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I've got a puzzle using dominoes for all here at Puzzling. I came up with it while playing around a few days ago, and I know it can be solved. It may not be a new idea, but to the best of my knowledge, no one else has posed it.

Using the standard set of 28 double-six dominoes, form a square frame of 9 by 9 squares with a 5 by 5 square gap in the middle (one domino being one square wide and two squares long). The dominoes must fit together in a brickwork-style interlocking pattern, such that the frame consists of an inner square frame of 12 dominoes and an outer frame of 16 dominoes.

Now, the catch:

All of the 2 by 2 squares of four adjacent domino-halves must have a pip count equal to 12.

Example

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    $\begingroup$ I've added an image which I think summarizes your puzzle. If not, please undo the edit. $\endgroup$
    – BmyGuest
    Feb 21, 2018 at 9:17
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    $\begingroup$ One thing is not 100% clear to me: If I shift a 2x2 with a pips count of 12 by one square (i.e. half its size) does it still needs to have a pips count of 12 ?? $\endgroup$
    – BmyGuest
    Feb 21, 2018 at 9:18
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    $\begingroup$ I've also linked an image of the standard set of double-six dominoes, for people who may not know what it looks like (for instance, me). $\endgroup$
    – Xenocacia
    Feb 21, 2018 at 9:18
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    $\begingroup$ @BmyGuest - I reckon yes, since the frame has an odd dimension and therefore cannot be dissected into 2x2 squares. $\endgroup$
    – Jaap Scherphuis
    Feb 21, 2018 at 10:11
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    $\begingroup$ @JaapScherphuis This was also my thining (hence the 3rd green square in my illustration), but I wanted to check back. $\endgroup$
    – BmyGuest
    Feb 21, 2018 at 16:04

2 Answers 2

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There was one idea I had that made constructing a solution a lot easier.

I decided to restrict myself to solutions where every pair of adjacent squares with one from the inner frame and one from the outer frame, adds up to 6. This automatically makes all 2x2 squares add up to 12.
The corner 2x2 squares are then forced to use two pairs of digits, arranged diagonally. Each corner must use a different pair of digits, otherwise you have a duplicate domino. All that is left is to fill in the edges of the frame. I hardly needed any backtracking to complete those sides.

Here is the solution I came up with.

enter image description here

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  • $\begingroup$ Nice! I also used this restriction when solving, but I didn't want to give too much away and spoil the fun. Now I am wondering if it can be solved without this added restriction.... Also, how many solutions might there be? No expectation for an answer there, just thinking out loud (in text lol). $\endgroup$
    – user15859
    Feb 25, 2018 at 23:00
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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.

enter image description here

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