\section{Introduction}
A linear \textbf{single change covering design} (\sccd($v,k,t,b$)) $(V,\mathcal{L})$ is a $v$-set $V$ and an ordered list of blocks $\mathcal{L}= (B_1,B_2,...,B_b)$ of size $k$  where every $t$-set, $T$, must occur in at least one block. Additionally each pair of consecutive blocks differs by exactly one element. A \textbf{circular singe change covering design} (CSCCD($v,k,t,b$)) is a \sccd ~where the first and last blocks also differ by one element. If no \dccd($v,k,t,b')$ exists for all $b' < b$ then a \dccd($v,k,t,b$) is called \textbf{{minimal}} 


The earliest discussion of \sccd s was in 1968 and focused on applications to efficient computing \cite{Nelder}. The calculation of $AA^T$ for some matrix requires processing all pairs of rows. In 1968 a limited number of rows could be stored in core memory at one time and updating RAM from a tape was slow. Therefore, visiting all pairs with the smallest number of accesses to the tape improves running time significantly. Similar constraints apply today with respect to CPU cache v.s. RAM access \cite{Tape_Drive, Core_Memory}. 

Gower, Preece, Constable, Zhang, Yucas, Wallis, McSorley and Phillips substantially developed the theory in the 1990s. For a summarized history please see \cite{Chafee_sccd} and \cite{inside}. Today we know constructions for all minimum designs of block size 2,3 and 4 as well as four classes modulo sixteen for block size 5. We also have difference constructions for circular \sccd ~for families with arbitrary $k$ \cite{Chafee_sccd}. 

This previous research only considered changing a single element. This prompts the question,``Could we instead change other fixed numbers of elements between blocks". We call such a design an $m$-change covering design. When $m=2$, these are called \textbf{double change covering designs (DCCD)}. These designs would prove useful in component testing for balancing costs of both testing a configuration and changing the components between tests. Given enough known $m$-change designs, one may choose the most cost efficient testing method. 


Preece et al. introduce the concept of linear double change covering designs and produce an example of a \dccd(9,3,2,12). Nirosh Gangoda Gamachchige and John McSorley begin a formal exploration of double change covering designs in Gangoda Gamachchige's M.Sc and Ph.D. these \cite{gangoda_gamachchige_double-change_2014, gangoda_gamachchige_double-change_2017}. In the M.Sc, Gangoda Gamachchige gives a construction for {\rm tight~CDCCD}($v,3,2,b$) developed using Bose and Skolem sequences on STS. He also gives constructions for {\rm CDCCD}($v,3,2,b$) when $v\equiv 0,2,4,5 \pmod{6}$. During his Ph.D. he examines the regularity of elements in a \dccd.
He finds that for a fixed $k$, linear \dccd($3k-5,k,2,2k-4)$ are minimal and if $2k-3=p^n$ for some prime $p\neq 2$, and $n\geq 1$ then $v-3k-5$ is the smallest $v$ where a linear \dccd($v,k,2,b)$ is minimal.
He shows fixed elements do not exist in economical linear \dccd($v,k,2,b)$ for $v\geq 3k-6$.
He shows that a minimal {\rm CDCCD}($v,k,2,b$) has a fixed element if and only if $v=2k-3$.
He shows all \dccd($v,2,2,b$) are minimal when $v\geq 6$
He shows there are no fixed elements in minimal \dccd($v,4,2,b$) for $v\geq 6$.
He constructs a minimal \dccd($v,4,2,b$) for $v=6,...,14$.   
He gives some alternative constructions that take minimal {\rm CDCCD}($v,4,2,b)$ to minimal {\rm CDCCD}($v',4,2,b')$ and ends by giving some general constructions for {\rm CDCCD}($v,4,2,b$) that are not necessarily minimal. 

In this paper we further explore linear and circular double change covering designs and discuss their constructions using expansion sets and 1-factorizations of complete graphs and difference families. 

\section{Background}

We will be using some graph theory in this work.  A \textbf{{perfect matching} or {1-factor}} of a graph $G$ is a subset of edges $M$ such that every vertex $v\in V$ is adjacent to exactly one edge $e \in M$. A partition of the edges of a graph into perfect matchings is a \textbf{{1-factorization}}. Each color in Figure~\ref{Figure: tikz K_6 1-factor} is a 1-factor in a 1-factorization of $K_6$. A 1-factorization of $K_n$ exists for all even $n$\cite{Factors_and_Factorizations}.

\begin{figure}[H]
\begin{center}
\begin{tikzpicture}
    \graph[circular placement, radius=2cm,
           empty nodes, nodes={circle,draw}]
    {
     \foreach \x in {a,...,f} {
        \foreach \y in {\x,...,f} {
          \x -- \y;
        };
      };
     
     
     
     
     
    };
    
   
   
    \draw [-,cap=rect,line width=1mm,draw=orange] (a)--(d);
    \draw [-,cap=rect,line width=1mm,draw=orange] (b)--(f);
    \draw [-,cap=rect,line width=1mm,draw=orange] (e)--(c);
  
    \draw [-,cap=rect,line width=1mm,draw=magenta] (b)--(d);
    \draw [-,cap=rect,line width=1mm,draw=magenta] (a)--(c);
    \draw [-,cap=rect,line width=1mm,draw=magenta] (e)--(f);
    
    \draw [-,cap=rect,line width=1mm,draw=green] (c)--(d);
    \draw [-,cap=rect,line width=1mm,draw=green] (b)--(e);
    \draw [-,cap=rect,line width=1mm,draw=green] (a)--(f);
    
    \draw [-,cap=rect,line width=1mm,draw=black] (e)--(d);
    \draw [-,cap=rect,line width=1mm,draw=black] (a)--(b);
    \draw [-,cap=rect,line width=1mm,draw=black] (c)--(f);
    
    \draw [-,cap=rect,line width=1mm,draw=cyan] (f)--(d);
    \draw [-,cap=rect,line width=1mm,draw=cyan] (a)--(e);
    \draw [-,cap=rect,line width=1mm,draw=cyan] (b)--(c);
   
    \foreach \x [count=\idx from 0] in {a,...,f} {
      \pgfmathparse{90 + \idx * (360 /6)}
      \node at (\pgfmathresult:2.0cm) [circle,draw=black,fill=white, line width=0.75mm] {\x};
      };

    \end{tikzpicture}
    \end{center}
\caption{1-factorization of $K_6$}
\label{Figure: tikz K_6 1-factor
\end{figure}


Let $(G,+,0)$ be a finite group of order $v$. For $D\subseteq G$ let $\partial D = \{d_1-d_2 : d_1 \neq d_2 \in D\}$. 
A  \textbf{($v,k,\lambda$)-{difference family}} is a collection $\{D_1,...,D_c\}$ of $k$-subsets of $G$, $|G|=v$, where $\partial D_1 \cup ... \cup \partial D_c=\lambda G\backslash\{0\}$. A difference family with $c=1$ is a $(v,k,\lambda)-${difference set}.

\begin{example} \label{Example: (7,3,1)-difference set} \label{Example: (13,3,1)-difference family}
$D= \{1,3,4\}$ is a (7,3,1)-difference set in $\mathbb{Z}_7$.  

$D = \big\{\{0,1,4\}, \{0,2,8\}\big\}$ is a (13,3,1)-difference family over $\mathbb{Z}_{13}$.
\end{example}


\section{Double Change Covering Designs}

A \textbf{double change covering design} (\dccd($v,k,t,b$)) $(V,\mathcal{L})$ is a $v$-set $V$ and an ordered list of blocks $\mathcal{L}= (B_1,B_2,...,B_b)$ of size $k$  where every $t$-set must occur in at least one block. Each pair of consecutive blocks differs by exactly two elements. A \dccd~ is \textbf{circular} if the first and last blocks also differ by two elements. The double change property requires that $v\geq k+2$ if $b>1$. An element $x\in V$ is \textbf{introduced} in a block if $x\in B_i$ and $x \not\in B_{i-1}$. As no block precedes $B_1$ in a linear \dccd~ every element is introduced here. A $t$-set is \textbf{covered} in $B_i$ if at least one element in the $t$-set was introduced in $B_i$.

