text:
\section{Introduction}
We consider finite undirected graphs
without loops or multiple edges.
For a graph $G$, we denote by $V(G)$ and $E(G)$
its vertex and edge sets, respectively.
An \textit{edge-coloring} of a graph $G$
is a mapping $color:E(G) \rightarrow \mathbb{C}$,
where $\mathbb{C}$ is a set of colors.
Then, the triple $(G, \mathbb{C}, color)$
is called an \textit{edge-colored graph}.
We often abbreviate an edge-colored graph $(G, \mathbb{C}, color)$ as $G$.
Note that
an edge colored graph is not necessary to be proper,
where
distinct red edges may have a common end vertex.

\subsection{%
Heterochromatic (or rainbow) spanning trees}
An edge-colored graph $G$ is said to be \textit{heterochromatic}%
\footnote{%
A heterochromatic graph is also said to be
\textit{rainbow}, \textit{multicolored}, \textit{totally multicolored},
\textit{polychromatic}, or \textit{colorful},
and so on.
}
if no two edges of $G$ have the same color,
that is,
$color(e_i) \ne color(e_j)$
for any two distinct edges $e_i$ and $e_j$ of $G$.
As far as I know,
there are three topics about heterochromatic graphs:
the Anti-Ramsey problem
introduced by Erd\H{o}s et al.
\cite{Erdos1975Anti-RamseyTheorems},
rainbow connection problems
introduced by Chartrand et al.
\cite{Chartrand2008RAINBOWGRAPHS},
and heterochromatic subgraph problems,
(see the surveys
\cite{Fujita2010RainbowSurvey}
\cite{Li2013RainbowSurvey}
\cite{Kano2008MonochromaticSurvey}
).
This paper focuses
on heterochromatic subgraph problems.

We denote by $\omega(G)$ the number of components of a graph $G$.
Given an edge-colored graph $G$ and a color set $R$,
we define $E_{R}(G) = \{ e \in E(G) ~|~ color(e) \in R \}$.
For simplicity,
we denote the graph
$(V(G), E(G) \setminus E_R(G))$ by $G-E_R(G)$,
and also denote
$E_{\{c\}}(G)$ by $E_c(G)$ for a color $c$.

Akbari \& Alipour
\cite{Akbari2007MulticoloredGraphs}
and Suzuki
\cite{Suzuki2006AGraph}
independently presented
a necessary and sufficient condition
for edge-colored graphs
to have a heterochromatic spanning tree.

\begin{theorem}[%
Akbari and Alipour
\cite{Akbari2007MulticoloredGraphs},
Suzuki
\cite{Suzuki2006AGraph}]
An edge-colored graph $G$
has a heterochromatic spanning tree
if and only if
\begin{equation*}
\omega(G-E_R(G)) \le |R|+1
\text{~~~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
\label{thm_Suzuki2006AGraph_1}
\end{theorem}

Suzuki
\cite{Suzuki2006AGraph}
proved the following theorem by using Theorem
\ref{thm_Suzuki2006AGraph_1}.

\begin{theorem}[Suzuki \cite{Suzuki2006AGraph}]
An edge-colored complete graph $G$ of order n
has a heterochromatic spanning tree
if $|E_c(G)| \le n/2$ for any color $c \in \mathbb{C}$.
\label{thm_Suzuki2006AGraph_2}
\end{theorem}

The complete graph $K_n$ has $(n-1)n/2$ edges,
thus the condition of Theorem
\ref{thm_Suzuki2006AGraph_2}
is equivalent to that
\begin{equation*}
\frac{|E_c(G)|}{|E(G)|}(n-1) \le 1
\text{~~~~~ for any color } c \in \mathbb{C}.
\end{equation*}

We can regard $|E_c(G)|/|E(G)|$ as
the probability of a color $c$
appearing in $G$.
The term ``Heterochromatic'' means that
any color appears once or zero times.
Thus,
we can interpret Theorem \ref{thm_Suzuki2006AGraph_2}
as saying that
if each color probability is at most $1/(n-1)$ in $G$
then $G$ has a spanning tree $T$
such that
each color probability is $1/(n-1)$ or $0$ in $T$.

\subsection{$f$-Chromatic spanning trees and forests}
The term ``Heterochromatic'' means that
any color appears at most once.
Suzuki
\cite{Suzuki2013AForests}
generalized ``once'' to a mapping $f$
from a given color set $\mathbb{C}$
to the set $\mathbb{Z}_{\ge 0}$
of non-negative integers,
and defined \textit{$f$-chromatic} graphs
as follows.

\begin{definition}[Suzuki \cite{Suzuki2013AForests}]
Let $G$ be an edge-colored graph.
Let $f$ be a mapping from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
$G$ is said to be \textit{$f$-chromatic}
if $|E_c(G)| \le f(c)$ for any color $c \in \mathbb{C}$.
\label{def_Suzuki2013AForests}
\end{definition}

Fig. \ref{fig_180825_1} shows an example
of an $f$-chromatic spanning tree
of an edge-colored graph.
For the color set $\mathbb{C} = \{1,2,3,4,5,6,7 \}$,
a mapping $f$ is given as follows:
\begin{gather*}
f(1)=3,
f(2)=2,
f(3)=3,
f(4)=0,
f(5)=0,
f(6)=1,
f(7)=2.
\end{gather*}
The left edge-colored graph
has the right $f$-chromatic spanning tree,
where each color $c$ appears at most $f(c)$ times.

\fig{0.85\textwidth}{!}
{fig_180825_1.pdf}
{An $f$-chromatic spanning tree
of an edge-colored graph.}
{fig_180825_1}

Suzuki
\cite{Suzuki2013AForests}
presented
the following necessary and sufficient condition
for edge-colored graphs to have
an $f$-chromatic spanning forest
with exactly $m$ components.

\begin{theorem}[Suzuki \cite{Suzuki2013AForests}]
Let $G$ be an edge-colored graph of order $n$.
Let $f$ be a mapping from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
Let $m$ be a positive integer such that $n \ge m$.
$G$ has
an $f$-chromatic spanning forest
with exactly $m$ components
if and only if
\begin{equation*}
\omega(G-E_R(G)) \le m+\sum_{c \in R}f(c)
\text{~~~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
\label{thm_Suzuki2013AForests_1}
\end{theorem}

Suzuki
\cite{Suzuki2013AForests}
proved the following Theorem
by using Theorem
\ref{thm_Suzuki2013AForests_1}.

\begin{theorem}[Suzuki \cite{Suzuki2013AForests}]
Let $G$ be an edge-colored graph of order $n$.
Let $f$ be a mapping from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
Let $m$ be a positive integer such that $n \ge m$.
If $|E(G)|>\binom{n-m}{2}$ and
\begin{equation*}
\frac{|E_c(G)|}{|E(G)|}(n-m) \le f(c)
\text{~~~~~ for any color } c \in \mathbb{C},
\end{equation*}
then $G$ has an $f$-chromatic spanning forest
with exactly $m$ components.
\label{thm_Suzuki2013AForests_2}
\end{theorem}

A heterochromatic graph is
an $f$-chromatic graph with $f(c) = 1$
for every color $c$.
Thus, these two theorems include
Theorem
\ref{thm_Suzuki2006AGraph_1} and
Theorem
\ref{thm_Suzuki2006AGraph_2}.
In this paper,
we will further generalize these theorems
and study color probability distributions
of edge-colored complete graphs
and its spanning trees.

\section{Main results}
In this paper,
I propose a \textit{$(g,f)$-chromatic graph}
as an edge-colored graph
where each color $c$ appears
at least $g(c)$ times and at most $f(c)$ times.
I also present a necessary and sufficient condition
for edge-colored graphs
to have a $(g,f)$-chromatic spanning forest
with exactly $m$ components.
(Theorem \ref{thm_180815_1}).
Using this criterion,
I show that
an edge-colored complete graph $G$ has
a spanning tree
with a color probability distribution
``similar'' to that of $G$
(Theorem \ref{thm_180913_1}).
Moreover, I conjecture that
an edge-colored complete graph $G$
of order $2n$ $(n \ge 3)$
can be partitioned into
$n$ edge-disjoint spanning trees
such that each has a color probability distribution
``similar'' to that of $G$
(Conjecture \ref{conj_180910_1}).

\subsection{%
$(g,f)$-Chromatic spanning trees and forests}
We begin with the definition
of a \textit{$(g,f)$-chromatic} graph.

\begin{definition}
Let $G$ be an edge-colored graph.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
$G$ is said to be \textit{$(g,f)$-chromatic}
if $g(c) \le |E_c(G)| \le f(c)$
for any color $c \in \mathbb{C}$.
\label{def_180815_1}
\end{definition}

Fig. \ref{fig_180825_2} shows
a $(g,f)$-chromatic spanning tree
of an edge-colored graph.
For the color set $\mathbb{C} = \{1,2,3,4,5,6,7 \}$,
mappings $g$ and $f$ are given as follows:
\begin{gather*}
g(1)=1,
g(2)=1,
g(3)=2,
g(4)=0,
g(5)=0,
g(6)=1,
g(7)=0,\\
f(1)=3,
f(2)=2,
f(3)=3,
f(4)=0,
f(5)=0,
f(6)=1,
f(7)=2.
\end{gather*}
The left edge-colored graph
has the right $(g,f)$-chromatic spanning tree,
where each color $c$ appears
at least $g(c)$ times and at most $f(c)$ times.

\fig{0.85\textwidth}{!}
{fig_180825_1.pdf}
{A $(g,f)$-chromatic spanning tree
of an edge-colored graph.}
{fig_180825_2}

We will see more examples.
First, we suppose that
$g$ and $f$ are given as follows:
\begin{gather*}
g(1)=3,
g(2)=1,
g(3)=3,
g(4)=0,
g(5)=0,
g(6)=1,
g(7)=2,\\
f(1)=3,
f(2)=2,
f(3)=3,
f(4)=0,
f(5)=0,
f(6)=1,
f(7)=2.
\end{gather*}
Then, the left edge-colored graph
in Fig. \ref{fig_180825_2}
has no $(g,f)$-chromatic spanning trees,
because $g(1)+g(2)+ \cdots +g(7)$ exceeds $7$,
the size of a spanning tree of the graph.

