In this section, we characterize a set of jump values *S*^{*} to a circulant graph *C*(*n*, *S*^{*}) admitting a restricted triangulation (Theorem 2.8). Also, we define the ‘smallest’ cardinality set of jump values *S*(*n*) such that *C*(*n*, *S*(*n*)) still admits a restricted triangulation. We give a characterization of a circulant graph *C*(*n*, *S*) that can be redrawn to admit a restricted triangulation (Corollary 2.13). These results are then applied to determine the convex skewness of the circulant graphs *G* in section (2.2).

#### Lemma 2.1

*If T*_{n} is a restricted triangulation of a circulant graph G = *C*(*n*, *S*), *then S contains D*(*T*_{n}).

#### Proof

Suppose *T*_{n} is a restricted triangulation of *G* = *C*(*n*, *S*). Let *d* ∈ *D*(*T*_{n}). By definition of span of edges, there is an edge *v*_{i}v_{j} in *T*_{n} such that *v*_{i}v_{j} = *v*_{i}v_{i± d} where *d* = min{|*j* − *i*|, *n* − |*j* − *i*|}. Since *T*_{n} is a restricted triangulation of *G*, then *E*(*T*_{n}) ⊂ *E*(*G*). Hence, *v*_{i}*v*_{i ± d} ∈ *E*(*G*). Thus, *d*∈ *S* (by definition of *S*) □

The following proposition proves that any circulant graph *C*(*n*, *S*) does not admit a restricted triangulation unless 1, 2 ∈ *S*.

#### Proposition 2.2

*Let n* ≥ 4 *be a natural number*. *Suppose the circulant graph C*(*n*, *S*) *admits a restricted triangulation*. *Then* 1, 2 ∈ *S*.

#### Proof

Suppose that *C*(*n*, *S*) is a circulant graph. Let *T*_{n} be a restricted triangulation of *C*(*n*, *S*).

It is well known that any triangulation on a set of point in the plane includes the boundary of the convex hull. For our case, *v*_{i}v_{i+1} ∈ *E*(*T*_{n}) for each *i* = 0, 1, …, *n* − 1 which yields 1 ∈ *D*(*T*_{n}) and then 1 ∈ *S* by Lemma 2.1.

Every maximal outer planar graph (as a special case, triangulation on a convex polygone) has at least two vertices of degree 2 (see [12]). Hence, if *v*_{i+1} is a vertex of degree 2 in *T*, then the diagonal edge *v*_{i}v_{i+2} must be in *E*(*T*_{n}) because *v*_{i+1} is incident just on two edges of *T*_{n} which are the boundary edges *v*_{i}v_{i+1} and *v*_{i+1}*v*_{i+2}. Thus, 2 ∈ *D*(*T*_{n}) since 2 is the distance between *v*_{i} and *v*_{i+1}. Then 2 ∈ *S* by Lemma 2.1. □

It is not difficult to verify that the converse of Proposition 2.2 is not true. For instance, *G*_{1} = *C*(12, *S*_{1}) where *S*_{1} = {1, 2, 3} does not admit any restricted triangulation, while each of *G*_{2} = *C*(12, *S*_{2}) and *G*_{3} = *C*(12, *S*_{3}) admits a restricted triangulation with *S*_{2} = {1, 2, 4} and *S*_{3} = {1, 2, 3, 5}. Note that |*S*_{2}| = |*S*_{1}| while *S*_{3} = *S*_{1} ∪ {5}. Hence, the question arises: what conditions on *S* do guarantee that *C*(*n*, *S*) admits a restricted triangulation and what is the *smallest* cardinality |*S*| of the set of jump values *S* for which *C*(*n*, *S*) admits a restricted triangulation?

The following proposition proves that the circulant graph *C*(*n*, *S*_{n}) admits a restricted triangulation where *S*_{n} is a set of ascending values {*a*_{1}, *a*_{2}, …, *a*_{s}} in which *a*_{1} = 1, *a*_{2} = 2, *a*_{s} =
$\begin{array}{}\lfloor \frac{n}{2}\rfloor \end{array}$ and *a*_{i+1} − *a*_{i} ∈ *S*_{n} for each *i* ∈ {1, 2, …, *s* − 1}.

#### Proposition 2.3

*Let n* ≥ 4 *be a natural number*. *Suppose S*_{n} is a set of ascending values {1, 2, *a*_{3}, …, *a*_{s}} *in which a*_{s} =
$\begin{array}{}\lfloor \frac{n}{2}\rfloor \end{array}$ *and S*_{n} has the property a_{i+1} − *a*_{i} ∈ *S*_{n} for each i ∈ {1, 2, …, *s* − 1}. *Then the circulant graph C*(*n*, *S*_{n}) *admits a restricted triangulation*.

#### Proof

Suppose that *C*(*n*, *S*_{n}) is a circulant graph. Since *a*_{s} ∈ *S*_{n} is a jump value, then *v*_{0}*v*_{as} and *v*_{n − as} *v*_{0} are two edges in *C*(*n*, *S*_{n}). Thus, we can define *G*_{1} and *G*_{2} to be two subgraphs of *C*(*n*, *S*_{n}), induced by the vertices *v*_{0}, *v*_{1}, *v*_{2}, …, *v*_{as} and *v*_{n − as}, *v*_{n − as+1}, …, *v*_{n − 1}, *v*_{0}, respectively. See Figure 1.

Fig. 1 *Q*_{aji+2}(*a*_{i}) on *G*_{1}

We shall construct the triangulation *T*_{G1} of *G*_{1}, and then *G*_{2} can be triangulated in a similar way.

(∗) It is important to mention that in the following argument the property *a*_{i+1} − *a*_{i} ∈ *S*_{n}, for each two consecutive values *a*_{i} and *a*_{i+1} in *S*_{n}, and 1, 2 ∈ *S*_{n}, are basic tools to construct *T*_{G1}.

