Extra-factorial sum: a graph-theoretic parameter in Hamiltonian cycles of complete weighted graphs

TL;DR

This paper explores the extra-factorial sum, a graph-theoretic parameter related to Hamiltonian cycles in complete weighted graphs, providing proofs, tutorial explanations, and new unpublished results.

Contribution

It offers a tutorial presentation of the extra-factorial sum with proofs and introduces new unpublished lemmas related to Hamiltonian cycles in complete weighted graphs.

Findings

The extra-factorial sum relates to the arithmetic mean of Hamiltonian cycle lengths.

It depends on the number of vertices n and involves factorial terms.

New lemmas extend the understanding of Hamiltonian cycle properties.

Abstract

A graph-theoretic parameter, in a form of a function, called the extra-factorial sum is discussed. The main results are presented in ref. [1] (Nastou et al., Optim Lett, 10, 1203-1220, 2016) and the reader is strongly advised to study the aforementioned paper. The current work presents subject matter in a tutorial form with proofs and some newer unpublished results towards the end (lemma six extension and lemma seven). The extra-factorial sum is relevant to Hamiltonian cycles of complete weighted graphs $WH_n$ with $n$ vertices and is obtained for each edge of $WH_n$. If this sum is multiplied by $1 / (n - 2)$ then it gives directly the arithmetic mean of the sum of lengths $l_i$ of all Hamiltonian cycles that traverse a selected edge $e_q$. The number of terms in this sum is a factorial proven to be $(n - 2)!$ which signifies that its value depends on $n$. Using the extra-factorial sum, the arithmetic mean of the sum of the squared lengths of $(n - 1)! / 2$ Hamiltonian cycles of $WH_{n}$ can be obtained as well.

Tables

Table 1. Table 1: The process of creating T i n + 1 superscript subscript 𝑇 𝑖 𝑛 1 T_{i}^{n+1} with the Steps Insert Vertex Algorithm (SIVA).

Input: $T_i^n$ $\left(n\ge 3\right)$ and vertex X outside the cycle.
$\to$	Step 1: copy $T_i^n$ and vertex X ($n$ times).
$\to$	Step 2: for each copy $T_i^n$ (step 1), remove a different $e_q$.
$\to$	Step 3: for each $T_i^n$ (step 2), add a copy of X.
$\to$	Step 4: for each $i$th copy (step 3), add a new pair $(e_{q1},e_{q2})$ so that each broken edge is relinked with X.
Output: $n$ new $T_i^{n+1}$ $\left(1\le i\le n\right)$ with $\left(n+1\right)$ edges and $\left(n+1\right)$ vertices.

Equations

$$\frac{[(n-1)-1]!}{2}{e}_{T^{n-1}}=\frac{(n-1)!}{2}$$

$$\frac{(n_{old}-1)!}{2}{e}_{T^{n_{old}}}=\frac{(n_{new}-1)!}{2}$$

$$n(n+1)(n+2)(n+3)\cdots (n+k-1)=\frac{(n+k-1)!}{2}$$

$$[(n-1)-2]!(e_{T^{n-1}}-1)=(n-2)!$$

$$[(n-1)-2]!(e_{T^{n-1}}-1)=(n-3)!(n-2)=(n-2)!$$

$$(n_{old}-2)!(e_{T^{n_{old}}}-1)=(n_{new}-2)!$$

$$\begin{array}{rl}& 2[(n-1)-3]!(e_{T^{n-1}}-2)=\\ & 2[(n-1)-3]![(n-1)-2]=\\ & 2(n-4)!(n-3)=\\ & 2(n-3)!\end{array}$$

$$\sum _{i=1}^{n(n-1)/2}w(e_i)=\sum _{i=1}^{(n-2)!}{l}_i^{AB}$$

$$\begin{array}{rl}\sum _{i=1}^{(n-2)!}{l}_i^{AB}=& \\ & (n-2)!X_1+(n-3)!X_2+2(n-3)!X_3\end{array}$$

$$\begin{array}{rl}{X}_1=& w(e_q^{AB})\\ X_2=& \sum _{q_1=1}^{(n-2)}w(e_{q_1}^{AX})+\sum _{q_2=1}^{(n-2)}w(e_{q_2}^{BY})\\ X_3=& \sum _{q=1}^{(n-2)(n-3)/2}w(e_q^{XY})\end{array}$$

$$\begin{array}{rl}(!)\sum _{i=1}^{(n-2)!}{l}_i^{AB}& =(n-2)w(e_q^{AB})+\\ & +[\sum _{q_1=1}^{(n-2)}w(e_{q_1}^{AX})+\sum _{q_2=1}^{(n-2)}w(e_{q_2}^{BY})]+\\ & +2[\sum _{q=1}^{(n-2)(n-3)/2}w(e_q^{XY})]\end{array}$$

