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

``Geometric probing considers problems of determining a geometric structure or
some aspect of that structure from the results of a mathematical or physical
measuring device, a probe.''~\cite[Page 1]{DBLP:journals/algorithmica/Skiena89}
Many probing tools have been studied in the literature such as finger probes~\cite{DBLP:journals/jal/ColeY87},
hyperplane (or line) probes~\cite{Dobkin:1986:PCP:12130.12174,DBLP:journals/ipl/Li88a},
diameter probes~\cite{DBLP:journals/ijrr/RaoG94},
x-ray probes~\cite{DBLP:journals/siamcomp/EdelsbrunnerS88,Gardner92},
histogram (or parallel x-ray) probes~\cite{MeijerSkiena96},
half-plane probes~\cite{DBLP:journals/jal/Skiena91} and
composite probes~\cite{DBLP:journals/amai/BrucksteinL91,DBLP:journals/ipl/Li88a} to name a few.
For example, diameter probes measure the width of the polygon at a certain direction; such measurements in all directions yield the diameter function of the polygon.

A geometric probing problem can be considered as a {\em reconstruction} problem. Can one reconstruct an object given a set of probes?
For diameter probes this is not the case: there exists a class of (curved) \emph{orbiform}  shapes such as the celebrated \emph{Reuleaux triangle} that have the same diameter function as the circle.
What is more surprising is that this is not the case even for polygonal shapes: there are uncountable families of polygons with the same diameter function, where the polygons  need not be regular, nor does their diameter function need to be constant~\cite{DBLP:journals/ijrr/RaoG94}.
In this paper we show, that an alternative probing device called an  \emph{$\omega$-wedge} yields a function that is free from these  drawbacks.

The $\omega$-wedge, first studied by Bose et al.~\cite{DBLP:journals/comgeo/BoseCSS16},  is the closed set of
points contained between two rays that are emanating from a single point, the \emph{apex} of the wedge. The angle $\omega$ formed by the two rays is such that $0 < \omega < \pi$.
A single probe of a convex $n$-gon $P$
is {\em valid} when
$P$ is inside the wedge and both rays of the wedge are tangent to
$P$, see Figure~\ref{fig:w-probe}a.\footnote{In~\cite{DBLP:journals/comgeo/BoseCSS16} probing with an $\omega$-wedge
is defined for a wider class of  convex objects. Since here we focus on the objects being convex polygons, we restrict our definition accordingly.}
A valid probe returns the coordinates of the apex
and of the two points of contact between the wedge and the polygon.
Using this tool,
a convex $n$-gon can be reconstructed using between $2n-3$ and $2n+5$ probes, depending on the value of $\omega$
and the number of
\emph{narrow vertices} (vertices whose internal angle is at most $\omega$) in $P$~\cite{DBLP:journals/comgeo/BoseCSS16}.
As an $\omega$-wedge rotates around $P$,
the locus of positions of the apex of the $\omega$-wedge
describes a curve called an \emph{$\omega$-cloud}, see Figure~\ref{fig:w-probe}c.
The $\omega$-cloud
finds applications in diverse geometric algorithms~\cite{abellanas2011coverage,ALOUPIS2010115,moslehi2017separating}.

\begin{figure}[ht]
\begin{minipage}{0.32\linewidth}
\centering
\includegraphics{wedge-1}
\\
(a)
\end{minipage}
\begin{minipage}{0.32\linewidth}
\centering
\includegraphics{narrow-pivot-dir}
\\
(b)
\end{minipage}
\begin{minipage}{0.32\linewidth}
\centering
\vspace{-1em}
\includegraphics[page=4,scale = 0.75]{narrow-pivot}
\\
(c)
\end{minipage}

\caption{A convex polygon $P$ (filled with gray color), and (a) A minimal $\omega$-wedge; (b) Narrow vertex $u$ of $P$,
wedges $W_\ell(u)$ and $W_r(u)$ (shaded with
rising  and
falling tiling pattern, respectively) and their directions $\dirl{u}$ and $\dirr{r}$ (dashed lines);
(c) The $\omega$-cloud $\Omega$ of $P$: the arcs (bold curved lines), pivots (filled disk marks), and all the supporting circles (light thin curved lines).}

\label{fig:w-probe}
\end{figure}

The $\omega$-cloud can be seen as a generalization of
the diameter function introduced by
Rao and Goldberg~\cite{DBLP:journals/ijrr/RaoG94}.
In their paper,
a diameter probe consists of
two parallel calipers turning around a convex object $P$ in the
plane. The function that returns the distance between these calipers as they
turn around $P$ is called a \emph{diameter function}.
Rao and Goldberg~\cite{DBLP:journals/ijrr/RaoG94}
show that two different
convex polygons can have the same diameter function, and the function need not be a constant.
 This implies that
recovering the exact shape and orientation of a convex $n$-gon given only its
diameter function is not always possible and additional information is required.
An $\omega$-wedge can be seen as
two non-parallel calipers turning around a convex object $P$ in the plane.
As we prove in this paper, two different convex
 polygons cannot have the same (non-constant) $\omega$-cloud function, as opposed to the case of Rao and Goldberg's tool.
This clearly shows the advantage of  the $\omega$-cloud against the latter tool.

Related to our probing method is also the method of Fleischer and Wang~\cite{DBLP:conf/isaac/FleischerW09}.
In their method, a convex $n$-gon $P$ is placed inside a circle of
radius $1$. A camera that sees only the silhouette of $P$ can be placed
anywhere on the circle. The angle with which the camera sees $P$ together
with the position of the camera is a probe of $P$.  Let $\alpha$ be the
largest angle of $P$. They prove that if no two camera tangents overlap,
then $\left\lceil \frac{3\pi}{\pi-\alpha}\right\rceil$ probes are necessary and
sufficient. Otherwise, approximately $\left\lceil
\frac{4\pi}{\pi-\alpha}\right\rceil$ probes
are sufficient. In our method, the apex of the $\omega$-wedge that turns
around $P$ can be seen as a camera. Thus our method is a variant of
theirs because instead of fixing the circle on which the camera can move, we fix
the angle from which $P$ can be seen by the camera.

In this paper, we prove that no two convex polygons have the same $\omega$-cloud (see Theorem~\ref{thm:uniqueness} in Section~\ref{sec:props}).
We further consider a harder, but more natural problem of reconstructing a polygon from its \emph{maximal} $\omega$-cloud, which is a variant of an $\omega$-cloud where the consecutive arcs with the same supporting circle are merged in one arc (see Section~\ref{subsec:max-cloud}).
We give two conditions under which the maximal $\omega$-cloud uniquely determines $P$:
(i) if $\omega$ with $\omega < \pi$ is fixed/given,
or (ii) without fixing $\omega$, but with a guarantee that $\omega < \pi/2$.
We show that if neither of the conditions (i), (ii) hold, then $P$ may not be unique.
Finally, we show that, when conditions  (i) or (ii) hold, the polygon $P$ can be reconstructed
in $O(n)$ time and  $O(1)$ working space in addition to the input,
where $n$ is the number of arcs in the input $\omega$-cloud (see Section~\ref{sec:reconstr}).

The results in this paper differ from the ones in  Bose et al.~\cite{DBLP:journals/comgeo/BoseCSS16}
in the following ways.
First, we consider a different probing method, where
only one piece of information is available for each probe instead of three,
namely the coordinates of the apex of the $\omega$-wedge.
Second,
we prove a fundamental property of the $\omega$-cloud, that
no two convex polygons have the same
$\omega$-cloud, and thus $\omega$-clouds yield a more advantageous probing method than diameter probes.
Third, we derive a list of interesting properties of the $\omega$-cloud, which were not known before,
and  that allow us to provide a simple algorithm to reconstruct
the polygon from its maximal $\omega$-cloud.
Note as well, that most of our results are more general than the ones from Bose et al.~\cite{DBLP:journals/comgeo/BoseCSS16}
since the results hold for any $0 < \omega < \pi$, instead of the more restricted range: $0 < \omega \leq \pi/2$.

\section{Properties of the $\omega$-cloud}\label{sec:props}

Let $P$ be an $n$-vertex convex polygon in $\mathbb{R}^2$.
For any vertex $v$ of $P$, let $\ang{v}$ be the internal angle of $P$ at $v$.
Let $\omega$ be a fixed angle with $0 < \omega < \pi$.
Consider an \emph{$\omega$-wedge} $W$; recall that it is the set of points contained between two rays $a_\ell$ and $a_r$ emanating from the same point $q$ (the \emph{apex} of $W$),
such that the angle between the two rays is exactly $\omega$. See Figure~\ref{fig:w-probe}a.
We call the ray $a_\ell$ (resp., $a_r$) that bounds
$W$ from the left (resp., right) as seen from $q$,
 the \emph{left} (resp., \emph{right}) \emph{arm} of $W$.
We say that an $\omega$-wedge $W$ is \emph{minimal} for $P$ if $P$ is contained in $W$ and the arms of $W$ are tangent to $P$.
The \emph{direction} of $W$ is given by the bisector ray of the two arms of $W$.
Note that for each direction, there is a unique minimal $\omega$-wedge.

\begin{definition}[$\omega$-cloud~\cite{bmss11}]
\label{def:w-cloud}
The \emph{$\omega$-cloud} of $P$ is the locus of the apices of all minimal $\omega$-wedges for $P$.
\end{definition}

Let $\Omega$ denote the $\omega$-cloud of $P$.
We define an \emph{arc} $\Gamma$ of $\Omega$ as a maximal connected portion of $\Omega$ such that for every point of $\Gamma$ the corresponding minimal $\omega$-wedge is combinatorially the same, i.e., its left and right arm touch the same pair of vertices of $P$.
Due to the inscribed angle theorem,  an arc of the $\omega$-cloud is a circular arc,
and $\Omega$ consists of a circular sequence of circular arcs, where each
pair of consecutive arcs shares an endpoint, see also~Bose et al.~\cite{bmss11} for more details.
We refer to such a sequence as a circular arc sequence.
The \emph{supporting circle} of a circular arc is the circle containing the arc.
The points on $\Omega$ that are the intersection of two consecutive arcs of $\Omega$
are called \emph{pivots}.
Note that if $\omega \geq \pi/2$, two consecutive arcs of the
$\omega$-cloud can lie on the same supporting circle.
In this case we call the pivot separating them a \emph{hidden pivot}. For example, points $b, d, f$ in Figure~\ref{fig:j-lous}a are hidden pivots for the polygon $abcdef$, when $\omega = 5/6$.
The total number of pivots (including hidden ones) is between $n$ and $2n$~\cite{bmss11}.

A vertex $v$ of $P$ is called \emph{narrow} if
the angle $\ang{v}$
is at most $\omega$.
Note that a pivot of $\Omega$ coincides with a vertex of $P$ if and only if that vertex is narrow.
In this case, we also call such pivot \emph{narrow}.
If $\ang{v} < \omega$, we call $v$ (the vertex or the pivot) \emph{strictly narrow}.
The portion of $\Omega$ between
two points $s,t \in \Omega$, unless explicitly stated otherwise,
is the portion of $\Omega$
one encounters when traversing $\Omega$ from $s$ to $t$ clockwise, excluding $s$ and $t$.
We denote this portion of $\Omega$ by $\Omega_{st}$.
The \emph{angular measure} of an arc $\Gamma$ of $\Omega$ is the angle spanned by $\Gamma$, measured from the center of its supporting circle;
the angular measure of a \emph{portion} of an arc is defined similarly.
For two points $s,t$ on $\Omega$, the \emph{total angular measure} of $\Omega$ from $s$ to $t$, denoted as
$\turnValue{s,t}$,
is the sum of the angular measures of all arcs in $\Omega_{st}$ (including the at most two non-complete arcs of $\Omega$).