Let *a*_{i}, *a*_{i+1} be two consecutive vertices in *S*_{n} and *a*_{i+1} − *a*_{i} = *a*_{ji} ∈ *S*_{n} for some *a*_{ji} ∈ {1, 2, *a*_{3}, …, *a*_{i}} (since *S*_{n} is a set of ascending values). Let *v*_{0} *v*_{ai} *v*_{ai+1} … *v*_{ai+1} *v*_{0} be a convex (*a*_{ji}+2)-gon in *G*_{1} denoted by *Q*_{aji+2}(*v*_{ai}). Clearly
$\begin{array}{}{G}_{1}={\cup}_{i=1}^{s-1}{Q}_{{a}_{{j}_{i}}+2}({v}_{{a}_{i}}).\end{array}$ First, we prove *Q*_{aji+2}(*v*_{ai}) can be triangulated by edges of *C*(*n*, *S*_{n}) for each *i* = 1, …, *s* − 1.

Note that *a*_{1} = 1, *a*_{2} = 2 and *a*_{3} ∈ {3, 4} (otherwise, if *a*_{3} = 5 then *a*_{3} − *a*_{2} = 5 − 2 = 3 ∉ *S*_{n}, a contradiction). It is easy to see that *Q*_{3}(*v*_{1}) ( = *v*_{0}*v*_{1}*v*_{2}*v*_{0}) is triangulated by *v*_{0}*v*_{1}*v*_{2}*v*_{0} and when *a*_{3} = 3 we have *Q*_{3}(*v*_{2}) ( = *v*_{0}*v*_{2}*v*_{3}*v*_{0}) is triangulated by *v*_{0}*v*_{2}*v*_{3}*v*_{0}, and also when *a*_{3} = 4 we have that *Q*_{4}(*v*_{2}) ( = *v*_{0}*v*_{2}*v*_{3}*v*_{4}*v*_{0}) is triangulated by the boundary edges *v*_{0}*v*_{2}*v*_{3}*v*_{4}*v*_{0} together with the diagonal edge *v*_{2}*v*_{4}.

Now we obtain a triangulation for *Q*_{aji+2}(*v*_{ai}), *i* = 3, …, *s* − 1.

Let *A*_{ai+1}(*v*_{ai}) = *v*_{0}*v*_{1} … *v*_{ai} *v*_{0}, be a convex (*a*_{i}+1)-gon in *G*_{1}. Clearly, *A*_{3}(*v*_{2}) ( = *v*_{0}*v*_{1}*v*_{2}*v*_{0}) is a triangle that is trivially triangulated by *T*_{1} = *v*_{0}*v*_{1}*v*_{2}*v*_{0}. If *i* = 3, then either *a*_{3} = 3 and *A*_{4}(*v*_{3}) ( = *v*_{0}*v*_{1}*v*_{2}*v*_{3}*v*_{0}) is a quadrilateral that is triangulated by *T*_{2} = *v*_{0}*v*_{2} ∪ *v*_{0}*v*_{1}*v*_{2}*v*_{3}*v*_{0}, or *a*_{3} = 4 and *A*_{5}(*v*_{4}) ( = *v*_{0}*v*_{1}*v*_{2}*v*_{3}*v*_{4}*v*_{0}) is a pentagon triangulated by *T*_{2} = *v*_{0}*v*_{2}*v*_{4} ∪ *v*_{0}*v*_{1}*v*_{2}*v*_{3}*v*_{4}*v*_{0}.

Let *B*_{aji+1}(*v*_{ai}) = *Q*_{aji+2}(*v*_{ai}) − *v*_{0}. Then *B*_{aji+1}(*v*_{ai}) = *v*_{ai}*v*_{ai+1} … *v*_{ai+aji} *v*_{ai} (since *a*_{i+1} − *a*_{i} = *a*_{ji}) which is equivalent to the convex polygon *v*_{0}*v*_{1} … *v*_{aji}v_{0} (by considering *a*_{i} = 0). Thus, *B*_{aji+1}(*v*_{ai}) is equivalent *A*_{aji+1}(*v*_{aji}).

If *i* = 3, we have *a*_{4} − *a*_{3} = *a*_{j3} ∈ *S*_{n} for some *a*_{j3} ∈ {1, 2, *a*_{3}}. Then, *B*_{aj3+1}(*a*_{3}) ( = *v*_{a3}*v*_{a3+1} … *v*_{a4} *v*_{a3} = *v*_{a3}*v*_{a3+1} … *v*_{a3+aj3} *v*_{a3}) is equivalent to *A*_{aj3+1}(*a*_{j3}) ( = *v*_{0}*v*_{1} … *v*_{aj3}*v*_{0}) which is triangulated by *T* ∈ {*T*_{1}, *T*_{2}}. Then, *B*_{aj3+1}(*a*_{3}) can be triangulated by *T*′ equivalent to *T*. Hence, *T*_{3} = *T*′ ∪ *v*_{a3}*v*_{0}*v*_{a4} is a triangulation of *Q*_{aj3+2}(*v*_{a3}).

Recursively, if we have *i* = *s* − 1 then *a*_{s} − *a*_{s − 1} = *a*_{js − 1} ∈ *S*_{n} for some *a*_{js − 1} ∈ {1, 2, *a*_{3}, …, *a*_{s − 1}}. Then, *B*_{ajs − 1+1}(*v*_{as − 1}) is equivalent to *A*_{ajs − 1+1}(*a*_{js − 1}) which is triangulated by *T* ∈ {*T*_{1}, *T*_{2}, *T*_{3}, …, *T*_{s − 2}} (depending on the value of *a*_{js − 1} ∈ {1, 2, *a*_{3}, …, *a*_{s − 1}}). Then, *B*_{ajs − 1+1}(*v*_{as − 1}) can be triangulated by *T*′ equivalent to *T*. Hence, *T*_{s − 1} = *T*′ ∪ *v*_{as − 1}*v*_{0}*v*_{as} is a triangulation of *Q*_{ajs − 1+2}(*v*_{as − 1}).

