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Characterization of some convergent bivariate subdivision schemes with nonnegative masks
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Applied and Computational Harmonic Analysis www.elsevier.com/locate/acha
Letter to the Editor
Characterization of some convergent bivariate subdivision schemes with nonnegative masks Li Cheng Vocational and Technical College, Institute of Nonlinear Analysis and Department of Mathematics, Lishui University, 323000 Lishui, China
a r t i c l e
i n f o
Article history: Received 2 October 2018 Received in revised form 29 August 2019 Accepted 25 September 2019 Available online xxxx Communicated by Ming-Jun Lai MSC: 65D17 26A18 39B12 Keywords: Convergence Nonnegative mask Bivariate subdivision scheme
a b s t r a c t Knowing that the convergence of a multivariate subdivision scheme with a nonnegative mask can be characterized by whether or not some finite products of row-stochastic matrices induced by this mask have a positive column. However, the number of those products is exponential with respect to the size of matrices. For nonnegative univariate subdivision, this problem is completely solved. Thus, the convergence in this case can be checked in linear time with respect to the size of a square matrix. This paper will demonstrate the necessary and sufficient conditions for the convergence of some nonnegative bivariate subdivision schemes by means of the so-called connectivity of a square matrix, which is derived by a given mask. Moreover, the connectivity can be examined in linear time with respect to the size of this matrix. © 2019 Elsevier Inc. All rights reserved.
1. Introduction Bivariate subdivision schemes are efficient iterative procedures for generating recursively discrete functions that are defined on finer and finer grids of points in R2 . Additionally, they are useful for surface modelling in computer aided geometric design (CAGD) as well as the animation industry. Let Z2 be the integer lattice. A subdivision scheme is defined by a fixed, finitely supported, real sequence (mask) {a(i, j)} := {a(i, j) : (i, j) ∈ Z2 }. Given an initial finite sequence of data values, v 0 = {v 0 (i, j)}, a subdivision scheme with a mask {a(i, j)} defines a sequence of values v k (i, j) recursively by the rule
v k (i, j) =
v k−1 (m, n)a(i − 2m, j − 2n).
(m,n)∈Z2
E-mail addresses: [email protected], [email protected]. https://doi.org/10.1016/j.acha.2019.09.004 1063-5203/© 2019 Elsevier Inc. All rights reserved.
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This scheme is said to be convergent if, for each v 0 , there exists a continuous function f such that sup |f (
lim
k→∞ (i,j)∈Z2
i j , ) − v k (i, j)| = 0 2k 2k
and f ≡ 0 for at least one v 0 . Let H be the hat function defined by H(y) = 1 − |y|, if |y| ≤ 1 and 0 for all other y. For (x1 , x2 ) ∈ R2 we set ψ(x1 , x2 ) = H(x1 ) · H(x2 ). Using v k (i, j) we get a “polygon” f k (x1 , x2 ) = (m,n)∈Z2 v k (m, n)ψ(2k x1 − m, 2k x2 − n). Clearly, f k (m/2k , n/2k ) = v k (m, n) and therefore the convergence of the subdivision scheme is equivalent to the uniform convergence of f k . On the other hand, write a1 (i, j) = a(i, j) and ak (i, j) = k−1 (m, n)a(i − 2m, j − 2n). In particular, taking v 0 (i, j) = δ0 (i, j), where δ0 (0, 0) = 1 and m,n∈Z2 a δ0 (i, j) = 0 if (i, j) = (0, 0), one has f k (x1 , x2 ) = (m,n) ak (m, n)ψ(2k x1 − m, 2k x2 − n). Therefore, the convergence of the subdivision scheme is equivalent to the uniform convergence of
ak (m, n)ψ(2k x1 − m, 2k x2 − n).
