Grids with tiles of two colors and connectors

Question

1. Make a square grid with an even side.

2. Make a continuous trail which passes through all small squares; the trail ends where it started.

Where there is an angle you can put a red or a blue tile.

3. On the small squares with no tiles place connectors to ensure the continuity of the trail remains intact.

4. The connectors can be bicolor or tricolor. If we connect two red tiles then the connector has a blue color. If we connect two blue tiles, the connector has a red color. And if we connect a red and a blue tile the connector has a green color (tricolor).

5. The number of red tiles has to be equal to the number of blue tiles.

The challenge of the puzzle is, when an even side square grid is given, to place the maximum number of tiles with the least number of tricolor connectors. All conditions given above are mandatory.

If the given grid is 14x14, find the maximum number of tiles with the least tricolor connectors. Your answer must contain a grid with a continuous trail, as well as a grid with red and blue tiles and with the appropriate connectors.

Answer 1

My path:

Is something of a spiral.

My grid:

Is a checkerboard missing eight tiles of each color.

Generalization:

Continuation of the spiral provides solutions for all grids of size $n=4k+2$ with $n+2$ tiles missing. Shown here is the path for a $54\times 54$ grid with $56$ missing tiles.

Answer 2

Here is one solution (with colours 1 and 2, and connectors indicated by the double line):

1--2  1--2  2--1  2--1  1--2  1--2  1--2
║  |  |  ║  ║  |  |  ║  ║  |  |  |  |  ║
║  1--2  ║  ║  2--1  ║  ║  1--2  1--2  ║
║        ║  ║        ║  ║              ║
1--2  1--2  2--1  2--1  1--2  2========2
   |  |        |  |        |  |
2--1  2========2  1========1  1========1
|                                      ║
1--2  1=====1  2--1  2--1  2--1  2--1  ║
   |  |     |  |  |  |  |  |  |  |  |  ║
2--1  2--1  2--1  2--1  2--1  2--1  2--1
|        |
1--2  1--2  1--2  1--2  1--2  1--2  1--2
   |  |     |  |  |  |  |  |  |  |  |  ║
2--1  2=====2  1--2  1--2  1--2  1--2  ║
|                                      ║
1--2  2=====2  1--2  1--2  1--2  1--2  ║
   |  |     |  |  |  |  |  |  |  |  |  ║
2--1  1--2  1--2  1--2  1--2  1--2  1--2
|        |
1--2  2--1  2--1  2--1  2--1  2--1  2--1
   |  |     |  |  |  |  |  |  |  |  |  ║
2--1  1=====1  2--1  2--1  2--1  2--1  ║
║                                      ║
║  2--1  2--1  2--1  2--1  2--1  2--1  ║
║  |  |  |  |  |  |  |  |  |  |  |  |  ║
2--1  2--1  2--1  2--1  2--1  2--1  2--1

This has 24 missing tiles and 0 green connectors, which I strongly believe is optimal. Note that it clearly generalizes to a (4n+2)×(4n+2) solution that has 8n missing tiles, and I believe this is optimal.

Surprisingly, it seems that the (4n)×(4n) problem is harder, because I can achieve only 4n missing tiles, but when my solution has (n mod 2) green connectors... For instance, here is my 12×12 solution, which has 1 green connector at the top-right corner.

1--2  1--2  1--2  1--2  1--2  1--2
║  |  |  |  |  |  |  |  |  |  |  ║
║  1--2  1--2  1--2  1--2  1--2  ║
║                                ║
1--2  1=====1  2--1  2--1  2--1  ║
   |  |     |  |  |  |  |  |  |  ║
2--1  2--1  2--1  2--1  2--1  2--1
|        |
1--2  1--2  1--2  1--2  1--2  1--2
   |  |     |  |  |  |  |  |  |  ║
2--1  2=====2  1--2  1--2  1--2  ║
|                                ║
1--2  2=====2  1--2  1--2  1--2  ║
   |  |     |  |  |  |  |  |  |  ║
2--1  1--2  1--2  1--2  1--2  1--2
|        |
1--2  2--1  2--1  2--1  2--1  2--1
   |  |     |  |  |  |  |  |  |  ║
2--1  1=====1  2--1  2--1  2--1  ║
║                                ║
║  2--1  2--1  2--1  2--1  2--1  ║
║  |  |  |  |  |  |  |  |  |  |  ║
2--1  2--1  2--1  2--1  2--1  2--1

I am unsure whether it is possible to achieve 4n missing tiles with no green connector, because I can achieve (8n−4) missing tiles and 0 green connectors, so there is no obvious parity issue.