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\title{\bf General Restriction of (s,t)-Wythoff's Game}

\author{Wen An Liu\thanks{This work is supported by the National
Natural Science Foundation of China under Grants 11171368 and 11171094.}\\
\small College of Mathematics and Information Science\\[-0.8ex]
\small Henan Normal University\\[-0.8ex]
\small Xinxiang, P. R. China\\
\small \tt{liuwenan@126.com}\\
\and
Haiyan Li\\
\small College of Mathematics and Information Science\\[-0.8ex]
\small Henan Normal University\\[-0.8ex]
\small Xinxiang, P. R. China\\[-0.8ex]
\small \texttt{lihaiyan1107@126.com} \\
 }

\date{\dateline{Dec 11, 2012}{}{}\\
\small Mathematics Subject Classifications: 91A46, 68R05}

\begin{document}
\maketitle

\begin{abstract}
A.S. Fraenkel introduced a new $(s,t)$-Wythoff's game which is a
generalization of both Wythoff's game and $a$-Wythoff's game.
 Four new models of a restricted version of $(s,t)$-Wythoff's game,
Odd-Odd $(s,t)$-Wythoff's Game, Even-Even $(s,t)$-Wythoff's Game,
 Odd-Even $(s,t)$-Wythoff's Game and Even-Odd $(s,t)$-Wythoff's Game, are investigated.
 Under normal or mis\`{e}re play convention, all $P$-positions of these four models are
given for arbitrary integers $s,t\geq 1$. For Even-Even $(s,t)$-Wythoff's Game,
 the structure of $P$-positions is given by recursive characterizations in term of mex function.
 For other models,
 the structures of $P$-positions are of algebraic form, which permit to
decide in polynomial time whether or not a given game position $(a,b)$ is a $P$-position.
% keywords are optional

  \bigskip\noindent \textbf{Keywords:} impartial combinatorial game; normal play convention; mis\`{e}re play
convention; $P$-position; $(s,t)$-Wythoff's game
\end{abstract}




\section{Introduction}


By game we mean a combinatorial game; we restrict our attention to
classical impartial games. There are two conventions: in
\emph{normal play convention}, the player first unable to move is
the loser (his opponent is the winner); in \emph{mis\`{e}re play
convention}, the player first unable to move is the winner (his
opponent is the loser). The positions from which the previous player
can win regardless of the opponent's moves are called
\emph{$P$-positions} and those from which the next player can win
regardless of the opponent's moves are called
\emph{$N$-positions}. The theory of such games can be found in
\cite{Berl1982, Bou1905,Con1976,Fraen2004}.

Throughout this paper, we use the following notations.

(1) By $Z^{\geq m}$ we denote the set of all integers not less than
$m$, i.e., $Z^{\geq m}=\{x\geq m| x \mbox{ is an integer}\}$. Let
$Z^{even}=\{2n|n\in Z^{\geq 0}\}$, $Z^{odd}=\{2n+1|n\in Z^{\geq
0}\}$.

(2) For any set $U\subseteq Z^{\geq 0}$, by $\mbox{mex}(U)$ we denote the
\emph{Minimum EXcluded value} of $U$, i.e., the smallest
nonnegative integer not in $U$. In particular,
$\mbox{mex}(\emptyset)=0$.

(3) By $\lfloor x\rfloor$ we denote the largest integer $\leq x$.

(4) We use the notation $(x_1,y_1)\rightarrow (x_2,y_2)$ if there is
a legal move from  $(x_1,y_1)$ to $(x_2,y_2)$.


\subsection{Wythoff's game}
Wythoff's game is played with two heaps of tokens. Each
player can either remove any number of tokens from a single heap
(\emph{Nim rule}) or remove the same number of tokens from both heaps
(\emph{Wythoff's rule}). All $P$-positions of Wythoff's game under normal
play convention were given in \cite{Wyth1907}. All $P$-positions of
Wythoff's game under mis\`{e}re play convention were determined
in \cite{Fraen2002}.



\subsection{Extension of Wythoff's game}
In many papers devoted to variations of Wythoff's game, new rules
are adjoined to the original ones. Such variations are called
\emph{extensions}.

As an example, \textit{$a$-Wythoff's game} was investigated in
\cite{Fraen1982}: Given an integer $a\geq 1$ and two heaps
of finitely many tokens. Two rules of moves are allowed.

(\emph{Nim Rule}) Take any positive number of tokens from a single heap, possibly the entire heap.

(\emph{General Wythoff's Rule}) Take tokens from both
heaps, $k>0$ tokens from one heap, and $\ell>0$ tokens from
the other, and $\mid k-\ell\mid <a,$ where $a>0$ is a fixed integer parameter.

A.S. Fraenkel \cite{Fraen1998} introduced a new \emph{$(s,t)$-Wythoff's game}. Given two parameters $s,t\in Z^{\geq 1}$ and two heaps
of finitely many tokens. There are two types of moves:

(\textit{Nim Rule}) Take any positive number of tokens from a single heap, possibly the entire heap.

(\textit{More General Wythoff's Rule}) Take tokens from
both heaps, $k>0$ from one heap and $\ell>0$ from the other, and
\begin{equation}\label{def-constraint-1}
0< \ k \leq \ell<sk+t.
\end{equation}

In \cite{Fraen1998}, the author gave the following results: Denote by $\mathscr{P}_{s,t}$
the set of all $P$-positions of $(s,t)$-Wythoff's game under normal play convention. Then
$\mathscr{P}_{s,t}= \bigcup\limits_{n=0}^{\infty}\{(A_{n},B_{n})\}$,
where for $n\geq 0$,
\begin{equation}\label{(s,t)-conclusion}
\left\{\begin{array}{l}
A_{n}=\mbox{mex}\{A_{i},B_{i}|0\leq i<n\},\\
B_{n}=sA_{n}+tn.
\end{array}\right.
\end{equation}



It is worth to mention that Wythoff's game is a special case
$s=t=1$ in $(s,t)$-Wythoff's game, and $a$-Wythoff's game is
a special case $s=1$ and $t=a$ in $(s,t)$-Wythoff's game. Thus
$(s,t)$-Wythoff's game is a generalization of both Wythoff's
game and $a$-Wythoff's game.

Under normal play convention, the set
$\mathscr{P}_{1,a}$ of all $P$-positions of $a$-Wythoff's game
and the set $\mathscr{P}_{1,1}$ of all $P$-positions of Wythoff's game are given by letting ($s=1$ and $t=a$) and $s=t=1$ in
Eq. (\ref{(s,t)-conclusion}), respectively (see \cite{Fraen1982,Wyth1907}).

Under mis\`{e}re play convention, all $P$-positions of $a$-Wythoff's game
were given in \cite{Fraen2002}. All $P$-positions of $(s,t)$-Wythoff's game
were determined in \cite{Liu2012}, for all integers $s,t\geq 1$.

Other examples of extensions of Wythoff's game were given in
\cite{Connell1959,FrOz1998, Fraen2005,Fraen2009,Sun2009}.

\subsection{Restriction of $(s,t)$-Wythoff's game}

There are a few papers where only subsets of Wythoff's moves are
allowed (see \cite{DuGr2009,DuFraen2009,Fraen1991}). Such variations are called
\emph{restrictions} of Wythoff's game.

We now introduce a new \emph{General Restriction of
$(s,t)$-Wythoff's Game}. Let $S_h$, $S_v$, $D_1$ and $D_2$ be subsets
of $Z^{\geq 0}$. Given two parameters $s,t\in Z^{\geq 1}$ and two heaps
of finitely many tokens. One of the heaps is designated as ``first heap"
and the other as ``second heap" throughout this game. By $(x,y)$ we denote a position of present game,
where $x$ and $y$ denote the numbers of tokens in the first and the
second heaps, respectively. There are three types of moves.

(\emph{Horizontal Move}) A player chooses the first heap and takes
$k\in (\{w|0<w\leq x\}\cap S_h)$ tokens, i.e.,
\begin{equation}\label{move-def-horizontal}
(x,y)\rightarrow (x-k,y)\mbox{ and } k\in (\{w|0<w\leq x\}\cap
S_h).\end{equation}
In this case, we call that $(x,y)$ is moved to $(x-k,y)$ in \emph{horizontal direction}, and $k$ is called \emph{horizontal distance}.

(\emph{Vertical Move}) A player chooses the second heap and takes
$\ell\in (\{z|0<z\leq y\}\cap S_v)$ tokens, i.e.,
\begin{equation}\label{move-def-vertical}
(x,y)\rightarrow (x,y-\ell)\mbox{ and } \ell\in (\{z|0<z\leq y\}\cap
S_v).\end{equation}
In this case, we call that $(x,y)$ is moved to $(x,y-\ell)$ in \emph{vertical direction}, and $\ell$ is called \emph{vertical distance}.

(\emph{Extended Diagonal Move}) A player takes tokens from both
heaps, $k\in (\{w|0<w\leq x\}\cap D_1)$ from the first heap and
$\ell\in (\{z|0<z\leq y\}\cap D_2)$ from the second heap, and
\begin{equation}\label{general-def-constraint}
0 \leq |\ell-k| \ <(s-1)\lambda+t, \ \ \lambda=\min\{k,\ell\}\in
Z^{\geq 1}.
\end{equation}
In this case, we call that $(x,y)$ is moved to $(x-k,y-\ell)$ in \emph{extended diagonal direction},
$k$ and $\ell$ are called \emph{extended diagonal distance}.

\begin{remark}\label{Remark1} Note that Eq. (\ref{def-constraint-1}) is
equivalent to
\begin{equation}\label{def-constraint-3}
0 \leq \ell-k \ <(s-1)k+t, \ \ k\in Z^{\geq 1},
\end{equation}
and Eq. (\ref{def-constraint-3}) is equivalent to Eq.
(\ref{general-def-constraint}).
\end{remark}

\begin{remark}\label{Remark2} $(s,t)$-Wythoff's game introduced by A.S.
Fraenkel in \cite{Fraen1998} is equivalent to $S_h=S_v=D_1=D_2=Z^{\geq 0}$
in General Restriction of $(s,t)$-Wythoff's Game.
Also $a$-Wythoff's game investigated in \cite{Fraen1982} is equivalent to
$S_h=S_v=D_1=D_2=Z^{\geq 0}$, $s=1$ and $t=a$ in General Restriction of $(s,t)$-Wythoff's Game.
Wythoff's game investigated in \cite{Wyth1907} is equivalent to
$S_h=S_v=D_1=D_2=Z^{\geq 0}$, $s=1$ and $t=1$ in General Restriction of $(s,t)$-Wythoff's Game.
\end{remark}

\begin{remark}\label{Remark3} In \cite{DuGr2009}, the authors investigated the case of Wythoff's game,
where ``horizontal distance" , ``vertical distance" and ``diagonal
distance" are bounded by a given positive integer $R$. This problem is equivalent to
$S_h=S_v=D_1=D_2=\{n\leq R|n\in Z^{\geq 0}\}$ and $s=t=1$ in General Restriction of
$(s,t)$-Wythoff's Game. The set of all $P$-positions of this game under normal play convention
was determined in \cite{DuGr2009}.
\end{remark}

\begin{remark}\label{Remark4} In \cite{DuGr2009}, the authors presented the following
problems:

(1) One can investigate the case of Wythoff's game, where only
``diagonal distance" is bounded. This problem is equivalent to
$S_h=S_v=Z^{\geq 0}$, $D_1=D_2=\{n\leq R|n\in Z^{\geq 0}\}$ ($R$ is
a fixed positive integer) and $s=t=1$ in General
Restriction of $(s,t)$-Wythoff's Game.

(2) One can investigate the case of Wythoff's game, where ``horizontal distance" and ``vertical distance" are bounded, but ``diagonal
distance" is infinite. This problem is equivalent to $S_h=S_v=\{n\leq
R|n\in Z^{\geq 0}\}$ ($R$ is a fixed positive integer),
$D_1=D_2=Z^{\geq 0}$ and $s=t=1$ in General Restriction of $(s,t)$-Wythoff's Game. Under normal play convention, the
set of all $P$-positions of this game was given in \cite{Liu2011}.

(3) One can investigate the bounded version of $a$-Wythoff's game,
where ``horizontal distance" and ``vertical distance" are bounded, but ``diagonal
distance" is infinite. This problem is equivalent to
$S_h=S_v=\{n\leq R|n\in Z^{\geq 0}\}$ ($R$ is a fixed positive
integer), $D_1=D_2=Z^{\geq 0}$ and $s=1$, $t=a$ in General Restriction of $(s,t)$-Wythoff's Game;
If ``horizontal distance" , ``vertical distance" and ``diagonal
distance" are bounded, then this problem is equivalent to $S_h=S_v=D_1=D_2=\{n\leq R|n\in
Z^{\geq 0}\}$ ($R$ is a fixed positive integer), $s=1$ and $t=a$ in
General Restriction of $(s,t)$-Wythoff's Game.
\end{remark}

\subsection{Our results}

For all extensions and restrictions of Wythoff's game, our main goal
is to find characterizations of $P$-positions, which
almost always differs from the original Wythoff's sequence (see
\cite{DuFraen2010,Liu2011}). In this paper, four models of General
Restriction of $(s,t)$-Wythoff's Game are investigated. Let us now briefly present the content of this paper.

