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\begin{document}
\title[Consensus List Colorings of Graphs and Physical Mapping of DNA]
{Consensus List Colorings of Graphs and \\Physical Mapping of DNA}
\author{N. V. R. Mahadev}
\address{Department of Computer Science, Fitchburg State
College}
\email{nmahadev@fsc.edu}
\author{Fred S. Roberts}
\address{Department of Mathematics and DIMACS, Rutgers
University}
\email{froberts@dimacs.rutgers.edu}
\thanks{The second author thanks the National 
Science Foundation for its support under grants NSF-SBR-9709134 and 
NSF-DBI-9982983 to 
Rutgers University and the Alfred P. Sloan Foundation for a grant to
Rutgers University.}


%\date{draft of August 28, 2002}

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\date{}

\keywords{list coloring, physical mapping, interval graph, fragment 
overlap graph.}
%\pagenumbering{roman}

\begin{abstract}
In graph coloring, one assigns a color to each vertex of a graph so
that neighboring vertices get different colors. We shall talk about a
bioconsensus problem relating to graph coloring and discuss the
applicability of the ideas to the DNA physical mapping problem. In
many applications of graph coloring, one gathers data about the
acceptable colors at each vertex.  A list coloring is a graph coloring
so that the color assigned to each vertex belongs to the list of
acceptable colors associated with that vertex.  We consider the
situation where a list coloring cannot be found.  If the data
contained in the lists associated with each vertex are made available
to individuals associated with the vertices, it is possible that the
individuals can modify their lists through trades or exchanges until
the group of individuals reaches a set of lists for which a list
coloring exists. We describe several models under which such a
consensus set of lists might be attained. In the physical mapping
application, the lists consist of the sets of possible copies of a
target DNA molecule from which a given clone was obtained and trades
or exchanges correspond to correcting errors in data.

\end{abstract}

\maketitle

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\section{Introduction}

    An old problem arising in decision theory, voting theory, and the
    social sciences is the {\em consensus problem}: How do we combine
    individual opinions into a decision by a group? This widely
    studied problem has a large associated literature. In recent
    years, methods developed in the social sciences have started to
    get wide use in the biological sciences. Use of consensus methods
    in biology is termed in this volume {\em bioconsensus}. In this
    paper, we introduce and develop some new models of consensus in a
    graph theory setting and discuss the application of these concepts
    in the context of the problem of physical mapping of DNA.

The new consensus models are concerned with graph coloring. Here, we
seek to assign a color to each vertex of a graph so that vertices
joined by an edge get different colors. Graph coloring is an old
subject with many important applications.  While graph theory has
widespread uses in molecular biology in particular and in other areas
of biology in general, graph coloring is not widely used. However, it
plays a role in the study of DNA physical mapping, as we shall
observe. The interesting and important variant called list coloring,
closely connected to applications, arises when the color assigned to a
vertex in a graph must be chosen from a list of acceptable colors
associated with that vertex.  In this paper, we consider the situation
when such a coloring cannot be found.  We introduce and study various
consensus-achieving models by which the individuals associated with
vertices might change the lists associated with their vertices so that
a coloring using acceptable colors might be achieved. These models are
applied to the physical mapping problem in the context of handling
certain kinds of errors in data.

    We begin with some basic definitions and motivations.  Suppose
$G = (V,E)$ is a graph.  A {\em coloring} is an assignment of a color
(herein a positive integer) $f(x)$ to each vertex $x$ in $V$ so that
$$\{x,y\} \in E \rightarrow f(x) \neq f(y).$$

\noindent 
The smallest number of colors in a graph coloring of $G$ is called the {\em
chromatic number} and is denoted $\chi(G)$.  Graph colorings arise in a
variety of applications. For example, in channel assignment, $V$ is a
set of transmitters, an edge means interference, and the color is the
channel assigned to the transmitter.  In traffic phasing, $V$ is a set
of individuals or cars or groups that request use of a facility such
as a computer, traffic intersection, or room; an edge means
interference; and the color is the time the facility is assigned to
the individual, car, or group.  For a variety of other applications
of graph coloring, see \cite{Roberts} \cite{RobertsRainbows}.

In the next section, we review the study of physical mapping and
describe the role of graph coloring in that study.  We introduce list
coloring in Section~\ref{Sec:ListColoring}. The following three
sections describe the three consensus models for list coloring and the
final section gives concluding remarks and describes future directions
for research.

\section{How Graph Coloring Enters into Physical Mapping}

This section will begin with an elementary introduction to physical
mapping. It closely follows Setubal and Meidanis \cite{SetMei}, which
is a good source of additional details. A human chromosome is a DNA
molecular consisting of about $10^8$ base pairs. A {\em physical map}
is defined to be a piece of DNA that tells us the location of certain
markers along the molecule. The {\em markers} are typically small, but
precisely defined, sequences. To build physical maps, we begin by
making several copies of the DNA molecule to be mapped, the so-called
{\em target DNA}. Then we break up each copy into {\em fragments}
through the use of {\em restriction enzymes}, with different enzymes
used for different copies. We obtain overlap information about the
fragments and use this information to obtain the mapping.

How do we obtain the overlap information? In general, each fragment is
still too long to be sequenced, so we cannot obtain overlap
information as is done in the process called {\em fragment assembly},
i.e., by sequencing and comparing fragments. The method used instead
is called {\em fingerprinting}. We shall describe the fingerprinting
method known as hybridization. Another commonly used method is known
as {\em restriction site analysis}.

