On approximation of asymmetric separators of the ncube
 Jan P Boroński^{1, 2}Email author and
 Marian Turzański^{3}
https://doi.org/10.1186/1687181220122
© Borońński and Turzańński; licensee Springer. 2012
Received: 1 February 2011
Accepted: 4 January 2012
Published: 4 January 2012
Abstract
A new combinatorial result intertwined with the Brouwer fixed point theorem for the ncube is given. This result can be used for any map (f_{1}, ..., fn): [0, 1]^{ n } → [0, 1]^{ n } to approximate the components of the set {(x_{1}, . . . , x_{ n }) ∈ [0, 1]^{ n } : f_{ i }(x_{1}, . . . , x_{ n }) = x_{ i }} that separate the ncube between the i th opposite faces. Equivalently, for maps g : [0, 1]^{ n } → ℝ such that g(x)g(y) ≤ 0 for any x ∈ {0} × [0, 1]^{n1}and y ∈ {1} × [0, 1]^{n1}, one can use the algorithm to approximate the components of g^{1}(0) that separate [0, 1]^{ n } between {0} × [0, 1]^{n1}and {1} × [0, 1]^{n1}. The methods are based on an earlier result of P. Minc and the present authors and relate to results of several other authors such as Jayawant and Wong, Kulpa and Turzański, and Gale.
Mathematics Subject Classification (2000): Primary 54H25; 5404; Secondary 55M20; 54F55.
Keywords
1. Introduction
In [1] Minc and the present authors described combinatorial methods that allow approximation of connected symmetric separators of the nsphere and ncube. The symmetric separators arise in the context of the BorsukUlam antipodal theorem and a theorem of Dyson for the 2sphere [2, 3]. The purpose of the present paper is to show how the results from [1] can be extended to the setting of asymmetric separators and the Brouwer fixed point theorem for the ncube. The classic result of L.E.J. Brouwer says that the ndimensional cube I^{ n } = [0, 1]^{ n } has the fixed point property; that is, for any map^{a} f : I^{ n } → I^{ n }, there is x ∈ I^{ n } such that f(x) = x. There are many important applications of the Brouwer's theorem such as, for example, those concerning existence of solutions for differential equations [4], or equilibrium strategies in multiperson games relating to market problems in economics [5]. This is why computability of fixed points became an important theme in the fixed point theory. The first fixed point algorithm was given by Scarf [6]. Soon after, other were given by Eaves [7] and Todd [8] (see, for example, [9, 10] for a comprehensive treatment of this subject with applications). There are also several combinatorial equivalents of Brouwer's theorem. The best known is probably Sperner's lemma [11] on coloring vertices of a barycentric subdivision of an nsimplex. Some other transfer the fixed point problem to the scenario of board games, such as Hex [12] or Chess [13].
In the present paper, in Theorem 3.1, we formulate yet another combinatorial result that implies the Brouwer fixed point theorem. Its baby version can be formulated as follows.
 1.
for every simplex σ ∈ S there is an edge [v, u] such that f (v) f (u) ≤ 0;
 2.
∪S separates I^{ n } between {0} × I^{n1}and {1} × I^{n1}.
The above theorem implies the Brouwer fixed point theorem in the following way. If (f_{1}, . . . , f_{ n }): I^{ n }→ I^{ n }is a map and $\mathcal{X}$ is a polyhedral complex with I^{ n } as its underlying space, then each g_{ i }(x_{1}, . . . , x_{ n }) = f_{ i }(x_{1}, . . . , x_{ n }) x_{ i } satisfies the assumptions of Theorem 1.1 and one can find C_{ i }, an approximation of a component of ${g}_{i}^{1}\left(0\right)$, that separates I^{ n } between the i th opposite (n  1)dimensional faces. By EilenbergOtto theorem (see [14]) ${\bigcap}_{i=1}^{n}{C}_{i}$ is nonempty and approximates a fixed point of f.
