#  Parallel edges in ribbon graphs and interpolating behavior of partial-
duality polynomials

Qiyao Chen  College of Mathematics, Hunan University, 410082 Changsha, China
<EMAIL_ADDRESS>and  Yichao Chen  College of Mathematics, Hunan
University, 410082 Changsha, China<EMAIL_ADDRESS>
(Date: ©)

###### Abstract.

Recently, Gross, Mansour and Tucker introduced the partial-twuality
polynomials. In this paper, we find that when there are enough parallel edges,
any multiple graph is a negative answer to the problem 8.7 in their paper
[European J. Combin. 95 (2021), 103329]: Is the restricted-orientable partial-
Petrial polynomial of an arbitrary ribbon graph even-interpolating$?$ In
addition, we also find a counterexample to the conjecture 8.1 of Gross,
Mansour and Tucker: If the partial-dual genus polynomial is neither an odd nor
an even polynomial, then it is interpolating.

###### Key words and phrases:

partial dual, partial Petrial, even-interpolating, interpolating

This work is supported by the NNSFC (Grant No. 11471106) and the JSSCRC( Grant
No. 2021530).

##  1\. Introduction

We assume that the readers are familiar with the basic knowledge of
topological graph theory. The reader is referred to [GMT21a] for the
explanation of all terms not defined here.

Let $G^{*|_{A}}$ ( $G^{\times|_{A}}$ ) be the partial dual (partial Petrial)
of $G$ with respect to $A\subset E(G)$. Denote by $v(G)$, $e(G)$, $f(G)$ and
$c(G)$ the number of vertices, edges, faces and connected components of $G,$
respectively. For $\bullet\in\\{\times,*,\times*,*\times,*\times*\\}$, Gross,
Mansour, and Tucker [GMT21a] introduced the  partial-$\bullet$ polynomial for
the ribbon graph $G$, i.e.,

$~{}^{\partial}{{\mathscr{E}}}^{\bullet}_{G}(z)=\sum\limits_{A\subseteq
E(G)}z^{{\mathfrak{eu}}[G^{\bullet|_{A}}]},$

where ${\mathfrak{eu}}[G^{\bullet|_{A}}]$ represents the Euler-genus of
$G^{\bullet|_{A}}$.

They also introduced the restricted orientable partial-$\times$ polynomial of
$G$ by enumerating Euler-genus only over edge-subsets $A\subseteq E(G)$ such
that $G^{\times|_{A}}$ is orientable. We recall that partial duality was
introduced by Chmutov in [Chm09]. In [EM12], Ellis-Monaghan and Moffatt
extended the partial-duality to include the Wilson dual, the Petrie dual, and
the two kinds of triality operators. In [AE19], Abrams and Ellis-Monaghan
called the five operators twualities. We may refer the reader to [EM13, GMT20,
GMT21a, GMT21b] for more background about partial-duality and partial Petrial.

A subdivision of $G$ is obtained by replacing an edge $e=uv$ of $G$ by a path
$uwv,$ and a proper edge is an edge with two different ends. The contraction
on edge $e$ is denoted by $G/e$, and we denote by $G-e$ the ribbon graph
obtained from $G$ by deleting the ribbon $e$. In [GMT20], Gross, Mantour, and
Tucker proved a subdivided edge recursion for partial-$*$.

######  Theorem 1.1.

[GMT20] Given a ribbon graph $G$ and a ribbon $e$. Let $K$ be a subdivision of
$G$, then

(1.1)
$~{}^{\partial}{{\mathscr{E}}}^{*}_{K}(z)=\begin{cases}2~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z),&\text{if
$e$ is a cut ribbon, }\\\
~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)+2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G-e}(z),&\text{if
$e$ is non-separating.}\\\ \end{cases}$

Gross, Mantour, and Tucker also give subdivided edges and parallel edges
recursions for partial-$\times$ in [GMT21a].

######  Theorem 1.2.

[GMT21a] Let $G$ be a ribbon graph with an e-type pp ribbon $e$, and let
$G+e^{\prime}$ be the ribbon graph obtained by adding a ribbon $e^{\prime}$
parallel to a ribbon $e$ in $G$. Thus

(1.2) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+e^{\prime}}(z)$
$\displaystyle=$
$\displaystyle(1+2z)~{}^{\partial}{{\mathscr{E}}}^{\times}_{G/e}(z)+(z^{2})~{}^{\partial}{{\mathscr{E}}}^{\times}_{G-e}(z)$

######  Theorem 1.3.

[GMT21a] Let $G$ be a ribbon graph with a ribbon $e$, and let $H$ be obtained
from $G$ by subdividing $e$ into edges $e_{1}$ and $e_{2}$. Then

(1.3) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{\times}_{H}(z)$
$\displaystyle=$ $\displaystyle
2~{}~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}(z).$

The support of a polynomial $f(z)=\displaystyle\sum_{i=0}^{n}a_{i}z^{i}$ is
the set $\\{i|a_{i}\neq 0\\}$. If $supp(f(z))$ is an integer interval $[m,n]$
of all integers from $m$ to $n$, inclusive, we call $f(z)$ an interpolating
polynomial. The size of integer interval $[m,n]$ is the number of elements of
$[m,n].$ If $supp(f(z))$ is the set of all even natural numbers in an integer
interval, then $f(z)$ is even-interpolating. If the terms of the non-zero
coefficient of the polynomial are even (odd) degree, we call it an even (odd)
polynomial.

Conjecture 8.1 and Problem 8.7 in their paper [GMT21a] state that

######  Conjecture 1.4.

If the partial-$*$ polynomial $~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)$ is
neither an odd nor an even polynomial, then it is interpolating.

######  Problem 1.5.

Is the restricted-orientable partial-$\times$ polynomial of an arbitrary
ribbon graph $G$ even-interpolating$?$

In this paper, we apply the operations of adding parallel edges and
subdividing edges to get some counter-examples of Conjecture 8.1 and Problem
1.2 [GMT21a]. We first disprove Conjecture 1.1 by finding an infinite family
of ribbon graphs as counterexamples. Then we answer the Problem 1.2 by proving
the restricted orientable partial polynomial of any multiple ribbon graph with
enough parallel edges ( we also have a tight lower bound for this) are not
even-interpolating.