(1.1)
(m,n)∈Z2
In this paper, when we say the subdivision scheme converges to ϕ, we mean (1.1) converges to ϕ, which is also equivalent to the scheme with δ0 (i, j) converging to ϕ. Cavaretta, Dahmen and Micchelli present a comprehensive discussion of this subject in [1]. The characterization of convergent subdivision schemes with the finitely mask can be found in [6,7,12]. Let E = {(0, 0), (1, 0), (0, 1), (1, 1)}. Theorem 1.1. A subdivision scheme associated with a fixed finitely supported real sequence (mask) {a(i, j) : (i, j) ∈ Z2 } converges if and only if
a(i + 2m, j + 2n) = 1, ∀ (i, j) ∈ Z2
(1.2)
(m,n)∈Z2
and lim
sup
k→∞ (i,j)∈Z2 , (e1 ,e2 )∈E
|ak (i, j) − ak (i − e1 , j − e2 )| = 0.
(1.3)
The first condition (1.2) is called sum rule and is easy to check. However, the second one is rather difficult to verify and can be determined by whether the so-called joint spectral radius of some square matrices is less than 1 (see [1,5,7]). However, by a result in [11] the decision problem of joint spectral radius is generally NP-hard. In 2005 the convergence of nonnegative univariate subdivision had been completely characterized (see [10,13]). The result is as follows. Theorem 1.2. Let {a(i) : i = 0, ..., N } be a nonnegative mask, which satisfies a(0), a(N ) = 0. Then, the univariate subdivision scheme associated with this mask converges if and only if 1) i a(2i) = i a(2i + 1) = 1 and 0 < a(0), a(N ) < 1, and 2) the greatest common divisor of {i : a(i) = 0} is 1, i.e. gcd{i : a(i) = 0} = 1. In this paper, we focus on the sufficient and necessary conditions for the convergence of the bivariate subdivision schemes with nonnegative masks. We begin with the following observation. For a square matrix B = {B(i, j)}1≤i,j≤N , we should define a directed graph G(Γ, K), where Γ = {1, ..., N } is the set of vertexes and the set of edges is given by K = {(i, j) : i, j ∈ Γ, B(i, j) = 0}.
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Definition 1.3. Let B = {B(i, j)}1≤i,j≤N be a square matrix and G(Γ, K) be the associated directed graph. The matrix B is connected if the directed graph G(Γ, K) is connected, i.e., for some vertex j ∈ Γ and any i ∈ Γ \ {j }, there exists a directed path from j to i. As pointed out in [14], roughly speaking, in the univariate case the convergence of subdivision schemes with nonnegative masks is equivalent to the connectivity of a square matrix. Remark 1.4. Let B 2 (m, n) = 0 for some 1 ≤ m, n ≤ N . Hence, N
B(m, j) · B(j, n) = 0.
j=1
So, there is at least one 1 ≤ τ ≤ N such that B(m, τ ) · B(τ, n) = 0. It follows from the above definition that we have edges (n, τ ) and (τ, m). In other words, in G(Γ, K) there exists a path from n to m via τ : n −→ τ −→ m. Consequently, if B p (m, n) = 0 for some p, there is a path from n to m in G(Γ, K). In particular, if B p (m, n) = 0 for all 1 ≤ m ≤ N , then B is connected. Denote the support of a mask by Ω = {(i, j) : a(i, j) = 0} and [Ω] the convex cover of Ω. Our main result of this paper is as follows: Theorem 1.5. Let the support Ω of a nonnegative mask {a(i, j)} in R2 be convex, i.e. Ω = [Ω] ∩Z2 , and satisfy the sum rule (1.2). The corresponding bivariate scheme converges if and only if the matrix A is connected, where A is a square matrix given by A((i, j), (m, n)) = a(−i + 2m, −j + 2n), (i, j), (m, n) ∈ [Ω] ∩ Z2 . The contents of the paper are as follows: Section 2 gives a brief view of some commonly used notations such as unimodular matrix, and a number of lemmas which lead to our main result, being stated as Theorem 1.5. Section 3 devotes to the proof of Theorem 1.5. The proof splits into three cases according to the difference k−1 values of (λ1 , λ2 ), where (λ1 , λ2 ) = l=0 2l (δ1,l , δ2,l ) for any k ∈ N, (δ1,0 , δ2,0 ), ..., (δ1,k−1 , δ2,k−1 ) ∈ E. In Section 4 we present a bivariate example for our main result; An algorithm to check the connectivity of matrix A is also given there. 