Inspired by Polyomino Z pentomino and rectangle packing into rectangle
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×b, find the smallest area larger rectangle that copies of a×b plus at least one W-pentomino will tile. Example shown, with the 1×1, you can tile a 3×3 as follows.
Now we don't need to consider 1×1 any longer as we have found the smallest rectangle tilable with copies of W plus copies of 1×1.
There are at least eight more solutions.
Answer
First, a generalizable solution for 1×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+1×3n rectangle (left)
Odd n fit in a 2n+1×4n rectangle (right)
Computer research shows that this is optimal for 1×8, 1×9, 1×10, 1×12 and 1×14.
This is a way to tile
a 6x3 rectangle with Ws and 1x2s:
This is optimal for this a×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:
and a rather large one for 1x5:
Continuing down the line we have (AFAIK) minimal tilings for 1x6:
for 1x7 one which looks like a Star Wars fighter:
for 1x8:
for 1x10:
for 2x3:
for 2x5:
for 2x11:
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×n I don't know yet. It does not lead to solutions for 2×7; for example, a rounded rectangle of sizes 46×74 and 32×102 cannot be tiled by W pentominos.
Moving on to the a=3 case, one for 3x4:
The solution for 3×4 is generalizable (but not necessarily minimal) to other 3×n (n not divisible by 3). Here is a 'recipe' for this, a 3×5 solution, which also explains the parameters xi and yi necessary for the construction:
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 (x0=3) and a sharp end at the right (y0=1). It turns out that when the area formed by x1 and y1 is tiled by vertical rectangles, this only works when y0=1 and x0=3 (horizontal, it's the other way around); symmetry dictates it suffices to tackle just the vertical case. In the 3×4 case, x1 is odd and y1 is even; here, it's the other way around. They cannot have the same parity.
The only other rules for xi and yi are divisibility rules for both a(=3) and b(=n). It seems to work just like magic:
a|x1
a|x2
a|x3−1
a|x4−1
a|x5
b|x1+3
b|x2+1
b|x3
b|x4
b|x5+1
a|y1+1
a|y2
a|y3−1
b|y1
b|y2+1
b|y3
The size of the solution is given by (∑3i=0yi+3)×(∑5i=0xi+2). To construct a solution for a given a×b, just choose the smallest (non-zero) xi and yi which satisfy the equations; for x1 and y1 you have to check their parity.
For example, in the 3×7 case the smallest solution for x1 is 18 and for y1 = 14, but they're both even and we have to settle for x1=39 (or y1=35 but that will give a larger solution). This is the result:
The solution for 3×10 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 28×55 solution for 3×5 is not the smallest one. The one below is smaller and is generalizable for 3×n where n is odd but not divisible by 3, but it's usually larger than the corresponding one for the previous family.
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