We now show that a piecewise linear approximation that results from applying Algorithm 3.1 can also be modeled with the generalized incremental method. We first prove two lemmata that are used afterward to prove the main result of this section.
Lemma 4.1
Let \(\mathcal {S} = \{S^0, \ldots , S^{2^d-1}\}\) be a refinement of a simplex S by Algorithm 3.1 with \(\mathcal {V}(S) = \{\bar{x}_0, \ldots , \bar{x}_d \}\). Then, each simplex of the subset of the corner sub-simplices \(\mathcal {S}' = \{S^{i_0}, \ldots , S^{i_{d}}\}\) of \(\mathcal {S}\) contains a vertex of the simplex S, i.e., \(\bar{x}_j \in \mathcal {V}(S^{i_j})\) for all \(j = 0, \ldots , d\). Moreover, for each pair of simplices \(S^{i_j}, S^{i_k} \in \mathcal {S}'\) the midpoint \(m_{jk}\) of the edge with endpoints \(\bar{x}_j\) and \(\bar{x}_k\) is contained in both vertex sets of the simplices.
Proof
First, the identity \({{\,\mathrm{id}\,}}\in {{\,\mathrm{Sym}\,}}_d\) always fulfills the conditions from Line 9 of Algorithm 3.1. Let \(S^{i_j}\) be the simplex that is constructed using the starting vertex \(v_0 = \frac{1}{2} (\bar{x}_0 + \bar{x}_j)\) and \(\tau = {{\,\mathrm{id}\,}}\), where \(j = 0, \ldots , d\). Due to the telescoping sum in Line 11, it follows that
$$\begin{aligned} v_j = \underbrace{\underbrace{\underbrace{\frac{1}{2} (\bar{x}_0 + \bar{x}_j)}_{v_0} + \frac{1}{2} (\bar{x}_1 - \bar{x}_0)}_{v_1} + \frac{1}{2} (\bar{x}_2 - \bar{x}_1)}_{\vdots } + \cdots + \frac{1}{2} (\bar{x}_j - \bar{x}_{j-1}) = x_j \end{aligned}$$
(5)
is contained in the vertex set of \(S^{i_j}\).
Furthermore, due to the telescoping sum in (5), we can rewrite Line 11 as
$$\begin{aligned} v_l \leftarrow \frac{1}{2} (\bar{x}_0 + \bar{x}_j) + \frac{1}{2} (\bar{x}_{l} - \bar{x}_{0}) = \frac{1}{2} (\bar{x}_{j} + \bar{x}_{l}) \end{aligned}$$
for \(\tau = {{\,\mathrm{id}\,}}\). Since \(m_{jk} = \frac{1}{2} (\bar{x}_{j} + \bar{x}_{k})\), we conclude that the vertices \(v_k\) and \(v_j\) that occur during the construction of \(S^{i_j}\) and \(S^{i_k}\), respectively, are equal to \(m_{jk}\). \(\square \)
Lemma 4.2
Let \(\mathcal {S}'' = \{S^0, \ldots , S^{2^d-1-(d+1)}\}\) be a refinement of a simplex S by Algorithm 3.1 without the \(d+1\) corner sub-simplices of \(\mathcal {S}'\) as in Lemma 4.1. Then, the union of all simplices of \(\mathcal {S}''\) is a (convex) polytope and the triangulation of \(\mathcal {S}''\) has an ordering with the properties (O1) and (O2).
