattern 1, 0, 0, 1, 0, 0, 1, 0, 0 ..., meaning that n MUST be either a multiple of 3 or one less than a multiple of 3.

Also, since creating loops requires matching numbers in pairs, and each number appears n times in the sets under consideration, n must also be even.

Taking these in combination, the only values of n where such a partition might be possible are n that either a multiple of six or two more than a multiple of six.

This gives 2, 6, 8, 12, 14, 18, 20, 24, 26, and 30 as the first 10 values of n that might work.

Its fairly trivial to show that such a partition exists for n = 2, and I am assuming you have already found such a partition for n = 6. I am interested in seeing if n = 8 can produce such a partition without breaking the existing parition for n =6, so I am going to see if I can partition ranks 7 and 8 of the n = 8 set to meet the requirements.

Also, I have done some thought on partitioning sets of dominoes into longer loops, and discovered the following pattern for modding the triangular numbers by 4: 1, 3, 2, 2, 3, 1, 0, 0, 1, 3, 2, 2, 3, 1, 0, 0... , which gives possible values of n of multiples of 8. Modding triangular numbers by 5 gives the pattern 1, 3, 1, 0, 0, 1, 3, 1, 0, 0... or values of n equal to 10k or 10k +4 for integer k.

I will report back if I find a partition of the n = 9 set's ranks 7 and 8.

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