Inspired by Polyomino T hexomino and rectangle packing into rectangle See also series Tiling rectangles with F pentomino plus rectangles and Tiling rectangles with Hexomino plus rectangle #1

Next puzzle in this series: Tiling rectangles with Heptomino plus rectangle #4

...up to 100 to come... I'll post them a few at a time. Why is this first one #3: Numbering is as per my heptomino data file and will skip rectifiable and uninteresting heptominoes. Some of them will be posed as no-computer hand-tiling only puzzles.

The goal is to tile rectangles as small as possible with the given heptomino, in this case number 3 of the 108 heptominoes. 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 $2\times 6$ 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 87 more but lots of them can be found by 'expansion rules'. I considered component rectangles of width 1 through 11 and length to 31 but my search was far from complete.

List of known sizes:

  • Width 1: Lengths 1 to 20, 22 to 25, 29 to 30
  • Width 2: Lengths 2 to 18, 22 to 24, 29 to 31
  • Width 3: Lengths 3 to 8, 14 to 15
  • Width 4: Lengths 4 to 25, 27, 29 to 31
  • Width 5: Lengths 7 to 8
  • Width 7: Length 8
  • Width 8: Lengths 9 to 10

Many of them could be tiled by hand fairly easily.


1 Answer 1


Let's get this party started with $1 \times 2$:

enter image description here

(that wasn't too obvious, see the edit ...)

and $2 \times 2$, which also works for $2 \times 4$:

enter image description here

Like here, there are again some generalizable solutions for $1 \times n$. Which one is smaller depends on $n$:

Left: if $n-1$ is divisible by 7, size: $(n+1) \times (n+5)$.
If $2n-1$ is divisible by 7, size: $(n+1) \times (2n+5)$.
If $3n-1$ is divisible by 7, size: $(n+1) \times (3n+5)$. (and so on:
If $kn-1$ is divisible by 7, size: $(n+1) \times (kn+5)$. )

Right: if $n$ is divisible by 7, size: $(n+1) \times (n+7)$. Otherwise, the size is $(n+1) \times 8n$.

enter image description here

They can probably be extended to $2 \times n$ ($n$ odd) in a similar way as with the hexomino puzzle. For example, here is $2 \times 7$:

enter image description here

  • $\begingroup$ 1x2 is optimal, as is 2x2 (also the 2x4 that gives you), also the 2x7 and the 1x7 and the 1x6. Your notation "n|7k+1, size: n+1x7k+6" is not clear to me, it appears to not follow standard rules of precedence of operators, and has a redundant "1x" which I don't follow. $\endgroup$ Commented Jun 9, 2018 at 19:43
  • $\begingroup$ $n|7k+1$ means $n$ divides $7k+1$. The $\times$ is only used to multiply the x and the y dimension, it never means multiplying numbers. It therefore has a lower precedence than addition. $\endgroup$
    – Glorfindel
    Commented Jun 9, 2018 at 19:45
  • $\begingroup$ I would have understood "n | (7k+1)" and "(n+1) x (7k+6)" a bit more easily. I still have trouble with "n divides (7k+1)" as I still can't get from there to the size of the component rectangle... and I can see the rectangle in the diagram... $\endgroup$ Commented Jun 9, 2018 at 20:00
  • $\begingroup$ Yeah, I've gotten better at that. Let me make an edit ... $\endgroup$
    – Glorfindel
    Commented Jun 9, 2018 at 20:01
  • $\begingroup$ OK I follow now. Eg if I decode that for '1x9' I get '10x32' which works... but there is a better generalisation for that case, ie one that results in a 10x10... and in 10x17 for the 1x16, and so on. These appear to be optimal for larger cases too, as well as for some cases where (3*2n-1) is divisible by 7, eg it works for both 1x30 and 1x15 (both size 10x31) $\endgroup$ Commented Jun 9, 2018 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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