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\begin{document} 
\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 
(2005), \#G02} 
\vskip 50pt 
\begin{center}
{\bf PARTIAL NIM}
\vskip 20pt
{\bf Chu-Wee Lim}\\
{\smallit Department of Mathematics, University of California Berkeley,
Berkeley, CA 94720-3840, USA}\\
{\tt limchuwe@math.berkeley.edu} 
\end{center}

\vskip 30pt
\centerline{\smallit Received: 10/19/04, Revised: 3/16/05,
 Accepted: 4/11/05,
Published: 4/20/05}
\vskip 30pt 

\centerline{\bf Abstract}


\noindent

In this paper, we offer a complete strategy for a variant of Nim which
is partial. Such a strategy can be construed although its canonical form
is exponentially complex. Variations of Nim are common in the literature
(see [1], [3]  and [4] for instance), but as far as we know, 
this variant of Partial Nim has not been explored before.

We shall assume that the reader is familiar
with the basics of combinatorial game theory. See [2] for an excellent
explanation of this theory.

\pagestyle{myheadings}
\markright{\smalltt INTEGERS: \smallrm 
ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 (2005),
\#G02\hfill}

\thispagestyle{empty} 
\baselineskip=15pt 
\vskip 30pt 

\section*{\normalsize 1. Introduction}

The game of Partial Nim is played as follows: we begin with several piles
of stones; some piles of which are colored Red, some of
which are colored bLue, while others are colored grEen. At each turn,
players Left and Right select a pile and remove a certain number of stones
from it, subjected to the following conditions.

\begin{enumerate}
\item In a bLue pile, Left may only remove an odd number of stones while
Right may only remove an even number.
\item Conversely, in a Red pile, Right may only remove an odd number of
stones while Left may only remove an even number.
\item In a grEen pile, there are no restrictions, just like conventional
Nim.
\end{enumerate}

The player who is unable to make a move loses. A simple mnemonic
for the above cases is as follows: bLue piles are advantageous to Left,
while Red piles are advantageous to Right. 

In the following sections, we shall describe
a complete strategy for this game.
The crux lies in the Lexicographical Ordering Principle (Theorem 2).

\vskip 30pt
\section*{\normalsize 2. Preliminaries}

First, the grEen piles easily simplify to a nim value $*n$, and by symmetry
we only need to look at the bLue piles.

\noindent{\bf Definition 1} : Let $A_n$ be the game of partial Nim with
a bLue pile of $n$ stones.

The canonical form of $A_n$ may be determined recursively. The first few
values of $A_n$ are:
\begin{align*}
A_0 &= 0 \\
A_1 &= 1 \\
A_2 &= \{ A_1 | A_0 \} = \ 1|0\\
A_3 &= \{ A_2, A_0 | A_1 \} = \frac 1 2
\end{align*}

\noindent{\bf Theorem 1} : For $n\ge 1$, we have:\\
(i) $A_{2n-1} = \frac 1 {2^{n-1}}$.\\
(ii) $A_{2n} = \{ 1\ |\ 0, A_{2n-2}\}$, which is canonical for $n>1$.\\
(iii) $A_2 > A_4 > A_6 > \dots $.

\noindent
{\it Proof.} The proof is by induction. Suppose the above 
holds for all $n\le m$, $m\ge 2$. Then
$$A_{2m+1} = \{ A_{2m}, A_{2m-2}, \dots, A_0\ |\ A_{2m-1}, A_{2m-3}, 
\dots, A_1 \}.$$
By induction hypothesis, each $A_{2i-1}$ on the right is $\frac 1 {2^{i-1}}$
so Right's best move is to $A_{2m-1} = \frac 1 {2^{m-1}}$. On the left, 
$A_{2i}$ ($i=1,\dots,m$) reverses back to 0 since $A_{2m+1}$ is 
positive. Hence $A_{2m+1} = \{0\ |\ 2^{-(m-1)}\} = \frac 1 {2^m}$.

For the even case, Left's best move is to $A_1=1$, while for Right's moves,
the induction hypothesis gives $A_2 > A_4 > \dots > A_{2m}$ so her best move
is to either $A_{2m}$ or $A_0 = 0$. Hence, 
$$A_{2m+2} = \{1 \ |\ 0, A_{2m}
\} \le \{ 1\ |\ 0, A_{2m-2}\} = A_{2m}.$$ Note that equality cannot 
hold since Right has a winning strategy in $A_{2m+2} - A_{2m}$ by taking
$A_{2m+2} \to A_{2m}$ leaving 0.