######  Remark 1.6.

Throughout the paper, we will use the rotation projection [GT87] of $G$
instead of the ribbon graph $G$ itself (See Figure 2.1 \- Figure 4.2).

##  2\. The parallel edge recursion for partial- $*$ polynomials

In this section, we derive a recursion for the partial-$*$ polynomials. Given
two disjoint ribbon graphs $G_{1}$ and $G_{2}$, we let $G_{1}\vee G_{2}$
denote the join of $G_{1}$ and $G_{2}$. The complement of $A$ in $E(G)$ is
$A^{c}=E(G)-A$.

######  Theorem 2.1.

[Mof12] Let $G$ be a ribbon graph and $A\subseteq E(G)$, then

(2.1) $\displaystyle{\mathfrak{eu}}(G^{*|_{A}})$ $\displaystyle=$
$\displaystyle 2c(G)+e(G)-f(A)-f(A^{c}).$

######  Theorem 2.2.

Let $G$ be a ribbon graph with a proper ribbon $e$, let $G+e_{1}$ be the
ribbon graph obtained by adding parallel ribbon $e_{1}$ to the ribbon $e$ as
shown in Figure 2.1. Then

(2.2) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G+e_{1}}(z)$
$\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)+2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G/e}(z).$

Figure 2.1.

###### Proof.

Let $A\subseteq E(G+e_{1}),\ A_{1}\subseteq E(G),\ A_{2}\subseteq E(G/e).$ It
is easy to see that $e(G+e_{1})=e(G/e)+2=e(G)+1.$ There are three cases.

$\mathbf{Case~{}1}:$ Suppose that $e_{1}\in A$ and $e\in A,$ and let
$A_{1}=A-e_{1}$. Since the ribbon graph $A_{1}$ is obtained from $A$ by
deleting the multiple edge $e_{1}$, which decreases the number of faces by 1,
we have

(2.3) $\displaystyle f(A)=f(A_{1})+1.$

We use the fact that the complement of $A$ in $E(G+e_{1})$ is equal to the
complement of $A_{1}$ in $E(G),$ then

(2.4) $\displaystyle f(A^{c})=f(A_{1}^{c}).$

Thus, by Theorem 2.1,

(2.5) $\displaystyle{\mathfrak{eu}}((G+e_{1})^{*|_{A}})$ $\displaystyle=$
$\displaystyle 2c(G+e_{1})+e(G+e_{1})-f(A)-f(A^{c})$ $\displaystyle=$
$\displaystyle 2c(G)+e(G)+1-f(A_{1})-1-f(A_{1}^{c})\quad\text{by
(\ref{A1})~{}and~{}(\ref{A2}) }$ $\displaystyle=$
$\displaystyle{\mathfrak{eu}}(G^{*|_{A_{1}}}).$

$\mathbf{Case~{}2}:$ Assume that $e_{1},e\in A^{c},$ and let $A=A_{1}$.
Obviously,

(2.6) $\displaystyle f(A)=f(A_{1}).$

Because $A^{c}_{1}$ is obtained from $A^{c}$ by deleting the multiple edge
$e_{1}$, which decreases the number of faces by 1. Therefore,

(2.7) $\displaystyle f(A^{c})=f(A^{c}_{1})+1.$

Hence

(2.8) $\displaystyle{\mathfrak{eu}}((G+e_{1})^{*|_{A}})$ $\displaystyle=$
$\displaystyle 2c(G+e_{1})+e(G+e_{1})-f(A)-f(A^{c})\quad\text{by (\ref{d0}) }$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+1-f(A_{1})-f(A_{1}^{c})-1\quad\text{by
(\ref{A3})~{}and~{}(\ref{A4}) }$ $\displaystyle=$
$\displaystyle{\mathfrak{eu}}(G^{*|_{A_{1}}}).$

Let $\mathscr{A}=\\{A|e_{1},e\in A\\}\cup\\{A|e_{1},e\in A^{c}\\},$
$\mathscr{A}_{1}=\\{A_{1}|e\in A_{1}\\},$ $\mathscr{A}_{2}=\\{A_{1}|e\in
A^{c}_{1}\\},$ then

(2.9)
$\displaystyle\sum\limits_{A\in\mathscr{A}}z^{{\mathfrak{eu}}\big{(}(G+e_{1})^{*|_{A}}\big{)}}$
$\displaystyle=$
$\displaystyle\sum\limits_{A_{1}\in\mathscr{A}_{1}}z^{{\mathfrak{eu}}\big{(}G^{*|_{A_{1}}}\big{)}}+\sum\limits_{A_{1}\in\mathscr{A}_{2}}z^{{\mathfrak{eu}}\big{(}G^{*|_{A_{1}}}\big{)}}\quad\text{by
(\ref{dm1})~{}and~{}(\ref{dm2}) }$ $\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z).$

$\mathbf{Case~{}3}:$ Let $e_{1}\in A^{c},$ $e\in A$, and we let $A_{2}=A/e$.
Recall that $A_{2}$ is obtained from $A$ by contracting the proper edge $e$
and $A^{c}_{2}$ is obtained from $A^{c}$ by contracting the proper edge
$e_{1}$. Since the contraction does not change the number of faces, we have

(2.10) $\displaystyle f(A)=f(A_{2}),\quad f(A^{c})=f(A_{2}^{c}).$

Therefore,

(2.11) $\displaystyle{\mathfrak{eu}}((G+e_{1})^{*|_{A}})$ $\displaystyle=$
$\displaystyle 2c(G+e_{1})+e(G+e_{1})-f(A)-f(A^{c})\quad\text{by (\ref{d0}) }$
$\displaystyle=$ $\displaystyle
2c(G/e)+e(G/e)+2-f(A_{2})-f(A_{2}^{c})\quad\text{by (\ref{A5}) }$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}((G/e)^{*|_{A_{2}}})+2.$

