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\section{Further ideas}
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\title[Conjugation-free geometric groups of arrangements II]{A conjugation-free geometric presentation of fundamental groups of arrangements II: Expansion and some properties}

\author[Eliyahu, Garber, Teicher]{Meital Eliyahu$^1$, David Garber and Mina Teicher}

\stepcounter{footnote}
\footnotetext{Partially supported by the Israeli Ministry of Science and Technology.}

\address{Meital Eliyahu, Department of Mathematics, Bar-Ilan University, 52900 Ramat-Gan, Israel}
\email{eliyahm@macs.biu.ac.il}

\address{David Garber, Department of Applied Mathematics, Faculty of
  Sciences, Holon Institute of Technology, 52 Golomb st., PO
  Box 305, 58102 Holon, Israel}
\email{garber@hit.ac.il}

\address{Mina Teicher, Department of Mathematics, Bar-Ilan University, 52900 Ramat-Gan, Israel}
\email{teicher@macs.biu.ac.il}

\begin{abstract}
A {\em conjugation-free geometric presentation} of a fundamental group
is a presentation with the natural topological generators $x_1, \dots, x_n$
and the cyclic relations:
$$x_{i_k}x_{i_{k-1}} \cdots x_{i_1} = x_{i_{k-1}} \cdots x_{i_1} x_{i_k} = \cdots = x_{i_1} x_{i_k} \cdots x_{i_2}$$
with no conjugations on the generators.

We have already proved in \cite{EGT} that if the graph of the arrangement is a disjoint union of cycles, then its fundamental group has a conjugation-free geometric presentation.
In this paper, we extend this property to arrangements whose graphs are a disjoint union of cycle-tree graphs.

Moreover, we study some properties of this type of presentations for a fundamental group of a line arrangement's complement. We show that these presentations satisfy a completeness property in the sense of Dehornoy, if the corresponding graph  of the arrangement has no edges. The completeness property is a powerful property which leads to many nice properties concerning the presentation (such as the left-cancellativity of the associated monoid and yields some simple criterion for the solvability of the word problem in the group).
\end{abstract}

\keywords{conjugation-free presentation, fundamental group, complete presentation, complemented presentation\\
\indent  {\it MSC2010:} Primary: 14H30; Secondary: 32S22, 57M05, 20M05, 20F05.}

\maketitle

\section{Introduction}
The fundamental group of the complement of plane curves is a very important topological invariant, which can be also computed for line arrangements. We present here some applications of this invariant.

Chisini \cite{chisini}, Kulikov \cite{Kul,Kul2} and Kulikov-Teicher \cite{KuTe} have used the fundamental group of complements of branch curves of
generic projections in order to distinguish between connected components of the moduli space of smooth projective surfaces, see also \cite{FrTe}.

Moreover, the Zariski-Lefschetz hyperplane section theorem (see \cite{milnor})
states that:
$$\pi_1 (\C\P^N \setminus S) \cong \pi_1 (H \setminus (H \cap S)),$$
where $S$ is a hypersurface and $H$ is a generic 2-plane.
Since $H \cap S$ is a plane curve, the fundamental groups of complements of curves
can be used also for computing the fundamental groups of complements of hypersurfaces in $\C\P^N$.

A different need for computations of fundamental groups is to
obtain more examples of Zariski pairs \cite{Z1,Z2}. A pair of plane
curves is called {\it a Zariski pair} if they have the
same combinatorics (to be exact: there is a degree-preserving bijection between the set of irreducible
components of the two curves $C_1,C_2$, and there exist regular neighbourhoods of the curves $T(C_1),T(C_2)$
such that the pairs $(T(C_1),C_1),(T(C_2),C_2)$ are homeomorphic and the homeomorphism respects the bijection
above \cite{AB-CR}), but their complements in $\P^2$ are not homeomorphic.
For a survey, see \cite{ABCT}.

It is also interesting to explore new finite non-abelian groups which
serve as fundamental groups of complements of plane curves in general;
see for example \cite{AB,AB1,Deg,Z1}.

\medskip

An {\it affine line arrangement} in $\C^2$ is a union of copies of $\C^1$ in $\C^2$. Such an arrangement is called {\em real} if the defining equations of all its lines can be written with real coefficients, and {\em complex} otherwise. Note that the intersection of a real arrangement with the natural copy of $\R^2$ in $\C^2$ is an arrangement of lines in the real plane, called the {\it real part} of the arrangement.

Similarly, a {\it projective line arrangement} in $\C\PP^2$ is a union of copies of $\C\PP^1$ in $\C\PP^2$. Note that the realization of the MacLane configuration \cite{maclane} is an example of a complex arrangement; see also \cite{ABC,Ry}.

For real and complex line arrangements $\mathcal L$, Fan \cite{Fa2} defined a graph $G(\mathcal L)$ which is associated to its multiple points (i.e., points where more than two lines are intersected). We give here its version for real arrangements: Given a real line arrangement $\mathcal L$, the graph $G(\mathcal L)$ of multiple points lies on the real part of $\mathcal L$. It consists of the multiple points of $\mathcal L$, with the
segments between the multiple points on lines which have at least two multiple points. Note that if the arrangement consists of three
multiple points on the same line, then $G(\mathcal L)$ has three vertices on the same line (see Figure \ref{graph_GL}(a)).
If two such lines happen to intersect in a simple point (i.e., a point where exactly two lines are intersected), it is ignored
(i.e., the lines do not meet in the graph). See another example in Figure \ref{graph_GL}(b) (note that Fan's definition gives a graph different from the graph defined in \cite{JY,WY}).

\begin{figure}[!ht]
\epsfysize 4cm
\centerline{\epsfbox{graph_GL.eps}}
\caption{Examples for $G(\mathcal L)$}\label{graph_GL}
\end{figure}

\medskip

In \cite{EGT} we introduce the notion of a {\it conjugation-free geometric presentation} of the fundamental group of an arrangement:
\begin{defn}
Let $G$ be a fundamental group of the affine or projective complements of some line arrangement with $n$ lines. We say that $G$ has {\em a conjugation-free geometric presentation} if $G$ has a presentation with the following properties:
\begin{itemize}
\item In the affine case, the generators $\{ x_1,\dots, x_n \}$ are the meridians of lines at some far side of the arrangement, and therefore the number of generators is equal to $n$.
\item In the projective case, the generators are the meridians of lines at some far side of the arrangement except for one, and therefore the number of generators is equal to  $n - 1$.
\item In both cases, the relations are of the following type:
$$x_{i_k} x_{i_{k-1}} \cdots x_{i_1} = x_{i_{k-1}} \cdots x_{i_1} x_{i_k} = \cdots = x_{i_1} x_{i_k} \cdots x_{i_2},$$
where $\{ i_1,i_2, \dots , i_k \} \subseteq \{1, \dots, m \}$ is an increasing subsequence of indices,
where $m=n$ in the affine case and $m=n-1$ in the projective case. Note that for $k=2$, we get the usual commutator.
\end{itemize}

\end{defn}

Note that in usual geometric presentations of the fundamental group, most of the relations have conjugations.

\medskip

The importance of this family of arrangements is that the fundamental group can be read directly from the arrangement or equivalently from its incidence lattice (where the {\em incidence lattice} of an arrangement is the partially-ordered set of non-empty intersections of the lines, ordered by
inclusion, see \cite{OT}) without any computation. Hence, for this family of arrangements, the incidence lattice determines the fundamental group of the complement (this is based on Cordovil \cite{cor}, too).

\medskip

We start with the easy fact that there exist arrangements whose fundamental groups have no conjugation-free geometric presentation: The fundamental group of the Ceva arrangement (also known as the {\it braid arrangement}, appears in Figure \ref{ceva}) has no conjugation-free geometric presentation (see \cite{EGT}).

\begin{figure}[!ht]
\epsfysize 4cm
\centerline{\epsfbox{ceva.eps}}
\caption{Ceva arrangement}\label{ceva}
\end{figure}

Note also that if the fundamental groups of two arrangements $\cL_1,\cL_2$  have conjugation-free geometric presentations and the arrangements intersect transversally, then the fundamental group of $\cL_1 \cup \cL_2$ has a conjugation-free geometric presentation, too. This is due to the important result of Oka and Sakamoto \cite{OkSa}:
\begin{thm}\label{OS_thm} {\bf (Oka-Sakamoto)} Let $C_1$ and $C_2$ be algebraic plane curves in $\C ^2$.
Assume that the intersection $C_1 \cap C_2$
consists of distinct $d_1 \cdot  d_2$ points, where $d_i \ (i=1,2)$ are the
respective degrees of $C_1$ and $C_2$. Then:
$$\pi _1 (\C ^2 - (C_1 \cup C_2)) \cong \pi _1 (\C ^2 -C_1) \oplus \pi _1 (\C ^2 -C_2)$$
\end{thm}

The main result of \cite{EGT} is:
\begin{prop}
The fundamental groups of following family of arrangements have a conjugation-free geometric presentation: a real arrangement $\mathcal L$, where $G(\mathcal L)$ is a disjoint union of cycles of any length, and the multiplicities of the multiple points are arbitrary.
\end{prop}

\medskip

In this paper, we continue the investigation of the family of arrangements whose fundamental groups have conjugation-free geometric presentations in two directions.
First, we extend this property to real arrangements whose graphs are a disjoint union of cycle-tree graphs, where an example for a cycle-tree graph is presented in Figure \ref{cycle-tree-intro}
(see Definition \ref{def-cycle-tree} below).

\begin{figure}[!ht]
\epsfysize 3cm
\centerline{\epsfbox{cycle-tree.eps}}
\caption{An example of a cycle-tree graph}\label{cycle-tree-intro}
\end{figure}

In the second direction, we study some properties of this type of presentations for a fundamental group of a line arrangement's complement. We prove:
\begin{prop}\label{prop-complete}
Let $\cL$ be a real arrangement whose fundamental group has a conjugation-free geometric presentation and its graph $G(\cL)$ has no edges. Then the presentation of the corresponding monoid is complete (and complemented).
\end{prop}

The completeness property is a powerful property which leads to many nice properties concerning the presentation (such as the left-\break cancellativity of the associated monoid and yields some simple criterion for the solvability of the word problem in the group and for Garside groups).

\medskip

The paper is organized as follows. In Section \ref{line_addition}, we prove that arrangements whose graphs are a disjoint union of cycle-tree graphs have a conjugation-free geometric presentation of the fundamental group of the complement. In Section \ref{sec-complemented}, we prove that conjugation-free geometric presentations are complemented presentations. Section \ref{sec-complete} deals with complete presentations, and includes the proof of Proposition \ref{prop-complete}.