Since,
$\begin{array}{}{G}_{1}={\cup}_{i=1}^{s-1}{Q}_{{a}_{{j}_{i}}+2}({v}_{{a}_{i}}),\end{array}$ then let
$\begin{array}{}{T}_{{G}_{1}}={\cup}_{i=1}^{s-1}{T}_{i}\end{array}$ be the triangulation of *G*_{1}. In a similar way we obtain *T*_{G2}. Thus, *T*_{n} = *T*_{G1} ∪ *T*_{G2} ∪ {*v*_{a}_{s}v_{n − as}}, where *a*_{s} ≠ *n* − *a*_{s}; otherwise {*v*_{as}v_{n − as}} = ∅ is a restricted triangulation of *G*. □

#### Corollary 2.4

*Let n* ≥ 4 *be a natural number*. *Suppose G* = *C*(*n*, *S*) *is a circulant graph*. *Then G admits a restricted triangulation if one of the following conditions hold*.

*S* = {1, 2, *a*_{3}, *a*_{4}, …,
$\begin{array}{}\lfloor \frac{n}{2}\rfloor \end{array}$} *where*
$\begin{array}{}{a}_{i}^{\prime}\end{array}$*s are consecutive odd values*.

*S* = {1, 2, *a*_{3}, *a*_{4}, …,
$\begin{array}{}\lfloor \frac{n}{2}\rfloor \end{array}$} *where*
$\begin{array}{}{a}_{i}^{\prime}\end{array}$ *are consecutive even values*.

#### Proof

In both cases *a*_{1} = 1, *a*_{2} = 2 and
$\begin{array}{}\lfloor \frac{n}{2}\rfloor \end{array}$ are in *S*. Further, for each *a*_{i+1}, *a*_{i} ∈ *S*_{n}, *a*_{i+1} − *a*_{i} ∈ {1, 2} ⊆ *S*_{n}. Then the circulant graph *C*(*n*, *S*) admits a restricted triangulation by Proposition 2.3. □

Now, we shall define a smallest size circulant graph that admits a restricted triangulation. Before proceeding, we present some ingredients that will be used to prove the main results in this section.

–

When *n* is even number, then *n* can be written as *n* = *t*.2^{r′} for some positive natural number *r*′ and some positive odd natural number *t*.

–

Let *S*_{α} be a set of jump values obtaining by next algorithm where *α* =
$\begin{array}{}\lfloor \frac{t}{3}\rfloor \end{array}$ and *β* =
$\begin{array}{}\lfloor \frac{t}{3}\rfloor \end{array}$ with *t* =
$\begin{array}{}\left\{\begin{array}{ll}n,& n\text{\hspace{0.17em}is odd ;}\\ t,& n=t{.2}^{{r}^{\prime}}.\end{array}\right.\end{array}$

–

Let *S*_{1} = {1, 2, 4, 8, …, *c*} with *c* = 2^{r} where
$\begin{array}{}r=\left\{\begin{array}{ll}0,& n\text{\hspace{0.17em}is odd ;}\\ {r}^{\prime}-1,& t=1;\\ {r}^{\prime},& t\ge 3.\end{array}\right.\end{array}$

–

If we have a set *B* = {*b*_{1}, *b*_{2}, …, *b*_{k}} of positive integers, then define *a*.*B* to be {*a*.*b*_{1}, *a*.*b*_{2}, …, *a*.*b*_{k}} where *a* ≥ 1 is an integer number.

#### Algorithm (*A*)

If *α* = 0, then let *S*_{α} = {*β*} and Stop. If *α* = 1, then let *S*_{α} = {1, *β*} and Stop. If 2 ≤ *α* ≤ 4, then let *S*_{α} = {1, 2, *α*, *β*} and Stop. Otherwise, let *L* = 0 be the number of starting level. Let *S*(*L*) = {1, 2, *β*}, *a*_{0} = *α*, and let *L* = *L*+1.

2- If *a*_{0} is odd and divisible by 3, then let *a*_{1, L} =
$\begin{array}{}\frac{{a}_{0}}{3}\end{array}$ and *a*_{2, L} = 2.*a*_{1, L}. Otherwise, let *a*_{1, L} = ⌊
$\begin{array}{}\frac{{a}_{0}}{2}\end{array}$⌋ and *a*_{2, L} = ⌈
$\begin{array}{}\frac{{a}_{0}}{2}\end{array}$⌉.

Let *S*(*L*) = {*a*_{1, L}, *a*_{2, L}, *a*_{0}}. Let
$\begin{array}{}{S}_{\alpha}=\bigcup _{l=0}^{L}S(l).\end{array}$

If 3 ∈ {*a*_{1, L}, *a*_{2, L}} and each of *α* and
$\begin{array}{}\lfloor \frac{\alpha}{2}\rfloor \end{array}$ is odd and not divisible by 3, then do the following:

Let *L* = 0 be the number of starting level. Let *S*(*L*) = {1, 2, 3, *α* − 3, *α*, *β*}, *a*_{0} = *α* − 3, and let *L* = *L*+1.

If *a*_{0} is odd and not divisible by 3, then let *a*_{1, L} = *a*_{2, L} = *a*_{0} − 3. If *a*_{0} is odd and divisible by 3, then let *a*_{1, L} =
$\begin{array}{}\frac{{a}_{0}}{3}\end{array}$ and *a*_{2, L} = 2.*a*_{1, L}. Otherwise, *a*_{1, L} = *a*_{2, L} =
$\begin{array}{}\frac{{a}_{0}}{2}\end{array}$.

Let *S*(*L*) = {*a*_{1, L}, *a*_{2, L}, *a*_{0}}. Let
$\begin{array}{}{S}_{\alpha}=\bigcup _{l=0}^{L}S(l).\end{array}$

If *a*_{1, L} ≤ 3, then arrange *S*_{α} to be a set of ascending values and stop. Otherwise, let *a*_{0} = *a*_{1, L} and *L* = *L*+1 and then repeat Step (ii).

If *a*_{1, L} ≤ 4, then arrange *S*_{α} to be a set of ascending values and stop. Otherwise, let *a*_{0} = *a*_{1, L} and *L* = *L*+1 and then repeat Step (2).

