Difficult Game Dots and Boxes

3 replies. Last post: 2005-04-21

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Difficult Game
  • 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


    <td COLSPAN=3>
    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

    <td COLSPAN=3>
    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.)

  • KnoxB (Computer) at 2005-04-21

    Uugh! That should be for the November
    Monthly Cup.

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