We give a stronger (but at the same time more technical) version of the above result in Theorem 3.1, and in Section 4, we show how along with Theorem 4.1 it can be used to approximate a connected separating component of the set of zeros of an arbitrary map f : I^{ n } → ℝ, which assumes opposite signs on some two opposite (n  1)faces of I^{ n }. The case when n = 2 was already considered in [15]. The methods used in the proof of Theorem 3.1 are based on those introduced in [1] where, in connection with the BorsukUlam antipodal theorem, it was shown how to approximate a connected separator of the nsphere S^{ n }(or I^{ n }), invariant under the antipodal map. Any such separator was corresponding to a component of f^{1}(0), with f : S^{ n } → ℝ (or f : I^{ n }→ ℝ) an odd map (related combinatorial results can be found in [2, 16]). However, the methods of [1] were dealing only with symmetric separators and are insufficient in the case of arbitrary separators. First, unlike in the case of symmetric separators and odd maps, if a map f : I^{ n }→ R satisfies the condition f({1} × I^{n1}) ⊆ [0, ∞) and f({0} × I^{n1}) ⊆ (∞, 0] for some i = 1, . . . , n, there may be no unique connected separator of I^{ n }in f^{1}(0). Clearly, f^{1}(0) may consist of several disjoint separating components, none of which needs to be symmetric. Second, the algorithms in [1] were making use of the fact that the symmetric component of f^{1}(0) is the separating omponent, when f is odd. Therefore, if a subcollection of the triangulation approximated a component of f^{1}(0) and, at the same time, was symmetric, this was sufficient to determine that it separated S^{ n }(or I^{ n }). This is why one is forced to develop new combinatorial criteria for arbitrary separators in I^{ n }. In Section 4 we furnish such a computer implementable criterion that allows isolating those subcollections of the triangulation, approximating a component of f^{1}(0), that separate I^{ n } from those that do not.
2. Preliminaries
For a collection of sets $\mathcal{K}$, by ${\mathcal{K}}^{*}$ we will denote the union of all its elements. π_{ i } : [0, 1]^{ n } → [0, 1] will denote the projection onto the i th coordinate. ${I}_{i}^{+}$ and ${I}_{i}^{}$ will denote the i th opposite (n  1)dimensional faces of I^{ n }, that is ${I}_{i}^{+}={\pi}_{i}^{1}\left(1\right)$and ${I}_{i}^{}={\pi}_{i}^{1}\left(0\right)$. C separates I^{ n } (or is a separator of I^{ n }) between ${I}_{i}^{+}$ and ${I}_{i}^{}$ if for any $x\in {I}_{i}^{+}\backslash C$,$y\in {I}_{i}^{}\backslash C$, there are U, V , distinct components of I^{ n }\C, such that x ∈ U and y ∈ V. A map $g:{\mathcal{X}}^{*}\to \mathbb{R}$ is piecewise linear if given {σ_{ j } : j = 1, . . . , N}, a triangulation of ${\mathcal{X}}^{*}$, for every j the restriction of g to the simplex σ_{ j } is linear, that is $g\left({\sum}_{i=1}^{k}{\lambda}_{i}{a}_{i}\right)={\sum}_{i=1}^{k}{\lambda}_{i}g\left({a}_{i}\right)$ where a_{1}, . . . , a_{ k } are the vertices spanning σ_{ j } and λ_{ i } ≥ 0 with ${\sum}_{i=1}^{k}{\lambda}_{i}=1$ (see [17]).
We will heavily rely on the following inductive procedure introduced by Minc and the two authors in [1]. Let $\mathcal{X}$ be a polyhedral complex such that ${\mathcal{X}}^{*}={\left[0,1\right]}^{n}$. Let $\mathcal{V}\left(\mathcal{X}\right)$ and $\mathcal{E}\left(\mathcal{X}\right)$ denote the collections of vertices and edges, respectively. Suppose $f:\mathcal{V}\left(\mathcal{X}\right)\to \mathbb{R}$ is a function. Let ${\mathcal{E}}_{f}$ be the collection of those edges $e=\u27e8u,v\u27e9\in \mathcal{E}\left(\mathcal{X}\right)$ that f (u) f (v) ≤ 0. Let $\mathcal{P}\subset \mathcal{X}$ be the collection of polytopes in $\mathcal{X}$ of dimension n. For any $e\in {\mathcal{E}}_{f},$ $\mathcal{C}\left(e\right)$ is defined by induction.

Let ${\mathcal{C}}_{0}\left(e\right)$ be the collection of those $P\in \mathcal{P}$ that contain e.

Suppose ${\mathcal{C}}_{i1}\left(e\right)$ has been defined. Define ${\mathcal{C}}_{i}\left(e\right)$ to be the collection of those $P\in \mathcal{P}$ such that the intersection $P\cap {\mathcal{C}}_{i1}{\left(e\right)}^{*}$ contains an edge from ${\mathcal{E}}_{f}$ or a vertex from f^{1} (0).