Finally, $A_{2m+2}$ is in canonical form since the Right option $A_{2m}$
is confused with 0 and does not reverse to 1 (which is incomparable to
$A_{2m+2}$).

\noindent
{\bf Corollary 1} : For an integer $n\ge 1$,\\
(i) $A_{2n}$ has a confusion interval of $(0,1)$.\\
(ii) At the end points, $A_{2n}$ is confused with both 0 and 1. \\
(iii) $A_{2n}$ has a temperature of $\frac 1 2$.

\noindent
{\it Proof.} (i) and (iii) follow from induction and the recursive formula
in Theorem 1(ii), while (ii) follows from the fact that $A_{2n}$ is
a first-player win (hence confused with 0) and $A_{2n}-1 = \{0\ |\ -1, 
A_{2n-2}\}$ is a first-player win as well.

Let us centralize the game $A_{2n}$ so that its mean is 0.

\noindent
{\bf Definition 2} : For $m\ge 1$, let $[m]$ denote the game $\frac 1 2
- A_{2m}$ and $[-m]$ denote the game $-[m]$. More generally, we define
$$[m_1, m_2, \dots, m_r] = [m_1] + [m_2] + \dots + [m_r]$$
for non-zero integers $m_1, m_2, \dots, m_r$. 

Theorem 1 immediately gives a recursive definition of $[m]$:
$$ [1] = \pm \frac 1 2, \quad [m+1] = \left\{\frac 1 2, [m]\ \left|\ 
-\frac 1 2 \right.\right\}.$$
It is clear that the difficulty of handling $[m]$ in sums of games increases
exponentially as $m$ gets larger. Expressing $[m_1, \dots, m_r]$ in 
canonical form is 
probably out of the question. Furthermore, we shall see later in Section 4
that the differences between these sums are extremely small, typically
about tiny $\tn_1$. Hence comparison between the games appears rather
difficult.

\vskip 30pt
\section*{\normalsize 3. Strategy}

Immediately, Theorem 1 and Corollary 1 give the following.

\noindent{\bf Lemma 1} : Let $m$ be a non-zero integer.\\
(i) The confusion interval of $[m]$ is the closed
interval $-\frac 1 2 \le x \le +\frac 1 2$.\\
(ii) The temperature of each $[m]$ is $\frac 1 2$.\\
(iii) $\dots [-3] < [-2] < [-1] = [1] < [2] < [3] < \dots$.

The next step is to analyze the pair $[i,j]$.

\noindent{\bf Lemma 2} : Let $i,j$ be non-zero integers. Then the game
$[i,j]$ is infinitesimal.

\noindent{\it Proof.}  Suppose $i,j>0$. Then Lemma 1(iii) gives
$[i,j] = [i] + [j] > [1] + [1] = 0$. So it suffices to show $[i,j]\not > \e$
for any positive number $\e$. Indeed, in the game $[i,j] - \e$, Right has
a winning strategy if she starts: take $[i]\to -\frac 1 2$ and leave
$[j] - (\frac 1 2 + \e)$ which is negative since $\frac 1 2 + \e$ lies
outside the confusion interval of $[j]$.

On the other hand, suppose we have $i>j>0$ and the game $[i,-j] = [i]-[j]$.
Then $[i]>[j]$ by Lemma 1(iii), while in the game $[i]-[j]-\e$,
Right takes $[i] \to -\frac 1 2$ as before to secure a win.

\noindent{\bf Theorem 2 (Lexicographical Ordering Principle)} : Suppose we 
have positive integers $m_1 \le m_2 \le \dots \le m_r$ and $n_1 \le n_2 \le
\dots \le n_r$. Then
$$ [m_1, m_2, \dots, m_r] < [n_1, n_2, \dots, n_r] $$
if the $r$-tuples satisfy $(m_i)_{i=1}^r < (n_i)_{i=1}^r$ in lexicographical
ordering, i.e. if there exists $1\le i\le r$, such that $m_j = n_j$ for all 
$1\le j<i$ but $m_i < n_i$. 

\noindent{\it Proof.}  First, the theorem holds for $r=1$ and for 
$m_1 = m_2 = \dots = m_r = 1$ (cf. Lemma 1(iii)). With these simple cases
in mind, we now apply induction on the sum $r + \sum_i (m_i + n_i)$.

