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A rectangular room has a floor tiled with tiles of two shapes: 1×4 and 2×2. The tiles completely cover the floor of the room, and no tile has been damaged, or cut in half. One day, a heavy object is dropped on the floor and one of the tiles is cracked. The handyman removes the damaged tile and goes to the storage to get a replacement. But he finds that there is only one spare tile, and it is of the other shape. Can he rearrange the remaining tiles in the room in such a way that the spare tile can be used to fill the hole?

Source of the puzzle Math.SE

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    $\begingroup$ The way the question is written, it seems like if there exists any size and shape room where the tiles can't be rearranged, then the answer should be No. It's not hard to come up with such a room. $\endgroup$ – Todd Wilcox Jan 11 '18 at 21:10
  • $\begingroup$ @Todd I would argue that since this is a more mathematical question, you would have to differentiate between the cases in which it is possible and impossible. $\endgroup$ – boboquack Jan 11 '18 at 23:39
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The answer is:

No

Suppose we colour the floor of the room under the tiles like so:

Dots at the intersections of every second row and column

extending up to the edge of the grid.

Then:

Each 2x2 tile covers 1 black dot (an odd number), while each 1x4 tile covers either 0 or 2 dots (an even number)

So:

Changing one tile for the other would change the parity of the number of dots it is possible to cover, however the dots don't change, so this is impossible.

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  • $\begingroup$ Awesomely elegant proof. $\endgroup$ – Phylyp Jan 11 '18 at 4:24
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    $\begingroup$ Why do I get the feeling OP has actually just broken their kitchen floor and was hoping somebody could find them a cheap way to fix it... $\endgroup$ – Bilkokuya Jan 11 '18 at 13:19
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    $\begingroup$ Important note: The dots are not part of the tiles, so they don't move with the tiles as you try to rearrange. They are part of the floor below (this avoids some messy argumentation and confusion). $\endgroup$ – Arthur Jan 12 '18 at 10:02
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One room is very small. On the top are two tiles 2x2 each side by side and below them is one 1x4 tile, so the room is rectangular 3 vertical x 4 horizontal. Like this:

  1   2   3   4
+---+---+---+---+
|       |       | 1
+       +       +
|       |       | 2
+---+---+---+---+     
|               | 3
+---+---+---+---+

Break any one of the three tiles and you can see by inspection that there is no rearrangement with the new tile of the other shape that can make a rectangle.

This is one counterexample so the answer to the question, in general, is no.

Of course it does not address the infinite number of other possibilities of other tile rooms. But all you need is one counterexample.

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    $\begingroup$ So why isn't the answer "perhaps, depending on the room and the quantities of tiles"? $\endgroup$ – boboquack Jan 12 '18 at 23:31
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The answer is:

No.

Reason:

Tiles are glued in place. To rearrange them would require destroying all the tiles and the handyman will be left with only a single tile.

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    $\begingroup$ +1 for the idea but what if the handyman used a heat gun on the tiles to melt the glue and very carefully lift up each tile? The answer would still be no but the reason would not be valid :) $\endgroup$ – ElPedro Jan 11 '18 at 19:48
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Yes (with a cheat):

There is nothing in the puzzle saying the spare tile is not to be cut, or even that the tiles to be rearranged are not to be cut (only that they had not been cut before).

or even:

no tile has been cut in half. It could have been cut in, say, quarters.

or:

He can rearrange the tiles so that the floor is no longer rectangular, then the parity argument does not hold. The easy way is to let parts of tiles extend to the jar below the door.

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  • $\begingroup$ "No tile has been damaged" "It could have been cut in, say, quarters." - I think that might just damage the tile. $\endgroup$ – theonetruepath Jan 11 '18 at 10:49
  • $\begingroup$ @theonetruepath "no tile has been damaged, or cut in half" - apparently, the author considers deliberate, careful premeditated cutting different from damaging. But I admitted my answer is cheating :-) $\endgroup$ – Radovan Garabík Jan 11 '18 at 11:23
  • $\begingroup$ Cutting tiles is not nearly as big a cheat as changing the size of the room! (Or if you're not changing the size of the room, explain how you can let some tiles extend under the door, while keeping the whole floor covered otherwise.) $\endgroup$ – Mr Lister Jan 11 '18 at 12:37
  • $\begingroup$ @MrLister While I could try to muse upon various door frame geometries or reusing part of the old tile, and came up with something that could be justifiable, I'll just admit that I have not considered this and you have a good point :-) $\endgroup$ – Radovan Garabík Jan 11 '18 at 14:32
  • $\begingroup$ Move the walls, I like it! And yes, I think you're right, it didn't explicitly deny cutting tiles when rearranging... $\endgroup$ – ilkkachu Jan 12 '18 at 9:30
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The answer is no, and not only for rectangular shapes, but any shape of room, with the assumption that there exists a tiling.And the proof is as follows:

Embedding the room in a big chessboard, you can number the rows and columns. You first consider rows and subtract the total number of squares in rows with odd index from those with even index. Now, for the given tiling you observe that neither 2x2 nor 1x4 put vertically contribute to this number. So only 1x4 put horizontally contribute, either 4 or -4. From this it is immediate that the number of horizontal 1x4 is well defined mod2. You do the same for columns and you get the number of vertical 1x4 mod2. Adding, the total number of 1x4 tiles is well defined mod2, and no such rearrangement exists

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  • $\begingroup$ Nice proof, and welcome to Puzzling! Why don’t you take a look at the tour whole you’re here? One thing you might notice if you look around is that some answers are hidden in spoiler blocks, which are made by prefixing critical lines with >! - if you have time, adding these to your answer might help in case another visitor comes and doesn’t want the answer given away immediately. Good luck, and happy puzzling! $\endgroup$ – boboquack Nov 1 at 6:57
  • $\begingroup$ thanks, i didn't know how to do this! $\endgroup$ – saret Nov 1 at 10:07

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