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Rectangular Rearrangement Resolution (Posted on 2015-03-20) Difficulty: 3 of 5
A rectangular floor is covered entirely by 2×2 and 1×4 tiles. One tile accidentally gets completely broken.
Each of the available spare tiles are corresponds to the other type.

Can that the floor be completely covered by rearranging the tiles?
Justify your answer.

*Assume that there are an unlimited number of spare tiles.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re: further Comment 5 of 5 |
(In reply to further by Charlie)

No, there is no minimum width and length for a rectangular room such that above that critical value you can assuredly find a rearrangement.


Consider a room with width 4n and an odd length.  It cannot be tiled using entirely 2x2 pieces.  Tile it using the maximum number of 2X2 pieces, with the remaining pieces being 1X4.  Now, if one of the 1x4 tiles break, we cannot find a successful rearrangement using the spare 2x2's, as we are already using the maximum possible number of 2X2's.

Less obviously, consider a rectangle that is 2m by 2n, where m and n are odd.  I think that it cannot be tiled entirely with 1x4 pieces (although I might be wrong on this point).  Tile it using the maximum number of 1x4 pieces, with the remaining pieces being 2x2.  Now, if one of the 2x2 tiles break, we cannot find a successful rearrangement using the spare 1x4's, as we are already using the maximum possible number of 1x4's.

Edited on March 20, 2015, 1:51 pm
  Posted by Steve Herman on 2015-03-20 13:38:32

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