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It is possible to produce the following arrangement in a standard game of Tetris (if you happen to get the right order of pieces and you place them appropriately).

enter image description here

Notice that all of the pieces are red, meaning they all came from the same kind of tetromino. For each of the 7 tetrominoes (I, L, J, S, Z, O, T), show how this exact arrangement can be produced where red corresponds to that tetromino, or show that it is impossible.

Just to be clear about the mechanics:

  • A row is cleared when every cell in that row is occupied.
  • When a row is cleared, everything above shifts down one row. There is no "gravity" or chain reactions.
  • You should be able to place each block directly from above. There should be no rotating or shifting at the last second.

If there are partial answers, I will accept the answer with the greatest number of tetrominoes solved, and earliest if there is a tie.

Good luck!

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It looks like

The Z might be impossible. The only way to construct the bottom line (empty line with the single square on the right) is to build some kind of base, put a standing Z on top, and then vanish every row below the top square of the standing Z.

This works all nice and fine,

except that there's always an odd number of empty squares below the line that the top square of the standing Z is on: the three other squares of the Z necessarily end up in that area. Therefore it's impossible to vanish those rows without adding anything to the target horizontal line (which was supposed to be empty), so you have to put another standing Z higher up, and again you're faced with the exact same problem.

There may, of course, be something obvious I'm missing.

For the J:

This seems to work: http://harddrop.com/fumen/?m115@vhU+NJhwBhuBHoBHnBHmBHlBNkBkkBklBkmBknBDaB?+bBmVBPjBrcBOcBmYBpjBAAA

With the L, there are some problems that are very similar to the Z case,

but choosing the correct square of the L as the remaining one, those can be overcome: http://harddrop.com/fumen/?m115@vhZ6NJniBnhBngBnfBifBBhBGjBHoBHnBHmBHlBNkB?kkBklBkmBknBDbBKaBCWBKcBmiBydBCYBpjBAAA

Here's the T:

http://harddrop.com/fumen/?m115@vhU9NJnnBnmBnlBnkBNkBElBEmBEnBEoBKfBqgBKiB?dgB9hBdjBlVBFXBlYBpjBAAA (EDIT: 16 move solution)

And finally, the S:

http://harddrop.com/fumen/?m115@vhevNJniBnhBngBnfB6jBOkBEgBEhBEiBJaBCoBHnB?HmBHlBmnByiBDhBskB6gBvaBnYBnXBHSBHRBDLBxMBTOBTP?BpjBAAA

And for completeness' sake, the solutions already found by others:

I-piece:

(slightly quicker than @athin's solution): http://harddrop.com/fumen/?m115@vhZhRJhvBDpBhvByxBhpBhrBpoBHoBHnBHmBHlBNkB?kkBklBkmBknBJeBDaBDbBDcBDdBhVBhXBpjBAAA

O-piece:

Like @jafe says, it's impossible to get an odd number of O-piece squares on any row.

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  • $\begingroup$ Your solutions are all correct. Since you've solved the most tetrominoes, I will accept your answer. $\endgroup$ – Riley Aug 13 '18 at 0:12
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Just to make sure, is this a valid answer for I Tetromino?

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  • $\begingroup$ Yes, this is valid. $\endgroup$ – Riley Aug 12 '18 at 14:07
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    $\begingroup$ By the way, I don't think you need to put that in a spoiler. The fact that it's a link already hides the solution. $\endgroup$ – Riley Aug 12 '18 at 14:42
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Just to point out the obvious one:

O is impossible since every row always has an even number of reds.

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  • $\begingroup$ Yes, you got the easiest one :P $\endgroup$ – Riley Aug 12 '18 at 14:10

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