Overview of Shelling for 2-Manifold Surface Reconstruction Based on 3D Delaunay Triangulation

Abstract

Recently, methods have been proposed to reconstruct a 2-manifold surface from a sparse cloud of points estimated from an image sequence. Once a 3D Delaunay triangulation is computed from the points, the surface is searched by growing a set of tetrahedra whose boundary is maintained 2-manifold. Shelling is a step that adds one tetrahedron at once to the growing set. This paper surveys properties that helps to understand the shelling performances: shelling provides most tetrahedra enclosed by the final surface, but it can “get stuck” or block in unexpected cases.

Appendices

Appendix 1: Proofs for Prerequisites of Star-Shape Shelling

1.1 Proof of Lemma 6

First we show that there are points $\mathbf {x}'\in \mathring{\Delta }_x$ and $\mathbf {y}'\in \mathbf {cx}'{\setminus } |A|$. Since |A| is a closed set in $\mathbb {R}^3$, there is $\epsilon > 0$ such that $B(\mathbf {y},\epsilon )\cap |A|=\emptyset$. Let $\tau \in \mathbb {R}$ be such that $\mathbf {x}-\mathbf {c}= \tau (\mathbf {y}-\mathbf {c})$. We have $\tau > 1$ since $\mathbf {y}$ is between $\mathbf {x}$ and $\mathbf {c}$. If $\mathbf {x}\in |\partial \Delta _x|$, there is a point $\mathbf {x}'$ very close to $\mathbf {x}$ such that $\mathbf {x}'\in \mathring{\Delta }_x$ and $\mathbf {y}'= \mathbf {c}+(\mathbf {x}'-\mathbf {c})/\tau \in B(\mathbf {y},\epsilon )$. If $\mathbf {x}\notin |\partial \Delta _x|$, we set $\mathbf {x}'=\mathbf {x}$ and $\mathbf {y}'=\mathbf {y}$.

Second we show that there are tetrahedron $\Delta _y\in T{\setminus } A$ and $\mathbf {x}''\in \mathring{\Delta }_x$ such that $\mathbf {y}''= \mathbf {c}+(\mathbf {x}''-\mathbf {c})/\tau \in \mathring{\Delta }_y$. Since |T| is convex and includes $\{\mathbf {c},\mathbf {x'}\}$, $\mathbf {y'}\in |T|$. Since $\mathbf {y'}\notin |A|$, there is a tetrahedron $\Delta _y\in T{\setminus } A$ such that $\mathbf {y}'\in \Delta _y$. If $\mathbf {y}'\in |\partial \Delta _y|$, there is a point $\mathbf {y}''$ very close to $\mathbf {y}'$ such that $\mathbf {y}''\in \mathring{\Delta }_y$ and $\mathbf {x}''= \mathbf {c}+\tau (\mathbf {y}''-\mathbf {c})\in \mathring{\Delta }_x$.

Last we find $\tilde{\mathbf {x}}$ and $\tilde{\mathbf {y}}$. Since $\mathring{\Delta }_x$ and $\mathring{\Delta }_y$ are open, we define $B'_x= B(\mathbf {x}'',\tau \epsilon ')$ and $B'_y= B(\mathbf {y}'',\epsilon ')$ such that $B'_x\subset \mathring{\Delta }_x$ and $B'_y\subset \mathring{\Delta }_y$. Let V be the union for every edge $e\in c(T)$ of the plane(s) that include(s) both e and $\mathbf {c}$. Since we are in a non-degenerate case, e and $\mathbf {c}$ are not collinear. Thus V is a finite union of planes and we cannot have $B'_x\subseteq V$. There is $\tilde{\mathbf {x}}\in B'_x{\setminus } V\subseteq \mathring{\Delta }_x$. This implies that $\mathbf {c}\tilde{\mathbf {x}}$ does not intersect the edges in c(T) and $\tilde{\mathbf {y}}= \mathbf {c}+(\tilde{\mathbf {x}}-\mathbf {c})/\tau \in B'_y\subseteq \mathring{\Delta }_y$.

1.2 Proof of Lemma 7

First we show that the set of the triangles $t_i\in c(T)$ intersected by $(\mathbf {xy})$ are such that $\mathbf {x}_{\lambda _i}= (\mathbf {xy})\cap t_i$ and the $\lambda _i$ are distinct. If $\mathbf {x}_\lambda$ is in a triangle $t_\lambda \in c(T)$, $t_\lambda \cap (\mathbf {cx})$ is a line segment $\mathbf {x}_{\lambda '_0}\mathbf {x}_{\lambda '_1}$ (indeed the intersection of two closed convexes is a closed convex). Furthermore, $\lambda '_0=\lambda '_1$ (otherwise, $(\mathbf {xy})$ is in the plane of $t_\lambda$ and intersects an edge of $t_\lambda$) and we obtain $\mathbf {x}_\lambda = (\mathbf {xy})\cap t_\lambda$. Let $\lambda _i\in \mathbb {R}$ such that $\mathbf {x}_{\lambda _i}\in t_i$. Now we show that $\lambda _i=\lambda _j$ implies $i=j$. Since $(\mathbf {xy})$ does not intersect the edges of $t_i$, $\mathbf {x}_{\lambda _i}\in \mathring{t}_i$. Thus $t_i= t_j$ (Lemma 5), i.e. $i= j$.

Second we note that there are at least two triangles $t_i$, i.e. $k\ge 1$. Indeed, $\mathbf {x}\in \mathring{\Delta }_x$ and $\Delta _x\in T$ imply that $(\mathbf {xy})$ intersects at least two triangles in $\partial \Delta _x\subset c(T)$.

Third the $\lambda _i$ are ordered such that $\lambda _i< \lambda _{i+1}$ and we show that there is a tetrahedron $\Delta _i\in T$ such that $t_i\in \partial \Delta _i$, $t_{i+1}\in \partial \Delta _i$ and $\mathbf {x}_\lambda \in \Delta _i$ iff $\lambda \in [\lambda _i,\lambda _{i+1}]$. Let $\mathbf {m}_i= \mathbf {x}_{({\lambda _i}+{\lambda _{i+1}})/2}$. Since |T| is convex, there is a tetrahedron $\Delta _i\in T$ such that $\mathbf {m}_i\in \Delta _i$. Since $\mathbf {m}_i$ is not in a triangle of c(T), $\mathbf {m}_i\in \mathring{\Delta }_i$. Therefore $\Delta _i\cap (\mathbf {xy})$ is a line segment $\mathbf {x}_{\lambda '_0}\mathbf {x}_{\lambda '_1}$ such that $\lambda '_0\ne \lambda '_1$. This implies that $\mathbf {x}_\lambda \in \Delta _i$ iff $\lambda \in [\lambda '_0,\lambda '_1]$. Furthermore $\mathbf {x}_{\lambda '_\epsilon }\in |\partial \Delta _i|$ (indeed $\mathbf {x}_{\lambda '_\epsilon }\notin \mathring{\Delta }_i$ which is open). Thus there are distinct triangles $t_j$ and $t_l$ in $\partial \Delta _i$ such that $\mathbf {x}_{\lambda '_0}\in t_j$ and $\mathbf {x}_{\lambda '_1}\in t_l$ and $\lambda '_0= \lambda _j$ and $\lambda '_1= \lambda _l$. Since $\lambda _j< \lambda < \lambda _l$ implies $\mathbf {x}_\lambda \in \mathring{\Delta }_i$, which in turn implies $\mathbf {x}_\lambda \notin t_m$ (Lemma 5), we have $j+1=l$. Since $\mathbf {m}_i\in \mathbf {x}_{\lambda _j}\mathbf {x}_{\lambda _{j+1}}$, we have $j= i$. Thus $t_i\in \partial \Delta _i$, $t_{i+1}\in \partial \Delta _i$, $\lambda '_0= \lambda _i$ and $\lambda '_1= \lambda _{i+1}$.

