Interesting Game Dots and Boxes
2 replies. Last post: 2005-03-10
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Knox (Computer) at 2005-03-04
I thought it might be nice if people occasionally posted
some good games with commentary. To start this off, I’m
posting the game between Knox and Oord2000 from the current
championship (3rd, level 1). In this game, both sides made
a premptive sacrifice. In fact, there were three premptive
sacrifices in the game as well as three one-box sacrifices.
All of the sacrifices except possibly a relatively early
one-box sac were the best plays.
<table CELLPADDING=5 BORDER=0>
<tr ALIGN=LEFT><th WIDTH=120>Knox<th WIDTH=120>Oord
Click Here
1. e3-f32. e4-f4
Position after 2nd move
3. a4-b44. a3-b3
Position after 4th move
5. d3-d46. c3-c4
Position after 6th move
7. d4-e48. b3-c3
9. c4-d4
Position after 9th move
<td COLSPAN=3>
Since Knox wants an even number of long chains, it
sacrifices one box to divide up the board into an even
number of regions. As a practical matter, Knox can
apply nimstring (a way to get “control”) when the board
divides up quickly so Knox should do well here.
10. c3-d3, d2-d3
<td COLSPAN=3> The majority of responses win at nimstring.
Knox has almost no means of discriminating between
various moves that win at nimstring. It will prefer
winning nimstring moves that break up quads and avoid
those that create them. Also, it prefers free moves to
those that give away boxes. For the next three plays,
the choice was simply random among the free winning
nimstring moves. [If Knox is losing nimstring, it will
play the move with the fewest free responses that win at
nimstring.]
11. e4-e512. c4-c5
Position after 12th move
13. f1-f214. d1-d2
Position after 14th move
15. e2-e3
<td COLSPAN=3>I was able to get the perfect play
results from here on out via D. Wilson’s analysis program.
Right now oord is winning by a single box.
16. d1-e1
Position after 16th move
<td COLSPAN=3>Now Knox is winning by 1 box!
This turned out to be the critical error as Knox kept
the 1 box advantage throughout the rest of the game
(both sides played perfectly from here on out!). Any
of the following moves would have won by 1 box with perfect
play: a4-a5, a5-b5, b5-c5, b6-c6, c6-d6, a2-a3, b2-b3,
a2-b2, b2-c2, b1-b2, b1-c1, c1-c2, c1-d1. Instead of
extending the chain with d1-e1, oord needed to try to
establish the correct parity in the rest of the board.
17. d5-d6! (Don’t ask me why!)
Knox switched from nimstring mode to search mode here.
It turned out to be the perfect time to switch because
this moves loses at nimstring! Curiously, Knox evaluated
the game as being in oord’s favor. Strangely,
every response by oord now loses by
exactly one box with perfect play.
18. b5-b6
19. c5-c6
Position after 19th move
<td COLSPAN=3>
Now Knox correctly evaluates the game as being 1 box
in its favor. Any of a4-a5, a5-b5, d6-e6, e5-e6, e6-f6,
or f5-f6 would also have won by 1 box.
20. b2-b3
Position after 20th move
<td COLSPAN=3>
I didn’t expect this. c1-c2 creating a 4’th chain seems
like such an obvious response. But oord must have already
planned to meet c1-c2 by sacrificing this chain. Instead
of b2-b3, I was thinking he would sac 1 box with c2-c3.
The winning response to this (after taking the box) is a2-b2.
21. c1-c2
<td COLSPAN=3>
Necessary as it is the only way to win. Knox can now see
to the end of the game. Hence, it already “knows” it is
going to win by at least 1 box.
22. b2-b3
Position after 22nd move — A nice pre-emptive sac.
<td WIDTH=120>23. b2-c2, c2-d2, c1-d1, a1-b1
<td COLSPAN=3>
The take-all/leave two decision is not an easy one.
Leaving two loses by 1 box though this is far from
obvious. After taking the entire chain, the long
chain parity in the rest of game favors oord. Knox
is going to have to eventually make a return pre-emptive
sacrifice to win this game. The two winning choices
after taking the chain are to either sac the three
chain in the top-central or (Knox’s choice) to play
one of the equivalent moves a1-a2 or a1-b1 which threaten
to create a fourth chain.
24. a2-b2
Position after 24th move
<td COLSPAN=3>
This one box sac is necessary to stop Knox from
creating a fourth chain and hence, stops Knox
from switching the parity to its favor.
25. b1-c1, e1-e2
Position after 25th move
<td COLSPAN=3>
Since the parity is in oord’s favor, Knox must make
a premptive sac on this 4-chain or on the 3-chain
at the top-middle.
<td WIDTH=130>26. d2-e2, d3-e3, e3-e4, f3-f4, d5-e5
Position after 26th move
<td COLSPAN=3>
The take-all and the leave two options both lose by
1 box! If you take all, then you should either sac
the 3-chain (as oord just did) or play a5-b5. If you
leave two, then the opponent should take these two
and either offer the 3-chain or play a5-b5. It seems
strange that a5-b5 is just as good as the premptive
sac — the premptive sac seems necessary in both cases
because the parity is not what is wanted.
<td WIDTH=120>27. d4-d5, c5-d5, c6-d6, a5-b5
Position after 27th move
<td COLSPAN=3>
Taking all is necessary. Knox wants two more chains
while oord wants one. The chain in the upper-left
cannot be stopped while the one in the upper-right
cannot be forced. Hence, oord is going to get his
one long chain (of length 4) and he will take this
at the end. But to stop the chain in the upper-right,
oord is going to have to sac 1 box. But Knox’s a5-b5
creates a second short chain in the upper-left so that
Knox will get two more boxes in the short chain swaps.
These three boxes plus the two that Knox is ahead by
are just enough to overcome oord’s final 4-chain.
28. e5-f5
29. f4-f5, f5-f6
Position after 29th move
<td COLSPAN=3>
It’s all over now. Just count up the boxes in
every other chain to determine how many more boxes
you are going to get. Knox wins 13-12.