3 ms·

Just realized that in a 2d puzzle you have only 4 corner pieces, but in a higher-dimensional (say, 20d) puzzle, almost all pieces would be corner pieces.

pfortuny 6y ago
Why? Are you not missing the “interior”?
- tgb 6y ago
  I agree, don't think it works out. Consider a 3x3...x3 puzzle (with N dimensions). There are 2^N corners but 3^N pieces so the corner-to-piece ratio decreases exponentially in N. However, what does "corner" mean in an N dimensional puzzle? I was taking it to mean that it has N sides without pegs or holes. But if you took it to mean at least 2 sides without pegs or wholes, then indeed the ratio would increase (I think there are 3^N - 2^N - 1 of those) with N. But I think this is the less-satisfactory way to generalize "corner" pieces and these really should be thought of as edge pieces.
  - bo1024 6y ago
    Ah, this is a nice point, it does work for the number of edge pieces. In this case (3x3x...x3) every piece but one is an edge piece! More generally, with k pieces on a side, there are k^N pieces total, and (k-2)^N internal pieces, so in the limit, the ratio of internal to total is (1-2/k)^N -> e^{-2N/k}. So if the number of dimensions gets to be bigger the grid size, almost all pieces are "edge" (i.e. outer) pieces. Even if N=k/2, something like 70% of the pieces are edge pieces.
mfenniak 6y ago
In a 2d, rectangular puzzle, with boring die-cut pieces, you have only 4 corner pieces. :-) Puzzles at higher difficulties are not rectangular, don't have straight sides, or have straight & corner-looking pieces in the middle. My favorite difficult puzzles are Stumpcraft -- this small business makes wooden, beautiful smelling puzzles with a laser-cutting technique. They're very difficult with intricate and intentionally misleading pieces and geometries. https://www.stumpcraft.com/ https://www.stumpcraft.com/
bo1024 6y ago
Ha. That sounded close to me but it's not right, I think the reason is that the corners are very small. Suppose a perfect grid in a hypercube. In d dimensions we have 2^d corner pieces. But if there are k pieces per side, then k^d total pieces (for example 10^d). If we consider pieces adjacent to a corner, including a diagonal, then these are little cubes of size 2^d sitting at the corners, so a total of (2^d)(2^d) = 4^d pieces close to a corner. In general taking little cubes out of the corners isn't enough to get most of the pieces. If we think about the diagonals of length k sqrt(d), we've only cut those a little shorter.

pfortuny 6y ago

Why? Are you not missing the “interior”?

tgb 6y ago
I agree, don't think it works out. Consider a 3x3...x3 puzzle (with N dimensions). There are 2^N corners but 3^N pieces so the corner-to-piece ratio decreases exponentially in N. However, what does "corner" mean in an N dimensional puzzle? I was taking it to mean that it has N sides without pegs or holes. But if you took it to mean at least 2 sides without pegs or wholes, then indeed the ratio would increase (I think there are 3^N - 2^N - 1 of those) with N. But I think this is the less-satisfactory way to generalize "corner" pieces and these really should be thought of as edge pieces.
- bo1024 6y ago
  Ah, this is a nice point, it does work for the number of edge pieces. In this case (3x3x...x3) every piece but one is an edge piece! More generally, with k pieces on a side, there are k^N pieces total, and (k-2)^N internal pieces, so in the limit, the ratio of internal to total is (1-2/k)^N -> e^{-2N/k}. So if the number of dimensions gets to be bigger the grid size, almost all pieces are "edge" (i.e. outer) pieces. Even if N=k/2, something like 70% of the pieces are edge pieces.

mfenniak 6y ago

In a 2d, rectangular puzzle, with boring die-cut pieces, you have only 4 corner pieces. :-) Puzzles at higher difficulties are not rectangular, don't have straight sides, or have straight & corner-looking pieces in the middle. My favorite difficult puzzles are Stumpcraft -- this small business makes wooden, beautiful smelling puzzles with a laser-cutting technique. They're very difficult with intricate and intentionally misleading pieces and geometries. https://www.stumpcraft.com/ https://www.stumpcraft.com/

bo1024 6y ago

Ha. That sounded close to me but it's not right, I think the reason is that the corners are very small. Suppose a perfect grid in a hypercube. In d dimensions we have 2^d corner pieces. But if there are k pieces per side, then k^d total pieces (for example 10^d). If we consider pieces adjacent to a corner, including a diagonal, then these are little cubes of size 2^d sitting at the corners, so a total of (2^d)(2^d) = 4^d pieces close to a corner. In general taking little cubes out of the corners isn't enough to get most of the pieces. If we think about the diagonals of length k sqrt(d), we've only cut those a little shorter.