text:
\section{Introduction}

~~~~The distance spectral radius of graphs have been studied extensively.
S. Bose, M. Nath and S. Paul \cite{B.S} determined the unique graph with maximal distance
spectral radius among graphs without a pendant vertex.
A. Ilic \cite{I.A} attained the unique graph whose distance spectral radius is maximum
among $n$-vertex trees with perfect matching and fixed maximum degree.
W. Ning, L. Ouyang and M. Lu \cite{N.W} characterized the graph with minimum distance spectral radius
among trees with given number of pendant vertices.
For more about the distance spectra of graphs see the survey \cite{A.M} as well as the references therein.

The least distance eigenvalues of connected graphs have been also studied.
H. Q. Lin \cite{L.H.Q.1} gave an upper bound on the least distance eigenvalue and
characterized all the connected graphs with the least distance eigenvalue
in $[-1-\sqrt{2}, a]$, where $a$ is the smallest root of $x^3-x^2-11x-7=0$ and $a\in (-1-\sqrt{2},-2)$.
H. Y. Lin and B. Zhou \cite{L.H.Y} attained the trees with the least distance eigenvalues
in $ [ -3-\sqrt{5},-2-\sqrt{2}  ] $ and the unicyclic and bicyclic graphs with least distance
eigenvalues in $(-2-\sqrt{2},-2.383)$.
G. L. Yu \cite{Y.L.G} introduced all the graphs with the least distance
eigenvalue in $[-2.383,0]$.

The complement of graph $G=( V( G ) ,E( G ) )$ is denoted by $G^c=( V( G^c ) ,E( G^c ) )$,
where $V( G^c ) =V( G ) $ and $E( G^c ) = \{ xy\notin E(G):x,y\in V( G ) ,\} $.
Y. X. Fan, F. F. Zhang and Y. Wang \cite{F.Y.Z} determined the connected graph with the minimal least eigenvalue
among all complements of trees.
G. S. Jiang, G. D. Yu, W. Sun and Z. Ruan \cite{J.G.S} gave the graph with the minimal least eigenvalue
among all graphs whose complements are connected and have only three pendent vertices.
S. C. Li and S. J. Wang \cite{L.S.C} introduced the unique connected graph whose least signless Laplacian
eigenvalue attains the minimum in the set of the complements of all trees.
G. D. Yu, Y. Z. Fan and M. L. Ye \cite{Y.G.D} achieved the unique graph which minimizes the least signless Laplacian eigenvalue
among all connected complements of unicyclic graphs.

Currently there is very little research about the distance eigenvalues of complements of graphs.
H. Q. Lin and S. Drury \cite{L.H.Q.2}
characterized the unique graphs whose distance spectral radius have maximum and minimum among all complements of trees,
and the unique graphs whose least distance eigenvalue have maximum and minimum among all complements of trees.
R. Qin, D. Li, Y. Y. Chen and J. X. Meng \cite{Q.R}
determined the unique graph which has maximum distance spectral radius
among all complements of unicyclic graphs and the unique graph which has maximum least distance eigenvalue
among all complements of unicyclic graphs of diameter three.

Let $G$ be a connected simple graph with the vertex set $V( G ) = \{ v_1,v_2,\cdots ,v_n  \} $.
Then the adjacency matrix of $G$ is $A( G ) =( a_{ij} ) _{n\times n}$,
where $a_{ij}=1$ if $v_i$ is adjacent to $v_j$, and $a_{ij}=0$ otherwise.
In this paper, we observe the relations between $D(G^c)$ and $A(G)$ and use them
to determine the unique graph whose distance spectral radius attains maximum and minimum
among all complements of graphs of diameter greater than three, respectively.
Furthermore, we also characterize the unique graph whose least distance eigenvalue attains maximum and minimum
among all complements of graphs of diameter greater than three, respectively.

\section {The distance spectral radius of the complements of graphs of diameter greater than three}
~~~~The below Lemma 2.1 reflects the relationship of $D(G^c)$ and $A(G)$.  \vskip 3mm

{\noindent \bf Lemma 2.1.}  {\it Suppose that $G$ is a simple graph on $n$ vertices whose diameter $d(G)$ is greater than three.
Then we have
\begin{enumerate}
	\setlength{\parskip}{0ex}
	\item[\rm (I.)]
	when $d( G ) > 3$, $D( G^c ) =J_n-I_n+A( G )$.
	\item[\rm (II.)]
	when $d( G ) =3$, $D( G^c )\geq J_n-I_n+A( G )$.
\end{enumerate} }

{\noindent \bf Proof.}
When $d(G)>3$, for any two vertices $u$ and $v$ of $G$,
there must exist the vertex $w$ of $G$ which is adjacent to neither $u$ nor $v$.
Thus $d_{G^c}(u,v)=2$ if $u$ is adjacent to $v$ in $G$,
and $d_{G^c}(u,v)=1$ otherwise.
This shows that $D(G^c)=J_n-I_n+A(G)$.

When $d(G)=3$, take two arbitrary vertices $u$ and $v$ of $G$.
Suppose that all vertices in $G\setminus \{u, v\}$ are adjacent to either $u$ or $v$.
Then $d_{G^c}(u,v)=3$ if $u$ is adjacent to $v$,
and $d_{G^c}(u_i,u_j)=1$ otherwise.
So We assume that there exists the vertex $w$ in $G\setminus \{u, v\}$ which is adjacent to neither $u$ nor $v$.
Then $d_{G^c}(u,v)=2$ if $u$ is adjacent to $v$, and $d_{G^c}(u,v)=1$ otherwise.
This shows that $D(G^c)\ge J_n-I_n+A(G)$. $\Box$ \vskip 3mm

In this section using the relations between $D(G^c)$ and $A(G)$ stated in Lemma 2.1
we determine the unique graph whose distance spectral radius attains maximum and minimum
among all complements of graphs of diameter greater than three, respectively.
\vskip 3mm

Suppose $G$ is a connected simple graph with
the vertex set $V(G) =\{ v_1,v_2,\cdots ,v_n \}$.
Let $x=( x_1,x_2,\cdots ,x_n ) ^T$ be an eigenvector of $D( G )$ with respect to the eigenvalue $\rho$,
where $x( v_i ) =x_i$ ($i=1,2,\cdots ,n$).
Then we have
\begin{equation}
  \rho x_i=\sum_{v_j\in V( G )}{d_{ij}x_j}.
\end{equation}

Suppose that $G$ is a connected simple graph.
In this paper we always assume that its complement $G^c$ is also connected.

Let the vertex $u$ connect the $s$ vertices of the complete graph $K_{n-2}$ and $v$  connect other $t$ ($=n-2-s$) vertices of $K_{n-2}$.
We denote by $H(s,t)$ the resulting graph.
\vskip 3mm

{\noindent \bf Lemma 2.2.} {\it Suppose that $G$ is a simple graph of diameter greater than three on $n$ vertices,
and let $H(s,t)$ be the graph defined above.
Then $\lambda_1(G^c) < \lambda _1(H^c(s,t))$.}
\vskip 3mm

{\noindent \bf Proof.}  Since $d( G ) >3$,
there must be two vertices $u$ and $v$ of $G$ such that $d_G( u,v ) >3$.
Clearly, the neighbours $N_G(u)$ and $N_G(v)$ of vertices $u$ and $v$ in the graph $G$ satisfy $N_G(u)\cap N_G(v)=\emptyset$.
Set $W=V(G)\backslash (N_G(u)\cup N_G(v)\cup \{u,v\})$.
Suppose that $s$ and $t$ are two positive integers such that $s\geq |N_G(u)|$, $t\geq |N_G(v)|$ and $s+t=n-2$.
Connecting all pairs of vertices of $G$ but $u$ and $v$,
connecting $u$ with $s-|N_G(u)|$ vertices of $W$ and connecting $v$ with other $t-|N_G(v)|$ vertices of $W$.
Obviously, the resulting graph is isomorphic to $H(s,t)$.

Let $x$ be the unit Perron vector of $D( G^c )$ with respect to $\lambda_1(G^c)$.
That is, each entry of $x$ is positive and $\|x\|=1$.
Note that $d( G ) >3$ and $d( H(s,t) ) =3$. From Lemma 2.1 we have
\begin{equation}
\begin{split}
\lambda_1(G^c) &=x^TD(G^c)x\\
&=x^T( J_n-I_n ) x+x^TA( G ) x\\
&<x^T( J_n-I_n ) x+x^TA( H(s,t) ) x\\
&\le x^TD( H^c(s,t) ) x.
\nonumber
\end{split}
\end{equation}
By Rayleigh's theorem we know $\lambda _1(H^c(s,t)) \ge x^TD( H^c(s,t) ) x$.
Then
$\lambda_1(G^c) < \lambda _1(H^c(s,t))$.
$\Box$ \vskip 3mm

Suppose that two graphs $G$ and $H$ are isomorphic. Then we write $G\cong H$.\vskip 3mm

{\noindent \bf Lemma 2.3.} {\it Let $H(s,t)$ be the graph as above.
Then
$\lambda _1(H^c(s,t)) \le \lambda _1(H^c(\lfloor\frac{n}{2}-1\rfloor, \lceil\frac{n}{2}-1\rceil))$
with equality if and only if
$H(s,t)\cong H(\lfloor\frac{n}{2}-1\rfloor, \lceil\frac{n}{2}-1\rceil)$.} \vskip 3mm

{\noindent \bf Proof.}
Let $k=\lambda _1(H^c(s,t))$.
Set $x$ to be the Perron vector of $D(H^c(s,t))$ with respect to $k$.
By the symmetry of $H^c(s,t)$ all the vertices in $N_{H(s,t)}(u)$ correspond to the same value $x_1$
and all the vertices in $N_{H(s,t)}(v)$ correspond to the same value $x_2$.
Let $x(u)=x_u$ and $x(v)=x_v$.
Then from the eigen-equation (1) we have
$$ \left\{ \begin{array}{l}
	kx_u=2sx_1+tx_2+x_v,\\
    kx_1=2x_u+2( s-1 ) x_1+3tx_2+x_v,\\
    kx_2=x_u+3sx_1+2( t-1 ) x_2+x_v,\\
    kx_v=x_u+sx_1+2tx_2.
\end{array}\right. $$
We can transform the above equations into a matrix equation $( kI_4-D ) x'=0$,
where $x'=( x_u,x_1,x_2,x_v ) ^T$ and
$$D=\left( \begin{matrix}
	0&		2s&		t&		1\\
	2&		2( s-1 )&		3t&		1\\
	1&		3s&		2( t-1 )&		2\\
	1&		s&		2t&		0\\
\end{matrix} \right).$$
Let $\phi _{s,t}( \lambda ) =\det ( I_4\lambda -D ) $. Then
\begin{equation}
\begin{split}
\phi _{s,t}( \lambda ) = & \lambda ^4+( -2s-2t+4 ) \lambda ^3+( -9s-9t-5st+3 ) \lambda ^2\\
&+ ( -12s-12t-4st-4 ) \lambda +( -4s-4t-4 ).
\nonumber
\end{split}
\end{equation}

Therefore, we obtain
$\phi _{s,t}( \lambda ) -\phi _{s-1,t+1}( \lambda ) =\lambda ( s-t-1 ) ( 5\lambda +4 )$.
Since the path $P_2$ of order 2 is an induced subgraph of $H^c(s,t)$,
$D(H^c(s,t))$ contains $D(P_2)$ as a principal submatrix.
Whereas $\lambda _1(P_2)=1$,
by Interlacing theorem we attain $\lambda _1( H^c(s,t))>1$.
Without loss of generality we assume $s\le t$.
We can compute out that $\phi _{s,t}( \lambda ) -\phi _{s-1,t+1}( \lambda ) < 0$ if $\lambda>1$.
Then $\lambda _1(H^c(s,t)) > \lambda _1( H^c(s-1,t+1))$.
Note that $s+t=n-2$. We finally obtain that
$\lambda _1(H^c(s,t)) \le \lambda _1(H^c(\lfloor\frac{n}{2}-1\rfloor, \lceil\frac{n}{2}-1\rceil))$.  $\Box$ \vskip 3mm

Combining Lemmas 2.2 and 2.3 we obtain the following main result. \vskip 3mm

{\noindent \bf Theorem 2.4.} {\it
Let $G$ be a simple graph of diameter greater than three on $n$ vertices.
Then $\lambda_1(G^c)< \lambda_1(H^c(\lfloor\frac{n}{2}-1\rfloor, \lceil\frac{n}{2}-1\rceil))$.} \vskip 3mm