2. Preliminaries and lemmas We begin with some notations. For a given mask {a(i, j)} let Q ⊂ Z2 be a finite quadrat satisfying [Ω] ∩ Z2 ⊆ Q. Let N = |Q| be the cardinality of the set Q. We define that for every = (1 , 2 ) ∈ E, the N × N matrix A , whose entries are given by A ((i, j), (m, n)) = a(−i + 1 + 2m, −j + 2 + 2n), (i, j), (m, n) ∈ Q. Clearly, if the mask {a(i, j)} is nonnegative and satisfies the sum rule (1.2), then A is row-stochastic matrix for all ∈ E. It is known (see e.g. [9]) that, among others, if the subdivision scheme with nonnegative mask {a(i, j)} is convergent, then for sufficient large p each product Ae1 · · · Aep has a positive column, e1 , ..., ep ∈ E. Later we will use this fact. A unimodular matrix M is a square matrix with integer entries having determinant 1 or −1. Let M2 be the set of 2 × 2 unimodular matrices, namely,
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M2 = {M : M is a 2 × 2 matrix with integer entries and | det M | = 1}. According to the properties of unimodular matrices, we note that the transformation of masks under a unimodular matrix does not affect the convergence and the divergence of the corresponding subdivision scheme. As a result, there holds Lemma 2.1. Let {a(i, j)} be a finite mask in R2 that satisfies the sum rule (1.2). Further, let b(i, j) = a((i, j)M ) for any given M ∈ M2 . Then, {b(i, j)} satisfies the sum rule. Moreover, the convergence behavior of the subdivision schemes associated with {a(i, j)} and {b(i, j)}, respectively, are the same. Proof. The mask {b(i, j)} satisfies the sum rule, because for any given M ∈ M2 and (i, j) ∈ Z2 there holds
b(i + 2m, j + 2n) =
(m,n)∈Z2
a((i + 2m, j + 2n)M ) = 1.
(m,n)∈Z 2
To show the second assertion we use induction to verify bl (i, j) = al ((i, j)M ), l = 1, 2, ... Therefore, if (1.3) in Theorem 1.1 holds for {a(i, j)}, so does {b(i, j)}, and vice versa. 2 For any given (σ, τ ) ∈ Z2 the translational mask is denoted as {b(i, j) : b(i, j) = a(i +σ, j +τ )}. Hence, the support of {b(i, j)} is Ω − (σ, τ ). We also find that the translation of a mask does not affect the convergence of the corresponding subdivision scheme, and can be described as follows. Lemma 2.2. Let {a(i, j)} be a finite mask in R2 and satisfy the sum rule (1.2). Further, let b(i, j) = a(i + σ, j + τ ) for any given (σ, τ ) ∈ Z2 . Then, {b(i, j)} satisfies the sum rule. Moreover, the convergence behavior of the subdivision schemes associated with {a(i, j)} and {b(i, j)} respectively are the same. Proof. Clearly, {b(i, j)} satisfies the sum rule. To show the second assertion we use induction to obtain l−1 l−1 bl (i, j) = al (i + i=0 2i σ, j + i=0 2i τ, ), l = 1, 2, .... Therefore, if (1.3) in Theorem 1.1 holds for {a(i, j) : α ∈ Z2 }, so does {b(i, j)}, and vice versa. 