\sccd~ and \dccd~ can optimize costs in component testing \cite{sccd}. For example, consider testing a device where running a test will cost \$10 and swapping components between tests costs \$1. Using the covering design with 7 blocks, {(1,2,3), (1,4,7), (1,5,6), (3,5,7), (3,4,6), (2,4,5), (2,6,7)} would cost \$91 to run, a SCCD(7,3,2,10) would cost \$112, and a DCCD(7,3,2,7) would cost \$85. If instead, it costs \$1 to run a test and \$5 to switch components these tests cost \$112, \$67, and \$77 respectively. \cite{sccd}


\begin{theorem}
    There are at least $g_1(v,k,t)=\ceiling{\frac{\binom{v}{t}-\binom{k}{t}}{2\binom{k-2}{t-1}+\binom{k-2}{t-2}}}+1$ blocks in a linear \dccd~ and $g_2(v,k,t)=\ceiling{\frac{\binom{v}{t}}{2\binom{k-2}{t-1}+\binom{k-2}{t-2}}}$ blocks in a {\rm CDCCD}.
\end{theorem}
\begin{proof}
    The first block of a linear \dccd~ will cover $\binom{k}{t}$ $t$-sets. Every other block will have two elements introduced. The number of $t$-sets that contain $i$ of the two introduced elements and $t-i$ of the remaining $k-2$ points is $\binom{2}{i}\binom{k-2}{t-i}$. Hence $\sum\limits_{i=1}^{t}\binom{2}{i}\binom{k-2}{t-i}$ $t$-sets are introduced in each block after the first. There are $\binom{v}{t}$ $t$-sets to cover, so $\binom{v}{t}\leq \binom{k}{t}+(b-1)\bigg(\sum\limits_{i=1}^s \binom{2}{i}\binom{k-2}{t-i}\bigg)$. In the circular case, each block covers at most $\sum\limits_{i=1}^t\binom{2}{i}\binom{k-2}{t-i}$ and the calculation is similar. 
\end{proof}

We will be focusing on strength $t=2$ designs for this paper so will drop the parameter $t$ and write \dccd($v,k,b$). If no \dccd($v,k,b')$ exists for all $b' < b$ then a \dccd($v,k,b$) is called \textbf{{minimal}}.  A (circular) \dccd($v,k,b$) is \textbf{{economical}} if it has $\ceiling{g'_1(v,k,2)}$ $\big(\ceiling{g'_2(v,k,2)}\big)$ blocks and \textbf{{tight}} if it is economical and $g_1(v,k,2)$ $\big(g_2(v,k,2)\big)$ is an integer. Any economical \dccd ~is minimum. In a tight \dccd ~every pair is covered exactly once. For some $v,k$ combinations tight \dccd ~can not exist because $g_1$ or $g_2$ do not have integer values, but economical designs can exist. An example of a {\rm CDCCD}(7,3,7) is given in Table~\ref{Table: dccd(7,3,2)}. The blocks are those of $STS(7)$. It is tight as either a \dccd ~or {\rm CDCCD}.

The elements $U_i = x \in B_i \cap B_{i+1}$ form the $i_{th}$ \textbf{unchanged subset}. Since no block proceeds $B_1$ or follows $B_b$ in linear \dccd~ we may consider any $k-2$ subset to be the unchanged subset $U_0$ and $U_b$ respectively. However, circular \dccd~ must maintain the double change between the first and last block, so $U_0=U_b= B_1 \cap B_b$ so we loose any free choice we had. For example the linear \dccd(7,3,7) has the unchanged subsets $U_1 = \{0\}, U_2 = \{4\}, U_3 = \{3\}, U_4 = \{6\}, U_5 = \{6\}, U_6 = \{5\}$ where $U_0$ can be any of $\{0\},\{1\},$ of $\{2\}$ and $U_7$ can be any of $\{1\},\{3\},$ or $\{5\}$. The {\rm CDCCD}(7,3,7) has the same $U_1,...,U_6$ however, $U_0 = U_7 =\{1\}$. If we can partition $V$ into a collection of unchanged subsets, we say the design has an \textbf{expansion set}. In Table~\ref{Table: dccd(7,3,2)}, only the linear \dccd(7,3,7) has an expansion set indicated by carrots, $\land$.

\begin{table}[H]
\centering
\begin{tabular}{lllllll} 
\hspace{.5em}B$_1$    & B$_2$    & B$_3$    & B$_4$    & B$_5$    & B$_6$    & B$_7$    \\
\hspace{.5em}0$^*$ & 0     & 2$^*$ & 0$^*$ & 1$^*$ & 5$^*$ & 5     \\
\hspace{.5em}1 & 4$^*$ & 4     & 6$^*$ & 6     & 6     & 3$^*$ \\
$_{\land}$2$^*$$_{\land}$  & 5$^*$$_{\land}$ & 3$^*$$_{\land}$ & 3     & 4$^*$$_{\land}$ & 2$^*$$_{\land}$ & 1$^*$$_{\land}$
\end{tabular}
\caption{A {\rm tight~CDCCD}(7,3,7)}
\label{Table: dccd(7,3,2)}\label{Table: dccd(7,3,2,7)}
\end{table}







\subsection{Expansion Set Recursion}

Preece et al. use expansion sets to construct a \sccd($v+1,k,b$) from a \sccd($v,k,b$) \cite{tsccd}. This has been extended to CSCCD \cite{circular}. We can do something similar with double change covering designs by utilizing 1-factorizations of a complete graph.

\begin{theorem}\label{Theorem: dccd(v+1,k,2) increase}
Let $(V,\mathcal{B})$ be a \text{\dccd}($v,k,b$) with $v$ an odd multiple of $k-2$ and expansion set. Then a \dccd($v^*, k,b^*$), ($V^*,\mathcal{B}^*$) exists for $v^*=v+\frac{v}{k-2}+1$ and $b^*= b+\frac{v}{k-2}\frac{v+k-2}{2k-4}$. Moreover, if ($V,\mathcal{B}$) is tight, economic or circular then ($V^*,\mathcal{B}^*$) is too. 
\end{theorem}

\begin{proof}
Let ($V,\mathcal{B})$ be a \dccd($v,k,b$) with the expansion set $\mathcal{E} = \{U_{i_j} : 1 \leq j \leq l = \frac{v}{k-2}\}$. Let $\{F_j\}_1^l$ be a 1-factorization of $K_{l+1}$ on set $X$ disjoint from $V$ \cite{Factors_and_Factorizations}. 

For each edge $e\in F_j$ let $B_{j,e}= U_{i_j}\cup e$ for each $j$. Insert all the $B_{j,e}$ in any order at each expansion location ${i_j} \in \mathcal{E}$. These insertions maintain the double change property. All the pairs that were covered in $(V,\mathcal{B})$ are still covered in $\mathcal{B}$. Since each pair of elements from $X$ is in exactly one 1-factor all the pairs in $X$ are covered. The blocks inserted at expansion set locations ensure each pair of elements $\{x,y\}$, $x\in X$, $y\in V$ is covered. So $(V^*,\mathcal{B}^*)$ is a \dccd($v^*,k,b^*$) with $v^*v+\frac{v}{k-2}+1$ and $b^*=b+\frac{v(v+k-2)}{2(k-2)^2}$. 