We denote by $G'$ the connected graph obtained from $G$ by deleting an edge of $G$ which are not adjacent. \vskip 3mm

{\noindent \bf Lemma 2.5.} {\it
Suppose that $G$ is a simple graph of diameter greater than three on $n$ vertices, and $G'$ is above.
Then $\lambda_1(G^c)\le \lambda_1(G'^c)$.} \vskip 3mm

{\noindent \bf Proof.}
Let $y$ be the unit Perron vector of $D(G'^c)$ with respect to $\lambda_1(G'^c)$.
Note that $d(G')>d(G)>3$.
From Lemma 2.1 we have
\begin{equation}
\begin{split}
\lambda_1(G'^c) &=y^TD(G'^c)y\\
&= y^T( J_n-I_n ) y+y^TA(G') y\\
&\le y^T( J_n-I_n ) y+y^TA(G) y\\
&= y^TD(G^c) y.
\nonumber
\end{split}
\end{equation}
By Rayleigh's theorem, $\lambda _1(G^c) \ge y^TD(G^c ) y$,
and so $\lambda_1(G^c)\ge \lambda _1(G'^c)$. $\Box$ \vskip 3mm

By repeatedly applying Lemma 2.5 we can prove that the result is true. \vskip 3mm

{\noindent \bf Lemma 2.6.} {\it
Suppose that $G$ is a simple graph of diameter greater than three on $n$ vertices,
and $T$ is a spanning tree of $G$.
Then $\lambda_1(G^c)\ge \lambda_1(T^c)$
with equality if and only if $G\cong T$.}
\vskip 3mm

{\noindent \bf Lemma 2.7.} (\cite{L.H.Q.2} )  {\it
Let $P_n$ be the path of order $n\ge 4$.
If $T$ is not isomorphic to the star graph $K_{1,n-1}$,
then $\lambda_1(T^c)\ge \lambda_1(P_n^c)$ with equality if and only if $T\cong P_n$.}
\vskip 3mm

Combining Lemmas 2.5, 2.6 and 2.7 we attain the following main result.
\vskip 3mm

{\noindent \bf Theorem 2.8.} {\it
Let $G$ be a simple graph of diameter greater than three on $n$ vertices.
Then $\lambda_1(G^c)\ge \lambda_1(P_n^c)$
with equality if and only if $G\cong P_n$.}

