On Convex Graphs Having Plane Spanning Subgraph of Certain Type

TL;DR

This paper investigates the conditions under which the complete graph minus a certain subgraph admits a g-angulation, providing necessary and sufficient criteria for various sizes of the subgraph.

Contribution

It establishes new necessary and sufficient conditions for the existence of g-angulations in graphs formed by removing specific subgraphs from complete graphs.

Findings

Characterization of subgraphs F with |E(F)| ≤ n-1 for g-angulation

Conditions for placing F in K_n when |E(F)| ≥ n to achieve g-angulation

Extension of previous results to broader classes of subgraphs

Abstract

Motivated by a result of [17], we determine necessary and sufficient conditions on $F\/$ with $|E(F)| \leq n-1\/$ for which $K_n - F\/$ admits a $g$-angulation. For $|E(F)| \geq n\/$, we investigate the possibility of placing $F\/$ in $K_n\/$ such that $K_n -F\/$ admits a $g$-angulation for certain families of graphs $F\/$.

Equations

$$v_0{v}_2{v}_4\dots v_{n-3}{v}_{n-1}{v}_1{v}_3{v}_5\dots v_{n-4}{v}_{n-2}{v}_0.$$

$$v_0{v}_2{v}_4\dots v_{n-4}{v}_{n-2}{v}_1{v}_3{v}_5\dots v_{n-5}{v}_{n-1}{v}_{n-3}{v}_0.$$

Full text

1 Introduction and Preliminary

By a geometric graph we mean a graph whose edges are straight line segments. By a convex graph, we mean a geometric graph whose vertices are in convex position.

Let $S\/$ be a set of $n\/$ points in general position in the plane. A $g$*-angulation * of $S\/$ is a plane graph in which each face interior to the convex hull of $S\/$ is a $g$-cycle. A *convex $g$-angulation * is a $g$-angulation on $S\/$ of $n\/$ points in convex position in the plane. We say that $G_{n}$ is a $g$-angulation of a graph $G(V,E)$ if $E(G_{n})\subseteq E$. In particular, the $g$-angulation is a triangulation if $g=3$.

The triangulation existence problem is the following: On a given graph $G$, decide whether there exists a triangulation of $G$. This problem is NP-complete (see [14]).  This article extends the problem of an article [17] on triangulability of convex geometric graphs to the $g$-angulation existence problem of a convex geometric graph $G\/$ by considering a spanning subgraph $F_{n}$ of $K_{n}$ with $G=K_{n}-F_{n}$ is the convex graph obtained from $K_{n}\/$ by deleting the set of edges of $F_{n}\/$.

To decide whether $G\/$ admits a $g$-angulation or not, we first characterize the forbidding configurations for any possible $g$-angulation of $K_{n}$. A configuration of $n$ vertices in convex position (vertices of $K_{n}$) having a common edge with each $g$-angulation of $K_{n}$, is a forbidding configuration for $g$-angulations of $K_{n}$.

Determining the smallest size of the forbidding configurations for $g$-angulations of $K_{n}$ is a natural Tur$\acute{a}$n-type question, as it is equivalent to determining the largest size of a convex geometric graph does not contain a $g$-angulation. We answer for the “forbidding configurations for $g$-angulations” question not only by determining their smallest size which is $n-g+1\/$, but rather giving a complete characterization of all those forbidding configurations of size at least $n-g+1\/$ and at most $n-1\/$.

We present the characterizations ${\cal F}_{n,g}(^{*})\/$ and ${\cal J}_{n,g}(^{*}_{\beta})\/$ ($\beta\in\{1,2,\ldots,2g-3\}\/$) that forbidding $g\/$-angulations of $K_{n}\/$.

By Theorem 2, Propositions 1, 2, and 3, we show that $K_{n}-{\cal J}_{n,g}(^{*})\/$ and $K_{n}-{\cal J}_{n,g}(^{*}_{\beta})\/$ admits no $g\/$-angulation respectively.

If $|E(F_{n})|\leq n-g+1\/$, we show by Theorem 3 that $K_{n}-F_{n}\/$ admits a $g$-angulation if and only if $F_{n}\neq{\cal F}_{n,g}(^{*})\/$. If $n-g+2\leq|E(F_{n})|\leq n-1\/$, we show by Theorem 4 that $K_{n}-F_{n}\/$ admits a $g\/$-angulation if and only if $F_{n}\neq{\cal J}_{n,g}(^{*}_{\beta})\/$.

For the case where $F_{n}\/$ has at least $n\/$ edges, it seems difficult to obtain a characterization on $F_{n}\/$ such that $K_{n}-F_{n}\/$ admits a triangulation. For this we confine our attention to seek for the possibility of arranging certain families of graphs $F_{n}\/$ as a convex geometric graph such that $K_{n}-F_{n}\/$ admits a $g$-angulation. If such a configuration exists for $F_{n}\/$, then we say that $F_{n}\/$ is potentially $g$-angulable in $K_{n}\/$. Potentially $g$-angulable graphs are considered in Section 5 where we (i) determine precisely the value of $n\/$ for which the $n\/$-cycle is potentially $g$-angulable in $K_{n}\/$ (Theorem 5),  and (ii) characterize all $2\/$-regular graphs which are potentially $g$-angulable in $K_{n}\/$ (Theorem 6).

The potentially $g$-angulable problem is extended to the regular case in Section 6 where we characterize all $3\/$-regular graphs which are potentially $4$-angulable in $K_{n}\/$ (Theorem 7).

Throughout, we shall adopt the following notations. Unless otherwise stated, the vertices of a convex complete graph $K_{n}\/$ will be denoted by $v_{0},v_{1},\ldots,v_{n-1}\/$ in cyclic clockwise ordered. Also, unless otherwise specified, any operation on the subscript of $v_{i}\/$ is reduced modulo $n\/$.

Lemma 1

Suppose $g\geq 3\/$ and $t\geq 2\/$ are natural numbers and assume that $n=g+t(g-2)\/$. Let $F\/$ be a subgraph of a convex complete graph $K_{n}\/$. Assume that $F\/$ has at most $n-1\/$ edges and having no boundary edge of $K_{n}\/$. Then $K_{n}-F\/$ has an edge of the form $v_{j}v_{j+g-1}\/$.

Proof: If the lemma is not true, then it implies that $v_{j-g+1}v_{j},v_{j}v_{j+g-1}\in E(F)\/$, and recursively, this implies that $F\/$ is a spanning subgraph of $K_{n}\/$ with minimum vertex-degree at least $2\/$. But this implies that $|E(F)|\geq n\/$, a contradiction.

2 $g$-angulable graphs

Theorem 1

:* Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 0$ is any natural number. Let $F\/$ be a subgraph of the complete convex graph $K_{n}$. Suppose $F\/$ contains no boundary edge of $K_{n}\/$ and $|E(F)|\leq n-g\/$. Then $K_{n}-F\/$ admits a $g$-angulation.*

Proof: We prove this by induction on $t\/$.

The result is trivially true if $t=0\/$. If $t=1\/$, then $n=2(g-1)\/$ and $F\/$ has at most $g-2\/$ edges. Hence $K_{n}-F\/$ has a vertex, say $v_{i}\/$ which is adjacent to every other vertex in $K_{n}-F\/$ and this means that $v_{i}v_{i+g-1}\/$ together with the boundary edges of $K_{n}\/$ form a $g\/$-angulation of $K_{n}-F\/$.

Assume that $t\geq 2\/$ and the result is true for all natural numbers $t^{\prime}\/$ where $t^{\prime}<t\/$. Since $g\geq 3\/$, by Lemma 1, $K_{n}-F\/$ contains an edge of the form $v_{i}v_{i+g-1}\/$. By relabeling the vertices of $K_{n}-F\/$, if necessary we may assume that $i=0\/$.

If, for some $j\in\{1,2,\ldots,g-2\}\/$, $v_{j}\/$ is adjacent to every other vertex in $K_{n}-F\/$, then $v_{j}v_{j+kg-2k+1}\/$ where $k=1,2,\ldots,t\/$ together with the boundary edges of $K_{n}\/$ form a $g\/$-angulation of $K_{n}-F\/$.

Hence we assume that, for each $j\in\{1,2,\ldots,g-2\}\/$, $v_{j}\/$ is incident with at least one edge of $F\/$. Then the subgraph obtained from $K_{n}-F\/$ by deleting all the vertices $v_{1},v_{2},\ldots,v_{g-2}\/$ is a convex graph of the form $K_{m}-F^{\prime}\/$ where $m=g+(t-1)(g-2)\/$ and $F^{\prime}=F-\{v_{1},v_{2},\ldots,v_{g-2}\}\/$. Moreover, $F^{\prime}\/$ has at most $m-g\/$ edges (since there are at least $g-2\/$ edges of $F\/$ incident to the vertices $v_{1},v_{2},\ldots,v_{g-2}\/$).

By induction $K_{m}-F^{\prime}\/$ admits a $g\/$-angulation $G^{\prime}\/$. As such, $G^{\prime}\cup\{v_{0}v_{1}v_{2}\cdots v_{g-1}\}\/$ is a $g\/$-angulation for $K_{n}-F\/$. This completes the proof.