Note that, *S*_{α} that obtained by Algorithm (*A*) is a set of ascending values 1, 2, *a*_{i − 1}, *a*_{i}, *a*_{i+1}, …, *α*, *β*.

#### Lemma 2.5

*Suppose S*_{α} = {*a*_{1}, *a*_{2}, …, *α*, *β*} *is a set obtained by Algorithm (A)*. *Then* 1, 2 ∈ *S*_{α} and a_{i+1} − *a*_{i} ∈ *S*_{α} *for each a*_{i+1}, *a*_{i} belong to S_{α}.

#### Proof

By Algorithm (*A*) step (1), we get 1, 2 ∈ *S*_{α}. Suppose that, *a*_{i}, *a*_{i+1} are two consecutive values belonging to *S*_{α}.

If *a*_{i+1} = *a*_{0} is an odd and divisible by 3 number, then by Algorithm (*A*) Step (2),
$\begin{array}{}{a}_{i}={a}_{2,L}=\frac{2.{a}_{i+1}}{3}.\end{array}$ Thus,
$\begin{array}{}{a}_{i+1}-{a}_{i}={a}_{i+1}-\frac{2.{a}_{i+1}}{3}=\frac{{a}_{i+1}}{3}={a}_{i-1}\in {S}_{\alpha}\end{array}$ (where
$\begin{array}{}{a}_{i-1}={a}_{1,L}=\frac{{a}_{i+1}}{3}\in {S}_{\alpha}).\end{array}$

If 3 ∈ *S*_{α} and *a*_{i+1} = *a*_{0} is an odd and not divisible by 3 number, then by Algorithm (*A*) Step (4), *a*_{i} = *a*_{i+1} − 3. Thus, *a*_{i+1} − *a*_{i} = 3 ∈ *S*_{α}.

Otherwise, by Algorithm (*A*) Step (2),
$\begin{array}{}{a}_{i}=\lceil \frac{{a}_{i+1}}{2}\rceil (\text{where\hspace{0.17em}}\lceil \frac{{a}_{i+1}}{2}\rceil ={a}_{2,L}\in {S}_{\alpha}).\end{array}$ Hence,
$\begin{array}{}{a}_{i+1}-{a}_{i}={a}_{i+1}-\lceil \frac{{a}_{i+1}}{2}\rceil =\lfloor \frac{{a}_{i+1}}{2}\rfloor ={a}_{i-1}\in {S}_{\alpha}\end{array}$ (where
$\begin{array}{}{a}_{i-1}=\lfloor \frac{{a}_{i+1}}{2}\rfloor ={a}_{1,L}\in {S}_{\alpha}).\end{array}$ In case when,
$\begin{array}{}\lceil \frac{{a}_{i+1}}{2}\rceil =\lfloor \frac{{a}_{i+1}}{2}\rfloor =\frac{{a}_{i+1}}{2},\end{array}$ then we have
$\begin{array}{}{a}_{i}=\frac{{a}_{i+1}}{2}.\end{array}$ Thus,
$\begin{array}{}{a}_{i+1}-{a}_{i}={a}_{i+1}-\frac{{a}_{i+1}}{2}=\frac{{a}_{i+1}}{2}={a}_{i}\in {S}_{\alpha}.\end{array}$

#### Theorem 2.6

*Let S*(*n*) = *S*_{1} ∪ *c*.*S*_{α}, *where n* ≥ 4. *Then C*(*n*, *S*(*n*)) *admits a restricted triangulation*.

#### Proof

Suppose that *C*(*n*, *S*(*n*)) is a circulant graph and let *α*, *β*, *c* and *t* is defined as above. We shall construct a restricted triangulation *T*_{n} of *C*(*n*, *S*(*n*)).

When *n* is even and *t* = 1, *S*(*n*) = *S*_{1} (In this case, *β* = 1 and then *S*_{α} = {1} and then *c*. *S*_{α} = {*c*} and *c* ∈ *S*_{1}). Then let $\begin{array}{}{T}_{n}=\bigcup _{i=0}^{r}\{{v}_{j{2}^{i}}{v}_{(j+1){2}^{i}},j=0,1,\dots ,\frac{n}{{2}^{i}}-1\}.\end{array}$ The circulant graph when *n* = 8 is depicted in Figure 2.

Fig. 2 *n* = 8 = 2^{3}, *t* = 1, *r* = *r*′ − 1 = 2, *T*_{8} = {*v*_{j} *v*_{(j+1)}, *j* = 0, …, 7} ∪ {*v*_{2j} *v*_{2(j+1)}, *j* = 0, …, 3} ∪ {*v*_{4j} *v*_{4(j+1)}, *j* = 0, 1}.

When *n* is even and *t* ≥ 3, *S*(*n*) = *S*_{1} ∪ *c*.*S*_{α}. In this case, and also when *n* is odd (which means *t* = *n*), we have *t*-gon which is induced by the vertices *v*_{0}, *v*_{c}, *v*_{2.c}, …, *v*_{(t − 2).c}, *v*_{(t − 1).c} (the shaded part in Figure 3).

We shall triangulate this *t*-gon by *T*′ which is obtained by one of the following three cases depending on *S*_{α}. According to *S*_{1}, let $\begin{array}{}T=\bigcup _{i=0}^{r}\{{v}_{j{2}^{i}}{v}_{(j+1){2}^{i}},j=0,1,\dots ,\frac{n}{{2}^{i}}-1\}\end{array}$ which triangulates unshaded part, between *n*-gon and *t*-gon, in Figure 3.

Fig. 3 *n* = 56 = 7.2^{3}, *t* = 7, *r* = *r*′ = 3, *T*_{56} = {*v*_{j} *v*_{(j+1)}, *j* = 0, …, 55} ∪ {*v*_{2j} *v*_{2(j+1)}, *j* = 0, …, 27} ∪ {*v*_{4j} *v*_{4(j+1)}, *j* = 0, …, 13} ∪ {*v*_{8j} *v*_{8(j+1)}, *j* = 0, …, 6}.