\section {The least distance eigenvalue of the complements of graphs of diameter greater than three}
~~~~In this section using the relations between $D(G^c)$ and $A(G)$ declared in Lemma 2.1
we determine the unique graph whose least distance eigenvalue attains maximum and minimum
among all complements of graphs of diameter greater than three, respectively. \vskip 3mm

Let $T(a,b)$ denote the tree obtained from the path $P_2$ of order $2$ by appending $a$ vertices to one vertex of $P_2$ and $b$ vertices to the other.
We denote by $T_1(a,b)$ the tree obtained from $P_3$ of order $3$ by appending $a$ vertices to one end vertex of $P_3$ and $b$ vertices to the other.
Let $T_2(a,b)$ be the tree obtained from $T(a,b)$ by appending an additional pendent edge to the group of $a$ pendent vertices of $T(a,b)$.
Clearly, $d(T(a+1,b))+1=d(T_1(a,b))=d(T_2(a,b))=4$. \vskip 3mm

{\noindent \bf Lemma 3.1.} (\cite{L.H.Q.2} ) {\it
Let $T(a+1,b)$, $T_1(a,b)$ and $T_2(a,b)$ be three trees of order $n$ ($=a+b+3$) as above.
Then we have
$$\lambda_n(T_1^c(a,b)\ge \lambda_n(T_2^c(a,b)))>\lambda_n(T^c(a+1,b)).$$
The equality holds if and only if $T_1(a,b)\cong T_2(a,b)$.} \vskip 3mm

Suppose that $G$ is a simple graph of diameter greater than three with
the vertex set $V(G) =\{ v_1,v_2,\cdots ,v_n\}$ ($n\ge 7$).
Let $x=\left( x_1,x_2,\cdots ,x_n \right) ^T$ be an eigenvector of $D\left( G^c \right) $ with respect to $\lambda _n(G^c)$,
where $x\left( v_i \right) =x_i$ ($i=1, 2,\cdots, n$).
Write $V_+=\left\{ v_i\in V\left( G^c \right): x_i>0 \right\} $,
$V_-=\left\{v_i\in V\left( G^c \right): x_i<0 \right\} $
and $V_0=\left\{ v_i\in V\left( G^c \right): x_i=0 \right\} $.
Let $|V_+\cup V_0|=p$ and $|V_-|=q$.
Without loss of generality in what follows we assume that $p\ge q$.
Note that $p+q=n\ge 7$. We have $p\ge 4$. \vskip 3mm

{\noindent \bf Lemma 3.2.} {\it Suppose that $G$ is a simple graph of diameter greater than three on $n\ge 7$ vertices.
If $q=1$ then $\lambda_n(G^c)\ge \lambda_n(T^c(n-3,1))$.} \vskip 3mm

{\noindent \bf Proof.}
Note that $q=1$. We let $V_-=\{v\}$.
Since $d(G)>3$, there must exist either the path $P_4=v\widetilde{u}_1\widetilde{u}_2\widetilde{u}_3$ or the path $P_5=u_1u_2vu_4u_5$.
Now we distinguish two cases as follows.

{\bf Case 1.} Suppose that there is the path $P_4=v\widetilde{u}_1\widetilde{u}_2\widetilde{u}_3$.

In this case deleting all edges in $G\setminus \{v\}$ except $\widetilde{u}_1\widetilde{u}_2$
and $\widetilde{u}_2\widetilde{u}_3$ and connecting all pairs of vertices
which are not adjacent between the vertex $v$ and all vertices of $G\setminus \{v,\widetilde{u}_1,\widetilde{u}_2,\widetilde{u}_3\}$ in $G$.
Thus we obtain the resulting graph which is isomorphic to the graph $T_1(n-3,2)$.
From the above argument we know that $x^T A(G)x=\sum_{v_iv_j\in V(G)}x_ix_j\ge \sum_{v_iv_j\in V(T_1(n-4,1))}x_ix_j=x^T A(T_1(n-4,1))x$.

{\bf Case 2.} Suppose that there is the path $P_5=u_1u_2vu_4u_5$.

In this case deleting all edges in $G\setminus \{v\}$ except $u_1u_2$ and $u_4u_5$ and connecting all pairs of vertices
which are not adjacent between the vertex $v$ and all vertices of $G\setminus \{u_1,u_2,v,u_4,u_5\}$ in $G$.
Thus we obtain the resulting graph which is isomorphic to the graph $T_2(n-4,1)$.
From the above argument we know that $x^T A(G)x=\sum_{v_iv_j\in V(G)}x_ix_j\ge \sum_{v_iv_j\in V(T_2(n-4,1))}x_ix_j=x^T A(T_2(n-4,1))x$.

Set $x$ to be the unit eigenvector of $D(G^c)$ with respect to $\lambda_n(G^c)$.
Note that $d(G)> 3$ and $d(T_1(n-3,2))=4$.
From Lemma 2.1 we have
\begin{equation}
	\begin{split}
		\lambda_n(G^c)&=x^TD(G^c)x\\
		&=x^T(J_n-I_n)x+x^TA(G)x\\
		&\ge x^T(J_n-I_n)x+x^TA(T_1(n-4,1))x\\
		&=x^TD(T_1^c(n-4,1))x.
		\nonumber
	\end{split}
\end{equation}
By Rayleigh's theorem we attain that
$\lambda_n(T_1^c(n-4,1))\le x^TD(T_1^c(n-4,1))x$.
Then we have $\lambda_n(G^c)\ge \lambda_n(T_1^c(n-4,1))$.

Similarly, we can determine that $\lambda_n(G^c)\ge \lambda_n(T_2^c(n-4,1))$.

By Lemma 3.1, $\lambda_n(T_1^c(n-4,1)\ge \lambda_n(T_2^c(n-4,1)))>\lambda_n(T^c(n-3,1))$.
From the above argument we have $\lambda_n(G^c)> \lambda_n(T^c(n-3,1))$. $\Box$ \vskip 3mm

Let $B_1(p,q)$ be the graph obtained from the complete bipartite graph $K_{p,q}$ by deleting the edge $uv$.
Suppose $u$ and $w$ are two vertices of the partition $U$ and $v$ belongs to the partition $V$.
Deleting all edges of $B_1(p,q)$ which are incident to $w$ except $wv$ we denote by $B_2(p,q)$ the resulting graph.
Clearly, $d(B_1(p,q))+1=d(B_2(p,q))=4$.

Suppose that $S$ is a subset of $V(G)$.
Then we denote by $G[S]$ the subgraph of $G$ induced by $S$. \vskip 3mm

{\noindent \bf Lemma 3.3.} {\it Suppose that $G$ is a simple graph of diameter greater than three on $n\ge 7$ vertices.
If $q\ge 2$ then we have $\lambda_n(G^c)\ge \lambda_n(B_2^c(p,q))$.} \vskip 3mm

{\noindent \bf Proof.} Set $x$ to be the unit eigenvector of $D(G^c)$ with respect to $\lambda_n(G^c)$.
Deleting all edges in $G[V_-]$ and $G[V_+\cup V_0]$ of $G$ we denote by $G'$ the resulting bipartite graph.
If $G'$ is connected then since $d(G')\geq d(G)>3$, there must be two vertices $u$ and $w$ such that $d_{G'}(u,w)=4$.
Let $P=uu_1u_2vw$ be the path between $u$ and $w$.
Then $u$ and $w$ are in the same partition, say $u$ and $w$ are both contained in $V_+\cup V_0$.
Without loss of generality assume that $x(u)\ge x(w)$.
Deleting all edges which are incident to $w$ except $vw$ and connecting all pairs of vertices
between $(V_+\cup V_0)\setminus \{w\}$ and $V_-$ which are not adjacent except $u$ and $v$ in $G'$.
Obviously, the resulting graph is isomorphic to the graph $B_2(p,q)$.
From the above construction we know that $x^T A(G)x=\sum_{v_iv_j\in E(G)}x_ix_j\ge \sum_{v_iv_j\in E(B_2(p,q))}x_ix_j=x^T A(B_2(p,q))x$.

So we can assume that $G'$ is not connected.
Since $G$ is connected, $G'$ must have one nontrivial component, that is, it contains at least one edge.
Now we distinguish two cases as follows.

{\bf Case 1.} $G'$ has at least two nontrivial components.

Suppose two edges $\tilde{u}\tilde{v}$ and $\tilde{u}'\tilde{v}'$ belong to two distinct nontrivial components.
Without loss of generality we assume that $x(\tilde{u})\ge x(\tilde{u}')\ge 0$.
Deleting all edges which are incident to $\tilde{u}'$ except $\tilde{u}'\tilde{v}'$ and connecting all pairs of vertices
between $(V_+\cup V_0)\setminus \{\tilde{u}'\}$ and $V_-$ which are not adjacent except $\tilde{u}$ and $\tilde{v}'$ in $G'$.
Obviously, the resulting graph is isomorphic to the graph $B_2(p,q)$.
From the above construction we know that $x^T A(G)x=\sum_{v_iv_j\in E(G)}x_ix_j\ge \sum_{v_iv_j\in E(B_2(p,q))}x_ix_j=x^T A(B_2(p,q))x$.

{\bf Case 2.} $G'$ has exactly one nontrivial component.

If $G'$ has exactly one isolated vertex $\bar{w}$,
then since $d(G)>3$, there must be two vertices $\bar{u}$ and $\bar{v}$ which are not adjacent in $G'$.
Without loss of generality we assume that $x(\bar{w})\ge 0$.
Connecting all pairs of vertices between $(V_+\cup V_0)\setminus \{\bar{w}\}$ and $V_-$ which are not adjacent
except $\bar{u}$ and $\bar{v}$ and connecting $\bar{w}$ and $\bar{v}$ in $G'$.
Obviously, the resulting graph is isomorphic to the graph $B_2(p,q)$.
From the above argument we know that $x^T A(G)x=\sum_{v_iv_j\in E(G)}x_ix_j\ge \sum_{v_iv_j\in E(B_2(p,q))}x_ix_j=x^T A(B_2(p,q))x$.

So we assume that $G'$ has at least two isolated vertices $\hat{w}$ and $\hat{w}'$.
Without loss of generality we assume that $x(\hat{w})\ge 0$.
Suppose $x(\hat{v})< 0$ in the edge $\hat{u}\hat{v}$.
We denote by $G''$ the graph obtained from $G'$ by connecting all pairs of vertices
between $(V_+\cup V_0)\setminus \{\hat{w}'\}$ and $V_-\setminus \{\hat{w}'\}$ which are not adjacent except $\hat{w}$ and $\hat{v}$.
Furthermore, connect $\hat{w}'$ and $\hat{v}$ if $x(\hat{w}')\ge 0$, and connect $\hat{w}'$ and $\hat{u}$ otherwise in $G''$.
Clearly, the resulting graph is isomorphic to the graph $B_2(p,q)$.
From the above construction we know that $x^T A(G)x=\sum_{v_iv_j\in E(G)}x_ix_j\ge \sum_{v_iv_j\in E(B_2(p,q))}x_ix_j=x^T A(B_2(p,q))x$.

Note that $d(G)> 3$ and $d(B_2(p,q))=4$.
From Lemma 2.1 and the above arguments we have
\begin{equation}
\begin{split}
\lambda_n(G^c)&=x^TD(G^c)x\\
&=x^T(J_n-I_n)x+x^TA(G)x\\
&\ge x^T(J_n-I_n)x+x^TA(B_2(p,q))x\\
&=x^TD(B_2^c(p,q))x.
\nonumber
\end{split}
\end{equation}
By Rayleigh's theorem we obtain that
$\lambda_n(B_2^c(p,q))\le x^TD(B_2^c(p,q))x$. Therefore, we have
$\lambda_n(G^c)\ge \lambda_n(B_2^c(p,q))$. $\Box$ \vskip 3mm

{\noindent \bf Lemma 3.4.} {\it
Let $B_2(p,q)$ and $B_1(p,q)$ be two graphs as above.
Then we have $\lambda_n(B_1^c(p,q))<\lambda_n(B_2^c(p,q))<-3$.
} \vskip 3mm

{\noindent \bf Proof.}
Let $\lambda_n$ be the least eigenvalue of $D(B_2^c(p,q))$.
Set $x$ to be an eigenvector of $D(B_2^c(p,q))$ with respect to $\lambda_n$.
By the symmetry of $B_2^c(p,q)$ all vertices in $\{V_+\cup V_0\}\setminus \{u,w\}$
correspond to the same value $x_1$ and
all the vertices in $V_-\setminus \{v\}$ correspond to the same value $x_2$.
Set $w$ to be the only one vertex of $N_T(v)\setminus \{u\}$.
Let $x(u)=x_u$, $x(v)=x_v$ and $x(w)=x_w$.
Then from the eigen-equation (1) we obtain
$$ \left\{ \begin{array}{l}
	\lambda_nx_u=x_v+x_w+(p-2)x_1+2(q-1)x_2,\\
    \lambda_nx_v=x_u+2x_w+2(p-2)x_1+(q-1)x_2,\\
    \lambda_nx_w=x_u+2x_v+(p-2)x_1+(q-1)x_2,\\
    \lambda_nx_1=x_u+2x_v+x_w+(p-3)x_1+2(q-1)x_2,\\
    \lambda_nx_2=2x_u+x_v+x_w+2(p-2)x_1+(q-1)x_2.
\end{array}\right.$$
We can transform the above equation into a matrix equation
$(\lambda_nI_5-D_{B_2^c})x'=0$, where $x'=(x_u, x_v, x_w, x_1, x_2)$ and
$$D_{B_2^c}=\left( \begin{matrix}
	0&1&1&p-2&2(q-1)\\
	1&0&2&2(p-2)&q-1\\
	1&2&0&p-2&q-1\\
	1&2&1&p-3&2(q-1)\\
	2&1&1&2(p-2)&q-2
\end{matrix} \right).$$
Let $\varphi_{p,q}(\lambda)=det(I_5\lambda-D_{B_2^c})$.
Then we get
\begin{equation}
\begin{split}
\varphi_{p,q}(\lambda)=
&{\lambda}^{5}- \left( q-5+p \right) {\lambda}^{4}\\
&- \left( 3\,pq+4\,p+q-10 \right) {\lambda}^{3}\\
&- \left( 8\,pq+6\,p-4\,q-8 \right) {\lambda}^{2}\\
&- \left( pq+10\,p-8 \right) \lambda+3\,pq-6\,p-2\,q+4.
\nonumber
\end{split}
\end{equation}
Similarly we have
\begin{equation}
\begin{split}
\phi_{p,q}(\lambda)=&det(I_4\lambda-D_{B_1^c})\\
=&{\lambda}^{4}+ \left( -q+4-p \right) {\lambda}^{3}\\
&+ \left( -8\,pq+2\,p+2\,q+4 \right) {\lambda}^{2}\\
&+ \left( -14\,pq+6\,p+6\,q \right) \lambda-5\,pq+2\,p+2\,q.
\end{split}
\end{equation}
By the above two equations we get
\begin{equation}
\begin{split}
\varphi_{p,q}(\lambda)-(\lambda+1)\phi_{p,q}(\lambda)=
&\left( 5\,pq-5\,p-2\,q+2 \right) {\lambda}^{3}\\
&+ \left( 14\,pq-14\,p-4\,q+4 \right) {\lambda}^{2}\\
&+ \left( 18\,pq-18\,p-8\,q+8 \right) \lambda+8\,pq-8\,p-4\,q+4.
\nonumber
\end{split}
\end{equation}

Since the path $P_4$ of order 4 is an induced subgraph of $B_2^c(p,q)$ and $B_1^c(p,q)$,
$D(P_4)$ is a principal submatrix of $D(B_2^c(p,q))$ and $D(B_1^c(p,q))$.
Whereas $\lambda_4(P_4)<-3$,
by Interlacing theorem we attain $\lambda_n(B_2^c(p,q))<-3$ and $\lambda_n(B_1^c(p,q))<-3$.
Note that $p\ge 4$ and $q\ge 2$.
We can compute out that
$\varphi_{p,q}(\lambda)-(\lambda+1)^2\phi_{p,q}(\lambda)>0$
when $\lambda<-3$.
This implies that $\lambda_n(B_2^c(p,q))>\lambda_n(B_1^c(p,q))$.
$\Box$ \vskip 3mm

{\noindent \bf Lemma 3.5.} {\it Let $B_1(p,q)$ and $T(n-3,1)$ be two graphs of order $n$ ($=p+q$) as above.
Then we have $\lambda_n(B_1^c(p,q))<\lambda_n(T^c(n-3,1))<-3$.} \vskip 3mm

{\noindent \bf Proof.}
Let $\lambda_n$ be the least eigenvalue of $D(T^c(n-3,1))$.
Set $x$ to be an eigenvector of $D(T^c(n-3,1))$ with respect to $\lambda_n$.
By the symmetry of $T^c(n-3,1)$ all vertices in $N_T(u)\setminus \{v\}$
correspond to the same value $x_1$.
Let $x(u)=x_u$, $x(v)=x_v$ and $x(w)=x_w$.
Then from the eigen-equation (1) we obtain
$$ \left\{ \begin{array}{l}
	\lambda_nx_u=3x_v+2(n-3)x_1+x_w,\\
	\lambda_nx_v=3x_u+(n-3)x_1+2x_w,\\
	\lambda_nx_1=2x_u+x_v+(n-4)x_1+x_w,\\
	\lambda_nx_w=x_u+2x_v+(n-3)x_1.
\end{array}\right.$$
We can transform the above equation into a matrix equation
$(\lambda_nI_4-D_{T^c})x'=0$, where $x'=(x_u, x_v, x_1, x_w)$ and
$$D_{T^c}=\left( \begin{matrix}
	0&3&2(n-3)&1\\
	3&0&n-3&2\\
	2&1&n-4&1\\
	1&2&n-3&0
\end{matrix} \right).$$
Let $\psi_{p,q}(\lambda)=det(I_4\lambda-D_{T^c})$.
Then we get
\begin{equation}
\begin{split}
\psi(\lambda)=
{\lambda}^{4}+ \left( -n+4 \right) {\lambda}^{3}+ \left( 4-6\,n
 \right) {\lambda}^{2}+ \left( -6\,n-8 \right) \lambda-12.
\end{split}
\end{equation}
Note that $n=p+q$. From the equations (2) and (3) we obtain
\begin{equation}
\begin{split}
		\phi_{p,q}(\lambda)-\psi(\lambda)=
		&\left( -8\,pq+8\,p+8\,q \right) {\lambda}^{2}\\