Each $B_{j,e}$ only covers new pairs and does not effect introductions in preexisting blocks, so if ($V,\mathcal{B}$) was tight or economical so is $(V^*,\mathcal{B}^*)$. The block insertions do not affect the double change between the first and last block, so if $(V,\mathcal{B})$ was circular so is ($V^*,\mathcal{B}^*$). 
\end{proof}

For example consider the linear \dccd(7,3,7) and expansion set in Table~\ref{Table: dccd(7,3,2,7)}. We may insert four new blocks at each expansion set location that consist of the unchanged subset and the vertices of an edge in a 1-factor of $K_8$ in Figure~\ref{Figure: K8} to construct a \dccd(13,3,35) as seen in Table~\ref{Table: DCCD(15,3,2,3)}.

\begin{figure}[H]
    \centering
    \begin{tikzpicture}
        \node[circle,draw=black,fill=white,line width=0.75mm] (15) at (0:0) {15};
        \pgfmathparse{90 + (0 * (360 /7))}
        \node[{minimum size=8mm, circle,draw}] (8) at (\pgfmathresult:2.7) {8};
        \pgfmathparse{90 + (1 * (360 /7))}
        \node[{minimum size=8mm, circle,draw}] (9) at (\pgfmathresult:2.7) {9};
        \pgfmathparse{90 + (2 * (360 /7))}
        \node[{minimum size=8mm, circle,draw}] (10) at (\pgfmathresult:2.7) {10};
        \pgfmathparse{90 + (3 * (360 /7))}
        \node[{minimum size=8mm, circle,draw}] (11) at (\pgfmathresult:2.7) {11};
        \pgfmathparse{90 + (4 * (360 /7))}
        \node[{minimum size=8mm, circle,draw}] (12) at (\pgfmathresult:2.7) {12};
        \pgfmathparse{90 + (5 * (360 /7))}
        \node[{minimum size=8mm, circle,draw}] (13) at (\pgfmathresult:2.7) {13};
        \pgfmathparse{90 + (6 * (360 /7))}
        \node[{minimum size=8mm, circle,draw}] (14) at (\pgfmathresult:2.7) {14};
       
       
       
       
       
       
        {
        \draw [-,cap=rect,line width=1mm,draw=black] (8)--(9);
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        \draw [-,cap=rect,line width=1mm,draw=black] (12)--(15);
        
        \draw [-,cap=rect,line width=1mm,draw=black] (8)--(10);
        \draw [-,cap=rect,line width=1mm,draw=black] (11)--(14);
        \draw [-,cap=rect,line width=1mm,draw=black] (12)--(13);
        \draw [-,cap=rect,line width=1mm,draw=black] (9)--(15);
        
        \draw [-,cap=rect,line width=1mm,draw=black] (8)--(11);
        \draw [-,cap=rect,line width=1mm,draw=black] (9)--(10);
        \draw [-,cap=rect,line width=1mm,draw=black] (12)--(14);
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        \draw [-,cap=rect,line width=1mm,draw=black] (8)--(15);
        \draw [-,cap=rect,line width=1mm,draw=black] (9)--(14);
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        \draw [-,cap=rect,line width=1mm,draw=black] (8)--(12);
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        \draw [-,cap=rect,line width=1mm,draw=black] (8)--(13);
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        \draw [-,cap=rect,line width=1mm,draw=black] (8)--(14);
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        {
        \draw [-,cap=rect,line width=1mm,draw=colour1] (8)--(9);
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        {
        \draw [-,cap=rect,line width=1mm,draw=colour2] (8)--(10);
        \draw [-,cap=rect,line width=1mm,draw=colour2] (11)--(14);
        \draw [-,cap=rect,line width=1mm,draw=colour2] (12)--(13);
        \draw [-,cap=rect,line width=1mm,draw=colour2] (9)--(15);
        }
        
        {
        \draw [-,cap=rect,line width=1mm,draw=colour3] (8)--(11);
        \draw [-,cap=rect,line width=1mm,draw=colour3] (9)--(10);
        \draw [-,cap=rect,line width=1mm,draw=colour3] (12)--(14);
        \draw [-,cap=rect,line width=1mm,draw=colour3] (13)--(15);
        }
        
        {
        \draw [-,cap=rect,line width=1mm,draw=colour4] (8)--(15);
        \draw [-,cap=rect,line width=1mm,draw=colour4] (9)--(14);
        \draw [-,cap=rect,line width=1mm,draw=colour4] (10)--(13);
        \draw [-,cap=rect,line width=1mm,draw=colour4] (11)--(12);
        }

        {
        \draw [-,cap=rect,line width=1mm,draw=colour5] (8)--(12);
        \draw [-,cap=rect,line width=1mm,draw=colour5] (13)--(14);
        \draw [-,cap=rect,line width=1mm,draw=colour5] (9)--(11);
        \draw [-,cap=rect,line width=1mm,draw=colour5] (10)--(15);
        }
        
        {
        \draw [-,cap=rect,line width=1mm,draw=colour6] (8)--(13);
        \draw [-,cap=rect,line width=1mm,draw=colour6] (9)--(12);
        \draw [-,cap=rect,line width=1mm,draw=colour6] (10)--(11);
        \draw [-,cap=rect,line width=1mm,draw=colour6] (14)--(15);
        }
        
        {
        \draw [-,cap=rect,line width=1mm,draw=colour7] (8)--(14);
        \draw [-,cap=rect,line width=1mm,draw=colour7] (9)--(13);
        \draw [-,cap=rect,line width=1mm,draw=colour7] (10)--(12);
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        }
        
       
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            \node at (\pgfmathresult:2.7 cm) [circle,draw=black,fill=white, minimum size=8mm,line width=0.75mm] {\x};
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    \end{tikzpicture}
\caption{K$_8$}
\label{Figure: K8}
\end{figure}

\begin{table}[H]
    \centering
    \setlength{\tabcolsep}{.5\tabcolsep}
    \begin{tabular}{llllllllllllllllll}
    
    \tiny{B$_1$} & \tiny{B$_{2}$} & \tiny{B$_{3}$} & \tiny{B$_{4}$} & \tiny{B$_{5}$} & \tiny{B$_{6}$} & \tiny{B$_{7}$} & \tiny{B$_{8}$} & \tiny{B$_{9}$}  & \tiny{B$_{10}$}  & \tiny{B$_{11}$} & \tiny{B$_{12}$} & \tiny{B$_{13}$} & \tiny{B$_{14}$} & \tiny{B$_{15}$} & \tiny{B$_{16}$} & \tiny{B$_{17}$} & \tiny{B$_{18}$} 
    \\ 
   
    \textcolor{colour1}{8} & \textcolor{colour1}{10} & \textcolor{colour1}{11} & \textcolor{colour1}{12} & 0$^*$ 
    & {0}  & {0}  & {0}  & {0} & 0 
    & \textcolor{colour3}{8}  & \textcolor{colour3}{9}  & \textcolor{colour3}{12} & \textcolor{colour3}{13} & 2$^*$ 
    & \textcolor{colour4}{8}  & \textcolor{colour4}{9} & \textcolor{colour4}{10} \\
   
    \textcolor{colour1}{9} & \textcolor{colour1}{14} & \textcolor{colour1}{13} & \textcolor{colour1}{15} & 1$^*$ 
    & \textcolor{colour2}{8}  & \textcolor{colour2}{11} & \textcolor{colour2}{12} & \textcolor{colour2}{9} & 4$^*$
    & {4}  & {4}  & {4} & {4}  & 4 
    &\textcolor{colour4}{15} & \textcolor{colour4}{14} & \textcolor{colour4}{13}  
     \\
   
    {2} & {2}  & {2}  & {2} & 2 
    & \textcolor{colour2}{10} & \textcolor{colour2}{14} & \textcolor{colour2}{13} & \textcolor{colour2}{15} & 5$^*$ 
    & \textcolor{colour3}{11} & \textcolor{colour3}{10} & \textcolor{colour3}{14} & \textcolor{colour3}{15} & 3$^*$ 
    & {3}  & {3} & {3}
    \\ \hline
   
    \tiny{B$_{19}$} & \tiny{B$_{20}$} & \tiny{B$_{21}$}  & \tiny{B$_{22}$} & \tiny{B$_{23}$} & \tiny{B$_{24}$} & \tiny{B$_{25}$} & \tiny{B$_{26}$} & \tiny{B$_{27}$} & \tiny{B$_{28}$} & \tiny{B$_{29}$} & \tiny{B$_{30}$} & \tiny{B$_{31}$} & \tiny{B$_{32}$} & \tiny{B$_{33}$} & \tiny{B$_{34}$} & \tiny{B$_{35}$} \\
   
    \textcolor{colour4}{11} & 0$^*$ & 1$^*$ 
    & \textcolor{colour5}{8}  & \textcolor{colour5}{13} & \textcolor{colour5}{9}  & \textcolor{colour5}{10} & 5$^*$ 
    & {5}  & {5}  & {5}  & {5} & 5 
    & \textcolor{colour7}{8}  & \textcolor{colour7}{9}  & \textcolor{colour7}{10} & \textcolor{colour7}{11} \\
   
    \textcolor{colour4}{12} & 6$^*$ & 6 
    & {6}  & {6}  & {6} & {6} & 6 
    & \textcolor{colour6}{8}  & \textcolor{colour6}{9}  & \textcolor{colour6}{10} & \textcolor{colour6}{14} & 3$^*$ 
    & \textcolor{colour7}{14} & \textcolor{colour7}{13} & \textcolor{colour7}{12} & \textcolor{colour7}{15} \\
   
    {3}  & 3 & 4$^*$ 
    & \textcolor{colour5}{12} & \textcolor{colour5}{14} & \textcolor{colour5}{11} & \textcolor{colour5}{15} & 2$^*$ 
    & \textcolor{colour6}{13} & \textcolor{colour6}{12} & \textcolor{colour6}{11} & \textcolor{colour6}{15} & 1$^*$ 
    & {1}  & {1}  & {1}  & {1} 
    
    \end{tabular}
    \caption{A \dccd(15,3,35) built from the \dccd(7,3,2) and $K_8$}
    \label{Table: DCCD(15,3,2,3)}\label{Table: DCCD(15,3,2,35)} \label{Table: DCCD(15,3,35}
\end{table}


We note that every new element $x\in X \backslash V$ in Theorem~\ref{Theorem: dccd(v+1,k,2) increase} is always removed one block after it is introduced. Consequently, $x$ is never in an unchanged subset so $\mathcal{B}^*$ never has an expansion set.

\subsection{Difference Methodology}

We can construct {\rm CDCCD}($v,k,b$) using a construction similar to difference sets and families. McSorley proved that $\{0,1,...,k-1\}$ is a base block of a CSCCD$(2k-1,k,2k-1)$ over $\mathbb{Z}_{2k-1}$ in \cite{circular}. This theorem is not stated in terms of difference methods but is equivalent to a difference method construction. In \cite{Chafee_sccd} we use difference methods to construct tight~CSCCD$(2c(k-1)+1,k,c^2 (2 k - 2) + c + 1)$ for all $c\geq 1, k\geq 2$. Here we extend these methods to {\rm CDCCD}.

Every element in the tight~CSCCD$(2k'-1,k',2k'-1)$ is introduced exactly once. If we replace $x\in \mathbb{Z}_{2k'-1}$ in each block with $2x$ and $2x+1$ in $\mathbb{Z}_{4k'-2}$ to produce a list of $2k'-1$ blocks $B_i$ of size $k=2k'$ from $\mathbb{Z}_{2k-2}$ the design is double change. The pairs $\{2x,2x+1\}$ are covered when introduced other pairs $\{y,z\} \subset \mathbb{Z}_{2k-2}$  are covered in the block where $\{\floor{\frac{y}{2}},\floor{\frac{z}{2}}\}$ are covered in the CSCCD. Thus,


\begin{theorem}\label{Theorem: tdcccd(4k-2,2k,2,2k-1) exist}
\label{Theorem: tdcccd(4k-2,2k,2k-1) exist}
\label{Theorem: tdcccd(2k-2,k,2,k-1) exist, k=2k'}
\label{Theorem: tdcccd(2k-2,k,k-1) exist, k=2k'}
A {\rm tight~CDCCD}$(2k-2,k,k-1)$ exists for all $k\geq 2$ if and only if $k$ is even.
\end{theorem}
\begin{proof}
Suppose that $k$ is odd, then $g_2(2k-1,k,2)=\ceiling{\frac{\binom{2k-2}{2}}{2\binom{k-2}{1}+\binom{k-2}{0}}}=k-1$. The discussion before this Theorem constructs these for $k$ even. We know show that $k$ odd cannot exist. Suppose there exists a point, $x$, in every block. There are $2(k-1)=v$ introductions so at least one point $y$ is introduced at least twice, so the pair $x,y$ is covered more then once, a contradiction to tightness. So no point may be in every block. However, there are not enough introductions for any point to be introduced more then once, so each point is introduced exactly once. 

When a point is introduced it covers $k-1$ pairs and each time it persists it covers $2$ more pairs. Thus each point is in $r$ blocks where $k-1+(r-1)2\geq v-1$ so we find $r\geq \frac{k-2}{2}$. Now when $r$ is odd, $r\geq \ceiling{\frac{k-2}{2}}$ which forces repeated pairs, a contradiction. Thus tight~CSCCD(2k-2,k,k-1) do not exist when $k$ is odd.
\end{proof}
For example the {\rm tight~CDCCD}(6,4,3) in Table~\ref{Table: tdcccd(6,4,2,3)} is constructed from the tight~CSCCD(3,2,3) to its left. 


\begin{table}[H]
\centering
\begin{tabular}{llllllllll}
B'$_1$ & B'$_2$ & B'$_3$   &&&&&B$_1$ & B$_2$ & B$_3$\\
0 & 2$^*$ &2            &&&&& 0 & 4$^*$ & 4 \\
1$^*$ & 1 & 0$^*$       &&&&&  1 & 5$^*$ & 5 \\
&&&&&&&                        2$^*$ & 2 & 0$^*$ \\
&&&&&&&                        3$^*_{\land}$ & 3$_{\land}$ & 1$^*_{\land}$
\end{tabular}
\caption{A tight~CSCCD(3,2,3) and tight~CSCCD(6,4,3)}
\label{tdcccd(6,4,2,3)}
\label{Table: tdcccd(6,4,2,3)} \label{Table: tscccd(3,2,2,3)}
\end{table}

\begin{proposition}
A {\rm CDCCD}($2k-2,k,k-1)$ has an expansion set if and only if $k=4$.
\end{proposition}
\begin{proof}
We have $v=2k-2=2(k-2)+2$, so $(k-2)|v$ if and only if $k=3,4$. When $k=3$ the \dccd~ cannot exist as $k$ is odd. When $k=4$ the {\rm CDCCD}(6,4,3) has an expansion set as shown in Table~\ref{Table: tdcccd(6,4,2,3)}.
\end{proof}

Using this in conjunction with the 1-factorization of $K_4$ on points $\{a,b,c,d\}$ and Theorem~\ref{Theorem: dccd(v+1,k,2) increase} we build the {\rm CDCCD}(10,4,9) seen in Table~\ref{Table: tdcccd(10,4,2,9)}.