In {\em hybridization}, we start with the set of fragments obtained
from each copy of the DNA molecule. The fragments are then replicated
using the method known as {\em cloning}. We end up with a collection
of thousands of clones, many corresponding to the same fragment. (We
shall be sloppy about the distinction between clones and fragments.)
Now the idea is to associate with each clone (fragment) information
that helps us uniquely describe it (as fingerprints do). This is
accomplished by checking whether certain small sequences called {\em
probes} bind (hybridize) to clones (fragments). The subset of probes
that do bind is called the clone's {\em fingerprint}. By comparing
fingerprints, we try to determine whether two clones overlap. The
principle used is that two clones that share part of their
fingerprints are likely to have come from overlapping regions of the
target DNA. It should be noted that overlap information does not tell
us the location of the corresponding fragment on the target DNA, only
something about the relative order of the clones (fragments),
information that was lost during the breakup.

It is important to emphasize that there are many errors in
hybridization data and this observation is an important motivation for
the application of list coloring consensus methods that we shall be
investigating. Still following Setubal and Meidanis, we can identify
various sources and kinds of errors. A probe might fail to bind where
it should or it may bind where it shouldn't. Humans can mis-read or
mis-record experimental results. During the cloning process, it is
possible for two pieces of the target DNA to join and then be
replicated as if they were a single clone. (Such clones are called
{\em chimeric clones} and as many of 60\% of the clones might be
chimeric.) Probes can also bind along more than one site along the
target DNA. (The method of {\em Sequence Tagged Site} or STS was
designed to avoid this.) Finally, the data obtained is often
incomplete. (Here, pooling methods can help.)

A key tool in physical mapping is the interval graph model. Let $F$ be
a family of sets. The {\em intersection graph} corresponding to $F$
has the sets in $F$ as its vertices and an edge between two sets if
and only if they overlap. A graph is called an {\em interval graph} if
it arises this way when the family $F$ consists of intervals on the
real line. Interval graphs have a long history and many applications
in both the social and biological sciences, as well as in many other
areas. For a discussion of interval graphs and their applications, see
for example
\cite{Fishburn},\cite{Golumbic},\cite{Roberts76},\cite{Roberts78}. There
are good algorithms for recognizing an interval graph and for
constructing a ``map" of intervals on the line that have the
corresponding intersection pattern.

From the overlap information, we create a {\em fragment overlap graph}
as follows. The vertices are the fragments (clones) and the edges mean
that the fragments overlap. If we had complete and error-free
information about fragment overlapping, then the resulting graph would
be an interval graph. If it is an interval graph, we can use the
corresponding ``map" of intervals to get information about the
relative order of the fragments on the target DNA molecule and begin
to obtain a physical map. However, the information we have is not
complete and error-free and the fragment overlap graph might not be an
interval graph. There are several approaches to handling this problem.
We describe three approaches that start by labeling each clone
(fragment) with identification of the copy of the target molecule from
which it came. This label can be thought of as a color. If two clones
come from the same copy of the target molecule, they cannot overlap,
and so the labeling gives us a graph coloring of the fragment overlap
graph. Later we shall discuss what to do if there are errors in the
labeling and we don't have a graph coloring. The possibility that
errors in the data lead to the
fragment overlap graph not being an interval graph is given a lot of
attention in the literature. The possibility that errors in
the labeling might lead to our not having a graph coloring is not
given this kind of attention. It will be our main emphasis in this
paper. However, we first consider what to do if the fragment overlap
graph is not an interval graph.

Let us consider first the case where the only type of error we have
made is to omit overlaps -- the {\em Case of False Negatives}.  Then
we might seek to add edges to the fragment overlap graph, enough to
obtain an interval graph.  We have to do this in such a way that the
labeling we have remains a graph coloring.  This may not be
possible. If it is, we can use the resulting graph to obtain a
candidate physical mapping.  If there are several ways to add edges,
we can look for the smallest number of added edges.  It turns out that
the problem of whether there is a way to add edges so as to obtain an
interval graph for which the coloring remains a coloring is
NP-hard. (See \cite{FHW},\cite{GGKS},\cite{GKS} for a discussion of
the NP-hardness of various physical mapping problems and \cite{KST}
for linear time algorithms for the simpler problem of transforming a
given graph into a proper interval graph with the smallest number of
edges. A {\em proper interval graph} is one where no interval properly
contains another.)

A second case is the {\em Case of False Positives} where the only type
of error we have made is to include overlaps when they are not
actually present. Then we might seek to delete edges from the fragment
overlap graph to obtain an interval graph with the labeling remaining
a graph coloring.  The latter requirement is trivially satisfied since
removing edges always maintains the coloring property.  It is natural
to seek the smallest number of edges to remove.  This is also an
NP-hard problem.

If both kinds of errors might occur, we can think of having a set of
edges $E_1$ on a vertex set $V$ that are definitely there (overlaps
that are definitely known to occur) and a set of edges that are
definitely not there. The latter defines a set $E_2$ of edges on $V$
that might be there, and we have $E_1 \subset E_2$. We have the same
coloring on both graphs $(V,E_1)$ and $(V,E_2)$. We then ask if there
is a set of edges $E$ so that $E_1 \subset E \subset E_2$ and so that
$(V,E)$ is an interval graph and the coloring remains a coloring for
$(V,E)$. The latter requirement is automatically satisfied. The
problem of determining if we can satisfy the former, known as the {\em
interval sandwich problem}, is also NP-hard \cite{GS}.

