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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Restricted triangulation on circulant graphs

Niran Abbas Ali
• Corresponding author
• Department of Mathematics, Universiti Putra Malaysia, 43400, Serdang Malaysia
• Department of Mathematics, College of Science, Al Mustansiriyah University, 10052 Filastin Street, Baghdad Iraq
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• Other articles by this author:
/ Hazim Michman Trao
• Department of Mathematics, Universiti Putra Malaysia, 43400, Serdang Malaysia
• Department of Mathematics, College of Science, Al Mustansiriyah University, 10052 Filastin Street, Baghdad Iraq
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• Other articles by this author:
Published Online: 2018-04-18 | DOI: https://doi.org/10.1515/math-2018-0033

## Abstract

The restricted triangulation existence problem on a given graph decides whether there exists a triangulation on the graph’s vertex set that is restricted with respect to its edge set. Let G = C(n, S) be a circulant graph on n vertices with jump value set S. We consider the restricted triangulation existence problem for G. We determine necessary and sufficient conditions on S for which G admitting a restricted triangulation. We characterize a set of jump values S(n) that has the smallest cardinality with C(n, S(n)) admits a restricted triangulation. We present the measure of non-triangulability of KnG for a given G.

Keywords: Triangulation; Circulant Graph

MSC 2010: 32C25

## 1 Introduction

A graph is an ordered pair G = (V, E), where V is a set of vertices, and E is a set of edges. An order of G is the number of its vertices while a size of G is the number of its edges. A graph is called geometric, if its edges are straight-line segments.

A triangulation Tn of a finite set of points V in the plane is a maximally connected, straight-line planar graph with vertex set V. Each bounded face is a triangle, and the triangulation includes the boundary of the convex hull.

Let n ≥ 3. A circulant graph G = C(n, S) is a graph on the vertex set V(G) = {v0, v1, …, vn − 1} such that each vertex vi is adjacent to vertices vi±a where i = 0, 1, …, n − 1 and the subscript index i ± a is reduced modulo n for all aS. That is, vivj is an edge of C(n, S) if and only if |ji| ∈ S or n − |ji| ∈ S. A set S ⊆ {1, 2, …, ⌊ n/2 ⌋} is called a set of jump values of G = C(n, S). When discussing circulant graphs, we will often assume that the vertices are the corners of a regular n-gon, labeled in clockwise order and the edges are straight line segments. Hence, circulant graphs can be considered as geometric graphs.

Circulant graphs include the family of cycles C(n, {1}) and the family of complete graphs Kn = C(n, {1, 2, …, ⌊ n/2⌋}). Clearly, when G1 = C(n, S1) and G2 = C(n, S2) are two circulant graphs such that |S1| < |S2|, then the size of G1 is smaller than the size of G2 (i.e., |E(G1)| < |E(G2)|). Hence, for simplicity we shall say that G1 = C(n, S1) is a smaller size circulant graph than G2 = C(n, S2) when |S1| < |S2|.

Let E be some set of edges spanned by V. We say that a triangulation Tn of V is restricted with respect to E if E(Tn) ⊆ E. The restricted triangulation existence problem, on a given graph G(V, E), is to decide whether there exists a triangulation of V that is restricted with respect to E. This problem was proven to be NP-complete (see [1, 2]). In section 2, we solve this problem for a certain geometric graph – a circulant graph. We give a characterization of a circulant graph G to admit a restricted triangulation.

Another related problem is beginning with a problem presented by Micha Perles on DIMACS Workshop on Geometric Graph Theory in 2002, which asks to determine the largest possible number h(n) such that every geometric graph on n vertices with at least $\begin{array}{}\left(\genfrac{}{}{0em}{}{n}{2}\right)-h\left(n\right)\end{array}$ edges has a non-crossing Hamiltonian path which has been studied by Černý et al. [3]. The authors in [4] motivated by Perles problem proved that if Fn is a subgraph of a convex complete graph Kn, where Fn contains no boundary edges of Kn and |E(Fn)| ≤ n − 3, then KnFn admits a restricted triangulation.

In section 3, we characterize a circulant graph G as a subgraph of a convex complete graph Kn and |E(G)| ≤ L(n) such that KnG allows a restricted triangulation.

We obtain a set of jump values S(n) = {a1, a2, …, a} that has the smallest cardinality |S(n)| such that C(n, S(n)) admits a restricted triangulation. That is, C(n, S(n)) is the smallest size circulant graph admitting a restricted triangulation.

Let vivj be an edge in a convex graph G, the distance between vi and vj in G is the length of a shortest (vi, vj)-path in G. A span of vivj is defined as the distance between its two end points vi and vj. In other words, a natural number d = min{|ji|, n − |ji|} such that vivj can be written as vivi+d or vivid, is a span of vivj. Hence, we can see that H = C(n, {d}) is a circulant graph in which vivjE(H) and each edge in H has a span d. Let E(Tn) be a set of edges of Tn and D(Tn) be a set of spans of all edges in E(Tn).

