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2 edition of Dictionary of painting found in the catalog.
Dictionary of painting
Martin L. Wolf
by Martin L. Wolf
Published 1958 by Peter Owen in London .
Painting -- Dictionaries.
Statement by Martin L. Wolf ; with an introduction by Eric Partridge.
Pagination viii, 335 p. ;
Art definition: Art consists of paintings, sculpture, and other pictures or objects which are created | Meaning, pronunciation, translations and examples. A Dictionary of the Art of Printing. William Savage. Longman, Brown, Page - An Act for the Encouragement of Learning, by vesting the copies of printed books in the authors or purchasers of such copies during the times therein mentioned," and by another act 5/5(2).
A work of art made from paint applied to canvas, wood, paper, or another support (noun). Related: Ad Reinhardt. Abstract Painting. André Derain. London Bridge. Andy Warhol. Campbell's Soup Cans. Andy Warhol. Double Elvis. Andy Warhol. Gold Marilyn Monroe. Andy Warhol. Self-Portrait. Barnett Newman. The Voice. A painting either on a wall or on a surface to be attached to a wall. Newsprint: Newspaper stock used for sketching, preliminary drawings and printing. Oil paint: Paint made by mixing ground pigment with oil (usually linseed oil) as a binder. Oil pastel: Oil based crayons. Palette: The surface used to dispense and mix paint on. Paper Mache.
Land O'Lakes, FL: Dealer's Choice Books, Falk, Peter Hastings. Dictionary of Signatures & Monograms of American Artists: From the Colonial Period to the Mid 20th Century. Madison, CT: Sound View Press, Goldstein, Franz. Buy a cheap copy of Dictionary of Art and Artists: book. Free shipping over $
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Dictionary of painting by Martin L. Wolf Download PDF EPUB FB2
A resource for both art and cultural studies, the Dictionary serves as a unique guide to all the visual arts: painting, sculpture, architecture, photography, drawing, printmaking, as well as the decorative arts.5/5(3). The Dictionary ranges A resource for both art and cultural studies, the Dictionary serves as a unique guide to all the visual arts: painting, sculpture, architecture, photography, drawing, printmaking, as well as Dictionary of painting book decorative arts/5(4).
Covering Western art from the ancient world to the present day, the Dictionary provides a careful balance of fact and critical appraisal ranging across painting, sculpture, drawing, and the applied arts.
Over entries provide the reader with information on a wide range of people and subjects, including artists, critics, collectors, dealers, patrons, museums, art schools, materials, and techniques/5(6). The Oxford Dictionary of Art is the unrivalled one-volume guide to the art of the Western world.
It provides a careful balance of fact and critical appraisal, ranging across painting, sculpture, and the graphic arts from classical times to the present. In addition, the book is thoroughly cross-referenced, so readers can easily find additional information on any topic of interest.
Well-written, filled with fascinating historical background and penetrating insight, The Oxford Dictionary of American Art and Artists is an essential one-volume resource for art Author: Ann Lee Morgan.
A noted scholar and historian of art, he was the author of several works, including The Illustrated Dictionary of Symbols in Eastern and Western Art (Westview Press, ).Cited by: This book is a wonderful resource for any student of art history.
To learn how symbols function, recur and morph across time and in the hands of different artists is an absolute pleasure. Hall's fluid, unfussy writing style makes this reference book eminently readable too/5(49).
Get this from a library. The dictionary of art. [Jane Turner;] -- Online ed. provides access to the entplus articles of Grove's Dictionary of art (, 34 vols.) with constant additions of new material and updates to the text, plus extensive image links. Welcome to Benezit Dictionary of Artists A comprehensive and definitive resource for artists' biographies, published since Includesbiographies as well as auction records, exhibition histories, and o images of artists' signatures and stamps of sale.
Welcome to Oxford Art Online. The gateway to Oxford's art reference works, including the peer-reviewed, regularly updated Grove® Dictionary of Art and the Benezit Dictionary of Artists; Includes overarticles that span ancient to contemporary art and architecture, as well as o images of works of art, structures, plans, and artist signatures.
This unique and authoritative reference work contains around 2, clear and concise entries on all aspects of modern and contemporary art. Its impressive range of terms includes movements, styles, techniques, artists, critics, dealers, schools, and galleries. There are biographical entries for artists worldwide from the beginning of the 20th century through to today's globalized art world.
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Boasting well over 6, contributors from 12 countries, the Dictionary offers its readers authoritative and comprehensive global coverage. A resource for both art and cultural studies, the Dictionary serves as a unique guide to all the visual arts: painting, sculpture, architecture, photography, drawing, printmaking, as well as the decorative arts.
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Covering Western art from the ancient Greeks to the present day, this best-selling and authoritative dictionary is more wide-ranging than any comparable reference work. It contains over 2, clear and concise entries on styles and movements, materials and techniques, and museums and galleries.
It also includes biographical entries for artists, critics, collectors, dealers, and patrons, with. Author. Michael Clarke, author. Michael Clarke is Director of the National Gallery of Scotland, and was awarded the CBE in His books include The Tempting Prospect: A Social History of English Watercolours (British Museum Press).
Deborah Clarke was previously an Assistant Keeper at Manchester City Art Galleries. Art Dictionary Fine Art Terms, Online Fine Art Dictionary collection of world star online Fine Art Terms Dictionaries Dictionary of Art Terms Dictionary of Art Terms, Greco-Roman Global Glossary - Classics Technology Center, Department of Fine Arts, Okanagan University College, Art and Architecture Thesaurus, All About Jewels, Encyclopedia of Symbols ArtLex - Art Dictionary used in discussing.
Modernist Venezuelan architect Carlos Raúl Villanueva designed the master plan and major buildings for the Ciudad Universitaria de Caracas, integrating art, architecture, and landscape. In the main auditorium, precast concrete panels produce brilliant patterns of sunlight and shade, and its interior includes Alexander Calder's distinctive installation of acoustic "floating clouds.".
The Oxford Dictionary of Art and Artists Oxford paperback reference: Editor: Ian Chilvers: Publisher: Oxford University Press, ISBN: X, Export. The Benezit Dictionary of Artists (in French, Bénézit: Dictionnaire des peintres, sculpteurs, dessinateurs et graveurs) is an extensive publication of bibliographical information on painters, sculptors, designers and engravers created primarily for art museums, auction houses, historians and dealers.
It was published by Éditions Gründ in Paris but has been sold to Oxford University Press. the art, process, or business of producing books, newspapers, etc., by impression from movable types, plates, etc. Dictionary Of Subjects And Symbols In Art book. Read 13 reviews from the world's largest community for readers.
It is clearly written, well arranged, an /5.The American Heritage Dictionary of the English Language, Fifth Edition: Fiftieth Anniversary Printing.
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Bosnian Islamic Center of Saint Louis
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A statement from a meeting of the Mufti of the Islamic Community in BiH on the upcoming elections
09/26/2018 Islamic Topics, News
Sarajevo, September 26, 2018 (MINA) – The Mufti of the Islamic Community in BiH, at a working meeting in Sarajevo on September 26, 2018, discussed, among other things, the current situation in society and in the context of the upcoming elections. At a meeting with Reis-l-Ulema Hussein ef. Kavazović and deputy reisu-l-ulema Hussein ef. Mufti's Smajić analyzed the role of the Islamic community in the electoral process and the challenges that its imams, bearers of religious authority and members meet.
Particular concern is expressed by the fact that political parties from the region, some officials of neighboring states and their institutions, are involved in possible electoral scams.
In addition, the Islamic Community in Bosnia and Herzegovina has the knowledge that in the smaller BiH entity it is organized in the votes of Bosniak returnees, and that this phenomenon is particularly pronounced in Srebrenica and other zones of committed genocide.
Therefore, the mufti appealed primarily to Muslims, but also to all citizens, not to be subjected to pressures and not to trade their rights and freedoms at the same time by inviting all domestic and international institutions involved in the conduct of elections to do everything that political corruption would not disguise the electoral process. Members of the Mufti Council of Bosnia and Herzegovina once again reminded the members of the Community of the great price of their freedom which they paid as a nation and as citizens of Bosnia and Herzegovina. They emphasized that all the refugees and migrants we meet in these cities in our cities should remind us of what is happening to the nations that lose their freedom and an efficient state that can worry about them.
They especially condemned the hate speech against Islam and Muslims that could be heard even before the official start of the election campaign, as well as attempts to drag the Islamic community and its authorities into election campaigns.
The Islamic Community expects all political subjects to offer concrete solutions to numerous social, economic and life problems of citizens in their election campaigns and during the campaign.
Mufti once again urged all members and members of the Islamic Community to exercise their civil rights and to vote on their own conscience and Islamic morale. They specifically called Bosniaks from Bosnia and Herzegovina's RS entity, if they did not vote by post, together with all other citizens on election day and voted in the places from which they were expelled.
The Mufti sent a message to the bearers of religious authority, above all the Imam, Hatibi and Muderis, not to succumb to political pressure, and to freely and dignistly continue their mission of encouraging good and deterrence from evil, clearly insisting on the promotion of universal Islamic values and morals in our society.
They appealed that citizens, and above all members of the Islamic community, are not subject to populism and to choose those who favor the general good of personal interests. They urged the Muslims to give their voice to those for whom they are convinced that they will preserve the faith, the traditional values of our nation and its state.
Mufti emphasized that Muslims can not choose those who are in politics solely for the sake of achieving their ambitions, which show fake care or behind other goals, as this is the closest way to removing our people from a real political scene.
It is necessary, the attitude is more of a mufti, to choose those who are worthy, responsible, ready for dialogue, credible and committed to helping and building, and not to destroying and confusing the conflict; those who will protect the life, wealth and honor of every human being; who will fight hatred and exclusion, and strive for truth and justice. Trust should be given to those who unambiguously strengthen the statehood of Bosnia and Herzegovina and who wish for every citizen of her country a good and equal right in every part of her territory.
Mufti invite all citizens regardless of their affiliation not to support those political options that sow discouragement and encourage divisions among people and nations. They have also threatened to block election results in the event that certain political options do not achieve their political goals. The state and the people must not be hostages of any political ambitions, and the mufti hope that the election results will be implemented without delay.
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Current Research and Scholarly Interests Inherited mutations in the RUNX1 gene cause a platelet disorder and increased risk of blood cancers. However, it is still unclear what actually causes progression to cancer in these patients. Using genetic editing, I am investigating how RUNX1 mutations contribute to disease.


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African Bank Targets 25 000 Customers A Month
African Bank has moved to set a realistic target for its customer base, due to South Africa's lacklustre economic environment, and has revised it down to 1.7 million customers in 2021 compared with a previous goal of reaching 2.5 million.
African Bank's chief executive, Busani Maluleke, said on Tuesday: "We made the change, because the initial target had been set in 2016, and 2016 was a different time.
"Many of you would agree that we live in a very different South Africa than what we were in 2016. The fact that our gross domestic product growth is pedestrian at best, the political uncertainty, the increasing levels of competition, all of these things together have led us to reconsider what is a much more of a realistic target" noted Maluleke.
He indicated the group's customer base had declined for the first two years after the bank's curatorship, as it had adopted a conservative risk appetite, which meant that the group could not lend to the same customers as it previously did.
"We are proud this has begun to turn around, and we have seen an increase in our customer numbers, and we expect this to accelerate with the launch of the MyWorld account," explained Maluleke.
The group said the MyWorld account, which was launched in May 2019, signalled a new phase in the bank's growth.
African Bank's group executive for digital and transactional banking, George Roussos, who was also at the event, said 100 000 MyWorld accounts had been opened since the launch.
"Our aspirations for 2020 are a good 200percent growth at least. We are aiming at around 25 000 customers a month. We will outperform for 2020," said Roussos.
The group showed the financial highlights for the year included the 11 percent increase in operating profit to R1.15 billion after tax.
African Bank reported a 115 percent growth in retail deposits to R2.4bn from R1.1bn in 2018. Net profit after tax for the year was up 13 percent to R1.15bn from R1.02bn in 2018.
However, the group said its net promoter scores (NPS), which measured customer experience, was disappointing. The group said that its customer NPS had declined to 39 from 43 in 2018, due to the launch of the group's digital channels.
"We have learnt lessons. With anything new that you do there will be teething problems. You will learn a lot. We are learning a lot and we are making the changes that are necessary to give our customers a better customer experience, and we expect to report a better ratio next year," Maluleke said.
African Bank has introduced voice analytics software that helps to mine and analyse audio data, detecting emotion, tone and stress in a customer's and consultant's voice. The company said 29 data scientists, who were enrolled in its DataDistillery@AB in January, would join the company in January 2020 to help in the use of data science. -BusinessReport
Unfunded SAA To Pay 50% Of Salaries
Business Confidence Ticks Up From 20-Year Low
Hey Buddy!, I found this information for you: "African Bank Targets 25 000 Customers A Month". Here is the website link: http://www.theweekly.co.za/african-bank-targets-25-000-customers-a-month/. Thank you.


