text:
\begin{document}

\title{Applying Skolem Sequences to Gracefully Label New Families of Triangular Windmills}
\author{Ahmad H. Alkasasbeh \footnote{ahmad84@mun.ca} \hspace{0.5in} Danny Dyer \footnote{dyer@mun.ca} \hspace{0.5in} Nabil Shalaby \footnote{nshalaby@mun.ca}\\
Department of Mathematics and Statistics \\
Memorial University of Newfoundland \\
St. John's, Newfoundland \\
A1C 5S7 Canada}
\date{}
\maketitle
\begin{abstract}

\noindent A function $f$ is a \textit{graceful labelling} of a graph $G=(V,E)$ with $m$ edges if $f$ is an injection $f:V\mapsto \{0,1,2,\dots,m\}$ such that each edge $uv \in E$ is assigned the label $|f(u)-f(v)|$, and no two edge labels are the same. If a graph G has a graceful labelling, we say that $G$ itself is graceful.\par
In this paper, we proved Rosa's conjecture for a new family of triangular cacti: Dutch windmills of any order with three pendant triangles.

\end{abstract}

{\bf Keywords:} Graceful Labellings; Skolem-type sequences.

\newpage
\section{Introduction}

In 1963, Ringel \cite{ringel} posed the following problem, given an arbitrary tree $T$ with $m$ edges. Can $K_{2m+1}$ be decomposed into $2m+ 1$ copies of $T$ ?\par
In 1967, Rosa introduced a new type of graph labelling called a $\beta-${\it valuation}. Later in, 1972, Golomb renamed $\beta$-valuation as a {\it graceful labelling}, as it is still known today. In \cite{rosa1} Rosa introduced the definition of graceful labelling and proved that Ringel's conjecture is valid if every tree has a graceful labelling.\par

In \cite{Bermond}, Bermond studied the graceful labelling of Dutch windmills. Dutch windmills are a subtype of triangular cacti, connected graphs whose blocks are all triangles.\par
In \cite{rosa}, Rosa stated the following conjecture:

\begin{enumerate}
\item all triangular cacti with $n\equiv 0$ or $1\ ($mod$\ 4)$ are graceful,
\item all triangular cacti with $n\equiv 2$ or $3\ ($mod$\ 4)$ are nearly graceful.
\end{enumerate}

In 1989, Moulton \cite{mou} proved Rosa's conjecture for a triangular snake, a type of triangular cactus whose block cutpoint graph is a path. Rosa, in \cite{rosa}, recommended using Skolem-type sequences to label various families of triangular cacti. In 2012, Dyer, Payne, Shalaby, and Wicks \cite{dyer} verified Rosa's conjecture for a new class of triangular cacti: Dutch windmills with at most two pendant triangles by using Skolem-type sequences as Rosa suggested. Gallian, in his survey (A Dynamic Survey of Graph Labeling) \cite{Gallian}, mentioned that a proof for all triangular cacti seems hopelessly difficult.\par
In this paper, we develop a system in which we add pendant triangles (in all possible configurations) to a family of gracefully labelled graphs to get new gracefully labelled families. We verify Rosa's conjecture for a new family of triangular cacti: Dutch windmills of any order with three pendant triangles, an extension of the results in \cite{dyer}. We can conclude this work with the following theorem.

\begin{theorem}
Every Dutch windmill with at most three pendant triangles is graceful or near graceful.
\end{theorem}

\section{Definitions and Preliminaries}
Essential definitions are introduced in this section, as well as previously--found results required to prove further results.
\subsection{Skolem and Langford Sequences}\label{skolemintro}

The following definitions came from the Handbook of Combinatorial Designs \cite{crc}.\\

\rm A {\it Skolem sequence} of order $n$ is a sequence $S=\left(s_{1},s_{2},\ldots,s_{2n}\right)$ of $2n$ integers satisfying these conditions:

\begin{enumerate}

\item for every $k\in\left\{1,2,\ldots,n\right\},$ there exist exactly two elements $s_{i},s_{j}\in S$ such that $s_{i}=s_{j}=k;$

\item if $s_{i}=s_{j}=k,$ with $i < j,$ then $j-i=k$.

\end{enumerate}

\noindent For example, $S_4=(1,1,4,2,3,2,4,3)$ or equivalently $\left\{\left(1,2\right),\left(4,6\right),\left(5,8\right),\left(3,7\right)\right\}$ is a Skolem sequence of order $4$.\par
If we have a Skolem sequence $\{(a_i,b_i)\}_{i=1}^n,$ then $i$ is called a \textit{pivot} of a Skolem sequence if $b_i+i \leq 2n.$\par
\begin{theorem} \cite{skolem}
A Skolem sequence of order $n$ exists if and only if $n\equiv0,1({\rm mod}\ 4)$.
\end{theorem}
{\rm A {\it hooked Skolem sequence} of order $n$ is a sequence $hS=\left(s_{1},s_{2},\ldots,s_{2n+1}\right)$ of $2n+1$ integers satisfying these conditions:

\begin{enumerate}

\item for every $k\in\left\{1,2,\ldots,n\right\},$ there exist exactly two elements $s_{i},s_{j}\in hS$ such that $s_{i}=s_{j}=k;$

\item if $s_{i}=s_{j}=k,$ with $i < j,$ then $j-i=k;$

\item $s_{2n}=0.$

\end{enumerate}}

\noindent For example, $hS_2=(1,1,2,0,2)$ is a hooked Skolem sequence of order $2$.\par
If we have a hooked Skolem sequence $\{(a_i,b_i)\}_{i=1}^n,$ then $i$ is called a \textit{pivot} of a hooked Skolem sequence if $2n \neq b_i+i \leq 2n+1.$\par
\begin{theorem}\cite{okeefe}
A hooked Skolem sequence of order $n$ exists if and only if $n\equiv2,3\ ({\rm mod}\ 4)$.
\end{theorem}

In $1897$, Heffter stated two difference problems  \cite{heffter1}. Heffter's first difference problem is can a set $\{1,\ldots, 3n\}$ be partitioned into $n$ ordered triples $(a_i,b_i,c_i)$ with $1\leq i\leq n$ such that $a_i+b_i=c_i$ or $a_i+b_i+c_i\equiv 0\;(\textup{mod}\;6n+1)$? If such a partition is possible, then $\{\{0,a_i+n,b_i+n\}|1\leq i\leq n\}$ will be the base blocks of a $CSTS(6n+1)$. Construction \ref{ds} gives a solution to Heffter's first difference problem.

\begin{construction}\cite{skolem}{\label{ds}}
From a Skolem sequence or a hooked Skolem sequence of order $n$, construct the pairs $(a_i,b_i)$ such that $b_i-a_i=i$ for
 $1\leq i\leq n$. The set of all triples $(i,a_i+n,b_i+n)$ for $1\leq i\leq n$ is a solution to the Heffter first difference problem.
 These triples yield the base blocks for a $CSTS(6n+1)$: $\{0,a_i+n,b_i+n\}$, $1\leq i\leq n$. Also,
 $\{0,i,b_i+n\}$, $1\leq i\leq n$ is another set of base blocks of a $CSTS(6n+1)$.
\end{construction}

{\label{baseblocks}}Let $S_4=(1,1,4,2,3,2,4,3)$ be a Skolem sequence of order $4$ yields the pairs $\{(1,2),(4,6),(5,8),(3,7)\}$. These pairs yield in turn the triples $\{(1,5,6),(2,8,10),(3,9,12), (4,7,11)\}$, forming a solution to the first Heffter problem.
These triples yield the base blocks for two $CSTS(25)s$:
\begin{enumerate}
\item $\{0,5,6\},\{0,8,10\},\{0,9,12\},$ and $\{0,7,11\}(\textup{mod}\;25)$;
\item $\{0,1,6\},\{0,2,10\},\{0,3,12\},$ and $\{0,4,11\}(\textup{mod}\;25)$.
\end{enumerate}

A \textit{Langford sequence} of defect $d$ and order $l$ is a sequence $L=\left(l_{1},l_{2},\cdots,l_{2l}\right)$ which satisfies these conditions:
\begin{enumerate}
	\item for every $k\in\left\{d,d+1,\ldots,d+l-1\right\},$ there exist exactly two elements $l_{i},l_{j}\in S$ such that $l_{i}=l_{j}=k$;
\item if $l_{i}=l_{j}=k,$ with $i < j,$ then $j-i=k$.
\end{enumerate}

A \textit{hooked Langford sequence} of defect $d$ and order $l$ is a sequence $L=\left(l_{1},l_{2},\cdots,l_{2l+1}\right)$ which satisfies these conditions:
\begin{enumerate}
	\item for every $k\in\left\{d,d+1,\ldots,d+l-1\right\},$ there exist exactly two elements $l_{i},l_{j}\in S$ such that $l_{i}=l_{j}=k$;

\item if $l_{i}=l_{j}=k,$ with $i < j,$ then $j-i=k;$
\item $l_{2m}=0.$

\end{enumerate}
\noindent For example, $(4,2,3,2,4,3)$ is a Langford sequence with $d=2$ and $l=3$ and $(8,4,7,3,6,4,3,5,8,7,6,0,5)$
 is a hooked Langford sequence with $d=3$ and $l=6$.

The necessary and sufficient conditions for the existence of (hooked) Langford sequences are given in Theorem~\ref{langfordexistence}.
\begin{theorem}\cite{simpson}\label{langfordexistence} \hspace{1cm}
\begin{enumerate}
\item A Langford sequence of order $m$ and defect $d$ exists if and only if
	\begin{enumerate}
		\item $m \geq 2d-1$
		\item $m \equiv 0 \text{ or } 1 \pmod 4$ and $d$ is odd, or
		\item $m \equiv 0 \text{ or } 3 \pmod 4$ and $d$ is even.
	\end{enumerate}
\item A hooked Langford sequence of order $m$ and defect $d$ exists if and only if
	\begin{enumerate}
		\item $m(m-2d+1)+2 \geq 0$
		\item $m \equiv 2 \text{ or } 3 \pmod 4$ and $d$ is odd, or
		\item $m \equiv 1 \text{ or } 2 \pmod 4$ and $d$ is even.
	\end{enumerate}
\end{enumerate}
\end{theorem}
Though they can be thought of as a natural generalization of Skolem sequences, Langford sequences have also been classically used to build new Skolem sequences by concatenating sequences of appropriate order, possibly interlacing hooks, if required.

\begin{lemma}\label{skla} If a (hooked) Skolem sequence of order $d-1$ exists, and a (hooked) Langford sequence of order $l$ and defect $d,$ then a (hooked) Skolem sequence of order $N=l+d-1$ exists. In particular, a new Skolem sequence of order $N$ is obtained by concatenating a Skolem sequence with a Langford sequence or by interlacing a hooked Skolem sequence and hooked Langford sequence. A new hooked Skolem sequence of order $N$ is obtained by concatenating a Skolem sequence with a hooked Langford sequence or a hooked Skolem sequence and a Langford sequence. \end{lemma}

Let $hS_{2}=(1,1,2,0,2)$ be a hooked Skolem sequence and \newline $hL_{3}^{6} = (8,4,7,3,6,4,3,5,8,7,6,0,5)$ be a hooked Langford sequence. Now, if we take the reverse of the hooked Skolem sequence and by interlacing $hL_{3}^{6}$ and the reverse of $hS_{2},$ then we obtain a new Skolem sequence \newline $S_{8}=(8,4,7,3,6,4,3,5,8,7,6,2,5,2,1,1)$ of order $8.$

\subsection{Graceful Labellings and Triangular Cacti}
Let $G=(V,E)$ be a graph with $m$ edges. Let $f$ be a labelling defined from $V(G)$ to $\{0,1,2,\dots,m\}$ ($f:V\mapsto \{0,1,2,\dots,m\}$) and let $g$ be the induced edge labelling defined from $E(G)$ to $\{1,2,\dots,m\}$ by $g(uv)=|f(u)-f(v)|,$ for all $uv \in E.$ The labelling $f$ is said to be \textit{graceful}, if $f$ is an injective mapping and $g$ is a bijection. If a graph $G$ has a graceful labelling, then it is graceful. \par

Let $G=(V,E)$ be a graph with $m$ edges. Let $f$ be a labelling defined from $V(G)$ to $\{0,1,2,\dots,m+1\}$ ($f:V\mapsto \{0,1,2,\dots,m+1\}$) and let $g$ be the induced edge labelling defined from $E(G)$ to $\{1,2,\dots,m-1,m+1\}$ by $g(uv)=|f(u)-f(v)|,$ for all $uv\in E.$ The labelling $f$ is said to be \textit{near graceful}, if $f$ is an injective mapping and $g$ is a bijection. If a graph $G$ has a near graceful labelling, then it is near graceful.\par

\begin{figure}[htbp]
\begin{center}
\scalebox{0.75}{
\begin{tikzpicture}
    \node[shape=circle,draw=black] (0) at (0,0) { 0};
    \node[shape=circle,draw=black] (3) at (2,0) { 3};
   		    \node[shape=circle,draw=black] (1) at (1,2) { 1};

    \path [-] (0) edge node[left,above] { $3$} (3);
    \path [-](0) edge node[left,above] { $1\ $} (1);
		\path [-] (1) edge node[left,above] { $\ 2$} (3);

		 \node[shape=circle,draw=black] (0) at (4,0) {0};
    \node[shape=circle,draw=black] (1) at (6,0) {1};
		 \node[shape=circle,draw=black] (6) at (3,2) {6};
    \node[shape=circle,draw=black] (5) at (7,2) {5};
		 \node[shape=circle,draw=black] (3) at (5,3) {3};

    \path [-] (0) edge node[left,above] {$\ 6$} (6);
    \path [-](6) edge node[left,above] {$3$} (3);
		\path [-] (3) edge node[left,above] {$2$} (5);
		  \path [-] (5) edge node[right,above] {$4\ $} (1);
    \path [-](1) edge node[left,above] {$1$} (0);
\end{tikzpicture}
}
\caption{Graceful labelling of $K_{3}$ and near graceful labelling of $C_{5}$}\label{c}
\end{center}
\end{figure}

\begin{comment}
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}
\draw [thick, -] (0,0) -- (1,0) -- (0.5,1) -- (0,0);

\node [below] at (0,0) {\tiny $0$};
\node [below] at (1,0) {\tiny $3$};
\node [above] at (0.5,1) {\tiny $1$};

\node [above] at (0.5,0) {\tiny $3$};
\node [left] at (0,0.5) {\tiny $1$};
\node [right] at (1,0.5) {\tiny $2$};

\draw[fill] (0,0) circle  [radius=.7pt];\draw[fill] (1,0) circle  [radius=.7pt];\draw[fill] (0.5,1) circle  [radius=.7pt];

\draw [thick, -] (3,0) -- (4,0) -- (4,0.7) -- (3.5,1) -- (3,0.7) -- (3,0);

\node [below] at (3,0) {\tiny $0$};
\node [below] at (4,0) {\tiny $1$};
\node [right] at (4,0.7) {\tiny $5$};
\node [above] at (3.5,1) {\tiny $3$};
\node [left] at (3,0.7) {\tiny $6$};

\node [above] at (3.5,0) {\tiny $1$};
\node [left] at (3,0.35) {\tiny $6$};
\node [right] at (4,0.35) {\tiny $4$};
\node [above] at (3.25,.82) {\tiny $3$};
\node [above] at (4,0.82) {\tiny $2$};

\draw[fill] (3,0) circle  [radius=.7pt];\draw[fill] (4,0) circle  [radius=.7pt];\draw[fill] (4,0.7) circle  [radius=.7pt];\draw[fill] (3.5,1) circle  [radius=.7pt];\draw[fill] (3,0.7) circle  [radius=.7pt];

\end{tikzpicture}
\caption{Graceful labelling of $K_{3}$ and near graceful labelling of $C_{5}$}\label{c}
\end{center}
\end{figure}
\end{comment}

In this paper, we will gracefully label some new families of triangular cacti and introduce some definitions and results related to triangular cacti.\par
{\rm A {\it triangular cactus} is a connected graph whose blocks are all triangles ($K_3$)}. {\rm A triangular cactus that has the property of all its blocks having a common vertex is said to be a \textit{regular Dutch windmill}, and the blocks will be called \textit{vanes}.} A {\it pendant triangle} is defined as a block that is added to any triangular cacti.\par
In this paper, we discuss the gracefulness of several families of triangular cacti. The necessary conditions for gracefulness and near gracefulness of general triangular cacti are as follows:
\begin{theorem}\cite{dyer}\label{necessarycacti}
Let $G$ be a triangular cactus with $n$ blocks. Then,
\begin{enumerate}
\item if $G$ is graceful, then $n \equiv 0 \text{ or } 1 \pmod 4$, and,
\item if $G$ is near graceful, then $n \equiv 2 \text{ or } 3 \pmod 4$.
\end{enumerate}\end{theorem}

\section{Dutch Windmills with Three pendant Triangles}\label{sectiondutch}

In order to verify Rosa's conjecture for a new family of triangular cacti, namely Dutch windmills of any order with three pendant triangles, we will use Langford sequences to obtain Skolem and hooked Skolem sequences of considerable sizes. This technique was introduced in \cite{dyer}. We categorize all such cacti into one of eleven types which we call Type $(a)$, Type $(b)$,\ldots, Type $(k)$ and then gracefully label each type.
\begin{figure}[htbp]
\begin{center}
\begin{picture}(300,300)
\linethickness{1pt}

\put(1,140){$(a)$}
\put(150,140){$(b)$}
\put(1,1){$(c)$}
\put(150,1){$(d)$}

\put(1,150){\begin{tikzpicture}[scale=1.1]
\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (85:1.931cm) node[draw=blue,shape=circle] (v9) {};
\path (110:1.931cm) node[draw=blue,shape=circle] (v10){};

\path (160:1.931cm) node[draw=red,shape=circle] (v11) {};
\path (180:2cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v5) edge[red] (v11)
(v5) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v9) edge[blue] (v3)
(v10) edge[blue] (v3)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(150,150){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (45:1.931cm) node[draw=blue,shape=circle] (v9) {};
\path (60:1.931cm) node[draw=blue,shape=circle] (v10){};

\path (160:1.931cm) node[draw=red,shape=circle] (v11) {};
\path (180:2cm) node[draw=red,shape=circle] (v12) {};

\path (0:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (-15:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v5) edge[red] (v11)
(v5) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v9) edge[blue] (v2)
(v10) edge[blue] (v2)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(1,11){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (85:1.931cm) node[draw=blue,shape=circle] (v9) {};
\path (110:1.931cm) node[draw=blue,shape=circle] (v10){};

\path (345:1.931cm) node[draw=red,shape=circle] (v11) {};
\path (0:2cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v1) edge[red] (v11)
(v1) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v9) edge[blue] (v3)
(v10) edge[blue] (v3)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(150,11){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (30:2.631cm) node[draw=blue,shape=circle] (v9) {};
\path (40:2.332cm) node[draw=blue,shape=circle] (v10) {};

\path (160:1.931cm) node[draw=red,shape=circle] (v11) {};
\path (180:2cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v5) edge[red] (v11)
(v5) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v13) edge[blue] (v9)
(v13) edge[blue] (v10)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\end{picture}

\label{twopendantfig}
\end{center}
\end{figure}

\begin{figure}[htbp]
\begin{center}
\begin{picture}(510,510)
\linethickness{0.9pt}

\put(25,435){$(e)$}
\put(220,435){$(f)$}

\put(25,305){$(g)$}
\put(220,305){$(h)$}

\put(25,185){$(i)$}
\put(220,185){$(j)$}

\put(120,50){$(k)$}

\put(25,445){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (45:1.931cm) node[draw=blue,shape=circle] (v9) {};
\path (60:1.931cm) node[draw=blue,shape=circle] (v10){};

\path (345:1.931cm) node[draw=red,shape=circle] (v11) {};
\path (0:2cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v1) edge[red] (v11)
(v1) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v9) edge[blue] (v2)
(v10) edge[blue] (v2)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(220,445){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (30:1.931cm) node[draw=blue,shape=circle] (v9) {};
\path (45:1.732cm) node[draw=blue,shape=circle] (v10) {};

\path (-15:1.931cm) node[draw=red,shape=circle] (v11) {};
\path (-30:2cm) node[draw=red,shape=circle] (v12) {};

\path (0:2.131cm) node[draw=green,shape=circle] (v13) {};
\path (15:1.932cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v1) edge[red] (v11)
(v1) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v1) edge[blue] (v9)
(v1) edge[blue] (v10)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(25,315){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (45:1.931cm) node[draw=blue,shape=circle] (v9) {};
\path (60:1.931cm) node[draw=blue,shape=circle] (v10){};

\path (15:2.931cm) node[draw=red,shape=circle] (v11) {};
\path (30:2.732cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v13) edge[red] (v11)
(v13) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v9) edge[blue] (v2)
(v10) edge[blue] (v2)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(220,315){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (30:2.631cm) node[draw=blue,shape=circle] (v9) {};
\path (40:2.332cm) node[draw=blue,shape=circle] (v10) {};

\path (0:2.931cm) node[draw=red,shape=circle] (v11) {};
\path (15:2.732cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v13) edge[red] (v11)
(v13) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v13) edge[blue] (v9)
(v13) edge[blue] (v10)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(25,195){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (30:2.631cm) node[draw=blue,shape=circle] (v9) {};
\path (40:2.332cm) node[draw=blue,shape=circle] (v10) {};

\path (345:1.931cm) node[draw=red,shape=circle] (v11) {};
\path (0:2cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v1) edge[red] (v11)
(v1) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v13) edge[blue] (v9)
(v13) edge[blue] (v10)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(220,195){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (30:2.631cm) node[draw=blue,shape=circle] (v9) {};
\path (40:2.332cm) node[draw=blue,shape=circle] (v10) {};

\path (20:3.331cm) node[draw=red,shape=circle] (v11) {};
\path (30:3.632cm) node[draw=red,shape=circle] (v12) {};

\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v9) edge[red] (v11)
(v9) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v13) edge[blue] (v9)
(v13) edge[blue] (v10)
(v9) edge[blue] (v10);
\end{tikzpicture}}

\put(120,60){\begin{tikzpicture}[scale=1.1]

\path (0:0cm) node[draw,shape=circle] (v0) {};
\path (0:1cm) node[draw,shape=circle] (v1) {};
\path (40:1cm) node[draw,shape=circle] (v2) {};
\path (92:1cm) node[draw,shape=circle] (v3) {};
\path (132:1cm) node[draw,shape=circle] (v4) {};
\path (174:1cm) node[draw,shape=circle] (v5) {};
\path (214:1cm) node[draw,shape=circle] (v6) {};
\path (266:1cm) node[draw,shape=circle] (v7) {};
\path (306:1cm) node[draw,shape=circle] (v8) {};

\path (3:2.631cm) node[draw=blue,shape=circle] (v9) {};
\path (-12:2.332cm) node[draw=blue,shape=circle] (v10) {};
\path (27:2.631cm) node[draw=red,shape=circle] (v11) {};
\path (40:2.732cm) node[draw=red,shape=circle] (v12) {};
\path (15:1.931cm) node[draw=green,shape=circle] (v13) {};
\path (30:1.732cm) node[draw=green,shape=circle] (v14) {};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v3)
(v3) -- (v4)
(v4) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v0) -- (v7)
(v7) -- (v8)
(v8) -- (v0)

(v14) edge[red] (v11)
(v14) edge[red] (v12)
(v11) edge[red] (v12)

(v1) edge[green] (v13)
(v1) edge[green] (v14)
(v13) edge[green] (v14)

(v13) edge[blue] (v9)
(v13) edge[blue] (v10)
(v9) edge[blue] (v10);
\end{tikzpicture}}
\end{picture}
\end{center}
\caption{All types of Dutch windmills of order $7$ with three pendant triangles.}\label{threependantfig}

\end{figure}

\newpage
\begin{theorem}\label{dutchthrm1}\cite{Bermond} Let $G$ be a dutch windmill with $n$ blocks. If there exists a Skolem sequence of order $n$, then $G$ is graceful.\end{theorem}

\begin{pf}Let $G$ be a dutch windmill with $n$ blocks. Let $S_{n}$ be a Skolem sequence of order $n$ of the form $\left(a_{i},b_{i}\right)$ where $a_{i}$ and $b_{i}$ are the subscripts of $i$ with $a_{i} < b_{i}, i=1,2,\ldots,n.$ These pairs give $n$ base blocks which are $\{0,a_{i}+n,b_{i}+n\}$, $1\leq i\leq n.$
\begin{enumerate}
	\item We can obtain three types of differences from $\{0,a_{i}+n,b_{i}+n\}$ as follows:
	\begin{enumerate}
		\item $A=\left\{(b_{i}+n)-(a_{i}+n)\right\}=\left\{1,2,\ldots,n\right\}$;
		\item $B=\left\{(a_{i}+n)-(0)\right\}$;
		\item $C=\left\{(b_{i}+n)-(0)\right\}$. Then, $A\cup B \cup C=\left\{1,2,\ldots,3n\right\}$.
	\end{enumerate}
	\item The base blocks $\{0,a_i+n,b_i+n\}_{i=1}^n$ give the vertex labels $\left\{0,1,2,\ldots,3n\right\},$ where $0$ is repeated $n$ times, but it is a common vertex. The base blocks formed by Skolem sequences present a graceful labelling.\end{enumerate}\end{pf}

\begin{theorem}\label{dutchthrm2}\cite{Bermond} Let $G$ be a Dutch windmill with $n$ blocks. If there exists a hooked Skolem sequence of order $n$, then $G$ is near graceful.\end{theorem}
\begin{pf}Let $G$ be a dutch windmill with $n$ blocks. Let $hS_{n}$ be a hooked Skolem sequence of order $n$ of the form $\left(a_{i},b_{i}\right)$ where $a_{i}$ and $b_{i}$ are the subscripts of $i$ with $a_{i} < b_{i}, i=1,2,\ldots,n.$ These pairs give $n$ base blocks which are $\{0,a_{i}+n,b_{i}+n\}$, $1\leq i\leq n.$
\begin{enumerate}
	\item We can obtain three types of differences from $\{0,a_{i}+n,b_{i}+n\}$ as follows:
	\begin{enumerate}
		\item $A=\left\{(b_{i}+n)-(a_{i}+n)\right\}=\left\{1,2,\ldots,n\right\}$;
		\item $B=\left\{(a_{i}+n)-(0)\right\}$;
		\item $C=\left\{(b_{i}+n)-(0)\right\}$. Then, $A\cup B \cup C=\left\{1,2,\ldots,3n-1,3n+1\right\}$.
	\end{enumerate}
	\item The base blocks $\{0,a_i+n,b_i+n\}_{i=1}^n$ give the vertex labels \newline $\left\{0,1,2,\ldots,3n-1,3n+1\right\},$ where $0$ is repeated $n$ times, but it is a common vertex. The base blocks formed by a hooked Skolem sequences present a near graceful labelling.\end{enumerate}\end{pf}

\par In the (near) graceful labelling by (hooked) Skolem sequences, we can use the base blocks of the form $\{0,a_i+n,b_i+n\}_{i=1}^n$ or $\{0,i,b_i+n\}_{i=1}^n$. They give two different vertex labels, but the same edge labels.

For instance, taking the base blocks from the example in Section~\ref{skolemintro} will give us the gracefully labelled dutch windmill of order $4$ shown in Figures~\ref{different3}$(a),(b),$ and $(c).$ The gracefully labelled Dutch windmills in $3(a)$ are formed by base blocks of the form $\left\{0,a_i+n,b_i+n\right\}$; the base blocks of the form $\left\{0,i,b_i+n\right\}$ in $3(b)$; and finally by a mixed set of base blocks that come from both forms in $3(c)$. We will use this technique in Chapter 3 when creating graceful labellings for Type $(k)$ dutch windmills.
\\
\\
\begin{figure}[htbp]
\begin{center}
\begin{picture}(260,260)
\linethickness{0.9pt}
\put(0,148){$(a)$}
\put(155,148){$(b)$}
\put(78,0){$(c)$}

\put(0,150){\begin{tikzpicture}[scale=1.4]

\path (0:0cm) node[draw,shape=circle] (v0) {$0$};
\path (20:1.5cm) node[draw,shape=circle] (v1) {\tiny $5$};
\path (70:1.5cm) node[draw,shape=circle] (v2) {\tiny $6$};
\path (110:1.5cm) node[draw,shape=circle] (v3) {\tiny $8$};
\path (160:1.5cm) node[draw,shape=circle] (v4) {\tiny $10$};
\path (200:1.5cm) node[draw,shape=circle] (v5) {\tiny $9$};
\path (250:1.5cm) node[draw,shape=circle] (v6) {\tiny $12$};
\path (290:1.5cm) node[draw,shape=circle] (v7) {\tiny $7$};
\path (340:1.5cm) node[draw,shape=circle] (v8) {\tiny $11$};
\draw
(v0) -- (v1) node[pos=.6,below] {\tiny $5$}
(v0) -- (v2) node[pos=.6,right] {\tiny $6$}
(v0) -- (v3) node[pos=.5,left] {\tiny $8$}
(v0) -- (v4) node[pos=.6,below] {\tiny $10$}
(v0) -- (v5) node[pos=.5,below] {\tiny $9$}
(v0) -- (v6) node[pos=.7,right] {\tiny $12$}
(v0) -- (v7) node[pos=.5,right] {\tiny $7$}
(v0) -- (v8) node[pos=.5,below] {\tiny $11$}
(v2) -- (v1) node[pos=.4,below] {\tiny $1$}
(v3) -- (v4) node[pos=.4,below] {\tiny $2$}
(v5) -- (v6) node[pos=.3,below] {\tiny $3$}
(v7) -- (v8) node[pos=.6,below] {\tiny $4$};
\end{tikzpicture}}

\put(155,150){\begin{tikzpicture}[scale=1.4]

\path (0:0cm) node[draw,shape=circle] (v0) {\tiny $0$};
\path (20:1.5cm) node[draw,shape=circle] (v1) {\tiny $1$};
\path (70:1.5cm) node[draw,shape=circle] (v2) {\tiny $6$};
\path (110:1.5cm) node[draw,shape=circle] (v3) {\tiny $2$};
\path (160:1.5cm) node[draw,shape=circle] (v4) {\tiny $10$};
\path (200:1.5cm) node[draw,shape=circle] (v5) {\tiny $3$};
\path (250:1.5cm) node[draw,shape=circle] (v6) {\tiny $12$};
\path (290:1.5cm) node[draw,shape=circle] (v7) {\tiny $4$};
\path (340:1.5cm) node[draw,shape=circle] (v8) {\tiny $11$};
\draw
(v0) -- (v1) node[pos=.6,below] {\tiny $1$}
(v0) -- (v2) node[pos=.6,right] {\tiny $6$}
(v0) -- (v3) node[pos=.5,left] {\tiny $2$}
(v0) -- (v4) node[pos=.6,below] {\tiny $10$}
(v0) -- (v5) node[pos=.5,below] {\tiny $3$}
(v0) -- (v6) node[pos=.7,right] {\tiny $12$}
(v0) -- (v7) node[pos=.5,right] {\tiny $4$}
(v0) -- (v8) node[pos=.5,below] {\tiny $11$}
(v2) -- (v1) node[pos=.4,below] {\tiny $5$}
(v3) -- (v4) node[pos=.4,below] {\tiny $8$}
(v5) -- (v6) node[pos=.3,below] {\tiny $9$}
(v7) -- (v8) node[pos=.6,below] {\tiny $7$};
\end{tikzpicture}}

\put(78,5){\begin{tikzpicture}[scale=1.4]

\path (0:0cm) node[draw,shape=circle] (v0) {\tiny $0$};
\path (20:1.5cm) node[draw,shape=circle] (v1) {\tiny $1$};
\path (70:1.5cm) node[draw,shape=circle] (v2) {\tiny $6$};
\path (110:1.5cm) node[draw,shape=circle] (v3) {\tiny $2$};
\path (160:1.5cm) node[draw,shape=circle] (v4) {\tiny $10$};
\path (200:1.5cm) node[draw,shape=circle] (v5) {\tiny $9$};
\path (250:1.5cm) node[draw,shape=circle] (v6) {\tiny $12$};
\path (290:1.5cm) node[draw,shape=circle] (v7) {\tiny $7$};
\path (340:1.5cm) node[draw,shape=circle] (v8) {\tiny $11$};
\draw
(v0) -- (v1) node[pos=.6,below] {\tiny $1$}
(v0) -- (v2) node[pos=.6,right] {\tiny $6$}
(v0) -- (v3) node[pos=.5,left] {\tiny $2$}
(v0) -- (v4) node[pos=.6,below] {\tiny $10$}
(v0) -- (v5) node[pos=.5,below] {\tiny $9$}
(v0) -- (v6) node[pos=.7,right] {\tiny $12$}
(v0) -- (v7) node[pos=.5,right] {\tiny $7$}
(v0) -- (v8) node[pos=.5,below] {\tiny $11$}
(v2) -- (v1) node[pos=.4,below] {\tiny $5$}
(v3) -- (v4) node[pos=.4,below] {\tiny $8$}
(v5) -- (v6) node[pos=.3,below] {\tiny $3$}
(v7) -- (v8) node[pos=.6,below] {\tiny $4$};
\end{tikzpicture}}
\end{picture}
\caption{Three different gracefully labelled Dutch windmills of order 4.}\label{different3}
\end{center}
\end{figure}

\par Since $x-y=(x+c)-(y+c)$ for any $x,y$, the following lemma is straightforward.
\begin{lemma}\label{lemma2}If we add a constant $c$ to each element of any base block $\left\{x,y,z\right\},$ which results in $\left\{x+c,y+c,z+c\right\}$, then the differences between the elements of $\left\{x+c,y+c,z+c\right\}$ will be the same.\end{lemma}

In a previous example, we consider the triple $\{0,8,10\}$ and add $2$ to each element to obtain $\{2,10,12\}$. This gives the blocks $\{0,5,6\},\{2,10,12\},\{0,9,12\},$ and $\{0,7,11\}.$ Consequently, we have obtained a graceful labelling for a Dutch windmill of order three with one pendant triangle.\par
Here we will present an example of how to label Figure~\ref{threependantfig}(b) by using a hooked Skolem sequence.
{\label{threepivotsexample}}Let $hS_{7}=(7,4,6,3,5,4,3,7,6,5,1,1,2,0,2)$ be a hooked Skolem sequence of order $7.$ This yields the pairs\newline  $\{(11,12),(13,15),(4,7),(2,6),(5,10),(3,9),(1,8)\}$. This sequence gives the base blocks of the form $\left\{0,a_{i}+n, b_{i}+n\right\}$ as follows:\newline $A=\{(0,18,19),(0,20,22),(0,11,14),(0,9,13),(0,12,17),(0,10,16),(0,8,15)\}.$\newline The above sequence $hS_{7}$ has three pivots, which are $1,4,$ and $7.$ We obtain a new set of base blocks by pivoting as follows:\newline $A'=\{(1,19,20),(0,20,22),(0,11,14),(4,13,17),(0,12,17),(0,10,16),(7,15,22)\}.$\par
We will use the set $A'$ to label our graph as follows: label the central vertex with $0$. Then label all the vanes by these base blocks containing $0.$ Finally, label the pendant triangles with the base blocks corresponding to the pivots. The blocks  $\left\{1,19,20\right\}$ and $\left\{7,15,22\right\}$ each intersect $\left\{0,20,22\right\}$ at a single distinct element, namely $20$ for the first block and $22$ for the second block. The block $\left\{4,13,17\right\}$ intersects $\left\{0,12,17\right\}$ at a single element, namely $17.$

\par To label dutch windmills with $n$ blocks that have three pendant triangles, we will use Skolem Skolem sequences of order $n$ with at least three pivots. Since our Skolem sequence $S$ has three distinct pivots $r,s,t,\  1 \leq r,s,t \leq n$ we will obtain a new set of base blocks $A'$. Then $A'$ contains the new triples $\{r,a_{r}+n+r,b_{r}+n+r\},\{s,a_{s}+n+s,b_{s}+n+s\},\{t,a_{t}+n+t,b_{t}+n+t\}.$ Exactly one of $r,a_{r}+n+r,$ or $b_{r}+n+r$ must appear in one of the other triples and this is the label where we attach the pendant triangle; (the elements of the triples in the set $A'$ must be labelled from ($\left\{0,1,2,\ldots,3n-1,3n+1\right\}$) $\left\{0,1,2,\ldots,3n\right\}$, and these values can not be more than ($3n+1$) $3n$), and it is the same for $s$ and $t$. In the same way, the same idea works for hooked Skolem sequences. Then, by either Theorem~\ref{dutchthrm1} or Theorem~\ref{dutchthrm2}, the Skolem sequence give the vertex labels of the gracefully labelled dutch windmill or near gracefully labelled dutch windmill.

\par In the following lemma, $\lambda$ represents a letter between $a$ and $k.$
\begin{lemma}\label{bound}If a (hooked) Skolem sequence of order $n$ exists that gives a (near) graceful labelling of a Type $(\lambda)$ dutch windmill with $n$ blocks, then there exists $N(n)=3n + 1$ such that for $m \geq N(n),$ there exists a (near) graceful labelling of Type $(\lambda)$ dutch windmill with $m$ blocks. \end{lemma}
\begin{pf}Let $G$ be a Type $(\lambda)$ dutch windmill with $n$ blocks (near) gracefully labelled by $S_{n},$ a (hooked) Skolem sequence of order $n$. By Theorem~\ref{langfordexistence}, a (hooked) Langford sequence exists with defect $d=n+1.$ Then by Lemma~\ref{skla}, we obtain a new (hooked) Skolem sequence of order $m,$ $S_{m}$ where $m \geq 3n + 1=N(n)$. Then, since $S_n$ had the pivot structure needed to (near) gracefully label a Type $(\lambda)$ dutch windmill with $n$ blocks, $S_m$ will yield the structure needed to (near) gracefully label a Type $(\lambda)$ dutch windmill with $m$ blocks.
\end{pf}

\subsection{Type $(a)$}
For Type $(a)$, when $n \leq 5$, there are not enough triangles to form a Type $(a)$ dutch windmill. When $n=6$, see the near graceful labelling in Figure~\ref{taypean6}.
\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[scale=1.2]

\path (0:0cm) node[draw,shape=circle] (v0) {\tiny $17$};
\path (20:1cm) node[draw,shape=circle] (v1) {\tiny $13$};
\path (60:1cm) node[draw,shape=circle] (v2) {\tiny $1$};
\path (120:1cm) node[draw,shape=circle] (v5) {\tiny $0$};
\path (160:1cm) node[draw,shape=circle] (v6) {\tiny $11$};
\path (70:2cm) node[draw,shape=circle] (v11) {\tiny $16$};
\path (50:2cm) node[draw,shape=circle] (v12) {\tiny $3$};
\path (110:2cm) node[draw,shape=circle] (v13) {\tiny $5$};
\path (130:2cm) node[draw,shape=circle] (v14) {\tiny $19$};
\path (90:0.9cm) node[draw=none,shape=circle] (v15){};

\path (-60:1cm) node[draw,shape=circle] (v16) {\tiny $10$};
\path (-120:1cm) node[draw,shape=circle] (v17) {\tiny $7$};

\path (-110:2cm) node[draw,shape=circle] (v18){\tiny $15$};
\path (-130:2cm) node[draw,shape=circle] (v19) {\tiny $6$};
\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v0) -- (v5)
(v5) -- (v6)
(v6) -- (v0)
(v2) -- (v11)
(v2) -- (v12)
(v11) -- (v12)
(v5) -- (v13)
(v5) -- (v14)
(v13) -- (v14)
(v0) -- (v16)
(v0) -- (v17)
(v16) -- (v17)
(v17) -- (v18)
(v17) -- (v19)
(v18) -- (v19);
\end{tikzpicture}
\caption{A near gracefully labelled Type $(a)$ dutch windmill with $6$ blocks}\label{taypean6}
\end{center}
\end{figure}

\begin{lemma}\label{typea}
Any Type $(a)$ dutch windmill with at least $6$ blocks is graceful or near graceful.\end{lemma}
\begin{pf}We begin by constructing a (hooked) Skolem sequence which we will use to label a Type $(a)$ Dutch windmill of order 7. Consider the (hooked) Skolem sequence, $hS_{7}=\left(3,4,7,3,2,4,2,5,6,7,1,1,5,0,6\right).$ This sequence gives triples of the form $\left(0,a_{i}+n, b_{i}+n\right)$ as follows: $\left(0,18,19\right),\left(0,12,14\right),\newline \left(0,8,11\right),\left(0,9,13\right),\left(0,15,20\right),\left(0,16,22\right),\left(0,10,17\right).$ The above sequence $hS_{7}$ has three pivots, which are $1,3,$ and $4.$ We convert these triples by pivoting as follows: $\left(1,19,20\right),\left(0,12,14\right),\left(3,11,14\right),\left(4,13,17\right),\left(0,15,20\right),\left(0,16,22\right),\newline \left(0,10,17\right).$ These triples near-gracefully label the Dutch windmill with $7$ blocks.\par

We have an order $7$ hooked Skolem sequence. By Theorem~\ref{langfordexistence}, a hooked Langford sequence of order $m$ with $d=8$ exists for $m \geq 15$. By Lemma~\ref{skla}, we obtain an associated Skolem sequence of order $n,$ and by Lemma~\ref{bound}, we obtain a gracefully labelled Type $(a)$ dutch windmill with $n$ blocks for $n \geq 22.$ Likewise, By Theorem~\ref{langfordexistence}, a Langford sequence of order $m$ with $d=8$ exists for $m \geq 15$. By Lemma~\ref{skla}, we obtain an associated hooked Skolem sequence of order $n$, and by Lemma~\ref{bound}, we obtain a near gracefully labelled Type $(a)$ dutch windmill with $n$ blocks for $n \geq 22.$ Now, we can see from the above steps and Theorem~\ref{necessarycacti} that a Type $(a)$ dutch windmill of order $n \geq 22$ with three pendant triangles is graceful or near graceful.\par

Skolem and hooked Skolem sequences with three pivots of order $8 \leq n \leq 21$ are given in Table~\ref{ttypea}. By using the same construction as $hS_{7}$ with the given pivots, then we have the labellings for Type $(a)$ dutch windmill with $n$ blocks for $8 \leq n \leq 21$. In this table and all subsequent tables, we will represent $10$ by $A$, $11$ by $B$, $12$ by $C$, and so on.\par
Therefore, for $n\geq 6,$ a Type $(a)$ dutch windmill with $n$ vanes is graceful or near graceful, as required.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $8$ & $4857411568723263$ & $1,2$ and $4$ \\
\hline $9$ & $759242574869311368$ & $2,4$ and $5$ \\
\hline $10$ & $A853113598A7426249706$ & $5,8$ and $A$ \\
\hline $11$ & $B68527265A8B7941134A309$ & $2,7$ and $B$ \\
\hline $12$ & $A8531135C8A6B9742624C79B$ & $3,6$ and $A$ \\
\hline $13$ & $B97D4262479B6CA8D5311358AC$ & $3,7$ and $B$ \\
\hline $14$ & $CA75311357EACDB864292468EBD09$ & $4,6$ and $C$ \\
\hline $15$ & $FDB9753EC3579BDF6A84CE64118A202$ & $4,9$ and $F$ \\
\hline $16$ & $9FDBG864292468BDFECAG75311357ACE$ & $3,4$ and $F$ \\
\hline $17$ & $FDB9H64282469BDF8GECAH75311357ACEG$ & $4,5$ and $9$ \\
\hline $18$ & $GEC9753113579ICEGHFDA8642B2468AIDFH0B$ & $4,6$ and $9$ \\
\hline $19$ & $IGEC9753113579JCEGIHFDA8642B2468AJDFH0B$ & $7,9$ and $C$ \\
\hline $20$ & $JHFDB9753CK3579BDFHJICEGA86411K468A2E2IG$ & $1,6$ and $9$ \\
\hline $21$ & $KIGEC9753113579LCEGIKAJHFDB8642A2468LBDFHJ$ & $5,A$ and $K$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(a)$.}
\label{ttypea}
\end{center}
\end{table}
\end{pf}

\subsection{Type $(b)$}
 For $n \leq 4$, there are not enough triangles to form Type $(b)$ dutch windmill.
\begin{lemma}\label{typeb}
Any Type $(b)$ dutch windmill with at least $5$ blocks is graceful or near graceful.
\end{lemma}
\begin{pf}
For Type $(b)$, consider $S_{5}=\left(4,1,1,5,4,2,3,2,5,3\right).$ This sequence has three pivots, which are $1,2$, and $4$. This gives us a graceful labelling of a Type $(b)$ dutch windmill with $5$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea}, by using Lemma~\ref{bound} with $S_{5}$ to obtain a (near) graceful labelling of a Type $(b)$ dutch windmill with $n$ blocks for $n \geq 16$. Table~\ref{ttypeb} provides Skolem and hooked Skolem sequences with three pivots of order $6 \leq n \leq 15,$ each of which gives a (near) graceful labelling of a Type $(b)$ dutch windmill with $n$ blocks.
\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $6$ & $5611453643202$ & $1,3$ and $5$ \\
\hline $7$ & $746354376511202$ & $1,4$ and $7$ \\
\hline $8$ & $3723258476541186$ & $1,3$ and $5$ \\
\hline $9$ & $759242574869311368$ & $1,4$ and $5$ \\
\hline $10$ & $2529115784A694738630A$ & $2,4$ and $5$ \\
\hline $11$ & $B68527265A8B7941134A309$ & $5,7$ and $B$ \\
\hline $12$ & $A8531135C8A6B9742624C79B$ & $5,6$ and $A$ \\
\hline $13$ & $B97D4262479B6CA8D5311358AC$ & $1,6$ and $7$ \\
\hline $14$ & $CA75311357EACDB864292468EBD09$ & $3,4$ and $C$ \\
\hline $15$ & $FDB9753EC3579BDF6A84CE64118A202$ & $6,9$ and $F$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(b)$.}
\label{ttypeb}
\end{center}
\end{table}
\end{pf}

\subsection{Type $(c)$}
 For $n \leq 4$, there are not enough triangles to form Type $(c)$ dutch windmill.
\begin{lemma}\label{typec}
Any Type $(c)$ dutch windmill with at least $5$ blocks is graceful or near graceful.\end{lemma}
\begin{pf}For Type $(c)$, consider $S_{5}=\left(3,5,2,3,2,4,5,1,1,4\right).$ This sequence has three pivots, which are $1,2$, and $3$. This gives us a graceful labelling of a Type $(c)$ dutch windmill with $5$ blocks. Following the method of Lemma~\ref{typea}, we use $S_{5}$ to obtain a (near) graceful labelling of a Type $(c)$ dutch windmill with $n$ blocks for $n \geq 16$. Table~\ref{ttypec} provides Skolem and hooked Skolem sequences with three pivots of order $6 \leq n \leq 15,$ each of which gives a (near) graceful labelling of a Type $(c)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $6$ & $6451146523203$ & $1,5$ and $6$ \\
\hline $7$ & $746354376511202$ & $4,6$ and $7$ \\
\hline $8$ & $3723258476541186$ & $1,4$ and $5$ \\
\hline $9$ & $372329687115649854$ & $2,3$ and $7$ \\
\hline $10$ & $36232A768119574A85409$ & $1,2$ and $3$ \\
\hline $11$ & $35232B549A841167B98A607$ & $1,2$ and $3$ \\
\hline $12$ & $3A232C78119AB7685C49654B$ & $1,2$ and $3$ \\
\hline $13$ & $CA8531135D8AC6B9742624D79B$ & $2,4$ and $6$ \\
\hline $14$ & $3B932A211DE9B6CA485647D5E8C07$ & $4,6$ and $A$ \\
\hline $15$ & $ECA8642D2468ACEFB953D735119B70F$ & $4,5$ and $E$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(c)$.}
\label{ttypec}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(d)$}
For $n \leq 4$, there are not enough triangles to form a Type $(d)$ dutch windmill. For $n=5,$ a graceful labelling of the $5-$vane dutch windmill appears in \cite{rosa}. When $n=6$, see the near graceful labelling in Figure~\ref{taypedn6}.
\begin{figure}[H]
\begin{center}
\scalebox{0.65}{
\begin{tikzpicture}
    \node[shape=circle,draw=black] (19) at (0,0) { $19$};
    \node[shape=circle,draw=black] (0) at (2,0) { $0$};
    \node[shape=circle,draw=black] (17) at (4,0) { $17$};
    \node[shape=circle,draw=black] (1) at (6,0) { $1$};
    \node[shape=circle,draw=black] (16) at (8,0) { $16$} ;
	   \node[shape=circle,draw=black] (6) at (10,0) { $6$};

		    \node[shape=circle,draw=black] (5) at (1,2) { $5$};
		    \node[shape=circle,draw=black] (11) at (3,2) { $11$};
		    \node[shape=circle,draw=black] (13) at (5,2) { $13$};
        \node[shape=circle,draw=black] (3) at (7,2) { $3$};
		    \node[shape=circle,draw=black] (9) at (9,2) { $9$};

    \path [-] (19) edge node[left,above] { $19$} (0);
    \path [-](19) edge node[left,above] { $14\ $} (5);
		\path [-] (5) edge node[left,above] { $\ 5$} (0);

    \path [-] (0) edge node[left,above] { $11\ $} (11);
    \path [-](0) edge node[left,above] { $17$} (17);
		\path [-] (17) edge node[left,above] { $\ 6$} (11);

		\path [-] (17) edge node[left,above] { $16$} (1);
    \path [-](17) edge node[left,above] { $4\ $} (13);
		\path [-] (1) edge node[left,above] { $\ 12$} (13);

    \path [-] (1) edge node[left,above] { $2\ $} (3);
    \path [-](1) edge node[left,above] { $15$} (16);
		\path [-] (3) edge node[left,above] { $\ 13$} (16);

	\path [-] (16) edge node[left,above] { $7\ $} (9);
    \path [-](16) edge node[left,above] { $10$} (6);
		\path [-] (6) edge node[left,above] { $\ 3$} (9);

 \node[shape=circle,draw=black] (10) at (5,-2) { $10$};
 \node[shape=circle,draw=black] (2) at (7,-2) { $2$};
\path [-] (2) edge node[left,above] { $\ 1$} (1);
    \path [-](1) edge node[left,above] { $9\ $} (10);
		\path [-] (2) edge node[left,above] { $8$} (10);
\end{tikzpicture}
}
\caption{A near gracefully labelled Type $(d)$ dutch windmill with $6$ blocks}\label{taypedn6}
\end{center}
\end{figure}

\begin{lemma}\label{typed}
Any Type $(d)$ dutch windmill with at least $5$ blocks is graceful or near graceful.\end{lemma}

\begin{pf}For Type $(d)$, consider $hS_{7}=\left(3,4,7,3,2,4,2,5,6,7,1,1,5,0,6\right).$ This sequence has three pivots, which are $1,2$, and $3$. This gives us a near graceful labelling of a Type $(d)$ dutch windmill with $7$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea}, by using Lemma~\ref{bound} with $hS_{7}$ to obtain a (near) graceful labelling of Type $(d)$ dutch windmill with $n$ blocks for $n \geq 22$. Table~\ref{ttyped} provides Skolem and hooked Skolem sequences with three pivots of order $8 \leq n \leq 21,$ each of which gives a (near) graceful labelling of a Type $(d)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $8$ & $4857411568723263$ & $1,4$ and $5$ \\
\hline $9$ & $759242574869311368$ & $1,5$ and $7$ \\
\hline $10$ & $A853113598A7426249706$ & $3,8$ and $A$ \\
\hline $11$ & $B68527265A8B7941134A309$ & $2,5$ and $7$ \\
\hline $12$ & $A8531135C8A6B9742624C79B$ & $6,8$ and $A$ \\
\hline $13$ & $B97D4262479B6CA8D5311358AC$ & $4,6$ and $7$ \\
\hline $14$ & $CA75311357EACDB864292468EBD09$ & $1,3$ and $A$ \\
\hline $15$ & $FDB9753EC3579BDF6A84CE64118A202$ & $5,B$ and $F$ \\
\hline $16$ & $9FDBG864292468BDFECAG75311357ACE$ & $4,B$ and $D$ \\
\hline $17$ & $FDB9H64282469BDF8GECAH75311357ACEG$ & $2,6$ and $9$ \\
\hline $18$ & $GEC9753113579ICEGHFDA8642B2468AIDFH0B$ & $3,6$ and $9$ \\
\hline $19$ & $IGEC9753113579JCEGIHFDA8642B2468AJDFH0B$ & $1,5$ and $E$ \\
\hline $20$ & $JHFDB9753CK3579BDFHJICEGA86411K468A2E2IG$ & $6,D$ and $F$ \\
\hline $21$ & $KIGEC9753113579LCEGIKAJHFDB8642A2468LBDFHJ$ & $6,E$ and $G$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(d)$.}
\label{ttyped}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(e)$}
 For $n \leq 3$, there are not enough triangles to form Type $(e)$ dutch windmill. For $n=4,$ a graceful labelling of the $4-$vane dutch windmill appears in \cite{rosa}.
\begin{lemma}\label{typee}
Any Type $(e)$ dutch windmill with at least $4$ blocks is graceful or near graceful.\end{lemma}

\begin{pf}For Type $(e)$, consider $S_{5}=\left(2,4,2,3,5,4,3,1,1,5\right).$ This sequence has three pivots, which are $1,2$, and $3$. This gives us a graceful labelling of a Type $(e)$ dutch windmill with $5$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea}, by using Lemma~\ref{bound} with $S_{5}$ to obtain a (near) graceful labelling of a Type $(e)$ dutch windmill with $n$ blocks for $n \geq 16$. Table~\ref{ttypee} provides Skolem and hooked Skolem sequences with three pivots of order $6 \leq n \leq 15,$ each of which gives a (near) graceful labelling of a Type $(e)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $6$ & $5611453643202$ & $3,4$ and $5$ \\
\hline $7$ & $746354376511202$ & $1,6$ and $7$ \\
\hline $8$ & $3723258476541186$ & $1,2$ and $3$ \\
\hline $9$ & $572825967348364911$ & $2,3$ and $7$ \\
\hline $10$ & $36232A768119574A85409$ & $2,3$ and $6$ \\
\hline $11$ & $B68527265A8B7941134A309$ & $4,6$ and $B$ \\
\hline $12$ & $A8531135C8A6B9742624C79B$ & $4,5$ and $6$ \\
\hline $13$ & $39B32D258A9C5B6784DA647C11$ & $4,8$ and $B$ \\
\hline $14$ & $DB964E1146C9BDA8537E35C8A7202$ & $3,6$ and $B$ \\
\hline $15$ & $CE3693BF262DC97EAB1187F4D5A4805$ & $1,7$ and $B$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(e)$.}
\label{ttypee}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(f)$}
For $n \leq 3$, there are not enough triangles to form Type $(f)$ dutch windmill. For $n=4,$ a graceful labelling of the $4-$vane dutch windmill appears in \cite{rosa}.
\begin{lemma}\label{typef}
Any Type $f$ dutch windmill with at least $4$ blocks is graceful or near graceful.\end{lemma}
\begin{pf}For Type $(f)$, consider $S_{5}=\left(2,4,2,3,5,4,3,1,1,5\right).$ This sequence has three pivots, which are $1,3$, and $4$. This gives us a graceful labelling of a Type $(f)$ dutch windmill with $5$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea}, by using Lemma~\ref{bound} with $S_{5}$ to obtain a (near) graceful labelling of a Type $(f)$ dutch windmill with $n$ blocks for $n \geq 16$. Table~\ref{ttypef} provides Skolem and hooked Skolem sequences with three pivots of order $6 \leq n \leq 15,$ each of which gives a (near) graceful labelling of a Type $(f)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $6$ & $6451146523203$ & $2,5$ and $6$ \\
\hline $7$ & $746354376511202$ & $5,6$ and $7$ \\
\hline $8$ & $3723258476541186$ & $4,5$ and $7$ \\
\hline $9$ & $746394376825291158$ & $2,6$ and $7$ \\
\hline $10$ & $A869117468A4973523205$ & $2,3$ and $A$ \\
\hline $11$ & $378392A2768B5946A54110B$ & $4,5$ and $9$ \\
\hline $12$ & $2529115B86AC947684B3A73C$ & $4,6$ and $9$ \\
\hline $13$ & $9BD3753AC957B82D2A46C84116$ & $2,7$ and $9$ \\
\hline $14$ & $E7D6C5B47654A8EDCB2928A311309$ & $5,6$ and $7$ \\
\hline $15$ & $3C9382B2E1198CDFAB5647E546AD70F$ & $4,5$ and $B$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(f)$.}
\label{ttypef}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(g)$}
For $n \leq 3$, there are not enough triangles to form Type $(g)$ dutch windmill. For $n=4,$ a graceful labelling of the $4-$vane dutch windmill appears in \cite{rosa}.
\begin{lemma}\label{typeg}
Any Type $(g)$ dutch windmill with at least $4$ blocks is graceful or near graceful.
\end{lemma}
\begin{pf}For Type $(g)$, consider $S_{5}=\left(3,4,5,3,2,4,2,5,1,1\right).$ This sequence has three pivots, which are $2,3$, and $4$. This gives us a graceful labelling of a Type $(g)$ dutch windmill with $5$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea},  by using Lemma~\ref{bound} with $S_{5}$ to obtain a (near) graceful labelling of a Type $(g)$ dutch windmill with $n$ blocks for $n \geq 16$. Table~\ref{ttypeg} provides Skolem and hooked Skolem sequences with three pivots of order $6 \leq n \leq 15,$ each of which gives a (near) graceful labelling of a Type $(g)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $6$ & $6451146523203$ & $1,2$ and $4$ \\
\hline $7$ & $746354376511202$ & $1,4$ and $5$ \\
\hline $8$ & $4857411568723263$ & $1,2$ and $5$ \\
\hline $9$ & $759242574869311368$ & $1,4$ and $7$ \\
\hline $10$ & $2529115784A694738630A$ & $1,4$ and $5$ \\
\hline $11$ & $B68527265A8B7941134A309$ & $2,5$ and $B$ \\
\hline $12$ & $A8531135C8A6B9742624C79B$ & $5,6$ and $8$ \\
\hline $13$ & $B97D4262479B6CA8D5311358AC$ & $1,4$ and $6$ \\
\hline $14$ & $CA75311357EACDB864292468EBD09$ & $1,3$ and $C$ \\
\hline $15$ & $FDB9753EC3579BDF6A84CE64118A202$ & $5,6$ and $F$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(g)$.}
\label{ttypeg}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(h)$}
 For $n \leq 3$, there are not enough triangles to form Type $(h)$ dutch windmill. For $n=4,5$ a graceful labelling of the $4-$vane and $5-$vane dutch windmill appears in \cite{rosa}.
\begin{lemma}\label{typeh}
Any Type $(h)$ dutch windmill with at least $4$ blocks is graceful or near graceful.
\end{lemma}
\begin{pf}
For Type $(h)$, consider $hS_{6}=\left(4,5,3,6,4,3,5,1,1,6,2,0,2\right)$. This sequence has three pivots, which are $1,3$, and $4$. This gives us a near graceful labelling of a Type $(h)$ dutch windmill with $6$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea}, again by using Lemma~\ref{bound} with $hS_{6}$ to obtain a (near) graceful labelling of a Type $(h)$ dutch windmill with $n$ blocks for $n \geq 19$. Table~\ref{ttypeh} provides Skolem and hooked Skolem sequences with three pivots of order $7 \leq n \leq 18,$ each of which gives a (near) graceful labelling of a Type $(h)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline

\hline $7$ & $746354376511202$ & $3,4$ and $5$ \\
\hline $8$ & $1157468543763282$ & $3,4$ and $5$ \\
\hline $9$ & $736931176845924258$ & $1,3$ and $7$ \\
\hline $10$ & $5262854A674981137A309$ & $1,4$ and $6$ \\
\hline $11$ & $635A37659B117A842924B08$ & $5,6$ and $7$ \\
\hline $12$ & $52426549B86CA711983B73AC$ & $4,5$ and $6$ \\
\hline $13$ & $86272C568B75D9A11C34B394AD$ & $1,5$ and $8$ \\
\hline $14$ & $2726AD1176C4EBA458D935C3B8E09$ & $4,6$ and $7$ \\
\hline $15$ & $8C3473D48117ACEFB96D52A2659BE0F$ & $2,5$ and $C$ \\
\hline $16$ & $637A3E6FC7D52A2G58BEC9FD48114B9G$ & $2,5$ and $7$ \\
\hline $17$ & $962D2G5649754EFHD78CAGB1138E3FACHB$ & $4,5$ and $D$ \\
\hline $18$ & $5D9485E4F1198BDG6HIAEC6FB7232A3G7CH0I$ & $8,9$ and $E$ \\

\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(h)$.}
\label{ttypeh}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(i)$}
 For $n \leq 3$, there are not enough triangles to form Type $(i)$ dutch windmill. For $n=4,5$ a graceful labelling of the $4-$vane and $5-$vane dutch windmill appears in \cite{rosa}. When $n=6$, see the near graceful labelling in Figure~\ref{taypein6}.
\begin{figure}[H]
\begin{center}
\scalebox{0.55}{
\begin{tikzpicture}
    \node[shape=circle,draw=black] (19) at (0,0) { $19$};
    \node[shape=circle,draw=black] (0) at (2,0) { $0$};
    \node[shape=circle,draw=black] (17) at (4,0) { $17$};
    \node[shape=circle,draw=black] (1) at (6,0) { $1$};
    \node[shape=circle,draw=black] (16) at (8,0) { $16$} ;

		    \node[shape=circle,draw=black] (5) at (1,2) { $5$};
		    \node[shape=circle,draw=black] (11) at (3,2) { $11$};
		    \node[shape=circle,draw=black] (13) at (5,2) { $13$};
        \node[shape=circle,draw=black] (3) at (7,2) { $3$};

    \path [-] (19) edge node[left,above] { $19$} (0);
    \path [-](19) edge node[left,above] { $14\ $} (5);
		\path [-] (5) edge node[left,above] { $\ 5$} (0);

    \path [-] (0) edge node[left,above] { $11\ $} (11);
    \path [-](0) edge node[left,above] { $17$} (17);
		\path [-] (17) edge node[left,above] { $\ 6$} (11);

		\path [-] (17) edge node[left,above] { $16$} (1);
    \path [-](17) edge node[left,above] { $4\ $} (13);
		\path [-] (1) edge node[left,above] { $\ 12$} (13);

    \path [-] (1) edge node[left,above] { $2\ $} (3);
    \path [-](1) edge node[left,above] { $15$} (16);
		\path [-] (3) edge node[left,above] { $\ 13$} (16);

 \node[shape=circle,draw=black] (9) at (1,-2) { $9$};
 \node[shape=circle,draw=black] (8) at (3,-2) { $8$};
\path [-] (8) edge node[left,above] { $1$} (9);
    \path [-](0) edge node[left,above] { $9\ $} (9);
		\path [-] (0) edge node[left,above] { $\ 8$} (8);

 \node[shape=circle,draw=black] (7) at (4,-2) { $7$};
 \node[shape=circle,draw=black] (10) at (6,-1) { $10$};
\path [-] (17) edge node[left] { $\ \ 10$} (7);
    \path [-](7) edge node[left,below] { $3$} (10);
		\path [-] (10) edge node[right] { $7$} (17);

\end{tikzpicture}
}
\caption{A near gracefully labelled Type $(i)$ dutch windmill with $6$ blocks}\label{taypein6}
\end{center}
\end{figure}

\begin{lemma}\label{typei}
Any Type $(i)$ dutch windmill with at least $4$ blocks is graceful or near graceful.\end{lemma}

\begin{pf}For Type $(i)$, consider $hS_{7}=\left(7,4,6,3,5,4,3,7,6,5,1,1,2,0,2\right).$ This sequence has three pivots, which are $4,5$, and $6$. This gives us a near graceful labelling of a Type $(i)$ dutch windmill with $7$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea}, by using Lemma~\ref{bound} with $hS_{7}$ to obtain a (near) graceful labelling of a Type $(i)$ dutch windmill with $n$ blocks for $n \geq 22$. Table~\ref{ttypei} provides Skolem and hooked Skolem sequences with three pivots of order $8 \leq n \leq 21,$ each of which gives a (near) graceful labelling of a Type $(i)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $8$ & $7536835726248114$ & $1,6$ and $7$ \\
\hline $9$ & $975386357946824211$ & $2,6$ and $7$ \\
\hline $10$ & $A853113598A7426249706$ & $2,3$ and $8$ \\
\hline $11$ & $B68527265A8B7941134A309$ & $4,8$ and $B$ \\
\hline $12$ & $A8531135C8A6B9742624C79B$ & $4,6$ and $8$ \\
\hline $13$ & $CA8531135D8AC6B9742624D79B$ & $4,6$ and $A$ \\
\hline $14$ & $DB964E1146C9BDA8537E35C8A7202$ & $3,9$ and $B$ \\
\hline $15$ & $ECA8642D2468ACEFB953D735119B70F$ & $3,A$ and $C$ \\
\hline $16$ & $9FDBG864292468BDFECAG75311357ACE$ & $2,6$ and $9$ \\
\hline $17$ & $FDB9H64282469BDF8GECAH75311357ACEG$ & $3,D$ and $F$ \\
\hline $18$ & $IGECA8642H2468ACEGIFD9753BH357911DF0B$ & $6,7$ and $I$ \\
\hline $19$ & $IGEC9753113579JCEGIHFDA8642B2468AJDFH0B$ & $2,5$ and $E$ \\
\hline $20$ & $JHFDB9753CK3579BDFHJICEGA86411K468A2E2IG$ & $6,F$ and $J$ \\
\hline $21$ & $KIGEC9753113579LCEGIKAJHFDB8642A2468LBDFHJ$ & $6,G$ and $K$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(i)$}
\label{ttypei}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(j)$ }
 For $n \leq 3$, there are not enough triangles to form Type $(j)$ dutch windmill. For $n=4$ a graceful labelling of the $4-$vane dutch windmill appears in \cite{rosa}.
\begin{lemma}\label{typej}
Any Type $(j)$ dutch windmill with at least $4$ blocks is graceful or near graceful.\end{lemma}

\begin{pf}For Type $(j)$, consider $S_{5}=\left(2,3,2,5,3,4,1,1,5,4\right).$ This sequence has three pivots, which are $1,2$, and $3$. This gives us a graceful labelling of a Type $(j)$ dutch windmill with $5$ blocks. The rest of the proof is analogous to the steps taken to prove Lemma~\ref{typea},  by using Lemma~\ref{bound} with $S_{5}$ to obtain a (near) graceful labelling of a Type $(j)$ dutch windmill with $n$ blocks for $n \geq 16$. Table~\ref{ttypej} provides Skolem and hooked Skolem sequences with three pivots of order $6 \leq n \leq 15,$ each of which gives a (near) graceful labelling of a Type $(j)$ dutch windmill with $n$ blocks.

\begin{table}[H]
\begin{center}
\begin{tabular}{|r|l|l|}
\hline $n$ & Skolem and hooked Skolem sequence & Pivots \\
\hline
\hline $6$ & $2326351146504$ & $1,2$ and $3$ \\
\hline $7$ & $232437546115706$ & $2,3$ and $4$ \\
\hline $8$ & $3753811576428246$ & $1,3$ and $5$ \\
\hline $9$ & $759242574869311368$ & $1,2$ and $7$ \\
\hline $10$ & $A853113598A7426249706$ & $1,3$ and $8$ \\
\hline $11$ & $35232B549A841167B98A607$ & $2,4$ and $5$ \\
\hline $12$ & $A8531135C8A6B9742624C79B$ & $3,6$ and $8$ \\
\hline $13$ & $A8D536C358A6B97D42C2479B11$ & $3,4$ and $A$ \\
\hline $14$ & $7A8C53E7358ADB9C6411E469BD202$ & $3,4$ and $A$ \\
\hline $15$ & $DB964F1146E9BD7CA853F735E8AC202$ & $4,5$ and $B$ \\
\hline
\end{tabular}
\caption{Skolem and hooked Skolem sequences with three pivots for Type $(j)$.}
\label{ttypej}
\end{center}
\end{table}

\end{pf}

\subsection{Type $(k)$}
Figure~\ref{taypekn} represents a portion type $(k)$ dutch windmill. For this type we can label all the triangles by the triples of the form $\left(0,a_{i}+n,b_{i}+n\right)$ where $1 \leq i \leq n,$ using the method of the previous sections with the exception of the three pendent triangles $2,3$ and $4$. Let triangle $1$ be labelled by $\left(0,a_{j}+n,b_{j}+n\right)$, we proceed under the assumption that the remaining triangles can be labelled by pivoting technique. By the pivoting we label triangle $2$ by $\left(k,a_{k}+n+k,b_{k}+n+k\right),$ where $b_{j}+n=a_{k}+n+k$ or $a_{j}+n=a_{k}+n+k$. We will consider $b_{j}+n=a_{k}+n+k$ at vertex $c$ in Figure~\ref{taypekn}. Again by pivoting we label triangle $3$ by $\left(l,a_{l}+n+l,b_{l}+n+l\right).$ We will consider $b_{k}+n+k=a_{l}+n+l,$ at vertex $a$, and hence vertex $b$ is labelled $k$. Note that $1 \leq j,k,l \leq n,$ and all are distinct. If we piviot triangle $4$ we label it by the triple $\left(s,a_{s}+n+s,b_{s}+n+s\right).$ However $k \neq s$ and $min\left(a_{s}+n+s,b_{s}+n+s\right) > n \geq k,$ since this is impossible we must abandon the pivoting method. Instead we will add some constant $c$ to obtain the triple $\left(c,a_{s}+c+n,b_{s}+c+n\right),$ and we will use triples of the form: $\left(0,i,b_{i}+n\right)$ to avoid any conflicts created. By Lemma~\ref{lemma2} adding any constant gives the same differences. (This is a generalization of the idea of pivoting.) The same approach works for hooked Skolem sequences.\par For this type, we will introduce the sequences and the corresponding triples to indicate forms of the triples we use.

\begin{figure}[H]
\begin{center}
\begin{tikzpicture}[scale=1.4]
\path (0:0cm) node[draw,shape=circle] (v0) {\tiny $0$};
\path (65:1cm) node[draw,shape=circle] (v1) {\tiny $c$};
\path (115:1cm) node[draw,shape=circle] (v2) {};
\path (75:2cm) node[draw,shape=circle] (v11) {\tiny $a$};
\path (55:2cm) node[draw,shape=circle] (v12) {\tiny $b$};
\path (101:2.3cm) node[draw,shape=circle] (v3) {};
\path (90:2.8cm) node[draw,shape=circle] (v4) {};
\path (38:2.8cm) node[draw,shape=circle] (v5) {};
\path (50:3.1cm) node[draw,shape=circle] (v6) {};

\node (zz) at (0,0.6) {$1$};
\node (zz) at (0.7,1.5) {$2$};
\node (zz) at (0,2.4) {$3$};
\node (zz) at (1.8,2) {$4$};

\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v1) -- (v11)
(v11) -- (v12)
(v12) -- (v1)
(v11) -- (v3)
(v3) -- (v4)
(v4) -- (v11)
(v12) -- (v5)
(v5) -- (v6)
(v6) -- (v12);
\end{tikzpicture}
\caption{Illustration for Type $(k)$ dutch windmill labelling}\label{taypekn}
\end{center}
\end{figure}

For $n \leq 3$, there are not enough triangles to form Type $(k)$ dutch windmill. For $n=4$, a graceful labelling of the $4$-vane dutch windmill appears in \cite{rosa}.
\begin{lemma}\label{typek}
Any Type $(k)$ dutch windmill with at least $5$ blocks is graceful or near graceful.\end{lemma}

\begin{pf}For Type $(k)$, consider $S_{5}=\left(2,3,2,5,3,4,1,1,5,4\right).$ This sequence gives triples of the form $\left(0,a_{i}+n,b_{i}+n\right)$ as follows: $\left(0,12,13\right),\left(0,6,8\right),\newline \left(0,7,10\right),\left(0,11,15\right),\left(0,9,14\right).$ The above sequence $S_{5}$ has three pivots, but its does not work for this case then we will use mixed forms of the triples to gracefully label a Type $(k)$ dutch windmill with $5$ blocks and the triples are:\newline  $\left(1,13,14\right),\left(1,7,9\right),\left(3,10,13\right),\left(0,11,15\right),\left(0,5,14\right).$ The first and the third triples are formed by pivoting the elements, the second triple is formed by adding $1$ to each element, the fourth triple is formed by base block of the form $\left(0,a_{i}+n,b_{i}+n\right)$ and the fifth triple is formed by base block of the form $\left(0,i,b_{i}+n\right).$ This gives us a graceful labelling of a Type $(k)$ dutch windmill with $5$ blocks.\par
We have an order $5$ Skolem sequence $S_{5}$. By Theorem~\ref{langfordexistence}, a hooked Langford sequence of order $m$ with $d=6$ exists for $m \geq 11$. By Lemma~\ref{skla}, we obtain an associated Skolem sequence of order $n,$ and by Lemma~\ref{bound}, we obtain a gracefully labelled Type $(k)$ dutch windmill with $n$ blocks for $n \geq 16.$ Likewise, By Theorem~\ref{langfordexistence}, a Langford sequence of order $m$ with $d=6$ exists for $m \geq 11$. By Lemma~\ref{skla}, we obtain an associated hooked Skolem sequence of order $n$, and by Lemma~\ref{bound}, we obtain a near graceful labelled Type $(k)$ dutch windmill with $n$ blocks for $n \geq 16.$ Now, we can see from the above steps and Theorem~\ref{necessarycacti} that a Type $(k)$ dutch windmill of order $n \geq 16$ with three pendant triangles is graceful or near graceful.\par

Skolem and hooked Skolem sequences with corresponding triples of order $6 \leq n \leq 15$ are given in Table~\ref{ttypek}, where the triples in bold are formed in an unusual way (i.e formed by one of the following forms: $\left(i,a_{i}+i+n,b_{i}+i+n\right),$ $\left(c,a_{i}+c+n,b_{i}+c+n\right),$ for some $c,$ or $\left(0,i,b_{i}+n\right)$) and the rest of the triples are formed by the typical $\left(0,a_{i}+n,b_{i}+n\right)$ construction. By using a similar structure as $S_{5}$ with the given triples, then we have the labellings for Type $(k)$ dutch windmill with $n$ blocks for $6 \leq n \leq 15$.\par
Therefore, for $n\geq 5,$ a Type $(k)$ dutch windmill with $n$ vanes is graceful or near graceful, as required.

\begin{table}[H]

\begin{center}

\scalebox{0.65}{
\begin{tabular}{|r|p{8cm}|p{12cm}|}
\hline $n$ & Skolem and hooked Skolem sequence & Triples \\
\hline

\hline $6$ & $2326351146504$ & $\left(\textbf{1,14,15}\right),\left(\textbf{1,8,10}\right),\left(\textbf{3,11,14}\right),\left(0,15,19\right),\left(0,12,17\right),
\left(\textbf{0,6,16}\right).$  \\

\hline $7$ & $746354376511202$ & $\left(\textbf{0,1,19}\right),\left(\textbf{0,2,22}\right),\left(\textbf{4,15,18}\right),\left(\textbf{7,11,20}\right),\left(0,12,17\right),\left(0,10,16\right),\newline \left(7,15,22\right).$  \\

\hline $8$ & $3753811576428246$ & $\left(\textbf{1,15,16}\right),\left(0,20,22\right),\left(\textbf{2,5,14}\right),\left(0,19,23\right),\left(\textbf{5,16,21}\right),
\left(0,18,24\right), \newline \left(0,10,17\right),\left(0,13,21\right).$  \\

\hline $9$ & $736931176845924258$ & $\left(\textbf{7,8,23}\right),\left(\textbf{0,2,25}\right),\left(\textbf{3,14,17}\right),\left(0,20,24\right),\left(0,21,26\right),
\left(0,12,18\right), \newline  \left(\textbf{7,17,24}\right),\left(0,19,27\right),\left(0,13,22\right).$  \\

\hline $10$ & $5262854A674981137A309$ & $\left(\textbf{1,25,26}\right),\left(0,12,14\right),\left(0,26,29\right),\left(\textbf{1,5,22}\right),
\left(0,11,16\right),\left(\textbf{6,19,25}\right), \newline \left(0,20,27\right),\left(0,15,23\right),\left(\textbf{0,9,31}\right),\left(0,18,28\right).$  \\

\hline $11$ & $635A37659B117A842924B08$ & $\left(0,22,23\right),\left(0,28,30\right),\left(0,13,16\right),\left(0,27,31\right),
\left(\textbf{7,12,26}\right),\left(\textbf{6,18,24}\right),\newline \left(\textbf{7,24,31}\right),\left(\textbf{0,8,34}\right),\left(0,20,29\right),\left(0,15,25\right),
\left(0,21,32\right).$  \\

\hline $12$ & $A8531135C8A6B9742624C79B$ & $\left(0,17,18\right),\left(0,29,31\right),\left(0,16,19\right),\left(0,28,32\right),\left(\textbf{6,11,26}\right),\left(\textbf{6,30,36}\right), \newline \left(0,27,34\right),\left(\textbf{8,22,30}\right),\left(\textbf{0,9,35}\right),\left(0,13,23\right),\left(0,25,36\right),\left(0,21,33\right).$  \\

\hline $13$ &$3113692D286C7B9A5847D54CBA$  & $\left(\textbf{5,6,21}\right),\left(\textbf{2,22,24}\right),\left(0,14,17\right),\left(0,32,36\right),\left(0,30,35\right),\left(\textbf{6,24,30}\right), \newline \left(0,26,33\right),\left(0,23,31\right),\left(0,19,28\right),\left(0,29,39\right),\left(0,27,38\right),\left(0,25,37\right), \newline \left(\textbf{0,13,34}\right).$   \\

\hline $14$ & $B36A349C647BDAE9578C25211D80E$ & $\left(0,38,39\right),\left(\textbf{2,37,39}\right),\left(\textbf{2,5,21}\right),\left(0,20,24\right),\left(0,31,36\right),\left(0,17,23\right), \newline \left(0,25,32\right),\left(0,33,41\right),\left(\textbf{0,9,30}\right),\left(0,18,28\right),\left(\textbf{11,26,37}\right),\left(0,22,34\right), \newline \left(0,27,40\right),\left(0,29,43\right).$  \\

\hline $15$ & $11FB479242DE76B9CF86A35D3E85C0A$ & $\left(\textbf{2,3,19}\right),\left(0,23,25\right),\left(\textbf{3,40,43}\right),\left(0,20,24\right),\left(0,38,43\right),\left(0,29,35\right), \newline  \left(0,21,28\right),\left(0,34,42\right),\left(\textbf{9,31,40}\right),\left(0,36,46\right),\left(\textbf{0,11,30}\right),\left(0,32,44\right), \newline \left(0,26,39\right),\left(0,27,41\right),\left(0,18,33\right).$  \\
\hline

\end{tabular}
}
\caption{Skolem and hooked Skolem sequences with the triples for Type $(k)$.}\label{ttypek}
\end{center}
\end{table}
\end{pf}

\begin{comment}
\\
\\
\begin{figure}[htbp]
\begin{center}
\begin{picture}(260,260)
\linethickness{0.9pt}
\put(0,148){$(a)$}
\put(155,148){$(b)$}
\put(78,0){$(c)$}

\put(0,150){\begin{tikzpicture}[scale=1.4]
\path (0:0cm) node[draw,shape=circle] (v0) {\tiny $0$};
\path (65:1cm) node[draw,shape=circle] (v1) {\tiny $c$};
\path (115:1cm) node[draw,shape=circle] (v2) {$$};
\path (75:2cm) node[draw,shape=circle] (v11) {\tiny $a$};
\path (55:2cm) node[draw,shape=circle] (v12) {\tiny $b$};
\path (101:2.3cm) node[draw,shape=circle] (v3) {$$};
\path (90:2.8cm) node[draw,shape=circle] (v4) {$$};
\path (38:2.8cm) node[draw,shape=circle] (v5) {$$};
\path (50:3.1cm) node[draw,shape=circle] (v6) {$$};