## 1.1 Importance of the triangulations and circulant graphs

Circulant graphs are an important class of interconnection networks in parallel and distributed computing. It can be used in the design of local area networks (see [5, 6]). On the other hand, computing a triangulation on a given graph has several important applications in different areas such as nondense matrix computations [7], database management [8] and artificial intelligence [9]. Moreover, triangulations are used in many areas of engineering and scientific applications such as finite element methods, approximation theory, numerical computation, computer-aided geometric design, computational geometry, etc. (see [10, 11]).

## 2 Restricted triangulation on G

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 Tn is a restricted triangulation of a circulant graph G = C(n, S), then S contains D(Tn).

#### Proof

Suppose Tn is a restricted triangulation of G = C(n, S). Let dD(Tn). By definition of span of edges, there is an edge vivj in Tn such that vivj = vivi± d where d = min{|ji|, n − |ji|}. Since Tn is a restricted triangulation of G, then E(Tn) ⊂ E(G). Hence, vivi ± dE(G). Thus, dS (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 Tn 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, vivi+1E(Tn) for each i = 0, 1, …, n − 1 which yields 1 ∈ D(Tn) 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 vi+1 is a vertex of degree 2 in T, then the diagonal edge vivi+2 must be in E(Tn) because vi+1 is incident just on two edges of Tn which are the boundary edges vivi+1 and vi+1vi+2. Thus, 2 ∈ D(Tn) since 2 is the distance between vi and vi+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, G1 = C(12, S1) where S1 = {1, 2, 3} does not admit any restricted triangulation, while each of G2 = C(12, S2) and G3 = C(12, S3) admits a restricted triangulation with S2 = {1, 2, 4} and S3 = {1, 2, 3, 5}. Note that |S2| = |S1| while S3 = S1 ∪ {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, Sn) admits a restricted triangulation where Sn is a set of ascending values {a1, a2, …, as} in which a1 = 1, a2 = 2, as = $\begin{array}{}⌊\frac{n}{2}⌋\end{array}$ and ai+1aiSn for each i ∈ {1, 2, …, s − 1}.

#### Proposition 2.3

Let n ≥ 4 be a natural number. Suppose Sn is a set of ascending values {1, 2, a3, …, as} in which as = $\begin{array}{}⌊\frac{n}{2}⌋\end{array}$ and Sn has the property ai+1aiSn for each i ∈ {1, 2, …, s − 1}. Then the circulant graph C(n, Sn) admits a restricted triangulation.

#### Proof

Suppose that C(n, Sn) is a circulant graph. Since asSn is a jump value, then v0vas and vnas v0 are two edges in C(n, Sn). Thus, we can define G1 and G2 to be two subgraphs of C(n, Sn), induced by the vertices v0, v1, v2, …, vas and vnas, vnas+1, …, vn − 1, v0, respectively. See Figure 1.

Fig. 1

Qaji+2(ai) on G1

We shall construct the triangulation TG1 of G1, and then G2 can be triangulated in a similar way.

(∗) It is important to mention that in the following argument the property ai+1aiSn, for each two consecutive values ai and ai+1 in Sn, and 1, 2 ∈ Sn, are basic tools to construct TG1.

Let ai, ai+1 be two consecutive vertices in Sn and ai+1ai = ajiSn for some aji ∈ {1, 2, a3, …, ai} (since Sn is a set of ascending values). Let v0 vai vai+1vai+1 v0 be a convex (aji+2)-gon in G1 denoted by Qaji+2(vai). Clearly $\begin{array}{}{G}_{1}={\cup }_{i=1}^{s-1}{Q}_{{a}_{{j}_{i}}+2}\left({v}_{{a}_{i}}\right).\end{array}$ First, we prove Qaji+2(vai) can be triangulated by edges of C(n, Sn) for each i = 1, …, s − 1.

Note that a1 = 1, a2 = 2 and a3 ∈ {3, 4} (otherwise, if a3 = 5 then a3a2 = 5 − 2 = 3 ∉ Sn, a contradiction). It is easy to see that Q3(v1) ( = v0v1v2v0) is triangulated by v0v1v2v0 and when a3 = 3 we have Q3(v2) ( = v0v2v3v0) is triangulated by v0v2v3v0, and also when a3 = 4 we have that Q4(v2) ( = v0v2v3v4v0) is triangulated by the boundary edges v0v2v3v4v0 together with the diagonal edge v2v4.

Now we obtain a triangulation for Qaji+2(vai), i = 3, …, s − 1.

Let Aai+1(vai) = v0v1vai v0, be a convex (ai+1)-gon in G1. Clearly, A3(v2) ( = v0v1v2v0) is a triangle that is trivially triangulated by T1 = v0v1v2v0. If i = 3, then either a3 = 3 and A4(v3) ( = v0v1v2v3v0) is a quadrilateral that is triangulated by T2 = v0v2v0v1v2v3v0, or a3 = 4 and A5(v4) ( = v0v1v2v3v4v0) is a pentagon triangulated by T2 = v0v2v4v0v1v2v3v4v0.