In Section 3, we define the first model, \emph{Odd-Odd $(s,t)$-Wythoff's Game},
which is equivalent to $S_h=S_v=D_1=D_2=Z^{odd}$ in General Restriction of $(s,t)$-Wythoff's Game.
Under normal play convention and for all $s,t\in Z^{\geq 1}$, the set of all $P$-positions is given by $\bigcup\limits_{m=0}^{\infty}\bigcup\limits_{n=0}^{\infty}\{(2m,2n)\}$;
Under mis\`{e}re play convention and for all $s,t\in Z^{\geq 1}$, the set of all $P$-positions is given by
$$\{(0,2p+1),(2p+1,0)|p\in Z^{\geq0}\}\cup\bigcup\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\{(2m,2n)\}.$$
The structures of $P$-positions are of algebraic form (Theorems \ref{OOW-normal} and \ref{OOW-misere}), which permit to
decide in polynomial time whether or not a given game position $(a,b)$ is a $P$-position.

In Section 4, we define the second model, \emph{Even-Even $(s,t)$-Wythoff's Game},
which is equivalent to  $S_h=S_v=D_1=D_2=Z^{even}$ in General Restriction of $(s,t)$-Wythoff's Game.
Under normal or mis\`{e}re play convention, and for all $s,t\geq 1$,
 the sets of all $P$-positions are given by recursive characterizations in term of mex function (Theorems \ref{EEW-normal} and \ref{EEW-misere}).

In Section 5, we define the third model, \emph{Odd-Even $(s,t)$-Wythoff's Game},
which is equivalent to ($S_h=D_1=Z^{odd}$ and $S_v=D_2=Z^{even}$) in
General Restriction of $(s,t)$-Wythoff's Game.
Under normal or mis\`{e}re play convention, and for all $s,t\in
Z^{\geq 1}$, the sets of all $P$-positions are given by algebraic
characterizations (Theorems \ref{OEW-normal-s=t=1}, \ref{OEW-misere-s=t=1}, \ref{OEW-normal-s+t>2}
and \ref{OEW-misere-s+t>2}), which provide polynomial time procedures.

In Section 6, we define the fourth model, \textit{Even-Odd $(s,t)$-Wythoff's Game},
which is equivalent to ($S_h=D_1=Z^{even}$ and $S_v=D_2=Z^{odd}$) in
 General Restriction of $(s,t)$-Wythoff's Game.
Under normal or mis\`{e}re play convention, and for all $s,t\in
Z^{\geq 1}$, the sets of all $P$-positions are given by explicit
formulas (Corollaries \ref{EOW-normal-s=t=1}, \ref{EOW-misere-s=t=1}, \ref{EOW-normal-s+t>2} and \ref{EOW-misere-s+t>2}),
which provide polynomial time procedures.

\section{Preliminaries}
Given any game $\Gamma$, we say informally that a $P$-position is
any position $u$ of $\Gamma$ from which the \textit{Previous} player
can force a win, that is, the opponent of the player moving from
$u$. An $N$-position is any position $v$ of $\Gamma$ from which the
\textit{Next} player can force a win, that is, the player who moves
from $v$. The set of all $P$-positions of $\Gamma$ is denoted by
$\mathscr{P}$, and the set of all $N$-positions of $\Gamma$ is
denoted by $\mathscr{N}$. Denote by $Option(u)$ all options
of $u$, i.e., the set of all positions that can be reached in one
move from $u$. It follows from Fraenkel \cite{Fraen2004} that
\begin{equation}\label{def-NP}
\begin{array}{ll}
u\in \mathscr {P}\Longleftrightarrow Option(u)\subseteq \mathscr{N},\\
u\in \mathscr{N}\Longleftrightarrow Option(u)\cap
 \mathscr{P}\neq\emptyset.
\end{array}
\end{equation}

In order to better understand the legal moves of General Restriction of
$(s,t)$-Wythoff's Game, we define the following notations.

By $Option_{h}(x,y)$ we denote the set of all positions that can be reached in one
move in ``horizontal direction" from a position $(x,y)$;

By $Option_{v}(x,y)$ we denote the set of all positions that can be reached in one
move in ``vertical direction" from a position $(x,y)$;

By $Option_{e}(x,y)$ we denote the set of all positions that can be reached in one
move in ``extended diagonal direction" from a position $(x,y)$.

It is obvious that for any position $(x,y)$,

(I) $Option(x,y)=Option_{h}(x,y)\cup Option_{v}(x,y)\cup Option_{e}(x,y)$;

(II) $Option_{h}(x,y)$, $Option_{v}(x,y)$ and $Option_{e}(x,y)$ are pairwise disjoint.


\begin{example}\label{example1} Eq. (\ref{def-NP}) can be used to check whether or not a given game position $(a,b)$ is a $P$-position.
 We consider General Restriction of $(s,t)$-Wythoff's Game under normal play convention and $s=t=2$, where $S_h=S_v=D_1=D_2=Z^{\geq 0}$.
 Then $u=(1,4)$ is a $P$-position.
\end{example}

\begin{proof}
 By $\mathscr{P}$ and $\mathscr{N}$ we denote the sets of all $P$-positions and all $N$-positions, respectively.
 It is obvious that $(0,0)$ is a $P$-position, i.e., $(0,0)\in \mathscr{P}$.

 (1) The positions $(0,1),(0,2),(0,3),(0,4)$ are $N$-positions.
 In fact, fix $m\in\{1,2,3,4\}$ and let $w=(0,m)$, one can move $(0,m)\rightarrow (0,0)$ by
 taking $m$ tokens in the vertical direction. Thus $(0,0)\in Option_v(w)$,
 i.e., $Option_v(w)\cap \mathscr{P}\neq \emptyset$. By Eq. (\ref{def-NP}), $(0,m)$ is an $N$-position.

(2) The position $(1,0)$ is an $N$-position. For $w=(1,0)$, one can move $(1,0)\rightarrow (0,0)$ by
 taking $1$ tokens in the horizontal direction. Thus $(0,0)\in Option_h(w)$,
 i.e., $Option_h(w)\cap \mathscr{P}\neq \emptyset$. By Eq. (\ref{def-NP}), $(1,0)$ is an $N$-position.

 (3) The positions $(1,1),(1,2),(1,3)$ are $N$-positions.
 In fact, fix $m\in\{1,2,3\}$ and let $w=(1,m)$. For $w=(1,m)$, one can move $(1,m)\rightarrow (0,0)$ by
 taking $k=1$ token from the first heap and $\ell=m$ token from the second heap. Note that Eq. (\ref{general-def-constraint}) is true:
 $$|\ell-k|=m-1<1+2=(s-1)\lambda+t, \lambda=k=1.$$

 (4) $Option_e(1,4)=\{(0,1),(0,2),(0,3)\}$. For $w=(1,4)$, one can move $(1,4)\rightarrow (0,m)$ with $1\leq m\leq 3$ by
 taking $k=1$ token from the first heap and $\ell=m$ tokens from the second heap, and $|\ell-k|=m-1<1+2=(s-1)\lambda+t, \lambda=k=1$.

 (5) It is obvious that $Option_h(1,4)=\{(0,4)\}$, $Option_v(1,4)=\{(1,m)|0\leq m\leq 3\}$. Thus
 $$\begin{array}{rcl}
 Option(1,4)&=&Option_{h}(1,4)\cup Option_{v}(1,4)\cup Option_{e}(1,4)\\
 &=&\{(0,4)\}\cup \{(1,m)|0\leq m\leq 3\}\cup \{(0,1),(0,2),(0,3)\}.\end{array}$$
 It follows form (1), (2) and (3) that $Option(1,4)\subseteq \mathscr{N}$. By Eq. (\ref{def-NP}),
 the position $(1,4)$ is a $P$-position.
 \end{proof}

\begin{proposition}\label{Property-NP} (\cite{DuFraen2010}, Characterization of the $P$-positions of an impartial acyclic game)
The sets of $P$- and $N$-positions of any impartial acyclic game (like Wythoff's game)
are uniquely determined by the following two properties:

$\bullet$ Any move from a $P$-position leads to an $N$-position (stability property of the $P$-positions).

$\bullet$ From any $N$-position, there exists a move leading to a $P$-position (absorbing property of the $P$-positions).

\end{proposition}
\begin{proof} See Proposition 1 in \cite{DuFraen2010}. \end{proof}

\section{Odd-Odd  $(s,t)$-Wythoff's Game}

  In this section, we introduce a new \emph{Odd-Odd  $(s,t)$-Wythoff's
 Game} (Denoted by OOW). Let $S_h$, $S_v$, $D_1$ and $D_2$ be subsets
of $Z^{\geq 0}$. Given two parameters $s,t\in Z^{\geq 1}$ and two heaps
of finitely many tokens. One of the heaps is designated as ``first heap"
and the other as ``second heap" throughout this game. By $(x,y)$ we denote a position of present game,
where $x$ and $y$ denote the numbers of tokens in the first and the
second heaps, respectively. Two rules of moves are allowed:

(\emph{Odd-Odd Nim Rule}) A player chooses one heap and takes an
arbitrary \textit{odd} number $k$ of tokens.

(\emph{Odd-Odd More General Wythoff's Rule}) A player takes tokens
from both heaps, \textit{odd} $k>0$ tokens from the first heap,
\textit{odd} $\ell>0$ tokens from the second heap, and
\begin{equation}\label{odd-odd-def-constraint}
0 \leq |\ell-k| \ <(s-1)\lambda+t, \ \ \lambda=\min\{k,\ell\}\in
Z^{\geq 1}.
\end{equation}

Obviously, OOW is equivalent to $S_h=S_v=D_1=D_2=Z^{odd}$
 in General Restriction of $(s,t)$-Wythoff's Game.

By the definition of OOW, the positions $(x,y)$ and $(y,x)$ are
equivalent, i.e., both $(x,y)$ and $(y,x)$ are $P$-positions, or are
$N$-positions. Theorems \ref{OOW-normal} and \ref{OOW-misere} will give the sets of all
$P$-positions of OOW under normal or mis\`{e}re play convention, respectively.
The corresponding winning strategies are also presented.


We define a function $\delta_n$ for $n\in Z^{\geq 0}$:
$$\delta_n=\left\{\begin{array}{ll}
0, &\mbox{if } n \mbox{ is even,}\\
1, &\mbox{if } n \mbox{ is odd.}
\end{array} \right.$$

\begin{theorem}\label{OOW-normal} By $\mathscr{P}_1$ we denote the set of all $P$-positions of OOW. Then for all $s,t\in Z^{\geq 1}$,
$$\mathscr{P}_1=\bigcup\limits_{m=0}^{\infty}\bigcup\limits_{n=0}^{\infty}\{(2m,2n)\}.$$
\end{theorem}

\begin{proof} Let $\mathcal
{M}_{1}=\bigcup\limits_{m=0}^{\infty}\bigcup\limits_{n=0}^{\infty}\{(2m,2n)\}$.
It suffices to show two things:

\textit{Fact A.} No options of a position in $\mathcal {M}_{1}$ can
be in $\mathcal {M}_{1}$.

\textit{Fact B.} Any position not in $\mathcal {M}_{1}$ can land in
a position in $\mathcal {M}_{1}$.

\vspace{0.5cm}
\noindent{\textit{Proof of Fact A}.} Let $(x,y)\in \mathcal
{M}_{1}$ be a position. Suppose that $(x,y)\rightarrow
(x',y')\in\mathcal {M}_{1}$. By the definition of $\mathcal
{M}_{1}$,  $x$, $y$, $x'$ and $y'$ are even.
Thus both $x-x'$ and $y-y'$ are even. This contradicts the rules of moves of OOW.

\vspace{0.5cm}
\noindent{\textit{Proof of Fact B}.} Let $(x,y)\notin
\mathcal{M}_{1}$ be a position. In this case, at least one of $x$
and $y$ is odd, i.e., $(\delta_x,\delta_y)=(0,1)$ or $(1,0)$ or
$(1,1)$. Thus we
can move $(x,y)\rightarrow (x-\delta_x,y-\delta_y)\in \mathcal{M}_{1}$ by taking one token from an odd-size heap.