$$\begin{array}{rl}(!)\sum _{i=1}^{(n-2)!}{l}_i^{AB}\frac{1}{(n-2)}& =\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AB}}{(n-3)!}\frac{1}{(n-2)}=\\ & =\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AB}}{(n-2)!}\end{array}$$

$$\frac{(!){\sum }_{i=1}^{(n-2)!}{l}_i^{AB}}{(n-2)}=\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AB}}{(n-2)!}=\frac{25+27}{2}=26$$

$$\begin{array}{rl}& \frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AC}}{(n-2)!}=25.5,\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AD}}{(n-2)!}=24.5\\ & \frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{BC}}{(n-2)!}=24.5,\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{BD}}{(n-2)!}=25.5\\ & \frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{CD}}{(n-2)!}=26\end{array}$$

$$\begin{array}{rl}& (n-2)!e_1^{AB}\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AB}}{(n-2)!}+\\ & +(n-2)!e_2^{AC}\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AC}}{(n-2)!}+\\ & +(n-2)!e_3^{AD}\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{AD}}{(n-2)!}+\\ & +(n-2)!e_4^{BC}\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{BC}}{(n-2)!}+\\ & +(n-2)!e_5^{BD}\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{BD}}{(n-2)!}+\\ & +(n-2)!e_6^{CD}\frac{{\sum }_{i=1}^{(n-2)!}{l}_i^{CD}}{(n-2)!}=\\ & =e_1^{AB}\sum _{i=1}^{(n-2)!}{l}_i^{AB}+e_2^{AC}\sum _{i=1}^{(n-2)!}{l}_i^{AC}+e_3^{AD}\sum _{i=1}^{(n-2)!}{l}_i^{AD}+\\ & +e_4^{BC}\sum _{i=1}^{(n-2)!}{l}_i^{BC}+e_5^{BD}\sum _{i=1}^{(n-2)!}{l}_i^{BD}+e_6^{CD}\sum _{i=1}^{(n-2)!}{l}_i^{CD}\end{array}$$

$$\begin{array}{rl}& \frac{l_1+l_2+l_3+\cdots +l_{(n-1)!/2}}{(n-1)!/2}=\\ & =n\frac{w(e_1)+w(e_2)+\cdots +w(e_{n(n-1)/2})}{n(n-1)/2}\end{array}$$

$$\begin{array}{rl}& \frac{l_1+l_2+l_3+\cdots +l_{(n-1)!/2}}{(n-1)!/2}=\\ & =\frac{(n-2)![w(e_1)+w(e_2)+\cdots +w(e_{n(n-1)/2})]}{(n-1)!/2}=\\ & =\frac{(n-2)!}{(n-1)!/2}[w(e_1)+w(e_2)+\cdots +w(e_{n(n-1)/2})]=\\ & =\frac{n}{n(n-1)/2}[w(e_1)+w(e_2)+\cdots +w(e_{n(n-1)/2})]\end{array}$$

Full text

Authors to whom all correspondence should be addressed; e-mails: farmermath70@gmail.com1 and nek.valous@nct-heidelberg.de3

Consider a complete graph $H_{n}$ with $n$ vertices. Such a graph has always $n\times(n-1)/2$ edges, since for each pair of vertices $(x,y)$ with $x\neq y$, exists only one edge $e_{q}$ with $1\leq q\leq n\times(n-1)/2$ that connects them. Additionally, each $e_{q}$ can have a weight $w(e_{q})\in$ $\mathbb{R}$ which makes $H_{n}$ a weighted graph $WH_{n}$. Fig. 1 shows a $WH_{6}$. In a graph $H_{n}$, every closed walk $T_{i}$ with $1\leq i\leq(n-1)!/2$ that traverses each vertex exactly once starting and ending at the same vertex is called a Hamiltonian cycle RefB1 . Each $T_{i}$ is comprised of $e_{T}$ edges and $n$ vertices $(e_{T}=n)$. The $i$th $T_{i}$ with $n$ vertices is called $T_{i}^{n}$. The length $l_{i}$ of a $T_{i}^{n}$ is obtained by summing the relevant $w(e_{q})$. Fig. 1 also shows a $T_{i}^{6}$: ACEFDBA which originates from $WH_{6}$ and has $l_{i}=65.1$. The number of $T_{i}^{n}$ is given by $(n-1)!/2$ RefB1 , e.g. $WH_{6}$ has 60 $T_{i}^{6}$.

