An excellent game from the 5th championship Dots and Boxes
1 replies. Last post: 2005-10-22
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KnoxB (Computer) at 2005-10-22
In this well played game, Knox managed to get the correct
parity but Scot was still able to pull out a narrow 1-box
victory. Of special note is Scot's 24'th and 26'th moves.
Here Scot made a very subtle pointless looking play that
was the only way to win followed by a very pretty quad
sacrifice with several free plays still available.
KnoxBScot1. a2-a32. c2-c33. c5-c64. a1-b15. c1-c26. b3-c37. c3-c48. a5-b59. c5-c4
Knox splits the board. As in KnoxB game with Michael posted earlier, the larger region is too large for KnoxB to make sense of but the situation is not so bad here; after any two more moves are made in the larger region, KnoxB will be able to make use of nimstring to try to get control. But until then, KnoxB's plays will be mostly random (it will play in the larger region and it will try to avoid/breakup quads). If getting into a position where KnoxB plays mostly random seems like a weakness, you are right! This kind of split
where the largest region is too large usually doesn't happen (I've never seen it happen to the original version of Knox but there is no reason why it couldn't, though it is less likely.) but even when it doesn't, the opening strategies of both versions of Knox are rather primitive compared to the top players which is why they can beat Knox and KnoxB. But at least these programs do have a real opening strategies which is more than I can say about any of the other
dots-and-boxes programs I'm aware of.
10. a4-a5
Scot wants to be able to force the needed parity on the left side. He wants an odd number of long chains and the right region will probably form 1 or 2 long chains (2 seems more likely). If one long chain forms on the right, he would need 0 or 2 on the left. Two on left seems impossible to force due to the lack of space, hence he plays to make
zero becomes a real possibility – to get 0, he needs two quads on the left. The upper quad can now be avoided only by a sacrifice. If instead two long chains form on the right, then Scot wants one long chain on the left.
11. e2-e312. a1-a2
Now Scot plays so that the bottom-left quad can be avoided only by a sac. (Note: though it was unintentional, Scot's move is in some sense an anti-knox move. After one more play in the large region, Knox can make use of nimstring.)
13. f1-f214. c4-c515. b6-c6
This was KnoxB's first play using its nimstring strategy. KnoxB is now winning nimstring meaning that it can eventually force control. Of course this doesn't guarantee that KnoxB is winning nor is it necessarily best to get control even when you can force it.
16. d3-e317. c2-d2
To use control to win a game, you need to have long enough chains around. Also, quads force the controlling player to give up 4 boxes to maintain control. So if you can force control, you should avoid quads and try to make the chains as long as possible. Hence, when there is choice among winning nimstring moves, Knox prefers those that break up
(or avoids forming) quads and those that extend long chains. In the current position most moves win at nimstring; Knox chose c2-d2 because it breaks up the quad in the right region – after c1-d1, the only way to stop the quad would be to sacrifice a box. Also, it is now very likely that a long chain will form in the lower part of the right region.
I obtained the perfect play results from here on out using D.Wilson's analysis program. Currently, Scot is winning by 1 box. (I didn't know this as the game was being played.)
18.a3-b3
One of the two winning plays (the other is b1-c1). The lower quad is now assured. Now the play a3-a4 would ensure a quad in the upper-left while the sac play (b3-b4 or b4-b5) would turn the upper-left into a long chain of length five.
19.b4-b5
Sacing a box to create a long chain is the only winning nimstring play but actually KnoxB switches to search mode and played the same move anyway. Knox “thinks” it is winning by 1 box (it is losing by one box).
20.a4-b4,b1-c1!!
The only winning play! (Why? I have no idea. Was this a great play or just luck?)
21.d5-e522.c5-d5!
On its move, Knox still incorrectly evaluates the position as 1 box in its favor. It looks like Scot is trying to get a second long chain on the right side starting at the top. Also, Scot's move has the potential to form a quad in the middle right which might be useful for defensive purposes if he doesn't get two long chains on the right.
Note: KnoxB's best practical alternatives seem to be 21.e3-f3 which is defeated only by a5-a6/a6-b6 and 21.d3-d4 which is defeated only by d5-d6 and sacrificing the 4-loop. Of course KnoxB couldn't even determine the correct evaluation let alone determine the number of optimal replies to each move. Also, Knox has no ability to apply any
psychological factors such as estimating which optimal replies would be harder for the opponent to find.
