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If you have hitherto conceal'd this sight,
Let it be tenable in your silence still.
And whatsoever else shall hap to-night,
Give it an understanding, but no tongue.
-Shakespeare, Hamlet

This is an arrangement puzzle. There are eight pieces, representing all of the distinct ways that one, two, and three holes can be placed in four spots on a 4x1 rectangle.

enter image description here

You will lay four of these down horizontally next to each other to form a square. You will then lay the remaining four vertically on top of that square. You may rotate or flip any of the pieces before putting them down. The aim of the puzzle is to cover up all of the holes on the bottom layer with the pieces on the top layer. In other words, none of the holes in the top pieces should overlap with the holes in the bottom pieces.

enter image description here

Example: The left picture shows the bottom layer consisting of four pieces placed horizontally. The middle picture shows the remaining four pieces placed vertically. The vertical pieces are then placed on top of the horizontal pieces; the right picture shows the result. In this case, three of the bottom holes are still uncovered. This is not a successful arrangement.

Please present or somehow describe a successful arrangement. There may be many solutions.

Citation: I stumbled across this puzzle online years ago and I unfortunately forget where I saw it. While I can't take credit for the idea of the puzzle, the write-up and art is mine.

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One possible arrangement is:

 ABAA 3-in
 ABBA 2-out
 BABA 2-alt
 AAAB 3-out
 1221
 iaio
 ndnu
  j t
Where A represents holes going in the horizontal direction, and B represents holes going in the vertical direction. To verify that this arrangement works, I've labeled the pieces as follows:
o - - - : 1-out
- o - - : 1-in
o o - - : 2-adj
o - o - : 2-alt
o - - o : 2-out
- o o - : 2-in
o o o - : 3-out
o o - o : 3-in

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A solution has already been given, but for completeness I'll list all solutions.

The puzzle has symmetries, so some solutions can be considered equivalent. To simplify things you can look at the location of the strip with alternating holes (i.e. _O_O ).

  1. We can assume it lies in the layer of horizontal strips, because we can turn over any solution, swapping the layers.
  2. We can also assume that it lies in the top half of the layer, because we can rotate the solution 180 degrees if not.
  3. Lastly we can assume that it has its holes on the right ( _O_O instead of O_O_ ) because inverting a solution (replacing holes by non-holes and vice versa) also produces a solution.
With the above in mind, there are $16$ solutions - $8$ with the alternating strip on the top row, $8$ with it on the second row.

 .O.O   .O.O   .O.O   .O.O   .O.O   .O.O   .O.O   .O.O
 ..O.   O..O   .OO.   OO.O   .O..   O.OO   O...   OOO.
 O...   .OOO   ...O   O...   OOO.   OOO.   O.OO   ..O.
 O..O   .OO.   O..O   .OO.   O..O   .OO.   .OOO   ...O

 OOO.   O..O   .OO.   ...O   .OOO   O...   O.OO   ..O.
 .O.O   .O.O   .O.O   .O.O   .O.O   .O.O   .O.O   .O.O
 O..O   .OO.   O..O   .OO.   O..O   .OO.   .O..   OO.O
 O.OO   .O..   OO.O   O.OO   ..O.   ..O.   O...   OOO.

There is actually another symmetry that relates the solutions with the alternating strip in the top row with those that have the strip in the second row. In any solution you can swap the first two rows and swap the last two rows, and then do the same for the first two and last two columns. This will also always result in a solution. So you could say that there are really only $8$ solutions.

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  • $\begingroup$ Very cool! How did you go about finding all of the solutions? $\endgroup$ – jamisans Aug 24 '18 at 2:12
  • $\begingroup$ @jamisans By computer. I once wrote a generic solver for edge-matching puzzles. It is flexible enough that it can solve this puzzle too, using 8 tiles that each have two orientations, each with with four "edges" that have to differ from the "edges" of the tiles in the other layer. $\endgroup$ – Jaap Scherphuis Aug 24 '18 at 6:17

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