Clearly, ${\mathcal{C}}_{i1}\left(e\right)\subset {\mathcal{C}}_{i}\left(e\right)$ and there is an integer q ≥ 0 such that ${\mathcal{C}}_{q}\left(e\right)={\mathcal{C}}_{q+1}\left(e\right).$ For the first such number q (e) set $\mathcal{C}\left(e\right)={\mathcal{C}}_{q\left(e\right)}\left(e\right)$. Note that $e\subset \mathcal{C}{\left(e\right)}^{*}$, and $\mathcal{C}{\left(e\right)}^{*}$ is connected.
3. Combinatorial theorem on separators of I^{ n }between opposite faces
Let χ be a polyhedral complex such that ${\mathcal{X}}^{*}={\left[0,1\right]}^{n}$. Note that $\mathcal{X}$ can be subdivided to give a triangulation of I^{ n }, without introducing new vertices [18], and consequently, every function $f:\mathcal{V}\left(\mathcal{X}\right)\to \mathbb{R}$ has a piecewise linear extension g : I^{ n } → ℝ. The following result is of purely combinatorial nature.
 1.
each $C\in \mathcal{C}\left(d\right)$ contains an edge from ${\mathcal{E}}_{f}$,
 2.
$\mathcal{C}{\left(d\right)}^{*}$ separates I^{ n } between ${I}_{i}^{+}$ and ${I}_{i}^{}$ and
 3.
for any other such ${d}^{\prime}\in \mathcal{L}$ with $\mathcal{C}{\left({d}^{\prime}\right)}^{*}$ satisfying ( 1)( 2), either $\mathcal{C}\left({d}^{\prime}\right)\cap \mathcal{C}\left(d\right)=\varnothing $ or $\mathcal{C}\mathsf{\text{(}}{d}^{\prime}\mathsf{\text{)=}}\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$.
Proof. Without loss we can assume that i = 1. Let g : I^{ n } → ℝ be a piecewise linear extension of f. Then, g is continuous and g(v) = f(v) for any $v\in \mathcal{V}\left(\mathcal{X}\right)$.
Claim 3.1.1. If $K\in \mathcal{X}$, then g(r) = 0 for some r ∈ K if and only if there is an edge $\left[w,v\right]\in \mathcal{E}\left(\mathcal{X}\right)$ such that $w,v\in K\cap \mathcal{V}\left(\mathcal{X}\right)$ and f (w) f (v) ≤ 0.
Proof of claim 3.1.1. First suppose $K\in \mathcal{X}$ is such that there are vertices $w,v\in K\cap \mathcal{V}\left(\mathcal{X}\right)$, and f (w) f (v) ≤ 0. Then, either f(w)f(v) = 0 or f(w)f(v) < 0. In the first case, clearly g(w) = f(w) = 0 or g(v) = f(v) = 0. Otherwise there must be r ∈ [u, v] such that g(r) = 0. For the converse, suppose g(r) = 0 for some r ∈ K. Then, $r={\sum}_{i=1}^{k}{\lambda}_{i}{a}_{i}$, where a_{ o }, . . . , a_{ k } are vertices of K spanning a simplex σ ⊆ K, and λ_{ i } ≥ 0 with ${\sum}_{i=1}^{k}{\lambda}_{i}=1$. Therefore, $0=g\left(r\right)={\sum}_{i=1}^{k}{\lambda}_{i}g\left({a}_{i}\right)={\sum}_{i=1}^{k}{\lambda}_{i}f\left({a}_{i}\right)$. Clearly, there is l such that f(a_{ l }) = 0, or there are a_{ j }, a_{ t } such that f(a_{ j })f(a_{ t }) < 0. □
Claim 3.1.2. $g\left({I}_{i}^{+}\right)\subseteq \left[0,+\infty \right),g\left({I}_{i}^{}\right)\subseteq \left(\infty ,0\right]$.
Proof of claim 3.1.2. Similarly to the proof of claim 3.1.1, this follows from the fact that if a_{ o }, . . . , a_{ k } spans a simplex σ and g (a_{ i }) ≥ 0 (g(a_{ i }) ≤ 0) for every i, then ${\sum}_{i=1}^{k}{\lambda}_{i}g\left({a}_{i}\right)\ge 0\left({\sum}_{i=1}^{k}{\lambda}_{i}g\left({a}_{i}\right)\le 0\right)$ for λ_{ i } ≥ 0. Consequently, g(σ) ⊆ [0,+∞) (g(σ) ⊆ (∞, 0]) for any such σ. □
Now, consider the following decomposition of the ncube.
