# What is the largest, compact connected-network polyomino for these tiles?

This is a graphical arrangement puzzle. Consider the following 36 tiles: Each tile has up to 8 connected directions and each tile has an initial up side. (The edge to whose centre the central pentagram points is the upper edge.)

These tiles have to be arranged into an arbitrary, single polyomino with a single restriction:

No tile may be placed such that a (brown) connection points directly at another tile that isn't pointing back.
Note that it is possible to have a connection pointing at "nothing" (on the edge of the polyomino), but if it points at another tile, then that tile must have a connection pointing back.

The resulting polyomino is scored like this:

• Dissect the polyomino (however you wish) into smaller rectangles such that each tile is part of exactly one such rectangle.

• The number of inner edges in these rectangles - i.e. all edges which connect two tiles of the rectangle - are counted, and that number is squared.

• If any tile was rotated from its original position, a value has to be subtracted from your score: This value is the square of the total number of 90° rotations used (compared to the original tile-set.) For example, if one tile was rotated clockwise by 90° and another by 180°, then the total number of 90° rotations is 3, and the penalty is $$3^2=9$$.

• Bonus score: Solutions which include all 36 tiles in a square are ranked by the number of 'separate networks' they form - the less networks the better, i.e. the ultimate solution additionally forms a single "brown" network.

## Find the highest possible score for this puzzle.

(The tile-arrangement has to be shown.)

## Example of a valid arrangement (only 9 tiles): As no tile was rotated, the score of this set is $$7^2 + 1^2 + 0^2 = 50$$ ( A 2x3 rectangle has 7 inner edges, a 2x1 rectangle has 1 inner edge, a single tile has no inner edge. )

## Example of an invalid arrangement (only 9 tiles): The bottom-right corner tile points towards the central tile, which isn't pointing back.

## 2nd Example of an invalid arrangement (only 4 tiles): Both C4 and A4 point at a tile (the one diagonal below) which is not pointing backwards. (That they two 'link' with each other via the corner is not important.)

• The most optimal solution would be to place all 36 into a single rectangle (of any dimensions), with no rotations. There is no guarantee that such a solution exists, but it is likely. One can write a program to find such a solution. Sep 22, 2015 at 17:52
• @ghosts_in_the_code Sure, but the puzzle is to find it - or the next best if it is not possible. Sep 22, 2015 at 18:12
• The simplest program to find the perfect single rectangle would be about ten lines long, but the sun would die out before it tried all 7*36! possibilities. Sep 22, 2015 at 18:45
• @Kevin It's much more than that, remember the tiles can be rotated. so, there is a factor of $4^{36}$ there. The death of the sun is tomorrow considering that. Sep 22, 2015 at 19:25
• well at the point where I stopped because I have to also work sometimes when I'm at work.... (I think) I easily had it 5x5 and messing around with the pieces to put it at 5x6. So I'm fairly certain the answer is going to be 5x6 + the rest. (It gets complicated fast from there to try and fit it to 6x6 so I doubt that solution exists, but I'm not there yet) - will post it tonight as coordinates. Sep 22, 2015 at 20:36

# 6x6 solution with 3 rotations: 3591

6x6 square = 1 bounding rectangle with 60 inner edges
3 total 90° rotations

$60^2 - 3^2 = 3591$ (single network) Thanks again to Sleafar for the image!

• Nice, looks like it's time to brute-force the rest, to see if there is a better option. Sep 26, 2015 at 11:25
• Very nice! I'll accept this unless a better solution comes about.... Sep 26, 2015 at 12:58

## 6x6 solution with rotations: 3479

$60^2 - 11^2 = 3479$ (single network) ## Inkscape image

Not part of the answer, but I have created an Inkscape image which should make handling the tiles way easier. Download it from here:

http://imgh.us/polyomino.svg

For snap to work configure it like in the screenshot below (the three activated buttons near the right border): • Thanks for making a start. As I've had to change the scoring to avoid the trivial case, I've updated your values. Sep 23, 2015 at 6:26
• Does E6 and E1 work though? they aren't linked (non continuous) Sep 23, 2015 at 18:35
• @Spacemonkey These is only "a single restriction" in the question, and it doesn't forbid this AFAICT. Sep 23, 2015 at 19:04
• @Sleafar you are correct, I read the OP anew. I must have imagined that restriction while looking at the polyomino examples Sep 23, 2015 at 20:01

# 6x6 solution with rotations: 3404

6x6 square = 1 bounding rectangle with 60 inner edges
14 total 90° rotations

$60^2 - 14^2 = 3404$ A big thanks to Sleafar for the Inkscape image! It made my life a lot easier when playing around with this.

• Well done! The first to find a square - and manually (and before the end of the universe ;c) ) I wonder if anybody could beat it with less rotations.. Sep 24, 2015 at 6:19
• As there could be more 6x6 solutions - even with the same point-score - I've added a bonus-score rule: If two 6x6 squares have the some point-score, the one with lesser 'separate' networks is superior. Sep 24, 2015 at 6:27
• Seems like a reasonable tie-breaker! Sep 24, 2015 at 12:47
• @BmyGuest 6*6 is not the only solution left. 4*9 with 8 rotations would have a higher score as well. Sep 24, 2015 at 18:24