Remark 1

:* The result in Theorem 1 is tight with respect to $|E(F)|\/$ since deleting $n-g+1\/$ edges from $K_{n}\/$ does not guarantee that the resulting graph admits a $g\/$-angulation. This follows directly from the next result in the next section.*

3 Graphs with at most $n-g+1\/$ edges

This section presents a characterization of ${\cal F}_{n,g}(^{*})\/$, that shares any possible $g$-angulation of $K_{n}$ with at least one edge. We show that $G\/$ admits no $g$-angulation when $G=K_{n}-{\cal F}_{n,g}(^{*})\/$, $G\/$ admits a $g$-angulation if and only if $F_{n}\neq{\cal F}_{n,g}(^{*})\/$.

Throughout, we let $d_{G}(v)\/$ denote the degree of $v\/$ in the graph $G\/$.

Definition 1

Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 1$ is any natural number. Let ${\cal F}_{n,g}(^{*})\/$ denote a convex geometric graph with no isolated vertices and having $n-g+1\/$ edges such that

(i) for any $v_{i}v_{j}\/$ in ${\cal F}_{n,g}(^{*})\/$, $|j-i|\equiv 1\ ({\rm mod}\ (g-2))\/$,

(ii) whenever $d(v_{i})\geq 2\/$, then $v_{j}v_{j-g+1}\/$ is an edge of ${\cal F}_{n,g}(^{*})\/$ for $j=i+1,\ldots,i+g-2\/$,

(iii) the neighbor $v_{i}\/$ of any pendant vertex satisfies the condition that $v_{j}v_{j-g+1}\/$ is an edge of ${\cal F}_{n,g}(^{*})\/$ for $j=i+1,\ldots,i+g-2\/$,

(iv) for any two pendant vertices $v_{r},v_{s}\/$ such that whenever $v_{r}v_{i}\/$ and $v_{s}v_{j}\/$ crosse each other in ${\cal F}_{n,g}(^{*})\/$, then $|i-j|\leq g-2\/$.

It is easy to see that in the definition, conditions (ii) and (iii) imply that all the pendant vertices of ${\cal F}_{n,g}(^{*})\/$ are in consecutive order. As such we may label the vertices of ${\cal F}_{n,g}(^{*})\/$ so that $d(v_{i})=1\/$ if and only if $i=k,k+1,\ldots,n-1\/$ where $k\/$ is a natural number with $g-2\leq k\leq n-2(g-1)\/$ if $t\geq 2\/$, and $k=0\/$ if $t=1\/$.

Examples of a geometric graph ${\cal F}_{n,g}(^{*})\/$ are depicted in Figure1 (a), (b), (c) and (d). In Figure1(a), ${\cal F}_{6,4}(^{*})$ with $t=1\/$, in (b) ${\cal F}_{12,4}(^{*})\/$ with minimum value of $k=g-2=2\/$, in (c) ${\cal F}_{12,4}(^{*})$ with maximum value of $k=n-2(g-2)=6$, and in (d) ${\cal F}_{12,4}(^{*})$ satisfies condition (iv) of Definition1.

Theorem 2

Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 1$ is any natural number. Then $K_{n}-{\cal F}_{n,g}(^{*})\/$ admits no $g$-angulation.

Proof:

We first observe that, when $t=1\/$, we have $n=2g-2\/$ and since ${\cal F}_{n,g}(^{*})\/$ consists of $g-1\/$ edges of the form $v_{j}v_{j+g-1}\/$, $j=0,1,\ldots,g-2\/$, $K_{n}-{\cal F}_{n,g}(^{*})\/$ admits no $g$-angulation.

Hence assume that $t\geq 2\/$.

Suppose $K_{n}-{\cal F}_{n,g}(^{*})\/$ admit a $g\/$-angulation $G_{n}\/$. Note that any $g$-angulation on a set of points in convex position has a diagonal $v_{q}v_{q+g-1}\/$.

Let $F_{m}={\cal F}_{n,g}(^{*})-\{v_{q+1},\ldots,v_{q+g-2}\}\/$ where $m=n-g+2\/$.

By the definition of ${\cal F}_{n,g}(^{*})\/$, we see that $d_{{\cal F}_{n,g}(^{*})}(v_{i})=1\/$ for each $i=q+1,q+2,\ldots,q+g-2$. It is readily checked that $F_{m}\/$ is of the form ${\cal F}_{m,g}(^{*})\/$ where $m=g+(t-1)(g-2)\/$. By induction, $K_{m}-F_{m}\/$ admits no $g$-angulation, which is a contradiction with $G_{n}$ is a $g$-angulation for $K_{n}-{\cal F}_{n,g}(^{*})\/$.

This completes the proof.

Theorem 3

:* Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 1$ is any natural number. Suppose $F_{n}\/$ is a subgraph of the convex complete graph $K_{n}\/$ such that $|E(F_{n})|\leq n-g+1\/$ and $F_{n}\/$ contains no boundary edges of $K_{n}\/$. Then $K_{n}-F_{n}\/$ admits a $g$-angulation unless $F_{n}={\cal F}_{n,g}(^{*})\/$.*

Proof: In view of Theorem 2, we assume that $F_{n}\neq{\cal F}_{n,g}(^{*})\/$. Also, we may assume that $F_{n}\/$ contains no isolated vertices (otherwise $K_{n}-F_{n}\/$ admits a $g$-angulation).

We prove the result by induction on $t\/$. The result is clearly true for $t=1\/$. Assume that $n=g+t(g-2)\/$ with $t\geq 2$ and the result is true for all convex graphs $K_{m}-F_{m}\/$ where $m=g+t^{\prime}(g-2)\/$ with $t^{\prime}\leq t-1$.

By Lemma 1, $K_{n}-F_{n}\/$ contains an edge of the form $v_{q}v_{q+g-1}\/$ which is not an edge of $F_{n}\/$.

Let $H\/$ be the convex graph obtained from $K_{n}-F_{n}\/$ by deleting the vertices $v_{q+1},\ldots,v_{q+g-2}$. Clearly, $H\/$ is of the form $K_{m}-F_{m}\/$ where $m=g+(t-1)(g-2)\/$.

If $F_{m}\neq{\cal F}_{m,g}(^{*})\/$, then $K_{m}-F_{m}\/$ admits a $g$-angulation (by induction) and this implies that $K_{n}-F_{n}\/$ admits a $g$-angulation. Hence assume that $F_{m}={\cal F}_{m,g}(^{*})\/$. As such, the vertices $v_{q+1},\ldots,v_{q+g-2}$ are pendant vertices in $F_{n}\/$ (otherwise $|E(F_{m})|\leq m-g\/$, a contradiction with assumption $F_{m}={\cal F}_{m,g}(^{*})\/$).

There are two cases to consider. Let $S=\{q+1,\ldots,q+g-2\}\/$.

Case (1): $m=2g-2\/$.

For each $i\in S\/$, $v_{i}\/$ is adjacent either (i) to $v_{i+g-1}\/$ or else (ii) to $v_{i-g+1}\/$.

If there exist $i,j\in S\/$ such that

$i<j\/$ and $v_{i}\/$ is adjacent to $v_{i+g-1}\/$ and $v_{j}\/$ is adjacent to $v_{j-g+1}\/$ $(\ast)\/$

then $v_{i}v_{i-g+1}\/$ and $v_{j}v_{j+g-1}\/$ are the diagonals of a required $g\/$-angulation.

Hence assume that no $i,j\in S\/$ satisfy the condition $(\ast)\/$. Let $i\/$ be the largest integer in $S\/$ such that $v_{i}\/$ is adjacent to $v_{i-g+1}\/$. This implies that $v_{j}\/$ is adjacent to $v_{j-g+1}\/$ if $j<i\/$, and $v_{j}\/$ is adjacent to $v_{j+g-1}\/$ otherwise. But this means that $F_{n}\/$ is ${\cal F}_{n,g}(^{*})\/$, a contradiction.

On the other hand, if there exists no $i\in S\/$ such that $v_{i}\/$ is adjacent to $v_{i-g+1}\/$, then $v_{j}\/$ is adjacent to $v_{j+g-1}\/$ for all $j\in S\/$. But again $F_{n}\/$ is is ${\cal F}_{n,g}(^{*})\/$, a contradiction.

Case (2): $m\geq 3g-4\/$.

Suppose there is an $i\in S\/$ such that $v_{i}\/$ is adjacent to a vertex $v_{j}\/$ in $F_{m}\/$ with $|j-i|\not\equiv 1\ ({\rm mod}\ (g-2))\/$. Then the diagonals $v_{i}v_{i+r(g-2)+1}\/$, $r=1,2,\ldots,t\/$ together with the boundary edges of $K_{n}\/$ yield a $g\/$-angulation of $K_{n}-F_{n}\/$.

Hence assume that for any $i\in S\/$, $v_{i}\/$ is adjacent to a vertex $v_{j}\/$ in $F_{m}\/$ such that $|j-i|\equiv 1\ ({\rm mod}\ (g-2))\/$.

(i) Suppose both $v_{q}$ and $v_{q+g-1}$ are non-pendent vertices in ${\cal F}_{m,g}(^{*})\/$.

