Tuesday, 29 December 2015

geometry - Tiling rectangles with F pentomino plus rectangles


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×b, find the smallest area larger rectangle that copies of a×b plus at least one F-pentomino will tile. Example shown, with the 1×1, you can tile a 3×3 as follows.


F plus 1x1


Now we don't need to consider 1×1 any longer as we have found the smallest rectangle tilable with copies of F plus copies of 1×1.


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



Here are a couple of 1×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 and 15 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:

enter image description here

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 1×10, 1×18 and 1×20 (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×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+1×3n rectangle (left)
Odd n fit in a 2n+1×4n rectangle (right)
enter image description here



1×3:




8×6=48
enter image description here



1×4:



6×6=36
enter image description here



1×5:




11×10=110
enter image description here



1×6:



8×8=64
enter image description here



1×7:




20×9=180
enter image description here



1×8:



22×9=198
enter image description here



1×9:




13×12=156
enter image description here



1×10:



20×11=220
enter image description here



1×11:




13×28=364
enter image description here



1×12:



17×17=289
enter image description here



1×13:




15×25=375
enter image description here



1×14:



16×16=256
enter image description here



1×15:




17×25=425
enter image description here



1×16:



18×18=324
enter image description here



1×18:




19×28=532
enter image description here



1×20:



21×30=630
enter image description here





Some minimal solutions for 2×2k+1; first 2×3:




10×11=110
enter image description here



2×5:



18×20=360
enter image description here



2×9:




30×30=900
enter image description here



2×11 (a nice one, but IMHO equally nice as 2×9):



36×36=1296
enter image description here



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





Another one for 3×4:



14×17=238
enter image description here



and for 3×5:



22×30=660
enter image description here






This night, I found a new one for 2×7 by combining the 2×9/2×11 padding with the inner pattern for the 1×16. This is a hand-made solution so I'm not sure it's minimal:



38x38 = 1444
enter image description here



In general, the generalizable solutions for 1×(10k+4) resp. 1×(10k+6) give rise to solutions for 2×(10k+9) resp. 2×(10k+11); the 2×7 solution mentioned above is basically a subdivision of the 2×21 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)×(3n+3).

enter image description here



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