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The six-pack is a thing of the past. Beverages of the future will use the seven-pack format. But how will the mighty spacemen of the future manage to ship the Slurm seven-pack efficiently in rectangular space crates?

Below is a schematic of the seven-pack. How can you completely fill a rectangle using seven-packs of this shape? It's ok to turn a seven-pack upside down, giving you a mirror image of the below shape. The arrangement requiring the smallest rectangular space crate wins.

enter image description here

Whimmy wham wham wozzle, dudes!


Stated simply, can you arrange an unlimited but finite number of the pictured shape and its mirror image so that they perfectly tessellate a rectangle?

Hint:

Remember that since the rectangle contains an integer number of seven-packs then M x N must be a multiple of 7. Furthermore, since 7 is prime at least one of the two dimensions of the rectangle must be a multiple of 7.

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    $\begingroup$ Don't worry, you can leave overlaps in the pattern and the delivery boy will drink the extras. $\endgroup$ – kaine Sep 30 '14 at 19:41
  • $\begingroup$ I assume empty spaces are not allowed? $\endgroup$ – Rob Watts Sep 30 '14 at 20:53
  • $\begingroup$ Both correct. I left the grid lines just to emphasize the dimensions of the shape. No empty spaces, no overlaps, no rearranging the position of the seventh can except, as I say, as a result of turning the whole seven-pack upside down. No funny business. Just a perfect n x m rectangular grid in which each space is occupied by exactly one can. $\endgroup$ – Matt Malone Sep 30 '14 at 21:05
  • $\begingroup$ What about rotation? I assume that's allowed, but you have not explicitly said so. $\endgroup$ – Rob Watts Sep 30 '14 at 21:12
  • $\begingroup$ Yes, rotation is fine. $\endgroup$ – Matt Malone Sep 30 '14 at 21:17
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I have a 14x14 (28 case) rectangle. It is completely symmetrical. Someone beat that.

28 Slurm

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  • $\begingroup$ This is, according to my sources, the smallest possible solution. Well done! $\endgroup$ – Matt Malone Oct 1 '14 at 3:36
  • $\begingroup$ Nice. Obviously at least one dimension needs to be a multiple of seven, and seven itself doesn't work. I'd tried placing tiles for 14 but couldn't get a good pattern so I was going to try to investigate residues from 21 or 28. Conceivably there may be a 21xN solution with less area than the 14x14 one. $\endgroup$ – supercat Oct 1 '14 at 3:43
  • $\begingroup$ The big thing was after placing four cases down on the bottom, i noticed that i could put a fifth that got me 14 and also was 90° off of the first one. Figured I'd try working the symmetry and after getting the edges done, the center just fell into place. After posting, i tried to shorten it to 14x10 or 14x8, but nothing was working. Guess i know why now. $\endgroup$ – Joel Rondeau Oct 1 '14 at 3:53
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    $\begingroup$ Congratulations! What did you draw this with? $\endgroup$ – 355durch113 Oct 1 '14 at 4:20
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    $\begingroup$ @Grantwalzer, My guess is "Excel\ Font section\ Draw Border" (that is alt H B W) $\endgroup$ – Rafe Oct 1 '14 at 12:41
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Here is an alternate solution:

Alternate solution

A computer search revealed 96 solutions for a 14x14 square, and no smaller solution.

The 96 solutions break down into 3 alternate ways to arrange the 4 center pieces, x2 for clockwise/counterclockwise arrangement, and 4 pairs of 2 pieces that can be flipped. That gives $3 * 2 * 2^4 = 96$ solutions.

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  • $\begingroup$ Did you deliberately choose to break the rotational symmetry or was that an accident? $\endgroup$ – Peter Taylor Oct 6 '14 at 9:15
  • $\begingroup$ No. I picked one solution from the 96 solutions I found. I didn't look carefully, I though it was symmetrical. $\endgroup$ – Florian F Oct 6 '14 at 9:17
  • $\begingroup$ 96? Yours differs from Joel Rondeau's in three of the four pairs going in from the middle of the edges (4x4 square with two opposite corners removed: two solutions, which are rotations of each other, so 2^4 = 16 configurations) and in the middle 4 blocks. Are there six configurations for the middle 4 blocks, or do you also have some which differ in the corners? $\endgroup$ – Peter Taylor Oct 6 '14 at 9:21
  • $\begingroup$ Answer updated. $\endgroup$ – Florian F Oct 6 '14 at 9:30

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