Then, let *T*_{n} = *T* ∪ *T*′ or *T*_{n} = *T*′ be a triangulation to *C*(*n*, *S*(*n*)) when *n* is even with *t* ≥ 3 or when *n* is odd, respectively.

Case (1)

When *α* = $\begin{array}{}\frac{t-2}{3}\end{array}$, then *t* = 2*β*+*α* (since *β* = *α*+1). Let △ = *v*_{0}*v*_{α}*v*_{α+β}*v*_{0} be a triangle. Then there are three (*α* + 1)-gons *G*_{1}, *G*_{2} and *G*_{3} induced by the vertices *v*_{0}, *v*_{c}, *v*_{2.c}, …, *v*_{α.c}; *v*_{(α).c}, *v*_{(α+1).c}, …, *v*_{2}*α*.*c* and *v*_{(α+β).c}, *v*_{(2α+2).c}, …, *v*_{(3α+1).c}, respectively. See Figure 4.

Fig. 4 *n* = 164, *t* = 41, *r* = 2, *α* = 13, and *β* = 14.

Case (2)

When *α* = $\begin{array}{}\frac{t-1}{3}\end{array}$, then *t* = 2*α*+*β* (since *β* = *α* + 1). Let △ = *v*_{0}*v*_{α}*v*_{2α}*v*_{0} be a triangle. Then there are three (*α* + 1)-gons *G*_{1}, *G*_{2} and *G*_{3}, induced by the vertices *v*_{0}, *v*_{c}, *v*_{2.c}, …, *v*_{α.c}; *v*_{(α).c}, *v*_{(α+1).c}, …, *v*_{2}*α*.*c* and *v*_{(2α).c}, *v*_{(2α + 1).c}, …, *v*_{(3α).c}, respectively. See Figure 5.

Fig. 5 *n* = 152, *t* = 19, *r* = 3, *α* = 6, and *β* = 7.

Case (3)

When *α* = $\begin{array}{}\frac{t}{3}\end{array}$, then *t* = 3*α*. There are three (*α* + 1)-gons *G*_{1}, *G*_{2} and *G*_{3} induced by the vertices *v*_{0}, *v*_{c}, *v*_{2.c}, …, *v*_{α.c}; *v*_{(α).c}, *v*_{(α+1).c}, …, *v*_{2}*α*.*c* and *v*_{2}*α*.*c*, *v*_{(2α+1).c}, …, *v*_{(3α−1).c}, *v*_{0}, respectively. See Figure 6.

In each case, *G*_{1}, *G*_{2} and *G*_{3} are of the same order. Suppose that *R* is an *r*-gon in *G*_{i}, for some *i* ∈ {1, 2, 3} such that *V*(*R*) = {*v*_{h}, *v*_{h+ai}, *v*_{h}+*a*_{i+1}, …, *v*_{h+ai+1}} for some *v*_{h} ∈ *V*(*G*_{i}) and two consecutive jump values *a*_{i} and *a*_{i+1} in *S*_{α}.

Without loss of generality assume that *R* is a subgraph of *G*_{1} and *h* = 0.

By Lemma 2.5, we have 1, 2 ∈ *S*_{α} and *a*_{i+1} − *a*_{i} ∈ *S*_{α} for each *a*_{i+1}, *a*_{i} belonging to *S*_{α}. Hence, we can use the argument (∗) of Proposition 2.3 to show that *R* can be triangulated by *T*_{r}. Moreover, similar as in the proof of Proposition 2.3 we can consider *G*_{1} as a finite union of such polygons. Then we can assume that *T*_{G1} is a finite union of the restricted triangulation of these polygons.

Obtain *T*_{G2} (on *G*_{2}) and *T*_{G3} (on *G*_{3}) by “rotating” the edges of *T*_{G1}, see Figures 4, 5 and 6. Let *T*′ = *T*_{G1} ∪ *T*_{G2} ∪ *T*_{G3} ∪ △ for case (1) and case (2), and let *T*′ = *T*_{G1} ∪ *T*_{G2} ∪ *T*_{G3} for case (3).

Fig. 6 *n* = 120, *t* = 15, *r* = 3, and *α* = 5.

This completes the proof.

□

#### Definition 2.7

*Let* *n* ≥ 4 *and* *t* *be defined as above*, *and let* *a*_{k} *be a natural number such that* ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* ≤ *a*_{k} ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋. *Then define* *S*^{*} *to be a set of ascending natural numbers such that*,

*S*^{*} = {1, 2, *a*_{3}, *a*_{4}, …, *a*_{k}},

*there is* {*x*, *z*} ⊂ *S*^{*} *such that* *n* = *x* + *y* + *z where* *y* ∈ {0, *a*_{i}} *for some* *i* ∈ {1, …, *k*},

*a*_{i+1} − *a*_{i} ∈ *S*^{*} *for each* *i* = 1, …, *k* − 1.

It is clear that, *S*^{*} = *S*_{n} when *a*_{k} = ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋. Our main result, Theorem 2.8, proves the sufficient and necessary condition for a circulant graph *C*(*n*, *S*) to admit a restricted triangulation.

#### Theorem 2.8

*Suppose* *G* = *C*(*n*, *S*) *is a circulant graph*. *G* *admits a restricted triangulation if and only if* *S* *contains* *S*^{*}.

#### Proof

Let *G* have a restricted triangulation *T*_{n}. We shall prove that *S* contains *S*^{*}. By Lemma 2.1, it is enough to prove *S*^{*} = *D*(*T*_{n}).

Arrange the spans of *D*(*T*_{n}) to be ascending values.

Since, the circulant graph *C*(*n*, *D*(*T*_{n})) admits *T*_{n}, then by Proposition 2.2, we have {1, 2} ⊆ *D*(*T*_{n}).

Let |*D*(*T*_{n})| = *s*, then *d*_{s} is the maximum span in *D*(*T*_{n}).