Next, in Fig. \ref{fig_180826_1},
we suppose that
$g$ and $f$ are given as follows:
\begin{gather*}
g(1)=0,
g(2)=2,
g(3)=2,
g(4)=0,
g(5)=0,
g(6)=1,
g(7)=0,\\
f(1)=3,
f(2)=2,
f(3)=3,
f(4)=0,
f(5)=0,
f(6)=1,
f(7)=2.
\end{gather*}
Then, in the left edge-colored graph,
any subgraph having $g(2)$, $g(3)$, and $g(6)$ edges
colored with $2$, $3$, and $6$, respectively,
contains the right subgraph,
which has a cycle.
Thus, the left graph
has no $(g,f)$-chromatic spanning trees.

\fig{0.85\textwidth}{!}
{fig_180826_1.pdf}
{The mapping $g$ forces us to use a cycle.}
{fig_180826_1}

The following is the main theorem,
which gives a necessary and sufficient condition
for edge-colored graphs
to have a $(g,f)$-chromatic spanning tree
as a corollary.

\begin{theorem}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$
such that $g(c) \le f(c)$ for any $c \in \mathbb{C}$.
Let $m$ be a positive integer
such that $n \ge m + \sum_{c \in \mathbb{C}}g(c)$.
$G$ has
a $(g,f)$-chromatic spanning forest
with exactly $m$ components
if and only if
\begin{equation*}
\omega(G-E_R(G)) \le
\min
\{~
m+\sum_{c \in R}f(c),~~
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
~\}
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
\label{thm_180815_1}
\end{theorem}

This theorem is proved in Section \ref{proof22}.
Note that
the size of a spanning forest
with exactly $m$ components of $G$ is $n-m$.
If $G$ has a $(g,f)$-chromatic spanning forest
with exactly $m$ components,
then the size of the forest
is at least $\sum_{c \in \mathbb{C}}g(c)$.
Thus,
the condition $n \ge m + \sum_{c \in R}g(c)$
is necessary.

We see the above last example again.
Let $G$ be the left graph in
Fig. \ref{fig_180826_1}.
$G$ has no $(g,f)$-chromatic spanning trees.
Thus, by Theorem \ref{thm_180815_1},
\begin{equation*}
\omega(G-E_R(G)) >
\min
\{~
1+\sum_{c \in R}f(c),~~
8-\sum_{c \in \mathbb{C} \setminus R}g(c)
~\}
\text{~~~ for some } R \subseteq \mathbb{C}.
\end{equation*}
Actually,
for $R = \{ 1,4,5,7 \}$,
$G-E_R(G)$ is the right graph in
Fig. \ref{fig_180826_1}
and we have
\begin{equation*}
\omega(G-E_R(G)) = 4,~~
1+\sum_{c \in R}f(c) = 6,~~
8-\sum_{c \in \mathbb{C} \setminus R}g(c) = 3.
\end{equation*}

We can prove the following theorem
by using Theorem \ref{thm_180815_1}.

\begin{theorem}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
Let $m$ be a positive integer such that $n \ge m$.
If $|E(G)|>\binom{n-m}{2}$ and
\begin{equation*}
g(c) \le \frac{|E_c(G)|}{|E(G)|}(n-m) \le f(c)
\text{~~~~~ for any color } c \in \mathbb{C},
\end{equation*}
then $G$ has a $(g,f)$-chromatic spanning forest
with exactly $m$ components.
\label{thm_180826_1}
\end{theorem}

This theorem is proved in Section \ref{proof23}.
Note that
an $f$-chromatic graph is
a $(g,f)$-chromatic graph with $g(c) = 0$
for any color $c$,
and $\omega(G-E_R(G)) \le n$
for any $R \subseteq \mathbb{C}$
since the number of components of any subgraph
of a graph of order $n$ is at most $n$.
Thus,
Theorem
\ref{thm_180815_1}
and
\ref{thm_180826_1}
include
Theorem
\ref{thm_Suzuki2013AForests_1}
and
\ref{thm_Suzuki2013AForests_2}.

\subsection{%
Color probability distributions
of edge-colored graphs}
\label{sec_Cpd}

We call $|E_c(G)|/|E(G)|$
the \textit{color probability}
of a color $c$
in an edge-colored graph $G$.
The \textit{color probability distribution} of $G$
is the sequence of the color probabilities.
Does a given edge-colored complete graph $G$ have
a spanning tree
with the same color probability distribution
as that of $G$ ?
Fig. \ref{fig_180830_1} shows an example.

\fig{0.75\textwidth}{!}
{fig_180830_1.pdf}
{An edge-colored complete graph
having a spanning tree
with the same color probability distribution
as that of it.}
{fig_180830_1}

Let $G$ and $T$ be the left and right graph
in Fig. \ref{fig_180830_1}, respectively.
Then,
\begin{gather*}
|E_1(G)|=3, |E_2(G)|=3, |E_3(G)|=6, |E_4(G)|=3,
|E(G)|=15,\\
|E_1(T)|=1, |E_2(T)|=1, |E_3(T)|=2, |E_4(T)|=1,
|E(T)|=5.
\end{gather*}
Thus, both color probability distributions are
$(0.2, 0.2, 0.4, 0.2)$
(Fig. \ref{fig_180830_2}).

\fig{0.75\textwidth}{!}
{fig_180830_2.pdf}
{Color probability distributions
of the graphs in Fig. \ref{fig_180830_1}.}
{fig_180830_2}

Fig. \ref{fig_180830_3} shows another example.
In the left complete graph $G$ of order $n=6$
has no spanning trees
with the same color probability distribution
as that of $G$,
because the number of colors in $G$ is $7$
exceeding the size of a spanning tree of $G$.

\fig{0.75\textwidth}{!}
{fig_180830_3.pdf}
{An edge-colored complete graph
having no spanning trees
with the same color probability distribution
as that of it,
but having a spanning tree
with a color probability distribution
similar to that of it.}
{fig_180830_3}

In other words,
for any spanning tree $T$ of $G$,
\begin{equation*}
\frac{|E_c(G)|}{|E(G)|} \ne \frac{|E_c(T)|}{|E(T)|}
\text{~~ for some color } c \in \mathbb{C},
\end{equation*}
that is,
\begin{equation*}
|E_c(T)| \ne \frac{|E_c(G)|}{|E(G)|}|E(T)|
= \frac{|E_c(G)|}{|E(G)|}(n-1)
\text{~~ for some color } c \in \mathbb{C}.
\end{equation*}

However,
the right spanning tree has
a color probability distribution
\textit{similar} to that of $G$
(see Fig. \ref{fig_180830_4}),
in the sense that
the following condition holds:
\begin{equation*}
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rfloor
\le
|E_c(T)|
\le
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rceil
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}

\fig{0.85\textwidth}{!}
{fig_180830_4.pdf}
{Color probability distributions
of the graphs in Fig. \ref{fig_180830_3}.}
{fig_180830_4}

In general,
let $G$ be an edge-colored complete graph
of order $n$,
and set
\begin{equation*}
g(c) =
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rfloor
\text{~~and~~}
f(c) =
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rceil
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
Then, by Theorem \ref{thm_180826_1},
$G$ has a $(g,f)$-chromatic spanning tree $T$.
By the definition \ref{def_180815_1},
$g(c) \le |E_c(T)| \le f(c)$
for any color $c \in \mathbb{C}$.
Thus, the following theorem holds.

\begin{theorem}
Any edge-colored complete graph $G$ of order $n$
has a spanning tree
with a color probability distribution
similar to that of $G$,
that is,
$G$ has a spanning tree $T$
such that
\begin{equation*}
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rfloor
\le
|E_c(T)|
\le
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rceil
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
\label{thm_180913_1}
\end{theorem}

\begin{remark}
Theorem \ref{thm_180913_1} is equivalent
to Theorem \ref{thm_180826_1} with $m=1$
for any edge-colored complete graph $G$
of order $n$.
\label{rem_180922_1}
\end{remark}
\begin{proof}
Theorem \ref{thm_180913_1} follows
from Theorem \ref{thm_180826_1}
by the above argument.

Let $G$ be an edge-colored graph of order $n$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
Suppose that Theorem \ref{thm_180913_1} holds.
Then,
$G$ has a spanning tree $T$
with a color probability distribution
similar to that of $G$.

If $g$,$f$, and $G$ satisfy the condition in Theorem \ref{thm_180826_1},
then we have
\begin{equation*}
g(c) \le
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rfloor
\text{ and }
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rceil
\le
f(c)
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}

Hence,
$T$ satisfies that for any color $c \in \mathbb{C}$,
\begin{equation*}
g(c) \le
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rfloor
\le
|E_c(T)|
\le
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rceil
\le f(c).
\end{equation*}

Therefore,
$T$ is a $(g,f)$-chromatic spanning tree of $G$.
\end{proof}

From Theorem \ref{thm_180913_1},
we can get the following theorem,
proved in Section \ref{proof26}.

\begin{theorem}
An edge-colored complete graph $G$ of order $n$
has a spanning tree
with the same color probability distribution
as that of $G$
if and only if
$|E_c(G)|$ is an integral multiple of $n/2$
for any color $c \in \mathbb{C}$.
\label{thm_180830_1}
\end{theorem}

\subsection{%
Spanning tree
decomposition conjectures}
In 1996,
Brualdi and Hollingsworth
\cite{Brualdi1996MulticoloredGraphs}
presented the following conjecture.

\begin{conjecture}[Brualdi and Hollingsworth
\cite{Brualdi1996MulticoloredGraphs}]
A properly edge-colored complete graph
$K_{2n}$ $(n \ge 3)$
with exactly $2n-1$ colors
can be partitioned into
$n$ edge-disjoint heterochromatic spanning trees.
\label{conj_Brualdi1996MulticoloredGraphs_1}
\end{conjecture}

Brualdi and Hollingsworth
\cite{Brualdi1996MulticoloredGraphs}
proved that
a properly edge-colored complete graph
$K_{2n} (n \ge 3)$
with exactly $2n-1$ colors
has two edge-disjoint heterochromatic spanning trees.
Krussel, Marshall, and Verrall
\cite{Krussel2000SpanningK2n}
proved that the graph
has three edge-disjoint heterochromatic spanning trees.
Kaneko, Kano, and Suzuki
\cite{Kaneko2006TwoGraphs}
proved that
a properly edge-colored complete graph
$K_n (n \ge 5)$
(not necessary with exactly $n-1$ colors)
has two edge-disjoint heterochromatic spanning trees.
Akbari and Alipour
\cite{Akbari2007MulticoloredGraphs}
proved that
an edge-colored complete graph $G$
(not necessary to be proper)
of order $n ( \ge 5)$
has two edge-disjoint heterochromatic spanning trees
if $|E_c(G)| \le n/2$ for any color $c \in \mathbb{C}$.
Carraher, Hartke, and Horn
\cite{Carraher2016Edge-disjointGraphs}
proved that
an edge-colored complete graph $G$
of order $n ( \ge 1000000)$
has at least $\lfloor n/(1000 \log n) \rfloor$
edge-disjoint heterochromatic spanning trees
if $|E_c(G)| \le n/2$ for any color $c \in \mathbb{C}$.
Horn
\cite{Horn2018RainbowOne-factorizations}
proved that
there exist positive constants $\epsilon, n_0$
so that
every properly edge-colored complete graph
$K_{2n} (2n \ge n_0)$
with exactly $2n-1$ colors
has at least $\epsilon n$ edge-disjoint
heterochromatic spanning trees.