Similarly, formula (2.11) also holds for the case: $e\in A^{c},$ and $e_{1}\in
A.$

Let $\overline{\mathscr{A}}=\\{A|e_{1}\in A^{c},e\in A\\}\cup\\{A|e\in
A^{c},e_{1}\in A\\},$ and then

(2.12)
$\displaystyle\sum\limits_{A\in\overline{\mathscr{A}}}z^{{\mathfrak{eu}}\big{(}(G+e_{1})^{*|_{A}}\big{)}}$
$\displaystyle=$ $\displaystyle
2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G/e}(z)\quad\text{by (\ref{dm4}) }$

Combining cases 1-3, we obtain

$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G+e_{1}}(z)$
$\displaystyle=\sum\limits_{A\in\mathscr{A}}z^{{\mathfrak{eu}}[(G+e_{1})^{*|_{A}}]}+\sum\limits_{A\in\overline{\mathscr{A}}}z^{{\mathfrak{eu}}[(G+e_{1})^{*|_{A}}]}$
$\displaystyle=2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G/e}(z)+~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z).$

∎

Figure 2.2.

######  Corollary 2.3.

Let $G$ be a ribbon graph with a proper ribbon $e$, and let
$G+\sum^{n}_{i=1}e_{i}$ be the ribbon graph obtained by adding $n$ parallel
ribbons $e_{1}$, $e_{2}$, $\cdots$, $e_{n}$ to the ribbon $e$ as shown in
Figure 2.2. Then

(2.13)
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G+\sum^{n}_{i=1}e_{i}}(z)$
$\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)+(2^{n+1}-2)z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G/e}(z).$

###### Proof.

By Theorem 2.2,

(2.14)
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G+\sum^{n}_{i=1}e_{i}}(z)$
$\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G+\sum^{n-1}_{i=1}e_{i}}(z)+2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{(G+\sum^{n-1}_{i=1}e_{i})/e_{n-1}}(z).$

Recall that the ribbons $e$, $e_{1}$, $e_{2}$, $\cdots$, $e_{n-2}$ in
$(G+\sum^{n-1}_{i=1}e_{i})/e_{n-1}$ are all loops, and the ribbon graph $G/e$
is obtained from $(G+\sum^{n-1}_{i=1}e_{i})/e_{n-1}$ by deleting $n-1$ loops.
Thus, from Proposition 3.2 in [GMT20],

(2.15)
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{(G+\sum^{n-1}_{i=1}e_{i})/e_{n-1}}(z)$
$\displaystyle=$ $\displaystyle
2^{n-1}~{}^{\partial}{{\mathscr{E}}}^{*}_{G/e}(z).$

(2.14), (2.15) and (2.2) clearly imply (2.13).

∎

Recall that the Tutte polynomial ([Tut54]) and Bollobas-Riordan polynomial
([Bol02]) have deletion-contraction recursions, as well as the role of series-
parallel graphs in the foundations of matroids([Bry71]).

######  Remark 2.4.

As pointed out by the anonymous referee, subdividing an edge is adding a
parallel edge in the dual ribbon graph. Thus Theorem 2.2 can be seen as the
dual form of Theorem 1.1.

Let $C_{n}$ and $K_{n}$ denote a $n$-cycle and a complete graph with $n$
vertices, respectively. A graph is series parallel ([AVJ99]), if its
2-connected components can be generated by repeatedly adding a parallel edge
and subdividing an edge, starting with $K_{2}$. As an example, we apply
Theorem 1.1 and Theorem 2.2 to calculate the partial -$*$ polynomial for a
series parallel graph.

Figure 2.3.  Figure 2.4.

######  Example 2.1.

Figure 2.3 shows the series parallel graphs $H,G$ and $G_{1}.$ It’s clear that
$G$ is isomorphic to $H$ with ribbon $e_{1}$ subdivided once, and
$G_{1}=H-e_{1}$. Moreover, $H$ is obtained from $G_{1}$ by adding a parallel
edge $e_{1}$ to the ribbon $e_{2}$, hence, we have

(2.16) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)$
$\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{H}(z)+2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{1}}(z)\quad\text{by~{}(\ref{1027})}$
$\displaystyle=$
$\displaystyle(2z^{2}+1)~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{1}}(z)+2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{1}/e_{2}}(z)\quad\text{by
~{} (\ref{dm0})}$

Since $G_{1}$ is isomorphic to $G_{1}/e_{2}$ with ribbon $e_{3}$ subdivided
once, let $G_{2}=(G_{1}/e_{2})-e_{3}$ as in Figure 2.4, it follows that

(2.17) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{1}}(z)$
$\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{1}/e_{2}}(z)+2z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{2}}(z)\quad\text{by
~{} (\ref{1027})}$

Combining (2.16) and (2.17), we have

(2.18) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)$
$\displaystyle=$
$\displaystyle(4z^{2}+1)~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{1}}(z)-4z^{4}~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{2}}(z).$

Similarly, we have

(2.19) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{1}}(z)$
$\displaystyle=$
$\displaystyle(4z^{2}+1)~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{2}}(z)-4z^{4}~{}^{\partial}{{\mathscr{E}}}^{*}_{K_{3}}(z)\quad\text{by
~{} (\ref{2402})},$

and

(2.20) $\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G_{2}}(z)$
$\displaystyle=$
$\displaystyle(4z^{2}+1)~{}^{\partial}{{\mathscr{E}}}^{*}_{K_{3}}(z)-4z^{4}~{}^{\partial}{{\mathscr{E}}}^{*}_{K_{2}}(z)\quad\text{by
~{} (\ref{2402})},$

where $K_{3}$ is obtained by adding a parallel edge and subdividing an edge to
$K_{2}$ once. It’s obvious that
$~{}^{\partial}{{\mathscr{E}}}^{*}_{K_{3}}(z)=6z^{2}+2$, and
$~{}^{\partial}{{\mathscr{E}}}^{*}_{K_{2}}(z)=2.$ Therefore, combining (2.18),
(2.19) and (2.20), we have

$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)$ $\displaystyle=$
$\displaystyle(32z^{6}+40z^{4}+12z^{2}+1)~{}^{\partial}{{\mathscr{E}}}^{*}_{K_{3}}(z)-(48z^{8}+332z^{6}+4z^{4})~{}^{\partial}{{\mathscr{E}}}^{*}_{K_{2}}(z)$
$\displaystyle=$ $\displaystyle 96z^{8}+240z^{6}+144z^{4}+30z^{2}+2.$

##  3\. Some counterexamples to the conjecture 1.1.

In this section, we find some infinite families of ribbon graphs as
counterexamples to Conjecture 1.1.

