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Important question, can this game be solved, like ordinary tic-tac-toe. Is there a solution that always leads to draw of defeat.

by deletes 13y ago

lucb1e 13y ago
Regardless of whether you can build a perfect AI on commodity hardware, though I haven't played this yet, it promises to be a much more interesting game than the normal tic-tac-toe! Harder to draw a lot on napkins though, but then again, a single game takes much longer.
- chii 13y ago
  use 9 napkins - problem solved ;)
  - ronaldx 13y ago
    if you use napkins then you can also move the boards around ;)
k2enemy 13y ago
Zermelo's theorem[1] says "yes," although I don't know which it is (between draw, p1 win, p1 lose). [1] https://en.wikipedia.org/wiki/Zermelo%27s_theorem_%28game_theory%29 https://en.wikipedia.org/wiki/Zermelo%27s_theorem_%28game_th...
- CJefferson 13y ago
  In fact, it can't be a p1 lose, because the extra piece p1 gets can never be a disadvantage (you can prove this more carefully). The question is then is it a p1 win, or draw? A similar situation arises in the game 'hex', except in that game there is no draw, so it has to be a p1 win!
  - Someone 13y ago
    "because the extra piece p1 gets can never be a disadvantage (you can prove this more carefully)" Please do so, as I do not see that this is trivially true. I see two tricky cases: - your opponent must play on the board where p1 is placed, and that board would have room iff p1 weren't present => Addition of p1 gives him a 'move anywhere' move. - you must play on the board where p1 is placed, and that board would have room iff p1 weren't present. The normal argument 'move anywhere and assume that that 'anywhere' is where your first move went, and you just played p1' does not help here, as changing the first move may change where your opponent's first move could have gone. Except from an exhaustive search, I do not see how to prove that you can prevent either case.
    - CJefferson 13y ago
      Yes, you are right. Actually, the fact you force your opponent isn't the problem (I don't think) but the fact that the presence of that piece might later give p2 a free move, where previously they wouldn't means it doesn't work. Sorry, that's what I get for thinking I didn't need to figure out all the details carefully!

Important question, can this game be solved, like ordinary tic-tac-toe. Is there a solution that always leads to draw of defeat.

by deletes 13y ago

Important question, can this game be solved, like ordinary tic-tac-toe. Is there a solution that always leads to draw of defeat.

lucb1e 13y ago

Regardless of whether you can build a perfect AI on commodity hardware, though I haven't played this yet, it promises to be a much more interesting game than the normal tic-tac-toe! Harder to draw a lot on napkins though, but then again, a single game takes much longer.

chii 13y ago
use 9 napkins - problem solved ;)
- ronaldx 13y ago
  if you use napkins then you can also move the boards around ;)

k2enemy 13y ago

Zermelo's theorem[1] says "yes," although I don't know which it is (between draw, p1 win, p1 lose). [1] https://en.wikipedia.org/wiki/Zermelo%27s_theorem_%28game_theory%29 https://en.wikipedia.org/wiki/Zermelo%27s_theorem_%28game_th...

CJefferson 13y ago
In fact, it can't be a p1 lose, because the extra piece p1 gets can never be a disadvantage (you can prove this more carefully). The question is then is it a p1 win, or draw? A similar situation arises in the game 'hex', except in that game there is no draw, so it has to be a p1 win!
- Someone 13y ago
  "because the extra piece p1 gets can never be a disadvantage (you can prove this more carefully)" Please do so, as I do not see that this is trivially true. I see two tricky cases: - your opponent must play on the board where p1 is placed, and that board would have room iff p1 weren't present => Addition of p1 gives him a 'move anywhere' move. - you must play on the board where p1 is placed, and that board would have room iff p1 weren't present. The normal argument 'move anywhere and assume that that 'anywhere' is where your first move went, and you just played p1' does not help here, as changing the first move may change where your opponent's first move could have gone. Except from an exhaustive search, I do not see how to prove that you can prevent either case.
  - CJefferson 13y ago
    Yes, you are right. Actually, the fact you force your opponent isn't the problem (I don't think) but the fact that the presence of that piece might later give p2 a free move, where previously they wouldn't means it doesn't work. Sorry, that's what I get for thinking I didn't need to figure out all the details carefully!