Based on these previous results,
I conjecture the following
as a generalization of Conjecture
\ref{conj_Brualdi1996MulticoloredGraphs_1}.

\begin{conjecture}
An edge-colored complete graph $G$
of order $2n$ $(n \ge 3)$
can be partitioned into
$n$ edge-disjoint spanning trees
$T_1, T_2, \ldots, T_n$
such that each has
a color probability distribution
is similar to that of $G$,
that is,
each $T_i$ satisfies that
\begin{equation*}
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(2n-1)
\right\rfloor
\le
|E_c(T_i)|
\le
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(2n-1)
\right\rceil
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
\label{conj_180910_1}
\end{conjecture}

Since $|E(G)|=2n(2n-1)/2$
for the complete graph $G$ of order $2n$,
we have
\begin{equation*}
\frac{|E_c(G)|}{|E(G)|}(2n-1)
= \frac{|E_c(G)|}{n}.
\end{equation*}
Thus, this conjecture implies that
$E_c(G)$ can be partitioned into
$n$ almost equal parts.
Fig. \ref{fig_180910_1}
shows an edge-colored complete graph $G$ of order $6$
and its partition into three edge-disjoint
spanning trees $T_1$, $T_2$, and $T_3$.
In this example,
\begin{gather*}
|V(G)|=2n=6, |E(G)|=15,\\
|E_1(G)|=7, |E_2(G)|=4,
|E_3(G)|=2, |E_4(G)|=2,\\
|E_1(T_1)|=3, |E_2(T_1)|=1,
|E_3(T_1)|=1, |E_4(T_1)|=0,\\
|E_1(T_2)|=2, |E_2(T_2)|=2,
|E_3(T_2)|=0, |E_4(T_2)|=1,\\
|E_1(T_3)|=2, |E_2(T_3)|=1,
|E_3(T_3)|=1, |E_4(T_3)|=1.
\end{gather*}
Thus, $E_c(G)$ is partitioned
into three almost equal parts for each color $c$,
and each $T_i$ $(1 \le i \le 3)$ satisfies that
\begin{equation*}
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(2n-1)
\right\rfloor
\le
|E_c(T_i)|
\le
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(2n-1)
\right\rceil
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
Hence, each $T_i$
has a color probability distribution
similar to that of $G$.

\fig{1.0\textwidth}{!}
{fig_180910_1.pdf}
{An example of Conjecture \ref{conj_180910_1}.}
{fig_180910_1}

By the same argument in the proof of
Remark \ref{rem_180922_1},
we can show that
Conjecture \ref{conj_180910_1} is equivalent
to the following proposition.

\begin{conjecture}
Let $G$ be an edge-colored complete graph
of order $2n$ $(n \ge 3)$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
If for any color $c \in \mathbb{C}$,
\begin{equation*}
g(c) \le \frac{|E_c(G)|}{n} \le f(c),
\end{equation*}
then $G$ can be partitioned into
$n$ edge-disjoint $(g,f)$-chromatic spanning trees.
\label{conj_180916_1}
\end{conjecture}

\section{Proofs}
\label{Proofs}
In this section,
we will prove
Theorem \ref{thm_180815_1},
Theorem \ref{thm_180826_1}, and
Theorem \ref{thm_180830_1}.
In order to prove Theorem \ref{thm_180815_1},
we will use
Lemma \ref{lem_180923_1} and \ref{lem_180923_2},
which will be proved in Section
\ref{proof31} and \ref{proof32}, respectively.
In order to prove Theorem \ref{thm_180826_1},
we will use almost trivial Lemma \ref{lem_180923_3},
which was proved by Suzuki
\cite{Suzuki2013AForests}.

\begin{lemma}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ be a mapping
from $\mathbb{C}$ to $\mathbb{Z}_{\ge 0}$.
$G$ has
a $(g,g)$-chromatic forest
if and only if
\begin{equation*}
\omega(G-E_R(G)) \le
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
\label{lem_180923_1}
\end{lemma}

Note that,
this lemma requires
the forest neither
to be a spanning forest nor
to have a fixed number of components.

\begin{lemma}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$
such that $g(c) \le f(c)$ for any $c \in \mathbb{C}$.
Let $m$ be a positive integer.
$G$ has
a $(g,f)$-chromatic spanning forest
with exactly $m$ components
if and only if
$G$ has both
an $f$-chromatic spanning forest
of size at least
$\sum_{c \in \mathbb{C}}g(c)$
with exactly $m$ components,
and
a $(g,g)$-chromatic forest.
\label{lem_180923_2}
\end{lemma}

Note that,
the $f$-chromatic spanning forest
and the $(g,g)$-chromatic forest
may be different in Lemma \ref{lem_180923_2}.

\begin{lemma}
\begin{equation*}
|E(G)| \le \binom{|V(G)|-\omega(G)+1}{2}
\text{~~ for any graph } G.
\end{equation*}
\label{lem_180923_3}
\end{lemma}

\subsection{Proof of Lemma \ref{lem_180923_1}}
\label{proof31}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ be a mapping
from $\mathbb{C}$ to $\mathbb{Z}_{\ge 0}$.

First,
we prove the necessity.
Suppose that
$G$ has a $(g,g)$-chromatic forest $F$.
By Definition \ref{def_180815_1},
$|E_c(F)|=g(c)$ for any color $c$.
For any $R \subseteq \mathbb{C}$,
the graph $(V(G),E_{\mathbb{C} \setminus R}(F))$
is a spanning forest of $G-E_R(G)$.
Thus,
\begin{align*}
\omega(G-E_R(G))
&\le \omega((V(G),E_{\mathbb{C} \setminus R}(F)))\\
&=|V(G)|-|E_{\mathbb{C} \setminus R}(F)|\\
&= |V(G)|-\sum_{c \in \mathbb{C} \setminus R}|E_c(F)|\\
&= n-\sum_{c \in \mathbb{C} \setminus R}g(c).
\end{align*}

Next,
we prove the sufficiency.
Suppose that
\begin{equation*}
\omega(G-E_R(G)) \le
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
Set $m=n-\sum_{c \in \mathbb{C}}g(c)$.
Then,
\begin{equation*}
n-m=\sum_{c \in \mathbb{C}}g(c)
= \sum_{c \in R}g(c)+\sum_{c \in \mathbb{C} \setminus R}g(c)
\text{~~~ for any } R \subseteq \mathbb{C},
\end{equation*}
that is,
\begin{equation*}
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
= m+\sum_{c \in R}g(c)
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
Thus, we have
\begin{equation*}
\omega(G-E_R(G)) \le
m+\sum_{c \in R}g(c)
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
Hence, by Theorem \ref{thm_Suzuki2013AForests_1},
$G$ has a $g$-chromatic spanning forest $F$
with exactly $m$ components.
By Definition \ref{def_Suzuki2013AForests},
$|E_c(F)| \le g(c)$ for any color $c \in \mathbb{C}$.
On the other hand,
we have
\begin{equation*}
\sum_{c \in \mathbb{C}}|E_c(F)|
= |E(F)|
= n-m
= n-(n-\sum_{c \in \mathbb{C}}g(c))
= \sum_{c \in \mathbb{C}}g(c).
\end{equation*}
Thus, $|E_c(F)| = g(c)$ for any color $c \in \mathbb{C}$.
Therefore,
by Definition \ref{def_180815_1},
$F$ is a $(g,g)$-chromatic forest of $G$.

\subsection{Proof of Lemma \ref{lem_180923_2}}
\label{proof32}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$
such that $g(c) \le f(c)$ for any $c \in \mathbb{C}$.
Let $m$ be a positive integer.

First,
we prove the necessity.
Suppose that
$G$ has
a $(g,f)$-chromatic spanning forest $F$
with exactly $m$ components.
By Definition \ref{def_180815_1},
$g(c) \le |E_c(F)|$ for any color $c \in \mathbb{C}$.
Thus,
$\sum_{c \in \mathbb{C}}g(c)
\le \sum_{c \in \mathbb{C}}|E_c(F)|
= |E(F)|$.
Hence,
$F$ is an $f$-chromatic spanning forest
of size at least
$\sum_{c \in \mathbb{C}}g(c)$
with exactly $m$ components of $G$.
Since $F$ is a $(g,f)$-chromatic forest,
$F$ contains some $(g,g)$-chromatic forest,
which is also a $(g,g)$-chromatic forest in $G$.

Next,
we prove the sufficiency.
Suppose that
$G$ has both
an $f$-chromatic spanning forest
of size at least
$\sum_{c \in \mathbb{C}}g(c)$
with exactly $m$ components,
and
a $(g,g)$-chromatic forest $F_g$.
Let $F_f$ be an $f$-chromatic spanning forest
of size at least
$\sum_{c \in \mathbb{C}}g(c)$
with exactly $m$ components of $G$
such that
it has the maximum number of edges of $F_g$.