If $v_{q+1}\/$ is not adjacent to $v_{q-g+2}\/$ in $F_{n}\/$, then obtained by deleting all the vertices $v_{q-g+3},\ldots,v_{q}\/$ is of the form $K_{m}-F^{\prime}_{m}\/$ where $m=n-g+2\/$, $|E(F^{\prime}_{m})|\leq m-g\/$ (because $d_{F_{n}}(v_{q})\geq 2\/$) and $F^{\prime}_{m}\/$ contains no boundary edges of $K_{m}\/$. By Theorem 1, $K_{m}-F^{\prime}_{m}\/$ has a $g\/$-angulation which together $v_{q-g+1}v_{q-g+2}\cdots v_{q+1}\/$ form a $g\/$-angulation for $K_{n}-F_{n}\/$.

If $v_{q+1}\/$ is adjacent to $v_{q-g+2}\/$ in $F_{n}\/$, then $v_{q+1}\/$ is not adjacent to $v_{q+g-2}\/$ in $F_{n}\/$. In this case, we consider the convex graph obtained by deleting the vertices $v_{q+2},v_{q+3},\ldots,v_{q+g-1}\/$ and apply similar argument before to conclude that $K_{n}-F_{n}\/$ admits a $g\/$-angulation.

(ii) Suppose only one of $v_{q}$ or $v_{q+g-1}$ is a non-pendent vertex in ${\cal F}_{m,g}(^{*})\/$.

We can assume without loss of generality that $v_{q}$ (since we can relabel the vertices of $K_{n}-F_{n}\/$). Then $v_{q+g-1}\/$ is a pendant vertex in $F_{n}\/$. If for some $i\in S\/$, $v_{i}\/$ is not adjacent to $v_{i-g+1}\/$ in $F_{n}\/$, then by the method similar to case in (i), we see that $K_{n}-F_{n}\/$ admits a $g\/$-angulation. On the other hand, if $v_{i}\/$ is adjacent to $v_{i-g+1}\/$ in $F_{n}\/$ for all $i\in S\/$, then $F_{n}\/$ is ${\cal F}_{n,g}(^{*})\/$, a contradiction.

(iii) Suppose both $v_{q}$ and $v_{q+g-1}$ are pendent vertices in ${\cal F}_{m,g}(^{*})\/$.

If $i\in S\/$, we let $v_{s_{i}}$ be the neighbor of $v_{i}$ in $F_{n}\/$.

Suppose there exist $i,j\in S\/$ such that

$i<j\/$ and $v_{i}v_{s_{i}}\/$ and $v_{j}v_{s_{j}}\/$ crosse in $F_{n}\/$ with $s_{j}-s_{i}>g-2\/$ $(\star)\/$

Let $H_{1}\/$ and $H_{2}\/$ denote the convex subgraphs of $K_{n}-F_{n}\/$ induced by the vertices $v_{s_{i}+a},\ldots,v_{s_{j}},\ldots,v_{i},\ldots,v_{j}\/$ and $v_{j},\ldots,v_{s_{i}},\ldots,v_{s_{i}+a}\/$ respectively where $a\in\{1,\ldots,g-2\}\/$ with $|s_{i}+a-j|\equiv 1({\rm mod}\ (g-2))\/$. Then $H_{1}\/$ and $H_{2}\/$ each admits a $g\/$-angulation (since $v_{i}$ and $v_{j}$ is adjacent to every other vertex of $H_{1}\/$ and $H_{2}\/$ respectively) which together yields a $g\/$-angulation for $K_{n}-F_{n}\/$.

Hence we assume that no $i,j\in S\/$ satisfy the condition $(\star)\/$.

Suppose that for some $i\in S\/$, $v_{s_{i}-j}v_{s_{i}-j+g-1}\notin E(F_{n})$ for some $j\in\{1,\ldots,g-2\}\/$.

Then the subgraph obtained by deleting $g-2\/$ vertices $v_{s_{i}-j+1},\ldots,v_{s_{i}},\ldots,\linebreak v_{s_{i}-j+g-2}\/$ (from $K_{n}-F_{n}\/$) is of the form $K_{m}-F^{\prime}_{m}\/$ where $m=n-g+2$ with $|E(F^{\prime}_{m})|\leq m-g\/$ (because $d_{F_{n}}(v_{s_{i}})\geq 2\/$) and $F^{\prime}_{m}\/$ contains no boundary edges of $K_{m}\/$. By Theorem 1, $K_{m}-F^{\prime}_{m}\/$ has a $g\/$-angulation which together with $v_{s_{i}-j}\cdots v_{s_{i}}\cdots v_{s_{i}-j+g-1}\/$ form a $g\/$-angulation for $K_{n}-F_{n}\/$.

Hence we assume that for any $i\in S\/$, $v_{s_{i}-j}v_{s_{i}-j+g-1}\/$ is an edge in $F_{n}\/$ for any $j\in\{1,\ldots,g-2\}\/$.

Suppose there is a pendant vertex $v_{r}\/$ such that $r\not\in S\/$ and the edge $v_{r}v_{s}\/$ (incident to $v_{r}\/$) crosses $v_{i}v_{s_{i}}\/$ ($i\in S\/$) with $|s-s_{i}|>g-2\/$, then again a $g\/$-angulation of $K_{n}-F_{n}\/$ can be constructed as in the previous case (where the condition $(\star)\/$ is satisfied).

Hence, for any pendant vertex $v_{r}\/$ where $r\not\in S\/$, $v_{r}v_{s}\/$ does not crosse $v_{i}v_{s_{i}}\/$ with $|s-s_{i}|>g-2\/$ for any $i\in S\/$. But this implies that $F_{n}\/$ is ${\cal F}_{n,g}(^{*})\/$, a contradiction.

This completes the proof.

4 $n-g+1+\mu\/$ edges

We present in this section a characterization of ${\cal J}_{n,g}(^{*}_{\beta})\/$ (where $\beta\in\{1,2,\ldots,2g-3\}\/$) of size at most $n-1\/$ that shares any possible $g$-angulation of $K_{n}\/$ by at least one edge. We show that $G=K_{n}-F_{n}\/$ admits a $g$-angulation if and only if $F_{n}\neq{\cal J}_{n,g}(^{*}_{\beta})\/$.

Definition 2

Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 1$ is any natural number. Let ${\cal J}_{n,g}(^{*}_{1})\/$ denote a convex geometric graph with $n\/$ vertices and $n-g+1+\mu\/$ edges such that ${\cal J}_{n,g}(^{*}_{1})-\{e_{1},e_{2},\ldots,e_{\mu}\}={\cal F}_{n,g}(^{*})\/$ for some $\mu$ edges $e_{1},e_{2},\ldots,e_{\mu}$. Here $1\leq\mu\leq g-2$.

Proposition 1

$K_{n}-{\cal J}_{n,g}(^{*}_{1})\/$* admits no $g$-angulation for any natural number $n=g+t(g-2)\/$ with $t\geq 1$.*

Proof: Suppose on the contrary that $K_{n}-{\cal J}_{n,g}(^{*}_{1})\/$ admits a $g$-angulation $G_{n}\/$. Since ${\cal J}_{n,g}(^{*}_{1})$ contains $\mu$ edges $\{e_{1},e_{2},\ldots,e_{\mu}\}$ such that ${\cal J}_{n,g}(^{*}_{1})-\{e_{1},e_{2},\ldots,e_{\mu}\}={\cal F}_{n,g}(^{*})\/$, it follows that $K_{n}-{\cal F}_{n,g}(^{*})\/$ admits $G_{n}\/$, which is a contradiction.

This completes the proof.

Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 0$ is any natural number. Let $F\/$ denote a subgraph of $K_{n}\/$.

• •

A pair of vertices $\{v_{r},v_{s}\}\/$ is called an $\alpha\/$*-pair * in $F\/$ if whenever pair of edges $v_{i}v_{r},v_{j}v_{s}\/$ (with $|r-i|\equiv 1({\rm mod}\ (g-2))\/$ and $|s-j|\equiv 1({\rm mod}\ (g-2))\/$), then $v_{i}v_{r}\/$ crosses $v_{j}v_{s}\/$ with $|i-j|>g-2\/$.

• •

A vertex $v_{i}\in V(F)$ is called a $g$*-angulable vertex * in $F\/$ if whenever edge $v_{i}v_{j}\/$ (if there exists) in $F\/$ satisfies that $|j-i|\not\equiv 1\ ({\rm mod}\ (g-2))\/$.

Definition 3

Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 2$ is any natural number. Let ${\cal J}_{n,g}(^{*}_{2})\/$ denote a convex geometric graph with $n\/$ vertices and $n-g+1+\mu\/$ edges (where $1\leq\mu\leq g-2\/$) with no $g\/$-angulable vertex in $K_{n}-{\cal J}_{n,g}(^{*}_{2})\/$ such that

(i) if $\{v_{r},v_{s}\}\/$ is an $\alpha\/$-pair of vertices in ${\cal J}_{n,g}(^{*}_{2})\/$ (which occurs at most once), then $v_{r},v_{s}\/$ are adjacent in ${\cal J}_{n,g}(^{*}_{2})\/$ with $|r-s|\equiv 1({\rm mod}\ (g-2))\/$, and

(ii) whenever $v_{i}v_{i+g-1}\not\in E({\cal J}_{n,g}(^{*}_{2}))$ and $\{v_{i+1},\ldots,v_{i+g-2}\}\/$ contains non-pendant vertex (which is a vertex of an $\alpha\/$-pair or a vertex in the neighbor of a vertex of an $\alpha\/$-pair in case ${\cal J}_{n,g}(^{*}_{2})\/$ has an $\alpha\/$-pair), then ${\cal J}_{n,g}(^{*}_{2})-\{v_{i+1},\ldots,v_{i+g-2}\}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{1})\/$ (where $m=n-g+2\/$).