  \node[legend_isps] (zz) at (0,0.6) {$1$};
 \node[legend_isps] (zz) at (0.7,1.5) {$2$};
 \node[legend_isps] (zz) at (0,2.4) {$3$};
 \node[legend_isps] (zz) at (1.8,2) {$4$};

\draw
(v0) -- (v1)
(v1) -- (v2)
(v2) -- (v0)
(v1) -- (v11)
(v11) -- (v12)
(v12) -- (v1)
(v11) -- (v3)
(v3) -- (v4)
(v4) -- (v11)
(v12) -- (v5)
(v5) -- (v6)
(v6) -- (v12);
\end{tikzpicture}}

\put(155,150){\begin{tikzpicture}[scale=1.4]

\end{tikzpicture}}

\put(78,5){\begin{tikzpicture}[scale=1.4]

\end{tikzpicture}}
\end{picture}
\caption{Illustration for Type $(k)$ dutch windmill labelling}\label{taypekn}

\end{center}
\end{figure}
\end{comment}

\begin{comment}
\begin{figure}[H]
\begin{center}
\scalebox{0.75}{
\begin{tikzpicture}
    \node[shape=circle,draw=black] (0) at (0,0) {0};
    \node[shape=circle,draw=black] (12) at (2,0) {12};
    \node[shape=circle,draw=black] (1) at (4,0) {1};
    \node[shape=circle,draw=black] (11) at (6,0) {11};

		    \node[shape=circle,draw=black] (9) at (1,2) {9};
		    \node[shape=circle,draw=black] (8) at (3,2) {8};
		    \node[shape=circle,draw=black] (3) at (5,2) {3};

    \path [-] (0) edge node[left,above] {$12$} (12);
    \path [-](9) edge node[left,above] {$3$} (12);
		\path [-] (12) edge node[left,above] {$4$} (8);

    \path [-] (8) edge node[left,above] {$7$} (1);
    \path [-](3) edge node[left,above] {$2$} (1);
		\path [-] (3) edge node[left,above] {$8$} (11);

		 \path [-] (0) edge node[left,above] {$9$} (9);
    \path [-](1) edge node[left,above] {$11$} (12);
		\path [-] (1) edge node[left,above] {$10$} (11);

 \node[shape=circle,draw=black] (2) at (2,4) {2};
 \node[shape=circle,draw=black] (7) at (4,4) {7};
\path [-] (8) edge node[left,above] {$6$} (2);
    \path [-](8) edge node[left,above] {$1$} (7);
		\path [-] (2) edge node[left,above] {$5$} (7);
\end{tikzpicture}
}
\caption{A gracefully labelled Dutch windmill of Type $(k)$ where $n=4$.}
\end{center}
\end{figure}
\end{comment}

\section{Some remarks}

In this paper, we proved Rosa's conjecture for a new family of triangular cacti: Dutch windmills of any order with three pendant triangles. This result, combined with those of Dyer, et al \cite{dyer}, gives the following theorem.

\begin{theorem}
Every Dutch windmill with at most three pendant triangles is graceful or near graceful.
\end{theorem}
\textbf{Question:} Can we gracefully label a dutch windmill consisting of $m$ triangles and $n$ $4$-cycles, all joined at a common point?

\par Langford sequences have been classically used to build new Skolem sequences. In this paper, we used this technique to gracefully label dutch windmills with three pendant triangles. However, in our construction, the Langford sequence implicitly contains a pivot. Thus we can in many cases label the dutch windmills with four pendant triangles, for dutch windmills of a large order.\par
\begin{theorem} There exists $M$ such that for every dutch windmill of order $m > M$ with exactly four pendant triangles where one pendant is attached to a vane containing no other pendants is graceful or near graceful.\end{theorem}
For example, Let $S_{8}=(4,8,5,7,4,1,1,5,6,8,7,2,3,2,6,3)$ be a Skolem sequence and $L_{9}^{17} = (24,17,21,22,18,14,11,19,25,23,10,20,9,16,13,15,12,11,\newline 17,14,10,9,18,21,24,22,19,13,12,16,15,20,23,25)$ be a Langford sequence. Now, if we take the Skolem sequence $S_{8}$ and the Langford sequence $L_{9}^{17},$ then we obtain a new Skolem sequence \newline $S_{25}=(4,8,5,7,4,1,1,5,6,8,7,2,3,2,6,3,24,17,21,22,18,14,11,19,25,23,10,\newline 20,9,16,13,15,12,11,17,14,10,9,18,21,24,22,19,13,12,16,15,20,23,25)$ of order $25.$ If we take that triples obtained from $S_{25}$ and pivot the triples $1,2,4$ and $11$ we can gracefully label the Dutch windmill of order $25$ with four pendant triangles.

\par Furthermore using the Langford sequences technique to gracefully label triangular cacti, we obtained the following theorem.
\begin{theorem}\label{newtheorem}
Let $G$ be a graph on $m$ edges that can be (near) gracefully labelled. Let $x$ be a vertex of $G$ that obtains the label zero under some (near) graceful labelling.
\begin{enumerate}
\item If $G$ is near gracefully labelled we can obtain a new graceful labelled graph $G^{\ast}$ of size $m+3(l+1)$ by adding an $l+1$ triangular vanes at the vertex $x$.

\item If $G$ is gracefully labelled we can obtain a new graceful labelled graph $G^{\ast}$ of size $m+3l$ edges by adding an $l$ triangular vanes at the vertex $x$.

\item If $G$ is gracefully labelled we can obtain a new near graceful labelled graph $G^{\ast}$ of size $m+3l$ by adding an $l$ triangular vanes at the vertex $x$.

\item If $G$ is near gracefully labelled we can obtain a new near graceful labelled graph $G^{\ast}$ of size $m+3(l+1)$ by adding an $l+1$ triangular vanes at the vertex $x$.

\end{enumerate}
 \end{theorem}
 In \cite {linek} Linek and Jiang studied $p$-extended Langford sequences. Here we will present the definition of a $p$-extended Langford sequence because we will use it in the proof of Theorem~\ref{newtheorem}.\par
A \textit{$p-$extended Langford sequence} of defect $d$ and $m$ differences is a sequence $S=\left(s_{1},s_{2},\cdots,s_{2m+1}\right)$ which satisfies these conditions:
\begin{enumerate}
\item for every $k\in\left\{d,d+1,\ldots,d+m-1\right\},$ there exist exactly two elements $s_{i},s_{j}\in S$ such that $s_{i}=s_{j}=k$;

\item if $s_{i}=s_{j}=k,$ with $i < j,$ then $j-i=k$;

\item $s_{p}=0$, where $p=1,2,\ldots,2m+1$.
\end{enumerate}

Now we will present the Proof of Theorem~\ref{newtheorem}.\\
\begin{pf}
We present a proof for the first case. The proof of the other three cases will be similar.\par
Let $G$ be a near gracefully labelled graph and the vertex labels for the near graceful graph $G$ are a subset of $\left\{0,1,2,\ldots,m+1\right\}$ and the edge labels are exactly $\left\{1,2,\ldots,m-1,m+1\right\}.$

\par Form an $p$-extended Langford sequence ($ELS$) with defect $d=m+2$ and the differences $d,d+1,\cdots,d+l-1$ and the extended position at the place $p=l_{2l+1}-(m-1).$ Following the constructions in \cite{linek1}, form an $p$-extended. Form a new sequence $ELS^{\ast}$ with all of the elements from $ELS$ along with the element $m$ at the position $k$ and $l_{2l+2}.$ Then from ($ELS^{\ast}$) we can construct triples of the form $D=\{0,a_{i}+d+l-1,b_{i}+d+l-1\}, 1 \leq i \leq d+l-1$. These base blocks give the vertex labels for $l+1$ triangles, where 0 is repeated $l+1$ times, as a common vertex.
We obtain the vertex labels from the set $\left\{0,d+l,d+l+1,\ldots,d+3l-1\right\},$ where $0$ is a common vertex and edge labels exactly are $\left\{d+l,d+l+1,\ldots,d+3l-1\right\} \cup \left\{m\right\}.$\par
The base blocks formed by the $ELS^{\ast}$ sequence with the original labelling of the graph $G$ give a graceful labelling for a new graph $G^{\ast}$, formed by attaching $l+1$ triangles at $0$ on the graph $G$, to get a gracefully labelled graph $G^{\ast}$.

For the second statement, we use the same technique as in the first with $L$ as a Langford sequence with defect $d=m+1$ and order $l$. For the third statement, we use the same technique as in the first with $hL$ as a hooked Langford sequence with defect $d=m+1$ and order $l$. For the fourth statement, we use the same technique as in the first with $W$ as a hooked extended Langford sequence with defect $d=m+2$ and order $l$ and use $W^{\ast}$.

\end{pf}

\par Let $G=C_{5}$ be a nearly gracefully labelled graph with the edge labels are exactly $\left\{1,2,3,4,6\right\},$ see Figure~\ref{c}. Let $ELS_{7}^{13}=(11,14,15,16,17,\newline 18,19,7,8,9,10,11,12,13,7,14,8,15,9,16,10,17,0,18,12,19,13)$ be an extended Langford sequence and $MELS_{7}^{13} = (11,14,15,16,17,18,19,7,8,9,10,11,\newline 12,13,7,14,8,15,9,16,10,17,\textbf{5},18,12,19,13,\textbf{5})$ be a modified extended Langford sequence.
\par This sequence gives triples of the form $\left(0,a_{i}+19, b_{i}+19\right)$ as follows: $\left(0,42,47\right),\left(0,27,34\right), \left(0,28,36\right),\left(0,29,38\right),\left(0,30,40\right),\left(0,20,31\right),\left(0,32,44\right),\newline\left(0,33,46\right),\left(0,21,35\right),\left(0,22,37\right),\left(0,23,39\right),\left(0,24,41\right),\left(0,25,43\right),\left(0,26,45\right).$
The base blocks formed by the $MELS_{7}^{13}$ sequence with the original labelling of the graph $G=C_{5}$ give a graceful labelling for a new graph $G^{\ast}$, formed by attaching $14$ triangles at $0$ on the graph $G$, to get gracefully labelled graph $G^{\ast}$.
\newline
\par In 1989, Moulton \cite{mou} proved Rosa's conjecture for a triangular snake, a type of triangular cactus whose block cutpoint graph is a path.\par\textbf{ Question:} Can we use Skolem type sequences to gracefully label triangular snakes? Can we use them to gracefully label triangular snakes with pendant triangles?
\section*{Acknowledgement}
Author Danny Dyer acknowledges research grant support from NSERC.

config: arxiv


TITLE: Show that if $\ 7|5a-2$ then $\ 49|a^2-5a-6\ $
QUESTION [1 upvotes]: Show that if $\ 7|5a-2$ then  $\ 49|a^2-5a-6\ $ , ($\ a$ is  positive  integer)
My work:
$7|5a-2 \Rightarrow\ 49|35a-14a,49a^2 \Rightarrow\ 49|14a^2+14 \Rightarrow\ 42a^2+42a,49a^2+49a\  \Rightarrow\ 49|7a^2+7a$
And I stopped here!!
I tried to make the coefficient of $a^2$ equal to 1,but no success...

REPLY [1 votes]: Hint $\ {\rm mod}\ 7\!:\  5a\!-\!2\equiv 5(\color{#c00}{a\!+\!1})\,$ and $\,a^2\!-\!5a\!-\!6\equiv a^2\!+\!2a\!+\!1\equiv (\color{#c00}{a\!+\!1})^2$