Z = {x ∈ I^{ n } : g(x) = 0}, N = {x ∈ I^{ n } : g(x) < 0}, P = {x ∈ I^{ n } : g(x) > 0}. Clearly Z separates I^{ n } between P and N. Let Z_{1}, . . . , Z_{ p } be the components of Z. It is well known that if X is a connected, locally connected and unicoherent space then any closed set separating X contains a connected subset separating X ([[19], p. 195], cf. [[20], p. 429, Theorem 1.(vi)]). Since Z is closed and separates I^{ n }, by unicoherence of I^{ n }, there must be q such that Z_{ q } separates I^{ n } between N and P. Consequently, Z_{ q } separates I^{ n } between ${I}_{i}^{+}$ and ${I}_{i}^{}$ by claim 3.1.2.
Consider $\mathcal{S}$ a subcollection of $\mathcal{K}$ such that $K\in \mathcal{S}$ if, it has nonempty intersection with Z_{ q }, that is $\mathcal{S}=\left\{K\in \mathcal{X}:K\cap {Z}_{q}\ne \varnothing \right\}$. Clearly ${Z}_{q}\subseteq {\mathcal{S}}^{*}$ and therefore ${\mathcal{S}}^{*}$ separates I^{ n } between ${I}_{i}^{+}$ and ${I}_{i}^{}$.
Now, let $\mathcal{L}$ be a subcollection of $\mathcal{E}\left(\mathcal{X}\right)$ such that ${\mathcal{L}}^{*}\subseteq {I}_{j}^{\epsilon}$ for some j ≠ 1 and ε ∈ {+, }, and ${\mathcal{L}}^{*}$ is an arc with endpoints $a\in {I}_{1}^{+}$ and $b\in {I}_{1}^{}$, that is ${\mathcal{L}}^{*}=\left[a,b\right]$. Since Z_{ q } ∩ Bd (I^{ n }) separates Bd (I^{ n }) between ${I}_{i}^{+}$ and ${I}_{i}^{}$, we conclude there is z ∈ [a, b] ∩ Z_{ q }. Additionally, there is $d\in \mathcal{L}$ such that z ∈ d. By claim 3.1.1 $d\in {\mathcal{E}}_{f}\cap \mathcal{L}$, and since d ∩ Z_{ q } ≠ ∅ therefore $d\in \mathcal{S}$.
Claim 3.1.3. $S\subseteq \mathcal{C}\left(d\right)$.
By definition of $\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$, for any $T\in {\mathcal{S}}_{2}$ and for any $\stackrel{\u0303}{T}\in {\mathcal{S}}_{1}$, we must have that whenever $s\in \mathcal{E}\left(\mathcal{X}\right)$ and $s\subseteq T\cap \stackrel{\u0303}{T}$ then $s\notin {\mathcal{E}}_{f}$. Otherwise T would be in ${\mathcal{S}}_{1}$. Therefore, $\left(T\cap \stackrel{\u0303}{T}\right)\cap {Z}_{q}=\varnothing $, by claim 3.1.1. Consequently, there is a partition of Z_{ q } into two disjoint sets ${Z}_{q}\cap {\mathcal{S}}_{1}^{*}$ and ${Z}_{q}\cap {\mathcal{S}}_{2}^{*}$. Since both are closed, we obtain a contradiction with connectedness of Z_{ q }. □
Now, property (1) is an immediate consequence of the definition of $\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$. Since ${\mathcal{S}}^{*}{\subseteq}^{*}\mathcal{C}\left(d\right)$, (2) easily follows from the fact that ${\mathcal{S}}^{*}$ separates I^{ n } between ${I}_{i}^{+}$ and ${I}_{i}^{}$. Now, suppose ${d}^{\prime}\in \mathcal{L}$ is another edge with $\mathcal{C}{\left({d}^{\prime}\right)}^{*}$ satisfying (1)(2). If $\mathcal{C}\left({d}^{\prime}\right)\cap \mathcal{C}\left(d\right)\ne \varnothing $, then there is K such that $K\in {\mathcal{C}}_{j}\left(d\right)$ and $K\in {\mathcal{C}}_{p}\left({d}^{\prime}\right)$ for some j and p. Consequently, ${\bigcup}_{i=0}^{j}{\mathcal{C}}_{i}\left(d\right)\subseteq {\bigcup}_{i=p}^{q\left({d}^{\prime}\right)}{\mathcal{C}}_{i}\left({d}^{\prime}\right)$ and ${\bigcup}_{i=j}^{q\left(d\right)}{\mathcal{C}}_{i}\left(d\right)\subseteq {\bigcup}_{i=p}^{q\left({d}^{\prime}\right)}{\mathcal{C}}_{i}\left({d}^{\prime}\right)$, by definition of $\mathcal{C}\left({d}^{\prime}\right)$. Clearly $\mathcal{C}\left(d\right)\subseteq \mathcal{C}\left({d}^{\prime}\right)$. Similarly $\mathcal{C}\left({d}^{\prime}\right)\subseteq \mathcal{C}\left(d\right)$. That justifies (3) and completes the proof. □