2 Next we focus on the connectivity of the matrix A (see [2]). Lemma 2.3. Let the nonnegative finite mask {a(i, j)} satisfy the sum rule (1.2). If the matrix A is connected, then for any M ∈ M2 and any finite convex set Γ ⊂ Z2 , i.e., [Γ] ∩ Z2 = Γ, such that [ΩM ] ∩ Z2 ⊆ Γ, the matrix B defined by B((i, j), (m, n)) = a(−(i, j)M −1 + 2(m, n)M −1 ), ∀ (i, j) (m, n) ∈ Γ is row-stochastic and connected with some (m, n) ∈ ([ΩM ] ∩ Z2 ) \ ∂[ΩM ]. About the connectivity of products of matrices can also be found in [8]. Recently, we established a new characterization of convergent multivariate subdivision schemes with nonnegative masks [3]. Let Γ ⊂ Z2 be finite and the direct sum Γk , k ∈ N, be defined by Γk = Γ + 2Γ + · · · + 2k−1 Γ. In the bivariate case this result is as follows: Lemma 2.4. The bivariate subdivision scheme with a nonnegative mask {a(i, j)}, whose support is Ω and that satisfies the sum rule (1.2), converges if and only if for any k ∈ N, (δ1,0 , δ2,0 ), ..., (δ1,k−1 , δ2,k−1 ) ∈ E k−1 and (λ1 , λ2 ) = l=0 2l (δ1,l , δ2,l ), the inclusion relations for any nonempty sets T and T of Q
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T − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk 2 ∩ Z ⊆ T and ∩ Z2 ⊆ T 2k 2k imply T ∩ T = ∅. In the remainder of this section, we continue to study the properties for the connectivity of the matrix A and the convexity of the support Ω. Lemma 2.5. Let the nonnegative finite mask a(i, j) satisfy the conditions of Theorem 1.5 and Ω be convex. If the matrix A is connected, then, for some suitable unimodular matrix M ∈ M2 and some integer (m, n), there holds (m, n) + E ⊆ ΩM . Proof. Denote Ω(1 , 2 ) = Ω ∩ {2Z2 + (1 , 2 )}, for (1 , 2 ) ∈ E and |Ω(1 , 2 )| to be the number of the elements in the set Ω(1 , 2 ). By Theorem 1.3 of [2], the connectivity of the matrix A implies that there are at least one inner integer in [Ω] and at most one Ω(1 , 2 ), with |Ω(1 , 2 )| = 1. Moreover, if |Ω(1 , 2 )| = 1, then the only one element of Ω(1 , 2 ) belongs to [Ω]o . In other words, this element is an inner integer in [Ω]. Hence, there are at least 7 integers in the convex cover [Ω]. With this in mind, let us prove that for some suitable unimodular matrix M ∈ M2 and some (m, n) ∈ Z2 there holds (m, n) + E ⊂ ΩM . We observe first the following simple fact: let a convex quadrilateral, whose four extreme points are integer, be given. If except those four integers there are no any integers on this quadrilateral, then this quadrilateral is parallelogram, that under some matrix M ∈ M2 is (m, n) + E for some (m, n) ∈ Z2 . Indeed, we may suppose that the four extreme points are (0, 0), a, b and c. The four edges are [(0, 0), a], [a, c], [c, b] and [(0, 0), b]. Thus the determinant of ab is ±1. Consequently, M =: ab is a matrix in M2 . So aM −1 = (1, 0) and bM −1 = (0, 1). Since the convex quadrilateral contains no any other integers we must have cM −1 = (1, 1). Hence, [(0, 0), a, b, c] = [E]M . We begin with the construction of such a convex quadrilateral. Let a, b, c, d be four elements of Ω, that belong to different sets Ω(1 , 2 ). Hence, each pair of these four elements is not congruent modulo 2. We observe the convex cover [a, b, c, d]. Clearly, [a, b, c, d] ∩ Z2 ⊂ Ω. This set is either a convex quadrilateral (whose extreme points are a, b, c, d) or a triangle, whose extreme points are three of these four elements (say a, b and c). The element d lies in [a, b, c]. We may assume that any edge of [a, b, c, d] contains only two integers, namely, the extreme points. Indeed, if, for example, the edge [a, b] contains an integer x, that is different from a and b, then x ≡ a (mod 2) or x ≡ b (mod 2). Let x ≡ a (mod 2). We replace [a, b, c, d] by [x, b, c, d]. The set [x, b, c, d] has less integers than [a, b, c, d] and [x, b, c, d] ⊂ [a, b, c, d]. Moreover, these four elements belong to different sets Ω(1 , 2 ). Under this assumption let integer y ∈ [a, b, c, d] and be different from a, b, c, d. Hence, y is congruent exact one of these four integers modulo 2 (say b). In case [a, b, c, d] is