\section{Adding a line through a single point preserves the conjugation-free geometric presentation}\label{line_addition}

We start with the following obvious observation, which is based on the Oka-Sakamoto decomposition theorem (see Theorem \ref{OS_thm} above):
\begin{obs}\label{obs_line_add}
Let $\cL$ be an arrangement whose fundamental group has a conjugation-free geometric presentation.
Let $L$ be a line which intersects $\cL$ transversally. Then
$\cL \cup L$ is also an arrangement whose fundamental group has a conjugation-free geometric presentation.
\end{obs}

In this section, we prove the following proposition, which is the next step:

\begin{prop}\label{prop_line_add}
Let $\cL$ be a real arrangement whose affine fundamental group has a conjugation-free geometric presentation.
Let $L$ be a line not in $\cL$, which passes through one intersection point $P$ of $\cL$. Then
$\cL \cup L$ is also an arrangement whose affine fundamental group has a conjugation-free geometric presentation.
\end{prop}

\begin{proof}
We can assume that the point $P$ is the leftmost and lowest point of the arrangement $\cL$ and all the intersection points of the line $L$ (except for $P$) are to the left of all the intersection points of the arrangement $\cL$ (except for $P$). We can also assume that the highest line in $\cL$ (with respect to the global numeration of the lines) passes through $P$. See Figure \ref{fig_L_union_l} for an illustration, where the arrangement $\cL$ is in the dashed rectangle.

\begin{figure}[!ht]
\epsfysize 4cm
\centerline{\epsfbox{fig_L_union_l.eps}}
\caption{An illustration of the real part of $\cL \cup L$}\label{fig_L_union_l}
\end{figure}

The above assumption is due to the following: First, one can rotate a line that participates in only one multiple point as long as it does not unite with a different line (by Results 4.8 and 4.13 of \cite{GTV}). Second, moving a line that participates in only one multiple point over a different line (see Figure \ref{triangle-line}) is permitted in the case of a triangle due to a result of Fan \cite{Fa2} that the family of configurations with $6$ lines and three triple points is connected by a finite sequence of smooth equisingular deformations. Moreover, by Theorem
4.11 of \cite{GTV}, one can assume that the point $P$ is the leftmost point of the arrangement $\cL$.

\begin{figure}[!ht]
\epsfysize 3cm
\centerline{\epsfbox{triangle-line.eps}}
\caption{Moving a line that participates in only one multiple point over a different line}\label{triangle-line}
\end{figure}

Let $n$ be the number of lines in $\cL$ and let $m$ be the multiplicity of $P$ in $\cL$.
So, the list of Lefschetz pairs of the arrangement $\cL$ is $$([a_1,b_1],[a_2,b_2],\dots,[a_{q-1},b_{q-1}],[1,m]),$$
where the Lefschetz pair $[1,m]$ corresponds to the point $P$ (for the theory used here for computing the fundamental group of the complements of arrangements, see \cite{EGT,GaTe,vK,BGT1}).
Since we have that $\pi_1(\C^2-\cL)$ has a conjugation-free geometric presentation, then we know that all the conjugations in the relations
induced by the van Kampen Theorem \cite{vK} (see also \cite{EGT}) can be simplified.

Now, let us deal with the arrangement $\cL \cup L$. By our assumptions, its list of Lefschetz pairs is (we write in small brackets the name of the corresponding point):
$$([a_1+1,b_1+1]_{(p_1)},[a_2+1,b_2+1]_{(p_2)},\dots,[a_{q-1}+1,b_{q-1}+1]_{(p_{q-1})},$$
$$[1,m+1]_{(p_q)},[m+1,m+2]_{(p_{q+1})},[m+2,m+3]_{(p_{q+2})},\dots, [n,n+1]_{(p_{q+(n-m)})} ).$$

\medskip

We start with the relations induced from intersection points on the line $L$. We first choose a set of $n+1$ generators of the fundamental group of its complement corresponding to its lines, namely $\{x_1,\dots,x_{n+1}\}$.
By the Moishezon-Teicher algorithm \cite{GaTe,BGT1} (see also \cite{EGT}), we now compute the skeletons corresponding to the points on the line $L$ (i.e. the points $p_j$, where $q \leq j \leq q+n-m$). Note that:
\begin{tiny}$$ \Delta \langle a_1+1,b_1+1 \rangle \Delta \langle a_2+1,b_2+1 \rangle \cdots \Delta \langle a_{q-1}+1,b_{q-1}+1 \rangle = \Delta \langle 2,n+1 \rangle \Delta ^{-1} \langle 2,m+1 \rangle,$$\end{tiny}
\hspace{-5pt}since given an arrangement, the multiplication of all the halftwists based on its Lefschetz pairs is equivalent to a unique halftwist of all the lines. By this observation, we get the skeletons in Figure \ref{bm_L_union_l}.

\begin{figure}[!ht]
\epsfysize 2cm
\centerline{\epsfbox{bm_L_union_l.eps}}
\caption{The skeletons of the points $p_{q+j}$ where\break $0 \leq j \leq n-m$}\label{bm_L_union_l}
\end{figure}

Hence, we get the following relations:

\medskip

\noindent
For the point $p_q$:
\begin{eqnarray*}
x_{n+1} x_n\cdots x_{n-m+2} x_1 & = & x_1 x_{n+1} x_n\cdots x_{n-m+2}=\\
&=& \cdots = x_n\cdots x_{n-m+2} x_1 x_{n+1}
\end{eqnarray*}
For the points $p_j$, where $q+1 \leq j \leq q+n-m$:
$$x_{n-m+2-j} x_1=  x_1 x_{n-m+2-j}$$
These relations are obviously without conjugations.

\medskip

Now, we move to the relations induced from points appearing in the original arrangement $\cL$.
The only change in the level of the Lefschetz pairs is an addition of one index in all the pairs, due to the line $L$.
Therefore, the induced braid monodromy and the relations will be changed by adding $1$ to every index, i.e.,
if we have a relation which involves the generators $x_{i_1},\dots, x_{i_k}$, then after adding the line, we have the same relation but with
generators $x_{i_1+1},\dots, x_{i_k+1}$, respectively.

Now, we know that the fundamental group of $\cL$ has a conjugation-free geometric presentation;
hence we have that by a simplification process, one can reach a presentation without conjugations.
If we imitate the simplification process of the presentation of the fundamental group of $\cL$ for the presentation of the fundamental group of $\cL \cup L$,
the cases in which we need to use the relations induced from the point $P$ are the relations that have been simplified by using the relations induced from $P$ before adding the line.
As above,  the  original relations induced from $P$ are:
\begin{eqnarray*}
R_p: \quad x_n x_{n-1}\cdots x_{n-m+1} &=& x_{n-1}\cdots x_{n-m+1} x_n =\\
& = & \cdots = x_{n-m+1} x_n \cdots x_{n-m+2},
\end{eqnarray*}
while the new ones are:
\begin{eqnarray*}
\tilde{R}_p: \quad x_{n+1} x_n\cdots x_{n-m+2} x_1 & = & x_1 x_{n+1} x_n\cdots x_{n-m+2}=\\
& = & \cdots = x_n\cdots x_{n-m+2} x_1 x_{n+1}.
\end{eqnarray*}

We can divide the relations induced from $\cL$ before adding the line $L$ into two subsets:
\begin{enumerate}
\item Relations that during the simplification process contain the subword $x_{n-m+2}^{-1} \cdots x_{n-1}^{-1} x_n x_{n-1} \cdots x_{n-m+2}$.
\item Relations that do not contain the above subword during its simplification process.
\end{enumerate}

For the second subset, the simplification process will be identical before adding the line $L$ and after it, since all the other relations induced by $\cL$ have not been changed by adding the line $L$ (except for adding 1 to the indices).

For the first subset, let us denote the relation by $R$.  Except for applying the relations induced from $P$, the rest of simplification process is identical to the one before adding the line (again, except for adding 1 to the indices). The only change is in the step of applying $R_p$. In this step, before adding the line $L$, the generator $x_1$ has not been involved in $R_p$, but after adding the line $L$, it appears in $\tilde{R}_p$. Hence, for applying $\tilde{R}_p$, we have to conjugate relation $R$ by $x_1$, and using the commutative relations which $x_1$ is involved in, we can diffuse $x_1$ into the relation $R$, so we can use the relation $\tilde{R}_p$ instead of $R_p$.

Hence, we can simplify all the conjugations in all the relations, so we have a conjugation-free geometric presentation, as needed.
\end{proof}

\begin{rem}
Note that adding a line which closes a cycle in $\cL$ might not preserve the conjugation-free geometric presentation property. For example, adding a line to an arrangement of 5 lines which creates the Ceva arrangement (see Figure \ref{ceva}) is not an action which preserves the conjugation-free geometric presentation property.
\end{rem}

Hence, we can extend the family of arrangements whose fundamental groups have a conjugation-free geometric presentation. We start with the following definition:

\begin{defn}\label{def-cycle-tree}
A {\em cycle-tree graph} is a graph which consists of a cycle, where each vertex of the cycle can be a root  of a tree, see Figure \ref{cycle-tree}.
It is possible that there exist some vertices also in the middle of an edge of the cycle or the trees.

\begin{figure}[!ht]
\epsfysize 3cm
\centerline{\epsfbox{cycle-tree.eps}}
\caption{An example of a cycle-tree graph}\label{cycle-tree}
\end{figure}

\end{defn}

\begin{cor}
Let $\cL$ be a real line arrangement whose graph is a disjoint union of cycle-tree graphs. Then the fundamental group of $\cL$ has a conjugation-free geometric presentation.
\end{cor}

\begin{proof}
We start by proving that a real arrangement whose graph is a cycle-tree graph has a fundamental group which has a conjugation-free geometric presentation.
We already have from \cite{EGT} that a real arrangement whose graph is a cycle has a fundamental group which has a conjugation-free geometric presentation. By Proposition \ref{prop_line_add}, adding a line which is either transversal to an arrangement or passes through one intersection point, preserves the property that the fundamental group has a conjugation-free geometric presentation. One can easily construct an arrangement whose graph is a cycle-tree graph from an arrangement whose graph is a cycle by inductively adding a line which is either transversal to the arrangement or passes through one of its intersection points. Hence, we get that an arrangement whose graph is a cycle-tree graph has a fundamental group which has a conjugation-free geometric presentation.

In the next step, using the theorem of Oka and Sakamoto \cite{OkSa} (see Theorem \ref{OS_thm} above), we can generalize the result from the case of one cycle-tree graph to the case of a disjoint union of cycle-tree graphs.
\end{proof}

\section{Complemented presentations}\label{sec-complemented}

A {\em semigroup presentation} $(\mathcal S,\mathcal R)$ consists of a nonempty set $\mathcal S$ and a family of pairs of nonempty words
$\mathcal R$ in the alphabet $\mathcal S$. The corresponding monoid $(\mathcal S,\mathcal R)$ is
$\langle \mathcal S | \mathcal R \rangle ^+ \cong (\mathcal S^*/\equiv _{\mathcal R}^+)$.

