4-DominoBrane Exclusive

Question

Here is a smaller puzzle from the set of domino puzzles, I call DominoBrane. This uses 4-Dominoes (all the normal dominoes, minus any with a 5 or a 6).

1. Each domino must connect, as in the normal game, to it's neighbour [0:3][3:3][3:1] etc and make a full circuit of all the 4-dominoes.

2. At each square, there must be no ambiguity as to which is the connected domino (eg, for a 4 there will be only one adjacent domino with a 4)

3. The dominoes must be constrained to a 5x6 grid.

Here is an example question, showing the path of the chain (or circuit).

╔═══╦═══╦═══╦═══╦═══╗
║   ║ 4 ║   ║ 4 ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║ 3 ║   ║ 3 ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║   ║ 2 ║   ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║ 1 ║   ║ 1 ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║ 0 ║   ║ 0 ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║ 3 ║   ║ 4 ║   ║
╚═══╩═══╩═══╩═══╩═══╝

And it's unique solution.. Note that there it is trivial to construct the circuit because each domino has only one place to go, following rule 2.

╔═══╦═══════╦═══════╗
║ 4 – 4   4 – 4   0 ║
║   ╠═══════╬═══╦═|═╣
║ 1 ║ 3   3 – 3 ║ 0 ║
╠═|═╬═|═════╣   ║   ║
║ 1 ║ 3   2 ║ 1 ║ 2 ║
║   ╠═════|═╬═|═╬═|═╣
║ 1 – 1   2 ║ 1 ║ 2 ║
╠═══╬═══════╣   ║   ║
║ 0 – 0   0 – 0 ║ 2 ║
║   ╠═══════╬═══╩═|═╣
║ 3 – 3   4 – 4   2 ║
╚═══╩═══════╩═══════╝ 

To clarify rule 2, as @Magma eloquently puts it: “For each square in the grid there is exactly one adjacent square which has the same number but is not part of the same domino.”

If one doesn't follow rule 2, there are multiple solutions, such as the following:

╔═══════╦═══════╦═══╗   
║ 2   4 – 4   4 – 4 ║
╠═|═╦═══╬═══════╣   ║   
║ 2 ║ 3 – 3   3 ║ 1 ║ 
║   ║   ╠═══╦═|═╬═|═╣   
║ 2 ║ 2 – 2 ║ 3 ║ 1 ║  <-- But both 2s on the left have 2 possible 
╠═|═╬═══╣   ║   ║   ║  connections, making it harder to work out.
║ 2 ║ 1 – 1 ║ 1 – 1 ║  <-- And here there are problems with the 1s
║   ║   ╠═══╩═══╬═══╣   
║ 0 ║ 0 – 0   0 – 0 ║  <-- Again a difficulty with the 0s.
╠═|═╩═══╬═══════╣   ║   
║ 0   3 – 3   4 – 4 ║
╚═══════╩═══════╩═══╝   

The puzzle: 4-DominoBrane #1, first published here

Following the rules above, find the solution where the numbers match the grid as follows (a blank means 'any number':

╔═══╦═══╦═══╦═══╦═══╗
║   ║   ║ 0 ║   ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║ 1 ║   ║ 1 ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║   ║ 2 ║   ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║ 3 ║   ║ 3 ║   ║
╠═══╬═══╬═══╬═══╬═══╣
║ 2 ║   ║ 4 ║   ║ 4 ║
╠═══╬═══╬═══╬═══╬═══╣
║   ║ 0 ║   ║ 0 ║   ║
╚═══╩═══╩═══╩═══╩═══╝

If you solve it, I would really love to know how!
You will be the first person to solve any 4-DominoBrane.

There is exactly one solution to this puzzle.

Be careful to follow rule #2 - there are many more solutions without that constraint.

Some starters

I may be wrong, so don't take my word for it!

Corners are gone through only one way.

Following the circuit, numbers always come in exactly 1 run of 4 (where the double is) and exactly 1 run of 2 (where the other two tiles join).

Rule #2 is your friend

Look where some numbers cannot go. For instance, can 1 be placed in the bottom two rows?

Work on establishing domino edges, chain directions, and numbers.

Even if you cannot yet place their ends, you can work out exactly which pieces will be needed in a sub-chain once there is a double, and that eliminates their use elsewhere.

Just like all other 'missing pieces' grids, start where there's more information!

Possible next step for @Bass after pondering diagram 1

I may be wrong but if you place a 3 in row-5, col-4 then one cannot use the 2 at row-3, col-3 into a 4-sequence, and the 2 sequence is already used in the bottom left.

Answer 1

First, let's make some general observations.

1. Because of how the dominos connect, as we follow the loop, we will encounter every digit in exactly two runs. Those runs will have lengths 2 and 4.

2. The loop can go through a corner in only one way.

So we start by noticing that the bottom zeroes must be part of the same run. (The other run of zeroes is at the top.)

Since the path must continue from the zeroes into both bottom corners, we know that the 0-2 and 0-4 pieces will get used there. (Exact positions unknown as of yet.)

This means that the top run of zeroes has the 0-1 and 0-3 pieces, which creates a run of 3s at the top, so the other 3s must belong to the same run.

Further examining the top 0, we notice that it must belong to the 0-3, which can only be horizontal. If it were the 0-1, then we would either break rule 2, or create a dead end at one of the top corners.

Also, the square between the ones cannot belong to the same domino as the 2: the 2-0 was already used at the bottom left, and a 2-1 would break rule 2.

Drawing all this out (using the CtC pencil mark scheme: centre means "must be one of these numbers", corner means "must be in one of these squares"), we have

Continuing from here, we notice that whichever way we connect the path at the bottom half, we will always

• use up the 3-4, and
• exit the bottom half at the leftmost and rightmost squares.
• also, the leftmost square in row 4 cannot be a 3.

.. and this is where I got stuck. No logical argument seemed to lead anywhere, and everything was always ambiguous, so after an hour or so, I just gave up and bifurcated. Sorry about that. If there was a logical way forward, I'd be happy to hear it.