(⋆) Let *t* be defined as before. Suppose that *d*_{s} < ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c*. Without loss of generality assume that, *c* = 1 and *d*_{s} = ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ − 1. Thus, *d*_{s} = $\begin{array}{}\frac{t-b}{3}\end{array}$ − 1, *b* ∈ {0, 1, 2} and then 3*d*_{s} + *b* + 3 = *t*. Let *R* be a convex *r*-gon induced by the vertices *v*_{ds}, *v*_{2ds}, *v*_{3ds}, *v*_{3}*d*_{s+1}, …, *v*_{3ds+b+1}, *v*_{3ds+b +2}, *v*_{t} (recall that, *v*_{t} = *v*_{3ds+b+3}). Suppose that *T*_{r} = *T*_{n}(*R*) is a subgraph of *T*_{n} that triangulates *R* and let **e** be an edge in *T*_{r} such that the span of **e** is the maximum with respect to *D*(*T*_{r}). Then:

–

either **e** = *v*_{2ds}*v*_{t} which yields that its span *t* − 2*d*_{s} ∈ *D*(*T*_{r}) ⊂ *D*(*T*_{n}); but *t*-2*d*_{s} > *d*_{s} + 1 (by above assumption, *t* − 2*d*_{s} = *d*_{s} + *b* + 3 > *d*_{s} + 1), which is a contradiction with maximality of *d*_{s} ∈ *D*(*T*_{n});

–

or, **e** = *v*_{d}_{s}v_{3}*d*_{s} which yields that 3*d*_{s} − *d*_{s} = 2*d*_{s} ∈ *D*(*T*_{r}) ⊂ *D*(*T*_{n}); but 2*d*_{s} > *d*_{s}, which also contradicts the maximality of *d*_{s} ∈ *D*(*T*_{n}).

Hence, *R* is not triangulated by *T*_{n} which is a contradiction with *C*(*n*, *S*) admitting a restricted triangulation *T*_{n}. Thus, *d*_{s} ≥ ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c*.

Now, in order to check property (ii) of *S*^{*} we have to consider two cases:

Case (1)

If *d*_{s} = $\begin{array}{}\frac{n}{2}\end{array}$, then *n* = 2.*d*_{s}. Hence let *x* = *z* = *d*_{s} and *y* = 0.

Case (2)

If ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* ≤ *d*_{s} ≤ $\begin{array}{}\frac{n-1}{2}\end{array}$, then *T*_{n} contains a triangle △ = *v*_{h}*v*_{h+ds}*v*_{h+ds+di}*v*_{h} for some *d*_{i} ∈ *D*(*T*_{n}) and *v*_{h} ∈ *V*(*G*). The span of the edge *v*_{h+ds+di}*v*_{h} ∈ *E*(*T*_{n}) is *n* − (*d*_{s}+*d*_{i}) ∈ *D*(*T*_{n}) (by definition of the span of edge). Hence let *x* = *d*_{s}, *y* = *d*_{i} and *z* = *n* − (*d*_{s}+*d*_{i}), then *n* = *x* + *y* + *z*.

To check the property (iii) of *S*^{*}, assume that *d*_{i} and *d*_{i+1}, *i* ∈ {1, 2, …, *s* − 1} are two consecutive spans in *D*(*T*_{n}).

If *d*_{s} = $\begin{array}{}\frac{n}{2}\end{array}$, then let *T*_{1} and *T*_{2} be two subgraphs of *T*_{n} induced by the vertices *v*_{0}, *v*_{1}, *v*_{2}, …, *v*_{x}; *v*_{x}, *v*_{x+1}, …, *v*_{x}+*y* − 1, *v*_{0} respectively (*x* = $\begin{array}{}\frac{n}{2}\end{array}$). Let *R* be a polygon induced by the vertices *v*_{j}, *v*_{j}+*d*_{i}, …, *v*_{j}+*d*_{i+1} where *v*_{j} ∈ *V*(*T*_{i}) and *v*_{j} *v*_{j+di} and *v*_{v}*v*_{j+di+1} are two diagonals in *T*_{i} for some *i* ∈ {1, 2}.

If ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* ≤ *d*_{s} ≤ $\begin{array}{}\frac{n-1}{2}\end{array}$, then let *T*_{1}, *T*_{2} and *T*_{3} be three subgraphs of *T*_{n} induced by the vertices *v*_{0}, *v*_{1}, *v*_{2}, …, *v*_{x}, *v*_{x}, *v*_{x+1}, …, *v*_{x+y}, and *v*_{x+y}, *v*_{x+y+1}, …, *v*_{x+y+z−1}, *v*_{0} respectively. Let *R* be a polygon induced by the vertices *v*_{j}, *v*_{j+di}, …, *v*_{j}+*d*_{i+1} where *v*_{j} ∈ *V*(*T*_{i}) and *v*_{j}*v*_{j+di} and *v*_{j}*v*_{j+di+1} are two diagonals in *T*_{i} for some *i* ∈ {1, 2, 3}.

Assume without loss of generality that *v*_{j}*v*_{j+di} and *v*_{j}v_{j+di+1} are two diagonals in *T*_{1}. Then, *j* < *j* + *d*_{i} < *j* + *d*_{i+1} (by definitions of *T*_{1} and *R*). It is clear that, *v*_{j} *v*_{j+di} … *v*_{j}+*d*_{i+1} *v*_{j} ⊂ *T*_{1}. Let *T*_{r} = *T*_{1}(*R*) be a subgraph of *T*_{1} that triangulates *R* with the boundary edges *v*_{j} *v*_{j+di} … *v*_{j}+*d*_{i+1} *v*_{j} of *R*.

Suppose that *v*_{j+di}*v*_{j+di+1} ∉ *E*(*T*_{r}). Then there is *d*_{i} < *d* < *d*_{i+1} such that *v*_{j}v_{j}_{+d} ∈ *E*(*T*_{r}). This implies that, *d* ∈ *D*(*T*_{r}) ⊂ *D*(*T*_{n}), a contradiction with *d*_{i} and *d*_{i+1} are two consecutive spans in *D*(*T*_{n}).