Figure 3.1. $G+\sum^{n}_{i=1}\bar{e}_{i}$ (left), $G$ (middle) and $G/e_{1}$
(right)

Let $G$ and $G+\sum^{n}_{i=1}\bar{e}_{i}$ be the ribbon graphs of Figure 3.1.
Clearly, $e(A)+e(A^{c})=e(G)$ and
${\mathfrak{eu}}(G^{*|_{A}})={\mathfrak{eu}}(G^{*|_{A^{c}}})$. Note that
$e(G)=7$, and for each edge set with the number of edges greater than 3 in
$G$, we can find the complement with the number of edges less than 4.

  1. (1)

$G$ has partial-$*$ polynomial $8z^{2}+48z^{4}+32z^{5}+40z^{6}$. By the
previous analysis, we need to show the partial-$*$ polynomial of dualizing
edge set with the number of edges less than 4 is
$4z^{2}+24z^{4}+16z^{5}+20z^{6}$. The $z^{2}$ term corresponds to dualizing
none, or twisted edge $e$, or all edges in the same $C_{3}$ (2 choices) and
all are isomorphic to $G$. The $z^{4}$ terms comes from dualizing one edge in
the same $C_{3}$ (6 choices), or a pair in the same $C_{3}$ (6 choices), or
one edge in the same $C_{3}$ and $e$ (6 choices), or a pair in the same
$C_{3}$ and $e$ (6 choices). Dualizing one edge of $C_{3}$ and two edges of
the other $C_{3}$ produces $z^{5}$ and $z^{6}$ (18 choices). Where dualizing
$e_{1}$, $e_{3}$, and one edge of $C_{3}$ ($e_{4}$, $e_{5}$, $e_{6}$ ) (3
choices), or $e_{5}$, $e_{6}$, and one edge of $C_{3}$ ($e_{1}$, $e_{2}$,
$e_{3}$ ) (3 choices) produces $z^{5}$, the remaining 12 choices produces
$z^{6}$. Dualizing one edge in each $C_{3}$ and $e$ produces $z^{5}$ and
$z^{6}$ (9 choices). Where dualizing $e_{2}$, $e_{4}$, and $e$ produces
$z^{5}$, the remaining 8 choices produces $z^{6}$. Dualizing one edge in each
$C_{3}$ (9 choices) also produces $z^{5}$.

  2. (2)

$G/e_{1}$ has partial-$*$ polynomial $8z^{6}+16z^{5}+32z^{4}+8z^{2}.$ For each
edge set with the number of edges greater than 3 in $G/e_{1},$ we can find the
complement with the number of edges less than 3. Moreover, for the edge set
whose number of edges is 3, the edge set with twisted edge $e$ is the
complement of the edge set without $e$. Therefore, we need to show the
partial-$*$ polynomial of dualizing edge set with the number of edges not
greater than 2, and edge set without $e$ where the number of edges is 3 are
$4z^{6}+8z^{5}+16z^{4}+4z^{2}$. The $z^{2}$ terms comes from dualizing none,
or $e$, or $C_{3}$, or $C_{2}$ and all are isomorphic to $G/e_{1}$. The
$z^{4}$ terms comes from dualizing one edge in $C_{3}$ and $C_{2}$ (5
choices), or a pair in $C_{3}$ (3 choices), or $e$ and one edge in $C_{3}$ and
$C_{2}$ (5 choices), or all edges in $C_{2}$ and one edge in $C_{3}$ (3
choices). Dualizing one edge of $C_{3}$ and one edge of $C_{2}$ produces
$z^{5}$ and $z^{6}$ (6 choices). Where dualizing $e_{3}$, $e_{5}$, $e_{6}$,
and $e_{2}$, $e_{5}$, $e_{6}$ produces $z^{5}$, the remaining 4 choices
produces $z^{6}$. Dualizing one edge in $C_{3}$ and one edge in $C_{2}$ (6
choices) also produces $z^{5}$.

Thus by Corollary 2.3, we have

$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G+\sum^{n}_{i=1}\bar{e}_{i}}(z)$
$\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{*}_{G}(z)+(2^{n+1}-2)z^{2}~{}^{\partial}{{\mathscr{E}}}^{*}_{G/e_{1}}(z)$
$\displaystyle=$
$\displaystyle(2^{n+4}-16)z^{8}+(2^{n+5}-32)z^{7}+(2^{n+6}-24)z^{6}$
$\displaystyle+32z^{5}+(2^{n+4}+32)z^{4}+8z^{2}.$

It is quite obvious that the polynomial above is neither an odd nor an even
polynomial, it’s also not interpolating.

######  Remark 3.1.

The anonymous referee also pointed out that we can take lots of joins of $G$
with itself to get counterexamples, since

$(z^{6}+z^{5}+z^{4}+z^{2})^{n}=z^{2n}+nz^{2n+2}+nz^{2n+3}+\cdots.$

##  4\.  A solution to the Problem 1.2.

In this section, we discuss the effect of adding multiple edges on the
restricted orientable partial-$\times$ polynomial. The restricted orientable
partial-$\times$ ribbon graph of $G$ is the orientable ribbon graph
$G^{\times|_{A}}$. For short, we denote the restricted orientable
partial-$\times$ polynomial of ribbon graph $G$ by
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}|_{O}(z)$.

