\documentclass[12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{enumerate}
\usepackage{xspace}
\usepackage{pstricks}
\usepackage{pst-tree}
\usepackage{subfigure}
%\usepackage[small,bf]{caption}
%\usepackage[subfigure21]{ccaption}
\usepackage[thmmarks,amsmath]{ntheorem}

% INTEGERS SETTINGS
\textwidth= 6.25in
\textheight= 9.0in
\topmargin = -10pt
\evensidemargin=10pt
\oddsidemargin=10pt
\headsep=25pt
\parskip=10pt
\font\smallit=cmti10
\font\smalltt=cmtt10
\font\smallrm=cmr9

% Add 30pt at the start of each section and use normalsize

\makeatletter
\renewcommand\section{\@startsection {section}{1}{\z@}%
                                   %{-3.5ex \@plus -1ex \@minus
                                   %-.2ex}%
                                   {-30pt}
                                   {2.3ex \@plus.2ex}%
                                   {\normalfont\normalsize\bfseries}}
\renewcommand\subsection{\@startsection{subsection}{2}{\z@}%
                                     {-3.25ex\@plus -1ex \@minus -.2ex}%
                                     {1.5ex \@plus .2ex}%
                                     {\normalfont\normalsize\bfseries}}
\makeatother


% OWN SETTINGS

% Theorems and such things
%\setlength{\theorempreskipamount}{\parskip}
%\setlength{\theorempostskipamount}{\parskip}

\theorembodyfont{\normalfont}
\theoremsymbol{{\small\ensuremath{\square}}}


\newtheorem{theorem}{Theorem}
\newtheorem{corr}{Corollary}
\newtheorem{lemma}{Lemma}
\newtheorem{observation}{Observation}

\theoremheaderfont{\it}
\theoremstyle{nonumberplain}
%\theoremstyle{plain}

\newenvironment{proof}
{\par\noindent{\textit{Proof }}}
{}

% Common symbols
\newcommand{\Pp}{\ensuremath{\mathcal{P}}\xspace}
\newcommand{\Np}{\ensuremath{\mathcal{N}}\xspace}

% Counters

\newcounter{saved}
\newcounter{rephrase}

% Float placement

\renewcommand{\topfraction}{.9}
\renewcommand{\bottomfraction}{.9}
\renewcommand{\textfraction}{.1}

% Comments during writing

\newcommand{\hnote}[1]%
{\marginpar{\raggedright\footnotesize{\textsf{#1}}}}
\newcommand{\pnote}[1]%
{\marginpar{\raggedright\footnotesize{\sffamily\textsl{#1}}}}

\newcommand{\hcomment}[1]{{\textsf{#1}}\xspace}
\newcommand{\pcomment}[1]{{\sffamily\textsl{#1}}\xspace}

\begin{document}
\vspace*{-60pt}
\centerline{\smalltt INTEGERS:
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4
(2004), \#G04}
\vskip 30pt

\begin{center}
{\bf NIM WITH A MODULAR MULLER TWIST}
\vskip 20pt
{\bf Hillevi Gavel}\\
{\smallit Department of Mathematics and Physics, M\"alardalen
University, V\"aster{\aa}s, Sweden}\\
{\tt hillevi.gavel@mdh.se}\\
\vskip 10pt
{\bf Pontus Strimling}\\
{\smallit Department of Mathematics and Physics, M\"alardalen
University, V\"aster{\aa}s, Sweden}\\
{\tt pontus.strimling@mdh.se}\\ 
\end{center}
\vskip 30pt
\centerline{\smallit Received: 11/20/03, Revised: 8/11/04, 
Accepted: 9/24/04, Published: 9/28/04 }
\vskip 30pt

\centerline{\bf Abstract}

\noindent
We study Nim with the additional rule that each player can put a
constraint on the next player's move. This is called a \emph{Muller
twist}. The constraints are of the form ``the number taken must not be
equivalent to \emph{some numbers} modulo~$n$''. A solution for a large
group of such games is given.

