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My young daughter and I love doing jigsaw puzzles together. The other day, we had a quite difficult one: 7 by 7 pieces!

We started as usual by splitting the pieces into 2 stacks: the edge pieces on one side, the middle pieces on the other side.

Before putting the edges together, we noticed that there were almost the same number of pieces in each stack:

Our 7x7 puzzle

I wondered then... our puzzle collection is huge! We have all the possible sizes, from 1x2 to 100x100 pieces.

Can we find one (or even more?) in our collection containing exactly the same number of edges pieces and middle pieces?

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  • $\begingroup$ Just to be 100% clear, could you include your definitions of middle and edge pieces? For example, if you have a 2x3 puzzle, would you define all of the pieces to be edge pieces? $\endgroup$
    – NeedAName
    Commented Aug 19, 2015 at 15:29
  • $\begingroup$ Yes, for a 2xN puzzle, all pieces would be edge pieces. $\endgroup$
    – Thomas
    Commented Aug 19, 2015 at 20:24

3 Answers 3

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What you're looking for are numbers $X$ and $Y$ where $2X + 2(Y-2) = (X-2)*(Y-2)$. This is generalized; a few possible (and practical) solutions would be:

  • $5$ by $12$, with $30$ of both edge and middle pieces.
  • $6$ by $8$, with $24$ of both edge and middle pieces.
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    $\begingroup$ There's no possibility with two odd numbers, since the left side of the equation would come out to an even number (always), while the right side would come out to an odd number (always). The closest you can get is within one, which the example fulfills (24 side pieces, 25 middle pieces). $\endgroup$
    – Bailey M
    Commented Aug 19, 2015 at 15:35
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    $\begingroup$ You're right, the next possible solution would be $10008$ by $858654$. But I don't think the OP wants more solutions than your two. $\endgroup$
    – user14478
    Commented Aug 19, 2015 at 15:44
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    $\begingroup$ Assuming you could scan about 10 pieces a second, it could take up to 27 years just to find the four corner pieces! $\endgroup$
    – Gordon K
    Commented Aug 19, 2015 at 15:48
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    $\begingroup$ @LuxxMiner Actually you don't need a program, The equation is equivalent to $(x-4)(y-4)=8$ Then we can factorize 8, and find only two incongruent solutions. $\endgroup$
    – Rohcana
    Commented Aug 19, 2015 at 17:57
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    $\begingroup$ @LuxxMiner I figured out your problem, you are having a good ol' integer overflow issue. $10008 \times 858654 \equiv 3474640 \pmod {2^{31}}$ and $2 \times (10008 + 858654 - 2) = 1737320$ which is half the previous figure. I guess you didn't use python then. :) $\endgroup$
    – Rohcana
    Commented Aug 19, 2015 at 23:25
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For an n x m puzzle, the number of edge pieces, E, is defined by

$2n+2(m-2)$

and the number of middle pieces, M, is then defined by

$nm-E$

If we want to find a puzzle size for which E=M, then we are looking for the case where

$nm-E=E \implies nm=2E \implies nm=4n+4m-8$

I wrote a quick nested for loop to search for solutions and found the pairs

$n=5, m=12$ and $n=6,m=8$
(In either case, swapping $n$ and $m$ is irrelevant i.e. $m=12,n=5$ leads to the same result)

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  • $\begingroup$ you can enclose math functions with $ symbols to format them to look nicer. $\endgroup$
    – Kingrames
    Commented Aug 19, 2015 at 15:43
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    $\begingroup$ And Bailey beat me by a mile. Stupid captcha! How am I supposed to know what pancakes look like, I'm just a robot! $\endgroup$
    – NeedAName
    Commented Aug 19, 2015 at 15:43
  • $\begingroup$ I can't tell if I did it correctly; the machine/browser I'm on right now won't render the formatted equations $\endgroup$
    – NeedAName
    Commented Aug 19, 2015 at 15:45
  • $\begingroup$ They look great. $\endgroup$
    – Kingrames
    Commented Aug 19, 2015 at 15:48
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Our goal is to find such $n$ and $m$ that $nm = 2(n-2)(m-2)$. They can't be equal of course.

Assuming $n<m$, then $\dfrac n{n-2} > \sqrt2$, so $n<8$.

Then again, $2(n-2) > n$, therefore $4<n<8$.

The equality $\frac{2n-4}n = \frac m{m-2}$ can be modified as $\frac{n-4}n = \frac2{m-2}$, so $\frac{2n}{n-4}$ must be an integer equal to $m-2$, making 8 divisible by $n-4$. The latter can be 1 or 2, so either $n=5, m=12$ or $n=6, m=8$.

Disclaimer: I know I could just try seeing if it can be done for all 3 possible values of $n$ after narrowing it down, but didn't feel like it.

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    $\begingroup$ Alternatively, rearrange $nm=2(n-2)(m-2)$ to $(n-4)(m-4)=8$ and compare with the factorizations $(\pm 1)\cdot(\pm 8)$ and $(\pm2)\cdot(\pm 4)$ of $8$, leading to the positive solutions $(5,12)$ and $(6,8)$ $\endgroup$
    – Hagen von Eitzen
    Commented Aug 19, 2015 at 20:28

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