		&+\left( -14\,pq+12\,p+12\,q+8 \right) \lambda-5\,pq+2\,p+2\,q+12.
\nonumber
\end{split}
\end{equation}

Since the Path $P_4$ of order $4$ is an induced subgraph of $T^c(n-3,1)$,
$D(T^c(n-3,1))$ contains $D(P_4)$ as a principal submatrix.
Whereas $\lambda_4(P_4)<-3$, we have $\lambda_n(T^c(n-3,1))<-3$.
Recall that $p\ge 4$ and $q\ge 2$.
Therefore, we can compute out that $\phi_{p,q}(\lambda)-\psi(\lambda)<0$ if $\lambda<-3$.
Thus, by Lemma 3.4 we get $\lambda_n(T^c(n-3,1))>\lambda_n(B_1^c(p,q))$. $\Box$ \vskip 3mm

{\noindent \bf Lemma 3.6.} {\it
Let $B_1(p,q)$ be the graph as above.
Then we have
$$\lambda_n(B_1^c(p,q))\ge \lambda_n\left(B_1^c\left(\left\lceil \frac{n}{2}\right\rceil,\left\lfloor \frac{n}{2}\right\rfloor\right)\right).$$
The equality holds if and only if
$B_1(p,q)\cong B_1(\lceil\frac{n}{2}\rceil,\lfloor \frac{n}{2}\rfloor)$.}\vskip 3mm

{\noindent \bf Proof.}
By the above equation (2) we obtain
$$\phi_{p,q}(\lambda)-\phi_{p-1,q+1}(\lambda)=\left( 8\,p-8\,q-8 \right) {\lambda}^{2}+ \left( 14\,p-14\,q-14
\right) \lambda+5\,p-5\,q-5.$$
Without loss of generality we assume that $p> q$.
By computation we obtain that $\phi_{p,q}(\lambda)-\phi_{p-1,q+1}(\lambda)\ge 0$ if $\lambda<-3$.
Thus, by Lemma 3.4 we have $\lambda_n(B_1^c(p,q))> \lambda_n(B_1^c(\lceil \frac{n}{2}\rceil,\lfloor \frac{n}{2}\rfloor))$.
$\Box$ \vskip 3mm

Combining Lemmas 3.2-3.6 we have the following main result. \vskip 3mm

{\noindent \bf Theorem 3.7.} {\it
Let $G$ be a simple graph of diameter greater than three on $n\ge 7$ vertices.
Then we have $$\lambda_n(G^c)>
\lambda_n\left(B_1^c\left(\left\lceil \frac{n}{2}\right\rceil,\left\lfloor \frac{n}{2}\right\rfloor\right)\right).$$} \vskip 3mm

Let $x$ be the unit eigenvector of $D(G^c)$ with respect to $\lambda_n(G^c)$.
Let $G'$ denote the connected graph obtained from $G$ by deleting an edge in $V_+\cup V_0$ or $V_-$
or connecting one pair of vertices between $V_+\cup V_0$ and $V_-$ which are not adjacent such that $d(G')>3$.
Clearly, $x^TA(G)x\geq x^TA(G')x$.\vskip 3mm

{\noindent \bf Lemma 3.8.} {\it Suppose that $G$ is a simple graph of diameter greater than three on $n\ge 7$ vertices.
Then $\lambda_n(G^c)\geq \lambda_n(G'^c)$.} \vskip 3mm

{\noindent \bf Proof.}
Let $x$ be the unit eigenvector of $D(G^c)$ with respect to $\lambda_n(G^c)$.
Note that $d(G)>3$ and $d(G')>3$.
From Lemma 2.1 we have
\begin{equation}
	\begin{split}
		\lambda_n(G^c)&=x^TD(G^c)x\\
		&= x^T(J_n-I_n)x+x^TA(G)x\\
		&\ge x^T(J_n-I_n)x+x^TA(G')x\\
		&=x^TD(G'^c)x.
		\nonumber
	\end{split}
\end{equation}
By Rayleigh's theorem,
$\lambda_n(G'^c)\le x^TD(G'^c)x$, and so $\lambda_n(G^c)\ge \lambda_n(G'^c)$. $\Box$ \vskip 3mm

From Lemma 3.8 we obtained that $\lambda_n(G^c)\le \lambda_n(\check{G}^c)$ if $\check{G}$ is obtained from $G$ by
connecting one pair of vertices in $V_+\cup V_0$ or $V_-$ which are not adjacent
or deleting an edge between $V_+\cup V_0$ and $V_-$.
Clearly, $x^TA(G)x\leq x^TA(\check{G})x$.\vskip 3mm

Let $K_{n-2}$ be a complete graph of order $n-2$.
We denote by $L'$ the graph by deleting an edge $wu$ of $K_{n-2}$ and appending a vertex $w'$ to $w$ and a vertex $v$ to $u$.
Clearly, $d(L')=4$.
Let $L''$ donote the graph by deleting an edge $w'u'$ of $K_{n-2}$ and appending a path of order $2$ to $u'$.
\vskip 3mm

{\noindent \bf Lemma 3.9.} {\it Suppose that $G$ is a simple graph of diameter greater than three on $n\ge 7$ vertices.
If $q=1$ then $\lambda_n(G^c)\le \lambda_n(L'^c)$.} \vskip 3mm

{\noindent \bf Proof.} Since $d(G)>3$, there must be two vertices $u_1$ and $u_5$ such that $d_G(u_1,u_5)=4$.
Let $V_-=\{v\}$ and $P=u_1u_2u_3u_4u_5$.
Now we distinguish four cases as follows.

{\bf Case 1.} Suppose that $v=u_3$ in $G$.

Without loss of generality we assume that $x(u_1)\ge x(u_5)$.
We denote by $G'$ the graph obtained from $G$ by deleting all edges which are incident to $v$ except $vu_2$,
deleting all edges which are incident to $u_5$ except $u_5u_4$ and appending them to $u_1$.
Connecting all pairs of vertices of $V(G)\setminus \{v,u_5\}$ which are not adjacent except $u_2$ and $u_4$ in $G'$.
Thus, the resulting graph is isomorphic to the graph $L'$.

{\bf Case 2.} Suppose that $v=u_2$ in $G$.

If $x(u_1)\ge x(u_5)$, we denote by $\widetilde{G}$ the graph obtained from $G$ by deleting all edges which are incident to $v$ except $vu_1$,
deleting all edges which are indicent to $u_5$ except $u_5u_4$ and appending them to $u_1$.
Connecting all pairs of vertices of $V(G')\setminus \{v,u_5\}$ which are not adjacent except $u_1$ and $u_4$ in $\widetilde{G}$.
Obviously, the resulting graph is isomorphic to the graph $L'$.

So we assume that $x(u_1)<x(u_5)$.
We denote by $\widetilde{G}'$ the graph obtained from $G$ by deleting all edges which are incident to $v$ except $vu_3$,
deleting all edges which are incident to $u_1$ and appending them to $u_5$.
Connecting all pairs of vertices of $V(\widetilde{G}')\setminus \{v,u_1\}$ and connecting $u_1$ and $u_5$.
Clearly, the resulting graph is isomorphic to the graph $B_2(p,q)$.

{\bf Case 3.} Suppose that $v=u_1$ in $G$.

If $x(u_2)\ge x(u_5)$, we denote by $\overline{G}$ obtained from $G$ by deleting edges which are incident to $v$
except $vu_2$ and deleting all edges which are incident to $u_5$ except $u_5u_4$.
Connecting all pairs of vertices of $V(\overline{G})\setminus \{v,u_5\}$ which are not adjacent except $u_2$ and $u_4$ in $\overline{G}$.
Clearly, the resulting graph is isomorphic to the graph $B_2(p,q)$.

So we assume that $x(u_2)<x(u_5)$.
We denote by $\overline{G}'$ obtained from $G$ by deleting edges which are incident to $v$ except $vu_2$
and deleting all edges which are incident to $u_2$ except $u_2u_3$ and $u_2v$ and appending them to $u_5$.
Connecting all pairs of vertices of $V(\overline{G}')\setminus \{v,u_2\}$ which are not adjacent except $u_3$ and $u_5$ in $\overline{G}'$.
Then the resulting graph is isomorphic to the graph $L''$.

{\bf Case 4.} Suppose that $v$ is adjacent to $u_3$ in $G$.

Without loss of generality assume that $x(u_1)\ge x(u_5)$.
We denote by $\hat{G}$ obtained from $G$ by deleting edges which are incident to $v$ except $vu_3$,
deleting all edges which are incident to $u_5$ and appending them to $u_1$ and connecting $u_5$ and $u_1$.
Connecting all pairs of vertices of $V(\hat{G})\setminus \{v,u_5\}$ except $u_1$ and $u_3$ in $\hat{G}$.
Thus, the resulting graph is isomorphic to the graph $B_2(p,q)$.

By the above arguments we obtain that the following facts.
In other cases, by deleting some edges of $G$ which are incident to $v$ we can pick out a path is isomorphic to the above four constructions.

By repeatedly applying Lemma 3.8 we can verify that
$\lambda_n(G^c)\le \lambda_n(L'^c)$ or $\lambda_n(G^c)\le \lambda_n(L''^c)$ with equality if and only if $G\cong L'$ or $G\cong L''$.

Let $\lambda_n$ be the least eigenvalue of $D(L'^c)$.
Set $x$ to be the eigenvector of $D(L'^c)$ with respect to $\lambda_n$.
By the symmetry of $L'^c$ all the vertices in $(V_+\cup V_0)\setminus \{u,v,w,w'\}$
correspond to the same value $x_1$.
Let $x(u)=x_{u}$, $x(v)=x_{v}$, $x(w)=x_w$ and $x(w')=x_{w'}$.
By the equation (1) we have
$$ \left\{ \begin{array}{l}
	\lambda_nx_u=2x_v+x_w+x_{w'}+2(p-4)x_1,\\
    \lambda_nx_v=2x_u+x_w+x_{w'}+(p-4)x_1,\\
    \lambda_nx_w=x_u+x_v+2x_{w'}+2(p-4)x_1,\\
    \lambda_nx_{w'}=x_u+x_v+2x_w+(p-4)x_1,\\
    \lambda_nx_1=2x_u+x_v+2x_w+x_{w'}+2(p-5)x_1.
\end{array}\right.$$
We can transform the above equation into a matrix equation
$(\lambda_nI_5-D_{L'^c})x'=0$, where $x'=(x_u, x_v,x_w, x_{w'}, x_1$) and
$$D_{L'^c}=\left( \begin{matrix}
	0&2&1&1&2(p-4)\\
    2&0&1&1&p-4\\
    1&1&0&2&2(p-4)\\
    1&1&2&0&p-4\\
    2&1&2&1&2(p-5)
\end{matrix} \right).$$
Let $\Psi(\lambda)=det(I_5\lambda-D_{L'^c})$. Then we get
\begin{equation}
\begin{split}
\Psi(\lambda)=det(I_5-D_{L'^c})
=&{\lambda}^{5}- \left( 2\,n-10 \right) {\lambda}^{4}\\
&- \left( -28+10\,n\right) {\lambda}^{3}-10\,n{\lambda}^{2}- \left( -4\,n+48 \right)\lambda.
\end{split}
\end{equation}
Similarly, we have
\begin{equation}
\begin{split}
\Psi'(\lambda)=det(I_5-D_{L''^c})
=&{\lambda}^{5}- \left( 2\,n-10 \right) {\lambda}^{4}- \left( -28+10\,n
 \right) {\lambda}^{3}\\
&- \left( 10+8\,n \right) {\lambda}^{2}- \left(
103-15\,n \right) \lambda+14\,n-70.
\nonumber
\end{split}
\end{equation}
From the above equations we get
$$\Psi(\lambda)-\Psi'(\lambda)=\left( -2\,n+10 \right) {\lambda}^{2}+ \left( -11\,n+55 \right)
\lambda-14\,n+70.$$

Since the path $P_5$ of order $5$ is an induced subgraph of $L'$ and $L''$,
$D(L'^c)$ and $D(L''^c)$ contain a principal submatrix $D(P_5)$.
Whereas $\lambda_5(P_5)<-5$, by Interlacing theorem we have $\lambda_n(L'^c)<-5$ and $\lambda_n(L''^c)<-5$.
Therefore, we can compute out that $\Psi(\lambda)-\Psi'(\lambda)<0$ if $\lambda<-5$ and $n\ge 7$.
This implies that $\lambda_n(L''^c)\le \lambda_n(L'^c)$.

Thus, by the above arguments we know $\lambda_n(G^c)\le \lambda_n(L'^c)$.
$\Box$ \vskip 3mm

Let $K_p$ and $K_q$ be two complete graphs of order $p$ and $q$, respectively.
We denote by $L(p,q)$ the graph by deleting an edge $wu$ of $K_p$ and connecting $u$ and $v$ of $K_q$.
Clearly, $d(L(p,q))=4$. \vskip 3mm

{\noindent \bf Lemma 3.10.} {\it Suppose that $G$ is a simple graph of diameter greater than three on $n\ge 7$ vertices.
If $q\ge 2$ then we have $\lambda_n(G^c)\le \lambda_n(L^c(p,q))$.} \vskip 3mm

{\noindent \bf Proof.}
Note that $G$ is a connected graph.
There must be two vertices $u$ of $V_+\cup V_0$ and $v$ of $V_-$ which are adjacent in $G$.
Since $d(G)>3$, we without loss of generality assume that there exists a vertex $w$ of $V_+\cup V_0$ which are not adjacent to $u$ in $G$.
Connecting all pairs of vertices in $V_+\cup V_0$ and $V_-$ which are not adjacent
except $w$ and $u$ and deleting all edges between $V_+\cup V_0$ and $V_-$ except $uv$ in $G$.
Obviously, the resulting graph is isomorphic to the graph $L(p,q)$.

By repeatedly applying Lemma 3.8 we can verify that the result is true. $\Box$ \vskip 3mm

{\noindent \bf Lemma 3.11.} {\it Let $L'$ and $L(p,q)$ be two graphs as above.
Then we have $\lambda_n(L'^c)<\lambda_n(L^c(p,q))$.} \vskip 3mm

{\noindent \bf Proof.}
Let $\lambda_n$ be the least eigenvalue of $D(L^c(p,q))$.
Set $x$ to be the eigenvector of $D(L^c(p,q))$ with respect to $\lambda_n$.
By the symmetry of $L^c(p,q)$ all the vertices in $(V_+\cup V_0)\setminus \{u,w\}$
correspond to the same value $x_1$ and
all the vertices in $V_-\setminus \{v\}$ correspond to the same value $x_2$.
Let $x(u)=x_{u}$, $x(v)=x_{v}$ and $x(w)=x_w$.
By the equation (1) we have
$$ \left\{ \begin{array}{l}
	\lambda_nx_u=2x_v+x_w+2(p-2)x_1+(q-1)x_2,\\
	\lambda_nx_v=2x_u+x_w+(p-2)x_1+2(q-1)x_2,\\
	\lambda_nx_w=x_u+x_v+2(p-2)x_1+(q-1)x_2,\\
	\lambda_nx_1=2x_u+x_v+2x_w+2(p-3)x_1+(q-1)x_2,\\
	\lambda_nx_2=x_u+2x_v+x_w+(p-2)x_1+2(q-2)x_2.
\end{array}\right.$$
We can transform the above equation into a matrix equation
$(\lambda_nI_5-D_{L^c})x'=0$, where $x'=(x_u, x_v,x_w, x_1, x_2)$ and
$$D_{L^c}=\left( \begin{matrix}
	0&2&1&2(p-2)&q-1\\
	2&0&1&p-2&2(q-1)\\
	1&1&0&2(p-2)&q-1\\
	2&1&2&2(p-3)&q-1\\
	1&2&1&p-2&2(q-2)
\end{matrix} \right).$$
Let $\Phi_{p,q}(\lambda)=det(I_5\lambda-D_{L^c})$. Then we get
\begin{equation}
	\begin{split}
		\Phi_{p,q}(\lambda)=
		&{\lambda}^{5}- \left( 2\,q-10+2\,p \right) {\lambda}^{4}\\
		&- \left( -3\,pq+16\,p+16\,q-40 \right) {\lambda}^{3}\\
		&- \left( -18\,pq+44\,p+50\,q-74\right) {\lambda}^{2}\\
		&- \left( -30\,pq+45\,p+63\,q-53 \right) \lambda+12\,pq-10\,p-22\,q+2.
	\end{split}
\end{equation}
From the equations (4) and (5) we have $$\Phi_{p,q}(\lambda)-\Psi(\lambda)= \left( 2\,p-6 \right) {\lambda}^{2}+ \left( 11\,p-33 \right)
\lambda+14\,p-42.$$

Since the path $P_5$ of order $5$ is an induced subgraph of $L^c(p,q)$,
$D(L^(p,q))$ contains $D(P_5)$ as a principal submatrix.
Whereas $\lambda_5(P_5)<-5$, by Interlacing theorem we get $\lambda_n(L^c(p,q))<-5$.
Recall that $p\ge 4$ and $q\ge 2$.
Therefore, we can compute out that $\Phi_{p,q}(\lambda)-\Psi(\lambda)> 0$ if $\lambda<-5$.
Thus, by Lemma 3.9 we obtain that $\lambda_n(L'^c)<\lambda_n(L^c(p,q))$. $\Box$ \vskip 3mm

{\noindent \bf Lemma 3.12.} {\it Let $L(p,q)$ be the graph as above.
Then we have $\lambda_n(L^c(p,q))\le \lambda_n(L^c(\lceil \frac{n}{2}\rceil,\lfloor\frac{n}{2}\rfloor))$.} \vskip 3mm

{\noindent \bf Proof.}
Note that $n=p+q$. By the equation (5) we have
\begin{equation}
	\begin{split}
\Phi_{p,q}(\lambda)-\Phi_{p-1,q+1}(\lambda)=
&\left( 3\,pq-6\,p-6\,q+12 \right) {\lambda}^{3}+ \left( 18\,pq-34\,p-
40\,q+74 \right) {\lambda}^{2}\\
&+ \left( 30\,pq-49\,p-67\,q+101 \right)