Proof
Alternatively to the vertex description, we can describe the simplex S by its half-space representation
$$\begin{aligned} S = \left\{ x \in {\mathbb {R}}^d \, :\, a_j^\top x \le b_j \, \text { with } a_j \in {\mathbb {R}}^d \, \text { and } b_j \in {\mathbb {R}}\, \text { for } j = 0, \ldots , d \right\} . \end{aligned}$$
(6)
We now describe the set \(\mathcal {S}''\) by adding the inequalities that separate all corner sub-simplices from the set S as in (6). For a vertex of S exactly d inequalities in (6) are tight. Due to Lemma 4.1, the vertex set of the corner sub-simplex \(S^{i_j} \in \mathcal {S}'\) consists of the vertex \(\bar{x}_j\) and all midpoints \(m_{jk}\) with \(k = 0, \ldots , d\). Let \(a_j^\top x \le b_j\) be the inequality that is not tight for \(\bar{x}_j\). Naturally, it is tight for all other vertices of S and it follows that
$$\begin{aligned} a_j^\top m_{jk} = \frac{1}{2} \, a_j^\top (\bar{x}_k + \bar{x}_j) = \frac{1}{2} (b_j + a_j^\top \bar{x}_j). \end{aligned}$$
(7)
Moreover, since the red refinement also delivers a triangulation of S, where all interiors of the sub-simplices are disjunct, we can describe \(S^{i_j}\) by substituting \(a_j^\top x \le b_j\) with \(a_j^\top x \le \frac{1}{2} (b_j + a_j^\top \bar{x}_j)\) in (6). Therefore, by separating all corner sub-simplices, we obtain the half-space description
$$\begin{aligned} \mathcal {S}'' =&\Bigg \{ x \in {\mathbb {R}}^d \, :\, a_j^\top x \le b_j, \ a_j^\top x \ge \frac{1}{2} (b_j + a_j^\top \bar{x}_j)\\&\text { with } a_j \in {\mathbb {R}}^d \, \text { and } b_j \in {\mathbb {R}}\, \text { for } j = 0, \ldots , d \Bigg \}. \end{aligned}$$
It follows that the union of all simplices of \(\mathcal {S}''\) is a convex polytope.
The consistency of the triangulation of S translates to the triangulation of \(\mathcal {S}''\), because only the corner sub-simplices are omitted. It is shown by [6] that a triangulation with the property that for each nonempty subset \(\tilde{\mathcal {S}} \subsetneq \mathcal {T}\) of a triangulation \(\mathcal {T}\) there exist simplices \(S_1 \in \tilde{\mathcal {S}}\) and \(S_2 \in \mathcal {T}{\setminus } \tilde{\mathcal {S}}\) such that \(S_1, S_2\) have d common vertices, always yields a triangulation with the properties (O1) and (O2). Therefore, we only have to prove that the triangulation of \(\mathcal {S}''\) has this property. To this end, we consider a nonempty subset \(\tilde{\mathcal {S}}\) of \(\mathcal {S}''\). Each facet of a d-dimensional simplex consists of d vertices of the simplex. Let \(\tilde{\mathcal {S}}_\text {F}\) be the set of all facets of the simplices of \(\tilde{\mathcal {S}}\), where the simplices are described by the convex hull of its vertices. Then, each facet in \(\tilde{\mathcal {S}}_\text {F}\) is either a facet of \(\mathcal {S}''\) (in the sense of convex hulls) or a common facet of two simplices of \(\tilde{\mathcal {S}}\). This is due to the consistency of the triangulation. Since \(\tilde{\mathcal {S}} \subsetneq \mathcal {S}''\), however, there must be a facet in \(\tilde{\mathcal {S}}_\text {F}\) that is a common facet of two simplices \(S_1, S_2\) such that \(S_1 \in \tilde{\mathcal {S}}\) and \(S_2 \in \mathcal {S}'' {\setminus } \tilde{\mathcal {S}}\). \(\square \)
Endowed with Lemmas 4.1 and 4.2, we are now ready to prove the main result of this section.
Theorem 4.3
Let \(\mathcal {T}\) be a triangulation with the properties (O1) and (O2). Then, any triangulation \(\mathcal {T}'\) obtained by applying Algorithm 3.1 to \(\mathcal {T}\) maintains the properties (O1) and (O2).