4. Algorithm approximating connected separators of I^{ n }
Suppose $\mathcal{K}$ is a partition of I^{ n } into k^{ n } congruent ncubes, all with side length equal to $\frac{1}{k}$. In this section we shall furnish a computer implementable criterion for the union of a subcollection of $\mathcal{K}$ to separate I^{ n } between some two opposite faces. Suppose $\mathcal{S}\subseteq \mathcal{K}$ and we want to determine if ${\mathcal{S}}^{*}$ separates I^{ n } between ${I}_{i}^{+}$ and ${I}_{i}^{}$.
Let $K\in \mathcal{K}\backslash \mathcal{S}$ be an ncube. G is a jface of K if dim (G) = j and G = K ∩ L for some K, $L\in \mathcal{K}$. We will define Comp $\left(K,\mathcal{S}\right)$ by induction. Let Comp_{1} $\left(K,\mathcal{S}\right)$ consists of K and all those cubes L in $\mathcal{K}\backslash \mathcal{S}$ such that K ∩ L is an (n  1)face. Suppose Comp_{ p }$\left(K,\mathcal{S}\right)$ has already been defined and let Comp_{p+1}$\left(K,\mathcal{S}\right)$ consists of all cubes in Comp_{ p }$\left(K,\mathcal{S}\right)$, and all those cubes R in $\mathcal{K}\backslash \mathcal{S}$ for which there is a cube L ∈ Comp_{ p }$\left(K,\mathcal{S}\right)$ such that L ∩ R is an (n  1)face. Since $\mathcal{K}$ consists of only finite number of cubes Comp_{ q }$\left(K,\mathcal{S}\right)$ = Comp_{q+1}$\left(K,\mathcal{S}\right)$ for some natural number q. Let q(K) be the first such number and let Comp $\left(K,\mathcal{S}\right)=\mathsf{\text{Com}}{\mathsf{\text{p}}}_{q\left(K\right)}\left(K,\mathcal{S}\right)$.
Proof. If the condition (4.1) is not satisfied, then clearly ${\mathcal{S}}^{*}$ does not separate I^{ n } between ${I}_{i}^{+}$ and ${I}_{i}^{}$. Namely, Comp ${\left(K,\mathcal{S}\right)}^{*}$ for some K contains a connected set, disjoint with ${\mathcal{S}}^{*}$, intersecting both ${I}_{i}^{+}$ and ${I}_{i}^{}$ in a nonempty set. For the converse, by contradiction suppose that the condition (4.1) is satisfied but ${\mathcal{S}}^{*}$ does not separate I^{ n } between ${I}_{i}^{+}$ and ${I}_{i}^{}$. Let A be a connected component of ${I}^{n}\backslash {\mathcal{S}}^{*}$ intersecting both ${I}_{i}^{+}$ and ${I}_{i}^{}$ in a nonempty set. Let $\mathcal{R}$ be a subcollection of $\mathcal{K}\backslash \mathcal{S}$ such that ${\mathcal{R}}^{*}$ is connected and $\mathcal{A}\subseteq {\mathcal{R}}^{*}$. Without loss of generality, we can assume that $\mathcal{R}$ is a minimal such collection. We will obtain a contradiction showing, by induction, that for any two cubes in $\mathcal{R}$ if their intersection is an mface, then m ∉ {0, . . . , n  2}. Suppose K, $L\in \mathcal{R}$ are two cubes such that K ∩ L ≠ ∅ but L ∉ Comp $\left(K,\mathcal{S}\right)$. Since K ∩ L must be an mface, for some m < n, we must have that K ∩ L is an mface with m < n  1. Suppose m = n  2, then there are exactly 2^{2}  2 other cubes sharing this mface. Let T be any of those two cubes. Then, T ∩ K and T ∩ L are (n  