\lambda+12\,pq-10\,p-22\,q+2.
\nonumber
\end{split}
\end{equation}
Recall that $p>q\ge 2$. By computation we obtain that $\Phi_{p,q}(\lambda)-\Phi_{p-1,q+1}(\lambda)>0$ if $\lambda< -5$.

Thus, by Lemma 3.11 we have $\lambda_n(L^c(p,q))< \lambda_n(L^c(\lceil \frac{n}{2}\rceil,\lfloor\frac{n}{2}\rfloor))$. $\Box$ \vskip 3mm

Combining Lemmas 3.8-3.12 we obtain the following main result.
\vskip 3mm

{\noindent \bf Theorem 3.13.} {\it
Suppose that $G$ is a simple graph of diameter greater than three on $n\ge 7$ vertices.
Then we have
$$\lambda_n(G^c)\le \lambda_n\left(L^c\left(\left\lceil \frac{n}{2}\right\rceil,\left\lfloor\frac{n}{2}\right\rfloor\right)\right).$$}

redpajama_set_name: RedPajamaArXiv


text:
A vöröstejű kígyógomba (Mycena sanguinolenta) a kígyógombafélék családjába tartozó, Eurázsiában és Észak-Amerikában honos, főleg fenyvesekben élő, nem ehető gombafaj.

Megjelenése
A vöröstejű kígyógomba kalapja 0,5-1,8 cm széles; alakja fiatalon kúpos, domború vagy harangszerű, idősen laposan, néha felkanyarodó szélekkel kiterül. Felszíne fiatalon hamvas, később csupasz. Széle áttetszően barázdált. Színe változatos; halványbarna, szürkésbarna, a közepén sötét vörösbarna.

Húsa vékony, nem túl törékeny; színe halványvöröses. Sérülésre vöröses tejnedvet választ ki. Szaga és íze nem jellegzetes.

Közepesen ritkás állású 13-17 lemeze keskenyen a tönkhöz nőtt vagy kis foggal lefutó. Színük fehéres vagy szürkés, élük vörösbarna vagy borvörös.

Tönkje 2-6 cm magas és 0,1-0,15 cm vastag. Alakja karcsú, egyenes vagy hajlott, üreges, törékeny. Felszíne fiatalon hamvas, később a csúcsán finoman szőrözött, lejjebb sima. Színe bézsbarna vagy barna, borvörös árnyalattal. Tövét fehér micélium borítja.

Spórapora fehér. Spórája gyümölcsmag alakú, sima, amiloid, mérete 8-10 x 4-6 µm.

Hasonló fajok
A lombos fákon élő vérző kígyógomba hasonlít hozzá.

Elterjedése és termőhelye
Eurázsiában és Észak-Amerikában honos. Magyarországon ritka.

Fenyvesekben, ritkábban bükkösökben fordul elő, általában nyirkos, mohás élőhelyeken; a korhadó növényi törmeléket, tűleveleket bontja. Ősszel terem.

Nem ehető.

Kapcsolódó cikkek
Magyarország nagygombafajainak listája

Források
Vöröstejű kígyógomba - Mycena sanguinolenta Miskolci Gombász Egyesület
Mycena sanguinolenta The Mycenas of Northern Europe

Gombafajok
Kígyógombafélék
Magyarország gombái
Európa gombái
Észak-Amerika gombái
Ázsia gombái

redpajama_set_name: RedPajamaWikipedia


text:
sarna.net
wiki |
Tag Archives: Ursus
Rabid Coyote vs. Ursus – The Second Line Bodyguards
While not as adaptable or flexible as OmniMechs, the Clans as a whole still see the indelible value of standard BattleMech technology. Even in their limited configuration state, when equipped with their venerable technology, Clan second line BattleMechs provide some of the most solid, dependable, and fun designs in the game. From the well-known Mad Cat Mk II to the iconic Kodiak and the re-imagined IIC lines, Clan second line designs range all the way from amazing to completely laughable.
Second line `Mechs are generally used for defensive operations, with some exceptions, and as such generally use less expensive and more stable parts and technology. Standard Engines are commonly found in second line designs, as well as standard chassis as opposed to Endo Steel construction. Ferro-Fibrous Armor, however, still appears to be the normal standard even with second line designs.
Today I want to take a look at two second line BattleMechs that were designed a few years, and couple thousand light years, apart from each other. These two `Mechs were designed to fill similar roles in their respective Clans' toumans, and each Clan went about the job a different way. In the case of Clan Coyote, they needed a lighter and faster design that could deliver a good punch standing alongside heavier units while still possessing enough speed to pursue attackers. For Clan Ghost Bear, the need for a strong and dependable second line design was apparent from the moment they arrived in the Inner Sphere and began taking and holding worlds.
Even with some of the differences in need, Clans Coyote and Ghost Bear ended up designing and producing similar BattleMechs that fit the needs of each Clan near perfectly. Let's dig into the histories, similarities, differences, and ultimately a comparative match-up, of the Rabid Coyote and the Ursus. For the purposes of this article, we are going to look at just the standard variants of both BattleMechs.
Rabid Coyote
Clan Coyote created the Rabid Coyote out of a need to support their highly successful assault class designs, such as the Canis and Savage Coyote. The larger `Mechs lacked a speed that was necessary for pursuing fleeing foes and suffered from the ability to engage faster enemies that could redeploy and attack from new angles.
During the time period following the Great Refusal, in the upheaval and scramble for new resources that followed the vacuum of Clan Smoke Jaguar's demise, Clan Coyote did not fare well. Clan Coyote, once at the top of Clan heap, now found itself on the edge of irrelevance. Even the introduction of Advanced Tactical Missile technology, solely a product of Clan Coyote, was not enough to catapult the Clan back into the spotlight.
The Khans needed something to help their touman, help their warriors on the battlefield, and help stabilize their decline. The future of the Clan was riding on the success of their next venture.
Designed as a bodyguard unit for heavier `Mechs, the Rabid Coyote was abjectly rejected upon arrival by the Warriors to which it was assigned. The concept that a BattleMech should play bodyguard and support to another BattleMech was one that Coyote Trueborns especially despised.
Regardless of the reception, the Rabid Coyote appears to be here to stay. Field tests and trials by fire have been positive for the new design, and Clan Coyote is committed to keeping the BattleMech in its ranks.
Armaments and Capabilities
Likely, at least in part, to further highlight their new technology, the Rabid Coyote's main armament is a single ATM 12 set in its left torso. The ATM has one ton of ammo for each type of ATM missile. This main weapon is backed up by four Medium Pulse Lasers, two in each arm, and an Electronic Countermeasure Suite (ECM) located near the cockpit. Ample armor protection, eight and a half tons of Ferro-Fibrous Armor, for a 55 ton BattleMech is provided, and battlefield staying power is backed up by its use of a standard engine, all mounted on an Endo Steel chassis. 12 Double Heat Sinks allow the Rabid Coyote to remain fairly cool when utilizing its entire arsenal.
This design on paper is as solid as they come. Perhaps a bit slow for a Clan Medium, in its intended role the 5/8 speed works out just fine, especially when compared to the speeds of the BattleMechs in whose company the Rabid Coyote is most commonly found.
Rolling off the production in 3059, seven years before the Rabid Coyote appeared in Clan Coyote's touman, the Ursus was a triumphant accomplishment for Khan Bjorn Jorgensson and for Clan Ghost Bear. The first Clan BattleMech produced completely within the Inner Sphere, the Ursus was designed to fill a painful gap in Clan Ghost Bear's second line forces.
The Ursus was designed somewhat with a "less is more" mentality that enforced its purpose. Using simple technologies like a standard chassis and engine, the design is dependable and able to take a lot of punishment for a fifty ton Medium `Mech. Received well by the touman as a whole, the majority of the first production runs found assignment to units primarily along the Draconis Combine border and saw action against both DCMS and Clan Nova Cat forces.
While not specifically created as a bodyguard unit, as in the case of the Rabid Coyote, the Ursus still found use in the role as it was soon assigned to duty alongside larger and slower second line BattleMechs like the Kodiak and Grizzly. In this role, the uncharacteristically slower speed of the Ursus did not feel at all like a liability, and the machine continued to shine as a defensive unit. Many commanders now consider the Ursus to be a quintessential companion to slower Heavy and Assault second line BattleMechs.
Click on the picture to visit IronWind Metals and get your own Ursus.
Armament and Capabilities
Thanks in part to its smaller engine, and the use of eight and a half tons of Ferro-Fibrous Armor, the Ursus packs an arsenal of weaponry that outclasses almost any other Clan second line Medium and many OmniMechs of its weight class. Its main armaments are on its right arm, which houses an Extended Range Large Laser and two Medium Pulse Lasers. These are backed up at long and medium ranges, respectively, by an LRM 10 on its left arm and an SRM 6 mounted in its center torso. Adding a final touch to the brawler nature of the BattleMech, each side torso sports an Extended Range Medium Laser, adding to the design's ability to engage effectively and decisively at medium range. Finally, the entire BattleMech is protected by an integrated Electronic Countermeasure Suite mounted in its right torso. 16 Double Heat Sinks allow the Ursus to stay just about as cool as the Rabid Coyote when using its vast array of weapons.
At first glance, it just doesn't seem like it is possible to fit all of that into a fifty ton chassis, but there it is. The secret to the Ursus is in its speed. 4/6 is painfully slow even for some Clan Heavy `Mechs, but in its intended role as a defensive and bodyguard unit, the Ursus just plain does not need to worry about being speedy.
Each of these BattleMechs represents a different take on what are essentially similar battlefield roles. The Rabid Coyote was additionally designed to emphasize pursuit after a successful defense, and the Ursus was designed to maximize overall defensive capability and sturdiness supporting larger `Mechs. Both designs are most commonly seen in the company of heavier designs, providing escort and support for the larger machines.
Both `Mechs' incorporation of ECM suites makes them invaluable on the battlefield in the company heavier units that are not commonly equipped with such countermeasures.
Also, it is highly unlikely that either of these two BattleMechs have ever seen each other on the battlefield. First, they are designed for the same role, not opposing roles, which means it would be more than exceedingly rare to see one or the other in an attacking force where the other is present among the defending forces. Mostly, though, is the fact that the Rabid Coyote is deployed by a Clan that is located totally in the Clan Homeworlds, and the Ursus is deployed by a Clan that is now completely located in the Inner Sphere. The Ursus was only ever produced in the Inner Sphere and was never transported to Clan space during the short time between its initial production and the Ghost Bears' exodus from Clan space.
IronWind Metals has Rabid Coyote models, too!
While Clan Wolf did obtain the design from their Coyote friends some time before their ejection from Clan space, eventually deploying their own variant during the Jihad, it is still highly unlikely for the two BattleMechs to have met on the field of batter. But we can imagine what a heads up engagement between these two BattleMechs might have looked like.
I'll give the long range game to the Ursus, even if slightly. Its ER Large Laser and LRM 10 are not affected by the Rabid Coyote's ECM Suite in the same way that the Ursus' ECM affects the Rabid Coyote's ATM 12 launcher. Fortunately for the Rabid Coyote because of its speed advantage over the slower Ghost Bear `Mech, the long range game should not last very long.
In the medium range, both BattleMechs enter each other's real danger zones. The Rabid Coyote adds four Medium Pulse Lasers into the mix, where as the Ursus adds two Medium Pulse Lasers and 2 ER Medium Lasers. If accuracy becomes an issue, the Rabid Coyote has the edge at this point. Its greater movement based defense also helps to further mitigate the non-Pulse Laser part of the Ursus' battery. However, a few luckier hits at shorter ranges with the ER Large Laser could keep the Ursus well in the fight.
It is at short range that the Rabid Coyote might finally turn the tables on the Ursus in a convincing manner. The High Explosive missiles on its ATM 12 will begin to exact a heavy toll on the Ursus, which does gain the use of its regular SRM 6 at this range.