A graph-theoretic problem is introduced which is defined as follows: for selected $e_{q}$ in $WH_{n}$, the aim is to find the sum of lengths $l_{i}$ of the $T_{i}^{n}$ that traverse $e_{q}$. Furthermore, it will be proved that each $e_{q}$ has a number of $T_{i}^{n}$ that traverse it which is $(n-2)!$. The Hamiltonian path problem is NP-complete RefJ1 and summing the $l_{i}$ of the $(n-2)!$ $T_{i}^{n}$ that traverse $e_{q}$ is computationally expensive. On the other hand, the arithmetic mean of this sum can be obtained and this new function is defined as the extra-factorial sum RefDef . This is a fraction with numerator the sum of $l_{i}$ of the $(n-2)!$ $T_{i}^{n}$ traversing $e_{q}$ and denominator the value $(n-3)!$. The parameter can be obtained directly without the need of computing the numerator and denominator separately. It will be proven that the extra-factorial sum of any $e_{q}$, if multiplied by $1/(n-2)$, yields the arithmetic mean of $l_{i}$ of the $(n-2)!$ $T_{i}^{n}$ traversing $e_{q}$. A direct deduction is that for any $WH_{n}$, the extra-factorial sum can be obtained for each $e_{q}$. This graph-theoretic parameter can be presented in a 2D Cartesian chart with the $y$-axis showing the values of the extra-factorial sum in the interval $(-\infty,+\infty)$ and the $x$-axis showing the ranked $e_{q}$ based on the extra-factorial sum values. Such a curve is a visualization of the distribution of $l_{i}$ in relation to the edges, and is predominantly a qualitative measure for comparing different $WH_{n}$.

Lemma 1. A Hamiltonian cycle generator GT is a $T_{i}^{n}$ in which if a vertex is added then this cycle can create new $T_{i}^{n+1}$ with $(n+1)$ edges and $(n+1)$ vertices. After creating the new $T_{i}^{n+1}$, GT is replaced by its child cycles.

Proof. If vertex X is added to $T_{i}^{n}$ then for each $e_{q}$ there is a unique pair ($e_{q1}$, $e_{q2}$) with common X that can break the edges, thus creating the new $T_{i}^{n+1}$. In this way, $n$ new $T_{i}^{n+1}$ can be created with $(n+1)$ edges and $(n+1)$ vertices each. $\blacksquare$

With an initial cycle generator $T_{i}^{8}=$ ABCDEFGHA and vertex X outside the cycle, then new $T_{i}^{8+1}$: $T_{1}^{8+1}$, $T_{2}^{8+1}$, $T_{3}^{8+1}$,$T_{4}^{8+1}$, $T_{5}^{8+1}$, $T_{6}^{8+1}$, $T_{7}^{8+1}$, $T_{8}^{8+1}$ are created (Fig. 2). Each new $T_{i}^{n+1}$ traverses the vertices of GT including X. In essence, each $e_{q}$ of GT (AB, BC, CD, DE, EF, FG, GH, HA) breaks in order to create new $T_{i}^{n+1}$. For example, the creation of AXBCDEFGHA requires the breaking of AB and the addition of AX and XB. The initial ABCDEFGHA ceases to exist after the creation of the new $T_{i}^{n+1}$. The obtained $T_{i}^{8+1}$ are equal to the $e_{q}$ of $T_{i}^{8}$, hence $T_{i}^{8}$ is defined as a GT. The process of creating $T_{i}^{n+1}$ is described in Table 1 with the Steps Insert Vertex Algorithm (SIVA). The inputs are $T_{i}^{n}$ and vertex X outside the cycle and the output is the $n$ new $T_{i}^{n+1}$ with $(n+1)$ edges and $(n+1)$ vertices RefJ2 .

Corollary 1. Every $T_{i}^{n}$ belongs to generation $(n-3)$ defined as $G=(n-3)$. The $T_{i}^{n}$ belonging to the same generation have the same number of edges and vertices. Since every $T_{i}^{n}$ has same number of edges and vertices then minimum is $n=3$. $T_{i}^{3}$ cannot be a GT therefore the generation is defined as $G=(n-3)=0$. The $T_{i}^{n}$ with $G=0$ is the initial $T_{1}^{3}$.

Henceforth, the edges of $H_{n}$ or $WH_{n}$ are denoted by $e_{n}$ and the edges of $T_{i}^{n}$ by $e_{T}^{n}$. When SIVA is applied to $T_{i}^{n}$ ($n\geq 3$) then this is defined as the $GT$ with $G=(n-3)$ that can create $n$ new $T_{i}^{n+1}$ with $G+1$ having $(n+1)$ vertices and $(n+1)$ edges. SIVA can be extended by creating $(n-1)!/2$ $T_{i}^{n}$ that exist in $H_{n}$ with $n$ vertices and $n(n-1)/2$ edges $(n\geq 3)$. Using SIVA to count $T_{i}^{n}$ offers the opportunity to count only the $T_{i}^{n}$ for which certain categorization criteria may apply. For $H_{3}$, the relationship $[n(n-1)/2]=e_{n}=n=3$ is valid; this means that $H_{3}$ has the additional property of being a $T_{1}^{3}$ with $G=0$, since for $T_{i}^{n}$ the number of vertices is equal to the number of edges.