Each point $x$ in the interior of an arc corresponds to a unique minimal $\omega$-wedge $W(x)$
with direction $\dir{x}$.
Let $u$ be a pivot of $\Omega$.
If $u$ is not strictly narrow,
$u$ also corresponds to a unique minimal $\omega$-wedge $W(u)$ with  direction  $\dir{u}$.
Otherwise,
$u$ corresponds to a closed interval of directions $[\dirl{u}, \dirr{u}]$,
where the angle between $\dirl{u}$ and $d_r(u)$ equals $\omega - \ang{u}$. See Figure~\ref{fig:w-probe}b.
Let $W_\ell(u)$ and $W_r(u)$ denote the minimal $\omega$-wedges with apex at $u$ and directions respectively $d_\ell(u)$ and $d_r(u)$.
For points $x$ on $\Omega$ that are not strictly narrow pivots, we define $\dirr{x}$ and $\dirl{x}$ both to be equal to $\dir{x}$, and  $W_\ell(x)$ and $W_r(x)$
equal to $W(x)$.

The now give a crucial property of the $\omega$-cloud that is the basis of the other properties.

\begin{lemma}
\label{lemma:ang-measure}
Let $s$ and $t$ be two points on $\Omega$ such that
$\Omega_{st}$ contains no narrow pivots.
Then the angle between $\dirr{s}$ and $\dirl{t}$ is $\turnValue{s,t}/2$.
\end{lemma}
\begin{proof}
Suppose first that $\Omega_{st}$ is a single arc, see Figure~\ref{fig:left-right-points}a.
The angle $\alpha$ between $\dirr{s}$ and $\dirl{t}$ equals the angle $\beta$ between the left arms of the two minimal $\omega$-wedges corresponding to these directions, which in turn equals angle $\beta'$.
By the inscribed angle theorem, $\beta'$ is half the central angle spanning the arc $\Omega_{st}$, which, by definition of the angular measure, equals $\turnValue{s,t}/2$.

Now suppose that $\Omega_{st}$ consists of several arcs, which are either arcs of $\Omega$ or (at most two) connected portions of those arcs.
By assumption, none of the pivots separating these arcs are narrow (although $s$ and $t$ may be, see the definition of $\Omega_{st}$).
Consider the change of direction
of the minimal $\omega$-wedge as its apex moves from $s$ to $t$. After traversing an arc with endpoints $u$ and $v$,
by the above observation, the direction changes by exactly $\turnValue{u,v}/2$.
Since none of the
pivots on $\Omega_{st}$ are narrow, there is no change in direction as the $\omega$-wedge passes through each pivot.
Therefore the total change in direction for the $\omega$-wedge is the sum of the changes induced by the traversed arcs, that is, $\turnValue{s,t}/2$.
\end{proof}

\begin{corollary}
  \label{cor:total-measure}
  The sum of the angular measures of all arcs of $\Omega$
  is $2(2\pi-\sum_{v \in S}(\omega-\ang{v}))$, where $S$ is the set of all
  narrow vertices of $P$.
  In particular,
 $P$ has no strictly narrow vertices if and only if
  the sum of angular measures of the arcs of $\Omega$ is $4\pi$.
\end{corollary}

\begin{proof}
Choose a point $x$ in the interior of an arc of  $\Omega$ and consider the change in the direction of the minimal $\omega$-wedge as its apex traverses the whole $\Omega$ and returns back to $x$. Since the minimal $\omega$-wedge at $x$ is unique,  the total angle the direction has turned is $2\pi$.
The narrow pivots break $\Omega$ in a number of connected components, for which Lemma~\ref{lemma:ang-measure} applies.
The total angular measure of the arcs of $\Omega$ is the sum of angular measures of these components.
In a narrow pivot $u$,
the minimal $\omega$-wedge turns from the position when its left arm coincides with the polygon edge incident to $u$ and following $u$ in the clockwise direction, to the position when its right arm coincides with the polygon edge
preceding $u$. See Figure~\ref{fig:w-probe}b.
Thus the angle the direction of the minimum $\omega$-wedge turned, while its apex has stayed in $u$, equals $\omega-\ang{u}$.
Summing up such  deficit for all narrow pivots, we obtain that the total turn of the $\omega$-wedge corresponding to the arcs of $\Omega$ is
$2\pi - \sum_{v \in S}(\omega-\ang{v})$. Lemma~\ref{lemma:ang-measure} as applied to the components of $\Omega$ as separated by the narrow pivots,  implies that
the total angular measure of all arcs of $\Omega$ equals $2(2\pi - \sum_{v \in S}(\omega-\ang{v}))$.
\end{proof}

Since each arm of any minimal $\omega$-wedge touches $P$, and since $P$ is convex,  Lemma~\ref{lemma:ang-measure} implies the following.

\begin{corollary}
\label{cor:max-measure}
For any arc $\Gamma$ of $\Omega$,
 the angular measure of $\Gamma$ is at most $2(\pi-\omega)$.
\end{corollary}

The following is a simple fact directly implied by the definition of the $\omega$-cloud.

\begin{lemma}
\label{lemma:normal-pivot}
Let $u$ be a non-narrow pivot of $\Omega$.  Let $x$ be the second point of intersection between the supporting circles of the two arcs of $\Omega$ adjacent to $u$ (the first point of intersection is $u$ itself).  The minimal $\omega$-wedge with the apex at $u$ touches $x$ with one of its arms. \end{lemma}

\begin{figure}[ht]
\begin{minipage}{0.32\linewidth}
\centering
\includegraphics{ang-measure}
\vspace{0.9em}
\\
(a)
\end{minipage}
\begin{minipage}{0.32\linewidth}
\centering
\includegraphics[page=2,scale = 0.83]{narrow-pivot}
\\
(b)
\end{minipage}
\begin{minipage}{0.32\linewidth}
\centering
\includegraphics[page=1, scale = 0.83]{narrow-pivot}
\\
(c)
\end{minipage}
\caption{
(a) Illustration for the proof of Lemma~\ref{lemma:ang-measure}, the case when $\Omega_{st}$ is a single arc. Vertices of $P$ are marked as filled disks.
(b) Point $x$ in the interior of an arc of $\Omega$,
wedge $W(x)$ with its direction $\dir{x}$,
points $x_\ell$ and $x_r$. (c) Narrow pivot $u$,
wedges $W_\ell(u)$ and $W_r(u)$,  points $u_\ell$ and~$u_r$.}
\label{fig:left-right-points}
\end{figure}

\textbf{Characterization of narrow pivots.}
Let $x$ be a point on $\Omega$. It corresponds to two  minimal $\omega$-wedges $W_\ell(x)$, $W_r(x)$ (which coincide if $x$ is not a narrow pivot).
Consider the open ray of the  right arm of $W_\ell(x)$.
A small neighborhood of its origin is in the interior of the region bounded by $\Omega$,
 thus the ray intersects $\Omega$ at least once.
Among the points of this intersection, let $x_\ell$ be the one closest to $x$. Define the point $x_r$ analogously for the left arm of $W_r(x)$.
See Figure~\ref{fig:left-right-points}b,c.

\begin{lemma}
\label{lemma:turns}
(a) Neither $\Omega_{x_{\ell}x}$ nor $\Omega_{xx_r}$ contains a narrow pivot.
(b) If $x$ is a narrow pivot, then $\turnValue{x_\ell,x} = \turnValue{x,x_r} = 2(\pi - \omega)$.
(c) If $x$ is not narrow,
then either $\turnValue{x,x_r} = 2(\pi - \omega)$, or $x_r$ is the first narrow pivot following $x$ in the clockwise direction. A symmetric statement holds for $x_\ell$.
\end{lemma}

\begin{proof}
We prove
the statement for $x$ and $x_r$;
 symmetric arguments apply for $x_\ell$ and $x$.

\textit{(a)} If $\Omega_{xx_r}$ contained a narrow pivot $u$, then $u$ had to lie inside the wedge
$W(x)$ (because it would be a vertex of $P$). However, in that case
the left arm of $W(x)$ should have intersected $\Omega$ before $x_r$, contradicting to the definition of $x_r$.

\textit{(b)} By Lemma~\ref{lemma:ang-measure}, $\turnValue{x,x_r}$ is twice the angle between $\dirr{x}$ and $\dirl{x_r}$.
Thus we need to show that the latter is $\pi - \omega$.
By definition, $x_r$ lies on the line $\ell$ through the left arm
of the wedge $W_r(x)$. Since $x$ is narrow, $\ell$ passes through an edge of $P$.
Now consider $W_\ell(x_r)$, the leftmost minimal $\omega$-wedge corresponding to $x_r$.
Its right arm is aligned with $\ell$, as this is the leftmost possible line through $x_r$
with $P$ entirely to its left. See Figure~\ref{fig:left-right-points}c.
Therefore the angle between the bisectors of $W_r(x)$ and $W_\ell(x_r)$ is $\pi-\omega$, and $\turnValue{x,x_r}$ is $2(\pi-\omega)$.

\textit{(c)} Let $x$ be not a narrow pivot. We suppose $\turnValue{x,x_r} \neq 2(\pi - \omega)$, and will show that $x_r$ is a narrow pivot of $\Omega$. By the above observation that there can be no narrow pivots between $x$ and $x_r$, this will imply the claim.
Suppose $x_r$ is not a narrow pivot.
Then there is a vertex $v \neq x_r$ of $P$,
 where the left arm of $W(x)$ touches $P$.
Then the right arm of $W_\ell(x_r)$ must be aligned with $\ell$, which would mean $\turnValue{x,x_r} = 2(\pi - \omega)$.
A contradiction.
\end{proof}

The above lemma characterizes narrow pivots in terms of the position of the corresponding minimal $\omega$-wedges.
Now we characterize them in terms of the adjacent arcs and their measures, the information that can be used by
the reconstruction algorithm.

\begin{lemma}
\label{lemma:narrow-char}
Let $u$ be a pivot of $\Omega$, and let $v$ and $w$ be the points on $\Omega$ such that
$\turnValue{v,u} = \turnValue{u, w} = 2(\pi - \omega)$. \\
 (a) If pivot $u$ is narrow, then
the supporting circles of all the arcs of $\Omega_{vw}$ pass through $u$.\\
(b) Pivot $u$ is narrow, if at least one of the following conditions is satisfied:
(i) $\Omega_{vu}$  consists of a single arc;  (ii) there is an  arc $\Gamma$ of
$\Omega_{vu}$ that is not incident to $u$, such that the supporting circle of $\Gamma$ contains $u$.
\\
 A symmetric statement holds for $\Omega_{uw}$.
\end{lemma}

\begin{proof}
(a) Suppose that $u$ is a narrow pivot.
Recall the definition of $u_\ell, u_r$ from above, see also Figure~\ref{fig:left-right-points}c.
 By
Lemma~\ref{lemma:turns}b the points $v$ and $w$ coincide with
$u_\ell$ and $u_r$, respectively. See Figure~\ref{fig:circles}. The minimal $\omega$-wedge, as its apex traverses $\Omega_{u_{\ell}u}$, is always touching $u$ with its left arm.
Indeed, when its apex is at $u_\ell$, its left arm is aligned with the line through $u$ and $u_\ell$, and thus with the edge $e_\ell$ of $P$ incident to $u$ and preceding it in clockwise order.
When the apex
reaches $u$, its direction is $\dirl{u}$ and its
right arm is touching $e_\ell$. Thus every placement of the minimal $\omega$-wedge in the considered interval was touching $u$ with its left arm. Symmetrically, every minimal $\omega$-wedge
 with the apex at any point between $u$ and $u_r$ is touching $u$ with its right arm.
The claim is implied.

(b) Suppose first that  $\Omega_{vu}$ consists of a single arc $\Gamma$.
We show that any minimal $\omega$-wedge corresponding to this arc must pass through both $u$ and $v$, which means that they both are vertices of $P$, and thus narrow pivots.  To see this, pick any point $y$ on $\Gamma$, and any point
$z$ on the supporting circle of $\Gamma$ outside $\Gamma$. See Figure~\ref{fig:b}. The quadrilateral $zvyu$ is inscribed in the circle, thus the sum of the angle at $z$ and the one at $y$ is $\pi$ (the angles $\beta$ and $\gamma$ in  Figure~\ref{fig:b}). The former angle is half the angular measure of $\Gamma$ (the angle $\beta$ in Figure~\ref{fig:b} is half the angle $\alpha$), i.e., it is $\pi-\omega$. Therefore the angle at $y$ is $\omega$, i.e., the minimal $\omega$-wedge
with the apex $y$ is passing through $u$ and $v$.

\begin{figure}
\begin{minipage}{0.49\linewidth}
\centering
\includegraphics[page=1]{narrow-char-proof-1}
\\ (a)
\end{minipage}
\begin{minipage}{0.49\linewidth}
\centering
\includegraphics[page=2]{narrow-char-proof-1}
\\ (b)
\end{minipage}

\caption{Illustration for the proof of Lemma~\ref{lemma:narrow-char}b,  the case when the portion of $\Omega$ between $v$ and $u$ consists of several  arc.}
\label{fig:narrow-char}
\end{figure}

Suppose  now that $\Omega_{vu}$ contains at least two arcs, and suppose that there exists arc  $\Gamma \in \Omega_{vu}$ that is not incident to $u$, but whose supporting circle $C$ passes through $u$, see Figure~\ref{fig:narrow-char}.
Let $\Gamma'$ be the arc of $\Omega_{vu}$ incident to $u$, and let $C'$ be the supporting circle of $\Gamma'$.
Let $u'$ be the most clockwise endpoint of $\Gamma$.
It is enough to show that pivot $u'$ corresponds to the turn of the $\omega$-wedge around $u$: indeed, this would imply
that $u$ coincides with a vertex of $P$, and thus it is a narrow pivot.

Assume for the sake of contradiction that the above does not hold.
Suppose first that the pivot $u'$ is not narrow.
Observe that the
wedge $W(u')$ cannot intersect arc $\Gamma$ with its arms.
Thus the arms of $W(u')$ touch circle $C$ in the apex $u'$ and two points $v_1, v_3$, that are the two vertices of $P$ that correspond to the arc $\Gamma$. See Figure~\ref{fig:narrow-char}a.
The two vertices $v_2,v_4$ corresponding to the arc $\Gamma'$
 must lie inside $W(u')$.
By construction, $v_1,v_2,v_3,v_4$ must be a  subsequence of the sequence of vertices of the polygon $P$
in clockwise order.
However, by the assumption of the lemma, both $C$ and $C'$ pass through $u$ and $u'$,
and thus points  $v_1,v_2,v_3,v_4$ (in this order)  are not in convex position, contradicting to $P$ being convex.

Suppose now that the pivot $u'$ is narrow. The minimal $\omega$-wedge as its apex traverses arc $\Gamma$
 turns around $u'$ and some vertex $v_1$ of $P$. See Figure~\ref{fig:narrow-char}b.
The minimal $\omega$-wedge $W(u)$ of $u$ must contain $u'$ since it is a vertex of $P$.
For example, in Figure~\ref{fig:narrow-char}b the right arm of this wedge passes through $u'$.
However, in this case one of the vertices of $P$ that correspond to the arc
$\Gamma'$ is outside $W(u')$, see the unfilled circle mark in Figure~\ref{fig:narrow-char}b. A contradiction.

The proof of item (b) for $\Omega_{uw}$ is analogous.
\end{proof}

\begin{figure}
\centering
\includegraphics{narrow-single-arc}
\caption{ Proof of Lemma~\ref{lemma:narrow-char}b, the case when the portion of $\Omega$ between $v$ and $u$ is a single arc.}
\label{fig:b}
\end{figure}

\begin{observation}
\label{obs:narrow}
If $u$ is a narrow pivot followed by a non-narrow pivot, then the condition (ii) of item (b) of Lemma~\ref{lemma:narrow-char} for $\Omega_{uw}$ is satisfied,
and
the supporting circle of the arc following $w$ does not pass through $u$.
\end{observation}

\begin{proof}
Let $e = uu'$ be the edge of $P$ adjacent to $u$ and following it in the clockwise direction.
Consider the  motion of the minimal $\omega$-wedge as its apex goes from $u$ to $w$. In the beginning, it rotates around $u$ and $u'$.
The first pivot $w'$ the apex reaches corresponds to the change of one of those points.
If $u$ is changed, then the right arm of the wedge must be collinear with $e$. In this case $w' = u' = w$, and this is a narrow pivot, which contradicts the assumption of this observation.
 If $u'$ is changed, then  the condition (ii) of item (b) of Lemma~\ref{lemma:narrow-char}
is satisfied for $\Omega_{uw}$  and the supporting circle of the arc following pivot $w'$ is passing through $u$ and some other vertex of $P$.

Let $\Gamma'$ be the arc following $w$. Since $\turnValue{u,w} = 2(\pi-\omega)$, the minimal $\omega$-cloud with the apex at $w$
has its right arm collinear with $e$.  If the supporting circle of $\Gamma'$ was passing through $u$, then the minimal $\omega$-wedge $W(a)$
 with the apex at any point $a$ in the interior of  $\Gamma'$ would touch $u$  with its right arm. But then the point $u'$ is outside
 $W(a)$, yielding a contradiction.
\end{proof}

\textbf{Characterization of hidden pivots.}
Recall that a pivot is called hidden if its incident arcs have the same supporting circle.

\begin{lemma}
\label{lemma:hidden-pivots}
Let $u$ be a hidden pivot of $\Omega$, let $\Gamma_\ell$ and $\Gamma_r$ be the two arcs of $\Omega$ incident to $u$, and let $v$ and $w$ be the other endpoints of $\Gamma_\ell$ and $\Gamma_r$, respectively.
Then $v$, $u$, and $w$ are all narrow and each of the arcs $\Gamma_\ell$, $\Gamma_r$ has angular measure $2(\pi-\omega)$.
\end{lemma}

\begin{proof}
Since $u$ is a hidden pivot,  $\Gamma_\ell$ and $\Gamma_r$ are supported by the same circle $C$. Consider the minimal $\omega$-wedge as its apex traverses $\Gamma_\ell$. Its arms are touching two vertices of $P$, both lying on $C$.
Since $u$ is a hidden pivot,  the vertex touched by the left arm of the wedge is $u$ (otherwise, there would be no possibility to switch from $\Gamma_\ell$ to $\Gamma_r$ at point $u$).
Let $a$ be the vertex of $P$ touched by the right arm of the wedge. When the apex of the wedge reaches $u$, the wedge becomes $W_\ell(u)$, and its right arm is passing through $a$ and $u$. If polygon $P$ had a vertex between $a$ and $u$, that vertex must lie outside $W_\ell$, which is impossible. Thus $a$ is actually $v$. Therefore, $\turnValue{v,u} = \turnValue{a,u}$, which by
Lemma~\ref{lemma:turns}b equals $2(\pi-\omega)$. A symmetric argument for $u$, $v$, and $\Gamma_r$ completes the proof.
\end{proof}

We proceed with the main result of this section.
\begin{theorem}
\label{thm:uniqueness}
For a circular arc sequence  $\Omega$, there is at most
 one convex polygon $P$ and one angle $\omega$ with $0 < \omega< \pi$ such that  $\Omega$ is the $\omega$-cloud of $P$.
\end{theorem}

\begin{proof}

We prove that the angle $\omega$ is unique. Suppose for the sake of contradiction that it is not the case, i.e., there are two distinct angles $\omega, \omega'$, for which
$\Omega$ is the $\omega$-cloud and the $\omega'$-cloud of some polygons.   Assume without loss of generality that $\omega < \omega'$.

Suppose first that the total angular measure of $\Omega$ is  $4\pi$. By Corollary~\ref{cor:total-measure}, it does not have any narrow pivots.
Fix any pivot $u$ of $\Omega$. Lemma~\ref{lemma:turns}c  uniquely determines the placement of the minimal $\omega$- and $\omega'$-wedge with the apex at point $u$:
 The two arms of the $\omega$-wedge must pass through the two points on $\Omega$ at distance $2(\pi-\omega)$ from $u$; analogously for the $\omega'$. By this construction, the  $\omega'$-wedge must enclose the  $\omega$-wedge.
But both wedges must touch the intersection point of the two supporting circles of the arcs incident to $u$ due to Lemma~\ref{lemma:normal-pivot}.
This is possible only if the left or the right arms of the two wedges coincide. But then by Lemma~\ref{lemma:turns}c, as applied to the $\omega$-wedge, the point on $\Omega$ at the distance $2(\pi-\omega)$ is a narrow pivot of $\Omega$. This contradicts our assumption.

Suppose now that the total angular measure of $\Omega$ is less than $4\pi$. By Corollary~\ref{cor:total-measure}, there are narrow pivots in $\Omega$. Moreover, if a pivot is narrow for $\omega$, it must be narrow for $\omega'$.
If all the pivots in $\Omega$ are narrow for $\omega$, they are also all narrow for $\omega'$, moreover, they are exactly the vertices of $P$, but then the same polygon $P$ has the $\omega$-cloud and $\omega'$-cloud coinciding with $\Omega$, which contradicts the definition of the $\omega$-cloud. Therefore, not all the pivots of $\Omega$ are narrow for $\omega$.

 Let $u$  be a narrow pivot for $\omega$ followed by a non-narrow pivot for $\omega$; it exists because of our above observation that not all pivots of $\Omega$ are narrow for $\omega$. Let $u, w, u', w'$ be the points of $\Omega$ with $\turnValue{u,v} = \turnValue{v,w} = 2(\pi-\omega)$ and $\turnValue{u',v} = \turnValue{v,w'} =
2(\pi-\omega')$.
By Lemma~\ref{lemma:narrow-char}, the supporting circles of  all the arcs
in $\Omega_{uw}$ pass through $u$. Since $\omega < \omega'$, point $w'$ lies in the interior of $\Omega_{uw}$. Thus the arc $\Gamma$ that follows $w'$ must have the supporting circle passing through $u$.
But by Observation~\ref{obs:narrow}, the supporting circle of $\Gamma$ does not pass through $u$.
This is a contradiction.

We showed that the angle $\omega$ is unique. We now show that the polygon $P$, for which $\Omega$ is the $\omega$-cloud, is unique as well.
Observe that $\Omega$ must have at least two arcs.
 Lemmas~\ref{lemma:narrow-char} and~\ref{lemma:hidden-pivots} uniquely identify the narrow pivots of $\Omega$, which are the narrow vertices of $P$ (including hidden narrow pivots).
 By Lemma~\ref{lemma:turns}a, the portion of $\Omega$ between any two narrow pivots has total angular measure at least
$2(\pi-\omega)$, and thus the components of $P$ as defined by excluding all narrow vertices  are uniquely determined by Lemma~\ref{lemma:turns}b. In particular, for each such component  Lemma~\ref{lemma:turns}b gives the minimal $\omega$-wedge $W$ with the apex at some point $x$ in that component. Let $\Gamma$ be an arc that contains or is incident to $x$, and $C$ be the supporting circle of $\Gamma$. Intersecting $W$ with $C$ gives the two vertices of $P$, that are tangent to the arms of the $\omega$-wedge as its apex traverses $\Gamma$.
\end{proof}

\subsection{Maximal $\omega$-cloud}
\label{subsec:max-cloud}
Given the $\omega$-cloud $\Omega$ of $P$, let the \emph{ maximal $\omega$-cloud} of $P$, denoted ${\Omega^*}$, be the result of merging all the pairs of consecutive arcs in $\Omega$ that have same supporting circle; equivalently ${\Omega^*}$ can be seen as the result of removing all the hidden pivots from $\Omega$.

The maximal $\omega$-cloud is a natural modification of the $\omega$-cloud for the reconstruction task, see Section~\ref{sec:reconstr}; it reflects the situation when as an input we are given a locus of the apices of all the minimal $\omega$-wedges without any additional information,
i.e., without the coordinates of the hidden pivots.
The following  two statements are corollaries of Lemma~\ref{lemma:hidden-pivots}.

\begin{lemma}
\label{cor:circular-cloud}
The maximal $\omega$-cloud of $P$ is a circle $C$ if and only if
$k = \pi/(\pi-\omega)$ is an integer and $P$ is a regular $k$-gon inscribed in $C$.
\end{lemma}
\begin{proof}
Suppose the maximal $\omega$-cloud ${\Omega^*}$ of $P$ is a circle $C$. Since all arcs of
${\Omega^*}$ of $P$ are supported by the circle $C$, all the pivots of $\Omega$ are hidden. Applying Lemma~\ref{lemma:hidden-pivots} to each pivot, we obtain that the pivots of $\Omega$ are exactly the vertices of $P$, and each arc has measure $2(\pi-\omega)$. Therefore the number of vertices of $P$ is $2\pi/(2(\pi-\omega))$, and the claim follows.
The other direction of the claim directly follows from
the definition of an  $\omega$-cloud.
\end{proof}

\begin{lemma}
\label{lemma:concat}
 An arc of ${\Omega^*}$ is greater than $2(\pi-\omega)$ if and only if this arc is a concatenation of
 at least two  co-circular arcs of $\Omega$ separated by hidden pivots, each of measure $2(\pi-\omega)$.
\end{lemma}

Lemmas~\ref{cor:circular-cloud} and~\ref{lemma:concat}
together with Theorem~\ref{thm:uniqueness} imply a similar uniqueness result for  ${\Omega^*}$:

\begin{theorem}
\label{thm:uniqueness-max}
 Given an angle $\omega$, and a circular arc sequence  ${\Omega^*}$ of at least two arcs, there is at most
 one convex polygon $P$ such that  ${\Omega^*}$ is the maximal $\omega$-cloud of $P$.
\end{theorem}

Another uniqueness result, which is useful
for our reconstruction algorithm, follows from   Theorem~\ref{thm:uniqueness} and the
fact that for $\omega < \pi/2$ the $\omega$-cloud and the maximal $\omega$-cloud coincide.

\begin{corollary} [of Theorem~\ref{thm:uniqueness}]
For a circular arc sequence  $\Omega$, there is at most
 one convex polygon $P$ and one angle
 $\omega< \pi/2$ such that  $\Omega$ is the maximal $\omega$-cloud of $P$.
\end{corollary}

We now show that the condition that $\omega< \pi/2$ is necessary for the above statement.

\begin{proposition}
\label{prop:non-unique}
There is a circular arc sequence such that it is the maximal $\omega$-cloud of $P$ and the maximal $\omega'$-cloud of $P'$ for distinct  angles $\omega, \omega'$ and polygons $P,P'$.
\end{proposition}
\begin{proof}
Let $P$ be a triangle with vertices $a,c,e$,  such that $\angle ace = \pi/3$,  $\angle cea = 16/45\pi$,
and $\angle cae = 14/45\pi$. See Figure~\ref{fig:j-lous}a.  Let $\omega = 2/3\pi$, and let $\Omega$ be
the $\omega$-cloud of $P$; it consists of three circular arcs $\Gamma_0, \Gamma_1,$ and $\Gamma_2$; note that the supporting circles of these arcs are pairwise distinct. Since $\Omega$ has no hidden pivots, $\Omega$ is also the maximal $\omega$-cloud of $P$. Let $b$ (resp.,
$d$ and $f$) be the midpoint of $\Gamma_0$ (resp.,  of $\Gamma_1$ and of $\Gamma_2$), and let $P'$ be the hexagon $abcdef$.
Then $\angle abc = \angle cde =  \angle efa =  \angle bcd = 2/3 \pi$, $ \angle fab = 29/45 \pi$, and  $ \angle fab = 29/45 \pi$.
Let $\omega' = 5/6\pi$, and note that all the angles at the vertices of $P'$ are smaller than $\omega'$.
It is easy to verify that $\Omega$ is the maximal $\omega'$-cloud of $P'$.
\end{proof}

\begin{figure}
\begin{minipage}{0.49\linewidth}
\centering
\includegraphics{j-lous-examp}
\\
(a)
\end{minipage}
\begin{minipage}{0.49\linewidth}
\centering
\includegraphics[page=5]{narrow-pivot}
\\
(b)
\end{minipage}
\caption{(a) A circular arc sequence of $3$ arcs, that is the maximal $2/3\pi-$cloud for triangle $ace$ and the maximal $5/6-$cloud for hexagon $abcdef$. (b) Illustration for the proof of Lemma~\ref{lemma:reconstr-no-narrow}.}
\label{fig:j-lous}

\end{figure}

\begin{remark}
The example in the proof of Proposition~\ref{prop:non-unique} can be continued by subdividing each arc in half and considering $11/12\pi$-cloud for the obtained $12$-gon, and so on.
\end{remark}

\section{Reconstructing $P$ from its maximal \texorpdfstring{$\omega$}{omega}-cloud}\label{sec:reconstr}
In this section we let
${\Omega^*}$ be a circular arc
 sequence.
Our goal is to reconstruct the convex polygon $P$ for which ${\Omega^*}$ is the maximal $\omega$-cloud for some angle $\omega$, or determine that no such polygon exists.

Note that if ${\Omega^*}$ is a single arc, i.e., it is a circle $C$, then $P$ is not unique. By Lemma~\ref{cor:circular-cloud}, it is a regular $\pi/(\pi-\omega)$-gon inscribed in $C$. The position of its vertices on $C$ is impossible to identify given only ${\Omega^*}$ and $\omega$.
Therefore we assume that ${\Omega^*}$ has at least two arcs.

First, in Section~\ref{sec:reconstr-given-w}, we consider $\omega$ to be given, and such that $0 < \omega < \pi$; in this case the maximal $\omega$-cloud of $P$ may differ from its $\omega$-cloud.
Afterwards, in Section~\ref{sec:reconstr-without-w}, we consider the setting where $\omega$ is not known, but it is known that  $0 < \omega < \pi/2$; in this case the two variations of the $\omega$-cloud coincide, and we use $\Omega$ instead of $\Omega^*$ to denote the input.

\subsection{ An  \texorpdfstring{$\omega$}{omega}-aware reconstruction algorithm}
\label{sec:reconstr-given-w}

Here we assume that $\omega$ is given, and that $0 < \omega < \pi$.
The main difficulty in the reconstruction task is caused by  strictly narrow vertices, as the turn of the minimal $\omega$-wedge at those vertices is not reflected in the $\omega$-cloud, see Corollary~\ref{cor:total-measure}.
In our reconstruction algorithm, we first find all the non-hidden
strictly narrow pivots of ${\Omega^*}$, and then
treat the connected portions of ${\Omega^*}$ between those pivots separately.
Lemma~\ref{lemma:reconstr-no-narrow} gives a procedure to process such a portion.

For two points $u$ and $v$ on ${\Omega^*}$, let $P_{uv}$ be the union of the edges and vertices of $P$ touched by the arms of the minimal $\omega$-wedge as its apex traverses $\Omega^*_{uv}$. Since $\Omega^*_{uv}$ is connected, when the apex of the wedge traverses $\Omega^*_{uv}$, for each of its two arms, the point of contact with $P$ traverses a connected portion of $P$; thus  $P_{uv}$ consists of at most two connected portions of $P$. Note that it is possible that one of the portions is a single vertex.

\begin{lemma}
\label{lemma:reconstr-no-narrow}
Given a portion  $\Omega^*_{uv}$ of ${\Omega^*}$ that
does not contain any strictly narrow pivots,
and the direction $\dirr{u}$ of the rightmost minimal $\omega$-wedge $W_r(u)$ of $u$,
the portion $P_{uv}$ of $P$ that corresponds to $\Omega^*_{uv}$
can be reconstructed
in time linear in the number
of arcs in $\Omega^*_{uv}$. The procedure requires $O(1)$ working space in addition to the input.
\end{lemma}
\begin{proof}
 Let  $\Gamma$ be the arc of $\Omega^*_{uv}$ incident to $u$, let $u'$ be the other endpoint of $\Gamma$,  and let $C$ be the supporting circle of $\Gamma$.
 See Figure~\ref{fig:j-lous}b.
 By knowing the value of $\omega$ and the direction
$\dirr{u}$,
we
determine the wedge $W_r(u)$.
The intersection between the wedge $W_r(u)$ and the circle  $C$
determines the
two vertices of $P$ touched by the minimal $\omega$-wedge as its apex traverses arc $\Gamma$.
In Figure~\ref{fig:j-lous}b, these two points are $u$ and $p$.
The direction of the leftmost minimal $\omega$-wedge at $u'$, $W_\ell(u')$, is   $\dirr{u} + \turnValue{u, u'}/2$ due to Lemma~\ref{lemma:ang-measure}.
If $u'$ is inside $\Omega^*_{uv}$, then $u'$ is not a strictly narrow vertex, and thus
there is a unique  minimal $\omega$-wedge $W(u')$
at $u'$, $W(u') = W_\ell(u')$.
Therefore, for each arc of $\Omega^*_{uv}$ we find the pair of vertices of $P$ that induces that arc.
Moreover,  by visiting  the pivots of $\Omega^*_{uv}$ one by one, we find the vertices of each of the two chains of $P_{uv}$ ordered clockwise.
To avoid double-reporting vertices of $P$, we keep the startpoints of the two chains, and whenever one chain reaches the startpoint of the other one, we stop reporting the points of the former chain.

This procedure visits the pivots of $\Omega^*_{uv}$ one by one, and performs $O(1)$ operations at each pivot, namely, finding the intersection between a given wedge and a given circle. No additional information needs to be stored.
\end{proof}

\subsubsection{Reconstruction algorithm.} As an input, we are given an angle  $\omega$, $0 < \omega < \pi$,
and a circular arc sequence ${\Omega^*}$ which is not a single circle.
We now describe an algorithm to check if ${\Omega^*}$ is the $\omega$-cloud of some convex polygon $P$,
and to return $P$ if this is the case.
It consists of two passes through ${\Omega^*}$, which are detailed below.
During the first pass we compute a list $S$ of all strictly narrow vertices of $P$ that are not hidden pivots. With each such vertex $u$,
we store the supporting lines of the two edges of $P$ incident to $u$. In the second pass we use this list to reconstruct $P$.

\textbf{First pass.}
We iterate through the pivots of ${\Omega^*}$ (recall that these are not hidden by the definition of ${\Omega^*}$).
For the currently processed pivot $u$, we maintain the point $v$ on ${\Omega^*}$ such that $\turnValue{v,u} = 2(\pi-\omega)$.
If pivot $u$ is narrow, we jump to the point on ${\Omega^*}$ at the distance $2(\pi-\omega)$ from $u$.
Moreover, if $u$ is strictly narrow, we add  $u$ to the list $S$.
If $u$ is not narrow, we process the next pivot of ${\Omega^*}$. We now give the details.

Let $\Gamma$ be the arc of ${\Omega^*}$ incident to $u$ and following it in clockwise direction. Let $\Gamma_r$ be
the arc following $\Gamma$, and $C_r$ be the supporting circle of $\Gamma_r$.
We consider several cases depending on the angular measure $|\Gamma|$ of $\Gamma$:

\begin{itemize}
\item[(a)]
$|\Gamma| < 2(\pi-\omega)$. See Figure~\ref{fig:first-pass}.
\begin{itemize}
\item[(i)]
Circle $C_r$ passes through $u$, see Figure~\ref{fig:first-pass}a. Then $u$ is narrow by Lemma~\ref{lemma:narrow-char}b.
By tracing ${\Omega^*}$, find the point $w$ on it such that $\turnValue{u,w} = 2(\pi-\omega)$.
In case $u$ is strictly narrow (i.e., $\angle vuw < \omega$), add $u$ to the list $S$ with the lines through $vu$ and $uw$ (these are the intended lines due to Lemma~\ref{lemma:turns}).
Set $v$ to be $u$, and $u$ to be $w$ (regardless the later condition).

\item[(ii)]
Circle  $C_r$ does not pass through $u$, see Figure~\ref{fig:first-pass}b. Then $u$ is not narrow
by Lemma~\ref{lemma:narrow-char}a.
Set $u$ to be the other endpoint of $\Gamma$, and update  $v$ accordingly.
\end{itemize}
\begin{figure}
\begin{minipage}{0.49\linewidth}
\centering
    \includegraphics[page=7,scale = 0.9]{narrow-pivot}
    \\
    (a)
\end{minipage}
\hfill
\begin{minipage}{0.49\linewidth}
\centering
    \includegraphics[page=6,scale = 0.9]{narrow-pivot}
    \\
    (b)
\end{minipage}
    \caption{First pass of the $\omega$-aware algorithm: pivot $u$ is (a) strictly narrow, (b) non-narrow}
    \label{fig:first-pass}
\end{figure}

\item[(b)] $|\Gamma| = 2(\pi-\omega)$. Then $u$ is narrow by Lemma~\ref{lemma:narrow-char}b.
Let $w$ be the other endpoint of $\Gamma$. Update $S$, $v$, and $u$ as in item a(i).

\item[(c)] $|\Gamma| = 2t(\pi-\omega)$ for some integer $t>1$. Then $\Gamma$ is in fact multiple arcs separated by hidden pivots, see Lemma~\ref{lemma:hidden-pivots} and Corollary~\ref{cor:max-measure}.  Let $p$ be the other endpoint of $\Gamma$.
Let $w$ and $w'$ be the points on $\Gamma$ such that $\turnValue{u,w} = 2(\pi-\omega)$ and $\turnValue{w',p} = 2(\pi-\omega)$.
 Update $S$ as in item a(i). Set $u$ to be $p$, and $v$ to be $w'$.
\item[(d)] Otherwise, stop and report that ${\Omega^*}$ is not the maximal $\omega$-cloud of any polygon.
\end{itemize}

Note that before starting the above procedure,
we need to find the starting positions $v$ and $u$.
We choose $v$ arbitrarily and traverse ${\Omega^*}$ until we reach the corresponding position of $u$ (i.e., $\turnValue{v,u} = 2(\pi-\omega)$).
Since we already have traversed the portion of ${\Omega^*}$ between $v$ and $u$,
in order to perform exactly one pass through ${\Omega^*}$, we will have to finish the above procedure as soon as
the pointer $u$ has reached the initial position of $v$ (not the initial position of $u$).
However, this is still enough for creating the complete list $S$. Indeed,
by Lemma~\ref{lemma:turns}b, there is at most one narrow pivot between $v$ and $u$.
There is no such narrow pivot if and only if the left arm of the  minimal $\omega$-wedge $W_r(v)$ and the right arm of $W_\ell(u)$ coincide, and in this case
there is nothing to add to $S$.
If there is such a narrow pivot $w$, then neither $u$ nor $v$ is narrow.
By Lemma~\ref{lemma:turns}c,
both the left arm of the wedge $W(v)$ and the right arm of $W(u)$ pass through $w$.  Thus $w$ can be found
 as their intersection.

\textbf{Second pass.} In case list $S$ is empty, we
apply the procedure of Lemma~\ref{lemma:reconstr-no-narrow} to the whole ${\Omega^*}$.
In particular,  as both the start and the endpoint,
we take the point $x$ with which we completed the first pass of the algorithm; the point  $x'$ such that $\turnValue{x',x} = 2(\pi-\omega)$ is already known from the first pass.
Then $\dirr{x} = \dir{x}$ is the direction of the  minimal $\omega$-wedge with the apex at $x$ and the right arm passing through $x'$.

Suppose now that list $S$ contains $k$ vertices.  They  subdivide ${\Omega^*}$ into $k$ connected portions that are free from strictly narrow non-hidden pivots. Each portion is treated as follows:
\begin{itemize}
\item
 If it is a single maximal arc of measure $2t(\pi-\omega)$, we simply separate it by $t-1$ equidistant points, and those points are exactly the vertices of the portion of $P$ corresponding to the considered component of ${\Omega^*}$, see Lemma~\ref{lemma:hidden-pivots}.

\item Otherwise it is a portion free from any strictly narrow pivots. We process this portion by the procedure of Lemma~\ref{lemma:reconstr-no-narrow}.
\end{itemize}

Correctness and complexity of
this algorithm
are analyzed in the following theorem.

\begin{theorem}
\label{thm:algo-given-w}
Given an angle  $\omega$ such that
$0 < \omega < \pi$, and a circular arc sequence ${\Omega^*}$ of $n$ arcs with $n>1$,
there is an algorithm to check if ${\Omega^*}$ is the maximal $\omega$-cloud of some
convex polygon $P$,
and to return $P$ if this is the case.
 The algorithm works in $O(n)$ time, making two passes through the input, and it
uses $O(1)$ working space. The constants in big-Oh depend only on $\omega$.
\end{theorem}

\begin{proof}
Consider the first pass of the algorithm.
Cases a(i), b, and c are exactly the cases where $u$ is a narrow pivot, see Lemmas~\ref{lemma:narrow-char} and \ref{lemma:hidden-pivots}.
The case d corresponds to an impossible situation
by Lemma~\ref{lemma:hidden-pivots} and Corollary~\ref{cor:max-measure}.
In case a(i), we update $u$ to be $w$, thus we are not processing the pivots between $u$ and $w$; Lemma~\ref{lemma:turns} guarantees
we do not miss any narrow pivot.
Correctness of the second pass is  implied by Lemmas~\ref{lemma:hidden-pivots} and \ref{lemma:turns}a.

Each pivot of ${\Omega^*}$ is visited exactly once during the first pass (it is either checked as the pivot $u$,  traversed in the case a(i), or skipped in the case c). The second pass as well visits each pivot once. The time spent during each such visit is $O(1 + t)$, where $t$ is the number of hidden pivots on the maximal arc adjacent to the visited pivot.
The storage required by the algorithm is the storage required for the list $S$, i.e., it is $O(k)$.

Observe finally that by Lemma~\ref{lemma:turns}, for any pair $u,v$ of narrow pivots of ${\Omega^*}$, $\turnValue{u,v} \geq 2(\pi-\omega)$. By Corollary~\ref{cor:total-measure} the total angular measure of the arcs of  ${\Omega^*}$ is at most $4\pi$. Thus the total number of narrow pivots  of ${\Omega^*}$  (including the hidden pivots) is at most $\floor{2\pi/(\pi-\omega)}$, which is a constant if $\omega$ is fixed. This completes the proof.
\end{proof}

\begin{remark}
It is not difficult to modify the above algorithm to work for the easier setting, where the input is the proper $\omega$-cloud of $P$
instead of its maximal version.
\end{remark}

\subsection{An  \texorpdfstring{$\omega$}{omega}-oblivious reconstruction algorithm}\label{sec:reconstr-without-w}
In this section we assume that $\omega < \pi/2$. Thus there are no hidden pivots in
the $\omega$-cloud of $P$, and therefore the input sequence equals the (not maximal) $\omega$-cloud
 $\Omega$.
 We now give the reconstruction algorithm, which is the main result of this section.

\begin{theorem}
\label{thm:vertex-only}
Given a circular arc sequence $\Omega$, there is an algorithm that finds the convex
polygon $P$ such that $\Omega$ is the $\omega$-cloud of $P$ for some angle $\omega$ with $0 < \omega < \pi/2$, if such a polygon exists.
Otherwise, it reports that such a polygon does not exist.
 The algorithm works in $O(n)$ time, making two passes through the input, and it
uses $O(1)$ working space.
\end{theorem}

\begin{proof}
If $\Omega$ is a single arc, i.e., it is a circle, we return the negative answer, as a circle cannot be an $\omega$-cloud for $\omega < \pi/2$ by Lemma~\ref{cor:circular-cloud}. Below we assume that $\Omega$ has at least two arcs.

If the total angular measure of $\Omega$ is $4\pi$, then no pivot is strictly narrow due to Corollary~\ref{cor:total-measure}. Starting from any pivot $u$ of $\Omega$ and applying Lemma~\ref{lemma:normal-pivot},
we find the placement of one of the arms of the minimal $\omega$-wedge with the apex $u$. Since no pivot is strictly narrow, Lemma~\ref{lemma:turns}c determines whether it is the right or the left arm. This gives us the angle and the placement of the wedge. We can perform the procedure of Lemma~\ref{lemma:reconstr-no-narrow} to reconstruct the polygon.

If the total  angular measure of $\Omega$ is less than $4\pi$, there must be a strictly narrow pivot. We perform a pass through $\Omega$ to find a
maximal sequence of arcs whose supporting circles contain
the same pivot $u$, and such that the sequence starts at $u$.
If we found such sequence of at least two arcs, the distance between $u$ and the endpoint of this sequence is $2(\pi-\omega)$ due to Lemma~\ref{lemma:narrow-char} and Observation~\ref{obs:narrow}. This determines the value of $\omega$. We run the  $\omega$-aware reconstruction procedure of Section~\ref{sec:reconstr-given-w}. If there is no such sequence of at least two arcs, then by Observation~\ref{obs:narrow}  all pivots must be narrow, but  since $\omega< \pi/2$, the polygon $P$ in that case must be a line segment. Whether $\Omega$ is an $\omega$-cloud of a line segment for some $\omega$ can be checked in one pass through $\Omega$.

We can combine finding the above maximal sequence with counting the total angular measure of $\Omega$ in one pass; this pass requires $O(1)$ working  storage. The claim then follows from the Theorem~\ref{thm:algo-given-w}.
\end{proof}

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It's been a century since one has held a Grimoire. There has been a prophecy and rumors saying that one has been discovered underneath the world of Yeïthum. That when found will grant unimaginable power.
Unexpectedly, young wolf Zärion gets into trouble. When he gets framed for having a part in the prophecy.
He's to travel to the forgotten Kingdom, Neithìum and retrieve the book and return it to the believer of the prophecy. Failure to do so will result in immense punishment.
While on his journey, Zärion runs into some complications. That results in unexpected discoveries.
- My first short at a tragic story. First person narrative from Zärion's point of view.
For future reference, the word is spelled "grimoire", not "girmoire".
Thanks for catching that silly mistake.
This page was created in 0.5188 seconds.


Develop confident linguists, who appreciate other cultures with this course, based closely around the IB's desired learner profile. This text caters for Language Acquisition - Group 2 - students learning Spanish as a second language at Standard and Higher levels. It provides a coherent course with carefully crafted tasks to develop all aspects of language learning, including clear explanations, thorough practice of grammar and a starter unit helps to bridge gap from pre-16 exams into the distinctive requirements of the IB Diploma. Each copy includes a free Audio CD providing the tracks for the listening exercises. - Builds language skills through carefully crafted tasks and grammar practice - Improves exam performance with activities for all aspects of IB Spanish assessment - Promotes global citizenship and an appreciation of Hispanic Culture through culturally appropriate stimulus material (including a particular emphasis on the Americas) Also available: vocabulary available online at Vocab Express, an interactive learning application.


Whether looking for a seasonal camp or a cozy year round home, this property is for you! Sitting on just a little over an acre, with a large deck and screened gazebo, you can enjoy the quiet area with views of South Bay of Lake Champlain. Lots of opportunities for all, especially the outdoor enthusiasts! With a short walk to Lake Champlain, you can enjoy beautiful sunsets from the pier in the South Bay! Motivated Sellers! Easy to Show!


Halls Creek lies on the edge of the Great Sandy Desert, at the junction of the Great Northern Highway and Duncan Road in the north west of the State. Halls Creek is the town closest to Purnululu National Park, the park of the famous Bungle Bungles. Also nearby is Wolfe Creek Crater, the second largest meteorite crater in the world. Tours of the crater and the Bungle Bungles are available from the town.
Visit the site of the first goldrush in Western Australia following the discovery in 1885. You can still find gold in the area, particularly around the old town site which is 16 kilometres east of Halls Creek and also features ruins of mud brick buildings and a small cemetery. Mineral exploration continues on a large scale in the area. The town is also the hub of a vast cattle industry.
Russian Jack, a goldmining character from the turn of the century who typified the hardy brand of early Halls Creek pioneers, has been immortalised in the town's main street. The bronze statue depicts him in the act of pushing his sick friend in a wheelbarrow over 300 kilometres to Halls Creek in search of medical attention.
Halls Creek services the tourist market with a hotel, motel, caravan park and two roadhouses.
Drive in or fly over?
There are two ways to encounter the Bungle Bungle: by four-wheel drive, or by air. For advanced four-wheel drivers the Bungle Range offers a perfect wilderness adventure and an unforgettable camping site. You'll find the turnoff to Purnululu National Park 250 kilometres – or around five hours – south of Kununurra, or 109 kilometres north of Halls Creek -Â around four hours drive.
Both camping spots there – Walardi or Kurrajong -Â have simple camping facilities. Most people camp there for a few days at least, so pack plenty of supplies! Take advantage of some the spectacular walks in the area. If you have a couple of days, go exploring in Picaninny Creek and Gorge, a spectacular walk not for the faint-hearted. Another wonderful walk is Echidna Chasm, on the northern side of the park.
The characteristics of the park fall neatly into two areas, north and south. The south, where the Bungle Bungle towers are, is best surveyed in the morning for full light-show effect. The north shows its best face in the afternoon.
The best time of year to visit the Bungle Bungle Range is well into the dry season, from April to October. Extreme weather makes access roads impassable outside of these times, and that's when it's time to take to the air. Some believe that flying offers the best Bungle Bungle experience regardless of the time of year. Scenic flights leave from Halls Creek or Kununurra, and gave an ideal view of this mystical landscape, sweeping effortlessly over the earthen towers of russet, black and moss-green.
The Bungle Bungle range is a stunning collection of beehive style geological formations arrayed in alternating bands of orange, black and green, in Purnululu National Park, 250 kilometres south of Kununurra, in the Kimberley, Western Australia. The area has been inscribed on the World Heritage List in 2003 for its outstanding natural heritage values.
While the reef is ancient, dating back hundreds of millions of years, locals kept it a secret and so the Bungle Bungle or Purnululu (meaning sandstone in the local Aboriginal Kija tongue) was not discovered by the wider world until the 1980s.
The hypnotically beautiful formations were formed by 20 million years of erosion and river movements. They are a combination of sandstone, and conglomerates (pebbles and boulders fixed together by finer material). Back when active faults were shaping the landscapes these sedimentary formations were dumped onto the Ord Basin, some 375 million years ago. The dome shapes are caused by natural erosion and uplift over 20 million years. The darker bands are more porous rock, which lets water move through and creates a dark algal growth. The domes are more fragile than they look. While the sandstone is held in place by overlying rock it is easily eroded, as revealed by the rounded tops.
The land is also rich in Aboriginal culture, being the site of indigenous art and sacred burial grounds. Wallabies, euros, and more than 130 bird species dwell in the park, including rainbow bee-eaters and flocks of beautiful wild budgerigars.
The Bungle Bungle Range, in Purnululu National Park, is one of the most fascinating geological landmarks in Western Australia.
Although the Bungle Bungle Range was extensively used by Aboriginal people during the wet season, when plant and animal life was abundant, few Europeans knew of its existence until the mid 1980s. The area has been a national park since 1987 and its unique appearance has captured the public imagination.
The park, which is open only between April and September, offers a remote wilderness experience, with few facilities and no accommodation; visitors must carry in all food and water. You can camp at Walardi and Kurrajong campsites, but it's essential to be well equipped.
In the Kija Aboriginal language 'purnululu' means sandstone. The name Bungle Bungle comes either from the corruption of an Aboriginal name for the area, or from a misspelling of one of the common Kimberley grasses found here – bundle bundle grass.
The Bungle Bungle Range rises up to 578 metres above sea level, and stands 200 to 300 metres above a woodland and grass-covered plain, with steep cliffs on the western face. It's cut by deep gullies and breaks up into complex areas of ridges and domes, with prominent orange and black or grey bands.
Virtually every visitor to Purnululu asks the same question – how did this remarkable landscape come about?
Well, the distinctive beehive-shaped towers of the Bungle Bungle are made up of sandstone (rocks formed by the consolidation of sand grains) and conglomerates (rocks composed mainly of pebbles and boulders and cemented together by finer material). These sedimentary formations were deposited into the Ord Basin 375 million to 350 million years ago, when active faults were altering the landscape.
The landforms seen today were produced by uplift and erosion during the past 20 million years. One of the most obvious features of the sandstone is the alternating orange and black or grey banding.
The darker bands were formed where moisture seeped through to the rock surface, promoting a dark algal growth. The orange bands are on harder rock which has been stained by iron and manganese.
While the geology of the Bungle Bungle is significant, the area is also rich in Aboriginal art, and there are also many burial sites.
The park is home to more than 130 bird species, including rainbow bee-eaters and flocks of budgerigars. The nailtail wallaby and the euro – a type of kangaroo – live around the base of the range, while the short-eared rock-wallaby is thought to live on top.


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Albert said...
Not a major factor--THE factor. I'll have some analysis of all this as soon as I can crunch the numbers if you're interested.
Tuesday, November 06, 2018 10:01:00 PM
It appears county clerk's retention percentages are inaccurate because he is using the same vote total for each race as the denominator which of course is not correct.
As for the 2020 class, get ready because several of you will be on the menu. Aurujo? Portman? Hooks? Raines? Sheahan? The guy who enjoined Toni's soda tax?
It's a real shame that the retention class voted to give the Cook County Democratic Party $40,000 after the Party stated they were going after Coghlan. Spineless. In effect, the retention class financed the Party mailers responsible for this travesty.
class of 2020 should take note of all this. Toni will be coming after some of you. When you decide on a political consultant, be sure you don't pick one who is also working with Toni behind your backs.
Aurujo is low-hanging fruit and an obvious choice. But the real question will be whether Injustice Watch (and Toni Preckwinkle) have the nerve to target female judges, particularly those of color. Two prime targets for 2020 remain: Megan Goldish (who, like Coghlan, is currently a defendant in a civil rights lawsuit stemming from her days as a Cook County ASA) and Jackie Portman, who graced the front page of the Sun-Times and Injustice Watch pages for her handling of the HOPE Court. Coghlan might be a one-offer. But if additional judges get taken out in 2020, then welcome to the Brave New World.
My long-time desire to become a judge has lost a lot of steam over the last two years, however, it was still there - until now. Not interested in working for Tim Evans or living in fear of the idiots that run the Cook County Democratic Party. Thank you, Jack, for posting your analysis and allowing commentators to post thoughtful perspectives and opinions. I am truly done and it is a relief.
2020 is going to be a train wreck in slow motion, driven by clowns whose hair is on fire. Rumors are already spiraling about all-female slates, comprised of Irish, Latina and Black aspiring jurists and how they plan to run against the machine. But the reality is that there is no machine. What there is a bunch of political hacks who will take your money, put you on a mailer and then do nothing else. If you want to win in this game, play for yourself. And don't delude yourself: the Supreme Court will only appoint women because women win and Theis is going into a 2022 retention cycle herself (assuming Injustice Watch and Toni don't come gunning for her too) and wants the appearance of strength on all fronts. Sorry boys!
Anon 11/8 @3:56 p.m. -- The mailer seems to have been pretty darned effective against Coghlan. Just sayin'....
Tim Evans' days are numbered. What did he to to stand up for Coghlan and put out a counter-narrative more closely related to the truth about the attack on Coghlan? Not a thing. Silence. Not a shred of support for one of his judges who was being unfairly attacked and maligned in the media.
The election for Chief Judge is in September 2019 and Tim has a female opponent — a serious opponent. Adios, Tim. Good luck winning without a retention class to threaten for votes.
If the full circuits had any brass (which they do not have) they would vote Coghlan into one of the associate judgeships that are open.
They should understand that united they will stand and divided they will fall.
That would be funny as hell...
Friday, November 09, 2018 9:51:00 PM
The public has spoken and Judge Coghlan failing to survive retention will never get put back by the Supreme's or any other judicial authority. Unlikely but he could try in a future election.
Saturday, November 10, 2018 9:17:00 AM
I believe that according to the Supreme Court rules, the full circuits can only select associate judges from the list of candidates who have submitted an application for associate judge. The full circuits are free to do write-ins but the attorneys at least have to be from the "long-list" of associate judge candidates certified by the Director of the AOIC.
This is Judge Dredd.
2020 is nigh. And the patriarchy will show you gals who is the dominant gender.
Will Albert be providing an analysis of this?
Recently saw Retired Judge Gloria Chevere was on brochure for supporting organization to dump Matt Coghlan. I know she is a regular contributor to this website with an obvious agenda. It's no surprise to see her name on the list of supporters. Their movement was started by injustice watch and a pac who's officers are suing judge Coghlan, they could not be more biased and obvious in their motives. No surprise they had the support of a retired judge who was photographed tanning in her backyard during court hours and is vigilant in the way people should dress in her courtroom.
Inspired by the candidacies of Judges Fernandez, Ortiz and Judge-Elect Silva, La Raza Latina will be running a countywide slate of Hispanic female candidates. And no, they will not be using Gloria Chevere or Louie Arroyo as campaign consultants.
Yes, I'm working on it. This one was like no other and there's a lot to sort through.
what does the fact that Judge Chevere is a supporter of the pac have to do with the fact that the Appellate Court twice overturned Coghlan's decision not to grant a rehearing for a defendant who said he'd confessed to murder after being tortured by detectives working for disgraced Chicago police Lt. Jon Burge and tends to sentence minorities more harshly on drug possession cases? What obvious agenda does she have besides getting rid of racist judges?
Tuesday, November 13, 2018 3:06:00 PM
Her obvious motivation was to prioritize eliminating racists so that slacker do-nothings like herself would have time to continue to do nothing and eventually retire fully vested despite having done absolutely nothing for years. Crook County at its best.
Why is anyone entitled to be retained? From my analysis it appears at least 2-3 more should not have been. The default shouldn't be retention. If you're a judge who can't get to work on time, manage a courtroom, behave with civility, and treat all litigants fairly you shouldn't be on the bench.
Thursday, November 15, 2018 10:33:00 AM
When you play the game of thrones you either win or die. Good luck 2020 Retention Class . . . looks like you're going to need it.
Megan Goldish has a HUGE target on her back. Best spend those big bucks on a real retention campaign, girlfriend.
@ anonymous on Wednesday, November 14, 2018 3:13:00 PM, so you have a problem with prioritizing getting rid of racists as oppose to slackers?
Nice try. Lloyd Brooks appointed countywide.
If by serious opponent you mean Sandra Ramos, try again. She has way too many skeletons and is thought of poorly by the rest of the judiciary.
Sandra Ramos for Chief Judge? Please.

redpajama_set_name: RedPajamaCommonCrawl


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Guy is a senior camera operator/editor in the VPI Production team with an awesome can do attitude. Nothing is too much trouble for this man!
So within the last 6 months, I have been actively filming in Mumbai India, Budapest Hungary, Wiesbaden Germany and Madrid Spain.
Q. What's been your biggest challenge across these countries with respect to obtaining quality film output?
The main challenge I frequently come across, which may also seem the most obvious, is language barrier. Whether filming in the most glamorous suburbs in Europe or in downtrodden India, language issues will always be equally difficult. To obtain high quality film often isn't only down to technical skill-set, but also social mindset. To direct cutaways, being able to carefully place microphones and lights in the right locations with confidence, and finally set the scene for the interview itself, all takes communication to make it perfect. Body language and gesturing is key, but also being talkative and comforting to make the interviewee feel at ease.
Q. And with the more isolated in home interviews you've done, what kind of cultural challenges have you faced for obtaining quality output?
The biggest cultural challenge I've recently faced, occurred when I was filming in-homes in the busy business district of Mumbai. London is noisy, busy, and often disruptive on occasion when filming, however this is drastically minimal compared to Mumbai. Even when inside, with all the windows bolted, shutters locked, doors closed, the excessive honking of overzealous rickshaws passing by caused big noise issues, in obtaining that perfect audio. Though with the best choice of microphone and placement of this, background noise can be kept to a minimum.
Q. What's been your favourite field experience to date and why?
My favourite field experience to date would be a week spent in Rio De Janeiro, Brazil. Shooting for Unilever, where the main topic of filming was cleaning and products surrounding this, it was also my first major film shoot and the furthest I have ever travelled to date. The reasoning behind why it was my favourite experience was down to the hospitality and generosity of each person I met. While doing Vox pops, asking passers by to participate, every single person was curious and ultimately interested in taking part. The laid back café culture in the city of Rio made it an excellent location to do Vox pops. Although I could not speak a word of Portuguese apart from the occasional 'Olá' or 'Obrigado!', each individual was completely open to my lack of linguistics and often was keen to engage me in English to the best of their ability.
Q. Where would you most like VPI to send you next for work?
If I could choose to be sent anywhere for work, it would have to be a location in Latin America. After seeing previous Vox Pops International projects which have been filmed in either Argentina or Columbia, it has given the impression that South America is full of rich culture, beautiful cities and interesting people. From those I know who have either visited for work or pleasure, both have described it as a dream location. From all of the good buzz around this, fingers crossed I'll be there next!
This entry was posted in and tagged Fieldwork, Filming. Bookmark the permalink.


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Hurt Morpeth teacher to return to UK
See Robson Green's message of support for villagers trying to save their local pub
Heroic efforts combating the Beast are recognised
Farmers, 4x4 owners and community leaders who helped Northumberland overcome the Beast from the East have been thanked for their efforts.
Friday, 27th April 2018, 12:12 pm
Updated Friday, 27th April 2018, 12:16 pm
Mike and Jamie Hope from Wooler. Picture by Jane Coltman
Northumbria Police invited them to a get-together in Belford to acknowledge their vital roles in getting the county moving again following the heavy snowfall in late February.
It was also an opportunity for the indivuduals involved to share stories of digging cars out of the snow, helping to deliver medical supplies and prescriptions and organising emergency food and accommodation for those left stranded.
The winter heroes and representatives from Northumbria Police who attended the event in Belford. Picture by Jane Coltman
Among those who helped was Rod Smith, of Beal Farm, who worked long stints to clear the A1 and the link road to Holy Island despite the tidal causeway being ice-bound.
"That was just the tip of the iceberg if you pardon the pun," said Rod. "The snow was at hedge height on the mainland and people just couldn't get off Holy Island.
"I spent 10 hours one day digging it out but I think that's something a lot of the farmers and others did.
"Communities have drifted apart over the years but I think this has highlighted that there are a lot of people out there who will help out in a crisis. Every down-side has an up-side and that has been bringing the community back together."
Richard White, who works for Spours, helped to clear the A1. Picture by Jane Coltman
Andrew Reed, of East Ancroft Farm helped to keep the Berwick-Wooler road open and the minor roads around Allerdean, Cheswick and Goswick.
"I spent a lot of time pulling out cars which had tried to find an alternative route to the A1," he said. "A number of them were foreign visitors and they were genuinely grateful."
He also helped someone travelling to hospital for dialysis to get through the drifts.
Of the get-together, he added: "It was a nice gesture to get that acknowledgement and recognition. If it hadn't been for people in the countryside there would have been a bigger mess. People could have lost lives."
Daniel Spours, from Twizell Farm, helped clear roads and deliver medicine. Picture by Jane Coltman
Father-and-son Mike and Jamie Hope, of Wooler garage, had a similar story to tell.
"There were cars stranded at Barmoor South Moor for 11 hours," he recalled. "We had to dig our way eight miles to help them."
They also delivered medication and even supplied shepherds with special vehicle tracks so they could access their livestock.
He was also full of praise for the community spirit shown.
Andrew Reed, from East Ancroft. Picture by Jane Coltman
"It was absolutely fantastic," he said. "Lots of people wanted to help in any way they could."
Rod Smith from Beal. Picture by Jane Coltman
Rev Matthew Knox and Julie Riley, of Berwick Deserves Better, helped organise food and accommodation. Picture by Jane Coltman
Geoff Frater, from Boulmer, was among those who helped. Picture by Jane Coltman
Government closes tax loophole on second homes
Chance to pay tribute to family GP
Northumberland swimmer aiming to raise awareness of hidden disabilities
Ashington woman develops cervical cancer after test incorrectly recorded as normal


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7 of the Best Happy Birthday Songs (Traditional and Funny)
Compu-Smart
I search endlessly to find you the best, funniest and greatest "Happy Birthday" songs for all ages (plus fun birthday facts).
"Happy Birthday to You" is a classic, but it's not the only song you can sing at a birthday party. Check out these great song options!
Image by profivideos from Pixabay
There's one thing in life that we all have in common, and that's a birthday—a day to celebrate once a year, whether we like it or not! In most English-speaking countries, it's traditional to sing "Happy Birthday to You" to honour the person celebrating a birthday. In fact, the "Happy Birthday" song is thought to be the most frequently sung melody in the world. It's not the only option, though. Below are some of the best birthday songs in existence.
The Top 7 Birthday Songs
"Happy Birthday to You" (Traditional)
"Happy Birthday, Mr. President" by Marilyn Monroe
"Happy Birthday" by Stevie Wonder
"Happy Birthday" by Altered Images
"Happy Birthday We Love You" by Daniel Dancer
"Happy Birthday" by the Chipmunks
"The Happy Happy Birthday Song" by The Arrogant Worms
1. "Happy Birthday to You!" (Traditional)
The "Happy Birthday" song was written by the sisters Patty and Mildred Hill, who were both American composers and teachers. This song was written in 1893 and initially called "Good Morning to All" for kindergarten children to sing. Now, it has become the most popular song ever written, and it's also played in many different countries and languages around the world.
The music publishing company Warner Chapel Music Incorporation owns the copyright to the song "Happy Birthday to You."
"Happy Birthday to You" is the highest-earning song of all time.
2. "Happy Birthday, Mr. President" by Marilyn Monroe
On May 19, 1962, Marilyn Monroe made an appearance at Madison Square Garden in New York, USA, especially to sing "Happy Birthday, Mr. President" for President John F. Kennedy (although, technically, it was not his actual birth-date. He was born on May 29).
After Monroe's performance, JFK joined her on stage and said, "I can now retire from politics after having had 'Happy Birthday' sung to me in such a sweet, wholesome way." The dress that she wore to the event was sold for a world record $1.26 million.
A rare image of Marilyn Monroe on her 30th birthday in the back of a car.
Marilyn Monroe's real name was Norma Jeane Mortenson.
Her birthday was on May 21, 1926.
She was an award-winning American actress, singer, icon and sex symbol. Marilyn's well-known films include Some Like It Hot, How to Marry a Millionaire, The Seven Year Itch, Gentlemen Prefer Blonds and Diamonds Are a Girls Best Friend.
3. "Happy Birthday" by Stevie Wonder
Stevie Wonder's hit songs include "You Are the Sunshine of My Life," "Superstition," "I Just Called to Say I Love You" and, of course, "Happy Birthday," which he wrote himself. He has performed this song for many notable figures, including King Carl XVI Gustaf and Oprah Winfrey (for her 50th birthday).
Stevie Wonder singing "Happy Birthday."
Stevie Wonder's real name is Stevland Hardaway Judkins.
He celebrates his birthday on May 13, 1950.
He is a blind American singer-songwriter, multi-instrumentalist and record producer who has recorded numerous hits and won many awards for his work.
4. "Happy Birthday" by Altered Images
This is an alternative "Happy Birthday" song by the 1980s Scottish post-punk rock band Altered Images. This song was their biggest hit at the time. Their other hits include "Don't Talk to Me About Love" and "See Those Eyes."
Lead singer Clare Grogan celebrates her birthday on March 17.
Clare also starred in the Scottish romantic comedy film Gregory's Girl and the science fiction comedy series Red Dwarf.
5. "Happy Birthday We Love You" by Daniel Dancer
When Daniel Dancer's daughter turned 11 in 2004, he realised he was tired of the same old birthday song, so he decided to write another one. The video above is a recording of the song with images to go with the words.
This song has become a special feature of Dancer's popular "Art for the Sky" artist in residency program (see artforthesky.com) that he leads at schools across the country. At the final assembly, before revealing the school's creation in the project film, he invites kids who had a birthday that week to come up on stage and share the song. Everyone always loves it!
6. "Happy Birthday" by the Chipmunks
This birthday song, sung by the squeaky-voiced Chipmunks, is especially popular with children.
The Chipmunks band was "born" in 1958.
The animated cartoon premiered in 1961.
7. "The Happy Happy Birthday Song" by The Arrogant Worms
This is a good choice for a hilarious birthday song, with lyrics like the following:
Now you're one year older!
Your life still isn't over!
You did not accomplish much
But you didn't die this year
I guess that's good enough
This song was released in 1995.
It appeared on the album C'est Cheese.
If you're sad about getting older, don't despair! Cheer up with these funny birthday jokes.
Image by Cucu Petronela from Pixabay
The most effective way to remember your wife's birthday is to forget it once.
The secret of staying young is to live honestly, eat slowly and always lie about your age.
Inside every older person is a younger person wondering what happened.
Birthdays are good for you. Statistics show that the people who have the most live the longest.
It's nice to be here. At my age, it's nice to be anywhere.
Forget about the past, because you can't change it. Forget about the future, because you can't predict it. Forget about the present, because I didn't get you one!
A diplomat is a man who always remembers a woman's birthday but never remembers her age.
Were any famous men or women born on your birthday? No, only babies.
Some birthdays have special significance, especially the multiples of 10.
Special and Traditional Birthdays
The most significant date-of-birth digit changes—for example, one's 1st, 10th, 20th, 30th, 50th or 100th birthday and certain "coming of age" years—are often accompanied by more legal privileges and responsibilities. In Western nations, these traditionally happen at the ages of 12, 15, 16, 18, 19 and 21, 30, 40, 50, 60, 70, 80, 90 and 100.
In most legal systems, one becomes a legal adult on a particular birthday (one's 18th or 21st). At different ages, one gains more rights and responsibilities, such as being able to vote, purchase alcohol or tobacco, legally buy lottery tickets, get married or obtain a vehicle driving license. One also becomes eligible for the military draft or voluntary enlistment at particular ages.
Traditional Birthday Gifts by Age
Just like weddings, birthdays are traditionally celebrated with gifts which vary in different countries. The following traditions are for both the UK and the USA, unless otherwise noted.
1st birthday: Cotton, UK; paper, USA
5th: Wood
10th: Tin
15th: Crystal
20th: China
25th: Silver
30th: Pearl
35th: Coral, UK; jade, USA
40th: Ruby
45th: Sapphire
50th: Gold
75th birthday: Diamond
More Fun Birthday Facts
If you're the Queen of England or another notable figure, you might celebrate your birthday on more than one occasion. The Queen's official birthday is on April 21st, which is marked by a 2nd public celebration in June and celebrated in London by a ceremony called "Trooping the Colour," also known as the "Queen's Birthday Parade."
Many adults avoid celebrating their birthday because it reminds them how old they're getting. Some people cannot afford to have any celebrations at all, and there are some people in deprived and Third World countries who have no idea what date their actual birthday is or how old they are.
On the day of your birth, you're generally thrown a party by family and friends who give you gifts in celebration of the blessed event. It is also customary for people to treat you very special on your birthday.. Of course, some would rather have a quiet time on their birthdays.
Celebrating one's birthday is not universal. Jehovah's Witnesses don't celebrate birthdays because they consider the origins of this celebration to be pagan festivals, much like Christmas and Easter.
Remember to sing one of the great birthday songs listed above to your pet!
Image by annmariephotography from Pixabay
Don't Forget Your Pet's Birthday!
Pets don't know when their birthday is, but they'll soon know it's a very special day for them, especially when you give them their favourite treats, extra attention, a new toy to play with, and—of course—a rendition of "Happy Birthday." If you don't know your pet's exact birthday, make one up.
© 2008 Compu-Smart
Umm on March 03, 2020:
Boris Johnson said we should "sing Happy Birthday twice whilst washing your hands to ensure you have exposed your hands to soap and water for long enough to kill the Coronavirus/Covid-19 germs", But which version was he taliking about!!?? LOL
Nell Rose from England on June 18, 2019:
They are great! love the Chipmonks! lol!
Compu-Smart (author) from London UK on June 12, 2019:
Hi Brenda, just send them this link. https://holidappy.com/party-planning/The-Best-Birt...
Brenda Mclaren on May 19, 2019:
How can I get this happy birthday song from altered images for a friend to send on messanger
jasmine on July 05, 2014:
ya to birthday song lagaty hi nhi ha toba toba
David on June 16, 2014:
Happy birthday Abby Elliot! You're such a beautiful and great actress. xoxo
Earth Mother on April 22, 2014:
Today is Earth Day. Happy Earth Day to you ALL!.
Kos on March 14, 2014:
Happy birthday to the best and nicest girl in the whole wide world. Her name is Helen Louise Manning and I miss her so much. I just hope she still remembers me and is having a wonderful time today. Sorry I can't be with you.
Kos X
Jano on March 06, 2014:
Happy birthday and congratulations to Japan's Misao Okawa who was born on 5th March 1898 and is the world's oldest living person.
She celebrated her 116th birthday yesterday.
She said the secret to a long life is eating sushi daily and getting at least eight hours sleep each day.
b-boy on February 10, 2014:
2day is my birthday and i got a new iphone AND ipad and im sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo APPY:)
Buggsy on January 10, 2014:
APPY BIRTHDAY TO MY BEAUTIFUL BUNNY RABBIT, TRUMP. I LOVE YOU LOADSSSSSSSS :)
Mercedes Londono on January 03, 2014:
Happy birthday to you my dear lovely brother, Paulo. Missing you loads. X
Tone/KOS on December 31, 2013:
Happy Birthday to thee one and only, Helen Louise Manning {Davies}, along time friend whoim I miss and would love to see again someday very soon!! HAPPY NEW YEAR Too!! XOXO :)
Sarah B. on December 20, 2013:
Happy Birthday, my beautiful sky. apologies for the loss of your beloved pet bandit. You are such a beautiful person and positive person. Im so proud of you.
jay on December 18, 2013:
happy birthday to my all time best friend "nawaf" and may all your wishes and dreams really come to live dearie xoxoxoxo
conor on December 17, 2013:
happy birthday granddad. i just want to tell you i love you so much. your the BEST!!!!
pretty little gleek on December 15, 2013:
today is't my birthday again but im still back to still listen to these! 8 best happy birthday songs!
sharifa on November 16, 2013:
HI. AM SHARIFA. HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY BIRTHDAY TO ALL MY DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV V IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII LLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL FRIENDS
Cindy2001 on October 20, 2013:
Omg,,the chimp monks are so adorable. I gonna ask my mom to buy me one,,or two,,or three lol!!
Ziggy on October 18, 2013:
Happy Birthday Barry, aka smurfy. I hope you have at least 100 more :)
Luke Davies on October 18, 2013:
TODAY IS MY BIRTHDAY AND I LOVED ALL THE VERY FUNNY JOKES AND SONGS. OHH, AND DONT FORGET TO BUY ME A PRESENT IF YOU KNOW ME!! : )
Larry on September 28, 2013:
Happy belated birthday to Google who has just turned 15:)
olivia on September 26, 2013:
Yàaaaaaaaaaaaaa aaàaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa I agreeeeeeeeee
HolidayGiftIdea on August 23, 2013:
This is a great hub. Keep them original.
sneha on June 22, 2013:
bakwaas
uf on November 18, 2012:
faltoooooooooooooooooooooooooooooooooooooooooooooooooo
nazim on November 17, 2012:
call this number bangladesh 01814708354
kavi on October 22, 2012:
happy bufday to all if you born as human n on your birthday............
Mahzia on August 28, 2012:
Happy b day little sis she is 6 now
me on July 01, 2012:
happy birthday to u
u stepped in some poo
dont waste it just tate it
alex on June 30, 2012:
My brother's birthday is coming up and im goin to show it to him..... wonder if he will understand it since he is 3?
barbara on May 31, 2012:
LOVE IT .. THAT IS A GREAT BIRTHDAY CARD ,,,
brian on February 09, 2012:
I really like this song
Ned on February 07, 2012:
Ok we get it
Keke on January 20, 2012:
My birthday is tomorrow. I will b thirteen!!!!!!
chini on January 08, 2012:
ARSLAN [1ST ONE] AND PRIYANKA[LAST ONE]
BOTH ARE CORRECT
ITS REALLY FALTU..
Patrick Mediana on December 28, 2011:
Happy Birthday ....!!!!
Reanza Veanica Mediana !!!!!!!!!!!
Onnesha Akhter Nova on November 24, 2011:
thank u vry much!!! i have loved the songs!!!!!!!!!!!!
Augusto on November 07, 2011:
My bday is todayyy novemmber 7th yay...guess it was yesterday but it still counts cuz I didn't sleep.
COWS on November 05, 2011:
COOOOL I LOVE MO?nERGKB
Mishka on October 30, 2011:
Happy Birth Day
faith on September 27, 2011:
AVINASH on September 26, 2011:
HAPPY BIRTH DAY ARCHANA & YOGESH ..
Kutie_pie512@yahoo.com on September 23, 2011:
Your birthday is hear
Lets all give a cheer
i hope its a hit...
Cause your Older Than S%#@ :)
Emaill me :D
- Torii
Andromeda Smith on August 26, 2011:
i remember i made one up while my school was singing happy bday to you to some people.
It goes like this;
you live in a loo
you look like a monkey
and you smell like poo poo
lina on May 13, 2011:
happy birthday lina
Cindy Bernard on April 13, 2011:
ewin on March 14, 2011:
coolio!
tae ann on January 29, 2011:
oh love it
jean dolfo on December 12, 2010:
wow nice songs......
k@ri on December 09, 2010:
Cute hub, I liked the song by Altered Images. I never heard that one before.
William F. Torpey on December 09, 2010:
Nice hub. I remember watching Marilyn Monroe singing "Happy Birthday" to President Kennedy. The anti-Kennedy crowd raised questions about the event and creating rumors that Kennedy and Marilyn had an affair.
JE$$!E on December 08, 2010:
that wassss sooooooo entertaining great job!!!!! :)
dakota austin on October 19, 2010:
Hi,my name is Dakota and i whet to the fair on 10/18/2010 and i had fun.And my conson whet to.
Joelle on September 30, 2010:
Happy birthday to you:~)
Denise Handlon from North Carolina on July 31, 2010:
Hi, I came across this hub while researching material for a hub on the internet. Great job. Very entertaining. Loved the footage of M.M. w/ the Pres.
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This is an arrangement for brass quintet and organ with optional timpani. Handels oratorio Samson contains several great movements that trumpet players play as solo or ensemble pieces at recitals apart from the complete oratorio. Besides Awake the Trumpets Lofty Sound, there is Let Their Celestial Concerts and one of the most popular trumpet and voice duets Let the Bright Seraphim.