Let Baji+1(vai) = Qaji+2(vai) − v0. Then Baji+1(vai) = vaivai+1vai+aji vai (since ai+1ai = aji) which is equivalent to the convex polygon v0v1vajiv0 (by considering ai = 0). Thus, Baji+1(vai) is equivalent Aaji+1(vaji).

If i = 3, we have a4a3 = aj3Sn for some aj3 ∈ {1, 2, a3}. Then, Baj3+1(a3) ( = va3va3+1va4 va3 = va3va3+1va3+aj3 va3) is equivalent to Aaj3+1(aj3) ( = v0v1vaj3v0) which is triangulated by T ∈ {T1, T2}. Then, Baj3+1(a3) can be triangulated by T′ equivalent to T. Hence, T3 = T′ ∪ va3v0va4 is a triangulation of Qaj3+2(va3).

Recursively, if we have i = s − 1 then asas − 1 = ajs − 1Sn for some ajs − 1 ∈ {1, 2, a3, …, as − 1}. Then, Bajs − 1+1(vas − 1) is equivalent to Aajs − 1+1(ajs − 1) which is triangulated by T ∈ {T1, T2, T3, …, Ts − 2} (depending on the value of ajs − 1 ∈ {1, 2, a3, …, as − 1}). Then, Bajs − 1+1(vas − 1) can be triangulated by T′ equivalent to T. Hence, Ts − 1 = T′ ∪ vas − 1v0vas is a triangulation of Qajs − 1+2(vas − 1).

Since, $\begin{array}{}{G}_{1}={\cup }_{i=1}^{s-1}{Q}_{{a}_{{j}_{i}}+2}\left({v}_{{a}_{i}}\right),\end{array}$ then let $\begin{array}{}{T}_{{G}_{1}}={\cup }_{i=1}^{s-1}{T}_{i}\end{array}$ be the triangulation of G1. In a similar way we obtain TG2. Thus, Tn = TG1TG2 ∪ {vasvnas}, where asnas; otherwise {vasvnas} = ∅ 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.

1. S = {1, 2, a3, a4, …, $\begin{array}{}⌊\frac{n}{2}⌋\end{array}$} where $\begin{array}{}{a}_{i}^{\prime }\end{array}$s are consecutive odd values.

2. S = {1, 2, a3, a4, …, $\begin{array}{}⌊\frac{n}{2}⌋\end{array}$} where $\begin{array}{}{a}_{i}^{\prime }\end{array}$ are consecutive even values.

#### Proof

In both cases a1 = 1, a2 = 2 and $\begin{array}{}⌊\frac{n}{2}⌋\end{array}$ are in S. Further, for each ai+1, aiSn, ai+1ai ∈ {1, 2} ⊆ Sn. 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.2r for some positive natural number r′ and some positive odd natural number t.

• Let Sα be a set of jump values obtaining by next algorithm where α = $\begin{array}{}⌊\frac{t}{3}⌋\end{array}$ and β = $\begin{array}{}⌊\frac{t}{3}⌋\end{array}$ 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 S1 = {1, 2, 4, 8, …, c} with c = 2r 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 = {b1, b2, …, bk} of positive integers, then define a.B to be {a.b1, a.b2, …, a.bk} where a ≥ 1 is an integer number.

#### Algorithm (A)

1. If α = 0, then let Sα = {β} and Stop. If α = 1, then let Sα = {1, β} and Stop. If 2 ≤ α ≤ 4, then let Sα = {1, 2, α, β} and Stop. Otherwise, let L = 0 be the number of starting level. Let S(L) = {1, 2, β}, a0 = α, and let L = L+1.

2. 2- If a0 is odd and divisible by 3, then let a1, L = $\begin{array}{}\frac{{a}_{0}}{3}\end{array}$ and a2, L = 2.a1, L. Otherwise, let a1, L = ⌊ $\begin{array}{}\frac{{a}_{0}}{2}\end{array}$⌋ and a2, L = ⌈ $\begin{array}{}\frac{{a}_{0}}{2}\end{array}$⌉.

3. Let S(L) = {a1, L, a2, L, a0}. Let $\begin{array}{}{S}_{\alpha }=\bigcup _{l=0}^{L}S\left(l\right).\end{array}$

4. If 3 ∈ {a1, L, a2, L} and each of α and $\begin{array}{}⌊\frac{\alpha }{2}⌋\end{array}$ is odd and not divisible by 3, then do the following:

1. Let L = 0 be the number of starting level. Let S(L) = {1, 2, 3, α − 3, α, β}, a0 = α − 3, and let L = L+1.

2. If a0 is odd and not divisible by 3, then let a1, L = a2, L = a0 − 3. If a0 is odd and divisible by 3, then let a1, L = $\begin{array}{}\frac{{a}_{0}}{3}\end{array}$ and a2, L = 2.a1, L. Otherwise, a1, L = a2, L = $\begin{array}{}\frac{{a}_{0}}{2}\end{array}$.