The proof is completed.\end{proof}

\begin{remark}\label{remark5} Given a game $\Gamma$. Let $\mathcal{M}$ be the set of all $P$-positions
of game $\Gamma$. The following facts are true:

\emph{Fact 1}. No options of a position in $\mathcal{M}$ can be in
$\mathcal{M}$.

\emph{Fact 2}. Any position not in $\mathcal{M}$ can land in a
position in $\mathcal{M}$ by a legal move.

We will determine the sets of all $P$-positions
of the games investigated in this paper, respectively. In all proofs,
the validity of \emph{Fact 1} and \emph{Fact 2} will be proved.
The method of the proofs is the same, though the proofs themselves vary greatly.
\end{remark}

\begin{theorem}\label{OOW-misere} By $\mathscr{P}_2$ we denote the set of all $P$-positions of OOW
under mis\`{e}re play convention. Then for all $s,t\in Z^{\geq 1}$,
$$\mathscr{P}_2=\{(0,2p+1),(2p+1,0)|p\in Z^{\geq0}\}\cup\bigcup\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\{(2m,2n)\}.$$
\end{theorem}

\begin{proof} Let
$$\begin{array}{rcl}
\mathcal {M}_{2}&=&\{(0,2p+1),(2p+1,0)|p\in Z^{\geq0}\}\cup\bigcup\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\{(2m,2n)\},\\
\mathcal{M}'_{2}&=&\{(0,2p+1),(2p+1,0)|p\in Z^{\geq 0}\},\\
\mathcal{M}''_{2}&=&\bigcup\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\{(2m,2n)\}.\end{array}$$

\noindent{\textit{Proof of Fact 1.}} Let $(x,y)$ be a
position in $\mathcal{M}_{2}$. For $(0,2p+1)\in \mathcal{M}'_{2}$,
 $(0,2p+1)\rightarrow (x',y')\in \mathcal{M}''_{2}$ is impossible since $x'>0$;
  $(0,2p+1)\rightarrow (0,2q+1) (q<p)$ is also impossible, since $2(p-q)$ is even.

For $(2m,2n)\in \mathcal{M}''_{2}$, $(2m,2n)\rightarrow (0,2p+1)$
(or $(2p+1,0)$) is impossible, since $2m$ (or $2n$) is even;
$(2m,2n)\rightarrow (2m',2n')\in \mathcal{M}''_{2}$ is also
impossible, since $2(m-m')$ is even,
which contradicts the rules of moves of OOW.

\vspace{0.5cm}
\noindent{\textit{Proof of Fact 2.}} Let $(x,y)$ with $x\leq
y$ be a position not in $\mathcal{M}_{2}$.

If $(x,y)=(0,2v)$ for some $v\in Z^{\geq 1}$, we move
$(x,y)=(0,2v)\rightarrow (0,2v-1)$ by taking one token from the heap
of size $2v$. If $(x,y)=(0,0)$ then next player wins without doing
any thing.


If $(x,y)=(2m,2n+1)$ for some $m,n\in Z^{\geq 1}$ and $n\geq m$, we
move $(2m,2n+1)\rightarrow(2m,2n)$, by taking one token from the
heap of size $2n+1$.


If $(x,y)=(2u+1,w)$ for some $u\in Z^{\geq 0}$ and $w\geq 2u+1$. We
need to consider two subcases:

(i) $u=0$. In this case, $w\geq 1$. If $w$ is odd, we move
$(2u+1,w)=(1,w)\rightarrow (0,w)\in \mathcal{M}'_{2}$; If $w$ is
even, we move $(2u+1,w)=(1,w)\rightarrow (0,w-1)\in
\mathcal{M}'_{2}$, by taking one token from each heap.

(ii) $u>0$. In this case, $w\geq 2u+1\geq 3$. If $w$ is odd, we move
$(2u+1,w)\rightarrow (2u,w-1)\in \mathcal{M}''_{2}$. If $w$ is even,
thus we move $(2u+1,w)\rightarrow (2u,w)\in \mathcal{M}''_{2}$.

The proof is completed.\end{proof}



\section{Even-Even  $(s,t)$-Wythoff's Game}

 In this section, we introduce a new \emph{Even-Even  $(s,t)$-Wythoff's
 Game} (Denoted by EEW). Let $S_h$, $S_v$, $D_1$ and $D_2$ be subsets
of $Z^{\geq 0}$. Given two parameters $s,t\in Z^{\geq 1}$ and two heaps
of finitely many tokens. One of the heaps is designated as ``first heap"
and the other as ``second heap" throughout this game. By $(x,y)$ we denote a position of present game,
where $x$ and $y$ denote the numbers of tokens in the first and the
second heaps, respectively. Two rules of moves are allowed:

(\emph{Even-Even Nim Rule}) A player chooses one heap and takes an
arbitrary \emph{even} number $k>0$ of tokens.

(\emph{Even-Even More General Wythoff's Rule}) A player takes
tokens from both heaps, \emph{even} $k>0$ tokens from the first
heap, \emph{even} $\ell>0$ tokens from the second heap and
\begin{equation}\label{even-even-def-constraint}
0 \leq |\ell-k| \ <(s-1)\lambda+t, \ \ \lambda=\min\{k,\ell\}\in
Z^{\geq 1}.
\end{equation}

Obviously, EEW is equivalent to $S_h=S_v=D_1=D_2=Z^{even}$
 in General Restriction of $(s,t)$-Wythoff's Game.


\begin{remark}\label{Notation 1}
The ``symmetric" notation $\{x,y\}$ for unordered pairs of
non-negative integers is used whenever the positions $(x,y)$ and
$(y,x)$ are equivalent, i.e., both $(x,y)$ and $(y,x)$ are
$P$-positions, or are $N$-positions.
\end{remark}

\begin{example}\label{example2} We consider EEW under normal play convention. Fix two integers $s=t=1$. It is obvious that
$(0,1)$ is a $P$-position, $(1,0)$ is also a $P$-position. Thus we use $\{0,1\}$ to denote two positions $(0,1)$ and $(1,0)$, i.e.,
$\{0,1\}=\{(0,1),(1,0)\}$. Generally, by the definition of EEW, two  positions $(x,y)$ and $(y,x)$ are equivalent, i.e., both $(x,y)$
and $(y,x)$ are $P$-positions, or are $N$-positions. Thus we use $\{x,y\}$ to denote two positions $(x,y)$ and $(y,x)$, i.e.,
$\{x,y\}=\{(x,y),(y,x)\}$.
\end{example}



Theorems \ref{EEW-normal} and \ref{EEW-misere} will give the sets
of all $P$-positions of EEW under normal or mis\`{e}re
play convention, respectively. The corresponding winning strategies are also presented.
Before the main results, we define
two sequences and give some properties in Lemma \ref{lemma 1}.

We define two sequences $A_n$ and $B_n$ for $n\in Z^{\geq 0}$ and all
$s,t\in Z^{\geq 1}$:
\begin{equation}\label{even-even-A-B-def}
\left\{\begin{array}{l}A_n=\mbox{mex}\{A_i,A_i+1,B_i,B_i+1|0\leq i<n\},\\
B_n=sA_n+(t+\delta_t)n.\end{array}\right.
\end{equation}
Tables 1 and 2 list the first few values of $A_n$
and $B_n$ for $s=t=1$ and $s=t=2$, respectively.
\begin{center}
Table 1.\quad The first few values of $A_n$ and $B_n$ for $s=t=1$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 &13 &14\\
\hline $A_{n}$&0&2 &6& 8 & 12 & 16 & 18& 22 & 24& 28 &32 & 34 &38 &42 &44 \\
 \hline $B_{n}$& 0 & 4 & 10 & 14& 20 & 26 & 30 &36& 40 & 46 &52 & 56& 62& 68& 72  \\
 \hline
 \end{tabular}
 \end{center}

 \begin{center}
Table 2.\quad The first few values of $A_n$ and $B_n$ for $s=t=2$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 &11 &12 &13 &14 \\
\hline $A_{n}$&0&2 &4& 8 & 10 & 14& 16 & 18 & 20& 24 &26 &30 &32 &34 &36 \\
 \hline $B_{n}$&0& 6 & 12  & 22 & 28 & 38 & 44 &50& 56 & 66 &72 &82 &88 &94 &100\\
 \hline
 \end{tabular}
\end{center}\vspace{0.5cm}

\begin{lemma}\label{lemma 1} Let $\{A_n\}^{\infty}_{n=0}$ and
$\{B_n\}^{\infty}_{n=0}$ be defined by Eq. (\ref{even-even-A-B-def}). We have the following properties:

(I) Both $A_n$ and $B_n$ are even for $n\geq 0$.

(II) Both $A_{n}$ and $B_n$ are strictly increasing sequences for
$n\geq 0$.

(III) $B_n>A_n+1>A_n$ for $n\geq 1$.

(IV) Let \begin{equation}\label{even-even-A-B}
\left\{\begin{array}{l}A=\bigcup\limits_{n=1}^{\infty} \{A_n\}\cup \bigcup\limits_{n=1}^{\infty}\{A_n+1\},\\
B=\bigcup\limits_{n=1}^{\infty}\{B_n\}\cup \bigcup\limits_{n=1}^{\infty}\{B_n+1\}.\end{array}
\right.\end{equation} Then $A$, $B$ are complementary with respect
to $Z^{\geq2}$, i.e., $A\cup B=Z^{\geq 2}$
and $A\cap B={\emptyset}$.
\end{lemma}

\begin{proof} (I) Note that $t+\delta_t$ is even for $t\in
Z^{\geq 1}$. We proceed by induction on $n$. Obviously, $A_0=B_0=0$,
$A_1=2$ and $B_1=sA_1+(t+\delta_t)$ are even. Suppose $m<n$, both
$A_m$ and $B_m$ are even. We now show that $A_n$ is even, and then
$B_n=sA_n+(t+\delta_t)n$ is even.

Indeed, suppose that $A_n$ is odd. Let $k=A_n$ and
$S=\{A_i,A_i+1,B_i,B_i+1|0\leq i<n\}$. By Eq.
(\ref{even-even-A-B-def}), the fact $k=\mbox{mex}(S)$ implies that $k\notin S$
and $k-1\in S$.

By the hypothesis of induction, $A_i$ and $B_i$ are even for all
$i\in\{0,1,\cdots,n-1\}$. Note that the facts $k-1\in S$ and $k-1$ is even imply that
$k-1\neq A_i+1$ or $B_i+1$. If there exists an integer
$i_0\in\{0,1,\cdots,n-1\}$ such that $k-1=A_{i_0}$ or $B_{i_0}$,
then $k\in S$. This contradicts $k\notin S$.

(II) By the definition of $A_{n}$ and \emph{mex} property, $A_{n}$
is strictly increasing sequence. Also $B_{n}$ is strictly increasing
sequence. Indeed, for $m>n$,
$$B_{m}-B_{n}=s(A_{m}-A_{n})+(t+\delta_t)(m-n)>0.$$

(III) Note that $t+\delta_t\geq 2$ for $t\in Z^{\geq 1}$. By Eq.
(\ref{even-even-A-B-def}), we have $B_n=sA_n+(t+\delta_t)n\geq
A_n+2n>A_n+1>A_n$, for $n\in Z^{\geq 1}$.

(IV) In fact, $A\cup B=Z^{\geq 2 }$ follows from the \textit{mex}
property and $A_0=0$, $B_0=0$, $A_1=2$. Suppose $A\cap B\neq
{\emptyset}$. It follows from (I) that $A_m+1\neq B_n$ and $A_m\neq
B_n+1$, thus the only possibility is $A_m=B_n$ for two integers
$m,n\in Z^{\geq 1}$. If $m>n$, then $A_m$ is \emph{mex} of a set
containing $B_n=A_m$, a contradiction. If $m\leq n$, then by (II) we
have
$B_n=sA_n+(t+\delta_t)n\geq sA_m+(t+\delta_t)m>A_m$, another contradiction.