Lemma 2. For $H_{3}$ and $k$ vertices $(k\geq 1)$ outside the graph, $H_{3}$ is a $T_{1}^{3}$ (initial $GT$). The $[(3+k)-1]!/2$ $T_{i}^{n}$ of the new $H_{k+3}$ are counted one-by-one when the new $k$ vertices are added to $H_{3}$ iteratively using SIVA.

Proof. A graph $H_{3}$ with $n=e_{n}$ is also a $T_{1}^{3}$. SIVA starts from $T_{1}^{3}$ and iterates $k$ times to introduce, in each repetition, a new vertex in every $T_{i}^{3+j}$ with $G=[(3+j)-3]$, for $1\leq j\leq k$. Fig. 3 shows $H_{3}=\textbf{ABC}$ and two new vertices X and Y $(k=2)$; ABC is also a $T_{1}^{3}=\textbf{ABCA}$.

Applying SIVA results in adding X into ABCA, which leads to the creation of three new $T_{i}^{4}$ ($G=1$): $T_{1}^{4}=\textbf{AXCBA}$, $T_{2}^{4}=\textbf{ACXBA}$, and $T_{3}^{4}=\textbf{AXBCA}$. Dotted edges represent breaking locations for creating the new cycles. After inserting X, ABC is converted to $H_{4}=\textbf{ABCX}$. Continuing along, SIVA adds Y to each $T_{i}^{4}$ $(1\leq i\leq 3)$ of the first generation ABCX. After adding Y, twelve new $T_{i}^{5}$ ($G=2$) are created ($1\leq i\leq 12$) (Fig. 4).

The new $T_{i}^{5}$ are created when SIVA is applied to each $T_{i}^{4}$ of ABCX. After inserting Y, $H_{4}$ is converted to $H_{5}=\textbf{ABCXY}$ (Fig. 5). In the general case (starting with $n=3$), when adding a new vertex to $H_{n}$ then SIVA places the vertex to every $T_{i}^{n}$, for $1\leq i\leq[(n-1)!/2]$. After adding the $k$th vertex, new $H_{n+k}$ are created with $(n+k)$ vertices and $[(n+k)(n+k-1)]/2$ edges, for $(n+k)>3$ and $k\geq 1$. This new graph incorporates the set of $[(n+k)-1]!/2$ $T_{i}^{n+k}$. Each $T_{i}^{n+k}$ belongs to $G=(n+k-3)$ with $(n+k)$ vertices and $(n+k)$ edges. Hence, the following recursive function is obtained:

[TABLE]

Alternatively, this can be expressed as:

[TABLE]

where $n_{new}-n_{old}=1$. This algebraic relationship states that the product of $(n_{old}-1)!/2$ of the $T_{i}^{n_{old}}$ by $e_{T}^{n_{old}}$ in each $T_{i}^{n_{old}}$ yields $(n_{new}-1)!/2$ of the $T_{i}^{n_{new}}$; these are created if $H_{n}$ increases its vertices by $1$ and its edges by $n$. Validity can be proved algebraically as a function of $n$, when SIVA increases the vertices of $H_{n}$ by $k$. The new $T_{i}^{5}$ ($G=2$) as a function of $n$, for $n=3$, are $(1\times n)\times(n+1)=12$. The term $(1\times n)$ are the $T_{i}^{4}$ ($G=1$) and the term $(n+1)$ are the $e_{T}^{4}$ in each $T_{i}^{4}$ ($G=1$). With the addition of vertices and starting with $n=3$, the new $T_{i}^{n+k}$ tends exactly to:

[TABLE]

This concludes the proof. $\blacksquare$

Lemma 3. The number of $T_{i}^{n}$ in $H_{n}$ $(n\geq 3)$ which traverse selected $e_{q}$ is $(n-2)!$.

