On existence of boards that be covered by every free tetromino

There is a board which can be covered by each of five free tetrominoes:

However, it's not simply-connected (has a hole). I wonder if there is a simply-connected board with the same property.

Solution 1:

I give here some statements that determine the properties of a figure that is divided into four-celled one-type polyminoes - both squares and zigzags. These statements, in my opinion, give the direction of proof of the non-existence of such a figure without internal voids, which is somewhat cumbersome. I think that there exists a simple and beautiful proof, but, unfortunately, I have not found it yet.

Preliminary definitions

A figure is a finite set of cells of the field.

A row is a set of all cells, located in one column or one line.

A segment is a finite connected set of cells of one row. The length of the segment is equal to the number of its cells.

A maximum segment (MF-segment) is a segment, all cells of which belong to the figure, and the endpoints are bounded by non-elements of the figure.

A maximum empty segment (ME-segment) is a segment, all cells of which do not belong to the figure, but the ends are bounded by the cells of the figure.

A strip is the set of all cells of the figure in the same row. The strip consists of MF-segments of one row. The length of the strip is the sum of the lengths of MF-segments and ME-segments that are in the row of the strip.

A border strip is a strip, which is the boundary left, right, bottom or top. Zigzag will be called vertical if its height is 3, otherwise it is horizontal.

In addition, we will distinguish S-and Z-zigzags.

Let the figure be divided into four-celled one-type polyminoes - both squares and zigzags. Taking into account that an entire number of sides of squares fit on each MF-segment, one can obtain the following assertions.

Properties of the figure

P1. Each MF-segment has an even length.

P2. Each cell of the boundary strip corresponds to the adjacent cell of the neighboring strip. Each MF-segment of the boundary strip corresponds to the adjacent segment of the neighboring strip.

Evidence. The squares whose cells form MF-segments of the boundary strip can be filled with 1x2-rectangles perpendicular to the row of the boundary strip. Such filling gives the necessary correspondence.

P3. The left and right cells of the upper and lower rows can not belong to the vertical zigzag.

Evidence. a). Let the left cell of the figure in the top row be the top cell of the vertical S-zigzag. Then the upper left horizontal segment consists of one upper cell of this S-zigzag and an even number of upper cells of horizontal Z-zigzags. There is a contradiction with P1.

b). Let the left cell of the figure in the top row be the upper cell of the vertical Z-zigzag. Then it is the left upper cell of the square and, therefore, part of the segment of the second line to the left of the side of this square must contain an even number of cells. This is possible only if the left cell of the second row is simultaneously the upper cell of the vertical Z-zigzag and the square. Carrying out further similar arguments for the following lines, we come to the conclusion that in this case the figure must contain an infinite number of lines.

P5, a). Each boundary strip consists of a single segment.

b). Each boundary strip is formed by cells of the same type, either S- or Z-zigzags.

c). The left and right strips are formed by cells of vertical zigzags.

d). The upper and lower strips are formed by cells of horizontal zigzags

P6. The length of the third from any edge of the strip is odd.

P7. The third row from any edge of the figure contains an ME-segment of odd length.

The assertions P5, P6 follow from the structure of the figure possessing the properties P1-P4. The assertion of P7 follows from P1, P5a and P6.

Thus, the left cell of the figure in the top row is a cell of a horizontal zigzag. For the right cell, the assertion is valid by virtue of symmetry. P4. The upper and lower cells of the right and left rows belong to the vertical zigzags. The assertion follows from P3 when the figure is rotated by $\pi/2$.

Then we can only consider a figure whose left cell of the bottom row is a cell of the horizontal Z-zigzag (for the S-zigzag reasoning is similar with respect to the mirror image of the figure).

To obtain the final result on the basis of the above statements, it is necessary to prove that the stripes of the figure framing the empty segments must be connected. It's as if it's obvious, but the formal proof seems cumbersome.

Addition

An example of a figure formed by a simple composition of two simple figures is given, each of which is formed by a broken and a closed sequence of one-type zigzags. In this case, the inner (small) figure consists of 4 zigzags and has inside one empty cell. The outer (large) figure consists of 12 zigzags and has inside an empty 5x5 square, in which a small figure is placed. In the example, I showed that with simple jointing of figures one can use their mirror images, and the types of zigzags in different parts can be different

Solution 2:

Something like this may work to prove a figure that can be tiled by both squares and skew tetrominoes must have holes. (This builds on some ideas from Yog Urts answer.)

We can always find a cycle of squares where they are connected when they are overlapped by the same tile in the skew tiling. Trough a series of transformations (which I'll describe below), all such cycles can be reduced to the minimum cycle which surrounds only one cell. Because transformations can only reduce the area of the interior by an even amount, it follows the original area of the interior of the inside must have been odd, and therefor cannot be tiled by any tetrominoes.

There are a few "holes" in the argument (maybe somebody can see how they can be removed?):

• The cycle exists, but for the transformations to be possible, the cycle of squares needs to be tileable by skew tetrominoes as well (I mean tileable on its own). This needs to be shown.
• There may be more transformations than the three I list below.
• It needs to be shown the transformations are always possible unless we have the base figure.

Here are the transformations.

If this is the original cycle:

The first transformation is bringing in an edge, that reduces the interior area by 2.

We can apply it again to make the figure look like this:

The second transformation is flipping a corner, that reduces the interior area by 4:

The third transformation is removing a double edge, leaving the interior area the same:

And after a few more transformations we end up at the base figure: