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HomeNewsMaynard James Keenan Says "Crippling Second Guessing" Delayed New Album
Maynard James Keenan Says "Crippling Second Guessing" Delayed New Album
September 3, 2019 News
On Friday, September 1st, BBC Radio 1's Rock Show With Daniel P Carter focused on Tool's new album Fear Inoculum and included an array of original interviews around the subject. Members of Bring Me the Horizon, Slipknot, Prophet of Rage, Mastodon and more spoke about Tool's impact on them as artists and music lovers, but the centerpiece of the episode were the chats with the band's own Maynard James Keenan and Justin Chancellor. The singer and bass player, respectively, discussed the new album — the band's first in 13 years — and the group's creative process.
Keenan was particularly blunt, as is his way. Talking about why Fear Inoculum took so long to be completed and released, the Tool, A Perfect Circle and Puscifer frontman did a little psychoanalysis of the band.
"I think a lot of it just that age where you want it to be right and we've had some success in the past and the fear of this thing coming out and not being accepted — the fear that it's not as good as it can be — that can be detrimentally crippling," he explained, as reported by the PRP. "Probably in if I had to 'psychology 101' [it], I would have to say, 'Well, yeah, that's why it would take 13 years to write something, because you're paranoid that it's not gonna be the best that it can be and then you second guess every single step that you make, when it was probably good enough — I shouldn't say good enough — it was fantastic 8 years ago.
"But then the crippling second guessing of yourself sets in and that psychology and that spiral you get in, it can be extremely daunting. And you can actually not even feel it happening. All of a sudden you wake up and it's 13 years later. The hard part is accepting the fact that maybe you're not as important as you think you are and you should probably just get on with it."
Keenan also discussed his relative lack on involvement in the earlier stages of the songwriting process and whether that is more of a recent development or the way the band has always functioned. "I would say way more so now," he said. "Because you have four strong personalities all fighting for space. So over the years I've had to just step out of the room just to let them fight it out. Because just to have another opinion in the room; If you think it's gridlocked now, to have a fourth arrogant person in the room is not gonna help anybody. So, you have to understand when there's too many chefs and just step out. Let them get there and then step in and do your magic when they're done."
Tool recently announced a big North American fall tour with Killing Joke — see the full itinerary here.
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Hindsight is always 20/20, but in the case of the '90s, that vision is deeply impaired. Today, metal fans like the remember the '90s as a creatively-fertile time when even the whitest of dudes could […]
5 Great Metal Second Albums
They call it the sophomore slump. Your band killed it with your debut — you guys had your whole lives to that point to make that album, after all. You won a growing legion of […]
Tool's Catalog Finally Available via Digital Service Providers
On Monday, July 29th, Tool had the internet buzzing with two massive announcements: 1) the title of their forthcoming new album, Fear Inoculum, and 2) the news that on Friday, August 2nd — a.k.a. today! […]


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\section{Introduction}

We use Bondy and Murty \cite{BM} for terminology and notations not
defined here. Let $G$ be a simple graph with $n$ vertices and $m$
edges. The {\it cyclomatic number} of a connected graph $G$ is
defined as $c(G)=m-n+1$. A graph $G$ with $c(G)=k$ is called a {\it
$k$-cyclic graph}. In particular, for $c(G)=0,1$ or $2$ we call $G$
a tree, unicyclic or bicyclic graph, respectively. Denote by
$\Delta$ the maximum degree of a graph. The eigenvalues
$\lambda_{1}, \lambda_{2},\ldots, \lambda_{n}$ of the adjacency
matrix $A(G)$ of $G$ are said to be the eigenvalues of the graph
$G$. The $energy$ of $G$ is defined as
$$E=E(G)=\sum_{i=1}^{n}|\lambda_{i}|.$$

For several classes of graphs it has been demonstrated that the
energy exceeds the number of vertices (see, \cite{G1}). In 2007,
Nikiforov \cite{N} showed that for almost all graphs,
$$E=\left(\frac{4}{3\pi}+o(1)\right)n^{3/2}.$$
Thus the number of graphs $G$ satisfying the condition $E(G)<n$ is
relatively small. In \cite{GR}, a connected graph $G$ of order $n$
is called $hypoenergetic$ if $E(G)<n$. For hypoenergetic graphs with
$\Delta \leq 3$, we have the following well known results.

\begin{lem}\cite{GLSZ}\label{lem1.1}
There exist only four hypoenergetic trees with $\Delta \leq 3$,
dipicted in Figure \ref{fig1}.
\end{lem}
\begin{figure}[ht]
\centering
  \setlength{\unitlength}{0.05 mm}%
  \begin{picture}(1592.3, 330.9)(0,0)
  \put(0,0){\includegraphics{fig1}}
  \put(40.00,60.93){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$S_1$\strut}}
  \put(335.53,52.15){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$S_3$\strut}}
  \put(727.62,49.22){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$S_4$\strut}}
  \put(1286.50,37.52){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$W$\strut}}
  \end{picture}%
 \caption{The hypoenergetic trees with maximum degree at most $3$.} \label{fig1}
\end{figure}

\begin{lem}\cite{N2}\label{lem1.2}
Let $G$ be a graph of order $n$ with at least $n$ edges and with no
isolated vertices. If $G$ is quadrangle-free and $\Delta (G)\leq 3$,
then $E(G)>n$.
\end{lem}

The present authors first in \cite{LM1} showed that complete
bipartite graph $K_{2,3}$ is the only hypoenergetic graph among all
unicyclic and bicyclic graphs with $\Delta \leq 3$, and then
recently they obtained the following general result:

\begin{lem}\cite{LM2}\label{lem1.3}
Complete bipartite graph $K_{2,3}$ is the only hypoenergetic
connected cycle-containing (or cyclic) graph with $\Delta \leq 3$.
\end{lem}

Therefore, all connected hypoenergetic graphs with maximum degree at
most 3 have been characterized.

\begin{lem}\cite{LM2}\label{lem1.4}
$S_1, S_3, S_4, W$ and $K_{2,3}$ are the only $5$ hypoenergetic
connected graphs with $\Delta \leq 3$.
\end{lem}

In \cite{MKG} Majstorovi\'{c} et al. proposed the following
conjecture, which is the second half of their Conjecture 3.7.

\begin{con}\cite{MKG}\label{con1.5}
There are exactly four connected graphs $G$ with order $n$ and
$\Delta \leq 3$ for which the equality $E(G)=n$ holds, which are
dipicted in Figure \ref{fig2}.
\end{con}
\begin{figure}[ht]
\centering
  \setlength{\unitlength}{0.05 mm}%
  \begin{picture}(1800.2, 437.5)(0,0)
  \put(0,0){\includegraphics{fig2}}
  \put(105.89,117.23){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$S_2$\strut}}
  \put(413.44,71.07){\fontsize{11.76}{14.11}\selectfont \makebox(371.9, 82.6)[l]{$K_{2,2}$\strut}}
  \put(919.87,93.07){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$Q$\strut}}
  \put(1388.26,37.52){\fontsize{11.76}{14.11}\selectfont \makebox(371.9, 82.6)[l]{$K_{3,3}$\strut}}
  \end{picture}%
  \caption{All connected graphs with maximum degree at most $3$ and $E=n$.} \label{fig2}
\end{figure}

In this paper, we will prove this conjecture.

\section{Main results}

The following results are needed in the sequel.

\begin{lem}\cite{DS}\label{lem2.1}
If $F$ is an edge cut of a graph $G$, then $E(G-F)\leq E(G)$, where
$G-F$ is the subgraph obtained from $G$ by deleting the edges in
$F$.
\end{lem}

\begin{lem}\cite{DS}\label{lem2.2}
Let $F=[S, V\setminus S]$ be an edge cut of a graph $G$ with vertex
set $V$, where $S$ is a nonempty proper subset of $V$. Suppose that
$F$ is not empty and all edges in $F$ are incident to one and only
one vertex in $S$, i.e., the edges in $F$ form a star. Then
$E(G-F)<E(G)$.
\end{lem}

\begin{lem}\cite{BP}\label{lem2.3}
The energy of a graph can not be an odd integer.
\end{lem}

In the following we first show that Conjecture \ref{con1.5} holds
for trees, unicyclic and bicyclic graphs, respectively. Then we show
that Conjecture \ref{con1.5} holds in general.

Let $F$ be an edge cut of a connected graph $F$. If $G-F$ has
exactly two components $G_1$ and $G_2$, then we denote $G-F=G_1+G_2$
for convenience. The following lemma is needed.

\begin{lem}\label{lem2.4}
Let $F$ be an edge cut of a connected graph $G$ of order $n$ such
that $G-F=G_1+G_2$. If $E(G_1)\geq |V(G_1)|$, $E(G_2)\geq |V(G_2)|$
and either at least one of the above inequalities is strict or the
edges in $F$ form a star or both, then $E(G)>n$.
\end{lem}

\noindent {\it Proof.}  If $E(G_1)>|V(G_1)|$ or $E(G_2)>|V(G_2)|$, then by Lemma
\ref{lem2.1}, we have
$$E(G)\geq E(G-F)=E(G_1)+E(G_2)>|V(G_1)|+|V(G_2)|=n.$$
Otherwise by Lemma \ref{lem2.2}, we have
$$E(G)>E(G-F)=E(G_1)+E(G_2)\geq |V(G_1)|+|V(G_2)|=n,$$
which completes the proof. \hfill \rule{4pt}{7pt}

The result Lemma \ref{lem2.4} is easy but useful in our proofs.

\begin{thm}\label{thm2.5}
$S_2$ and $Q$ are the only two trees $T$ with order $n$ and $\Delta
\leq 3$ for which the equality $E(T)=n$ holds.
\end{thm}

\noindent {\it Proof.}  Let $T$ be a tree with $n$ vertices and $\Delta \leq 3$. From
Table 2 of \cite{CDS}, we know that $S_2$ and $Q$ are the only two
trees with $\Delta \leq 3$ and $n\leq 10$ for which the equality
$E=n$ holds. By Lemma \ref{lem2.3}, we may assume that $n\geq 12$ is
even. We will prove that $E(T)>n$.

We divide the trees with $\Delta \leq 3$ into two classes: {\bf
Class 1} contains the trees $T$ that have an edge $e$, such that
$T-e=T_1+T_2$ and $T_1, T_2\not\cong S_1, S_3, S_4, W$. {\bf Class
2} contains the trees $T$ in which there exists no edge $e$, such
that $T-e=T_1+T_2$ and $T_1, T_2\not\cong S_1, S_3, S_4, W$, i.e.,
for any edge $e$ of $T$ at least one of components of $T-e$ is
isomorphic to a tree in $\{S_1, S_3, S_4, W\}$.

\noindent {\bf Case 1.} $T$ belongs to Class 1. Then there exists an
edge $e$ such that $T-e=T_1+T_2$ and $T_1, T_2\not\cong S_1, S_3,
S_4, W$. Hence by Lemmas \ref{lem1.1} and \ref{lem2.2}, we have
$E(T)>E(T-e)=E(T_1)+E(T_2)\geq |V(T_1)|+|V(T_2)|=n$, which completes
the proof.

\noindent {\bf Case 2.} $T$ belongs to Class 2. Consider the center
of $T$. There are two subcases: either $T$ has a (unique) center
edge $e$ or a (unique) center vertex $v$.

\noindent {\bf Subcase 2.1.} $T$ has a center edge $e$. The two
fragments attached to $e$ will be denoted by $T_1$ and $T_2$, i.e.,
$T-e=T_1+T_2$.

Without loss of generality, we assume that $T_1$ is isomorphic to a
tree in $\{S_1, S_3, S_4$, $W\}$.

If $T_1$ is isomorphic to a tree in $\{S_1, S_3, S_4\}$, then it is
easy to see that $n\leq 11$, which is a contradiction.

If $T_1\cong W$ and it is attached to the center edge $e$ through
the vertex of degree 2, then it is easy to see that $T$ must be the
tree as given in Figure \ref{fig3} (a) or (b). By direct computing,
we have that $E(T)=12.61708>12=n$ in the former case while
$E(T)=14.91128>14=n$ in the latter case. If $T_1\cong W$ and it is
attached to the center edge $e$ through a pendent vertex, see Figure
\ref{fig3} (c). Since $T$ belongs to Class 2, deleting the edge $f$,
we then have that $T_2\cup e$ is isomorphic to a tree in $\{S_1,
S_3, S_4, W\}$, which contradicts to the fact that $e$ is the center
edge of $T$.
\begin{figure}[ht]
\centering
  \setlength{\unitlength}{0.05 mm}%
  \begin{picture}(1993.4, 2404.1)(0,0)
  \put(0,0){\includegraphics{fig3}}
  \put(833.42,2146.42){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$e$\strut}}
  \put(273.06,2139.18){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$e$\strut}}
  \put(239.24,1897.64){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(a)\strut}}
  \put(768.21,1897.64){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(b)\strut}}
  \put(1599.09,2185.07){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$e$\strut}}
  \put(1746.76,2158.63){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$T_2$\strut}}
  \put(1486.12,1920.44){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(c)\strut}}
  \put(1492.61,2186.23){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$f$\strut}}
  \put(1445.24,1713.09){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$v$\strut}}
  \put(386.34,1663.54){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$v$\strut}}
  \put(366.56,1169.11){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$v$\strut}}
  \put(1441.29,1152.01){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$v$\strut}}
  \put(452.42,511.23){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$v$\strut}}
  \put(1472.93,546.83){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{$v$\strut}}
  \put(373.73,1411.12){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(d)\strut}}
  \put(1413.60,1381.63){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(e)\strut}}
  \put(397.46,683.32){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(f)\strut}}
  \put(1441.29,792.66){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(g)\strut}}
  \put(409.33,58.36){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(h)\strut}}
  \put(1441.72,37.52){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(i)\strut}}
  \end{picture}%
  \caption{The graphs in the proof of Theorem \ref{thm2.5}. }\label{fig3}
\end{figure}

\noindent {\bf Subcase 2.2.} $T$ has a center vertex $v$. If $v$ is
of degree 2, then the two fragments attached to it will be denoted
by $T_1$ and $T_2$. If $v$ is of degree 3, then the three fragments
attached to it will be denoted by $T_1$, $T_2$ and $T_3$.

Let $v_i$ be the adjacent vertex of $v$ in $T_i$. Denote
$T-vv_1=T_1+T'_2$. Since $T$ belongs to Class 2, either $T_1$ or
$T'_2$ is isomorphic to a tree in $\{S_1, S_3, S_4, W\}$.

\noindent {\bf Subsubcase 2.2.1.} $T'_2$ is isomorphic to a tree in
$\{S_1, S_3, S_4, W\}$.

Clearly $T'_2\not \cong S_1$. If $T'_2\cong S_3$ or $S_4$, then it
is easy to see that $n\leq 7$, which is a contradiction. If
$T'_2\cong W$ and $v$ is of degree $3$, then it is easy to see that
$n\leq 10$, which is a contradiction. If $T'_2\cong W$ and $v$ is of
degree $2$, i.e., $N(v)=\{v_1,v_2\}$. Consider $T-vv_2$, since $T$
belongs to Class 2, we have that $T_1\cup vv_1$ is isomorphic to a
tree in $\{S_1, S_3, S_4, W\}$. By the fact that $v$ is the center
of $T$, we have that $T_1\cup vv_1\cong W$, and so $n=13$, which is
a contradiction.

\noindent {\bf Subsubcase 2.2.2.} $T_1$ is isomorphic to a tree in
$\{S_1, S_3, S_4, W\}$.

If $T_1\cong S_1$, then it is easy to see that $n\leq 4$, which is a
contradiction.

If $T_1\cong S_3$ and $v_1$ is of degree 2 in $T_1$, then it is easy
to see that $n\leq 10$, which is a contradiction. If $T_1\cong S_3$
and $v_1$ is a pendent vertex in $T_1$, denote by $u$ the unique
adjacent vertex of $v_1$ in $T_1$. Since $T$ belongs to Class 2,
deleting the edge $uv_1$, we then have that $T'_2\cup vv_1$ is
isomorphic to a tree in $\{S_1, S_3, S_4, W\}$, and so $n\leq 9$,
which is a contradiction.

If $T_1\cong S_4$ or $T_1\cong W$ and $v_1$ is of degree 2 in $T_1$,
then by the facts that $T$ belongs to Class 2, $v$ is the center of
$T$ and $n$ is even, it is not hard to obtain that $T_2$, $T_3$ must
be isomorphic to a tree in $\{S_1, S_3, S_4, W\}$, and at least one
of $T_2$ and $T_3$ is isomorphic to a tree in $\{S_4, W\}$, and if
$T_2$ ($T_3$, respectively) is isomorphic to $W$, then $v_2$ ($v_3$,
respectively) is of degree 2 in $T_2$ ($T_3$, respectively). Hence
there are 6 such trees, as given in Figure \ref{fig3} (d), (e), (f),
(g), (h) and (i). The energy of these trees are 12.72729 ($>12=n$),
12.65406 ($>12=n$), 16.81987 ($>16=n$), 16.77215 ($>16=n$), 19.18674
($>18=n$) and 23.38426 ($>22=n$), respectively.

If $T_1\cong W$ and $v_1$ is a pendent vertex in $T_1$, denote by
$u$ the unique adjacent vertex of $v_1$ in $T_1$. Since $T$ belongs
to Class 2, deleting the edge $uv_1$, we then have that $T'_2\cup
vv_1$ is isomorphic to a tree in $\{S_1, S_3, S_4, W\}$, which
contradicts to the fact that $v$ is the center vertex of $T$. The
proof is thus complete. \hfill \rule{4pt}{7pt}

From Table 1 of \cite{CDS}, we know that $K_{2,2}$ is the only
connected graph of order 4 with $\Delta \leq 3$ and $E=4$. From
Tables 1 and 2 of \cite{CP}, we know that $K_{3,3}$ is the only
connected cycle-containing graph of order 6 with $\Delta \leq 3$ and
$E=6$.

\begin{thm}\label{thm2.6}
$K_{2,2}$ is the only unicyclic graph with $\Delta \leq 3$ for which
the equality $E=n$ holds.
\end{thm}

\noindent {\it Proof.}  Let $G\not\cong K_{2,2}$ be a unicyclic graph of order $n$ with
$\Delta \leq 3$. It is sufficient to show that $E(G)>n$. By Lemmas
\ref{lem1.2} and \ref{lem2.3}, we can assume that $n\geq 8$ is even
and $G$ contains a quadrangle $C=x_1x_2x_3x_4x_1$. We distinguish
the following four cases:

\noindent {\bf Case 1.} There exists an edge $e$ on $C$ such that
the end vertices of $e$ are of degree 2.

Without loss of generality, we assume that $d(x_1)=d(x_4)=2$. Let
$F=\{x_1x_2,x_4x_3\}$, then $G-F=G_1+G_2$, where $G_1\cong S_2$ and
$G_2$ is a tree of order at least $6$ since $n\geq 8$. Since
$\Delta(G)\leq 3$, $G_2$ can not be isomorphic to $W$ or $Q$.
Therefore we have $E(G_1)=|V(G_1)|$ and $E(G_2)>|V(G_2)|$ by Lemma
\ref{lem1.1} and Theorem \ref{thm2.5}. It follows from Lemma
\ref{lem2.4} that $E(G)>n$.

\noindent {\bf Case 2.} There exist exactly two nonadjacent vertices
$x_i$ and $x_j$ on $C$ such that $d(x_i)=d(x_j)=2$.

Without loss of generality, we assume that $d(x_2)=d(x_4)=2$,
$d(x_1)=d(x_3)=3$. Let $y_3$ be the adjacent vertex of $x_3$ outside
$C$. Then $G-x_3y_3=G_1+G_2$, where $G_1$ is a unicyclic graph and
$G_2$ is a tree. Notice that $E(G_1)\geq |V(G_1)|$ by Lemma
\ref{lem1.3}. If $G_2\not\cong S_1,S_3,S_4,W$, then we have
$E(G_2)\geq |V(G_2)|$ by Lemma \ref{lem1.1} and so $E(G)>
E(G-x_3y_3)\geq n$ by Lemma \ref{lem2.4}. Therefore we only need to
consider the following four subcases.

\noindent {\bf Subcase 2.1.} $G_2\cong S_1$. Let
$F=\{x_2x_3,x_3x_4\}$, then $G-F=G'_1+G'_2$, where $G'_2\cong S_2$
and $G'_1$ is a tree of order at least $6$ since $n\geq 8$. If
$G'_1\cong W$, then $n=9$, which is a contradiction. Otherwise, it
follows from Lemmas \ref{lem1.1} and \ref{lem2.4} that $E(G)>n$.

\noindent {\bf Subcase 2.2.} $G_2\cong S_3$. Then $G$ must have the
structure as given in Figure \ref{fig4} (a) or (b). In the former
case, $G-y_3z=G'_1+G'_2$, where $G'_1$ is a unicyclic graph and
$G'_2\cong S_2$. It follows from Lemmas \ref{lem1.4} and
\ref{lem2.4} that $E(G)>n$. In the latter case,
$G-\{x_1x_2,x_4x_3\}=G'_1+G'_2$, where $G'_2$ is the tree of order
$5$ containing $x_3$ and $G'_1$ is a tree of order at least $3$. By
Lemma \ref{lem1.1} and Theorem \ref{thm2.5}, we have
$E(G'_2)>|V(G'_2)|$. If $G'_1\not\cong S_3,S_4,W$, then we have
$E(G)>n$ by Lemmas \ref{lem1.1} and \ref{lem2.4}. Since
$\Delta(G)\leq 3$, $G'_1$ can not be isomorphic to $S_4$ or $W$. If
$G'_1\cong S_3$, then $G$ must be the graph as given in Figure
\ref{fig4} (c). By choosing the edge cut $\{x_1x_2,x_1x_4\}$, we can
similarly obtain that $E(G)>n$.
\begin{figure}[ht]
\centering
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  \end{picture}%
  \caption{The graphs in the proof of Theorem \ref{thm2.6}. }\label{fig4}
\end{figure}

\noindent {\bf Subcase 2.3.} $G_2\cong S_4$. Then $G$ must have the
structure as given in Figure \ref{fig4} (d). Let
$F=\{x_2x_3,x_3x_4\}$, then $G-F=G'_1+G'_2$, where $G'_2$ is the
tree of order $5$ containing $x_3$ and $G'_1$ is a tree of order at
least $4$. By Lemma \ref{lem1.1} and Theorem \ref{thm2.5}, we have
$E(G'_2)>|V(G'_2)|$. If $G'_1\not\cong S_4,W$, then we have $E(G)>n$
by Lemmas \ref{lem1.1} and \ref{lem2.4}. If $G'_1\cong S_4$, then
$n=9$, which is a contradiction. If $G'_1\cong W$, then $G$ must be
the graph as given in Figure \ref{fig4} (e). By choosing the edge
cut $\{x_1x_2,x_3x_4\}$, we can similarly obtain that $E(G)>n$.

\noindent {\bf Subcase 2.4.} $G_2\cong W$. Then $G$ must have the
structure as given in Figure \ref{fig4} (f) or (g). In the former
case, $G-y_3z=G'_1+G'_2$, where $G'_1$ is a unicyclic graph and
$G'_2$ is a tree of order $6$. It follows from Lemmas \ref{lem1.4}
and \ref{lem2.4} that $E(G)>n$. In the latter case,
$G-\{x_2x_3,x_3x_4\}=G'_1+G'_2$, where $G'_2$ is the tree of order
$8$ containing $x_3$ and $G'_1$ is a tree of order at least $4$. If
$G'_1\not\cong S_4,W$, then we have $E(G)>n$ by Lemmas \ref{lem1.1}
and \ref{lem2.4}. If $G'_1\cong S_4$, then $G$ must be the graph as
given in Figure \ref{fig4} (h). By choosing the edge cut
$\{x_1x_2,x_1x_4\}$, we can similarly obtain that $E(G)>n$. If
$G'_1\cong W$, then $n=15$, which is a contradiction.

\noindent {\bf Case 3.} There exists exactly one vertices $x_i$ on
$C$ such that $d(x_i)=2$.

Without loss of generality, we assume that $d(x_1)=2$. Let
$F=\{x_1x_4,x_2x_3\}$, then $G-F=G_1+G_2$, where $G_1$ is the tree
of order at least $3$ containing  $x_1$ and $G_2$ is a tree of order
at least $4$. Since $\Delta(G)\leq 3$, $G_1, G_2$ can not be
isomorphic to $S_4$, $W$ or $Q$. If $G_1\not\cong S_3$, then we have
$E(G)>n$ by Lemmas \ref{lem1.1}, \ref{lem2.4} and Theorem
\ref{thm2.5}. If $G_1\cong S_3$, then
$G-\{x_1x_2,x_2x_3\}=G'_1+G'_2$, where $G'_1$ is the tree of order
at least $5$ containing $x_1$ and $G'_2\cong S_2$. If $G'_1\not\cong
W$, then we have $E(G)>n$ by Lemmas \ref{lem1.1} and \ref{lem2.4}.
If $G'_1\cong W$, then $n=9$, which is a contradiction.

\noindent {\bf Case 4.} $d(x_1)=d(x_2)=d(x_3)=d(x_4)=3$.

Let $F=\{x_1x_4,x_2x_3\}$, then $G-F=G_1+G_2$, where $G_1$ and $G_2$
are trees of order at least $4$ and it is easy to see that $G_1,
G_2$ can not be isomorphic to $S_4$, $W$ or $Q$. So it follows from
Lemmas \ref{lem1.1}, \ref{lem2.4} and Theorem \ref{thm2.5} that
$E(G)>n$. The proof is thus complete. \hfill \rule{4pt}{7pt}

\begin{thm}\label{thm2.7}
There does not exist any bicyclic graph with $\Delta \leq 3$ for
which the equality $E=n$ holds.
\end{thm}

\noindent {\it Proof.}  Let $G$ be a bicyclic graph of order $n$ with $\Delta \leq 3$.
We know that $E(G)\neq n$ for $n=4$ or $6$. By Lemmas \ref{lem1.2}
and \ref{lem2.3}, we may assume that $n\geq 8$ is even and $G$
contains a quadrangle. Then we will show that $E(G)> n$.

If the cycles in $G$ are disjoint, then it is clear that there
exists a path $P$ connecting the two cycles in $G$. For any edge $e$
on $P$, we have $G-e=G_1+G_2$, where $G_1$ and $G_2$ are unicyclic
graphs. By Lemma \ref{lem1.3}, we have $E(G_1)\geq |V(G_1)|$ and
$E(G_2)\geq |V(G_2)|$. Therefore we have $E(G)>n$ by Lemma
\ref{lem2.4}. Otherwise, the cycles in $G$ have two or more common
vertices. Then we can assume that $G$ contains a subgraph as given
in Figure \ref{fig5} (a), where $P_1,P_2,P_3$ are paths in $G$. We
distinguish the following three cases:
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  \put(2436.02,1115.69){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$y$\strut}}
  \put(2097.16,705.40){\fontsize{10.69}{12.82}\selectfont \makebox(187.8, 75.1)[l]{$G_1$\strut}}
  \put(776.21,121.49){\fontsize{10.69}{12.83}\selectfont \makebox(112.7, 75.1)[l]{(p)\strut}}
  \put(1714.02,616.06){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$u$\strut}}
  \put(1721.02,190.86){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$v$\strut}}
  \put(1517.78,415.05){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$x$\strut}}
  \put(1949.44,483.88){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$y$\strut}}
  \put(1885.34,37.75){\fontsize{10.69}{12.83}\selectfont \makebox(112.7, 75.1)[l]{(q)\strut}}
  \put(1614.21,88.17){\fontsize{10.69}{12.82}\selectfont \makebox(187.8, 75.1)[l]{$G_1$\strut}}
  \put(660.70,561.21){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$u$\strut}}
  \put(664.78,198.95){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$v$\strut}}
  \put(481.66,331.26){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$x$\strut}}
  \put(827.42,448.23){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$y$\strut}}
  \put(2256.32,2014.89){\fontsize{10.69}{12.83}\selectfont \makebox(112.7, 75.1)[l]{(i)\strut}}
  \put(2220.40,2396.36){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$P$\strut}}
  \put(1856.00,2396.56){\fontsize{10.69}{12.82}\selectfont \makebox(187.8, 75.1)[l]{$P_1$\strut}}
  \put(2191.74,2108.17){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$v$\strut}}
  \put(2428.52,2506.37){\fontsize{10.69}{12.83}\selectfont \makebox(187.8, 75.1)[l]{$e_3$\strut}}
  \put(2249.14,2240.90){\fontsize{10.69}{12.83}\selectfont \makebox(187.8, 75.1)[l]{$e_2$\strut}}
  \put(2152.26,2699.89){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$u$\strut}}
  \put(2155.24,1887.89){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$u$\strut}}
  \put(1948.62,1720.52){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$x$\strut}}
  \put(2198.58,1702.14){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$z$\strut}}
  \put(2160.40,1536.63){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$v$\strut}}
  \put(1623.19,1803.95){\fontsize{10.69}{12.82}\selectfont \makebox(112.7, 75.1)[l]{$y$\strut}}
  \end{picture}%
  \caption{The graphs in the proof of Theorem \ref{thm2.7}.} \label{fig5}
\end{figure}

\noindent {\bf Case 1.}  At least one of $P_1$, $P_2$ and $P_3$, say
$P_2$ has length not less than $3$.

Let $e_1$ and $e_2$ be the edges on $P_2$ incident with $u$ and $v$,
respectively. Then $G-\{e_1, e_2\}=G_1+G_2$, where $G_1$ is a
unicyclic graph and $G_2$ is a tree of order at least $2$. It
follows from Lemma \ref{lem1.3} that $E(G_1)\geq |V(G_1)|$. If
$G_2\not\cong S_3, S_4, W, S_2, Q$, then we have $E(G_2)> |V(G_2)|$
by Lemma \ref{lem1.1} and Theorem \ref{thm2.5}, and so $E(G)>n$ by
Lemma \ref{lem2.4}. Hence we only need to consider the following
five subcases.

\noindent {\bf Subcase 1.1.} $G_2\cong S_3$. Then $G$ must have the
structure as given in Figure \ref{fig5} (b) or (c). In either case,
$G-\{e_2,e_3\}=G'_1+G'_2$, where $G'_1$ is a unicyclic graph and
$G'_2\cong S_2$. Obviously, $G'_1\not\cong K_{2,2}$. Then
$E(G'_1)>|V(G'_1)|$ by Lemma \ref{lem1.3} and Theorems \ref{thm2.6}.
Since $E(G'_2)=|V(G'_2)|$, we have $E(G)>n$ by Lemma \ref{lem2.4}.

\noindent {\bf Subcase 1.2.} $G_2\cong S_4$. Then $G$ must have the
structure as given in Figure \ref{fig5} (d). Obviously,
$G-\{e_3,e_4\}=G'_1+G'_2$, where $G'_1$ is a unicyclic graph which
is not isomorphic to $K_{2,2}$ and $G'_2\cong S_2$. Similar to the
proof of Subcase 1.1, we have $E(G)>n$.

\noindent {\bf Subcase 1.3.} $G_2\cong W$. Then $G$ must have the
structure as given in Figure \ref{fig5} (e), (f) or (g). Obviously,
$G-\{xy,yz\}=G'_1+G'_2$, where $G'_1$ is a unicyclic graph which is
not isomorphic to $K_{2,2}$ and $G'_2$ is a tree of order $5$ or
$2$. Similarly, we can obtain that $E(G)>n$.

\noindent {\bf Subcase 1.4.} $G_2\cong S_2$. Since $G_1$ is a
unicyclic graph, if $G_1\not\cong K_{2,2}$, then we can similarly
obtain that $E(G)>n$. If $G_1\cong K_{2,2}$, then $n=6$, which is a
contradiction.

\noindent {\bf Subcase 1.5.} $G_2\cong Q$. Then $G$ must have the
structure as given in Figure \ref{fig5} (h) or (i). In the former
case, $G-\{e_3, e_4\}=G'_1+G'_2$, where $G'_2$ is a path of order
$4$ and $G'_1$ is a unicyclic graph which is not isomorphic to
$K_{2,2}$. Similarly, we can obtain that $E(G)>n$. In the latter
case, $G-\{e_2, e_3\}=G'_1+G'_2$, where $G'_2$ is a tree of order
$5$ and $G'_1$ is a unicyclic graph which is not isomorphic to
$K_{2,2}$. Similarly, we can obtain that $E(G)>n$.

\noindent {\bf Case 2.} All the paths $P_1$, $P_2$ and $P_3$ have
length $2$.

We assume that $P_1=uxv$, $P=uzv$ and $P_2=uyv$. Let $F=\{uy,vy\}$,
then $G-F=G_1+G_2$, where $G_1$ is a unicyclic graph and $G_2$ is a
tree. It follows from Lemma \ref{lem1.3} that $E(G_1)\geq |V(G_1)|$.
If $G_2\not\cong S_1, S_3, S_4, W$, then we have $E(G_2)\geq
|V(G_2)|$ by Lemma \ref{lem1.1} and so $E(G)>n$ by Lemma
\ref{lem2.4}. Hence we only need to consider the following four
subcases.

\noindent {\bf Subcase 2.1.} $G_2\cong S_1$. Let $F'=\{uy,zv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2\cong S_2$ and $G'_1$ is a tree of
order at least $6$ since $n\geq 8$. It is easy to see that $G'_1$
can not be isomorphic to $Q$ or $W$. Therefore we have
$E(G'_1)>|V(G'_1)|$ and $E(G'_2)=|V(G'_2)|$ by Lemma \ref{lem1.1}
and Theorem \ref{thm2.5}. It follows from Lemma \ref{lem2.4} that
$E(G)>n$.

\noindent {\bf Subcase 2.2.} $G_2\cong S_3$. Then $G$ must have the
structure as given in Figure \ref{fig5} (j). Let $F'=\{uy,zv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2$ is the path of order $4$
containing $y$ and $G'_1$ is a tree of order at least $4$ since
$n\geq 8$. Clearly, $G'_1$ can not be isomorphic to $S_4$, $Q$ or
$W$. Similar to the proof of Subcase 2.1, we have $E(G)>n$.

\noindent {\bf Subcase 2.3.} $G_2\cong S_4$. Then $G$ must have the
structure as given in Figure \ref{fig5} (k). Let $F'=\{uy,zv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2$ is the tree of order $5$
containing $y$ and $G'_1$ is a tree of order at least $3$. Clearly,
$G'_1$ can not be isomorphic to $S_4$ or $W$. If $G'_1\not\cong
S_3$, then we can similarly obtain that $E(G)>n$. If $G'_1\cong
S_3$, then $G$ must be the graph as given in Figure \ref{fig5} (l).
By choosing the edge cut $\{uy,uz,xv\}$, we can also obtain that
$E(G)>n$.

\noindent {\bf Subcase 2.4.} $G_2\cong W$. Then $G$ must have the
structure as given in Figure \ref{fig5} (m). Let $F'=\{uy,zv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2$ is the tree of order $8$
containing $y$ and $G'_1$ is a tree of order at least $3$. Clearly,
$G'_1$ can not be isomorphic to $S_4$ or $W$. If $G'_1\cong S_3$,
then $n=11$, which is a contradiction. If $G'_1\not\cong S_3$, then
we can similarly obtain that $E(G)>n$.

\noindent {\bf Case 3.} One of the paths $P_1$, $P_2$ and $P_3$ has
length $1$, and the other two paths have length $2$.

Without loss of generality, we assume that $P=uv$, $P_1=uxv$ and
$P_2=uyv$. Let $F=\{uy,vy\}$, then $G-F=G_1+G_2$, where $G_1$ is a
unicyclic graph and $G_2$ is a tree. Similarly, if $G_2\not\cong
S_1, S_3, S_4, W$, then we have $E(G)>n$. Hence we also need to
consider the following four subcases.

\noindent {\bf Subcase 3.1.} $G_2\cong S_1$. Let $F'=\{uy,uv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2\cong S_2$ and $G'_1$ is a tree of
order at least $6$ since $n\geq 8$. Since $\Delta(G)\leq 3$, $G'_1$
can not be isomorphic to $Q$ or $W$. Similar to the proof of Subcase
2.1, we have $E(G)>n$.

\noindent {\bf Subcase 3.2.} $G_2\cong S_3$. Then $G$ must have the
structure as given in Figure \ref{fig5} (n). Let $F'=\{uy,uv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2$ is the path of order $4$
containing $y$ and $G'_1$ is a tree of order at least $4$ since
$n\geq 8$. Clearly, $G'_1$ can not be isomorphic to $S_4$ or $W$.
Similarly, we have $E(G)>n$.

\noindent {\bf Subcase 3.3.} $G_2\cong S_4$. Then $G$ must have the
structure as given in Figure \ref{fig5} (o). Let $F'=\{uy,uv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2$ is the tree of order $5$
containing $y$ and $G'_1$ is a tree of order at least $3$. Clearly,
$G'_1$ can not be isomorphic to $S_4$ or $W$. If $G'_1\not\cong
S_3$, then we can similarly obtain that $E(G)>n$. If $G'_1\cong
S_3$, then $G$ must be the graph as given in Figure \ref{fig5} (p).
By choosing the edge cut $\{xu,xv\}$, we can similarly obtain that
$E(G)>n$.

\noindent {\bf Subcase 3.4.} $G_2\cong W$. Then $G$ must have the
structure as given in Figure \ref{fig5} (q). Let $F'=\{uy,uv,xv\}$,
then $G-F'=G'_1+G'_2$, where $G'_2$ is the tree of order $8$
containing $y$ and $G'_1$ is a tree of order at least $2$. Clearly,
$G'_1$ can not be isomorphic to $S_4$ or $W$. If $G'_1\cong S_3$,
then $n=11$, which is a contradiction. If $G'_1\not\cong S_3$, then
we can similarly obtain that $E(G)>n$. The proof is thus complete.
\hfill \rule{4pt}{7pt} \\[3mm]
\noindent{\bf Proof of Conjecture \ref{con1.5}:} Let $G$ be a
connected graph of order $n$ with $\Delta \leq 3$. Clearly, if $G$
is isomorphic to a graph in $\{S_2, Q, K_{2,2},K_{3,3}\}$, then
$E(G)=n$. We will prove that $E(G)\neq n$ if $G\not\cong S_2$, $Q$,
$K_{2,2}$ or $K_{3,3}$ by induction on the cyclomatic number $c(G)$.
It follows from Theorems \ref{thm2.5}, \ref{thm2.6} and \ref{thm2.7}
that the result holds for $c(G)\leq 2$. Let $k\geq 3$ be an integer.
We assume that the result holds for $c(G)<k$. Now let $G$ be a graph
with $c(G)=k\geq 3$. We will show that $E(G)\neq n$.

By Lemma \ref{lem2.3}, the result holds if $n$ is odd. By the fact
that $K_{3,3}$ is the only connected cycle-containing graph of order
6 with $\Delta \leq 3$ and $E=6$, we know that the result holds for
$n\leq 6$. So in the following we assume that $n\geq 8$ is even. In
our proof we will repeatedly make use of the following claim:

\noindent {\bf Claim 1.} {\it Let $F$ be an edge cut of $G$ such
that $G-F=G_1 + G_2$ with $c(G_1), c(G_2)<k$. If $G_1, G_2\not\cong
S_1, S_3, S_4, W$ or $K_{2,3}$ and either the edges in $F$ form a
star or at least one of $G_1$ and $G_2$ is not isomorphic to $S_2,
Q$ or $K_{2,2}$, then we are done. }

\noindent {\it Proof.}  By Lemma \ref{lem1.4}, we have $E(G_1)\geq |V(G_1)|$ and
$E(G_2)\geq |V(G_2)|$. Clearly, $G_1, G_2\not\cong K_{3,3}$. If
$G_i\not\cong S_2, Q$ or $K_{2,2}$, then by induction hypothesis, we
have $E(G_i)\neq |V(G_i)|$. Therefore we have $E(G)>n$ by Lemma
\ref{lem2.4}. \hfill \rule{4pt}{7pt}

In what follows, we use $\hat{G}$ to denote the graph obtained from
$G$ by repeatedly deleting the pendent vertices. Clearly,
$c(\hat{G})=c(G)$. Denote by $\kappa'(\hat{G})$ the edge
connectivity of $\hat{G}$. Since $\Delta(\hat{G})\leq 3$, we have
$1\leq \kappa'(\hat{G})\leq 3$. Therefore we only need to consider
the following three cases.

\noindent {\bf Case 1.} $\kappa'(\hat{G})=1$.

Let $e$ be a cut edge of $\hat{G}$. Then $\hat{G}-e$ has exactly two
components, say, $H_1$ and $H_2$. It is clear that $c(H_1)\geq 1$,
$c(H_2)\geq 1$ and $c(H_1)+c(H_2)=k$. Consequently, $G-e$ has
exactly two components $G_1$ and $G_2$ with $c(G_1)\geq 1$,
$c(G_2)\geq 1$ and $c(G_1)+c(G_2)=k$, where $H_i$ is a subgraph of
$G_i$ for $i=1,2$. If neither $G_1$ nor $G_2$ is isomorphic to
$K_{2,3}$, then we are done by Claim 1. Otherwise, without loss of
generality, we assume that $G_1\cong K_{2,3}$. Then $G$ must have
the structure as given in Figure \ref{fig6} (a). Now, let
$F=\{e_1,e_2\}$. Then $G-F=G'_1+G'_2$, where $G'_1\cong K_{2,2}$ and
$G'_2=G_2\cup e$. Therefore we have that $c(G'_2)=k-2\geq 1$ and
$G'_2\not\cong K_{2,2}, K_{2,3}$, and so we are done by Claim 1.
\begin{figure}[ht]
\centering
   \setlength{\unitlength}{0.05 mm}%
  \begin{picture}(2240.7, 1075.8)(0,0)
  \put(0,0){\includegraphics{fig6}}
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  \put(392.02,902.47){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_1$\strut}}
  \put(389.07,678.47){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_2$\strut}}
  \put(770.16,596.85){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$G_2$\strut}}
  \put(1168.43,876.79){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_3$\strut}}
  \put(1538.69,950.67){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_1$\strut}}
  \put(1528.28,669.79){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_2$\strut}}
  \put(1813.06,609.31){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$G_2$\strut}}
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  \put(343.14,248.57){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_2$\strut}}
  \put(216.17,362.62){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_3$\strut}}
  \put(40.00,205.28){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_4$\strut}}
  \put(1094.30,444.50){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_1$\strut}}
  \put(1109.38,197.32){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_2$\strut}}
  \put(977.61,358.38){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_3$\strut}}
  \put(802.09,202.38){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_4$\strut}}
  \put(1839.02,453.55){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_1$\strut}}
  \put(1821.93,131.44){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_2$\strut}}
  \put(1719.44,360.82){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_3$\strut}}
  \put(1541.83,210.79){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_4$\strut}}
  \put(321.18,72.87){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(c)\strut}}
  \put(1092.30,50.91){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(d)\strut}}
  \put(1832.73,37.52){\fontsize{11.76}{14.11}\selectfont \makebox(124.0, 82.6)[l]{(e)\strut}}
  \end{picture}%
  \caption{The graphs in the proof of Case 1 and Subcase 2.1 of Conjecture \ref{con1.5}.} \label{fig6}
\end{figure}

\noindent {\bf Case 2.} $\kappa'(\hat{G})=2$.

Let $F=\{e_1, e_2\}$ be an edge cut of $\hat{G}$. Then $\hat{G}-F$
has exactly two components, say, $H_1$ and $H_2$. Clearly,
$c(H_1)+c(H_2)=k-1\geq 2$.

\noindent {\bf Subcase 2.1.} $c(H_1)\geq 1$ and $c(H_2)\geq 1$.
Therefore, $G-F$ has exactly two components $G_1$ and $G_2$ with
$c(G_1)\geq 1$, $c(G_2)\geq 1$ and $c(G_1)+c(G_2)=k-1$, where $H_i$
is a subgraph of $G_i$ for $i=1,2$. If $G_1, G_2\not\cong K_{2,3}$
and at least one of $G_1$ and $G_2$ is not isomorphic to $K_{2,2}$,
then we are done by Claim 1. If at least one of $G_1$ and $G_2$ is
isomorphic to $K_{2,3}$, say $G_1\cong K_{2,3}$. Then $G$ must have
the structure as given in Figure \ref{fig6} (b). Now, let
$F'=\{e_2,e_3,e_4\}$, then $G-F'=G'_1+G'_2$, where $G'_1\cong
K_{2,2}$ and $G'_2=G_2\cup e_1$. Therefore we have that
$c(G'_2)=k-3$ and $G'_2\not\cong K_{2,2}, K_{2,3}$, and so we are
done by Claim 1. If $G_1,G_2\cong K_{2,2}$, then $G$ must be the
graph as given in Figure \ref{fig6} (c), (d) or (e). Let $F'=\{e_1,
e_3, e_4\}$, then $G-F'=G'_1+G'_2$, where $G'_1\cong S_2$ and
$G'_2\not\cong K_{2,2}$ is a unicyclic graph. Hence we are done by
Claim 1.

\noindent {\bf Subcase 2.2.} One of $H_1$ and $H_2$, say $H_2$ is a
tree. Therefore, $G-F$ has exactly two components $G_1$ and $G_2$
with $c(G_1)=k-1$ and $c(G_2)=0$, where $H_i$ is a subgraph of $G_i$
for $i=1,2$. Since $k-1\geq 2$, $G_1\not\cong S_2,Q,K_{2,2}$. If
$G_1\not\cong K_{2,3}$ and $G_2\not\cong S_1,S_3,S_4,W$, then we are
done by Claim 1. So we assume that this is not true. We only need to
consider the following five subsubcases.

\noindent {\bf Subsubcase 2.2.1.} $G_2\cong S_1$. Let
$V(G_2)=\{x\}$, $e_1=xx_1$ and $e_2=xx_2$. It is clear that
$d_{G_1}(x_2)=1$ or $2$. If $d_{G_1}(x_2)=1$, let
$N_{G_1}(x_2)=\{y_1\}$ (see Figure \ref{fig7} (a), where $y_1$ may
be equal to $x_1$). Let $F'=\{e_1, x_2y_1\}$. Then $G-F'=G'_1+G'_2$,
where $G'_1$ is a graph obtained from $G_1$ by deleting a pendent
vertex and $G'_2\cong S_2$. Therefore, $c(G'_1)=k-1\geq 2$. If
$G'_1\not\cong K_{2,3}$, then we are done by Claim 1. Otherwise,
$n=7$, which is a contradiction.
\begin{figure}[ht]
\centering
    \setlength{\unitlength}{0.05 mm}%
  \begin{picture}(2348.1, 3013.8)(0,0)
  \put(0,0){\includegraphics{fig7}}
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  \put(383.52,2841.44){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$x_1$\strut}}
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  \end{picture}%
  \caption{The graphs in the proof of Subcase 2.2 of Conjecture \ref{con1.5}.} \label{fig7}
\end{figure}

If $d_{G_1}(x_2)=2$, let $N_{G_1}(x_2)=\{y_1,y_2\}$ (see Figure
\ref{fig7} (b), where one of $y_1$ and $y_2$ may be equal to $x_1$).
Let $F'=\{e_1, x_2y_1, x_2y_2\}$. Then $G-F'=G'_1+G'_2$, where
$G'_1$ is a graph obtained from $G_1$ by deleting a vertex of degree
$2$ and $G'_2\cong S_2$. Therefore, $c(G'_1)=k-2\geq 1$. If
$G'_1\not\cong K_{2,2}, K_{2,3}$, then we are done by Claim 1.
Otherwise, $n=6$ or $7$, which is a contradiction.

\noindent {\bf Subsubcase 2.2.2.} $G_2\cong S_3$. If $e_1$, $e_2$
are incident with a common vertex in $G_2$, then $G$ must have the
structure as given in Figure \ref{fig7} (c). Similar to the proof of
Subsubcase 2.2.1, we can obtain that there exists an edge cut $F'$
such that $G-F'=G'_1+G'_2$ satisfying that $c(G'_1)=k-1$ if
$d_{G_1}(x_2)=1$ or $c(G'_1)=k-2$ if $d_{G_1}(x_2)=2$ and $G'_2$ is
a path of order $4$. If $G'_1\not\cong K_{2,3}$, then we are done by
Claim 1. Otherwise $n=9$, which is a contradiction.

If $e_1$, $e_2$ are incident with two different vertices in $G_2$,
then $G$ must have the structure as given in Figure \ref{fig7} (d)
or (e). Let $F'=\{e_2, e_3\}$, then $G-F'=G'_1+G'_2$, where
$G'_1=G_1\cup e_1$ and $G'_2\cong S_2$. Therefore we have that
$c(G'_1)=k-1\geq 2$ and $G'_2\not\cong K_{2,3}$, and so we are done
by Claim 1.

\noindent {\bf Subsubcase 2.2.3.} $G_2\cong S_4$. If $e_1$, $e_2$
are incident with a common vertex in $G_2$, then $G$ must have the
structure as given in Figure \ref{fig7} (f). Similar to the proof of
Subsubcase 2.2.1, we can obtain that there exists an edge cut $F'$
such that $G-F'=G'_1+G'_2$ satisfying that $c(G'_1)=k-1$ if
$d_{G_1}(x_2)=1$ or $c(G'_1)=k-2$ if $d_{G_1}(x_2)=2$ and $G'_2$ is
a tree of order $5$. If $G'_1\not\cong K_{2,3}$, then we are done by
Claim 1. Otherwise $G$ must be the graph as given in Figure
\ref{fig7} (h) or (i). In the former case let $F''=\{e_1,e_3,e_4\}$
while in the latter case let $F''=\{e_1,e_3,e_4,e_5\}$. Then
$G-F''=G''_1+G''_2$, where $G''_1\cong K_{2,2}$, $G''_2$ is a tree
of order $6$ and $G''_2\not\cong Q$. Therefore we are done by Claim
1.

If $e_1$, $e_2$ are incident with two different vertices in $G_2$,
then $G$ must have the structure as given in Figure \ref{fig7} (g).
Let $F'=\{xy, yz\}$, then $G-F'=G'_1+G'_2$, where $G'_1=G_1\cup
\{e_1,e_2\}$ and $G'_2\cong S_2$. Therefore we have that
$c(G'_1)=k-1\geq 2$ and $G'_2\not\cong K_{2,3}$, and so we are done
by Claim 1.

\noindent {\bf Subsubcase 2.2.4.} $G_2\cong W$. If $e_1$, $e_2$ are
incident with a common vertex in $G_2$, then $G$ must have the
structure as given in Figure \ref{fig7} (j). Similar to the proof of
Subsubcase 2.2.1, we can obtain that there exists an edge cut $F'$
such that $G-F'=G'_1+G'_2$ satisfying that $c(G'_1)=k-1$ if
$d_{G_1}(x_2)=1$ or $c(G'_1)=k-2$ if $d_{G_1}(x_2)=2$ and $G'_2$ is
a tree of order $8$. If $G'_1\not\cong K_{2,3}$, then we are done by
Claim 1. Otherwise, $n=13$, which is a contradiction.

If $e_1$, $e_2$ are incident with two different vertices in $G_2$,
then $G$ must have the structure as given in Figure \ref{fig5} (e),
(f) or (g) ($e_1$, $e_2$ may be incident with a common vertex in
$G_1$). Let $F'=\{xy, yz\}$, then $G-F'=G'_1+G'_2$, where $G'_2$ is
the tree of order $5$ or $2$ containing $y$. Clearly,
$c(G'_1)=k-1\geq 2$ and $G'_1\not\cong K_{2,3}$. Therefore we are
done by Claim 1.

\noindent {\bf Subsubcase 2.2.5.} $G_1\cong K_{2,3}$ and
$G_2\not\cong S_1, S_3, S_4, W$. It is easy to see that $G$ must
have the structure as given in Figure \ref{fig7} (k). Let
$F'=\{e_1,e_3,e_4,e_5\}$. Then $G-F'=G'_1+G'_2$, where $G'_1\cong
S_2$ and $G'_2$ is a tree of order at least $6$ since $n\geq 8$. It
is easy to see that $G'_2$ can not be isomorphic to $W$ or $Q$.
Therefore we are done by Claim 1.

\noindent {\bf Case 3.} $\kappa'(\hat{G})=3$.

Noticing that $\Delta (\hat G)\leq 3$ and $\Delta (G)\leq 3$, we
obtain that $G=\hat G$ is a connected $3$-regular graph. Hence we
have $n+k-1=m=\frac{3}{2}n$, i.e., $n=2k-2$. Since $n\geq 8$, we
have $k\geq 5$.

Let $F=\{e_1, e_2, e_3\}$ be an edge cut of $G$. Then $G-F$ has
exactly two components, say, $G_1$ and $G_2$. Clearly,
$c(G_1)+c(G_2)=k-2\geq 3$. Let $c(G_1)\geq c(G_2)$. If $c(G_2)\geq
3$, then we are done by Claim 1. Hence we only need to consider the
following three subcases.

\noindent {\bf Subcase 3.1.} $c(G_2)=0$ and $c(G_1)=k-2$. Let
$|V(G_2)|=n_2$. Then we have $3n_2=\sum_{v\in
V(G_2)}d_{G}(v)=2(n_2-1)+3=2n_2+1$. Therefore, $n_2=1$, i.e.,
$G_2=S_1$. Let $V(G_2)=\{x\}$, $e_1=xx_1$, $e_2=xx_2$ and
$e_3=xx_3$. Let $N_{G_1}(x_2)=\{y_1, y_2\}$ (see Figure \ref{fig8}
(a)). Let $F'=\{e_1, e_3, x_2y_1, x_2y_2\}$. Then $G-F'=G'_1+ G'_2$,
where $G'_2\cong S_2$ and $G'_1$ is a graph obtained from $G_1$ by
deleting a vertex of degree $2$. Therefore, $c(G'_1)=k-3\geq 2$. If
$G'_1\not\cong K_{2,3}$, then we are done by Claim 1. If $G'_1\cong
K_{2,3}$, then $n=7$, which is a contradiction.
\begin{figure}[ht]
\centering
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  \put(1139.65,623.10){\fontsize{11.76}{14.11}\selectfont \makebox(206.6, 82.6)[l]{$e_1$\strut}}
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  \caption{The graphs in the proof of Case 3 of Conjecture \ref{con1.5}.} \label{fig8}
\end{figure}

\noindent {\bf Subcase 3.2.} $c(G_2)=1$ and $c(G_1)=k-3$. Let
$|V(G_2)|=n_2$. Then we have $3n_2=\sum_{v\in
V(G_2)}d_{G}(v)=2n_2+3$. Therefore, $n_2=3$, i.e., $G_2$ is a
triangle. If $G_1\not\cong K_{2,3}$, then we are done by Claim 1. If
$G_1\cong K_{2,3}$, then $G$ must be the graph as given in Figure
\ref{fig8} (b). Let $F'=\{e_1, e_4, e_5, e_6\}$. Then $G-F'=G'_1+
G'_2$, where $G'_1\cong S_2$ and $G'_2$ is a bicyclic graph which is
not isomorphic to $K_{2,3}$. Then we are done by Claim 1.

\noindent {\bf Subcase 3.3.} $c(G_2)=2$ and $c(G_1)=k-4$. Let
$|V(G_2)|=n_2$. Then we have $3n_2=\sum_{v\in
V(G_2)}d_{G}(v)=2(n_2+1)+3=2n_2+5$. Therefore, $n_2=5$. If neither
$G_1$ nor $G_2$ is isomorphic to $K_{2,3}$, then we are done by
Claim 1. Otherwise, we assume that $G_2\cong K_{2,3}$ (similar for
$G_1\cong K_{2,3}$). Then $G$ must have the structure as given in
Figure \ref{fig8} (c). Let $F'=\{e_1, e_2, e_4, e_5\}$. Then
$G-F'=G'_1+ G'_2$, where $G'_2\cong K_{2,2}$ and $G'_1$ is a
$(k-4)$-cyclic graph which is not isomorphic to $K_{2,3}$. Then we
are done by Claim 1. The proof is thus complete. \hfill \rule{4pt}{7pt}

redpajama_set_name: RedPajamaArXiv


text:
Q: QSqlDatabase::addDatabase() check if host exists Currently I'm working on a C++ application using the qt framework to read a specific database from a mysql server or a sqlite3 file. I have the main UI to display the data and a login-dialog to connect to the database.
The dialog checks if the filename is valid and a sqlite3 database and now i need this check for a given IP address cause otherwise the program gets blocked.
Is is possible to check if the IP address in the network exists and maybe to check if the server is available?

redpajama_set_name: RedPajamaStackExchange


Advanced phones boxy design of BlackBerry Passport, equipped with touch screen and physical keyboard. BlackBerry introduced in separate events in three different locations.
The realistic and strange shape of BlackBerry Passport smartphone has a square finished design.
After a lot of information and pictures leak, the BlackBerry release date has been set. The telephone company "Strawberry Black" send event invitations on the same day in three different areas of Canada, UK and Dubai.
BlackBerry paid attention for their looks similar like the LG Vu combines a touchscreen and a full qwerty keyboard. The wide 4.5 inch display with 1440×1440 resolution to 453 ppi pixel density higher than the M8 One or Galaxy S5. The keyboard is equipped with physical attractiveness on Passport but was reduced to 3 rows. This design of the keys is bigger, perfectly flat and a change compared to the previous keyboard BlackBerry.
As a high-end device, the Blackberry smartphone Passport equipped with powerful configuration 2.2 GHz quad-core Snapdragon processor 800. It has 3GB of RAM, 32 GB of internal memory, expandable with a memory card. Together, the BlackBerry also focuses on the entertainment features integrated 13 megapixel camera with optical image stabilization and front camera of 2 megapixel.
Blackberry Passport includes the connectivity of LTE network, Wi-Fi standard, Bluetooth 4.0, GPS and NFC. 3450 mAh high capacity battery is also equipped with high expectations on new BlackBerry phones. Dimensions are 128×90.3×9.3 mm and weight of 194 gm.
BlackBerry smartphone with box styling, combined with the traditional QWERTY keyboard, but is designed more simply with just 3 lines of buttons.
After John Chen officially unveiled the Passport with traditional qwerty keyboard this product has caused the attention of community technology. With a square shape, unique design and a qwerty keyboard with 3 rows, BlackBerry Passport deserve to be one of the phones most anticipated in the fall of this year. You will have a better experience when reading text, surf the web and view the document.
The emphasis on the form of the Passport not only in design, 3 row qwerty keypad also possesses many interesting features, such as stroking up and down to browse, swipe right to delete from, or stroking the keys over to implementation of punctuation, numbers.
BlackBerry Passport Cost in India is around Rupees 35,000 to 40,000.


Discussion in 'Transformers News and Rumors' started by Matt Booker, Jul 29, 2010.
Haven't seen this reported anywhere, figured I'd post this.
This morning at Walmart I picked up Prime and Banzaitron.
They also had Payload, Seaspray, and the bigger Powercore Combiners packs.
My tfw search skills have atrophied.
At least it's good to hear Walmart has something at last.
Deluxes were still Wall O Mudflap though.
How's the Prime look in person?
Not bad. I'm really thinking of a friend of mine touch it up, but the unpainted chunks that look garish in pictures look okay in person.
It's got a lot of detail that really needs a wash to bring it out, but it's not as bad as it looked.
Searching the News and Rumor RSS feed is another good option (especially if you read RSS feeds with Google Reader).
If it says released, there's a good chance that there's been a news story about it.


text:
Q: How to share variable across files in different directory I have a variable defines in one .cpp file.
(file1.cpp)
int N;
....
N =3;

(directory/file2.cpp)  ---->  file2.cpp is in another directory.
extern int N;

cout << n << endl ;    -----> It is printing 0.

It should hv printed 3..right ? What is the mistake I am doing..Please let me know. Directory Structure is:
(main Directory)
file1.cpp   directory

(directory)
file2.cpp

Thanks,
Avinash

A: In one file this variable should be define as is:
  int N;

in others as extern (it is better to use header file to include such declaration), i.e.:
  extern int N;

The directory structure is not important, just cpp files must be compiled together to allow linker resolving connections.

A: To make it work you can use following approach:
Header file which provides the declarations so that other translation units can easily import it:
// dir1/file1.hpp
extern int N;

Implementation file with variable definition:
// dir1/file1.cpp
int N = 3;

Example client which uses the variable:
// main.cpp
#include "dir1/file1.hpp"

#include <iostream>

int main () {
    std::cout << N << std::endl;
    return 0;
}

Note that include guards are omitted in dir1/file1.hpp


text:
Age-specific strategies are needed when caring for older individuals with HIV
A new article highlights the differences between older and younger adults living with HIV, and offers age-specific strategies on how to provide care.
Older individuals differ from younger patients with HIV in many ways--for example, they tend to have an increased risk for late or missed diagnoses, differing side effects from antiretroviral therapies, and increased comorbidities. Only a few existing treatment guidelines make distinctions when it comes to care by age, however.
By examining findings from clinical research, investigators have now offered easy-to-follow approaches for providing care to older people living with HIV. "Our article recommends an overall strategy in which various tools are used to provide information on the likely course of HIV for an individual, which clinicians can then use to guide discussions on personalized, person-centered care," said Dr. Jonathan Appelbaum, co-author of the Journal of the American Geriatrics Society article. One tool that may be especially useful is the Veterans Aging Cohort Study (VACS) Index, which can help distinguish between those who are aging well and those who may appear older than their chronological age.
"Our article highlights the need for more research on screening, evaluation, and treatment in older adults infected with HIV. As the number of older people living with HIV grows, we need scientifically sound data to help prepare our clinicians and healthcare systems for the unique needs and expectations of this group," said co-author Dr. Aroonsiri Sangarlangkarn.
Related Hiv Articles:
Defective HIV proviruses reduce effective immune system response, interfere with HIV cure
A new study finds defective HIV proviruses, long thought to be harmless, produce viral proteins and distract the immune system from killing intact proviruses needed to reduce the HIV reservoir and cure HIV.
1 in 7 people living with HIV in the EU/EEA are not aware of their HIV status
Almost 30,000 newly diagnosed HIV infections were reported by the 31 European Union and European Economic Area (EU/EEA) countries in 2015, according to data published today by ECDC and the WHO Regional Office for Europe.
Smoking may shorten the lifespan of people living with HIV more than HIV itself
A new study led by researchers at Massachusetts General Hospital finds that cigarette smoking substantially reduces the lifespan of people living with HIV in the US, potentially even more than HIV itself.
For smokers with HIV, smoking may now be more harmful than HIV itself
HIV-positive individuals who smoke cigarettes may be more likely to die from smoking-related disease than the infection itself, according to a new study published in the Journal of Infectious Diseases.
Patients diagnosed late with HIV infection are more likely to transmit HIV to others
An estimated 1.2 million people live with HIV in the United States, with nearly 13 percent being unaware of their infection.
The Lancet HIV: New HIV infections stagnating at 2.5 million a year worldwide
A major new analysis from the Global Burden of Disease 2015 study, published today in The Lancet HIV journal, reveals that although deaths from HIV/AIDS have been steadily declining from a peak in 2005, 2.5 million people worldwide became newly infected with HIV in 2015, a number that hasn't changed substantially in the past 10 years.
NIH scientists discover that defective HIV DNA can encode HIV-related proteins
Investigators from the National Institutes of Health have discovered that cells from HIV-infected people whose virus is suppressed with treatment harbor defective HIV DNA that can nevertheless be transcribed into a template for producing HIV-related proteins.
Study examines risk of HIV transmission from condomless sex with virologically suppressed HIV infection
Among nearly 900 serodifferent (one partner is HIV-positive, one is HIV-negative) heterosexual and men who have sex with men couples in which the HIV-positive partner was using suppressive antiretroviral therapy and who reported condomless sex, during a median follow-up of 1.3 years per couple, there were no documented cases of within-couple HIV transmission, according to a study appearing in the July 12 issue of JAMA, an HIV/AIDS theme issue.
HIV vaccine design should adapt as HIV virus mutates
Researchers from UAB, Emory and Microsoft demonstrate that HIV has evolved to be pre-adapted to the immune response, worsening clinical outcomes in newly infected patients.
Charlie Sheen's HIV disclosure may reinvigorate awareness, prevention of HIV
Actor Charlie Sheen's public disclosure in November 2015 that he has the human immunodeficiency virus (HIV) corresponded with the greatest number of HIV-related Google searches ever recorded in the United States, according to an article published online by JAMA Internal Medicine.
More Hiv News and Hiv Current Events
Related Hiv Reading:

redpajama_set_name: RedPajamaCommonCrawl


text:
Tag: Reading
Guess Who Reading Discussion Prompt for Craft of Prose
In this reading discussion prompt, students are asked to consider what elements contribute to our understanding of character in this "Guess Who" game in which students draw a character who appears in chapters 3–12 of Harry Potter and the Goblet of Fire and then describe them by answering a series of questions. They will use their descriptions to come up with three clues about the character: one is a concrete detail about the character's appearance, the second is to identify a scene in which they appear, and the third is a literary craft element that helps reveal their character.
Fall 2016 Required Texts & Course Descriptions
ENG 2015: Poetry Workshop
American poet C.D. Wright once wrote: "If I wanted to understand a culture, my own for instance . . . I would turn to poetry first. For it is my confirmed bias that the poets remain the most 'stunned by existence,' the most determined to redeem the world in words." In this course, we will hold poetry to this noble standard, as an amplifier for the voices in our culture and an invocatory rendering of our world. In doing so, I'll ask you to not only read and write poetry but also begin to look at your surroundings as a poet would. This requires close examination of images, scrutiny of your thoughts and feelings about subject matter, and consideration for other points of view. Additionally, you will be asked to think deeply about language, in terms of its meanings, its sounds, its rhythms, and its forms. You should bring to this class a hard work ethic supported by curiosity and generosity. As a means of introduction to the craft of poetry, students will submit original poems for workshop, a collaborative discussion about writing techniques and their effects on readers. In addition to workshop, you will be asked to engage with the writing of contemporary poets, to read like a writer would. I've chosen Poetry magazine as our required text so that you will have a lens through which to examine the current landscape of American poetry and to see that even today poets are still trying to "redeem the world in words."
Hirsch, Edward. A Poet's Glossary. Houghton Mifflin Harcourt, 2014. ISBN: 978-0151011957.
Poetry magazine student subscription, available at https://www.poetryfoundation.org/poetrymagazine/subscribe
ENG 2016OL: Online Prose Workshop
This course will focus on the prose forms of the short story and personal essay, and emphasize drafting and revision. Students will respond to published prose and write their own pieces for workshop, a collaborative discussion about the effects of a writer's choices on readers. You should bring to this class a hard work ethic supported by curiosity and generosity. We will base our discussions on how texts work rather than what they mean, after Francine Prose's ideal of "reading like a writer." We will use Megan Mayhew Bergman's collection of short stories, Almost Famous Women, and Zadie Smith's collection of essays, Changing My Mind, as a touchstone for learning writing skills and discovering genre conventions. My approach to teaching writing is founded on the belief that our writing skills must be practiced and cultivated, and that one must continually challenge one's aesthetics, habits, and concerns throughout one's writing life in order to write anything of consequence to one's readers and, perhaps more importantly, one's self.
Bergman, Megan Mayhew. Almost Famous Women. Scribner, 2015. 256 pages. ISBN: 978-1476788814.
Smith, Zadie. Changing My Mind: Occasional Essays. Penguin, 2010. 320 pages. ISBN: 978-0143117957.
ENG 2031: Craft of Prose
In Italo Calvino's Invisible Cities, Marco Polo tells Kublai Khan that "the city must never be confused with the words that describe it." In this course, we will shoot high and aim to write so richly and uniquely about our fictional worlds that they will be rendered in our readers' imaginations as palpable, the words and places indistinguishable, symbiotic, "real." We will take as our lodestars a number of texts, including Calvino's Invisible Cities, a short story collection by Centenary's fall 2016 visiting author Megan Mayhew Bergman, and one of the Harry Potter novels. We will translate these immersive reading experiences into writing skills through discussion, exercises, and workshop. Several times throughout the semester, students will turn in original writing for workshop, a collaborative discussion about writing techniques and their effects on readers, and later revise one of the pieces using the comments received in workshop. Additionally, we will play host to Centenary's Fall 2016 visiting writer, Megan Mayhew Bergman, and prepare accordingly. You should bring to this class a hard work ethic supported by curiosity and generosity. We will base our discussions on how texts work rather than what they mean, after Francine Prose's ideal of "reading like a writer." My approach to teaching writing is founded on the belief that our writing skills must be practiced and cultivated, and that one must continually challenge one's aesthetics, habits, and concerns throughout one's writing life in order to write anything of consequence to one's readers and, perhaps more importantly, one's self.
Texts and Supplies
Bergman, Megan Mayhew. Almost Famous Women. Scribner, 2015. 256 pages. ISBN: 978-1476788814.*
Calvino, Italo. Invisible Cities. Harcourt, 1978. 165 pages. ISBN: 978-0156453806. +
Everett, Percival. I Am Not Sidney Poitier. Graywolf Press, 2009. 234 pages. ISBN: 978-1555975272. +
Rowling, J.K. Harry Potter and the Goblet of Fire. Scholastic Paperbacks, 2002. 752 pages. ISBN: 978-0439139601.*
Russell, Karen. Vampires in the Lemon Grove: And Other Stories. Vintage, 2014. +
Writing Journal
ENG 3099: Special Topics: Literary Editing & Publishing
"Editing, like writing, is fundamentally about composing a world," Peter Gizzi writes in his essay "On the Conjunction of Editing and Composition." In this course, students will learn how this act of composition takes place, from submissions to printing, by reading first-hand accounts of editors in the profession and through practical application. This reading intensive course will challenge you to read like an editor rather than a reader, writer, or critic, and ask you to consider how literary magazines contribute to literary culture. You will be exposed to many different types of editing styles, and you will be asked to begin to cultivate your own approach to editing a literary magazine or journal while being introduced to all the skillsets needed to create a publication. We will use literary magazines as a case study for all of our publishing inquiries, but we will likewise touch upon the book-publishing process and market. The class will include an investigation into the history of literary magazines; editorial meetings in which students will evaluate and debate sample pieces; papers that analyze literary magazines, editorial roles, and the state of contemporary publishing; and a final editorial project in which student groups will "compose a world" through a mock literary magazine by developing its mission, design, and content. In many ways, this course acts as a kind of introductory practicum for students interested in pursuing future publishing opportunities as editors, production editors, and as writers.
Lupton, Ellen. Thinking with Type, 2nd edition. Princeton Architectural Press, 2010.
Kurowski, Travis. Paper Dreams: Writers and Editors on the American Literary Magazine. Atticus Books, 2013.
Art supplies, paper, etc. and whatever else you may need to create your final editorial project.
"No River, No Bridge: Writing Poetry, Writing Nonfiction" Handout and Exercise
Ermakov, Dimitri (1846-1916).
I taught a class titled "No River, No Bridge: Writing Poetry, Writing Nonfiction" at Appomattox Regional Governor's School's Writers' Fest yesterday. We discussed hybridity of poetry and nonfiction, as well as the ways in which poetry can take on and explore "truth" using essayistic strategies without the requirement of "fact." Several students came up afterward to show me their exercises, and one told me that she felt like it was one of the best things she's written. The class ranged from high school students to adults.
In this course we will examine works of poetry and creative nonfiction in order to open up these genres to hybrid works. We will discuss features that are shared by the genres, how to bend the rules, and look at some examples before doing an exercise.
With over forty students in a one-hour class, we had these objectives:
Examine definitions of genre and the elements therein.
Think about the function and aims of each genre.
Read excerpts of texts that bridge the gap between poetry and nonfiction.
Write some assays—yes, some attempts—that use both poetry and essay strategies.
Examine definitions of genre and the elements therein. We looked at definitions of essays and lyric essays and Edward Hirsch's definitions of line, lineation, lyric, prose poem, and stanza from A Poet's Glossary. We looked at these definitions because they seek to define and, therefore, draw boundaries. We needed to know these definitions in order to explode these boundaries. In particular, we looked at Hirsch's definition of line, in which he quotes Longenbach:
"Poetry is the sound of language organized in lines," James Longenbach asserts in The Art of the Poetic Line (2008). "More than meter, more than rhyme, more than images or alliteration or figurative language, line is what distinguishes our experience of poetry as poetry, rather than some other kind of writing."
And compared it to the prose poem definition:
A composition printed as prose that names itself poetry. The prose poem takes advantage of its hybrid nature—it avails itself of the elements of prose (what Dryden called "the other harmony of prose") while foregrounding the devices of poetry. . . . Baudelaire used prose poems to rebel against the straitjacket of classical French versification. He dreamed of creating "a poetic prose, musical without rhyme or rhythm, supple and jerky enough to adapt to the lyric movements of the soul, to the undulations of reverie, to the somersaults of conscience." Baudelaire's prose poems—along with Rimbaud's Les Illuminations (1886) and Malarmé's Divagations (1897)—created a mixed musical form (part social, part transcendental) that has been widely and internationally practiced in the twentieth century. "There is no such thing as prose," Malarmé insisted in 1891. "There is the alphabet, and then there are verses which are more or less closely knit, more or less diffused. So long as there is a straining toward style, there is versification."
We thought about these paradoxes, discussed why lineation makes a poem a poem, as well as considered some alternative definitions. We then examined Hirsch's definition of "lyric"—
The lyric poem immerses us in the original waters of consciousness, in the awareness, the aboriginal nature, of being itself. . . . The lyric, which offers us a supposed speaker, a person to whom we often assign the name of the author, shades off into the dramatic utterance
—in juxtaposition with a discussion of point of view in essay. Think about the function and aims of each genre. When I think about the intersections and differences of poetry and nonfiction, it's helpful for me to illustrate the two genres with metaphors.
Poem as a river. With all of their musical elements, I think of poems as a river—something fluid, flowing, that rushes and eddies, deltas, and empties out to sea. Poems are one of our oldest forms of literature and therefore seem natural to our landscape of language.
Essay as a bridge. The essay, an analytical form, builds upon its subject piece by piece, element by element, fact by fact, experience by experience. We might think about this strategy as similar to building pillars with blocks, hammering planks together, supporting and trussing.
We might use these metaphors to understand hybrid works—nonfiction prose poems, lyric essays—better. Sometimes a poem is a river we want to cross; we see and hear the rushing water, but we want to examine it from a sturdy vantage point where we stay safe and dry. Some music comes into the poem, but we rely on the exposition that essays provide. Adversely, we could be in the water, prone to the rapids, splashed in the face, paddling fiercely. Just trying to stay afloat in the overflow of language, of experience. There are times, however, where we might pass under the shadow of a bridge, have insights into reason, into conquering this flow, bridging one experience to another. Read excerpts of texts that bridge the gap between poetry and nonfiction.
"Say It, Say It Anyway You Can" by Vievee Francis (two versions—the prose poem version from Rattle and the lineated version that appeared in her collection Horse in the Dark). Let's take a look at this poem, which started off as a prose poem and then became a lineated poem. Discuss the poet's possible reasoning for lineating the poem. What are the effects of the two forms?
An excerpt from Citizen by Claudia Rankine and an excerpt from Bluets by Maggie Nelson. Talk to me a little bit about what these prose forms borrow from poetry. Are they poetry? (How were they marketed? How were they identified?) Is it enough to say something is poetry? Nonfiction? Are these true hybrids? How would you categorize it? Does it need to be categorized?
"No River, No Bridge" Exercise
Select an everyday object that's in your house. Ex. a television, a pair of jeans, a can of kidney beans. (1 min.)
Now describe the object using all of your senses, with only concrete details. Do not use any metaphors or other figurative language. (2 min.)
Imagine how the object was made. Write a step-by-step description of its fabrication. (3 min.)
Imagine the harvesting, mining, or creation of the raw materials that went into making this object. (i.e. the shearing of sheep for your wool sweater or the mining of silicone to make your iPhone). Describe. (2 min.)
Now talk a little bit about how you came to own this object (and talk about not remembering how you got it, if that's true.) Tell us the backstory. Or, what connotations does the object have? Do you have specific memories associated with it? (2 min.)
Now describe the object using only metaphors. ("My pair of jeans is muddy water reflecting sky, a bird with clipped wings . . . ") (2 min.)
Share and discuss: How has the object changed in each iteration (concrete description, origin story, deconstruction, entrance, emotional connection, metaphorical naming)? Which form (poetry or essay) better conveys these approaches? Why? How can we combine these strategies?
"Fame Makes a Man Take Things Over" Exercise
Drawing of Stage Door Johnnies (1894)
Class: Writing Out of the Ordinary
Genre: Creative nonfiction
Readings: A packet of persona poems and dramatic monologues
1. Pick a celebrity, sports star, cartoon or comic book character, product mascot (ex. Count Chocula, the Geico gecko, etc.) or newsworthy individual (Octomom, Charles Manson, etc.).
2. Create a mundane problem for that character or person. (Kobe Bryant can't open a jelly jar. Elvis Presley can't fit into his old slacks. Speedy Gonzalez gets stuck in a mouse trap.)
3. Free write for ten minutes in the voice of that character as they're attempting to resolve the problem. What concerns them? Are they worried about their public image? How does this problem relate to bigger problems for them? What sorts of language do they use? Are they thinking about the problem at hand or something else? Where are they at? More specific questions: What are they wearing? What kind of jelly is Kobe Bryant trying to get into? Strawberry or grape? Who set the trap for Speedy? Has Elvis tried dieting? (Hint: You don't have to answer these specific questions, but be sure to take leaps like this with your own characters.)
The Shape of Words
In Design as Art, Bruno Munari writes, "Not only does each letter of a word have a shape of its own, but all its letters taken together give shape to the word." When we read, we read the shape of the word if the word is familiar to us. If the word is unfamiliar, we must read the shape of each letter, attributing sound to each one and combining them. We use this amalgamated sound to attempt to locate meaning. We troll our memories for previous usage. We look at context. If we're scholars, we might even recognize roots and stems. In effect, in all of these investigative processes, we look to the history of the word—personal, textual, or etymological—in order to understand its present meaning.
Sometimes when I'm tired or sick or otherwise impaired, however, I look at words and they're only shapes; they mean nothing. It makes reading more akin to viewing visual art, a Rothko perhaps. I look at the whole page, the shape, and the rivers of white across the lines. I see the words as bodies, as segmented insects. Regardless of whether or not I view the shape of the letters or words, meaning becomes untethered to those shapes. They become tangible. I care more for the vessel than what it contains.
While this as a permanent condition could be problematic, the transient aphasia provides me with a means to reboot language, the way time seems new and fresh on those rare and brief occasions when I forget how old I am. It reminds one that language is a symbol rather than an incarnation and that, in order for language to mean, I must be present. I must supply language with meaning. Reading is a transaction rather than a gift of information.
Words need me as much as I need them. When I realize this, suddenly the axis shifts again, and those shapes suddenly have definition.
On Journaling: a Revelation
Codex Manesse, fol. 383r, Meister Konrad von Würzburg
Sometimes discussions with my students lead me to articulate things about writing that I've fathomed but haven't been able to put into words. For instance, in Intro to Creative Writing this morning, my students were discussing Joan Didion's "On Keeping a Notebook" and how they could use journaling as a foundation for their creative writing. Because of the students' provocative questions and Didion's exegesis on the practice—"I imagine . . . that the notebook is about other people. But of course it is not. . . . our notebooks give us away, for however dutifully we record what we see around us, the common denominator of all we see is always, transparently, shamelessly, the implacable 'I.'"—I was able to finally say what I've often felt in keeping my writing journal:
Journaling is an exercise in being a character.
In keeping a journal, I distance myself from the self that appears in my writing. (Think Dante the writer vs. Dante the character.) This not only allows me to receive critical input without feeling as if I'm under attack but it also, and perhaps more importantly, gives me the opportunity to view my own writing as a reader would.
Maybe other writers out there have come to this realization, but the idea, and the way I was able to say it, surprised me in its clarity. Maybe this too is a product of journaling: I have distanced myself enough so that I am allowed to be surprised by myself, a moxie I find quite germane to the writing of the lyric.