\pagestyle{myheadings}
\markright{\smalltt INTEGERS: 
\smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4 (2004),
\#G04\hfill}

\thispagestyle{empty}
\baselineskip=15pt
%\vskip 30pt 
%Unnecessary, since I've added 30pt in the definition of section.

\section{Introduction}

An interesting twist on a combinatorial game can be obtained if one
allows the players to put constraints on each other's moves. This is
called ``introducing a \emph{Muller Twist}'', after Blaise Muller who
invented the game \emph{Quarto\textsuperscript{\textregistered}}, a
game where each player chooses the piece the next player should
place. This game and several others are described in an
article~\cite{oddeven} by F.~Smith and P.~St\u{a}nic\u{a}.

A move in a game with a Muller twist consists of a ``physical'' move
and a choice of a restriction. A position consists of a ``physical''
position and a restriction.  Imagine a chess game where the other
player determines which kind of piece you are allowed to move or
points out a square that you must not move to, or a bridge game where
your opponent determines which suite you should play.  These would all
be Muller twist games.

Adding a Muller twist to a single game may obviously make it more
entertaining to play.  It may also make the mathematical analysis of
the game more challenging.  In this article we will study the game of
Nim with restrictions on the number of sticks that can be taken,
aiming to solve as large a group of such games as possible.


\section{The game Nim and twists thereupon}

Nim is an important and well understood combinatorial game, see
\cite{vinway} and~\cite{ONAG}. It is commonly played with sticks
divided into piles. In each move a player may pick as many sticks as
he wants to, but only from one pile. The two players take turns and
when a player cannot make a move (because there are no sticks left) he
loses.  It can be shown \cite{vinway,ONAG} that a position is a
winning position for the player leaving it (a \emph{\Pp-position}) if
the Nimsum of the pile sizes is~0; otherwise, the position is a
winning position for the person receiving it (an \emph{\Np-position}).
The Nimsum $\oplus$ of a set of numbers is computed by bitwise Xor on
the binary representation of the numbers, that is, addition modulo~2
without carry. So
\[
9\oplus3=(2^3+2^0)\oplus(2^1+2^0) =
2^3+2^1+(\underbrace{2^0\oplus2^0}_0)=10
\]
The key observation is (see~\cite{ONAG}) the following lemma, which we
present without proof:

\begin{lemma}\label{l:nimsum}
\mbox{}
\begin{itemize}
\item
If a set of numbers have Nimsum~0 and you change one of them, then
the new set of numbers will have a Nimsum differing from~0.
\item
If a set of numbers have Nimsum differing from~0, then there is a way to decrease at
least one of the numbers so that the new Nimsum is~0.
\end{itemize}
\end{lemma}


\subsection{Odd-or-even-Nim}\label{s:eoon}
\emph{Odd-or-even-Nim} is analyzed in F.~Smith and
P.~St\u{a}nic\u{a}'s article~\cite{oddeven}.  In this twisted form,
the previous player tells the next player if she should take an odd or
an even number of sticks.  Smith and St\u{a}nic\u{a} show that the
\Pp-positions are the positions satisfying one of three properties:
\begin{itemize}
\item
Nimsum~0, restriction even
\item
Nimsum~1, restriction even
\item
Nimsum~0, all piles are even and restriction odd.
\end{itemize}
The proof is elegant, but provides no intuition for why the results
look like they do.  In order to gain better understanding, we looked
closer at the game structure and generalized the Muller twist.


``Odd'' or ``even'' tells if a number is equivalent to 1 or to~2
modulo~2. A related twist is to play modulo~3 instead. Then there are
three classes of numbers, equivalent to 1, 2 or~3, and the player can
disallow one of them. So the player leaving the position says ``the
number of sticks taken must \emph{not} be equivalent to~$x$
modulo~3''.  An example of a game of this form is given in
Figure~\ref{f:demospel}.