Dehornoy \cite{Deh1} has defined the notion of a {\it complemented presentation} of a semigroup:

\begin{defn}
A semigroup presentation $(\mathcal S,\mathcal R)$ is called
{\em complemented} if, for each $s \in \mathcal S$, there is no relation $s \ldots = s \ldots$ in $\mathcal R$ and, for $s, s' \in \mathcal S$,
there is at most one relation $s \ldots = s' \ldots$ in $\mathcal R$.
\end{defn}

Our type of presentations satisfies this property as follows:
\begin{lem}
 A conjugation-free geometric presentation is a complemented presentation.
\end{lem}

\begin{proof}
Any pair of lines intersect exactly once, hence their corresponding generators appear as prefixes in exactly one relation.
Since there are no conjugations, this is their unique appearance as a pair of prefixes.
\end{proof}

\begin{rem} \ \\
\begin{enumerate}
\item This property does not hold for presentations of fundamental groups in general (due to the conjugations in the relations).
\item This property does not hold in the homogeneous minimal presentations introduced by Yoshinaga \cite{Yos}.
\end{enumerate}
\end{rem}

\section{Complete presentations}\label{sec-complete}

In this section, we will study which cases of conjugation-free geometric presentations are also {\em complete} in the sense of Dehornoy \cite{Deh2}.
The completeness property is a very important and powerful property.  In Section \ref{complete_back},
we supply some background on this property and some of its applications. In Section \ref{result}, we present our results in this direction.

\subsection{Background on complete presentations}\label{complete_back}

We follow the survey of Dehornoy \cite{Deh3}.
We start by defining the notion of a {\it word reversing}:

\begin{defn}
For a semigroup presentation $(\mathcal S,\mathcal R)$ and\break $w,w' \in ({\mathcal S} \cup {\mathcal S}^{-1})^*$,
{\em $w$ reverses to $w'$ in one step}, denoted by $w \curvearrowright _{\mathcal R} ^1 w'$,
if there exist a relation $sv' = s'v$ of $\mathcal R$ and $u,u'$ satisfying:
$$ w = us^{-1}s'u' \ {\rm and} \  w'= uv'v^{-1}u'.$$
We say that {\em $w$ reverses to $w'$ in $k$ steps}, denoted by $w \curvearrowright _R ^k w'$, if there exist words $w_0, \dots ,w_k$
satisfying $w_0 = w, w_k = w'$ and $w_i \curvearrowright _R ^1 w_{i+1}$ for each $i$.
The sequence $(w_0, \dots ,w_k)$ is called {\em an $R$-reversing sequence from $w$ to $w'$}.

We write $w \curvearrowright w'$, if $w \curvearrowright _{\mathcal R} ^k w'$ holds for some $k \in \mathbb{N}$.
\end{defn}

\begin{defn}
A semigroup presentation $(\mathcal S,\mathcal R)$ is called {\em complete} if, for all words $w,w' \in \mathcal S^*$:
$$w \equiv ^+_{\mathcal R} w'  \quad \Rightarrow \quad  w^{-1}w' \curvearrowright _{\mathcal R} \varepsilon .$$
where $\varepsilon$ is the empty word.
\end{defn}

In the next definition, we define the {\it cube condition}, which is a useful tool for verifing the completeness property.

\begin{defn}
Let $(\mathcal S, \mathcal R)$ be a semigroup presentation, and\break $u, u', u'' \in \mathcal S^*$. We say that $(\mathcal S, \mathcal R)$ satisfies the {\em cube condition for $(u, u', u'')$} if:
$$u^{-1} u'' u''^{-1} u' \curvearrowright_{\mathcal R} v' v^{-1} \qquad \Rightarrow \qquad
(u v')^{-1} (v u') \curvearrowright_{\mathcal R} \varepsilon .$$

For $X \subseteq \mathcal S^*$, we say that $(\mathcal S, \mathcal R)$ satisfies {\em the cube condition on $X$} if it satisfies the cube condition for every triple $(u, u', u'')$ where $u, u', u'' \in X$.
\end{defn}

\begin{figure}[!ht]
\epsfysize 3cm
\centerline{\epsfbox{cube.eps}}
\caption{An illustration of the cube condition}\label{cube}
\end{figure}

\begin{defn}
A semigroup presentation $(\mathcal S, \mathcal R)$ is said to be {\em homogeneous} if there exists an $\equiv^+_{\mathcal R}$-invariant mapping $\lambda: \mathcal S^* \to \mathbb{N}$
satisfying, for $s \in \mathcal S$ and $w \in \mathcal S^*$, $$\lambda(s w) > \lambda(w).$$
\end{defn}

A typical case of a homogeneous presentation is where all relations in $\mathcal R$ preserve the length of words, i.e., they have the form
$v'=v$ where $v'$ and $v$ have the same length.

Dehornoy \cite{Deh2} has proved the following result:
\begin{prop}
Assume that $(\mathcal S, \mathcal R)$ is a homogeneous semigroup presentation.  Then
$(\mathcal S, \mathcal R)$ is complete if and only if it satisfies the cube condition on $\mathcal S$.
\end{prop}

The next definition is needed for introducing an operation used in an equivalent condition for the cube condition:

\begin{defn}
For a complemented semigroup presentation $(\mathcal S, \mathcal R)$ and $w, w' \in \mathcal S^*$, the {\em $\mathcal R$-complement of $w'$ in $w$, denoted $w \backslash w'$}, (``$w$ under $w'$''), is the unique word $v' \in \mathcal S^*$ such that $w^{-1} w'$ reverses to $v'v^{-1}$ for some $v \in \mathcal S^*$, if such a word exists.
\end{defn}

Dehornoy \cite{Deh2} has proved that the cube condition is equivalent to some expression involving the complement operation:
\begin{prop}
Assume that $(\mathcal S, \mathcal R)$ is a complemented semigroup presentation. Then, for all words $u, u', u'' \in \mathcal S^*$, the following are equivalent:
\begin{enumerate}
\item $(\mathcal S, \mathcal R)$ satisfies the cube condition on $\{u, u', u'' \}$.
\item either $(u \backslash u') \backslash (u \backslash u'')$ and
$(u' \backslash u) \backslash (u' \backslash u'')$ are $\mathcal R$-equivalent or they are not defined,
and the same holds for all permutations of $u, u', u''$.
\end{enumerate}
\end{prop}

Here, we survey some important consequences and applications arising
from the completeness property.

\begin{prop}[\cite{Deh2}, Proposition 6.1]
Every monoid that admits a complete complemented presentation is left-cancellative (i.e., $xy=xz \Rightarrow y=z$).
\end{prop}

\begin{prop}[\cite{Deh2}, Proposition 6.10]
Assume that $(\mathcal S, \mathcal R)$ is a complete semigroup presentation. If $(\mathcal S,\mathcal R)$ is
complemented, then the monoid $\langle \mathcal S | \mathcal R \rangle^+$ admits least common multiples.
\end{prop}

\subsection{Completeness of conjugation-free geometric presentations} \label{result}
In this section, we prove that a conjugation-free geometric presentation is complete if its corresponding graph has no edges:

\begin{prop}
Let $\cL$ be a real arrangement whose fundamental group has a conjugation-free geometric presentation and its graph $G(\cL)$ has no edges. Then the presentation of the corresponding monoid is complete (and complemented).
\end{prop}

\begin{proof}
It is obvious that the conjugation-free geometric presentations are homogeneous (since all the words in the same relation are of the same length).  Hence, we prove this proposition by verifying the equivalent version of the cube condition (for any triple $(u,u',u'') \in ({\mathcal S^*})^3$, the words $(u \backslash u') \backslash (u \backslash u'')$ and $(u' \backslash u) \backslash (u' \backslash u'')$ are $\mathcal R$-equivalent) case-by-case.

\medskip

\begin{enumerate}
\item[Case 1:] The three generators correspond to three lines $\ell_i,\ell_j,\ell_k$ intersecting in three simple points, see Figure \ref{case1}.

\begin{figure}[!ht]
\epsfysize 2cm
\centerline{\epsfbox{case1.eps}}
\caption{Case 1}\label{case1}
\end{figure}

In this case, the relations induced by the three simple points are: $[x_i,x_j]=[x_i,x_k]=[x_j,x_k]=e$, where $x_i,x_j,x_k$ are the generators of
the lines $\ell_i,\ell_j,\ell_k$ respectively.
So, we have:
$$(x_i \backslash x_j) \backslash (x_i \backslash x_k) = x_k = (x_j \backslash x_i) \backslash (x_j \backslash x_k),$$
which are indeed $\mathcal R$-equivalent.

By symmetry, this holds to any permutation of $x_i,x_j,x_k$ as needed.

\medskip

\item[Case 2:] The three generators correspond to three lines $\ell_i,\ell_j,\ell_k$ passing through the same multiple point, see Figure \ref{case2}. For this case, we have two subcases: in the first case, the corresponding lines appear consecutively in the intersection point. In the second case, the corresponding lines appear separately in the intersection point.

\begin{figure}[!ht]
\epsfysize 3cm
\centerline{\epsfbox{case2.eps}}
\caption{Case 2}\label{case2}
\end{figure}

\medskip

\begin{enumerate}
\item[Case 2a:] {\it The lines appear consecutively in the intersection point:} Without loss of generality, we can assume that the multiple point has multiplicity 4, see Figure \ref{case2}(a). Hence, the relations induced by this multiple point are:
    $$x x_k x_j x_i = x_k x_j x_i x = x_j x_i x x_k = x_i x x_k x_j,$$
    where $x_i,x_j,x_k,x$ are the generators of the lines $\ell_i,\ell_j,\ell_k,\ell$ respectively.
    Hence, we have:
    $$(x_i \backslash x_j) \backslash (x_i \backslash x_k) = e = (x_j \backslash x_i) \backslash (x_j \backslash x_k),$$
    which are indeed $\mathcal R$-equivalent.

    Any other permutation of $x_i,x_j,x_k$ yields $e$ in both sides of the condition, so the condition is satisfied for any permutation.

\medskip

\item[Case 2b:] {\it The lines do not appear consecutively in the intersection point:} Without loss of generality, we can assume that the multiple point has multiplicity 6, see Figure \ref{case2}(b). Hence, the relations induced by this multiple point are:
    \begin{eqnarray*}
    z x_k y x_j x x_i & = & x_k y x_j x x_i z = y x_j x x_i z x_k = x_j x x_i z x_k y =\\
    & =& x x_i z x_k y x_j = x_i z x_k y x_j x,
    \end{eqnarray*}
    where $x_i,x_j,x_k,x,y,z$ are the generators of the lines $\ell_i$,$\ell_j$, $\ell_k$,$\ell_1$,$\ell_2$,$\ell_3$ respectively.
    Hence, we have:
    $$(x_i \backslash x_j) \backslash (x_i \backslash x_k) = e = (x_j \backslash x_i) \backslash (x_j \backslash x_k),$$
    which are indeed $\mathcal R$-equivalent.

    Any other permutation of $x_i,x_j,x_k$ yields $e$ in both sides of the condition, so the condition is satisfied for any permutation.

\end{enumerate}

\end{enumerate}

\medskip

Hence, we have verified the equivalent version of the cube condition ($(u \backslash u') \backslash (u \backslash u'')$ and $(u' \backslash u) \backslash (u' \backslash u'')$ are $\mathcal R$-equivalent) for any triple of generators $u, u', u''$ in any case that the graph has no edges, so we are done.
\end{proof}

Hence, we have the following corollary:
\begin{cor}
Let $\cL$ be a real arrangement whose fundamental group has a conjugation-free geometric presentation and its graph $G(\cL)$ has no edges. Then, the corresponding monoid is cancellative and has least common multiples.
\end{cor}

\begin{rem}
The condition that the graph has no edges is essential, since if we take a line arrangement whose graph contains an edge, and its fundamental group has a conjugation-free geometric presentation, we can find a triple of generators for which the cube condition is not satisfied anymore.
\end{rem}

\section*{Acknowledgments}
We would like to thank an anonymous referee of our previous paper \cite{EGT} for pointing us out the possible connection between the conjugation-free geometric presentations and Dehornoy's complete presentations.
We owe special thanks to Patrick Dehornoy for many discussions.

We wish to thank Tadashi Ishibe for pointing us a wrong consequence from the published version of this paper, and consequently it weakens our result concerning complete presentations.

redpajama_set_name: RedPajamaArXiv


text:
Joseph of Arimathea
Glastonbury abounds in Myths and Legends. There are the Arthurian and Celtic stories and that have inspired contemporary pilgrims interested in goddess and pagan views. Running beside these this is another stream of what might be called Christian myths and legends.
There is of course the great Abbey, which has its own myths and history. Running beside this are a number of intriguing tales relating to Joseph of Arimathea. As with so many legends, there is little or no factual
of these tales, and they were certainly amplified and encouraged by the monks of the Middle Ages in order to increase the fame and popularity of the great Abbey. But we need to remember that, as with all myths and legends, there is always spark of eternal truth lying behind them, even though that truth may have been embellished in order to make it more readily understandable to lay listeners.
In this paper, we will look at some aspects of these stories.
We do not have a concrete historical information on Joseph, but medieval interest in genealogy raised claims that Joseph was a relative of Jesus. specifically, Mary's uncle, or according to some genealogies, Joseph's uncle. He was said to have owned the grave in which Jesus was buried and to have had close links with his nephew. The following legends are told about his activities.
A merchant – Joseph was said to have been a trader in metals and in the course of his activities, visited England – to the tin mines in Cornwall and the lead mines in the Mendip Hills of Somerset.
The young Jesus – there is an old Cornish story that relates how. on one of his journeys:
"Joseph of Arimathea came in a boat to Cornwall, and brought the child Jesus with him,
and the latter taught him how to extract the tin and purge it of its '.
Whilst in England, Jesus and Joseph travelled to Glastonbury and there young Jesus is said to have helped build, with his own hands, the original church of daub and wattle, the vetusta ecclesia - dedicated to Mary, his mother. in what is now the Abbey grounds.
The Crucifixion – During the crucifixion, Joseph gathered drops of blood and sweat from the dying Jesus and these were contained in two glass 'cruets'. On his death, Jesus was buried in a grave owned by Joseph.
Pilgrimage -after the death of Jesus, Joseph gather together a small band of pilgrims, said to have included Mary, the mother of Jesus, and the pilgrims set off to travel across Europe eventually ending up in Glastonbury.
Glastonbury – There is an interesting legend of Joseph's arrival in Glastonbury. After his long journey, he was weary and on a hill"
'He set his walking staff on the ground to sleep, it miraculously took root, leafed out and blossomed as the "Glastonbury Thorn".
The hill on which the staff was grounded is now known as Wearyall Hill and the thorn, and its modern daughters, blooms at Christmas time; to honour the birth of Jesus.
Joseph had brought with him the two 'cruets', one red and one white, with the blood and sweat of Jesus and these are said to be buried somewhere in Glastonbury. Some legends say that he also brought with him, the cup of the Last Supper the holy Chalice and this also is buried somewhere – one of the places is claimed to be the foot of the Tor in what is now the garden of Chalice Well. This magical chalice Is also sometimes claimed to be the Holy Grail – the object of the search of Arthur and the Knights of the round table.
Joseph settled with his disciples and built 12 huts around the church that has been built by the young Jesus, the vetusta ecclesia. This original church was preserved when the great abbey was built but was later destroyed by the great fire. The stories of the origins of the Abbey church have led to its claim of being the oldest above-ground Christian church in the world.
Many see Josephs' arrival as the ushering in of the Age of Pisces and the birth of Christianity in England.
Wwhether Jesus's mother Mary, was one of the disciples that settled in Glastonbury is not clear. But the ancient original church was held in high regard by the monks, and the legends relating to it, led to the concept of Glastonbury Abbey being the most sacred land in England, and hence the great poem hymn of William Blake - Jerusalem – 'and did those feet in ancient times etc.''

redpajama_set_name: RedPajamaCommonCrawl


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text:
Q: Get all https://bitbucket.org/alceawisteria/ostr/issues
and then display them below in the current HTML document.
Is this possible with js?
(If not, how can it be done otherwise ?)
Tried to implement approaches from the "GitHub issue widget" code to no avail.

A: This solves the issue via PHP link scraping
<?php
require 'simple_html_dom.php';
$dom = new DOMDocument;
@$dom->loadHTML($html);
$links = $dom->getElementsByTagName('a');
$url = 'https://bitbucket.org/alceawisteria/ostr/issues/';
$html = file_get_contents($url);
$dom = new DOMDocument();
@$dom->loadHTML($html);
$xpath = new DOMXPath($dom);
$nodes = $xpath->query('//a[@class="execute"]');
foreach ($nodes as $node){
    echo $link->nodeValue;
    echo " getAttribute('href');
    echo ">";
    echo $node-> getAttribute('href');
    echo "", '
';
}
?>

redpajama_set_name: RedPajamaStackExchange


It would be perfectly natural if your reaction to the title of this blog post was something along the lines of, "A Certified Financial Planner telling me I need a comprehensive financial plan…surprise, surprise!" Maybe you've been told by other financial 'professionals' that you needed to buy their product only to find years later that you were sold a bag of goods you didn't need.
Well, I ask that you keep an open mind because I really do believe that having a personalized comprehensive financial plan MUST become the norm in our society. Why?
Let's look at this graph of what life tends to look like for many people. Everyone's situation is going to be different, obviously, but this graph looks at major life events that result in a large financial commitment.
The higher the lines go, the bigger the financial need you have at that particular moment of your life. For example, getting married typically unleashes a number of new financial needs in a relatively short number of years.
What's clear about the above graph is that there are a number of different financial needs we will face during our lifetimes. Those needs differ depending on what stage of life we are in, but no matter the age, there's almost always some sort of life event happening that needs our financial resources.
As a society, we are not doing a good job of preparing for our major financial needs. The consequences of this go well beyond dollars and cents. Money problems are one of the leading causes of divorce in America. Likewise, young adults are entering adult life after college with crippling levels of student loan debt, which impacts their plans for getting married and buying a home.
There are any number of reasons why this is all happening, but I think the overriding issue is what we saw in the graphic on the previous page – there are a lot of moving parts in life and trying to manage our finances can be overwhelming and complex.
"Personalized Comprehensive Financial Planning" is a mouthful to say but important to understand. There are two words that I'd like to focus on.
Personalized: Every person and family is different. While the general path of our lives may look similar, there are always nuances which make each family's situation unique. Many big financial companies offer financial solutions to bits and pieces of our financial lives but not the whole picture, and oftentimes they take a cookie-cutter approach to solving our problems.
A Certified Financial Planner, on the other hand, works with you one-on-one to understand the big picture of your life. We take into account what life stage you are in, your family makeup and the financial resources you have. Your plan is tailored to your unique situation, and you get the attention to detail that you deserve so you can rest knowing that all of your unique needs are being addressed.
Comprehensive: Have you ever had a chance to vacation at an all-inclusive resort? You buy your airline ticket, reserve your room at the hotel, and you're done. No worries about where you are going to eat or what you're going to do. Just a truly relaxing vacation with everything taken care of for you while you're at the resort.
Working with a Certified Financial Planner is like going to the all-inclusive resort. We take a comprehensive approach to identifying and prioritizing all of your life goals and look for ways to protect you in the event something unexpected happens. The plan you get equips you to start tackling your goals proactively. Best of all, we help you put your plan into action and monitor your progress for you, increasing your chances of success.
As my crude graphic demonstrates below, working with a Certified Financial Planner helps you untangle all of the moving parts of your financial life and then organizes everything into a PLAN.
Ideally, the person you work with will explain everything to you in plain English. Managing your entire financial picture is complex but it's not rocket science. Having a plan that is simple and easy to understand will help you stay committed to it over the course of your life.
At Honey Lake Advisors, we work with clients on an ongoing basis to keep your plan fresh and updated for any life changes, while continually monitoring your progress. That way, you always have a trusted advisor to turn to when you need one.


The exhibition will be open to the public from 29th September 2018 - 26th January 2019.
29th September 2018 - 26th January 2019.
Supported by the Heritage Lottery Fund, Arts Council England and Norwegian Embassy in London.
Bruce Castle Museum and Haringey Archive's collection of Gestetner artefacts include the fantastically illustrated house publications, which record the life of the factory in the 1950s and 1960s underscoring the companies close roots to Tottenham.
The exhibition will include a retrospective of Alt Går Bra's mimeographed works over the last four years. This section of the exhibition will centre on Alt Går Bra's book The Mimeograph, A Tool for Radical Art and Political Contestation, a compilation of texts by contemporary scholars worldwide dedicated to mimeography and the first known publication dedicated to mimeographic works. Alt Går Bra will show its mimeographed books, publications and prints, along with newly produced work for this exhibition.


Crustacean Man, real name unknown, was a mutant who was designated to become a Robot Fighter. Experimented on and trained by 1-A as a predecessor to Magnus, he showed mental instability and was forced to live in the Phobos Asylum until he was killed by Malevelents.
This article has no notable quotes listed.


The Country Music Hall of Fame and Museum is the gateway into the extraordinary history of country music. This 37 million dollar complex houses impressive artifacts, interactive exhibits, and prestigious awards of the country music greats of yesterday and today. The venue occupies an entire city block, and is the single best way to appreciate the heritage and impact of country music today. There are several exciting ways to explore the Country Music Hall of Fame, each offering unique benefits. Tour the museum on your own or purchase the audio tour to hear the music of your favorite stars, interact with incredible exhibits, and learn more about the amazing performers who have helped to make this genre what it is today.
When you step into the Country Music Hall of Fame and Museum, you will be embarking on a trip through decades of country music history in this architecturally-stunning building. A towering wall of honor features displays of every gold and platinum record to hit the country music charts, as well as exhibits dedicated to the music trailblazers and legends who have been inducted into the Hall of Fame. The 40,000-square-foot museum also features rare recordings, memorabilia, and special traveling exhibits.
Enjoy admission to the Country Music Hall of Fame & Museum on a two-hour tour of this museum that serves as the gateway into the extraordinary history of country music. This museum and hall of fame is the best way to appreciate the heritage and impact of country music today, all inside an architecturally stunning building.
Not only will you have access to the entire museum with this option, but you will also take a guided tour of Historic RCA Studio B, the only historic studio tour in Nashville. This studio is one of the world's most important recording studios. More than 35,000 songs have been brought to life including more than 1,000 top ten American hits, and over 150 Elvis Presley recordings.
Take in one of the largest popular music museums in the world, along with a Celebrity Audio Tour. Your audio guide will take you through the different eras of country music, and your audio guide will be joined by others such as Charlie Daniels, Vince Gill, Dolly Parton, and more, who will provide behind-the-scenes stories, personal memories, and more!
What is included in the admission price for Country Music Hall of Fame and Museum?
Includes all of the exhibits.
Approximately how long could a customer plan to spend at Country Music Hall of Fame and Museum?
Guests spend about 2 hours here.
There is a gift shop and eateries onsite.
Wheelchairs are available on a first-come first-serve basis.
Is Country Music Hall of Fame and Museum handicap accessible?
For what ages is Country Music Hall of Fame and Museum appropriate?
Is parking available at Country Music Hall of Fame and Museum? Is it free?
Parking is available downtown for various prices.
"The staff was knowledgeable and accommodating"
I'm not even a country music fan, my wife and brother-in-law are. But, i spent more time in the museum then they did. The staff was knowledgeable and accommodating, they were familiar with the vme tickets, they were able to provide additional information regarding the area, and were very professional.
"This is well worth the day."
This is well worth the day. You may go at your own pace, great for school age kids because their are touch screen computers, kids can pull up records, look at the old or new record label,and listen to a little diddy, i took my kindergartner and he had a blast. Not engaging enough for kids under 5, even though i took a 4 and 2 year old as well. I recommend when you come out you go around the corner and eat lunch at jacks bbq!!!!!!!!!!!!!!!
"I would recommend this to people that are going to visit Nashville."
The reba exhibit was wonderful. It is great to see and hear about singers and actors family history and how they get started in the business. It was great to see the older style of country music outfits and to hear some older country songs. The carrie underwood exhibit was amazing as well. I would recommend this to people that are going to visit nashville.
No disappointments here. There was enough of old time stars and new ones! Interesting exhibits and the videos throughout gave me a chance to sit and rest and still see and learn.
Tip: Buying tickets ahead of time through vacations made easy certainly made it easy. No waiting in long lines.


text:
Home / Shop / Books / Acting & Auditioning
A Dream of Passion: The Development of the Method
$16.00 (as of July 16, 2019, 5:56 pm)	& FREE Shipping. Details $12.88
{"main_aff_id":"safeavon-20","aff_ids":[{"country":"com","aff_id":"safeavon-20"}]}
Categories: Acting & Auditioning, Books
"The definitive source book on acting."—Los Angeles Times
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"A must for young actors—for old ones, too, for that matter."—Paul Newman
"An exploration of the creative process that will reward all who are interested in the nature of inspiration."—Library Journal
"An important cultural document."—Booklist
Plume Books
POXM-0452261988, XM-0452261988
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40, hundredths-inches, 790, hundredths-inches, 25, Hundredths Pounds, 520, hundredths-inches
Dog Sees God: Confessions of a Teenage Blockhead – Acting Edition (Acting Edition for Theater Productions) Paperback – January 1, 2006
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We Have Always Lived in a Castle (Play Script) (Acting Edition for Theater Productions)
On Golden Pond. (Acting Edition for Theater Productions) Paperback – January, 1998
Five Women Wearing the Same Dress (Acting Edition for Theater Productions) Paperback – January, 1998
August: Osage County (Acting Edition for Theater Productions) Paperback – November 10, 2009
How I Learned to Drive – Acting Edition (Acting Edition for Theater Productions) Paperback – October 1, 1997
Harvey Paperback – August 20, 2014
Stop Kiss – Acting Edition (Acting Edition for Theater Productions) 33173rd Edition


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text:
Q: Proper way to concatenate a tuple with keys and values I have a tuple that is populated with elements from a directory:
{'C:/foo' : ['bar1', 'bar2'], 'C:/foo2' : ['bar']}

and I am trying to concatenate each key with its values:
C:/foo/bar1
C:/foo/bar2
C:/foo2/bar

I thought this would work:
[''.join('%s/%s' % (d.keys(), value)) for value in d.values()]

But it's just concatenating the key with the value in the same way as:
for k,v in d.items():
  print(k'/'v)

Any help in this area would be greatly appreciated! If I am going at this the wrong way please tell me.
Thanks!

A: You can use os.path.join to build up paths in this manner.
import os
d = {'C:/foo' : ['bar1', 'bar2'], 'C:/foo2': ['bar']}
for k,v in d.items():
    for value in v:
        print(os.path.join(k,value))

Output
C:/foo2\bar
C:/foo\bar1
C:/foo\bar2

Then to make sure the slashes are all the correct way you can use os.path.normpath
for k,v in d.items():
    for value in v:
        print(os.path.normpath(os.path.join(k,value)))

Output
C:\foo2\bar
C:\foo\bar1
C:\foo\bar2

To be clear, you should prefer using os.path because it will robustly handle path manipulation in an operating system specific manner. In this case ensuring the correct number of slashes (including escaping if necessary), and which types of slashes (forward vs backward).

A: You first need to iterate through the dictionary, and then iterate through the lists.
You can do this in a simple list comprehension, without using a for loop.
files_dict = {'C:/foo' : ['bar1', 'bar2'], 'C:/foo2' : ['bar']}
filenames = [dir+"/"+file for dir, files in files_dict.items() for file in files]

Output:
['C:/foo/bar1', 'C:/foo/bar2', 'C:/foo2/bar']

A: There is a typo in your tuples list, a colon is missing:
a = {'C:/foo' : ['bar1', 'bar2'], 'C:/foo2': ['bar']}

for key in a.keys():
    vals = a[key]
    for v in vals:
        concat = key + '/' + v
        print concat


1.It adopts mechatronics control equipment and PLC combined control steam boiler/hot water boiler/2 ton boiler/3 ton boiler boilers, steam boilers, heater.
hot water heating boiler, steam boiler,SZS gas fired . Diesel Fired Steam Boilers, China 1 to 20 Ton/Hr Low Pressure Gas Fired Steam Boiler .


Ok... this was originally meant to be a Desert Eagle and it looked pretty good... but it didn't shoot very well at all, that's why I didn't call it a Desert Eagle. In the future (next week) i will make a replica that does shoot. Some features: 30 - 40+ ft range True trigger 8 round mag Sounds really cool CREDIT TO TRAUTS, SUBSCRIBE TO HIM!!! I USED HIS MECHANISM FOR THIS GUN. A VIDEO WILL BE UP THIS WEEK AND INSTRUCTIONS ON THE WEEKEND!!!
Replicas... what can I say? Some people like them, some people hate them. I'm starting to get sick of them, after I post my F2000, I'm going to take a break and make a war kit. It will have: Sniper Assault Rifle Shotgun Sidearm For my sniper, it will definitely be a slingshot. The shotgun will have multiple bullets per shot, and the assault rifle will have good range and a high ROF. I'm not really sure if I should make the sidearm oodamo or not. Do you guys think I should follow through with this idea, or should I just make replicas? If I do this, I will certainly go back to replicas after, but this will come first. Tell me what you guys think I should do.
Ok, so I went pro.... now what ?
Hi guys, just wanted to let you know that I don't think I'll be on ibles anymore, there are just more important things in my life than K'nex. Sorry if this looks like i'm just trying to get publicity too, I'm not. And if you think I'm fake, then I must be the best faker in the world. PS. If you're interested in buying some of my K'nex or all of it, please let me know. I have about 15 pounds, and if anyone is interested in buying I might do a full piece count.
Hey guys, or whoever is reading this and even slightly cares. I haven't made a gun in a really long time, and I haven't really responded to the last 300-400 comments. Now that it's the holiday break, I'm going to have a lot more spare time. So this is what I feel like doing, making a new gun. I want your ideas for it, because I'm really not that creative.. So what do you guys want me to make? Also add me on Xbox Live; TH3 BR0WN B0Y. The "0's" are zeros.
Anybody wanna buy about 20lbs of knex? I'm selling for $100 plus shipping.
I just made a chat room for us here on Ibles. KI has had one for a while and nobody has bothered to make on for ibles. Here's the site. This is mainly just an experiment to see how many people use this, but if a lot start to then I will be on fairly often.
I realize it separates the bullet from the rest, but how does this work? Could somebody make me a diagram?
Almost. Here is a prototype turreted NAR. Discuss.
ITS FINISHED!!!!!! CHECK IT OUT HERE!!!!!!
Am I the only one that can do this? Have any of you guys noticed this?
Any of you guys want to learn how to do the Rubik's Cube?
This is my M4 replica. The features: -Removeable mags -Modified Motaboi crane stock -4 attachement rails -Iron sights -True trigger -pin guide -40-50' range -Fake charging handle -New magazine -Trigger gaurd It shoots blue rods, external mag pusher. So, whadaya think? should I post?
Hey guys, it's MotaBoi and im just saying, I'm back. Kind of, I still don't think I'll be making anything with k'nex anymore or what I will be posting. I wrote to the moderator and he told me I could have my account back if I wrote 2 apologies and you guys accepted them. One to Kevin, and one to the community. Here's my apology to the community, and I will send a private message to Keven about my apology. I wrote these before and I asked the moderator I talked to to post them for me, but he didnt really do anything about it. Hey guys, it's me, MotaBoi. As some of you may know kNeXFreek has recently come back to instructables and a few of us decided to play a practical joke on him by changing our display picture to something very offensive to him. I know some of you must be mad at me for doing this and I'm sorry. Why I'm apologizing to you guys like this now is because I need you to all forgive me so that I can have my account back. If you guys can forgive me and want me to be able to be back on ibles, please say so so that I can have my account back PS. If you have any questions about my guns or anything, post them to my OB so that I know, and spread the word!
UPDATE: I have posted a link to this on a mythbusters sugest a myth forum. Cross your fingers everyone! I'm sure most of you have seen at least 1 or 2 episodes of Mythbusters, but wouldn't it be cool if one of our knex guns or other creations were actually on the show. Post a comment saying your name if you want to sign this and then, in a month, I will submit this to the Mythbusters themselves. Every week I will update the count. Remember, the more people, the more likely it will actually be on the Mythbusters. I just need a few more and I'll post it on their comment board. People so far 222222222539 TheDunkis Knexsuperbuilderfreak MrMoonatic1 Tombuckey Qwerty2008 Killerboy RMConstruction Kairah Axiys zx Falco33 Knex mad sprout_less Shadowman39 billythebob NYPA nekie81 bartboy chopstx knex gun builder Killer~Safecracker Mr.McCoy bounty1012 Selezoina Big Z logic boy Motaboi lioneatr Rockstar1000000 Rustymia Bounty1012 Cupcake43 Rec0n MotaBoi danbaha cunningham4 ceaserama H14T4TCH1 ElvenChild Update: so far there are 39 people who signed this petition. Keep it up!!
IF YOU ARE A KNEXER AND YOU HAVE A PS3 AND PSN, ADD ME AND POST YOUR USERNAME IN THE COMMENTS!! ALSO TELL ME IF YOU PLAY MW2, I DO AND IM LEVEL 50 ON 2ND PRESTIGE ON MY OLD ACCOUNT, AND LEVEL 18 ON 10TH PRESTIGE ON MY HACKED ACCOUNT. I ONLY PLAY ON MY HACKED ACCOUNT. MY USERNAME IS: A_HUNGRY_APE JUST LEAVE A COMMENT WITH YOUR USERNAME, IF YOU PLAY MW2 AND YOUR ONLINE LEVEL!!!!!!!
Knexer quiz- CLOSED! ALL ANSWERS FOUND!
Knexer quiz. You may have seen a few of these before, done by Lemonie and Jayefuu, but I just couldnt resist making one for the knexers! I do ask one thing of you though, if you know more than one of them, please only answer one, to give others a chance, thanks :-) Every picture represents a well-known knexer, can you guess them right? :-) Sorry guys, but there are no prizes :-( The first picture is an example on how to solve these, it says hiyadudez, and the answer would look like this: Hill + Yard sale + Dudel + (ez) Now some may be quite difficult, but some may be very easy. I will say it again to be clear, I ask one thing of you, if you know more than one of them, please only answer one, to give others a chance, thanks :-) 1. Hill + Yard sale + Dudel + (z) 2. Where Dutch people live + World War 1 + Lord of the rigs (Congrats Knex mad) 3. Shad + Mow + Man + (39) (Congrats Shadowman39! 4. Oblivion rollercoaster + iTunes (Congrats LaserJunkie) 5. Goo + Train (Quite easy but congrats Motaboi) 6. Iron + Man (Congrats Knexfan9182) 7. Killer whale + Safety hat + Nutcracker (Congrats Knex gun builder) 8. Knee + Exilim + Free to dance + (K) (Congrats Perkinator) 9. Chocolate log + Ice cubes + Buoy (Congrats Dsman195276) 10. Love + Down + Eye (Congrats Killer~safecracker 11. Good luck!
Here's this week's run of total views. Sorry I missed the last two runs over Christmas, was too busy having fun and didn't have the script with me where I was. We have LOTS of new members since the last time I ran it, about 40 more I think! And lots more in the high totals, Fungus Amungus and Randofo have joined the group so next week we'll be able to see if they're gaining on Canida and whether Fungus is getting enough weekly views to stay ahead of Scoochmaroo. Instructabler 09/12/2010 06/01/2011 Gain canida 4,016,751 4,133,770 117,019 fungus amungus 3,810,955 scoochmaroo 3,396,097 3,566,430 170,333 randofo 2,737,169 Honus 2,472,615 2,503,181 30,566 Kiteman 1,733,654 1,760,269 26,615 Plasmana 1,603,873 1,631,315 27,442 Tool Using Animal 1,565,973 1,586,400 20,427 T3h_Muffinator 1,080,291 zieak 1,006,130 1,036,709 30,579 killerjackalope 956,702 989,201 32,499 gmjhowe 952,161 975,903 23,742 drinkmorecoffee 909,894 Weissensteinburg 851,790 SaskView 827,172 840,127 12,955 lemonie 777,977 793,879 15,902 Brennn10 679,755 ChrysN 575,263 593,019 17,756 mikey77 567,052 582,631 15,579 jeff-o 525,555 540,726 15,171 PKM 519,124 527,697 8,573 laxap 458,068 478,370 20,302 mikeasaurus 440,136 aeray 402,540 belsey 386,587 400,703 14,116 depotdevoid 368,485 377,391 8,906 Jayefuu 344,693 351,698 7,005 ModMischief 325,951 330,818 4,867 scraptopower 290,296 317,326 27,030 Spl1nt3rC3ll 277,916 280,333 2,417 KentsOkay 243,351 245,855 2,504 nmcclana 236,457 AngryRedhead 206,707 219,817 13,110 Doctor What 215,596 219,715 4,119 rimar2000 170,374 182,030 11,656 BrittLiv 151,895 170,049 18,154 Ninzerbean 163,897 168,016 4,119 comodore 154,701 156,851 2,150 Lithium Rain 148,100 151,967 3,867 yokozuna 136,331 145,008 8,677 Kaelessin 140,385 144,778 4,393 Hiyadudez 130,115 134,884 4,769 Lynne Bruning 131,330 133,761 2,431 bertus52x11 123,686 127,354 3,668 The Jamalam 120,388 121,719 1,331 RavingMadStudios 100,547 105,709 5,162 thermoelectric 101,427 104,441 3,014 uniqueutopia 97,394 100,581 3,187 JamesRPatrick 94,541 Goodhart 89,846 92,295 2,449 DJ Radio 88,004 90,497 2,493 janw 79,364 87,283 7,919 jen7714 57,733 64,459 6,726 jen7714 57,733 64,459 6,726 MichelMoermans 61,419 63,651 2,232 Killer%7ESafeCracker 58,893 61,050 2,157 nelson8815 55,564 Re-design 54,114 Shadowman39 45,771 48,570 2,799 TNEN 46,104 46,832 728 kelseymh 45,354 46,203 849 Sunkicked 43,805 44,695 890 beckervdejazz.com 40,844 42,220 1,376 ynze 15,664 41,336 25,672 The Ideanator 36,036 39,299 3,263 Berkin 37,659 38,822 1,163 wizgirl 38,030 mman1506 37,123 EmmettO 35,441 Biggsy 34,344 knuckel 32,043 33,935 1,892 artificialintelligence 26,396 30,363 3,967 nickodemus 27,906 28,827 921 TSC 24,535 26,906 2,371 MotaBoi 24,657 25,633 976 jwystup 21,387 25,524 4,137 KoffeeKommando 25,191 cbm104 24,882 peguiono 23,772 24,316 544 m6233555m 23,626 23,965 339 Jimmy Proton 12,837 23,398 10,561 kcls 20,841 22,929 2,088 lilyfrancis 20,950 22,689 1,739 Knex_Gun_Builder 21,624 22,063 439 WARLOCK76 19,349 21,330 1,981 plane phanatic 20,283 20,945 662 CrayfishYAY 20,126 20,756 630 Derin 20,190 20,698 508 steveastrouk 19,237 19,998 761 possum888 18,499 19,233 734 tombuckey 17,016 17,723 707 zascecs 16,698 16,888 190 Tornado96 15,654 16,755 1,101 knexsuperbulderfreak 13,562 14,010 448 floris2burn 13,243 13,799 556 Legofanatic 12,363 13,127 764 uberdum05 12,300 12,626 326 qazwsx755 11,244 11,926 682 wtmi.podbean.com 10,949 NachoMahma 10,655 10,771 116 masterochicken 10,540 10,758 218 FrozenStar 10,242 10,474 232 acidbass 9,115 9,478 363 cdawisconsin 9,295 zurichko 8,744 lizzyastro 7,738 8,309 571 Kryptonite 6,568 6,769 201 Sorunome 5,690 6,596 906 cherishcherub 5,701 5,942 241 moocowdog 4,773 5,115 342 zero.gx 4,847 4,991 144 drpepper1234 4,595 jamesdude 3,983 4,070 87 slithien 3,535 3,789 254 RNB 2,858 3,369 511 MegaMetal8 3,319 wareneutron 2,513 2,802 289 tbcross 2,513 2,748 235 cj81499 2,464 2,508 44 solarblade90 2,109 2,310 201 Jag56 1,871 1,988 117 Flannel UK 413 1,311 898 shadon 1,100 1,210 110 Randomguy65 941 1,099 158 dombeef 739 788 49 AwesomeSwordGuy 532 727 195 Fr-Pa-Co 191 i like cheese 50 66 16 LivStoleYourPie 56 megametal8 3,150 -3,150 This forum topic is a printout of the results of a Perl script I run weekly. To be included in it, join this group. The Perl script visits the profile page of each member of the group and extracts their Total Views then saves them to a file that can be opened by me in Excel. I then post the last 3 columns so everyone can compare their weekly gain with their friends. While you can easily see this data for you on the You page, this script allows people to compare their success/gain with their friends. I hand out patches as achievements when members of the group reach 50k, 100k, 250k, 500k and 1m views.
Total Views Count - 20/01/2011 - UPDATED!


International equity and fixed income investment managers are taking greater control of the FX trades they need to execute to settle purchases and sales of non-dollar securities as a result of increased client scrutiny of FX transactions. What tools and techniques are buyside firms using to get best execution in FX? How are theymeeting the challenge of trading restricted currencies in emerging markets? How best can traders use the armory of pretrade analytical systems, brokeralgorithms, RFQ tools and TCA services?


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…at Lafayette College, that is. Last February I had the opportunity to join fellow members of the American Friends of Lafayette (yes, that's a thing, and you can—and should—join by visiting this website) in the "reserved seating" section (that's the second row, folks) of the absolutely packed Colton Chapel on the campus of Lafayette College in Easton, Pennsylvania. We were settled into the pews to hear Sarah Vowell discuss her New York Times bestseller, Lafayette in the Somewhat United States.
Lafayette College is, obviously, an apt place to hear someone expound upon the merits of Marie-Joseph Paul Yves Roch Gilbert du Motier, Marquis de Lafayette (say that five times fast). The college was founded in 1826 on the heels of the Marquis' "Farewell Tour" of the United States, and aptly named for him "in memory and out of respect for the signal services rendered by General Lafayette in the great cause of freedom" (from the Charter of Lafayette College, Article I). Lafayette College isn't the only place in the United States named for or out of respect of the Marquis—in her book, Vowell notes that "nowadays, Lafayette is a place, not a person," and she makes an appropriate observation. According to Lafayette scholar (and AFofL president) Alan Hoffman, over 80 localities in the United States have been named in honor of the Marquis, with Fayetteville, North Carolina being the very first in 1783.
Place and landscape are interesting lenses through which to examine Lafayette's relationship with the United States. Though his final resting place is in Paris, Lafayette is even buried in American soil–from Bunker Hill, as a matter of fact–collected in 1825.
The next morning was beautiful. Everything around me was new to me, the room, the bed draped in delicate mosquito curtains…the strange new beauty of the landscape outside my windows, the luxuriant vegetation—all combined to produce a magical effect.
Vowell describes how Lafayette, in 1777, in his enthusiasm and naiveté, wasn't entirely sure what he was looking at, but loved it. Lafayette surveyed scenes outside of his window that were, true to Vowell's title, somewhat the United States—Congress declared independence from Great Britain a year earlier, but Parliament had yet to recognized it. Similarly, the very physical landscape Lafayette described was, well, schizophrenic. Huger's Georgian-inspired South Carolina home where the ultimate patriot awoke that morning was nestled near a town named after—who else–King George III, yet Georgetown was a hotbed of patriotism that contributed to the infamy of everyone's favorite Revolutionary rascal Mel Gibson Francis Marion.
Vowell's text weaves us through the rest of Lafayette's tenure in the somewhat-but-becoming-ever-closer-to-actually-being United States, all the way to up to Lafayette's participation in the Siege of Yorktown, the somewhat-last-battle of the American Revolution.
But Lafayette's relationship with the somewhat United States didn't end there. In 1824 at the invitation of President James Monroe, the Marquis, now the last surviving general of the American Revolution, embarked on a "Farewell Tour" of the country he so passionately helped establish. A bona fide rock star, Lafayette arrived in New York to an estimated 80,000 screaming fans.
The Marquis de Lafayette by Ary Scheffer, 1824.
All of this is to say nothing about Lafayette's influential views on abolition, women's rights, and other important reforms that, if he didn't support directly, he was eventually called down to support (e.g., Evelyn Wotherspoon Wainwright's 1920 pleading to a statue of Lafayette in D.C.'s Lafayette Square to "Speak, Lafayette, dead these hundred years but still living in the hearts of the American people.") As Vowell argues in Lafayette in the Somewhat United States, the American people, and the rights and liberties they enjoy, would not be ours without a little help from our friends (see what I did there?)–one friend, specifically—the Marquis de Lafayette, and his constant contributions to the somewhat, ever growing, ever discoursing, ever revolutionary United States. I encourage you to check out Sarah Vowell's book, and learn more about why Lafayette is permanently fixed on the landscape of the United States of America—and in the hearts and minds of its citizens. No somewhat about it.
Robert P. Hay, "The American Revolution Twice Recalled: Lafayette's Visit and the Election of 1824," Indiana Magazine of History, Volume 69, Issue 1, pp 43-62.
Sarah Vowell, Lafayette in the Somewhat United States, New York: Riverhead Books, 2015.


Sobukwe Day is observed and commemorated during the month of February since the death of our founding President Robert Mangaliso Sobukwe on the year of 1978. This year will be marking the 40th Death Annivesary of Sobukwe.
Searching for the voice of SOBUKWE!!!


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Independence, Freedom, Islamic Republic: Talking About a Revolution, and its Promises
Annalisa Perteghella
The Islamic Republic, whose survival nobody would have betted on, still lives on. No matter what John Bolton predicted last year – "the Islamic Republic will not last until its 40th birthday" – or what common sense suggested in the early days of the revolution, when very few people thought it would have lasted more than six months.
The failure in predicting the survival of the revolution only matches the failure in predicting its very same outbreak. Besides then-US president Jimmy Carter praising Iran as an "island of stability in one of the most troubled areas of the world" just a year before the revolution began, we have CIA reports clearly admitting the failure of intelligence to grasp what was going on in Iran at that time: "I would suggest even Ayatollah Khomeini didn't realize how well his force was moving along", said admiral Stansfield Turner, then-Director of the CIA, in an interview on February 4, 1979.
It is hard to overstate the importance of the Iranian revolution, the last successful revolution of the 20th century, whose significance extended far beyond Iran, with an impact on the Middle East and the broader Islamic world. The outcome of the revolution was manifold: from the spurring of opposition movements throughout the Muslim world – often imbued with radicalism – to an effort to produce Islamic alternatives to failed state institutions with the aim of creating a new civil society.
And many were its promises. The revolution's main slogan "Independence, Freedom, Islamic Republic" perfectly summarises the major ideals in whose name it was fought, which were social and political, as well as religious. The Islamic Republic – this unique state conformation bringing together the imperative of "Islamic government" with the French Fifth Republic institutional structure – is maybe the most successful fulfillment of the revolutionary promises.
But scarcely a year after the revolution the state of affairs was in flagrant contradiction with the hopes that had been raised in its early days. The very imposition of an Islamic Republic was the result of the purging of all other revolutionary voices. The ideals of justice were hijacked by the hyper-speculative economy set up by the founders, which went from being potential embodiments of the ideal of redistributive justice to mechanisms for enriching parasitic groups. This, together with the nationalisation and expropriation efforts carried out by the state as well as the emigration of a large proportion of the middle class, led to an increase in poverty. While failing to close the gap between rich and poor, this led to the emergence of new elites. A sullen atmosphere descended upon the effervescence of the early days of the revolution, with the nihilist reinterpretation of the ideology of martyrdom taking young Iranians' lives on the Iraq-Iran war front. The secularization of religion went hand in hand with an increasingly tense social situation and an economy that was in regression, in the context of a revolution that had gone adrift, together with the hopes it had embodied.
And yet, while the revolution had gone off course, the Islamic Republic survived. It survived by adjusting to the new reality and by putting the nezam, the "system", first. Contrary to outsiders' perceptions of the Republic as a monolith, it survived by opening – and closing – its gates to the different voices of Iran's vast political panorama. It survived by adapting the "neither East nor West" revolutionary slogan to the need of opening up to both East and West in order to endure. While the emphasis on "independence" has never gone lost, Iran has been able to compromise when the survival of the Republic was at stake.
Forty years later, mixed conclusions can be drawn from the revolution. The signals of crisis are manifold. Not only the economic crisis openly acknowledged by the government but, once again, the legitimacy crisis which has much to do with the slow but inevitable passing away of the old generation of revolutionaries, while the majority of the Iranian population – which is composed of under-40s – has not lived the experience of the revolution. Yet Iranian society is alive. Despite duplicitous US appeals to the Iranian population, aimed at showing a fake and dangerous sense of empathy, what is sure is that Iranians do not seem to crave another revolution, either domestically born or instigated from outside.
The desire for change is ultimately a desire for a better life. The joy of the July 2015 celebrations for the nuclear deal appear in stark contrast with today's sense of hopelessness and disillusion among vast sectors of society, who see "the doors to the outside world closing again". What Iran has tried to show the world in these 40 years is that it is a normal country, made up of people and common aspirations. Too often the outside world has mistaken the people for the system, failing to understand that a country is both its people and its leaders.
And while there is a serious debate inside the country over "whether the Islamic Republic can survive in its present form", it's probably time for the outside world to acknowledge that, while the revolution is long gone, what was once looked at as a fragile experiment has evolved into a robust reality. Let's move on from there.
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text:
Q: No output after running a program testing different loop structures This is about a program aimed to generate a series of continuous integers from 1 to 1000 which sum up to 1000. My function, While1000, works but if I try to modify this code to use other loop structures to get the same output it does not work. My modified code blocks when using do-while and for loops. I get no response when I run them in Eclipse. No syntax error or Exception warning.
Here are the different versions of the code:
public class While1000 {

    public static void main(String args[]) {
        int i, j, k, sum = 0;
        for (i = 1; i <= 1000; i++) {
            j = i;
            sum = i;
            while (sum < 1000)
                sum += ++j;

            if (sum == 1000) {
                for (k = i; k <= j; k++)
                    System.out.print(k + " ");
                System.out.println();
            }
        }
    }

}

public class dowhile1000 {
    public static void main(String args[]) {
        int i, j, k, sum = 0;
        for (i = 1; i <= 1000; i++) {
            do {
                j = i;
                sum += ++j;
            }
            while (sum >= 1000);
            if (sum == 1000) {
                for (k = i; k <= j; k++)
                    System.out.print(k + " ");
                System.out.println();
            }

        }
    }
}

public class For1000 {
    public static void main(String args[]) {
        int i, j, k, sum = 0;
        for (i = 1; i <= 1000; i++) {
            for (j = i; sum >= 1000; ++j) {
                sum += j;
            }

            if (sum != 1000) continue;
            for (k = i; k <= j; k++)
                System.out.print(k + " ");
            System.out.println();
        }
    }
}

The right output should be:
28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52

55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70

198 199 200 201 202

1000

A: *

*When you are in a loop you have to re-initialization your variable

*The stop condition are important

dowhile1000
for (i = 1; i <= 1000; i++) {
    --> j = i;   // Shouldn't be initialization do-while
    --> sum = 0; // Should be initialization to 0 after each iteration
    do {
        sum += ++j;
    }
    --> while (sum < 1000); // Keep adding to the sum until is greater than 999
    if (sum == 1000) {
        for (k = i + 1; k <= j; k++)
            System.out.print(k + " ");
        System.out.println();
    }

}

For1000
for (i = 1; i <= 1000; i++) {
    --> sum = 0; // Should be initialization to 0 after each iteration
    --> for (j = i; sum < 1000; ++j) { // Keep adding to the sum until is greater than 999
        sum += j;
    }

    if (sum != 1000) continue;
    for (k = i; k < j; k++)
        System.out.print(k + " ");
    System.out.println();
}

