Numerical notation

Some further notation will help us describe the possibilities a little better.

(0)
If neither player has a legal move in a given position, then we denote that position by

$\displaystyle 0=\{\ \vert\ \}=\{\emptyset \vert\emptyset \},
$

where $ \emptyset$ denotes the empty set. In the game of Domineering, this occurs if there is only one square remaining:

\begin{picture}(50,50)
\put(0,0){\framebox (10,10)}
% put(10,0)\{ framebox(10,10...
...\}
% put(30,0)\{ framebox(10,10)\}
% put(0,10)\{ framebox(10,10)\}
\end{picture}
In this position, the player to move loses.

(1)
If Left has exactly one legal move in a given position, and Right has none, then we denote that position by

$\displaystyle 1=\{0 \vert\ \}=\{\{\emptyset \vert\emptyset \} \vert\ \}.
$

In the game of Domineering, this occurs if there are two or three consecutive squares in a horizonal row remaining (and any number of isolated squares):

\begin{picture}(50,50)
\put(0,0){\framebox (10,10)}
\put(10,0){\framebox (10,10)...
...)}
% put(30,0)\{ framebox(10,10)\}
% put(0,10)\{ framebox(10,10)\}
\end{picture}
In this position, Left wins no matter who plays first.

(-1)
If Right has exactly one legal move in a given position, and Left has none, then we denote that position by

$\displaystyle -1=\{\ \vert \}=\{\ \vert\ \{\emptyset \vert\emptyset \} \}.
$

In the game of Domineering, this occurs if there are two or three consecutive squares in a vertical row remaining (and any number of isolated squares):

\begin{picture}(50,50)
\put(0,0){\framebox (10,10)}
\put(0,10){\framebox (10,10)...
...)}
% put(30,0)\{ framebox(10,10)\}
% put(0,10)\{ framebox(10,10)\}
\end{picture}
In this position, Right wins no matter who plays first.

(*)
If Left and Right each have exactly one legal move in a given position, then we denote that position by

$\displaystyle *=\{\ 0\vert \}=\{\{\emptyset \vert\emptyset \} \vert\ \{\emptyset \vert\emptyset \} \}.
$

In the game of Domineering, this occurs if

\begin{picture}(50,50)
\put(0,0){\framebox (10,10)}
\put(0,10){\framebox (10,10)...
...0)\}
\put(10,0){\framebox (10,10)}
% put(0,10)\{ framebox(10,10)\}
\end{picture}
In this position, whoever plays first wins.



David Joyner 2007-09-03