Remark 2

Suppose $v_{i}v_{i+g-1}\not\in E({\cal J}_{n,g}(^{*}_{2}))$ and either all vertices in $\{v_{i+1},\linebreak\ldots,v_{i+g-2}\}\/$ are pendant (in case ${\cal J}_{n,g}(^{*}_{2})\/$ has no $\alpha\/$-pair) or $\{v_{i+1},\ldots,v_{i+g-2}\}\/$ contains a non-pendant vertex but neither a vertex of an $\alpha\/$-pair nor a vertex in the neighbor of a vertex of an $\alpha\/$-pair (in case ${\cal J}_{n,g}(^{*}_{2})\/$ has an $\alpha\/$-pair), then it is easy to see that ${\cal J}_{n,g}(^{*}_{2})-\{v_{i+1},\ldots,v_{i+g-2}\}\/$ is ${\cal J}_{m,g}(^{*}_{2})$ where $m=n-g+2\/$. See Figure 12.

An example of a geometric graph ${\cal J}_{n,g}(^{*}_{2})\/$ with and without $\alpha\/$-pair is depicted in Figure 2 (a) and (b) respectively. In Figure 2 (a), $\{v_{6},v_{10}\}\/$ is the $\alpha\/$-pair in ${\cal J}_{14,5}(^{*}_{2})\/$.

Proposition 2

$K_{n}-{\cal J}_{n,g}(^{*}_{2})\/$* admits no $g$-angulation for any natural number $n=g+t(g-2)\/$ with $t\geq 2$.*

**Proof:**We prove this by induction on $n\/$.

Consider the case $t=2\/$. Here $n=3g-4\/$.

Lemma 1 asserts the existence of an edge $v_{q}v_{q+g-1}\/$ in $K_{n}-{\cal J}_{n,g}(^{*}_{2})\/$. Choose $q\/$ such that $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ contains non-pendant vertex. By definition ${\cal J}_{n,g}(^{*}_{2})-\{v_{q+1},\ldots,v_{q+g-2}\}=F_{m}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{1})\/$, $K_{m}-F_{m}\/$ admits no $g\/$-angulation by Theorem 2 and Proposition 1. Here $m=n-g+2\/$.

Assume on the contrary that $K_{n}-{\cal J}_{n,g}(^{*}_{2})\/$ admits a $g$-angulation $G_{n}\/$. Then $G_{n}\/$ does not contain the edge $v_{q}v_{q+g-1}\/$ (otherwise this implies that $K_{m}-F_{m}\/$ admits $g\/$-angulation, a contradiction).

Since $t=2\/$, $G_{n}\/$ has only two diagonal edges. Clearly at least one of these diagonal edges, say $e_{1}\/$ is incident to a vertex in $\{v_{q+1},\ldots,v_{q+g-2}\}\/$. This is clearly not possible if all vertices in $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ are non-pendant. Hence assume that $e_{1}\/$ is a incident to a pendant vertex in $\{v_{q+1},\ldots,v_{q+g-2}\}\/$. That is $e_{1}=v_{j}v_{r}\/$ where $j\in\{q+1,\ldots,q+g-2\}\/$. This means that $v_{j}v_{s}\/$ is an edge of ${\cal J}_{n,g}(^{*}_{2})\/$ (where $|r-s|=g-2\/$) and $v_{s}\/$ is a non-pendant vertex in ${\cal J}_{n,g}(^{*}_{2})\/$ (otherwise $v_{s}\/$ is an isolated vertex in $F_{m}={\cal J}_{n,g}(^{*}_{2})-\{v_{q+1},\ldots,v_{q+g-2}\}\/$, a contradiction).

Let $v_{p}v_{p+g-1}\/$ be the second diagonal edge of $G_{n}\/$ (where $p>j\/$). It is easy to see that $v_{s}\/$ is one of the vertex in $\{v_{p+1},\ldots,v_{p+g-2}\}\/$ (see for example Figure 3). But this means that $v_{j}\/$ is an isolated vertex in ${\cal J}_{n,g}(^{*}_{2})-\{v_{p+1},\ldots,v_{p+g-2}\}\/$ (a contradiction).

Now suppose $t\geq 3\/$.

Assume on the contrary that $K_{n}-{\cal J}_{n,g}(^{*}_{2})$ admits a $g$-angulation $G_{n}\/$. Note that any $g$-angulation on a set of points in convex position has a diagonal $v_{q}v_{q+g-1}\/$.

Let $F_{m}={\cal J}_{n,g}(^{*}_{2})-\{v_{q+1},\ldots,v_{q+g-2}\}\/$ where $m=n-g+2\/$. Then $G_{n}-\{v_{q+1},\ldots,v_{q+g-2}\}\/$ is a $g$-angulation for $K_{m}-F_{m}\/$.

*Case (1): * ${\cal J}_{n,g}(^{*}_{2})\/$ contains no $\alpha\/$-pair.

(i) Suppose $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ contains non-pendent vertex. Then by the definition of ${\cal J}_{n,g}(^{*}_{2})\/$, $F_{m}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{1})\/$. By Theorem 2 or Proposition 1 respectively, $K_{m}-F_{m}\/$ admits no $g$-angulation, a contradiction.

(ii) Suppose all vertices in $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ are pendant vertices. Then $F_{m}={\cal J}_{m,g}(^{*}_{2})\/$. By induction, $K_{m}-F_{m}\/$ admits no $g$-angulation, again a contradiction that $G_{n}$ is a $g$-angulation for $K_{n}-{\cal J}_{n,g}(^{*}_{2})\/$.

*Case (2): * ${\cal J}_{n,g}(^{*}_{2})\/$ contains an $\alpha\/$-pair.

If $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ contains a vertex of an $\alpha\/$-pair or a vertex in the neighbor of a vertex of an $\alpha\/$-pair, then (by definition of ${\cal J}_{n,g}(^{*}_{2})\/$,) $F_{m}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{1})\/$. If $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ contains neither a vertex of an $\alpha\/$-pair nor a vertex in the neighbor of a vertex of an $\alpha\/$-pair, then $F_{m}\/$ is ${\cal J}_{m,g}(^{*}_{2})\/$.

In any case, by Theorem 2, Proposition 1 or by induction, $K_{m}-F_{m}\/$ admits no $g$-angulation, a contradiction that $G_{n}$ is a $g$-angulation for $K_{n}-{\cal J}_{n,g}(^{*}_{2})\/$.

This completes the proof.

Definition 4

Suppose $g\geq 4\/$ and $\gamma\geq 3\/$ are natural numbers and let $n=g+t(g-2)\/$ where $t\geq 3\/$ is any natural number. Let ${\cal J}_{n,g}(^{*}_{\gamma})\/$ denote a convex geometric graph with $n\/$ vertices and $n-g+1+\mu\/$ edges (where $1\leq\mu\leq g-2\/$) with no $g$-angulable vertex and having only $\gamma-1\/$ $\alpha\/$-pairs of vertices such that

(i) whenever $\{v_{r},v_{s}\}\/$ is an $\alpha\/$-pair of vertices, then $v_{r},v_{s}\/$ are adjacent in ${\cal J}_{n,g}(^{*}_{\gamma})\/$ with $|r-s|\equiv 1({\rm mod}\ (g-2))\/$, and

(ii) whenever $v_{i}v_{i+g-1}\not\in E({\cal J}_{n,g}(^{*}_{\gamma}))$ and $\{v_{i+1},\ldots,v_{i+g-2}\}\/$ contains a vertex of an $\alpha\/$-pair or a vertex in the neighbor of a vertex of an $\alpha\/$-pair, then ${\cal J}_{n,g}(^{*}_{\gamma})-\{v_{i+1},\ldots,v_{i+g-2}\}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{\beta})\/$ where $m=n-g+2\/$ and $\beta\leq\gamma-1\/$.

An example of a convex geometric graph ${\cal J}_{n,g}(^{*}_{3})\/$ is depicted in Figure 4. In Figure 4 (a), $\{v_{9},v_{12}\}$ and $\{v_{13},v_{16}\}$ are two $\alpha\/$-pairs in ${\cal J}_{18,4}(^{*}_{3})\/$, in (b) $\{v_{13},v_{16}\}$ is the $\alpha\/$-pair in the convex geometric graph ${\cal J}_{16,4}(^{*}_{2})\/$ which obtained from ${\cal J}_{18,4}(^{*}_{3})\/$ by deleting two vertices $v_{7}\/$ and $v_{8}\/$ (with $v_{7}\/$ is a neighbor of a vertex ($v_{12}\/$) of the $\alpha\/$-pair $\{v_{9},v_{12}\}\/$). For more instance see Figure 13.