\addvspace{-2cm}
\section{List Coloring} \label{Sec:ListColoring}
In many applications of graph coloring, there is an extra
complication.  There are some acceptable colors for each vertex and
the color assigned to a vertex must be chosen from the set of
acceptable colors.  For instance, in channel assignment, we specify a
set of acceptable channels and in traffic phasing a set of acceptable
times.  In physical mapping, we might lose or inaccurately record
information about which copy of the target DNA molecule a given clone
came from, and only know that the copy it came from belongs to one of
a set of possible copies.

To formalize this idea, we
let $S(x)$ denote a nonempty set of integers assigned to vertex $x$ of
graph $G$ and call $S$ a {\em list assignment} for $G$.  We seek a
graph coloring $f$ so that for every $x \in V$, we have $f(x) \in
S(x)$.  Such a coloring is called a {\em list coloring} for $(G,S)$.
If a list coloring exists, we say that $(G,S)$ is {\em list
colorable}.  List colorings were introduced independently by Vizing
\cite{Vizing} and Erd\"{o}s, Rubin, and Taylor \cite{ERT} and there
have been a very large number of papers about this subject in the past
decade.  The recent flurry of activity concerning list colorings is
surveyed in the papers \cite{Tuza} and \cite{Kratochvil}. Earlier
surveys are in \cite{Alon} and \cite{JT}.

Since it is NP-complete to determine if a graph is colorable in at
most $k$ colors if $k \geq 3$, it is easy to see that it is
NP-complete to determine if there is a list coloring for $(G,S)$ if
$|\cup S(x)| \geq 3$.  (It is even NP-complete for special cases such
as bipartite graphs -- see \cite{JS}, \cite{Kubale}. See \cite{Tuza} for other
special cases that are NP-complete.)  If $|\cup S(x)| = 2$, then it
is easy to see that it is polynomial to determine if $(G,S)$ has a
list coloring (see \cite{Vizing} and \cite{ERT}).


In physical mapping, in the Case of False Negatives, we might now ask:
Given $(G,S)$, can we add edges to $G$ to obtain an interval graph $H$
so that $(H,S)$ is list colorable? The answer is always ``no" if
$(G,S)$ is not list colorable. Adding edges makes it harder to obtain
a list coloring.

However, the comparable question in the Case of False Positives is
interesting: Given $(G,S)$, can we subtract edges from $G$ to obtain
an interval graph $H$ so that $(H,S)$ is list colorable? If so, what
is the smallest number of edges we can remove to accomplish this?

The Sandwich question is also interesting. Given graphs $G_1 =
(V,E_1)$ and $G_2 = (V,E_2)$, with $E_1 \subset E_2$, and given a list
assignment $S$ on $V$, is there an interval graph $H = (V,E)$ with
$E_1 \subset E \subset E_2$ so that $(H,S)$ is list colorable?

While these questions are all interesting, we defer answers to
them. Instead, we shall ask what is perhaps a more fundamental
question, one that doesn't involve changing edges. Namely, we shall
simply think of making modifications in the assignment $S$ to obtain
an assignment
$S^\prime$ with $(G,S^\prime)$ list colorable.

To put our problem in a social choice or consensus context, we think
of vertices as corresponding to individuals. It is this point of view
we shall take in what follows. If $(G,S)$ has no list coloring, some
individuals represented by vertices may have to make sacrifices by
exchanging or expanding their lists in order for a list coloring to
exist. In physical mapping, the motivation is a bit different. It is
based on the idea that there could be errors in the data about the
possible copies of target DNA a clone came from. In the next three
sections, we introduce and discuss three models for how individuals
might change their lists. We think of these models as providing
procedures for the group of individuals corresponding to the vertices
to reach a consensus about a list coloring. In the physical mapping
context, we think of these as providing procedures for modifying the
sets of possible copies associated with a given clone with as few
changes as possible.

In what follows, we use the notation $K_n$ for the complete graph with
$n$ vertices and $I_n$ for its complement, the graph with $n$ vertices
and no edges. If $G$ and $H$ are graphs with disjoint vertex sets,
then $G+H$ denotes the graph whose vertex set is the union of the
vertex sets of $G$ and $H$ and whose edge set is the union of the edge
sets of $G$ and $H$ plus all edges joining a vertex in $G$ to a vertex
in $H$.

\section{The Adding Model}\label{Sec:Adding}

In the first consensus model, we allow each individual to add one
color from the set of colors already used, i.e., from $\cup S(x)$.
For example, in the channel assignment application, a person may add
one acceptable channel, and in the traffic phasing application, a
person may add one acceptable time.  In the physical mapping
application, this is the case where a possible copy number might have
been omitted from the list associated with a clone. Note that allowing
the addition of more than one color does not add any flexibility since
a person can always add only the color that he/she actually uses.

We say that $(G,S)$ is {\em $p$-addable} if we can identify $p$
distinct vertices $x_1, x_2, \ldots , \linebreak[0] x_p$ in $G$ and
(not necessarily distinct) colors $c_1, c_2, \ldots , c_p$ in $\cup
S(x)$ so that if $S^\prime(x_i) = S(x_i) \cup \{c_i\}, i = 1, 2,
\ldots , p$, and $S^\prime(u) = S(u)$ otherwise, then $(G,S^\prime)$ is
list-colorable.  It is straightforward to observe that $(G,S)$ is
$p$-addable for some $p$ iff
\begin{equation}
|\cup\{S(x): x \in V\}| \geq \chi(G). \label{eq:Weak}
\end{equation}
Necessity follows from the observation that if $(G,S)$ is $p$-addable,
then there is a graph coloring using colors from $\cup S(x)$.
Sufficiency follows since Equation~\ref{eq:Weak} implies that we can
find a coloring $f$ using colors in $\cup S(x)$.  We then let $c_i$ be
$f(x_i)$ for all vertices $x_i$ in $G$.

If $k \geq 3$, it is NP-complete to determine if $\chi (G) \leq k$, so
it follows that if $|\cup S(x)| \geq 3$, it is NP-complete to
determine if $(G,S)$ is $p$-addable for some $p$.  On the other hand,
since there is a polynomial algorithm to determine whether or not
$\chi (G) \leq 2$, there is a polynomial algorithm to determine
whether or not $(G,S)$ is $p$-addable for some $p$ in the case where
$|\cup S(x)| = 2$.

One way to measure how hard it is to reach consensus in problems such
as channel assignment or traffic phasing or physical mapping is to determine the smallest
number of ``individuals" who have to add an acceptable choice, i.e., the
smallest $p$ such that $(G,S)$ is $p$-addable.  If such a $p$ exists,
we define $I(G,S)$ to be $p$ and otherwise we say $I(G,S)$ is
undefined.  We call $I(G,S)$ the {\em inflexibility}. The higher it
is, the more inflexible the group is.  Note of course that $I(G,S) = 0$
iff $(G,S)$ is 0-addable iff $(G,S)$ is list colorable.

\subsection{Complete Bipartite Graphs}

To illustrate the notions of $p$-addability and inflexibility and give
the combinatorial flavor of the arguments, we note that complete
bipartite graphs have played an important role in the history of list
colorings and so we give a few sample results about these graphs in
this subsection. Let $K(n,m)$ denote the graph with two ``partite"
classes of vertices, $n$ vertices in one class $A$ and $m$ in the
other $B$, and all vertices in class $A$ joined to all vertices in
class $B$ by edges.

Consider $K(10,10)$. Define $S$ as follows. On the 10 vertices of
class $A$, use for the sets $S(x)$ the ten 2-element subsets of
$\{1,2,3,4,5\}$, and similarly for the 10 vertices of class
$B$. Suppose $f(x)$ is a list coloring obtained after additions change
list assignment $S$ into list assignment $S^\prime$. Suppose $f$ uses
$r$ colors on $A$ and $s$ colors on $B$. Then of course $r+s \leq 5.$
Now $\binom{5-r}{2}$
%$(\stackrel{5-r}{{\scriptstyle 2}})$
%$\left( \begin{array}{c} 5-r \\ 2 \end{array} \right)$
 sets on $A$ don't use these $r$ colors and so at least
$\binom{5-r}{2}$
%$(\stackrel{5-r}{{\scriptstyle 2}})$
%$\left( \begin{array}{c} 5-r \\ 2 \end{array} \right)$
 sets on $A$ need colors added. Similarly, at least
$\binom{5-s}{2}$ sets on B
%$(\stackrel{5-s}{{\scriptstyle 2}})$
%$\left( \begin{array}{c} 5-s \\ 2 \end{array} \right)$ sets in $B$
need colors added. This number of additions will work since all other
sets on $A$ have one of the $r$ colors and similarly for $B$. It
follows that
$$I(K(10,10),S) \leq \binom{5-r}{2} + \binom{5-s}{2}$$

%$$I(K(10,10)) \leq (\stackrel{5-r}{{\scriptstyle 2}}) + 
%(\stackrel{5-s}{{\scriptstyle 2}}).$$

%\left( \begin{array}{c} 5-r \\ 2 \end{array} \right) +
%\left( \begin{array}{c} 5-s \\ 2 \end{array} \right).$$
%\binom{5-r}{2} + \binom{5-s}{2}.$$

In fact, it is easy to show that one has equality here for $r = 3$ and
$s = 2$, thus giving us $I(K(10,10),S) = 4.$

A similar argument shows that
$I(K(\binom{m}{2},\binom{m}{2},S)$
%$I(K(\left( \begin{array}{c} \lfloor (\stackrel{m}{{\scriptstyle
%2}})\rfloor \\ 2 \end{array} \right), \left( \begin{array}{c} \lfloor
%(\stackrel{m}{{\scriptstyle 2}}) \rfloor \\ 2 \end{array} \right) )$
%
%$I(K(\left( \begin{array}{c} \lfloor m \rfloor \\ 2 \end{array}
%\right), %\left( \begin{array}{c} \lfloor m \rfloor \\ 2 \end{array}
%\right) )$
for an appropriate list assignment $S$ is given by $\lfloor {m/2}
\rfloor \lfloor {m/2 - 1} \rfloor$ if $m$ is even and $\lfloor {m/2}
\rfloor^2$ if $m$ is odd.

To give another sample result, consider $K(7,7)$ and the special case
where all sets $S(x)$ have the same size, a common assumption in
practice and theory. In particular, consider the case where all
$|S(x)| = 3$. If $|\cup S(x)| = 6$, then we shall show that
$(K(7,7),S)$ is 1-addable. Consider the seven 3-element sets $S(x)$
for $x$ in partite set $A$. A simple combinatorial argument shows that
there are integers $i$ and $j$ in $\{1,2,\ldots,6\}$ so that at most
one of these seven sets does not contain either $i$ or $j$. Add $i$ to
the set missing $i$ and $j$ if there is one, obtaining $S^\prime$. Let
$f(x)$ be $i$ or $j$ for $x$ in $A$. Then for every $y$ in partite set
$B$, $S(y) = S^\prime (y)$ has a third element different from $i$ and
$j$ and we can define $f(y)$ to be that element. Hence, $f$ is a list
coloring for $(G,S^\prime)$. We conclude that $(K(7,7),S)$ is
1-addable.

