Tiling rectangles with U pentomino plus rectangles

Question

The goal is to tile rectangles as small as possible with the U 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 U-pentomino will tile. Example shown, with the $1\times 1$, you can tile a $2\times 3$ 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 U plus copies of $1\times 1$.

There are at least 6 more solutions. I tagged it 'computer-puzzle' but you can certainly work some of these out by hand. The larger ones might be a bit challenging.

Glorfindel · Accepted Answer · 2018-05-17T11:13:36.357

6

Here is a way to tile a

6x13 = 78

rectangle with U pentominoes and 1x4 rectangles, which is an improvement over @athin's 9x10 solution:

As a bonus, here are two suboptimal solutions, one of which is asymmetric:

link to two 11x8 = 88 solutions

For 1x5:

12x20 = 240

for 1x6:

14x24 = 336

and for 3x4:

19x40 = 760

edited May 17 '18 at 11:13

answered Apr 19 '18 at 21:20

Glorfindel

28,033
9
96
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Yup that's optimal – theonetruepath Apr 20 '18 at 06:21
@theonetruepath I̶ (my program) found two more, I believe we've reached the "6 more solutions" now. – Glorfindel May 17 '18 at 10:08
... make that three, actually. So I guess this question counts as solved as well. – Glorfindel May 17 '18 at 11:14
Yup I haven't reached 19x40 yet, all others minimal. – theonetruepath May 17 '18 at 12:00
FYI, my program takes a fixed rectangle size (e.g. 3x4) and then enumerates all 'boards' in increasing order. Larger rectangle sizes are evaluated much faster, and for this particular pentomino + rectangles with width > 1 there's another trick: you know that the U pentominos always come in pairs. – Glorfindel May 17 '18 at 12:04
Yeah I took a shortcut to do them all simultaneously. I use a 'shape file' with all the pentominoes and all rectangles from 1x1 through 10x10, then I solve for each outer rectangle in size order, using one pentomino and one small rectangle at a time, with no optimisations. I have eight copies of the program doing all eight pentominoes plus all rectangles to 10x10 running simultaneously, plus another eight doing the remaining rectangles up to 15x15, requiring no input from me. They are all set to go to area 6000 or (more likely) until I stop them. All running on a 10 core (20 thread) CPU. – theonetruepath May 17 '18 at 13:43

score 3 · Answer 2 · answered Apr 18 '18 at 17:59

3

Beside Riley's 2 solutions:

$ 1 \times 4 $

Area: $10 \times 9 = 90$

$ 2 \times 3 $

Area: $10 \times 7 = 70$

answered Apr 18 '18 at 17:59

athin

34,177
4
70
221

Both nice solutions, the $2\times 3$ is optimal but there is a smaller area for $1\times 4$ – theonetruepath Apr 18 '18 at 18:05

score 2 · Answer 3 · answered Apr 17 '18 at 03:40

2

Here's two to get this one started

$3\times 4$ using a $1\times 2$

.

$7\times 4$ using a $1\times 3$

.

answered Apr 17 '18 at 03:40

Riley

14,424
4
39
77

Both optimal yes – theonetruepath Apr 17 '18 at 03:43

Tiling rectangles with U pentomino plus rectangles

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