3. Let S(L) = {a1, L, a2, L, a0}. Let $\begin{array}{}{S}_{\alpha }=\bigcup _{l=0}^{L}S\left(l\right).\end{array}$

4. If a1, L ≤ 3, then arrange Sα to be a set of ascending values and stop. Otherwise, let a0 = a1, L and L = L+1 and then repeat Step (ii).

5. If a1, L ≤ 4, then arrange Sα to be a set of ascending values and stop. Otherwise, let a0 = a1, L and L = L+1 and then repeat Step (2).

Note that, Sα that obtained by Algorithm (A) is a set of ascending values 1, 2, ai − 1, ai, ai+1, …, α, β.

#### Lemma 2.5

Suppose Sα = {a1, a2, …, α, β} is a set obtained by Algorithm (A). Then 1, 2 ∈ Sα and ai+1aiSα for each ai+1, ai belong to Sα.

#### Proof

By Algorithm (A) step (1), we get 1, 2 ∈ Sα. Suppose that, ai, ai+1 are two consecutive values belonging to Sα.

If ai+1 = a0 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 $\begin{array}{}{a}_{i-1}={a}_{1,L}=\frac{{a}_{i+1}}{3}\in {S}_{\alpha }\right).\end{array}$

If 3 ∈ Sα and ai+1 = a0 is an odd and not divisible by 3 number, then by Algorithm (A) Step (4), ai = ai+1 − 3. Thus, ai+1ai = 3 ∈ Sα.

Otherwise, by Algorithm (A) Step (2), $\begin{array}{}{a}_{i}=⌈\frac{{a}_{i+1}}{2}⌉\left(\text{where\hspace{0.17em}}⌈\frac{{a}_{i+1}}{2}⌉={a}_{2,L}\in {S}_{\alpha }\right).\end{array}$ Hence, $\begin{array}{}{a}_{i+1}-{a}_{i}={a}_{i+1}-⌈\frac{{a}_{i+1}}{2}⌉=⌊\frac{{a}_{i+1}}{2}⌋={a}_{i-1}\in {S}_{\alpha }\end{array}$ (where $\begin{array}{}{a}_{i-1}=⌊\frac{{a}_{i+1}}{2}⌋={a}_{1,L}\in {S}_{\alpha }\right).\end{array}$ In case when, $\begin{array}{}⌈\frac{{a}_{i+1}}{2}⌉=⌊\frac{{a}_{i+1}}{2}⌋=\frac{{a}_{i+1}}{2},\end{array}$ 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) = S1c.Sα, 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 Tn of C(n, S(n)).

When n is even and t = 1, S(n) = S1 (In this case, β = 1 and then Sα = {1} and then c. Sα = {c} and cS1). Then let $\begin{array}{}{T}_{n}=\bigcup _{i=0}^{r}\left\{{v}_{j{2}^{i}}{v}_{\left(j+1\right){2}^{i}},j=0,1,\dots ,\frac{n}{{2}^{i}}-1\right\}.\end{array}$ The circulant graph when n = 8 is depicted in Figure 2.

Fig. 2

n = 8 = 23, t = 1, r = r′ − 1 = 2, T8 = {vj v(j+1), j = 0, …, 7} ∪ {v2j v2(j+1), j = 0, …, 3} ∪ {v4j v4(j+1), j = 0, 1}.

When n is even and t ≥ 3, S(n) = S1c.Sα. In this case, and also when n is odd (which means t = n), we have t-gon which is induced by the vertices v0, vc, v2.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 Sα. According to S1, let $\begin{array}{}T=\bigcup _{i=0}^{r}\left\{{v}_{j{2}^{i}}{v}_{\left(j+1\right){2}^{i}},j=0,1,\dots ,\frac{n}{{2}^{i}}-1\right\}\end{array}$ which triangulates unshaded part, between n-gon and t-gon, in Figure 3.

Fig. 3

n = 56 = 7.23, t = 7, r = r′ = 3, T56 = {vj v(j+1), j = 0, …, 55} ∪ {v2j v2(j+1), j = 0, …, 27} ∪ {v4j v4(j+1), j = 0, …, 13} ∪ {v8j v8(j+1), j = 0, …, 6}.

