$\begingroup$Is there any way to calculate how many border-pieces a puzzle has, without knowing its width-height ratio? I guess it's not even possible, but I am trying to be sure about it.
Thanks for your help!
BTW you might want to know that the puzzle has 3000 pieces.
$\endgroup$56 Answers
$\begingroup$Obviously, $w\cdot h=3000$, and there are $2w+h-2+h-2=2w+2h-4$ border pieces. Since $3000=2^3\cdot 3\cdot 5^3$, possibilities are \begin{eqnarray}(w,h)&\in&\{(1,3000),(2,1500),(3,1000),(4,750),(5,600),(6,500),\\&&\hphantom{\{}(8,375),(10,300),(12,250),(15,200),(20,150),(24,125)\\ &&\hphantom{\{}(25,120),(30,100),(40,75),(50,60),(h,w)\},\end{eqnarray}
Considering this, your puzzle is probably $50\cdot60$ (I've never seen a puzzle with $h/w$ or $w/h$ ratio more than $1/2$), so there are $216$ border pieces. This is only $\frac{216\cdot100\%}{3000}=7.2\%$ of the puzzle pieces, which fits standards.
$\endgroup$$\begingroup$Admitting it's a regular grid of 3000 pieces, no more, no less, there aren't that many possibilities for the size of the borders, as they must be a divisor of 3000.
However, even if you don't know the ratio, it's usually safe to assume it's between 1:1 and 2:1, as most puzzles are nice rectangles (using a very-non-mathematical notion of nice, by which I mean, most pictures, paintings, sheets of paper, screens, etc...).
This leaves us with the following possibilities:
- 50x60 -> 2x48 + 2x58 = 212 border pieces and 4 corners
- 40x75 -> 2x38 + 2x74 = 224 border pieces and 4 corners
- 30x100 is already far from a 2:1 ratio but would yield 2x28 + 2x98 = 252 border pieces and 4 corners.
Assume that the puzzle is a rectangle of $a\times b\ $ cm$^2$, and the pieces can be idealized as rectangles of $c\times d\ $ cm$^2$. Then $m={a\over c}$ pieces border along an $a$-side and $n={b\over d}$ pieces along a $b$-side. You have told us that $m\cdot n=3000$, and you want to know the number $N:=2m+2n-4$. We don't have enough information to determine $N$. By the AGM-inequality one has ${m+n\over 2}\geq \sqrt{mn}$. This implies the estimate $N\geq216$.
$\endgroup$$\begingroup$Think of this like a perfect square. Example say the puzzle is 100 pieces, that's 10X10 grid, so that puzzle would have 40 boarder pieces approximately. There is no way to figure exactly the width and height. So going by that using a 300X300 grid equaling 3000 pieces, you best answer is approximately 1200 for all four sides of the boarder. If you want a realistic answer, just Google some 3000 piece puzzles and find many similar ones then get your best answer.
$\endgroup$$\begingroup$You could apply the Monte Carlo method. Pick a random piece (that is, with an uniform distribution) and note if it is a border piece or not. Repeat until the ratio $$k = \frac{\# \text{noticed border pieces}}{\#\text{total}}$$ converges (up to the desired precision). Then $3000 \cdot k$ will be your answer (or generally $nk$ for $n$-piece puzzle).
Good luck! ;-)
$\endgroup$1$\begingroup$500 piece has 88 border pieces. 750 piece has 108 border pieces. 1000 piece has 128 border pieces. These numbers hold true for standard rectangle puzzles.
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