Proof. Counting the $T_{i}^{n}(n\geq 3)$ can be done using SIVA but with some differences comparing to Lemma 2: $e_{q}$ does not break. Specifically, if $e_{q}$ is selected then $k=n-3$ and the first $T_{1}^{3}$ is created with the two vertices of $e_{q}$ and any one vertex of $H_{n}$. Then SIVA iterates by inserting the remaining $k$ vertices for $n=(3+k)$. In each iteration, it is imperative that $e_{q}$ does not break. This variation is visualized in Fig. 6 and Fig. 7 for $G=1$ and $G=2$, respectively. For $H_{5}$ (Fig. 5) and selected $e_{q}^{AB}$ that links A and B, in order to count the $T_{i}^{n}$ that always traverse $e_{q}^{AB}$, only the $T_{i}^{n-1}$ traversing $e_{q}^{AB}$ are counted recursively. The procedure begins with $H_{5}$ and $e_{q}^{AB}$; then three of the five vertices are selected such that $e_{q}^{AB}$ is always traversed: these three vertices form an $H_{3}$ and a $T_{1}^{3}$. SIVA is applied to the edges ($e_{q}^{AC}$, $e_{q}^{BC}$) of $T_{1}^{3}$ except $e_{q}^{AB}$; hence a $T_{1}^{4}$ ($G=1$) is created. For $T_{1}^{4}=\textbf{ABCXA}$ and $T_{2}^{4}=\textbf{ABXCA}$, SIVA adds the fifth and final vertex of $H_{5}$ without breaking $e_{q}^{AB}$. In the general case of $H_{n}$, the algorithm proceeds till all remaining vertices are added. Hence, a set of $T_{i}^{n}$ is created that always traverse $e_{q}^{AB}$. The number of $T_{i}^{n}$ is $(n-2)!$ which is proven by induction: this is valid for $H_{3}$ since the cycles that traverse $e_{q}$ are $(n-2)!=1$. Similarly for $H_{4}$ since $(n-2)!=2$. Therefore:

[TABLE]

The term $(n-2)!$ expressing the $T_{i}^{n}$ that traverse $e_{q}$ can be obtained using SIVA, for each $T_{i}^{n-1}$ traversing $e_{q}$ without breaking it. The fact that $e_{q}$ does not break is denoted by the second term $(e_{T^{n-1}}-1)$: this expresses the edges in $T_{i}^{n-1}$ allowed to break in an one-by-one fashion for creating the new $T_{i}^{n}$. Overall, the product defines the $T_{i}^{n-1}$ traversing $e_{q}$ by the value $(e_{T^{n-1}}-1)$ from each $T_{i}^{n-1}$ excluding $e_{q}$. Since $e_{T^{n}}=n$, this can be written as:

[TABLE]

Alternatively, this can be expressed as:

[TABLE]

where $n_{new}-n_{old}=1$. This expresses the $T_{i}^{n-1}$ in $H_{n-1}$ that always traverse $e_{q}$ by the edges in $T_{i}^{n-1}$ excluding $e_{q}$; this equals to the $T_{i}^{n}$ in $H_{n}$ that traverse $e_{q}$. $\blacksquare$

An edge intersects or not any other edge; if $e_{q}^{AB}$ intersects A and B (Fig. 5) then it intersects an edge adjacent to $e_{q}^{AB}$, meaning that they share the same vertex. Hence, for every edge intersecting $e_{q}^{AB}$: $e_{q}^{AX}$, $e_{q}^{BY}$ where $X,Y\neq(A\parallel B)$, e.g. edges intersecting AB are: AY, AX, AC and BY, BX, BC.

Lemma 4. The number of $T_{i}^{n}$ in $H_{n}$ $(n\geq 3)$ which traverse a selected pair of adjacent edges is $(n-3)!$.

Proof. Counting the $T_{i}^{n}$ ($n\geq 3$) that satisfy the lemma is proven through Lemma 3. For example, in $H_{5}$ (Fig. 5) edges $e_{q}^{AY}$ and $e_{q}^{YX}$ with common Y are selected. Initially, it is observed that there are no $T_{i}^{5}$ that traverse $e_{q}^{AX}$ and no edge capable of linking Y with vertices other than A and X, therefore $e_{q}^{AX}$, $e_{q}^{YB}$, and $e_{q}^{YC}$ have to be removed. For $H_{n}$ $(n\geq 4)$ and pair of adjacent edges (e.g. $e_{q}^{AY}$ and $e_{q}^{YX}$), the edge that links the vertices of the selected adjacent edges is removed. This means $e_{q}^{AX}$ and any other edge that links Y with any other vertex excluding $e_{q}^{YK}$, where $K\neq(A\parallel B)$. After the removal of edges, Y can be removed temporarily since for all $T_{i}^{5}$, A and B are always linked. The graph resulting from the removal of Y is an $H_{4}$ with $(n-1)=4$ vertices. In this new $H_{4}$, the number of $T_{i}^{4}$ that traverse $e_{q}^{AB}$ is equal to the number of $T_{i}^{5}$ traversing the adjacent $e_{q}^{AY}$ and $e_{q}^{YX}$ of the initial $H_{5}$. Previously, it was proven that the number of $T_{i}^{n}$ that traverse a selected edge is $(n-2)!$. Consequently, the number of $T_{i}^{n}$ that traverse any selected pair of adjacent edges is $(n-1)-2!=(n-3)!$. $\blacksquare$

Lemma 5. The number of $T_{i}^{n}$ in $H_{n}$ ($n\geq 4$) which traverse a selected pair of non-adjacent edges is $2(n-3)!$.