We will study several variations of this game, varying the modulo used
and the number of forbidden moves. We will solve those games where all
combinations of up to a given number of options are allowed as
restrictions, and also some games where not all combinations may be
used.


\begin{figure}[t]
\psset{linewidth=.5pt,nodesep=2pt,levelsep=1cm}
\pstree{\Tr{$\phantom{^{\not\equiv2}}(4,2,1)^{\not\equiv1}$}}
{\Tr{$\phantom{^?}(2,2,1)^?$}
 \Tr{$\phantom{^?}(1,2,1)^?$}
 \Tr{$\phantom{^?}(4,1)^?$}}
\hfil
\pstree{\Tr{$\phantom{^{\not\equiv2}}(4,2,1)^{\not\equiv1}$}}
{\pstree{\Tr{$\phantom{^{\not\equiv2}}(2,2,1)^{\not\equiv1}$}}
 {\pstree{\Tr{$\phantom{^{\not\equiv2}}(2,1)^{\not\equiv1}$}}
  {\pstree{\Tr{$\phantom{^{\not\equiv2}}(1)^{\not\equiv1}$}}
   {\Tr{$\dagger$}}}
  \pstree{\Tr{$\phantom{^{\not\equiv2}}(2,1)^{\not\equiv2}$}}
  {\pstree{\Tr{$\phantom{^{\not\equiv2}}(1,1)^{\not\equiv1}$}}
   {\Tr{$\dagger$}}}}
 \Tn
 \Tn}
\caption{A player is passed three piles containing 4, 2 and 1~sticks
together with the restriction ``the number taken must not be
equivalent to~1 modulo~3''. That means that taking 1 or 4~sticks is
forbidden, so the player has to take 2 or 3~sticks from one of the
piles. There are three ways to do this. If the player chooses to take
2~from the largest pile, and hands over the position with the
restriction ``not~1'', the other player has to take all the sticks in
one of the two 2-stick piles. Then, there are two ``sensible''
constraints that can be put. (Forbidding~3, which is not possible
anyway, seems pointless.)  Both constraints make it possible for the
first player to reduce the game to a set of one-stick piles, and
setting the restriction ``not~1'' makes this a terminal position, a
losing position for the other player. Thus, $(4,2,1)$ with the
restriction ``not~1'' is a winning position for the player being
handed that position; an \Np-position.  Furthermore, going to
$(2,2,1)$ and setting the restriction ``not~1'' is the \emph{only} way
to win. One can check this by investigating the other options in the
same way as used with this one.}\label{f:demospel}
\end{figure}


\subsection{Modular one-blocking Nim}

In our first generalized version of odd-or-even-Nim, which we will
call \emph{modular one-blocking Nim}, the restriction is of the form
``the number of sticks taken must not be equivalent to~$x$
modulo~$n$''.

Our analysis of this game turned up a surprise: The modulo used
actually does not matter! The only important thing is whether the
restriction allows you to take \emph{exactly} one stick or not.

\begin{theorem}\label{t:pposooe}\setcounter{rephrase}{\value{theorem}}
In modular one-blocking Nim the \Pp-positions are characterized by
having either of two properties:
\begin{enumerate}
\item\label{P1}
Positions where the Nimsum of the quotients of the pile sizes 
divided by~2 is~0, and where the restriction is ``the number of
sticks taken must not be equivalent to~1 modulo~$n$''.
\item\label{P2}
Positions where the Nimsum of the quotients is~0 and where all the
remainders of the pile sizes divided by~2 are~0 (that is,
all the pile sizes are even), any restriction.
\end{enumerate}