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\section{Introduction}
\noindent In this paper, we consider only simple and finite graphs. We use $V(G)$ for the vertex set and $E(G)$ for the edge set of a graph $G$. The neighborhood, $N_G(v)$ or shortly $N(v)$ of a vertex $v$ of $G$ is the set of all vertices adjacent to $v$ in $G$. For further graph-theoretic terminology and notation, we refer Bondy and Murty \cite{MR2368647} and Hammack $et$ $al.$\cite{MR2817074}.
\par A Kotzig array \cite{kotzig1971magic}, $KA(a,b)$ is an $a\times b$ array in which every row contains each of the integers $0,1,\cdots, b-1$ exactly once and the sum of the entries in each column is the same constant $c = \frac{a(b - 1)}{2}$. The following theorem can be obtained from \cite{MR3013201}.
\par In 2003, Miller {\cite{MR1999203}} introduced the concept of distance magic labeling of a graph $G$, which is injective function $l : V(G)\rightarrow \{1,2,\cdots,|V(G)|\}$ such that for any $u$ of $G$, the weight $w_G(u)$ of $u$ is a constant $\mu$, where $w_G(u)$ is the sum of labels of all neighbors of $u$. A graph $G$ that admits such a labeling $l$ is called a distance magic graph, or shortly, a $dmg$.
\par The following results provide some necessary condition for distance magicness of regular graphs.
\begin{theorem}\label{oddregular}
	\textnormal{\cite{MR1999203,Rao2004,vilfred}} No $r$-regular graph with $r$-odd can be a distance magic graph.
\end{theorem}
\begin{theorem}\label{2mode4}
	\textnormal{\cite{MR2259699}} Let $EIT(a, r)$ be an equalized tournament with an even number $a$ of teams and $r\equiv 2 \mod 4$. Then $a\equiv 0 \mod 4$.
\end{theorem}
\noindent A simple argument on parity, can prove that an odd regular graph is not distance magic \cite{MR1999203} and therefore, $G=K_{3,3}$ given in Figure \ref{Fig5.1}, is not a distance magic. That is, it is not possible to find a labeling from $V(G)$ to $\{1,2,...,6\}$, with all vertices have same constant weight. But the weights of all vertices, are  a unique constant, when the existing set $\{1,2,\cdots,6\}$ is replaced by certain sets $S_1,S_2$ and $S_3$ with $|S_1|=|S_2|=|S_3|=6$ (see Figure 1).
\begin{figure}[ht]
	\centering
	\includegraphics[width=120mm]{Fig5_1}
	\caption{$S_1,$ $S_2$ and $S_3$-magic graph $G$}
	\label{Fig5.1}
\end{figure}
\par Motivated by the above fact, Godinho et al. \cite{MR3743955} defined the concept of $S$-magic labeling of a graph and introduce certain measure namely, the distance magic index to predict the nature of distance of magicness of a graph.  \begin{definition}
	\textnormal{\cite{MR3743955}} Let $G$ be a graph on $n$ vertices and let $S$ be a set of any $n$ positive integers. Then $G$ is said to be \textit{$S$-magic} if there exists a bijection $l$ from $V(G)$ to $S$ such that for any vertex $u$ of $G$, $\sum_{v\in N(u)} l(v)$ is a constant $\mu'$ and $\mu'$ is the $S$-magic constant.
\end{definition}
\begin{definition}
	\textnormal{\cite{MR3743955}} Let $G$ be a graph on $n$ vertices and let $S$ be a set of $n$ positive integers with the largest number $\eta(S)$. Define $i(G)$ to be the infimum of all $\eta(S)$'s, where the infimum runs overall $S$ for which $G$ admits $S$-magic labeling. Then the \textit{distance magic index} of $G$, $\theta(G)=i(G)-n$.
\end{definition}
In 2015, Godinho et al. \cite{godinho2015s} defined the concept of  fractional domination number and its relationship with $S$-magic graphs.
\begin{definition}\cite{godinho2015s}
    Let $G = (V,E)$ be a graph without isolated vertices. A function $g : V \rightarrow [0, 1]$ is called a total dominating function of $G$ if $g(N(v)) = \sum_{u \in N(v)}  g(u) \ge 1$ for all $v \in V$. The fractional total domination number $\gamma_{ft}(G)$ is defined to be $min\{|g| : g$ is a total dominating function of $G\}$, where $|g| = \sum_{v \in V} g(v)$.
    \end{definition}

\begin{theorem}\textnormal{{\cite{MR1999203,kotlar}}}     Let $1 \le n_1 \le n_2 \le n_3 \le n_4$ and $s_{i} = \sum_{i = 1}^{4} n_i$. Suppose $G$ is either $K_{n_1,n_2}$ or $K_{n_1,n_2,n_3}$ or $K_{n_1,n_2,n_3, n_4}$. There exists a distance magic labeling of $G$ if and only if the following conditions hold.
    \begin{enumerate}
    \item[$(a)$] $n_2 \ge 2$,
    \item[$(b)$] $n(n + 1) \equiv 0$ (mod $2p)$, where $n = s_{p} = |V (K_{n_{1},,...,n_{p}})|$, and
    \item[$(c)$] $\displaystyle \sum_{j=1}^{s_{i}} (n - j + 1) \ge \frac{n(n+1)i}{2p}$ for $1 \le i \le p$,
    \end{enumerate}
    where $p=2$ if $G$ is complete bipartite graph, and $p=3$ if $G$ is complete tripartite graph, and $p=4$ if $G$ is complete four-partite graph.
 \end{theorem}
 \begin{lemma} \textnormal{ \cite{godinho2015s})}
     The complete r-partite graph $G = K_{m_1,m_2,...,n_r}$ is $S$-magic with labeling f if and only if the sum of the labels of all the vertices in any two partite sets are equal.
 \end{lemma}
 \begin{theorem} \textnormal{\cite{MR3743955}}
Let $G$ be the complete bipartite graph $K_{n_{1}, n_{2}}$, where $2 \le n_{1} \le n_{2}$ and $n_{1} + n_{2} = n$. Then\\
    $\displaystyle \theta(G)$ = $\begin{cases}
    0 & \text{ if } n(n+1) \ge 2n_{2}(n_{2} + 1) \textnormal{\it ~and~} n\equiv 0 \text{ or } 3 (\text{mod~} 4)\\
    1 & \text{ if } n(n+1) \ge 2n_{2}(n_{2} + 1) \textnormal{\it ~and~} n\equiv 1 \text{ or } 2 (\text{mod~} 4)\\
     \bigl\lceil \frac{|n(n+1) - 2n_{2}(n_{2} + 1)|}{2n_
     {1}} \bigr\rceil & \text{ if } n(n+1) < 2n_{2}(n_{2} + 1).
    \end{cases}$
 \end{theorem}
    \begin{theorem} \textnormal{\cite{godinho2015s}}
       If a graph $G$ admits an $S$-magic labeling $f$ with magic constant $k$, then $k = \frac{\alpha}{\gamma_{ft}(G)}$, where $\alpha = \sum_{i \in S}i$.
    \end{theorem}
    \begin{corollary} \textnormal{\cite{godinho2015s}}
    If $G$ is $S$-magic, then the smallest $S$-magic constant corresponds to the $S$-magic labeling for which $  \sum_{i \in S}i$ is minimum.
    \end{corollary}
    \par Let $G$ be a graph on $n$ vertices. It is clear that $G$ is distance magic if and only if $\theta(G) = 0$. On the other hand, if $G$ is not $S$-magic, for any $S$ with $|V(G)|$ positive integers, then $\theta(G)$ is not finite. Godinho et al.\cite{MR3743955} identified a parameter for a general graph $G$, defined by $g(x) = \frac{1}{2}[(2\delta(n+x)-\delta^2+\delta)-\Delta(\Delta+1)]$, which can be effectively used to determine the distance magic index of general graphs. From \cite{MR3743955}, one can see that if $g(0)<0$, then $G$ is not a distance magic graph and hence $\theta(G)\geq 1$.
The following result gives a lower bound for $\theta(G)$.
\begin{theorem}\label{godinho}
	\textnormal{\cite{MR3743955}} If $G$ is a graph of order $n$ such that $g(0)<0$, then $\theta(G)\geq \biggl\lceil \frac{|g(0)|}{\delta}\biggr\rceil$.
\end{theorem}
\par However, if $g(0)\geq 0$, there is no result on $G$ to argue whether for that graph, $\theta(G)=0$ or $\theta(G) \geq 1$. Note that if $G$ is an $r$-regular graph with $n$ vertices, then $g(0)>0$ and hence Theorem \ref{godinho} does predict any possible $\theta(G)$.
\section{Distance magic index of complete-tripartite graphs}
In this section, we prove the distance magicness and the magic index of the complete tripartite graph $K_{n_1, n_2, n_3}$, where $V=\bigcup_{i=1}^3 V_i$ and $|V_i|=n_i$ so that $n_1 + n_2 + n_3 = |V|=n$ (say) and the edge set $E =\{uv: u\in V_i,v\in V_j, 1\leq i,j\leq 3 \textnormal{~and~} i\ne j\}$ with $|E |=n_1n_2 + n_2n_3 + n_1n_3$.    Also when $l$ is a mapping from the vertex set of $K_{n_1, n_2, n_3}$ to $S$, where $S$ is set of positive integers with $|S|=|V(K_{n_1, n_2, n_3})|$, we say label set of $V_i$ is $\{l(v) : v\in V_i\}$, which is denoted by $L_i$ and  we use $s(L_i)$ to denote the sum of labels of all the elements in $L_i$, where $i=1,2,3$. Without loss of generality, one can restrict $2 \le n_{1} \le n_{2} \le n_{3}$. For any $i<j$, we denote $\zeta_i^j = \sum_{\gamma=i}^{j} \gamma$. Now, by using the above notation and terminology, we first rewrite the third condition in Theorem 1.3\cite{MR1999203, kotlar} for the complete tripartite graph $K_{n_1, n_2, n_3}$
 \begin{equation}\label{1a}
     \zeta_{n - n_1 + 1}^{n} = \sum_{j=1}^{n_{1}} (n-j+1) = \frac{n_{1}}{2} (2n - n_{1} + 1) \ge \frac{n(n+1)}{6} = \frac{\zeta_1^{n}}{3}
 .\end{equation}
 and,
 \begin{equation*}
      \zeta_{n_3 + 1}^{n} = \sum_{j=1}^{n_{1}+n_{2}} (n-j+1) = \frac{n_{1}+n_{2}}{2} (2n - n_{1} - n_{2} + 1) \ge \frac{n(n+1)}{3} = \frac{2\zeta_1^{n}}{3}
 .\end{equation*}
 Since,
 \begin{equation}\label{2a}
     \frac{n_{1}+n_{2}}{2} (2n - n_{1} - n_{2} + 1) + \frac{n_{3}(n_{3}+1)}{2} = \frac{\zeta_1^{n}}{3} + \frac{2\zeta_1^{n}}{3} = \frac{n(n+1)}{2} = \zeta_1^{n}
     .\end{equation}
     we get,
    \begin{equation}\label{3a}
        3\zeta_1^{n_3} \le \zeta_1^{n}
    .\end{equation}
    From the above equations \eqref{1a} and \eqref{3a}, one can write the following theorem.
    \begin{theorem}
        $K_{n_{1}, n_{2}, n_{3}}$ is distance magic if and only if the following conditions hold.
    \begin{enumerate}
    \item[$1)$] $n_{2} \ge 2$,
    \item[$2)$] $2\zeta_1^n \equiv 0\mod 6$,
    \item[$3)$] $3\zeta_{n-n_1+1}^{n} \ge \zeta_1^{n}$ and,
    \item[$4)$] $3\zeta_{1}^{n_3} \le \zeta_1^{n}$.
    \end{enumerate}
    \end{theorem}
    Note that  Theorem [3.1]  is a rewritten from Theorem [1.3] for the particular case of complete tripartite graphs, $K_{n_{1}, n_{2}, n_{3}}$. Therefore, from above theorem, one can classify $K_{n_{1}, n_{2}, n_{3}}$ into the following five cases.
    \begin{enumerate}
        \item[$i)$] $3\zeta_{n-n_1+1}^{n} \ge  \zeta_1^{n} \ge 3\zeta_{1}^{n_3}   \text{ and }2\zeta_1^{n}  \equiv 0 \text{ mod } 6$.
        \item[$ii)$] $\zeta_1^{n} > 3\zeta_{n-n_1+1}^{n}$ and $\zeta_1^{n} \ge 3\zeta_{1}^{n_3} $.
        \item[$iii)$] $3\zeta_{n-n_1+1}^{n} < \zeta_1^{n}  < 3\zeta_{1}^{n_3}   $.
        \item[$iv)$]  $3\zeta_{n-n_1+1}^{n} \ge  \zeta_1^{n} \ge 3\zeta_{1}^{n_3}  \text{ and } 2\zeta_1^{n}  \equiv 2 \text{ mod } 6$.
        \item[$v)$]  $\zeta_1^{n} \le 3\zeta_{n-n_1+1}^{n}$ and $\zeta_1^{n} < 3\zeta_{1}^{n_3} $.
    \end{enumerate}

