Difficult Game Dots and Boxes
3 replies. Last post: 2005-04-21
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Knox (Computer) at 2005-04-21
The following game between Knox and Pako decided the
December monthly cup. Using D.Wilson’s analysis program,
I found that the game contained a run of at least six
consecutive mistakes! Futhermore, I do not understand
why they are mistakes; they look like good moves. The
correct evaluations of some of these positions are not at
all what they seem they should be. If nothing else, this
game demonstrates that Dots and Boxes can be really hard.Pako vs. Knox
<table CELLPADDING=5 BORDER=0>
<tr ALIGN=LEFT><th WIDTH=120>Knox<th WIDTH=120>Oord
Click Here
1. b1-c12. b1-b2
Position after 2nd move
3. d2-d34. e4-f4
Position after 4th move
5. c2-d26. d1-d2
Position after 6th move
<td COLSPAN=3>
Knox is trying to create a region in the lower-right, and
wants to split the rest of the board to get three regions.
7. e2-e38. b3-b4
Position after 8th move
<td COLSPAN=3>
Knox doesn’t think it can split up board favorably so it tries to create
a defensive quad (or force Pako to sac a box to avoid the quad).
9. b3-c310. d3-d4
Position after 9th move
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Pako breaks up the potential quad. Again Knox is trying
to create a separate region in the lower-right.
11. d1-e112. c5-d5
Position after 12th move
<td COLSPAN=3> Knox still doesn’t like the prospects
of splitting up the board and again tries for a quad.
13. c4-d414. f4-f5
Position after 14th move
15. d5-d6
<td COLSPAN=3>I was able to get the perfect play results from here on
out via D. Wilson’s analysis program. Right now Pako is winning by a
single box. Also, there is nothing profound about Knox’s 14’th move.
Knox wasn’t able to find anything constructive to do. In such a case,
it looks at the free moves (i.e. those that don’t give away any boxes),
throws away a couple of those that it “thinks” would give the opponent
some direct immediate help towards creating the desired parity, and
then randomly picks one of the remaining moves. Knox could face this
exact position 15 times in a row and play a different move each time.
16. b4-b5?
Position after 16th move
<td COLSPAN=3>Now Pako is winning by 5 boxes.
(Knox switched to search mode here and evaluated the position
as being very strongly in Pako’s favor).
The two optimal plays are 16.f2-f3 and the 1 box sac 16.e4-e5 (or the
equivalent e5-f5). The unique winning response to this sac is the
two box sac d4-d5!
17. d4-e4?
Position after 17th move
<td COLSPAN=3>
Now the winning margin is back down to a single box.
The two optimal plays (which would maintain the 5 box margin)
are e3-f3 and the more interesting 1 box sac c4-c5 (d4-d5).
The unique optimal response to c4-c5 is c6-d6.
18. c3-c4
Position after 18th move
<td COLSPAN=3>
Since Knox is not ahead in boxes already taken (0 to 0), this premptive
sacrifice must be losing against best play, yet there is a logic to it.
There doesn’t seem to be any other way for Knox to force an odd number
of long chains. Hence even though this must be losing, it seems like
it should keep the margin down and make it tougher for Pako. Yet it
is not the best; this sac increases Pako’s winning margin to 3 boxes.
The optimal lines of play here are extremely difficult and complex.
They involve multiple consecutive sacrifices. At one point, both
the take all and leave two options are optimal and furthermore,
both optimal continuations leads to the identical sequence of
positions but with the two players being on opposite sides. And
to top it off, these optimal sequences reach another take all/leave
two option where both choices are optimal and furthermore, both optimal
continuations yet again lead to the identical positions but with the
two players on opposite sides. Bizarre! Ready boys and girls? :-)
There are two optimal moves here 18.f3-f4 and 18.f2-f3.
(note: if no options are given for a move, then that is the
unique optimal move)
(A) 18.f3-f4 19.e2-f2 20.the sac c3-c4 (or equivalent moves)
The optimal continuations are
(a1) 21. take all, a4-b4
(a2) 21. take all, sac c1-c2 22. take all, a3-a4
(a3) 21. leave two. now either 22.a4-b4 or 22.sac c1-c2 23.take all,
a3-a4
(B) 18.f2-f3 19. sac e2-f2(e3-f3)The optimal continuations are
(b1) 20.take (e3-f3/e2-f2), sac c3-c4 21. take all, sac c1-c2
22.take all and leave two are both optimal.
(b2) 20.take (e3-f3/e2-f2), sac d2-e2
Now both the take all and leave two choices are optimal. The two
optimal sequences start with 21. take all, sac c3-c4 and 21. leave two
22. sac c3-c4.
Now regardless of which player makes the c3-c4 (or equivalent) sac, the
optimal response is to take all and sac c1-c2. In response to the c1-c2
sac, both the take all and leave two options are optimal. The two
optimal continuations are move A: take all, a2-a3 and move A: leave two,
move B: a2-a3.