Last we assume that a tetrahedron $\Delta \in T$ meets $(\mathbf {xy})\cap \Delta \ne \emptyset$ and show that $\Delta$ is a $\Delta _i$. The proof is similar as in the third step above. We have $(\mathbf {xy})\cap \Delta = \mathbf {x}_{\tilde{\lambda }_0}\mathbf {x}_{\tilde{\lambda }_1}$ such that $\tilde{\lambda }_0\le \tilde{\lambda }_1$ and $\mathbf {x}_{\tilde{\lambda }_\epsilon }$ is in a triangle $\tilde{t_\epsilon }\in \partial \Delta$. Since $(\mathbf {xy})$ does not intersect the $\Delta$ edges, $(\mathbf {xy})$ intersects $\mathring{\tilde{t_\epsilon }}$, which in turn implies that $\tilde{\lambda }_0\ne \tilde{\lambda }_1$. Now there are distinct triangles $t_j$ and $t_l$ such that $\tilde{t_0}= t_j$, $\tilde{t_1}= t_l$, $\tilde{\lambda }_0= \lambda _j$ and $\tilde{\lambda }_1= \lambda _l$. Since $\tilde{\lambda }_0< \lambda < \tilde{\lambda }_1$ implies $\mathbf {x}_\lambda \in \mathring{\Delta }$, which in turn implies $\mathbf {x}_\lambda \notin t_m$ (Lemma 5), we have $l=j+1$.

1.3 Proof of Theorem 7

Let $O\subseteq T$ be $\mathbf {c}$-star-shaped, a triangle $t\in \partial O$, a tetrahedron $\Delta \in O$ such that $t\in \partial \Delta$, and show that $\Delta$ and $\mathbf {c}$ are in the same t half-space. If $t\in \partial T$, both $\Delta$ and $\mathbf {c}$ are in the convex set T (indeed T is Delaunay) and thus there are in the same half-space of t. If $t\notin \partial T$, there is a tetrahedron $\Delta '\in T{\setminus } O$ such that $t=\Delta \cap \Delta '$. Let $\mathbf {b}$ be the barycentre of t. Since $\mathbf {c}$ is not in the t plane, $\mathring{\Delta }\cap \mathbf {bc}\ne \emptyset$ or $\mathring{\Delta '}\cap \mathbf {bc}\ne \emptyset$. Assume (reductio ad absurdum) that $\mathring{\Delta '}\cap \mathbf {bc}\ne \emptyset$. Since O is $\mathbf {c}$-star-shaped, $\mathbf {bc}\subseteq |O|$. Therefore there is $\Delta ''\in O$ such that $\mathring{\Delta '}\cap \Delta ''\ne \emptyset$. Lemma 5 implies that $\Delta '= \Delta ''\in O$, which contradicts $\Delta '\in T{\setminus } O$.

Assume that $O\subseteq T$ is not $\mathbf {c}$-star-shaped and show that there is a tetrahedron $\Delta \in O$ and a triangle $t\in \partial O\cap \partial \Delta$ such that $\Delta$ and $\mathbf {c}$ are not in the same t half-space. Since O is not $\mathbf {c}$-star-shaped, there is a point $\mathbf {x}'\in |O|{\setminus }\{\mathbf {c}\}$ such that $\mathbf {cx}'$ is not included in |O|. Thus there is a point $\mathbf {y}'\in \mathbf {cx}'{\setminus } |O|$. Now we use Lemma 6 with $A= O$. Let $\Delta _x\in O$ be a tetrahedron such that $\mathbf {x}'\in \Delta _x$. Thus there are tetrahedron $\Delta _y\in T{\setminus } O$ and points $\mathbf {x}\in \mathring{\Delta _x}$ and $\mathbf {y}\in \mathring{\Delta _y}\cap \mathbf {cx}$ such that $(\mathbf {cx})$ does not intersect the edges in c(T). Then we obtain a tetrahedron series $\Delta _i$ covering $\mathbf {xy}$ as described by Lemma 7. There are j and m such that $\Delta _j= \Delta _x\in O$ and $\Delta _m= \Delta _y\in T{\setminus } O$. Since $\mathbf {x}_0= \mathbf {x}\in \Delta _j$ and $\mathbf {x}_1= \mathbf {y}\in \Delta _m$, $j\le m$ and $\lambda _m\le 1$. Therefore there is l such that $j\le l< m$, $\Delta _l\in O$, $\Delta _{l+1}\in T{\setminus } O$ and $\lambda _{l+1}\le \lambda _m$. Let $\Delta = \Delta _l$ and the triangle $t= \Delta _l\cap \Delta _{l+1}$. Note that $\mathbf {x}_\lambda \in \Delta$ iff $\lambda \in [\lambda _l,\lambda _{l+1}]$, $\mathbf {x}_\lambda \in t$ iff $\lambda =\lambda _{l+1}$, $\lambda _{l+1}\le 1$, $\mathbf {c}= \mathbf {x}_\mu$ with $\mu > 1$ ($\mathbf {x}_1$ is between $\mathbf {x}_0$ and $\mathbf {c}$). Thus $\Delta$ and $\mathbf {c}$ are in different half-spaces of t.

Appendix 2: Proofs for Star-Shape Shelling

1.1 Proof of Lemma 12

We change the coordinate basis such that $\mathbf {v}_1^T= \begin{pmatrix} 0&0&0\end{pmatrix}$, $\mathbf {v}_2^T= \begin{pmatrix} 1&0&0\end{pmatrix}$, $\mathbf {v}_3^T= \begin{pmatrix} 0&1&0\end{pmatrix}$, $\mathbf {v}_4^T= \begin{pmatrix} 0&0&1\end{pmatrix}$. Thus

$$\begin{aligned} H'_1= & {} \{ (x_2,x_3,x_4)^T\in \mathbb {R}^3, x_2+x_3+x_4\ge 1\} \nonumber \\ H'_2= & {} \{ (x_2,x_3,x_4)^T\in \mathbb {R}^3, x_2\le 0\} \nonumber \\ H'_3= & {} \{ (x_2,x_3,x_4)^T\in \mathbb {R}^3, x_3\le 0\} \nonumber \\ H'_4= & {} \{ (x_2,x_3,x_4)^T\in \mathbb {R}^3, x_4\le 0\}. \end{aligned}$$

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Now we see that $(x_2,x_3,x_4)\in H'_1\cap H'_2\cap H'_3\cap H'_4$ is impossible.

Assume that $\mathbf {c}\in H'_1\cap H_2\cap H_3\cap H_4$. We have $\mathbf {v}_1\in H_2\cap H_3\cap H_4$ (indeed $\Delta = H_1\cap H_2\cap H_3\cap H_4$). Since $H_2\cap H_3\cap H_4$ is convex and includes both $\mathbf {c}$ and $\mathbf {v}_1$, $\mathbf {cv}_1\subset H_2\cap H_3\cap H_4$. Let plane $\pi _1= H_1\cap H'_1$. Thus $\mathbf {cv}_1\cap \pi _1\subseteq H_1\cap H_2\cap H_3\cap H_4\cap H'_1= \Delta \cap \pi _1= t_1$. We obtain $\mathbf {cv}_1\cap \pi _1\subseteq \mathbf {cv}_1\cap t_1$. Since $\mathbf {c}$ and $\mathbf {v}_1$ are in different $t_1$ half-spaces, $\mathbf {cv}_1\cap \pi _1\ne \emptyset$ and we obtain $\mathbf {cv}_1\cap t_1\ne \emptyset$.