Proposition 3

$K_{n}-{\cal J}_{n,g}(^{*}_{\gamma})\/$* admits no $g$-angulation for any natural number $n=g+t(g-2)\/$ with $t\geq 3$.*

Proof: We prove this by induction on $n\/$.

Consider the case $t=3\/$. Here $n=4g-6\/$.

Assume on the contrary that $K_{n}-{\cal J}_{n,g}(^{*}_{\gamma})$ admits a $g$-angulation $G_{n}\/$. Note that $G_{n}\/$ has a diagonal $v_{q}v_{q+g-1}\/$ for some $q\/$.

Let $F_{m}={\cal J}_{n,g}(^{*}_{\gamma})-\{v_{q+1},\ldots,v_{q+g-2}\}\/$ where $m=n-g+2\/$. Then $G_{n}-\{v_{q+1},\ldots,v_{q+g-2}\}\/$ is a $g$-angulation of $K_{m}-F_{m}\/$.

Let $\{v_{r},v_{s}\}\/$ be an $\alpha$-pair in ${\cal J}_{n,g}(^{*}_{\gamma})\/$ and let $v_{i}v_{r}\/$ and $v_{j}v_{s}\/$ the edges incident to $v_{r}\/$ and $v_{s}\/$ respectively. Here $|i-r|\equiv 1({\rm mod}\ (g-2))\/$ and $|j-s|\equiv 1({\rm mod}\ (g-2))\/$. Since $|i-j|>g-2\/$ and $|r-s|=g-1\/$.

Since $m=n-g+2\/$ then $t=2\/$. Hence, there is no any $\alpha$-pair in $F_{m}\/$.

Thus, $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ contains a vertex of $v_{i},v_{j},v_{r},v_{s}\/$.

By definition of ${\cal J}_{n,g}(^{*}_{\gamma})\/$, $F_{m}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{\beta})\/$ with $\beta=1,2\/$.  By Theorem 2 or Proposition 1 or else Proposition 2, $K_{m}-F_{m}\/$ admits no $g$-angulation, a contradiction.

Now suppose $t\geq 4\/$.

Assume on the contrary that $K_{n}-{\cal J}_{n,g}(^{*}_{\gamma})$ admits a $g$-angulation $G_{n}\/$. Then $G_{n}\/$ has a diagonal $v_{q}v_{q+g-1}\/$.

Let $F_{m}={\cal J}_{n,g}(^{*}_{\gamma})-\{v_{q+1},\ldots,v_{q+g-2}\}\/$ where $m=n-g+2\/$. Then $G_{n}-\{v_{q+1},\ldots,v_{q+g-2}\}\/$ is a $g$-angulation for $K_{m}-F_{m}\/$.

If $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ contains a vertex of an $\alpha\/$-pair or a vertex in the neighbor of a vertex of an $\alpha\/$-pair, then (by definition of ${\cal J}_{n,g}(^{*}_{\gamma})\/$,) $F_{m}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{\beta})\/$ where $m=n-g+2\/$ and $\beta\leq\gamma-1\/$. If $\{v_{q+1},\ldots,v_{q+g-2}\}\/$ contains neither a vertex of an $\alpha\/$-pair nor a vertex in the neighbor of a vertex of an $\alpha\/$-pair, then $F_{m}\/$ is ${\cal J}_{m,g}(^{*}_{\gamma})\/$.

In any case, by Theorem 2, Proposition 1, Proposition 2 or by induction, $K_{m}-F_{m}\/$ admits no $g$-angulation, a contradiction that $G_{n}$ is a $g$-angulation for $K_{n}-{\cal J}_{n,g}(^{*}_{\gamma})\/$ (see for example Figure 12).

This completes the proof.

An example of a convex geometric graph ${\cal J}_{n,g}(^{*}_{\gamma})\/$ is depicted in Figure 5. In Figure 5 (a), ${\cal J}_{18,6}(^{*}_{8})\/$ has $\{v_{10},v_{15}\}\/$, $\{v_{10-\ell},v_{15-\ell}\}\/$ and $\{v_{6-\ell},v_{1-\ell}\}\/$ as $\alpha$-pairs where $\ell=1,2,3\/$, in Figure 5 (b) ${\cal J}_{18,6}(^{*}_{9})\/$ has $\{v_{7+\ell},v_{12+\ell}\}\/$ and $\{v_{16+\ell},v_{3+\ell}\}\/$ as $\alpha$-pairs where $\ell=0,1,2,3\/$.

Theorem 4

Suppose $g\geq 3\/$ is a natural number and let $n=g+t(g-2)\/$ where $t\geq 1$ is any natural number. Suppose $F_{n}\/$ is a subgraph of the convex complete graph $K_{n}\/$ such that $n-g+2\leq|E(F_{n})|\leq n-1\/$ and $F_{n}\/$ contains no boundary edges of $K_{n}\/$. Then $K_{n}-F_{n}\/$ admits a $g\/$-angulation unless $F_{n}\/$ is ${\cal J}_{n,g}(^{*}_{\beta})\/$ for some $\beta\in\{1,2,\ldots,2g-3\}\/$.

Proof: In view of Propositions 1, 2 and 3, we assume that $F_{n}\neq{\cal J}_{n,g}(^{*}_{\beta})\/$ for any $\beta\geq 1\/$.

The case $g=3\/$ has been treated in [17]. Hence we assume that $g\geq 4\/$.

We prove the result by induction on $t\/$.

Suppose $t=1\/$. If $K_{n}-F_{n}\/$ contains an edge of the form $v_{j}v_{j+g-1}\/$, then $v_{j}v_{j+g-1}\/$ together with the boundary edges of $K_{n}\/$ is a $g\/$-angulation of $K_{n}-F_{n}\/$.

On the other hand, if $v_{j}v_{j+g-1}\/$ is not an edge of $K_{n}-F_{n}\/$ for any $j=0,1,\ldots,g-2\/$, (that is, $v_{j}v_{j+g-1}\/$ is an edge of $F_{n}\/$), then $F_{n}\/$ contains ${\cal F}_{n,g}(^{*})\/$ as a subgraph. This contradicts the assumption that $F_{n}\neq{\cal J}_{n,g}(^{*}_{1})\/$.

Now, assume that $t\geq 2$ and the result is true for all convex graphs $K_{m}-F_{m}\/$ where $m=g+t^{\prime}(g-2)\/$, $t^{\prime}\leq t-1$.

By Lemma 1, $K_{n}-F_{n}\/$ contains an edge of the form $v_{j}v_{j+g-1}\/$ which is not an edge of $F_{n}\/$. By relabeling if necessary, we may take $j=0$.

Now delete the set of vertices $v_{1},\dots,v_{g-2}$ from $K_{n}-F_{n}\/$. Let $K_{m}-F_{m}\/$ denote the resulting convex graph. Here $m=g+(t-1)(g-2)\/$.

If $|E(F_{m})|\leq m-g$, then $K_{m}-F_{m}\/$ admits a $g$-angulation by Theorem 1. Clearly, this $g\/$-angulation gives rise to a $g$-angulation for $K_{n}-F_{n}\/$.

Hence we assume that $|E(F_{m})|\geq m-g+1$.

Suppose $F_{m}\/$ is neither ${\cal F}_{m,g}(^{*})\/$ nor ${\cal J}_{m,g}(^{*}_{\beta})\/$ for any $\beta\/$. By Theorem 3 or by induction $K_{m}-F_{m}\/$ admits a $g$-angulation. Again this $g\/$-angulation gives rise to a $g$-angulation for $K_{n}-F_{n}\/$.

Hence assume that $F_{m}\/$ is either ${\cal F}_{m,g}(^{*})\/$ or ${\cal J}_{m,g}(^{*}_{\beta})\/$ for some $\beta\in\{1,2,\ldots 2g-3\}\/$.

Suppose that $\{v_{1},\dots,v_{g-2}\}\/$ contains no vertex of an $\alpha\/$-pair in $F_{n}\/$.

If $F_{m}={\cal F}_{m,g}(^{*})\/$, then $F_{n}\/$ is either ${\cal F}_{n,g}(^{*}_{1})\/$ or ${\cal J}_{n,g}(^{*}_{2})\/$. Either case is a contradiction.

Hence assume that $F_{m}={\cal J}_{m,g}(^{*}_{\beta})\/$. If $\{v_{1},\dots,v_{g-2}\}\/$ contains a neighbor of a vertex of an $\alpha\/$-pair in $F_{n}\/$, then $F_{n}={\cal F}_{n,g}(^{*}_{\gamma})\/$ ($\gamma=\beta+1\/$). If $\{v_{1},\dots,v_{g-2}\}\/$ contains no neighbor of a vertex of an $\alpha\/$-pair in $F_{n}\/$, then $F_{n}={\cal F}_{n,g}(^{*}_{\beta})\/$. Either case is a contradiction.

Now assume that $\{v_{r},v_{s}\}\/$ is an $\alpha\/$-pair in $F_{n}\/$ and $v_{r}\in\{v_{1},\dots,v_{g-2}\}\/$.

If $v_{r},v_{s}\/$ are adjacent and $|r-s|\equiv 1({\rm mod}\ (g-2))\/$ then $F_{n}={\cal J}_{n,g}(^{*}_{\beta})\/$ for some $\beta\in\{2,\ldots,2g-3\}\/$, a contradiction. Hence either $v_{r},v_{s}\/$ are non-adjacent or else $|r-s|\not\equiv 1({\rm mod}\ (g-2))\/$ .