In my estimation of the two BattleMechs, if the Ursus can prolong the longer range engagement, keeping to long range and the far end of medium range, for as long as possible, it will keep the upper hand. However, once the Rabid Coyote is able to make use of its superior speed and close to short range, the power of ATM 12 system will likely overpower the Ursus' ability to answer back as effectively.
I know that at this point you might be thinking that I'm going to avoid outright declaring a winner in this fight, and you would be right to think so. Both of these BattleMechs are well armed and armored for their weight class, and in a medium weight, non-striker defensive role, they are both perfect in their own ways.
While not as widespread as other designs, both the Ursus and Rabid Coyote are available in certain places should you be looking for a flavorful addition to a unit.
The Ursus has found its way, in the form of the Ursus 2 variant, into the touman of Clan Hell's Horses, and the original variant also traveled with many of the Ghost Bear Clusters gifted to the Republic of the Sphere after the Jihad. Of course, you can almost always come up with a reason to have any BattleMech in any force, especially a mercenary unit, and I would not be surprised to see the Ursus scattered about very thinly in the forces of the Draconis Combine, Clan Nova Cat, and even Clan Wolf. Salvage is a beautiful thing.
The Rabid Coyote only regularly appears among Clan Wolf forces in the Inner Sphere, but the Homeworld Clans Coyote and Cloud Cobra both field the design in large numbers. It is also not unheard of to see the design fielded by Clan Burrock and the Dark Caste. Like the Ursus, the design may have spread in very limited numbers, as salvage, from Clan Wolf and into the forces of Clans Jade Falcon, Ghost Bear, and Hell's Horses, as well as even possibly into the LAAF.
If you play any of these factions, or you just really want an awesome and defensive Clan tech BattleMech, and you have the means, I highly recommend picking either one (or both!) up.
If you want to know more about either the Ursus or Rabid Coyote, check out the reference materials listed below:
BattleTech Technical Readout: 3060 on BattleCorps, DriveThruRPG, or Amazon.
Want to see me compare two other BattleMechs? Suggest a pairing in the comments, and if it strikes my fancy I might just write it!
This entry was posted in Miniatures, Reviews and tagged Alpha Strike, BattleTech, Clan Coyote, Clan Ghost Bear, IronWind Metals, Rabid Coyote, Ursus on 2016/01/20 by Dave Martin.
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Tangent Space to Fiber Product of Transverse Mappings I'm trying to show two things: that for two smooth transverse mappings $f:X \to Z$ and $g:Y \to Z$, the map $h:W \to Z\times Z$, $h'(x,y)(v_x,v_y) = (f'(x)(v_x), g'(y)(v_y))$ is surjective (I see that it happens to be the case that this is true for vector spaces, but I'm frustrated that I can't prove it in general), and the tangent space $T_{(x,y)}W$ where $W$ consists of $(x,y) \in X \times Y$ such that $f(x) = g(y)$ is contained in the kernel of the map $\phi:T_xX \times T_yY \to T_zZ$ defined by $\phi(v_x,v_y) = f'(x)(v_x) - g'(y)(v_y)$. Any help would be appreciated.

A: The idea in your last comment works indeed. And local is enough, as being an embedded submanifold is a local property. That is, it is enough to show that for any $(x,y)\in X\times Y$ with $f(x)=g(y)$, there is an open neighborhood $(x,y)\in U\subset X\times Y$ such that $W\cap U$ is an embedded submanifold of $U$.
Now, suppose $f(x_0)=g(y_0)=z_0$, and let $(V,\phi)$ be a coordinate chart of $Z$ around $z_0$. Write $U:=f^{-1}(V)\times g^{-1}(V)\subset X\times Y$, and define$$k:U\to\mathbb{R}^k,\quad (x,y)\mapsto \phi(f(x))-\phi(g(y)).$$ By construction, we have $W\cap U=k^{-1}(0).$ It follows from transversality that $0$ is a regular value of $k$, and the claim follows from the implicit function theorem.

redpajama_set_name: RedPajamaStackExchange


text: Q: Why can we assign both energy and particle number to each state in grand canonical emsemble? The system in grand canonical emsemble together with the surrounding reservoir is isolated, thus have conserved particle number $N$. However, the system itself only has fixed average $\langle N\rangle$, and the quantum mechanical operator $\hat{N}$ necessarily does not commute with the Hamiltonian. Why can we assign both energy and particle number to each state?


text:
A determined woman works with a hardened boxing trainer to become a professional.
Physically Afflicted Character
All TerritoriesDomesticArgentinaAustraliaAustriaBelgiumBrazilBulgariaChileCzech RepublicDenmarkEgyptFinlandFranceGermanyGreeceHong KongHungaryIcelandItalyJapanLatviaLithuaniaMexicoNetherlandsNew ZealandNorwayPolandPortugalRomaniaRussia/CISSingaporeSlovakiaSouth KoreaSpainSwedenSwitzerlandTaiwanThailandTurkeyUnited Arab EmiratesUnited KingdomUruguay
DistributorWarner Bros.
Opening$179,953
Budget$30,000,000
Release Date (Wide)Dec 15, 2004 (Jan 28, 2005) - Jun 30, 2005
GenresDrama Sport
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Dec 20, 2004 Monday 22 $24,640 -56.9% - 8 $3,080 $260,047 6 false
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Feb 23, 2005 Wednesday 4 $623,152 -0.6% +18.4% 2,105 $296 $56,896,013 71 false
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Mar 1, 2005 Tuesday 3 $968,411 +12.2% +54.4% 2,125 $455 $66,683,367 77 false
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Mar 3, 2005 Thursday 3 $911,278 +4.8% +49.1% 2,125 $428 $68,464,587 79 false
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Mar 5, 2005 Saturday 5 $3,837,721 +67.8% +7.5% 2,350 $1,633 $74,588,914 81 false
Mar 6, 2005 Sunday 5 $2,011,094 -47.6% +7.8% 2,350 $855 $76,600,008 82 false
Mar 7, 2005 Monday 5 $531,306 -73.6% -38.5% 2,350 $226 $77,131,314 83 false
Mar 8, 2005 Tuesday 4 $611,193 +15% -36.9% 2,350 $260 $77,742,507 84 false
Mar 9, 2005 Wednesday 4 $620,330 +1.5% -28.7% 2,350 $263 $78,362,837 85 false
Mar 10, 2005 Thursday 4 $538,014 -13.3% -41% 2,350 $228 $78,900,851 86 false
Mar 11, 2005 Friday 6 $1,393,920 +159.1% -39% 2,375 $586 $80,294,771 87 false
Mar 12, 2005 Saturday 6 $2,448,133 +75.6% -36.2% 2,375 $1,030 $82,742,904 88 false
Mar 13, 2005 Sunday 6 $1,311,303 -46.4% -34.8% 2,375 $552 $84,054,207 89 false
Mar 14, 2005 Monday 6 $442,176 -66.3% -16.8% 2,375 $186 $84,496,383 90 false
Mar 15, 2005 Tuesday 6 $502,244 +13.6% -17.8% 2,375 $211 $84,998,627 91 false
Mar 16, 2005 Wednesday 6 $471,221 -6.2% -24% 2,375 $198 $85,469,848 92 false
Mar 17, 2005 Thursday 6 $452,407 -4% -15.9% 2,375 $190 $85,922,255 93 false
Mar 18, 2005 Friday 8 $1,095,620 +142.2% -21.4% 2,210 $495 $87,017,875 94 false
Mar 19, 2005 Saturday 8 $1,874,634 +71.1% -23.4% 2,210 $848 $88,892,509 95 false
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Mar 23, 2005 Wednesday 8 $414,435 +2.3% -12.1% 2,210 $187 $91,140,621 99 false
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Mar 25, 2005 Friday 10 $841,093 +107.5% -23.2% 1,710 $491 $92,387,058 101 false
Mar 26, 2005 Saturday 10 $1,078,215 +28.2% -42.5% 1,710 $630 $93,465,273 102 false
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Mar 28, 2005 Monday 10 $331,476 -50.4% -12.1% 1,710 $193 $94,464,754 104 false
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Mar 30, 2005 Wednesday 10 $233,465 -5.6% -43.7% 1,710 $136 $94,945,494 106 false
Mar 31, 2005 Thursday 10 $250,440 +7.3% -38.2% 1,710 $146 $95,195,934 107 false
Apr 1, 2005 Friday 12 $464,069 +85.3% -44.8% 1,405 $330 $95,660,003 108 false
Apr 2, 2005 Saturday 12 $661,391 +42.5% -38.7% 1,405 $470 $96,321,394 109 false
Apr 3, 2005 Sunday 12 $340,969 -48.4% -49% 1,405 $242 $96,662,363 110 false
Apr 4, 2005 Monday 12 $98,390 -71.1% -70.3% 1,405 $70 $96,760,753 111 false
Apr 5, 2005 Tuesday 12 $116,490 +18.4% -52.9% 1,405 $82 $96,877,243 112 false
Apr 6, 2005 Wednesday 12 $116,368 -0.1% -50.2% 1,405 $82 $96,993,611 113 false
Apr 7, 2005 Thursday 12 $130,082 +11.8% -48.1% 1,405 $92 $97,123,693 114 false
Apr 8, 2005 Friday 14 $213,007 +63.7% -54.1% 857 $248 $97,336,700 115 false
Apr 9, 2005 Saturday 14 $352,612 +65.5% -46.7% 857 $411 $97,689,312 116 false
Apr 10, 2005 Sunday 14 $136,400 -61.3% -60% 857 $159 $97,825,712 117 false
Apr 11, 2005 Monday 14 $61,580 -54.9% -37.4% 857 $71 $97,887,292 118 false
Apr 12, 2005 Tuesday 14 $68,223 +10.8% -41.4% 857 $79 $97,955,515 119 false
Apr 13, 2005 Wednesday 14 $61,445 -9.9% -47.2% 857 $71 $98,016,960 120 false
Apr 14, 2005 Thursday 14 $63,497 +3.3% -51.2% 857 $74 $98,080,457 121 false
Apr 15, 2005 Friday 17 $179,739 +183.1% -15.6% 902 $199 $98,260,196 122 false
Apr 16, 2005 Saturday 16 $308,263 +71.5% -12.6% 902 $341 $98,568,459 123 false
Apr 17, 2005 Sunday 17 $140,182 -54.5% +2.8% 902 $155 $98,708,641 124 false
Apr 19, 2005 Tuesday 14 $58,449 +7.3% -14.3% 902 $64 $98,821,545 126 false
Apr 20, 2005 Wednesday 14 $59,438 +1.7% -3.3% 902 $65 $98,880,983 127 false
Apr 21, 2005 Thursday 14 $62,327 +4.9% -1.8% 902 $69 $98,943,310 128 false
Apr 22, 2005 Friday 18 $86,490 +38.8% -51.9% 455 $190 $99,029,800 129 false
Apr 24, 2005 Sunday 19 $76,735 -47.5% -45.3% 455 $168 $99,252,599 131 false
Apr 27, 2005 Wednesday 17 $26,015 +4.1% -56.2% 455 $57 $99,328,024 134 false
Apr 28, 2005 Thursday 17 $24,435 -6.1% -60.8% 455 $53 $99,352,459 135 false
Apr 30, 2005 Saturday 23 $87,261 +84% -40.3% 235 $371 $99,487,154 137 false
May 1, 2005 Sunday 26 $40,098 -54% -47.7% 235 $170 $99,527,252 138 false
May 2, 2005 Monday 18 $14,866 -62.9% -39.1% 235 $63 $99,542,118 139 false
May 3, 2005 Tuesday 18 $16,769 +12.8% -32.9% 235 $71 $99,558,887 140 false
May 4, 2005 Wednesday 18 $16,932 +1% -34.9% 235 $72 $99,575,819 141 false
May 5, 2005 Thursday 20 $15,241 -10% -37.6% 235 $64 $99,591,060 142 false
May 6, 2005 Friday 23 $72,535 +375.9% +52.9% 390 $185 $99,663,595 143 false
May 7, 2005 Saturday 22 $112,755 +55.4% +29.2% 390 $289 $99,776,350 144 false
May 8, 2005 Sunday 21 $75,048 -33.4% +87.2% 390 $192 $99,851,398 145 false
May 9, 2005 Monday 17 $27,613 -63.2% +85.7% 390 $70 $99,879,011 146 false
May 10, 2005 Tuesday 15 $31,430 +13.8% +87.4% 390 $80 $99,910,441 147 false
May 11, 2005 Wednesday 17 $28,275 -10% +67% 390 $72 $99,938,716 148 false
May 12, 2005 Thursday 17 $29,317 +3.7% +92.4% 390 $75 $99,968,033 149 false
May 13, 2005 Friday 20 $53,219 +81.5% -26.6% 345 $154 $100,021,252 150 false
May 14, 2005 Saturday 19 $90,782 +70.6% -19.5% 345 $263 $100,112,034 151 false
May 15, 2005 Sunday 19 $44,347 -51.2% -40.9% 345 $128 $100,156,381 152 false
May 16, 2005 Monday 18 $11,454 -74.2% -58.5% 345 $33 $100,167,835 153 false
May 17, 2005 Tuesday 18 $13,595 +18.7% -56.7% 345 $39 $100,181,430 154 false
May 18, 2005 Wednesday 18 $11,479 -15.6% -59.4% 345 $33 $100,192,909 155 false
May 19, 2005 Thursday 19 $10,383 -9.5% -64.6% 345 $30 $100,203,292 156 false
May 20, 2005 Friday 24 $22,190 +113.7% -58.3% 110 $201 $100,225,482 157 false
May 23, 2005 Monday 21 $9,956 -53.1% -13.1% 110 $90 $100,293,675 160 false
May 24, 2005 Tuesday 21 $8,066 -19% -40.7% 110 $73 $100,301,741 161 false
May 25, 2005 Wednesday 20 $7,947 -1.5% -30.8% 110 $72 $100,309,688 162 false
May 26, 2005 Thursday 21 $9,262 +16.5% -10.8% 110 $84 $100,318,950 163 false
May 27, 2005 Friday 22 $13,136 +41.8% -40.8% 75 $175 $100,332,086 164 false
May 28, 2005 Saturday 22 $20,225 +54% -45.3% 75 $269 $100,352,311 165 false
May 29, 2005 Sunday 22 $16,086 -20.5% -24.2% 75 $214 $100,368,397 166 false
Monday 22 $11,250 -30.1% +13% 75 $150 $100,379,647 167 false
May 31, 2005 Tuesday 18 $5,835 -48.1% -27.7% 75 $77 $100,385,482 168 false
Jun 1, 2005 Wednesday 19 $5,436 -6.8% -31.6% 75 $72 $100,390,918 169 false
Jun 2, 2005 Thursday 19 $5,034 -7.4% -45.6% 75 $67 $100,395,952 170 false
Jun 3, 2005 Friday 22 $7,529 +49.6% -42.7% 59 $127 $100,403,481 171 false
Jun 4, 2005 Saturday 22 $11,648 +54.7% -42.4% 59 $197 $100,415,129 172 false
Jun 5, 2005 Sunday 22 $7,657 -34.3% -52.4% 59 $129 $100,422,786 173 false
Jun 6, 2005 Monday 20 $2,949 -61.5% -73.8% 59 $49 $100,425,735 174 false
Jun 7, 2005 Tuesday 20 $3,181 +7.9% -45.5% 59 $53 $100,428,916 175 false


text:
Biography: historical, political & military
Crossing the Line : A Bluejacket's Odyssey in World War II EPUB
by Kernan Alvin B. Kernan
Part of the Yale Library of Military History series
In this memoir of life aboard aircraft carriers during World War II, Alvin Kernan combines vivid recollections of his experience as a young enlisted sailor with a rich historical account of the Pacific war. Kernan served in many battles and was aboard the Hornet when it was sunk by torpedoes in the Battle of the Santa Cruz Islands.
"One of the most arresting naval autobiographies yet published."—Sir John Keegan