We will prove that
$F_f$ is the desired $(g,f)$-chromatic spanning forest
with exactly $m$ components of $G$
by contradiction.

Suppose that
$F_f$ is not a $(g,f)$-chromatic spanning forest
with exactly $m$ components of $G$.
Then, since $F_f$ is $f$-chromatic
but not $(g,f)$-chromatic,
we may assume that for some color, say color $1$,
$|E_1(F_f)| \le g(1)-1$.

Since $F_g$ is $(g,g)$-chromatic,
$|E_1(F_g)|=g(1)$.
Thus, $|E_1(F_f)| < |E_1(F_g)|$.
Hence, $E_1(F_g) \setminus E_1(F_f) \ne \emptyset$.
Let $e$ be an edge in $E_1(F_g) \setminus E_1(F_f)$.
Adding the edge $e$ to $F_f$,
we consider the resulting graph
$(V(F_f),E(F_f) \cup \{ e \})$
denoted by $F_f^+$.
Since $F_f$ is $f$-chromatic and $e \notin E_1(F_f)$,
we have
\begin{equation*}
|E_c(F_f^+)|=
\begin{cases}
|E_c(F_f)|+1 \le g(c) \le f(c)
& \text{if $c=1$},\\
|E_c(F_f)| \le f(c)
& \text{if $c \ne 1$}.
\end{cases}
\end{equation*}
Thus, $F_f^+$ is also an $f$-chromatic
spanning subgraph of $G$.

If the edge $e$ connects two distinct components
of $F_f$ in $F_f^+$,
then $F_f^+$ is an $f$-chromatic spanning forest
with exactly $m-1$ components of $G$.
Since $F_g$ is $(g,g)$-chromatic,
$|E(F_g)| = \sum_{c \in \mathbb{C}}g(c)$.
Since $|E(F_f)| \ge \sum_{c \in \mathbb{C}}g(c)$,
we have
\begin{equation*}
|E(F_f^+)|=|E(F_f)|+1 \ge \sum_{c \in \mathbb{C}}g(c) +1
= |E(F_g)|+1 > |E(F_g)|.
\end{equation*}
Thus,
$E(F_f^+) \setminus E(F_g) \ne \emptyset$.
Let $e'$ be an edge in $E(F_f^+) \setminus E(F_g)$.
Then, we have
\begin{gather*}
\omega(F_f^+ - e')=\omega(F_f^+)+1=m,\\
|E(F_f^+ - e')|=|E(F_f^+)|-1
=|E(F_f)| \ge \sum_{c \in \mathbb{C}}g(c),
\end{gather*}
where $F_f^+ - e'$ denotes the graph
$(V(F_f^+),E(F_f^+) \setminus \{ e' \})$.
Hence,
since $F_f^+$ is an $f$-chromatic spanning forest
of $G$,
$F_f^+ - e'$ is an $f$-chromatic spanning forest
of size at least
$\sum_{c \in \mathbb{C}}g(c)$
with exactly $m$ components of $G$.
Recall that
$e \in E(F_g)$ and $e' \notin E(F_g)$.
Then,
$F_f^+ - e'$, namely,
$(V(F_f),(E(F_f) \cup \{ e \}) \setminus \{ e' \})$
has more edges of $F_g$ than $F_f$,
which is a contradiction
to the maximality of $F_f$.

Therefore, we may assume that
the both endpoints of $e$
are contained in one component of $F_f$.
Then,
$\omega(F_f^+)=\omega(F_f)=m$ and
$F_f^+$ has exactly one cycle $C$,
which contains $e$.
Since $F_g$ has no cycles,
$C$ has some edge $e' \notin E(F_g)$.
Then, $F_f^+ - e'$ is a forest and
\begin{gather*}
\omega(F_f^+ - e')=\omega(F_f^+)=m,\\
|E(F_f^+ - e')|=|E(F_f^+)|-1
=|E(F_f)| \ge \sum_{c \in \mathbb{C}}g(c).
\end{gather*}
Thus,
since $F_f^+$ is an $f$-chromatic spanning subgraph
of $G$,
$F_f^+ - e'$ is an $f$-chromatic spanning forest
of size at least
$\sum_{c \in \mathbb{C}}g(c)$
with exactly $m$ components of $G$.
Recall that
$e \in E(F_g)$ and $e' \notin E(F_g)$.
Then,
$F_f^+ - e'$, namely,
$(V(F_f),(E(F_f) \cup \{ e \}) \setminus \{ e' \})$
has more edges of $F_g$ than $F_f$,
which is a contradiction
to the maximality of $F_f$.

Consequently,
$F_f$ is the desired $(g,f)$-chromatic spanning forest
with exactly $m$ components of $G$.

\subsection{Proof of Theorem \ref{thm_180815_1}}
\label{proof22}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$
such that $g(c) \le f(c)$ for any $c \in \mathbb{C}$.
Let $m$ be a positive integer
such that $n \ge m + \sum_{c \in \mathbb{C}}g(c)$.

First,
we prove the necessity.
Suppose that
$G$ has
a $(g,f)$-chromatic spanning forest $F$
with exactly $m$ components.
Since $F$ is a $(g,f)$-chromatic forest,
$F$ contains some $(g,g)$-chromatic forest.
Thus, by Lemma \ref{lem_180923_1}, we have
\begin{equation*}
\omega(G-E_R(G)) \le
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
On the other hand,
since $F$ is a $(g,f)$-chromatic spanning forest
with exactly $m$ components of $G$,
$F$ is an $f$-chromatic spanning forest
with exactly $m$ components of $G$.
Thus, by Theorem \ref{thm_Suzuki2013AForests_1},
we have
\begin{equation*}
\omega(G-E_R(G)) \le m+\sum_{c \in R}f(c)
\text{~~~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
Therefore,
\begin{equation*}
\omega(G-E_R(G)) \le
\min
\{~
m+\sum_{c \in R}f(c),~~
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
~\}
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}

Next,
we prove the sufficiency.
Suppose that
\begin{equation}
\omega(G-E_R(G)) \le
\min
\{~
m+\sum_{c \in R}f(c),~~
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
~\}
\text{~~~ for any } R \subseteq \mathbb{C}.
\label{eq_180925_1}
\end{equation}

By (\ref{eq_180925_1}), we have
\begin{equation*}
\omega(G-E_R(G)) \le m+\sum_{c \in R}f(c)
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
Thus, by Theorem \ref{thm_Suzuki2013AForests_1},
$G$ has an $f$-chromatic spanning forest $F$
with exactly $m$ components of $G$.
By our assumption
that $n \ge m + \sum_{c \in \mathbb{C}}g(c)$,
we have
\begin{equation*}
|E(F)| = n-m \ge \sum_{c \in \mathbb{C}}g(c).
\end{equation*}
Thus, $F$ is
an $f$-chromatic spanning forest
of size at least
$\sum_{c \in \mathbb{C}}g(c)$
with exactly $m$ components.

On the other hand,
by (\ref{eq_180925_1}), we have
\begin{equation*}
\omega(G-E_R(G)) \le n-\sum_{c \in \mathbb{C} \setminus R}g(c)
\text{~~~ for any } R \subseteq \mathbb{C}.
\end{equation*}
Thus, by Lemma \ref{lem_180923_1},
$G$ has a $(g,g)$-chromatic forest.

Therefore,
by Lemma \ref{lem_180923_2},
$G$ has
a $(g,f)$-chromatic spanning forest
with exactly $m$ components.

\subsection{Proof of Theorem \ref{thm_180826_1}}
\label{proof23}
Let $G$ be an edge-colored graph of order $n$.
Let $g$ and $f$ be mappings from $\mathbb{C}$
to $\mathbb{Z}_{\ge 0}$.
Let $m$ be a positive integer such that $n \ge m$.
Suppose that $|E(G)|>\binom{n-m}{2}$ and
\begin{equation}
g(c) \le \frac{|E_c(G)|}{|E(G)|}(n-m) \le f(c)
\text{~~~~~ for any color } c \in \mathbb{C}.
\label{eq_180926_1}
\end{equation}
Then, since $\sum_{c \in \mathbb{C}}|E_c(G)|=|E(G)|$,
we have
\begin{equation}
\sum_{c \in \mathbb{C}}g(c)
\le
\sum_{c \in \mathbb{C}}\frac{|E_c(G)|}{|E(G)|}(n-m)
= n-m,
\text{ that is, }
n \ge m + \sum_{c \in \mathbb{C}}g(c).
\label{eq_180926_2}
\end{equation}

We will prove that
$G$ has a $(g,f)$-chromatic spanning forest
with exactly $m$ components
by contradiction.

Suppose that
$G$ has no $(g,f)$-chromatic spanning forests
with exactly $m$ components.
By (\ref{eq_180926_2}) and our assumption,
we can apply Theorem \ref{thm_180815_1} to $G$
and we have
\begin{equation*}
\omega(G-E_R(G)) >
\min
\{~
m+\sum_{c \in R}f(c),~~
n-\sum_{c \in \mathbb{C} \setminus R}g(c)
~\}
\text{~~~ for some } R \subseteq \mathbb{C}.
\end{equation*}

That is,
$\omega(G-E_R(G)) \ge m+\sum_{c \in R}f(c) +1$ or
$\omega(G-E_R(G)) \ge
n-\sum_{c \in \mathbb{C} \setminus R}g(c) +1$
for some $R \subseteq \mathbb{C}$.
We denote $G-E_R(G)$ by $G'$.

\begin{claim}
\begin{equation*}
\omega(G') \ge m+1
\text{~ and ~}
\omega(G') \ge n+1- \frac{|E(G')|}{|E(G)|}(n-m)
\end{equation*}
\label{clm_180926_1}
\end{claim}
\begin{proof}
First, we suppose that
$\omega(G') \ge m+\sum_{c \in R}f(c)+1$
for some $R \subseteq \mathbb{C}$.
Since $f(c) \ge 0$ for any color $c$,
$\omega(G') \ge m+\sum_{c \in R}f(c)+1 \ge m+1$.

