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.

KnoxOordClick Here1. 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-c39. c4-d4

Position after 9th move

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 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-e3I 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

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!!

Position after 17th move

The only winning move! (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-b619. c5-c6

Position after 19th move

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

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

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. 23. b2-c2, c2-d2, c1-d1, a1-b1

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

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

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.26. d2-e2, d3-e3, e3-e4, f3-f4, d5-e5

Position after 26th move

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.

27. d4-d5, c5-d5, c6-d6, a5-b5

Position after 27th move

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-f529. f4-f5, f5-f6

Position after 29th move

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.

Great Analysis. Too bad i lost, but i'm glad i found some nice moves.