Assume that $\mathbf {c}\in H'_1\cap H'_2\cap H_3\cap H_4$. The tetrahedron $\Delta = \mathbf {v}_1\mathbf {v}_2\mathbf {v}_3\mathbf {v}_4$ is split in two tetrahedra $\Delta _1= \mathbf {v}_1\tilde{\mathbf {v}}_2\mathbf {v}_3\mathbf {v}_4$ and $\Delta _2= \tilde{\mathbf {v}}_1\mathbf {v}_2\mathbf {v}_3\mathbf {v}_4$ using $\tilde{\mathbf {v}}_2= \tilde{\mathbf {v}}_1= \mathbf {v}= (\mathbf {v}_1+\mathbf {v}_2)/2$. Let $\tilde{H_1}$ and $\tilde{H_2}$ be the half-spaces of the triangle $\mathbf {v}\mathbf {v}_3\mathbf {v}_4$ such that $\mathbf {v}_1\in \tilde{H_1}$ and $\mathbf {v}_2\in \tilde{H_2}$. Note that $\Delta _2$ is like $\Delta$ except that the vertex $\mathbf {v}_1$ is replaced by $\tilde{\mathbf {v}}_1$ and the half-space $H_2$ is replaced by $\tilde{H_2}$: $\Delta _2= H_1\cap \tilde{H_2}\cap H_3\cap H_4$ ($t_1$ does not change, the half-spaces $H_3$ and $H_4$ do not change since their boundary planes include $\tilde{\mathbf {v}}_1$, $\mathbf {v}_1$ and $\mathbf {v}_2$ that are collinear). Assume $\mathbf {c}\in \tilde{H_2}$ (the other case $\mathbf {c}\in \tilde{H_1}$ is similar). We have $\mathbf {c}\in H'_1\cap \tilde{H_2}\cap H_3\cap H_4$ (indeed, $\mathbf {c}\in H'_1\cap H_3\cap H_4$ and $\mathbf {c}\in \tilde{H_2}$). According to the second assertion of the lemma applied to $\Delta _2$, $\mathbf {c}\tilde{\mathbf {v}}_1\cap t_1\ne \emptyset$. Thus $\mathbf {cv}\cap (t_1\cup t_2)\ne \emptyset$.

Appendix 3: Non-Shellable Star Shapes in Degenerate Cases

Here we show that there are star shapes O that are non-shellable in degenerate cases. More precisely, we find $\mathbf {h}$-star shapes O such that $\partial O$ has a singular vertex and $\mathbf {h}$ is in a plane of a triangle in $\partial O$. Since $\partial O$ is a 2-manifold if O is shellable (Theorem 4), we see that such an O is non-shellable.

For example, $O= \{ \mathbf {habc}, \mathbf {hdef} \}$ is a $\mathbf {h}$-star shape such that $\partial O$ has a singular vertex $\mathbf {h}$. Figure 18 shows another example O that is star-shaped with respect to a point $\mathbf {h}\notin |\partial O|$:

$$\begin{aligned} O= S_1\cup \{ \mathbf {bcef}, \mathbf {bcgd}\}\hbox { where } S_1= \mathbf {ab}\times \mathbf {d}\hbox {-}\mathbf {e}\hbox {-}\mathbf {f}\hbox {-} \mathbf {g}\hbox {-}\mathbf {d}.\nonumber \\ \end{aligned}$$

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Note that $\mathbf {h}$ is in the plane of the triangle $\mathbf {bcg}\in \partial O$ and $\mathbf {b}$ is a singular vertex of $\partial O$. Now we show that O can be star-shaped.

Fig. 18

A non-shellable star shape O with respect to $\mathbf {h}$ such that $\mathbf {h}\notin |\partial O|$ (right): O is the union of $\mathbf {ab}\times \mathbf {d}\hbox {-}\mathbf {e}\hbox {-} \mathbf {f}\hbox {-}\mathbf {g}\hbox {-}\mathbf {d}$ (left) and $\{ \mathbf {bcef}, \mathbf {bcgd} \}$ (middle)

Property 2

Let $S_2= \mathbf {bc}\times \mathbf {d}\hbox {-}\mathbf {e}\hbox {-} \mathbf {f}\hbox {-}\mathbf {g}\hbox {-}\mathbf {d}$ and $\mathbf {h}\in |S_1|\cap (\mathbf {bc})$. If both $S_1$ and $S_1\cup S_2$ are convex, O is $\mathbf {h}$-star-shaped.

Proof

Let $\mathbf {x}\in |O|$ and show that $\mathbf {xh}\subseteq |O|$. We have $\mathbf {x}\in |S_1|$ or $\mathbf {x}\in \mathbf {bcef}$ (or similarly $\mathbf {x}\in \mathbf {bcgd}$). Since $S_1$ is convex and $\mathbf {h}\in |S_1|$, $\mathbf {x}\in |S_1|$ implies $\mathbf {xh}\subseteq |S_1|\subseteq |O|$.

Now we assume $\mathbf {x}\in \mathbf {bcef}$ and show $\mathbf {xh}\cap \mathbf {bcde}= \mathbf {xh}\cap \mathbf {bce}$. Since $\mathbf {h}\in (\mathbf {bc})$ and $\mathbf {x}\in \mathbf {bcef}$, $\mathbf {hx}$ and $\mathbf {bcef}$ are in the same half-space of $\mathbf {bce}$. Since $\mathbf {bcde}$ is in the other half-space of $\mathbf {bce}$, $\mathbf {hx}\cap \mathbf {bcde}$ is included in the plane of $\mathbf {bce}$. We obtain $\mathbf {xh}\cap \mathbf {bcde}= \mathbf {xh}\cap \mathbf {bce}$ and similarly $\mathbf {xh}\cap \mathbf {bcfg}= \mathbf {xh}\cap \mathbf {bcf}$.

Since $S_1\cup S_2$ is convex and $\mathbf {h}\in |S_1\cup S_2|$, $\mathbf {xh}\subseteq |S_1\cup S_2|= |O\cup \{\mathbf {bcde},\mathbf {bcfg} \}|$. Thus

$$\begin{aligned} \mathbf {xh}= \mathbf {xh}\cap |O\cup \{ \mathbf {bcde}, \mathbf {bcfg} \}|= \mathbf {xh}\cap (|O|\cup \mathbf {bce}\cup \mathbf {bcf}).\nonumber \\ \end{aligned}$$

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Since $\mathbf {bce}\subseteq |O|$ and $\mathbf {bcf}\subseteq |O|$, we obtain $\mathbf {xh}\subseteq |O|$. $\square$

Appendix 4: Proof of Theorem 10

Lemma 18

Assume that $A\subseteq T^\infty$ and $B\subseteq T{\setminus } A$. Then $c(A)\cap c(B)$ is a simplicial complex in $\mathbb {R}^3$. Furthermore, $c(A)\cap c(B)= c(\partial A)\cap c(\partial B)$.

