# How to calculate the number of pieces in the border of a puzzle?

Is there any way to calculate how many border-pieces a puzzle has, without knowing it's width-height ratio? I guess it's not even possible, but I am trying to be sure about it.

BTW you might want to know that the puzzle has 3000 pieces.

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Assuming the puzzle is some kind of Rectangle or a Square, you can find the number of border pieces only if you know the size of the full puzzle as well as the size of the small pieces(assuming all small pieces are of the same size). –  lsp Mar 22 '13 at 11:39
The problem is, you can't even be sure your puzzle has 3000 pieces. It could very well have 3008, for example, but be advertised as 3000. And let's hope you're speaking of a puzzle with regular square-shaped pieces with nothing fancy –  Joubarc Mar 22 '13 at 13:00
@Joubarc The last reference is just brilliant! :-) –  dtldarek Mar 22 '13 at 13:04
The Wikipedia page has more examples, but I found that one the most impressive. Its border pieces are also a nice illustration of ways puzzle makers can mess with the part count. –  Joubarc Mar 22 '13 at 13:09
As @Joubarc: says most puzzles I have seen have somewhat more pieces than the number on the box. But roughly speaking $4 \sqrt {\text{pieces}}$ is a lower limit and the real answer won't be too much greater. –  Ross Millikan Mar 22 '13 at 13:17

Obviously, $w\cdot h=3000$, and there are $2w+h-2+h-2=2w+2h-4$ border peaces. 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.

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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 $N\geq216$.

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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! ;-)

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I really want to vote this up, but can't in good conscience since it's not (meant to be) helpful. :-) The other answers aren't all that helpful either, due to the issues raised in the comment. –  half-integer fan Mar 22 '13 at 14:30