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This actually isn't true for modern variants of Tetris. A correct implementation of the random number generator requires that the 7 pieces be shuffled into a ba
by zeta0134 9y ago
This actually isn't true for modern variants of Tetris. A correct implementation of the random number generator requires that the 7 pieces be shuffled into a bag and then drawn randomly, ensuring an even distribution, and disallowing long runs without a desired piece. (Maximum of 12.) The bag is re-shuffled every 7 blocks in the sequence.
https://tetris.wiki/Tetris_Guideline https://tetris.wiki/Tetris_Guideline
https://tetris.wiki/Random_Generator https://tetris.wiki/Random_Generator
Thus, in a game of Tetris adhering to these official guidelines, the longest sequential run of S and Z pieces would be 4, in the following sequence:
[5 Random Blocks] [Z and S in any order]
[Z and S in any order] [5 Random Blocks]
I'd be very curious to see a proof of an impossible sequence which adheres to the official rules, as that would be quite a feat indeed!
- anvandare 9y agoIn that case it would probably still be possible, but the proof would have to be found by example. It seems like an advanced version of the Eight Queens puzzle: which sequence of bags guarantees a losing board state?
- ethan_g 9y ago>I'd be very curious to see a proof of an impossible sequence which adheres to the official rules, as that would be quite a feat indeed! It sounds like, given enough piece previews, you can play without ever losing: http://tetris.wikia.com/wiki/Playing_forever http://tetris.wikia.com/wiki/Playing_forever
- vacri 9y ago> [5 Random Blocks] [Z and S in any order] [Z and S in any order] [5 Random Blocks] Any dab hands at regex out there? I think this is the regex to match, but something feels wrong about it (([^sz]){5}[sz]{2})([sz]{2}([^sz]){5})