text:
\section{Introduction}
\label{Introduction}

Our terminology  follows Oxley \cite{oxl2}. We say that a matroid $M$ {\it uses} an element $e$ or a set $Z$ of elements if $e \in E(M)$ or $Z \subseteq E(M)$. Let $N$ be a matroid. A matroid $M$ with $|E(M)| \geq 2$ is {\it $N$-connected} if, for every pair of distinct elements $e,f$ of $E(M)$, there is a minor of $M$ that is isomorphic to $N$ and uses $\{e,f\}$.

We will assume, unless otherwise stated, that the matroids discussed here have at least two elements. Note that $U_{1,2}$-connectivity coincides with the usual notion of connectivity for matroids. Hence, relying on a well-known inductive property of matroid connectivity \cite{wtt}, we have that if $M$ is $U_{1,2}$-connected, $e \in E(M)$, and $|E(M)| \geq 3$, then $M \backslash e$ or $M / e$ is $U_{1,2}$-connected. Our first theorem shows that $U_{1,2}$ is the only connected matroid with this property.

\begin{theorem}\label{minor_induction}
Let $N$ be a matroid. If, for every $N$-connected matroid $M$ with $|E(M)| > |E(N)|$ and, for every $e$ in $E(M)$, at least one of $M \backslash e$ or $M / e$ is $N$-connected, then $N$ is isomorphic to one of $U_{1,2}$, $U_{0,2}$, or $U_{2,2}$.
\end{theorem}

One attractive property of matroid connectivity is that elements can be assigned to components.
We say that a matroid $N$ has the {\it transitivity property} if, for every matroid $M$ and every triple $\{e,f,g\} \subseteq E(M)$, if $e$ is in an $N$-minor with $f$, and $f$ is in an $N$-minor with $g$, then $e$ is in an $N$-minor with $g$.  Let $M(\mathcal{W}_2)$  be the rank-$2$ wheel. In Section \ref{components}, we prove the following result.

\begin{theorem}\label{transitivity3}
The only   matroids with the transitivity property are $U_{1,2}$ and $M(\mathcal{W}_2)$.
\end{theorem}

On combining the last two theorems, we get the following result, which indicates how special the usual matroid connectivity is.

\begin{corollary}
\label{combine}
Let $N$ be a matroid with the transitivity property such that whenever $M$ is an $N$-connected matroid, $e \in E(M)$, and $|E(M)| > |E(N)|$, at least one of $M\backslash e$ and $M/e$ is $N$-connected. Then $N \cong U_{1,2}$.
\end{corollary}

The concept of $N$-connectivity can also convey interesting information when $N$ is disconnected, as the next result indicates.

\begin{theorem}\label{loop and coloop}
A matroid $M$ is $U_{0,1} \oplus U_{1,1}$-connected if and only if every clonal class of $M$ is trivial.
\end{theorem}

The paper is structured as follows. In the next section, we recall Cunningham and Edmonds's decomposition theorem for connected matroids that are not $3$-connected, which is a basic tool in our proofs. Sections \ref{3-connected matroids}, \ref{connected matroids}, and \ref{disconnected} treat the cases of $N$-connected matroids when $N$ is $3$-connected, connected, and disconnected, respectively. In particular, we prove Theorems \ref{minor_induction}, and \ref{transitivity3} in Section  \ref{components} and Theorem~\ref{loop and coloop} in Section~\ref{disconnected}.
Finally, in Section \ref{larger}, we consider what can be said when every set of three elements occurs in some minor.
Moss \cite{tm1} showed that $3$-connected matroids can be characterized as those in which every set of four elements is contained in a minor isomorphic to a member of $\{\mathcal{W}^2, \mathcal{W}^3, \mathcal{W}^4, M(\mathcal{W}_3), M(\mathcal{W}_4), Q_6\}$.

\section{Preliminaries}
\label{preliminaries}

The concept of $N$-connectivity is closely related to roundedness, which is exemplified by Bixby's \cite{bix} result that if $e$ is an element of a $2$-connected non-binary matroid $M$, then $M$ has a $U_{2,4}$-minor using $e$. Formally, let $t$ be a positive integer and let $\mathcal{N}$ be a class of matroids. A matroid $M$ {\it has an $\mathcal{N}$-minor} if $M$ has a minor isomorphic to a member of $\mathcal{N}$. Seymour \cite{sey2} defined  $\mathcal{N}$ to be {\it $t$-rounded} if, for every $(t+1)$-connected matroid $M$ with an $\mathcal{N}$-minor and every subset $X$ of $E(M)$ with at most $t$ elements, $M$ has an $\mathcal{N}$-minor using $X$.
Thus Bixby's result shows that $\{U_{2,4}\}$ is $1$-rounded. Seymour \cite{sey} extended this result as follows.

\begin{theorem}\label{nonbinary 2 elts}
Let $M$ be a $3$-connected matroid having a $U_{2,4}$-minor, and let $e$ and $f$ be  elements of $M$. Then $M$ has a $U_{2,4}$-minor using $\{e,f\}$.
\end{theorem}

The connectivity function $\lambda_M$ of a matroid $M$ is defined for every subset $X$ of $E(M)$ by $\lambda_M(X) = r(X) + r(E(M) - X) - r(M)$;
equivalently,
$\lambda_M(X) = r(X) + r^*(X) - |X|.$
For disjoint subsets $A,B$ of $E(M)$, define $\kappa_M(A,B) = \min \{ \lambda_M(X) : A \subseteq X \subseteq E(M) - B\}$.

\begin{lemma}\label{connectivity_function}
If $N$ is a minor of $M$ and $A,B$ are disjoint subsets of $E(N)$, then $\kappa_N(A,B) \leq \kappa_M(A,B)$.
\end{lemma}

Next we give a brief outline of Cunningham and Edmonds's   decomposition  \cite{cthesis}  of matroids that are $2$-connected but not $3$-connected.  More complete details can be found in \cite[Section 8.3]{oxl2}. First recall that when $(X,Y)$ is a $2$-separation of a connected matroid $M$, we can write $M$ as $M_X \oplus_2 M_Y$ where $M_X$ and $M_Y$ have ground sets $X \cup p$ and $Y \cup p$.  A {\it matroid-labeled tree} is a tree $T$ with vertex set $\{ M_1, M_2, \ldots, M_n \}$ such that each $M_i$ is a matroid and, for distinct vertices $M_j$ and $M_k$, the sets $E(M_j)$ and $E(M_k)$ are disjoint if $M_j$ and $M_k$ are non-adjacent, whereas if $M_j$ and $M_k$ are joined by an edge $e$, then $E(M_j) \cap E(M_k) = \{e\}$, and $\{e\}$ is not a separator in either $M_j$ or $M_k$.

When $f$ is an edge of a matroid-labeled tree $T$ joining vertices $M_i$ and $M_j$, if we contract the edge $f$, we obtain a new matroid-labeled tree $T/ f$ by relabeling the composite vertex that results from this contraction as $M_i \oplus_2 M_j$, with every other vertex retaining its original label.

A {\it tree decomposition} of a $2$-connected matroid $M$ is a matroid-labeled tree $T$ such that if $V(T) = \{M_1, M_2, \ldots, M_n \}$ and $E(T) = \{ e_1, e_2, \ldots, e_{n-1} \}$, then

\begin{enumerate}[label=(\roman*)]
\item $E(M) = (E(M_1) \cup E(M_2) \cup \cdots \cup E(M_n)) - \{e_1, e_2, \ldots, e_{n-1} \}$;
\item $|E(M_i)| \geq 3$ for all $i$ unless $|E(M)| < 3$, in which case, $n = 1$ and $M_1 = M$; and
\item the label of the single vertex of $T / \{ e_1, e_2, \ldots, e_{n-1} \}$ is $M$.
\end{enumerate}

We call the members of $\{ e_1, e_2, \ldots, e_{n-1} \}$ {\it basepoints} since each member of this set is the basepoint of a $2$-sum when we construct $M$. Cunningham and Edmonds  (in \cite{cthesis}) proved the following (see also \cite[Theorem~8.3.10]{oxl2}).

\begin{theorem}
\label{cet}
Let $M$ be a $2$-connected matroid. Then $M$ has a tree decomposition $T$ in which every vertex label that is not a circuit or a cocircuit is $3$-connected, and there are no adjacent vertices that are both labeled by circuits or are both labeled by cocircuits. Moreover, $T$ is unique up to relabeling of its edges.
\end{theorem}

The tree decomposition $T$ whose existence is guaranteed by the last theorem is called the {\it canonical tree decomposition} of $M$. Although circuits and cocircuits with at most three elements are $3$-connected matroids, when we refer to a {\it $3$-connected vertex}, we shall mean one with at least four elements. Clearly, for each edge $p$ of $T$, the graph $T\backslash p$ has two components. Thus $p$ induces a partition of $V(T)$ and a corresponding partition $(X_p,Y_p)$ of $E(M)$. The latter partition is a
 $2$-separation of $M$;  we say that it is {\it displayed} by the edge $p$. Moreover, $M= M_{X_p} \oplus_2 M_{Y_p}$  where $M_{X_p}$ and $M_{Y_p}$ have ground sets   $X_p \cup p$ and $Y_p \cup p$, respectively. We shall refer to this 2-sum decomposition of $M$ as having been  {\it induced} by the edge $p$ of $T$.

We shall frequently use the following well-known result, which appears, for example, as \cite[Lemma~2.15]{jt}.

\begin{lemma}\label{path in tree}
Let $M_1$ and $M_2$ label distinct vertices in a tree decomposition $T$ of a connected matroid $M$. Let $P$ be the path in $T$ joining $M_1$ and $M_2$, and let $p_1$ and $p_2$ be the edges of $P$ meeting $M_1$ and $M_2$, respectively. Then $M$ has a minor that uses $(E(M_1) \cup E(M_2)) \cap E(M)$ and is isomorphic to the $2$-sum of $M_1$ and $M_2$, with respect to the basepoints $p_1$ and $p_2$.
\end{lemma}

We will  often use the next result,  another consequence of  Theorem~\ref{cet}.

\begin{lemma}\label{replace_basepoint}
Let $(X,Y)$ be a $2$-separation displayed by an edge $p$ in a $2$-connected matroid $M$. Suppose $y \in Y$. Then $M$ has, as a minor, the matroid $M_X(y)$ that is obtained from $M_X$ be relabeling $p$ by $y$. In particular, let $N$ be a $3$-connected minor of $M$ with $|E(N)| \geq 4$ and $|E(N) \cap Y| \leq 1$. If $|E(N) \cap Y| = 1$, let $y \in E(N) \cap Y$; otherwise let $y$ be an arbitrary element of $Y$. Then $M_X(y)$ has $N$ as a minor.
\end{lemma}

Let $T$ be the canonical tree decomposition of a $2$-connected matroid $M$, and let $M_0$ label a vertex of $T$. Let $p_1, p_2, \ldots, p_d$ be the edges of $T$ that meet $M_0$. For each $p_i$, let $(X_i, Y_i)$ be the $2$-separation of $M$ displayed by $p_i$, where $M_0$ is on the $X_i$-side of the $2$-separation. For each $i$, let $y_i \in Y_i$. Then, by repeated application of Lemma~\ref{replace_basepoint}, we deduce that $M$ has, as a minor, the matroid that is obtained from $M_0$ by relabeling $p_i$ by $y_i$ for all $i$ in $\{1,2,\ldots,d\}$. We denote this matroid by $M_0(y_1,y_2,\ldots,y_d)$ and call it a {\it specially relabeled} $M_0$-minor of $M$.