a convex quadrilateral, we replace [a, b, c, d] by [a, y, c, d]. If [a, b, c, d] is a triangle and y ≡ d (mod 2), then in the segment [y, d] there is an integer y that is not congruent d modulo 2. However, y must be congruent exact one of a, b or c modulo 2 (say c). We replace [a, b, c, d] by [a, b, y , d]. Thus, beginning with [a, b, c, d] we obtain a convex cover deduced by four elements of Ω, that belong to different sets Ω(1 , 2 ). Moreover, this new convex cover is a subset of [a, b, c, d] and contains less integers than [a, b, c, d]. Repeat the above steps we get finally a convex cover [a , b , c , d ], that is a subset of [a, b, c, d] and contains exact four integers a , b , c and d . Those four integers belong to different sets Ω(1 , 2 ). Clearly, [a , b , c , d ] ∩ Z2 ⊂ Ω. If this set is a quadrilateral, we have nothing more to do as above shows. Otherwise, [a , b , c , d ] is a triangle, whose extreme points are three of these four elements (say a , b and c ). The element d lies in [a , b , c ]. We define three half-lines L1 , L2 , L3 beginning at d through a , b , c , respectively. We remember that Ω is
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convex and contains at least 7 integers. Let x ∈ Ω be different from a , b , c , d . Moreover, x does not lie on L1 , L2 and L3 . Without loss of generality, we may assume that x lies between L1 and L2 . On the other hand, x must be congruent to one of a , b , c and d modulo 2. Put x1 :=
x + {a , b , c , d } ∩ Z2 . 2
It is easy to see that, if x1 = d , the set [a , b , x, d ] is the desired convex quadrilateral. If x1 = d , the convexity of Ω implies x1 ∈ Ω. Obviously, x1 lies also between L1 and L2 and x1 ∈ / [a , b , c ]. We replace x by x1 and repeat the above process until x1 = d . It remains to show the existence of such an x. Indeed, there are at least three integers in Ω \ {a, b , c , d }. If all integers from Ω \ {a , b , c , d } lie on L1 ∪ L2 ∪ L3 , then at most one of these three half-lines contains no any those integers because there is maximal one set Ω(1 , 2 ) with |Ω(1 , 2 )| = 1. Let L1 and L2 contain y1 , y2 ∈ Ω \ {a , b , c , d }, respectively. The convexity of Ω implies that we must have integers z1 ∈ [a , y1 ] ∩ Ω and z2 ∈ [b , y2 ] ∩ Ω such that z1 ≡ d (mod 2) and z2 ≡ d (mod 2). Clearly, (z1 + z2 )/2 is again an integer and belongs to Ω. However, (z1 + z2 )/2 ∈ / L1 ∪ L2 ∪ L3 . Therefore, we have always such an x. 2 We also have the following fact: Lemma 2.6. Let the nonnegative finite mask a(i, j) satisfy the conditions of Theorem 1.5 and Ω be convex. If the matrix A is connected, then one of the extreme points of [(m, n) + E] is an inner point of [Ω]M . Proof. From Lemma 2.5 we know [(m, n) +E] ⊆ ΩM for some M ∈ M2 . On the other hand, the connectivity of A implies [Ω]o ∩ Z2 = ∅ (see Theorem 1.3 of [2]). Hence, [Ω]o M ∩ Z2 = ∅. We may assume (m, n) = (0, 0). Let (ε1 , ε2 ) ∈ [Ω]o M ∩ Z2 . If (ε1 , ε2 ) is an extreme point of [E], we have nothing more to do. Otherwise, for all (δ1 , δ2 ) ∈ [E] and 0 ≤ t < 1 the vertex (e1 , e2 ) = (ε1 , ε2 ) + (δ1 − ε1 , δ2 − ε2 )t is an inner point of [Ω]M because [E] ⊆ [Ω]M . In case of εi = 0 or 1, i = 1, 2, we choose ei = δi = 0 or 1, respectively. Denote η = maxi,εi >1 (εi − 1)/εi if there is εi > 1, otherwise η = 0; and ζ = maxi,εi <0 |εi |/(1 + |εi |) if there is εi < 0, otherwise ζ = 0. Finally, t = max{η, ζ}. Since (ε1 , ε2 ) is not an extreme point of [E], we must have 0 < t < 1. Our (e1 , e2 ) is given by ei =
1,
εi ≥ 1,
0,
εi ≤ 0
and our (δ1 , δ2 ) by δi =
εi , εi +
εi = 0 or 1, ei −εi t ,
otherwise.