Let $r\ge 2$ and $G = [n_1, \dots, n_r, -m_1, \dots, -m_r]$. We first prove
that $G\ge 0$. If any $m_i$
and $n_j$ are equal, we may cancel both and apply the induction hypothesis
for $r-1$. So we assume that $m_i \ne n_j$ for any $i,j$. Now if Right has
the first move in $G$, she has the following options:

{\it Case 1} : Take $[n_i] \to -\frac 1 2$, so the right incentive
$\Delta^R = G - G^R = \frac 1 2 + [n_i]$. Then the best choice for Right 
is $[n_r]$. Left may
 counter this move by taking $[m_r] \to \frac 1 2$ to produce
$$ [n_1, n_2, \dots, n_{r-1}, -m_1, -m_2, \dots, -m_{r-1}]$$
which is non-negative by induction hypothesis. So Left wins in this case.

{\it Case 2} : Take $-[m_i] \to -[m_i-1]$ for some $m_i > 1$. Now we
still have $$(m_1, \dots, m_{i-1}, m_i-1, m_{i+1}, \dots, m_r) < (n_1, 
\dots, n_r),$$
so the resulting game from $G$ is positive by induction hypothesis. 
Left still wins.

{\it Case 3} : Take $-[m_i] \to -\frac 1 2$, so $\Delta^R = \frac 1 2
-[m_i]$. Since the smallest $[m_i]$ is $[m_1]$, Right would do best by
moving $G$ to $[n_1, \dots, n_r, -m_2, \dots, -m_r] -\frac 1 2.$
If $m_2 < n_2$ as well, then Left may counter with $[n_1] \to \frac 1 2$
and reduce $G$ to
$$ [n_2, n_3, \dots, n_r, -m_2, -m_3, \dots, -m_r],$$
which is $\ge 0$ by induction hypothesis. Otherwise, $m_2 > n_2$ and
Left responds with $[-m_2] \to \frac 1 2$ leaving
$$ [n_1, n_2, \dots, n_r, -m_3, \dots, -m_r] = [n_1, n_2, \dots, n_r,
-1, -1, -m_3, \dots, -m_r].$$
Since $m_1 \ge 1$ and $m_2 > n_2 \ge 1$, we have reduced the sum
$\sum (m_i + n_i)$, and induction applies. 

The above shows $G \ge 0$. To complete the proof for $G > 0$, note that
$$[m_1, \dots, m_{r-1}, m_r] < [m_1, \dots, m_{r-1}, m_r + 1] \le 
[n_1, \dots, n_r].$$

It is natural now to ask for a comparison between $[m_1, \dots, m_r]$
and $[n_1, \dots, n_s]$ in general. Suppose $r+s$ is even and $r>s$.
Then we may add pairs of $[1,1] = 0$ to the second game until both
games have equal number of terms. The Lexicographical Ordering
Principle gives us $[m_1, \dots, m_r] > [n_1, \dots, n_s]$. Hence, w
have a more general result

\noindent
{\bf Corollary 2} : Suppose $m_1 \le m_2 \le \dots \le m_r$ and
$n_1 \le n_2 \le \dots \le n_s$ are positive integers, and $r<s$ are
of the same parity. Assume further that at most one $m_i$ (resp. $n_j$)
is equal to 1. Then
$$ [m_1, m_2, \dots, m_r] < [n_1, n_2, \dots, n_s].$$

In summary, we get the following ordering:
\begin{align*}
0 = [1,1] < [1,2] < [1,3] < [1,4] < \dots < [1,m] < & \dots \\
          < [2,2] < [2,3] < [2,4] < \dots < [2,m] < & \dots \\
                  < [3,3] < [3,4] < \dots < [3,m] < & \dots \\
\vdots\qquad & \vdots\qquad \\
   < [1,2,2,2] < [1,2,2,3] < [1,2,2,4] < [1,2,2,5] < & \dots \\
               < [1,2,3,3] < [1,2,3,4] < [1,2,3,5] < & \dots \\
\vdots\qquad & \vdots\qquad \\
< [1,2,2,2,2,2] < [1,2,2,2,2,3] < [1,2,2,2,2,4] < & \dots\\
\vdots\qquad & \vdots\qquad 
\end{align*}

Next, we shall compare the games $[m_1, \dots, m_r]$ and $[n_1, \dots,
n_s]$ when $r$ and $s$ are of different parity. Again, assume 
$m_1 \le \dots \le m_r$ and $n_1 \le \dots \le n_s$, and that at most
one $m_i$ (resp. $n_j$) is equal to 1. As before, let $G = [m_1, \dots, m_r,
-n_1, \dots, -n_s]$ be the difference. Since $[m,n]$ is infinitesimal for
all $m,n\ne 0$, $G$ has confusion interval $(-\frac 1 2, +\frac 1 2)$. 
The only uncertainty lies in the endpoints $-\frac 1 2, +\frac 1 2$.