\begin{proof}
First we must show that if we are given a position of type~\ref{P1}
or~\ref{P2}, then we have to return a position from the complement set
(if any move at all is possible):
\begin{enumerate}
\item[\ref{P1}.]  If the Nimsum of the quotients is~0 and the
restriction is ``not~1'', then we have to take at least \emph{two}
sticks. Taking two or more sticks from a pile changes the quotient,
which means that the Nimsum of the quotients will change too
(Lemma~\ref{l:nimsum}), so we will hand over a position where the
Nimsum is not~0.
\item[\ref{P2}.]  
If the Nimsum is~0 and all the remainders are~0 too, then we have to
change a quotient when we take some sticks, even if we take just one
and thus cause the Nimsum to change.
\end{enumerate}
Second, we must show that if we are given a position from the
complement set, then it is always possible to make a move and return a
position of type~\ref{P1} or~\ref{P2}:
\begin{enumerate}
\item
If the Nimsum of the quotients is not~0, then there is at least one
way to change the quotients to make it so. There are two different
ways to give the chosen pile the desired quotient -- one where the
remainder is~0 and one where it is~1. The restriction is only able to
block one of these alternatives. We will certainly be able to hand
over a position of type~\ref{P1}. If none of the remainders are~1, we
can hand over type~\ref{P2} instead.
\item
If the Nimsum is~0, at least one remainder is~1 and we are allowed to
take just one stick, then we can take this remainder and keep the
Nimsum~0. We can either hand over a position of type~\ref{P1}
or~\ref{P2} (the latter if just one of the remainders in the original
position was~1).
\end{enumerate}
\end{proof}
\end{theorem}

If we look at the game in Figure~\ref{f:demospel}, we find that the
position $(4,2,1)^{\not\equiv1}$ has the quotients $(2,1,0)$. The
Nimsum of the quotients is $2 \oplus 1\oplus 0 = 3 \neq 0$, so this
position is indeed an \Np-position. When the player moves to
$(2,2,1)^{\not\equiv1}$ the quotients are changed to $(1,1,0)$, and $1
\oplus 1 \oplus 0 = 0$. The remainders are $(0,0,1)$, so the player
has to block~1 to make this a \Pp-position. The given move is the only
way for the first player to win the game, as stated.


\begin{figure}
\centering
\setlength{\arraycolsep}{.2em}
\subfigure[Constraint ``not~1''. (In odd-or-even-Nim: constraint
even.)]% 
{$
\begin{array}{c|cc|cc|cc|c}
&0&1&2&3&4&5&6\\
\hline
0&\Pp&\Pp&\Np&\Np&\Np&\Np&\Np\\
1&\Pp&\Pp&\Np&\Np&\Np&\Np&\Np\\
\hline
2&\Np&\Np&\Pp&\Pp&\Np&\Np&\Np\\
3&\Np&\Np&\Pp&\Pp&\Np&\Np&\Np\\
\hline
4&\Np&\Np&\Np&\Np&\Pp&\Pp&\Np\\
5&\Np&\Np&\Np&\Np&\Pp&\Pp&\Np\\
\hline
6&\Np&\Np&\Np&\Np&\Np&\Np&\Pp
\end{array}
$}
\hfil
\subfigure[Constraint anything except ``not~1''. (In odd-or-even-Nim:
constraint odd.)]% 
{$
\begin{array}{c|cc|cc|cc|c}
&0&1&2&3&4&5&6\\
\hline
0&\Pp&\Np&\Np&\Np&\Np&\Np&\Np\\
1&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
\hline
2&\Np&\Np&\Pp&\Np&\Np&\Np&\Np\\
3&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
\hline
4&\Np&\Np&\Np&\Np&\Pp&\Np&\Np\\
5&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
\hline
6&\Np&\Np&\Np&\Np&\Np&\Np&\Pp
\end{array}
$}
\caption{\Np- and \Pp-positions when playing modular one-blocking Nim with two
piles. On the matrix we have superimposed a grid. Positions within the
same ``box'' of the grid have the same quotients when divided by~2. The
remainders give the position within the box. In~(b) we can note that
the \Pp-positions are the ones where both numbers are even, that is,
where both remainders are~0.}\label{f:ooen}
\end{figure}


The \Pp-positions in modular one-blocking Nim are the same as in
odd-or-even-Nim (if we regard ``not~1'' as the same thing as ``even''
and everything else as ``odd''). An illustration is given in
Figure~\ref{f:ooen}. The equivalence of the two sets follows from the
following observations:

The quotient when dividing a number by~2 is what you get if you cut
out the least significant bit, while the remainder is that bit.  If
the Nimsum of the quotients is~0, then the Nimsum of the original
numbers will have~0s in all positions except the least significant
one, that may be 0 or~1. That is, the Nimsum of the original numbers
will be 1 or~0.  The \Pp-positions of odd-or-even-Nim were described
as
\begin{itemize}
\item
Nimsum~0, restriction \emph{even} 
\item
Nimsum~1, restriction \emph{even}
\item
Nimsum~0, all numbers even, restriction \emph{odd}
\end{itemize}
The first two kinds are the same as type~\ref{P1}, and the last one is
type~\ref{P2}. (If all the numbers are even, the least significant bits
will be~0, so if the Nimsum of the quotients is~0, so is the Nimsum of
the original numbers.)


\section{Further generalization}

\newcommand{\blap}{$k$-blocking modular Nim\xspace} Even further
generalizations of odd-or-even-Nim can be made. In the previous
section, we blocked \emph{one} modular option.  We will now consider a
game where a greater number of options can be blocked.  As a start, we
will express Theorem~\ref{t:pposooe} in a more general way:

\setcounter{saved}{\value{theorem}}
\setcounter{theorem}{\value{rephrase}}
\addtocounter{theorem}{-1}

\begin{theorem}[rephrased]
In modular one-blocking Nim the \Pp-positions are the positions where
the Nimsum of the quotients of the pile sizes divided by~2 is~0 and
where all the remainders are smaller than the smallest amount that can
be taken according to the constraint.
\end{theorem}

\setcounter{theorem}{\value{saved}}

The reason that the theorem involves division by \emph{two} is that we
can force the adversary to take at least two sticks, but not more. If
we block~1, taking 2 is an option. If we block~2, taking just one is
possible. We can call~2 ``the smallest unavoidable number''.


\subsection{Blocking a fixed number of options}
Our next generalized set of games will be called \emph{$k$-blocking
modular Nim}.  Here, a constraint consists of exactly $k$~different
subconstraints of the kind ``you must not take a number equivalent
to~$x$ modulo~$n$''. All the $\binom{n}{k}$~possible combinations of
subconstraints may be used. So in 4-blocking modular Nim with
modulo~10, one possible restriction is ``you can not take a number
equivalent to 1, 3, 6 or~8 modulo~10''.


\begin{lemma}
If the constraint consists of $k$~subconstraints, then you can force
the adversary to take at least $k+1$~sticks. (That is when the set of
subconstraints is ``not~1'', ``not~2'', \dots, ''not~$k$''.) We will
call this the \emph{strictest constraint}.
\end{lemma}


\begin{theorem}
\label{t:pontus}
In $k$-blocking modular Nim the \Pp-positions are the
positions where
\begin{itemize}
\item 
the Nimsum of the quotients of the pile sizes when divided by~$k+1$
is~0 
\end{itemize}
and
\begin{itemize}
\item
all remainders are smaller than the smallest number of sticks
that can be taken according to the given constraint.
\end{itemize}

\begin{proof}
Starting with the \Pp-positions:
\begin{itemize}
\item%[P1:]
If all the remainders are smaller than the smallest number of sticks
that can be taken, then the move will change one of the quotients, so
the Nimsum of the quotients will not be~0 (Lemma~\ref{l:nimsum}).
\end{itemize}
Continuing with the \Np-positions:
\begin{itemize}
\item%[P2:]
If the Nimsum of the quotients is not~0, then one can make it~0 by
changing one of the quotients (Lemma~\ref{l:nimsum}). There are
$k+1$~different pile sizes that have the desired quotient, and the
constraint is only able to block $k$~of them. Having made the
Nimsum~0, we can choose a constraint that is bigger than all the
remainders. (If we choose the strictest constraint, the one that
forces the adversary to take at least $k+1$~sticks, all the remainders
will be smaller than that, as they can not exceed~$k$.)

If the Nimsum of the quotients is~0, but at least one of the
remainders is at least as big as the smallest number of sticks we may
take, then we can decrease that remainder keeping the quotient
constant. The Nimsum will still be~0, and we can choose a constraint
strict enough to leave a \Pp-position.