######  Example 4.1.

Let $D_{n}$ be the dipole ribbon graph in the sphere. Clearly, if one twists
any proper subset of the edges, the resulting ribbon graph is non orientable,
there will be circuits of lenght two containing only one twisted edge. Thus
the only orientable partial -$\times$ duals are $D_{n}$ and $D_{n}^{\times}$.
The latter has $1$ or $2$ faces depending on whether $n$ is odd or even and
hence has orientable genus $\frac{n-1}{2}$ or $\frac{n-2}{2}$. Furthermore, we
have $~{}^{\partial}{{\mathscr{E}}}^{\times}_{D_{n}}|_{O}(z)=1+z^{n-1}$, when
$n$ is odd;
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{D_{n}}|_{O}(z)=1+z^{n-2}$, when $n$
is even. Thus, when $n\geq 5,$ the restricted-orientable partial-$\times$
polynomial of $D_{n}$ is not even-interpolating.

######  Lemma 4.1.

Let $G$ be a ribbon graph, then
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}|_{O}(z)=~{}^{\partial}{{\mathscr{E}}}^{\times}_{G^{\times|_{A}}}|_{O}(z).$

###### Proof.

Let $A_{1}\subseteq E(G)$, and $A\subseteq E(G)$. For each orientable ribbon
graph $G^{\times|_{A_{1}}}$, there exists $B\subseteq E(G^{\times|_{A}})$ such
that the orientable ribbon graph $(G^{\times|_{A}})^{\times|_{B}}$ and
$G^{\times|_{A_{1}}}$ are the same.

∎

Lemma 4.1 shows that it suffices to consider the case that $G$ is an
orientable ribbon graph.

######  Lemma 4.2.

Let $G$ be a ribbon graph, then the number of the restricted orientable
partial-$\times$ ribbon graphs of $G$ is $2^{v(G)-1}$.

###### Proof.

By [GMT21a], we know that the proportion of partial-$\times$ duals of G that
are orientable is $\frac{1}{2^{\beta(G)}}$ . There are $2^{e(G)}$ spanning
subgraphs in $G$, thus, the number of the restricted orientable
partial-$\times$ ribbon graphs of $G$ is

$\displaystyle\frac{2^{e(G)}}{2^{\beta(G)}}$ $\displaystyle=$
$\displaystyle\frac{2^{e(G)}}{2^{e(G)-v(G)+1}}=2^{v(G)-1}.$

∎

Let $e$ be a ribbon of $G$, and let $A$ be a subset of $E(G)$, we define
${\mathfrak{eu}}^{0}(G^{\times|_{A}})$ and
${\mathfrak{eu}}^{1}(G^{\times|_{A}})$ as $e\in A$ and $e\notin A$ of the
Euler-genus of orientable ribbon graph $G^{\times|_{A}}$, respectively.
Similarly, we let $f^{0}(G^{\times|_{A}})$ and $f^{1}(G^{\times|_{A}})$,
respectively, denote the number of faces of orientable ribbon graph
$G^{\times|_{A}}$ with $e\in A$ and $e\notin A$. And let
$f^{0}_{max}(G)=max\\{f^{0}(G^{\times|_{A}})|e\in A\\}$,
$f^{1}_{min}(G)=min\\{f^{1}(G^{\times|_{A}})|e\notin A\\}$.

######  Theorem 4.3.

Let $G$ be a ribbon graph with a proper ribbon $e$, and for even $n$, let
$G+\sum^{n}_{i=1}e_{i}$ be the ribbon graph obtained by adding $n$ parallel
ribbons $e_{1}$, $e_{2}$, $\cdots$, $e_{n}$ to the ribbon $e$ (see Figure
2.2). Then the polynomial
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
not even-interpolating for sufficiently large $n$.

###### Proof.

Suppose that
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
even-interpolating. We need to analysis the number of faces for orientable
ribbon graphs $G^{\times|_{A}}$ and
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}}$.

If $e\notin A$, and $G^{\times|_{A}}$ is an orientable ribbon graph, then
$f(G^{\times|_{A}})=f^{1}(G^{\times|_{A}})$. We let $A^{\prime}=A$, thus
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}}$ is also an orientable ribbon
graph, and

(4.1) $\displaystyle f((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}})$
$\displaystyle=$ $\displaystyle
f^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}})$ $\displaystyle=$
$\displaystyle f^{1}(G^{\times|_{A}})+n,$

because $G^{\times|_{A}}$ is obtained from
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}}$ by deleting $n$ multiple
ribbons $e_{1}$, $e_{2}$, $\cdots$, $e_{n}$, and deleting a multiple ribbon
will reduce one face.

Otherwise, put $e\in A_{0}$, suppose that the ribbon graph
$G^{\times|_{A_{0}}}$ is an orientable ribbon graph, then
$f(G^{\times|_{A_{0}}})=f^{0}(G^{\times|_{A_{0}}})$. Due to $e_{i}\cup e$ is a
2-cycle in $G+\sum^{n}_{i=1}e_{i}$, let
$A^{\prime}_{0}=A_{0}\cup\sum^{n}_{i=1}e_{i}$, thus
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}}$ is also an orientable
ribbon graph. Since the ribbon graph $G^{\times|_{A_{0}}}$ is obtained from
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}}$ by deleting $n$ twisted
multiple ribbons $e_{1}$, $e_{2}$, $\cdots$, $e_{n}$, and the deletion of the
even number of twisted ribbons does not change the number of faces, it follows
that

(4.2) $\displaystyle f((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})$
$\displaystyle=$ $\displaystyle
f^{0}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})$ $\displaystyle=$
$\displaystyle f^{0}(G^{\times|_{A_{0}}}).$

According to the relationship between $f^{1}(G^{\times|_{A}})$ and
$f^{0}(G^{\times|_{A_{0}}})$, we have the following two cases.