The proof is completed.\end{proof}

\begin{theorem}\label{EEW-normal} By $\mathscr{P}_3$ we denote the set of
all $P$-positions of EEW under
normal play convention. Then for all $s,t\in Z^{\geq 1}$,
$$\mathscr{P}_{3}=\bigcup\limits_{i=0}^{\infty}\left\{
\begin{array}{ll}
\{A_i,B_i\},\{A_i,B_i+1\},\\
\{A_i+1,B_i\},\{A_i+1,B_i+1\}
\end{array}
\right\},$$
where $A_n$ and $B_n$ are defined by Eq. (\ref{even-even-A-B-def}).
\end{theorem}

\begin{proof} Before we give the proof of Theorem \ref{EEW-normal}, Tables 3 and 4 list the first few values of $A_n$ and $B_n$,
which show us how to determine $\mathscr{P}_3$ by using Theorem \ref{EEW-normal}:
\begin{center}
Table 3.\quad The first few values of $A_n$ and $B_n$ for $s=t=1$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 &13 &14\\
\hline $A_{n}$&0&2 &6& 8 & 12 & 16 & 18& 22 & 24& 28 &32 & 34 &38 &42 &44 \\
\hline $A_{n}+1$&1&3 &7& 9 & 13 & 17 & 19& 23 & 25& 29 &33 & 35 &39 &43 &45 \\
 \hline $B_{n}$& 0 & 4 & 10 & 14& 20 & 26 & 30 &36& 40 & 46 &52 & 56& 62& 68& 72  \\
 \hline $B_{n}+1$& 1 & 5 & 11 & 15& 21 & 27 & 31 &37& 41 & 47 &53 & 57& 63& 69& 73  \\
 \hline
 \end{tabular}
 \end{center}
For $s=t=1$, it follows from Table 3 that
$$\mathscr{P}_3=\left\{\begin{array}{l}(0,0),(0,1),(1,0),(1,1);\\
(2,4),(2,5),(3,4),(3,5);(4,2),(5,2),(4,3),(5,3);\\
(6,10),(6,11),(7,10),(7,11);(10,6),(11,6),(10,7),(11,7);\cdots\end{array}\right\}$$

 \begin{center}
Table 4.\quad The first few values of $A_n$ and $B_n$ for $s=t=2$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 &11 &12 &13 &14 \\
\hline $A_{n}$&0&2 &4& 8 & 10 & 14& 16 & 18 & 20& 24 &26 &30 &32 &34 &36 \\
\hline $A_{n}+1$&1&3 &5& 9 & 11 & 15& 17 & 19 & 21& 25 &27 &31 &33 &35 &37 \\
 \hline $B_{n}$&0& 6 & 12  & 22 & 28 & 38 & 44 &50& 56 & 66 &72 &82 &88 &94 &100\\
  \hline $B_{n}+1$&1& 7 & 13  & 23 & 29 & 39 & 45 &51& 57 & 67 &73 &83 &89 &95 &101\\
  \hline
 \end{tabular}
\end{center}
For $s=t=2$, it follows from Table 4 that
$$\mathscr{P}_3=\left\{\begin{array}{l}(0,0),(0,1),(1,0),(1,1);\\
(2,6),(2,7),(3,6),(3,7);(6,2),(7,2),(6,3),(7,3);\\
(4,12),(4,13),(5,12),(5,13);(12,4),(13,4),(12,5),(13,5);\cdots\end{array}\right\}$$


We now give the proof of Theorem \ref{EEW-normal}. Let
$$\mathcal{M}_{3}=\bigcup\limits_{i=0}^{\infty}\left\{
\begin{array}{ll}
\{A_i,B_i\},\{A_i,B_i+1\},\\
\{A_i+1,B_i\},\{A_i+1,B_i+1\}
\end{array}
\right\}.$$

\noindent{\textit{Proof of Fact 1.}} Let $(x,y)$ with $x\leq
y$ be a position in $\mathcal{M}_{3}$. It follows from (III) of Lemma \ref{lemma 1}
that there exists an integer $n\geq 0$ such that $x=A_n$ or $A_n+1$, and
$y=B_n$ or $B_n+1$.

Suppose that $(x,y)\rightarrow (x',y)\in \mathcal{M}_{3}$ with
$x'=A_m$ or $A_m+1$, by Even-Even Nim Rule. Then the fact $x'\leq
x-2<A_n$ implies that $m<n$. Thus $(x',y)\notin \mathcal{M}_{3}$, a
contradiction. Suppose that $(x,y)\rightarrow (x,y')\in
\mathcal{M}_{3}$ with $y'=B_m$ or $B_m+1$, by Even-Even Nim
Rule. Then $y'\leq y-2<B_n$ implies that $m<n$. Thus $(x,y')\notin
\mathcal{M}_{3}$, another contradiction.

Suppose that $(x,y)\rightarrow (x',y')\in \mathcal{M}_{3}$ by
Even-Even More General Wythoff's Rule. Then $x-x'>0$ is
even and $y-y'>0$ is even. It follows from (I) and (III) of Lemma \ref{lemma 1} that
$k=x-x'=A_n-A_m$, $\ell=y-y'=B_n-B_m$ and $m<n$. Thus
$$0<k\leq \ell=s(A_n-A_m)+(t+\delta_t)(m-n)\geq sk+t,$$
 which contradicts Eq. (\ref{even-even-def-constraint}).


\vspace{0.5cm}
\noindent{\textit{Proof of Fact 2.}} Let $(x,y)$ be a
position not in $\mathcal{M}_{3}$. Without loss of generality,
assume that $0\leq x\leq y$.

If $x=0$ or $1$, we move $(x,y)\rightarrow (x,\delta_y)\in
\mathcal{M}_{3}$. This move is legal, since $y-\delta_y$ is even and
$(x,y)\notin \mathcal{M}_{3}$ implies that $y>1\geq \delta_y$.

If $x\geq 2$ then the integer $x$ appears exactly once in exactly
one of $A$ and $B$, since $A$ and $B$ are complementary with respect
to $Z^{\geq 2}$ (Lemma \ref{lemma 1}, (IV)). Therefore, we have one of the
following two cases: (i) $x=B_n$ or $x=B_n+1$, (ii) $x=A_n$ or
$x=A_n+1$, for some $n\geq 1$.

\textit{Case (i)}: $x=B_n$ or $x=B_n+1$, $n\geq 1$. We
move $$(x,y)\rightarrow (x,A_n+\delta_y)\in \mathcal{M}_{3},$$ i.e.,
we take $y-A_n-\delta_y$ tokens from the heap of $y$ tokens. It
follows from (I) and (III) of Lemma \ref{lemma 1} that $$y\geq x\geq B_n>A_n+1\geq
A_n+\delta_y$$ and $y-A_n-\delta_y$ is even. Thus the above move is
legal.

\textit{Case (ii)}: $x=A_n$ or $x=A_n+1$, $n\geq 1$. In
this case, we have $y>B_n+1$ or $x\leq y<B_n$.

(1) $y>B_n+1$. We move
$$(x,y)\rightarrow (x,B_n+\delta_y)\in \mathcal{M}_{3},$$
i.e., we take $y-B_n-\delta_y$ tokens from the heap of $y$ tokens.
This ia a legal move, since $y>B_n+1\geq B_n+\delta_y$ and
$y-B_n-\delta_y$ is even.

(2) $x\leq y<B_n$. We distinguish the following two subcases: $x\leq
y<sA_n+t+\delta_t$ or $sA_n+t+\delta_t\leq y<B_n$:

(2.1) $x\leq y<sA_n+t+\delta_t$. We move
$$(x,y)\rightarrow (x-A_n,\delta_y)\in \mathcal{M}_{3},$$
since $x-A_n=0$ or $1$, and $\delta_y=0$ or $1$. This move is legal:

1) $k=A_n$ is even;

2) $\ell=y-\delta_y$ is even;

3) $y\geq x$ implies that $\ell=y-\delta_y\geq A_n=k$. Note that
$y<sA_n+t+\delta_t$ and $sA_n+t+\delta_t$ is even, so $y\leq
sA_n+t+\delta_t-2+\delta_y$. Hence,
$$\begin{array}{rcl}|\ell-k|&=&y-\delta_y-A_n\\
&\leq &(s-1)A_n+t+\delta_t-2\\
&<&(s-1)A_n+t.\end{array}$$


(2.2) $sA_n+t+\delta_t\leq y<B_n$. Put
$m=\lfloor\frac{y-sA_n-\delta_y}{t+\delta_t}\rfloor.$ We move
$$(x,y)\rightarrow (x-A_n+A_m,B_m+\delta_y)\in \mathcal{M}_{3},$$
since $x-A_n=0$ or $1$, and $\delta_y=0$ or $1$. This move is legal:

(a) $k=A_n-A_m>0$ and $k$ is even. Firstly, we show that $0\leq
m<n$. Note that $y-sA_n\geq t+\delta_t\geq 2> \delta_y,$ so
$\frac{y-sA_n-\delta_y}{t+\delta_t}>0$. Thus
$m=\lfloor\frac{y-sA_n-\delta_y}{t+\delta_t}\rfloor\geq 0$. On the
other hand, the facts $y<B_n$ and $B_n$ is even imply that $y-\delta_y<B_n$.
Thus $$y-sA_n-\delta_y<B_n-sA_n=(t+\delta_t)n,$$ and
$$m=\lfloor\frac{y-sA_n-\delta_y}{t+\delta_t}\rfloor \leq
\frac{y-sA_n-\delta_y}{t+\delta_t}<n.$$ It follows from (I) and (II) of
Lemma \ref{lemma 1} that $k=A_n-A_m>0$ and $k$ is even.


(b) $\ell=y-B_m-\delta_y>0$ and $\ell$ is even. By the definition of
$m$, we have $m\leq \frac{y-sA_n-\delta_y}{t+\delta_t}$ and
 $$\begin{array}{lcl}y&\geq &(t+\delta_t)m+sA_n+\delta_y\\
&=&B_m+\delta_y+s(A_n-A_m)\\
&>&B_m+\delta_y.\end{array}$$ Thus $\ell=y-B_m-\delta_y>0$ and
$\ell$ is even.

(c) $|\ell-k|<(s-1)k+t$. By the definition of $m$, we have
$m>\frac{y-sA_n-\delta_y}{t+\delta_t}-1$ and
$y<(t+\delta_t)(m+1)+sA_n+\delta_y.$ Thus $$\begin{array}{rcl}\ell&=&y-B_m-\delta_y\\
&<&(t+\delta_t)(m+1)+sA_n-sA_m-(t+\delta_t)m\\
&=&s(A_n-A_m)+t+\delta_t.\end{array}$$ We note that $y-B_m-\delta_y$
and $s(A_n-A_m)+t+\delta_t$ are even,
so $$\begin{array}{rcl}\ell&=&y-B_m-\delta_y\\
&\leq& s(A_n-A_m)+t+\delta_t-2\\
&<&s(A_n-A_m)+t;\end{array}$$ On the other hand, $y-B_m-\delta_y\geq
s(A_n-A_m)\geq A_n-A_m$ by virtue of (b).
Therefore, $|\ell-k|<(s-1)k+t$.

The proof is completed.\end{proof}

\begin{theorem}\label{EEW-misere} By $\mathscr{P}_4$ we denote the set of all
$P$-positions of EEW under mis\`{e}re
play convention. Then for all $s,t\in Z^{\geq 1}$,

$$\mathscr{P}_{4}=\bigcup\limits_{i=0}^{\infty}\left\{
\begin{array}{ll}
\{E_i,H_i\},\{E_i,H_i+1\},\\
\{E_i+1,H_i\},\{E_i+1,H_i+1\}
\end{array}
\right\},$$ where $E_{n}$ and $H_{n}$ are given by the following two
cases:

(A) If $s\neq 1$ or $t>2$, then for $n\geq 0$,
\begin{equation}\label{even-even-Misere-2}
\left\{\begin{array}{l}E_n=\mbox{mex}\{E_i,E_i+1,H_i,H_i+1|0\leq i<n\},\\
H_n=sE_n+(t+\delta_t)n+2.\end{array}\right.
\end{equation}

(B) If $s=1$ and $t\in \{1,2\}$, then $E_{0}=H_{0}=4$ and for $n\geq 1$,
\begin{equation}\label{even-even-Misere-1}
\left\{\begin{array}{l}E_n=\mbox{mex}\{E_i,E_i+1,H_i,H_i+1|0\leq i<n\},\\
H_n=E_n+2n.\end{array}\right.
\end{equation}
\end{theorem}

\begin{proof} Before we give the proof of Theorem \ref{EEW-misere}, Tables 5 and 6 list the first few values of $E_n$
and $H_n$, which show us how to determine $\mathscr{P}_4$ by using Theorem \ref{EEW-misere}:

\begin{center}
Table 5.\quad The first few values of $E_n$ and $H_n$ for $s=t=2$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 &11 &12 &13 &14\\
\hline $E_{n}$&0&4 &6& 8 & 10 & 14& 16 & 20 & 22& 26 &28 &32 &34 &36 &38\\
 \hline $H_{n}$&2&12 & 18  & 24 & 30 & 40 & 46 &56& 62 & 72 &78 &88 &94 &100 &106\\
 \hline
 \end{tabular}
 \end{center}
For $s=t=2$, it follows from Table 5 that
$$\mathscr{P}_4=\left\{\begin{array}{l}(0,2),(0,3),(1,2),(1,3);(2,0),(3,0),(2,1),(3,1);\\
(4,12),(4,13),(5,12),(5,13);(12,4),(13,4),(12,5),(13,5);\\
(6,18),(6,19),(7,18),(7,19);(18,6),(19,6),(18,7),(19,7);\cdots\end{array}\right\}$$

\begin{center}
Table 6.\quad The first few values of $E_n$ and $H_n$ for $s=t=1$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14\\
\hline $E_{n}$&4&0 &6& 8 & 12 & 16 & 18& 22 & 24& 28 & 32 &34 &38 &42 &44\\
 \hline $H_{n}$& 4& 2 & 10 & 14& 20 & 26 & 30 &36& 40 & 46 &52 &56 &62 &68 &72 \\
 \hline
 \end{tabular}
\end{center}
For $s=t=1$, it follows from Table 6 that
$$\mathscr{P}_4=\left\{\begin{array}{l}(4,4),(4,5),(5,4),(5,5);\\
(0,2),(0,3),(1,2),(1,3);(2,0),(3,0),(2,1),(3,1);\\
(6,10),(6,11),(7,10),(7,11);(10,6),(11,6),(10,7),(11,7);\cdots\end{array}\right\}$$