By our assumption (\ref{eq_180926_1}),
\begin{align*}
\sum_{c \in R}f(c)
&\ge
\sum_{c \in R}\frac{|E_c(G)|}{|E(G)|}(n-m)
= \frac{n-m}{|E(G)|}\sum_{c \in R}|E_c(G)|
= \frac{n-m}{|E(G)|}|E_R(G)|\\
&= \frac{n-m}{|E(G)|}(|E(G)|-|E(G')|)
= n-m-\frac{|E(G')|}{|E(G)|}(n-m).
\end{align*}
Thus, we have
\begin{align*}
\omega(G')
&\ge m+\sum_{c \in R}f(c)+1\\
&\ge m+ n-m-\frac{|E(G')|}{|E(G)|}(n-m) +1
= n+1- \frac{|E(G')|}{|E(G)|}(n-m).
\end{align*}

Next, we suppose that
$\omega(G') \ge
n-\sum_{c \in \mathbb{C} \setminus R}g(c) +1$
for some $R \subseteq \mathbb{C}$.
By (\ref{eq_180926_2}),
$\sum_{c \in \mathbb{C}}g(c) \le n-m$.
Thus, we have
\begin{align*}
\omega(G')
\ge n-\sum_{c \in \mathbb{C} \setminus R}g(c) +1
\ge n-\sum_{c \in \mathbb{C}}g(c) +1
\ge n-(n-m)+1
= m+1.
\end{align*}

By our assumption (\ref{eq_180926_1}),
\begin{equation*}
\sum_{c \in \mathbb{C} \setminus R}g(c)
\le
\sum_{c \in \mathbb{C} \setminus R}\frac{|E_c(G)|}{|E(G)|}(n-m)
= \frac{|E_{\mathbb{C} \setminus R}(G)|}{|E(G)|}(n-m)
= \frac{|E(G')|}{|E(G)|}(n-m).
\end{equation*}
Thus, we have
\begin{align*}
\omega(G')
&\ge n-\sum_{c \in \mathbb{C} \setminus R}g(c) +1\\
&\ge n-\frac{|E(G')|}{|E(G)|}(n-m) +1
= n+1- \frac{|E(G')|}{|E(G)|}(n-m).
\end{align*}
\end{proof}

By Claim \ref{clm_180926_1},
\begin{equation*}
n-\omega(G')+1 \le \frac{|E(G')|}{|E(G)|}(n-m).
\end{equation*}
Since $n \ge \omega(G')$,
$n-\omega(G')+1 \ge 1$,
that is, $n-\omega(G')+1 \ne 0$.
Thus,
\begin{align*}
|E(G)|
&\le
\frac{n-m}{n-\omega(G')+1}|E(G')|.
\intertext{
Since $|V(G')|=|V(G)|=n$,
by Lemma \ref{lem_180923_3},}
|E(G)|
&\le
\frac{n-m}{n-\omega(G')+1}
\binom{|V(G')|-\omega(G')+1}{2}\\
&\le
\frac{n-m}{n-\omega(G')+1}
\times \frac{(n-\omega(G')+1)(n-\omega(G'))}{2}\\
&=
\frac{(n-m)(n-\omega(G'))}{2}.
\intertext{
By Claim \ref{clm_180926_1},
$\omega(G') \ge m+1$.
Thus,}
|E(G)|
&\le
\frac{(n-m)(n-(m+1))}{2}
= \binom{n-m}{2},
\end{align*}
which contradicts to our assumption
that $|E(G)|>\binom{n-m}{2}$.

Therefore,
$G$ has a $(g,f)$-chromatic spanning forest
with exactly $m$ components.

\subsection{Proof of Theorem \ref{thm_180830_1}}
\label{proof26}
If an edge-colored complete graph $G$ of order $n$
has a spanning tree $T$
with the same color probability distribution
as that of $G$,
that is,
\begin{equation*}
\frac{|E_c(G)|}{|E(G)|} = \frac{|E_c(T)|}{|E(T)|}
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
then
\begin{equation*}
|E_c(G)| = \frac{|E_c(T)||E(G)|}{|E(T)|}
= \frac{|E_c(T)|n(n-1)/2}{n-1}
= \frac{|E_c(T)|n}{2}
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
Thus, since $|E_c(T)|$ is an integer,
$|E_c(G)|$ is an integral multiple of $n/2$.

Next, let $G$ be
an edge-colored complete graph $G$ of order $n$.
For any color $c \in \mathbb{C}$,
we suppose that
$|E_c(G)| = k_c \times n/2$
for some $k_c \in \mathbb{Z}_{\ge 0}$.
By Theorem \ref{thm_180913_1},
$G$ has a spanning tree $T$
such that
\begin{equation*}
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rfloor
\le
|E_c(T)|
\le
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rceil
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
Since $|E(G)|=n(n-1)/2$ and
$|E_c(G)| = k_c \times n/2$
($k_c \in \mathbb{Z}_{\ge 0}$),
we have
\begin{equation*}
k_c
=
\lfloor k_c \rfloor
=
\left\lfloor
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rfloor
\le
|E_c(T)|
\le
\left\lceil
\frac{|E_c(G)|}{|E(G)|}(n-1)
\right\rceil
=
\lceil k_c \rceil
=
k_c.
\end{equation*}
Thus, $|E_c(T)|=k_c$.
Then,
\begin{equation*}
\frac{|E_c(G)|}{|E(G)|}
= \frac{k_c \times n/2}{n(n-1)/2}
= \frac{k_c}{n-1}
= \frac{|E_c(T)|}{|E(T)|}
\text{~~ for any color } c \in \mathbb{C}.
\end{equation*}
Therefore,
the color probability distribution of $T$
is the same as that of $G$.

\bigskip
\noindent
\textbf{Acknowledgments}
This work was supported by
JSPS KAKENHI Grant Number 16K05254.


text:
Lions WR Kenny Golladay expected to play Week 3 vs. Cardinals
The Lions will get back their best offensive weapon this week.
By Jeremy Reisman@DetroitOnLion Sep 21, 2020, 8:17am EDT
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Photo by Christian Petersen/Getty Images
Chances are if you're a Detroit Lions fan and you're here, you could probably use some good news.
And while I can't sit here and tell you that the team is about to turn everything around and save the season, I can tell you that one of their most significant pieces on offense is about to return to the lineup after missing the first two weeks.
According to ESPN's Adam Schefter, Lions Pro Bowl receiver Kenny Golladay is expected to play next week against the Arizona Cardinals. Per Schefter, Golladay was close to playing this past Sunday against the Packers, but the Lions simply didn't want to risk aggravating the injury. Head coach Matt Patricia spoke vaguely about this a couple weeks ago.
"We're trying to prevent them from being long-term type of soft-tissue injuries," Patricia said. "So even if we get guys that are tight, that may be on the milder side, maybe a seven-, 10-, 14-day type stuff, we're trying to prevent the four-week type of injury there."
Golladay's return should help out a receiving corps that has struggled in his absence. Marvin Jones Jr. has just 78 yards in two games, while Marvin Hall only has 50. Rookie Quintez Cephus is the only receiver who really appears to have stepped up in Golladay's absence, hauling in 97 yards on six catches.
The Lions are particularly hurting in their ability to stretch the field like they did in 2019. After throwing for eight plays of 40+ yards in eight games last season (three of which went to Golladay), Matthew Stafford hasn't connected on a single one through two games.
Golladay is coming of a breakout year in which he led the league in touchdown catches (11) and set a career high with 1,190 yards.

redpajama_set_name: RedPajamaCommonCrawl


text:
Police Investigate Man For Criticising UKIP On Twitter
Officers from Cambridgeshire Police visit a man's home after he fact-checked a list of ten of the eurosceptic party's policies on Twitter.
Luke Macgregor / Reuters / Reuters
Cambridgeshire police have confirmed that they visited a man after he criticised UKIP on Twitter.
Michael Abberton posted a tweet last Monday in which he fact-checked ten controversial UKIP policies.
Officers came to his home on Saturday after receiving a complaint from an unknown party. The message had been retweeted hundreds of times.
Abberton gave an account of the incident on his blog, explaining how he was visited by two officers after he re-posted the list on the social network.
He claims the police told him he could not tweet about the visit.
This is the original tweet which prompted the police visit.
Michael Abberton @MichaelAbberton
10 Great Reasons to vote #UKIP. I don't know who made it - so I referenced it from official #UKIP websites
12:48 PM - 05 May 2014
Michael Abberton@MichaelAbbertonFollow
10 Great Reasons to vote #UKIP. I don't know who made it - so I referenced it from official #UKIP websites
01:48 PM - 5 May 14ReplyRetweetFavorite
Cambridgeshire Police confirmed to BuzzFeed that they had visited the man's home over the incident and didn't dispute his version of events.
According to a spokesperson the police were investigating a potential breach of the law that governs misrepresentation in election publications:
We were called with a complaint about a message on social media at about 12.40pm on Friday. Inquiries were made as to whether any offences had been committed under the Representation of the People Act but none were revealed and no further action was taken.
The police would not confirm whether they had been called by UKIP, or another member of the public. A spokesperson said:
We're always called by individuals, whether they represent an organisation or just themselves. We would not give any more information about the individual who made the call.


Anyone? Anyone? Bueller? This website really needs work!
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Aspiring to scuba my dream is to be so submerged in nature that i can't hardly breathe, if not for the tank. The beauty of diving so far with only luminous fish and coral around me is what drives me to learn about it.
Does thalos belong in this group?