    The resulting theorem determines the distance magic index of the complete tripartite graphs for some of the cases mentioned above.
    \begin{lemma}
        Let $K_{n_{1}, n_{2}, n_3}$ be a complete tripartite graph such that $$3\zeta_{n-n_1+1}^{n} \ge  \zeta_1^{n} \ge 3\zeta_{1}^{n_3}  \text{ and } 2\zeta_1^{n} \equiv 2 \mod 6.$$ Then $K_{n_1 + 1, n_2, n_3}$ is a distance magic graph and consequently, $\theta(K_{n_1 + 1, n_2, n_3}) = 0$.
    \end{lemma}
    \begin{proof}    Consider, $6\zeta_{n-n_1+1}^{n+1} = 3(n_1 + 1)(2(n+1) - (n_1+1) + 1)$,
       \begin{eqnarray*}
           3(n_1 + 1)(2(n+1) - (n_1+1) + 1) &=& 3n_1(2n - n_1 + 1) - 6n_1 + 6n\\ &\ge& n(n+1) - 6n_1 + 6n.
       \end{eqnarray*}
        Since $n_1 \le \frac{n}{3}$. Hence,
        \begin{eqnarray*}
            3(n_1 + 1)(2(n+1) - (n_1+1) + 1) &\ge& n(n+1) + 6n - 6n_1\\ &\ge& n(n+1) + 4n\\ &=& n^2 + 5n \\ &>& n^2 + 3n + 2\\ &=& (n+1)(n+2) = 2\zeta_{1}^{n+1}.
        \end{eqnarray*}
        Now consider $6\zeta_{1}^{n_3} = 3n_3(n_3+1)$,
        \begin{eqnarray*}
            3n_3(n_3 + 1) \le n(n+1) \le (n+1)(n+2) = 2\zeta_{1}^{n+1}.
        \end{eqnarray*}

       \noindent Therefore,
        \begin{eqnarray*}
            2\zeta_{1}^{n+1} = (n+1)(n+2)  \cong n(n+1) + 2(n+1) \text{ mod } 6 \cong 2n + 4 \text{ mod } 6.
        \end{eqnarray*}
        Since $2\zeta_{1}^{n} \cong 2$ mod $6$, $2n \cong 2$ mod $6$, we have,
        \begin{equation*}
            2\zeta_{1}^{n+1} \cong 0 \text{ mod } 6
        .\end{equation*}
        Hence by Theorem 3.1, $K_{n_1 + 1, n_2, n_3}$ is a distance magic graph.
        \end{proof}
    \begin{theorem}
       Let G be the complete tripartite graph $K_{n_{1}, n_{2}, n_{3}}$. Then,
\begin{enumerate}
    \item[$(i)$] If  $3\zeta_{n-n_1+1}^{n} \ge  \zeta_1^{n} \ge 3\zeta_{1}^{n_3}   \text{ and }2\zeta_1^{n}  \equiv 0 \text{ mod } 6$, then $\theta(G)=0$.

    \item[$(ii)$] If $\zeta_1^{n} > 3\zeta_{n-n_1+1}^{n}$ and $\zeta_1^{n} \ge 3\zeta_{1}^{n_3} $, then
    \begin{equation*}
        \theta(G) = \begin{cases}
         \biggl\lceil \frac{3\zeta_{1}^{n-n_1} - 2\zeta_{1}^{n}}{2n_{1}} \biggr\rceil & \text{ if } n - n_{1} \equiv 0 \text{ or } 3 \text{ (mod } 4)\\
        \biggl\lceil \frac{3\zeta_{1}^{n-n_1} - 2\zeta_{1}^{n} + 1}{2n_{1}}\biggr \rceil & \text{ if } n - n_{1} \equiv 1 \text{ or } 2 \text{ (mod } 4). \end{cases} .\end{equation*}
    \item[$(iii)$] If
   $3\zeta_{n-n_1+1}^{n} < \zeta_1^{n}  < 3\zeta_{1}^{n_3}$, then
    \begin{equation*}
        \theta(G) \ge  \biggl\lceil \frac{\zeta_{1}^{n_3} - \zeta_{n-n_1+1}^{n}}{n - n_{3}} \biggr\rceil.
    \end{equation*}
    \item[$(iv)$] If $3\zeta_{n-n_1+1}^{n} \ge  \zeta_1^{n} \ge 3\zeta_{1}^{n_3}  \text{ and } 2\zeta_1^{n}  \equiv 2 \text{ mod } 6$,
    \begin{equation*}
        1 \le \theta(G) \le n+1.
    \end{equation*}
    \item[$(v)$] If $\zeta_1^{n} \le 3\zeta_{n-n_1+1}^{n}$ and $\zeta_1^{n} < 3\zeta_{1}^{n_3},$ then
    \begin{equation*}
        \theta(G) =  \biggl\lceil \frac{\zeta_{1}^{n_3} - \zeta_{n-n_1+1}^{n}}{n_{1}} \biggr\rceil,  \text{ if } \frac{\zeta_{1}^{n_3} - \zeta_{n-n_1+1}^{n}}{n_{1}} \ge \theta(H)
    \end{equation*}
    \begin{equation*}
        \theta(G) \le \theta(H) + 1,  \text{ if } \frac{\zeta_{1}^{n_3} - \zeta_{n-n_1+1}^{n}}{n_{1}} < \theta(H),\textnormal{~~~~~~~~}
    \end{equation*}\\
    where $H \cong K_{n_{2}, n_{3}}$.

\end{enumerate}
 \end{theorem}
 \begin{proof}
Let $G=K_{n_{1}, n_{2}, n_{3}}$ be a complete tripartite graph, where  $2 \le n_{1} \le n_{2}\le n_{3}$ and $V =V_1\cup V_2\cup V_3$ with $|V_i|=n_i$, and  $|V | =n_1+n_2+n_3=n$~(say) and with the edge set $E =\{uv: u\in V_i,v\in V_j, 1\leq i,j\leq 3 \textnormal{~and~} i\ne j\}$. Note that $|E|=n_1n_2 + n_2n_3 + n_1n_3$.  For any vertex $v\in V_i$, since the neighborhood of $v$ is $V_j\cup V_k$, the degree of $v$ is  $n_j+n_k$, where $i,j,k$ are three distinct integers and $1 \leq i,j,k\leq 3$.\\
    \noindent \textbf{Proof of case $(i)$} \\ Suppose that $3\zeta_{n-n_1+1}^{n} \ge  \zeta_1^{n} \ge 3\zeta_{1}^{n_3} \text{ and}2\zeta_1^{n}  \equiv 0 \mod 6$. By using Theorem 3.1, we have $\theta(G) = 0$.

     \noindent \textbf{Proof of case $(ii)$} \\ Let $L_{i}$ denote the set of labels of the partition $V_{i}$, where $i=1,2,3$.
    Let the label sets of partitions $V_{1}$, $V_{2}$ and $V_{3}$ be $L_{1} = \{n - n_{1} + 1, \cdots, n\}$, $L_{2} = \{n_{3} + 1,\cdots, n - n_{1}\}$ and $L_{3} = \{1,\cdots, n_{3}\}$ respectively. Let $H$ be a subgraph of $G$ such that $H \cong K_{n_{2},n_{3}}$. Consider,
    \begin{eqnarray*}
        \frac{(n - n_{1})(n - n_{1} + 1)}{2} = \frac{n^{2} + n + n_{1}^{2} - 2nn_{1} - n_{1}}{2}\\
        = \frac{n^{2} + n}{2} - \frac{n_{1}(2n - n_{1} + 1)}{2}.
    \end{eqnarray*}
    Since $3n_{1}(2n - n_{1} +1) < n(n+1)$ and $3n_{3}(n_{3} + 1) \le n(n+1)$,
    \begin{eqnarray}\label{6}
        \frac{(n - n_{1})(n - n_{1} + 1)}{2} > \frac{n^{2} + n}{3} \ge n_{3}(n_{3} + 1).
    \end{eqnarray}
    Hence by Theorem 1.4,\\
    \begin{equation}\label{7}
        \theta(H) = \begin{cases}
    0 & \text{if}  n - n_{1} \equiv 0 \text{or} 3 (\mod 4)\\
    1 & \text{ if }  n - n_{1} \equiv 1 \text{ or } 2 (\mod 4).\\
    \end{cases} \\
    \end{equation}
    \vskip .2cm \noindent Hence by Theorem 1.4 \cite{MR3743955}, the graph $H \equiv K_{n_{2},n_{3}}$ is $S$-magic with $S = \{1, 2,\cdots, n-n_{1}\}$ if $\theta(H) = 0$ and with $S = \{1, 2,\cdots, n-n_{1}-1, n-n_{1}+1\}$ if $\theta(H) = 1$. Hence from this,
\begin{eqnarray}\label{8}
       h = \begin{cases} \frac{(n-n_{1})(n-n_{1}+1)}{4}, & \textnormal{if} n-n_{1} \equiv 0 \textnormal{ or} 3 (\mod 4)\\
        \frac{(n-n_{1})(n-n_{1}+1)+2}{4}, & \textnormal{if} n-n_{1} \equiv 1 \text{or} 2 (\mod 4).
        \end{cases}
\end{eqnarray}
    where $h$ is the magic constant of the subgraph $H \cong K_{n_{2}, n_{3}}$. Note that $h$ is an integer.\newline