Proof

This is Lemma 1 in [14] by replacing the set B by a single tetrahedron. The proof is very similar as the proof in Appendix A.1 of [14]. $\square$

We show that every vertex $\mathbf {v}\in c(\partial (A\cup B))$ meets $(\partial (A\cup B))_\mathbf {v}= \mathbf {v}\times \mathbf {x}_*$ where $\mathbf {x}_*$ is a cycle. Thanks to Lemma 15, $\partial (A\cup B)= (\partial A{\setminus }\partial B)\cup (\partial B{\setminus }\partial A)$. Thus $\mathbf {v}\in c(\partial A)$ or $\mathbf {v}\in c(\partial B)$. We also have

$$\begin{aligned} (\partial (A\cup B))_\mathbf {v}= ((\partial A)_\mathbf {v}{\setminus } (\partial B)_\mathbf {v})\cup ((\partial B)_\mathbf {v}{\setminus } (\partial A)_\mathbf {v}). \end{aligned}$$

(32)

There are three cases. If $\mathbf {v}\in c(\partial A){\setminus } c(\partial B)$, $(\partial B)_\mathbf {v}=\emptyset$. Thanks to Eq. 32, $(\partial (A\cup B))_\mathbf {v}= (\partial A)_\mathbf {v}$. Since $\partial A$ is a 2-manifold, there is a cycle $\mathbf {x}_*$ such that $(\partial (A\cup B))_\mathbf {v}= (\partial A)_\mathbf {v}= \mathbf {v}\times \mathbf {x}_*$. The proof is the same if $\mathbf {v}\in c(\partial B){\setminus } c(\partial A)$.

Now we consider the last case $\mathbf {v}\in c(\partial A)\cap c(\partial B)$. Since $\partial A$ is a 2-manifold,

$$\begin{aligned} \exists \mathbf {a}_i\in c(\partial A), (\partial A)_\mathbf {v}= \mathbf {v}\times \mathbf {a}_*, \mathbf {a}_*= \mathbf {a}_1\hbox {-}\cdots \mathbf {a}_m\hbox {-}\mathbf {a}_1, \end{aligned}$$

(33)

and $\mathbf {a}_*$ is a cycle. Since $\partial B$ is a 2-manifold,

$$\begin{aligned} \exists \mathbf {b}_i\in c(\partial B), (\partial B)_\mathbf {v}= \mathbf {v}\times \mathbf {b}_*, \mathbf {b}_*= \mathbf {b}_1\hbox {-}\cdots \mathbf {b}_n\hbox {-}\mathbf {b}_1, \end{aligned}$$

(34)

and $\mathbf {b}_*$ is a cycle. Since $c(\partial A)\cap c(\partial B)= c(A)\cap c(B)$ (Lemma 18) and $c(A)\cap c(B)$ is a 2-manifold with boundary, there are $\mathbf {c}_i\in c(\partial A)\cap c(\partial B)$ such that

$$\begin{aligned} (\partial A\cap \partial B)_\mathbf {v}= \mathbf {v}\times \mathbf {c}_*, \mathbf {c}_*=\mathbf {c}_1\hbox {-}\cdots \mathbf {c}_k, \end{aligned}$$

(35)

and there are two cases: $\mathbf {c}_*$ is a cycle and $k\ge 3$, or all $\mathbf {c}_i$ are distinct and $k\ge 2$. Since $(\partial A\cap \partial B)_\mathbf {v}\subseteq (\partial A)_\mathbf {v}$ and $(\partial A\cap \partial B)_\mathbf {v}\subseteq (\partial B)_\mathbf {v}$, we have $\mathbf {c}_*\subseteq \mathbf {a}_*$ and $\mathbf {c}_*\subseteq \mathbf {b}_*$. Assume (reductio ad absurdum) that $\mathbf {c}_*$ is a cycle. Since $\mathbf {c}_*$ is included in $\mathbf {a}_*$ and $\mathbf {b}_*$ that are also cycles, $\mathbf {a}_*= \mathbf {c}_*= \mathbf {b}_*$. This implies $(\partial A)_\mathbf {v}= (\partial B)_\mathbf {v}$. Thanks to Eq. 32, $(\partial (A\cup B))_\mathbf {v}=\emptyset$. This contradicts $\mathbf {v}\in c(\partial (A\cup B))$.

Now the $\mathbf {c}_i$ are distinct and $k\ge 2$. Since the $\mathbf {c}_i$ are distinct and $\mathbf {c}_*\subseteq \mathbf {a}_*$, $k\le m$. We shift the $\mathbf {a}_i$ such that $\mathbf {a}_i= \mathbf {c}_i$ if $1\le i\le k$. Similarly $k\le n$ and $\mathbf {b}_i= \mathbf {c}_i$ if $1\le i\le k$. Thanks to Eq. 32, $(\partial (A\cup B))_\mathbf {v}= \mathbf {v}\times \mathbf {x}_*$ where

$$\begin{aligned} \mathbf {x}_*= \mathbf {a}_k\hbox {-}\cdots \mathbf {a}_m\hbox {-}\mathbf {a}_1 \cup \mathbf {b}_k\hbox {-}\cdots \mathbf {b}_n\hbox {-}\mathbf {b}_1. \end{aligned}$$

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Since $\mathbf {a}_1= \mathbf {b}_1= \mathbf {c}_1$ and $\mathbf {a}_k= \mathbf {b}_k= \mathbf {c}_k$,

$$\begin{aligned} \mathbf {x}_*= \mathbf {c}_k\hbox {-}\mathbf {a}_{k+1}\hbox {-}\mathbf {a}_{k+2} \hbox {-}\cdots \mathbf {a}_m\hbox {-}\mathbf {c}_1\hbox {-}\mathbf {b}_n\hbox {-} \mathbf {b}_{n-1}\hbox {-}\cdots \mathbf {b}_{k+1}\hbox {-}\mathbf {c}_k. \end{aligned}$$

(37)

Then we show that $c(\mathbf {x}_*)$ has at least 3 distinct vertices. It has $\mathbf {c}_k$ and $\mathbf {c}_1$. Furthermore, $\mathbf {c}_1\ne \mathbf {c}_k$ since $k\ge 2$. Assume (reduction ad absurdum) that it has no other vertex. Thus $m=k=n$. This implies $(\partial A)_\mathbf {v}= (\partial B)_\mathbf {v}$, which is impossible (see above).

Last we show that the vertices in the vertex sequence $\mathbf {x}_*$ are distinct. We have $\mathbf {c}_1\ne \mathbf {c}_k$ since $k\ge 2$. We also have $\mathbf {c}_1\ne \mathbf {a}_i$ if $i>k$, since $\mathbf {c}_1= \mathbf {a}_1$ and the $\mathbf {a}_i$ are distinct (indeed, $\mathbf {a}_*$ is a cycle). Similarly, $\mathbf {c}_1\ne \mathbf {b}_i$ if $i>k$, $\mathbf {c}_k\ne \mathbf {a}_i$ and $\mathbf {c}_k\ne \mathbf {b}_i$ if $i> k$. Assume (reductio ad absurdum) that $\mathbf {a}_i=\mathbf {b}_j$ if $i> k$ and $j> k$. Thus $\mathbf {va}_i\in c(\partial A)\cap c(\partial B)$. Since $c(\partial A)\cap c(\partial B)$ is 2D pure (indeed, $c(\partial A)\cap c(\partial B)$ is a 2-manifold with boundary), there is a triangle $t\in (\partial A\cap \partial B)_\mathbf {v}$ such that $\mathbf {a}_i\in t$. This implies that $\mathbf {a}_i=\mathbf {a}_j$ with $i> k$ and $j\le k$ (impossible).