$\mathbf{Case~{}1}:$ If $f^{1}(G^{\times|_{A}})\geq
f^{0}(G^{\times|_{A_{0}}})$. By Euler’s formula,

(4.3) $\displaystyle{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})$ $\displaystyle=$
$\displaystyle 2c(G)+e(G)-v(G)-f^{0}(G^{\times|_{A_{0}}})$ $\displaystyle\geq$
$\displaystyle 2c(G)+e(G)-v(G)-f^{1}(G^{\times|_{A}})$ $\displaystyle=$
$\displaystyle{\mathfrak{eu}}^{1}(G^{\times|_{A}}),$ (4.4)
$\displaystyle{\mathfrak{eu}}^{0}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}(G^{\times|_{A_{0}}})\quad\text{by (\ref{A.1})}$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})+n,$
(4.5)
$\displaystyle{\mathfrak{eu}}^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}})$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{1}(G^{\times|_{A}})-n\quad\text{by (\ref{A.0})}$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{1}(G^{\times|_{A}}).$

Moreover, by (4.3)-(4.5), we get
${\mathfrak{eu}}^{0}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})\geq{\mathfrak{eu}}^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}})$.
So by the hypothesis that
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
even-interpolating, there is a subinterval
$[{\mathfrak{eu}}^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}}),{\mathfrak{eu}}^{0}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})]$
in
$supp(~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z))$.
Since there are only even numbers in
$[{\mathfrak{eu}}^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}}),{\mathfrak{eu}}^{0}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})]$
(the Euler-genus of an orientable ribbon graph is even), it follows that the
size of
$[{\mathfrak{eu}}^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}}),{\mathfrak{eu}}^{0}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})]$
is

(4.6)
$\displaystyle\frac{{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})-{\mathfrak{eu}}^{1}(G^{\times|_{A}})+n}{2}+1\quad\text{by
(\ref{A.5}) and (\ref{A.6}) }$

By Lemma 4.2, there are $2^{v(G)-1}$ orientable ribbon graphs in
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}}$. Even if the number of faces
of the restricted orientable partial-$\times$ ribbon graph of
$(G+\sum^{n}_{i=1}e_{i})$ are different, there are only $2^{v(G)-1}$ different
numbers. By (4.6), for sufficiently large $n$, we have

$\frac{{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})-{\mathfrak{eu}}^{1}(G^{\times|_{A}})+n}{2}+1\geq
2^{v(G)-1}+1.$

Hence, contradicting the assumption that
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
even-interpolating. We have proved the case 1.

$\mathbf{Case~{}2}:$ If $f^{1}(G^{\times|_{A}})<f^{0}(G^{\times|_{A_{0}}})$.
By Euler’s formula, we have
${\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})<{\mathfrak{eu}}^{1}(G^{\times|_{A}})$,
however, when $n$ is large enough, by (4.4)-(4.5), we still obtain
${\mathfrak{eu}}^{0}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{0}}})\geq{\mathfrak{eu}}^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}}})$.
The remainder of the proof is analogous to that in case 1 and so is omitted.

∎

Next, we give a lower bound for
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
not even-interpolating for the number $n$.

######  Theorem 4.4.

Let $G$ be a ribbon graph with a proper ribbon $e$, and let
$G+\sum^{n}_{i=1}e_{i}$ be the ribbon graph obtained by adding $n$ parallel
ribbons $e_{1}$, $e_{2}$, $\cdots$, $e_{n}$ to the ribbon $e$ (see Figure
2.2).

  1. (1)

If $~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}|_{O}(z)$ has only one term, and

$n\geq\begin{cases}4,&\text{if
$f^{0}_{max}(G+e_{1})=f^{0}(G^{\times|_{A}})+1$, }\\\ 3,&\text{if
$f^{0}_{max}(G+e_{1})=f^{0}(G^{\times|_{A}})-1$,}\\\ \end{cases}$

then $~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$
is not even-interpolating.

  2. (2)

If $~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}|_{O}(z)$ has more than one
term, and $n\geq
min\\{f^{0}_{max}(G+e_{1})-f^{1}_{min}(G)+4,f^{0}_{max}(G)-f^{1}_{min}(G)+4\\},$
then $~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$
is not even-interpolating.

###### Proof.

Suppose that
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
even-interpolating. We observe that the number of faces of
${\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$ is
$f^{0}_{max}(G+\sum^{n}_{i=1}e_{i})$, and the number of faces of
${\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i})$ is
$f^{1}_{min}(G+\sum^{n}_{i=1}e_{i})$. Let $e\in A_{0}$, $e\notin A_{1}$,
assume that both $G^{\times|_{A_{0}}}$ and $G^{\times|_{A_{1}}}$ are
orientable ribbon graphs, then
$f(G^{\times|_{A_{0}}})=f^{0}(G^{\times|_{A_{0}}})$,
$f(G^{\times|_{A_{1}}})=f^{1}(G^{\times|_{A_{1}}})$.

For Item (1), we have

(4.7) $\displaystyle
f^{1}(G^{\times|_{A_{1}}})=f^{0}(G^{\times|_{A_{0}}})=f^{1}_{min}(G)=f^{0}_{max}(G).$

Observe that $f^{0}_{max}(G+e_{1})=f^{0}_{max}(G)+1,$ or
$f^{0}_{max}(G+e_{1})=f^{0}_{max}(G)-1,$ which implies that we should consider
the following two cases.