We now give the proof of Theorem \ref{EEW-misere}. Let $$\mathcal {M}_{4}=
\bigcup\limits_{i=0}^{\infty}\left\{
\begin{array}{ll}
\{E_i,H_i\},\{E_i,H_i+1\},\\
\{E_i+1,H_i\},\{E_i+1,H_i+1\}
\end{array}
\right\},$$

and $E= \bigcup\limits_{n=0}^{\infty} \{E_{n}\}\cup \bigcup\limits_{n=0}^{\infty}\{E_{n}+1\}$, $H=
\bigcup\limits_{n=0}^{\infty} \{H_{n}\}\cup \bigcup\limits_{n=0}^{\infty}\{H_{n}+1\}$. Then we claim that:


{\bf Fact A.1} If $s\neq 1$ or $t>2$, then both $E_{n}$ and $H_{n}$ are
even for $n\in Z^{\geq 0}$, and both $E_{n}$ and $H_{n}$ are
strictly increasing sequences for $n\geq 0$. The proofs are similar
to ones of (I) and (II) of Lemma \ref{lemma 1}.

{\bf Fact A.2} If $s\neq 1$ or $t>2$, then $E\cup H=Z^{\geq 0}$ and
$E\cap H=\emptyset$. In fact, $E\cup H=Z^{\geq 0}$ follows from the
definition of \emph{mex}. Suppose $E\cap H\neq {\emptyset}$. It follows from 
Fact A.1 that $E_m+1=H_n$ and $E_m=H_n+1$ are impossible, thus there
exist two integers $m,n\in Z^{\geq 0}$ such that $E_m=H_n$. If $m>n$
then $E_{m}=\mbox{mex}\{E_{i},E_{i}+1,H_{i},H_{i}+1|0\leq i<m\}$,
which contradicts $E_{m}=H_{n}$; If $m\leq n$ then
$$\begin{array}{rcl}
H_{n}=sE_{n}+(t+\delta_t)n+2\geq sE_{m}+(t+\delta_t)m+2
>E_{m},\end{array}$$ which also contradicts $E_{m}=H_{n}$.

{\bf Fact B.1} If $s=1$ and $t\in \{1,2\}$, then both $E_{n}$ and $H_{n}$
are even for $n\in Z^{\geq 0}$, both $E_{n}$ and $H_{n}$ are
strictly increasing sequences for $n\geq 1$. Indeed,
$$E_{1}=0<6=E_{2}<8=E_{3}<12=E_{4}<16=E_{5}<\cdots$$ and
$$H_{1}=2<10=H_{2}<14=H_{3}<20=E_{4}<26=E_{5}<\cdots.$$

{\bf Fact B.2} If $s=1$ and $t\in \{1,2\}$, then $E\cup H=Z^{\geq 0}$ and
$E\cap H=\{4\}$. Its proof is similar to those of Fact A.2.

\vspace{0.5cm}
\noindent{\textit{Proof of Fact 1.}} Let $(x,y)$ with $x\leq
y$ be a position in $\mathcal {M}_{4}$. By the definition of
$\mathcal {M}_{4}$, there exists an integer $n\in Z^{\geq 0}$ such
that $x=E_n$ or $E_n+1$, and $y=H_n$ or $H_n+1$.

Suppose that $(x,y)\rightarrow (x',y)\in \mathcal {M}_{4}$ with
$x'=E_m$ or $E_m+1$, by Even-Even Nim Rule. Then the fact $x'\leq
x-2<E_n$ implies that $m<n$. Thus $(x',y)\notin \mathcal {M}_{4}$, a
contradiction. Similarly, $(x,y)\rightarrow (x,y')\in \mathcal
{M}_{4}$ is also impossible.


Suppose that $(x,y)\rightarrow (x',y')\in \mathcal {M}_{4}$ by
Even-Even More General Wythoff's Rule. Then $x-x'>0$ is
even and $y-y'>0$ is even. It follows from Facts A.1 and B.1 that
$k=x-x'=E_n-E_m$, $\ell=y-y'=H_n-H_m$ and $m<n$. Thus
$$0<k\leq
\ell=s(E_n-E_m)+(t+\delta_t)(m-n)\geq sk+t,$$ which contradicts Eq.
(\ref{even-even-def-constraint}).

\vspace{0.5cm}
\noindent{\textit{Proof of Fact 2.}} Let $(x,y)$ be a
position not in $\mathcal {M}_{4}$. Without loss of generality,
assume that $0\leq x\leq y$.
 By Facts A.2 and B.2, we have one
of the following two cases: (i) $x=H_n$ or $x=H_n+1$, (ii) $x=E_n$
or $x=E_n+1$, for some $n\geq 0$.

\textit{Case (i)}: $x=H_n$ or $x=H_n+1$, $n\geq 0$. In
this case, the fact $(x,y)\notin \mathcal {M}_{4}$ implies that $y>E_n+1$.
In fact, if $n=0$, then $y\geq x\geq H_0\geq E_0$, thus $y>E_0+1$;
For $n\geq 1$, if $y\leq E_n+1$, then it follows from Eqs (\ref{even-even-Misere-2}) and (\ref{even-even-Misere-1}) that
$$x\geq H_n\geq E_n+2>E_n+1\geq y,$$ a contradiction.

We move $$(x,y)\rightarrow (x,E_n+\delta_{y})\in \mathcal {M}_{4},$$
i.e., we take $y-E_n-\delta_{y}$ tokens from the heap of $y$ tokens.
Note that $y-E_n-\delta_{y}>1-\delta_{y}\geq 0$ and $y-E_n-\delta_{y}$ is
even, thus the above move is legal.

\textit{Case (ii)}: $x=E_n$ or $x=E_n+1$, $n\geq 0$. The fact $(x,y)\notin \mathcal {M}_{4}$ implies that
$y>H_n+1$ or $x\leq y<H_n$.

(1) $y>H_n+1$. We move $$(x,y)\rightarrow
(x,H_n+\delta_y)\in \mathcal {M}_{4},$$ i.e., we take
$y-H_n-\delta_y$ tokens from the heap of $y$ tokens. This ia a legal
move, since $y-H_n-\delta_y>1-\delta_{y}\geq 0$ and $y-H_n-\delta_y$ is even.

(2) $x\leq y<H_n$. We need to consider the situations
for (A) and for (B), respectively.

{\bf(2-A)} $s\neq 1$ or $t>2$. If $n=0$, then $0\leq x\leq y<2$ and hence $(x,y)=(0,0)$ or
$(0,1)$ or $(1,1)$. The next player wins without doing anything. It
remains to consider $n\geq 1$. We proceed by
distinguishing the following two subcases: (2-A-1) $x\leq
y<sE_n+t+\delta_t+2$; (2-A-2) $sE_n+t+\delta_t+2\leq y<H_n$.

(2-A-1) $x\leq y<sE_n+t+\delta_t+2$. In this subcase, we move
$$(x,y)\rightarrow (x-E_{n},2+\delta_{y})\in \mathcal {M}_{4},$$
since $x-E_n\in \{0,1\}=\{E_0,E_0+1\}$ and $2+\delta_{y}\in \{H_0,H_0+1\}$. Note that $sE_n+t+\delta_t$ is
even, so $y<sE_n+t+\delta_t+1+\delta_{y}$. This move is legal:

1) $k=E_n>0$ is even;

2) $\ell=y-2-\delta_{y}$ is even. Note that $y\geq x\geq E_n$ and
$E_n$ is even, thus $y-\delta_y\geq E_n$ and
$\ell=y-2-\delta_{y}\geq E_n-2\geq E_1-2>0;$

3) It is easy to see that

$$\begin{array}{rcl}
|\ell-k|&=&|y-2-\delta_y-E_n|\\
&<&(s-1)E_n+t+\delta_t-1\\
&\leq& (s-1)E_n+t.\end{array}$$

(2-A-2) $sE_n+t+\delta_t+2\leq y<H_n$. Put
\begin{equation}\label{even-even-def-m}
m=\lfloor\frac{y-sE_n-2-\delta_y}{t+\delta_t}\rfloor.\end{equation}
We move
\begin{equation}\label{even-even-move}
(x,y)\rightarrow (x-E_n+E_m,H_m+\delta_y)\in \mathcal {M}_{4},
\end{equation}
since $x-E_n+E_m\in \{E_m,E_m+1\}$ and $H_m+\delta_{y}\in \{H_m,H_m+1\}$. This move is legal:

(A-a) \emph{$k=E_n-E_m>0$ and $k$ is even}. Firstly, we show that $0\leq
m<n$. Note that $$y-sE_n-2\geq t+\delta_t\geq 2> \delta_y,$$ so
$\frac{y-sE_n-2-\delta_y}{t+\delta_t}>0$. Thus
$m=\lfloor\frac{y-sE_n-2-\delta_y}{t+\delta_t}\rfloor\geq 0$; On the
other hand, the fact $y<H_n$ implies that $y-\delta_y<H_n$. Thus
$$y-sE_n-2-\delta_y<H_n-sE_n=(t+\delta_t)n,$$ and
$$m=\lfloor\frac{y-sE_n-2-\delta_y}{t+\delta_t}\rfloor \leq
\frac{y-sE_n-2-\delta_y}{t+\delta_t}<n.$$ By Facts A.1 and A.2,
$k=E_n-E_m>0$ and $k$ is even.

(A-b) \emph{$\ell=y-H_m-\delta_y>0$ and $\ell$ is even}. By the definition
of $m$, we have $m\leq \frac{y-sE_n-2-\delta_y}{t+\delta_t}$ and
 $$\begin{array}{lcl}y&\geq &(t+\delta_t)m+sE_n+2+\delta_y\\
&=&H_m+\delta_y+s(E_n-E_m)\\
&>&H_m+\delta_y.\end{array}$$ Thus $\ell=y-H_m-\delta_y>0$ and
$\ell$ is even.

(A-c) \emph{$0\leq \ell-k<(s-1)k+t$}. By the definition of $m$, we have
$$m>\frac{y-sE_n-2-\delta_y}{t+\delta_t}-1,$$ i.e.,
$$y<(t+\delta_t)(m+1)+sE_n+2+\delta_y.$$
By Eq. (\ref{even-even-Misere-2}), we have
$H_m=sE_m+(t+\delta_t)m+2.$
Thus $$\begin{array}{rcl}\ell&=&y-H_m-\delta_y\\
&<&(t+\delta_t)(m+1)+sE_n-sE_m-(t+\delta_t)m\\
&=&s(E_n-E_m)+t+\delta_t.\end{array}$$
Note that $y-H_m-\delta_y$
and $s(E_n-E_m)+t+\delta_t$ are even, so
$$\begin{array}{rcl}y-H_m-\delta_y\leq s(E_n-E_m)+t+\delta_t-2<s(E_n-E_m)+t;\end{array}$$
On the other hand,
$$\begin{array}{rcl}\ell&=&y-H_m-\delta_y\geq s(E_n-E_m)\geq E_n-E_m=k\end{array}$$
by virtue of (A-b).
Therefore, $0\leq \ell-k<(s-1)k+t$.