Appendix 5: Proof of Theorem 12

First Theorem 17 details the splitting of a 2-sphere in c(T) by a cycle. Then we show Theorem 12 (splitting of $\partial S$ by the annulus N) by applying Theorem 17 to every cycle in $\partial N$.

Theorem 17

Let $L\subseteq T$ be a 3-ball and $\mathbf {b}_*\in c(\partial L)$ be a cycle. There are $B_i$ such that $B_1\cup B_2=\partial L$, $\partial B_1= \partial B_2= \mathbf {b}_*$, $c(B_1)\cap c(B_2)= c(\mathbf {b}_*)$, every $B_i$ is strongly connected.

Intuitively, the closed curve $\mathbf {b}_*$ splits the 2-sphere $\partial L$ in two disjoint and connected sets (2-balls) of triangles $B_1$ and $B_2$ that have the common boundary $\mathbf {b}_*$. This is a discrete version of the Jordan Theorem for a 2-sphere encoded by a simplicial complex.

Proof

Since the 3-ball $L\subseteq T$, $\partial L$ is a 2-sphere (Theorem 5). Let $\mathbf {m}\in |\partial L|$ that is neither a vertex nor in an edge of $c(\partial L)$. Let $\varphi$ be a homeomorphism such that $\varphi (|\partial L|{\setminus }\{\mathbf {m}\} )= \mathbb {R}^2$.

Let $G=(V,E)$ be the graph of the vertices and edges in $c(\partial L)$. Let $G^*= (V^*,E^*)$ be the adjacency graph of the triangles in $\partial L$, i.e. $V^*=\{ v(t), t\in \partial L \}$ and $E^*= \{ e(t_1,t_2), t_i\in \partial L, t_1\cap t_2\hbox { is an edge } \}$. Both G and $G^*$ have drawings (mappings) in $\mathbb {R}^2$ by $\varphi$. The drawing of G is $(\varphi (V), \varphi (E))$. The drawing of $v(t)\in V^*$ is the drawing of the barycentre of $t\in \partial L$. The drawing of $e(t_1,t_2)\in E^*$ is the drawing of a polygonal curve linking the barycentres of adjacent triangles $t_1$ and $t_2$ (the polygonal curve is included in $t_1\cup t_2$ and intersects the edge $t_1\cap t_2$). Let $F= \mathbf {b}_*$ and $F^*$ be the dual edges of F, i.e. $F^*= \{ e(t_1,t_2)\in E^*, t_1\cap t_2\in F\}$. Thanks to Proposition 4.6.1 in [6] (the plane duality theorem) and since G is connected and the drawings of G and $G^*$ are dual and F is a cycle, $F^*$ is a minimal cut of $G^*$.

Word “cut” means that the graph $(V^*,E^*{\setminus } F^*)$ is disconnected. Word “minimal” means that $(V^*,E^*{\setminus } F^*)$ becomes connected if we remove any edge from $F^*$. If we don’t remove an edge from $F^*$, $(V^*,E^*{\setminus } F^*)$ has exactly two connected components $\{ v(t), t\in B_1\}$ and $\{ v(t), t\in B_2\}$: $B_1\cup B_2= \partial L$, $B_1\cap B_2= \emptyset$, and every $B_i$ is non-empty strongly connected using the edges in $E^*{\setminus } F^*$.

Let edge $e\in \partial B_i$ and show $e\in \mathbf {b}_*$. Since $\partial L$ is a 2-manifold and the $B_i$ partition $\partial L$, there are triangles $t_i\in B_i$ such that $e= t_1\cap t_2$. If $e(t_1,t_2)\in E^*{\setminus } F^*$, $v(t_1)$ and $v(t_2)$ are in the same connected component of $(V^*,E^*{\setminus } F^*)$ (impossible). Thus $e(t_1,t_2)\in F^*$ and $e\in F= \mathbf {b}_*$.

Let edge $e\in \mathbf {b}_*$ and show $e\in \partial B_i$. We have distinct triangles $t_i\in \partial L$ such that $e= t_1\cap t_2$. Since $F=\mathbf {b}_*$, $e(t_1,t_2)\in F^*$. If $t_1\in B_1$ and $t_2\in B_1$ (reductio ad absurdum), we remove the edge $e(t_1,t_2)$ from $F^*$ and add it to $E^*{\setminus } F^*$ without connecting $B_1$ and $B_2$, i.e. F is not a minimal cut (impossible). Thus $t_1\in B_1$ and $t_2\in B_2$; we obtain $e\in \partial B_i$.

Since $\partial B_i= \mathbf {b}_*$, $c(\mathbf {b}_*)\subseteq c(B_1)\cap c(B_2)$. Now we show $c(B_1)\cap c(B_2)\subseteq c(\mathbf {b}_*)$. Since $B_1\cap B_2=\emptyset$, there are not triangles in $c(B_1)\cap c(B_2)$. Let edge $e\in c(B_1)\cap c(B_2)$; e is included in a triangle in $B_1$ and another in $B_2$ and thus $e\in \mathbf {b}_*$. Let vertex $\mathbf {v}\in c(B_1)\cap c(B_2)$. Since $\partial L$ is a 2-manifold, there is a triangle series $t_i\in (\partial L)_\mathbf {v}$ such that every $t_i\cap t_{i+1}$ is an edge. There is a $t_j\in B_1$ and $t_k\in B_2$. Therefore there is i such that $t_i\in B_1$ and $t_{i+1}\in B_2$. We see that the edge $e=t_i\cap t_{i+1}$ meets $e\in c(B_1)\cap c(B_2)$. Since $e\in \mathbf {b}_*$ (see above), $\mathbf {v}\in c(\mathbf {b}_*)$. $\square$

Now we do the proof of Theorem 12.

First we introduce notations by applying Theorem 17 to cycles $\mathbf {b}_*$ and $\mathbf {c}_*$. There are $B_i$ such that $B_1\cup B_2=\partial L$, $\partial B_1= \partial B_2= \mathbf {b}_*$, $c(B_1)\cap c(B_2)= c(\mathbf {b}_*)$ and every $B_i$ is strongly connected. There are $C_i$ such that $C_1\cup C_2=\partial L$, $\partial C_1= \partial C_2= \mathbf {c}_*$, $c(C_1)\cap c(C_2)= c(\mathbf {c}_*)$ and every $C_i$ is strongly connected.

Second we show that $N\subseteq B_2$ (similarly, $N\subseteq C_2$). Assume (reductio ad absurdum) that there are triangles $t_1\in N\cap B_1$ and $t_2\in N\cap B_2$. Since N is strongly connected and $N\subseteq B_1\cup B_2$, we can choose $t_1$ and $t_2$ such that $t_1\cap t_2$ is an edge. Therefore $t_1\cap t_2\in \partial B_i= \mathbf {b}_*$. However, an edge between two distinct triangles in N is $\mathbf {b}_i\mathbf {c}_i$ or $\mathbf {b}_{i+1}\mathbf {c}_i$ or $\mathbf {c}_{i+1}\mathbf {b}_i$ (contradiction). We obtain $N\subseteq B_2$.

