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

Question

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.

Answer 1

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! ;-)

Answer 2

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.

Answer 3

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.

Answer 4

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 $N\geq216$.