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Then the black stone's moves $ (1\ 0)\ $ and $\ (1\ 0)\ $
would stumble, and $\ (2\ 1)\ $ would have only two unoccupied
nearest neighbors. Thus, only one move $\ (2\ 1)\ $$\ (1\ 2)\ $ is left:
Thus, let me play $\ (0\ 3),$$\ (-\!1\ 3),$
Then the black stone's moves $ (1\ 0)\ $ and $\ (1\ 0)\ $
would stumble, and $\ (2\ 1)\ $ would have only two unoccupied
nearest neighbors. Thus, only one move $\ (2\ 1)\ $ is left:
Thus, let me play $\ (0\ 3),$
Then the black stone's moves $ (1\ 0)\ $ and $\ (1\ 0)\ $
would stumble, and $\ (2\ 1)\ $ would have only two unoccupied
nearest neighbors. Thus, only one move $\ (1\ 2)\ $ is left:
An even-move, also called a white move, in
position $\ P\ $$\ P:=(b\ X)\ $ is an arbitrary
$\ y\in\mathbb Z^2\setminus\{b\}.\ $ The resulting position is defined as $\ Q:=(b\ Y),\ $ where
$\ Y:=X\cup\{y\}.\ $ (Even-move $\ y\in X\ $ would be silly but legal).
An even-move, also called a white move, in
position $\ P\ $ is an arbitrary
$\ y\in\mathbb Z^2\setminus\{b\}.\ $ The resulting position is defined as $\ Q:=(b\ Y),\ $ where
$\ Y:=X\cup\{y\}.\ $ (Even-move $\ y\in X\ $ would be silly but legal).
An even-move, also called a white move, in
position $\ P:=(b\ X)\ $ is an arbitrary
$\ y\in\mathbb Z^2\setminus\{b\}.\ $ The resulting position is defined as $\ Q:=(b\ Y),\ $ where
$\ Y:=X\cup\{y\}.\ $ (Even-move $\ y\in X\ $ would be silly but legal).