The following result, which is straightforward to prove by repeated application of Lemma~\ref{connectivity_function}, is well known.

\begin{lemma}\label{replace_basepoint2}
Let $N$ be a $3$-connected matroid with $|E(N)| \geq 3$. Let $M$ be a $2$-connected matroid with canonical tree decomposition $T$. Then there is a unique vertex $M'$ of $T$ such that, for each edge $p$ of $T$,   the partition of $V(T)$ induced by $p$ has the vertex $M'$   on the same side as at least $|E(N)| - 1$ elements of $N$. Moreover, there is a specially relabeled $M'$-minor of $M$ that has $N$ as a minor.
\end{lemma}

\section{$3$-connected matroids}
\label{3-connected matroids}

Let $\mathcal{N}$ be a set of matroids. A matroid $M$ is {\it $\mathcal{N}$-connected} if, for every two distinct elements $e$ and $f$ of $M$, there is an $N$-minor of $M$ that uses $\{e,f\}$ for some $N$ in $\mathcal{N}$. A consequence of \cite[Proposition 4.3.6]{oxl2} is that
a matroid with at least three elements is $\{U_{1,3}, U_{2,3}\}$-connected if and only if it is connected. The first result in this section characterizes $U_{2,3}$-connected matroids.
One may hope for a characterization of $3$-connectivity in terms of $\mathcal{N}$-connectivity, but no such characterization exists. To see this, note that
if $M$ is $\mathcal{N}$-connected, then so is $M \oplus_2 M$.  A characterization of $3$-connectivity in terms of minors containing $4$-element sets, as opposed to the $2$-element sets currently under consideration, is given in \cite{tm1}.

\begin{proposition}\label{U23}
A matroid $M$ is $U_{2,3}$-connected if and only if $M$ is connected and simple.
\end{proposition}

\begin{proof}

Suppose $M$ is $U_{2,3}$-connected. Clearly $M$ is connected and simple. Conversely, if $M$ is connected and simple, and $e$ and $f$ are distinct elements of $M$, then $M$ has a circuit $C$ containing $\{e,f\}$ and $|C| \geq 3$. Hence $M$ has a $U_{2,3}$-minor using $\{e,f\}$, so $M$ is $U_{2,3}$-connected.
\end{proof}

\begin{corollary}\label{U13}
A matroid $M$ is $U_{1,3}$-connected if and only if $M$ is connected and cosimple.
\end{corollary}

We will describe $N$-connectivity for a $3$-connected matroid $N$ by first considering the case when $N$ is $U_{2,4}$.
We will refer to binary and non-binary matroids that label vertices of a canonical tree decomposition as {\it binary} and {\it non-binary vertices}.

\begin{theorem}
A matroid $M$ is $U_{2,4}$-connected if and only if $M$ is connected and non-binary, and, in the canonical tree decomposition  of $M$,
\begin{enumerate}[label=(\roman*)]
\item every binary vertex has at most one element that is not a basepoint; and
\item on every path between two binary vertices that each contain a unique element of $E(M)$, there is a non-binary vertex.
\end{enumerate}
\end{theorem}

\begin{proof}
Suppose $M$ is non-binary and connected, and the canonical tree decomposition $T$ of $M$ satisfies the above conditions. Suppose $e$ and $f$ are distinct elements of $M$. If $e$ and $f$ are in the same $3$-connected vertex $M_0$ of $T$, then, by (i), $M_0$ is non-binary. Thus, by Theorem~\ref{nonbinary 2 elts}, $M$ has a $U_{2,4}$-minor using $\{e,f\}$.

Next suppose $e$ belongs to a binary vertex $M_1$ of $T$, and $f$ belongs to a non-binary vertex $M_0$ of degree $d$. By Lemma~\ref{replace_basepoint}, $M$ contains a specially labeled $M_0$-minor $M_0(e, y_2, y_3, \ldots, y_d)$ using $\{e,f\}$. Similarly, let $e$ and $f$ belong to binary vertices $M_1$ and $M_2$, and let $M_0$ be a  non-binary vertex on the path between them in $T$. Then $M$ contains a specially labeled $M_0$-minor $M_0(e,f,y_3,y_4,\ldots,y_d)$.
Thus, by Theorem~\ref{nonbinary 2 elts}, $M$ has a $U_{2,4}$-minor  using $\{e,f\}$.

Suppose now that $M$ is $U_{2,4}$-connected. Clearly $M$ is non-binary and connected.
If a binary vertex   $M_1$ in $T$  contains two non-basepoints $e$ and $f$, then, by Lemma~\ref{replace_basepoint2},  a $U_{2,4}$-minor of $M$ using $\{e,f\}$ must be a minor of $M_1$; a contradiction.

Now suppose $e$ and $f$ are the unique non-basepoints of binary vertices $M_1$ and $M_2$, respectively, in $T$, and let $N$ be a $U_{2,4}$-minor of $M$ using $\{e,f\}$. By Lemma~\ref{replace_basepoint2}, $T$ has a nonbinary vertex $M_0$ such that, for  every edge $p$ of $T$, the partition of $V(T)$ induced by $p$ has $M_0$ on the same side as at least $|E(N)| - 1$ elements of $N$. Let $p_1$ be the edge incident with $M_0$ such that $M_1$ and $M_0$ are on opposite sides of the induced partition of $V(T)$. Then $M_2$  must be on the same side of this partition as $M_0$. Hence $M_0$ lies on the path in $T$ between $M_1$ and $M_2$.
\end{proof}

The last theorem can be generalized  as follows.

\begin{theorem}\label{general}
Let $N$ be a $3$-connected matroid with at least four elements. A matroid $M$ is $N$-connected if and only if $M$ is connected, has $N$ as a minor, and, in the canonical tree decomposition  of $M$,
\begin{enumerate}[label=(\roman*)]
\item every vertex that is not $N$-connected has at most one element that is not a basepoint; and
\item on every path between two vertices that are not $N$-connected and that each have   unique non-basepoints, there is an $N$-connected vertex.
\end{enumerate}
\end{theorem}

\section{Connected matroids}
\label{connected matroids}

In this section, we consider $N$-connected matroids when $N$ is connected but not $3$-connected.

\begin{theorem}\label{2wheel}

A matroid $M$ is $M(\mathcal{W}_2)$-connected if and only $M$ is connected and non-uniform.
\end{theorem}

\begin{proof}
If $M$ is $M(\mathcal{W}_2)$-connected, then it is clearly both connected and non-uniform.
To prove the converse, suppose $M$ is connected and non-uniform. We argue by induction that $M$ is $M(\mathcal{W}_2)$-connected. This is immediate if $|E(M)| = 4$, since $M(\mathcal{W}_2)$ is the unique $4$-element connected, non-uniform matroid. Assume it holds for $|E(M)| < n$ and let $|E(M)| = n > 4$. Distinguish two elements $x$ and $y$ of $E(M)$.

Suppose there is an element $e$ of $E(M) - \{x,y\}$ such that $M / e$ is disconnected. Then $M$ is the parallel connection, with basepoint $e$, of two matroids $M_1$ and $M_2$. Now $M \backslash e$ is connected. We may assume that it is uniform; otherwise, by the induction assumption, $M \backslash e$ and hence $M$ has an $M(\mathcal{W}_2)$-minor using $\{x,y\}$. Now $r(E(M_1) - e) + r(E(M_2) - e) - r(M \backslash e) = 1$.
Suppose each of $|E(M_1) - e|$ and $|E(M_2) - e|$ has at least two elements. Then $M \backslash e$ has a $2$-separation. Since $M \backslash e$ is uniform, it follows that $M \backslash e$ is a circuit or a cocircuit. In the latter case, $M$ is also a cocircuit; a contradiction. If $M \backslash e$ is a circuit, then $M$ is the parallel connection of two circuits, and $M$ is easily seen to have an $M(\mathcal{W}_2)$-minor using $\{ x,y \}$.

Now suppose that $|E(M_1) - e| = 1$. Thus $M$ has a circuit, $\{ e,f \}$ say, containing $e$. As $M \backslash e$ is uniform but $M$ is not, $r(M) \geq 2$, so $M \backslash e$ has a circuit containing $\{f,x,y\}$. It follows that $M$ has an $M(\mathcal{W}_2)$-minor with ground set $\{e,f,x,y\}$.

We may now assume that $M / e$ is connected for all $e$ in $E(M) - \{x,y\}$. Moreover, by replacing $M$ with $M^*$ in the argument above, we may also assume that $M \backslash e$ is connected for all such $e$. If $M \backslash e$ or $M / e$ is non-uniform, then, by the induction assumption, $M$ has an $M(\mathcal{W}_2)$-minor using $\{x,y\}$. Thus both $M \backslash e$ and $M / e$ are uniform. Let $r(M \backslash e) = r$. Then every circuit of $M \backslash e$ has $r+1$ elements. Since $M$ is not uniform, it has a circuit containing $e$ that has at most $r$ elements. Contracting $e$ from $M$ produces a rank-$(r-1)$ matroid having a circuit with at most $r-1$ elements. Since $M / e$ is uniform, this is a contradiction.
\end{proof}

We omit the straightforward proof of the next result.

\begin{lemma}\label{transitivity}
If $M$, $N$, and $N'$ are matroids such that $M$ is $N$-connected and $N$ is $N'$-connected, then $M$ is $N'$-connected.
\end{lemma}

If we wish to describe the class of $N$-connected matroids for a $3$-connected matroid $N$, it suffices to describe the $N$-connected matroids that are $3$-connected and then apply Theorem~\ref{general}. If $N$ is not $3$-connected, the task of describing $N$-connected matroids becomes harder, and we omit any attempt to provide a general theorem for $N$-connectivity in this case. We will instead give characterizations for two specific matroids that are not $3$-connected, namely $U_{1,4}$ and its dual $U_{3.4}$. We will use the following theorem of Oxley \cite{oxl1}.

\begin{theorem
\label{rank and corank 3}
Let $M$ be a $3$-connected matroid having rank and corank at least three,  and suppose that $\{x,y,z\} \subseteq E(M)$. Then $M$ has a minor isomorphic to one of $U_{3,6}, P_6, Q_6, \mathcal{W}^3$, or $M(K_4)$ that uses $\{x,y,z\}$.
\end{theorem}

\begin{proposition} \label{U14}
A $3$-connected matroid $M$ is $U_{1,4}$-connected if and only if either $M \cong U_{2,n}$ for some $n \geq 5$, or $M$ has rank and corank at least three.
\end{proposition}

\begin{proof}
Clearly if $n \geq 5$, then $U_{2,n}$ is $U_{1,4}$-connected. Now assume that $r(M) \geq 3$ and $r^*(M) \geq 3$. Suppose $\{x,y\} \subseteq E(M)$. Then, by Theorem~\ref{rank and corank 3}, $M$ has an $\mathcal{N}$-minor using $\{x,y\}$ where $\mathcal{N}$ is  $\{U_{3,6}, P_6, Q_6, \mathcal{W}^3, M(K_4) \}$. One easily checks that each member of $\mathcal{N}$ is $U_{1,4}$-connected. Hence, by Lemma~\ref{transitivity}, $M$ is $U_{1,4}$-connected.