\vspace{0.5cm}
{\bf(2-B)} $s=1$ and $t\in \{1,2\}$. Note that $x\leq y<H_n$. In this case, $n=0$ is impossible; If $n=1$
then $0=E_1\leq x\leq y<H_1=2$, i.e., $(x,y)=(0,0)$ or $(0,1)$, or
$(1,1)$. The next player wins without doing anything. It remains to
consider $n\geq 2$:


Put
\begin{equation}\label{even-even-def-m-B}
m=\lfloor\frac{y-E_n-\delta_y}{2}\rfloor.
\end{equation}
 We move
\begin{equation}\label{even-even-move-B}
(x,y)\rightarrow (x-E_n+E_m,H_m+\delta_y)\in \mathcal {M}_{4},
\end{equation}
 since $x-E_n=0$ or $1$, and $\delta_y=0$ or $1$. This move is legal:

(B-a) \emph{$k=E_n-E_m>0$ and $k$ is even}. Firstly, we show that $0\leq
m<n$. Note that if $y$ is even, we have $y\geq x \geq
E_n=E_n+\delta_y$; if $y$ is odd, we have $y\geq
E_n+1=E_n+\delta_y$. Thus $y\geq E_n+ \delta_y,$ i.e.,
$m=\lfloor\frac{y-E_n-\delta_y}{2}\rfloor\geq 0$; On the other hand,
the fact $y<H_n$ implies that $y-\delta_y<H_n$. Thus
$$y-E_n-\delta_y<H_n-E_n=2n,$$ and
$$m=\lfloor\frac{y-E_n-\delta_y}{2}\rfloor \leq
\frac{y-E_n-\delta_y}{2}<n.$$ By Fact B.1, $k=E_n-E_m>0$ and $k$ is
even.

(B-b) \emph{$\ell=y-H_m-\delta_y>0$ and $\ell$ is even}. By the definition
of $m$, we have $m\leq \frac{y-E_n-\delta_y}{2}$. It follows from Eq.
(\ref{even-even-Misere-1}) and $E_{0}=H_{0}=4$ that
\begin{equation}\label{even-even-def-Hn}
H_n=E_n+2n\mbox{ for }n\geq 0.
\end{equation}
Thus
 $$\begin{array}{lcl}y&\geq &2m+E_n+\delta_y\\
&=&H_m+\delta_y+E_n-E_m\\
&>&H_m+\delta_y.\end{array}$$
Hence $\ell=y-H_m-\delta_y>0$ and $\ell$ is even.

(B-c) $0\leq \ell-k<t$. By the definition of $m$, we have
$m>\frac{y-E_n-\delta_y}{2}-1$, i.e., $y<2(m+1)+E_n+\delta_y.$ By
Eq. (\ref{even-even-def-Hn}), we have
 $$\begin{array}{rcl}\ell&=&y-H_m-\delta_y\\
&<&2(m+1)+E_n-E_m-2m\\
&=&E_n-E_m+2.\end{array}$$
Note that $y-H_m-\delta_y$ and
$E_n-E_m+2$ are even, so $\ell=y-H_m-\delta_y\leq
E_n-E_m<E_n-E_m+t$; On the other hand, $\ell=y-H_m-\delta_y\geq
E_n-E_m=k$ by virtue of (B-b).
Therefore, $0\leq \ell-k<t$.

The proof is completed. \end{proof}



\section{Odd-Even  $(s,t)$-Wythoff's Game}

 In this section, we introduce a new \emph{Odd-Even $(s,t)$-Wythoff's
 Game} (denoted by OEW). Let $S_h$, $S_v$, $D_1$ and $D_2$ be subsets
of $Z^{\geq 0}$. Given two parameters $s,t\in Z^{\geq 1}$ and two heaps
of finitely many tokens. One of the heaps is designated as ``first heap"
and the other as ``second heap" throughout the game. By $(x,y)$ we denote a position of present game,
where $x$ and $y$ denote the numbers of tokens in the first and the
second heaps, respectively. Two rules of moves are allowed:

(\emph{Odd-Even Nim Rule}) A player chooses the first heap and
takes \textit{odd} $k>0$ tokens, or chooses the second heap and
takes \textit{even} $\ell>0$ tokens.

(\emph{Odd-Even More General Wythoff's Rule}) A player takes
tokens from both heaps, \textit{odd} $k>0$ tokens from the first
heap, \textit{even} $\ell>0$ tokens from the second heap, and
\begin{equation}\label{odd-even-def-constraint}
0 \leq |\ell-k| \ <(s-1)\lambda+t, \ \ \lambda=\min\{k,\ell\}\in
Z^{\geq 1}.
\end{equation}


Obviously, OEW is equivalent to $S_h=D_1=Z^{odd}$ and
$S_v=D_2=Z^{even}$ in General Restriction of $(s,t)$-Wythoff's Game. Therefore OEW
 is a restricted version of $(s,t)$-Wythoff's game.

\begin{remark}\label{remark6} OEW has no ``symmetry", i.e., two positions $(x,y)$ and
$(y,x)$ maybe not equivalent (see Remark \ref{Notation 1}). As an example, we consider OEW under normal play convention.
 Obviously, $(0,0)$ is a $P$-position. The position $(0,1)$ is a $P$-position,
 as the only possible move is by taking $1$ token from the second heap. But this is not a legal move.
 The position $(1,0)$ is an $N$-position as one can move $(1,0)$ to $(0,0)$ by taking $1$ tokens from the first heap.
 Thus two positions $(0,1)$ and $(1,0)$ are not equivalent.

 The position $(3,8)$ is an option of position $(10,8)$, as one can move $(10,8)$ to $(3,8)$ by taking $7$ tokens from the first heap.
But the position $(8,3)$ is not an option of position $(8,10)$,
as the move of taking $7$ tokens from the second heap is not a legal move.
\end{remark}

\begin{remark}\label{remark7} In OEW, two parameters $s\geq 1$ and $t\geq 1$ are positive integers.
 If $s=t=1$, then  Eq. (\ref{odd-even-def-constraint}) can not hold:
 $$|\ell-k|\geq 1=(s-1)\lambda+t,\lambda=\min\{k,\ell\}\in
Z^{\geq 1},$$
 i.e., Odd-Even More General Wythoff's Rule is invalid.
For ($s=1$ and $t>1$) or ($s>1$ and $t\geq 1$),
Odd-Even More General Wythoff's Rule is valid. Therefore, we will
give the results on $s=t=1$ and $s+t>2$, respectively. All
$P$-positions of OEW under normal or mis\`{e}re play convention are
given, and the corresponding winning strategies are also presented (Theorems \ref{OEW-normal-s=t=1}, \ref{OEW-misere-s=t=1}, \ref{OEW-normal-s+t>2}
and \ref{OEW-misere-s+t>2}).
\end{remark}


\subsection{All $P$-positions of  OEW: $s=t=1$}

\begin{theorem}\label{OEW-normal-s=t=1} Given two parameters $s=t=1$. By $\mathscr{P}_{5}$ we denote the set of all
$P$-positions of OEW under normal play convention. Then
$$\mathscr{P}_{5}=\bigcup\limits_{n=0}^{\infty}\{(2n,0), (2n,1), (2n+1,2), (2n+1,3)\}.$$
\end{theorem}

\begin{proof} Let $$\mathscr{W}=\bigcup\limits_{n=0}^{\infty}\{(2n,0), (2n,1), (2n+1,2),
(2n+1,3)\}.$$

\noindent{\textit{Proof of Fact 1.}} Let $(a,b)$ and
$(a',b')$ are two distinct positions of $\mathscr{W}$. It is easy to see that there exists no legal move such that
$(a,b)\rightarrow (a',b')$ or $(a',b')\rightarrow
(a,b)$.

\vspace{0.5cm}
\noindent{\textit{Proof of Fact 2.}} Let $(a,b)$ be a
position not in $\mathscr{W}$. It suffices to show that there
exists a legal move such that $(a,b)\rightarrow (a',b')\in \mathscr{W}$.


(2.1) $a=2n$ for some integer $n \in Z^{\geq 0}$. In this case,
the fact $(a,b)\notin \mathscr{W}$ implies that $b\geq 2$. We move
$$(a,b)\rightarrow (2n,\delta_b)\in \mathscr{W}.$$

(2.2)  $a=2n+1$ for some integer $n \in Z^{\geq 0}$. In this case,
the fact $(a,b)\notin \mathscr{W}$ implies that $b=0$, $b=1$ or $b\geq
4$:

$\bullet$ $b\in\{0,1\}$. We move $(a,b)=(2n+1,b)\rightarrow
(2n,b)\in \mathscr{W}$.

$\bullet$ $b\geq 4$. We move $(a,b)=(2n+1,b)\rightarrow (2n+1,2+\delta_b)\in \mathscr{W}$.

The proof is completed. \end{proof}

\begin{theorem}\label{OEW-misere-s=t=1} Given two parameters $s=t=1$. By $\mathscr{P}'_{5}$ we denote the set of all
$P$-positions of OEW under mis\`{e}re play convention. Then
$$\mathscr{P}'_{5}=\bigcup\limits_{n=0}^{\infty}\{(2n,2), (2n,3), (2n+1,0), (2n+1,1)\}.$$
\end{theorem}

\begin{proof} Let
$$\mathscr{M}=\bigcup\limits_{n=0}^{\infty}\{(2n,2), (2n,3),
(2n+1,0), (2n+1,1)\}.$$

\noindent{P\textit{roof of Fact 1.}} Let $(a,b)$ and
$(a',b')$ are two distinct positions of $\mathscr{M}$. It is easy to see that there exists no legal move such that
$(a,b)\rightarrow (a',b')$ or $(a',b')\rightarrow (a,b)$.

\vspace{0.5cm}
\noindent{\textit{Proof of Fact 2.}} Let $(a,b)$ be a
position not in $\mathscr{M}$. It suffices to show that there
exists a legal move such that $(a,b)\rightarrow (a',b')\in \mathscr{M}$.

(2.1) $a=2n+1$ for some integer $n \in Z^{\geq 0}$. In this case,
the fact $(a,b)\notin \mathscr{M}$ implies that $b\geq 2$. We move
$$(a,b)\rightarrow (2n+1,\delta_{b})\in \mathscr{M}.$$


(2.2)  $a=2n$ for some integer $n \in Z^{\geq 0}$. We distinguish the following two subcases: $n=0$ or $n\geq 1$.

(2.2.1) $n=0$. In this subcase, the fact $(a,b)\notin \mathscr{M}$ implies
that $b=0$ or $b=1$ or $b\geq 4$. It is obvious that $(0,0)$ and
$(0,1)$ are $N$-positions.  If $b\geq 4$, we move
$$(a,b)=(0,b)\rightarrow (0,2+\delta_{b})\in \mathscr{M}.$$

(2.2.2) $n\geq 1$. In this subcase, the fact $(a,b)\notin \mathscr{M}$ implies that $b=0$ or $b=1$ or $b\geq 4$.

$\bullet$ $b\in\{0,1\}$. We move $(a,b)=(2n,b)\rightarrow (2n-1,b)\in \mathscr{M}$.

$\bullet$ $b\geq 4$. We move $(a,b)=(2n,b)\rightarrow (2n,2+\delta_b)\in \mathscr{M}$.

The proof is completed. \end{proof}

\subsection{All $P$-positions of  OEW: $s+t>2$}


\begin{theorem}\label{OEW-normal-s+t>2} By $\mathscr{P}_{6}$ we denote the set of all
$P$-positions of OEW under normal play convention. Then for all $s,t\in Z^{\geq 1}$ with $s+t>2$,
$$\mathscr{P}_{6}=\bigcup\limits_{n=0}^{\infty}\{(A_n,B_n),(A_n,B'_n)\},$$
where for $n\geq 0$,
\begin{equation}\label{odd-even-normal-s+t>2}\left\{
\begin{array}{lll}
A_n=n,\\
B_n=\delta_n(sA_n+t+\delta_{s+t}),\\
B'_n=B_n+1.\\
\end{array}
\right.\end{equation}
\end{theorem}

\begin{proof} Before we give the proof of Theorem \ref{OEW-normal-s+t>2}, Tables 7 and 8 list the first few values of $A_n$ and $B_n$ for $s=t=2$, $s=2$ and $t=3$,
respectively, which show us how to determine $\mathscr{P}_6$ by using Theorem \ref{OEW-normal-s+t>2}.

\begin{center}
Table 7.\quad The first few values of $A_n$ and $B_n$ for $s=t=2$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 &11 &12 &13 &14\\
\hline $A_{n}$&0&1 &2& 3 & 4 & 5& 6 & 7 & 8& 9 &10 &11 &12 &13 &14\\
 \hline $B_{n}$&0&4 & 0& 8  & 0 &12 &0 & 16 &0& 20 & 0 &24 &0 &28 &0\\
 \hline $B'_{n}$&1&5 & 1  & 9 & 1 & 13 & 1 &17& 1 & 21 &1 &25 &1 &29 &1\\
 \hline
 \end{tabular}
 \end{center}
For $s=t=2$, it follows from Table 7 that
$$\mathscr{P}_6=\left\{\begin{array}{l}(0,0),(0,1),(1,4),(1,5),(2,0),(2,1),(3,8),(3,9),(4,0),(4,1),\\
(5,12),(5,13),(6,0),(6,1),(7,16),(7,17),(8,0),(8,1),\cdots\end{array}\right\}$$

\begin{center}
Table 8.\quad The first few values of $A_n$ and $B_n$ for $s=2$ and $t=3$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 &11 &12 &13 &14\\
\hline $A_{n}$&0&1 &2& 3 & 4 & 5& 6 & 7 & 8& 9 &10 &11 &12 &13 &14\\
 \hline $B_{n}$&0&6 & 0& 10  & 0 &14 &0 & 18 &0& 22 & 0 &26 &0 &30 &0\\
 \hline $B'_{n}$&1&7 & 1  & 11 & 1 & 15 & 1 &19& 1 & 23 &1 &27 &1 &31 &1\\
 \hline
 \end{tabular}
 \end{center}
For $s=2$ and $t=3$, it follows from Table 8 that
$$\mathscr{P}_6=\left\{\begin{array}{l}(0,0),(0,1),(1,6),(1,7),(2,0),(2,1),(3,10),(3,11),(4,0),(4,1),\\
(5,14),(5,15),(6,0),(6,1),(7,18),(7,19),(8,0),(8,1),\cdots\end{array}\right\}$$

We now give the proof of Theorem \ref{OEW-normal-s+t>2}. Let $\mathcal{W}=\bigcup\limits_{n=0}^{\infty}\{(A_n,B_n),(A_n,B'_n)\}$.