Assume without loss of generality that $r<s\/$. Let $i\/$ be the largest integer such that $v_{i}\/$ is adjacent to $v_{r}\/$ in $F_{n}\/$, and $j\/$ the smallest integer such that $v_{j}\/$ is adjacent to $v_{s}\/$ in $F_{n}\/$.

Let $k\/$ be an integer such that $i<k<j\/$ and $|k-s|\equiv 1({\rm mod}\ (g-2))\/$. Then $v_{r}\/$ (respectively $v_{s}\/$) is a $g\/$-angulable vertex in the convex subgraph induced by $v_{k},v_{k+1},\ldots,v_{s}\/$ (respectively $v_{s},v_{s+1},\ldots,v_{k}\/$). This yields a $g\/$-angulation for $K_{n}-F_{n}\/$.

This completes the proof.

5 Potentially $g$-angulable graphs

We now look at the possibility of placing a graph $F\/$ with $n\/$ vertices and $n\/$ edges in the convex complete graph $K_{n}\/$ so that $K_{n}-F\/$ admits a $g$-angulation. We shall confine our attention to the case where $F_{n}\/$ is a $2\/$-regular graph.

Definition 5

:* Let $K_{n}\/$ be a convex complete graph with $n\/$ vertices. $F\/$ is said to be potentially $g$-angulable if there exists a configuration of $F\/$ in $K_{n}\/$ such that $K_{n}-F\/$ admits a $g$-angulation.*

Theorem 5

:* Suppose $F_{n}\/$ is an $n\/$-cycle and $g\geq 4$ is a natural number such that $n=g+t(g-2)$. Then $F_{n}\/$ is potentially $g$-angulable if and only if $n\geq 5\/$.*

Proof: It is easy to see that $K_{n}-F_{n}\/$ admits no $g$-angulation if $n\leq 4\/$.

For $n=6\/$ suppose $F\/$ is a $6\/$-cycle. Let $F\/$ be of the form $v_{0}v_{2}v_{4}v_{1}v_{5}v_{3}v_{0}\/$, then $K_{6}-F\/$ admits a $g$-angulation for $g\in\{4,6\}$, when $g=4$ the diagonal is $v_{2}v_{5}\/$.

For the rest of the proof, we assume that $n\geq 5\/$, where $n\neq 6$.

When $n\/$ is odd, let $F_{n}\/$ takes the form

[TABLE]

When $n\/$ is even, let $F_{n}\/$ takes the form

[TABLE]

In both cases, the edges $v_{2}v_{3+i(g-2)}\/$, $i=1,\ldots,t\/$ together with the boundary edges $v_{0}v_{1}v_{2}\ldots v_{n-1}v_{0}\/$ is a $g$-angulation of $K_{n}-F_{n}\/$.

This completes the proof.

Theorem 6

:* Let $F_{n}\/$ be a $2\/$-regular graph with $n\/$ vertices and $g\geq 4$ is a natural number such that $n=g+t(g-2)$. Then $F_{n}\/$ is potentially $g$-angulable if and only if $n\geq 5\/$.*

Proof: If $F_{n}\/$ is connected, the result is true by Theorem 5. Hence we assume that $F_{n}\/$ is a union of disjoint cycles.

Let $C$ be a smallest cycle in $F_{n}$ and let $v_{x}v_{y}$ and $v_{y}v_{z}$ are two edges in $C$. Consider $F^{*}\/$ to be a union of disjoint cycles in $F_{n}-C$ and let $|V(C)|=p$.

(i) If $F^{*}$ is a $4$-cycle, Take $F^{*}=\{v_{1}v_{3}v_{2}v_{4}v_{1}\}\/$. Insert $v_{x}$, $v_{y}$ and $v_{z}$ of $C$ into edges $v_{4}v_{1},v_{1}v_{2},v_{2}v_{3}$ respectively. Then $K_{7}-F_{7}$ admits $7$-angulation with $t=0$. In case that $p=4$, insert the fourth vertex $v_{w}$ into the edge $v_{3}v_{4}$. Then $K_{8}-F_{8}$ admits $4$-angulation with two diagonal edges $v_{y}v_{3}$ and $v_{y}v_{4}$ and admits $5$-angulation with a diagonal edge $v_{y}v_{w}$.

(ii) If $F^{*}$ is not a $4$-cycle, then place $F^{*}$ on $K^{*}_{n-p}$ so that $F^{*}$ contains no boundary edge of $K^{*}_{n-p}$. Let $v_{1},v_{2},\ldots,v_{n-p}$ denotes the vertices of $K^{*}_{n-p}$. Insert $v_{x}$, $v_{y}$ and $v_{z}$ of $C$ into edges $v_{1}v_{2},v_{2}v_{3},v_{3}v_{4}$ respectively. In case that $p>3$ insert the rest of vertices of $C$, which are $p-3$ vertices, into the edges of the path $v_{4}v_{5}\cdots v_{n-p}v_{1}$ and place $C$ on $K_{n}$ so that $C$ contains no boundary edge of $K_{n}$.

Relabel the vertices of $K_{n}$ to be $u_{0},u_{1},u_{2},\ldots,u_{n-1}$ with $v_{y}=u_{0}$. Hence, we have $u_{0}u_{1+i(g-2)}$, $i=1,2,\dots,t$ together with the edges $u_{0}u_{1}u_{2}u_{3}\ldots u_{n-1}u_{0}\/$ is a $g$-angulation of $K_{n}-F_{n}\/$ (since $u_{0}$ is adjacent only to $u_{n-2}$ and $u_{2}$).

This completes the proof.

6 Regular graphs

In view of the results in the preceding section, it is natural to ask which regular graph is potentially $g$-angulable in $K_{n}\/$.

Problem:  Let $r\geq 3\/$ and $g\geq 3\/$ be two natural numbers and let $G\/$ be an $r\/$-regular graph with $n\/$ vertices where $n=g+(g-2)t\/$. It is true that there is a natural number $n_{0}(r,g)\/$ such that when $n\geq n_{0}(r,g)\/$, then $G\/$ is potentially $g$-angulable in the convex complete graph $K_{n}\/$?

We believe that the above problem is true. However we do not have a complete answer for this even when restricted to the case $r=3\/$. Nevertheless we offer the following special case of a $3\/$-regular graph which is well-known in the literature.

Suppose $n\/$ and $k\/$ are two integers such that $1\leq k\leq n-1\/$ and $n\geq 5\/$. The generalized Petersen graph $P(n,k)\/$ is defined to have vertex-set $\{a_{i},b_{i}:i=0,1,\ldots,n-1\}\/$ and edge-set $E_{1}\cup E_{2}\cup E_{3}\/$ where $E_{1}=\{a_{i}a_{i+1}:i=0,1,\ldots,n-1\}\/$, $E_{2}=\{b_{i}b_{i+k}:i=0,1,\ldots,n-1\}\/$ and $E_{3}=\{a_{i}b_{i}:i=0,1,\ldots,n-1\}\/$ with subscripts reduced modulo $n\/$. Edges in $E_{3}\/$ are called the spokes of $P(n,k)\/$.

Proposition 4

:* Suppose $1\leq k<n/2\/$ and $g\geq 4$ is a natural number such that $2n=g+t(g-2)$. Then the generalized Petersen graph $P(n,k)\/$ is potentially $g$-angulable in the convex complete graph $K_{2n}\/$ where $n\geq 5\/$.*

Proof: Let the vertices of $K_{2n}\/$ be denoted $v_{1},v_{2},v_{3},\ldots,v_{2n}\/$. We shall pack $P(n,k)\/$ on $K_{2n}\/$ so that $K_{2n}-P(n,k)\/$ admits a $g$-angulation.

Case (1):  $k=1\/$

$P(n,1)\/$ consists of two $n\/$-cycles $C=a_{0}a_{1}a_{2}\cdots c_{n-1}a_{0}\/$ and $C^{\prime}=b_{0}b_{1}b_{2}\cdots b_{n-1}b_{0}\/$ together with the edges $a_{i}b_{i}\/$, $i=0,1,2,\ldots,n-1\/$.

Place $C\/$ on $K_{2n}\/$ so that $C\/$ takes the form $v_{2}v_{4}v_{6}\cdots v_{2n-4}v_{2n-2}v_{2n}v_{2}\/$ and that $C^{\prime}\/$ takes the form $v_{1}v_{3}v_{5}v_{4}v_{7}\cdots v_{2n-3}v_{2n-1}v_{1}\/$.

Case (2):  $1<k<n/2\/$

$P(n,k)\/$ consists of two $n\/$-cycles $C=a_{0}a_{1}a_{2}\cdots c_{n-1}a_{0}\/$ and $C^{\prime}=\{b_{i}b_{i+k},\ i=0,1,2,\ldots,n-1\}\/$ together with the edges $a_{i}b_{i}\/$, $i=0,1,2,\ldots,n-1\/$.

