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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.

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3 Answers 3

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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

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  • $\begingroup$ Yup that's optimal $\endgroup$
    – theonetruepath
    Apr 20, 2018 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, 2018 at 10:08
  • $\begingroup$ ... make that three, actually. So I guess this question counts as solved as well. $\endgroup$
    – Glorfindel
    May 17, 2018 at 11:14
  • $\begingroup$ Yup I haven't reached 19x40 yet, all others minimal. $\endgroup$
    – theonetruepath
    May 17, 2018 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, 2018 at 12:04
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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

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  • $\begingroup$ Both nice solutions, the $2\times 3$ is optimal but there is a smaller area for $1\times 4$ $\endgroup$
    – theonetruepath
    Apr 18, 2018 at 18:05
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Here's two to get this one started

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

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enter image description here

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

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enter image description here

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  • $\begingroup$ Both optimal yes $\endgroup$
    – theonetruepath
    Apr 17, 2018 at 3:43

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