Proof. For $T_{i}^{n}$ ($n\geq 4$) and two selected non-adjacent edges, then SIVA creates $(e_{T}^{n}-2)=(n-2)$ new $T_{i}^{n+1}$ with $(n+1)$ vertices since the two non-adjacent edges do not break. Counting the $T_{i}^{n}$ that fulfill the lemma starts from $n=4$ since this is the minimum value of $n$ permitting the creation of $T_{i}^{n}$ that can have pairs of non-adjacent edges. Only $T_{1}^{4}$, $T_{2}^{4}$, and $T_{3}^{4}$ exist in $H_{4}$; from these and for every pair of non-adjacent edges (e.g. $e_{q}^{AB}$ and $e_{q}^{XC}$) only $T_{1}^{4}$ and $T_{2}^{4}$ traverse them, therefore counting starts from them (Fig. 8). SIVA creates the $T_{i}^{n}$ contained in each of the initial $T_{1}^{4}$ and $T_{2}^{4}$ without breaking $e_{q}^{AB}$ and $e_{q}^{XC}$. SIVA iterates $k$ times with $n=k-4$ for each $T_{1}^{4}$ and $T_{2}^{4}$ and creates $2[(4+k)-3]!$ new $T_{i}^{k+4}$ with $(4+k)$ vertices and $(4+k)$ edges. For $k=1$ and $n=4$, the expression $2[(n+k)-3]!=4$ is valid since for $T_{1}^{4}=\textbf{ABCDA}$ and $T_{2}^{4}=\textbf{ABDCA}$ with the addition of a new vertex, SIVA creates $(e_{T}^{n}-2)=(n-2)=2$ new $T_{i}^{5}$ that traverse $e_{q}^{AB}$ and $e_{q}^{XC}$. Given that $T_{1}^{4}$ and $T_{2}^{4}$ are two then $2(e_{T}^{n}-2)=2(n-2)=4$ new $T_{i}^{5}$ are created (Fig. 9).

The dotted lines (Fig. 9) denote the edges that break in order to add the new $Y$. The number of $T_{i}^{5}$ that fulfill the lemma is $2(n-3)!=4$. When SIVA adds a new vertex in $H_{5}$ (without breaking $e_{q}^{AB}$ and $e_{q}^{XC}$) then for $n=5$ and $k=1$, $2[(n+k)-3]!=12$ new $T_{i}^{6}$ are created traversing $e_{q}^{AB}$ and $e_{q}^{XC}$ since for each of the old $T_{i}^{5}$ three vertices break. This is expressed as: $4(e_{T}^{n}-2)=4(n-2)=12$, and in the general case:

[TABLE]

The constant expresses the $T_{i}^{4}$ where counting starts from and proceeds with SIVA $k$ times without breaking the non-adjacent edges. The term $(n-4)!$ expresses the child cycles originating from each initial $T_{i}^{4}$ in $H_{n-1}$. The term $(n-3)$ expresses the edges belonging to $T_{i}^{n-1}$ that break in an one-by-one fashion (except for the selected edges) in order for SIVA to insert a new vertex. $\blacksquare$

Consider the graph $WH_{5}$ (based on the graph $H_{5}$ of Fig. 5) with $w(e_{q}^{AB})=4$, $w(e_{q}^{AX})=15$, $w(e_{q}^{AC})=12$, $w(e_{q}^{AY})=0.5$, $w(e_{q}^{BC})=33$, $w(e_{q}^{BX})=-0.4$, $w(e_{q}^{BY})=15$, $w(e_{q}^{CY})=33$, and $w(e_{q}^{XY})=7$. This graph has $n(n-1)/2=10$ edges and $(n-1)!/2=12$ $T_{i}^{5}$. Each $T_{i}^{5}$ has $l_{i}$ equal to the sum of its consisting edge weights. Fig. 10 shows the $T_{i}^{5}$ into two groups; the first (top two rows) contains the cycles that traverse the selected $e_{q}^{AB}$ while the second (bottom two rows) the remaining that do not.

