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In Tetris, Mr. T loves performing Perfect Clears, when a piece clears all lines in the playing field. As you might guess, when it comes to tetriminos, he has a personal favorite: the undeniably cutest O piece. But a Perfect Clear using only O pieces is not terribly exciting, so Mr. T wants to use his second favorite tetrimino: the T piece.


Starting with an empty field, can Mr. T achieve a perfect clear using only T pieces? If so, how fast can he pull it off? Mr. T is not easily persuaded by smooth talk, so please justify your answer with a convincing argument.


The standard Tetris field has ten columns. The T tetrimino has three minos adjacent to a central mino in a T shape, as shown below.

OOO
 O
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  • $\begingroup$ Why is he called Mr. T when O is his favourite piece? $\endgroup$ – Glorfindel May 1 at 13:34
  • $\begingroup$ Maybe T for Tetris? :) $\endgroup$ – Omega Krypton May 1 at 13:35
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    $\begingroup$ @Glorfindel Just a bit of humor and a small reference to a character from a recent Tetris game. :) $\endgroup$ – noedne May 1 at 13:38
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    $\begingroup$ n̶o̶o̶,̶ ̶s̶c̶h̶e̶z̶o̶ ̶i̶s̶ ̶t̶h̶e̶ ̶c̶u̶t̶e̶s̶t̶ ̶b̶o̶y̶ $\endgroup$ – athin May 1 at 14:18
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    $\begingroup$ "I pity the fool who messes with the T Man" - Mr. T $\endgroup$ – CpILL May 1 at 23:49
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It can be done with

10 T tetrominoes (if you allow rotation – this is necessary because otherwise it's not possible to get as much 'squares' in the first row as in the second, which is a prerequisite for clearing the field).

Some explanation:

enter image description here

First, drop the three yellow ones and the orange one on the left (first picture); this clears the bottom line.
Then, drop the three light blue ones and the dark blue one on the left (second picture); this clears the second line.
The final two tetrominos (green) can be dropped (third picture) to clear the final two lines.

Since

the number of columns (10) is not divisible by 4, the total number of rows must be even. I don't see a possibility to do it in two rows, you'll need 'vertically' placed tetrominos and those always need three rows.

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    $\begingroup$ That's a great answer. I wasn't anywhere close to that. $\endgroup$ – Jaap Scherphuis May 1 at 13:30
  • $\begingroup$ "you'll need 'vertically' placed tetrominos" -> theorem. I managed to prove it. $\endgroup$ – Joshua May 1 at 15:37
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    $\begingroup$ The first spoiler block contains some reasoning why they are needed. $\endgroup$ – Glorfindel May 1 at 15:39
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A simple solution involves

10 T Tetrominos.

Stack the first five arranged the same direction (on the side of the T with the bottom facing either left or right. In my example, the bottom is facing the right). This eliminates the second row (shown with Xs through them). Next, flip the T 180 degrees so the bottom is facing the other direction and fill the gaps as such. This should immediately eliminate rows 3 and 4. The squares in row 5 fall down to row 1 and the sum of those squares eliminates the final row.

enter image description here

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    $\begingroup$ Welcome to Puzzling! I have placed parts of your answer in spoiler tags to avoid spoiling other solvers. Please take the tour! $\endgroup$ – noedne May 1 at 18:19
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    $\begingroup$ Regarding your solution, I should have specified that naive gravity was in play, as in most games. This certainly works under cascade gravity that some game modes have. $\endgroup$ – noedne May 1 at 18:21

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