Hence, *v*_{j+di}*v*_{j}_{+di+1} ∈ *E*(*T*_{r}) ⊂ *E*(*T*_{n}). Then *d*_{i+1} − *d*_{i} ∈ *D*(*T*_{n}).

Thus, *D*(*T*_{n}) = *S*^{*}. This completes the proof of the necessity.

To show the sufficiency, suppose that *S*^{*} ⊆ *S*. Then, *C*(*n*, *S*^{*}) is a subgraph of *C*(*n*, *S*).

Let △ = *v*_{0}*v*_{x}v_{x+y}*v*_{x+y+z} be a triangle (since, *x* + *y* + *z* = *n*). Clearly, *E*(△) ⊂ *E*(*G*). Then there are three polygons *G*_{1}, *G*_{2} and *G*_{3}, induced by the vertices *v*_{0}, *v*_{1}, *v*_{2}, …, *v*_{x}; *v*_{x}, *v*_{x+1}, …, *v*_{x+y} and *v*_{x+y}, *v*_{x+y+1}, …, *v*_{x+y+z} respectively, and *G*_{2} = ∅ where *y* ∈ {0, 1}.

Suppose that *R* is an *r*-gon in *G*_{i}, for some *i* ∈ {1, 2, 3} such that *V*(*R*) = {*v*_{h}, *v*_{h+ai}, *v*_{h+ai +1}, …, *v*_{h+ai+1}} for some *v*_{h} ∈ *V*(*G*_{i}) and two consecutive jump values *a*_{i} and *a*_{i+1} in *S*^{*}.

Without loss of generality assume that *R* is a subgraph of *G*_{1} and *h* = 0.

Since *S*^{*} = {1, 2, *a*_{3}, *a*_{4}, …, *a*_{k}} and satisfies that *a*_{i+1} − *a*_{i} ∈ *S*^{*} for each *i* = 1, …, *k* − 1, then we can use argument (∗) of Proposition 2.3 to show that *R* can be triangulated by *T*_{r}. Consider *G*_{1} as a finite union of such polygons. Then we can assume that *T*_{G1} is a finite union of the restricted triangulation of those polygons.

Obtain *T*_{G2} (on *G*_{2}) and *T*_{G3} (on *G*_{3}) in a similar way. Then *T*_{G1} ∪ *T*_{G2} ∪ *T*_{G3} is a triangulation of *C*(*n*, *S*^{*}). Since *C*(*n*, *S*^{*}) is a spanning subgraph of *G*, then *T*_{G1} ∪ *T*_{G2} ∪ *T*_{G3} is a triangulation of *G*.

This completes the proof. □

#### Corollary 2.9

*S*(*n*) is *S*^{*}.

#### Proof

By definition of *S*(*n*), either *S*(*n*) = *S*_{1} (when *n* is even and *t* = 1) or *S*(*n*) = *S*_{α} (when *n* is odd) or else *S*(*n*) = *S*_{1} ∪ *c*.*S*_{α}. By definition of *S*_{1}, we have that *S*_{1} is a set of ascending values and contains 1, 2; also *S*_{α} is a set of ascending values by Algorithm (*A*) and contains 1, 2 by Lemma 2.5. Thus *S*(*n*) is a set of ascending values containing 1, 2.

Now, let *ℓ* denote the cardinality of *S*(*n*).

When *S*(*n*) = *S*_{1}, *a*_{ℓ} = 2^{r} = $\begin{array}{}\frac{n}{2}\end{array}$. Let *x* = *z* = *a*_{ℓ} and *y* = 0 then we have *n* = *x* + *y* + *z*.

When *S*(*n*) = *S*_{α} or *S*(*n*) = *S*_{1} ∪ *c*.*S*_{α}, we have by definition of *β*, *a*_{ℓ} = ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c*. To get *n* = *x* + *y* + *z* we have three cases. When *α* = $\begin{array}{}\frac{t-2}{3}\end{array}$, *β* = *α* + 1 and *t* = 3.*α* + 2. Let *x* = *c*.*α*, *y* = *z* = *c*.*β* (since, *n* = *c*.*t*). When *α* = $\begin{array}{}\frac{t-1}{3}\end{array}$, *β* = *α* + 1 and *t* = 3.*α* + 1. Let *x* = *y* = *c*.*α*, *z* = *c*.*β*. When *α* = $\begin{array}{}\frac{t}{3}\end{array}$, let *x* = *y* = *z* = *c*.*α* (since *β* = *α*).

Let *a*_{i} and *a*_{i+1} be any two consecutive values in *S*(*n*). If *a*_{i}, *a*_{i+1} ∈ *S*_{1}, then *a*_{i} = 2^{i} and *a*_{i+1} = 2^{i+1} and then *a*_{i+1} − *a*_{i} = 2^{i+1} − 2^{i} = 2^{i} ∈ *S*_{1}. If *a*_{i} = 2^{r}, then *a*_{i} is the last value in *S*_{1} and the first in *c*.*S*_{α} (recall that, *c* = 2^{r}) and if *a*_{i}, *a*_{i+1} ∈ *S*_{α} then by Lemma 2.2, *a*_{i+1} − *a*_{i} ∈ *S*_{α}. Thus, *a*_{i+1} − *a*_{i} ∈ *S*(*n*) for each *a*_{i+1}, *a*_{i} ∈ *S*(*n*).

This completes the proof. □

#### Corollary 2.10

*Let* *n* ≥ 4 *be a natural number*. *Suppose* *G* = *C*(*n*, *S*) *is a circulant graph*. *Then* *G* *admits a restricted triangulation if one of the following conditions hold*.

*When* *n* *is odd*, *S* *contains* {2} *and all odd values* *a*_{i} ≤ *a*_{s} *where* ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ ≤ *a*_{s} ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋.

*When* *n* *is even*, *S* *contains* {1} *and all even values* *a*_{i} ≤ *a*_{s} *where* ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* ≤ *a*_{s} ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋.

#### Proof

In both cases *a*_{1} = 1, *a*_{2} = 2 belong to a set of ascending values *S*, and ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* ≤ *a*_{s} ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋ (when *n* is an odd natural number, ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* = ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉). Further, for each *a*_{i+1}, *a*_{i} ∈ *S*, *a*_{i+1} − *a*_{i} ∈ {1, 2} ⊂ *S*.