    \noindent \textbf{Subcase 1} Let $n - n_{1} \equiv 0$ or $3(\mod 4)$. Then from \eqref{7}, $\theta(H) = 0$. Thus $L_{1} = \{n - n_{1} + 1,\cdots, n\}$, $L_{2}$ is the set of labels of  $V_{2}$ in $H$ and $L_{3}$ is the set of all labels of $V_{3}$ in $H$. Now for the graph $G$ to be $S$-magic, $s(L_{1}) = h$. But $s(L_{1}) = \frac{n_{1}}{2}(2n - n_{1} + 1)$. Therefore, let
    \begin{eqnarray*}
    \centering
     \lambda &=& h - s(L_1) = \frac{(n-n_1)(n-n_1+1)}{4} - \frac{n_1}{2}(2n - n_1 + 1).\\
    \textnormal{Then, }\mbox{~~~~~~~~~} \lambda &=& h - \{s(L_1) + s(L_2) + s(L_3)\} + s(L_2) + s(L_3)\\
          &=& 3h - \frac{n(n+1)}{2}.
    \end{eqnarray*}
    \begin{equation}\label{9}
        \therefore \lambda = \frac{6h - n(n+1)}{2} =  \frac{3(n-n_{1})(n-n_{1}+1)-2n(n+1)}{4}
    .\end{equation}
    From equation \eqref{6}, $\lambda > 0$. Hence, let $\lambda = n_{1}q + r$, where $q \ge 0$ and $0 \le r < n_{1}$. Now, increase each label of $L_{1}$ by $q$ and keep $L_{2}$ and $L_{3}$. Then, $L_{1}^{'} = \{n - n_{1} + q + 1,\cdots, n + q\}$, $L_{2}^{'} = L_{2}$ and $L_{3}^{'} = L_{3}$.
    \vskip .2cm \noindent Therefore, $$s(L_{1}^{'}) = s(L_{1}) + n_{1}q.$$
   \noindent {\bf (i)} Suppose $r=0$. We can label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the labels $L_{1}^{'}$, $L_{2}^{'}$ and $L_{3}^{'}$ respectively. From equation \eqref{8} and \eqref{9}, it is clear that, $s(L_{1}^{'}) = s(L_{2}^{'}) = s(L_{3}^{'})$. Hence by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n+q$, the magic index, $\theta(G) = q$.\\
    {\bf (ii)} Suppose $r>0$, We replace the label $n + q - r + 1 \in L_{1}^{'}$ with the label $n + q - r + 1 + r = n + q + 1$. Hence the resulting labels are $L_{1}^{''} = \{n - n_{1} + q + 1,\cdots, n + q - r, n + q - r +2,\cdots, n + q + 1\}$, $L_{2}^{''} = L_{2}^{'}$ and $L_{3}^{''} = L_{3}^{'}$. Now $s(L_{1}^{''}) = s(L_{1}^{'}) + r = s(L_{1}) + n_{1}q + r$, $s(L_{2}^{''})=s(L_{2})$ and $s(L_{3}^{''}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the labels $L_{1}^{''}$, $L_{2}^{''}$ and $L_{3}^{''}$ respectively.
    \vskip .2cm \noindent From equation \eqref{8} and \eqref{9}, it is clear that $s(L_{1}^{''}) = s(L_{2}^{''}) =s(L_{3}^{''})$. Hence by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + q + 1$, the magic index, $\theta(G) = q + 1$. Therefore,
    \begin{equation*}
        \theta(G) = \biggl\lceil \frac{6h - n(n+1)}{2n_{1}} \biggr\rceil = \biggl\lceil \frac{3(n - n_{1})(n - n_{1} + 1) - 2n(n+1)}{4n_{1}} \biggr\rceil
    .\end{equation*}
    Since the subgraph $H$ is $S$-magic with $\theta(H) = 0$, by Corollary 1.1, $h$ is the minimum magic constant for the subgraph $H$. Therefore, $\theta(G)$ is minimum as it is a linear function of $h$. \newline
    \noindent \textbf{Subcase 2}. Let $n - n_{1} \equiv 1 $ or $2$ (mod $4)$. Hence from \eqref{7}, $\theta(H) = 1$. So, take $L_{1}^{'} = \{n-n_{1}, n-n_{1}+2,\cdots, n\}$, $L_{2}^{'}$ is the set of labels of $V_{2}$ in $H$ and $L_{3}^{'}$ is the set of labels of $V_{3}$ in $H$. Now if $G$ is $S$-magic, then by Lemma 1.1, we must have $s(L_{1}^{'}) = s(L_{2}^{'}) = s(L_{3}^{'})$. But $s(L_{1}^{'}) = s(L_{1}) - 1 = \frac{n_{1}}{2}(2n - n_{1} +1) - 1$. Hence, let
    \begin{eqnarray*}
        \lambda &=& h - s(L_{1}) + 1 = \frac{(n-n_{1})(n-n_{1}+1)+2}{4} - \bigl(\frac{n_{1}}{2}(2n - n_{1} + 1) -1\bigr)   \\
        \implies \lambda &=& h - s(L_{1}) + 1 - s(L_{2}) - s(L_{3}) + s(L_{2}) + s(L_{3}) \\ &=& 3h - \frac{n(n+1)}{2}.\\
       \textnormal{Thus,~~} \lambda &=& \frac{6h - n(n+1)}{2} =  \frac{3(n-n_{1})(n-n_{1}+1)-2n(n+1) + 6}{4}
    \end{eqnarray*}
    \\ \noindent  As in the previous subcase, from equation \eqref{6}, $\lambda > 0$. Hence, let $\lambda - 1 = n_{1}q + r - 1$, where $q \ge 0$ and $0 \le r < n_{1}$.
    \vskip .2cm \noindent  First, let us consider $q > 1$. Now  increase each label of $L_{1}^{'}$ by $q$ and keep $L_{2}^{'}$ and $L_{3}^{'}$ unchanged, we get $L_{1}^{''} = \{n - n_{1} + q, n - n_{1} + 2 + q,\cdots, n + q\}$, $L_{2}^{''} = L_{2}$ and $L_{3}^{''} = L_{3}$. Therefore, $s(L_{1}^{''}) = s(L_{1}) + n_{1}q - 1$. If $r = 0$, we can label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the labels $L_{1}^{''}$, $L_{2}^{''}$ and $L_{3}^{''}$ respectively. From equation \eqref{8} and \eqref{9}, we see that $s(L_{1}^{''}) = s(L_{2}^{''}) = s(L_{3}^{''})$. Hence by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + q$, the magic index, $\theta(G) = q$. Since $\lambda - 1 = n_{1}(q - 1) + n_{1} - 1$, we have the distance magic index, $$\theta(G) = q = \biggl\lceil \frac{\lambda - 1}{n_{1}} \biggr\rceil.$$

    \vskip .2cm \noindent {\bf (i)} Suppose $r$ is $1$. Then replace the label $n - n_{1} + q \in L_{1}^{''}$ with the label $n - n_{1} + q +1$. Hence the resulting labels are $L_{1}^{'''} = \{n - n_{1} + q + 1, n - n_{1} + q + 2, \dots, n + q\}$, $L_{2}^{'''} = L_{2}$ and $L_{3}^{'''} = L_{3}$. Therefore, $s(L_{1}^{'''}) = s(L_{1}) + n_{1}q$, $s(L_{2}^{'''}) = s(L_{2})$ and $s(L_{3}^{'''}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the labels $L_{1}^{'''}$, $L_{2}^{'''}$ and $L_{3}^{'''}$ respectively. From equation \eqref{8} and \eqref{9}, we see that $s(L_{1}^{'''}) = s(L_{2}^{'''}) = s(L_{3}^{'''})$. Hence, by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + q$, the distance magic index $$\theta(G) = q = \biggl\lceil \frac{\lambda - 1}{n_{1}} \biggr\rceil.$$

     \vskip .2cm \noindent {\bf (ii)} Suppose $r>1$.  Then replace the label $n - n_{1} + q \in L_{1}^{''}$ with the label $n - n_{1} + q +1$ and $n + q - r + 2 \in L_{1}^{''}$ with the label $n + q - r + 2 + r - 1 = n + q + 1$. Hence the resulting labels are $L_{1}^{''''} = \{n - n_{1} + 1 + q, n - n_{1} + 2 + q, \dots, n + q - r + 1, n + q - r + 3, ..., n + q + 1\}$, $L_{2}^{''''} = L_{2}^{'}$ and $L_{3}^{''''} = L_{3}^{'}$. Now $s(L_{1}^{''''}) = s(L_{1}^{'}) + r = s(L_{1}) + n_{1}q + r - 1$, $s(L_{2}^{''''}) = s(L_{2})$ and $s(L_{3}^{''''}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the labels sets $L_{1}^{''''}$, $L_{2}^{''''}$ and $L_{3}^{''''}$ respectively. From equation \eqref{8} and \eqref{9}, we see that  $s(L_{1}^{''''}) = s(L_{2}^{''''}) = s(L_{3}^{''''})$. Hence, by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + q + 1$, the distance magic index $\theta(G) = q + 1$. Since $\lambda - 1 = n_{1}q + r - 1$ and $r > 1$, $$\theta(G) = q + 1 = \biggl\lceil \frac{\lambda - 1}{n_{1}} \biggr\rceil.$$

    \noindent Now, if we take $q = 1$, then $\lambda  - 1 = n_{1} + r - 1$. Hence, we can not increase  $n - n_{1} \in L_{1}^{'}$ by 1 as $n - n_{1} + 1 \in L_{2}^{'} \cup L_{3}^{'}$. Hence, increase each label of $L_{1}^{'}-\{n-n_1\}$ by 1, and keep $L_{2}^{'}$ and $L_{3}^{'}$ unchanged and hence, we get, $L_{1}^{'''} = \{n - n_{1}, n - n_{1} + 3,\cdots, n + 1 \}$, $L_{2}^{'''} = L_{2}$ and $L_{3}^{'''} = L_{3}$. Therefore $$s(L_{1}^{'''}) = s(L_{1}) + n_{1} - 1.$$
     \vskip .2cm \noindent {\bf (i)} Suppose $r=0$. Then replace the label $n + 1 \in L_{1}^{'''}$ with $n + 2$. Hence the resulting labels are $L_{1}^{''''} = \{n - n_{1}, n - n_{1} + 3,\cdots, n, n + 2\}$, $L_{2}^{''''} = L_{2}$ and $L_{3}^{''''} = L_{3}$. Therefore, $s(L_{1}^{''''}) = s(L_{1}) + n_{1}$, $s(L_{2}^{''''}) = s(L_{2})$ and $s(L_{3}^{''''}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the labels $L_{1}^{''''}$, $L_{2}^{''''}$ and $L_{3}^{''''}$, respectively. From equation \eqref{8} and  equation \eqref{9}, it is clear that, $s(L_{1}^{''''}) = s(L_{2}^{''''}) = s(L_{3}^{''''})$. Hence by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + 2 = n + q + 1$, the magic index, $\theta(G) = 2 = q+1$. Since $\lambda - 1 = n_{1} + r - 1$, we have $\theta(G) = \lceil \frac{\lambda - 1}{n_{1}} \rceil$.\\
     \vskip .2cm \noindent {\bf (ii)} Suppose $r = 1$. Then replace the label $n - n_{1} \in L_{1}^{'''}$ with $n - n_{1} + 2$. Hence the resulting labels are $L_{1}^{''''} = \{n - n_{1} + 2, n - n_{1} + 3,\cdots, n + 1\}$, $L_{2}^{''''} = L_{2}$ and $L_{3}^{''''} = L_{3}$. Therefore, $s(L_{1}^{''''}) = s(L_{1}) + n_{1}$, $s(L_{2}^{''''}) = s(L_{2})$ and $s(L_{3}^{''''}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the label set $L_{1}^{''''}$, $L_{2}^{''''}$ and $L_{3}^{''''}$, respectively. From \eqref{8} and \eqref{9} it is clear that, $s(L_{1}^{''''}) = s(L_{2}^{''''}) = s(L_{3}^{''''})$. Hence by lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + 1 = n + q $, the magic index, $\theta(G) = 1 = q$. Since $\lambda - 1 = n_{1}$, $\theta(G) = \lceil \frac{\lambda - 1}{n_{1}} \rceil$.\\
     \vskip .2cm \noindent {\bf (iii)} Suppose that $r > 1$.  We replace the label $n - n_{1} \in L_{1}^{'''}$ with the label $n - n_{1} + 2$ and $n - r + 3 \in L_{1}^{'''}$ with $n - r + 3 + r - 1 = n + 2$. Hence the resulting label sets are $L_{1}^{''''} = \{n - n_{1} + 2, n - n_{1} + 3,\cdots, n - r + 2, n - r + 4,\cdots, n +2\}$, $L_{2}^{''''} = L_{2}$ and $L_{3}^{''''} = L_{3}$. Now $s(L_{1}^{''''}) = s(L_{1}) + n_{1} + r$, $s(L_{2}^{''''}) = s(L_{2})$ and $s(L_{3}^{''''}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the labels $L_{1}^{''''}$, $L_{2}^{''''}$ and $L_{3}^{''''}$ respectively. From \eqref{8} and \eqref{9} it is clear that, $s(L_{1}^{''''}) = s(L_{2}^{''''}) = s(L_{3}^{''''})$. Hence by lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + 2 = n + q + 1$, the magic index, $\theta(G) = 2 = q+1$. Since $\lambda - 1 = n_{1} + r - 1$, $\theta(G) = \lceil \frac{\lambda - 1}{n_{1}} \rceil$.\\