Suppose we draw the search tree of optimal positions here starting
at the position after the sac d2-e2 (in move 20). At move 21,
the tree will split into two branches, one for each of the two optimal
choices. The continuing optimal positions along both branches
will be identical but the two players will be on different sides
in the two branches. Then both branches will do a two-way split
at move A. The final positions at the end of the four branches will
all be identical but the players will be on differing sides depending
on how that particular position was reached.
Now if both players were omniscient gurus, Pako could force himself to
be on either side of the resulting optimal position. At move 21, Pako
could play take all so that he will face the second take all/leave two
decision and then forcibly choose whichever side he wanted. If instead
Pako were to choose leave two at move 21, then Knox would get the second
take all/leave two decision and could choose either side. But of course
neither player qualifies as an omniscient guru as this game more than
amply demonstrates.
This line is perhaps too bizarre for words.
19. take all, c1-c2?
Position after 19th move
<td COLSPAN=3>
Now Knox is winning by one box!
The take all decision was correct but the followup sacrifice wasn’t.
After taking the chain, every sacrifice loses by one
box, f2-f3 wins by three boxes, and all other non sacrifice plays win
by one box.
I must say I’m baffled as to best plays. Pako wants an even number
of long chains and there are an odd number on the board. Hence,
Pako’s premptive sacrifice should be correct and Pako has enough
boxes to be able to afford this. And how on earth can a move like
a2-a3 or b6-c6, which simply lets a chain grow, be a better move?
The correct evaluations of the resulting positions are not at all
what they appear to be and I don’t know why.
20. c1-c2,b2-b3?? (i.e. leave two)
Position after 20th move
<td COLSPAN=3>
The worse move of the game. This changes the result from a one box win
to a 5 box loss. Again I’m puzzled as to why this move is bad; it looks
correct. If Knox takes everything, then the parity of the long chains
favors Pako. By leaving two, the subsequent doublecross move (b2-c2)
will swap the parity back to Knox’s favor.
(note: If I had given Knox a little more time, it would have made
the correct choice and gone on to win. Of course I have to give it some
time limit and I don’t want to give it an unfair amount of time.)
21. b2-c2, d3-e3?
<td COLSPAN=3>
This drops the winning margin back down to a single box.
The best moves (after b2-c2) were b5-b6, a4-a5, and a2-a3.
This move was the last of the suboptimal plays.
Once again, I really don’t understand why Pako’s move drops the winning
margin by four boxes; it looks like a good move to me. Pako wants an
odd number of long chains (the doublecross he just made, b2-c2, swaps
the parity) and there are an even number on the board. There is no
place left to get another long chain so he makes a premptive sacrifice
on the shorter chain. How can this be bad?
22. take all, f2-f3
Position after 22nd move
<td COLSPAN=3>
Knox can now see to the end of the game and makes the unique best play.
The smoke has cleared and Pako had to be happy by this point.
Pako wants an odd number of long chains. There is a single long chain
of four boxes at the top and furthermore, both ends of this chain must
each grow by at least one box. Pako can stop the formation of any
additional long chains and thus guaranteeing that he will get the last
long chain which will contain at least six boxes and probably more.
The only thing Knox can do is to threaten to create an additional long
chain that will force Pako to sac a box or two to stop it. Hence,
the reason for Knox’s move f2-f3. Pako must respond with the two
box sac e3-f3 to stop the formation of the long chain.
But it is hard to see how Knox can force enough sacrificed boxes to
overcome that long chain at the end. Knox does manage to keep it
closer than it appears at a casual glance but Pako has a pretty
clear victory path and doesn’t have any difficult choices to make.
(note: Knox’s current move is the first of five consecutive unique
optimal plays! I.e. there is no other way to keep the winning
margin down so low).
23. e3-f3 24.take and a2-a3
Position after 24th move
25. a3-b3 26.a2-b2, a4-a5
Position after 26th move
27. b6-c628. b5-b6
Position after 28th move
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Instead of 27. b6-c6, an immediate 27.a4-b4 would also have worked.
The reason for 27. b6-c6 is to grow the last chain. After the move,
this end must grow by least two boxes instead of by at least one box.
29. a4-b430. take and d6-e6
Position after 30th move
31. e5-e630. take and d6-e6
Position after 32nd move
<td COLSPAN=3>
It’s all over now. Pako will get a single box plus the chain of seven.
These eight plus the five he already gives him a 13-12 victory. -
Knox (Computer) at 2005-04-21
Sorry but the column heading for the players moves
should be
<tr ALIGN=LEFT><th WIDTH=120>Pako<th WIDTH=120>Knox
(I made the pako vs knox file from the Knox vs Oord file
and forgot to change the column headings.)