On the other hand, we shall show that there is a list assignment $S$
for $K(7,7)$ with each $|S(x)| = 3, |\cup S(x)| = 7$, and so that
$(K(7,7),S)$ is not 0-addable. On partite set $A$, use the seven sets
$\{i,i+1,i+3\}, i = 1, 2, \ldots, 7$, where addition is modulo 7. Use
the same sets on partite set $B$. Suppose there is a list coloring
$f$. We shall show that $F = \{f(x): x \in A\}$ contains one of the
sets $\{i,i+1,i+3\}$. Since this set is $S(y)$ for some $y$ in partite
set $B$, there is no way to pick a color $f(y)$ from $S(y)$. Suppose
that $F$ has consecutive $i$ and $i+1$ for some $i$, including $i =
7$. Without loss of generality, assume that 1 and 2 are in $F$. If 4
is in $F$, then $\{1,2,4\}$ is in $F$. Suppose 4 is not in $F$. Since
for some $x$ in $A$, $S(x) = \{3,4,6\}$, we must have 3 or 6 in
$F$. Similarly, using $\{4,5,7\}$, we see that 5 or 7 is in $F$. If 3
and 5 are in $F$, then $\{2,3,5\}$ is contained in $F$. Other cases
are similar. The case where $F$ has no consecutive $i$ and $i+1$ is
also similar.


\subsection{Upper Bounds on $\mathbf{I(G,S)}$}

Since we can add colors to at most each vertex, it is obvious that
$I(G,S) \leq |V(G)|$ provided that $(G,S)$ is $p$-addable for some
$p$.  (In fact, it is not hard to prove the stronger result that if
$(G,S)$ is $p$-addable for some $p$, then $I(G,S) \leq n - \omega(G),$
where $\omega(G)$ is the size of the largest clique of $G$.)

\begin{them}\label{Main} There are $(G,S)$ such that $I(G,S)/|V(G)|$
is arbitrarily close to 1.
\end{them}

\noindent 
This theorem has the interpretation that there are
situations where almost everyone has to ``sacrifice'' by taking as
acceptable an alternative not on their initial list.  In physical
mapping, there are situations where essentially every list of copies
needs to be expanded.

\addvspace{-2em}
\begin{proof}  Let $G = G_m$ =
$I_{\binom{m}{2}}$.
%$I_{(\stackrel{{\scriptstyle m}}{2})}$ + $K_{m-1}$.
   Define $S$ as follows.  On
$I_{\binom{m}{2}}$,
%$I_{(\stackrel{{\scriptstyle m}}{2})}$,
 use all pairs from
$\{1,2,\ldots,m\}$ and on $K_{m-1}$ use the pairs $\{i,i+1\}, i = 1,
2, \ldots, m-1$.  We shall show that $I(G,S) =
\binom{m-1}{2}$.
%(\stackrel{m-1}{{\scriptstyle 2}})$.
  To show $\geq$, we note that no
matter what additions are made to the sets $S(x_i)$ for $x_i$ in
$K_{m-1}$, we still need $m-1$ colors to color $K_{m-1}$.  Then the
$\binom{m-1}{2}$
%$(\stackrel{m-1}{{\scriptstyle 2}})$
vertices of
$I_{\binom{m}{2}}$
%$I_{(\stackrel{{\scriptstyle m}}{2})}$
 whose sets consist of two of
these colors cannot be colored from their sets.  The inequality $\leq$
follows if we take $f(x)$ to be $i$ if $S(x) = \{i,i+1\}$ for $x$ in
$K_{m-1}$, $f(x) = m$ if $m \in S(x)$ for $x$ in
$I_{\binom{m}{2}}$,
%$I_{(\stackrel{{\scriptstyle m}}{2})}$,
 and otherwise add $m$ to set
$S(x)$ for $x$ in
% $I_{(\stackrel{{\scriptstyle m}}{2})}$
$I_{\binom{m}{2}}$, and take
$f(x) = m$.  Since $|V(G)| =
\binom{m}{2} + m-1$,
%(\stackrel{m}{{\scriptstyle 2}}) +
%m-1$,
 it is easy to show that the theorem holds.
\end{proof}
%\begin{flushright} Q.E.D. \end{flushright}

Note that the result still holds if we add the common requirement that
all sets $S(x)$ have the same cardinality $k$.  Take $m > k$ and use
$I_{\binom{m}{k}}$
%$I_{(\stackrel{{\scriptstyle m}}{k})}$
in place of
$I_{\binom{m}{2}}$
%$I_{(\stackrel{{\scriptstyle m}}{2})}$.
 On
$I_{\binom{m}{k}}$,
%$I_{(\stackrel{{\scriptstyle m}}{k})}$,
 use all $k$-element subsets from $\{1,2,\ldots,m\}$ and on $K_{m-1}$,
use the sets $\{i,i+1,\ldots,i+k-1\}, i = 1, 2, \ldots, m-1$, where
addition is modulo $m$. Then, $I(G,S) = \binom{m-1}{k}$.
%(\stackrel{m-1}{{\scriptstyle k}})$.