The number of $T_{i}^{5}$ that traverse $e_{q}^{AB}$ is $(n-2)!$. According to Lemma 4, for the $T_{i}^{5}$ that traverse $e_{q}^{AB}$ the number of edges that intersect this edge is $(n-3)!$ meaning that $e_{q}^{AY}$, $e_{q}^{AX}$, $e_{q}^{AC}$, $e_{q}^{BY}$, $e_{q}^{BX}$, and $e_{q}^{BC}$ appear twice in the $T_{i}^{5}$ subset that traverse $e_{q}^{AB}$. According to Lemma 5, for the $T_{i}^{5}$ that traverse $e_{q}^{AB}$ the number of edges that do not intersect this edge is $2(n-3)!$ meaning that $e_{q}^{YX}$, $e_{q}^{XC}$, $e_{q}^{YC}$ appear four times in the $T_{i}^{5}$ subset that traverse $e_{q}^{AB}$. The sum of $l_{i}$ of the $T_{i}^{5}$ that traverse $e_{q}^{AB}$ is: $\sum_{i=1}^{(n-2)!}l_{i}^{AB}=l_{1}+l_{2}+l_{3}+l_{4}+l_{5}+l_{6}=349.2$, where $T_{1}^{5}=\textbf{YCABXY}(l_{1}=55.6)$, $T_{2}^{5}=\textbf{YCBAXY}(l_{2}=92)$, $T_{3}^{5}=\textbf{YXCBAY}(l_{3}=59.5)$, $T_{4}^{5}=\textbf{YCXBAY}(l_{4}=52.1)$, $T_{5}^{5}=\textbf{YCXABY}(l_{5}=82)$, and $T_{6}^{5}=\textbf{YXCABY}(l_{6}=53)$.

Lemma 6. For $WH_{n}$ and selected edge (e.g. $e_{q}^{AB}$) there exist a unique summational graph $\sum WH_{n}^{AB}$ that corresponds to $e_{q}^{AB}$. Each edge in $WH_{n}$ corresponds to a unique $\sum WH_{n}$. For $e_{q}^{AB}$, the resulting $\sum WH_{n}^{AB}$ is a copy of the initial $WH_{n}$ where each weight is multiplied as follows: i) $w(e_{q}^{AB})$ multiplied by $(n-2)!$, ii) the weight of each edge intersecting $e_{q}^{AB}$ multiplied by $(n-3)!$, and iii) the weight of each edge not intersecting $e_{q}^{AB}$ multiplied by $2(n-3)!$. The sum of $w(e_{i})$ of $\sum WH_{n}$ corresponding to $e_{q}^{AB}$ is equal to the sum of $l_{i}$ of the $T_{i}^{n}$ that traverse $e_{q}^{AB}$:

[TABLE]

Proof. The sum of $l_{i}$ of the $T_{i}^{5}$ traversing $e_{q}^{AB}$ is visualized in $\sum WH_{5}^{AB}$ (Fig. 11): this shows a copy of $WH_{5}$ for $e_{q}^{AB}$ where $w(e_{q}^{AB})$ is multiplied by $(n-2)!=6$, the weight of each edge intersecting $e_{q}^{AB}$ by $(n-3)!=2$, and the weight of the remaining edges by $2(n-3)!=4$. The sum of edge weights in $\sum WH_{5}^{AB}$ is equal to the sum of $l_{i}$ of the $T_{i}^{5}$ that traverse $e_{q}^{AB}$.

This concludes the proof. $\blacksquare$

The relationship between the sum of $l_{i}$ of the $T_{i}^{n}$ ($WH_{n}$) and the sum of edge weights in $\sum WH_{n}^{AB}$ is given by RefJ2 :

[TABLE]

where $X_{1}$ is the weight of $e_{q}^{AB}$, $X_{2}$ is the sum of weights with $X,Y\neq(B\parallel A)$ for the edges that intersect $e_{q}^{AB}$ (their number is $2(n-2)$), and $X_{3}$ is the sum of weights with $X,Y\neq(B\parallel A)$ for the edges that do not intersect $e_{q}^{AB}$ (their number is $[n(n-1)]/2-[2(n-2)]-1=[(n-2)(n-3)]/2$). Overall:

[TABLE]

Since, the common term is $(n-3)!$ the extra-factorial sum of edge AB can be written as:

[TABLE]

The symbol (!) denotes the removal of the term $(n-3)!$. The extra-factorial sum for a selected edge multiplied by $1/(n-2)$ is equal to the arithmetic mean of $l_{i}$ of the $(n-2)!$ cycles that traverse that edge:

[TABLE]