\begin{table}[H]
\centering
\begin{tabular}{lllllllll}
0 & a$^*$ & c$^*$ & 4$^*$ & 4 & 4 & 4 & a$^*$ & b$^*$ \\
1 & b$^*$ & d$^*$ & 5$^*$ & 5 & 5 & 5 & d$^*$ & c$^*$ \\
2$^*$ & 2 & 2 & 2 & a$^*$ & b$^*$ & 0$^*$ & 0 & 0 \\
3$^*$ & 3 & 3 & 3 & c$^*$ & d$^*$ & 1$^*$ & 1 & 1
\end{tabular}
\caption{A {\rm tight~CDCCD}(10,4,9)}
\label{tdcccd(10,4,2,9)}
\label{Table: tdcccd(10,4,2,9)}
\end{table}




Adding a single new point to every block in the \dccd ~constructed in Theorem~\ref{Theorem: tdcccd(4k-2,2k,2k-1) exist} yields 

\begin{theorem}
A {\rm tight~CDCCD}$(2k-3,k,k-2)$ exists for all odd $k\geq 3$.
\end{theorem}



Beyond manipulating the CSCCD ~by doubling elements, we can directly construct difference sets for {\rm CDCCD}. Example~\ref{Example: dcccd(10,4,9), c=1} will help us gain some intuition for doing this.



\begin{example}\label{Example: dcccd(10,4,9), c=1}
Consider a {\rm CDCCD}(11,4,11). In Figure~\ref{Figure: K11} we show a base block $\{0,1,2,5\}$ and its successor after incrementing by $1$. The vertices only in the first block are highlighted in blue, the vertices only on the second block are highlighted in pink, and the vertices in the unchanged subset in black. We place a line between each introduced element and the other elements in the block, blue for the first block and pink for the second; these are the covered pairs and every difference in $\mathbb{Z}_{11}$ appears. Each subsequent block is obtained by rotating the previous block by one. After developing the blocks fully in $\mathbb{Z}_{11}$ all pairs will be covered. We also see pictorially that there will be double changes between consecutive blocks. Table~\ref{Table: dcccd(11,4,2,11)} shows the table form the {\rm CDCCD}(11,4,11). 


\begin{figure}[H]
    \centering
    \begin{tikzpicture}
        \pgfmathparse{90 + (0 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (0) at (\pgfmathresult:2.7) {0};
        \pgfmathparse{90 + (10 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (1) at (\pgfmathresult:2.7) {1};
        \pgfmathparse{90 + (9 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (2) at (\pgfmathresult:2.7) {2};
        \pgfmathparse{90 + (8 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (3) at (\pgfmathresult:2.7) {3};
        \pgfmathparse{90 + (7 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (4) at (\pgfmathresult:2.7) {4};
        \pgfmathparse{90 + (6 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (5) at (\pgfmathresult:2.7) {5};
        \pgfmathparse{90 + (5 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (6) at (\pgfmathresult:2.7) {6};
        \pgfmathparse{90 + (4 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (7) at (\pgfmathresult:2.7) {7};
        \pgfmathparse{90 + (3 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (8) at (\pgfmathresult:2.7) {8};
        \pgfmathparse{90 + (2 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (9) at (\pgfmathresult:2.7) {9};
        \pgfmathparse{90 + (1 * (360 /11))}
        \node[{minimum size=1mm, circle,draw}] (10) at (\pgfmathresult:2.7) {\tiny{10}};
       
       
       
       
       
       
       
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (0)--(2);
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (0)--(5);
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (1)--(2);
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (1)--(5);
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (2)--(5);
        
       
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (1)--(3);
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (1)--(6);
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (2)--(3);
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (2)--(6);
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (3)--(6);
        
        
       
       
       
       
       
       
        
        {
        \pgfmathparse{90 + (0 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (0) at (\pgfmathresult:2.7) {0};
        \pgfmathparse{90 + (10 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (1) at (\pgfmathresult:2.7) {1};
        \pgfmathparse{90 + (9 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (2) at (\pgfmathresult:2.7) {2};
        \pgfmathparse{90 + (6 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (5) at (\pgfmathresult:2.7) {5};
        }
        {
        \pgfmathparse{90 + (0 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (0) at (\pgfmathresult:2.7) {0};
        \pgfmathparse{90 + (6 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (5) at (\pgfmathresult:2.7) {5};
        \pgfmathparse{90 + (10 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (1) at (\pgfmathresult:2.7) {1};
        \pgfmathparse{90 + (9 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (2) at (\pgfmathresult:2.7) {2};
        \pgfmathparse{90 + (8 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=magenta, text=white}] (3) at (\pgfmathresult:2.7) {3};
        \pgfmathparse{90 + (5 * (360 /11))}
        \node[{minimum size=1mm, circle, draw, fill=magenta, text=white}] (6) at (\pgfmathresult:2.7) {6};
        }
        
    \end{tikzpicture}
\caption{Difference construction for \dccd(11,4,11)}
\label{Figure: K11}
\end{figure}
    
\end{example}

\begin{table}[H]
    \centering
    \begin{tabular}{lllllllllll}
    \small{B$_1$} & \small{B$_{2}$} & \small{B$_{3}$} & \small{B$_{4}$} & \small{B$_{5}$} & \small{B$_{6}$} & \small{B$_{7}$} & \small{B$_{8}$} & \small{B$_{9}$}  & \small{B$_{10}$}  & \small{B$_{11}$} \\
    0 & 3$^*$ & 3 & 3 & 6$^*$ & 6  & 6 & 9$^*$ & 9  & 9  & 1$^*$  \\
    1 & 1 & 4$^*$ & 4 & 4 & 7$^*$  & 7 & 7 & 10$^*$ & 10 & 10 \\
    2$^*$ & 2 & 2 & 5$^*$ & 5 & 5  & 8$^*$ & 8 & 8  & 0$^*$  & 0  \\
    5$^*$ & 6$^*$ & 7$^*$ & 8$^*$ & 9$^*$ & 10$^*$ & 0$^*$ & 1$^*$ & 2$^*$  & 3$^*$  & 4$^*$ 
    \end{tabular}
    \caption{A {\rm CDCCD}(11,4,11) }
    \label{Table: DCCCD(11,4,2,11)}\label{Table: dcccd(11,4,2,11)}\label{Table: dcccd(11,4,11)}
\end{table}









We can construct difference families with up to $5$ base blocks for all $k$. 

\begin{theorem}[{\rm CDCCD}~ difference families with $c$ initial blocks]
\label{Theorem: Initial c blocks for all k >= 3}

A {\rm tight~CDCCD}$(c(4k-6)+1,k,c^2(4k-6)+c)$ exists for all $k\geq 3$ when $c \leq 5$
\end{theorem}

\begin{proof}
To build each of the {\rm tight~CDCCD}$(c(4k-6)+1,k,c^2(4k-6)+c)$, let $A= \{0,1,...,(k-3)\}$ and consider the initial blocks labeled $B_{0,1},...,B_{0,c}$ in order, given in Table~\ref{Table: initial blocks for dcccd(c(4k-6)+1,k,c^2(4k-6)+c)}.

\begin{table}[H]
    \centering
    \begin{tabular}{|l|l|}
        \hline
        \textbf{$c$} & \textbf{Initial Blocks} \\ \hline
    
        1 & $A \cup \{(k-2),2(k-2)+1\}$               \\ \hline
        2 & $A \cup \{2(k-2),5(k-2)+2\}, A \cup \{(k-2),3(k-2)+1\}$               \\ \hline

        3 & $A \cup \{2(k-2),7(k-2)+3\}, A \cup \{3(k-2)+1,5(k-2)+2\},$ \\ & $A \cup \{(k-2),4(k-2)+2\}$              \\ \hline

