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  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.

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

  2. 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).

  3. 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.

April14

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  • $\begingroup$ I have improved my solution. $\endgroup$ Nov 14, 2021 at 2:58

2 Answers 2

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My path:

Is something of a spiral.

Path

My grid:

Is a checkerboard missing eight tiles of each color.

Grid

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.

Larger Grid

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  • $\begingroup$ Wow so my hypothesis for (4n+2)×(4n+2) was wrong. Do you think yours is optimal? Also, are you able to generalize your solution? $\endgroup$
    – user21820
    Nov 10, 2021 at 7:07
  • $\begingroup$ @user21820 My previous solution was not optimal. My new solution, which I believe is optimal, does extend to larger grids. $\endgroup$ Nov 14, 2021 at 2:57
  • $\begingroup$ Very nice, but I can't upvote it again. Though it's a pity it's off by 2 from 'expected' based on the (4n)×(4n) solution that you found earlier. $\endgroup$
    – user21820
    Nov 14, 2021 at 3:02
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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.

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  • $\begingroup$ Also note that the solution in the question has the same number of missing tiles as my solution but 3 green connectors compared to my 0, so it is highly unlikely that there is a parity obstruction. All this makes me suspect that it is possible to get a (4n)×(4n) solution with 4n missing tiles and 0 green connectors, but I could not figure out how. $\endgroup$
    – user21820
    Nov 8, 2021 at 16:47
  • $\begingroup$ The grid has to be 14 X 14, as shown in the example provided. No empty spaces. $\endgroup$ Nov 8, 2021 at 21:52
  • $\begingroup$ @VassilisParassidis: My diagram is for 14×14, look again. I did not draw the grid as it is unnecessary. Each number denotes a coloured tile. $\endgroup$
    – user21820
    Nov 8, 2021 at 21:56
  • 1
    $\begingroup$ 4n x 4n: image link $\endgroup$ Nov 9, 2021 at 1:42
  • $\begingroup$ @DanielMathias: Very nice! I didn't realize that attaching at alternating sides would eliminate the length-2 connectors except for the start and end. Would you mind if I include your solution in my answer and credit you? $\endgroup$
    – user21820
    Nov 9, 2021 at 6:33

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