Then, let Tn = TT′ or Tn = 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 = 2β+α (since β = α+1). Let △ = v0vαvα+βv0 be a triangle. Then there are three (α + 1)-gons G1, G2 and G3 induced by the vertices v0, vc, v2.c, …, vα.c; v(α).c, v(α+1).c, …, v2α.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 = 2α+β (since β = α + 1). Let △ = v0vαv2αv0 be a triangle. Then there are three (α + 1)-gons G1, G2 and G3, induced by the vertices v0, vc, v2.c, …, vα.c; v(α).c, v(α+1).c, …, v2α.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 = 3α. There are three (α + 1)-gons G1, G2 and G3 induced by the vertices v0, vc, v2.c, …, vα.c; v(α).c, v(α+1).c, …, v2α.c and v2α.c, v(2α+1).c, …, v(3α−1).c, v0, respectively. See Figure 6.

In each case, G1, G2 and G3 are of the same order. Suppose that R is an r-gon in Gi, for some i ∈ {1, 2, 3} such that V(R) = {vh, vh+ai, vh+ai+1, …, vh+ai+1} for some vhV(Gi) and two consecutive jump values ai and ai+1 in Sα.

Without loss of generality assume that R is a subgraph of G1 and h = 0.

By Lemma 2.5, we have 1, 2 ∈ Sα and ai+1aiSα for each ai+1, ai belonging to Sα. Hence, we can use the argument (∗) of Proposition 2.3 to show that R can be triangulated by Tr. Moreover, similar as in the proof of Proposition 2.3 we can consider G1 as a finite union of such polygons. Then we can assume that TG1 is a finite union of the restricted triangulation of these polygons.

Obtain TG2 (on G2) and TG3 (on G3) by “rotating” the edges of TG1, see Figures 4, 5 and 6. Let T′ = TG1TG2TG3 ∪ △ for case (1) and case (2), and let T′ = TG1TG2TG3 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 ak be a natural number such that$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ cak ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋. Then define S* to be a set of ascending natural numbers such that,

1. S* = {1, 2, a3, a4, …, ak},

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

3. ai+1aiS* for each i = 1, …, k − 1.

It is clear that, S* = Sn when ak = ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋. 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 Tn. We shall prove that S contains S*. By Lemma 2.1, it is enough to prove S* = D(Tn).

Arrange the spans of D(Tn) to be ascending values.

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

Let |D(Tn)| = s, then ds is the maximum span in D(Tn).

(⋆) Let t be defined as before. Suppose that ds < ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ c. Without loss of generality assume that, c = 1 and ds = ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ − 1. Thus, ds = $\begin{array}{}\frac{t-b}{3}\end{array}$ − 1, b ∈ {0, 1, 2} and then 3ds + b + 3 = t. Let R be a convex r-gon induced by the vertices vds, v2ds, v3ds, v3ds+1, …, v3ds+b+1, v3ds+b +2, vt (recall that, vt = v3ds+b+3). Suppose that Tr = Tn(R) is a subgraph of Tn that triangulates R and let e be an edge in Tr such that the span of e is the maximum with respect to D(Tr). Then:

• either e = v2dsvt which yields that its span t − 2dsD(Tr) ⊂ D(Tn); but t-2ds > ds + 1 (by above assumption, t − 2ds = ds + b + 3 > ds + 1), which is a contradiction with maximality of dsD(Tn);

• or, e = vdsv3ds which yields that 3dsds = 2dsD(Tr) ⊂ D(Tn); but 2ds > ds, which also contradicts the maximality of dsD(Tn).

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

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

• Case (1)

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

• Case (2)

If ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ cds$\begin{array}{}\frac{n-1}{2}\end{array}$, then Tn contains a triangle △ = vhvh+dsvh+ds+divh for some diD(Tn) and vhV(G). The span of the edge vh+ds+divhE(Tn) is n − (ds+di) ∈ D(Tn) (by definition of the span of edge). Hence let x = ds, y = di and z = n − (ds+di), then n = x + y + z.

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

If ds = $\begin{array}{}\frac{n}{2}\end{array}$, then let T1 and T2 be two subgraphs of Tn induced by the vertices v0, v1, v2, …, vx; vx, vx+1, …, vx+y − 1, v0 respectively (x = $\begin{array}{}\frac{n}{2}\end{array}$). Let R be a polygon induced by the vertices vj, vj+di, …, vj+di+1 where vjV(Ti) and vj vj+di and vvvj+di+1 are two diagonals in Ti for some i ∈ {1, 2}.

If ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ cds$\begin{array}{}\frac{n-1}{2}\end{array}$, then let T1, T2 and T3 be three subgraphs of Tn induced by the vertices v0, v1, v2, …, vx, vx, vx+1, …, vx+y, and vx+y, vx+y+1, …, vx+y+z−1, v0 respectively. Let R be a polygon induced by the vertices vj, vj+di, …, vj+di+1 where vjV(Ti) and vjvj+di and vjvj+di+1 are two diagonals in Ti for some i ∈ {1, 2, 3}.

Assume without loss of generality that vjvj+di and vjvj+di+1 are two diagonals in T1. Then, j < j + di < j + di+1 (by definitions of T1 and R). It is clear that, vj vj+divj+di+1 vjT1. Let Tr = T1(R) be a subgraph of T1 that triangulates R with the boundary edges vj vj+divj+di+1 vj of R.