\noindent
{\bf Theorem 3}: Let $G$ be as above, with $r<s$. Then\\
(i) the game $G$ is confused with $-\frac 1 2$; and \\
(ii) $G < +\frac 1 2$, unless we have the {\it exceptional case} of
$r=s-1$ and $(m_i)_{i=1}^r \ge (n_i)_{i=1}^r$ lexicographically.

\noindent
{\it Proof.} (i) Look at $G+\frac 1 2$. If Left starts, he wins easily by
taking $-[n_s] \to +\frac 1 2$ to produce $[m_1, \dots, m_r, -n_1, \dots,
-n_{s-1}] + 1$. Since the bracketed part has an even number of terms,
it is infinitesimal and hence $G + \frac 1 2 > 0$.

If Right starts and $r>0$, he takes $[m_r] \to -\frac 1 2$ and leaves the
game $[m_1, \dots, m_{r-1}, -n_1, \dots, -n_s]$ which is negative by
Corollary 2. On the other hand, if $r=0$, Right repeatedly takes any $-[n_j] 
\to -\frac 1 2$. Left then has no choice but to take a $-[n_{j'}] \to
+\frac 1 2$ or $[m_i] \to +\frac 1 2$. Since $r+s$ is odd, this exchange
favours Right.

(ii) Consider $G-\frac 1 2$. If Right starts, she wins by turning any
$[m_i]$ or $-[n_j]$ to $-\frac 1 2$. But if Left starts, he has the
following options.

{\it Case 1} : Take $[m_i] \to \frac 1 2$, producing $[m_1, \dots, \hat{m_i},
\dots, m_r, -n_1, \dots, -n_s]$ which is negative by Corollary 2. This move
is clearly bad.

{\it Case 2} : Left takes $[m_i] \to [m_i-1]$ for some $m_i>1$, which
Right counters by $[m_i - 1] \to -\frac 1 2$ and wins.

{\it Case 3} : Left's only chance is to take $[-n_j] \to \frac 1 2$. This
gives $\Delta^L = \frac 1 2 + [n_j]$ so the best choice is in $[-n_s]$.
The resulting game is
$$ [m_1, m_2, \dots, m_r, -n_1, -n_2, \dots, -n_{s-1}].$$
By the Lexicographical Ordering Principle and its Corollary, this game is 
negative (Right wins) {\it unless} $r = s-1$ and $(m_i)_{i=1}^r \ge 
(n_i)_{i=1}^r$ lexicographically; in which case it is $\ge 0$ and Left wins.
This proves the theorem.

Note that we have not taken into account the grEen part of partial Nim,
i.e. we need to compare $[m_1, \dots, m_r]$ and $*N$. This will be 
covered in the next section.

\noindent{\bf Examples} : Let us analyze the following games, where the
brackets () and (()) refer to bLue and Red piles respectively.
\begin{enumerate}
\item (3,3,9), ((2,1)). The odd piles sum up to $\frac 1 2 + \frac 1 2 
+ \frac 1 {2^4} - 1$. The confusion interval of ((2)) is $(-1,0)$, hence
the {\it first player wins}.

\item (1,2,6), ((3,3,3,5)). The odd piles add up to $-\frac 3 4$, while
$(2,6) = A_2 + A_4 = 1 + [-1,-3]$ is 1-ish. So {\it Left wins}.

\item (1), ((8,4,6)). This game is $1 - A_4 - A_2 - A_3 = -\frac 1 2 + 
[4,2,3]$. Since $[4,2,3]$ is confused with $\frac 1 2$, the {\it first
player wins}.

\item (1,2,6,10), ((3,5,5,4,4,8)). This game is $[2,2,4,-1,-3,-5] > 0$
since lexicographically $(1,3,5) < (2,2,4)$. {\it Left wins}.