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Southampton's Band Of Skulls fifth album release, Love Is All You Love, is the follow-up to 2016's By Default. The new album was produced by Richard X and was recorded in London's Miloco Studios, The Smoakstack studio in Nashville, and in the Southampton studio of Band Of Skulls founder members' Russell Marsden (guitar, vocals) and Emma Richardson (bass, vocals). Following the amicable departure of drummer Matt Hayward, Russell and Emma started the writing process for the record in 2017 and quickly accumulated a host of new songs. It was a free and experimental process mixing electronic and live instruments, working quickly, getting ideas down, then moving on.
Work then switched to The Toyshop studio at Miloco to collaborate with Richard X, known for his innovative production with artists from Pet Shop Boys to Goldfrapp to Sugababes, whose "Freak Like Me" was one of the standout singles of the 2000s. Richard X encouraged Band Of Skulls new spirit of experimentation, pushing the combination of live and electronic as mixing the two was creating interesting dynamics. Once the writing was complete the band decamped to Nashville to record the live element of the record with Eagles Of Death Metal's Julian Dorio on drums.
All these elements added a heavy punch and power to Love Is All You Love, especially on the unrelenting "Carnivorous" and short sharp shock of "That's My Trouble". "Cool Your Battles" is more widescreen, while "We're Alive" has a heavy dance pulse with a touch of melancholy. The song "Love Is All You Love" focuses on the groove and keeps the feel of when they tracked it live in Nashville. This song in particular seems to encapsulate the sentiment of the whole record.
"This record is tough and tender with a rebellious spirit", say Emma and Russell. "It is a positive record. It looks to what is possible and how a little love and human kindness can help bring about positive change. We drew on personal experiences, changes in our situation, and couldn't help but be affected by what has been going on in the world politically, culturally and environmentally. Attempting to overcome that sometimes powerless feeling by creating and connecting with people and making something new".


- Print with minimal noise levels.
- Work faster with One Touch Screen Print.
- Enjoy a Hi-Speed USB 2.0 interface.
- Print in less than 8.5 seconds with Ready Mode.
20 ppm, Time to first page (black, normal): 8.5 s.
Standard input capacity: 150 sheets.
Standard output capacity: 100 sheets.


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Home»Review»A Review of Elon Musk's Relationship With Venture Capitalist Shivon Zilis
A Review of Elon Musk's Relationship With Venture Capitalist Shivon Zilis
Shivon Zilis is a Canadian venture capitalist. She works in the artificial intelligence and technology fields and is a mother of nine. She has worked with Elon Musk in the past and has been accused of violating ethical boundaries. Nevertheless, she has defended her relationship with Musk.
Elon Musk's relationship with shivon zilis violates ethical boundaries
shivon zilis has nine children
Shevon zilis is an artificial intelligence expert
shivon zilis works for Tesla
There have been mixed reactions to Elon Musk's relationship with Neuralink executive Shivon Zilis, the co-CEO of the rocket-development company. Some experts have argued that the couple's relationship may have violated Neuralink's corporate ethics. Others say it was an unorthodox breach of professional conduct.
According to court documents obtained by Business Insider, Elon Musk and Zilis had twins last November. While the two haven't publicly spoken about the twins, their relationship has raised ethical questions in the workplace. In addition to the twins, the couple has nine other children together. Shivon Zilis is currently a director at Neuralink, where Elon Musk is co-CEO. She previously worked at Tesla for two years.
The relationship between Musk and Zilis is particularly concerning given that both are high-level executives. Zilis reportedly shares Musk's interest in AI and cryptocurrency. In addition to sharing his interests in artificial intelligence, the two have also tweeted about population issues. In May, Musk shared a graphic from the Wall Street Journal that showed a fall in U.S. fertility rates.
The relationship is also disturbing because Zilis works for Musk's Neuralink brain chip company. Zilis works as the director of special projects and operations at Neuralink, where the father was the founder. Several other colleagues said that she told colleagues that she and Musk conceived the twins using IVF. Five people with knowledge of the situation have confirmed the story.
Elon Musk is rumored to be the father of nine children, including twins he welcomed with his business partner, Shivon Zilis. The couple expanded their family with the help of a surrogate in November 2021, according to court documents. A judge in Austin, Texas, approved the arrangement in May, but the children's names are not yet known. Elon and Zilis did not respond to a request for comment.
The news comes as the couple, who are based in San Francisco, recently moved to Austin to raise their twins. While Zilis was living in San Francisco, the couple purchased a home in a gated Austin community before they had their twins. They then filed legal documents to change the names of the twins to reflect the father.
Elon Musk's twins were conceived in IVF, a procedure in which embryos are fertilized in the lab. Shivon Zilis is a senior executive at the company Neuralink, where she works as the director of operations and special projects. She told her colleagues that she conceived her children through IVF, but Reuters could not independently verify her story.
After graduating from Yale University, Shivon Zilis worked as a project director at Tesla until the company was acquired by Elon Musk. She is also on the board of OpenAI, another company co-founded by Musk. She was also mentioned in Forbes' list of 30 under thirty in venture capital. Her work is related to human-machine interfaces.
Zilis met Musk through OpenAI, a nonprofit that promotes artificial intelligence research and deployment. She served as the youngest member of OpenAI's board of directors. She later joined Tesla as a project director and deployed her AI expertise on the Autopilot and chip-design teams. Musk has rumored that Zilis will run Twitter after its acquisition by Alphabet.
Shivon Zilis is an artificial intelligence expert and an entrepreneur. She was born and raised in Canada. She studied at Yale University, where she played ice hockey. After college, she moved to California, where she worked in various fields, including the technology industry. She currently lives in Austin, Texas.
Zilis studied economics and philosophy at Yale University. She has worked at several companies, including IBM and Tesla. She was named a director of operations for Neuralink, a neurotechnology firm co-founded by Elon Musk. In 2015, she met Musk, and the two became friends. In the same year, she welcomed twins with Musk.
Zilis has a private Instagram account with only a few followers, but later removed it following news of her pregnancy. However, she has over seventy thousand followers on Twitter. Zilis is a Canadian-American. She is currently the director of operations for Neuralink, a neurotech start-up that was co-founded by Elon Musk and chairs by Elon Musk.
Previously, Zilis was an IBM researcher and worked on financial technologies for three years. After that, she joined Bloomberg Beta, a venture capital firm, where she led investments in machine learning and data. She also served as a board director for several AI startups. Ultimately, Zilis hopes to use AI for the good of society.
Shivon Zilis is a noted artificial intelligence expert and has been associated with several of Elon Musk's projects. She is currently the director of Neuralink, a company which develops implantable brain chips to improve cognitive function. Zilis and Musk have been married for five years, but Zilis has been secretive about her pregnancy.
Musk is the richest man on the planet. He is also a tech entrepreneur and investor. He is the CEO of SpaceX, Tesla, The Boring Company, and Neuralink, as well as an artificial intelligence expert. He is also married to Shivon Zilis, a Neuralink top rank holder.
Shivon Zilis is an artificial intelligence expert and has worked for IBM and Bloomberg Beta. In 2015, she was named to the Forbes 30 Under 30 list. She later became a director of OpenAI, Musk's artificial intelligence research nonprofit. Zilis has worked on projects for Tesla and Neuralink, and has been an active participant on Twitter.
Zilis previously worked as a Project-Director for Tesla, where she worked under Elon Musk. She currently works for Neuralink, another company that Musk founded in 2016. She was a member of the Yale University women's ice hockey team and was active in the Creative Destruction Lab. She was listed on Forbes' 30 under 30 list and has been recognized by LinkedIn as one of the world's most influential people under the age of 30.
Musk and Shivon have been married since October 2015. Their first child was born in March 2015, while their second child was delivered via surrogate. After the birth of their first child, the Musks kept their pregnancy a secret until March of this year, when a Vanity Fair reporter heard the lone cry of a baby upstairs. Musk then promoted Zilis to project director for Tesla's Neuralink company.
Zilis is a noted artificial intelligence expert. She has also been a top executive for Tesla's Neuralink firm, which develops implantable brain-machine interfaces. Zilis, who received a bachelor's degree from Yale University, worked as a Project Director for the company for five years. In addition to her work at Tesla, Zilis has served as a Fellow for the Creative Destruction Lab since 2015.
Zilis met Musk while working at Tesla, where she served as a Project Director. She is now working on a new company that uses neural networks to improve autonomous cars. Zilis met Musk through a mutual connection to OpenAI. The two married in 2015, and Musk has two children with Boucher.
After Zilis's pregnancy, Musk and Zilis have remained on good terms. They have continued to work together. Zilis helped Musk set up a meeting with the CEO of Synchron. The couple's second child was welcomed via surrogate in December 2021.
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text:
Big tobacco firms source of two-thirds of smuggled cigarettes, study finds
Illicit Firearms Trade
Illicit Tobacco & Alcohol Trade
Illicit Pharmaceuticals Trade
Illicit Goods Trade
Illicit Fuel Trade
Crime Types
Guillaume Goudreau
New research from academics at the University of Bath in the UK has revealed that big cigarette firms are continuing to facilitate the smuggling of tobacco while funding studies that routinely overestimate the size of the illicit trade in its products.
Published in BMJ journal Tobacco Control and based on leaked documents, a report from the university's Tobacco Control Research Group (TCRG) shows how the industry has gone to great lengths to undermine the Illicit Trade Protocol (ITP), a treaty signed in 2012 that aims to eliminate the illicit trade in tobacco products.
The establishment of the ITP involved the introduction of a reliable tracking mechanism that would allow for the origin of all tobacco products to be traced back through supply chains to their source should they be found being offered on the black market.
According to the leaked documents on which the Bath research is based, the world's four largest tobacco firms developed a plan to use front organisations and third party entities to promote their own track and trace system to governments across the globe, misleadingly giving the impression the system was independent of the industry.
The study also found the tobacco industry remains the source of roughly two-thirds of all smuggled cigarettes across the globe, suggesting tobacco firms are failing to fulfil their responsibilities under the ITP treaty, and may even be actively involved in the smuggling trade.
Commenting on the findings of the study, TCRG Director Professor Anna Gilmore said: "This has to be one of the tobacco industry's greatest scams: not only is it still involved in tobacco smuggling, but big tobacco is positioning itself to control the very system governments around the world have designed to stop it from doing so.
"The industry's elaborate and underhand effort involves front groups, third parties, fake news and payments to the regulatory authorities meant to hold them to account."
In March, former New York Mayor Michael Bloomberg committed to spending $20 million on founding a new international watchdog to monitor the deceptive practices employed by big tobacco companies.
Announcing his plans at the World Conference on Tobacco or Health in Cape Town, Bloomberg said: "Over the last decade tobacco control measures have saved nearly 35 million lives, but as more cities and countries take action, the tobacco industry is pushing to find new users, particularly among young people.
"We cannot stand by as the industry misleads the public in an effort to get more people hooked on its products — and this global watchdog will help us hold the industry accountable."
Malaysia needs new anti-smuggling task force to prevent massive tax losses, MICCI says
Multi-million dollar modern-day cigarette bootlegging gang arrested in New York
Record haul of 9.5 million smuggled cigarettes seized by customs officers in Western Australia
Global UN-backed crackdown on illicit tobacco comes into force in September
Polish police arrest five after seizing 39 million cigarettes during Europol-backed operation
Related Topics:_featuredbig tobaccocigarette smugglingTobacco Control Research Group
Five members of sham marriage gang arrested in Poland and Romania during Europol-backed operation
Police find major narcotunnel running across Mexican border to abandoned KFC restaurant
Crooked vendors exploiting flaw in eBay's feedback system to con buyers into purchasing bogus and dangerous items
Matúš Tóth
Buyers on eBay are being duped into purchasing substandard and counterfeit products due to a flaw in the online auction platform's seller feedback system, according to an investigation conducted by UK consumer group Which?
The watchdog found that dishonest vendors can take advantage of these flaws by linking positive reviews of genuine products manufactured by companies such as Apple and Samsung to fake and low-quality items.
Which? found that crooked sellers are able to link thousands of positive reviews to eBay listings they have nothing to do with.
The organisation discovered that real reviews can be associated with fake products that are potentially dangerous, such as counterfeit mobile phone chargers that can pose a fire risk.
Sellers are able to do this by using "product IDs" associated with genuine items when adding their products to eBay, subsequently benefitting from the positive reviews those items have attracted.
The system is intended to make the process of listing products on eBay quicker and easier by allowing sellers to pull information from similar items that have a linked product ID.
As part of its investigation, Which? purchased 20 bogus Apple and Samsung accessories such as chargers and USB cables that were supposed to be official and shared the same reviews as products manufactured by the two technology firms
Calling for online ecommerce platforms to be held accountable for flaws in their seller feedback systems that allow dishonest vendors to pull the wool over buyers' eyes, Head of Home Products and Services at Which? Natalie Hitchins said: "Our investigation has uncovered yet another example of online reviews being manipulated to mislead people.
"eBay's product review system is confusing for consumers and could even direct them towards counterfeit or dangerous products sold by unscrupulous sellers.
"Online reviews influence billions of pounds of consumer spending each year.
"The [UK Competition and Markets Authority] must now investigate how fake and misleading reviews are duping online shoppers, taking the strongest possible action against sites that fail to tackle the problem."
Responding to the findings of Which?'s investigation eBay said in a statement: "The research does not fully consider that there are distinctions between product reviews (which provide buyers with a holistic review of the same product), and seller feedback (which can be used to see specific reviews of a seller's performance and may reflect the item's condition)."
Earlier this month, Bloomberg reported that US politicians had called on lawmakers to hold ecommerce companies such as eBay and Amazon to account if they fail to prevent third-party vendors selling counterfeit or substandard products on their platforms.
Major 'lover boy' prostitution gang broken up by coalition of European law enforcement agencies
A Romanian human trafficking and prostitution network that used the "lover boy" method to entrap young women before forcing them into sex work has been broken up a coalition of European law enforcement agencies.
The lover boy method, also known as the "Romeo pimp" method, involves young men seducing victims with the objective of coercing them into prostitution.
Lover boy traffickers groom their victims to believe they have entered into a serious romantic relationship before using emotional, psychological and sometimes physical abuse to intimidate them into working in the sex services industry.
Investigators from Spain, Romania, the Czech Republic and several other European nations were involved in the operation that resulted in the dismantling of the gang, which is said to have groomed and exploited at least 10 young women by forcing them to work as prostitutes.
The operation resulted in the arrest of 14 people in Romania and Spain, the safeguarding of 10 trafficking victims, and the confiscation of a number of items, including a quantity of cash, jewellery, expensive vehicles and several electronic devices.
In total, the agencies taking part in the effort raided 16 properties in the Czech Republic, Romania and Spain.
Having groomed their victims, Romanian members of the network would develop manipulative dependent relationships with the young women they targeted before forcing them into sex work.
Once under the traffickers' control, victims would be abused and drugged before being sold onto other members of the network for as much as €6,000 ($6,632) each.
The women would then be moved between locations and countries on a regular basis as part of the gang's efforts to avoid the attention of police.
Profits made by the network were laundered through the purchase of property, expensive jewellery and high-value cars.
Ongoing investigations into the network's activities are focussed on the theory that it was working in cooperation with another gang.
Enquires have already resulted in the identification of more than 40 additional women who fell victim to the two criminal organisations.
In a statement, Europol said: "Europol facilitated the information exchange between the participating countries, provided coordination support and analysed operational information against Europol's databases to give leads to investigators.
"Europol conducted a financial analysis based on the information provided which highlighted the extension of the criminal activity of the group and the presence and flow of illicit profits to other jurisdictions."
Taking cocaine will not cure people struck down with the coronavirus, French government warns public
Authorities in France have been forced to inform the public that taking cocaine will not cure people infected with the coronavirus.
Taking to Twitter on Sunday, the French Ministry for Solidarity and Health told its followers that cocaine is not only ineffective when it comes to fighting the coronavirus, but is also a highly addictive drug that can cause serious harm to users' health.
The government department was seeking to counter fake news circulating on social media that taking the drug could cure or prevent the virus, including doctored news stories that appeared to confirm the drug's effectiveness at fighting the disease.
The ministry's Twitter post included a link to a government information page that provided further guidance on disinformation circulating about the coronavirus outbreak.
As well as encouraging those worried about the coronavirus to start taking cocaine, online trolls have also suggested that bleach can also help fight the disease.
In a post on Twitter that has attracted many thousands of engagements, @Jordan_Sather_ told his followers: "Would you look at that. Not only is chlorine dioxide (aka 'MMS') an effective cancer cell killer, it can wipe out coronavirus too.
"No wonder YouTube has been censoring basically every single video where I discuss it over the last year."
In August of 2019, the US Food and Drug Administration issued a warning about that dangers of consuming bleach, noting: "Drinking any… chlorine dioxide products can cause nausea, vomiting, diarrhoea, and symptoms of severe dehydration."
As well as warning about cocaine's inability to fight the coronavirus, the French government has also told members of the public that spraying bleach or alcohol on their bodies will not neutralise viruses they have already been infected with.
Elsewhere, US Vodka maker Tito's Homemade was last week forced to urge people not to make DIY hand sanitiser out of its products.
Responding to one of its customers who said they had done just that, the company said on Twitter: "Per the CDC [Centres for Disease Control and Prevention], hand sanitizer needs to contain at least 60% alcohol. Tito's Handmade Vodka is 40% alcohol, and therefore does not meet the current recommendation of the CDC. Please see attached for more information."
For its part, the World Health Organisation, which today officially categorised the coronavirus as a pandemic, has published a webpage dispelling misinformation about the disease, noting that the virus cannot be killed of avoided by taking a hot bath or using hand dryers.
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US politicians call for state action against Pornhub over allegations it hosted rape and child abuse videos
Californian border officers catch Mexican man with enough fentanyl to kill 1.2 million people
Tweets by @illicit_trade_