$\mathbf{Case~{}1}:$ $f^{0}_{max}(G+e_{1})=f^{0}_{max}(G)+1$. Note that
deleting an even number of twisted multiple edges does not change the number
of faces. Then, when $n$ is even, the maximum value of
$f^{0}(G+\sum^{n}_{i=1}e_{i})$ is the maximum value of $f^{0}(G)$, that is

(4.8) $\displaystyle f^{0}_{max}(G+\sum^{n}_{i=1}e_{i})$ $\displaystyle=$
$\displaystyle f^{0}_{max}(G).$

When $n$ is odd, and deleting $n-1$ twisted multiple edges does not change the
number of faces. We have

$\displaystyle f^{0}_{max}(G+\sum^{n}_{i=1}e_{i})$ $\displaystyle=$
$\displaystyle f^{0}_{max}(G+e_{1})$ $\displaystyle=$ $\displaystyle
f^{0}_{max}(G)+1.$

By Euler’s formula, it follows that

(4.10) $\displaystyle{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}_{max}(G)\quad\text{by~{}(\ref{tm88})}$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}(G^{\times|_{A_{0}}})\quad\text{by~{}(\ref{tm1})}$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})+n,$

for even $n$, and

(4.11) $\displaystyle{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}_{max}(G)-1\quad\text{by~{}(\ref{tm8})}$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}(G^{\times|_{A_{0}}})-1\quad\text{by~{}(\ref{tm1})}$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})+n-1,$

for odd $n$.

Let $A^{\prime}_{1}=A_{1}$. It is easy to see that
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{1}}}$ is an orientable ribbon
graph, and
$f((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{1}}})=f^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{1}}})$.
Clearly, for any ribbon subset $A^{\prime}_{1}$ of $G+\sum^{n}_{i=1}e_{i}$, we
have
$f^{1}((G+\sum^{n}_{i=1}e_{i})^{\times|_{A^{\prime}_{1}}})=f^{1}(G^{\times|_{A_{1}}})+n.$
Furthermore,

(4.12) $\displaystyle f^{1}_{min}(G+\sum^{n}_{i=1}e_{i})$ $\displaystyle=$
$\displaystyle f^{1}_{min}(G)+n,$

similarly, we have

(4.13) $\displaystyle{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i})$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{1}_{min}(G)-n\quad\text{by (\ref{tm111})}$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}(G^{\times|_{A_{0}}})-n\quad\text{by (\ref{tm1})}$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}}).$

Recall that
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
even-interpolating, so, there is a subinterval
$[{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i}),{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})]$
in
$supp(~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z))$.
Note that the Euler-genus of the orientable ribbon graph
$(G+\sum^{n}_{i=1}e_{i})^{\times|_{A}}$ is either smaller than
${\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i})$ or larger than
${\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$, so, there are only two
numbers ${\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i})$,
${\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$ in
$[{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i}),{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})]$.

If $n$ is odd, from (4.11) and (4.13), the size of
$[{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i}),{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})]$
is
$\frac{{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})+n-1-{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})}{2}+1$.
Otherwise, from (4.10) and (4.13), the size of
$[{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i}),{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})]$
is
$\frac{{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})+n-{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})}{2}+1$.
When the size of
$[{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i}),{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})]$
is greater than or equal to 3, it is contrary to the hypothesis that
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
even-interpolating. Therefore,
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
not even-interpolating for $n\geq 4$.

$\mathbf{Case~{}2}:$ $f^{0}_{max}(G+e_{1})=f^{0}_{max}(G)-1$. Similarly, when
$n$ is odd,

$\displaystyle f^{0}_{max}(G+\sum^{n}_{i=1}e_{i})$ $\displaystyle=$
$\displaystyle f^{0}_{max}(G+e_{1})$ $\displaystyle=$ $\displaystyle
f^{0}_{max}(G)-1.$ (4.15)
$\displaystyle{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}_{max}(G)+1\quad\text{by~{}(\ref{tm9})}$
$\displaystyle=$ $\displaystyle
2c(G)+e(G)+n-v(G)-f^{0}(G^{\times|_{A_{0}}})+1\quad\text{by~{}(\ref{tm1})}$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})+n+1,$

If $n$ is odd, from (4.15) and (4.13),
$\frac{{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})+n+1-{\mathfrak{eu}}^{0}(G^{\times|_{A_{0}}})}{2}+1$
is the size of
$[{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i}),{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})]$.
The remainder of the argument is analogous to that in case 1 and omitted.
Hence,
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z)$ is
not even-interpolating for $n\geq 3$.

For item (2), we shall adopt the same procedure as in the proof of case 1.

When $n$ is odd, by (4), we have
$f^{0}_{max}(G+\sum^{n}_{i=1}e_{i})=f^{0}_{max}(G+e_{1})$, and

(4.16) $\displaystyle{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$
$\displaystyle=$ $\displaystyle
2c(G+e_{1})+e(G+e_{1})+n-1-v(G+e_{1})-f^{0}_{max}(G+e_{1})$ $\displaystyle=$
$\displaystyle 2c(G+e_{1})+e(G+e_{1})-v(G+e_{1})-f^{0}_{max}(G+e_{1})+n-1$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{0}_{min}(G+e_{1})+n-1.$

When $n$ is even, by (4.8), we have
$f^{0}_{max}(G+\sum^{n}_{i=1}e_{i})=f_{max}^{0}(G)$, and

(4.17) $\displaystyle{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})$
$\displaystyle=$ $\displaystyle 2c(G)+e(G)+n-v(G)-f^{0}_{max}(G)$
$\displaystyle=$ $\displaystyle 2c(G)+e(G)-v(G)-f^{0}_{max}(G)+n$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{0}_{min}(G)+n.$

By (4.13), we have

(4.18) $\displaystyle{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i})$
$\displaystyle=$ $\displaystyle 2c(G)+e(G)+n-v(G)-f^{0}_{min}(G)-n$
$\displaystyle=$ $\displaystyle 2c(G)+e(G)-v(G)-f^{0}_{min}(G)$
$\displaystyle=$ $\displaystyle{\mathfrak{eu}}^{1}_{max}(G).$

Recall that there exists a subinterval
$[{\mathfrak{eu}}^{1}_{max}(G+\sum^{n}_{i=1}e_{i}),{\mathfrak{eu}}^{0}_{min}(G+\sum^{n}_{i=1}e_{i})]$
in
$supp(~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z))$.
By (4.16)-(4.18), there exist two subintervals
$[{\mathfrak{eu}}^{1}_{max}(G),$ ${\mathfrak{eu}}^{0}_{min}(G+e_{1})+n-1]$ and
$[{\mathfrak{eu}}^{1}_{max}(G),{\mathfrak{eu}}^{0}_{min}(G)+n]$ in
$supp(~{}^{\partial}{{\mathscr{E}}}^{\times}_{G+\sum^{n}_{i=1}e_{i}}|_{O}(z))$.
When