\item (1,3,12), ((5,5,8,4,2)). The resulting sum is $[4,2,1,-6] > 0$,
so {\it Left wins}.

\item (3,3,4,8), ((6,10,6)). The sum is $\frac 1 2 + [3,3,5,-2,-4]$.
Since removing $[5]$ gives $[2,4] < [3,3]$, the game $[3,3,5,-2,-4] > -\frac
1 2$ by Theorem 3, i.e. {\it Left wins}.

\item (3,3,8,8), ((6,10,6)). Similar to (5), but in $[3,3,5,-4,-4]$,
deleting the game $[5]$ results in the exceptional case of $[3,5] < [4,4]$.
So the {\it first player wins}.
\end{enumerate}

In canonical form, the game in example (6) is:
$$ 1, \{ 1 ||| 0 || 0 | -1 , \{0 | -1\} \}\ ||\ \{1 ||| 0 || 0 | -1, 
\{0 | -1\}\}\ |\ \{0 | -1, \{0|-1\}\}, \{0||0|-1, \{0|-1\}\}$$
according to David Wolfe's {\it Gamesman Toolkit}, which also verified
this game to be positive.

\vskip 30pt

\section*{\normalsize 4. Comparing $[m_1, \dots, m_r]$ With Infinitesimals}

For an even $r$, let us compare the games $[m_1, \dots, m_r]$ and 
$+_k = \{0 || 0 | -k \}$. Throughout
the rest of this article, for games $A$ and $B$, we will write $A \gg B$
if $A > n\cdot B$ for any positive integer $n$.

\noindent
{\bf Lemma 3} : Let $D_m = [m+1, -m]$. Then we have
$$ \tn_1 = D_1 \gg D_2 \gg D_3 \gg \dots \gg \tn_{1+\e}$$
for any number $\e>0$.

\noindent
{\it Proof.} The equality on the left is easy. The Lexicographical
Ordering Principle gives $D_{m} > i\cdot D_{m+1}$ for any positive integer 
$i$. For the rightmost inequality, it suffices to show $D_m \ge \tn_{1+\e}$
for any $m$ and number $\e>0$. 

Indeed, we wish to show that $G = [m]-[m+1]+ \tn_{1+\e} \le 0$. Suppose
Left starts. If he moves $[m] \to [m-1]$, Right can respond with $-[m+1]
\to -[m]$. On the other hand, if he takes $-[m+1] \to \frac 1 2$ or
$[m] \to \frac 1 2$, Right makes a move in the tiny component and wins.

\noindent{\bf Corollary 3} : The game $G=[m_1, \dots, m_r]$, if positive,
satisfies $\tn_{1+\e} < G < \tn_{1-\e}$ for any number $\e>0$.

\noindent{\it Proof.} If $r$ is odd, then $G$ is fuzzy. When $r$ is even,
the right inequality follows easily from the Lexicographic Ordering
Principle :
$$[m_1, \dots, m_r] = (r+2)[1,1] + [m_1, \dots, m_r] 
< \overset{r+1}{\overbrace{[1,2,\dots,1,2]}}
= (r+1)[1,2] < \tn_{1-\e}.$$
To prove the left inequality, we may append copies of $[1,1]$ or $[-1,-1]$ 
to $G$, such that in $G = [m_1, \dots, m_r]$, there are as many positive
$m_i$'s as negative $m_i$'s. In this way, we may rewrite $G$
as $[m_1, \dots, m_s] - [n_1, \dots, n_s] = \sum_{i=1}^s [m_i, -n_i]$, where 
$n_1 \le \dots \le n_s$, $m_1 \le \dots \le m_s$. Since $G$ is positive,
we may assume, with no loss of generality, that $m_1 > n_1$.
Thus, $[m_1, -n_1] \ge D_{n_1}$. As for $[m_i, -n_i]$
with $i>1$, if $m_i \ge n_i$ then $[m_i, -n_i] \ge 0$ and
dropping the term $[m_i, -n_i]$ gives a smaller $G$.
We may therefore assume
$m_i < n_i$ for all $i > 1$, whence by lemma 3,
$$G = [m_1, -n_1] - \sum_{i=2}^s [n_i, -m_i] > D_{m_1} - k\cdot D_{m_1 + 1} 
> \tn_{1+\e},$$
for some positive integer $k$. 

Now we may complete our analysis, by comparing $[m_1, \dots, m_r]$ and
$*N$.