Proof
For the following proof, we first show that there is an ordering of the sets \(\mathcal {S}'\) and \(\mathcal {S}''\) from Lemmas 4.1 and 4.2 with the properties (O1) and (O2). The second part of the proof merges these two orderings to obtain an overall ordering with the properties (O1) and (O2).
Let \(\mathcal {T}= \{S_1, \ldots , S_n\}\) and \(S_l\) be the simplex that has to be refined, while \(\bar{x}_0^{S_l}, \ldots , \bar{x}_d^{S_l}\) are its labeled vertices. We first consider \(d \ge 4\) and \(d = 2, 3\) afterward. The \(2^d\) simplices \(S_l^0, \ldots , S_l^{2^d-1}\), into which \(S_l\) is divided by the red refinement, have due to Lemma 4.1 the property that the corner sub-simplices contain the vertices of \(S_l\). Without loss of generality, let \(\bar{x}_j^{S_l} \in \mathcal {V}(S_l^j)\).
We first show that the set \(\mathcal {S}'\) of the corner sub-simplices has an ordering with the required properties. The corner sub-simplices \(S_l^j \in \mathcal {S}'\) yield a complete graph \(G=(V,E)\) with the simplices as the node set V and the midpoints \(m_{jk}\) as the edge set E connecting the simplices \(S_l^j\) and \(S_l^k\). Due to \(m_{jk} = m_{kj}\), we assume in the following for the notation \(m_{jk}\) that \(j < k\) holds. For each Hamiltonian path in G, we can use the path itself as an ordering of the simplices that correspond to the nodes of G. An edge connecting two consecutive nodes of the path corresponds to a common midpoint of two consecutive simplices. Therefore, the ordering naturally has property (O1), which indicates that two consecutive simplices have at least one common vertex. The ordering has the property O2 as well, which states that the last vertex of any simplex is equal to the first vertex of the next one: With two consecutive simplices \(S_l^j\) and \(S_l^k\) that correspond to two consecutive nodes of the Hamiltonian path, we only have to set \(\bar{x}_d^{S_l^j} = m_{jk}\) and \(\bar{x}_0^{S_l^k} = m_{jk}\).
Moreover, it follows from Lemma 4.2 that there is an ordering
$$\begin{aligned} \left( S_l^{i_0}, \ldots , S_l^{i_{2^d-1-(d+1)}}\right) \end{aligned}$$
(8)
of the simplex set \(\mathcal {S}''\) with the properties (O1) and (O2). Since the vertices of the simplices of \(\mathcal {S}''\) are the midpoints \(m_{jk}\), there must be two midpoints \(m_{jk}\) and \(m_{st}\) with
$$\begin{aligned} m_{jk} = \bar{x}_0^{S_l^{i_0}} \qquad \text {and}\qquad m_{st} = \bar{x}_d^{S_l^{i_{2^d-1-(d+1)}}}. \end{aligned}$$
We now link the orderings of \(\mathcal {S}'\) and \(\mathcal {S}''\) to obtain an overall ordering with the properties (O1) and O2. Let R be a Hamiltonian path in the sub-graph of G that consists of the vertices \(V {\setminus } \{0, j, k, s, t, d\}\). Such a path is always attainable, because any sub-graph of a complete graph is also complete. With \(j \not = k\) one of the following three cases applies to the node j: \(j = s\), \(j = t\), or \(j \not =s \wedge j \not = t\). With \(s \not = t\), we have the following three cases for the node s: \(s = j\), \(s = k\), or \(s \not =j \wedge s \not = k\). Consequently, due to \(j \not = k\) and \(s \not = t\) only the following five cases are possible for the nodes j, k, s, and t:
$$\begin{aligned} j = s \,\wedge \, k \not = t, \quad j = s \,\wedge \, k = t, \quad j \not = s \,\wedge \, k = t, \quad k = s, \quad j \not = s \,\wedge \, k \not = t. \end{aligned}$$
(9)
The case \(k = s\) is equivalent to the case \(j = t\), since the inverse of the ordering (8) also has the properties (O1) and (O2).