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text:
Q: How to view active query underlying cursor In pg_stat_activity I can see that a client is working its way through some query results using a cursor. But how can I see what the original query is?
pipeline=> select pid, query from pg_stat_activity where state = 'active' order by query_start;
  pid  |                                                         query
-------+--------------------------------------------------------------------------------------
  6734 | FETCH FORWARD 1000 FROM "c_109886590_1"
 26731 | select pid, query from pg_stat_activity where state = 'active' order by query_start;
(2 rows)

I see there is pg_cursors, but it is empty:
pipeline=> select * from pg_cursors;
 name | statement | is_holdable | is_binary | is_scrollable | creation_time
------+-----------+-------------+-----------+---------------+---------------
(0 rows)

The server is on AWS RDS.
pipeline=> select version();
                                                   version
--------------------------------------------------------------------------------------------------------------
 PostgreSQL 9.3.3 on x86_64-unknown-linux-gnu, compiled by gcc (GCC) 4.6.3 20120306 (Red Hat 4.6.3-2), 64-bit
(1 row)

A: You can't.
pg_cursors is backend-local. It doesn't show cursors that aren't part of the current connection.
PostgreSQL has no way to find out what query underlies a cursor from another session.
The only way I can think of to do this is using log analysis, with log_statement = all and a suitable log_line_prefix.

redpajama_set_name: RedPajamaStackExchange


Edgo has been through many changes since its inception in 1956, but the most important things about it have remained the same. Throughout our operating history, providing oilfield and support services to the oil and gas industry, we have combined the skills and tenacity that are required to ensure success in physically harsh and politically complex environments. We maintain and promote the pioneering spirit and resilience that helped us evolve into a highly diverse and successful multidisciplinary business and leading service provider to the oil and gas industry.
Our continuing success is founded on four key areas: the skills and expertise of our people; our ability to deliver fully managed programmes; effective local partnerships; and long-established roots in the communities we serve. These features, combined with a long-term commitment to serving customers and collaborating with partners in the oil and gas industry, have made Edgo what it is today.
In all of our operations, we are focused on the principle of maximising local content, and encouraging the development of indigenous skills and expertise. We demonstrate this through the training of local staff and investment in the communities where we work. This approach has always been central to our company ethos.