Place $C\/$ and $C^{\prime}\/$ on $K_{2n}\/$ so that $C\/$ takes the form $v_{2}v_{4}v_{6}\ldots v_{2n-2}v_{2n}v_{2}\/$ and $C^{\prime}\/$ takes the form $\{v_{i}v_{i+2k}\ :\ i=1,3,\ldots,2n-1\}\/$. The operations on the subscripts are reduced modulo $2n\/$.

In both cases: Let the spokes take the form  $v_{i}v_{i+3}\/$,  $i=0,2,4,\ldots,2n-2\/$. Here also the operations on the subscripts are reduced modulo $2n\/$.

In both cases: if $t=1$, then $K_{n}-F_{n}\/$ has a $g$-angulation whose diagonal is $v_{1}v_{g}$. If $t\geq 2$, consider the subgraph $H\/$ induced by the sets of vertices $\{v_{g},v_{g+1},\ldots,\linebreak v_{2n-(g-1)}\}\cup\{v_{0},v_{1}\}\/$. Since the vertex $v_{0}\/$ is not adjacent to every vertex in $H\/$ and $|V(H)|=2n-2(g-2)$, $K_{2n-2(g-2)}-H\/$ admits a $g$-angulation $G\/$. Then $G\cup\{v_{1}v_{2}\cdots v_{g}\}\cup\{v_{2n-(g-1)}v_{2n-(g-1)+1}\ldots v_{2n-1}v_{0}\}\/$ is a $g$-angulation for $K_{2n}-P(n,k)\/$.

This completes the proof.

In this part, we shall prove that $3\/$-regular graph with $n=4+2t\/$ vertices where $t\geq 2\/$ is potentially $4$-angulable.

Lemma 2

:* Let $G\/$ be a $3\/$-regular graph with $n=4+2t\/$ vertices where $t\geq 2\/$. Suppose $V(G)$ can be labeled as $v_{1},v_{2},\ldots,v_{n}$ such that*

(i) $v_{1}v_{a},v_{1}v_{b},v_{1}v_{c}\in E(G)$ where $a,b,c\/$ are distinct odd integers (different from $1\/$, and

(ii) $|i-j|\notin\{1,n-1\}$ whenever $v_{i}v_{j}\in E(G)$.

Then $G\/$ is potentially $4$-angulable.

Proof: Place the vertices $v_{1},v_{2},\ldots,v_{n}\/$ of $G\/$ in convex position and put them in clock wise order. Then $v_{1}v_{4+2k}\/$, $k=0,1,\ldots,t-1$ together with all boundary edges of $K_{n}\/$ is a $4$-angulation of $K_{n}-G$.

In the next lemma, we shall show that all $3\/$-regular graphs with at least $8\/$ vertices admit a labeling as described in Lemma 2 unless it is the cube $Q_{3}\/$ (on $8\/$ vertices) in which case it has a labeling as shown in Figure 10. If $Q_{3}\/$, with the given labeling, is placed on convex position (with clockwise order), then $v_{2}v_{5},v_{1}v_{6}\/$ together with the boundary edges of $K_{8}\/$ is a $4$-angulation of $K_{8}-Q_{3}$. This implies that all connected $3\/$-regular graphs are potentially $4\/$-angulable.

To facilitate the proof of Theorem 7, we shall need to consider $3\/$-regular graphs where double edges are allowed. By a $2\/$-cycle in a graph, written $uvu\/$, we mean two edges of the graph of the form $uv\/$ and $vu\/$.

Let $G\/$ be a $3\/$-regular graph on $n\/$ vertices and let $e=xy$ be an edge in $G$. Suppose $N_{G}(x)=\{x_{1},x_{2},y\}$ and $N_{G}(y)=\{y_{1},y_{2},x\}$. Let $G_{e}\/$ be the graph obtained from $G-e\/$ by replacing the paths $x_{1}xx_{2}\/$ and $y_{1}yy_{2}\/$ with the edges $x_{1}x_{2}\/$ and $y_{1}y_{2}\/$ respectively. Then $G_{e}\/$ is a $3\/$-regular graph with $n-2\/$ vertices.

In the case that $G\/$ has no $2\/$-cycle, it is easy to see that $G\/$ contains an edge $e\/$ such that $G_{e}\/$ has at most one $2\/$-cycle if $n\geq 6\/$.

Lemma 3

:* Let $G\/$ be a connected $3\/$-regular graph with $n\geq 8\/$ vertices having at most one $2\/$-cycle.*

(i) Suppose $G\/$ has no multiple edges. Then $V(G)\/$ admits a labeling as described in Lemma 2 unless $G\/$ is the cube which has labeling as shown in Figure 10.

(ii) Suppose $G\/$ has a $2\/$-cycle $uvu\/$. Then $V(G)\/$ can be labeled as $v_{1},v_{2},\ldots,v_{n}\/$ such that $u=v_{1},v=v_{a}\/$, $v_{1}v_{b}\in E(G)\/$ where $a,b\/$ are distinct odd integers different from $1\/$, and that $|i-j|\notin\{1,n-1\}$ whenever $v_{i}v_{j}\in E(G)$.

Proof: We prove this lemma by induction on $n\/$.

For $n=8\/$, we have checked that each cubic graph on $8\/$ vertices, except $Q_{3}\/$ cube admits a labeling on its vertices that satisfies the conditions (i) and (ii) (see Figure 11).

Let $G\/$ be a $3\/$-regular graph on $n\/$ vertices where $n\geq 10\/$.

**Case (1) ** $G$ has no $2\/$-cycle.

Let $e=xy$ be chosen such that $G_{e}\/$ has at most one $2\/$-cycle.

(1.1) $G_{e}$ has no $2\/$-cycle.

If $G_{e}\/$ is the cube $Q_{3}\/$, then $G\/$ is any one of the four $3\/$-regular graphs depicted in Figure 8. Each of these graphs has a labeling that satisfies condition (i) of the lemma.

(b) Suppose $v_{1}\not\in\{x_{1},x_{2},y_{1},y_{2}\}\/$. See Figure 15.

(b1) Suppose $\{x_{1},x_{2}\}\cap\{y_{1},y_{2}\}=\emptyset\/$.

Let $v\in\{x_{1},x_{2},y_{1},y_{2}\}\/$ be such that $v\/$ is not adjacent to $v_{n-2}\/$ in $G_{e}\/$. Without loss of generality, assume that $v=x_{1}\/$. In $G_{e}\/$, change the label for $v\/$ from $v_{i}\/$ to $v_{n-1}\/$. Extend this labeling on $G_{e}\/$ to a labeling of $G\/$ by assigning $x\/$ with $v_{i}\/$ and $y\/$ with $v_{n}\/$.

(b2) Suppose $\{x_{1},x_{2}\}\cap\{y_{1},y_{2}\}\neq\emptyset\/$.

Assume that $x_{2}=y_{2}\/$. Then $y_{1}\neq x_{1}\/$.

Suppose that $v_{n-2}\in\{x_{1},x_{2},y_{1}\}$.

When $v_{n-2}=x_{2}$, then in $G_{e}\/$, change the label of $x_{1}$ from $v_{i}\/$ to $v_{n-1}\/$. In $G\/$, assign $v_{i},v_{n}\/$ to $x,y\/$ respectively.

When $v_{n-2}\neq x_{2}$, then assume without loss of generality that $v_{n-2}=x_{1}\/$. In $G_{e}\/$, change the label for $y_{1}\/$ from $v_{i}\/$ to $v_{n}\/$ and change the label for $x_{2}\/$ from $v_{j}\/$ to $v_{n-1}\/$. Extend this labeling on $G_{e}\/$ to a labeling of $G\/$ by assigning $y\/$ with $v_{i}\/$ and $x\/$ with $v_{j}\/$.

Hence assume that $v_{n-2}\not\in\{x_{1},x_{2},y_{1}\}$.

Suppose $x_{2}=v_{n-3}$. Let $v\in\{x_{1},y_{1}\}\/$, say $v=y_{1}\/$ be such that $v\/$ is not labeled with $v_{3}\/$. Then in $G_{e}\/$, change the label for $v_{2}\/$ to $v_{n}\/$. In $G\/$, assign $v_{n-1},v_{2}\/$ to $x,y\/$ respectively.

Suppose $x_{2}\neq v_{n-3}$. Let $v\in\{x_{1},y_{1}\}\/$, say $v=y_{1}\/$ be such that $v\/$ is not labeled with $v_{n-3}\/$. Then in $G_{e}\/$, change the labels for $v_{n-2}\/$ to $v_{n-1}\/$. In $G\/$, assign $v_{n},v_{n-2}\/$ to $x,y\/$ respectively.

(1.2) $G_{e}$ has a $2\/$-cycle. See Figure 16.

Here we may assume that $x_{1}x_{2}x_{1}\/$ is the $2\/$-cycle in $G_{e}\/$. By induction, $G_{e}\/$ has a labeling $v_{1},v_{2},\ldots,v_{n-2}\/$ that satisfies condition (ii) of the lemma. There are two cases to consider.

(a) $\{x_{1},x_{2}\}\cap\{y_{1},y_{2}\}=\emptyset\/$.

We may assume without loss of generality that $x_{1}=v_{1}\/$ and $x_{2}=v_{a}\/$.