Keeping the cases in (9) in mind, we define the path
Please note that if \(t = d\) in (10a), we can again invert the ordering (8) such that \(t \not = d\) and \(k = d\). The same can be applied in case of \(k = d\) in (10c) and (10d). Moreover, any permutation of the labeling \(\bar{x}_i\) of the simplex \(S_l^{i_0}\), where \(i = 0, \ldots , d-1\) and of the simplex \(S_l^{i_{2^d-1-(d+1)}}\), where \(i = 1, \ldots , d\), is permissible. Thus, we can assume that \(m_{st} \not = m_{0d}\) and that \(m_{jk} \not = m_{0u}\) if \(m_{st} = m_{ud}\). This guarantees us that the path H is always a Hamiltonian path.
The path H corresponds to an ordering, where the vertices in H are the corner sub-simplices \(S_l^j \in \mathcal {S}'\). As showed above, this ordering has the properties (O1) and (O2). We now insert the ordering (8) of the simplex set \(\mathcal {S}''\) into the one of H after the simplex that corresponds to the vertex j. The union of \(\mathcal {S}'\) and \(\mathcal {S}''\) corresponds to the set of all \(2^d\) sub-simplices into which \(S_l\) is divided by the red refinement. Therefore, the resulting ordering covers all \(2^d\) sub-simplices. The merging of the two orderings of \(\mathcal {S}'\) and \(\mathcal {S}''\) via the midpoints \(m_{jk}\) and \(m_{st}\) finally leads to an overall ordering \((S_l^{r_0}, \ldots , S_l^{r_{2^d-1}})\) that has the properties (O1) and (O2).
In case of \(d = 2\), we use the ordering \((S_l^0, S_l^1, S_l^2, S_l^3)\), where \(S_l^0\), \(S_l^1\), and \(S_l^3\) are the three corner sub-simplices and \(S_l^2\) the remaining center sub-simplex, see again Fig. 1 for an illustration.
For \(d = 3\), we assume without loss of generality that the vertex \(\bar{x}_3\) of the simplex that has to be refined is contained in \(S_l^7\). We use the ordering \((S_l^0, S_l^1, S_l^2, S_l^{3}, S_l^{4}, S_l^{5}, S_l^{6}, S_l^7)\), where \(S_l^0\), \(S_l^1\), \(S_l^2\), and \(S_l^7\) are the four corner sub-simplices contained in \(\mathcal {S}'\) and \(S_l^{3}\)–\(S_l^{6}\) the four center sub-simplices of \(\mathcal {S}''\). Finally, the orderings of \(\mathcal {S}'\) and \(\mathcal {S}''\) are linked via \(m_{jk} = m_{02}\) and \(m_{st} = m_{27}\).
With these orderings for the refined simplex \(S_l\) and all dimensions \(d \ge 2\), we complete the proof as follows. We order the simplices of \(\mathcal {T}'\) as
$$\begin{aligned} (S_1, \ldots , S_{l-1}, S_l^{r_0}, \ldots , S_l^{r_{2^d-1}}, S_{l+1}, \ldots , S_n). \end{aligned}$$
(11)
Since the corner sub-simplices of \(S_l\) yield a complete graph, we can order them such that
$$\begin{aligned} \bar{x}_d^{S_{l-1}} = \bar{x}_0^{S_l^{r_0}} \qquad \text {and}\qquad \bar{x}_d^{S_l^{r_{2^d-1}}} = \bar{x}_0^{S_{l+1}} \end{aligned}$$
holds. Therefore, the simplices \(S_{l-1}\), \(S_l^{r_0}\) and \(S_l^{r_{2^d-1}}\), \(S_{l+1}\) are linked as required in O2. Altogether, due to the inheritance from \(\mathcal {T}\), we conclude that the ordering (11) of the simplices of \(\mathcal {T}'\) has the properties (O1) and (O2). \(\square \)