A: You are a bit sloppy with brackets and indentation. That does not improve the readability of your code.
dowhile1000: the do-while-loop and the while loop would be different, if the while condition could be false before entering the loop. That is not the case. So the do-while-condition has to be the same than the while condition. Your do-while condition is not true after the first run of the loop, so the loop terminates. And of course j and sum have to be intitialized before the while loop. j is reset with every loop run in your code.
For1000: A for loop is repeated as long as the condition in statement 2 is true. In the for-j-loop, the condition is never true when entering the loop, so the loop is never executed. sum has to be reset before every new j. Because j is incremented after the for condition is true, in the for-k-loop the test has to be k

public class While1000 {

    public static void main(String args[]) {

    int i, j, k, sum = 0;
    for (i = 1; i <= 1000; i++) {

        j = i;
        sum = i;

        while (sum < 1000)
        sum += ++j;

        if (sum == 1000) {
        for (k = i; k <= j; k++)
            System.out.print(k + " ");
        System.out.println();
        }
    }
    }
}

public class Dowhile1000 {

    public static void main(String args[]) {

    int i, j, k, sum = 0;
    for (i = 999; i <= 1000; i++) {

        j = i;
        sum = 0;

        do {
        sum += j++;
        } while (sum < 1000);

        if (sum == 1000) {
        for (k = i; k < j; k++)
            System.out.print(k + " ");
        System.out.println();
        }

    }
    }
}

public class For1000 {

    public static void main(String args[]) {

    int i, j, k, sum = 0;
    for (i = 1; i <= 1000; i++) {
        sum = 0;
        for (j = i; sum < 1000; ++j) {
        sum += j;
        }

        if (sum != 1000)
        continue;
        for (k = i; k < j; k++)
        System.out.print(k + " ");
        System.out.println();
    }
    }
}


Robert began his journey in the Las Vegas real estate industry in 2002 by investing in Las Vegas residential homes and condos. Based on several years of successful investing, Robert was approached by friends and family to share his expertise in real estate investing in the Las Vegas market. Robert used this opportunity to join longtime friend, business partner and fellow investor, Richard Crighton at Crighton Real Estate Holdings and obtained his Nevada Real Estate Broker/Salesman License. Since 2004 he has been assisting real estate investors in negotiating successful real estate transactions across the greater Las Vegas Valley. He has experience in, but not limited to, residential investing, commercial real estate advising, loss mitigation, rehabilitating apartment buildings and homes, high rise transactions, creative financing (AITD's, Contract of Sale, Owner Will Carry, etc). Robert has exercised his fundamental knowledge of loss mitigation, in combination with business acumen, to lead the charge in assisting homeowners with Loan Modifications and Short Sales far before the concept became a household name.
Angela plans to get her Massachusetts and Rhode Island salesperson license towards the end of 2014 and join Robert in at Sankey Real Estate in North Attleboro, MA. The two of them plan to build up a strong real estate team in both locations as time goes on. Let Angela Adams and The Adams Team at Rothwell Gornt Companies guide you through this ever changing real estate market.
Claudia Stack has resided in the Attleboro and North Attleboro area for the past twenty years and has come to truly love it. She is very familiar with the neighborhoods and school systems. Stack recently joined Sankey Real Estate and brings with her several years of real estate marketing experience in the residential and commercial markets. She is very excited to be using dynamic new marketing tools that are on the cutting-edge. Stack will help you to promote your property in a fresh and compelling way. She also has the knowledge to assist buyers with selecting the best home for their needs. Whether selling or buying, Stack has the know-how to best serve her clients.
Stack was born and raised in Providence, Rhode Island alongside eight siblings. She grew up in a twenty-room, historic Victorian home and has always been fascinated with all types of homes - old, new...and everything in between. Claudia graduated from Classical High School in 1984 and also attended Community College of Rhode Island and Rhode Island College.
In her spare time, Claudia is at her happiest spending time with her husband Rick, their twelve year old daughter, and their many friends and family. She also enjoys cooking and entertaining, biking, hiking and yoga.
My name is Robyn Kulis. I joined Sankey Realty in 2017. I have a huge passion for real estate, Architecture, Interior Design and Home Improvements.
I grew up in a small town, North Attleboro, MA and graduated from No. Attleboro High school. Then, I attended Community College Of Rhode Island. I have worked in the medical field for over 20 years. My career change was when I found my passion to help others and families with real estate and their search for reaching their goals.
Now, I enjoy going to work every day, meeting new people and helping clients find the right house to call home is my passion. In my free time, I enjoy spending time with family, going for walks with my dog (Stormie), meditation, reiki, reading and working out.
Let Robyn show you how hard she works for her clients to surpass all of their expectations.
Motivated Rhode Island Realtors Wanted!
We are in the process of setting up these profiles for Massachusetts real estate websites as well as Massachusetts forums etc. Once we have them set up we will link our profiles here forum views to keep up with our newest posts and blogs contributions.


For an outdoor enthusiast Cabo san Lucas on the southern tip of Mexico has a range of activities to choose from, whether you love the ocean or are more of a land lover. There are many adventures to be had, either by yourself or as part of an organized tour. Activities in the ocean include snorkeling with the impressive whale shark to going on a camel safari. An adventure awaits you regardless of what sort of outdoor adventure you are looking for. You can take a whale watching tour to see the amazing Grey whales in Magdalena Bay. These magnificent creatures use these waters off Baja California as a breeding ground and nursery. There are only two other places in the world where these whales have breeding grounds. Aside from whale watching you can dive in and snorkel alongside the massive Whale shark. Dolphins and rays can also be spotted in the Pacific waters off Cabo san Lucas.
Sportfishing and kayaking are also popular pursuits in the waters around Cabo san Lucas aside from scuba diving, snorkeling and swimming. You can usually rent equipment such as kayaks so don't worry if you don't own all the equipment that you may need.Big game fishing in the ocean is something that helped put Cabo san Lucas on the map, with big fish such as Marlin and Sailfin being caught often.
They even have a water sport called Flyboarding which has become very popular in the area. Another quite recent water sport is standup paddle. They teach lessons in this at the scenic Arch landmark.You can also jetski and waterski or even surf in the ocean. There are also surf lessons given if you have never surfed before but want to learn how. They also offer a variety of tours including glass bottom boat tours which will take you past beautiful landmarks such as Lands End, the Arch and Lovers Beach. On the way you can look down and view the prolific marine life of the Pacific Ocean. On land you can go hiking, camping or on an ecotour to view the beautiful wildlife. There is even an equestrian center which offers horseback riding on the beach and on trails among other equine activities.
For the motor enthusiasts there is off road racing that you can partake in or watch. There are also tours that you can take by either renting out a jeep or on an ATV. You can travel through the desert and catch views of the ocean along the way. They also offer outback adventure trips in an H2 Humvee that you can rent. For some fun and adventure you can even go zip lining in the canopy over the scenic canyon terrain of the Boca de Sierra. This is sure to get your heart racing. There is a golf course designed by pro golfers if golf is your sport.
No matter what your tastes are and what type of outdoor activity you prefer, Cabo san Lucas is sure to have something that you will enjoy doing.


Coming to Mondo Moxie this June the horror comedy Tremors.
Summary: Natives of a small isolated town defend themselves against strange underground creatures which are killing them one by one.
"Horror/comedies often tread too far to one side or the other of that fine line; Tremors walks it like a tightrope."
"Fast-moving, light-handed, assured, even witty at times, and filled with satisfying special effects, Tremors plays like a redneck "Dune.""
"With its excellent, offbeat cast, its sprightly pacing and its goofy tone, Tremors is the kind of movie that propels you out of the theater with a grin on your face"
"This is what a monster movie is supposed to be like, and it's terrific."
"A delightful throwback to such '50s and '60s films as "Them," "The Deadly Mantis" and "Attacks" of both "The Giant Leeches" and "The Crab Monsters.""
"Directed and co-written by Ron Underwood, Tremors maintains a good, steady tongue-in-cheek tone while working nicely as a suspense thriller."
"While liberally dosing the action with humor, Underwood is able to preserve an undertone of genuine menace and substantial suspense. His shooting style is clean and classical, distinguished by camera movements that emphasize the line of the action without becoming conspicuous in themselves."


I had a co-worker tell me today how amazed he was at my dealing with… everything. I have to be honest. Sentiments such as these are humbling and they make me pause. There's no way anyone can really deal with … everything well. God only knows I've had (and continue to have) my moments. But I'm doing the only thing I know how to do and that's to keep going. Negativity and anger really don't help. They just spew more ick into the world. You have to keep on even when the days aren't going well because that is the only way you'll get enough momentum to get out. And that's not something I'm regurgitating from a Hallmark card. I had to experience it again, the hard way.
On Tuesday of last week I was scared. After what seemed to be a normal Monday, Dave took a dip (as we were warned may happen) and spiked a fever due to infection. I realized how much the chemo could weaken his immune system and how quickly something he would normally deflect could bring him down. Fast. As the doctor wrote out prescriptions for antibiotics and a new course of pain management, I could tell by his tone that he was concerned and here's why. The pathology reports were still inconclusive as to what type of Lymphoma Dave was fighting. So, while that would have been nice to know, we were still in a holding pattern — swatting at the general idea quasi blind while waiting for specific direction on how to counter-attack the mutant cells. It is a circumstance that makes you weary.
A few days later on Friday I hit a low, and this time it had little to do with Dave's situation. Our eldest son, Brandon, was diagnosed with a dilated aorta that is most likely due to a connective tissue disorder. Watching him take the news from a room full of specialists in white coats was heart wrenching. In a matter of moments, what he identified as his life and his way was being completely re-arranged by the suspected misspelling of the FBN1 gene. The medical team knew of Dave's recent diagnosis and that just made the entire moment feel pitiful. How in the world was all of this happening? So for all of you out there who think I've been doing really well with… everything, here's the reality: I broke down. There were not enough tissues in that room and I was a complete sap.
But guess who was there for me when we got home… yep, you know it. It was Dave. He got me situated on the couch, brought me a mug of coffee, called all the grandparents with the news and took care of the kids after-school so I could take a nap. I'm sure that wore him out but it was a gift that I'll hang onto through all this nonsense. No matter what, we take care of each other. And in doing that, we can take on the world.
A few days have passed and we have gotten over last week's hurdles. Dave is feeling much better and is getting ready for another CT scan this Friday with chemo the following Monday. Brandon is amazing and realizes what a blessing early detection can be. Neither are giving up or giving in. They both are keeping on. When in Rome, right?
So why the change? Dave wanted to have a little bit of control should the chemo decide to temporarily make him reach for hats this fall. I realized that my hair would be better served helping someone else feel whole while going through treatment. So, with the help of my Dad for Dave and my friend Terry and Beautiful Lengths for me, we went short and sassy.