        4 & $ A \cup \{2(k-2),8(k-2)+4\}, 
        A \cup \{3(k-2)+1,5(k-2)+2\},$ \\ & $
        A \cup \{4(k-2)+2,7(k-2)+4\}, 
        A \cup \{(k-2),6(k-2)+3\}$    \\ \hline

        5 & $ A \cup \{2(k-2),6(k-2)+3\},
        A \cup \{3(k-2)  , 9(k-2)+4\},$ \\ & $ 
        A \cup \{4(k-2)+2, 7(k-2)+4\},
        A \cup \{5(k-2)+3, 8(k-2)+4\},$ \\ & $
        A \cup \{ (k-2)  ,10(k-2)+5\}$      \\ \hline

    \end{tabular}
    \caption{Families of base blocks for {\rm CDCCD}}
    \label{Table: initial blocks for dcccd(c(4k-6)+1,k,c^2(4k-6)+c)}
\end{table}


Each block contains $A$ in addition to two distinct elements. Additionally, $B_{0,c}$ contains $A \cup \{(k-2)\}$, which contains $A$ incremented by 1, thus $(B_{0,1},B_{0,2},...,B_{0,c}, B_{1,1},...,B_{1,c},...,B_{v,c})$ is double change.

Each $B_{0,i}= \{0,1,...,k-3,x,y\}$ covers differences $\pm \{y-x,x-k+3,...,x,y-k+3,...,y\}$, which will cover all the differences $\mathbb{Z}_{v}$. Developing in $\mathbb{Z}_v$ covers all pairs. 
\end{proof}

Example~\ref{Example: dcccd(10,4,9), c=1} is an example of Theorem~\ref{Theorem: Initial c blocks for all k >= 3} when $c=1$ and $k=4$. Let us examine what happens when $c=2$. 

\begin{example}\label{Example: dcccd(21,4,42) when c=2}
When $k=4$ and $c=2$ we construct the {\rm tight~CDCCD}(21,4,42) in Table~\ref{Table: tdcccd(21,4,2,42) highlighted}. In this Table we highlight every $x\in X$ such that the pair $(0,x)$ is covered.

\begin{table}[H]
\centering
\setlength{\tabcolsep}{.5\tabcolsep}
\begin{tabular}{lllllllllllllllllllll}
\small{B}\tiny{$_{0,1}$} & \small{B}\tiny{$_{0,2}$} & \small{B}\tiny{$_{1,1}$} & \small{B}\tiny{$_{1,2}$} & \small{B}\tiny{$_{2,1}$} & \small{B}\tiny{$_{2,2}$} & \small{B}\tiny{$_{3,1}$} & \small{B}\tiny{$_{3,2}$} & \small{B}\tiny{$_{4,1}$} & \small{B}\tiny{$_{4,2}$} & \small{B}\tiny{$_{5,1}$} & \small{B}\tiny{$_{5,2}$} & \small{B}\tiny{$_{6,1}$} & \small{B}\tiny{$_{6,2}$} \\ 

{0}      & {0}      & {5$^*$}  & {3$^*$}  & {3}      & {3}      & {3}      & {3}      & {8$^*$}  & {6$^*$}  & {6}      & {6}      & {6}      & {6}      \\
{1}      & {1}      & {1}      & {1}      & {6$^*$}  & {4$^*$}  & {4}      & {4}      & {4}      & {4}      & {9$^*$}  & {7$^*$}  & {7}      & {7}      \\
\textcolor{cyan}{4$^*$}  & \textcolor{cyan}{2$^*$}  & {2}      & {2}      & {2}      & {2}      & {7$^*$}  & {5$^*$}  & {5}      & {5}      & {5}      & {5}      & {10$^*$} & {8$^*$}  \\
\textcolor{cyan}{12$^*$} & \textcolor{cyan}{7$^*$}  & {13$^*$} & {8$^*$}  & {14$^*$} & {9$^*$}  & {15$^*$} & {10$^*$} & {16$^*$} & {11$^*$} & {17$^*$} & {12$^*$} & {18$^*$} & {13$^*$} \\
\hline 

\small{B}\tiny{$_{7,1}$} & \small{B}\tiny{$_{7,2}$} & \small{B}\tiny{$_{8,1}$} & \small{B}\tiny{$_{8,2}$} & \small{B}\tiny{$_{9,1}$} & \small{B}\tiny{$_{9,2}$} & \small{B}\tiny{$_{10,1}$} & \small{B}\tiny{$_{10,2}$} & \small{B}\tiny{$_{11,1}$} & \small{B}\tiny{$_{11,2}$} & \small{B}\tiny{$_{12,1}$} & \small{B}\tiny{$_{12,2}$} & \small{B}\tiny{$_{13,1}$} & \small{B}\tiny{$_{13,2}$} \\
{11$^*$} & {9$^*$}  & {9}      & {9}      & \textcolor{cyan}{9}      & {9}      & {14$^*$} & {12$^*$} & {12}     & {12}     & {12}     & {12}     & {17$^*$} & {15$^*$} \\

{7}      & {7}      & {12$^*$} & {10$^*$} & \textcolor{cyan}{10}     & {10}     & {10}     & {10}     & {15$^*$} & {13$^*$} & {13}     & {13}     & {13}     & {13}     \\

{8}      & {8}      & {8}      & {8}      & \textcolor{cyan}{13$^*$} & {11$^*$} & {11}     & {11}     & {11}     & {11}     & {16$^*$} & {14$^*$} & {14}     & {14}     \\

{19$^*$} & {14$^*$} & {20$^*$} & {15$^*$} & {0$^*$}  & {16$^*$} & {1$^*$}  & {17$^*$} & {2$^*$}  & {18$^*$} & {3$^*$}  & {19$^*$} & {4$^*$}  & {20$^*$} \\
\hline 

\small{B}\tiny{$_{14,1}$} & \small{B}\tiny{$_{14,2}$} & \small{B}\tiny{$_{15,1}$} & \small{B}\tiny{$_{15,2}$} & \small{B}\tiny{$_{16,1}$} & \small{B}\tiny{$_{16,2}$} & \small{B}\tiny{$_{17,1}$} & \small{B}\tiny{$_{17,2}$} & \small{B}\tiny{$_{18,1}$} & \small{B}\tiny{$_{18,2}$} & \small{B}\tiny{$_{19,1}$} & \small{B}\tiny{$_{19,2}$} & \small{B}\tiny{$_{20,1}$} & \small{B}\tiny{$_{20,2}$} \\

