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.
Let's take a 25 x 25 puzzle of a color gradient(also called color ramp and color progression). No two connections the same, even it is off by a millimeter. All the pieces are in a pile, with no way to identify the pieces. The fastest would be the first solution, assembling the border first.
Number 1, which is the fastest, would require you to sift through ...
Let's start by just attempting to complete the puzzle, and see how far I get:
WARNING! I didn't know how to effectively spoiler my whole answer, so it's not covered. read at your own risk!
Let's call The possible formations of the sides of a piece 0 for a straight edge, 1 for a single "out", 2 for a single "in", and 3 for the doubled in/out. This naming is ...
To a real-life problem I had to give a real-life answer:
But you asked for an actual tiling, without gaps, so here it is.
PS: there is a simpler pattern where pairs disassemble with a single translation:
The solved dominosa looks like this
At the start there is only one way to place an 88 and 38, this leaves only one way to place the 33 and just two possibilities for the 44. After that we try to place the 47 domino. There is only one 7 with three 4s for this domino. If it would make a 2x2 square with the 44 domino, we get a contradiction.
This means the 47 ...
Efficiency means that you handle each tile as little as possible, and that you make as few mistakes as possible. Ideally you want to pick up a tile and fit it to at least one other rather than put it down again unmatched. Incorrect fitting wastes time when correct matches are rejected and when you have to unpick the mistake.
From experience, the most ...
I'm not entirely sure that my interpretation of how the backwards timer works is entirely correct, but if it works like this:
Then I think I can get
Here's the diagram (pardon the mess):
Here's how we'd set the timers, and what would happen:
I have solved Ravensburger's Neuschwanstein castle (5000 pcs) some ten years ago.
It has an impressive size of 153 x 101 cm (see details on 5000 pcs puzzles).
Once the puzzle was solved, I used this Puzzle Conserver from Ravensburger:
I think nowadays a newer item replaced this, called Puzzle Glue & Go.
Note that this "glue" should be used on the ...
Don't use glue at all, just tightly "trap" the puzzle between a backing board (behind) an acrylic or glass sheet (front), and plastic/wood around the edges to stop the puzzle pieces falling out, secured to both the back and front.
I've never seen a professional one, but I've seen a few made by a moderately skilled DIY'er with a few lengths of Balsa wood and ...
Very partial solution: this was as far as I could get during my lunch break and I think I went wrong somewhere.
UPDATE 1: Nearly forgot to keep working on this...
UPDATE 2: Seems like @Wesley Situ already solved the whole thing, but I'm gonna see if I can finish this myself in my own time, just for fun.
Solved at last! I believe that the final answer is:
The completed jigsaw looks like this:
How can we find these words?
With the hints in mind we now see that in the three central 'pixels' of each tile edge that jigsaws into another tile we have:
So what does this mean? Well, realise that:
This satisfies the title since: