RE: A great win by Michael Dots and Boxes
1 replies. Last post: 2005-09-27Reply to this topic Return to forum
KnoxB (Computer) at 2005-09-27
UUggh!! The first few links are still incorrect!
It better work this time or else I quit. :-)
KnoxBMichael1. d2-d32. f2-f33. d5-d64. d4-e45. d3-d46. e4-f47. d1-d2
Since Knox wants an even number of long chains, it divides up the board into two disjoint regions. Usually, Knox has an advantage over humans when the board quickly divides up because it can then make use of nimstring to try to get control. But here the larger region is so large that it can't make use of nimstring (nor search). When this happens,
at best Knox can only make use of some weak heuristics. In the present position that means, don't divide up the big region into two largish regions and don't create any quads (it may be better to try to create a quad with the idea that this might reduce the available chain-making space and make it less likely for a third long chain to appear). Other than that, KnoxB's next three plays are random.
Position after the 8th move Played to get a long chain in the lower-right.
9. a2-b210. e4-e511. a3-a412. c4-d413. c2-d214. c4-c5
Michael's last three moves were played to get a second long chain. Knox's plays were largely random but as luck would have it, these moves resulted in an extremely complex position that is very difficult for humans to play correctly but present no problems for a computer. Perhaps Michael would have been better off not to allow so many free shots on the left side and instead do something that would make the left side more in his favor.
KnoxB switches over to search mode but its play probably wasn't any better than random. I was able to get the perfect play results from the current position on using D.Wilson's analysis program. Currently, Michael has a one box win under perfect play.
Michael makes a long chain and prevents this chain from
extending into the left region so that he can get another long chain on the left. The move makes intuitive sense but actually is not a winning move under perfect play; KnoxB is now winning by 1 box. The three winning plays are e5-e6, b1-c1, and b2-b3. (There are also 10 or so moves that are defeated only by one or two replies). The reasons may be beyond human understanding (they're certainly well beyond *my* understanding).
KnoxB doesn't find one the three winning replies – a4-b4, b2-c2, and a1-a2(a1-b1) – flipping the perfect play result back to a 1 box win for Michael. This move is probably the best losing move because it is the losing move with the minimum number of winning replies, only 1, and furthermore this reply is not at all obvious. Very few people could have won against Knox from this position.
A great play by Michael. The obvious play is to sacrifice one box in the lower-right in order to guarantee a long chain there but this actually loses (the winning replies would be a4-b4,b2-b3,b2-c2, and a1-b1/a1-a2). Michael said he couldn't find a winning path when making the sacrifice so he played to try to prevent KnoxB from creating another region in the top corner. This move is the *only* winning move and if Michael didn't find it, KnoxB would have found the win. As it is, the win is far from easy. (Also, there are several alternatives that are defeated only by one or two moves).
A good try because there are only three non-obvious winning replies. To get an idea of just how complex this position is, consider the winning replies to the following alternatives. 19.a2-a3 is defeated only by a4-a5 and by a5-b5; 19. a4-b4 is defeated only by b2-b3 and by e5-e6; 19. b1-c1 is defeated only by b1-b2 and by a4-a5; 19.f1-f2/e1-f1 is defeated only by b2-b3; and 19.a1-a2/a1-b1 is defeated only by e5-e6. This last alternative is particularly mysterious. Who would suspect that the only way to defeat the innocent looking a1-a2 is to play a move completely away from where all the action is and furthermore, a move that does the opposite of what it seems this player wants (it would seem that Michael would want to make the long chains as long as possible not shorten them.)
Superlative. The only other winning moves are a2-a3 and
d6-e6. None of these moves are obvious and I don't understand why these moves are the winning ones.
A good try. The next move in the lower-right will determine whether a long chain forms there or not. But …..
Good alternative moves are 21.b2-b3 which is defeated only by b2-c2 and 21.b1-c1 which is defeated only by a2-a3.
How does Michael do it? Michael finds the only move in the critical unresolved left region that wins (strangely, any free move in the upper right also wins). Resolving the lower-right region loses because the left region can then be resolved in an appropriate manner. Sacrificing one box to create a long chain in the lower-right loses only to b2-b3. Making a quad in the lower-right loses by 3 boxes to a2-a3 (and loses by one box to a3-b3 and to b3-c3).
Michael writes that instead of thinking in terms of a good move for himself, this is more like thinking in terms of “can I find a reply for you that kills my move.” Indeed, some of the refutation moves for the alternatives are hard to find. E.g. b2-b3 is defeated only by b3-c3; a3-b3 is defeated only by e1-f1; a5-a6/a6-b6 is defeated only by b1-c1; and a2-a3 is defeated only by e1-f1.
Michael says that with a possible loop/quad in the unresolved area, it is straightforward to see that giving a square to create a long chain in the lower-right is suicide. He then makes the curious comment that against a human, he would probably make that sacrifice even knowing that it is a losing move! The reason is that calculation then becomes alot easier and his opponent will make some error after the sacrifice 99.5% of the time.
Knox makes a long chain (making the quad in the lower-right is defeated only by b2-b3). Michael can't afford to resolve the lower-right either (the 1 box sac is still defeated only by b2-b3). Michael writes that he has been trying to guide the left into one bigger chain but simply can't force it.
Making a second chain by 25.b2-b3 is defeated by making the quad and making the quad is defeated (only) by b2-b3.
The only way to stop the left from being one really long chain is to sacrifice three boxes. So now the sacrifice to create a third long chain is worth it. In fact, it is the only way to win.
This countersac is the best move and gives Michael one last chance to screw up but it is very unlikely that he will make such a mistake.
Of course. The long chains are long enough that you should
without worry play it exactly as you would at the end of a game – decline the last two and get all but two of the next chain. Again the only way for KnoxB to stop the left side from becoming one really long chain is to sacrifice three boxes (leaving a quad and a long chain of length 3 on the left) and the three extra boxes are enough for Michael to win anyway.
KnoxB could also play a move connecting the upper-right corner dot to a neighbor, in order to force Michael to sacrifice a box (to avoid a long chain in the upper-right). But this has no effect on the final score because Michael is going to do this anyway.
This ends all the suspense. KnoxB takes the free box and the last free move. yielding the
position. All you have to do now is count up how many boxes you are going to get. Knox gets the lower-left corner box, then the second of the two-chains (1+2=3 more so far). Knox opens up the 3-chain. Michael takes it and opens up the quad. Knox declines the quad to get 5 boxes in the last chain. So KnoxB get 1+2+5=8 more plus the 4 already captured for a total of 12. That leaves 13 boxes and the victory for Michael.