{15}     & \textcolor{cyan}{15}     & {15}     & {15}     & {20$^*$} & {18$^*$} & \textcolor{cyan}{18}     & {18}     & {18}     & {18}     & {2$^*$}  & {0$^*$}  & {0}      & {0}      \\
{18$^*$} & \textcolor{cyan}{16$^*$} & {16}     & {16}     & {16}     & {16}     & {0$^*$}  & {19$^*$} & {19}     & {19}     & {19}     & \textcolor{cyan}{19}     & \textcolor{cyan}{3$^*$}  & \textcolor{cyan}{1$^*$}  \\
{14}     & \textcolor{cyan}{14}     & {19$^*$} & {17$^*$} & {17}     & {17}     & \textcolor{cyan}{17}     & {17}     & {1$^*$}  & {20$^*$} & {20}     & \textcolor{cyan}{20}     & {20}     & {20}     \\
{5$^*$}  & {0$^*$}  & {6$^*$}  & {1$^*$}  & {7$^*$}  & {2$^*$}  & \textcolor{cyan}{8$^*$}  & {3$^*$}  & {9$^*$}  & {4$^*$}  & {10$^*$} & \textcolor{cyan}{5$^*$}  & \textcolor{cyan}{11$^*$} & \textcolor{cyan}{6$^*$} 
\end{tabular}
\caption{A Tight {\rm CDCCD}(21,4,2,42)}
\label{tdcccd(21,4,2,42)}
\label{Table: tdcccd(21,4,2,42)}
\label{tdcccd(21,4,2,42) highlighted}
\label{Table: tdcccd(21,4,2,42) highlighted}
\end{table}
\end{example}

\begin{example} \label{Example: dcccd(13,3,2,26) pictorially}
We may also visualize
multiple base blocks pictorially as we did in Example~\ref{Example: dcccd(10,4,9), c=1}. Consider the base blocks $\{0,1,4\}$ and $\{0,2,7\}$ of a {\rm tight~CDCCD}(13,3,26). These blocks are highlighted light blue and dark blue respectively in Figure~\ref{Figure: tdcccd(13,3,2,26)} and the unchanged subset is highlighted black. All differences are covered in these two blocks, and in any two consecutive blocks. Figure~\ref{Figure: dcccd(13,3,2,26) Step 2} shows how we increment Figure~\ref{Figure: dcccd(13,3,2,26) Step 1} by one to move from block 1 to block 3. The light blue highlighting $B_1$ is incremented by 1 to form the light red $B_3$. 

\begin{figure}[H]
    \centering
    \begin{tikzpicture}
        \pgfmathparse{90 + (0 * (360 /13))}
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        \pgfmathparse{90 + (11 * (360 /13))}
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        \pgfmathparse{90 + (10 * (360 /13))}
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        \pgfmathparse{90 + (7 * (360 /13))}
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        \pgfmathparse{90 + (5 * (360 /13))}
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        \pgfmathparse{90 + (2 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (11) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (1 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (12) at (\pgfmathresult:2.65) {};
       
       
       
       
       
       
        
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (0)--(1);
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (0)--(4);
        \draw [-,cap=rect,line width=0.3mm,draw=cyan] (1)--(4);
        
        \draw [-,cap=rect,line width=0.3mm,draw=blue] (0)--(2);
        \draw [-,cap=rect,line width=0.3mm,draw=blue] (0)--(7);
        \draw [-,cap=rect,line width=0.3mm,draw=blue] (2)--(7);
        
        
        
       
       
        \foreach \x [count=\idx from 0] in {12,...,0} {
            \pgfmathparse{90 + ((\idx+1) * (360 /13))}
            \node at (\pgfmathresult:2.8 cm) [circle,draw=black,fill=white, minimum size=1mm,line width=0.2mm] {\x};
        };
        
        \pgfmathparse{90 + (0 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (0) at (\pgfmathresult:2.8) {0};
        \pgfmathparse{90 + (12 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (1) at (\pgfmathresult:2.8) {1};
        \pgfmathparse{90 + (9 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (4) at (\pgfmathresult:2.8) {4};
        
        \pgfmathparse{90 + (0 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (0) at (\pgfmathresult:2.8) {0};
        \pgfmathparse{90 + (12 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (1) at (\pgfmathresult:2.8) {1};
        \pgfmathparse{90 + (9 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (4) at (\pgfmathresult:2.8) {4};
        \pgfmathparse{90 + (11 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=blue, text=white}] (2) at (\pgfmathresult:2.8) {2};
        \pgfmathparse{90 + (6 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=blue, text=white}] (7) at (\pgfmathresult:2.8) {7};
        
        
    \end{tikzpicture}
\caption{$c=2$, $k=3$ {\rm tight~CDCCD}(13,3,2,26) initial blocks $B_1$ and $B_2$}
\label{Figure: dcccd(13,3,2,26) Step 1}\label{Figure: dcccd(13,3,2,26)} \label{Figure: tdcccd(13,3,2,26)}
\end{figure}

\begin{figure}[H]
    \centering
    \begin{tikzpicture}
        \pgfmathparse{90 + (0 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (0) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (12 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (1) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (11 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (2) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (10 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (3) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (9 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (4) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (8 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (5) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (7 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (6) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (6 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (7) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (5 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (8) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (4 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (9) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (3 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (10) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (2 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (11) at (\pgfmathresult:2.65) {};
        \pgfmathparse{90 + (1 * (360 /13))}
        \node[{minimum size=1mm, circle,draw}] (12) at (\pgfmathresult:2.65) {};
       
       
       
       
       
       
        
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (1)--(2);
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (1)--(5);
        \draw [-,cap=rect,line width=0.3mm,draw=magenta] (2)--(5);
        
        \draw [-,cap=rect,line width=0.3mm,draw=blue] (0)--(2);
        \draw [-,cap=rect,line width=0.3mm,draw=blue] (0)--(7);
        \draw [-,cap=rect,line width=0.3mm,draw=blue] (2)--(7);
        
        
        
       
       
        \foreach \x [count=\idx from 0] in {12,...,0} {
            \pgfmathparse{90 + ((\idx+1) * (360 /13))}
            \node at (\pgfmathresult:2.8 cm) [circle,draw=black,fill=white, minimum size=1mm,line width=0.2mm] {\x};
        };
        
        \pgfmathparse{90 + (0 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (0) at (\pgfmathresult:2.8) {0};
        \pgfmathparse{90 + (12 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (1) at (\pgfmathresult:2.8) {1};
        \pgfmathparse{90 + (8 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=cyan, text=white}] (5) at (\pgfmathresult:2.8) {5};
        
        \pgfmathparse{90 + (0 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=blue, text=white}] (0) at (\pgfmathresult:2.8) {0};
        \pgfmathparse{90 + (12 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=magenta, text=white}] (1) at (\pgfmathresult:2.8) {1};
        \pgfmathparse{90 + (8 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=magenta, text=white}] (5) at (\pgfmathresult:2.8) {5};
        \pgfmathparse{90 + (11 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=black, text=white}] (2) at (\pgfmathresult:2.8) {2};
        \pgfmathparse{90 + (6 * (360 /13))}
        \node[{minimum size=1mm, circle, draw, fill=blue, text=white}] (7) at (\pgfmathresult:2.8) {7};
    \end{tikzpicture}
\caption{$c=2$, $k=3$ {\rm tight~CDCCD}(13,3,2,26) $B_2$ and $B_3$ which is $B_1$ incremented by 1}
\label{Figure: dcccd(13,3,2,26) Step 2}
\end{figure}


\iffalse
Reading the light blue block first and then the dark blue block, we increment both blocks by one and repeat the colouring order. Note that taking every difference in B$_1$ and B$_2$ involving at least one introduced point produces the differences $1,2,...,12 \pmod{13}$. When we increment the base blocks we always have these differences in consecutive blocks. This produces the {\rm tight~CDCCD}(13,3,26) as seen in Table~\ref{Table: tdcccd(13,3,2,26)}. Note here that light blue and dark blue blocks increments to the pink and red blocks respectively.

\begin{table}[H]
    \centering
    \setlength{\tabcolsep}{.5\tabcolsep}
    \begin{tabular}{lllllllllllllllll}
    
    \tiny{\textcolor{cyan}{B$_1$}} & \tiny{\textcolor{blue}{B$_{2}$}} & \tiny{\textcolor{magenta}{B$_{3}$}} & \tiny{\textcolor{red}{$B_{4}$}} & \tiny{B$_{5}$} & \tiny{B$_{6}$} & \tiny{B$_{7}$} & \tiny{B$_{8}$} & \tiny{B$_{9}$}  & \tiny{B$_{10}$}  & \tiny{B$_{11}$} & \tiny{B$_{12}$} & \tiny{B$_{13}$} & \tiny{B$_{14}$} & \tiny{B$_{15}$} & \tiny{B$_{16}$} & \tiny{B$_{17}$}  \\
   
