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Here's a puzzle I've been thinking about for a while, but still don't have an answer to.

Lets say a friend is playing Tetris, but instead of the computer picking pieces at random you get to choose which pieces are added. (Unlike the computer you do not have to pick from a bag, you may choose any piece any time) Now since you are a good friend you obviously want to use this power to cause your friend to lose. And there are cases where you can force a loss on your opponent. However the my question is can you choose a static order of pieces that will force your opponent to lose no matter what moves they make?

For example if the board were only 3 squares tall you could force your opponent to lose by only choosing one variety of S or Z pieces and feeding your friend only that piece. However traditionally Tetris is played on a 10x20 board in which this strategy does not work.

So can you find a sequence of pieces that forces a loss? I would prefer to see shorter than longer solutions.

You may choose to ignore or allow complex moves like T spins. And I would love to see solutions that do both.

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It is

possible,

and the following strategy does it:

your (soon to be ex-) friend simply alternates between "S" and "Z" pieces.

See

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.55.8562&rep=rep1&type=pdf

for formal proof.

There is

an earlier publication addressing the same question, which is not available online; this is Richard Tucker's article "Tetris" in Eureka, the approximately annual journal of the Archimedeans, which is a student mathematical society at the University of Cambridge. It's in Eureka number 51 from 1992, it's two pages long, and the winning strategy Tucker proposes for the player's adversary is different from the one in the paper above: pick a positive irrational number and drop S and Z pieces in that ratio. (Full disclosure #1: Eureka is not a peer-reviewed scholarly journal. Full disclosure #2: I have not checked all the details of the argument in the article. Full disclosure #3: Richard Tucker is a friend of mine.)

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Here is a link to a Mathematics and Computer Science journal published by Kaitlyn M. Tsudura from Saint Mary's University in Halifax, Canada. It summarizes the results of studies published by John Bruztowski and Heidi Burgiel

http://euclid.trentu.ca/aejm/V4N1/Tsuruda.V4N1.pdf

The studies were basically to determine whether it was possible to make a Tetris game only winnable but they found that it is possible to make games un-winnable:

found some cases to show it is certain that after you play 69,600 S and Z tetrominoes, you will lose. It is also shown that all Tetris games will end because any game with arbitrary initial conditions, and a string of 127,000 alternating S and Z tetrominoes will force the game to end.

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All of this S/Z only talk is confusing to me. It seems like any number of Z/S pieces is trivial to solve, given an even width (which you have). Because, you just have to orient the pieces so that they are 3 high, two across. After you fill an entire row, the middle part of the tetromino dissappears, and the top row falls onto/into the bottom row. Becasuse the S/Z pieces are lengthwise, it seems that the top row and the bottom row complement each other perfectly, and all the Z/S pieces disappear.

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    $\begingroup$ If one row disappears, the rows above will fall down exactly one step regardless of the contents of the rows below. Blocks can remain hanging in mid-air without support from below. When a row of vertical S/Z pieces is completed and the middle removed, you will always have two rows of rubble left and they will not interweave. $\endgroup$ – Jaap Scherphuis Aug 4 '17 at 3:11

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