3
$\begingroup$

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 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 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:

U plus 1x1

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.

$\endgroup$
4
$\begingroup$

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:

enter image description here

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

link to two 11x8 = 88 solutions

For 1x5:

12x20 = 240

enter image description here

for 1x6:

14x24 = 336

enter image description here

and for 3x4:

19x40 = 760

enter image description here

$\endgroup$
  • $\begingroup$ Yup that's optimal $\endgroup$ – theonetruepath Apr 20 '18 at 6:21
  • $\begingroup$ @theonetruepath I̶ (my program) found two more, I believe we've reached the "6 more solutions" now. $\endgroup$ – Glorfindel May 17 '18 at 10:08
  • $\begingroup$ ... make that three, actually. So I guess this question counts as solved as well. $\endgroup$ – Glorfindel May 17 '18 at 11:14
  • $\begingroup$ Yup I haven't reached 19x40 yet, all others minimal. $\endgroup$ – theonetruepath May 17 '18 at 12:00
  • $\begingroup$ 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. $\endgroup$ – Glorfindel May 17 '18 at 12:04
3
$\begingroup$

Beside Riley's 2 solutions:

$ 1 \times 4 $

Area: $10 \times 9 = 90$
enter image description here

$ 2 \times 3 $

Area: $10 \times 7 = 70$
enter image description here

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

Here's two to get this one started

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

.

enter image description here

.

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

.

enter image description here

$\endgroup$
  • $\begingroup$ Both optimal yes $\endgroup$ – theonetruepath Apr 17 '18 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.