Tiling rectangles with Heptomino plus rectangle #7

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The goal is to tile rectangles as small as possible with the given heptomino, in this case number 7 of the 108 heptominoes (see example below). 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 of the given heptomino will tile.

Example with the $1\times 1$ you can tile a $2\times 5$ as follows:

Now we don't need to consider $1\times 1$ further as we have found the smallest rectangle tilable with copies of the heptomino plus copies of $1\times 1$.

I found only 7 more. I considered component rectangles of width 1 through 11 and length to 31 but my search may not be complete.

List of known sizes:

• Width 1: Lengths 1 to 5
• Width 2: Lengths 2, 3, 5

These could all be tiled by hand, of course the bigger ones will be challenging. I'm making this one a 'hand tiling only' puzzle. In other words, use a computer to do anything except look up or compute the arrangements.

• There seem to be an awful lot of these. Are they just going to continue indefinitely? (Looking at the history, the earliest ones attracted rather more upvotes and favourite-ings than more recent ones have.) There are 108 different heptominoes and I'm not sure we really want 108 "tiling rectangle with heptomino plus rectangles" puzzles... – Gareth McCaughan Jun 24 '18 at 23:53
• @GarethMcCaughan I think there are an awful lot of riddles. But this isn't my personal puzzle emporium so I let it slide. The real question is, are they that unpopular and/or without merit that they should simply not be here? I'm open to persuasion. – theonetruepath Jun 25 '18 at 0:48
• I'm not suggesting that they be deleted, or anything like that. I just don't look forward with particular joy to the next hundred puzzles that are exactly like this one but with a different choice of heptomino. And yes, there are too many bad riddles -- but they're all posted by different people, often newcomers to PSE, rather than one person posting a dozen riddles that all work the same way :-). – Gareth McCaughan Jun 25 '18 at 2:47
• Well I have some sympathy for your feeling here: I don't look forward to the work involved in posting 100 more very similar puzzles. Maybe instead I will 'condense' them in some way to concentrate the more interesting ones in far fewer puzzles. – theonetruepath Jun 25 '18 at 3:47

I found a solution for the $2\times5$. It obviously also works for $1\times5$.

$16\times16$:

Here is a better $1\times5$ solution.

$9\times16$:

• 2x5=16x16 is minimal, there is a smaller 1x5. Just 1x4 and 1x5 left now... You are leading by area found, with just that one... – theonetruepath Jun 24 '18 at 11:00
• @theonetruepath I found a better solution for the $1\times5$. – Jaap Scherphuis Jun 24 '18 at 11:28
• Yep the 1x5=9x16 is minimal. Lucky last, the 1x4. Hint: Its tiling is smaller than that of the 1x5. – theonetruepath Jun 25 '18 at 3:45
• Your total area is 256+144=400. Glorfindel was 325 so you get the nod... – theonetruepath Jun 26 '18 at 9:29

$1 \times 2$:

$3 \times 5 = 15$

$2 \times 2$:

$4 \times 10 = 20$

$1 \times 3$:

$7 \times 10 = 70$

and a variation on that theme which works for $2 \times 3$:

$13 \times 10 = 130$

This is about as hard as I can solve without using even pen and paper to think them out (obviously I'm using Excel to make the pictures). FWIW I'm working on an iPad app to help with tiling by hand, it's a good exercise to polish my Swift. the app works now (more or less), and helped me finding a solution for $1 \times 4$:

$9 \times 10 = 90$

• Yup 1x2=3x5 and 2x2=4x10 are minimal. – theonetruepath Jun 24 '18 at 7:44
• 1x3=7x10, 2x3=10x13 both minimal – theonetruepath Jun 24 '18 at 10:57
• Yup 1x4=7x10 is minimal. All done! – theonetruepath Jun 26 '18 at 9:28