"An honest story of collective courage, evocative, well-written, and fixed before the colors fade."—Kirkus Reviews
"[Kernan] recounts a wonderful and exciting American story about a poor farm boy from Wyoming who enlisted in the Navy. . . .[He] has written eight other books. I will go back and read them all."—John Lehman, Air & Space
"Details . . . make the moment vivid; that is what it was like, on the Hornet in its last hours."—Samuel Hynes, New York Times Book Review
Naval forces & warfare
Also in the Yale Library of Military History series | View all
Baghdad at Sunrise
War by Land, Sea, and Air
A Question of Command


American top vocalist John Legend brought his sweet sounds to the 2014 Grammy's stage.
American top vocalist John Legend brought his sweet sounds to the 2014 Grammy's stage.
Dressed in a white suit, the R&B crooner serenaded the crowd with his piano ballad "All of Me," which was recently certified gold.
While he didn't win Best R&B Album, he did take home his supermodel wife Chrissy Teigen.


I am in the process of moving and recently spent the weekend stripping wallpaper. I couldn't help but be fascinated by the layers of colour and decoration that I found. Who was the person who decided that deep pink was the colour of choice for the woodchip wallpaper they'd put in the front room? Aside from the obvious question of 'what were they thinking?', I found myself wondering more. When had they lived here? What were they like? As another layer gradually came off, there was a more decorative layer – flowers and leaf detail. Who put that there and when? What was their story? I finally got through to the bottom layer and discovered that the walls themselves had been painted at some point in a mustardy-yellow colour, complete with red & green spots, clearly added one by one, across the walls. What about those people? Why that colour combination? Were they the first ones to live there? Did they ever wonder about who would live in that tiny flat in the future? As a PhD student, my mind turned, as it often does, to my research. To the concept of ancestral time. As I peeled back through the layers of interior decoration and found myself wondering about the lives of the previous residents, I reflected on the many ways that we are impacted by and connected to previous generations in our everyday lives.
recalling a past when building upwards was the only option, staying within the safety of the city boundaries. The grandeur of the Georgian New Town, designed by James Craig in 1768, reminds me of the ways people with a vision for a better future can impact that future: the streets are largely unchanged and I sometimes walk along wondering whether James Craig and his colleagues knew that their work would still be having impact in the lives of the people of Edinburgh nearly 250 years later.
The past impacts the present. The past impacts the future. The present impacts the future. The idea of ancestral time grounds me in the knowledge that I am indeed connected to the generations who have gone before. That our society is connected to the generations who have gone before. And as such, we are also deeply connected to the generations who are still to come. Not just in the next ten, twenty or fifty years, but to the people whose lives will be playing out in the 2100s, 2200s, and beyond. It feels strange just writing those dates. Will someone be moving into 'my' flat in 2100, wondering why I chose a simple beige paint for my walls? Will there be people walking the streets of Edinburgh in 2214 looking at the old historic buildings – those which are being built today? What vision are we casting for a future 250 years from now? How are we living in a way that considers our impact on future generations?
When we connect ourselves to the past through the concept of ancestral time, we realise that ancestral time also connects us to the future. Ancestral Time provides me with a grounding for looking to the future with a bit more humility about my life today, and with a good deal more reflection on the responsibility and impact that we have for the future as well.
I was just reviewing Calvinism and its early beginnings in Geneva spreading in Europe and to the New World. How the past has impact today and the future. How Calvinism today is still related from its early Reformed theology. So what would Ancestral Time do with our theological future now that we are globalized? Thanks for your blog Alice and maybe see you some- Time?
An interesting thought – how theology and ideas are shaped through ancestral time as well. Definitely more to think about! Thanks for your comment.


The Hall is a popular and busy community hub for the arts in Bewdley. It is perfect for art classes, easy to get to and accessible for everyone. There is always lots going on, centrally located with a lovely community café, it is a meeting place for those interested in the arts, cinema or just to have a bit to eat. We are so pleased to be hosting our classes here, it is a natural fit for anyone wanting to learn to paint and draw in the Bewdley area.


The activity of EFC characterize all-Estonian team-work between voluntary associations, research establishments, local administrations and state cultural institutions. EFC co-ordinates national and international folklore events, regulates international exchange between folklore groups, mediates information, gives a methodological, organizational and scientific assistance, organizes festivals, seminars and conferences, arranges teaching on folklore. EFC has a leadership in organizing the main folklore event in Estonia – CIOFF® International Folklore Festival BALTICA.


One of the best feelings in this world is to laugh from the heart.
Laughter is good for body, mind and soul. Research shows that laughter does improve our health. In fact, you cannot feel stressed or sad when we are laughing.
When was the last time you Laugh Out Loud?
When we were a kid, we used to laugh whole day. As we grow, we forget it. We think that laughing is only for kids. No, it is not only for kids. It is for everyone who is alive.
Do you remember any of your friends who laugh from the heart?
When I think about laughter, it reminds me one of my friends, "Sumee". She knows how to laugh from the heart.
She was able to find humour everywhere. She can create a virtual funny world around her any time. She can laugh anytime. It makes people happy around her as well.
Almost everyone has at least one friend in their friend circle, who laughs more and makes others laugh as well. Appreciate him for spreading the happiness.
If you laugh more, you feel a sense of wellness. Laughter keeps you away from the stress and anxiety.
Laughter reduces the risk of heart attacks.
Laughter boosts the immune system. Doctors always said that a person with a positive set of attitude always laughs more and have fewer diseases in his lifetime.
So laugh more and live longer.
Plan a get together with old friends whenever possible. Celebrate little victories with friends. Plan a comedy movie or plan a trip. You already know when you meet them you laugh more.
Get a Pet if you like. They often do funny things. You can laugh more with them.
Spend time with playful people around you.
Try to find humour in every situation of life. If you cannot, think about your funniest friend. How he will be able to deal with the situation. Even thinking of him can make you laugh. Once you laugh, no problem seems too big.
Play with kids around you. Be one of them. I believe there is no use of grow up, if you are not a child sometimes.


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redpajama_set_name: RedPajamaGithub


Velocity digital optical Toslink cables from C2G are perfect for connecting your DVD,CD,satellite receiver,game console or other digital audio equipment for clear,full detailed sound. Velocity Toslink cables are manufactured from high quality PMMA optical fiber with a polished lens and gold plated collet to provide an accurate digital signal. The stylish PVC jacket adds flexibility and durability for years of listening pleasure.


text:
Il  fa parte di un gruppo di mausolei risalenti al tardo periodo Kofun (VII secolo), situati presso il villaggio di Asuka, nel bacino di Nara, in Giappone.

Takamatsuzuka è un tumulo di dimensioni piccole comprendente una piccola cella funeraria. Le pareti interne sono tutte ricoperte di stucco e di una decorazione pittorica affine alle tecniche cinesi e coreane.

Le raffigurazioni sono rappresentate in coincidenza con i punti cardinali nel rispetto della cosmologia cinese: e quindi una coppia di animali appare in ognuno dei quattro punti, come ad esempio la tartaruga e il serpente a nord affiancati da un gruppo di quattro uomini o di quattro donne. Sul soffitto sono rappresentate le costellazioni. Non particolarmente ricco appare il corredo funerario, composto da uno specchio metallico, da alcuni piattini d'argento e i resti di una spada di bronzo.

Incerta appare l'attribuzione della tomba: qualche esperto propende per un aristocratico giapponese, mentre altri pensano di ricondurre la struttura in onore di un esponente continentale, visti anche i molti richiami stilistici con le tombe coreane e l'affinità dei costumi con quelli cinesi dei Chin orientali.
Nell'ipotesi di una famiglia aristocratica giapponese, quella più accreditata è quella dell'imperatore Tenmu (673-686) e dei suoi figli:
 Principe Osakabe (? - 705),
 Principe Yuge (? - 699),
 Principe Takechi (654 - 696),

Secondo alcuni esperti, tra i quali l'archeologo Namio Egami, la civiltà Kofun fu influenzata da una forte invasione di cavalieri coreani caratterizzati da un forte accentramento di poteri politici e religiosi nelle mani di un capo. Ecco perché la tomba aristocratica di Takamatsuzuka è circondata da resti di tombe di cavalieri arricchite da corredi di armi.

Note

Bibliografia

Altri progetti

Collegamenti esterni
  Distretto di Nara : le tombe di Takamatsuzuka ;
  Decorazioni di Takamatsuzuka ;
  Storia e dipinti di Takamatsuzuka.

Siti archeologici del Giappone
Storia del Giappone


Our October 2018 basic box focused on ink. It included a 45ml bottle of Pebeo Graphic India Ink, a Grumbacher Academy round brush size 0, A full four piece set of Sakura Pigma markers (brush tip, .25mm micron tip, .45mm micron tip, and 1mm tip) in a beautiful sepia tone, and a Global Art Flexi-Sketch landscape 5x7" sketchbook. These supplies were specifically chosen to allow you to create beautiful ink illustrations. The 45ml bottle of Pebeo ink is extremely potent and will give you the ability to create hundreds of ink pieces. It washes out to beautiful grays with the addition of water. The sepia set from Sakura allows you to add a warm tone to your drawings and add detailed linework. The Grumbacher brush allows for finely detailed illustrations and a true command over the ink. Have fun and remember to tag your art with #sketchBox to be considered for our monthly contest. The winner gets their art printed on our boxes! This picture was taken by our talented subscriber @thehueinator - check out their art on instagram!


text:
How does ICT affect your life?
ICT has contributed a lot to change our everyday life such as letter to e-mail, market shopping to on-line shopping, classroom learning to e-learning, etc. This paper present's the effects of ICT as Home and Domestic Activities, Social Networking, Education, Health, Commerce, Banking, and Employment.
What does ICT mean?
How does ICT enhance learning?
ICTs can enhance the quality of education in several ways: by increasing learner motivation and engagement, by facilitating the acquisition of basic skills, and by enhancing teacher training. ICTs are also transformational tools which, when used appropriately, can promote the shift to a learner-centered environment.
What is the positive impact of ICT on education?
It enhances scope of education by facilitating mobile learning and inclusive education. It facilitates research and scholarly communication. Impact of ICT and its potential for the education field is manifold. It positively affects all the stakeholders of the education field.
How does ICT affect future employment?
ICTs connect people to jobs. ... ICTs also enable new, more flexible forms of employment and work: Online contracting uses ICT to increase access to work opportunities worldwide, mainly for smaller employers.
How does ICT affects your life as a student?
ICT sector plays an important role, notably by contributing to rapid technological progress and productivity growth. ... ICT may also be a significant motivational factor in students' learning, and can support students' engagement with collaborative learning. It has a great impact in our daily lives.