It’s easy to see that 0 ≤ δi ≤ 1, so (δ1 , δ2 ) belongs to [E]. As 0 < t < 1 and (e1 , e2 ) = (ε1 , ε2 ) + (δ1 − ε1 , δ2 − ε2 )t , the vertex (e1 , e2 ) is an inner point of [Ω]M and belongs to E. 2 3. Proof of the main result We are now in the position to prove Theorem 1.5. Proof of Theorem 1.5. At the beginning of Section 2 we have defined the matrices A for ∈ E and mentioned that for convergent subdivision scheme with a finite nonnegative mask there is p > 1 such that A1 · · · Ap has a positive column with any 1 , ..., p ∈ E (see [9] for the detail). As matrix A is a submatrix
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of A0 and row-stochastic, the product Ap must also have a positive column if the scheme is convergent. Consequently, A is connected (see Remark 1.4), which proves the necessity. To prove the sufficiency we will take advantage of Lemma 2.4. Let integer k ≥ 1 and (λ1 , λ2 ) = (1,0 , 2,0 ) + 2(1,1 , 2,1 ) + ... + 2k−1 (1,k−1 , 2,k−1 ). We have to show that (see Lemma 2.4), for subsets T and L of a finite set, L − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk 2 ∩ Z ⊆ T and ∩ Z2 ⊆ L 2k 2k
(3.1)
implies T ∩ L = ∅. Let us make some simplifications. We observe the first inclusion. Denote T1 − (λ1 , λ2 ) + Ωk Tj − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk ∩ Z2 = T1 , ∩ Z2 = T2 , ..., ∩ Z2 = Tj+1 , ... k k 2 2 2k Thus, Tj ⊆ T for all j ∈ N. Consequently, since T is a finite set there are ≥ 1 and τ ≥ 1 such that T = T+τ . In other words, with (λτ1 , λτ2 ) = (λ1 , λ2 ) + 2k (λ1 , λ2 ) + ... + 2k(τ −1) (λ1 , λ2 ) T − (λτ1 , λτ2 ) + Ωkτ ∩ Z2 = T . 2kτ Clearly, the last identity is again true if we replace τ by τ p for all p ≥ 1. Obviously, this consideration is also valid for the second inclusion of (3.1). Therefore, to finish the proof we need only to show: for subsets T, L of a finite set L − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk 2 ∩ Z = T and ∩ Z2 = L 2k 2k
(3.2)
implies T ∩ L = ∅. We divide the proof into three cases according to different values of (λ1 , λ2 ). Case 1. (λ1 , λ2 ) = (0, 0). We may write the first inclusion of (3.2) as T1 + Ω Tk−1 + Ω T +Ω ∩ Z2 = T1 , ∩ Z2 = T2 , ..., ∩ Z2 = T. 2 2 2 We remember (see [3]) that T ⊆ [Ω] ∩ Z2 . As Ω is convex, T ⊆ Ω. Now let (x1 , x2 ) ∈ T ⊆ Ω. Then ((x1 , x2 ) + (x1 , x2 ))/2 = (x1 , x2 ) ∈ T1 . Consequently, T ⊆ T1 ⊆ T2 ⊆ ... ⊆ T , i.e. T = T1 = ... = Tk−1 . We conclude that the first inclusion of (3.2) implies T +Ω ∩ Z2 = T. 2
(3.3)
