Tiling rectangles with F pentomino plus rectangles

Question

Inspired by Polyomino Z pentomino and rectangle packing into rectangle

Also in this series: 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 W pentomino plus rectangles

Tiling rectangles with X pentomino plus rectangles

The goal is to tile rectangles as small as possible with the F 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 F-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 F plus copies of $1\times 1$.

There are at least 17 more solutions. More expected. I tagged it 'computer-puzzle' but you can certainly work some of these out by hand. The larger ones might be a bit challenging.

Answer 1

Here are a couple of $1 \times n$ solutions, which I believe to be optimal. They seem to follow some pattern(s), some of which are generalizable (spoiler ahead):

the rectangles are at the edges of the board; often one rectangle per edge, sometimes more, and sometimes (the $n$ = 12, 13, 15 and 17 case) in the middle. The cases $n$ = 4, 6, 14 and 16 are extra 'nice' and can be generalized to $n=10k+4$ and $n=10k+6$ (though the solutions won't necessarily be optimal). Here's a nice visualization of the inductive step:



Note that these solutions also work for all $n$ not divisible by 5; e.g. two 1x7 rectangles fit in a single 1x14 rectangle, so we can just reuse that solution. There is a smaller one, though (see below).

The solutions for $1 \times 10$, $1 \times 18$ and $1 \times 20$ (all below) also seem to form some kind of generalizable family.

As @JaapScherphuis notes in the comment, there's a (probably non-optimal) generalizable solution for $1 \times n$, $n$ is even, rectangles, just like for the W-pentomino. By halving the rectangles, we can also obtain solutions for odd $n$, and the parts with just rectangles and no F-pentominos can be shortened.

Even $n$ fit in a $2n+1 \times 3n$ rectangle (left)
Odd $n$ fit in a $2n+1 \times 4n$ rectangle (right)

$1 \times 3$:

$8 \times 6 = 48$

$1 \times 4$:

$6 \times 6 = 36$

$1 \times 5$:

$11 \times 10 = 110$

$1 \times 6$:

$8 \times 8 = 64$

$1 \times 7$:

$20 \times 9 = 180$

$1 \times 8$:

$22 \times 9 = 198$

$1 \times 9$:

$13 \times 12 = 156$

$1 \times 10$:

$20 \times 11 = 220$

$1 \times 11$:

$13 \times 28 = 364$

$1 \times 12$:

$17 \times 17 = 289$

$1 \times 13$:

$15 \times 25 = 375$

$1 \times 14$:

$16 \times 16 = 256$

$1 \times 15$:

$17 \times 25 = 425$

$1 \times 16$:

$18 \times 18 = 324$

$1 \times 17$:

$19 \times 31 = 589$

$1 \times 18$:

$19 \times 28 = 532$

$1 \times 19$:

$21 \times 31 = 651$

$1 \times 20$:

$21 \times 30 = 630$

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Some minimal solutions for $2 \times 2k+1$; first $2 \times 3$:

$10 \times 11 = 110$

$2 \times 5$:

$18 \times 20 = 360$

$2 \times 7$:

$25 \times 30 = 750$

$2 \times 9$:

$30 \times 30 = 900$

$2 \times 11$ (a nice one, but IMHO equally nice as $2 \times 9$):

$36 \times 36 = 1296$

The last two are part of a generalizable solution, see the bottom of the post.

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Another one for $3 \times 4$:

$14 \times 17 = 238$

and for $3 \times 5$:

$22 \times 30 = 660$

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A while ago, I found a new one for $2 \times 7$ by combining the $2 \times 9$/$2 \times 11$ padding with the inner pattern for the $1 \times 16$. This is a hand-made solution which is not minimal (see above):

38x38 = 1444

In general, the generalizable solutions for $1 \times (10k+4)$ resp. $1 \times (10k+6)$ give rise to solutions for $2 \times (10k+9)$ resp. $2 \times (10k+11)$; the $2 \times 7$ solution mentioned above is basically a subdivision of the $2 \times 21$ one minus some extraoneous padding.

The length of the red line is $10k+4$ or $10k+6$; this makes the purple line $10k+9$ or $10k+11$ and the total solution $(3n+3) \times (3n+3)$.

Answer 2

This should be the minimal solution

Answer 3

I'll get the easy one out of the way: the answer for $1 \times 2$ is

a $3 \times 6$ rectangle. A tiling exists with two F pentominoes and four $1 \times 2$ rectangles. It can be is obtained by taking the $1 \times 1$ solution, reflecting it about the top edge, and joining together the pairs of $1 \times 1$ rectangles into $2 \times 2$ rectangles in an obvious way.

This is minimal because

the smallest dimension of the rectangle must be 3, because the diameter of the F pentomino is 3. Suppose there is only one F pentomino present. In this case, both dimensions must be odd: the pentomino covers an odd number of squares, and every $1 \times 2$ tile covers an even number, so the total number of squares in the grid must be odd. Suppose the smallest dimension is 3. It is obvious that the $3 \times 3$ case won't work. A little trial and error shows that the $3 \times 5$ case won't work either; effectively, we would have to extend the rectangle in the original diagram "upwards" to accomplish this, but a checkerboard coloring argument shows that one domino would have to cover two squares of the same color to accomplish this.

For two F pentominoes, one or both dimensions must be even by the same parity argument as above. This means that the three smallest possible grids are $3 \times 4$, $4 \times 4$, and $3 \times 6$ (which we know to exist). It is not possible to fit two F pentominoes into a $3 \times 4$ grid; and while it is possible to fit them into a $4 \times 4$ grid, the remaining space can't be filled with dominoes. Thus $3 \times 6$ is minimal.