For different $WH_{n}$, the extra-factorial sum can be visualized in a Cartesian chart. Two $WH_{14}$ ($G_{1}$ and $G_{2}$) are used as examples: $G_{1}$ has random edge weights and $G_{2}$ is a copy of $G_{1}$ with weights multiplied by $1/2$. The charts of the ranked extra-factorial sums of the 91 edges of each graph are shown in Fig. 13. The first edge has the smallest extra-factorial sum value, while the last (91st) has the largest. This demonstrates that even if the edges have different extra-factorial sums, the corresponding curves are identical. This is because $G_{2}$ originates from $G_{1}$ meaning that the graphs are linearly dependent. The similarity in the curves signifies that the cycle length distributions (corresponding to each edge) are also identical. Since the extra-factorial sum provides an overview of the subset of cycles that traverse each edge, future work can focus on absolute or relative similarity RefJ3 for two or more complete weighted graphs. Consider the graph $WH_{4}=\textbf{ABCD}$ (based on Fig. 5 but with no vertex Y and edges AY, XY, BY, CY, and where A corresponds to A, X to B, B to D, C to C) (Fig. 12).

The weights are: $w(e_{q}^{AB})=12$, $w(e_{q}^{AD})=7$, $w(e_{q}^{AC})=8$, $w(e_{q}^{BC})=4$, $w(e_{q}^{BD})=5$, and $w(e_{q}^{CD})=2$. The graph has $T_{1}^{4}=\textbf{ACBDA}$ with $l_{1}=24$, $T_{2}^{4}=\textbf{ABCDA}$ with $l_{2}=25$, and $T_{3}^{4}=\textbf{ABDCA}$ with $l_{3}=27$. The arithmetic means of $l_{i}$ of the $T_{i}^{4}$ that traverse each edge are:

[TABLE]

[TABLE]

If the weight of each edge (IJ) in $WH_{4}$ is multiplied by its own value $\frac{\sum_{i=1}^{(n-2)!}l_{i}^{IJ}}{(n-2)!}$, then a new graph $WH_{4}^{{}^{\prime}}$ results. According to Lemma 3, every edge appears $(n-2)!$ times in the $T_{i}^{4}$ of $WH_{4}^{{}^{\prime}}$. Therefore the sum of $l_{i}$ of the $T_{i}^{4}$ ($WH_{4}^{{}^{\prime}}$) is given by:

[TABLE]

Substituting the corresponding values, the sum becomes: $12(25+27)+8(24+27)+7(24+25)+4(24+25)+5(24+27)+2(25+27)$. This can be written as: $24(7+8+5+4)+25(7+4+2+12)+27(8+5+2+12)$ which is: $(576)+(625)+(729)$. Hence, the sum of $l_{i}$ of the $T_{i}^{4}$ ($WH_{4}^{{}^{\prime}}$) is equal to the sum of squared lengths of $T_{i}^{4}$ of the initial $WH_{4}$.

Lemma 7. The arithmetic mean of $l_{i}$ of the $T_{i}^{n}$ ($WH_{n}$ or $WH_{n}^{{}^{\prime}}$) is given by:

[TABLE]

Proof. According to Lemma 3, the following can be deduced:

[TABLE]

This concludes the proof. $\blacksquare$

In a similar framework, the arithmetic mean of the sum of $l_{i}$ of the $T_{i}^{n}$ that do not traverse $e_{q}$ can be obtained as well. It is interesting to note that these sums can be further analyzed in the context of a controlled change in edge weights, with the aim of analyzing and comparing two or more $WH_{n}$ using the extra-factorial sum curves. The process of computing the arithmetic mean of the sum of squared lengths of $T_{i}^{n}$ ($WH_{n}$) in combination with the inverse process (given $WH_{n}^{{}^{\prime}}$ to produce an exact corresponding $WH_{n}$ in which the sum of squared lengths of $T_{i}^{n}$ is equal to the sum of $l_{i}$ of the $T_{i}^{n}$ of the initial $WH_{n}^{{}^{\prime}}$) can provide solutions pertinent to the following problem: the existence or not of at least one $T_{i}^{n}$ (edge weights having positive and negative values) with negative length. The problem of creating a $WH_{n}$ from an initial $WH_{n}^{{}^{\prime}}$ is rather complex due to the fact that for each $WH_{n}^{{}^{\prime}}$ there may correspond more than one $WH_{n}$. Furthermore, an interesting application pertinent to signal processing RefJ4 is to study the behavior of the extra-factorial sum curves ($WH_{n}$) that have irrational edge weights (specifically trigonometric numbers) which are functions of time. Yet, another application is in Hopfield neural networks RefJ5 ; these are modelled as $WH_{n}$ and the extra-factorial sum may contribute to new methods of training the network, e.g. changing the weights in each iteration can incorporate the magnitude of the extra-factorial sum.