text:
\section{Introduction}

Variations on coloring problems in graph theory have involved many ways of
generating vertex colorings.  Without restrictions, the minimum number of
distinct colors needed to label the vertices of $G$ so that adjacent vertices
have different colors is the \textit{chromatic number} $\chi(G)$.  We consider
restricted colorings produced from weights on the edges and vertices.

Let $V(G)$ and $E(G)$ denote the vertex set and edge set of a graph $G$, and
let $\Gamma_G(v)$ denote the set of edges incident to a vertex $v$.  An
\textit{$S$-weighting} of a graph $G$ is a map $w\colon\, E(G)\to S$.  A
\textit{total $S$-weighting} is a map $w\colon\, E(G)\cup V(G)\to S$.  More
specifically, a $k$-weighting is an $S$-weighting with $S=\{1,\ldots,k\}$.
For a weighting $w$, let $\phi_w(v)=\sum_{e\in \Gamma_G(v)}w(e)$.
For a total weighting $w$, let $\phi_w(v)=w(v)+\sum_{e\in \Gamma_G(v)}w(e)$;
that is, each vertex is assigned the total of the weights it ``sees''.  A
weighting or total weighting $w$ is \textit{proper} if $\phi_w$ is a proper
coloring of $G$.  We seek proper $k$-weightings or proper total $k$-weightings
for small $k$.

\begin{conj}[\rm $1,2,3$-Conjecture; Kar\'onski--{\L}uczak--Thomason \cite{KLT}]
\label{123conj}
Every graph without isolated edges has a proper $3$-weighting.
\end{conj}

\begin{conj}[\rm  $1,2$-Conjecture; Przyby{\l}o--Wo\'zniak \cite{pw}]
\label{12conj}
Every graph has a proper total $2$-weighting.
\end{conj}

Toward the $1,2,3$-Conjecture, the original paper proved it for graphs with
chromatic number at most $3$.  Addario-Berry et al.~\cite{admrt} showed that
every graph without isolated edges has a proper $k$-weighting when $k=30$.
After improvements to $k=15$ in \cite{adr} and $k=13$ in \cite{wy}, Kalkowski,
Kar\'onski, and Pfender \cite{kkp} showed that every graph without isolated
edges has a proper $5$-weighting.  Toward the $1,2$-Conjecture, Przyby\l o and
Wo\'zniak \cite{pw} proved it for complete graphs and for graphs with chromatic
number at most $3$.  Kalkowski~\cite{K} showed that every graph has a proper
total $3$-weighting; furthermore, there is such a weighting with the edge
weights in one spanning tree fixed arbitrarily and the vertex weights chosen
from $\{1,2\}$.  Seamone~\cite{S} surveyed progress on these and related
problems.

List versions of the conjectures have been proposed.  A graph is
$k$-\textit{weight-choosable} if whenever each edge is given a list of $k$
available integers, a proper weighting can be chosen from the lists.
A graph is $(k,k')$-\textit{weight-choosable} if whenever each vertex has a
list of size $k$ and each edge has a list of size $k'$, a proper total
weighting can be chosen from the lists.

\begin{conj}[Bartnicki--Grytczuk--Niwczyk \cite{bgn}]\label{bgn}
Every graph without isolated edges is $3$-weight-choosable.
\end{conj}

\begin{conj}[Wong--Zhu \cite{wz}]\label{wz}
Every graph is $(2,2)$-weight-choosable. Every graph without isolated edges is
$(1,3)$-weight-choosable.
\end{conj}

These conjectures are stronger than the original conjectures, which concern
the special case where the lists consist of the smallest positive integers.

Wong, Yang, and Zhu~\cite{wyz} proved that the complete multipartite graph
$K_{n,m,1,1,...,1}$ is $(2, 2)$-weight-choosable and that complete bipartite
graphs other than $K_2$ are $(1, 2)$-weight-choosable.
Bartnicki, Grytczuk, and Niwczyk \cite{bgn} applied the Combinatorial
Nullstellensatz \cite{a} to prove Conjecture \ref{bgn}  for complete graphs,
complete bipartite graphs, and trees.  Wong and Zhu \cite{wz} applied the
Combinatorial Nullstellensatz to prove Conjecture \ref{wz} for complements of
linear forests; this includes complete graphs.  They also proved that every
tree with an even number of edges is $(1,2)$-weight-choosable.  Wong, Yang, and
Zhu~\cite{wyz} continued this approach, proving Conjecture~\ref{wz} for graphs
with maximum degree $3$.  Finally, Wong and Zhu~\cite{wz2} proved that every
graph is $(2,3)$-weight-choosable.

Our results use the ``Discharging Method'' and apply to sparse graphs.
Sparseness is imposed by bounding the \textit{maximum average degree} of $G$,
denoted ${\rm Mad}\,(G)$, which is the largest average degree among the subgraphs of
$G$: ${\rm Mad}\,(G)=\max_{H\subseteq G} \FR{2|E(H)|}{|V(H)|}$.  A consequence of our
results is that Conjectures~\ref{123conj} and \ref{12conj} hold for every graph
$G$ such that ${\rm Mad}\,(G)<8/3$.  However, that consequence is already known,
since every subgraph of a graph $G$ with ${\rm Mad}\,(G)<8/3$ has a vertex with degree
at most $2$, and therefore by induction is $3$-colorable.  The original papers
proved the conjectures for $3$-colorable graphs.

The novelty of our results is thus in the structure theorems proved by
discharging (which may be of use in solving other problems) and in the
reducibility theorems showing that various configurations cannot occur in
minimal counterexamples to the conjecture.

We note that proofs via the Combinatorial Nullstellensatz are nonconstructive,
in that the parameter space to be searched for the guaranteed weighting is
exponentially large.  Inductive proofs via discharging, such as ours, yield
polynomial-time algorithms to produce the solution, though implementation may
be complicated.  The original proofs of these conjectures for $3$-colorable
graphs are also constructive.

\medskip
Because our proofs for ${\rm Mad}\,(G)<8/3$ are fairly long, we first present in
Section~\ref{sec52} short proofs of the weaker results that the claims hold
when ${\rm Mad}\,(G)<5/2$.  The reducibility arguments in these proofs are used in
the stronger results.

To discuss both problems together, let {\it $j$-weighting} mean a $3$-weighting
when $j=3$ and a total $2$-weighting when $j=2$.  A graph is {\it $j$-bad} if
it has no proper $j$-weighting (and no isolated edge if $j=3$).  Configurations
forbidden from minimal $j$-bad graphs are {\it $j$-reducible}.

Our proofs of $j$-reducibility use the restriction of weights to values at most
$j$, so they do not extend to the list versions.  Also, unlike in most coloring
problems, vertices of degree $1$ do not immediately yield reducible
configurations, since the weight on a pendant edge affects whether its incident
edges are properly colored.

In Section~\ref{sec52} we first obtain some $3$-reducible configurations.
We next use discharging to show that every graph with average degree less than
$5/2$ contains a $3$-reducible configuration.  We assign each vertex initial
charge equal to its degree and then shift charge so that if no specified
configuration occurs, then every vertex has final charge at least $c$.
Since ${\rm Mad}\,(G)<c$ and $G'\subseteq G$ imply ${\rm Mad}\,(G')<c$ (by definition), this
proves that $G$ has a proper $3$-weighting when ${\rm Mad}\,(G)<5/2$.
In Section~\ref{sec52} we also use this method to prove that $G$
has a proper (total) $2$-weighting when ${\rm Mad}\,(G)<5/2$.  Both results use the
same discharging argument, although the sets of reducible configurations are
different.

When ${\rm Mad}\,(G)\ge 5/2$, no longer must $G$ have a configuration among those in
Section~\ref{sec52}.  We need additional reducible configurations to complete
an unavoidable set when ${\rm Mad}\,(G)<8/3$.  In Section~\ref{sec12} and
Section~\ref{sec123}, respectively, we complete the proofs of the
$1,2$-Conjecture and the $1,2,3$-Conjecture for graphs $G$ such that
${\rm Mad}\,(G)<8/3$.

\section{Reducible Configurations and ${\rm Mad}\,(G)<5/2$}\label{sec52}

In discharging arguments for sparse graphs, it is convenient to have concise
terminology for vertices satisfying degree constraints.

\begin{definition}\label{basic}
A vertex with degree $k$, at least $k$, or at most $k$ is a
\textit{$k$-vertex}, a \textit{$k^+$-vertex}, or a \textit{$k^-$-vertex},
respectively.  A \textit{$j$-neighbor} of $v$ is a $j$-vertex that is a
neighbor of $v$.

Write $N_G(v)$ for the neighborhood of $v$ in $G$ and $d_G(v)$ for its degree.
For $v\in V(G)$ and $U\subseteq N_G(v)$, let $[v,U]$ denote the set of edges
joining $v$ to $U$.

A weighting or total weighting $w$ \textit{satisfies} an edge $uv$ if
$\phi_w(u)\ne\phi_w(v)$, or equivalently if $\rw{u}{v}\ne\rw{v}{u}$, where we
define $\rw{x}{y}=\phi_w(x)-w(xy)$ when $x$ and $y$ are adjacent.

A {\em configuration} in a graph $G$ is a subgraph $C$ together with specified
degrees in $G$ for $V(C)$.  The {\em core} of the configuration is $E(C)$, and
the resulting {\em derived graph} is $G-E(C)$.
\end{definition}

Our simplest configurations consist of a vertex with specified degree and the
edges from it to certain neighbors of specified degrees.
We begin with a lemma that reduces the length of reducibility proofs:
$1$-neighbors are ``easier'' to handle than $2$-neighbors, so when we claim
that a configuration is reducible when a particular vertex has degree $1$ or
$2$, in the proof we may assume that it has degree $2$.

\begin{lem}\label{assume2}
If a vertex $z$ in a $2$- or $3$-reducible configuration $C$ has degree $1$ in
$C$ and is specified as a $2$-vertex of the full graph, then the configuration
$C'$ obtained from $C$ by instead specifying $z$ as a $1$-vertex (with
neighbor $v$) is also reducible.
\end{lem}
\begin{proof}
Let $H$ be a graph containing $C'$, and let $H'$ be the derived graph; $z$ is
isolated in $H'$.  Form $G$ by adding vertices $a$ and $b$ and edges $ab$ and
$bz$ to $H$.  Now $C$ arises in $G$, and its derived graph $G'$ arises from
$H'$ in the same way that $G$ arises from $H$.

If $H$ is a minimal $j$-bad graph, then $H'$ has a desired weighting.  Since
also the path $P_3$ has such a weighting, $G'$ has such a weighting.  Since $C$
is $j$-reducible, $G$ therefore also has such a weighting.  To
obtain the desired weighting of $H$, note that all edges are satisfied when $a$
and $b$ are deleted from the weighting of $G$, except possibly $zv$.

For $j=2$, the weight on $z$ is needed only to satisfy $zv$ in $H$ and can be
respecified so that $zv$ is satisfied.  For $j=3$, the edge $zv$ is satisfied
automatically since $d_H(v)>1$.
\end{proof}

Reducibility proofs may use some types of inferences many times.  The next
lemma enables us to express statements concisely and reduce repetitive language.
It can be stated in more generality, but for clarity we list just typical
situations in which we will use it.

\begin{lem}\label{choose}
Let $w$ be a partial $j$-weighting of a graph $G$ ($w$ is not specified
everywhere).  In the situations below, the weights on a set $S$ can be chosen
to satisfy the edges in a set $F$ if the weights on all the edges (or
vertices) incident to $F$ and not in $S$ are already known:

1) The edges of $F$ have a common endpoint $v$, incident to all edges of $S$
(possibly also $v\in S$ when $j=2$), and $|F|\le(j-1)|S|$.

2) $F$ consists of two edges, with $S$ a single edge joining them and $j=3$.
\end{lem}

\begin{proof}
Let $k=|S|$.  Since weights are chosen from $\{1,\ldots,j\}$, the sum
of $k$ weights has $1+(j-1)k$ possible values.  Each edge in $F$ uses that sum
in determining whether the values of $\phi$ differ at its endpoints.  Each edge
in $F$ thus forbids at most one value of the sum in a proper $j$-weighting.
There are at least $k(j-1)$ possible augmentations above the least value of the
sum, so when $k(j-1)\ge|F|$ the labels can be chosen to satisfy all of $F$.

Note that in (2) the weights on $F$ may be unspecified; the weight on an edge
does not affect whether it is satisfied.  Similarly, if $F=\{uv\}$, and $S$ is
a single edge incident to $v$ or is $v$ itself, and the weights of all other
items incident to $uv$ are known, then the weight on $S$ can be chosen in
$j-1$ ways to satisfy $F$.
\end{proof}

\begin{remark}\label{plan}
We use Lemma~\ref{choose} frequently in reducibility arguments, invoked without
mention in $2$-reducibility when we write ``choose $w(vz)$ to statisfy $vx$''
or ``choose $w(v)$ and $w(vz)$ to satisfy $vx$ and $vv'$''.  In
$3$-reducibility, a choice can satisfy more.  In $2$-reducibility we can choose
one weight to avoid one value, but in $3$-reducibility it can avoid two values.

Another method of satisfying an edge $uv$ is to create sufficient imbalance
between the contributions at $u$ and $v$ to guarantee that $\phi(u)\ne\phi(v)$
when the weighting is completed.  When we write ``Set $w(uv)=3$ to ensure
satisfying $vz$'', we mean that no way of choosing weights on the remaining
edges can produce $\phi(w)=\phi(v)$.  Saying that an edge is ``automatically
satisfied'' has a similar meaning.  For example, any edge joining a $1$-vertex
to a $3$-vertex is automatically satisfied for (total) $2$-weightings, while
putting weight $1$ at the $1$-vertex ensures satisfying the edge even when the
neighbor has degree $2$.

The figures for configurations show the core in bold; the derived graph $G'$ is
obtained by deleting the core.  Also, with $w'$ assumed to be a proper
$j$-weighting of $G'$, the label on an edge $e$ is $w'(e)$, and the label in a
circle at a vertex $x$ with one neighbor $u$ is $\rp{x}{u}$.  To satisfy $xu$,
the sum of the contributions at $u$ other than $w'(xu)$ must differ from
$\rp{x}{u}$.

The figures do not show cases where some of the specified vertices may be equal.
For instances where such equalities do not affect the validity of the written
argument, we make no comment about possible changes in the illustration.
\end{remark}

The task of proving reducibility for a configuration $C$ is the task of
modifying or extending a proper $j$-weighting $w'$ of the derived graph $G'$
to obtain a $j$-weighting $w$ of $G$ such that the edges in or incident to the
core of $C$ become satisfied, while the other edges of $G'$ remain satisfied.
If we do not change the weights on edges of $G'$ incident to the core, then
we do not change the fact that all edges of $G'$ not incident to the core
are satisfied.

With these preparations, we begin the reducibility arguments.  The first lemma
eliminates many degenerate cases of later configurations in which specified
vertices may be identical.

\begin{lem}\label{triangle}
The following configurations are both $2$-reducible and $3$-reducible:

(1) A $3$-cycle through two $2$-vertices and one $4^-$-vertex.

(2) A $3$-cycle through one $2$-vertex $z$ and two vertices that each may be
a $3$-vertex, a $4$-vertex with a $1$-neighbor, or a $5$-vertex with a
$1$-neighbor.  In addition, one neighbor of $z$ is allowed to be a $4$-vertex
with a $2$-neighbor other than $z$ (and no $1$-neighbor).
\end{lem}
\begin{proof}
When $G$ is a $3$-cycle, the weights can be chosen to produce colors
$\{3,4,5\}$ at the vertices, for either value of $j$.  Suppose $G\ne C_3$.
In each case, we extend a proper $j$-weighting $w'$ of a subgraph $G'$ obtain
by deleting the edges of the cycle.  The cases appear in Figure~\ref{figtri}.

For (1), let $v$ be the $3^+$-vertex on the cycle, and let $z$ and $z'$ be
the $2$-vertices.  To extend $w'$ to $w$, first set $w(zz')=1$, and also set
$w(v)=2$ if $j=2$.  This ensures satisfying $vz$ and $vz'$.  If $j=2$, then fix
$w(z)=1$, choose $w(vz)$ and $w(vz')$ to satisfy $\Gamma_{G'}(v)$, and choose
$w(z')$ to satisfy $zz'$.  If $j=3$, then require $w(vz)\ne w(vz')$ with
$w(vz)\in\{1,2\}$ and $w(vz')\in\{2,3\}$ to satisfy $zz'$.  There are three
choices for $w(vz)+w(vz')$, so we can choose them also to satisfy
$\Gamma_{G'}(v)$, since $d_{G'}(v)\le2$.

For (2), let $v$ and $v'$ be the other vertices of the triangle.
If $d_G(v)\ge4$, then let $u$ be a vertex of smallest degree in $N(v)-\{z\}$;
similarly define $u'\in N(v')$.  Form $G'$ from $G$ by deleting
$\{vz,v'z,vv'\}$ and the edges $vu$ and $v'u'$ (if they exist).
Figure~\ref{figtri} shows one of the possibilities at each of $v$ and $v'$.

We first ensure that $vz$ and $v'z$ will be satisfied by setting $w(vv')=j$
(and $w(z)=1$ if $j=2$); this will yield $\rw zv\le 3<4\le \rw vz$, since
$d_G(v)\ge3$.

For $d_G(v)=3$, choose $w(vz)$ (and $w(v)$ if $j=2$) to satisfy the one edge in
$\Gamma_{G'}(v)$.  For $d_G(v)\in\{4,5\}$ and $d_G(u)=1$, choose $w(vz)$ and
$w(vu)$ (and $w(v)$ if $j=2$) to satisfy $\Gamma_{G'}(v)$.  These cases have
extra flexibility, so that if all contributions to $\phi_w(v')$ are already
known, then $vv'$ can also be satisfied.

For $d_G(v)=4$ and $d_G(u)=2$, choose $w(vu)$ (and $w(u)$ if $j=2$) to satisfy
$\Gamma_{G'}(u)$, and then choose $w(vz)$ (and $w(v)$ if $j=2$) to satisfy
$vu$ and $\Gamma_{G'}(v)$.  In this case we do not satisfy $vv'$ using edges at
$v$.  Instead, we satisfy $vv'$ using one of the earlier cases at $v'$ after
$\phi_w(v)$ is known; this case is only allowed to occur at one of $\{v,v'\}$.
\end{proof}

\begin{figure}[h]
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\caption{Cases (1) and (2) for Lemma~\ref{triangle}\label{figtri}}
\end{figure}

\begin{lem}\label{3config}
The following configurations are $3$-reducible.

{\bf A}. A $2$-vertex or $3$-vertex having a $1$-neighbor.

{\bf B}. A $4^-$-vertex whose neighbors all have degree $2$.

{\bf C}. A $3$-vertex having two $2$-neighbors, one of which has a $2$-neighbor.

{\bf D}. A $4$-vertex having a $1$-neighbor and a $2^-$-neighbor.

{\bf E}. A $5^+$-vertex $v$ with
$3p_1+2p_2\ge d_G(v)$, where $p_i$ is the number of $i$-neighbors of $v$.
\end{lem}
\begin{proof}
Let $v$ be such a vertex in a graph $G$.  Let $U_i$ be the set of $i$-neighbors
of $v$.  Form the derived graph $G'$ as specified in Definition~\ref{basic}
(deleting the bold core), except that in addition any resulting isolated edges
are also deleted.  We show that a proper $3$-weighting $w'$ of $G'$ can
be used to obtain a proper $3$-weighting $w$ of $G$.

{\bf Case A:} {\it $d_G(v)\le3$ and $v$ has a $1$-neighbor $u$.}
As in Lemma~\ref{choose}, we can choose $w(uv)$ to satisfy the other edges at
$v$.  With $d_G(v)\ge2$, the edge $uv$ is automatically satisfied.

By Case {\bf A}, deleting the core in Cases {\bf B,C,D} leaves no isolated
edges.

\begin{figure}[h]
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\vspace{-.5pc}
\caption{Cases {\bf A, B, C} for Lemma~\ref{3config}\label{3fig52}}

\end{figure}
\vspace{-.5pc}

{\bf Case B:} {\it $d_G(v)\le 4$ and $U_2=N_G(v)$}.
Let $z$ and $z'$ be $2$-neighbors of $v$, with $N_G(z)=\{v,y'\}$ and
$N_G(z')=\{v,y\}$.  By Lemma~\ref{triangle}, $\{y,y'\}\cap\{z,z'\}=\varnothing$ (see
Figure~\ref{3fig52}B).  Let $G'=G-\{vz,vz'\}$.
If $d_G(v)=2$, then choose $w(vz)$ to satisfy $yz$ and $vz'$, and choose
$w(vz')$ to satisfy $y'z'$ and $vz$.
If $d_G(v)\in\{3,4\}$, then for $z\in N_G(v)$ with $zy\in E(G')$, choose
$w(vz)\in\{2,3\}-\{\rp{y}{z}\}$ to satisfy $yz$.  Since $d(v)\ge3\ge w'(zy)$,
such choices on all of $\Gamma_G(v)$ also satisfy $zv$.

{\bf Case C:} {\it $d_G(v)=3$ and $U_2=\{z,z'\}$, with $z$ having a
$2$-neighbor $y$.}
By Lemma~\ref{triangle}, we may assume $y\ne z'$.  Let $G'=G-\{vz,vz',zy\}$,
leaving $vx,z'y',yu\in E(G')$ (see Figure~\ref{3fig52}C).  Choose $w(vz)$ to
satisfy $zy$ and $vz'$, then $w(vz')$ to satisfy $z'y'$ and $vx$, and finally
$w(zy)$ to satisfy $yu$ and $vz$.

{\bf Case D:} {\it $d_G(v)=4$ and $N_G(v)=\{u,z,x,x'\}$ with $d_G(u)=1$
and $d_G(z)\le2$.}  By Lemma~\ref{assume2}, we may assume $d_G(z)=2$.  Let
$G'=G-\{vu,vz\}$, leaving $zy\in E(G')$ (see Figure~\ref{3fig52DE}D, where $y$
may be in $N_G(v)$).  When choosing $w(vz)$ to satisfy $zy$ and choosing
$w(vu)$ to satisfy $vz$, each has at least two possible values.  Hence they can
be chosen with three possible values for $w(vz)+w(vu)$, yielding a choice that
also satisfies $vx$ and $vx'$.

\begin{figure}[h]
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\vspace{-.5pc}
\caption{Cases {\bf D} and {\bf E} for Lemma~\ref{3config}\label{3fig52DE}}
\end{figure}
\vspace{-.5pc}

{\bf Case E:} {\it $d_G(v)\ge5$ and $3p_1+2p_2\ge d_G(v)$.}
For $z\in U_2$, let $y$ be the neighbor of $z$ in $G'$.  To satisfy $yz$ when
$y\notin U_2$ (see Figure~\ref{3fig52DE}E), we need $w(vz)\ne\rp yz$; there are
at least two such choices for $w(vz)$.  (If $y\in U_2$, then $yz$ is deleted in
$G'$; let $w(yz)=1$.  Now $w(vy)\ne w(vz)$ is needed to satisfy $yz$, leaving
three choices for $w(vy)+v(vz)$.)

Edges to $U_1$ are automatically satisfied, since $d_G(v)\ge2$.
For $z\in U_2$, the edge $zv$ will be satisfied, since $d_G(v)\ge5$ yields
$\rw vz\ge 4>3\ge w(zy)$.

It remains to satisfy $\Gamma_{G'}(v)$.
Let $\sigma=\sum_{e\in E(G)-E(G')} w(e)$.  We
need $\sigma\ne\phi_{w'}(x)-\phi_{w'}(v)$ when $x\in N_{G'}(v)$, so $\sigma$ must avoid
$d_G(v)-p_1-p_2$ values.  It suffices to show that there are $1+2p_1+p_2$
choices for $\sigma$, since we are given $2p_1+p_2\ge d_G(v)-p_1-p_2$.

Weights on edges from $v$ to $U_1$ have three choices.  Those to $U_2$ have at
least two choices, except that two such edges incident to neighboring
$2$-vertices instead have three choices for the sum of their two weights.
Starting with the smallest choices, we can make $2p_1+p_2$ augmentations to the
sum, always using choices that satisfy the constraints discussed earlier.
Hence there are enough choices for $\sigma$ to satisfy the final constraints.
\end{proof}

We now present a discharging argument to obtain an unavoidable set of
configurations.

\begin{lem}\label{52struct}
If a graph $G$ without isolated edges has average degree less than $5/2$,
then $G$ contains one of the following configurations.

{\bf A}. A $2$-vertex or $3$-vertex having a $1$-neighbor.

{\bf B}. A $4^-$-vertex whose neighbors all have degree $2$.

{\bf C}. A $3$-vertex having two $2$-neighbors, one of which has a $2$-neighbor.

{\bf D}. A $4$-vertex having a $1$-neighbor and a $2^-$-neighbor.

{\bf E}. A $5^+$-vertex $v$ with
$3p_1+p_2\ge 2d_G(v)-4$, where $p_i$ is the number of $i$-neighbors of $v$.

\end{lem}

\begin{proof}
We prove that a graph $G$ containing none of {\bf A-E} has average degree at
least $5/2$.  Give every vertex $v$ in $G$ initial charge $d_G(v)$.  Move
charge via the following rules:

\medskip
{\narrower

\noindent
(1) Each $4^+$-vertex gives $\FR32$ to each $1$-neighbor and
$\FR12$ to each $2$-neighbor.

\smallskip

\noindent
(2) Each $3$-vertex with a $2$-neighbor gives total $\frac{1}{2}$ to
its $2$-neighbors, split equally if it has two $2$-neighbors.

}
\medskip

\noindent
Let $\mu(v)$ denote the resulting charge at $v$; it suffices to check that
$\mu(v)\ge\FR52$ for all $v$.  For ${\bf Z}\in\{{\bf A,B,C,D,E}\}$, let
$\overline{{\bf Z}}$ mean ``configuration {\bf Z} does not occur in $G$''.

Case $d(v)=1$: By $\overline{{\bf A}}$, the neighbor of a $1$-vertex $v$ has degree at least
$4$, so $\mu(v)=\FR52$.

Case $d(v)=2$: By $\overline{{\bf A}}$ and $\overline{{\bf B}}$, $v$ has a $3^+$-neighbor and receives from
it $\FR14$ or $\FR12$.  If only $\FR14$, then $\overline{{\bf C}}$ implies that $v$ also
receives at least $\FR14$ from its other neighbor, so $\mu(v)\ge\FR52$.

Case $d(v)=3$: By $\overline{{\bf A}}$ and $\overline{{\bf B}}$, $v$ has no $1$-neighbor and at most two
$2$-neighbors.  Hence $v$ gives away $0$ or $\FR12$, and $\mu(v)\ge\FR52$.

Case $d(v)=4$: By $\overline{{\bf D}}$ and $\overline{{\bf B}}$, $v$ has at most one $1$-neighbor, has a
$2$-neighbor only if it has no $1$-neighbor, and has at most three
$2$-neighbors.  It loses at most $\FR32$, and $\mu(v)\ge\FR52$.

Case $d(v)\ge5$: $v$ gives $\FR32$ to each $1$-neighbor and $\FR12$ to each
$2$-neighbor.  By $\overline{{\bf E}}$,
$\mu(v)=d_G(v)-\FR12(3p_1+p_2)\ge d_G(v)-\FR12(2d_G(v)-5)=\FR52$.
\end{proof}

\begin{thm}{\label{aved}}
If $G$ has no isolated edge, and ${\rm Mad}\,(G)<\frac{5}{2}$, then $G$ has a proper
$3$-weighting.
\end{thm}
\begin{proof}
A minimal counterexample contains none of the configurations {\bf A-E} in
Lemma~\ref{3config}.  However, it has average degree less than $5/2$, and hence
it contains a configuration listed in Lemma~\ref{52struct}.  The lists are the
same except for {\bf E}.  Since a $5^+$-vertex $v$ satisfying $3p_1+p_2\ge
2d_G(v)-4$ also satisfies $3p_1+2p_2\ge d_G(v)$, every graph with ${\rm Mad}\,(G)<5/2$
contains a reducible configuration.
\end{proof}

For the $1,2$-Conjecture, we again begin with reducible configurations.
Isolated edges are now allowed, which eliminates some technicalities.
We will be able to use the unavoidable set obtained in Lemma~\ref{52struct},
but the list of $2$-reducible configurations is different.  The new technique
here is that in obtaining the proper total $2$-weighting of $G$ from such a
weighting of a subgraph $G'$, we may erase weights from some vertices and
recolor them.

Configuration {\bf B} in the next lemma is more general than is needed for
the $1,2$-Conjecture when ${\rm Mad}\,(G)<5/2$, but we will need its full generality
in the proof for ${\rm Mad}\,(G)<8/3$.

\begin{lem}\label{2config}
A minimal $2$-bad graph contains none of the following configurations.

{\bf A}. A $3^-$-vertex having a $1$-neighbor.

{\bf B}. A $4^-$-vertex having two $2^-$-neighbors.

{\bf C}. A $5^+$-vertex $v$ whose number of $2^-$-neighbors is at least
$(d(v)-1)/2$.
\end{lem}

\begin{proof}
In each case, we obtain a proper total $2$-weighting of $G$ from such a
weighting $w'$ of the derived graph $G'$.  Let $v$ be the specified vertex.
Since isolated edges have proper total $2$-weightings,
we may assume that any $1$-vertex in $G$ has a $2^+$-neighbor.  In the
extension arguments, we use Lemma~\ref{choose} frequently to choose labels.

{\bf Case A:} {\it $d(v)\le 3$, and $v$ has a $1$-neighbor $u$.}
For $d(v)=3$, let $N_{G'}(v)=\{x,x'\}$ (see Figure~\ref{2fig52}A).  Uncolor
$v$, and then choose $w(v),w(uv)\in\{1,2\}$ to satisfy $vx$ and $vx'$.
Now choose $w(u)$ to satisfy $uv$.  When $d(v)=2$, we only need $w(v)+w(uv)$
to avoid one value.

\begin{figure}[h]
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\caption{Cases {\bf A} and {\bf C} for Lemma~\ref{2config}\label{2fig52}}
\end{figure}

{\bf Case C:} {\it $d(v)\ge5$ and $v$ has at least $\FR{d(v)-1}2$
$2^-$-neighbors.}
Let $U$ be a set of $p$ such neighbors, where $p=\CL{\FR{d(v)-1}2}$, and let
$G'=G-[v,U]$ and $X=N_{G'}(v)$.  By Lemma~\ref{assume2}, for $z\in U$ we may
assume $d_G(z)=2$ and let $\{y\}=N_{G'}(z)$ (see Figure~\ref{2fig52}C).
Uncolor $z$.

By Lemma~\ref{choose}, we can choose the $p+1$ weights on
$\{v\}\cup\{vz\colon\, z\in U\}$ to satisfy $[v,X]$ in $G$, since $p+1\ge d(v)-p$.
Now choose $w(z)$ for $z\in U$ to satisfy $zy$.  Finally, since $d(v)\ge5$, we
have $\rw{v}{z}\ge 5>4\ge w(z)+w(zy)$, so $zv$ is automatically satisfied.

\smallskip
{\bf Case B:} {\it $v$ has two $2^-$-neighbors $z$ and $z'$.}
By Lemma~\ref{assume2}, we may assume $d_G(z)=d_G(z')=2$.
By Lemma~\ref{triangle}, we may assume $zz'\notin E(G)$.
Let $G'=G-\{vz,vz'\}$.  Since the edges of $G'$ incident to the core
must be satisfied in the extension to $G$, uncolor $v$, $z$, and $z'$.

{\bf Subcase 1:} {\it $d_G(v)\in\{3,4\}$}.
Let $X\!=N_G(v)-\{z,z'\}$ (see Figure~\ref{2configB}).  Let $a=\SM xX w'(vx)$.
If $a\ge 2$, then setting $w(v)=2$ ensures satisfying $vz$ and $vz'$.  Using
$|X|\le 2$, choose $w(vz)$ and $w(vz')$ to satisfy $\Gamma_{G'}(v)$.
Now choose $w(z)$ and $w(z')$ to satisfy $yz$ and $y'z'$, respectively.

Hence we may assume $a=1$, which requires $d_G(v)=3$.  If $w'(yz)=1$,
then set $w(vz')=2$ to ensure satisfying $vz$.  Next choose $w(z')$ to satisfy
$z'y'$, and then choose $w(v)$ and $w(vz)$ to satisfy $vz'$ and $vx$.
Finally, choose $w(z)$ to satisfy $yz$.

By symmetry, we may now assume $a=1$ and $w'(yz)=w'(y'z')=2$, as in the
middle in Figure~\ref{2configB}.  If $w'(y)=1$, then we can exchange $w'(y)$
and $w'(yz)$ with no effect on the satisfaction of any edge in $\Gamma_{G'}(y)$
except $yz$, thereby reaching the case in the preceding paragraph.  Hence by
symmetry we may also assume $w'(y)=w'(y')=2$.

Let $b=\rp xv$.  If $b=4$, then set $w(zv)=w(v)=w(vz')=2$ to satisfy $vx$ and
ensure satisfying $vz$ and $vz'$; then choose $w(z)$ and $w(z')$ to satisfy
$zy$ and $z'y'$, respectively.  If $b\ne4$, then set $w(v)=2$ and
$w(z)=w(zv)=w(vz')=w(z')=1$.  By $\overline{{\bf A}}$, we have $d_G(y)\ge 2$ and hence
$\rp yz>2$, so $yz$ is satisfied (similarly for $y'z'$).  These values also
satisfy $\Gamma_G(v)$.

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\caption{Case {\bf B} for Lemma~\ref{83red2}\label{2configB}}
\end{figure}

{\bf Subcase 2:} {\it $d_G(v)=2$.}
For this subcase, let $z_1=z$ and $z_2=z'$.  For $i\in\{1,2\}$, let
$\{y_i\}=N_{G'}(z_i)$, let $b_i=w'(y_iz_i)$, and let $a_i=\rp{y_i}{z_i}$, as on
the right in Figure~\ref{2config}.  To satisfy $y_iz_i$, fix $w(vz_i)=3-w(z_i)$
when $a_i$ is even and $w(vz_i)=w(z_i)$ when $a_i$ is odd.  We then must choose
$w(z_1)$ and $w(z_2)$ (and hence $w(vz_1)$ and $w(vz_2)$) to satisfy $vz_2$ and
$vz_1$.  We need $b_1+w(z_1)\ne w(v)+w(vz_2)$ and $b_2+w(z_2)\ne w(v)+w(vz_1)$.

When $a_1-a_2$ is even, set $w(v)=1$.  When $a_1$ and $a_2$ are both even,
using $w(vz_i)=3-w(z_i)$ converts the requirements to $b_1+w(z_1)\ne 4-w(z_2)$
and $b_2+w(z_2)\ne 4-w(z_1)$.  With two choices for both $w(z_1)$ and $w(z_2)$,
we can pick them so that $w(z_1)+w(z_2)\notin\{4-b_1,4-b_2\}$.  When $a_1$ and
$a_2$ are both odd, using $w(vz_i)=w(z_i)$ it suffices to choose
$w(z_1),w(z_2)\in\{1,2\}$ so that $w(z_1)-w(z_2)\notin\{1-b_1,b_2-1\}$.  Since
the difference can be any of the three values in $\{1,0,-1\}$, this also can be
done.

When $a_1$ and $a_2$ have opposite parity, we may assume that $a_1$ is even.
Now set $w(v)=3-b_1$ and $w(z_1)=w(z_2)=b_1$.  Using $w(vz_1)=3-w(z_1)$ and
$w(vz_2)=w(z_2)$, we have satisfied $vz_1$ because
$b_1+w(z_1)=2b_1\ne 3=w(v)+w(z_2)$, and we have satisfied $vz_2$ because
$b_2+w(z_2)=b_2+b_1\ne 6-2b_1=w(v)+3-w(z_1)$.
\end{proof}

\begin{thm}\label{52thm2}
If ${\rm Mad}\,(G)<5/2$, then $G$ has a proper total $2$-weighting.
\end{thm}
\begin{proof}
By Lemma~\ref{2config}, a minimal counterexample contains no configuration
listed there.  Since it has average degree less than $5/2$, it contains a
configuration listed in Lemma~\ref{52struct}.  Configurations {\bf A--D} are
all $2$-reducible, by {\bf A} and {\bf B} of Lemma~\ref{2config}.  Hence to
show that every graph with ${\rm Mad}\,(G)<5/2$ contains a $2$-reducible
configuration, it
suffices to show that a $5^+$-vertex $v$ satisfying $3p_1+p_2\ge 2d_G(v)-4$
also satisfies $2p_1+2p_2\ge d_G(v)-1$.  If the desired inequality fails, then
subtracting $2p_1+2p_2\le d_G(v)-2$ from the given inequality yields
$p_1\ge d_G(v)-2$.  Since $d_G(v)-2\ge(d_G(v)-1)/2$, the desired inequality
follows.
\end{proof}

\section{Proper Total $2$-weighting when ${\rm Mad}\,(G)<8/3$}\label{sec12}

A graph formed by adding a pendant edge at each vertex of a $3$-regular graph
has average degree $\FR52$.  It has no configuration in Lemma~\ref{2config},
since each $4$-vertex has one $1$-neighbor and three $4$-neighbors.  Further
$2$-reducible configurations will require multiple ``almost-reducible''
vertices.  We introduce two types.

\begin{definition}\label{bvert}
A \textit{$\beta$-vertex} is a  $3$-vertex having exactly one $2$-neighbor
and no $1$-neighbor.
A $\beta'$-vertex is a $2k$-vertex, where $k\geq 2$, having exactly $k-1$
neighbors of degree $1$ and no $2$-neighbor.  For $\gamma\in\{\beta,\beta'\}$,
a \textit{$\gamma$-neighbor} of $v$ is a $\gamma$-vertex in $N(v)$.
\end{definition}

We will show in Lemma~\ref{83red2} that various configurations involving such
vertices are $2$-reducible.  Theorem~\ref{83disch2} shows that these plus
the configurations in Lemma~\ref{2config} form an unavoidable set when
${\rm Mad}\,(G)<8/3$.  The argument would be shorter if adjacent $\beta'$-vertices of
degree $4$ formed a reducible configuration, but our usual method fails there.

\begin{example}\label{nonred}
Let $v$ and $v'$ be adjacent $\beta'$-vertices of degree $4$ in $G$, having
$1$-neighbors $u$ and $u'$, respectively.  As in Section~\ref{sec52}, the core
$F$ is $\{uv,vv',v'u'\}$, and $G'=G-F$.  A total $2$-weighting $w'$ of $G'$ may
assign labels as indicated in Figure~\ref{betafig}.  To extend $w'$, we need
$w(uv)+w(v)+w(vv')\in \{3,6\}$; hence these three weights must be equal.
Similarly, $w(u'v')+w(v')+w(vv')\in\{3,6\}$.  Since $w(vv')$ can take only one
value, we have forced $\phi_w(v)=\phi_w(v')$.  Hence no extension to a proper
total $2$-weighting is possible.

Another would-be-useful but non-reducible configuration consists of a
$\beta$-vertex $v$ whose $2$-neighbor $z$ has a $2$-neighbor $y$.
A total $2$-weighting $w'$ of $G'$ may assign labels as indicated in
Figure~\ref{betafig}.  The values of $\phi'_{w'}$ at the neighbors of $v$ other
than $z$ force $w(v)=w(vz)=1$.  Now satisfying $vz$ requires $w(z)=w(zy)$.
Similarly, satisfying $xy$ requires $w(y)=w(yz)$.  We conclude $w(z)=w(y)$,
but now $yz$ cannot be satisfied, since also $w(yx)=w(zv)$.
\qed
\end{example}

\vspace{-.5pc}
\begin{figure}[h]
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\caption{Non-reducible: adjacent $\beta'$-vertices, or $\beta$-vertex near
extra $2$-vertex\label{betafig}}
\end{figure}

\vspace{-.5pc}

Graphs formed by adding a pendant edge at each vertex of a $3$-regular graph
contain only configuration {\bf F} among those in Lemma~\ref{83red2}.  Although
its reducibility proof does not require the full flexibility of choosing
weights in it, Example~\ref{nonred} shows that the local argument cannot be
completed when a $\beta'$-vertex has only one $\beta'$-neighbor (of degree $4$).

Example~\ref{nonred} also shows that a $\beta$-vertex is not reducible,
even when its $2$-neighbor has another $2$-neighbor.  Nevertheless, when a
$\beta$-vertex appears in a minimal $2$-bad graph we can guarantee satisfying
all but one specified edge at that vertex.  This is useful when we can ensure
satisfying that edge, such as when its other endpoint has high degree.

\begin{lem}\label{beta2}
Let $v$ be a $\beta$-vertex with $2$-neighbor $z$ in a minimal $2$-bad graph
$G$.  Name vertices so that $N_G(v)=\{z,x,u\}$ and $N_G(z)=\{v,y\}$ (see
Figure~\ref{beta2fig}).  If $G-vz$ has a proper partial $2$-weighting $w'$ that
satisfies $\Gamma_{G}(x)$ and $\Gamma_{G}(y)$, then $G$ has a partial
$2$-weighting $w$ that satisfies the same edges other than $vu$, plus $vz$,
without changing any weights on $G-\{v,z\}$ except possibly on $yz$ and $y$.
\end{lem}

\begin{proof}
Let $G'=G-vz$.  By Lemma~\ref{2config}{\bf A}, $d(y)\geq 2$.  We want to choose
$w(v)$, $w(z)$, and $w(vz)$ to satisfy $\{xv,vz,zy\}$, leaving edges other than
$vu$ satisfied.

Let $a=\rp{y}{z}$.  If $a\ge4$, then setting $w(zv)=1$ ensures satisfying
$yz$, after which we choose $w(v)$ to satisfy $vx$ and $w(z)$ to satisfy $vz$.

If $a=3$ and $d_G(y)=3$, then $w'(y)=1$ ($y=u$ is allowed).  If $w'(yz)=2$,
then we can exchange the weights on $y$ and $yz$ and apply the previous case.
If $w'(yz)=1$, then setting $w(z)=w(zv)=1$ satisfies both $yz$ and $zv$, after
which we choose $w(v)$ to satisfy $vx$.

The remaining case is $d_G(y)=2$; let $N_G(y)=\{z,u'\}$ ($u'=u$ is allowed).
Uncolor $y$ and $yz$.  Setting $w(yz)=1$ and $w(v)=2$ ensures satisfying $zv$.
Now choose $w(vz)$ to satisfy $vx$, $w(y)$ to satisfy $yu'$, and $w(z)$ to
satisfy $yz$.
\end{proof}

\vspace{-1pc}
\begin{figure}[h]
\gpic{
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\caption{Configuration for Lemma~\ref{beta2}\label{beta2fig}}
\end{figure}

\vspace{-.5pc}

\begin{lem}\label{83red2}
The configurations below are $2$-reducible.

{\bf A}. A $3^-$-vertex having a $1$-neighbor.

{\bf B}. A $4^-$-vertex having two $2^-$-neighbors.

{\bf C}. A $5^+$-vertex $v$ whose number of $2^-$-neighbors is at least
$(d_G(v)-1)/2$.

{\bf D}. Two adjacent $\beta$-vertices.

{\bf E}. A $\beta$-vertex with a $\beta'$-neighbor.

{\bf F.} A $\beta'$-vertex of degree $4$ having two $\beta'$-neighbors of degree $4$.

{\bf G.} A $3$-vertex such that each neighbor is a $\beta$-vertex or is a
$\beta'$-vertex of degree $4$.

\end{lem}

\begin{proof}
Configurations {\bf A}--{\bf C} were shown to be $2$-reducible in
Lemma~\ref{2config}.  For {\bf D}--{\bf G}, as usual we consider a minimal
$2$-bad graph $G$ containing the specified configuration, and the derived graph
$G'$ is obtained by deleting the core, shown in bold in
Figures~\ref{fig2D}--\ref{fig2G}.  In each case we have a proper total
$2$-weighting $w'$ of $G'$ and produce a proper total $2$-weighting $w$ of $G$
by choosing weights on the deleted edges and on their endpoints, leaving all
other weights as in $w'$, with the possible exception of applying Lemma 3.3.
For each successive configuration, we know that the earlier configurations do
not occur in $G$.

\smallskip
\textbf{Case D:} {\it $v$ and $v'$ are adjacent $\beta$-vertices.}
As shown in Figure~\ref{fig2D}D, $v$ and $v'$ have degree $3$, with
$2$-neighbors $z$ and $z'$, respectively.  Let $N_G(v)=\{z,v',x\}$ and
$N_G(v')=\{z',v,x'\}$, also $N_G(z)=\{v,y\}$ and $N_G(z')=\{v,y'\}$ ($y=y'$
and/or $x=x'$ are allowed in the argument).  By Lemma~\ref{triangle}, we may
assume $z\ne z'$, $y\ne x$, and $y'\ne x'$.  Let $G'=G-\{zv,vv',v'z'\}$.

Consider first the degenerate case $zz'\in E(G)$, so $y=z'$ and $y'=z$.
This also handles the case $y=x'$ or $y'=x$ under appropriate relabeling.
Set $w(zz')=1$ and $w(vv')=2$ to ensure satisfying $zv$ and $z'v'$.
Set $w(z)=w(zv)=2$.  Now choose $w(v)$ to satisfy $zv$, choose $w(v')$ and
$w(v'z')$ to satisfy $vv'$ and $v'x'$, and choose $w(z')$ to satisfy $zz'$.

Hence we may assume that the vertices are distinct as on the left in
Figure~\ref{fig2D}D.  Let $a=w'(zy)$, $b=\rp{y}{z}$, $c=w'(vx)$, and
$d=\rp{x}{v}$.  Define $a',b',c',d'$ analogously using $y',z',v',x'$
In all subcases, set $w(vv')=2$.

{\bf Subcase 1:} {\it $d_G(y)=2$ (or $d_G(y')=2$).}  Let $N_G(y)=\{z,u\}$.
Uncolor $u$ and $yu$.  Treat $y=z'$ (which implies $u=v'$) as a special case.
When $y=z'$, set $w(zz')=w(z')=1$; in general, set $w(z)=w(z')=1$.  In both
cases, this ensures satisfying $vz$ and $v'z'$ (since $w(vv')=2$).  Now set
$w(v'z')=1$ when $y=z'$; otherwise choose $w(v'z')$ to satisfy $y'z'$.  In both
cases, next choose $w(v')$ to satisfy $v'x'$, then $w(v)$ and $w(vz)$ to
satisfy $vv'$ and $vx$.  Finally, choose $w(z)$ to satisfy $zz'$ when $y=z'$;
otherwise, choose $w(u)$ to satisfy $yu$ and then $w(yu)$ to satisfy the other
edge at $u$.

{\bf Subcase 2:} {\it $d_G(y)\ne2$}.  By $\overline{A}$, we may assume
$d(y)\ge3$.
If $c=2$ or $a=1$, then $w(vv')=2$ ensures satisfying $zv$.  Set $w(v')=2$
to guarantee satisfying $z'v'$.  Now choose $w(z'v')$ to satisfy $v'x'$, and
choose $w(z')$ to satisfy $z'y'$.  Next choose $w(zv)$ and $w(v)$ to satisfy
$vx$ and $vv'$.  Finally, choose $w(z)$ to satisfy $zy$.

We may therefore assume $c=1$ and $a=2$, and by symmetry $c'=1$ and $a'=2$.
We may also assume $w'(y)=w'(y')=2$, since otherwise we can switch weights on
$y$ and $yz$ (or on $y'$ and $y'z'$), which leaves the other edges at $y$ or
$y'$ satisfied and yields the subcase above.

With $d_G(y)\ge3$, we have $b\ge4$ (since $w'(y)=2$).  By symmetry,
$d_G(y')\ge3$ and $b'\geq 4$.  Now setting $w(z)=w(z')=1$ ensures satisfying
$\Gamma_G(z)$ and $\Gamma_G(z')$ (since $w(vv')=2$).  Finally, choose
$w(zv)+w(v)$ to avoid $d-2$ and and $w(z'v')+w(v')$ to avoid $d'-2$ (allowing
two choices for each sum) so that the sums are different.  This satisfies $vx$,
$v'x'$, and $vv'$.

\begin{figure}[h]
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\vspace{-1pc}
\caption{Cases {\bf D} and {\bf E} for Lemma~\ref{83red2}\label{fig2D}}
\end{figure}

\smallskip
\textbf{Case E:} {\it $v$ is a $\beta$-vertex with a $\beta'$-neighbor $v'$.}
Let $z$ be the $2$-neighbor of $v$, with $N_G(v)=\{x,z,v'\}$ and
$N_G(z)=\{y,v\}$.

If $d_G(v')\ge6$, then let $U$ be the set of $1$-neighbors of $v'$.  Set
$w(v')=2$ to guarantee satisfying $vv'$ (since now
$\rw{v'}{v}\ge 7>6\ge\rw{v}{v'}$).  Including $vv'$, the number
of remaining edges incident to $v'$ with unchosen weights equals the number of
edges from $v'$ to $N_G(v')-U$; by Lemma~\ref{choose}, we can choose these
weights to satisfy these edges.  The edges of $[v,U]$ are automatically
satisfied, regardless of the weights on $U$.  Finally, now that $w(vv')$ is
chosen, Lemma~\ref{beta2} allows us to complete $w$ to a proper
$2$-weighting of $G$.

We may therefore assume $d_G(v')=4$, as in Figure~\ref{fig2D}E.  Let  $u$ be
the $1$-neighbor of $v'$.  Let $G'=G-\{zv,vv',v'u\}$, leaving $v'x',v'x''\in
E(G')$.  Let $a=w'(zy)$, $b=w'(v'x')$, $b'=w'(v'x'')$, and $c=w'(vx)$.
The argument allows $y\in\{x',x''\}$.  Fix $w(u)=1$.

If $b+b'-c\geq 2$, then requiring $w(v)+w(zv)=3$ guarantees satisfying $v'v$.
Next choose $w(v'v)$ to satisfy $vx$, and choose $w(v')$ and $w(v'u)$ to satisfy
$v'x'$ and $v'x''$.  With $w(v)=3-w(zv)$, there are three choices for
$w(zv)+w(z)$, so we can choose $w(zv)$ and $w(z)$ with
$w(z)+w(zv)\ne \rp yz$ to satisfy $yz$ and $w(z)+a\ne 3-w(zv)+c+w(vv')$ to
satisfy $zv$.  We may therefore assume $b+b'-c\leq 1$.

If $c=2$ or $a=1$, then requiring $w(v)+w(vv')=3$ guarantees satisfying $zv$.
Now choose $w(zv)$ to satisfy $vx$ and then $w(z)$ to satisfy $zy$.
Finally, tentatively set $w(vv')=2$ and $w(v)=1$, and then choose $w(v')$
and $w(v'u)$ to satisfy $v'x'$ and $v'x''$.  If $vv'$ is not now satisfied,
then $w(v')+w(v'u)<4$.  Now exchange weights on $vv'$ and $v$ while increasing
$w(v')$ or $w(v'u)$ to satisfy $vv'$ and preserve the satisfaction of
$v'x'$ and $v'x''$.

Hence we may assume $c=1$ and $a=2$.  Since also $b+b'-c\leq 1$, we have
$b=b'=1$.  Tentatively set $w(v'u)=2$, and choose $w(v')$ and $w(vv')$ to
satisfy $v'x'$ and $v'x''$, with $w(v')\ge w(vv')$.  If $w(v')=2$, then
$vv'$ is automatically satisfied (since $c=1$), and Lemma~\ref{beta2}
completes the extension to $w$.  If $w(v')=1$ and the application of
Lemma~\ref{beta2} produces $w(zv)=w(v)=2$, then $vv'$ is not satisfied.
In this case, $\rp{x'}{v'},\rp{x''}{v'}=\{6,7\}$, and changing $w(v'u)$ to
$1$ satisfies $vv'$ while still satisfying $v'x'$ and $v'x''$.

\smallskip
\textbf{Case F:} {\it $v$ is a $\beta'$-vertex of degree $4$ with
$\beta'$-neighbors $z$ and $z'$ of degree $4$.} Let $N_G(v)=\{x,u,z,z'\}$.  Let
$u,y,y'$ be the $1$-neighbors of $v,z,z'$, respectively.  (see
Figure~\ref{fig2F}).

{\bf Subcase 1:} {$zz'\in E(G)$}.  In this case, we have a triangle of
$\beta'$-vertices having degree $4$.  Let $G'=G-\{vz,vz',zz',vu,zy,z'y'\}$.
Let $t$ and $t'$ be the remaining neighbors of $z$ and $z'$, respectively.
By symmetry, we need only consider two cases: $w'(zt)\ne w'(z't')$, and the
case $w'(zt)=w'(z't')=w'(vx)=c$.

If $w'(zt)\ne w'(z't')$, then by symmetry we may assume $w'(zt)=1$ and
$w'(z't')=2$.  Set $w(zz')=w(z')=w(z'v)=2$ and $w(zv)=w(v)=1$ to ensure
satisfying $zz'$ and $z'v$.  Now choose $w(vu)$ to satisfy $\Gamma_{G'}(v)$ and
choose $w(z'y')$ to satisfy $\Gamma_{G'}(z')$.  Finally, choose $w(z)$ and
$w(zy)$ to satisfy $zv$ and $\Gamma_{G'}(z)$.

In the other case, let $w(vu)=w(v)=w(vz)=w(vz')=a$.  Choose $a$ to satisfy
$vx$.  Let $w(zz')=3-a$; this ensures satisfying $vz$ and $vz'$.  With
$w(z')$ arbitrary, choose $w(z'y')$ to satisfy $z't'$, and finally choose
$w(z)$ and $w(zy)$ to satisfy $zz'$ and $zt$.

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\caption{Cases {\bf F1} and {\bf F2} for Lemma~\ref{83red2}\label{fig2F}}
\end{figure}

{\bf Subcase 2:} {\it $zz'\notin E(G)$}.  Let $G'=G-\{yz,zv,vu,vz',z'y'\}$.
Let $a=\rp{x}{v}$.  If $a\ne 6$, then requiring $w(v)+w(vz')=3$ and
$w(vz)+w(vu)=3$ satisfies $xv$, with $w(vz')$ and $w(vz)$ still choosable
freely.  Choose $w(zy)$, $w(z)$, and $w(vz)$ so that their sum avoids
$\{\rp{s}{z},\rp{t}{z}\}$, where $\{s,t\}=N_{G'}(z)$, and so that
$w(zy)+w(z)+w'(sz)+w'(tz)\ne 3-w(vz)+w(v)+w(vz')+w'(vx)$ (to satisfy $vz$).
Since $w(v)+w(vz')=3$, there are three constants for $w(zy)+w(z)+w(vz)$ to
avoid, so such a choice exists.  Finally, choose $w(vz')$, $w(z')$, and
$w(z'y')$ to satisfy $vz$ and $\Gamma_{G'}(z)$, again making their sum avoid
three known values.

Hence we may assume $a=6$.  Now choosing $w(v)=w(vu)=w(vz)$ guarantees
satisfying $vx$; let $b$ denote the value to be chosen for them.
Let $c$ be the total weight assigned by $w'$ to $\Gamma_{G'}(z')$.
If $c=2$, or if $c=3$ and $w(vx)=2$, then let $b=2$.  Otherwise, let $b=1$.
In either case, $vz'$ is guaranteed to be satisfied.  Finally, choose $w(z)$
and $w(zy)$ to satisfy $zs$ and $zt$, choose $w(vz')$ to satisfy $vz$, and
choose $w(z')$ and $w(z'y')$ to satisfy $\Gamma_{G'}(z')$.

\smallskip
\textbf{Case G:} {\it $v$ is a $3$-vertex with neighbors $z_1,z_2,z_3$ (where
$d_G(z_1)\ge d_G(z_2)\ge d_G(z_3)$) such that each is a $\beta$-vertex or is a
$\beta'$-vertex of degree $4$.}
For $i\in\{1,2,3\}$, let $y_i$ be the neighbor of $z_i$ with degree
$5-d_G(z_i)$.  When $d_G(z_i)=3$, let $x_i$ be the other neighbor of $z_i$ and
let $y'_i$ be the other neighbor of $y_i$.  When $d_G(z_i)=4$, let
$\{x_i,x'_i\}=N_G(z_i)-\{v,y_i\}$.

We first reduce to the case where $N_G(v)$ is independent.  Adjacent
$\beta$-vertices are forbidden by $\overline{{\bf D}}$.  Adjacent $\beta$- and
$\beta'$-vertices are forbidden by $\overline{{\bf E}}$.

The third possibility is that $z$ and $z'$ are adjacent $\beta'$-vertices
having a common $3$-neighbor $v$.  The situation is illustrated by deleting $u$
from the left graph in Figure~\ref{fig2F}.  Label the vertices as described
there, with $G'=G-\{vz,vz',zz',zy,z'y'\}$.  If $w'(zt)=2$ or $w'(vx)=1$, then
set $w(vz)=w(vz')=1$ and $w(z')=w(zz')=2$ to ensure satisfying $vz$ and $vz'$.
Choose $w(v)$ to satisfy $vx$, choose $w(z'y')$ to satisfy $z't'$, and
choose $w(z)$ and $w(zy)$ to satisfy $zt$ and $zz'$.

By symmetry, we may now assume $w'(zt)=w'(z't')=1$ and $w'(vx)=2$.  Also,
$w'(x)=2$, or we can switch the weights on $x$ and $vx$ to reach the case just
discussed.  Now, using $d(x)\ge3$ (since $x$ is a $\beta$- or $\beta'$-neighbor
of $v$), we have $\rp xv\ge4$.  Now set $w(zv)=w(v)=w(vz')=1$ and $w(zz')=2$
to satisfy $vx$ and ensure satisfying $vz$ and $vz'$.  Finally, set
$w(z')=1$, choose $w(z'y')$ to satisfy $z't'$, and choose $w(z)$ and $w(zy)$ to
satisfy $zt$ and $zz'$.

Hence we may assume that $N_G(v)$ is independent.  If two $\beta$-vertices in
$N_G(v)$ have a common $2$-neighbor, say $z_1$ and $z_2$ with common
$2$-neighbor $y$, then let $G'=G-\{vz_1,vz_2,z_1y,z_2y\}$.  Set
$w(vz_1)=w(vz_2)=2$ and $w(y)=1$ to ensure satisfying $z_1y$ and $z_2y$.
Now choose $w(v)$ to satisfy $vz_3$, choose $w(z_1)$ and $w(z_1y)$ to satisfy
$vz_1$ and $\Gamma_{G'}(z_1)$, and choose $w(z_2)$ and $w(z_2y)$ to satisfy
$vz_2$ and $\Gamma_{G'}(z_2)$.

Now $N_G(v)$ independent and the $2$-neighbors of $\beta$-neighbors of $v$ are
distinct.  The remaining cases are shown in Figure~\ref{fig2G}.  The argument
does not require the vertices on circles to be distinct.  Let Subcase $j$ be
the situation where $j$ neighbors of $v$ are $\beta'$-vertices.  In each
subcase, the deleted core consists of $\Gamma_G(v)$ and
$\{z_1y_1,z_2y_2,z_3y_3\}$.  To obtain $w$ from $w'$, we must satisfy these six
edges and six additional edges incident to them.  We have the freedom to choose
weights on the six deleted edges and their seven incident vertices.

We define operation $S_i$ to satisfy the edges in the $i$th ``branch'' when
$w(vz_i)$ has been specified.  If $d_G(z_i)=3$, then $S_i$ uses
Lemma~\ref{beta2} to choose $w(z_i)$, $w(z_iy_i)$, and $w(y_i)$ (plus possible
changes to weights on $y_iy'_i$ and $y'_i$ but not on $z_ix_i$ or $z_iv$) so
that $z_ix_i$, $z_iy_i$, and $y_iy'_i$ become satisfied.  If $d_G(z_i)=4$, then
$S_i$ chooses $w(z_i)$ and $w(z_iy_i)$ to satisfy $z_ix_i$ and $z_ix_i'$.
(When $d_G(z_i)=4$, automatically $z_iy_i$ is satisfied, and $w(y_i)$ is
irrelevant.)

\begin{figure}[h]
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    \special{pa 6388 967}%
    \special{pa 5828 406}%
    \special{fp}%
    \special{pa 6388 967}%
    \special{pa 5828 1527}%
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    \special{fp}%
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}
\vspace{-1pc}

\caption{Case {\bf G} for Lemma~\ref{83red2}\label{fig2G}}
\end{figure}

\textbf{Subcase 0:}
Set $w(v)=w(vz_1)=w(vz_2)=w(vz_3)=2$, and consider $i\in\{1,2,3\}$.
If $w'(x_iz_i)=1$, then $z_iv$ is automatically satisfied; apply $S_i$.
If $w'(x_iz_i)=2$, then $z_iy_i$ is automatically satisfied.  Set $w(z_i)=1$
to satisfy $z_iv$.  Choose $w(z_iy_i)$ to satisfy $z_ix_i$, and choose $w(y_i)$
to satisfy $y_iy'_i$.

\textbf{Subcase 1:}
Let $a=\rp{x_1}{z_1}$ and $a'=\rp{x_1'}{z_1}$.  When $w'(z_1x_1)=w'(z_1x'_1)=1$
and $\{a,a'\}=\{6,7\}$, set $w(z_3v)=w(z_2v)=2$.  Otherwise, set
$w(z_3v)=w(z_2v)=1$.  With $w(z_3v)$ and $w(z_2v)$ fixed, apply $S_2$ and
$S_3$.  Now choose $w(v)$ and $w(vz_1)$ to satisfy $vz_2$ and $vz_3$, with
$w(v)\le w(vz_1)$.

If we have set $w(z_3v)=w(z_2v)=2$, then $w'(z_1x_1)=w'(z_1x_1')=1$;
satisfy $vz_1$ by setting $w(z_1)=w(z_1y_1)=1$.  Since $\{a,a'\}=\{6,7\}$,
this also satisfies $\Gamma_{G'}(z_1)$.

If we have set $w(z_3v)=w(z_2v)=1$, then $w'(z_1x_1)+w'(z_1x_1')\ge3$ or
$\{a,a'\}\ne\{6,7\}$.  In the first case, $vz_1$ is automatically satisfied;
apply $S_1$.  In the second, $w(z_3v)=w(z_2v)=w'(z_1x_1)=w'(z_1x'_1)=1$ and
$\{a,a'\}\ne\{6,7\}$; choose $b\in\{6,7\}-\{a,a'\}$.  If $w(v)=1$, then $vz_1$
is automatically satisfied; apply $S_1$.  Otherwise, $w(v)=w(vz_1)=2$, since we
chose $w(v)\le w(vz_1)$.  Now choose $w(z_1)$ and $w(z_1y_1)$ with sum $b-3$.
This satisfies $vz_1$ and $\Gamma_{G'}(z_1)$.

\textbf{Subcase 2:}
Set $w(z_3v)=1$.  With $w(z_3v)$ fixed, apply $S_3$.
If $w'(z_1x_1)=2$, then setting $w(z_1v)=w(v)=1$ ensures satisfying $z_1v$
and $z_2v$; choose $w(z_2v)$ to satisfy $vz_3$ and apply $S_1$ and $S_2$.
By symmetry, we may thus assume $w'(z_ix_i)=w'(z_ix'_i)=1$ for $i\in\{1,2\}$.

If $w(z_3)+w'(z_3x_3)+w(z_3y_3)>3$, then setting $w(v)=w(vz_2)=w(vz_1)=1$
satisfies $vz_3$ and ensures satisfying $vz_2$ and $vz_1$; apply $S_2$
and $S_1$.  Hence we may also assume $w(z_3)+w'(z_3x_3)+w(z_3y_3)=3$.
Let $a_i=\rp{x_i}{z_i}$ and $a'_i=\rp{x_i'}{z_i}$, for $i\in\{1,2\}$.

If  $\{a_1,a'_1\}\neq \{5,6\}$, then set $w(vz_1)=w(v)=1$ and $w(vz_2)=2$.
Now $vz_3$ is satisfied and $vz_2$ is automatically satisfied; apply $S_2$.
Choose $b\in\{5,6\}-\{a_1,a'_1\}$.  Choose $w(z_1)$ and $w(z_1y_1)$ summing to
$b-2$.  This satisfies $vz_1$ and $\Gamma_{G'}(z_1)$.

By symmetry, we may thus assume $\{a_1,a'_1\}=\{a_2,a'_2\}=\{5,6\}$.  Let
$w(v)=2$ and $w(vz_i)=w(z_i)=w(z_iy_i)=2$ for $i\in\{1,2\}$ to satisfy all
remaining edges.

\textbf{Subcase 3:}
Set $w(v)=w(vz_1)=w(vz_2)=w(vz_3)=1$ to guarantee satisfying each $vz_i$.  Now
for each $i$ choose $w(y_iz_i)$ and $w(z_i)$ to satisfy $z_ix_i$ and $z_ix'_i$.
\end{proof}

Case {\bf F} and Case {\bf G} in Lemma~\ref{83red2} can be generalized.
If $v$ in the former case or $z_i$ in the latter is a $\beta'$-vertex of
any even degree, then the configuration remains $2$-reducible.  We omit this
since it is not needed to prove the $1,2$-Conjecture for ${\rm Mad}\,(G)<8/3$; the
more restrictive configurations in the lemma complete an unavoidable set.

\begin{lem}\label{83disch2}
If $G$ has average degree less than $8/3$, then $G$ contains one of the
following:

{\bf A}. A $3^-$-vertex having a $1$-neighbor.

{\bf B}. A $4^-$-vertex having two $2^-$-neighbors.

{\bf C}. A $5^+$-vertex $v$ whose number of $2^-$-neighbors is at least
$(d_G(v)-1)/2$.

{\bf D}. Two adjacent $\beta$-vertices.

{\bf E}. A $\beta$-vertex with a $\beta'$-neighbor.

{\bf F.} A $\beta'$-vertex of degree 4 having two $\beta'$-neighbors of degree
$4$.

{\bf G.} A $3$-vertex such that each neighbor is a $\beta$-vertex or is a
$\beta'$-vertex of degree $4$.
\end{lem}

\begin{proof}
We prove that a graph $G$ containing none of {\bf A-G} has average degree at
least $8/3$.  Give every vertex $v$ in $G$ initial charge $d_G(v)$.  Move
charge via the following rules:

\medskip
{\narrower

\noindent
(1) Each $1$-vertex takes $\FR53$ from its neighbor.
\smallskip

\noindent
(2) Each $2$-vertex takes $\FR23$ from one $3^+$-neighbor.
\smallskip

\noindent
(3) Each $3$-vertex with a $2$-neighbor takes $\FR16$ from each other neighbor.
\smallskip

\noindent
(4) Each $4$-vertex with a $1$-neighbor takes $\FR16$ from each other neighbor
not a $\beta'$-vertex.
\smallskip

}

\noindent
Let $\mu(v)$ denote the resulting charge at $v$; it suffices to check that
$\mu(v)\ge\FR83$ for all $v$.  For ${\bf Z}\in\{{\bf A},\ldots,{\bf G}\}$, let
$\overline{{\bf Z}}$ mean ``configuration {\bf Z} does not occur in $G$''.  Note that by
$\overline{{\bf B}}$, the vertex taking charge in Rule 3 or Rule 4 is a $\beta$-vertex or a
$\beta'$-vertex, respectively.

Case $d(v)=1$: By Rule 1, $v$ receives $\FR53$ and has final charge $\FR83$.

Case $d(v)=2$: By $\overline{{\bf A}}$ and $\overline{{\bf B}}$, $v$ gives no charge and receives $\FR23$
from a $3^+$-neighbor.

Case $d(v)=3$:
By $\overline{{\bf A}}$, $v$ has no $1$-neighbor.  If $v$ also has no $2$-neighbor, then by
$\overline{{\bf G}}$ at most two neighbors take charge $\FR16$ from it, so $\mu(v)\ge\FR83$.
If $v$ has a $2$-neighbor, then by $\overline{{\bf B}}$ it is a $\beta$-vertex, has only
one $2$-neighbor, and may give $\FR23$ to that $2$-neighbor.  By $\overline{{\bf D}}$ and
$\overline{{\bf E}}$, $v$ loses no other charge.  If $v$ does loses $\FR23$, then $v$ needs to
regain $\FR13$ and does so via Rule 3.

Case $d(v)=4$:
By $\overline{{\bf B}}$, $v$ has at most one $2^-$-neighbor.  If $v$ has no $1$-neighbor, then
$v$ loses at most $\FR23+\FR36$.  Hence in this case $\mu(v)>\FR83$.  If $v$
has a $1$-neighbor, then $v$ loses $\FR53$ to it and is a $\beta'$-vertex.
By $\overline{{\bf E}}$ and $\overline{{\bf F}}$, $v$ has no $\beta$-neighbor and at most one
$\beta'$-neighbor.  Hence it gives away no other charge and receives at least
$\FR26$ to reach $\mu(v)\ge \FR83$.

Case $d(v)\ge5$:
By $\overline{{\bf C}}$, $v$ has at most $\FR{d(v)-2}2$ $2^-$-neighbors.  If the inequality
is strict, then $v$ gives at most $\FR53\FR{d(v)-3}2$ to those vertices and
at most $\FR16$ to each other neighbor, yielding
$$
\mu(v)\ge d(v)-\FR53\FR{d(v)-3}2-\FR16\FR{d(v)+3}2
=\FR{d(v)}{12}+\FR94\ge \FR{32}{12}=\FR83.
$$
If $v$ has exactly $\FR{d(v)-2}2$ $2^-$-neighbors and at least one of them
is a $2$-vertex, then $d(v)\ge6$ and
$$
\mu(v)\ge d(v)-\FR53\FR{d(v)-4}2-\FR23-\FR16\FR{d(v)+2}2
=\FR{d(v)}{12}+\FR83-\FR16> \FR83.
$$
In the remaining case, $v$ is a $\beta'$-vertex with degree at least $6$.
By definition, $v$ has $\FR{d(v)-2}2$ $1$-neighbors and no $2$-neighbor.
By $\overline{{\bf E}}$, $v$ has no $\beta$-neighbor.  By Rules 3 and 4, $v$ gives charge
only to its $1$-neighbors.  Hence
$$
\mu(v)\ge d(v)-\FR53\FR{d(v)-2}2 =\FR{d(v)}{6}+\FR53\ge \FR83.
$$
\vspace{-3pc}

\end{proof}
\medskip

Since every configuration in Lemma~\ref{83disch2}
is listed in Lemma~\ref{83red2}, the following is proved.

\begin{thm}
The $1,2$-Conjecture holds for each graph $G$ such that ${\rm Mad}\,(G)<8/3$.
\end{thm}

\section{Proper $3$-weighting when ${\rm Mad}\,(G)<8/3$}\label{sec123}

For the discussion of proper $3$-weightings, again it will be helpful to have
notation for special types of vertices.  The definition of $\beta$-vertex is
the same as before, but instead of $\beta'$-vertices we introduce
$\alpha$-vertices and $\gamma$-vertices.

\begin{definition}
An \textit{$\alpha$-vertex} is a $2$ vertex with a $2$-neighbor.
A \textit{$\beta$-vertex} is a $3$-vertex with a $2$-neighbor.
A \textit{$\gamma$-vertex} is a $4$-vertex with a $1$-neighbor
or is a $3$-vertex with an $\alpha$-neighbor or two $2$-neighbors.
\end{definition}

\begin{lem}\label{83disch3}
If $G$ has average degree less than $8/3$, then $G$ contains one of the
following:

{\bf A}. A $2$-vertex or $3$-vertex having a $1$-neighbor.

{\bf B}. A $4^-$-vertex whose neighbors all have degree $2$.

{\bf C}. A $3$-vertex having an $\alpha$-neighbor and another $2$-neighbor.

{\bf D}. A $4$-vertex having a $1$-neighbor and a $2^-$-neighbor.

{\bf E}. A $5^+$-vertex $v$ with
$3p_1+2p_2\ge d(v)$, where $p_i$ is the number of $i$-neighbors of $v$.

{\bf F}. Two adjacent $\gamma$-vertices.

{\bf G}. A $3$-vertex with two $\gamma$-neighbors.

{\bf H}. A $6$-vertex or $7$-vertex having a $1$-neighbor and four
$\gamma$-neighbors.

{\bf I}. A $5$-vertex having a $1$-neighbor and three $\gamma$-neighbors.

{\bf J}. A $4$-vertex with $p+q+r\ge5$, where $p,q,r$ are its numbers of
$2$-neighbors, $\gamma$-neighbors, and $\alpha$-neighbors, respectively.

{\bf K}. A $\gamma$-vertex whose $3^+$-neighbors are all $\beta$-vertices.
\end{lem}

\begin{proof}
We prove that a graph $G$ containing none of {\bf A-K} has average degree at
least $8/3$.  Give every vertex $v$ in $G$ initial charge $d(v)$.
Move charge via the following rules:

\medskip
{\narrower

\noindent
(1) Each $1$-vertex takes $\FR53$ from its neighbor.
\smallskip

\noindent
(2) Each $\alpha$-vertex takes $\FR23$ from its $3^+$-neighbor.
\smallskip

\noindent
(3) Each $2$-vertex that is not an $\alpha$-vertex takes $\FR13$ from each
neighbor.
\smallskip

\noindent
(4) Each $\gamma$-vertex takes $\FR13$ from each $3^+$-neighbor that is not
a $\beta$-vertex.

}
\medskip

\noindent
Let $\mu(v)$ denote the resulting charge at $v$; it suffices to check that
$\mu(v)\ge\frac{8}{3}$ for all $v$.  For
${\bf Z}\in \{{\bf A},\ldots,{\bf K}\}$, let $\overline{{\bf Z}}$ mean ``configuration
${\bf Z}$ does not occur in $G$''.  Fix a vertex $v$ and let $p_i$, $q$, and
$r$ count its $i$-neighbors, $\gamma$-neighbors, and $\alpha$-neighbors,
respectively.

Case $d(v)=1$: By Rule 1, $v$ receives $\FR53$ and has final charge $\FR83$.

Case $d(v)=2$: By $\overline{{\bf A}}$, $v$ gives no charge; via Rule 2 or Rule 3, $v$
receives $\FR23$.

Case $d(v)=3$: By $\overline{{\bf A}}$ and $\overline{{\bf B}}$, $v$ has no $1$-neighbor and at most two
$2$-neighbors.

If $v$ has no $2$-neighbor, then $v$ is not a $\gamma$-vertex or a
$\beta$-vertex.  It loses nothing via Rule 2 or 3, and via Rule 4 it gives
$\FR13$ to each $\gamma$-neighbor.  By $\overline{{\bf G}}$, $v$ loses at most $\FR13$, and
$\mu(v)\ge\FR83$.

If $v$ has a $2$-neighbor, then $v$ is a $\beta$-vertex and loses nothing via
Rule 4.  By $\overline{{\bf B}}$ and $\overline{{\bf C}}$, $v$ gives at most $\frac{2}{3}$ to its
$2$-neighbors, with equality only if it is a $\gamma$-vertex.  In this case, by
$\overline{{\bf A}}$ and $\overline{{\bf B}}$, $v$ has a $3^+$-neighbor $u$.  By $\overline{{\bf F}}$, $u$ is not a
$\gamma$-vertex, and by $\overline{{\bf K}}$ not all choices of $u$ are $\beta$-vertices.
Hence from some $3^+$-neighbor $v$ receives $\FR13$, and $\mu(v)\ge\FR83$.

Case $d(v)=4$:
If $v$ has a $1$-neighbor, then $v$ is a $\gamma$-vertex.  By $\overline{{\bf D}}$ it has no
other $2^-$-neighbor and loses at most $\FR53$.  Its other neighbors are
$3^+$-vertices.  By $\overline{{\bf F}}$, none is a $\gamma$-vertex, and by $\overline{{\bf K}}$ they are not
all $\beta$-vertices.  Hence $v$ receives at least $\FR13$ via Rule 4, and
$\mu(v)\ge\FR83$.

If $v$ has no $1$-neighbor, then $v$ is not a $\gamma$-vertex.  By $\overline{{\bf J}}$,
$p+q+r\le 4$.  An $\alpha$-neighbor contributes to both $p$ and $r$.  Hence $v$
loses exactly $(p+q+r)/3$, yielding $\mu(v)\ge\FR83$.

Case $d(v)\ge5$:
The charge lost by $v$ is $\FR13(5p_1+p_2+q+r)$.  Hence $v$ is happy if
$5p_1+p_2+q+r\le 3d(v)-8$.  Also, configurations with $3p_1+2p_2\ge d(v)$
are forbidden, by $\overline{{\bf E}}$.  Hence if $v$ is not happy and is not in a
configuration forbidden by $\overline{{\bf E}}$, then
\begin{equation}\label{ineq}
5p_1+p_2+q+r\ge 3d(v)-7 \quad\textrm{and}\quad 3p_1+2p_2\le d(v)-1.
\end{equation}
Since $r\le p_2$ and $q\le d(v)-p_1$, the first inequality above yields
$4p_1+2p_2\ge 2d(v)-7$.  Eliminating $2p_2$ from the two inequalities then
yields $2d(v)-7-4p_1\le d(v)-1-3p_1$, which simplifies to $d(v)-6\le p_1$.
Since also $p_1\le \FL{(d(v)-1)/3}$, we obtain $d(v)\le8$.

If $p_1=0$, then substituting $q+r\le d(v)$ in (1) yields
$(d(v)-1)/2\ge 2d(v)-7$, which simplifies to $d(v)\le13/3$.  Hence we may
assume $p_1\ge1$.  We consider below the remaining unexcluded possibilities for
$(p_1,p_2,r,q)$.  In each case these are the choices allowed by (\ref{ineq}).

For $d(v)=5$, the remaining case is $(1,0,0,q)$ with $q\in\{3,4\}$, forbidden
by $\overline{{\bf I}}$.

For $d(v)=6$, the remaining case is $(1,1,1,4)$, forbidden by $\overline{{\bf H}}$.

For $d(v)=7$, the case $(1,p_2,r,q)$ requires $p_2\le1$, which yields
$5p_1+2p_2+q\le 12 <14$, so $v$ remains happy.  Hence the remaining case is
$(2,0,0,q)$ with $q\in\{4,5\}$, forbidden by $\overline{{\bf H}}$.

For $d(v)=8$, with $p_1\le 2$ and $3p_1+2p_2\le 7$, we have
$5p_1+2p_2\le 10$.  At most $16/3$ is lost, and hence $\mu(v)\ge\FR83$.
\end{proof}

The next lemma explains the role of $\gamma$-vertices.

\begin{lem}\label{gamma}
Let $v$ be a $\gamma$-vertex having a $3^+$-neighbor $x$.
Define $F\subseteq E(G)$ as follows:

$F=\{vu\}$ if $v$ is a $4$-vertex with $1$-neighbor $u$,

$F=\{vz,vz'\}$ if $v$ is a $3$-vertex with $2$-neighbors $z$ and $z'$,

$F=\Gamma_G(z)$ if $v$ is a $3$-vertex with $\alpha$-neighbor $z$.

\noindent
Given any weighting of $G-F$, weights in $\{1,2,3\}$ can be chosen on $F$ to
satisfy all edges in $F$ or incident to edges of $F$ except $vx$, without
changing the weights on edges not in $F$.
\end{lem}
\begin{proof}
Figure~\ref{figgamma} shows $F$ in bold; the weight on $vx$ is fixed.
When $v$ is a $4$-vertex, choose $w(vu)$ to satisfy the two edges from $v$
to $N_G(v)-\{x,u\}$.  When $v$ is a $3$-vertex with $2$-neighbors $z$ and
$z'$ having neighbors $y$ and $y'$ other than $v$, choose $w(vz)$ to
satisfy $vz'$ and $zy$, and choose $w(vz')$ to satisfy $vz$ and $z'y'$.
When $v$ is a $3$-vertex with $\alpha$-neighbor $z$ having neighbor $y$ other
than $v$, let $x'$ and $y'$ be the remaining neighbors of $v$ and $y$.
Choose $w(vz)$ to satisfy $vx'$ and $zy$, and choose $w(zy)$ to satisfy
$vz$ and $yy'$.
\end{proof}

\begin{figure}[h]
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    \hbox{\vrule depth0.828in width0pt height 0pt}%
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\caption{Three cases for Lemma~\ref{gamma}, with $F$ in bold\label{figgamma}}
\end{figure}

In employing Lemma~\ref{gamma}, the difficulty is ensuring that the edge $vx$
will be satisfied.  Generally, we will need to ensure that some edge is
satisfied regardless of the choice of weight on some incident edge.  When
$v$ is a $\gamma$-vertex, let $F_v$ denote the set of one or two edges
designated as $F$ in Lemma~\ref{gamma} (bold in Figure~\ref{figgamma}).

Like Lemma~\ref{triangle}, the next lemma excludes degenerate cases for the
reducibility proofs.

\begin{lem}\label{4cycle}
Let $z$ and $z'$ be $\beta$-vertices having respective $2$-neighbors $y$ and
$y'$ that are equal or adjacent.  The following cases lead to
$3$-reducible configurations:

(1) $zz'\in E(G)$.

(2) $z$ and $z'$ have a common neighbor $v$ with a $1$-neighbor $u$, such that
$d_G(v)\in\{4,5\}$.

\end{lem}
\begin{proof}
See Figure~\ref{fig4cyc} for these cases.

(1) Suppose $zz'\in E(G)$.  If $y=y'$, then Lemma~\ref{triangle} applies.
If $yy'\in E(G)$, then let $G'=G-\{zz',zy,yy',z'y'\}$.  Set $w(yy')=1$ to
ensure satisfying $zy$ and $z'y'$.  Set $w(zy)=1$.  Now choose $w(z'y')$ to
satisfy $yy'$ and $zz'$, and choose $w(zz')$ to satisfy $zx$ and $z'x'$, where
$N_G(z)=\{z',y,x\}$ and $N_G(z')=\{z,y',x'\}$ ($x=x'$ is allowed).

(2) By Case (1), we may assume $zz'\notin E(G)$, so $x\ne z'$ and $x'\ne z$.

(2a) If $y=y'$, then let $G'=G-\{vz,vz',zy,z'y',vu\}$.  Set $w(zy)=w(z'y')=1$
to ensure satisfying $zy$ and $z'y'$.  Next choose $w(zv)\in\{2,3\}$ to satisfy
$zx$ and choose $w(z'v)\in\{2,3\}$ to satisfy $z'x'$.  To ensure satisfying
$vz$ and $vz'$, we now choose $w(vu)$ to satisfy $\Gamma_{G'}(v)$ (if
$d_G(v)=5$) or $w(vu)\in\{2,3\}$ to satisfy $\Gamma_{G'}(v)$ (if $d_G(v)=4$).

(2b) If $yy'\in E(G)$, then let $G'=G-\{vz,vz',zy,z'y',yy',vu\}$.  Set
$w(yy')=1$ to ensure satisfying $zy$ and $z'y'$.
Set $w(zy)=1$ and $w(vz')=3$ to ensure satisfying $zv$.
Next choose $w(z'y')$ to satisfy $yy'$ and $z'x'$.
Two choices of $w(vz)$ will satisfy $zx$.  Along with the three choices
available for $w(vu)$
these choices can be made to satisfy $vz'$ and $\Gamma_{G'}(v)$.
\end{proof}

\begin{figure}[h]
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\caption{Three cases for Lemma~\ref{4cycle}\label{fig4cyc}}
\end{figure}

\begin{lem}\label{83red3}
The following configurations are $3$-reducible.

{\bf A}. A $2$-vertex or $3$-vertex having a $1$-neighbor.

{\bf B}. A $4^-$-vertex whose neighbors all have degree $2$.

{\bf C}. A $3$-vertex having an $\alpha$-neighbor and another $2$-neighbor.

{\bf D}. A $4$-vertex having a $1$-neighbor and a $2^-$-neighbor.

{\bf E}. A $5^+$-vertex $v$ with
$3p_1+2p_2\ge d(v)$, where $p_i$ is the number of $i$-neighbors of $v$.

{\bf F}. Two adjacent $\gamma$-vertices.

{\bf G}. A $3$-vertex with two $\gamma$-neighbors.

{\bf H}. A vertex $v$ such that $p_1+2q\ge d(v)$ and $p_1+q> 4$, where $p_1$
is the number of $1$-neighbors and $q$ is the number of $\gamma$-neighbors of
$v$.

{\bf I}. A $5$-vertex having a $1$-neighbor and three $\gamma$-neighbors.

{\bf J}. A $4$-vertex with (1) two $\alpha$-neighbors, (2) an $\alpha$-neighbor,
another $2$-neighbor, and a $\gamma$-neighbor, or (3) a $2$-neighbor and three
$\gamma$-neighbors.

{\bf K}. A $\gamma$-vertex whose $3^+$-neighbors are all $\beta$-vertices.
\end{lem}

\begin{proof}
Lemma~\ref{3config} shows that {\bf A-E} are $3$-reducible.  Let $G$ be a
minimal $3$-bad graph containing one of {\bf F-K}.  When $z$ is a
$\gamma$-vertex with a $3^+$-neighbor $v$ and $w(zv)$ has been chosen, the
phrase ``apply Lemma~\ref{gamma} to $z$'' means ``apply Lemma~\ref{gamma} to
choose weights on $F_z$ to satisfy $F_z$ and the edges incident to them other
than $zv$''.  In each case, we extend a proper $3$-weighting $w'$ of a proper
subgraph $G'$ of $G$ to a proper $3$-weighting $w$ of $G$.

The phrase ``Figure $n$ is accurate'' means that the vertices in the
illustration are known to be distinct, except possibly for
non-$\gamma$-vertices on circles, to which no edges of the core are adjacent.
There are three types of $\gamma$-vertices.  Let those of degree $4$ be
{\it $\gamma_4$-vertices}, those of degree $3$ with an $\alpha$-neighbor be
{\it $\gamma_{3a}$-vertices}, and those of degree $3$ with two $2$-neighbors be
{\it $\gamma_{3b}$-vertices}.

\smallskip
{\bf Case F:} {\it $v$ and $v'$ are adjacent $\gamma$-vertices.}
Let $G'=G-vv'-F_v-F_{v'}$.  By symmetry, the first subcase covers when $v$ or
$v'$ is a $\gamma_{3b}$-vertex.  By Lemma~\ref{triangle}, $v$ and $v'$ do
not have a common $2$-neighbor.

{\bf Subcase 1:} {\it $v$ is a $\gamma_{3b}$-vertex.}
Let $N_G(v)=\{v',z,z'\}$.  By Lemma~\ref{triangle}, we may assume
$zz',zv',z'v'\notin E(G)$, so Figure~\ref{figF12} is accurate.  Set $w(vv')=3$
to ensure satisfying  $vz$ and $vz'$.  Apply Lemma~\ref{gamma} to $v'$.
Choose $w(vz)$ to satisfy $\Gamma_{G'}(z)$.  Choose $w(vz')$ to satisfy $vv'$
and $\Gamma_{G'}(z')$.

\begin{figure}[h]
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\caption{Cases {\bf F1} and {\bf F2} for Lemma~\ref{83red3}\label{figF12}}
\end{figure}

{\bf Subcase 2:} {\it $v$ and $v'$ are both $\gamma_{3a}$-vertices.}
Let $z$ and $z'$ be the $\alpha$-neighbors of $v$ and $v'$, and let
$y$ and $y'$ be the $2$-neighbors of $z$ and $z'$.  By Lemma~\ref{4cycle},
Lemma~\ref{triangle}, and {\bf B}, Figure~\ref{figF12} is accurate.  Let $x$
and $x'$ be the remaining neighbors of $v$ and $v'$, respectively.

Choose $w(vz)$ to satisfy $zy$.  Now choose $w(v'z')$ to satisfy $vv'$ and
$z'y'$.  Next choose $w(vv')$ to satisfy $vx$ and $v'x'$.  Finally, choose
$w(zy)$ to satisfy $vz$ and $\Gamma_{G'}(y)$, and choose $w(z'y')$ to satisfy
$v'z'$ and $\Gamma_{G'}(y')$.

{\bf Subcase 3:} {\it $v$ and $v'$ are both $\gamma_{4}$-vertices.}
Let $N(v)=\{v',z_1,z_2,u\}$ and $N(v')=\{v,z_1',z_2',u'\}$, with
$d_G(u)=d_G(u')=1$.  For $i\in\{1,2\}$, let $a_i=w'(vz_i)$ and $b_i=\rp{z_i}v$;
similarly define $a_i'$ and $b_i'$ using $\{v',z_1',z_2'\}$.  Since
$d_G(u),d_G(u')=1$, Figure~\ref{figF34} is accurate.

If $a_1+a_2>a_1'+a_2'$, then set $w(uv)=3$ to ensure satisfying $vv'$.  Next
choose $w(vv')$ to satisfy $vz_1$ and $vz_2$, and choose $w(v'u')$ to satisfy
$v'z_1'$ and $v'z_2'$.

By symmetry, we may thus assume $a_1+a_2=a_1'+a_2'$.  If $b_1\ne a_2+4$, then
choose $w(vv')\in\{1,3\}$ to ensure satisfying $vz_1$.  Now choose
$w(v'u')$ to satisfy $v'z_1'$ and $v'z_2'$, and then choose $w(vu)$ to satisfy
$vz_2$ and $vv'$.  Hence by symmetry we may assume that each entry of
$(b_1,b_2,b_1',b_2')$ exceeds the corresponding entry of $(a_2,a_1,a_2',a_1')$
by exactly $4$.  Now setting $w(uv)=w(vv')=3$ and $w(v'u')=2$ completes the
extension.

\begin{figure}[h]
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\caption{Cases {\bf F3} and {\bf F4} for Lemma~\ref{83red3}\label{figF34}}
\end{figure}

{\bf Subcase 4:} {\it $v$ is a $\gamma_{3a}$-vertex and $v'$ is a
$\gamma_{4}$-vertex.}
Let $N_G(v)=\{v',z,u\}$, with $z$ being the $\alpha$-vertex.  Let $y$ be the
$2$-neighbor of $z$, with $N_G(y)=\{z,y'\}$.  Let $N_G(v')=\{v,x,x',u'\}$, with
$d_G(u')=1$.  By Lemma~\ref{triangle} and {\bf D}, Figure~\ref{figF34} is
accurate.  Let $a=w'(v'x)+w'(v'x')$, $b=w'(vu)$, and $c=w'(yy')$.

If $a\ne b$, then choose $w(u'v')\in\{1,3\}$ to ensure satisfying $vv'$.
Now choose $w(vv')$ to satisfy $v'x$ and $v'x'$, choose $w(vz)$ to satisfy
$vu$ and $yz$, and choose $w(zy)$ to satisfy $vz$ and the other edge at $y$.

If $a=b$, then set $w(u'v')=c$.  Now choose $w(vv')$ to satisfy $v'x$ and
$v'x'$.  Next choose $w(vz)$ to satisfy $vv'$, $yz$, and $uv$, which succeeds
because $vv'$ and $yz$ both are satisfied if and only if $w(vz)\ne c$.
Finally choose $w(zy)$ to satisfy $vz$ and the other edge at $y$.

{\bf Case G:} {\it $v$ is a $3$-vertex having two $\gamma$-neighbors $z$ and
$z'$.}
Let $N_G(v)=\{z,z',x\}$.  In each case, we will
let $G'=G-\{vz,vz'\}-F_z-F_{z'}$ (and $w'$ is a proper $3$-weighting of $G'$).
By {\bf F}, $zz'\notin E(G)$.  By Lemma~\ref{triangle}, neither $z$
nor $z'$ shares a $2$-neighbor with $v$ or has a $2$-neighbor adjacent to a
$2$-neighbor of $v$.  Hence Figures~\ref{figG12} and~\ref{figG3} are accurate.

{\bf Subcase 1:} {\it $z$ is a $\gamma_{3a}$-vertex.}
Let $N_G(z)=\{v,y,u\}$, with $y$ the $\alpha$-neighbor of $z$.  Let $y'$ be the
$2$-neighbor of $y$, with $N_G(y')=\{y,y''\}$.  Let $a=w'(vx)$, $b=w'(zu)$, and
$c=w'(y'y'')$, as shown on the left in Figure~\ref{figG12}.

If $b\ne a$, then choose $w(vz')\in\{1,3\}$ to ensure satisfying $vz$.
With $w(vz')$ known, apply Lemma~\ref{gamma} to $z'$.  Next choose $w(vz)$ to
satisfy $vx$ and $vz'$.  With $vz$ automatically satisfied and $w(vz)$ chosen,
it now suffices to apply Lemma~\ref{gamma} to $z$.

Hence we may assume $b=a$.  In this case, set $w(vz')=c$ and apply
Lemma~\ref{gamma} to $z'$.  Choose $w(vz)$ to satisfy $vx$ and $vz'$.
Now choose $w(zy)$ to satisfy $vz$, $zu$, and $yy'$; this succeeds because
$yy'$ and $vz$ both forbid $w(zy)=c$, so at most two choices for $w(zy)$ are
forbidden.  Finally, choose $w(yy')$ to satisfy $zy$ and $y'y''$.

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\vspace{-.5pc}

\caption{Cases {\bf G1} and {\bf G2}  for Lemma~\ref{83red3}\label{figG12}}
\end{figure}

{\bf Subcase 2:} {\it $z$ is a $\gamma_{3b}$-vertex.}
Let $N_G(z)=\{v,y_1,y_2\}$, with $N_G(y_1)=\{z,y_1'\}$ and
$N_G(y_2)=\{z,y_2'\}$.  By Subcase 1, we may assume that $z'$ is not a
$\gamma_{3a}$-vertex.

Suppose that $z'$ is a $\gamma_4$-vertex, with $1$-neighbor $u$, as in
Figure~\ref{figG12} in the middle.  Set $w(vz)=3$ to ensure satisfying
$zy_1$ and $zy_2$.  Now $w(uz')$ has two choices that satisfy $vz'$, and
$w(vz')$ has two choices that satisfy $vx$.  With at least three choices for
the sum $w(uz')+w(vz')$, the two edges in $\Gamma_{G'}(z')$ can also be
satisfied.  Now choose $w(zy_1)$ to satisfy $\Gamma_{G'}(y_1)$ and
$w(zy_2)$ to satisfy $zv$ and $\Gamma_{G'}(y_2)$.

Hence we may assume that $z'$ is also a $\gamma_{3b}$-vertex, as on the right
in Figure~\ref{figG12}.  If $\rp xv\ne6$, then set $w(vz)=w(vz')=3$.  This
satisfies $vx$ and also ensures satisfying $F_z$ and $F_{z'}$.  Choose
$w(zy_1)$ to satisfy $y_1y_1'$, and choose $w(zy_2)$ to satisfy $y_2y_2'$ and
$zv$.  Choose weights on $F_{z'}$ by the same method.

If $\rp xv=6$ and $w'(y_1y_1')\ne 3$, then set $w(vz)=2$ and $w(vz')=3$ to
satisfy $vx$ and ensure satisfying $zy_1$.  Choose $w(zy_1)\in\{2,3\}$ to
satisfy $y_1y_1'$ and ensure satisfying $zy_2$.  Now choose $w(zy_2)$ to
satisfy $y_2y_2'$ and $vz$.  Since setting $w(vz')=3$ ensures satisfying
$F_{z'}$, we can choose weights on $F_{z'}$ to finish as in the preceding
paragraph.

Hence we may assume that $\rp xv=6$ and (by symmetry) that all the edges of
$G'$ incident to the $2$-neighbors of $z$ and $z'$ have weight $3$ under $w'$.
Since we may assume by Subcase 1 that neither $z$ nor $z'$ is a
$\gamma_{3a}$-vertex, we can complete the extension by giving all the missing
edges weight $1$, unless $w'(vx)=1$.  In that case, just change $w(vz)$ to $3$
and choose $w(zy_i)\in\{2,3\}$ to satisfy $y_iy_i'$, for $i\in\{1,2\}$.

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\caption{Case {\bf G3}  for Lemma~\ref{83red3}\label{figG3}}
\end{figure}

{\bf Subcase 3:} {\it $z$ and $z'$ are both $\gamma_4$-vertices.}
Let $N_G(z)=\{v,y_1,y_2,u\}$ and $N_G(z')=\{v,y_1',y_2',u'\}$, with
$d_G(u)=d_G(u')=1$.  Let $a=w'(vx)$.  For $i\in\{1,2\}$, let $b_i=w'(zy_i)$ and
$b_i'=\rp{y_i}z$, as on the left in Figure~\ref{figG3}.

Let $b=b_1+b_2$.  If $b\ne a$, then choose $w(vz')\in\{1,3\}$ to ensure
satisfying  $vz$.  With $w(vz')$ fixed, choose $w(z'u')$ to satisfy
$\Gamma_{G'}(z')$.  Now choose $w(vz)$ to satisfy $vx$ and $vz'$, and then
choose $w(zu)$ to satisfy $\Gamma_{G'}(z)$ and complete the extension.

If $b_1'\ne b_2+4$, then choose $w(vz)\in\{1,3\}$ to ensure satisfying
$zy_1$.  Now restrict $w(u'z')$ to two choices that satisfy $vz'$, and restrict
$w(vz')$ to two choices that satisfy $vx$.  Having at least three choices
for the sum $w(u'z')+w(vz')$ allows also satisfying the edges of
$\Gamma_{G'}(z')$.  Finally, choose $w(uz)$ to satisfy $zy_2$ and $zv$ and
complete the extension.

By symmetry, the only remaining case is $b=a$, $b_1'=b_2+4$, and $b_2'=b_1+4$,
and similarly for $\Gamma_{G'}(z')$, as on the right in Figure~\ref{figG3}.
Now $w(vz)+w(zu)<4$ satisfies $zy_1$ and $zy_2$, and $w(uz)\ne w(vz')$
satisfies $vz$.  Similarly, $w(vz')+w(z'u')>4$
satisfies $z'y'_1$ and $z'y'_2$, and $w(u'z')\ne w(vz)$ satisfies $vz'$.  Set
$w(vz)=w(zu)=1$ and $w(z'u')=3$, and choose $w(vz')\in\{2,3\}$ to satisfy $vx$.

{\bf Case H:} {\it A vertex $v$ such that $p_1+2q\ge d(v)$ and $p_1+q>4$.}
Let $Z$ be the set of $\gamma$-neighbors of $v$.  Let $R$ be the set of edges
from $v$ to $1$-neighbors and to $Z$, shown bold in Figure~\ref{figHI3}.  By
{\bf F}, the set $Z$ is independent.  Form $G'$ from $G$ by deleting $R$ and
$F_z$ for each $z\in Z$.  Here $v$ and a $\gamma$-neighbor of $v$ play the
roles of $x$ and $v$ in Figure~\ref{figgamma}, respectively.

Let $R'$ be a set of $d(v)-p_1-q$ edges from $v$ to $Z$.  Assign weight $3$ to
all of $R-R'$.  Choose weights on $R'$ from $\{2,3\}$ to satisfy the
$d(v)-p_1-q$ edges in $\Gamma_{G'}(v)$.

Consider $vz$ with $z\in Z$.  Including the weights on $\Gamma_{G'}(v)$, the
sum of the weights on $\Gamma_G(v)-\{vz\}$ is now at least $3(q-1)+3p_1$, which
by hypothesis exceeds $9$.  At most three edges are incident to $vz$ at $z$, so
$vz$ is automatically satisfied, as are the edges from $v$ to $1$-neighbors.
Now the weights on $R$ are fixed; apply Lemma~\ref{gamma} to the vertices of
$Z$.

\begin{figure}[h]
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\caption{Cases {\bf H} and {\bf I} for Lemma~\ref{83red3}\label{figHI3}}
\end{figure}

{\bf Case I:} {\it A $5$-vertex $v$ having a $1$-neighbor and three
$\gamma$-neighbors.}
The argument of Case {\bf H} does not suffice here, since $p_1+q=4$.  Let $u$
be the $1$-neighbor of $v$, and let $z_1,z_2,z_3$ be its $\gamma$-neighbors.
As in {\bf H}, let $R=\{vu,vz_1,vz_2,vz_3\}$ (bold on the right in
Figure~\ref{figHI3}), and let $G'=G-R-\bigcup_i F_{z_i}$.  Let $a=\rp{x}{v}$,
and let $b=w'(vx)$.  By Lemma~\ref{4cycle}, we may assume that no two of the
$\gamma$-neighbors are $3$-vertices with a common $2$-neighbor, and by {\bf F}
they form an independent set.  Hence Figure~\ref{figHI3} is accurate.

If $a\ne12$, then put weight $3$ on all edges of $R$ to satisfy $vx$ and
ensure satisfying $\{vz_1,vz_2,vz_3\}$.  Finally, apply
Lemma~\ref{gamma} to each $z_i$.

If $a=12$, then set $w(vz_1)=2$ so that $vx$ is automatically satisfied.
Having specified $w(vz_1)$, apply Lemma~\ref{gamma} to $z_1$.  Now let
$c=\rho_w(z_1,v)$.  If $c\le 7$, or if $c=8$ and $b\ge2$, then set
$w(vz_2)=w(vz_3)=3$ to ensure satisfying $vz_1$.  If $c=9$, or if $c=8$ and
$b=1$, then set $w(vz_2)=w(vz_3)=1$ to ensure satisfying $vz_1$.  Next
apply Lemma~\ref{gamma} to $z_2$ and $z_3$.  Finally, choose $w(vu)$ to
satisfy $vz_2$ and $vz_3$.

\smallskip
{\bf Case J:} {\it A $4$-vertex with specified neighbors.}
As usual, by the prior lemmas and reducible configurations, Figure~\ref{figJ3}
is accurate.

{\bf Subcase 1:} {\it A $4$-vertex $v$ with $\alpha$-neighbors $z$ and $z'$.}
Let $N_G(z)=\{v,y\}$ and $N_G(z')=\{v,y'\}$, as in Figure~\ref{figJ3}.
By Lemma~\ref{triangle}, we may assume $\{y,y'\}\cap\{z,z'\}=\varnothing$.  By {\bf B},
we have $y\ne y'$ and $yy'\notin E(G)$.  Let $G'=G-\{vz,vz',zy,z'y'\}$.  At
least two choices for $w(vz)$ satisfy $zy$, and similarly two choices for
$w(vz')$ satisfy $z'y'$.  This yields at least three choices for
$w(vz)+w(vz')$, which is enough to satisfy $\Gamma_{G'}(v)$.  Finally, choose
$w(zy)$ to satisfy its two incident edges and $w(z'y')$ to satisfy its two
incident edges.

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\vspace{-.5pc}

\caption{Cases {\bf J1,J2,J3}  for Lemma~\ref{83red3}\label{figJ3}}
\end{figure}

{\bf Subcase 2:} {\it A $4$-vertex $v$ with an $\alpha$-neighbor $z$, another
$2$-neighbor $u$ (that is not an $\alpha$-vertex), and a $\gamma$-neighbor
$x$.}  Name the vertices (uniquely) so that $y',y,z,v,u,u'$ form a path in
order, and let $x'$ be the remaining neighbor of $v$, as in Figure~\ref{figJ3}.
Let $G'=G-\{yz,zv,vu,vx\}-F_x$.
Set $w(vx)=2$ to ensure satisfying $vu$.  Apply Lemma~\ref{gamma} to $x$.
At least two choices for $w(vu)$ satisfy $uu'$, and at least two choices for
$w(vz)$ satisfy $yz$.  With at least three choices for $w(vu)+w(vz)$, at least
one satisfies $vx$ and $vx'$.  Finally, choose $w(yz)$ to satisfy $vz$ and
$yy'$.

{\bf Subcase 3:} {\it A $4$-vertex $v$ with a $2$-neighbor $z$ and three
$\gamma$-neighbors $x_1,x_2,x_3$.}
Let $N(z)=\{v,y\}$, as in Figure~\ref{figJ3}.  Let
$G'=G-\Gamma_G(v)-\bigcup_iF_{x_i}$.

If $d(x_1)=3$, or if $d(x_1)=4$ and $\phi_{w'}(x_1)\le 4$, then the value
of $\rho_w(x_1,v)$ will be at most $7$, since when $d(x_1)=4$ there is only one
edge in $F_{x_1}$ (the edge incident to the $1$-neighbor of $x_1$).  Hence
setting $w(vx_2)=w(vx_3)=3$ and restricting $w(zv)$ to $\{2,3\}$ ensures
satisfying $vx_1$ and $vz$.  Apply Lemma~\ref{gamma} to $x_2$ and $x_3$.  Now
choose $w(zv)$ (in $\{2,3\}$) to satisfy $yz$, and then choose $w(vx_1)$ to
satisfy $vx_2$ and $vx_3$.  Finally, apply Lemma~\ref{gamma} to $x_1$.

By symmetry, we may now assume $d(x_i)=4$ and $\phi_{w'}(x_i)\ge5$ for all $i$.
Choose $w(vz)\in\{1,2\}$ to satisfy $zy$.  Set $w(vx_1)=2$ and
$w(vx_2)=w(vx_3)=1$ to ensure satisfying all edges incident to $v$.
Finally, apply Lemma~\ref{gamma} to each $x_i$.

\smallskip
{\bf Case K:} {\it $v$ is a $\gamma$-vertex whose $3^+$-neighbors are all
$\beta$-vertices.}  See Figure~\ref{figK3}.  Let $S$ be the set of neighbors of
$v$ whose degrees are not specified by the definition of $v$ being a
$\gamma$-vertex.  By $\{${\bf B,C,D}$\}$, all vertices of $S$ are
$3^+$-vertices.  By {\bf F}, they cannot be $\gamma$-vertices, so by the
hypothesis of this case, each vertex of $S$ has degree $3$, with one
$2$-neighbor and one $3^+$-neighbor other than $v$ (by {\bf A}).
By Lemmas~\ref{triangle} and~\ref{4cycle} and Cases {\bf B} and {\bf F},
Figure~\ref{figK3} is accurate in each subcase.

{\bf Subcase 1:} {\it $v$ is a $\gamma_{3b}$-vertex.}
Let $N_G(v)=\{z,z',u\}$, where $d_G(u)=3$.  Let $y$ be the $2$-neighbor of $u$.
Let $G'=G-\Gamma_G(v)-uy$.  Set $w(vu)=3$ to ensure satisfying all of
$\{vz,vz',uy\}$.  Choose $w(uy)$ to satisfy its two incident edges other than
$vu$.  Choose $w(vz)$ to satisfy the other edge at $z$, and choose $w(vz')$ to
satisfy $vu$ and the other edge at $z'$.

{\bf Subcase 2:} {\it $v$ is a $\gamma_4$-vertex.}
Let $N_G(v)=\{z_1,z_2,z_3,u\}$, with $d_G(u)=1$.  Let $y_i$ be the $2$-neighbor
of $z_i$.  Let $G'=G-\Gamma_G(v)-\{z_1y_1,z_2y_2,z_3y_3\}$.  For each $i$, set
$w(vz_i)=3$ to ensure satisfying $z_iy_i$, and choose $w(z_iy_i)$ to
satisfy its two incident edges other than $z_iy_i$.  Also $vz_1,vz_2,vz_3$
are satisfied.

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\caption{Cases {\bf K1,K2,K3}  for Lemma~\ref{83red3}\label{figK3}}
\end{figure}

{\bf Subcase 3:} {\it $v$ is a $\gamma_{3a}$-vertex.}
Let $u$ be the $\alpha$-neighbor of $v$ (with $N_G(u)=\{v,u'\}$).
By {\bf C}, $v$ does not have another $2$-neighbor.
Let $N_G(z)=\{v,x,y\}$ and $N_G(z')=\{v,x',y'\}$, with $d_G(y)=d_G(y')=2$.
By {\bf J}, each $\beta$-neighbor of $v$ is not a $\gamma$-vertex, which means
that the other neighbors of $y$ and $y'$ are $3^+$-vertices.

Let $G'=G-\Gamma_G(v)-\{uu',zy,z'y'\}$.  Let $a$ and $b$ be the weights under
$w'$ of the edges incident to $y$ and $z$ in $G'$, respectively.
Set $w(vz')=3$ to ensure satisfying $z'y'$, and choose $w(z'y')$
to satisfy the edges incident to $z'y'$ other than $vz'$.

If $a\le b$, then $zy$ is automatically satisfied.  Choose $w(zy)$
to satisfy $\Gamma_{G'}(y)$.  Now choose $w(vu)$ to satisfy $vz$ and $uu'$, and
then choose $w(vz)$ to satisfy $vz'$ and $zx$.  Finally, choose $uu'$ to
satisfy $vu$ and $u'u''$.

If $a>b$, then setting $w(zy)=1$ ensures satisfying the other edge at $y$
(since its other endpoint has degree at least $3$).  With $b\le2$ and
$w(vz')=3$, the edge $vz$ is automatically satisfied.  Now choose $w(vz)$
to satisfy the other edges at $z$, choose $w(vu)$ to satisfy $vz'$ and $uu'$,
and choose $w(uu')$ to satisfy its incident edges.
\end{proof}

\begin{thm}
Every graph $G$ with ${\rm Mad}\,(G)<\FR83$ has a proper $3$-weighting.
\end{thm}
\begin{proof}
It suffices to show that every configuration in the unavoidable set in
Lemma~\ref{83disch3} is shown to be $3$-reducible in Lemma~\ref{83red3}.  The
configurations are the same in the two lemmas except for {\bf H} and {\bf J}.

For {\bf H}, if $d(v)\in\{6,7\}$ and $v$ has a $1$-neighbor and four
$\gamma$-neighbors, then $p_1+2q\ge d(v)$ and $p_1+q>4$.  For {\bf J}, a
$4$-vertex $v$ with $p+q+r\ge 5$ must have an $\alpha$-neighbor.
If $v$ has another $\alpha$-neighbor, then {\bf J1} applies.
If $v$ has a $\gamma$-neighbor and another $2$-neighbor, then {\bf J2} applies.
Otherwise, all other neighbors are $\gamma$-neighbors (reducible by {\bf J3})
or all are $2$-neighbors (reducible by {\bf B}).
\end{proof}

Some of the $3$-reducible configurations in Lemma~\ref{83red3} are more general
than the configurations forced in Lemma~\ref{83disch3}.  Also, there are other
$3$-reducible configuration we have not used, such as (1) a $4$-vertex having
two $2$-neighbors and one $\gamma$-neighbor and (2) a more general version of
configuration {\bf H}.  This suggests that with more work this approach could
be pushed to prove the conclusion under a weaker restriction on ${\rm Mad}\,(G)$.

{\small

redpajama_set_name: RedPajamaArXiv