Dots and Boxes commentary. You can get the excellent book by Elwyn Berlekamp here.

The game is played by drawing a square grid of boxes. After that, players alternate connecting dots, as shown below (no diagonal lines). Whenever a square is completed, the player that drew the final line claims that box with their initials, and moves again. When all boxes have been claimed, the person with the most boxes wins.

I've introduced a new notation scheme. Any move will be either the lower edge (—) or left edge (|)of a given square. Thus, the moves already made are A2|, A4—, B1|, B2|, B3|, B4—, C1|, C2|, C4—, D2|, and D3|. Such a scheme, for a 5x5 grid, would use A-F and 1-6.


Stephen Kloder -- The winning move is A3_.
After player 2 takes the squares at A2 and A3 (otherwise p1 takes them
with no other change in gameplay), he can draw a line adjacent to A1
(A1_, A1|) or C1 (C1_, D1|). If p2 draws in A1, p1 can draw in D1 (or
vice versa). Thus in the next move p2 is left with no choice but to
fill A1 and draw a line somewhere in B or C, giving the other 6 squares
(and the game) to p1.
(Note that A2_ is similar, but falls to A3|).

Joseph DeVincentis -- The winning move is A3-.
This gives away the boxes A2 and A3. Opponent can take them, not take them,
or just take one of them, as he wishes, but then must make another move.

If that other move is anything other than A1-, A1|, C1-, or D1|, you
immediately claim 5 boxes for the win.

After his move, claim A2 and/or A3 if opponent refused to do so.

If opponent played A1- or A1| on his second move, you can now also claim A1,
and then play either C1- or D1|. Any move by opponent then allows you to
claim the remaining 6 boxes.

If opponent played C1- or D1| on his second move, you play either A1- or
A1|, giving away that box. Opponent can take it or not, but in either case
must then play some other move, which lets you take all 6 of the boxes in
the B and C columns.

If you make any other move first, opponent can make a 2 or 3 unit rectangle
in the first column, which forces 2 boxes to be closed with a single move,
and changes the parity of the game. Ex.: C1-, A1|, A2-, A3|. At this point
you may claim the boxes in the A column but then your final move lets
opponent claim all the others.

Joe and I played a game. I went first, and lost. Joe comments.

Ed Pegg Jr wrote:
> > 1. B1| C3| ......... A2| A3|
> > 2. B4_ A3_ ......... B2_ D3|
> > 3. C3_ ......... [C4_] C2|
> Resign... you have this one.

My strategy, initially, was based on the observation that in the QDB
phase of the game, since you can have at most 2 edges of each of the
9 squares filled in, you can have at most 18 "edge-sides" (where a
line drawn on the outer boundary of the grid has one edge-side and
an internal line has two edge-sides) before somebody has to draw the
third edge to a box.

Now, each QDB move-pair adds 2, 3, or 4 edge-sides, so I decided on
the strategy of playing the opposite number of edge-sides as you did
on each turn that we were still in QDB. This makes 6 edge-sides per
round, so by my 3rd play (if still in QDB) they'll all be gone.
However, on my third move (and maybe even on my second) I will be
trying to make there be an odd number of isolated groups of boxes
(groups that would be claimed on a single turn) so that when we fall
into normal dots-and-boxes, and have no "safe" moves left, I get to
claim the last, and thus, biggest group of boxes.

Now, there's a problem with this playing-to-18-edge-sides concept,
and that is that the 2 edge-sides for the middle box might not be
playable if too many of the adjacent boxes get two other edges played
to them first. So in my second move, I intentionally played one of
those and left a second one playable.

After you broke the pattern with your third move, there were either
3 or 4 free moves remaining, depending on which ones were taken in
the C1 and C2 boxes, and what looked like it would become 3 groups
of boxes, counting the single box I would take that turn as one of
them (or just 2, not counting that one). I was expecting to give away
to you some or all of the boxes in the lower right, only focusing
on not allowing you the kind of parity-swap involved in the solution
to the D&B puzzle you posted earlier this month, and ultimately
forcing you to play into the big group.

> I've decided a box move gets bracketed. [C3] might work, too, indicating
> the last line of a box + initials.

This works. Sometimes a single move can complete two adjacent boxes
(but only after a player has declined taking one of them for strategic
purposes!) and this could be shown with double brackets. Of course,
chains of box-taking moves can simply all be written out after each other.

The [C3] notation might work better placed after the normal way of
writing the move: C4_ [C3] C2|.

Ed again. In QDB, the interesting analysis starts very quickly. I like it.

James Allen

Thanks for the great website!! I check it out frequently. I design puzzles
myself and will try to look for one worthy enough for your site.

I'm writing about 3x3 QDB --- it turns out it's not so quick after all!
Second-player wins but sometimes his
only winning first move is to make a 3-of-4 and put the game in slow mode.
In fact, the only possible position after his first move which wins for him
and doesn't contain any 3-of-4 is:

*---*   *   *

*   *---*   *
*   *   *   *

*   *   *---*

or equivalent. (Note that the two moves at any corner are interchangeable.)

(All these comments assume my software is bug-free.)

Dan Hoey

I've written my own program, and it agrees with the results of James
Allen's that the second player Bertha wins 3x3 QDB, but only by
._. . . creating a three-sided box, except in the unique position
. ._. . shown at left. Depending on the first player Abby's move,
. | . . Bertha can guarantee herself five or six of the boxes. The
. . ._. unique position guarantees her five, and the only time it is
a suboptimal choice is when Abby's first move is taken in
two opposite corners--then Bertha should make a three in one of the
corners and end up with six boxes.

Bertha's tendency to make a three as soon as possible is illustrated
in Ed vs. Joe:

1. B1| C3| ......... A2| A3|
2. B4_ A3_ ......... B2_ D3|
3. C3_ ......... [C4_] C2|
Ed resigns.

In hindsight we see that Ed could have maintained control by playing
3. C2| C3_ to make the first three instead of Joe. My program says he
would then have gotten five boxes with optimal play, but that he could
have gotten six boxes with any of the sixteen moves
{C1|, C1_, C2_, D1|, D2|} {A2_, B2|},
C2_ {B1_, C1|},
or {C2_, D2|} {C3_, C4_}.

It's not surprising that Joe's second move could have won. He could
even have guaranteed six boxes with any of the twenty-four moves
{B1_, B2_, C1_, C2_, C3_, C4_, D1|, D2|, D3|} {A2_, B2|},
any two of {B1_,C1|,B2_},
or any two of {C3_,C4_,D3|}.

And from James's result, we know that Ed's second move could have won.
Ed would have had to add two sides to one of the boxes A1, A2, A3, B1,
or C3, guaranteeing himself five boxes with optimal play.

On Joe's first move, he could have guaranteed five boxes by adding two
sides to one of the boxes A1, A2, B3, or C3.