# Tiling rectangles with W pentomino plus rectangles

Also in this series: Tiling rectangles with F pentomino plus rectangles

Tiling rectangles with N pentomino plus rectangles

Tiling rectangles with T pentomino plus rectangles

Tiling rectangles with U pentomino plus rectangles

Tiling rectangles with V pentomino plus rectangles

Tiling rectangles with X pentomino plus rectangles

The goal is to tile rectangles as small as possible with the W pentomino. Of course this is impossible, so we allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one W-pentomino will tile. Example shown, with the $1\times 1$, you can tile a $3\times 3$ as follows.

Now we don't need to consider $1\times 1$ any longer as we have found the smallest rectangle tilable with copies of W plus copies of $1\times 1$.

There are at least eight more solutions.

## 2 Answers

First, a generalizable solution for $$1 \times n$$, $$n$$ is even. By halving the rectangles, we can also obtain solutions for odd $$n$$, and the parts with just rectangles and no W-pentominos can be shortened.

These can be used for F pentominos as well.
Even $$n$$ fit in a $$2n + 1 \times 3n$$ rectangle (left)
Odd $$n$$ fit in a $$2n + 1 \times 4n$$ rectangle (right)

Computer research shows that this is optimal for $$1 \times 8$$, $$1 \times 9$$, $$1 \times 10$$, $$1 \times 12$$ and $$1 \times 14$$.

This is a way to tile

a 6x3 rectangle with Ws and 1x2s:

This is optimal for this $$a \times b$$ because

to fill the gap at the 'bottom' of the W, you need at least two 1x2s (using another W to fill it leads to at least a 5x4 rectangle which is bigger. That fixes the right half of the rectangle; that a 5x3 rectangle is not possible is easy to see with a checkerboard argument, and for a 4x4 we'd need another 7 squares, which is odd, so another W, and the only position to place that W leaves two corner squares empty.

And here is a way to tile

a 4x7 rectangle with Ws and 1x3s:

I've found two more, one for 1x4:

(11x8 = 88)

and a rather large one for 1x5:

(15x12 = 180)

Continuing down the line we have (AFAIK) minimal tilings for 1x6:

(8x18 = 144)

for 1x7 one which looks like a Star Wars fighter:

(10x37 = 370)

for 1x8:

(17x24 = 408)

for 1x10:

(21x30 = 630)

for 2x3:

(7x10 = 70)

for 2x5:

(20x20 = 400)

for 2x11:

(38x38 = 1444)

Notice the 8 by 8 'rounded square' in the center of the 2x11 solution, wrapped by two layers of W pentominos before moving on to the rectangles. This is the same layout as the 2x5 solution, but there the rounded square has dimension 2 by 2, so it vanishes. Whether this solution type is generalizable for other $$2 \times n$$ I don't know yet. It does not lead to solutions for $$2 \times 7$$; for example, a rounded rectangle of sizes $$46 \times 74$$ and $$32 \times 102$$ cannot be tiled by W pentominos.

Moving on to the $$a = 3$$ case, one for 3x4:

(19x28 = 532)

The solution for $$3 \times 4$$ is generalizable (but not necessarily minimal) to other $$3 \times n$$ ($$n$$ not divisible by 3). Here is a 'recipe' for this, a $$3 \times 5$$ solution, which also explains the parameters $$x_i$$ and $$y_i$$ necessary for the construction:

$$28 \times 55 = 1540$$

First, concentrate on the shape formed by the W pentominos alone. Note that it can extend indefinitely from the top left in two directions, and for each end, we have two choices for a 'tail'; a blunt end as seen at the bottom ($$x_0 = 3$$) and a sharp end at the right ($$y_0 = 1$$). It turns out that when the area formed by $$x_1$$ and $$y_1$$ is tiled by vertical rectangles, this only works when $$y_0 = 1$$ and $$x_0 = 3$$ (horizontal, it's the other way around); symmetry dictates it suffices to tackle just the vertical case. In the $$3 \times 4$$ case, $$x_1$$ is odd and $$y_1$$ is even; here, it's the other way around. They cannot have the same parity.

The only other rules for $$x_i$$ and $$y_i$$ are divisibility rules for both $$a (= 3)$$ and $$b (= n)$$. It seems to work just like magic:

$$a |\, x_1$$
$$a |\, x_2$$
$$a |\, x_3 - 1$$
$$a |\, x_4 - 1$$
$$a |\, x_5$$
$$b |\, x_1 + 3$$
$$b |\, x_2 + 1$$
$$b |\, x_3$$
$$b |\, x_4$$
$$b |\, x_5 + 1$$
$$a |\, y_1 + 1$$
$$a |\, y_2$$
$$a |\, y_3 - 1$$
$$b |\, y_1$$
$$b |\, y_2 + 1$$
$$b |\, y_3$$

The size of the solution is given by $$(\sum_{i=0}^3 y_i + 3) \times (\sum_{i=0}^5 x_i + 2)$$. To construct a solution for a given $$a \times b$$, just choose the smallest (non-zero) $$x_i$$ and $$y_i$$ which satisfy the equations; for $$x_1$$ and $$y_1$$ you have to check their parity.

For example, in the $$3 \times 7$$ case the smallest solution for $$x_1$$ is 18 and for $$y_1$$ = 14, but they're both even and we have to settle for $$x_1 = 39$$ (or $$y_1 = 35$$ but that will give a larger solution). This is the result:

$$31 \times 70 = 2170$$

The solution for $$3 \times 10$$ is rather small compared to the rectangle size and my program was able to verify (after a few weeks of calculation) that it's the minimal solution.

Interestingly, the $$28 \times 55$$ solution for $$3 \times 5$$ is not the smallest one. The one below is smaller and is generalizable for $$3 \times n$$ where $$n$$ is odd but not divisible by 3, but it's usually larger than the corresponding one for the previous family.

$$35 \times 38 = 1330$$

• Yup. Two down, six to go. – theonetruepath Apr 15 '18 at 10:28
• 1x2, 1x3, 1x4, 1x5 all minimal. Four more that I've found so far. – theonetruepath May 9 '18 at 10:03
• I've found seven more, including a $1 \times 2k$ generalizable one. Will post them later today. – Glorfindel May 9 '18 at 17:12
• Your 1x6 is smaller than mine which means I have to go find a bug. 1x7 is the same, also 1x8 and 2x5. My program hasn't reached the others yet. – theonetruepath May 10 '18 at 9:16
• Heh. I noticed a strange one in mine, my initial 2x3 solution was 10x10, consisting of the currently displayed 10x7 solution + five 2x3 tiles on top of each other. It turns out I wrongfully discarded some possibilities. Luckily I spotted that before submitting :) – Glorfindel May 10 '18 at 9:21

@Glorfindel My program finally got to your generalised 1x2k for k=6 (1x12) and confirmed it is minimal yes...I guess it's likely that larger k might be minimal too.

• I upvoted, but just note, answering your own question is a shifty way to earn reputation... – Mr Pie Jul 1 '18 at 3:29