Third we show that $c(N)\cap c(B_1)= c(\mathbf {b}_*)$ (similarly, $c(N)\cap c(C_1)= c(\mathbf {c}_*)$). Since $c(\mathbf {b}_*)\subseteq c(B_1)$ and $c(\mathbf {b}_*)\subseteq c(N)\subseteq c(B_2)$ and $c(B_1)\cap c(B_2)= c(\mathbf {b}_*)$, we have

$$\begin{aligned} c(\mathbf {b}_*)\subseteq c(N)\cap c(B_1)\subseteq c(B_1)\cap c(B_2)\subseteq c(\mathbf {b}_*). \end{aligned}$$

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Thus $c(N)\cap c(B_1)= c(\mathbf {b}_*)$.

Fourth we show that $B_1\cap C_1=\emptyset$. Since $\partial B_1= \mathbf {b}_*$, there is a triangle $\tilde{t}= \mathbf {d}\mathbf {b}_0\mathbf {b}_1\in B_1$. Assume (reductio ad absurdum) that $\tilde{t}\in C_1$. Thus $\mathbf {b}_0\in c(C_1)$. Since $\mathbf {b}_0\in c(N)$, we have $\mathbf {b}_0\in c(C_1)\cap c(N)= c(\mathbf {c}_*)$, i.e. $\mathbf {b}_0$ is equal to a $\mathbf {c}_j$ (impossible). We see that every triangle $\tilde{t}= \mathbf {d}\mathbf {b}_0\mathbf {b}_1\in B_1{\setminus } C_1$. Assume (reductio ad absurdum) that there is a triangle $t\in C_1\cap B_1$. Since $B_1$ is strongly connected and $\tilde{t}\in B_1{\setminus } C_1$ and $t\in B_1\cap C_1$, we can find triangles $t'\in B_1{\setminus } C_1$ and $t''\in B_1\cap C_1$ such that $t'\cap t''$ is an edge. Therefore $t'\cap t''\in \partial C_1= \mathbf {c}_*\subseteq \partial N$. This implies that $t'\in N$ or $t''\in N$, which is impossible since $B_1\cap N\subseteq B_1\cap B_2= \emptyset$.

Fifth we show that $N\cup C_1= B_2$. Since $B_1\cap C_1=\emptyset$ and $C_1\subseteq B_1\cup B_2$, $C_1\subseteq B_2$. Thus $N\cup C_1\subseteq B_2$ and we only need to show that $t'''\in B_2$ implies $t'''\in N\cup C_1$. Since $B_2$ is strongly connected and $\emptyset \ne N\cup C_1\subseteq B_2$, there is a triangle series $t_i\in B_2$ such that $t_0\in N\cup C_1$ and $t_n=t'''$ and every $t_i\cap t_{i+1}$ is an edge that is not in $\partial B_2$. Since $N\cap C_1=\emptyset$,

$$\begin{aligned} \partial (N\cup C_1)= & {} (\partial N{\setminus }\partial C_1)\cup (\partial C_1{\setminus }\partial N) \nonumber \\= & {} ((\mathbf {b}_*\cup \mathbf {c}_*){\setminus }\mathbf {c}_*) \cup (\mathbf {c}_*{\setminus } (\mathbf {b}_*\cup \mathbf {c}_*)) \nonumber \\= & {} \mathbf {b}_*=\partial B_2. \end{aligned}$$

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Therefore every $t_i\cap t_{i+1}$ is not in $\partial (N\cup C_1)$. Now we have $t_{i+1}\in N\cup C_1$ if $t_i\in N\cup C_1$. Since $t_0\in N\cup C_1$, we obtain $t'''= t_n\in N\cup C_1$.

Last we show $c(B_1)\cap c(C_1)=\emptyset$ and conclude using $B=B_1$ and $C= C_1$. Since $C_1\subseteq B_2$ (see above), $c(B_1)\cap c(C_1)\subseteq c(B_1)\cap c(B_2)= c(\mathbf {b}_*)$. Furthermore $B_1\subseteq C_2$ (indeed, $B_1\cap C_1=\emptyset$ and $B_1\subseteq C_1\cup C_2$ imply $B_1\subseteq C_2$). Thus $c(B_1)\cap c(C_1)\subseteq c(C_2)\cap c(C_1)= c(\mathbf {c}_*)$. We obtain $c(B_1)\cap c(C_1)\subseteq c(\mathbf {b}_*)\cap c(\mathbf {c}_*)=\emptyset$.

Appendix 6: Proof of Theorem 14

We show that $\partial (A\cup B)$ is a connected 2-manifold that has the same genus as $\partial A$ (therefore they are homeomorphic). The principle of the proof is the following. Theorem 10 implies that $\partial (A\cup B)$ is a 2-manifold. The set $D= \partial A\cap \partial B$ is a 2-ball. Then we show that $\partial (A\cup B)$ is connected by studying the connectivity of $\partial A{\setminus } D$ and $\partial B{\setminus } D$ and since their closures include the 1-sphere $\partial D$. Last we use properties of Euler’s characteristic $\chi (X)$ of $X\subseteq c(T)$ to show that both $\partial (A\cup B)$ and $\partial A$ have the same genus. We remind that $\chi (X)$ is the number of the vertices plus number of the triangles minus the number of the edges in X, and thus, it is closely related to the genus of X (Sect. 2.6).

We start by the two following lemmas, which are useful to show that $\partial A{\setminus } D$ and $\partial B{\setminus } D$ are connected.

Lemma 19

Let $M\subseteq c(T)$ be a 2-manifold. Let $D\subset M$. Then $\partial (M{\setminus } D)=\partial D$ and $c(M{\setminus } D)\cap c(D)= c(\partial D)$.

Proof

First we show $\partial D= \partial (M{\setminus } D)$. Let $e\in c(M)$ be an edge. Since M is a 2-manifold, there are exactly two triangles $t_1$ and $t_2$ in M that includes e and $e= t_1\cap t_2$. Since $D\subseteq M$, $e\in \partial D$ iff $t_1\in D$ and $t_2\in M{\setminus } D$ iff $e\in \partial (M{\setminus } D)$.

If a simplex $\sigma \in c(\partial D)$, $\sigma \in c(D)$ and $\sigma \in c(\partial (M{\setminus } D))\subseteq c(M{\setminus } D)$. Therefore $\sigma \in c(M{\setminus } D)\cap c(D)$. Conversely, let $\sigma \in c(M{\setminus } D)\cap c(D)$ and show $\sigma \in c(\partial D)$. This means that there are triangles $t_1\in D$ and $t_2\in M{\setminus } D$ such that $\sigma \subseteq t_1\cap t_2$. Assume that $\sigma$ is an edge (case 1). Thus $\sigma = t_1\cap t_2$. Since $D\subseteq M$ and M is a 2-manifold, the only triangles in M that includes $\sigma$ are $t_1$ and $t_2$. Therefore there is only one triangle in D that includes $\sigma$ and we obtain $\sigma \in c(\partial D)$. Assume that $\sigma$ is a vertex (case 2). Since M is a 2-manifold, $M_\mathbf {v}$ is a triangle series $t'_i\in M$ such that every $t'_i\cap t'_{i+1}$ is an edge including $\mathbf {v}$. There is j and k such that $t'_j=t_1\in D$ and $t'_k=t_2\in M{\setminus } D$. Thus we can find i such that $t'_i\in D$ and $t'_{i+1}\in M{\setminus } D$ (or vice versa). Now vertex $\sigma$ is a face of the edge $e= t'_i\cap t'_{i+1}$ such that $e\in c(M{\setminus } D)\cap c(D)$ and we just showed (case 1) that $e\in c(\partial D)$. Therefore $\sigma \in c(\partial D)$. $\square$

Lemma 20

Assume that $M\subseteq c(T^\infty )$ is a connected 2-manifold. If a 2-ball $D\subseteq M$, $M{\setminus } D$ is connected.