text:
Andhra Pradesh Crime News Entertainment Gallery Hyderabad News India News
BJP Leader Slams Own Government In Karnataka
News Desk Mar 28, 2022
A senior BJP leader in Karnataka has slammed right-wing organisations' call to ban Muslim traders from temple premises as a retaliatory gesture to the hijab row. H Vishwanath, a member of Karnataka's legislative council, has also accused…
Bengal CM Mamata Banerjee targets BJP
Syed Mohiuddin Mar 23, 2022
Kolkata: Targeted by the BJP over the killing of eight people in Bengal's Birbhum, Chief Minister Mamata Banerjee today said "such incidents have happened in Gujarat and Rajasthan too" and assured action "in a fair manner". She also said…
Delhi CM Arvind Kejriwal Slams BJP
New Delhi, March 23: Delhi Chief Minister Arvind Kejriwal today attacked the BJP over the "postponement" of municipal elections, saying that AAP will quit politics if BJP gets these polls conducted timely and wins them. His remarks came…
Sonia Gandhi asks state party chiefs to resign after poll debacle
New Delhi, March 15: Heads have started to roll after the poll Debacle in five states as Congress interim president Sonia Gandhi has asked state party presidents to put in their papers. The party suffered massive defeats in five states…
Nepotism would not be allowed in BJP: PM Modi
New Delhi, March 15: In a strong message to BJP Members of Parliament, Prime Minister Narendra Modi on Tuesday said that nepotism would not be allowed in the party. Addressing the first meeting of the BJP parliamentary party after its…
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"Instacart covers 8 wards in DC and delivers groceries from four major stores: Costco, Harris Teeter, Safeway and Whole Foods"
Prince Of Petworth April 1, 2014 at 2:30pm
Photo by PoPville flickr user Mr.TinDC
From a press release:
"Instacart (www.Instacart.com), the only service that can deliver groceries from multiple local stores within an hour, announced today that it is adding delivery service from Safeway for the Washington, DC area.
Instacart also expanded its delivery area to include neighborhoods Seven Corners and Bailey's Crossroads in Falls Church, VA. These new areas have been added to Instacart's existing delivery area which includes all of D.C., nearby areas of Virginia (Arlington, Alexandria, and McLean), as well as neighborhoods of MD (Takoma Park, Chevy Chase, Bethesda, and Silver Spring).
In the roughly six weeks since the same-day grocery delivery service launched in the DC area, Instacart has grown its user base by 15 times, now covers 8 wards in DC and 7 neighborhoods in VA and MD, and delivers groceries from four major stores: Costco, Harris Teeter, Safeway and Whole Foods.
Originally launched in San Francisco and later expanded to Chicago and Boston, Instacart has disrupted the traditional grocery delivery space with its crowdsourced labor force dubbed "Personal Shoppers" who shop for and deliver grocery orders using their own vehicles in as little as one hour. This eliminates the need for costly infrastructure such as warehouses, trucks and full-time drivers. Instacart is not affiliated with the stores; its model is to connect customers with Personal Shoppers who shop on their behalf.
Customers can open an account at www.Instacart.com, and get free delivery on their first order of $35 or more.
"Safeway was a very popular request from our DC area customers and gives them a great range of options for groceries; it was a logical choice for a new store addition," said Instacart founder and CEO Apoorva Mehta. "We are very happy with the enormous growth we've had for our same-day grocery delivery service in the DC area; clearly the convenience of grocery delivery has resonated with this community of busy, hard-working people who are thrilled to do their grocery shopping online from their favorite stores and have everything delivered quickly, right to their doorstep."
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Internal Security situation in the country has vastly improved : Rajnath Singh
Rajnath Singh addresses Inaugural Session of DGP/IGP Conference in Gujarat
Gujarat , 20 Dec 2018
The Union Home Minister has said the Internal Security situation in the country has vastly improved, particularly in the North-East and the LWE affected States. Addressing the Inaugural Session of the Conference of Director Generals and Inspector Generals of Police at Kevadia in Narmada district of Gujarat today, Rajnath Singh noted with satisfaction the improving security scenario in the North-East, particularly the peaceful rollout of NRC in Assam including publication of the final draft NRC list on July 30, 2018."Security scenario in the North-East region has witnessed steady improvement in recent years. All parameters of violence have improved in 2018 in comparison to corresponding period of 2017. There is 17% decline in total incidents and 38% reduction in the casualties of civilians," said Shri Rajnath Singh. "AFSPA was selectively relaxed in Arunachal Pradesh and Meghalaya, inviting positive public response. Besides, to encourage the militants to join the mainstream, GoI has revised the Surrender-Cum-Rehabilitation Policy by enhancing financial benefits to the surrendering militants," he added.The Union Home Minister said India's security concerns have been conveyed to Myanmar even as efforts are on to harness the trade potential with the Eastern neighbour and beyond. "A number of our militant groups are operating from the territories of neighbouring countries, particularly Myanmar. Our concerns have been highlighted in the bilateral meetings with these countries. In order to plug the gaps on the Indo-Myanmar border, an Inter-Ministerial Committee headed by MHA has formulated a revised protocol for effective implementation of Free Movement Regime."
Shri Rajnath Singh said violence in the LWE affected areas continues to improve since the last eight years. "We achieved major operational successes in core areas of CPI/Maoist in Chhattisgarh, Jharkhand and Odisha. Coordination among different State Police Forces and CAPFs, and generation of regular actionable intelligence was among the main factors which led to an almost two-fold increase in the losses suffered by LWEs this year." "Reflecting on these successes and the upper hand of the government over the Maoists, after a comprehensive review the number of districts covered under SRE scheme were reduced from 126 to 90," he added.The Union Home Minister said the communal situation has also remained under control this year to date. "The communal situation during the current year remained under control, with a decrease of 12% in communal incidents as compared to the corresponding period in 2017," said Shri Rajnath Singh. "Misuse of Social Media for religious denigration also witnessed a significant decline during this year," he added.Shri Rajnath Singh terrorism in the hinterland has also been stamped out. "Law enforcement agencies, through concerted action coordinated by Intelligence Bureau, have managed to overcome the first tide of propaganda and mobilization by Islamic State in Iraq and Syria. Consequently, India witnessed very few incidents of IS inspired violence. In concerted counter-terrorism operations, Law enforcement agencies have arrested nearly 125 terrorist suspects as against 117 arrested in 2017," said Shri Rajnath Singh. "The new phase of Islamic State propaganda emanating from Afghanistan-Pakistan region is also being countered effectively," he added.
The Union Home Minister commended the National Investigation Agency (NIA) for clamping down on terror activities. "In countering terror, NIA is also doing very good work. A total number of 45 NIA Special Courts have been constituted in the States / Union Territories. The NIA, since its inception, has registered 183 cases till December, 2017, out of which, 135 cases have been charge sheeted. Trial has been concluded in 37 cases, out of which 35 cases have resulted in conviction. This is one of the highest conviction rates in the world by any Agency."The Union Home Minister said Pakistan is perpetrating terrorist activities on Indian soil from J&K to Punjab and the hinterland. Shri Rajnath Singh said the security situation in J&K continues to be vitiated even though the stone pelting incidents have declined. "Attempts by militants to infiltrate in large numbers, intermittent attacks by militants and efforts at local recruitment continue. Terror infrastructure in Pakistan and PoK remains in the form of training camps, launching pads and communication control stations. Financing from across for militants and separatists is a cause of concern. The separatists also exploit every possible situation to agitate the people to fan further anti-India sentiments which leads to law and order situation."The Union Home Minister directed the security establishment to check efforts to revive terrorism in Punjab. "In Punjab, recent terror incidents and related interdictions indicate concerted efforts on part of Pak-establishment and Sikh extremist elements based in Pakistan to revive terrorism in Punjab with active support of other Sikh radical/extremist entities based abroad, especially Europe and America.
Pak-establishment is also seeking to forge an understanding between Sikh extremist groups and Pak-based Islamist outfits as well as Kashmir-centric terrorist groups for targeting Indian interests," he said.Shri Rajnath Singh said our "economic installations and iconic institutions" are also being continuously targeted from across the Western border through espionage and cyberspace. "Pakistan continues to abet terrorism and during the period January 01, 2017 to November 29, 2018, 17 ISI backed espionage modules have been neutralized in the country resulting in the arrest of 25 espionage agents which include 02 Pakistanis. Various sensitive organizations, including the Indian Army, Air Force, Navy, ITBP, DRDO, BSF, CRPF, MEA, Airport Authority of India, Ministry of Petroleum and Natural Gas and the Indian Railways, etc. have been targeted by PIOs from across the border through spoofed / crank calls to elicit sensitive information," he said.Shri Rajnath Singh said the Government is simultaneously pursuing diplomatic efforts to stop terror emanating from Pakistan. "India is a Member of Financial Action Task Force (FATF), an inter-Governmental Body, which makes recommendations relating to Combating of Financing of Terrorism, Money Laundering, etc. Efforts on this platform by India have ensured that Pakistan is rightly put on grey list."The Union Home Minister said while increased training and sensitisation programmes to cover critical installations spread all over the country are being regularly undertaken and sensitisation of top scientists and senior managers is being done in strategic and sensitive installations and sectors, including Atomic, Space, Defence, Power, Oil & Gas, etc. the State Police need to be alert on this front.Shri Rajnath Singh advocated Public Private Partnership to boost cyber security. "What is important in Cyber domain is to harness tremendous talent lying outside the Government's domain.
In a move to create synergy,Industry associations such as Data Security Council of India (DSCI), NASSCOM, Cyber Forensic Labs, set up in certain States,have taken up tasks of awareness creation and training programmes on Cyber Crime investigation.These are welcome moves. Academia, like National Law School, Bangalore and NALSAR University of Law, Hyderabad are also engaged in conducting several awareness and training programmes on Cyber Laws and Cyber Crimes,".The Union Home Minister meanwhile asked officials to be cautious in their dealings with foreigners. "National Security has economic underpinning. In this area, two aspects become important. Firstly, interface of various foreign entities with the political and administrative structure increase manifold. Here, I would like to advice police and administration officials not to bypass prescribed procedure for interaction with officials of foreign missions which requires communication to be routed through Ministry of Home or External Affairs, except in cases where such contact is permitted according to the Protocol Hand Book," he said.Shri Rajnath Singh complimented the security establishment for continuously striving to make the country secure and safe. "It is undeniable that the Police forces give a sense of confidence to people and allay their fears over security. I also acknowledge that the Police Forces are working in difficult environment. I attribute the credit for India's success in the area of security to her Intelligence Agencies, State Police Forces and Central Armed Police Forces," he said.The Union Ministers of State for Home Affairs, Shri Kiren Rijiju and Shri Hansraj Gangaram Ahir are attending the Conference from 20th-22nd December, 2018, being held at Kevadia, the venue of Statue of Unity. The Prime Minister Shri Narendra Modi will also address the Police Chiefs and Senior Security Officers during the three-day Conference.
Tags: Rajnath Singh , Kiren Rijiju , Hansraj Gangaram Ahir
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