To prove the converse, assume that $M$ is $U_{1,4}$-connected. Since $r^*(U_{1,4})=3$, it follows that $r^*(M) \geq 3$. The required result holds if $r(M) \geq 3$. But, since $M$ is $3$-connected and $U_{1,4}$-connected, $r(M) \geq 2$. Moreover, if $r(M) = 2$, then $M \cong U_{2,n}$ for some $n \geq 5$.
\end{proof}

Duality gives a corresponding result for $U_{3,4}$-connectivity.

\begin{corollary} \label{U34}
A $3$-connected matroid $M$ is $U_{3,4}$-connected if and only if either $M \cong U_{n-2,n}$ where $n \geq 5$, or $M$ has rank and corank at least $3$.
\end{corollary}

Observe that this fails to fully characterize $U_{3,4}$-connectivity for if we let $M = M(K_{2,3})$, then $M$ is $U_{3,4}$-connected but none of the matroids in its canonical tree decomposition is $U_{3,4}$-connected. We can instead describe $U_{3,4}$-connectivity in terms of forbidden configurations of matroids in the canonical tree decomposition.

\begin{proposition}\label{U34+}
Suppose $M$ is not $3$-connected. Then $M$ is $U_{3,4}$-connected if and only if $M$ is connected and simple, and, in the canonical tree decomposition $T$ of $M$, there is no vertex of degree at most two   that is labeled by some $U_{2,n}$ such that its only neighbors in $T$ are cocircuits that use elements of $E(M)$.
\end{proposition}

\begin{proof}
Let $T$ be the canonical tree decomposition of $M$. Assume $M$ is $U_{3,4}$-connected. Then, by Lemma~\ref{transitivity},  $M$ is $U_{2,3}$-connected, so $M$ is connected and simple. Suppose that $T$ has a vertex $M_0$ whose degree $d$ is at most two such that $M_0$ is labeled by some $U_{2,n}$ and has its only neighbors $M_1,\dots, M_d$ labeled by cocircuits that use elements of $E(M)$. For each $i$ in $\{1,\dots,d\}$, suppose $f_i \in E(M_i) \cap E(M)$. Then $M$ can be obtained from a copy of $U_{2,n}$ using $\{f_1,\dots,f_d\}$ by, for each $i$,  adjoining some  matroid via parallel connection across the basepoint $f_i$. If $ d = 1$, let $f_2$ be an element of $M_0$ other than $f_1$. Clearly $M$ has no circuit using $\{f_1,f_2\}$ that has more than three elements.

Now assume that $M$ is connected and simple and that $T$ satisfies the specified conditions. Let $\{e,f\}$ be a subset of  $E(M)$ that is not contained in a $U_{3,4}$-minor.  Assume first that $e$ and $f$ belong to the same vertex $M_1$ of $T$. As $M$ is simple, $M_1$ is not a cocircuit. Now $M$ has a specially relabeled $M_1$-minor using $\{e,f\}$. Thus, by Corollary~\ref{U34},  $M_1 \cong U_{2,n}$ for some $n \ge 3$. Let $p$ be an edge of $T$ that meets $M_1$. Consider the $2$-sum $N_1 \oplus_2 N_2$ induced by $p$ where $\{e,f\} \subseteq E(N_1)$. Certainly $N_1$ has a circuit containing $\{e,f,p\}$, and $N_2$ has a circuit of size at least three containing $p$. Thus $M$ has a $U_{3,4}$-minor containing $\{e,f\}$; a contradiction.

We now know that $e$ and $f$ belong to distinct vertices $M_1$ and $M_2$ of $T$. Each edge $p$ of the path $P$ in $T$ joining $M_1$ and $M_2$ induces a 2-sum decomposition of $M$ into two matroids, $N_{1p}$ and $N_{2p}$. Moreover, an element $x_i$ of $E(N_{ip})$ is in a circuit of $N_{ip}$ of size at least three containing $p$ unless $x_i$ is parallel to $p$ in $N_{ip}$. Thus $e$ or $f$ is parallel to $p$ in $N_{1p}$ or $N_{2p}$, respectively. Let the edges of $P$, in order, be $p_1,p_2,\dots,p_k$ where $p_1$ meets $M_1$. We may assume that $e$ is parallel to $p_1$ in $N_{1p_1}$. Then the vertex $M_1$ of $T$ containing $e$ is a cocircuit.

Suppose $k \ge 3$. As no two adjacent vertices of $T$ are cocircuits, neither $e$ nor $f$ is parallel to $p_2$ in $N_{1p_2}$ or $N_{2p_2}$. Hence $M$ has a $U_{3,4}$-minor using $\{e,f\}$. This contradiction implies  that $k \in \{1,2\}$. Suppose $k = 2$. Then  $f$ is parallel to $p_2$ in $N_{2p_2}$. Thus $M_2$ is a cocircuit. Since $M$ has no $U_{3,4}$-minor using $\{e,f\}$, the vertex $M_3$ of $T$ that is adjacent to both $M_1$ and $M_2$ is  isomorphic to some $U_{2,n}$. By assumption, $M_3$ must have another neighbor in $T$ to which it is joined by the edge $q$, say. Then, for the 2-sum decomposition $Q_1 \oplus_2 Q_2$ of $M$ induced by $q$, there is a circuit of $Q_1$ containing $\{e,f,q\}$ and a circuit of $Q_2$ of size at least three containing $q$. Thus $M$ has a $U_{3,4}$-minor using $\{e,f\}$. This contradiction implies that $k = 1$. Then $M = N_{1p_1} \oplus_2 N_{2p_1}$. Thus the specially relabeled minor $N_{2p_1}(e)$ uses $\{e,f\}$. Now the canonical tree decomposition $T'$ of $N_{2p_1}(e)$ can be obtained from the component of $T\backslash p_1$ using $N_{2p_1}$ by replacing $M_2$ by $M_2(e)$. As $e$ and $f$ are contained in the same vertex of $T'$, we deduce from the second paragraph that $N_{2p_1}(e)$, and hence $M$, has a $U_{3,4}$-minor using $\{e,f\}$; a contradiction.
\end{proof}

\section{Disconnected matroids}
\label{disconnected}

We now turn our attention to $N$-connectivity where $N$ is disconnected. The following is essentially immediate.

\begin{proposition}\label{U_nn_basis}
Let $n$ be an integer exceeding one. A matroid $M$ is $U_{n,n}$-connected if and only if $M$ is simple with rank at least $n$.
\end{proposition}

Recall that elements $x$ and $y$ of a matroid $M$ are {\it clones} if the bijection on $E(M)$ that interchanges $x$ and $y$ but fixes every other element yields the same matroid.
Next we prove Theorem~\ref{loop and coloop}, showing that a matroid is $U_{0,1} \oplus U_{1,1}$-connected if and only if no element has a clone. The proof will use the well-known fact (see, for example, \cite{bs}) that two elements in a matroid are clones if and only if they are in precisely the same  cyclic flats.

\begin{proof}[Proof of Theorem~\ref{loop and coloop}]

Suppose every clonal class of $M$ is trivial and let $x$ and $y$ be distinct elements of $M$. Then $M$ has a cyclic flat $F$ that contains exactly one of $x$ and $y$, say $x$. In $M / (F - x)$, the element $x$ is a loop but $y$ is not. Thus $M$ has a $U_{0,1} \oplus U_{1,1}$-minor using $\{x,y\}$, so $M$ is $U_{0,1} \oplus U_{1,1}$-connected.

Conversely, assume $M$ is $U_{0,1} \oplus U_{1,1}$-connected, but $M$ has elements $x$ and $y$ that are in the same cyclic flats. Suppose that $M / C \backslash D \cong U_{0,1} \oplus U_{1,1}$ and $E(M / C \backslash D) = \{x,y\}$. Let $x$ be the loop of $M / C \backslash D$. Then $x \in \mathrm{cl}_M(C)$. Thus $y \in \mathrm{cl}_M(C)$, so $y$ is a loop in $M / C \backslash D$; a contradiction.
\end{proof}

Recall, for the next result, that an element is {\it free} in a matroid if it is not a coloop and every circuit that contains it is spanning.

\begin{theorem}
A matroid $M$ is $U_{1,2} \oplus U_{1,1}$-connected if and only if $M$ is loopless, has at most one coloop, and has at most one free element.
\end{theorem}

\begin{proof}
Clearly if $M$ is $U_{1,2} \oplus U_{1,1}$-connected, then it obeys the specified conditions.
Conversely, suppose $M$ is loopless, has at most one coloop, and has at most one free element.
Let $e$ and $f$ be elements of $M$. Suppose first that $M$ is disconnected. If $e$ and $f$ are in the same component, then they are in a $U_{1,2}$-minor of that component, so $M$ has a $U_{1,2} \oplus U_{1.1}$-minor  using $\{e,f\}$. If $e$ and $f$ are in different components, then one of these components is not a coloop. That component has a $U_{1,2}$-minor using $e$ or $f$. It follows that $M$ has a $U_{1,2} \oplus U_{1,1}$-minor  using $\{e,f\}$.

Now suppose $M$ is connected. Suppose  that $e$ is free in $M$. Then $f$ is in some non-spanning circuit, $C_f$. Choose $g$ in $C_f - f$. Contracting $C_f - \{f,g\}$ and deleting every other element of $M$
yields a $U_{1,2} \oplus U_{1,1}$-minor of $M$ using $\{e,f\}$.

Suppose neither $e$ nor $f$ is free in $M$. If there is a non-spanning circuit $C$ containing $\{e,f\}$, we can find a $U_{1,2} \oplus U_{1,1}$-minor by contracting every element of $C$ except $e$ and $f$, and deleting every other element except for one. Now suppose every circuit containing $\{e,f\}$ is spanning. Since $e$ is not free, there is a non-spanning circuit $C$ containing $e$.
Clearly $f \not\in \mathrm{cl}(C)$ otherwise $M|{\mathrm{cl}(C)}$ is a connected matroid of rank less than $r(M)$ so it contains a circuit containing $\{e,f\}$; a contradiction.
Therefore, after we contract all of $C$ except for $e$ and one other element, we see that $f$ will not be a loop. Thus we can find a $U_{1,2} \oplus U_{1,1}$-minor using $\{e,f\}$.
\end{proof}

\begin{corollary}
A matroid $M$ is $U_{1,2} \oplus U_{0,1}$-connected if and only if $M$ is coloopless and has at most one element that is in every dependent flat.
\end{corollary}

\section{$N$-connectivity as compared to connectivity}
\label{components}

Before proving Theorem~\ref{minor_induction}, we state and prove its converse.

\begin{proposition}
If $N \in \{U_{1,2} , U_{0,2} , U_{2,2} \}$, then, for every $N$-connected matroid $M$ with $|E(M)| \geq 3$ and for every $e$ in $E(M)$, at least one of $M \backslash e$ or $M / e$ is $N$-connected.
\end{proposition}

\begin{proof}
The result is immediate if $N \cong U_{1,2}$. By duality, it suffices to deal with the case when $N \cong U_{2,2}$. Suppose $M$ is $U_{2,2}$-connected, and $|E(M)| \geq 3$. By Proposition~\ref{U_nn_basis}, $M$ is simple with rank at least two. Therefore if $M$ is $U_{2,2}$-connected and $r(M) > 2$, we can delete any element $e$ of $M$ and still have an $N$-connected matroid. Observe that if $r(M) = 2$, then $M$ must be connected since it is simple. Therefore $M$ has no coloops, so $r(M \backslash e) = 2$ for all $e$ of $E(M)$. Thus $M \backslash e$ is $U_{2,2}$-connected.
\end{proof}

\stepcounter{theorem}
\begin{proof}[Proof of Theorem~\ref{minor_induction}]

First we consider the case when $N$ is connected. Then $N$ is $U_{1,2}$-connected. Thus, by Lemma~\ref{transitivity}, every $N$-connected matroid is $U_{1,2}$-connected and so is connected. Suppose $M$ is an $N$-connected matroid with $|E(M)| > |E(N)|$.