\addvspace{-5em}
\section{The Trading Model}\label{Sec:Trading}

A second consensus model considers the possibility of side agreements
among individuals to lead to increased flexibility.  In particular, it
allows the trade (or purchase) of colors from one individual's
acceptable set to another's.  In the first model, we paid no attention
to where the added colors in a color set come from.  In the new model,
we think of trades as taking place in a series of steps.  A {\em
trade} from $x$ to $y$ means that we remove color $c$ from $S(x)$ and
add it to $S(y)$, i.e., we define a new list assignment $S^\prime$ by
taking $S^\prime(x) = S(x) - \{c\}, S^\prime(y) = S(y) \cup \{c\},$
and otherwise $S^\prime(u) = S(u)$. In this new model, in the physical
mapping application, we think of the possibility that a label was
incorrectly recorded in a set of possible labels of another clone and
should be moved.

We shall ask how many trades are required in order to obtain a list
assignment that has a list coloring.  We say that $(G,S)$ is {\em
$p$-tradeable} if this can be done in $p$ trades.  For instance, if we
take the path $P_k$ consisting of $k$ vertices and define $S$ by
putting the set $\{1\}$ on all but the last vertex and the set $\{2\}$
on the last vertex, then if $k \geq 4$, $(P_k,S)$ is not $p$-tradeable
for any $p$.  There are not enough 2's available. This also shows that
Equation~\ref{eq:Weak} is not sufficient for $p$-tradeability for some
$p$.  The following problem, denoted as problem $(G, c_1, c_2, \ldots,
c_r)$, is clearly related to $p$-tradeability: Given graph $G$ and
positive integers $c_1, c_2, \ldots, c_r$, is there a graph coloring
of $G$ so that for $i = 1, 2, \ldots, r$, the number of vertices
receiving color $i$ is at most $c_i$?  If $c_i$ is the number of times
that $i$ occurs in some $S(x)$, then $(G,S)$ is $p$-tradeable for some
$p$ if and only if the problem $(G, c_1, c_2, \ldots, c_r)$ has a
positive answer. The problem $(G, c_1, c_2, \ldots, c_r)$ arises in
``timetabling'' applications in scheduling.  In \cite{deWerra97} and
\cite{CRM99}, it is shown that the problem $(G, c_1, c_2, \ldots,
c_r)$ is NP-complete, even for the class of line graphs of bipartite
graphs. Hansen, Hertz, and Kuplinsky \cite{HHK} study this problem for
the special case where all $c_i$ are the same. An edge coloring
version of the same problem was shown to be NP-complete even for
bipartite graphs with maximum degree 3 in \cite{Even}. The papers
\cite{deWerra97}, \cite{deWerra99}, \cite{CRM99} consider the problem
$(G, S, c_1, c_2, \ldots, c_r)$ where $S$ is a list assignment for the
graph $G$ and we ask if there is a list coloring for $(G,S)$ solving
problem $(G, c_1, c_2, \ldots, c_r)$.  It is shown that the problem
$(G, S, c_1, c_2, \ldots, c_r)$ can be solved in polynomial time when
$G$ is a union of vertex-disjoint cliques. Dror, et al. \cite{Dror}
study the problem $(G, S, c_1, c_2, \ldots, c_r)^*$ where $\sum c_i$
is the number of vertices of $G$ and each color $i$ must be used {\em
exactly} $c_i$ times. The former is a necessary condition for the
problem to have a positive answer.  Dror, et al. show that problem
$(G, S, c_1, c_2, \ldots, c_r)^*$ is NP-complete even if the graph is
a path or a union of paths with each $|S(v)| \leq 2$.  This shows that
the problem $(G, S, c_1, c_2, \ldots, c_r)$ is also NP-complete in
this case, since if $\sum c_i$ is the number of vertices, then problem
$(G, S, c_1, c_2, \ldots, c_r)^*$ has a positive answer if and only if
problem $(G, S, c_1, c_2, \ldots, c_r)$ does. Dror, et al. also show
that if the total number of colors in all of the lists is fixed (not
part of the input), the problem $(G, S, c_1, c_2, \ldots, c_r)^*$ is
solvable in polynomial time on paths and vertex-disjoint unions of
paths.  The special case where $|\cup S(x)| = 3$ had previously been
solved in \cite{Xu}.

If $|\cup S(x)| = s \geq 3$, it is clearly NP-complete to determine if
$(G,S)$ is $p$-tradeable for some $p$.  For, given $G$ and $s$, define
$S(x) = \{1,2,\ldots,s\}$ for all $x$ in $G$.  Then $\chi(G) \leq s$
iff $(G,S)$ is $p$-tradeable for some $p$ iff $(G,S)$ is
list-colorable.  However, we have the following observation.

\begin{prop}\label{Poly} If $|\cup S(x)| = 2$, then it is polynomial to
determine whether or not $(G,S)$ is $p$-tradeable for some $p$.
\end{prop}
\begin{proof} It suffices to consider the case where $G$ is connected.  Note
that if $(G,S)$ is $p$-tradeable for some $p$, then $G$ is
2-colorable, i.e., bipartite. By connectedness, if $G$ is bipartite,
there is a unique bipartition into classes $A$ and $B$.  Then it is
easy to see that the necessary and sufficient condition for
$p$-tradeability for some $p$ is that $G$ is bipartite with classes
$A$ and $B$ and there are at least $|A|$ 1's and at least $|B|$ 2's in
$\cup S(x)$ or vice versa.
\end{proof}
%\begin{flushright} Q.E.D. \end{flushright}