1)faces, K ∩ L ⊆ T and T must be in $\mathcal{S}$. A contradiction with the fact that (K ∩ L) ∩ A ≠ ∅ and therefore m ≠ n  2. Suppose we have already proved that m < n  i. We shall show that m ≠ n  (i + 1). Suppose otherwise, that is K ∩ L is an n  (i + 1)face, for some K, $L\in \mathcal{R}$. Then, there are 2^{i+1} 2 other cubes having this n  (i + 1)face in common. Let T be one of them such that T ∩ K is an (n  1)face. Then, T ∩ L is an (n  i)face and K ∩ L ⊆ T. Since (K ∩ L) ⊆ T and (K ∩ L) ∩ A ≠ ∅, therefore A ∩ T ≠ ∅ and $T\notin \mathcal{S}$. Consequently, T ∈ Comp $\left(K,\mathcal{S}\right)$ with T ∩ L an (n  i)face, which leads to a contradiction by an inductive step.
It follows that for any two K, $L\in \mathcal{R}$ we have L ∈ Comp $\left(K,\mathcal{S}\right)$. Consequently, $\mathcal{R}=\mathsf{\text{Comp(}}K\mathsf{\text{,}}\mathcal{S}\mathsf{\text{)}}$ for some K such that $K\cap {I}_{i}^{+}\ne \varnothing $ and Comp ${\left(K,\mathcal{S}\right)}^{*}\cap {I}_{i}^{}\ne \varnothing $. A contradiction that completes the proof. □
A collection of cubes in $\mathcal{K}$ and the collection of the faces of all dimensions of cubes in $\mathcal{K}$ forms a polyhedral complex with $\mathcal{K}$ as its generating collection. Denote this complex by $\mathcal{X}$.
 1.
each $C\in \mathcal{C}\left(d\right)$ contains an edge from ${\mathcal{E}}_{f}$,
 2.
$\mathcal{C}{\left(d\right)}^{*}$ is connected, and
 3.
$\mathcal{C}{\left(d\right)}^{*}$ separates I^{ n } between $x\in {I}_{1}^{+}$ and $y\in {I}_{1}^{}$ for each $x,y\in {I}^{n}\backslash \mathcal{C}{\left(d\right)}^{*}$.
Set $\mathcal{L}=\left\{\left[\frac{i}{k},\frac{i+1}{k}\right]\times \left\{0\right\}\times ...\times \left\{0\right\}:i=0,...,k1\right\}$, and notice that ${\mathcal{L}}^{*}$is a segment joining ${I}_{i}^{+}$ and ${I}_{i}^{}$. Therefore, $\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$ will be the desired collection satisfying (1)  (3) for some $d\in \mathcal{L}$.
Algorithm (outline)
Step 1. Add all elements of ${\mathcal{E}}_{f}\cap \mathcal{L}$ to List A.
Step 2. Repeat Step 3Step 11 until List A is empty.
Step 3. Pick an edge d from List A.
Step 4. Generate $\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$. Remove d from List A.
Step 5. Add all elements $K\in \mathcal{K}$ such that $K\cap {I}_{i}^{+}\ne \varnothing $ to List B.
Step 6. Repeat Step 7Step 9 until List B is empty.
Step 7. Pick a cube K from List B.
Step 8. Generate Comp (K, $\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$). Remove K from List B.
Step 9. If there is L ∈ Comp (K, $\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$) such that $L\cap {I}_{i}^{}\ne \varnothing $ then go back to Step 3.
Otherwise, go back to Step 7.
Step 10. List all elements from $\mathcal{C}\mathsf{\text{(}}d\mathsf{\text{)}}$ ($\mathcal{C}{\left(d\right)}^{*}$ is a separator).
Step 11. Go back to Step 3.
Endnote
^{a}By a map we will always mean a continuous function. Whenever continuity is not assumed we will use the term function instead.
Declarations
Acknowledgements
The authors are grateful to the referees for their careful reading of our manuscript and helpful comments that improved the paper.
Authors’ Affiliations
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