\noindent{\textit{Proof of Fact 1.}} By the definition of
function $\delta_{n}$, we have
$$B_n=\delta_n(sA_n+t+\delta_{s+t})=\delta_n(sn+t+\delta_{s+t})$$ is
even and $B'_n=B_n+1$ is odd.

Given $(A_n,B_n)\in\mathcal {W}$. Note that $(A_n,B_n)\rightarrow
(A_m,B'_m)\in\mathcal {W}$ is not a legal move, since
$\ell=B_n-B'_m$ is odd. Similarly, $(A_n,B'_n)\rightarrow
(A_m,B_m)\in\mathcal {W}$ is also impossible.

Given $(A_n,B_n)\in\mathcal {W}$ or $(A_n,B'_n)\in\mathcal {W}$.
Suppose that $(A_n,B_n)\rightarrow (A_m,B_m)\in\mathcal {W}$ or
$(A_n,B'_n)\rightarrow (A_m,B'_m)\in\mathcal {W}$. In both cases, we
have $m<n$, $k=A_n-A_m=n-m$ is odd, and $\ell=B_n-B_m$.

If $n$ is even and $m$ is odd then
$$\ell=B_n-B_m=0-(sm+t+\delta_{s+t})<0,$$ which is impossible; If $n$ is odd and $m$ is even then
$$\ell=B_n-B_m=(sA_n+t+\delta_{s+t})-0\geq sn\geq s(n-m)\geq k>0,$$ and
$$\begin{array}{rcl}
\ell&=&B_n-B_m=sA_n+t+\delta_{s+t}\\
&\geq& sn+t\geq s(n-m)+t=sk+t,
\end{array}$$ which contradicts Eq. (\ref{odd-even-def-constraint}).

\vspace{0.5cm}
\noindent{\textit{Proof of Fact 2.}} Let $(x,y)$ be a
position not in $\mathcal{W}$. We will show that there exists a
legal move such that $(x,y)\rightarrow (A_n,B_n)\in \mathcal{W}$ or
$(x,y)\rightarrow (A_n,B'_n)\in \mathcal{W}$.

Put $x=n=A_n$ for some integer $n\in {Z}^{\geq 0}$. We distinguish
two cases:

(2.1) $x=n=A_n$ is even. In this case, $B_n=0$, $B'_n=1$. The
fact $(x,y)\notin \mathcal{W}$ implies that $y\geq 2$.
 We move $(x,y)\rightarrow (A_n,B_n+\delta_y)\in \mathcal{W}$ by virtue of $\delta_y=0$ or $1$.

(2.2) $x=n=A_n$ is odd. In this case,
\begin{equation}\label{odd-even-x-odd-Bn}
B_n=sx+t+\delta_{s+t}\mbox{ is even},
\end{equation}
and \begin{equation}\label{x-odd-B'n} B'_n=B_n+1\mbox{ is odd}.
\end{equation}
The fact $(x,y)\notin \mathcal{W}$ implies that $y\geq B_n+2$ or
$0\leq y\leq B_n-1$.

(2.2.1) $y\geq B_n+2$. Now $y\geq B_n+2>B_n+\delta_y$ and
$y-B_n-\delta_y$ is even. We move
$$(x,y)\rightarrow (A_n,B_n+\delta_y)\in\mathcal{W},$$ by taking $y-B_n-\delta_y$ tokens from the second heap.

(2.2.2) $0\leq y\leq B_n-1$. We distinguish the following three
subcases: $y\in\{0,1\}$ or $2\leq y\leq x$ or $x+1\leq y\leq B_n-1$.

$\bullet$ $y\in\{0,1\}$. We move
$$(x,y)\rightarrow (x-1,\delta_y)=(A_{n-1},B_{n-1}+\delta_y)\in \mathcal {W},$$
 since $n-1$ is odd, $B_{n-1}=0$, and $y=B_{n-1}+\delta_y$.


$\bullet$ $2\leq y\leq x$. We move
$$(x,y)\rightarrow (x-y+1+\delta_y,\delta_y),$$
by taking $y-1-\delta_y$ tokens from the first heap and $y-\delta_y$
tokens from the second heap. Let $x-y+1+\delta_y=m$. We note that
$m=A_m$ is even and $B_m=0$. Note that $\delta_y=B_m$ if $y$ is even,
$\delta_y=B_m+1=B'_m$ if $y$ is odd. Thus
$$(x-y+1+\delta_y,\delta_y)=(A_m,B_m+\delta_y)\in \mathcal{W}.$$
This move is legal. Indeed,

 1) $k=y-1-\delta_y>0$ is odd;

 2) $\ell=y-\delta_y>0$ is even;

3) $0\leq |\ell-k|=1<(s-1)k+t$.

$\bullet$ $x+1\leq y\leq B_n-1$. We move
$(x,y)\rightarrow(0,\delta_y)\in \mathcal{W}$,
 by taking $x$ tokens from the first heap and $y-\delta_y$ tokens from the second heap.
 This move is legal. Indeed,

 1) $k=x$ is odd;

 2) $\ell=y-\delta_y$ is even;

 3) We note that $y\geq x+1$ implies that $\ell=y-\delta_y\geq x=k$.
 By Eq. (\ref{odd-even-x-odd-Bn}), $B_n-1$ is odd. The fact $y\leq B_n-1$ implies that $y\leq B_n-2+\delta_y$. Hence,
 $$\begin{array}{rcl}
 0\leq |\ell-k|&=&y-\delta_y-x\\
 &\leq& sx+t+\delta_{s+t}-2-x\\
 &=& (s-1)k+t+\delta_{s+t}-2\\
 &<&(s-1)k+t,\end{array}$$
by virtue of $k\geq 1$.

The proof is completed. \end{proof}

\begin{theorem}\label{OEW-misere-s+t>2} By $\mathscr{P}'_6$ we denote the set of all
$P$-positions of OEW under mis\`{e}re play convention. Then for all $s,t\in Z^{\geq 1}$
with $s+t>2$,
$$\mathscr{P}'_6=\bigcup\limits_{n=0}^{\infty}\{(E_n,H_n),(E_n,H'_n)\},$$
where $E_0=0$, $H_0=2$, $H'_0=3$ and for $n\geq 1$,
\begin{equation}\label{odd-even-misere-s+t>2}\left\{
\begin{array}{lll}
E_n=n,\\
H_n=(1-\delta_n)(sE_n-s+t+\delta_{s+t}),\\
H'_n=H_n+1.\\
\end{array}
\right.\end{equation}
\end{theorem}

\begin{proof} Before we give the proof of Theorem \ref{OEW-misere-s+t>2}, Tables 9 and 10 list the first few values of $E_n$, $H_n$ and $H'_n$
for $s=t=2$, $s=1$ and $t=2$, respectively, which show us how to determine $\mathscr{P}'_6$ by using Theorem \ref{OEW-misere-s+t>2}.
\begin{center}
Table 9.\quad The first few values of $E_n$, $H_n$ and $H'_n$ for $s=t=2$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 &11 &12 &13 &14\\
\hline $E_{n}$&0&1 &2& 3 & 4 & 5& 6 & 7 & 8& 9 &10 &11 &12 &13 &14\\
 \hline $H_{n}$&2&0 & 4& 0  & 8 &0 &12 &0 &16 & 0 & 20 &0 &24 &0 &28\\
 \hline $H'_{n}$&3 & 1 &5 & 1 &  9 & 1& 13 & 1 &17 &1& 21 &1 &25 &1 &29\\
 \hline
 \end{tabular}
 \end{center}

\begin{center}
Table 10.\quad The first few values of $E_n$, $H_n$ and $H'_n$ for $s=1,t=2$.\\[0.3cm]
\begin{tabular}{c|cccccccccccccccccc}
\hline
$n$&0&1 &2& 3 & 4 & 5 & 6 & 7 & 8 & 9 &10 &11 &12 &13 &14\\
\hline $E_{n}$&0&1 &2& 3 & 4 & 5& 6 & 7 & 8& 9 &10 &11 &12 &13 &14\\
 \hline $H_{n}$&2&0 & 5& 0  & 9 &0 &13 &0 &17 & 0 & 21 &0 &25 &0 &29\\
 \hline $H'_{n}$&3 & 1 &6 & 1 &  10 & 1& 14 & 1 &18 &1& 22 &1 &26 &1 &30\\
 \hline
 \end{tabular}
 \end{center}

We now give the proof of Theorem \ref{OEW-misere-s+t>2}. Let
$$\mathcal{G}=\{(0,2),(0,3)\}\cup\bigcup\limits_{n=1}^{\infty}\{(E_n,H_n),(E_n,H'_n)\}.$$

\noindent{\textit{Proof of Fact 1.}} By the definition of
function $\delta_{n}$, if $n$ is odd then $H_n=0$,
if $n\geq 1$ is even then $H_n=s(n-1)+t+\delta_{s+t}$ is even.
  Thus for $n\in Z^{\geq 1}$, $H_n$ is even and $H'_n=H_n+1$ is odd.

Note that $(0,2)$ or $(0,3)$ can not move to $(E_n,H_n)$
or $(E_n,H'_n)$ as $E_n>0$. If $n$ is odd, $H_n=0$ implies that $(E_n,H_n)$ or
$(E_n,H'_n)$ can not move to $(0,2)$ or $(0,3)$; If $n$ is even,
$E_n=n$ is even implies that $(E_n,H_n)$ or $(E_n,H'_n)$ can not
move to $(0,2)$ or $(0,3)$.

Given $(E_n,H_n)\in\mathcal {G}$. Note that $(E_n,H_n)\rightarrow
(E_m,H'_m)\in\mathcal {G}$ is not a legal move, since
$\ell=H_n-H'_m$ is odd. Similarly, $(E_n,H'_n)\rightarrow
(E_m,H_m)\in\mathcal {G}$ is also impossible.

Given $(E_n,H_n)\in\mathcal {G}$ or $(E_n,H'_n)\in\mathcal {G}$.
Suppose that $(E_n,H_n)\rightarrow (E_m,H_m)\in\mathcal {G}$ or
$(E_n,H'_n)\rightarrow (E_m,H'_m)\in\mathcal {G}$. In both cases, we
have $1\leq m<n$, $k=E_n-E_m=n-m$ is odd, and $\ell=H_n-H_m$.

If $n$ is odd and $m$ is even then
$$\ell=H_n-H_m=0-(s(m-1)+t+\delta_{s+t})<0,$$ which is impossible;
If $n$ is even and $m$ is odd, then
$$\ell=H_n-H_m=s(n-1)+t+\delta_{s+t}-0\geq s(n-m)+t>sk\geq k>0,$$ and
$$\begin{array}{rcl}
\ell&=&H_n-H_m\\
&=&sE_n-s+t+\delta_{s+t}\\
&\geq& s(n-1)+t\\
&=&s(n-m)+s(m-1)+t\\
&\geq& s(n-m)+t=sk+t,\end{array}$$
by virtue of $m\geq 1$, which contradicts Eq. (\ref{odd-even-def-constraint}).

\vspace{0.5cm}
\noindent{\textit{Proof of Fact 2.}} Let $(x,y)$ be a
position not in $\mathcal {G}$. It suffices to show that there
exists a legal move such that $(x,y)\rightarrow (E_n,H_n)\in
\mathcal {G}$ or $(x,y)\rightarrow (E_n,H'_n)\in \mathcal {G}$.