To show the property (2) of *S*^{*} we have to consider two cases.

Case (1)

When *n* is an odd natural number. Let *n* ≥ 7 (When *n* = 5, then *S* = {1,2} and clearly *C*(5, *S*) admits a restricted triangulation *T*_{5} = {*v*_{0}*v*_{1}*v*_{2}*v*_{3}*v*_{4}*v*_{0}} ∪ {*v*_{0}*v*_{2}, *v*_{0}*v*_{3}}). If ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ is an odd number, then ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ = *a*_{i} for some *i* ∈ {3, …, *s*}. If ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ is an even number, then ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ + 1 = *a*_{i} for some *i* ∈ {4, …, *s*}.

Whether ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ is odd or even, we have *n* − 2*a*_{i} is odd and 1 ≤ *n* − 2*a*_{i} ≤ *a*_{i}. Then *n* − 2*a*_{i} = *a*_{j} for some *j* ∈ {1, …, *i*}. Let *x* = *y* = *a*_{i}, *z* = *a*_{j}. Thus, *n* = *x* + *y* + *z*.

Case (2)

When *n* is an even natural number, we have *r* ≥ 1 and then *c*(≥ 2) is even. Hence, ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* is even and then ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ *c* = *a*_{i} for some *i* ∈ {2, …, *s*}. Now, we have *n* − 2*a*_{i} is even and either *n* − 2*a*_{i} = 0 or 2 ≤ *n* − 2*a*_{i} ≤ *a*_{i}. If *n* − 2*a*_{i} = 0, let *x* = *z* = *a*_{i}, *y* = 0. If 2 ≤ *n* − 2*a*_{i} ≤ *a*_{i}, then *n* − 2*a*_{i} = *a*_{j} for some *j* ∈ {2, …, *i*} and let *x* = *z* = *a*_{i}, *y* = *a*_{j}. Thus, *n* = *x* + *y* + *z*.

By Theorem 2.8, the circulant graph *C*(*n*, *S*) admitting a restricted triangulation.

□

#### Proposition 2.11

([13]). *Suppose* *G* = *C*(*n*, {*a*_{1}, *a*_{2}, …, *a*_{k}}) *and* *H* = *C*(*n*, {*b*_{1}, *b*_{2}, …, *b*_{k}}) *with* {*a*_{1}, *a*_{2}, …, *a*_{k}} = *q*.{*b*_{1}, *b*_{2}, …, *b*_{k}}, *where the multiplication is reduced modulo* *n* *and* *gcd*(*q*, *n*) = 1. *Then* *G* *is isomorphic to* *H*.

#### Example 2.12

*Let* *n* = 9 *and* *S*(*n*) = {1, 2, 3}.

*When* *S*_{1} = {2, 3, 4}, *then we have* *S*_{1} = {2, 6, 4} = 2.{1, 3, 2} = 2.*S*(*n*).

*When* *S*_{2} = {1, 3, 4}, *then we have* *S*_{2} = {8, 12, 4} = 4.{2, 3, 1} = 4.*S*(*n*).

The next corollary considers circulant graph *G* = *C*(*n*, *S*) admits a restricted triangulation when there is an integer *q* ≥ 1 with *gcd*(*q*, *n*) = 1 such that *S*^{*} ⊆ *q* ⋅ *S*.

#### Corollary 2.13

*let* *G* = *C*(*n*, *S*) *be a circulant graph*. *Suppose there is an integer* *q* ≥ 1 *with* *gcd*(*q*, *n*) = 1 *such that* *S*^{*} ⊆ *q*.*S* *or* *q*.*S*^{*} ⊆ *S* *where the multiplication is reduced modulo* *n*. *Then* *G* *has a configuration that admits a restricted triangulation*.

#### Proof

Suppose that, *q* ≥ 1 is an integer such that *gcd*(*q*, *n*) = 1 and *S*^{*} ⊆ *q*.*S* or *q*.*S*^{*} ⊆ *S*. Then, there is a set *S*′ ⊂ *S* such that *S*^{*} = *q*.*S*′ or *q*.*S*^{*} = *S*′. Thus, by Proposition 2.11, the subgraph *H* = *C*(*n*, *S*′) of *C*(*n*, *S*) is isomorphic to *C*(*n*, *S*^{*}). By Theorem 2.8, *H* = *C*(*n*, *S*′) admits a restricted triangulation. Thus, *G* has a configuration that admits a restricted triangulation. This completes the proof. □

## 2.1 An application

The skewness of a graph *G*, denoted *sk*(*G*), is the minimum number of edges to be deleted from *G* such that the resulting graph is planar. The convex skewness of a convex graph *G*, denoted *sk*_{c}(*G*) is the minimum number of edges to be removed from *G* so that the resulting graph is a convex plane graph (see [14]).

#### Proposition 2.14

*Let* *G* = *C*(*n*, *S*) *be a circulant graph and let* *q* ≥ 1 *be an integer such that* *gcd*(*q*, *n*) = 1. *Then* *sk*_{c}(*G*) = *E*(*G*) − (2*n* − 3) *if* *S*^{*} ⊆ *S*, *or* *qS*^{*} ⊆ *S*, *or* *S*^{*} ⊆ *qS*.

#### Proof

If *S*^{*} ⊆ *S*, then *G* = *C*(*n*, *S*) admits a restricted triangulation, by Theorem 2.8. If *qS*^{*} ⊆ *S* or *S*^{*} ⊆ *qS*, then *G* = *C*(*n*, *S*) admits a restricted triangulation, by Corollary 2.13.

It is known that, any triangulation *T* of a convex *n*-gon has 2*n* − 3 edges (*n* − 3 of them are non-boundary edges). If any new straight line segment is added to the triangulation, it will intersect with some non-boundary edge of *T*. Hence, we have *sk*_{c}(*G*) = *E*(*G*) − (2*n* − 3). □

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