Note that Knox never tries to determine the number of optimal replies and then use that information as a tie-breaker but there is no reason it couldn't when you get close enough to the end of the game. But when you get that close to the end, this may not provide any real benefit. Also, Knox often plays a move where the number of optimal
responses is minimal or near minimal. This seems to be an artifact of the “iterative deepening” search procedure.
23.e5-e6
KnoxB's move cuts off the top two squares which makes it pretty much impossible to force two long chains on the right. I.e. Knox is going to get the desired parity. Knox sees to the end of the game and finally realizes that it is losing by one box. (this was a mild surprise to me because the game seemed to me to be going Knox's way.)
24.e6-f6!!
Scot makes the only winning play! (f5-f6 is of course equivalent). This is a very subtle play. The reason is that the position on the right side is similar to something chessplayers call zugzwang – a german term for 'move-compulsion.' A player is said to be in zugzwang if every threat is defended against but any move you make
disturbs this situation, i.e. every move leaves your opponent with some threat that you cannot defend against. (this is called move-compulsion because you cannot pass your turn. By the rules of the game, you are compelled to move and are thus forced to ruin your own position.) In the current game, any move on the right side by either player besides Scot's move is poison, it gives your opponent
a strong response. It's a challenging exercise to find out how to defeat some of these moves (in particular, f2-f3, e3-e4, e4-e5, f4-f5, and e5-f5 are very tough to defeat). Considering that just about everything on the right is bad, then why not just play 24. a6-b6/a5-a6? This move actually loses! It is defeated only by e3-f3!
25.a6-b6
This is natural considering the previous comments. Now doesn't Scot have to make a poisonous move on the right?
26.a2-b2!!
No he doesn't! I gave a second “!” just for the beauty of the move. Scot doesn't want to be the first to play in the “bad” region but that doesn't mean that this sacrifice works. KnoxB has the choice between taking four boxes and moving first in the rest of the game or giving the four boxes to Scot and going second in the rest of the game. One of these options must be as good or better than the other option and hence must result in taking at least 1/2 of the remaining boxes. Hence, the only way this sacrifice can be a winning play is if under best play, either option leads to a 50-50 split of the remaining boxes giving Scot a one box victory due to the fact that he already has one box. And this is precisely the situation. After the quad is taken,
going second in the remaining region is worth exactly four boxes under best play which precisely cancels out the four boxes in the quad. As quad sacrifices go, this is about as nice as they come.
27.b2-b3,b2-c2,b1-b2,f2-f3
When both options lead to the same evaluation, Knox greedily takes the boxes. But when this evaluation is also a loss, Knox should really decline the boxes (here by b2-c2) under the “enough rope” principle. Since both options lose by the same amount, the only way you can win is if your opponent makes a mistake. Under which option does that seem more likely? The choice is between whether you want your opponent to move first or second in the remaining region. If your opponent moves first, then for each of your opponent's turns there will be more available moves for your opponent than if your opponent moved second. So make things slightly more complicated for your opponent by forcing him to go first, and to force him to go first in the remaining region you decline the boxes. This is sometimes called the “enough rope” principle, the phrase is “give your opponent enough rope to let him hang himself.”
I modified both versions of Knox so that they will now decline the boxes when faced with this situation. Not that it would have made any difference in this game; Scot certainly seemed to know what he was doing.
Note that this was the point at which Knox lost control. Knox could have maintained control by declining by the
boxes.
28.e3-e4
Yep. Either this or e4-e5 to get a central quad and ensure only one long chain on the right.
29.d1-e130.e4-f4
Either this or 30.e4-e5 finishes Knox off.
31.e4-e532.d1-d2
There is nothing to do now but count up boxes. Knox gets the next box plus the second 2-chain and then opens up the 6-chain attached to the loop. Maintaining control will net one box (give up the quad to get the 5-chain) which is fewer than the two boxes you have to give up in order to maintain control. Hence, Scot takes the entire 6-chain and opens the quad. Knox declines the quad to get the 5-chain. So Knox
gets 1 + 2 + 5 = 8 more boxes plus the 4 already captured for a total of 12. Scot gets the remaining 13 and hence wins. If the opponent has a low rating, then I would play this out at least to the point after the opponent makes his choice between taking the entire 6-chain or leaving the last two boxes of this chain. Who knows, your opponent might make a mistake. For a player of Scot's caliber however, the choice is trivial.