(4.19)
$\frac{{\mathfrak{eu}}^{0}_{min}(G)+n-{\mathfrak{eu}}^{1}_{max}(G)}{2}+1\geq
3,$

or

(4.20)
$\frac{{\mathfrak{eu}}^{0}_{min}(G+e_{1})+n-1-{\mathfrak{eu}}^{1}_{max}(G)}{2}+1\geq
3,$

there are at least three numbers in
$[{\mathfrak{eu}}^{1}_{max}(G),{\mathfrak{eu}}^{0}_{min}(G+e_{1})+n-1]$ or
$[{\mathfrak{eu}}^{1}_{max}(G),\\\ {\mathfrak{eu}}^{0}_{min}(G)+n]$
respectively. However, there are only two numbers
${\mathfrak{eu}}^{1}_{max}(G)$, ${\mathfrak{eu}}^{0}_{min}(G+e_{1})+n-1$ in
$[{\mathfrak{eu}}^{1}_{max}(G),{\mathfrak{eu}}^{0}_{min}(G+e_{1})+n-1]$ and
two numbers ${\mathfrak{eu}}^{1}_{max}(G)$, ${\mathfrak{eu}}^{0}_{min}(G)+n$
in $[{\mathfrak{eu}}^{1}_{max}(G),{\mathfrak{eu}}^{0}_{min}(G)+n]$, it is
impossible.

Combining (4.16)-(4.20), we finish the proof.

∎

######  Theorem 4.5.

[GMT21a] Let $G=G_{1}\vee G_{2}$. Then

$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}(z)$ $\displaystyle=$
$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{\times}_{G_{1}}(z)~{}^{\partial}{{\mathscr{E}}}^{\times}_{G_{2}}(z).$

######  Remark 4.6.

It is easy to see that Theorem 1.3 still holds for
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}|_{O}(z)$.

Now, let’s take two examples to illustrate the above results.

Figure 4.1. $C_{4}+\sum^{n}_{i=1}e_{i}$

######  Example 4.2.

Let $C_{m}$ be the planar ribbon $m$-cycle. Evidently,
$~{}^{\partial}{{\mathscr{E}}}^{\times}_{C_{2}}|_{O}(z)=2.$ Using the fact
that $C_{m}$ can be obtained from $C_{2}$ by subdividing a ribbon $m-2$ times,
we have

$\displaystyle~{}^{\partial}{{\mathscr{E}}}^{\times}_{C_{m}}|_{O}(z)$
$\displaystyle=$ $\displaystyle
2^{m-2}~{}^{\partial}{{\mathscr{E}}}^{\times}_{C_{2}}|_{O}(z)\quad\text{by
Theorem \ref{sub}}$ $\displaystyle=$ $\displaystyle 2^{m-1}.$

The ribbon graph $C_{m}+\sum^{n}_{i=1}e_{i}$ is obtained from $C_{m}$ by
attaching $n$ parallel edges $e_{1}$, $e_{2}$, $\cdots$, $e_{n}$ to the ribbon
$e$, as shown in Figure 4.1. It is clear that $f^{0}(C_{m}^{\times|_{A}})=2$.
Since the number of faces of orientable ribbon graph
$(C_{m}+e_{1})^{\times|_{B}}$ with $e\in B$ is 1, it follows that
$f^{0}_{max}(C_{m}+e_{1})=f^{0}(C_{m}^{\times|_{A}})-1=1$. By Theorem 4.4 (1),
when $n\geq 3,$ the restricted-orientable partial-$\times$ polynomial of
$C_{m}+\sum^{n}_{i=1}e_{i}$ is not even-interpolating.

Let $H=D_{n}\vee C_{2}$, then by Theorem 4.5 and Example 4.1, the restricted-
orientable partial-$\times$ polynomial of $H$ is not even-interpolating for
$n\geq 5$.

Figure 4.2. $G$ and $G+\sum^{n}_{i=1}\bar{e}_{i}$

######  Example 4.3.

Let $G$ and $G+\sum^{n}_{i=1}\bar{e}_{i}$ be the ribbon graphs in Figure 4.2.
When
$A\in\\{\\{e_{1},e_{3},e\\},\\{e_{2},e_{3},e\\},\\{e_{1},e_{3},e_{4}\\},\\{e_{2},e_{3},e_{4}\\}\\}$,
the number of faces of orientable ribbon graph $G^{\times|_{A}}$ is 1. When
$A\in\\{\\{e_{1},e_{2}\\},\\{e_{4},e\\},\\{e_{1},e_{2},e_{4},e\\},\emptyset\\}$,
the number of faces of orientable ribbon graph $G^{\times|_{A}}$ is 3. We
infer that $~{}^{\partial}{{\mathscr{E}}}^{\times}_{G}|_{O}(z)=4+4z^{2},$
$f^{0}_{max}(G)=3$, and $f^{1}_{min}(G)=1$. When
$B\in\\{\\{e_{1},e_{3},e,\bar{e}_{1}\\},\\{e_{2},e_{3},e,\bar{e}_{1}\\},$
$\\{e_{1},e_{2},e_{4},e,\bar{e}_{1}\\},\\{e_{4},e,\bar{e}_{1}\\}\\}$, we have
$f((G+\bar{e}_{1})^{\times|_{B}})=2$, thus,

$\displaystyle f^{0}_{max}(G+\bar{e}_{1})$ $\displaystyle=$ $\displaystyle
2~{}<~{}f^{0}_{max}(G).$

Therefore, by Theorem 4.4 (2), when $n\geq 2-1+4=5,$ the restricted-orientable
partial-$\times$ polynomial of $G+\sum^{n}_{i=1}\bar{e}_{i}$ is not even-
interpolating.

###  4.1. Acknowledgments

We are grateful to the anonymous referees for their valuable comments.

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