How does ICT skills make you better?
ICT Can Help Foster More Efficient Ways of Working Being able to use ICT systems effectively allows employees more time to concentrate on areas of their job role that require soft skills. For example, many pharmacies use robot technology to assist with picking prescribed drugs.
How can ICT make a difference in the teaching and learning process?
There is widespread belief that ICTs can and will empower teachers and learners, transforming teaching and learning processes from being highly teacher-dominated to student-centered, and that this transformation will result in increased learning gains for students, creating and allowing for opportunities for learners ...
What is ICT skill?
Information and communications technology (ICT) skills refer to one's ability to converse with people through various technologies. ... ICT skills could also include the ability to use older communication technologies such as telephones, radios, and televisions.
What is entry level 3 ICT?
The NCFE Functional Skills qualification in ICT at Entry Level 3 is an ideal qualification for learners wishing to develop practical, transferable skills in ICT in order to work confidently, effectively and independently in life.
What are good ICT skills?
The ability to use computer operating systems, to access software programmes and manage the basic functions of a computer.
Being able to confidently use core computer programmes to produce common digital information such as Word documents and PowerPoint presentations.
What are the four types of ICT?
Each of the different ICTs—print, audio/video cassettes, radio and TV broadcasts, computers or the Internet—may be used for presentation and demonstration, the most basic of the five levels. Except for video technologies, drill and practice may likewise be performed using the whole range of technologies..
What are the two types of ICT?
In business, ICT is often categorised into two broad types of product: Traditional computer-based technologies (things you can typically do on a personal computer or using computers at home or at work)
What is ICT strand?
ICT is a Senior High School strand that prepares you for a career in Information and Communication Technology. ... It equips you with advanced skills in computer systems, programming, creating web pages, and basic animation.
What is Vygotsky theory of social constructivism?
How does feminist theory differ from constructivist theories?
What is the difference between constructivism and social constructivism?
What is difference between realism and neorealism?
What is a constructivist worldview?
What is constructivism essay?
What is the difference between Interpretivism and constructivism?
What is gender constructivism?
What is constructivist grounded theory?
How tall is the average apartment building?
What age is most crucial for brain development?
Is Kant A constructivist?
What is constructivism in qualitative research?
Is John Dewey a constructivist?
What are the key components of constructivism?
What is constructivism in history?
Who is the proponent of constructivist theory?
What does Piaget's theory focus on?


text:
Pennsylvania COVID-19 Early Warning Monitoring Dashboard Update for Nov. 13 – Nov. 19: Case Increases Top 36,000; Percent Positivity at 11.1% and 63 Counties with Substantial Transmission
Governor Tom Wolf and Secretary of Health Dr. Rachel Levine today released a weekly status update detailing the state's mitigation efforts based on the COVID-19 Early Warning Monitoring System Dashboard, highlighting a seven-day case increase of 36,133 cases, statewide percent positivity of 11.1%, and a concerning 63 counties with substantial transmission status.
The update includes the following:
Level of community transmission as a basis for the recommendations for Pre-K to 12 schools to determine instructional models.
Data on cases among 5-18-year-olds.
Cases that reported visiting a business among potential locations where exposures may have occurred.
Updated travel recommendations.
The dashboard is designed to provide early warning signs of factors that affect the state's mitigation efforts. The data available on the early warning monitoring dashboard includes week-over-week case differences, incidence rates, test percent-positivity, and rates of hospitalizations, ventilations and emergency room visits tied to COVID-19. This week's update compares the period of November 13 – November 19 to the previous seven days, November 6 – November 12.
"Another week of significant increases of COVID-19 across Pennsylvania is a call to action," Gov. Wolf said. "We need all Pennsylvanians to take the steps they can to protect one another. We need Pennsylvanians to follow the most recent orders and wear a mask, social distance, avoid travel and gatherings and wash their hands. It is only by working together that Pennsylvanians can prevent the spread of the virus."
As of Thursday, November 19, the state has seen a seven-day case increase of 36,133 cases; the previous seven-day increase was 27,326 cases, indicating 8,807 more new cases across the state over the past week compared to the previous week.
The statewide percent-positivity went up to 11.1% from 9.6% last week. Every county in the state has a concerning percent positivity above five percent except for one county, Cameron County at 1.9 percent.
"This week's data, in terms of hospitalization increase, an increase in the use of ventilators, case increase and percent positivity are worrisome," Secretary of Health Dr. Rachel Levine said. "Latest models show we could run out of ICU beds within a week. We know COVID-19 does not discriminate and is affecting every county in the Commonwealth. It is affecting all Pennsylvanians, no matter your race, ethnicity, age, socioeconomic status or whether you live a rural, suburban or urban area."
Community Transmission
As of Friday's data, Adams, Allegheny, Armstrong, Beaver, Bedford, Berks, Blair, Bradford, Bucks, Butler, Cambria, Carbon, Centre, Chester, Clarion, Clearfield, Clinton, Columbia, Crawford, Cumberland, Dauphin, Delaware, Elk, Erie, Franklin, Fulton, Greene, Huntingdon, Indiana, Jefferson, Juniata, Lackawanna, Lancaster, Lawrence, Lebanon, Lehigh, Luzerne, Lycoming, McKean, Mercer, Mifflin, Monroe, Montgomery, Montour, Northampton, Northumberland, Perry, Philadelphia, Potter, Schuylkill, Snyder, Somerset, Tioga, Union, Venango, Washington, Westmoreland, Wyoming and York counties were in the substantial level of community transmission. The departments of Education and Health will speak with school district representatives in these counties to discuss the implications of this level of transmission.
For the week ending November 12, two counties were in the low level of transmission, two counties in the moderate level, with 63 with substantial transmission:
Low – Cameron, Sullivan
Moderate – Pike, Wayne
Substantial – Adams, Allegheny, Armstrong, Beaver, Bedford, Berks, Blair, Bradford, Bucks, Butler, Cambria, Carbon, Centre, Chester, Clarion, Clearfield, Clinton, Columbia, Crawford, Cumberland, Dauphin, Delaware, Elk, Erie, Fayette, Forest, Franklin, Fulton, Greene, Huntingdon, Indiana, Jefferson, Juniata, Lackawanna, Lancaster, Lawrence, Lebanon, Lehigh, Luzerne, Lycoming, McKean, Mercer, Mifflin, Monroe, Montgomery, Montour, Northampton, Northumberland, Perry, Philadelphia, Potter, Schuylkill, Snyder, Somerset, Susquehanna, Tioga, Union, Venango, Warren, Washington, Westmoreland, Wyoming, York
Cases Among 5-18-Year-Olds
The Department of Health is providing weekly data on the number of statewide cases of COVID-19 among 5 to 18-year-olds.
Throughout the pandemic, there have been 23,742 total cases of COVID-19 among 5 to 18-year-olds. Of that total, 3,937 occurred between November 13 – November 19. For the week of November 6 – November 12, there were 3,198 cases of COVID-19 among 5 to 18-year-olds.
Cases by demographic group is available on the DOH website.
Business Visits
The Department of Health is providing weekly data on the number of individuals who responded to case investigators that they spent time at business establishments (restaurants, bars, gym/fitness centers, salon/barbershops) and at mass gatherings 14 days prior to the onset of COVID-19 symptoms.
It is important to note that due to the recent number of cases, the department is prioritizing case investigations. In addition to the need for people to answer the call, the significant number of cases helps contribute to the low percentages in contact tracing data. All of this reinforces the need for Pennsylvanians to take steps to prevent the spread of COVID-19.
Of the 34,719 confirmed cases reported between November 8 and November 14, 10 percent (3,619) provided an answer to the question as to whether they spent time at a business establishment.
Of those who did provide an answer, 1.4 percent, or 483, answered yes, they visited a business establishment 14 days prior to onset of symptoms:
48 percent (233) of those who said yes reported going to a restaurant;
24 percent (109) of those who said yes reported going to some other business establishment;
15 percent (74) of those who said yes reported going to a bar;
14 percent (66) of those who said yes reported going to a gym/fitness center; and
7 percent (34) of those who said yes reported going to a salon/barbershop.
Of the 34,719 confirmed cases, 10.5 percent (3,628) answered the question as to whether they attended a mass gathering or other large event. Of the 10.5 percent, 16.3 percent (591) answered yes to whether they attended a mass gathering or other large event 14 days prior to onset of symptoms.
Compared to data reported on November 16, this week's data saw an increase for people who reported going to a gym/fitness center (14 percent vs. 11 percent last week) and going to a bar (15 percent vs. 12.5 percent last week),.The data saw a decrease for people who reported going a restaurant (48 percent vs. 53 percent last week) and to another business (23 percent vs. 26 percent last week). Numbers remained the same for those going to some a salon/barbershop (7 percent vs. 7 percent last week). The number of those who attended a mass gathering or other large event decreased from to 16.3 percent from 18.4 percent last week.
The numbers above highlight business settings and mass gatherings as possible sites for transmission. With less than half of those asked about what types of businesses they visited or if they attended a mass gathering responding to the question, the department is reminding Pennsylvanians that it is essential that people answer the phone when case investigators call and to provide full and complete information to these clinical professionals.
Last week, the Department of Health changed its travel guidance. Dr. Levine issued an order requiring anyone who visits from another state to have a negative COVID-19 test within 72 hours prior to entering the commonwealth.
If someone cannot get a test or chooses not to, they must quarantine for 14 days upon arrival in Pennsylvania.
Pennsylvanians visiting other states are required to have a negative COVID-19 test within 72 hours prior to their return to the commonwealth or to quarantine for 14 days upon return to Pennsylvania.
This order, which took effect on Friday, November 20, does not apply to people who commute to and from another state for work or medical treatment.
It is important that people understand that this order is in place to prevent the spread of COVID-19 in Pennsylvania. A concerning number of recent cases have been linked to travel, and if people are going to travel, we need them to take steps to protect themselves, their loved ones and their community, and that involves having either a negative test, or quarantining.
Gov. Wolf continues to prioritize the health and safety of all Pennsylvanians through the COVID-19 pandemic. Pennsylvanians should continue to take actions to prevent the spread of COVID-19, regardless of in what county they live. This includes wearing a mask or face covering anytime they are in public. COVID-19 has been shown to spread easily in the air and contagious carriers can be asymptomatic.


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How To: Differentiate between count & noncount noun in English
In this video a teacher shows the difference between count and noncount nouns. This is very important for good American English grammar.
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