Analogous to the definition of the inflexibility $I(G,S)$, we define
the {\em trade inflexibility} $I_t(G,S)$ to be the smallest $p$ so
that $(G,S)$ is $p$-tradeable, with $I_t(G,S)$ undefined if there is
no such $p$.  We note that if $I_t(G,S)$ is defined, then $I_t(G,S)
\leq |V(G)|$. For, suppose $S^\prime$ is obtained from $S$ by a
sequence of $p$ trades and $(G,S^\prime)$ has a list coloring $f$.
Then each $f(x)$ is in some $S(y)$ and the number of $x$'s for which
$f(x) = i$ is at most the number of times $i$ is in some $S(y)$. So,
we can arrange to trade the required number of $i$'s to sets assigned
to vertices $x$ for which $f(x) = i$.  There is at most one incoming
trade into each set this way.

The following theorem has the same interpretation as the analogous
theorem for $I(G,S)$.

\begin{them}\label{Main2} There are $(G,S)$ such that
$I_t(G,S)/|V(G)|$ is arbitrarily close to 1.
\end{them}

\begin{proof}
Suppose that $m > 3p+2$. Consider the graph $G = G(m,p) = K_{m-p} +
pI_{m-p}$, where $pI_{m-p}$ is $I_{m-p} + I_{m-p} + \ldots + I_{m-p}$,
with $p$ copies of $I_{m-p}$.  Define $S$ as follows. On $K_{m-p}$,
use the sets
$$\{i,i+1,m-p+1,m-p+2,\ldots,m\}$$

\noindent for $i = 1,2,\ldots,m-p-1$ and
$$\{m-p,1,m-p+1,m-p+2,\ldots,m\}.$$

\noindent On each copy of $I_{m-p}$, use the sets $\{i,i+1\}, i =
1,2,\ldots,m-p-1$, and $\{m-p,1\}$.

Let $f$ be a list coloring obtained after trades give a new list
assignment $S^\prime$.  Each $i, 1 \leq i \leq m-p,$ appears in two
sets $S(x)$ on $K_{m-p}$ and two sets $S(x)$ on each copy of
$I_{m-p}$, so in $2(p+1)$ sets in all.  Since $m > 3p+2$, and
therefore $m-qp > 2(p+1)$, any list coloring $f$ uses a color
different from $1,2,\ldots,m-p$ on each copy of $I_{m-p}$.  Thus, on
each such copy, $f$ uses a color in the set
$\{m-p+1,m-p+2,\ldots,m\}$.  It follows that none of these colors are
used by $f$ on $K_{m-p}$.  Thus, $f$ on $K_{m-p}$ must use the colors
$1,2,\ldots,m-p$ and none of these colors can be used by $f$ on any
copy of $I_{m-p}$.

We are left with $p$ colors to be used by $f$ on the $p$ copies of
$I_{m-p}$.  It follows that $f$ uses color $m-p+1$ on all vertices of
one copy of $I_{m-p}$, color $m-p+2$ on all vertices of a second copy
of $I_{m-p}, \ldots$, and color $m$ on all vertices of a $p^{th}$ copy
of $I_{m-p}$.  To obtain $S^\prime$, we must trade $m-p$ copies of
each of $m-p+1,m-p+2,\ldots,m$ to sets on copies of $I_{m-p}$. Hence,
we need a minimum of $p(m-p)$ trades.  This number suffices. Thus,
$$I_t(G,S)/|V(G)| = p(m-p)/(p+1)(m-p) \rightarrow 1$$

\noindent as $p \rightarrow \infty$.
\end{proof}
%\begin{flushright} Q.E.D. \end{flushright}

There might be some situations in which we only wish to allow trades
to go from a vertex to an adjacent (neighboring) vertex.  This might
occur in channel assignment, for example, if adjacency (interference)
corresponds to physical proximity.  Adjacency had been interpreted as
interference, which in turn could correspond to physical closeness. In
this case, we might only be willing or able to trade with a nearby
person.  It is not clear what this means in the physical mapping
application. We shall say that $(G,S)$ is {\em $p$-neighbor-tradeable}
if there is a sequence of $p$ trades, each from a vertex to an
adjacent one, that results in a list colorable list assignment. Let
$I_{t,n}(G,S)$ be the smallest $p$ so that $(G,S)$ is
$p$-neighbor-tradeable and be undefined if there is no such $p$.  In
contrast to the situation with $p$-tradeability, it is possible for
$I_{t,n}(G,S)$ to be larger than $|V(G)|$.  In fact,
$I_{t,n}(G,S)/|V(G)|$ can be arbitrarily large.  This is easy to see
and we shall indicate why below.

\section{The Exchange Model}\label{Sec:Exchange}

A variant on the trade model arises when we do not use one-way trades,
but instead use two-way exchanges.  Here, a color from $S(x)$ and a
color from $S(y)$ are interchanged at each step.  In physical mapping,
this might correspond to the case where labels of two clones are
transposed. We think of exchanges as taking place in a series of
steps.  Formally, an {\em exchange} between $x$ and $y$ means that we
find a color $c$ in $S(x)$ and a color $d$ in $S(y)$ and move $c$ to
$S(y)$ and $d$ to $S(x)$ by taking $S^\prime(x) = S(x) \cup \{d\} -
\{c\}, S^\prime(y) = S(y) \cup \{c\} - \{d\}$, and otherwise
$S^\prime(u) = S(u)$.