Suppose that vj+divj+di+1E(Tr). Then there is di < d < di+1 such that vjvj+dE(Tr). This implies that, dD(Tr) ⊂ D(Tn), a contradiction with di and di+1 are two consecutive spans in D(Tn).

Hence, vj+divj+di+1E(Tr) ⊂ E(Tn). Then di+1diD(Tn).

Thus, D(Tn) = 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 △ = v0vxvx+yvx+y+z be a triangle (since, x + y + z = n). Clearly, E(△) ⊂ E(G). Then there are three polygons G1, G2 and G3, induced by the vertices v0, v1, v2, …, vx; vx, vx+1, …, vx+y and vx+y, vx+y+1, …, vx+y+z respectively, and G2 = ∅ where y ∈ {0, 1}.

Suppose that R is an r-gon in Gi, for some i ∈ {1, 2, 3} such that V(R) = {vh, vh+ai, vh+ai +1, …, vh+ai+1} for some vhV(Gi) and two consecutive jump values ai and ai+1 in S*.

Without loss of generality assume that R is a subgraph of G1 and h = 0.

Since S* = {1, 2, a3, a4, …, ak} and satisfies that ai+1aiS* for each i = 1, …, k − 1, then we can use argument (∗) of Proposition 2.3 to show that R can be triangulated by Tr. Consider G1 as a finite union of such polygons. Then we can assume that TG1 is a finite union of the restricted triangulation of those polygons.

Obtain TG2 (on G2) and TG3 (on G3) in a similar way. Then TG1TG2TG3 is a triangulation of C(n, S*). Since C(n, S*) is a spanning subgraph of G, then TG1TG2TG3 is a triangulation of G.

This completes the proof. □

S(n) is S*.

#### Proof

By definition of S(n), either S(n) = S1 (when n is even and t = 1) or S(n) = Sα (when n is odd) or else S(n) = S1c.Sα. By definition of S1, we have that S1 is a set of ascending values and contains 1, 2; also Sα 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 denote the cardinality of S(n).

When S(n) = S1, a = 2r = $\begin{array}{}\frac{n}{2}\end{array}$. Let x = z = a and y = 0 then we have n = x + y + z.

When S(n) = Sα or S(n) = S1c.Sα, we have by definition of β, a = ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ 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 ai and ai+1 be any two consecutive values in S(n). If ai, ai+1S1, then ai = 2i and ai+1 = 2i+1 and then ai+1ai = 2i+1 − 2i = 2iS1. If ai = 2r, then ai is the last value in S1 and the first in c.Sα (recall that, c = 2r) and if ai, ai+1Sα then by Lemma 2.2, ai+1aiSα. Thus, ai+1aiS(n) for each ai+1, aiS(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.

1. When n is odd, S contains {2} and all odd values aias where$\begin{array}{}\frac{n}{3}\end{array}$⌉ ≤ as ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋.

2. When n is even, S contains {1} and all even values aias where$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ cas ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋.

#### Proof

In both cases a1 = 1, a2 = 2 belong to a set of ascending values S, and ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ cas ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋ (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 ai+1, aiS, ai+1ai ∈ {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 T5 = {v0v1v2v3v4v0} ∪ {v0v2, v0v3}). If ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ is an odd number, then ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ = ai 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 = ai for some i ∈ {4, …, s}.

Whether ⌈$\begin{array}{}\frac{n}{3}\end{array}$⌉ is odd or even, we have n − 2ai is odd and 1 ≤ n − 2aiai. Then n − 2ai = aj for some j ∈ {1, …, i}. Let x = y = ai, z = aj. 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 = ai for some i ∈ {2, …, s}. Now, we have n − 2ai is even and either n − 2ai = 0 or 2 ≤ n − 2aiai. If n − 2ai = 0, let x = z = ai, y = 0. If 2 ≤ n − 2aiai, then n − 2ai = aj for some j ∈ {2, …, i} and let x = z = ai, y = aj. 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, {a1, a2, …, ak}) and H = C(n, {b1, b2, …, bk}) with {a1, a2, …, ak} = q.{b1, b2, …, bk}, 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 S1 = {2, 3, 4}, then we have S1 = {2, 6, 4} = 2.{1, 3, 2} = 2.S(n).

When S2 = {1, 3, 4}, then we have S2 = {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*qS.

#### 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 skc(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 skc(G) = E(G) − (2n − 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 2n − 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 skc(G) = E(G) − (2n − 3). □

## 3 Kn − G admits a restricted triangulation

In this section we turn to another question: which circulant graph G on n vertices does satisfy that KnG admits a restricted triangulation and what is the largest size of G such that KnG still admits a restricted triangulation? We answered the first question by Corollary 3.1 and Corollary 3.3. We show that C(n, S(n)) is a smallest size circulant graph that admits a restricted triangulation, in order to answer the second question by Theorem 3.5. In what follows, let 𝓝 = {1, 2, …, ⌊ n/2⌋}.