Proof

Let vertices $\mathbf {v}$ and $\mathbf {v}'$ in $c(M{\setminus } D)$. Since they are in c(M) and M is connected, there is a path $\mathbf {v}_0\hbox {-}\mathbf {v}_1\hbox {-}\cdots \mathbf {v}_n\subseteq c(M)$ such that $\mathbf {v}_0=\mathbf {v}$ and $\mathbf {v}_n=\mathbf {v}'$. If this path has an edge which is not in $c(M{\setminus } D)$, we modify it to obtain another path such that all its edges are in $c(M{\setminus } D)$ (thus $M{\setminus } D$ is connected).

Let j be the smallest index i such that $\mathbf {v}_i\mathbf {v}_{i+1}\notin c(M{\setminus } D)$. We show that $\mathbf {v}_j\in c(\partial D)$. Since $\mathbf {v}_j\mathbf {v}_{j+1}\in c(M)= c(M{\setminus } D)\cup c(D)$, $\mathbf {v}_j\mathbf {v}_{j+1}\in c(D)$. If $j=0$, we have $\mathbf {v}_j\in c(M{\setminus } D)\cap c(D)= c(\partial D)$ (Lemma 19). If $j> 0$, $\mathbf {v}_{j-1}\mathbf {v}_j\in c(M{\setminus } D)$ and we have $\mathbf {v}_j\in c(M{\setminus } D)\cap c(D)= c(\partial D)$.

Let k be the greatest index i such that $\mathbf {v}_i\mathbf {v}_{i+1}\notin c(M{\setminus } D)$. Similarly, $\mathbf {v}_{k+1}\in c(\partial D)$.

Since D is a 2-ball, $\partial D$ is a cycle of edges $\mathbf {d}_i\mathbf {d}_{i+1}$. There are integers l and m such that the path $\mathbf {d}_l\hbox {-}\mathbf {d}_{l+1}\hbox {-}\cdots \mathbf {d}_m\subset \partial D$ meets $\mathbf {d}_l= \mathbf {v}_j$ and $\mathbf {d}_m= \mathbf {v}_{k+1}$. Thanks to Lemma 19, every edge $\mathbf {d}_i\mathbf {d}_{i+1}\in c(M{\setminus } D)$. Last we concatenate $\mathbf {v}_0\hbox {-}\cdots \mathbf {v}_j$ and $\mathbf {d}_l\hbox {-}\cdots \mathbf {d}_m$ and $\mathbf {v}_{k+1}\hbox {-}\cdots \mathbf {v}_n$ to obtain a path $\mathbf {v}_0\hbox {-}\cdots \mathbf {v}_n\subseteq c(M{\setminus } D)$ such that $\mathbf {v}_0=\mathbf {v}$ and $\mathbf {v}_n=\mathbf {v}'$. $\square$

Now we show Theorem 14.

First we show that $\partial (A\cup B)$ is a connected 2-manifold. Thanks to Theorem 10, $\partial (A\cup B)$ is a 2-manifold. Furthermore $\partial A{\setminus } D$ is connected and $\partial B{\setminus } D$ is connected (Lemma 20) and $\emptyset \ne c(\partial D)\subseteq c(\partial A{\setminus } D)\cap c(\partial B{\setminus } D)$ (thanks to Lemma 19). Thus $\partial (A\cup B)= (\partial A{\setminus } D)\cup (\partial B{\setminus } D)$ is connected.

Second we show that $c(D)\cap c(\partial (A\cup B))= c(\partial D)$. Lemma 19 using $M\in \{ \partial A, \partial B \}$ implies $c(D)\cap c(\partial A{\setminus } D)= c(\partial D)= c(D)\cap c(\partial B{\setminus } D)$. Therefore

$$\begin{aligned} c(D)\cap (c(\partial A{\setminus } D)\cup c(\partial B{\setminus } D))= c(\partial D). \end{aligned}$$

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We obtain the result using Lemma 15.

Third we show that $\chi (c(\partial (A\cup B)))= \chi (c(\partial A))$. Thanks to Lemma 15, $D\cup \partial (A\cup B)=\partial A\cup \partial B$. Since $\chi (X\cup Y)= \chi (X)+\chi (Y)-\chi (X\cap Y)$,

$$\begin{aligned}&\chi (c(D))+\chi (c(\partial (A\cup B)))-\chi (c(D)\cap c(\partial (A\cup B))) \nonumber \\&\quad = \chi (c(\partial A))+\chi (c(\partial B))-\chi (c(\partial A)\cap c(\partial B)). \end{aligned}$$

(41)

Furthermore, $\chi$ is known for 2-balls, cycles and 2-spheres:

$$\begin{aligned}&\chi (c(D))= \chi (c(\partial A)\cap c(\partial B))= 1, \end{aligned}$$

(42)

$$\begin{aligned}&\chi (c(D)\cap c(\partial (A\cup B)))= \chi (c(\partial D))=0, \end{aligned}$$

(43)

$$\begin{aligned}&\chi (c(\partial B))=2. \end{aligned}$$

(44)

Thus

$$\begin{aligned} 1+\chi (c(\partial (A\cup B)))-0=\chi (c(\partial A))+2-1. \end{aligned}$$

(45)

Now we see that $\partial A$ and $\partial (A\cup B)$ have the same genus.

Appendix 7: An Example for Theorem 15

First we remind how Z is constructed in [24] using shortened notations: a vertex is an integer between 1 and 12, the tetrahedron with vertices $\{ 1,2,3,4 \}$ is written 1.2.3.4, the edge with vertices $\{ 1,2 \}$ is written 1.2, $1.2\times 3\hbox {-}4\hbox {-}5\hbox {-}6$ is the tetrahedron set $\{1.2.3.4, 1.2.4.5, 1.2.5.6\}$, etc. Let V be the set of the vertices $1,2\cdots 12$. The 3D coordinates of the vertices in V are given in Table 1.

Table 1 3D coordinates of the 12 vertices in V

Then Z is defined as the disjoint union of three subsets of tetrahedra (Fig. 19). The center subset is

$$\begin{aligned} Z_0= \{ 1.2.3.4, 1.2.5.6, 2.3.6.7, 3.4.7.8, 4.1.8.5 \}. \end{aligned}$$

(46)

The top subset is the union of the tetrahedron cycles $Z_{2.10}$ and $Z_{4.12}$:

$$\begin{aligned} Z_{2.10}=2.10\times \hbox {5-6-7-3-1-5}, Z_{4.12}= \hbox {4.12}\times \hbox {7-8-5-1-3-7}.\nonumber \\ \end{aligned}$$

(47)

The bottom subset is the union of the tetrahedron cycles $Z_{1.9}$ and $Z_{3.11}$:

$$\begin{aligned} Z_{1.9}=1.9\times \hbox {5-6-2-4-8-5}, Z_{3.11}=\hbox {3.11}\times \hbox {7-8-4-2-6-7}.\nonumber \\ \end{aligned}$$

(48)

Fig. 19

Top views of the three subsets of Z. Left the top subset is the union of the tetrahedra cycles $Z_{2.10}$ and $Z_{4.12}$. Middle the center subset has five internal tetrahedra. Right the bottom subset is the union of the tetrahedra cycles $Z_{1.9}$ and $Z_{3.11}$. Note that vertices are sometimes duplicated

We obtain

$$\begin{aligned} Z= Z_0\cup Z_{1.9}\cup Z_{2.10}\cup Z_{3.11}\cup Z_{4.12}. \end{aligned}$$

(49)

Second we check that Z is included in a set of tetrahedra T such that c(T) is a 3D Delaunay triangulation.