We shall ask how many exchanges are required in order to obtain a list
assignment that has a list coloring.  We say that $(G,S)$ is {\em
$p$-exchangeable} if this can be done in $p$ exchanges.  Note that
$(G,S)$ is $p$-exchangeable for some $p$ iff it is $q$-tradeable for
some $q$. For, if $(G,S)$ is $p$-exchangeable, then since every
exchange can be broken into two trades, it is $2p$-tradeable.  If
$(G,S)$ is $p$-tradeable for some $p$, then, as we have observed
before, we can accomplish this with $p$ trades so that every vertex
receives a trade at most once.  Every vertex that receives a trade can
therefore trade back one of its colors without affecting the result.
So, this is $p$-exchangeable.  Thus, the complexity of determining
whether or not $(G,S)$ is $p$-exchangeable for some $p$ is the same as
that of determining whether or not $(G,S)$ is $p$-tradeable for some
$p$.

Analogous to the definition of trade inflexibility $I_t(G,S)$, we
define the {\em exchange inflexibility} $I_e(G,S)$ to be the smallest
$p$ so that $(G,S)$ is $p$-exchangeable, with $I_e(G,S)$ undefined if
there is no such $p$.  By the above reasoning, we note that $I_e(G,S)
\leq I_t(G,S)$ and so if $I_e(G,S)$ is defined, $I_e(G,S) \leq
|V(G)|$.

 In the example in the proof of Theorem~\ref{Main2}, we have
$I_e(G,S) = I_t(G,S)$.  The proof applies with only minor modification
at the end. Thus, we have the following theorem.

\begin{them}\label{Main3} There are $(G,S)$ such that
$I_e(G,S)/|V(G)|$ is arbitrarily close to 1.
\end{them}

This has the same interpretation as the analogous theorem for $I(G,S)$.

Just as with the trade model, there might be situations in the
exchange model in which we only allow exchanges to go between adjacent
vertices.  We define {\em $p$-neighbor-exchangeable} and
$I_{e,n}(G,S)$ analogously to the same concepts for tradeable.

One can show that $I_{e,n}(G,S)/|V(G)|$ can be arbitrarily large.  For
instance, consider the path $P_{2k+1}$ on $2k+1$ vertices with sets $S(x) =
\{1\}$ on the first $k+1$ vertices and sets $S(x) = \{2\}$ on the last
$k$ vertices.  The only way to color this path with colors from the
set $\cup S(x) = \{1,2\}$ is to alternate colors.  Thus, we must move
2's to the left and 1's to the right.  Doing this by a series of
exchanges between neighbors is analogous to changing the identity
permutation into another permutation by transpositions of the form $(i
\; \; i+1)$. The number of transpositions needed to do this is readily
established. According to a well-known formula (see for example
\cite{Jerrum}), the number of such transpositions required to switch
an identity permutation into the permutation $\pi$ of
$\{1,2,\ldots,n\}$ is given by
$$J(\pi) = |\{(i,j): 1 \leq i < j \leq n \; \; \& \; \; \pi(i) > \pi(j)\}.$$

\noindent Here, $\pi$ is the permutation
$$1 \; \; k+2 \; \; 2 \; \; k+3 \; \; 3 \; \; k+4 \; \ldots \; \; k \;
\; 2k+1 \; \; k+1$$

\noindent and $J(\pi) = k(k+1)/2$.  Hence,
$$I_{e,n}(P_{2k+1},S)/|V(P_{2k+1})| = k(k+1)/2(2k+1) \rightarrow \infty$$

\noindent as $k \rightarrow \infty$.  An analogous proof shows that
for this
example,
$$I_{t,n}(P_{2k+1},S)/|V(P_{2k+1})| \rightarrow \infty$$

\noindent as $k \rightarrow \infty$.

\section{Concluding Remarks}

We have presented three consensus procedures individuals might use to
modify their acceptable sets in order for the group to achieve a list
colorable situation.  So far, very little is known about these
procedures.  For instance, it would be very helpful to describe
conditions under which $(G,S)$ is $p$-tradeable or $p$-exchangeable
for some $p$, at least for special classes of $G$ and $S$.  Very
little is known about the values of or bounds for the parameters
$I(G,S)$, $I_t(G,S)$, $I_{t,n}(G,S)$, $I_e(G,S)$, and $I_{e,n}(G,S)$
for specific graphs or classes of graphs and specific list assignments
or classes of list assignments.  Also, we might ask about values of or
bounds for these parameters under the extra restriction that all sets
$S(x)$ have the same fixed cardinality.  Finally, we might ask for
good algorithms for finding optimal ways to modify list assignments so
that we obtain a list colorable assignment under the different
consensus models, again at least for special classes of graphs and
classes of list assignments.

Specifically motivated by the physical mapping problem, we have posed
the following questions. Given $(G,S)$, can we subtract edges from $G$
to obtain an interval graph $H$ so that $(H,S)$ is list colorable? If
so, what is the smallest number of edges we can remove to accomplish
this?

Also, given graphs $G_1 = (V,E_1)$ and $G_2 = (V,E_2)$, with $E_1
\subset E_2$, and given a list assignment $S$ on $V$, is there an
interval graph $H = (V,E)$ with $E_1 \subset E \subset E_2$ so that
$(H,S)$ is list colorable?

Of course, we also want to ask whether the consensus models are useful
in physical mapping, and to this point these ideas are preliminary and
have not been applied in practice.

\section*{Acknowledgement}
 The authors thank Paul Dreyer for his helpful comments.

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\end{document}