#### Corollary 3.1

Let G = C(n, S) be a circulant graph. Then KnG admits a restricted triangulation if one of the following conditions hold.

1. If there is S* such that SS* = ∅.

2. S contains all odd i of 𝓝 except {1, ⌊ n/2⌋}.

3. S contains all even i of 𝓝 except {2, ⌊ n/2⌋}.

#### Proof

First, it is well known that KnC(n, S) = C(n, 𝓝 − S).

Suppose that SS* = ∅. Then, 𝓝 − S contains S* which yields that the circulant graph C(n, 𝓝 − S) admits the restricted triangulation by Theorem 2.8, and this shows the property (1).

In (2), 𝓝 − S contains all even values of 𝓝 together with {1, ⌊ n/2⌋}. In (3), 𝓝 − S contains all odd values of 𝓝 together with {2, ⌊ n/2⌋}. By Corollary 2.4, C(n, 𝓝 − S) admits the restricted triangulation for both cases.

This completes the proof. □

#### Definition 3.2

([14]). Let Kn be a convex complete graph with n vertices. F is said to be potentially triangulable in Kn if there exists a configuration of F in Kn such that KnF admits a triangulation.

#### Corollary 3.3

Let G = C(n, S) be a circulant graph. Suppose S*q.(𝓝 − S) for some an integer q ≥ 1 with gcd(q, n) = 1. Then G is potentially triangulable in Kn.

#### Proof

Suppose that, q ≥ 1 is an integer such that gcd(q, n) = 1 and S*q.(𝓝 − S) for some S*. Then, there is a set S′ ⊂ 𝓝 − S such that S* = q.S′. Then by Proposition 2.11, the subgraph C(n, S′) of C(n, 𝓝 − S) is isomorphic to C(n, S*). By Theorem 2.8, C(n, S′) has a configuration that admits a restricted triangulation. Thus, C(n, 𝓝 − S) (= KnG) admits a restricted triangulation. This completes the proof. □

To answer the second part of the question, we shall determine the largest size L(n) of G for which KnG admits a restricted triangulation. Before proceeding, let first |E(C(n,S(n)))| = 𝓔 where |S(n)| = . Then we deduce that, $\begin{array}{}{\mathcal{E}}_{\ell }=\left\{\begin{array}{l}n\ell -\frac{n}{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t=1;\\ n\ell ,\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{otherwise.}\end{array}\right\\end{array}$

The next result shows that whenever C(n, S) admits a restricted triangulation then |S| ≥ |S(n)|. That is, C(n, S), is not a smaller size than C(n, S(n)).

#### Theorem 3.4

C(n, S(n)) is the smallest size circulant graph admitting a restricted triangulation if n ≥ 4.

#### Proof

Recall that, S(n) = S1cSα. By Theorem 2.6, C(n, S(n)) admits a restricted triangulation. Hence, we just show that S(n) is the smallest cardinality set for which the conclusion remains true.

Assume that C(n, S) is a circulant graph that admits a restricted triangulation where S = {b1, b2, b3, …, bs} is a set of ascending jump values to C(n, S).

By Theorem 2.8, S contains S*. Thus, 1, 2 ∈ S and ⌈$\begin{array}{}\frac{t}{3}\end{array}$⌉ ⋅ cbs ≤ ⌊$\begin{array}{}\frac{n}{2}\end{array}$⌋ and for any i ∈ {1, …, s − 1}, bi+1bi = bjS where j ∈ {1, …, i}.

In case when bi+1 is even we have, according to Sα (by Algorithm (A) step (2) where bi+1 = a0 is even) and S1 (by definition of S1), that the difference between the two consecutive values bi and bi+1 is always bi.

If j ∈ {1, …, i − 1}, then the number of values in S(n) is less than the number of values in S. If bi+1bi = biS, then the number of values in S(n) is equal to the number of values in S. Thus, |S| ≥ |S(n)|.

In case when bi+1 is odd, then bjbi and there are three cases which are depending only on Sα. (by definition of S1).

• Case (1)

bj ∈ {1, 2}.

In this case, the difference between the two consecutive values bi and bi+1 is at most 2. According to Algorithm (A), the difference between the two consecutive values is at most 2 only when bi = ⌊$\begin{array}{}\frac{{b}_{i+2}}{2}\end{array}$⌋ and bi+1 = ⌈$\begin{array}{}\frac{{b}_{i+2}}{2}\end{array}$⌉ or when bi = 3 and bi = 5. Then the number of values in S(n) is equal to or less than the number of values in S. Whereas Step (2) and Step (4) of Algorithm (A) give the difference between any two consecutive values being at least 3, which implies |S| ≥ |S(n)|.