    \noindent Now, take $q = 0$. Then $\lambda = r$. Since $\lambda > 0$, we have $r > 0$. If $r = 1$, we replace $n \in L_{1}^{'}$ with $n + 1$. Hence the resulting labels are $L_{1}^{v} = \{n - n_{1}, n - n_{1} + 2,\cdots, n - 1, n + 1\}$, $L_{2}^{v} = L_{2}$ and $L_{3}^{v} = L_{3}$. Therefore, $s(L_{1}^{v}) = s(L_{1}) + 1$, $s(L_{2}^{v}) = s(L_{2})$ and $s(L_{3}^{v}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the label sets $L_{1}^{v}$, $L_{1}^{v}$ and $L_{1}^{v}$, respectively. From \eqref{8} and \eqref{9}, it is clear that $s(L_{1}^{v}) = s(L_{3}^{v}) = S(L_{2}^{v})$. Hence by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + 1$, the magic index, $\theta(G) = 1 = q + 1$. Since $\lambda - 1 = r - 1$, $\theta(G) = \lceil \frac{\lambda - 1}{n_{1}} \rceil$.
    even
    If $r > 1$. we replace the label $n - r + 1 \in L_{1}^{'}$ with $n - r + 1 + r = n + 1$. Hence the resulting labels are $L_{1}^{v} = \{n - n_{1}, n - n_{1} + 2,\cdots,n - r, n - r + 2, \cdots, n + 1\}$, $L_{2}^{v} = L_{2}$ and $L_{3}^{v} = L_{3}$. Therefore, $s(L_{1}^{v}) = s(L_{1}) + r$, $s(L_{2}^{v}) = s(L_{2})$ and $s(L_{3}^{v}) = s(L_{3})$. We label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ with the label sets $L_{1}^{v}$, $L_{1}^{v}$ and $L_{1}^{v}$, respectively. From \eqref{8} and \eqref{9}, one can see that  $s(L_{1}^{v}) = s(L_{3}^{v}) = s(L_{2}^{v})$. Hence by Lemma 1.1, $G$ is $S$-magic. Since the highest label is $n + 1$, the magic index, $\theta(G) = 1 = q + 1$. Since $\lambda - 1 = r - 1$, we have $\theta(G) = \bigl\lceil \frac{\lambda - 1}{n_{1}} \bigr\rceil$.
    \begin{equation*}
        \therefore \theta(G) = \biggl\lceil \frac{6h - n(n+1) - 2}{2n_{1}} \biggr\rceil = \biggl\lceil \frac{3(n - n_{1})(n - n_{1} + 1) - 2n(n+1) + 2}{4n_{1}} \biggr\rceil
    .\end{equation*}
    Since the subgraph $H$ is $S$-magic with $\theta(H) = 1$, by Corollary 2.7.1, $h$ is the minimum magic constant for the subgraph $H$. Therefore, $\theta(G)$ is minimum as it is a linear function of $h$.\\
  \noindent \textbf{Proof of case} (iii). \\   Assume that $3\zeta_{n-n_1+1}^{n} < \zeta_1^{n}  < 3\zeta_{1}^{n_3}$. Then
    \begin{equation*}
        \frac{n_{1}(2n - n_{1} + 1)}{2} < \frac{n_{3}(n_{3} + 1)}{2}
    .\end{equation*}

    Hence, from equation (4), $g(0) < 0$. Therefore, by Lemma 1.2,
    \begin{equation}
        \theta(G) \ge \biggl\lceil \frac{|n_{1}(2n - n_{1} + 1) - n_{3}(n_{3}+1)|}{2(n - n_{3})} \biggr\rceil.
    \end{equation}

\begin{example} Consider the complete tripartite graph $K_{5,6,7}$. Now, by the case $(i)$ of Theorem 2.2. its distance magic index $0$ with magic constant $k = 114$ and it is illustrated in Figure 1.
    \begin{figure}[ht]
        \centering
        \includegraphics[scale=0.1]{k5,6,7.png}
        \caption{Example of case $(i)$, $G \cong K_{5, 6, 7}$, $\theta(G) = 0$, $k = 114$}
    \end{figure}
   On the other hand, Figure 2 gives an $S$-magic labeling of the complete tripartite graph $K_{3,8,9}$ and its distance magic index $7$ with magic constant $k = 154$. Note that by using case $(ii)$ of Theorem 2.2, one can say that the magic index of $K_{3, 8, 9}$ cannot be $0, 1, 2, 3, 4, 5$ and $6$,
       \begin{figure}[ht]
        \centering
    \includegraphics[scale=0.1]{k3,8,9.png}
        \caption{Example of case $(ii)$, $G \cong K_{3, 8, 9}$, $\theta(G) = 7$, $k = 154$}
    \end{figure}\\
    \end{example}
    \noindent \textbf{Proof of case $(iv)$}\\
    Here, $3\zeta_{n-n_1+1}^{n} \ge  \zeta_1^{n} \ge 3\zeta_{1}^{n_3}  \text{ and } 2\zeta_1^{n}  \equiv 2 \text{ mod } 6$. Hence by Theorem 1.3, $\theta(G) \ge 1$. By  Lemma 3.1, one can see that $\theta(K_{n_1 + 1, n_2, n_3}) = 0$.
    Thus, by removing the vertex with maximum label, say $u$, from $V_1$ and adding that label to any other vertex, say $u'$, of $V_1$ such that $u + u' > n$ will make $K_{n_1, n_2, n_3}$, $S$-magic. Since $u \le n + 1$, and $u' \le n$, the highest label, $u + u' \le 2n +1$. Therefore, $\theta(G) \le 2n + 1 - n = n +1$. Hence,
    $1 \le \theta(G) \le n + 1.$
    \newline \noindent But we have to prove the existence of vertices $u$ and $u'$ in $V_1$ such that $u + u' > n$.
    Suppose $u + u' \le n$. Let us consider $S(L_1)$ of graph $K_{n_1+1,n_2,n_3}$
        \begin{eqnarray*}
        s(L_1) &<& (n_1)u + u' \le (n_1-1)u + n\\
            &<& (n_1-1)\frac{n}{2} + n =
            \frac{n(n_1 + 1)}{2}\\
            &\le& \frac{n(n+3)}{6} < \frac{(n+1)(n+2)}{6}.
        \end{eqnarray*}
        Which is a contradiction, Since $s(L_1) = \frac{(n+1)(n+2)}{6}$.\\
    \textbf{Proof of case $(v)$}.\\
    In this case, we have, $\zeta_1^{n} \le 3\zeta_{n-n_1+1}^{n}$ and $\zeta_1^{n} < 3\zeta_{1}^{n_3} $. Suppose that the partitions $V_1$, $V_2$ and $V_3$ are labelled by the set of labels $L_{1} = \{n,\cdots, n - n_{1} + 1\}$, $L_{2} = \{n - n_{1},\cdots, n_{3} + 1\}$ and $L_{3} = \{n_{3},\cdots, 1\}$, respectively. Let $H \cong K_{n_{2}, n_{3}}$. Consider,
      \begin{eqnarray*}
        \frac{(n-n_{1})(n - n_{1} + 1)}{2} &=& \frac{n^2 + n}{2} - \frac{n_{1}(2n - n_{1} + 1)}{2}\\
        &\le& \frac{n^2 + n}{3} \\ &<& n_{3}(n_{3} + 1).
    \end{eqnarray*}
    Hence, by using Theorem 1.4, we have
    \begin{equation}
        \theta(H) = \biggl\lceil \frac{|(n - n_{1})(n - n_{1} + 1) - 2n_{3}(n_{3}+1)|}{2n_{2}} \biggr\rceil
    .\end{equation}
            Let $$\frac{|(n - n_{1})(n - n_{1} + 1) - 2n_{3}(n_{3}+1)|}{2} = n_{2}\mu + r, \quad 0 \le r < n_{2}$$
            Hence $H$ is $S$-magic with $L_{3}^{'} = L_{3}$ and\\
            $L_{2}^{'} =$ $\begin{cases}\{n_{3} + 1 + \mu, \cdots, n - n_{1} + \mu\}, & \text{ if } r = 0 \\
            \{n_{3} + 1 + \mu, \cdots, n - n_{1} + \mu - r, n - n_{1} + \mu - r + 2,\cdots, n - n_{1} + \mu + 1\}, & \text{ if } r > 0.\end{cases}$
            Let $L_{1} ^{'} = \{n - n_{1} + \mu + 2, \cdots, n + \mu + 1\}$. Implies, $s(L_{1}^{'}) = s(L_{1}) + n_even{1}(\mu+1)$.\\
           {\bf Subcase 1}.  Suppose $s(L_{1}^{'}) \le s(L_{3})$. Let $s(L_{3}) - s(L_{1}^{'}) = n_{1}\lambda + r_{1}$ where, $0 \le r_{1} < n_{1}$. This implies, $s(L_{3}) - s(L_{1}) = n_{1}\lambda + n_{1}(\mu+1) + r_{1}$. Now, increase each label of $L_{1}^{'}$ by $\lambda$ and hence, we get $L_{1}^{''} = \{n - n_{1} + \mu + \lambda + 2,\cdots, n + \mu + \lambda + 1\}$. If $r_{1} = 0$, we can label the vertices of the partition $V_{1}$, $V_{2}$ and $V_{3}$ by the label set $L_{1}^{''}$, $L_{2}^{'}$ and $L_{3}$, respectively. Since $s(L_{1}^{''}) = s(L_{2}^{'}) = s(L_{3})$, by Lemma 1.1, the graph $G$ is $S$-magic. Since the highest label is $n + \mu + \lambda + 1$, the magic index $\theta(G) = \mu + \lambda + 1$. If $r_{1} > 0$, replace $n + \mu + \lambda + 2 - r_{1} \in L_{1}^{''}$, with $n + \mu + \lambda + 2$. Thus the resulting label set $L_{1}^{'''} = \{n - n_{1} + \mu + \lambda + 2, \cdots, n + \mu + \lambda - r_{1}+1, n + \mu + \lambda + 3 - r_{1}, \cdots, n + \mu + \lambda + 2\}$. Hence, $s(L_{1}^{'''}) = s(L_{1}^{'}) + n_{1}\lambda + r_{1} = s(L_{2}^{'}) = s(L_{3})$. Therefore, by Lemma 1.1, we label the partitions $V_{1}$, $V_{2}$ and $V_{3}$ respectively by $L_{1}^{'''}$, $L_{2}^{'}$ and $L_{3}$ in such a way that the graph $G$ is an $S$-magic. Since the highest label is $n + \mu + \lambda + 2$, the magic index, $\theta(G) = \mu + \lambda + 2$. Therefore,
    \begin{equation*}
        \theta(G) = \biggl\lceil \frac{2h - n_{1}(2n-n_1+1)}{2n_{1}} \biggr\rceil = \biggl\lceil \frac{n_{3}(n_{3} + 1) - n_1(2n-n_1 + 1)}{2n_{1}} \biggr\rceil
    .\end{equation*}
   {\bf Subcase 2}. Suppose $s(L_{1}^{'}) > s(L_3)$. Hence the maximum highest possible label, is $n + \mu + 1$. Therefore, $\theta(G) \le \mu + 1 \le \theta(H) + 1$ and hence $\theta(G) \le \theta(H) + 1$.
    \end{proof}