    \textcolor{cyan}{0}$^*$ & \textcolor{blue}{0} & \textcolor{magenta}{1}$^*$ & \textcolor{red}{1} & 2$^*$ & 2 & 3$^*$ & 3 & 4$^*$ & 4 & 5$^*$ & 5 & 6$^*$ & 6 & 7$^*$ & 7 & 8$^*$ \\
   
    \textcolor{cyan}{1} & \textcolor{blue}{2}$^*$ & \textcolor{magenta}{2} & \textcolor{red}{3}$^*$ & 3 & 4$^*$ & 4 & 5$^*$ & 5 & 6$^*$ & 6 & 7$^*$ & 7 & 8$^*$ & 8 & 9$^*$ & 9   \\
   
    \textcolor{cyan}{4}$^*$ & \textcolor{blue}{7}$^*$ & \textcolor{magenta}{5}$^*$ & \textcolor{red}{8}$^*$ & 6$^*$ & 9$^*$ & 7$^*$ & 10$^*$ & 8$^*$ & 11$^*$ & 9$^*$ & 12$^*$ & 10$^*$ & 0$^*$ & 11$^*$ & 1$^*$ & 12$^*$ \\ \hline
   
   \tiny{B$_{18}$} & \tiny{B$_{19}$} & \tiny{B$_{20}$} & \tiny{B$_{21}$}  & \tiny{B$_{22}$} & \tiny{B$_{23}$} & \tiny{B$_{24}$} & \tiny{B$_{25}$} & \tiny{B$_{26}$}\\
   
    8 & 9$^*$ & 9 & 10$^*$ & 10$^*$ & 11$^*$ & 11 & 12$^*$ & 12 \\
   
    10$^*$ & 10 & 11$^*$ & 11 & 12$^*$ & 12 & 0$^*$ & 0 & 1$^*$ \\
   
    2$^*$ & 0 & 3 & 1 & 4 & 2 & 5 & 3 & 6 
    \end{tabular}
    \caption{Tight {\rm CDCCD}(13,3,26)}
    \label{Table: tdcccd(13,3,2,26)}\label{Table: tdcccd(13,3,26)}
\end{table}
\fi

\end{example}


\begin{proposition}
When $(k-2) | 2c+1 $, a {\rm CDCCD} ~constructed via Theorem~\ref{Theorem: Initial c blocks for all k >= 3} has an expansion set.
\end{proposition}

\begin{proof}
In Theorem~\ref{Theorem: Initial c blocks for all k >= 3} the unchanged subsets are all $A+i$, $i\in \mathbb{Z}_v$. When $(k-2)|2c+1$ then $(k-1)|v$. Taking $i\in \{0,k-2,2k-4,...\}$ partitions $\mathbb{Z}_v$ giving the
expansion set $\mathcal{E} = \bigcup\limits_{i=0}^{\frac{4ck-6c+1}{k-2}} \{ U_{ic(k-2)} \}$.
\end{proof}

\begin{corollary} \label{Corollary: Odd number of expansion sets}
If $\frac{2c+1}{k-2}$ is odd, then a {\rm tight~CDCCD}$(c(4k-6)+|\mathcal{E}|+2,k,c^2(4k-6)+c+|\mathcal{E}|\frac{|\mathcal{E}|+1}{2})$ exists.
\end{corollary}
\begin{proof}
The order of the expansion set is $|\mathcal{E}|=\frac{4ck-6c+1}{k-2}=4c+\frac{2c+1}{k-2}$. This is odd if and only if $\frac{2c+1}{k-2}$ is odd then we apply Theorem~\ref{Theorem: dccd(v+1,k,2) increase}. 
\end{proof}

\begin{corollary}
\label{Corollary: tdcccd(15,3,2,35) exists}
\label{Corollary: tdcccd(15,3,35) exists}
\label{Corollary: tdcccd(21,5,30) exists}
\label{Corollary: 11 additional tdcccd using recusion on diff families}

The following tight {\rm CDCCD}~ exist: {\rm CDCCD}(15,3,35), {\rm CDCCD}(21,5,30), {\rm CDCCD}(27,3,117), {\rm CDCCD}(55,7,135), {\rm CDCCD}(39,3,247), {\rm CDCCD}(105,9,364), {\rm CDCCD}(67,3,489), {\rm CDCCD}(77,5,418), {\rm CDCCD}(161,11,640), {\rm CDCCD}(63,3,651),  {\rm CDCCD}(253,13,1386)
\end{corollary}




\begin{proposition}
\label{Proposition: tdcccd(61,4,366) exists}
A {\rm tight~CDCCD}$(61,4,366)$ exists. 
\end{proposition}


\begin{proof}
Consider the initial blocks in order as shown in Table~\ref{Table: c=6, k=4 dcccd(61,4,366) example}. Each contains $A=\{0,1\}$ and $B_{0,c}$ contains $A$ and $A+1$ thus $(B_{0,1},B_{0,2},...,B_{0,c}, $ $B_{1,1},...,B_{1,c},...,B_{v,c})$ is double change. $B_{0,i}= \{0,1,x,y\}$ covers differences $\pm \{y-x,x-1,x,y-1,y\}$ and as $x,y$ range over values in Table~\ref{Table: c=6, k=4 dcccd(61,4,366) example} these cover all differences in $\mathbb{Z}_6$
\end{proof}


\begin{table}[H]
    \centering
    \begin{tabular}{|l|l|l|}
        \hline
        \textbf{$k$}  & \textbf{$c$} & \textbf{Initial Blocks} 
        \\ \hline
         4  & 6 & $ \{0, 1, 4, 19\}, \{0, 1, 6, 22\}, \{0, 1, 8, 25\},\{0, 1, 10, 48\}, \{0, 1, 12, 32\},$ \\
         && $\{0, 1, 2, 28\}$ \\ \hline
          
    \end{tabular}
    \caption{Base Blocks for a {\rm CDCCD}(61,4,366)}
    \label{Table: c=6, k=4 dcccd(61,4,366) example}
\end{table}


\section{Conclusion}
We have proved a recursion that uses 1-factorizations and expansion sets to construct \dccd($v+\frac{v}{k-2}+1,k,b+\frac{v}{k-2}\frac{v+k-2}{2k-4}$) from \dccd($v,k,b$) when $\frac{v}{k-2}$ is odd. 
We use the family of tight~CSCCD($2k'-1,k',2k'-1)$ to build {\rm tight~CDCCD}($2k-2,k,k-1$) and {\rm CDCCD}($2k-3,k,k-2$).
We use difference methods to construct {\rm tight~CDCCD}(61,4,366) and {\rm tight~CDCCD}($c(4k-6)+1,k,c^2(4k-6)+c$) for $k\geq 3$, $1\leq c\leq 5$. When possible we apply the recursion to these to construct eleven additional {\rm CDCCD}.

There are several directions to explore next. Can we develop a version of Theorem~\ref{Theorem: dccd(v+1,k,2) increase} that works when the size of the expansion set is even? 
Can we develop a version of Theorem~\ref{Theorem: dccd(v+1,k,2) increase} to build economic \dccd~ from tight \dccd~ as in \cite{Chafee_sccd}. 
Does there exist recursive construction that uses expansion sets and two \dccd~ to build tight \dccd~ or {\rm CDCCD}~ as in \cite{sccd,Chafee_sccd} respectively? 
McSorely proved  the number of blocks in minimum CSCCD~ as long as $v\leq 2k-1$ \cite{circular}. Can we solve the existence of minimum {\rm CDCCD}~ when $v\leq 2k-2$? Finally, can we find more linear \dccd~ beyond when $k=4$?



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