(Note: This situation only occurs after suboptimal play, as the player
giving the constraint could have chosen a stricter one, making the
position a \Pp-position instead.)
\end{itemize}
\end{proof}
\end{theorem}


An illustration of  \Np- and \Pp-positions in a game is given in
Figure~\ref{f:spel42}. 


\begin{figure}[t]
\centering
\setlength{\arraycolsep}{.2em}
\subfigure[The strictest constraint: ``not~1 and not~2''.]
{$
\begin{array}{c|ccc|ccc|c}
&0&1&2&3&4&5&6\\
\hline
0&\Pp&\Pp&\Pp&\Np&\Np&\Np&\Np\\
1&\Pp&\Pp&\Pp&\Np&\Np&\Np&\Np\\
2&\Pp&\Pp&\Pp&\Np&\Np&\Np&\Np\\
\hline
3&\Np&\Np&\Np&\Pp&\Pp&\Pp&\Np\\
4&\Np&\Np&\Np&\Pp&\Pp&\Pp&\Np\\
5&\Np&\Np&\Np&\Pp&\Pp&\Pp&\Np\\
\hline
6&\Np&\Np&\Np&\Np&\Np&\Np&\Pp
\end{array}$}
\hfill
\subfigure[``Not~1, not~3'' and ``not~1, not~4''. (That is,
constraints blocking~1 but not~2.)]
{$
\begin{array}{c|ccc|ccc|c}
&0&1&2&3&4&5&6\\
\hline
0&\Pp&\Pp&\Np&\Np&\Np&\Np&\Np\\
1&\Pp&\Pp&\Np&\Np&\Np&\Np&\Np\\
2&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
\hline
3&\Np&\Np&\Np&\Pp&\Pp&\Np&\Np\\
4&\Np&\Np&\Np&\Pp&\Pp&\Np&\Np\\
5&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
\hline
6&\Np&\Np&\Np&\Np&\Np&\Np&\Pp
\end{array}
$}
\hfill
\subfigure[Constraints ``not~2, not~3'', ``not~2, not~4'' and ``not~3,
not~4''. (That is, constraints not blocking~1.)]
{$
\begin{array}{c|ccc|ccc|c}
&0&1&2&3&4&5&6\\
\hline
0&\Pp&\Np&\Np&\Np&\Np&\Np&\Np\\
1&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
2&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
\hline
3&\Np&\Np&\Np&\Pp&\Np&\Np&\Np\\
4&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
5&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
\hline
6&\Np&\Np&\Np&\Np&\Np&\Np&\Pp
\end{array}
$}
\caption{Playing 2-blocking Nim modulo~4, and using two piles. 
``2-blocking'' means that it is the quotient and remainder when
dividing by three that is important. When using just two piles, the
Nimsum equals~0 if and only if the piles are equal, which means that
the \Pp-positions will be placed along the main
diagonal.}\label{f:spel42}
\end{figure}


\begin{corr}[push-them-upwards]
It is never an advantage to give any other constraint than the
strictest one, and sometimes it is a disadvantage.
\end{corr}


\subsection{Blocking a variable number of options}

A further generalizations of the game allows blocking of any number of
options, up to a certain limit~$k$. This does not change the analysis,
since it is never disadvantageous to use the maximum number~$k$ of
blockings.


\subsection{Blocking certain combinations of options}

We propose as an open problem to find the solution of the general
modular blocking Nim game, that is, when the set of allowed
restrictions is \emph{any} subset of the set of all possible
combinations of subrestrictions.