\par  Now, we discuss the distance magic index of complete multi-partite graphs and some related graphs. We denote $K(a,b)$ as the complete multi-partite graph with $b$ partitions and each partition having $a$ vertices in it, where $a\geq 1, b\geq 1$. Note that for a graph $G$, the graph $mG$ is the disjoint union of $m$ copies of the graph $G$. Recall that the lexicographic product~\cite{MR2817074} of two graphs $G$ and $H$ is a graph $G \circ H$ with the vertex set $V(G)\times V(H)$ and two vertices $(u,v)$ and $(u',v')$ are adjacent if and only if $u$ is adjacent to $u'$ in $G$, or $u=u'$ and $v$ is adjacent to $v'$ in $H$.
\begin{theorem}
	Let $h \geq 1.$ A Kotzig array $KA(2h, b)$ exists for any $b$;
	$KA(2h + 1, b)$ exists if and only if $b$ is odd.
\end{theorem}
In 2022, Froncek et al.\cite{AVPrajeesh}  introduced the notion of the lifted quasi-Kotzig arrays and used them to construct the quasimagic rectangles as a generalization of magic rectangles.
	\begin{definition} \cite{AVPrajeesh}
		For an odd $a$ and even $b$, a quasi-magic rectangle $QMR(a,b:d)$ is an $a\times b$ array with the entries $1,2,\dots,d-1,d+1,\dots, ab+1$, each appearing exactly once, such that the sum of each row is equal to a constant $\rho$ and the sum of each column is equal to a constant $\sigma$.\end{definition}
	They proved the existence of $QMR(a,b:d)$ for all possible values of $a$ and $b$ and $d=ab/2+1$. When $\gcd(a,b)=1$, and $d=ab/2+1$, they obtained a characterization for the existence of $QMR(a,b:d)$.
	\begin{theorem}\textnormal{\cite{AVPrajeesh}}\label{thm:main}
		{There exist quasimagic rectangles $QMR(a,2t:a t+1)$ for all odd $a$ and $t \geq 1$ with exceptions of $QMR(a,2:a+1)$ for $a \equiv 1(\bmod~4)$. In addition, when $\gcd(a, 2 t)=1$,  then the necessary condition is also sufficient.}
	\end{theorem}
	Note that it is certainly, one of the tools to determine the distance magic index of certain partite graphs.

\section{Quasimagic rectangle: A tool to determine the distance magic index}
\par  In this section, the distance magic indices of the graphs such as $mK(a,b)$, $m(C_{b}\circ\overline{K}_a)$ and $G\circ \overline{K}_a$, are computed, where $G$ is an arbitrary regular graph. These graphs already appear in the literature in the articles \cite{MR2848089,MR1999203,MR2533199}. These articles mainly discuss about the distance magicness of the above mentioned graphs. The existing results of  distance magicness of these graphs, are tabulated below for a quick reference. \\
\noindent Case 1: The distance magicness of $K(a,b)$ is given in Table \ref{Table1}.
\begin{table}[h]
	\centering
	\begin{tabular}{|c|c|c|}
		\hline
		values of $a$ & odd $b$'s & even $b$'s\\
		\hline
		even & $dmg$ \cite{MR1999203} & $dmg$ \cite{MR1999203} \\
		\hline
		odd & $dmg$ \cite{MR1999203} & not~  $dmg$ \cite{MR1999203}  \\
		\hline
	\end{tabular}
	\caption{}
	\label{Table1}
\end{table}
\\ \noindent Case 2: The distance magicness of $mK(a,b),m>1$ is given in Table \ref{Table2}.
\begin{table}[htbp]
 \centering
\begin{tabular}{|c|c|c|c|}
\hline values of $m$ & values of $a$ & odd $b$'s & even $b$'s\\ \hline
odd  & even & $dmg$ \cite{MR2533199}  &$dmg$\cite{MR2533199}\\ \hline
even & even  &  $dmg$ \cite{MR2533199}& $dmg$ \cite{MR2533199} \\	\hline
odd &  odd  &$dmg$ \cite{MR2533199}  &  not $dmg$ \cite{MR2533199}  \\ \hline
 even & odd &  not dmg if $b=4t+1$ \cite{MR2848089}, not dmg if $b=4t+3$ \cite{MR2533199}
& not $dmg$ \cite{MR2533199} \\ \hline
\end{tabular}
\caption{}
\label{Table2}
\end{table}
\newline
Case 3: The distance magicness of $m(C_b\circ\overline{K}_a)$ is given in Table \ref{Table3}.
\begin{table}[htbp]
	\begin{center}
		{\footnotesize \begin{tabular}{|c|c|c|c|}
				\hline
				values of $m$ & values of $a$ & even $b$'s & odd $b$'s\\
				\hline
				odd &  {even} &  $dmg$ \cite{MR2533199}& $dmg$ \cite{MR2533199} \\\hline
				even & even &  $dmg$ \cite{MR2533199} & $dmg$ \cite{MR2533199} \\
				\hline
				odd &  odd   & $dmg$ if $b=4t$, {not $dmg$} if $b=4t+2$  \cite{MR2533199}&  {$dmg$} \cite{MR2533199} \\ \hline
				even & odd & $dmg$ if $b=4t$, {not $dmg$} if $b=4t+2$ \cite{MR2533199}  &  {not $dmg$} \cite{MR2533199} \\
				\hline
		\end{tabular}}
	\end{center}
	\caption{}
	\label{Table3}
\end{table}
\newline
Case 4: If $G$ is an $r$-regular graph on $b$ vertices, the distance magicness of $G\circ\overline{K}_a$ is given in Table \ref{Table4}.
\begin{table}[h]
	\begin{center}
		{\footnotesize \begin{tabular}{ |c|c|c|c|c|}
				\hline
				values of $r$
				& values of $a$ & even $b$'s, $b=4t$ & even $b$'s, $b=4t+2$& odd $b$'s\\
				\hline
				even, $r=4s$ &  odd   & {not yet solved}& { not yet solved} & $dmg$ \cite{MR2848089}\\\hline
				even, $r=4s+2$ & odd & { not yet solved } &  {not $dmg$} \cite{MR2848089} &  $dmg$ \cite{MR2848089}\\hline
				odd & & {not $dmg$}& {not $dmg$}& no such graphs\\\hline
				even & even  & $dmg$\cite{MR1999203} &  $dmg$ \cite{MR1999203}
				&  $dmg$ \cite{MR1999203}
				\\\hline
				odd &even&  $dmg$ \cite{MR1999203}& $dmg$ \cite{MR1999203}& no such graphs \\
				\hline
		\end{tabular}}
	\end{center}
	\caption{}
	\label{Table4}
\end{table}
\begin{theorem}\label{index1hnp} For the graph $K(a,b)$, 	\\
	$\theta(K(a,b)) =$
	$\begin{cases}
	0 &   \text{for $a$ even
		or $a$ and $b$ both odd} \\
	1 &   \text{for $a$ odd and $b$ even except when $a \equiv 1 \bmod 4$ and $b=2$}.
	\end{cases}$
\end{theorem}
\begin{proof}
	If $a$ is even or $a$ and $b$ are both odd, then from Table 1, $\theta(K(a,b))=0$. Now, if  $a$ is odd and $b$ is even except when $a \equiv 1 \mod 4$ and $b=2$, then again from Table 1,   $\theta(K(a,b))\neq 0$.
	\par Now, if $a$ is odd and $b$ is even, then by using Theorem 2.4, construct quasimagic rectangle $QMR(a,b)$ from  the set $S = \{1,2,\cdots,\frac{ab}{2},\frac{ab}{2}+2,\cdots,ab+1\}$. Then, label $i^{th}$  partition of $K(a,b)$ by $i^{\textnormal{th}}$ column of $QMR(a,b)$, where $1\leq i \leq b$. Thus, if $\rho$ is the column sum of a $QMR(a,b)$, then $K(a,b)$ is $S$-magic graph with magic constant $\rho(b-1)$ and therefore, $\theta(K(a,b))$ is 1.
\end{proof}
\begin{theorem}\label{multi6}
	Let $a>1,b>1,m> 1$. Then\\
	$\theta(mK(a,b))$ =
	$\begin{cases}
	0 &   \text{for} a \text{is even $(or)$}
              ~mab \text{~is odd}\\
	1 &  \text{otherwise.}
	\end{cases}$
\end{theorem}
\begin{proof}
	From Table 2, it can be observed that when $m$ is even, and $a$ is odd or $m,a$ are odd and $b$ is even; we have $\theta(mK(a,b))\neq 0$. For the remaining cases $\theta(mK(a,b))= 0$.
	\par For all the cases, when $mK(a,b)$ is not distance magic, by using Theorem 2.4,  construct a $QMR(a,mb)$ with column sums $\rho$.
	Now, use it to label $m$ copies of the graph $K(a,b)$ by choosing $mb$ columns of $QMR(a,mb)$ to the $(mb)$-partitions of $mK(a,b)$.  Hence, an $S$-magic labeling of $mK(a,b)$ with $\mu' =\rho(b-1)$ is obtained. Therefore, $\theta(mK(a,b))=1.$
\end{proof}
\begin{theorem}\label{thm14}
	Let $m\geq 1, a>1$ and $b\geq3$.  Then\\
	\begin{equation*}
	\theta(m(C_{b}\circ \overline{K}_a) =
	\begin{cases}
	0 &   \text{if $a$ is~even $(or)$ $mab$ is odd, $(or)$ $a$ is odd, and $b\equiv 0 \textnormal{~mod~} 4$},\\
	1 &   otherwise.
	\end{cases}
	\end{equation*}
\end{theorem}
\begin{proof}