For instance, take bi+1 = 13 then bi ∈ {11, 12}. Take S ∈ {S1, S2, S3, S4, S5, S6} where S1 = {1, 2, 3, 5, 10, 12, 13}, S2 = {1, 2, 3, 6, 9, 12, 13}, S3 = {1, 2, 3, 5, 6, 11, 13}, S4 = {1, 2, 3, 6, 9, 11, 13}, S5 = {1, 2, 3, 5, 10, 11, 13}, or S6 = {1, 2, 3, 4, 7, 9, 11, 13}. Then the cardinality of S is 7. By Algorithm (A) step (2):

• either, bi = 7 and bi−1 = 6 and then Sα = {1, 2, 3, 6, 7, 13}. Thus, the cardinality of Sα is 6;

• or, bi+2 = 25 and then bi+1 = 13 and bi = 12. Then, Sα = {1, 2, 3, 6, 12, 13, 25}. In this case, Take S ∈ {S1, S2} where S1 = {1, 2, 3, 5, 10, 12, 13, 25}, and S2 = {1, 2, 3, 6, 9, 12, 13, 25}. Note that S is of cardinality 8 while Sα is of cardinality 7.

• Case (2)

j ∈ {3, …, i − 2}.

In Sα, this case is satisfied in step (4) of Algorithm (A) (when bi = a0 is odd and not divisible by 3). But in this case, Algorithm (A) states bi to be bi+1 − 3 which is always even. Then by step (2) when bi(= a0) is even we have that bi−1 = $\begin{array}{}\frac{{b}_{i}}{2}\end{array}$. That is, the difference between the two consecutive values bi and bi−1 is always bi−1 (since bi = 2bi−1). While, in S, we have either bibi−1 = bi−1 and then |S| = |S(n)| or bibi−1 = bkbi−1 and then |S| > |S(n)|. Thus, |S| ≥ |S(n)|.

For instance, take bi+1 = 23 and S = {1, 2, 3, 6, 8, 9, 17, 23}. By Algorithm (A) we have Sα = {1, 2, 3, 5, 10, 20, 23}. Note that S is of cardinality 8 while Sα is of cardinality 7.

• Case (3)

j = i − 1.

If bi+1 is not divisible by 3, then by Algorithm (A) step (2) we have bj = bi−1 = ⌊$\begin{array}{}\frac{{b}_{i+1}}{2}\end{array}$⌋ and bi = ⌈$\begin{array}{}\frac{{b}_{i+1}}{2}\end{array}$⌉ (where bi−1 = b1,L and bi = b2,L), and then the difference between the two consecutive values bi and bi−1 is 1. While in S, we have either bibi−1 = 1 and then |S| = |S(n)| or bibi−1 = bk ≠ 1 and then |S| > |S(n)|.

If bi+1 is divisible by 3, then in Sα by Algorithm (A) step (2) (where bi+1 = a0 is odd and divisible by 3). Then bi−1 = $\begin{array}{}\frac{{b}_{i+3}}{2}\end{array}$ and bi = 2.bi−1 (where bi−1 = a1,L and bi = a2,L). That is, the difference between the two consecutive values bi and bi−1 is always bi−1 itself (since bi = 2.bi−1). While in S, either bibi−1 = bi−1 and then |S| = |S(n)| or bibi−1 = bk for some k ∈ {1, 2, … i − 2} and then |S| > |S(n)|.

This completes the proof.

□

Let L(n) = $\begin{array}{}\left(\genfrac{}{}{0}{}{n}{2}\right)-{\mathcal{E}}_{\ell }\left(\text{where}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(\genfrac{}{}{0}{}{n}{2}\right)\end{array}$ is the size of Kn and 𝓔 is the size of C(n,S(n))). Then, we conclude that $\begin{array}{}L\left(n\right)=\left\{\begin{array}{ll}\frac{n\left(n-2\ell \right)}{2},& \text{t=1;}\\ \frac{n\left(n-2\ell -1\right)}{2},& \text{otherwise.}\end{array}\end{array}$

By Theorem 3.4, we have that 𝓔 is the smallest number of edges of a circulant graph that admits a restricted triangulation. The following result is to measure the non-triangulability of KnG.

#### Theorem 3.5

Let G = C(n, S) be a circulant graph. Then KnG admits no restricted triangulation if |E(G)| > L(n).

#### Proof

Let |E(G)| > L(n). Then |E(KnG)| = $\begin{array}{}\left(\genfrac{}{}{0}{}{n}{2}\right)-|E\left(G\right)|<\left(\genfrac{}{}{0}{}{n}{2}\right)-L\left(n\right)\end{array}$ = 𝓔. Then |𝓝 − S| < 𝓔. By Theorem 3.4, C(n, 𝓝 − S) (= KnG) admits no restricted triangulation.

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Accepted: 2018-01-24

Published Online: 2018-04-18

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 358–369, ISSN (Online) 2391-5455,

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