3 ms·

This is actually wrong. Some friends of mine proved that if you drop only S and Z pieces in an irrational ratio, the player is doomed irrespective of the dimens

• temporalparts 6y ago
  This is interesting! Can you link the proof?
  • moomin 6y ago
    I can’t, but it was published in Eureka (Cambridge Maths Society/Archimedean journal) somewhere between 1988 and 1993 I think. I don’t know how to get old copies but they were all typeset in TeX. The same people also proved (earlier) that with 3-ominos its a win for the player. Incidentally, if you’re into cool maths that isn’t part of any major research line, Eureka is a great read.
    • chriskanan 6y ago
      Proof: https://harddrop.com/wiki/A_deadly_piece_sequence https://harddrop.com/wiki/A_deadly_piece_sequence Paper: https://www.semanticscholar.org/paper/How-to-lose-at-tetris-Burgiel/11c12871bfa138fa8bb93a4e5dbcca36c5d214fa?p2df https://www.semanticscholar.org/paper/How-to-lose-at-tetris-...
    • gjm11 6y ago
      The proof that Tetris is a win for the computer was published in Eureka 51 (1992) by Richard Tucker. (Someone else has posted a link to another paper proving this, but it's from 1997; Richard got there first.) The proof that Tris is a win for the player was published in Eureka 51 (1991) by Adam Chalcraft. Richard and Adam definitely know one another and I would be unsurprised to learn that they talked to one another about this stuff, but the actual writeups in Eureka were by different people. (Quite possibly a substantial amount of the thinking was done collectively at meetings of the "Puzzles and Games Ring", a sort of sub-organization of the Archimedeans (the student mathematical society at the University of Cambridge, which publishes Eureka).
      • moomin 5y ago
        waves Might have known you’d have the exact details.
• amptorn 6y ago
  No. It's right. If only S pieces are provided, you can make lines for ever. If S and Z pieces are provided, you can't.