	If $a$ is even or $mab$ is odd, or $a$ is odd and $b \equiv 0 \bmod 4$, then from Table 3, $\theta(G)=0.$
	Now the remaining cases are given below. \\
	Case 1. $a$ is odd, $m$ is even, $b\equiv 2 \bmod 4.$\\
	Case 2. $a$ is odd, $m$ is odd, $b\equiv 2 \bmod 4.$\\
	Case 3. $a$ is odd, $m$ is even, $b$ is odd.
	\par Now, for all these cases, a $QMR(a,mb)$ is constructed by using Theorem 2.4, with individual column-sum $\rho$ and individual row-sum $\sigma$. The first $b$ columns of $QMR(a,mb)$ is used to label the first copy of  $m(C_{b}\circ \overline{K}_a)$, by labeling   the $i^{th}$ layer of $K_{a}$ of $C_{b}\circ \overline{K}_a$ using the $i^{th}$ column of $QMR(a,mb)$. Similarly, the next $b$ columns are used to label the vertices of the second copy  $C_{b}\circ \overline{K}_a$. If we proceed like this, one can label all copies of $m(C_{b}\circ \overline{K}_a)$, using the column of  $QMR(a,mb)$.  Thus, $m(C_{b}\circ \overline{K}_a)$ is $S$-magic with $\mu' = 2\rho$ and therefore, $\theta(m(C_{b}\circ \overline{K}_a))=1$.
\end{proof}
\begin{theorem}\label{thm15}
	Let $a\ge 1$, and $G$ be an $r$-regular graph on $b$ vertices. Then
	\begin{equation*}
	\theta(G\circ \overline{K}_a) =
	\begin{cases}
	1 &   \text{for $a$ and $r$ are odd except when $a \equiv 1 \bmod 4$ and $b=2$} \\
	1 &   \text{for $a$ is odd, $r\equiv 2 \bmod 4$, $b \equiv 2\bmod 4$},\\
	0 &   \text{for $a$ is even $(or)$ $a$ and $b$ are odd, and $r$ is even}.
	\end{cases}
	\end{equation*}

\end{theorem}
\begin{proof}
	Let $v_{1},v_2,\cdots,v_{b}$ be the vertices of $G$. Now, for any $j\in\{1,2,\cdots,b\}$, let $V_{j} = \{v_{j}^{i}: i\in\{1,2,\cdots, a\}\}$ be set the vertices of $G\circ \overline{K}_a$ that replace the vertex $v_{j}$ of $G$. In fact, the vertex set of $G\circ \overline{K}_a$ is $\displaystyle \bigcup_{j=1}^{b}V_{j}$ and $G\circ \overline{K}_a$ is a $(ra)$-regular graph on $ab$ vertices. \par Now consider the following cases:\\ Case 1. $a$ is even. From Table 4, we get $\theta(G\circ \overline{K}_a)=0$.\\
	Case 2. $a$ is odd. When $r$ is even, and $b$ is odd, from Table 4, we have
	$\theta(G\circ \overline{K}_a)=0$. On the other hand, if $r\equiv 2 \bmod 4$, and $b \equiv 2 \bmod 4$, then from Table 4,   $\theta(G\circ \overline{K}_a)\neq 0$. Finally, if $r$ is odd, then $b$ is even, and by Theorem \ref{oddregular}, $\theta(G\circ \overline{K}_a)\neq0$.\\
	\par Now, for both the cases, when $\theta(G\circ \overline{K}_a)\neq0$,  construct $QMR(a,b)$ using Theorem 2.4. Use the $j^{th}$ column of $QMR(a,b)$ to label the set of vertices, $V_{j}$, for all $j \in \{1,2,\cdots,b\}$. Hence, an $S$-magic labeling of $G\circ \overline{K}_a$, with  $\mu' = r\rho$ is obtained. Therefore, $\theta(G\circ \overline{K}_a)=1.$
\end{proof}
\begin{example}
		A quasimagic rectangle $QMR(3,10)$ is given below. Now, Figure \ref{Fig5.3} illustrates how one can use this $QMR$ to give $S$-magic labeling of the $9$-regular graph $G \circ \overline{K}_3$, where $S$ is the set of all entries of this $QMR$.  \vskip 0.12cm
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
		\hline
   22 & 23 & 5 &  20 & 1 & 10 &  12 & 6 & 30 & 31 \\\hline
		17 & 18 & 19 & 3 & 26 & 27 &  8 & 13 & 14 & 15 \\\hline
		9 & 7 & 24 & 25 & 21 & 11 & 28 & 29 & 4 & 2  \\\hline
\end{tabular}
 \end{center}
\vskip .2cm \noindent One can see that the column sums are 48 and row sums are 160 and moreover, $\theta(G\circ \overline{K}_3)=1.$
\begin{figure}[ht]
	\centering
	\includegraphics[width=120mm]{Fig5_3}
	\caption{An $S$-magic labeling of $G \circ \overline{K}_3$ using $QMR(3,10)$}
	\label{Fig5.3}
\end{figure}
 \end{example}

\section{Conclusion and scope}
In this paper,   the distance magic index of complete tri-partite graphs $K_{n_1,n_2,n_3}$ for all possible values of $n_1,n_2$ and $n_3$, are determined and  Table 5 gives the summary. However, the problem of determining the distance magic index of complete $r$-partite graphs $K_{n_1,n_2,\cdots, n_r}$ is still open, where $r>3$. In last section, the quasimagic rectangle $QMR$ is  identified as useful tool to determine the distance magic indices of certain regular graphs such as disjoint union of $m$ copies of $K(a,b)$, disjoint union of $m$ copies of $C_{b}\circ \overline{K}_a$ and $G \circ {K}_a$, where $G$ is an arbitrary regular graph.

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Ben Axler (Spruce Point Capital Management), Patricia Olasker (Davies), Sunny Puri (Anson Funds), David Salmon (Laurel Hill), moderated by Andrea Mandel-Campbell (NATIONAL Capital Markets)Is Your Company Safe?
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Co-sponsored by NUS (FASS Environment Cluster, and the Institute of Water Policy, Lee Kuan Yew School of Public Policy) and AONBenfield Asia Pte. Ltd.
The Environment Cluster in the National University of Singapore (NUS), Faculty of Arts and Social Sciences (FASS), is an initiative to foster interdisciplinary approaches in environment-related research. The Environment Cluster seeks to catalyze broader collaborative research efforts within the FASS, among different faculties in the NUS and with other international universities and research institutions.
Environmental issues are not only scientific and engineering problems but are also social, economic, cultural and political problems. For this reason, FASS has an important role in the field of environmental research. Environmental problems are generally complex, marked by effects that are economic, social, political and potentially even security-related, as well as the direct effects on the physical environment. . At the same time, political, social, economic and cultural factors drive many of the processes that create environmental problems. Recognizing such complexities, the FASS Environment Cluster seeks to bring together members from different disciplines to engage in meaningful and substantive research within the environmental field.
In order to facilitate such research, the Environment Cluster works to promote and organize the research capacity of the FASS. At the same time, the Environment Cluster also acts as the entry point for various other faculties and institutes for collaborative research efforts.
(5) Upland Ecosystems and Metropolitan Relationships.


