A Tetris Puzzle, with solution

This puzzle occurred to me whilst pondering the close packing possibilities of the seven tetrads from the classic puzzle video game, Tetris. I have never seen this puzzle before, so I assume I am the originator. However, since it seems like the type of question one might ask, I am not certain. If you have seen this puzzle or a similar puzzle involving Tetris pieces before, please /msg me and I will include references. Actually, this type of question is rather typical of polyomino explorations, so I would be not-at-all surprised if someone has posed this exact question before. (Update: QuietLight informs me that this problem was presented as extra credit in a 7th grade math class, many a year ago.)

Consider the seven Tetrads from Tetris. They are represented below in all their ASCII majesty:

 _    _        _    _        _    _      _ _
|_|  |_|_    _|_|  |_|      |_|  |_|_   |_|_| 
|_|  |_|_|  |_|_|  |_|_    _|_|  |_|_|  |_|_|
|_|    |_|  |_|    |_|_|  |_|_|  |_|

These seven tetrominoes represent all (connected) possibilities, up to rotation. If we were to allow reflection, then the "S" and "Z" shapes would be equivalent, as would the two "L" shapes. In this problem, to keep with the classic Tetris theme, we will allow rotation but not reflection of the pieces in this puzzle.

Now, these seven blocks have four squares each, for a total of 28 squares. The question is, can we fit all seven blocks (each block being used exactly once) into a 28-square grid of dimensions 7 by 4, so that all squares of the grid are covered by blocks, and so that no blocks overlap? Of course, we shall assume that the tetrads stay rectilinear with the grid's dimensions, and that no dissection of the grid or the blocks, or other such funny business is occurring.

If such a tiling exists, demonstrate it. If such a tiling does not exist, prove that it cannot be done.

Solution below.

Spoilers Below! No Peeking!

It turns out it is impossible to cover the 7x4 grid with the seven tetrads. A classic proof method from polyomino theory helps us here.

Consider our 7x4 grid:

 _ _ _ _ _ _ _

Now imagine if the grid were covered with a checkerboard pattern. Due to our ASCII limitations we will use X's and O's to simulate the checkerboard's usual red and white squares.

 _ _ _ _ _ _ _

Note that there are 14 X's and 14 O's in the "painted" grid above.

Now, when each tetrad is placed on the grid, some of its squares must be over gridboxes marked X, and some must be over boxes marked O. For each tetrad there are only two possibilities, depending on where the block ends up being placed in the grid:

 _    _        _    _        _    _      _ _
|X|  |X|_    _|X|  |X|      |X|  |X|_   |X|O| 
|O|  |O|X|  |X|O|  |O|_    _|O|  |O|X|  |O|X|
|X|    |O|  |O|    |X|O|  |O|X|  |X|


 _    _        _    _        _    _      _ _
|O|  |O|_    _|O|  |O|      |O|  |O|_   |O|X| 
|X|  |X|O|  |O|X|  |X|_    _|X|  |X|O|  |X|O|
|O|    |X|  |X|    |O|X|  |X|O|  |O|

Now note that for six out of the seven tetrads, each one takes exactly two X's and two O's. The exception is the "T" tetrad, which takes either three X's and one O, or three O's and one X.

Imagine that we place the "T" tetrad on the grid first, anywhere we think it will work. No matter where it is placed, we will have used up three of either the X's or the O's; without loss of generality, say the "T" tetrad is placed over three X's and one O. Therefore we have 11 X's and 13 O's remaining.

Since each of the six remaining tetrads MUST use two X's, it is only possible to place five more tetrads on the grid, using 10 of the 11 X's. When it comes time to place the final tetrad we will see that we have only one X remaining to be covered, and this sixth tetrad, like its five counterparts, needs to be placed over TWO X's.

Thus, no matter where the "T" tetrad is placed, its placement will guarantee that the remaining tetrads cannot all be fit into the grid. Thus all seven tetrads cannot be placed to exactly cover the grid. QED.

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