Put $x=n=E_n$ for some integer $n\in {Z}^{\geq 0}$.

(2.1) $x=n=0$. In this case, we have $y=0$ or $y=1$ or $y\geq 4$. It
is obvious that $(0,0)$ and $(0,1)$ are $N$-positions. If $y\geq 4$,
then we move $(0,y)\rightarrow (0,2+\delta_y)\in \mathcal {G}$ by
virtue of $\delta_y=0$ or $1$.

(2.2) $x=n>0$ is odd. In this case, $H_n=0$, $H'_n=1$. The fact
$(x,y)\notin \mathcal {G}$ implies that $y\geq 2$. We move
$(x,y)\rightarrow (E_n,H_n+\delta_y)\in \mathcal {G}$.

(2.3) $x=n>0$ is even. In this case,
\begin{equation}\label{odd-even-x-even-Hn}
H_n=sx-s+t+\delta_{s+t}\mbox{ is even},
\end{equation}
and \begin{equation}\label{x-even-H'n} H'_n=H_n+1\mbox{ is odd}.
\end{equation}
 The fact $(x,y)\notin \mathcal {G}$ implies that  $y\geq H_n+2$ or $0\leq
y\leq H_n-1$.

(2.3.1) $y\geq H_n+2$, $n\in Z^{\geq 1}$. We move $$(x,y)\rightarrow
(E_n,H_n+\delta_y)\in \mathcal {G},$$ which is legal, since
$y-H_n-\delta_y>0$ is even.

(2.3.2) $0\leq y\leq H_n-1$, $n\in Z^{\geq 1}$. We distinguish the
following three subcases: $y\in\{0,1\}$ or $2\leq y\leq x-1$ or $x\leq
y\leq H_n-1$.

$\bullet$ $y\in\{0,1\}$. We move
$$(x,y)\rightarrow(x-1,\delta_y)=(E_{n-1},H_{n-1}+\delta_y)\in \mathcal {G},$$
 since $n-1$ is odd, $H_{n-1}=0$, and $\delta_y=0$ or $1$.

$\bullet$ $2\leq y\leq x-1$. We move
$$(x,y)\rightarrow(x-y-1+\delta_y,\delta_y),$$ by taking
$y+1-\delta_y$ tokens from the first heap and $y-\delta_y$ tokens
from the second heap. Let $x-y-1+\delta_y=m$. Note that $m$ is
odd, $H_m=0$, and $\delta_y=H_m+\delta_y$. Obviously, $\delta_y=H_m$
if $y$ is even, $\delta_y=H_m+1=H'_m$ if $y$ is odd. Thus
$$(x-y-1+\delta_y,\delta_y)=(E_m,H_m+\delta_y)\in \mathcal {G}.$$
This move is legal. Indeed,

 1) $k=y-\delta_y+1>0$ is odd;

 2) $\ell=y-\delta_y>0$ is even and $\ell\geq k$;

 3) $0<|\ell-k|=1<(s-1)k+t$.

$\bullet$ $x\leq y\leq H_n-1$, $n\in Z^{\geq 1}$. We move
$$(x,y)\rightarrow (1,\delta_y)\in \mathcal {G},$$
 by taking $x-1$ tokens from the first heap and $y-\delta_y$ tokens from the second heap.
 If $y$ is even, $\delta_y=0=H_1$; If $y$ is odd, $\delta_y=1=H_1+1=H'_1$.
 Thus $(1,\delta_y)=(E_1,H_1+\delta_y)\in \mathcal {G}$. This move is legal. Indeed,

 1) $k=x-1$ is odd;

 2) $\ell=y-\delta_y$ is even;

 3) Note that $y\geq x$ implies that $$\ell=y-\delta_y\geq x>x-1=k.$$
 By Eq. (\ref{odd-even-x-even-Hn}), $H_n-1$ is odd, so the fact $y\leq H_n-1$ implies that $y\leq H_n-2+\delta_y$. Hence,
 $$\begin{array}{rcl}
 0\leq |\ell-k|&=&y-\delta_y-x+1\\
 &\leq& sx-s+t+\delta_{s+t}-2-x+1\\
 &=& (s-1)(k+1)-s+t+\delta_{s+t}-1\\
 &=& (s-1)k+t+\delta_{s+t}-2\\
 &<&(s-1)k+t,\end{array}$$
by virtue of $k\geq 1$.

The proof is completed. \end{proof}


\section{Conclusion}

Three new models, Odd-Odd $(s,t)$-Wythoff's Game, Even-Even
$(s,t)$-Wythoff's Game, Odd-Even $(s,t)$-Wythoff's Game, are investigated.
 Under normal or mis\`{e}re play convention, all $P$-positions of these three models are
given for all integers $s,t\geq 1$.

Similar to Odd-Even $(s,t)$-Wythoff's Game, we can
define the fourth model, \emph{Even-Odd $(s,t)$-Wythoff's Game}
(Denoted by EOW): Let $S_h$, $S_v$, $D_1$ and $D_2$ be subsets
of $Z^{\geq 0}$. Given two parameters $s,t\in Z^{\geq 1}$ and two heaps
of finitely many tokens. One of the heaps is designated as ``first heap"
and the other as ``second heap" throughout the game. By $(x,y)$ we denote a position of present game,
where $x$ and $y$ denote the numbers of tokens in the first and the
second heaps, respectively. Two rules of moves are allowed:

(\emph{Even-Odd Nim Rule}) A player chooses the first heap and
takes \emph{even} $k>0$ tokens, or chooses the second heap and
takes \emph{odd} $\ell>0$ tokens.

(\emph{Even-Odd More General Wythoff's Rule}) A player takes
tokens from both heaps, \emph{even} $k>0$ tokens from the first
heap, \emph{odd} $\ell>0$ tokens from the second heap, and
\begin{equation}\label{even-odd-def-constraint}
0 \leq |\ell-k| \ <(s-1)\lambda+t, \ \ \lambda=\min\{k,\ell\}\in
Z^{\geq 1}.
\end{equation}


Obviously, EOW is equivalent to $S_h=D_1=Z^{even}$ and
$S_v=D_2=Z^{odd}$ in General Restriction of $(s,t)$-Wythoff's Game.

 If $(x,y)$ is a $P$-position of OEW, then $(y,x)$ is a $P$-position of EOW.
 Thus we have

\begin{corollary}\label{EOW-normal-s=t=1} Given two parameters $s=t=1$. By $\mathscr{P}_{7}$ we denote the set of all
$P$-positions of EOW under normal play convention. Then
$$\mathscr{P}_{7}=\bigcup\limits_{n=0}^{\infty}\{(0,2n), (1,2n), (2,2n+1), (3,2n+1)\}.$$
\end{corollary}

\begin{corollary}\label{EOW-misere-s=t=1} Given two parameters $s=t=1$. By $\mathscr{P}'_{7}$ we denote the set of all
$P$-positions of  EOW under mis\`{e}re play convention. Then
$$\mathscr{P}'_{7}=\bigcup\limits_{n=0}^{\infty}\{(2,2n), (3,2n), (0,2n+1), (1,2n+1)\}.$$
\end{corollary}

\begin{corollary}\label{EOW-normal-s+t>2} By $\mathscr{P}_{8}$ we denote the set of all
$P$-positions of  EOW under normal play convention. Then for all $s,t\in Z^{\geq 1}$
with $s+t>2$,
$$\mathscr{P}_{8}=\bigcup\limits_{n=0}^{\infty}\{(B_n,A_n),(B'_n,A_n)\},$$
\textit{where for $n\geq 0$,}
\begin{equation}\label{even-odd-normal-s+t>2}\left\{
\begin{array}{lll}
A_n=n,\\
B_n=\delta_n(sA_n+t+\delta_{s+t}),\\
B'_n=B_n+1.\\
\end{array}
\right.\end{equation}
\end{corollary}

\begin{corollary}\label{EOW-misere-s+t>2} By $\mathscr{P}'_8$ we denote the set of all
$P$-positions of EOW under mis\`{e}re play convention. Then for all $s,t\in Z^{\geq 1}$
with $s+t>2$,
$$\mathscr{P}'_8=\{(2,0),(3,0)\}\cup\bigcup\limits_{n=1}^{\infty}\{(H_n,E_n),(H'_n,E_n)\},$$
where for $n\geq 1$,
\begin{equation}\label{even-odd-misere-s+t>2}\left\{
\begin{array}{lll}
E_n=n,\\
H_n=(1-\delta_n)(sE_n-s+t+\delta_{s+t}),\\
H'_n=H_n+1.\\
\end{array}
\right.\end{equation}
\end{corollary}

Recall that Wythoff's game is a special case $s=t=1$ in
$(s,t)$-Wythoff's game, and $a$-Wythoff's game is a special
case $s=1$ and $t=a$ in $(s,t)$-Wythoff's game. Thus
$(s,t)$-Wythoff's game is a generalization of both Wythoff's game
and $a$-Wythoff's game. Under normal play convention, the set of all
$P$-positions of $a$-Wythoff's game and the set of all
$P$-positions of Wythoff's game can be obtained by letting ($s=1$
and $t=a$) and $s=t=1$ in Eq. (\ref{(s,t)-conclusion}), respectively
(see \cite{Wyth1907},\cite{Fraen1982}).

Our results on OOW, EEW, OEW and
EOW are given for all integers $s\geq 1$ and $t\geq 1$. Thus the
corresponding results on ($s=1$ and $t=a$) or $s=t=1$ have
been obtained.

Given two integer $K\geq 1$ and $r\in \{0,1,2\cdots,K-1\}$.
We use notation $Z^{(r)}_K=\{Kn+r|n\in Z^{\geq 0}\}$.


\begin{open}\label{problem1} Let $S_h=S_v=D_1=D_2=Z^{(0)}_K$
in General Restriction of $(s,t)$-Wythoff's Game. How
to determine all $P$-positions under normal or mis\`{e}re play convention?

Note that $Z^{even}=Z^{(0)}_2$, thus Theorems \ref{EEW-normal} and \ref{EEW-misere} have settled the special case $K=2$ of this
problem. Can we generalize Theorems \ref{EEW-normal} and \ref{EEW-misere} from $K=2$ to an arbitrary integer $K\geq 3$?
\end{open}

\begin{open}\label{problem2} Let $S_h=S_v=D_1=D_2=Z^{(r)}_K$
in General Restriction of $(s,t)$-Wythoff's Game, where
$r\in\{0,1,2\cdots,K-1\}$ be a fixed integer. How to determine all
$P$-positions under normal or mis\`{e}re play convention?

Note that $Z^{odd}=Z^{(1)}_2$, thus Theorems \ref{OOW-normal} and \ref{OOW-misere}
have settled the special case $K=2$ and $r=1$ of this problem. Can
we generalize Theorems \ref{OOW-normal} and \ref{OOW-misere} from $K=2$ and $r=1$ to arbitrary integers $K\geq 3$ and $r\in \{0,1,2\cdots,K-1\}$?
\end{open}

\begin{open}\label{problem3} Let $S_h=D_1=Z^{(r_1)}_K$ and
$S_v=D_2=Z^{(r_2)}_K$ in General Restriction of
$(s,t)$-Wythoff's Game, where $r_1,r_2\in\{0,1,2\cdots,K-1\}$ and
$r_1\neq r_2$ be fixed integers. How to determine all $P$-positions under normal or mis\`{e}re play convention?

Note that $Z^{even}=Z^{(0)}_2$ and $Z^{odd}=Z^{(1)}_2$, thus
Theorems \ref{OEW-normal-s=t=1}, \ref{OEW-misere-s=t=1}, \ref{OEW-normal-s+t>2} and \ref{OEW-misere-s+t>2}
have settled the special case $K=2$ of this
problem. Can we generalize these results from ($K=2$, $r_1=1$ and
$r_2=0$) to arbitrary integers $K\geq 3$ and $r_1\neq r_2\in
\{0,1,2\cdots,K-1\}$?
\end{open}

\begin{open}\label{problem4} One can investigate the case of
$(s,t)$-Wythoff's game which is only restricted on Extended Diagonal Moves.
As an example, let $S_h=S_v=Z^{\geq 0}$ and $D_1=D_2=Z^{(r)}_K$ ($K$ is a fixed
positive integer, $r\in\{0,1,2\cdots,K-1\}$) in General
Restriction of $(s,t)$-Wythoff's Game. Can we obtain the
corresponding results?
\end{open}

{\bf Acknowledgements} The authors are grateful to the responsible
editor and the anonymous referees for their valuable comments and
suggestions, which have greatly improved the earlier version of this
paper. The research is supported by the National Natural Science
Foundation of China under Grants 11171368 and 11171094.


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