# Are tiny point-symmetric grids of two Pentominos impossible to solve?

The game of Pentomino is a tiling puzzle game played on a grid. A Pentomino piece is a two dimensional shape of five non-diagonally connected tiles. There are exactly 12 unique pieces (ignoring rotation/reflection). The goal of the game is to fill the board with the available pieces without overlapping.

This question concerns point-symmetric boards (not necessarily rectangular) of 10 spaces, i.e. with room for exactly two pieces.

Filling such a board is, of course, trivial when using the same piece twice.

Using the Y-piece twice on a point-symmetric hourglass board:
####
O#
OOOO

However, it seems impossible to do when using two different pieces. Is this true? How could this be proved?

This is not the case for larger boards or larger pieces (more than 5 tiles). Of course, the standard 6x10 board using 12 Pentomino pieces is also a counter example when using more than two pieces. Small counter examples:

Three 5-tile Pentomino pieces (V, P and L pieces) on a 3x5 grid:
vvvpp
vLppp
vLLLL

The 5-tile pieces P, T and F on an oval-shaped point-symmetric grid.
ppTF
pppTFFF
TTTF

Two 8-tile pieces on a 4x4 grid:
####
#OOO
#OOO
##OO
• A proof in either direction is "easy", just tedious: It's enough to check all possible ways of joining 2 pentominos together. There are 12 of them, each can be in at most 4*4=16 possible orientations (rotations by a multiple of 90 degrees, plus flips), and there are at most (5*4)^2=400 different ways to join two together (5 tiles to choose from and 4 possible directions in which it could be joined, for both pentominos), so there are at most 12*16*400=76800 cases to check. A good task for a computer program. Mar 7, 2020 at 16:28

Let's see what happens after you rotate the partition around the board's center. The region in which one part maps onto the other part is point symmetric around the board's center. Therefore, the region in which one part is mapped onto the same part is:

• also point symmetric with the same center
• itself partitioned into two point symmetric regions with the same center and
• those two regions have the same area

The smallest non-empty such region has an area of 4. There are three possible configurations (I tetromino, S/Z tetromino, O tetromino). To connect the outer partition of the I tetromino or the S/Z tetromino into a single polyomino you need four extra tiles. For the O tetromino you need five extra, and furthemore you can't connect both regions at once.

Therefore, there is no partition of a 10-tile point symmetric board into a pair of incongruent pentominoes, and only two partitions of 12-tile point symmetric boards into a pair of incongruent hexominoes, up to symmetry:

I:   S/Z:
@@@@ @@@.
.@@. @...
.... @@..
• This sounds right, but I do not understand it yet. In the first part, do you make any assumptions if the point of rotation is on a cell, between two cells or between four cells?
– mafu
Mar 8, 2020 at 9:15
• The same argument applies no matter what center you choose. No matter how you place the four tiles whose partition assignment the rotation preserves, you don't have enough extra tiles to connect them into a pair of pentominoes. Mar 8, 2020 at 9:27
• Can you briefly reiterate why the three condition points are mandatory?
– mafu
Mar 8, 2020 at 9:47