Let $v\in\{y_{1},y_{2}\}\/$, say $v=y_{1}\/$ be such that $v\/$ is not labeled with $v_{n-2}\/$. In $G_{e}\/$, change the label for $y_{1}\/$ from $v_{i}\/$ to $v_{n}\/$. Extend this labeling of $G_{e}\/$ to a labeling of $G\/$ by assigning the label $v_{n-1}\/$ and $v_{i}\/$ to $x\/$ and $y\/$ respectively.

(b) $\{x_{1},x_{2}\}\cap\{y_{1},y_{2}\}\neq\emptyset\/$.

(b1) $\{x_{1},x_{2}\}=\{y_{1},y_{2}\}\/$.

We may assume that $x_{1}=v_{1},x_{2}=v_{a}\/$ and $x_{3}=v_{b}\/$ where $x_{3}\/$ is the other neighbor of $x_{1}\/$ in $G_{e}\/$. To obtain a required labeling for $G\/$, we first change the label of a vertex in $G_{e}\/$ from $v_{c}\/$ to $v_{n}\/$ where $c\/$ is an odd integer and $c\not\in\{1,a,b\}\/$, and then assign $v_{n-1},v_{c}\/$ to $x,y\/$ respectively.

(b2) $x_{2}=y_{2}\/$ and $x_{1}\neq y_{1}\/$.

Then either (b2.1) $x_{2}=v_{1}\/$ and $x_{1}=v_{a}\/$ or (b2.2) $x_{1}=v_{1}\/$ and $x_{2}=v_{a}\/$.

(b2.1) In this case, $y_{1}=v_{b}\/$. In $G_{e}\/$, change the label of $y_{1}\/$ to $v_{n}\/$. Then extend this new labeling on $G_{e}\/$ to a required labeling of $G\/$ by assigning $v_{n-1},v_{b}\/$ to $x,y\/$ respectively.

(b2.2) In this case, we consider the label on $y_{1}\/$.

Suppose $y_{1}\neq v_{n-2}\/$. First change the label of a vertex in $G_{e}\/$ from $v_{c}\/$ to $v_{n}\/$ where $c\/$ is an odd integer and $c\not\in\{1,a,b\}\/$, and then assign $v_{n-1},v_{c}\/$ to $x,y\/$ respectively.

Suppose $y_{1}=v_{n-2}\/$. Then in $G_{e}\/$, if $a=3\/$, then change $v_{6}\/$ to $v_{n}\/$; if $a\neq 3\/$, then change $v_{2}\/$ to $v_{n}\/$. In any case, in $G\/$, label $x\/$ with $v_{n-1}\/$, and label $y\/$ with $v_{6}\/$ and $v_{2}\/$ respectively. This gives a required labeling for $G\/$.

**Case (2) ** $G$ has a $2\/$-cycle. See Figure 14.

Let $xyx\/$ be the $2\/$-cycle of $G\/$. Also, let $x_{1}\/$ (respectively $y_{1}\/$) be the other neighbor of $x\/$ (respectively $y\/$).

Here let $e=xy\/$. Then $G_{e}\/$ is a connected $3\/$-regular graph on $n-2\/$ vertices having at most one $2\/$-cycle.

(2.1) $G_{e}$ has no $2\/$-cycle. In this case, $x_{1}\/$ and $y_{1}\/$ are not adjacent in $G\/$.

If $G_{e}\/$ is the cube $Q_{3}\/$, then $G\/$ the $3\/$-regular graph depicted in Figure 10. which has a labeling that satisfies condition (ii) of the lemma.

Hence we assume that $G_{e}\/$ is not the cube $Q_{3}\/$.

By induction, $V(G_{e})$ can be labeled as $v_{1},v_{2},\ldots,v_{n-2}$ which satisfies condition (i) of Lemma 2. We shall use this labeling on $V(G_{e})\/$ to obtain a labeling for $G\/$ that satisfies the conditions of Lemma 2.

(a) Suppose $v_{1}\in\{x_{1},y_{1}\}\/$.

Assume without loss of generality that $x_{1}=v_{1}\/$ and $y_{1}=v_{a}\/$. In $G_{e}\/$, change the labels of $x_{1},y_{1}\/$ to $v_{n-1},v_{n}\/$ respectively. Extend this new labeling on $V(G_{e})\/$ to a required labeling of $G\/$ by labeling $x,y\/$ with $v_{1},v_{a}\/$ respectively.

(b) Suppose $v_{1}\not\in\{x_{1},y_{1}\}\/$.

(b1) $v_{n-2}\in\{x_{1},y_{1}\}\/$.

Suppose $x_{1}=v_{n-2}\/$.

If no vertex in $N_{G_{e}}(v_{n-2})-y_{1}\/$ is $v_{n-4}\/$, then in $G_{e}\/$, change the labels for $v_{1}\/$ to $v_{n}\/$, and interchange the labels of $v_{n-3}\/$ and $v_{n-2}\/$. Now, extend this new labeling on $G_{e}\/$ to a required labeling in $G\/$ by assigning $v_{1},v_{n-1}\/$ to $x,y\/$ respectively.

If some vertex in $N_{G_{e}}(v_{n-2})-y_{1}\/$ is $v_{n-4}\/$, then in $G_{e}\/$, change the labels for $x_{1},v_{1},v_{n-3}\/$ to $v_{n-1},v_{n},v_{n-2}\/$ respectively. Now, extend this new labeling on $G_{e}\/$ to a required labeling in $G\/$ by assigning $v_{1},v_{n-3}\/$ to $x,y\/$ respectively.

(b2) $v_{n-2}\not\in\{x_{1},y_{1}\}\/$.

Suppose the label of $x_{1}\/$ is $v_{i}\/$ where $i\/$ is odd. Then in $G_{e}\/$, change the label of $v_{1}\/$ to $v_{n}\/$. In $G\/$, assign $v_{1},v_{n-1}\/$ to $x,y\/$ respectively.

Hence assume that $x_{1}=v_{i},y_{1}=v_{j}\/$ are such that $i\/$ and $j\/$ are both even.

Suppose $2\in\{i,j\}$. Assume without loss of generality that $i=2$.

If $v_{4}\notin N_{G_{e}}(x_{1})-y_{1}\/$, then in $G_{e}$, interchange the labels of $v_{2}\/$ and $v_{3}\/$. Also, change the label of $v_{1}\/$ to $v_{n}\/$. Now, extend this new labeling on $G_{e}\/$ to a required labeling in $G\/$ by assigning $v_{1},v_{n-1}\/$ to $x,y\/$ respectively.

If $v_{4}\in N_{G_{e}}(x_{1})-y_{1}\/$, then in $G_{e}\/$, change the label of $x_{1}\/$ from $v_{2}\/$ to $v_{n-1}\/$. Also, change the label of $v_{1},v_{3}\/$ to $v_{n},v_{2}\/$ respectively. Now, extend this new labeling on $G_{e}\/$ to a required labeling in $G\/$ by assigning $v_{1},v_{3}\/$ to $x,y\/$ respectively.

Hence assume that $2\notin\{i,j\}$ and that $i<j$.

Let $u$ be the vertex in $G_{e}$ having the label $v_{i+1}$. Note that such a vertex exists because $i<n-2$.

To extend the labeling of $G_{e}\/$ to a required labeling for $G\/$, we first interchange the labels of $v_{i}\/$ and $v_{i+1}\/$ (in $G_{e}\/$). There are two cases to consider.

Now, if $u$ is not adjacent to a vertex with the label $v_{i-1}$ in $G_{e}$, then change the label for $v_{1}\/$ to $v_{n}\/$. In $G\/$, we assign $v_{1},v_{n-1}\/$ to $x,y\/$ respectively.

If $u$ is adjacent to a vertex with the label $v_{i-1}$ in $G_{e}$, then change the label for $v_{1},v_{i-1}\/$ to $v_{n-1},v_{n}\/$ respectively. In $G\/$, assign $v_{1},v_{i-1}\/$ to $x,y\/$ respectively.

(2.2) $G_{e}$ has a $2\/$-cycle.

In this case, $x_{1}x_{2}\in E(G)\/$. By induction, $G_{e}\/$ has a labeling $v_{1},v_{2},\ldots,v_{n-2}\/$ that satisfies condition (ii) of the lemma. To obtain a required labeling for $G\/$, we first change the label of the vertex $v\/$ (which is adjacent to $x_{1}\/$) in $G_{e}\/$ from $v_{b}\/$ to $v_{n}\/$, and then extend this new labeling to $G\/$ by assigning $v_{1},v_{n-1},v_{b}\/$ to $x,y,x_{1}\/$ respectively.

This completes the proof.

Theorem 7

Let $G\/$ be a $3\/$-regular graph with $n=4+2t\/$ vertices where $t\geq 2\/$. Then $G\/$ is potentially $4$-angulable.

Proof: Suppose $G\/$ is the cube $Q_{3}\/$ and has labeling as shown in Figure 10. Then $K_{n}-G\/$ admits a $4$-angulation that has the diagonal edges $v_{1}v_{6}\/$ and $v_{2}v_{5}\/$. Now suppose that $G\/$ is a cubic not $Q_{3}\/$. By Lemma 3, $V(G)\/$ admits a labeling as described in Lemma 2. Then $G\/$ is potentially $4$-angulable.