Misere dots

Are they just always counted for two "chains"?

In misere dots, anything is a chain (long chains, double crosses, triple crosses, even quadruple crosses with right shape you cannot cut in two double crosses: T4, quintuple crosses: X) except quads that just don't count... In the end, the player without control will play in a quad and the controlling player will just turn that into two double crosses.

I'm prepared to bet that in misere dots, it is a first player win.
And as most misere games, far enough for the ending position behave like normal games, that could shed some light on the outcome of normal dots and boxes.

So the player in control would want many quads but not long chains since they make it expensive to stay in control.

Not really, because:
- for quads they can cost the player in control at least 3 squares if other player attacks into them before they are complete
- for long chains, you can usually trade at least the four last squares (3+1) against the start of the chain. So it costs you squares only if it is longer than eight squares.

The real problem for the player in control is to decide how many chains he must get before starting to sacrifice so that it will not cost him more than 12 squares to keep the right parity. And he must also be assured that the rest of the board will not turn into another chain. Or be sure he will get another one for a disguised control with wrong parity.

That is true for the 5x5 but in general long chains seem to favour the player not in control.

It is a fun thing about making a difference if quads are complete or not.

http://www.trmph.com/dnb/board#5,b9d9f9g10j11h9a8a6b5d5e6e8f5h5i6i8k10k8j5c4c2b1d1f1g2g4h1k4

First player to move and win (non-misere).

If it's non-misere, why do you post it in a thread on misere dots?

If you haven't solved the puzzle yet, why do you start bitching at me? I was referring to the last post by Christian K.

Oh OK :-P

The tournament is over. Well done dryad.

I'd play d11, personally. But I agree with Taz, I wouldn't play it in a misere thread.

WP Dryad.

AGAIN!

http://www.littlegolem.net/jsp/forum/topic2.jsp?forum=110&topic=142

Don't really get your point, but here you are. :P

How does one defeat the "rotational symmetry" strategy when playing misère dots and boxes? I can see how to do it for the case where the board has odd side lengths (one just sacrifices the middle box), but I don't see how to escape from it otherwise.

I think one answer is that when 1st player will do a double-cross threat (that is a double-cross minus the closing move), 2nd player will not want to play the symmetric move because then 1st player can chose the parity.

See for example :
http://www.trmph.com/dnb/board#5,f11f1e10g2e8g4f7f5d7h5d9h3b9j3i2c10k2a10h7d5i8c4i10c2h11d1b5j7b3j9b1j11b7
here p2 should not play 34.j5 because of :
http://www.trmph.com/dnb/board#5,f11f1e10g2e8g4f7f5d7h5d9h3b9j3i2c10k2a10h7d5i8c4i10c2h11d1b5j7b3j9b1j11b7j5c8a8k4

but should play http://www.trmph.com/dnb/board#5,f11f1e10g2e8g4f7f5d7h5d9h3b9j3i2c10k2a10h7d5i8c4i10c2h11d1b5j7b3j9b1j11b7c8a8i4j5 instead.

Here it is easy because it is the end of the game, but 1st player can force p2 to take a difficult decision beforehand.

An example on 4x4:
http://www.trmph.com/dnb/board#4,d9f1e8e2d7f3b7h3f7d3g8c2h7i8b3a2e4e6c6g4c4g6b9h1c8i2b5h5

Ah, yes. Thanks!