\noindent
{\bf Theorem 4} : Consider the game $[m_1, \dots, m_r]$, where the $m_i$'s
are all non-zero integers.\\
(i) If $r$ is even, then $[m_1, \dots, m_r] + (*N)$ is fuzzy when $N>0$. \\
(ii) If $r$ is odd, then $-\frac 1 2 < [m_1, \dots, m_r] + (*N) < +\frac 1 2$
when $N>0$.
\\
In particular, we see that in partial Nim, $*1$ is already ``remote''.

\noindent
{\it Proof.} (i) Indeed, $G = [m_1, \dots, m_r]$, if positive, lies
strictly between $\tn_{1+\e}$ and $\tn_{1-\e}$. Since both of these bounds
are incomparable with $*N$, so is $G$.

\noindent
(ii) Consider $G = [m_1, \dots, m_r] +\frac 1 2 + (*N)$. If it were Left's
turn, he can easily secure a win by moving $m_r \to \frac 1 2$. On the
other hand, if it were Right's turn, he cannot afford to move $[m_i] \to
[m_i+1]$ (for some $m_i < -1$) or Left can win as before. Hence his only
chance is $m_i \to -\frac 1 2$, which results in the game
$$ [m_1, \dots, \hat{m_i}, \dots, m_r] + (*N)$$
which is a win for the next player by part (i). Hence, Right loses and
$G > 0$. Replacing $[m_1, \dots, m_r]$ by $[-m_1, \dots, -m_r]$, we get
the other inequality.

With this theorem, our analysis for partial Nim is complete. 

\noindent{\bf Examples} : Let us add $*N$ ($N>0$) to the games in the
previous example.

\begin{enumerate}
\item (3,3,9), ((2,1)), *N. The outcome is still fuzzy, i.e. {\it first 
player wins}.

\item (1,2,6), ((3,3,3,5)), *N. As before, {\it Left wins}.

\item (1), ((8,4,6)), *N. The game is now $-\frac 1 2 *N + [4,2,3] < 0$.
{\it Right wins}.

\item (1,2,6,10), ((3,5,5,4,4,8)), *N. The resulting game is 
$[2,2,4,-1,-3,-5] + *N$ which is fuzzy. {\it First player wins}.

\item (1,3,12), ((5,5,8,4,2)), *N. The game $[4,2,1,-6] + (*N)$ is
fuzzy - {\it first player wins}.

\item (3,3,4,8), ((6,10,6)), *N. The sum is now $\frac 1 2 *N + [3,3,5,-2,-4]$
which is positive. {\it Left wins}. 

\item (3,3,8,8), ((6,10,6)), *N. The game $\frac 1 2 * N + [3,3,5,-4,-4]$
is also positive - {\it Left wins}.

\end{enumerate}

\vskip 30pt
\section*{\normalsize 5. Other Variants of Partial Nim \& Future Work}

It is natural to consider exploring other variants of Partial Nim. For
example, what if we had altered the rules for this version: {\it in a bLue 
(resp. Red) pile, Left (resp. Right) can take any number of stones, while 
Right (resp. Left) can only take an odd number}?

This variant turns out to be easy.

\noindent
{\bf Theorem 5} : Let $B_n$ be this game for a bLue pile of $n$ stones. \\
(i) If $n=2m$, then $B_n =\ \up + \up^2 + \dots + \up^m$.\\
(ii) If $n=2m+1$, then $B_n = \ \up + \up^2 + \dots + \up^m +\ *$.

\noindent
{\it Proof.} Omitted, left as an exercise to the reader.

It may also be fruitful to 
examine partial variants of well-known games such as Kayles, Dawson's Chess
or Treblecross. We hope that these can yield interesting new results.

\vskip 30pt
\section*{\normalsize Acknowledgements}

The author would like to thank Professor Elwyn Berlekamp for his 
encouragement in getting the paper written, and for his excellent course
in Combinatorial Game Theory back in Fall 2002. 

\vskip 30pt
\section*{\normalsize References}

\renewcommand{\labelenumi}{[\theenumi]}
\begin{enumerate}
\item M. H. Albert and R. J. Nowakowski. Nim restrictions. {\it Integers},
4:Paper G1, 10pp. (electronic). 2004.

\item Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy. {\it Winning
ways for your mathematical plays, Volume 1}. A.K. Peters Ltd., second
edition, 2001.

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\end{enumerate}

\end{document}