Lemma 21

Thanks to Table 1, Z is included in a 3D Delaunay triangulation whose vertex set is V.

Proof

Let $\mathbf {c}_\Delta$ and $r_\Delta$ be the center and radius of the circumscribing sphere of the tetrahedron $\Delta$. Let

$$\begin{aligned} e= \min _{\Delta \in Z, \mathbf {u}\in V{\setminus } c(\Delta )} d^2(\mathbf {u}, \mathbf {c}_\Delta )-r^2_\Delta , \end{aligned}$$

(50)

where d is the Euclidean distance between two 3D points. We have $e= 0.5> 0$. Thus, for every $\Delta \in Z$, there is no $V{\setminus } c(\Delta )$ vertex in the (interior of the) sphere that includes the four $\Delta$ vertices. Therefore Z is included in a 3D Delaunay triangulation of V. $\square$

Third we show that Z is the union of four pipes and a 3-ball.

Lemma 22

We have $Z= P\cup S$ where

$$\begin{aligned} P=&p(5.9.1, 6.2.10)\cup p(7.11.3, 6.2.10) \nonumber \\&\cup \,p(5.9.1, 8.4.12)\cup p(7.11.3, 8.4.12), \end{aligned}$$

(51)

$$\begin{aligned} S=&\{ 1.2.3.4, 1.9.2.4, 2.10.1.3, 3.11.2.4,4.12.1.3 \} \end{aligned}$$

(52)

and S is a 3-ball.

Proof

Equation 15 is rewritten as

$$\begin{aligned} p(\mathbf {a}_0\mathbf {b}_0\mathbf {c}_0,\mathbf {a}_1\mathbf {b}_1\mathbf {c}_1) =&\mathbf {b}_0\mathbf {c}_0\times \mathbf {a}_0\hbox {-}\mathbf {a}_1\hbox {-}\mathbf {b}_1 \cup \{ \mathbf {a}_0\mathbf {a}_1\mathbf {b}_1\mathbf {c}_0 \} \nonumber \\&\cup \,\mathbf {b}_1\mathbf {c}_1\times \mathbf {c}_0\hbox {-}\mathbf {a}_0\hbox {-}\mathbf {a}_1. \end{aligned}$$

(53)

Thanks to Eq. 53,

$$\begin{aligned}&p(5.9.1, 6.2.10)= 9.1\times 5\hbox {-}6\hbox {-}2\cup \{ 5.6.2.1\}\cup 2.10\times 1\hbox {-}5\hbox {-}6 \nonumber \\&p(7.11.3, 6.2.10)= 11.3\times 7\hbox {-}6\hbox {-}2\cup \{7.6.2.3\}\cup 2.10\times 3\hbox {-}7\hbox {-}6 \nonumber \\&p(5.9.1, 8.4.12)= 9.1\times 5\hbox {-}8\hbox {-}4\cup \{ 5.8.4.1\}\cup 4.12\times 1\hbox {-}5\hbox {-}8 \nonumber \\&p(7.11.3, 8.4.12)= 11.3\times 7\hbox {-}8\hbox {-}4\cup \{ 7.8.4.3\}\cup 4.12\times 3\hbox {-}7\hbox {-}8. \end{aligned}$$

Now we see that

$$\begin{aligned} P= & {} (Z_0{\setminus }\{ 1.2.3.4\}) \nonumber \\&\cup \, 9.1\times 2\hbox {-}6\hbox {-}5\hbox {-}8\hbox {-}4\cup 2.10\times 1\hbox {-}5\hbox {-}6\hbox {-}7\hbox {-}3 \nonumber \\&\cup \, 11.3\times 2\hbox {-}6\hbox {-}7\hbox {-}8\hbox {-}4\cup 4.12\times 1\hbox {-}5\hbox {-}8\hbox {-}7\hbox {-}3 \nonumber \\= & {} (Z_0{\setminus }\{ 1.2.3.4\})\cup (Z_{1.9}{\setminus }\{ 1.9.2.4\}) \nonumber \\&\cup \,(Z_{2.10}{\setminus }\{ 2.10.1.3\})\cup (Z_{3.11}{\setminus }\{ 3.11.2.4\}) \nonumber \\&\cup \,(Z_{4.12}{\setminus }\{ 4.12.1.3\}) \nonumber \\= & {} Z{\setminus } S. \end{aligned}$$

(54)

and $S\subset Z$. Note that S is a 3-ball since it has a shelling started by the central tetrahedron 1.2.3.4 and the four other tetrahedra $\Delta _i$ are added to $O_i$ such that $c(\Delta _i)\cap c(O_i)= c(t_i)$ where $t_i\in \partial 1.2.3.4$. $\square$

Fourth, we show that $c(S)\cap c(P)$ is an annulus.

Lemma 23

We have $c(S)\cap c(P)= c(N)$ where

$$\begin{aligned} N= & {} \{\mathbf {9.2}.1,2.\mathbf {1.10}\}\cup \{3.\mathbf {2.11},\mathbf {10.3}.2\} \nonumber \\&\cup \, \{\mathbf {11.4}.3,4.\mathbf {3.12}\}\cup \{1.\mathbf {4.9},\mathbf {12.1}.4 \}. \end{aligned}$$

(55)

The vertices in bold fonts form edges in $\partial N$:

$$\begin{aligned} \partial N= 9\hbox {-}2\hbox {-}11\hbox {-}4\hbox {-}9\cup 1\hbox {-}10\hbox {-}3\hbox {-}12\hbox {-}1. \end{aligned}$$

(56)

Proof

Let $\sigma \in c(S)\cap c(P)$ be a simplex. Since vertices 5,6,7,8 are not in c(S), $\sigma$ does not have a vertex in $\{ 5,6,7,8\}$. According to Eq. 16, the only triangles in $\partial p(\mathbf {a}_0\mathbf {b}_0\mathbf {c}_0,\mathbf {a}_1\mathbf {b}_1\mathbf {c}_1)$ that does not have vertices $\mathbf {a}_0$ and $\mathbf {a}_1$ are $\{\mathbf {b}_0\mathbf {b}_1\mathbf {c}_0, \mathbf {b}_1\mathbf {c}_0\mathbf {c}_1\}$. Using Eq. 51 and $\mathbf {a}_0\in \{5,7\}$ and $\mathbf {a}_1\in \{6,8\}$, $\sigma$ is included is one of the triangles in $\tilde{N}$ where

$$\begin{aligned} \tilde{N}= & {} \{ 9.2.1,2.1.10 \}\cup \{ 11.2.3,2.3.10\} \nonumber \\&\cup \, \{ 9.4.1,4.1.12 \}\cup \{ 11.4.3,4.3.12\}. \end{aligned}$$

(57)

Therefore $c(S)\cap c(P)\subseteq c(\tilde{N})$. Conversely, $\tilde{N}\subseteq c(P)$ and $\tilde{N}\subseteq c(S)$ imply $c(\tilde{N})\subseteq c(S)\cap c(P)$. Last we check that $N= \tilde{N}$. $\square$

Thus $Z= P\cup S$ with $n= 4$ and $P= \cup _i P_i$ as in Theorem 15. Since Z is strongly non-shellable [24], $Z= P\cup S'$ and $S= S'$.