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Matthew Farraye, PA-C
Michael W. Geary, M.D
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Katherine M. Merra, M.D.
Deanna Packard, PA-C
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I had knee surgery under the care of Dr. McConville and his team. Everything went great from start to finish. He and his team are prompt, courteous and caring. Appointments were easy to schedule and were on-time. I was amazed at how easy and trouble free the entire process was. I would highly recommend Dr. McConville and his staff at So. Sh. Orthopedics. Excellent Care/Excellent Doctor/Excellent Facility! - Patient of Dr. McConville
The Connection Between Breast Cancer and Bone Loss
Second only to skin cancer, breast cancer is one of the most common cancers in women. It makes up about 30% of all new female cancer cases yearly. Accompanied by intense treatment and lifestyle changes, breast cancer can also adversely affect other body parts.
Ben's Story
In October 2015, Ben suffered a severe bite by a family dog. Ben's mom, Kristy, brought Ben to South Shore Hospital's Pediatric Emergency Department. As the team determined how to treat Ben's bite, Kristy noticed her son was flushed. The team confirmed he had a fever and saw that his hand was swelling. The Trauma team paged John Kadzielski, MD, an orthopedic surgeon and Director of the Division of Hand Surgery at South Shore Hospital, who was on call that day…
Caroline's Story
Positioning herself to perfectly execute a winning shot is something Caroline McCarthy does well. Last summer, however, she never could have imagined that a tie-breaking point in a game would land her in South Shore Hospital's emergency department with a severe wrist fracture….
Nick's Story
When young athletes get injured, they typically embark on a challenging journey that can often take them off the field for some time. Star student-athlete, Nick Hall, understands this journey all too well, when an ordinary practice at the start of his Senior football season at Cohasset High School ended with a torn ACL and meniscus…..
Kevin Connelly of Holbrook was a star hockey player in high school. But he suffered an injury at that time which required knee surgery.
It would be more than 30 years before the lingering effects of that high school injury would become an issue for Kevin….
Peter's Story
Norwell resident Peter Hasham returned to Michael Ayers, MD, for a total knee replacement at the end of January, after having two successful total hip replacements in 2012. "Before my first hip replacement, I remember being fearful, not certain what recovery would be like," Peter said. "However, because my experience was so amazing, I was actually excited to have my knee surgery, remembering how quickly I became pain-free after my hip surgeries," Peter said…..
Elsa's Story
The cold never bothered her anyway…but the hip pain did. Elsa Gould decided she had had enough of the agony associated with her arthritic right hip.
"It hurt even to lie down and try to sleep," she said…..
Tom's Story
Plymouth resident, Tom Silvia is no stranger to pain and endurance. In fact, the now active 68-year-old suffered from Polio at the age of six. Initially paralyzed from the neck down, Tom endured years of therapy, spent a week in an iron lung and worked his way from a wheelchair to crutches in a year's time…..
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April 29, 2012 General Ag News and Notes Post a Comment
California braces for a deadly stalker of citrus
LOS ANGELES >> The worst disease known to the citrus industry may have arrived in California on a bud of friendship.
A graft of pomelo — a symbol of good fortune and prosperity in many Asian cultures — was the likely source of the state's first documented case of huanglongbing, a citrus disease with no known cure, say researchers involved in the investigation. The suspected plant shoot, or budwood, was passed freely among San Gabriel Valley church friends who loved to garden and experiment with hybridization, according to residents.
Until a month ago, California was the last major citrus-growing region in the world to avoid a scourge that has decimated groves in China, Brazil and Florida. The disease arrived the way experts had long predicted: in a tree in a Southern California yard. Now, agriculture officials are scrambling to slow the disease's march north and save a $2 billion industry based in the Central Valley.
Authorities launched a massive containment effort involving quarantines, pesticides and public hearings when a lemon-pomelo tree in Mary Wang's lush Hacienda Heights yard tested positive for the disease on March 30. The sickly looking tree was quickly removed for study.
June 9, 2011 General Ag News and Notes Post a Comment
Hawaii officials looking for stinging caterpillar
Hawaii agriculture officials are asking for the public's help in spotting infestations of the stinging nettle caterpillar, which appears to have recently spread to Kauai.
The state Department of Agriculture said Wednesday Kauai residents may begin to see more of the bugs during the summer, the peak months for the species.
The Big Island, Maui, and Oahu already have established populations of the caterpillar, which carries a painful sting.
Last August, a Kauai plant nursery owner found one and turned it in to the agency's plant quarantine branch. The department has since found adult moths in Wailua, Kapaa and Kilauea.
The caterpillar is white and has a long stripe running down its back. Those allergic to the bug may have difficulty breathing or develop other serious symptoms after being stung.
Hawaii officials looking for stinging caterpillar – Hawaii News – Staradvertiser.com
November 25, 2010 General Ag News and Notes Post a Comment
Kona coffee beans, plants quarantined over pest
KAILUA-KONA (AP) – Coffee plants and unroasted beans from Hawaii's Big Island are being quarantined in hopes of preventing the spread of a crop-destroying pest from Kona farms to other islands.
The Hawaii Board of Agriculture unanimously approved the emergency quarantine Tuesday due to the coffee berry borer, which has been found in 21 West Hawaii farms but hasn't been seen on other islands.
The quarantine restricts the movement of coffee plants, plant parts, green beans and bags unless the items are treated with pesticides or heating methods to kill the beetle and its larvae, according to the Department of Agriculture.
"Movement of green beans is restricted unless it's fumigated," said Department of Agriculture spokeswoman Janelle Saneishi.
The beetle was first detected in West Hawaii-grown coffee beans in mid-September. Agriculture officials haven't yet determined how it arrived on the Big Island.
The quarantine could last up to a year. It doesn't apply to farmers who are sending green beans out of state.
October 25, 2010 General Ag News and Notes Post a Comment
Ag officials: Kona coffee facing quarantine
In response to the threat posed by the coffee berry borer, state agriculture officials are preparing to establish a quarantine on the transport of green coffee beans from South Kona.
The pest's presence was confirmed Sept. 8. Hawaii was one of the few remaining coffee-producing areas in the world that had not been infested by the bug, which has been known to cut crop production up to 20 percent.
Lyle Wong, plant industry administrator with the state Department of Agriculture, said Friday the Plants and Animals Advisory Committee would meet in a week or so on whether to recommend a quarantine be enacted.
He said a meeting was held Monday, but due to a failure to advertise it six days beforehand, another meeting must be called.
"What went before the advisory board was a proposal for quarantine of the whole Kona coast, but we will have to do it again," he said.
If the pest is deemed an "immediate emergency" and the committee passes the recommendation, it will go before the Department of Agriculture board for approval and implementation, Wong said.
A quarantine means that green, or non-roasted, coffee beans would have to be treated with heat or an insecticide before they could be shipped off island.
June 17, 2010 General Ag News and Notes Post a Comment
Barking mad – Star Advertiser
Pet owners run out of patience with the shortage of inspectors and longer waits at Honolulu Airport's quarantine station
The line of tired and weary pet owners can stretch out the door at the Honolulu Airport's animal quarantine office, and tempers occasionally grow testy.
In just the last three weeks, the anti-rabies quarantine station has seen as many as 60 pet owners per day trying to squeeze through a time window that used to be 12 hours a day.
Since the number of inspectors reviewing both applications and animals was cut to two from four in December and mandatory furlough days went into effect, pet owners now have 31/2 fewer hours to get their pets processed through the increasingly busy quarantine station below Gate 26.
Bo a no-go for Hawaii trip – PATRICK GAVIN | POLITICO
Sorry Bo. The first family may be vacationing in Hawaii this holiday season, but the first dog will be stuck in cold, snowy D.C.
The Hawaii Department of Agriculture said that the Portuguese water dog will not be allowed into the state thanks to strict anti-rabies quarantine rules.
Had the family elected to bring Bo, he would have had to either spend 120 days in quarantine or endure two rounds of rabies vaccinations and a 120-day waiting period.
The Honolulu Advertiser also notes that Bo "would have been subject to Hawai'i's sometimes contradictory leash laws. City ordinances require dogs to be leashed on Kailua Beach — and their owners to clean up their feces. But the State Department of Land and Natural Resources — which has jurisdiction over the ocean — allows dogs to swim in the water without leashes, Laura Stevens, DLNR's education and outreach coordinator, said today."
Bo a no-go for Hawaii trip – PATRICK GAVIN | POLITICO CLICK
September 30, 2009 General Ag News and Notes Post a Comment
Maui Meeting to Address Shortage of Agricultural Inspectors
In light of concerns about potential layoffs and/or furloughs facing the Hawaii Department of Agriculture, Mayor Tavares has organized a group of individuals to bring the most up-to-date information to various Maui communities. The Mayor feels this is an important investment of our time because the decisions that are ultimately made will impact all of us to varying degrees. A shortage of agricultural inspectors could pose a host of threats to our environment, agriculture industry, tourism, and public health and safety.
The group of presenters include Anna Mae Shishido – Maui County Supervisor of the Department of Agriculture's Plant Quarantine Branch, Warren Watanabe – Executive Director of the Maui County Farm Bureau, Teya Penniman – Manager of the Maui Invasive Species Committee, and myself.
The Kula Community Association has graciously agreed to host our group at its next meeting, which will be open to its entire membership and the community-at-large. It will be held at the Kula Elementary School cafeteria on Tuesday, October 6th, starting at 6:00 p.m.
Please join us if you can and feel free to invite your family, friends, neighbors, and co-workers to this very important meeting. We will provide you with the most current information and let you know how to stay informed about this issue and what you can do to help out.
Kuhea
Kuhea Paracuelles
Environmental Coordinator
200 South High Street, 9th Floor
Wailuku, Hawai`i 96793
E-mail: kuhea.paracuelles@co.maui.hi.us
East Bay Express – Shoddy Science
Shoddy Science
A national panel criticizes the USDA's scientific research on the light brown apple moth but affirms the agency's power to start another round of aerial spraying.
As expected, a panel from the National Academy of Sciences said on Monday that the government has the legal authority to embark on a massive new eradication effort against the light brown apple moth, thereby opening the door for another round of aerial pesticide spraying. But the panel also criticized the United States Department of Agriculture for engaging in shoddy science to substantiate its war on the moth.
The 21-page report came in response to petitions submitted by opponents of the government's extermination plans. They had asked the USDA to reclassify the light brown apple moth from being a major pest to one that could be easily controlled by farmers. Such a move would have prohibited aerial spraying or other major eradication efforts that the government is now planning.
Opponents believe the USDA and state officials have severely overstated the threats posed by the moth, and have noted that it has lived for more than one hundred years in Australia, New Zealand, and Hawaii without causing serious, sustained damage to crops or native plants and trees. The USDA, nonetheless, believes the moth will destroy large swaths of cropland throughout California and much of the southern United States. The agency also considers it a serious threat to native redwood and pine forests.
Industry fights for inspectors – Starbulletin.com
Agricultural groups fear state layoffs will backlog shipments
By Erika Engle
POSTED: 01:30 a.m. HST, Sep 08, 2009
Agricultural industry executives worry that Hawaii businesses will wither on the vine and incoming food will rot on the docks if the state goes through with massive layoffs of agriculture inspectors.
Plans call for laying off 50 of the state's 78 agriculture inspectors, 64 percent of that specialized work force.
Diminished inspection capacity could also cost hundreds of millions of dollars each year if additional invasive species get established, industry officials say.
State inspectors both certify products to be exported out of Hawaii and inspect food and plants being imported into the state.


Getting the Best Financial Advice – Who Influences You?
We all have influences and we all want to make sure we get the best financial advice. Other people often shape who we become, what we believe, and how we act. So, who influences me from a financial standpoint? I have had several influences. I'll share a few of them with you.
My Stepfather has been a huge influence on my financial life. He is very successful from a career and a financial standpoint. I credit him with my interest in finances and investing. He taught me how to create a budget, how to analyze stocks, and how to get set up with a brokerage account and a ROTH IRA.
I don't know how much he has, but I do know he is a self-made millionaire and became a millionaire at a fairly young age. I try to emulate his approach toward money as much as I can.
We often bounce ideas off one other for stock picks and investing strategies. I am blessed to have a close family member who can help me navigate the waters of finances. If you have someone in your family who is financially savvy, I'd highly recommend spending some time talking with them and learning from them. They can be an invaluable resource as you learn and grow financially.
All three have quality stories related to finance, saving, investing, and the stock market. If you are looking to analyze a stock or bond I've found that Morningstar is an excellent resource. There is a paid version of the site, but the free one works just fine.
If you like printed publications, Barron's is an excellent weekly paper. They also have a website which mirrors the printed version. You will gain access to it if you subscribe. It's a bit pricey but well worth it if you are looking to do in-depth analysis. Be aware that it is a rather advanced publication and isn't for the novice investor.
I've also had some influences in my life who have taught me what not to do financially. Most people you encounter in your daily life will fall into this category on some level. There is a good chance that many of your co-workers, friends, and family members will not be financially astute. Some will be worse off than others, but the reality is that in most instances financially successful individuals will be more of a rarity than the norm.
It's important to take a step back and observe what others around you are doing and how they are acting. I often notice my co-workers buying new cars every couple of years, going out to eat for lunch daily, and walking into work with all the latest tech gadgets. Sometimes it's easy to fall into the herd mentality and give into the temptation to keep up with your peers, but it can also be a recipe for financial hardship. Stay true to your beliefs and your goals.
So, who and what influences you? I'd love to hear some of your personal influences, how they've shaped your financial perspectives, and who inspires you. Comment below and share your story.
Do Investments Make Good Gifts?


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Through its monthly presentations, speakers, films and quarterly newsletters, ROCLA educates Rochester-area residents about the culture, economics, and politics of the countries of Central and South America, Mexico and the Caribbean, and the ways globalization, trade and drug war policies impact the lives of their people.
ROCLA Winter 2022-2023 Newsletter
ROCLA's Winter 2022-2023 Newsletter - Providing members and friends with an expanded quarterly newsletter that covers many more Latin American countries and timely issues such as climate change, migration and cultural change.
Prospects for the Pink Tide in Latin America in a Volatile World Context
Feb 8, 2023 7PM - Prospects for the Pink Tide in Latin America in a Volatile World Context with Roger Harris. Join our mailing list to receive Zoom meeting info.
Reform in Honduras and US Opposition
Dec 7, 2022 7PM - Reform in Honduras and US Opposition with Karen Spring. Join our mailing list to receive Zoom meeting info. Video now available!
Founded in 1973, the Rochester Committee on Latin America seeks to build bridges between the Rochester, NY community and the people of Latin America.
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Your vacation in Tulum as a cultural getaway
BY layer22
Tulum is a destination where culture plays a main role.
If you are interested in history, ancient civilizations, amazing archaeological sites and are also longing to enjoy a relaxing time among nature, you must choose your vacation in Tulum as a cultural getaway.
The Maya culture is alive. Hundreds of Maya communities are immersed in the jungle and have their own sustainable economies. Most of the hotel's employees are Maya and they speak their own language.
The Maya are proud people, they practice their traditions and replicate ancient recipes and rituals that are also available for you to enjoy. Their culture is characterized by monumental architecture, complex astronomical and mathematical systems, hieroglyphic writing and iconography, as well excellency in observation astronomy that guided their sustainable agricultural system and calendar.
The most impressive Maya archeological site is "Tulum" which was built on a cliff situated directly in front of the sea and facing the rising sun. An unbelievable scenario for breathtaking pictures, where you can imagine how ordinary life might have been when the Maya culture was booming around the 6th century A. D. This is the only Maya archaeological site built on the ocean.
Just north of the city of Tulum is the tallest archaeological site, named Cobá, where you can climb the 120 step temple which was the center of the civilization.
The Muyil archaeological site is located approximately 9.3 miles south of Tulum. Muyil is part of the Sian Ka'an Biosphere Reserve, with impressive structures and an amazing mangrove canal system that will lead you through one of the many ancient commercial paths.
The Sian Ka'an Biosphere Reserve starts where Tulum ends. It has dunes, lagoons, savannahs, beaches, and protected ecosystems where you can practice snorkeling, kayaking and bird watching, among other activities.
Sian Ka'an in the Mayan language, translates into "Where the sky is born". No words can describe the beauty of this unique tropical ecosystem, you must experience it.
If you are looking for a contemporary twist on culture, you cannot miss IK LAB, Tulum's top art gallery hosting artwork from artists from all over the world, directed by Peggy Guggenheim's great-grandson in a magical treehouse setting.
These are only some of the reasons for choosing your vacation in Tulum as a cultural getaway.
Adults Only Hotel in Tulum
Road Tulum-Boca Paila km7 | Zona Hotelera, Mexico, Tulum 77780
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