The following observations tell how far we have come with this
problem: 


\begin{observation}
The characterization of \Pp-positions in
Theorem~\ref{t:pontus}  holds even if we do not allow all
combinations of at most $k$~subconstraints, as long as the strictest
constraint, that is, the combination ``not~1, not~2, \dots, not~$k$''
is an option.
The same proof goes through. The important thing is that we are able to
use the strictest constraint, and thus force the adversary to change a
quotient.
\end{observation}







\begin{figure}[t]
\centering
\setlength{\arraycolsep}{.2em}
\subfigure[Constraint ``not~2, not~3'' (that is: the number taken must
be equal to~1 modulo~3).]
{$
\begin{array}{c|ccc|ccc|c}
&0&1&2&3&4&5&6\\
\hline
0&\Pp&\Np&\Np&\Pp&\Np&\Np&\Pp\\
1&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
2&\Np&\Np&\Pp&\Np&\Np&\Pp&\Np\\
\hline
3&\Pp&\Np&\Np&\Pp&\Np&\Np&\Pp\\
4&\Np&\Np&\Np&\Np&\Np&\Np&\Np\\
5&\Np&\Np&\Pp&\Np&\Np&\Pp&\Np\\
\hline
6&\Pp&\Np&\Np&\Pp&\Np&\Np&\Pp
\end{array}$
}
\hfil
\subfigure[Constraint ``not~1, not~3'' (the number taken must
be equal to~2 modulo~3).]
{$
\begin{array}{c|ccc|ccc|c}
&0&1&2&3&4&5&6\\
\hline
0&\Pp&\Pp&\Np&\Np&\Pp&\Np&\Np\\
1&\Pp&\Pp&\Np&\Np&\Pp&\Np&\Np\\
2&\Np&\Np&\Pp&\Np&\Np&\Pp&\Np\\
\hline
3&\Np&\Np&\Np&\Pp&\Np&\Np&\Pp\\
4&\Pp&\Pp&\Np&\Np&\Pp&\Np&\Np\\
5&\Np&\Np&\Pp&\Np&\Np&\Pp&\Np\\
\hline
6&\Np&\Np&\Np&\Pp&\Np&\Np&\Pp
\end{array}
$}
\caption{Playing blocking Nim modulo~3, each constraint blocking two
options, but the strictest combination ``not~1, not~2''
disallowed. The pattern is completely different from when the
strictest combination is allowed. If ``not~1, not~2'' is an option,
then we get the same pattern as in
Figure~\ref{f:spel42}.}\label{f:nimskum}
\end{figure}

\begin{observation}
The characterization of \Pp-positions in Theorem~\ref{t:pontus} does
\emph{not} hold if we \emph{do not} include the strictest constraint
``not~1, not~2, \dots, not~$k$'' as an option.
The parts ``the constraint is only able to block $k$~of them'' and
``If we choose the strictest constraint'' of the proof will not work
anymore.

If the grid size is determined by the number of constraints, so that
we divide by~$k+1$, then we are not always able to force the adversary
to change a quotient, since we need the strictest constraint to ensure
that.

If the grid size is determined by the lowest unavoidable number (less
than~\mbox{$k+1$}) the $k$~subconstraints may be able to block every
way to get the desired quotient.
\end{observation}

Figure~\ref{f:nimskum} shows an example of a game where the strictest
constraint is excluded.



\section{Conclusion}
There are surprisingly few Muller twist games solved. Some, besides
Odd-or-even-Nim, are described in~\cite{oddeven}, some others
in~\cite{HolReit1}. We hope that we have given inspiration for further
works in this field. We thank an anonymous referee and Kimmo~Eriksson
for many useful comments.


\begin{thebibliography}{1}

\bibitem{oddeven} F.~Smith, P.~St\u{a}nic\u{a},
\emph{Comply/Constrain Games or Games with a Muller Twist},
Integers~2, art.~G3, 2002.

\bibitem{vinway} E.R.~Berlekamp, J.H.~Conway, R.K.~Guy, \emph{Winning
Ways}, A~K~Peters, Ltd., 2001.

\bibitem{ONAG} J.H.~Conway, \emph{On Numbers and Games},
A~K~Peters, Ltd., 2001. 

\bibitem{HolReit1} A.~Flammenkamp, A.~Holshouser, H.~Reiter,
\emph{Dynamic One-Pile Blocking Nim}, Electron. J.
Combin., Volume~10(1), art.~N4, 2003.


\end{thebibliography}


\end{document}