Assume $N$ is simple. Then, by Proposition~\ref{U23} and Lemma~\ref{transitivity}, $N$, and hence $M$, is $U_{2,3}$-connected.
Let $M_1$ and $M_2$ be isomorphic copies of $M$ with disjoint ground sets. Pick arbitrary elements $g_1$ and $g_2$ in $M_1$ and $M_2$, and let $M_3$ be the parallel connection of $M_1$ and $M_2$ with respect to the basepoints $g_1$ and $g_2$, which we relabel as $g$ in $M_3$. Then one easily sees that $M_3$ is $N$-connected.
Let $e, f \in E(M_1) - g$. By assumption, we can remove all the elements of $E(M_1) - \{e,f,g\}$ from $M_3$ via deletion or contraction to obtain a matroid $M_4$ that is still $N$-connected. Since $M_4$ is $U_{2,3}$-connnected, it follows that $\{e,f,g\}$ is a triangle in $M_4$. Moreover,   $\{e,f\}$ is a series pair in $M_4$. However, neither $M_4 \backslash e$ nor $M_4 / e$ is $U_{2,3}$-connected since $M_4 \backslash e$ is disconnected, and $M_4 / e$ has $f$ and $g$ in parallel.
We deduce that $N$ is not simple. Dually, $N$ is not cosimple. The only uniform matroid that is neither simple nor cosimple is $U_{1,2}$, so either $N \cong U_{1,2}$, or $N$ is non-uniform.

Next we show that $N$ cannot be non-uniform. Suppose, instead, that $N$ is non-uniform. Then, as $N$ is connected, by Theorem~\ref{2wheel}, $N$ is $M(\mathcal{W}_2)$-connected.

Recall that $M$ is $N$-connected with $|E(M)| > |E(N)|$. Let $n = |E(N)| + 1$ and distinguish elements $e, f$ of $E(M)$. Let each of $M_1, M_2, \ldots, M_n$ be a copy of $M$  and let $e_i$ and $f_i$ be the elements of $M_i$ corresponding to $e$ and $f$. Let $M'$ be the parallel connection of $M_1, M_2, \ldots, M_n$ with respect to the basepoints $e_1,e_2,\ldots,e_n$ where  these elements are relabeled as $e$ in $M'$. By assumption, for each $M_i$, we can remove $E(M_i) - \{e,f_i\}$ from $M'$ in such a way that the resulting matroid $M''$ is $N$-connected. Since $M''$ is connected, it must be isomorphic to $U_{1,n+1}$, which is clearly not $M(\mathcal{W}_2)$-connected; a contradiction.
We conclude that $N$ cannot be non-uniform, and hence the theorem holds when $N$ is connected.

Next we consider the case when $N$ is disconnected, first showing the following.

\begin{sublemma}\label{sublemma}
If each element of $N$ is a loop or a coloop, then $N \cong U_{0,2}$ or $U_{2,2}$.
\end{sublemma}

Suppose $n \geq 3$ and let $N \cong U_{n,n}$. Let $M = U_{2,3} \oplus U_{n-2,n-2}$. Then $M$ is $N$-connected, but if $e$ is a coloop of $M$, then neither $M \backslash e$ nor $M / e$ has a $U_{n,n}$-minor. Therefore $N \not \cong U_{n,n}$; dually, $N \not \cong U_{0,n}$.

If $N = U_{0,1} \oplus U_{1,1}$, then let $M = M(K_4)$. By Theorem~\ref{loop and coloop}, $M$ is $N$-connected, but, for every $e$ of $E(M)$, both $M \backslash e$ and $M / e$ have nontrivial clonal classes and are therefore not $N$-connected. Now assume $N \cong U_{0,n} \oplus U_{m,m}$ for some $n  \geq  2$ and  $m \geq 1$. Then $U_{0,n+1} \oplus U_{m,m}$ is an $N$-connected matroid, say $M$. But if $e$ is a coloop, then neither $M \backslash e$ nor $M / e$ has an $N$-minor. On combining this contradiction with duality, we conclude that \ref{sublemma} holds.

Now assume that $N$ has $k+s$ components $N_1, N_2, \ldots, N_{k+ s}$ where those with at least two elements are $N_1, N_2, \ldots, N_{k}$. Then $k \ge 1$. For each $i$ in $\{1,2,\ldots,k\}$, choose an element $e_i$ of $N_i$ and relabel it as $p$. Let $M'$ be the parallel connection of $N_1, N_2, \ldots, N_{k}$ with respect to the basepoint $p$ where we take $M' = N_1$ if $k=1$. Let $N'$ be a copy of $N$ whose ground set is disjoint from $E(N)$, and let $n'_i$ be the component of $N'$ corresponding to $N_i$.
Let  $M_1 = N' \oplus M'$.
We   show next  that

\begin{sublemma}
\label{m1n}
$M_1$ is $N$-connected.
\end{sublemma}

Suppose  $\{e,f\} \subseteq E(M_1)$. Certainly $M_1$ has an $N$-minor using $\{e,f\}$ if   $\{e,f\} \subseteq E(N')$. Next suppose that $e \in E(M')$. Then, since $M'$ is a connected parallel connection, we see that, for each $i$ in $\{1,2,\ldots,k+s\}$, there is an $N_i$-minor of $M'$ using $e$. Thus, if $f \in E(N')$, say $f \in E(N'_j)$, then we can choose $i \neq j$ and get an $N$-minor of $M_1$ using $\{e,f\}$ unless $k= 1 = j$. In the exceptional case, $M'$ has an $N_2$-minor with ground set $\{e\}$ and again we get an $N$-minor of $M_1$ using $\{e,f\}$. We may now assume that $f \in E(M')$, say
$f \in E(N_j)$. Then $M'$ has an $N_j$-minor using $\{e,f\}$, so $M_1$ has an $N$-minor using $\{e,f\}$. Thus \ref{m1n} holds.

Since $M_1$ is $N$-connected, by assumption, we may delete or contract elements of $M_1$ until we obtain an $N$-connected matroid $M_2$ with $|E(M_2)| = |E(N)| + 1$. In particular, we may remove elements from $M'$ in $M_1$ until a single element  $g$  remains. Now choose $e$ in $E(N'_1)$. Then $M_2 \backslash e$ or $M_2/e$ is isomorphic to $N$. But both $M_2 \backslash e$ and $M_2/e$ have more one-element components than $N'$; a contradiction.
\end{proof}

Recall that we say that a matroid $N$ has the transitivity property if, for every matroid $M$ and every triple $\{e,f,g\} \subseteq E(M)$, if $e$ is in an $N$-minor with $f$, and $f$ is in an $N$-minor with $g$, then $e$ is in an $N$-minor with $g$. Clearly $N$ has the transitivity property if and only if $N^*$ has the transitivity property.

\begin{lemma}\label{parallel-delete}
Suppose $N$ is a  matroid having the transitivity property. Let $N'$ be     obtained from $N$ by adding an element $f$ in parallel to a non-loop element $e$ of $N$. Then there is an element $g$ of $E(N')$ such that $N' \backslash g$ is isomorphic to $N$ and has $\{e,f\}$ as a $2$-circuit. Moreover,   $g$ is in a $2$-circuit in $N$.
\end{lemma}

\begin{proof}
The transitivity property implies that $\{e,f\}$ is in an $N$-minor  of $N'$.  Since $r^*(N') > r^*(N)$,  there must be an element $g$ of $E(N') - \{e,f\}$ such that $N'\backslash g \cong N$.  Since we have introduced a new $2$-circuit in constructing $N'$, when we delete $g$, we must destroy a $2$-circuit.
\end{proof}

By the last lemma and duality, we obtain the following result.

\begin{corollary}
\label{pd2}
If $N$ is a matroid having the transitivity property, then $N$ has a component with more than one element.
\end{corollary}

The following elementary observation and its dual will be used repeatedly in the proof of Theorem~\ref{transitivity3}.

\begin{lemma}
\label{transit}
Suppose $N$ is a matroid with the transitivity property. Let $N_0$ be a component of $N$ with the largest number of elements. Suppose $f$ is added in parallel to an element $e$ of $N_0$. Let $N_0'$ and $N'$ be the resulting extensions of $N_0$ and $N$, respectively. Suppose $g \in E(N')$ such that $N'\backslash g \cong N$. Then $g \in E(N_0')$.
\end{lemma}

Recall that a set $S$ of elements of a matroid $M$ is a {\it fan} if $|S| \geq 3$ and there is an ordering $(s_1, s_2, \ldots, s_n)$ of the elements of $S$ such that, for all $i$ in $\{1,2, \ldots, n-2\}$,

\begin{enumerate}[label=(\roman*)]
\item $\{s_i, s_{i+1}, s_{i+2}\}$ is a triangle or a triad; and
\item when $\{s_i, s_{i+1}, s_{i+2}\}$ is a triangle, $\{s_{i+1}, s_{i+2}, s_{i+3}\}$ is a triad; and when $\{s_i, s_{i+1}, s_{i+2}\}$ is a triad, $\{s_{i+1}, s_{i+2}, s_{i+3}\}$ is a triangle.
\end{enumerate}

Note that the above extends the definition given in \cite{oxl2} by eliminating the requirement that $M$ be simple and cosimple. We shall follow the familiar practice here of blurring the distinction between a fan and a fan ordering.

\begin{lemma}\label{special_fan}
Let $(s_1, s_2, \ldots, s_n)$ be a fan $X$ in a matroid $M$ such that each of $\{s_1, s_2\}$ and $\{s_{n-1}, s_n\}$ is a circuit or a cocircuit. Then $X$ is a component of $M$.
\end{lemma}

\begin{proof}
By switching to the dual if necessary, we may assume that $\{s_1, s_2, s_3\}$ is a triangle of $M$. Thus $\{s_1, s_2\}$ is a cocircuit. Observe that $\{s_i : i \mathrm{~is~odd} \}$ spans $X$.
If $n$ is odd, this is immediate, and if $n$ is even, it follows from the fact that $\{s_{n-1}, s_n\}$ is a circuit in this case.
By duality, $\{s_i : i \mathrm{~is~even} \}$ spans $X$ in $M^*$. Hence $r(X) + r^*(X) \leq |X|$; that is, $\lambda(X) \leq 0$, so $X$ is a component of $M$.
\end{proof}

We define a {\it special fan} to be a fan $(s_1, s_2, \ldots s_k)$ such that $\{s_1, s_2\}$ is a cocircuit of $M$. We will now show that $U_{1,2}$ and $M(\mathcal{W}_2)$ are the only connected matroids with the transitivity property.

\stepcounter{theorem}
\begin{proof}[Proof of Theorem~\ref{transitivity3}]
 It is clear that $U_{1,2}$ has the transitivity property. By Theorem~\ref{2wheel}, two elements of $M$ are in an $M(\mathcal{W}_2)$-minor together if and only if they are in a connected, non-uniform component together. It follows that $M(\mathcal{W}_2)$ has the transitivity property.

Suppose that $N$ has the transitivity property.    Assume that $N$ is not isomorphic to $U_{1,2}$ or $M(\mathcal{W}_2)$.
Next we show the following.

\begin{sublemma}
\label{biggest}
Let $N_0$ be a largest component of $N$. Then $N_0$ is isomorphic to $U_{1,2}$ or $M({\mathcal W}_2)$.
\end{sublemma}

Assume that this assertion fails. Then, by Corollary~\ref{pd2}, $N_0$ has at least two, and hence at least three, elements. Take an element $e$ of $N_0$ and add an element $f$ in series with it. Let the resulting coextensions of $N_0$ and $N$ be $N'_0$ and $N'$, respectively. Then, by the transitivity property, $N'/a \cong N$ for some element $a$ of $E(N') - \{e,f\}$. Furthermore, by the dual of Lemma~\ref{transit}, $a \in E(N_0)$. We deduce that $N_0$ has a 2-cocircuit, say $\{a,b\}$. In $N_0$, add an element $c$ in parallel to $a$ to get $N_1$. Then, by transitivity and Lemma~\ref{transit}, there is an element $s_1$ of $E(N_1) - \{a,c\}$ such that $N_1 \backslash s_1 \cong N_0$. Since $N_1\backslash b$ has $\{a,c\}$ as a component, the component sizes of $N_1 \backslash b$ and $N_0$ do not match, so
 $s_1 \neq b$.
Thus $s_1 \in E(N) - \{a,b,c\}$, so $N_1 \backslash s_1$ has $\{c,a,b\}$ as a cocircuit. Next add an element $d$ to $N_1 \backslash s_1$, putting it in series with $c$. Let the resulting matroid be $N_2$. By the dual of Lemma~\ref{transit}, there is an element $s_2$ of $E(N_2) - \{c,d\}$ such that $N_2/d \cong N_0$.
Moreover, $s_2$ must be in a $2$-cocircuit of $N_2$, and $s_2$ is in a triangle in $N_2$ as $N_2 /s_2$ must have a $2$-circuit that is not present in $N_2$ since adding $d$ destroyed the 2-circuit $\{a,c\}$.
Now $s_2 \neq a$ since $N_2 / a$ has $\{c,d\}$ as a component.

Suppose $s_2 = b$. Then $b$ is in a $2$-cocircuit $\{b,e\}$ in $N_2$.
Moreover, $N_2$ has a triangle $T$ containing $b$. By orthogonality, $T = \{b,e,a\}$.
Then $(d,c,a,e)$ is a fan $X$ in $N_2 / b$ having $\{c,d\}$ as a cocircuit and $\{a,e\}$ as a circuit. By Lemma~\ref{special_fan}, $X = E(N_2 / b)$, so $N_0 \cong N_2 / b \cong M(\mathcal{W}_2)$; a contradiction.

We now know that $s_2 \neq b$, so $s_2 \not\in \{a,b,c,d\}$. Thus $N_0$ has  $(d,c,a,b)$ as a special fan.
Among all the special fans of $N_0$ and $N_0^*$, take one, $(a_1, a_2, \ldots, a_k)$, with the maximum number of elements. Then $k \geq 4$.
First assume  $\{a_{k-2}, a_{k-1}, a_k\}$ is a triad. Suppose $\{a_{k-1}, a_k\}$ is a $2$-circuit of $N_0$.  Then, by Lemma~\ref{special_fan}, the special fan is the whole component $N_0$.  As $N_0 \not \cong M({\mathcal W}_2)$, we see that $k \ge 6$.
Add an element $f$ in parallel to   $a_3$ to form a new matroid $N_0'$. Then $\{a_1, a_3\}$ is in an $N_0$-minor of $N_0'$, and so is $\{a_1, f\}$. By the  transitivity property, $N_0'$ has  $\{a_3, f\}$  in an $N_0$-minor. Since $N_0'$ has $\{a_3,f\}$ and $\{a_{k-1},a_k\}$ as its only 2-circuits, while $N_0$ has a single 2-circuit, we deduce that $N_0'\backslash a_k \cong N_0$. But every element of $N_0$ is in a cocircuit of size at most three, yet $f$ is in no such cocircuit of $N_0'\backslash a_k$; a contradiction.

It remains to deal with the cases when, in $N_0$, either $\{a_{k-2}, a_{k-1}, a_k\}$ is a triad and $\{a_{k-1}, a_k\}$ is not a circuit, or $\{a_{k-2}, a_{k-1}, a_k\}$ is a triangle. In these cases, add $a_0$ in parallel with $a_1$ to produce $N_3$. To obtain an $N_0$-minor of $N_3$ using $\{a_0, a_1\}$,   we must delete an element $z$ of $N_3$ that belongs to a $2$-circuit. Now $z$ is not in $\{a_2, a_3, \ldots, a_k\}$ as  none of these elements is  in a $2$-circuit, so $N_3 \backslash z$ is isomorphic to $N_0$ and has $(a_0, a_1, \ldots, a_k)$ as a special fan. This contradicts our assumption that a special fan    in $N_0$ or $N_0^*$ has at most $k$ elements. We conclude that \ref{biggest} holds.

\begin{sublemma}
\label{noone}
$N$  has no single-element component.
\end{sublemma}

To see this, let $N_0$ be a largest component of $N$. By \ref{biggest}, $N_0$ is isomorphic to $U_{1,2}$ or $M({\mathcal W}_2)$.  Assume that $N$ has a single-element component $N_1$ with $E(N_1) = \{a\}$. By replacing $N$ by its dual if necessary, we may assume that $a$ is a coloop of $N$. Let $c$ be an element  that is in a 2-cocircuit of $N_0$. Now let $N'$ be obtained from $N$ by adding an element $b$ so that $N'$ has $\{a,b,c\}$ as a triangle and $\{a,b\}$ as a cocircuit. Then, by the transitivity property, $N'\backslash g \cong N$ for some element $g$ not in $\{a,b\}$. By the choice of $N_0$, we deduce that $g$ must be in the same component $N_0'$ of $N'$ as $\{a,b,c\}$. Moreover, $g$ must be in a 2-cocircuit of $N'_0$. But $N'_0$ contains no such element. Hence \ref{noone} holds.

\begin{sublemma}
\label{iftwo}
$N$  has a single component of maximum size.
\end{sublemma}

Assume that this fails, and let $N_0$ and $N_1$ be components of $N$ of maximum size.
Let $\{a_i,b_i\}$ be a 2-circuit of $N_i$. Let $N'_i$ be obtained from $N_i$ by adding $c_i$ in series with $b_i$. Now take a copy of $U_{2,3}$ with ground set $\{c_0,z,c_1\}$ and adjoin $N'_0$ and $N'_1$ via parallel connection across $c_0$ and $c_1$, respectively. Truncate the resulting matroid to get $N_{01}$. Then $r(N_{01}) = r(N_0) + r(N_1) + 1$. Let $N'$ be obtained from $N$ by replacing  $N_0 \oplus N_1$ by $N_{01}$.
Now $N_{01} /c_0$ and $N_{01} /c_1$ have $(N_0 \oplus N_1)$-minors using $\{z,c_1\}$ and $\{z,c_0\}$, respectively. Hence $N'/c_0$ and $N'/c_1$ have $N$-minors using $\{z,c_1\}$ and $\{z,c_0\}$. Thus, by transitivity, $N'$ has an $N$-minor $\tilde{N}$ using $\{c_0,c_1\}$. As $r(N') = r(N) + 1$, there are elements $e$, $f$, and $g$ of $E(N') - \{c_0,c_1\}$ such that $\tilde{N} = N' / e \backslash f,g$. Now $N'/e$   must have two disjoint 2-circuits  that are not in $N'$. Thus $e \in E(N_{01})$. As $e \not\in \{c_0,c_1\}$, it follows that $N_0 \cong M({\mathcal W}_2) \cong N_1$ and, by symmetry, we may assume that $e = a_0$. But $N_{01}/a_0$ does not have an $(M({\mathcal W}_2) \oplus M({\mathcal W}_2))$-minor. Thus \ref{iftwo} holds.

By \ref{biggest} and \ref{iftwo}, $N$ has a single largest component $N_0$ and it is isomorphic to $M({\mathcal W}_2)$. As $N$ is disconnected, we may assume by duality that $N$ has a component  $N_1$ that is isomorphic to $U_{1,k}$ for some $k$ in $\{2,3\}$.  Now take a copy of $U_{2,3}$ with ground set $\{c_0,z,c_1\}$ and adjoin copies of $U_{2,k+1}$ via parallel connection across $c_0$ and $c_1$, letting the resulting matroid be $N_{01}$. Replacing $N_0 \oplus N_1$ by $N_{01}$ in $N$ to give $N'$, we see that $r(N') = r(N) + 1$. Moreover, $N'/c_0$ and $N'/c_1$ have $N$-minors using $\{c_1,z\}$ and $\{c_0,z\}$, respectively. But  $c_0$ and $c_1$ are the only elements $e$ of $N'$ such that $N'/e$ has    two disjoint 2-circuits  that are not in $N'$. Thus $N'$ has no $N$-minor using $\{c_0,c_1\}$. This contradiction completes the proof of the theorem.
\end{proof}

We conclude this section by proving Corollary~\ref{combine}, which demonstrates how two of the basic properties of matroid connectivity are enough to characterize it.

\begin{proof}[Proof of Corollary~\ref{combine}]
Assume that $N \not\cong U_{1,2}$. Then, by Theorem~\ref{minor_induction} and duality, we may assume that $N \cong U_{2,2}$. But $U_{2,2}$ does not have the transitivity property as the matroid $U_{1,2} \oplus U_{1,1}$ shows.
\end{proof}

\section{Three-element sets}
\label{larger}

The notion of $N$-connectivity defined here relies on sets of two elements. Sets of size three have already been an object of some study.
Seymour asked whether every $3$-element set in a $4$-connected non-binary matroid belongs to a $U_{2,4}$-minor but Kahn~\cite{kahn} and Coullard~\cite{cou} answered this question negatively. Seymour \cite{ps2} characterized the internally $4$-connected binary matroids that are $U_{2,3}$-connected, but the problem of completely characterizing  when every triple of elements in an internally $4$-connected   matroid is in a $U_{2,3}$-minor remains  open  \cite[Problem 15.9.7]{oxl2}.

For a  $3$-connected binary matroid $M$ having rank and corank at least three, Theorem~\ref{rank and corank 3} shows  that every triple of elements of  $M$ is in an $M(K_4)$-minor. The next result extends this theorem to connected binary matroids. As the proof, which is based on Lemma~\ref{replace_basepoint2}, is so similar to those appearing earlier, we omit the details.

\begin{proposition}
Let $M$ be a connected binary matroid. For every triple $\{x,y,z\} \subseteq E(M)$, there is an $M(K_4)$-minor using $\{x,y,z\}$ if and only if every matroid in the canonical tree decomposition of $M$ has rank and corank at least $3$.
\end{proposition}

redpajama_set_name: RedPajamaArXiv


What's the Big Deal? | Hangover Soap Co.
Have you ever used soap before? Vegas oddsmakers would say 'no'.
I'm not trying to imply that you're dirty (though, I have heard rumors). But those bars you buy at the grocery store are not soap.
Look closely at the packaging – yep, those nasty things are called "cleansing bars" or "beauty bars" because they legally can't be called soap.
Finding the right combination of oils is as much an art form as it is a science. After lots of testing and experimenting, we believe we've created the formula/recipe for the perfect soap.
Our bars will cleanse your skin without leaving it dry and chapped. And you can be certain that you WON'T be applying any nasty or harmful chemicals. Nothing but love and sweet ingredients to nourish and pamper your skin.
Kaolin Clay detoxifies by pulling impurities out of your skin, and provides a great 'slip' when shaving.
Activated Charcoal is like the 'Mob Boss' of ingredients, taking care of all sorts of skin problems such as acne, bug bites and oily skin. It can even help reduce pore size.


text:
Q: Certificate error in Crystal Reports After the last upgrade of Crystal Reports for Visual Studio, every time I open the report in design view I'm getting multiple certificate errors (see picture below). The errors comes from the SAP web site productupdates.sap.com, which indeed has certificate expired. But I cannot find any references to this site in my project. How to get rid of them? This doesn't happen on all computers, only on mine.
Posted this question in SAP forum - no answer.
Thank you.

A: https://productupdates.sap.com/ is expired.
See whether Crystal Reports->Check for Updates on Start Up is checked
Turning it off and re-opening the solution appears to stop the continuous certificate security alerts.

redpajama_set_name: RedPajamaStackExchange


The list of City Club amenities is extensive.
A full-service health club features cardiovascular & strength training equipment, cycling room and group exercise classes. Locker rooms include whirlpools, saunas, and steam room. Our staff of personal trainers are dedicated to helping you achieve your goals! Our fully-equipped Pilates and Yoga studio offers high quality One-on-One Sessions and Group Classes.
The outdoor recreation center features 13 rubico and hard surface tennis courts, a pro shop, and a 5,500 sq ft state of the art lazy river resort pool.
Our creative kitchen staff prepares award-winning cuisine, served in a casual Members Only restaurant, City Club Grill & Bar.
The elegant banquet rooms can host a wide variety of special events featuring distinctive catering and professional service for both social and business affairs. We understand that every event is unique, so we are delighted to fulfill special requests to create your customized event experience.
Riverspa, is our on-site day spa offering manicures, pedicures, massages, facials, body treatments, and more. City Club Members receive preferred pricing on all services.
The Carriage House Hotel provides the services and amenities of a boutique hotel within a stylish and intimate environment. Our superb location is within minutes of the major business and pleasure destinations in Acadiana. City Club Members receive preferred pricing on all services.
We haven't forgotten the kids! Because. We have a nursery with an enthusiastic staff for those children too young to enjoy the Club on their own. We also have youth fitness classes in our Kids' Club.
City Club at River Ranch strives to be the private club of choice in Lafayette, enhancing the lifestyles of its Members and their Guests by providing recreational, social, business, dining, and fitness services and amenities, proudly delivered by a professional team, which is dedicated to an uncompromising level of service for every experience. Belong to a place where you are sure to always receive a warm welcome and a fond farewell.


Meaning large port in French, Grand Port is on the east island in Mauritius. Mahebourg which is the main town in the Grand Port area was initially built by the Dutch but was later taken over by the French. As a result this acts as a constant reminder of the island's colonial past. Grand Port is the first port of Mauritius. Book your budget flight to Mauritius through Air Africa, to experience some of the island's well known beach areas such as Blue Bay which is not too far from Mahebourg.
Enjoy some diving and snorkeling!
It is an ideal place for diving and snorkeling while also being recognized as one of the best bathing spots in the area. Surrounded by buildings that harken to the colonial days and by Filao trees this exotic location is also good for some windsurfing and yachting. As if this was not enough enticement to visit the Grand Port area, it is also known to be the place where the first humans landed on the island and colonized it. This was back in 1598.
While holidaying in Grand Port also take in such places of interest as the Kestrel Valley. It is located between mountains. Once functioning as private hunting grounds it now primarily serves as a nature reserve for endemic birds such as the Mauritius Kestrel a rare bird of prey. Kestrel Valley offers wonderful nature hikes and trails through pristine woodlands. These guided excursions allow visitors to trek through rare native rainforests filled with meandering streams and rock pools for swimming. Certain areas have been closed off to visitors to maintain its untouched quality for the animals such as the Java deer, wild boar, monkeys and a variety of birds. Make sure to visit the museum and take in the artifacts exhibition and the visitor centre as well as to walk around the tropical garden. Take an evening stroll around Mahebourg's lovely waterfront with convenient benches located all around, for people to just sit, relax and reflect.
Also be sure to sample the rich and tasty Mauritian cuisine that reflects its past with a mix of cultural influences. The British tradition of afternoon tea drinking is adhered to in Grand Port as well. Thanks to Air Africa cheap flights to Grand Port, it is now all the more convenient to visit this lovely area in Mauritius.


text:
Q: How to end a process using C++? I'm trying to create a C++ program  which ends a process. So starting from scratch, I'm not too sure which direction I should be heading? I mean based on the research I've done the solution involves these steps (let me know if it's incorrect logic):

*

*Get process name (in my case its always the same one)

*Iterate through all processes and find matching name (assuming this process has a unique name)

*Get the id and terminate it
As I mentioned I don't have to worry about checking for duplicate process names since its assumed to be unique, but now I need guidance on what sort of library, classes etc I should be looking at?

A: Use the TerminateProcess function.

A: See EnumProcesses and research Handles to processes.
Pass an empty array of type pIDs to EnumProcesses, it will fill it up with the pID of all currently running processes so use a big array. This way, you can enumerate through all running processes.
Use OpenProcess to obtain the handle associated with the PID and call GetModuleFileNameEx to get its fully-qualified path to see if the one you are looking at is the one you want.
Then you can use the TerminateProcess on the handle if it is the one you want to end.


TLDR: On roads with multiple lanes in same direction where the speedlimit is 45mph or under, when cars are stopped you may pass them between the lanes riding at 15mph or under.
Lane filtering will be legal from 2019 May 14 until 2022 July 01 unless the provisions are renewed or established permanently.
Yea, I saw that and wondered how strictly the rules will be enforced, and how the typical overly aggressive driver will feel about it if they missed the change in the law?
I've never lived where it was legal, so I guess I'll have to start working on my technique.
Filtering has been legal in the UK for many years, but a large section of the car driving community dont like it because the bike is making progress and they are not.
I'm genuinely worried about it. I need to think about how and when I'll use it. I recognize the safety concern that has led to filtering being adopted. But cagers' road rage isn't so quantifiable as bikers getting rear ended. This is why the provision will be reevaluated and has automatic sunset.
It's great if used properly. I never get frustrated on a bike in traffic whilst I do in the car sometimes.
I especially like the bit in the law saying the inside lane has to be stationary. Over here most of the accidents now happen because the cages thinks the other lane is moving faster so they change lanes without looking.
Keeping the speed down is a good idea too. We have too many numpties on bikes going up the middle at 70 mph.
Let's face it. If a car is stationary it can't change lanes very quick and if you are only doing 15 mph you should be able to stop pretty quick if a car does pull out.
USSENTERNCC1701E, twag4 and oldjackbob like this.


The APIs given below are erroring out, the same code is running perfectly in 12.2.4. But in 12.2.6 it is throwing error.
Please suggest something for this. If there is any patch or any other solution for this.


text:
Russian Neo-Nazis 'Celebrate' Hitler's Birthday, Part Two: April 20, 2013
Publication: Eurasia Daily Monitor Volume: 10 Issue: 80
By: Richard Arnold
April 29, 2013 09:35 PM Age: 9 years
(Source: AP)
Adolf Hitler's birthday, traditionally a concern for authorities determined to prevent racist violence in Russia (see Part One of this two-part article in EDM, April 15), appears to have passed without much disruption in 2013. At the time of writing, there were relatively few reports of "celebrations" and no news accounts of racist violence directly connected to the Fuehrer's birth. This is possibly due to the increased media attention given to the regions of Dagestan and Chechnya in the wake of the arrest of the Dzokhar Tsarnaev, one of the alleged Boston Marathon bombers: the media had their attention elsewhere. Of course, the lack of overt "celebration" could also be due to more effective policing of the activities of Neo-Nazi groups in Russia, which the government has been pursuing with greater zeal since 2010. For instance, on April 12, the procurator of St. Petersburg added articles from Hitler's Mein Kampf that have appeared on the Russian social networking site Vkontate.ru to the list of materials banned under article 282 of the Russian constitution, which forbids incitement of racial or religious intolerance (http://ria.ru/incidents/20130412/932357545.html). However, given attempts by the international Far Right to present Russia as another "front" in the struggle for white civilization, a vengeance attack for the Boston bombings on Caucasians (referred to by their derogatory Russian epithet, cherny, meaning "black") in Russia by Neo-Nazi groups is not out of the question.
This is not to say that there was no violence this year whatsoever. On April 18, a fight broke out between fans of the soccer team from Rostov-on-Don and Caucasian fans supporting the FC Terek Grozny (Chechnya) team, allegedly started after the Rostov fans began chanting "Seig Heil!" (http://www.sova-center.ru/racism-xenophobia/news/racism-nationalism/2013/04/d26946/). As a result of the fight, two of the Terek Grozny fans were hospitalized with serious wounds to the face and head. In another case, in the center of St. Petersburg on April 21, three young men, one of whom had a shaved head and all of whom had bovver boots and black jackets, inflicted serious wounds on two people who were of "non-Slavic appearance" (http://www.sova-center.ru/racism-xenophobia/news/racism-nationalism/2013/04/d26948/). While such events are terrible, they are not the sort of systemic violence that induced Russian authorities to warn dark-skinned foreign students not to leave their dormitories on Hitler's birthday, a practice that started in 2002.
Others commemorated the day in a more peaceful (though still offensive) fashion. In St. Petersburg, fans of the "Zenit" team unveiled a banner on April 17 in a soccer match against Kuban FC, displaying a symbol popular with Neo-Nazis. On April 18, fans of the Voronezh side unveiled a banner with a Celtic Cross, another symbol popular with skinheads and the white internationalist movement. On April 21, in a match between CSKA Moscow and Spartak, both sets of soccer fans had banners supporting racism: the CSKA fans held up a Celtic Cross and the Spartak fans had a banner proclaiming: "support against tolerance" (http://www.sova-center.ru/racism-xenophobia/news/racism-nationalism/2013/04/d26947/). In Voronezh, on April 20, Neo-Nazis unveiled a banner on a bridge with a portrait of Adolf Hitler, a swastika, and the caption, "Happy Birthday, Hitler" (http://www.sova-center.ru/racism-xenophobia/news/racism-nationalism/2013/04/d26945/). The association of soccer with racism is well-established, and in 2010, a protest at the murder of Spartak Moscow fan Yegor Svidorov by people from the Caucasus allowed Neo-Nazis to organize protests that drew 5,500 on Manezhnaya (Manezhka) Square outside the Kremlin. While the association between soccer and racism is not constrained to Russia, other countries have taken steps to prevent it. The overall moderate tone of events is summed up by the fact that a photograph of a cake with swastika icing, baked by students in Kirov, attracted attention on the internet (http://www.pro-goroda.ru/sobytiya/den-rozhdeniya-gitlera-studenty-ispekli-tort-so-svastikoy-14039). If confectionary counts as news, then there is very little else to report.
The relative calm on Hitler's birthday in 2013 could reflect a change of strategy in the Russian Far Right—a move away from street violence toward political participation based around removing Putin from the Kremlin. The nationalists realize they are gaining in popularity as the only serious opposition to the regime, and some commentators argue that the next Duma will see a nationalist caucus making up as much as 10 percent of the votes (http://svpressa.ru/politic/article/66135/). Certainly this hypothetical change of strategy would be in line with the 2011 visit of skinhead gang leader Dmitrii Demushkin to Chechnya and embrace of Chechens as people with whom Russian nationalists share values (http://www.apn.ru/publications/article25954.htm). The presence of the nationalists in the protests against Putin (Alexei Navalny, whose slogan "stop feeding the Caucasus!" took pride of place in last year's protests, is an ideologue of Russian nationalism) suggests that they find elections a more promising route to power than direct street action. Indeed, the nationalists seem to be more concerned with planning for another "day of protest" against the Putin regime on the traditional holiday of May 1, than with celebrating Hitler's birthday. Protests are planned for at least three Russian cities—Volgograd, Astrakhan and St. Petersburg (http://www.dpni.org/articles/novostnaya/35574/)—under the banner of Imperial Russia (a black, yellow, and gold striped flag, which is a symbol of the far-right Movement Against Illegal Immigration—DPNI). The continued involvement of the nationalists in protests against Putin is, thus, clear. Hitler's birthday in 2013 may have passed off relatively peacefully, but that is no guarantee for other events this year.
The IMU Expansion in Afghanistan's Tak..Next
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Militant Leadership Monitor – July 2018
The July 2018 issue of Militant Leadership Monitor features a brief on Boko Haram's missing commander, Man Chari, as well as profiles of Tihama Resistance Brigade Commander Shaykh Abd al-Rahman Shuae Hajari; IS Russian Propaganda Master Abu Jihad; The New Ideologue-Warrior Leader of TTP, Mufti Noor Wali Mehsud; and Syrian Tiger Forces Commander... MORE

redpajama_set_name: RedPajamaCommonCrawl


Aldora offers one of the largest decorative glass selections for shower enclosures in North America. These patterns can make your shower enclosure a visual highlight of the bathroom.
Aldora shower design and fabrication team will make your custom shower enclosure a reality. Aldora can transfer your drawing into a sophisticated layout that shows all measurements, hinge notches, pull locations and a description of all hardware. Our knowledgeable customer service representative can walk you through your design and help with recommendations.
Aldora's fabrication department takes care to ensure that the glass components of your project are measured correctly, notched and drilled for hardware as specified and kept free from unsightly scratches. Your project will arrive complete along with the hardware requested.


More posts by Neil Jagdish Patel.
More posts by Seif Lotfy.
More posts by Gord Allott.
Axiom was founded by three engineers, co-workers, and friends: Neil Jagdish Patel, Seif Lotfy, and Gord Allott. We have worked together for over a decade, through multiple companies and on multiple products.
Since starting Axiom in 2016, we have worked to usher in a new generation of Cloud Appliances to the world that push User Experience, Performance, Functionality, and Design to a new level.
In our previous companies, our job was to build a brand new Desktop User Experience, or a Crash Reporting and User Analytics service. We always had to rely on supporting services that did not fit the product exactly, and as a result compromises would have to be made.
There is never a reason to do something that provides a worse version of what you want to make, there are always solutions even if it means making it yourself. And so, we never compromise on anything. Design, UX, Privacy, Security, anything.
We endeavour to always provide the best experience that we can no matter the complexity. You can see this in the way we built our installer, or the design of our interfaces, or the consideration around privacy and security of our products.
Our belief is that your data is your data. We design and build products that run directly inside your infrastructure, and ensure that your business data never leaves your firewall. Whether your goal is to run our services on public or private clouds, or even on bare metal, you can rest assured that we've taken every step to ensure that your data stays safely in your hands.
Even when designing our minimal backend services, such as those that would provide out-of-the-box SMS, voice calls, or emails. We have gone as far as to ensure that it's possible to easily replace our default implementations with your own (e.g. your own Twilio and SendGrid api keys).
It is our belief that nobody has any right to your data but you.
No matter how fancy the UI or how easy the installation, the most significant metric when using a data product is speed. How quickly did the page load? How fast did ingestion take place? How quickly were you able to see your data?. You should never need to wait more than a second for something to happen, no matter how complex it is.
Our architecture means we aren't struggling to scale our server backends to customer demands like most cloud services are; 99/100% spent on engineering is spent directly on the product. Providing you with a better experience, faster updates, and a more stable experience.
Privacy and security shouldn't come at the cost of high-maintenance. Many organisations have flocked to SaaS apps over the past few years because the alternative options would essentially require a dedicated team to run them.
So, when building our products, we set strict rules around ease of installation and maintenance. No terminals, no staring at blank screens, and no unsafe scripts. Instead, simple, guided installations make setup a breeze. Once you're ready to go, you know that automatic updates (with automatic rollback should anything go wrong) will always keep you fresh with the latest feature and security updates, All without you having to lift a finger.
We are excited to go on this journey, Watchly is just the beginning.


In the 1990s, hip-hop reached its creative peak. By the 2000s, it reached it commercial peak — bursting through to mainstream audiences like never before.
Let's look at the 10 artists who led the way.
By the way, this list mainly covers artists who dropped solo albums during the grading period. So no Andre 3000 or Pusha T.
Now let's highlight the names that led rap in the early stages of the 21st century.
Edd said: Hip hop's infatuation with "trap rap" in the 2000s put the Snowman firmly in the spotlight. After grinding independently for a few years, he broke out with Thug Motivation 101, solidifying himself as one of the decade's most recognizable voices.
Edd said: Unless you're new here, you know Nelly is far from one of my favorites. But even I can't hate on his phenomenal success in the 2000s, establishing the "hip-pop" genre that dominates airwaves today.
Edd said: While the homie Esco wasn't nearly as successful in the 2000s as he was the prior decade, he still remained one of the game's most influential MCs. He gave us quite a few strong albums during the time period as well.
Edd said: TI basically crowned himself King of the South in the early 2000s, and few challenged him for that throne. Southern sounds heavily influenced hip-hop at the beginning of the 21st century, and Tip led the charge.
Edd said: Luda may well be the MVP of 2000s. It's easy to write him off as a novelty act, but he was more than just a commercial cash cow. Ludacris sprinkled the decade with some of the era's memorable guest verses. Don't sleep on Luda, when he was at the top of his game, he was a force.
Edd said: From underdog to media mogul, 50 spent the first half of the decade dominating the boards and the boardroom. At their peak, 50 and his G Unit brethren were untouchable. Get Rich or Die Tryin' was one of the few albums to live up to its massive hype. By the end of the decade, 50 ran out of steam but you can't overlook his early success.
Edd said: Weezy is an artist who matured right before our eyes. From "wobbledy-wobbledy" Cash Money hypeman to full-fledged superstar in the late 00s, Wayne honed his game until he reached rap's pinnacle. High-profile guest spots and a deluge of mixtapes made him a household name.
Edd said: Love him, hate him, but lord knows no one had an impact on the game like Kanye West. His production skills launched some of rap's best records and inspired countless "soul beat" knockoffs. And his everyman rap style — and evolution into crazed rock star — made him rap's most polarizing figure. And, of course, he's blessed us with a mountain of classic tracks. That's all that counts.
Edd said: A lot of critics give Eminem much too hard of a time, crediting his success to his skin color. Anyone who says that hasn't listened to his discography. Just a few years after his debut, Em was mentioned in the same breath as rap's most legendary lyricists. The '00s became Shady's decade, dominating the charts, smashing sales records and delivering mind-bending bars that few of his peers could match.
Edd said: Jay is hip-hop's juggernaut, and the 2000s defined his legacy. From rap wars with legends to stints in the boardroom before heading back to the booth, Jay became the posterchild for rap success. And sure, he's clearly lost a step creatively in recent years, but he provided more than his share of classics throughout the decade. Jay really was the blueprint.
Now it's your turn. Who are your picks for the best of the 2000s?
P. Diddy ? Should be replaced with Jeezy.
Why would you hate him, he is a good rapper.
There are none of the good ones like Chance the rapper. Where are they at? Kayne and Jay-Z could never be consider good rappers by any means. Where is Kendrick Lamar?
50 Cent is #1, fight me.


Eclipse is an IDE tool that helps us to develop software. Compared to a text editor, eclipse provides many additional useful features to make the development of software easier and faster. Know eclipse better for better productivity.


We are seeing a new virus in the wild that is evading blocks on an EXE filename. It uses malformed MIME which prevents us from detecting the filename of the attachment.
If you decide to use the rule, you should create it in the "default" stream (and "base" realm if you are running CanIt-Domain-PRO).
This page was last modified on 9 May 2014, at 11:33.


text:
‹ Back to Management Team
Rajnish Kapur
President of India Operations
Rajnish Kapur is the President of DMI's India Operations. He is responsible for building, leading, and managing world class delivery capability for all divisions of DMI's business units. Rajnish joined DMI in 2013 and scaled the Indian team from its infancy into a vibrant organization, geared to help make DMI customers successful in their transformation endeavors.
Rajnish has over 25 years of experience in high-end technology consulting, services, and product development industry. He has worked with marquee customers such as, General Electric, Sabre, Amadeus, and TUI. Prior to DMI, Rajnish was a part of the leadership team at MakeMyTrip.com (NASDAQ: MMYT) and served as Chief Innovation Officer. At MakeMyTrip.com, he had a multifaceted role spanning technology, automation, product development, and customer service. Before joining MakeMyTrip, Rajnish was Executive Vice President at Kale Consultants Limited (now Accelya-Kale) where he was responsible for the Travel, Hospitality, and Logistics vertical.
He was also the Co-Founder and Chief Technology Officer of eTravelIndia.com, which started operations in the year 2000. Rajnish was the Founder, Chief Executive and Technology Officer of Cognosys Software, UK, a company he spun and incubated off from eTravelIndia in 2001. Prior to eTravelIndia.com, Rajnish spent 10 years with IIS Infotech Limited (now Sopra-Steria Group) both in India and the UK. He holds a Bachelor's degree in Engineering (Computer Science) from Bangalore University and B.A. (Hons) in Economics from Delhi University.


Wow! What a day! It's 7:30 pm and I am leaving kampong Cham and going Phnom Penh. It's been an eventful day.
It started early this morning in prayer and study looking over the Me kong river. Went from that to breakfast with Pastor Meng Hong. A good man and pastor with a huge heart. After breakfast we went to the hall where the pastors conference was already in progress. There were 54 pastors in attendance. It was a great day of interaction. I taught two leadership development sessions and then we had lunch and when we returned I taught on marriage and ministry. I found that the same relational issues exist everywhere in the world. It was quite funny.
While I was teaching, the team was working at the Hope center. They worked on such projects as painting the gates, creating a new walk way, and two different landscaping projects. Some of the team went to the market to prepare for the party tonight. Buying the clothes, backpacks, school supplies and toiletries. This is not an easy job. Buying for 31 orphans is definitely a major task. A big thanks goes out to signature dental who provided toothbrushes, and toothpaste for all the kids.
So the team and I all met back at the hotel and rushed to get ready for the party. This requires, getting all the elements organized all the gift bags filled all the roles disseminated. Big Props to autumn and Taylor and our Cambodian coordinator Kimmy.
We went to the hall we rented for the party, got the music going, blew up balloons hung up the decorations, set out the stations-coloring, nails/hair/make up for the girls to make them feel special. A balloon station where team members make balloon animals for the kids, a twister station, a painting station, a scarf making session. Just the biggest party you've ever seen.
They love it. Every year they just get so excited. It's like Christmas. We party, dance and then have the big gift give away.
This year I was only able to stay for the very first part so I left to go to phnom penh. I have an early morning meeting with the staff at new life and then a lunch meeting with the expat teams at new life. Looking forward to all of those.
God is so good. All the time! I miss my girls but love what God is doing. At least I have one of my girls with me. The team will finish the party tonight then have a soccer clinic tomorrow with the hope center children.
Then we are almost finished. It's gone fast. I can't say enough about this team. Each one of them have been superb in every way. Everything you want in a team.
Adrienne Jefferis – harvest Okc campus.
You should be proud of these committed people who have given their time and treasure to this cause and they have made a difference.
Every year when we drive up to the hope center the kids are in the front porch clapping and cheering smiling and so happy to see us, and we are equally as happy to see them.
This year there was a noticeable difference. Many of the older kids have grown significantly in the last year. They are so beautiful and respectful and treat us like family.
Every year as they greet us and sing songs and introduce themselves, I think to myself where would they be if they didn't have the hope center. Would some of them be dead. Died of disease or malnutrition, or lack of food? Would some of them be forever stuck in the cycle of poverty, having no education, or decent clothing or medical attention, and never be able to break out. Continuing the cycle into the next generation.
Would some be the victims of a heartless trafficker who thinks children are a commodity to be bought and sold for work or sexual perversion?
The truth is now they are safe, they are covered. They may not have parents who are alive but they have family in this hope center.
Thank God for Pastor Meng Hong and Vat dei for having the courage and strength to do this. It is our pleasure to help support them. It is our honor to come each year and have a party for the kids and distribute 3 new sets of clothes for each of them, to get them all of their school supplies, to bring them new toiletries. To get them bicycles to ride to school, to do upkeep and new projects around the home, and so much more like a washer and dryer, the installation of indoor bathrooms, an overhang so the kids can play in the shade, painting and reprinting the home, landscaping, sewing machine, to make clothes and to teach the kids.
Over these years we have supported them in a significant way by all of this and much more including money to help with ongoing expenses throughout the year.
The most fun thing though is what will happen tonight. Tonight we will throw our annual party for them. Something they don't normally get. We eat together, we dance, we play games, we give them all their new stuff as presents. It's truly like Christmas to them, and it's such a joy to us.
Thank you! From the bottom of my heart Thank You! If you have supported a missionary if you have given towards projects if you have prayed for us. If you have sent a family member to come. Thank you.
This is true religion, this is the heart of God. This is pleasing to the father.
The most important thing is that if these kids were not able to be placed in this home they would probably not know Jesus or have heard of him in any clear way.
This is great if it was just changing the life of one child but the truth is it is changing the life of generations.
Yesterday was phenomenal. I had the privilege of speaking in all the services at the central campus and the team was able to be a part of the opening of a brand new service and campus for new life.
They are reaching so many people. To give some perspective the average Cambodian church that's 10 yrs old or more averages 20 adults. They started this service with 108 adults. This church is truly going to another level.
The service last night was powerful. A huge praise celebration. I spoke on the power of praise and the place erupted in praise. god really touched lives. The team was praying for people and joining in the celebration. It was remarkable to see.
In the afternoon the team was able go to the dump and minister to some of the most impoverished people in phnom penh. It's hard for many west enters to fathom the depth of poverty in the developing world but it is good for everyone who can to see it and do something about it. We are doing what we can. Needless to say yesterday was a good day.
We are going out to kampong Cham today to our hope center there to spend 2 and 1/2 days with the orphans there. We will conduct classes, games, party's, work on the home.
It's a privilege to work for the kingdom.
Team did a great job with feeding program yesterday and ministry at next step houses last night.
The leaders here are like sponges soaking up every word and learning how to grow.
This church is growing and developing. Definitely going to another level. I'm so encouraged by these leaders and their pastors. God is so good and His church is alive and well.
The team is going this afternoon to the killing fields museum then they are leading the youth outreach tonight. Busy, impacting, fun, and exciting day.
of travel the team got on the bus and drive 4 more hours out to the province. Jet lagged and tired they jumped out of the bus, played games with the orphans at the hope center, talked to them and were just absolutely stellar.
If you've never done that than you wouldn't understand how incredibly taxing and challenging that would be, but we all agreed we were not going to let travel problems stop us from connecting with our hope center.
I'm so proud of this team. We were able to bless pastor Sokhan and Esther and their team with resources to help with the amazing job they do with the orphans.
This morning the team went with their feeding program team and went to the village to help with feeding village kids who are in desperate need. While I traveled early this morning to phnom penh to meet with pastor Jesse and their pastoral team. We had a great meeting. The presence of God was present and blessed us all. They have a team of world changers. This church is truly positioned to influence the nation.
Team is on their way back to the city and we will meet up in just a few minutes.
We will connect tonight with university students who have aged out of children's homes. They are orphans. New life has "next step" homes to provide housing and help while they attend university. Great ministry opportunity.
Thanks for the support. Continue to pray.
Arrived in Tai Pei about an hour ago. The long flight seemed short. Most of the team slept most of the way. We are so excited to finally be on this side of the world.
We will be here for about another 2 hrs then off to cambodia. Because of the lost time we will need to take off to the province almost immediately.
So pray for us as this will be a long and tiring day but well worth every second. We can't wait to see Pastor Sokhan and Esther and all the kids.
We are expecting a great time and to be able to help them significantly with the time we have. These pastors are so committed, it is our pleasure to come along side and support and resource.


text:
Evans plays down Tour chances
Aussie star defends reputation
Cadel Evans is remaining cautious about his chances of winning the 2009 Tour de France after second-place finishes in '07 and '08.
The Australian believes his Silence-Lotto team are improving, but says they need more funding if they are to challenge the likes of Astana.
In an interview with Cyclingnews.com, he said: "As a team we've come close in the last few years and for the budget we have, we provide good value for the sponsor.
"We've come second for the last two years and we've been getting better and better. Our team is a bit different to Lance's [Armstrong] team, who he used to co-own. Lotto is a government-owned team so they have other priorities than just the Tour; they're also obliged to promote Belgian cycling at all levels."
Yaroslav Popovych and Robbie McEwen have left the team, while Bernard Kohl has been banished. In come Sebastian Lang, Charly Wegelius and Thomas Dekker in a lineup that looks as though it's built around Evans' needs.
"Last year may have been my best chance to win the Tour [so far], but I had back luck, so we'll just have to wait and see what happens in 2009," he said.
Evans also explained his reaction to an incident in which he batted away the hand of a journalist on last year's Tour.
"What they didn't understand was that I had skin missing off the entire left hand side of my body and that every time someone touched me it was painful," he said.
"There's an amount of disrespect you can tolerate but at certain points it's just too much. I'd just lost yellow and probably the Tour. Then I've got someone tapping me on the shoulder again and again. I think my response wasn't so bad, considering. I know it wasn't a perfect situation."
For the full interview with Cadel Evans, go to Cyclingnews.com.


The taste is rounded, with bright citrus nuances, with hints of honey, vanilla and spices.
Olmeca Tezon Blanco is self-sufficient in its pure form, but can be the basis for the unique cocktails.
Olmeca Tezon Blanco - bright, rich taste of tequila super-premium 100% agave. Olmeca Tezon Blanco won a gold medal from the Institute of tasting drinks in Chicago (Beverage Testing Institute) in 2007.
A distinctive feature of this tequila is a unique way to get the juice of blue agave, it is based on the use of wheel-press, made from volcanic mineral tezontl. Another feature of the production of tequila Olmeca Tezon - double check the resulting alcohol. Firstly - a chemical laboratory analysis of each batch of distillate, and secondly - assessment drink two tasters, whose delicate scent captures the slightest deviation from the sample. Bottle Tezon - it is a work of art, each one numbered for tracking quality control beverage.
In our store you can buy tequila Olmeca Tezon Blanco, 0.75 L, price Olmeca Tezon Blanco, 0.75 L — 4 788 rub.. Producer tequila Olmeca. Delivery Olmeca Tezon Blanco, 0.75 L.


This is a placeholder page for Eakin David, which means this person is not currently on this site. We do suggest using the tools below to find Eakin David.
You are visiting the placeholder page for Eakin David. This page is here because someone used our placeholder utility to look for Eakin David. We created this page automatically in hopes Eakin David would find it. If you are not Eakin David, but are an alumni of Romulus High School, register on this site for free now.


text:
December 21, 2022 @ 10:00 pm - December 22, 2022 @ 1:00 am
« Team Trivia at Pale Fire
The Wreath Barn at Harmony Harvest Farm »
Bring your friends and show off your singing chops. Every Wednesday 10PM-1AM in Ruby's Lounge. 21+ Only.
December 21, 2022 @ 10:00 pm
Drinks, music
booking@clementinecafe.com
Ruby's Lounge
Harrisonburg, VA 22801 United States + Google Map
Movie Musical Magic
An event every day that begins at 6:30 pm, repeating until January 28, 2023
Movie Musical Magic A Musical Theatre Revue
One event on January 28, 2023 at 6:30 pm
January 28, 2023 @ 6:30 pm - 12:00 am


text:
/**
 *
 */
package model;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.ResultSetMetaData;
import java.sql.SQLException;
import java.sql.Statement;

/**
 * Title: Testdb Description: simple hello world db example of a standalone
 * persistent db application
 *
 * every time it runs it adds four more rows to sample_table it does a query and
 * prints the results to standard out
 *
 * Author: Karl Meissner karl@meissnersd.com
 */
public class Database {
    Connection conn; // our connection to the db - persist for life of program

    // we dont want this garbage collected until we are done
    public Database() throws Exception { // note more
					 // general
					 // exception

	// Load the HSQL Database Engine JDBC driver
	// hsqldb.jar should be in the class path or made part of the current
	// jar
	Class.forName("org.hsqldb.jdbcDriver");

	// connect to the database. This will load the db files and start the
	// database if it is not already running.
	// db_file_name_prefix is used to open or create files that hold the
	// state of the db.
	// It can contain directory names relative to the
	// current working directory
	conn = DriverManager.getConnection(
		"jdbc:hsqldb:file:../contas/bin/database/db;shutdown=true",
		"sa", "");
    }

    public void shutdown() throws SQLException {

	Statement st = conn.createStatement();

	// db writes out to files and performs clean shuts down
	// otherwise there will be an unclean shutdown
	// when program ends
	st.execute("SHUTDOWN");
	conn.close(); // if there are no other open connection
    }

    // use for SQL command SELECT
    public synchronized ResultSet query(String expression) throws SQLException {

	Statement st = null;
	ResultSet rs = null;

	st = conn.createStatement(); // statement objects can be reused with

	// repeated calls to execute but we
	// choose to make a new one each time
	rs = st.executeQuery(expression); // run the query

	// do something with the result set.
	return rs;
	// st.close();

	// NOTE!! if you close a statement the associated ResultSet
	// is closed too
	// so you should copy the contents to some other object.
	// the result set is invalidated also if you recycle an Statement
	// and try to execute some other query before the result set has been
	// completely examined.
    }

    // use for SQL commands CREATE, DROP, INSERT and UPDATE
    public synchronized void update(String expression) throws SQLException {

	Statement st = null;

	st = conn.createStatement(); // statements

	int i = st.executeUpdate(expression); // run the query

	if (i == -1) {
	    System.out.println("db error : " + expression);
	}

	st.close();
    } // void update()

    public static void dump(ResultSet rs) throws SQLException {

	// the order of the rows in a cursor
	// are implementation dependent unless you use the SQL ORDER statement
	ResultSetMetaData meta = rs.getMetaData();
	int colmax = meta.getColumnCount();
	int i;
	Object o = null;

	// the result set is a cursor into the data. You can only
	// point to one row at a time
	// assume we are pointing to BEFORE the first row
	// rs.next() points to next row and returns true
	// or false if there is no next row, which breaks the loop
	for (; rs.next();) {
	    for (i = 0; i < colmax; ++i) {
		o = rs.getObject(i + 1); // Is SQL the first column is indexed

		// with 1 not 0
		System.out.print(o.toString() + " ");
	    }

	    System.out.println(" ");
	}
    } // void dump( ResultSet rs )
}

redpajama_set_name: RedPajamaGithub


The Conduit Magazine is a leisure magazine covering West Dorset and South Somerset with regular garden features and lots of plant sales taking place in the area. The gardening column is written by Castle Gardens.
Great location for weddings and special events. Plus of course our May plant fair.
Superb Dorset manor house recently used for filming. Restaurant, shop and great gardens. Weddings and special events facilities.
Collection of traditional hardy perennials cottage garden plants and new introductions. Home made chutneys, jams and jellies from locally sourced fruits and vegetables.
Snape cottage garden an interesting garden to visit, also the home of "snapestakes", plant supports and obelisks.
Cottage garden perennials, especially hardy geraniums and good bee plants.