On the other hand, the matrix A is connected. Hence, we have (i, j) ∈ [Ω] ∩ Z2 = Ω such that for all fixed (m, n) ∈ Ω \ {(i, j)}, there is a path from (i, j) to (m, n). In particular, for a given (m, n) ∈ T , let the path from (i, j) to (m, n) be (i, j) −→ (m1 , n1 ) −→ (m2 , n2 ) −→ ... −→ (mp−1 , np−1 ) −→ (mp , np ) −→ (m, n). By the definition of a directed graph deduced from the matrix A, we then have (μ, ν), (μ1 , ν1 ), ..., (μp , νp ) in Ω satisfying
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(m + μ, n + ν) (mp + μp , np + νp ) (m1 + μ1 , n1 + ν1 ) = (mp , np ), = (mp−1 , np−1 ), ..., = (i, j). 2 2 2 We conclude from (3.3) that (mp , np ), ..., (m1 , n1 ), (i, j) ∈ T . The same is also true for L from the second inclusion of (3.2). We obtain that (i, j) ∈ L. Thus (i, j) ∈ T ∩ L = ∅. This concludes the first case. Case 2. (λ1 , λ2 ), 0 < λ1 , λ2 < 2k − 1. Remember that Ω is convex. Lemmas 2.1-2.3 and 2.5 allow us to suppose that E ⊂ Ω. On the other hand, the inclusions of (3.2) are still valid if we replace k by k for any ≥ 1. In the following proof we may choose an that is large enough. Let (β1 , β2 ) ∈ T and (λ1 , λ2 ) = (λ1 , λ2 ) + 2k (λ1 , λ2 ) + ... + 2k(−1) (λ1 , λ2 ) for a large enough . Because 0 < λ1 , λ2 < 2k − 1, the numbers λ1 and λ2 are between (2k − 1)/(2k − 1) and (2k − 2)(2k − 1)/(2k − 1). Noticing that T is finite, we have for a large enough and some γj ∈ E ⊂ Ω (binary expression) (β1 , β2 ) = 2k (0, 0) + (λ1 , λ2 ) −
k−1
2j γj .
j=0
Since
k−1 j=0
2j γj ∈ Ωk we get (0, 0) ∈
T − (λ1 , λ2 ) + Ωk ∩ Z2 = T. 2k
Clearly, the same holds also for L. Consequently, (0, 0) ∈ T ∩ L = ∅. Case 3. (λ1 , λ2 ) = (λ1 , 0) with 0 < λ1 < 2k − 1 or (λ1 , λ2 ) = (0, λ2 ) with 0 < λ2 < 2k − 1. We prove the assertion only for (λ1 , λ2 ) = (λ1 , 0) with 0 < λ1 < 2k − 1. k−1 Write (λ1 , 0) = j=0 2j (j , 0) where j = 0 or 1. Since 0 < λ1 < 2k − 1 there exists at least one j with j = 1 and one j with j = 0. Denote for 0 ≤ τ ≤ k − 1
(λ1 (τ ), 0) =
k−1
2j (j+τ , 0),
j=0
where j+τ is cyclic defined, i.e. j+τ = j+τ if 0 ≤ j + τ ≤ k − 1 and j+τ = j+τ −k if j + τ ≥ k. Let us observe the first identity of (3.2). This identity means T1 − (1 , 0) + Ω Tk−1 − (k−1 , 0) + Ω T − (0 , 0) + Ω ∩ Z2 = T1 , ∩ Z2 = T2 , ..., ∩ Z2 = T. 2 2 2 Thus, for any 0 ≤ τ ≤ k − 1 Tτ − (λ1 (τ ), 0) + Ωk ∩ Z 2 = Tτ , 2k where T0 = T . Similarly, the second identity of (3.2) implies Lτ − (λ1 (τ ), 0) + Ωk ∩ Z2 = Lτ . 2k It is easy to see that, if for some 0 ≤ τ ≤ k − 1 there holds Tτ ∩ Lτ = ∅ then T ∩ L = ∅. We know that Ω is convex and E ⊂ Ω by Lemma 2.5. Moreover, by using Lemma 2.6 we may assume that (0, 0) is an inner integer of Ω. At first, just like in Case 2, for a fixed k and large enough ∈ N, any (β1 , β2 ) ∈ Tτ with β2 ≤ 0 can be written as
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(β1 , β2 ) = 2k (0, 0) + (λ1 (τ ), 0) −
k−1
2j γj ,
9
(3.4)
j=0
for some γj ∈ Ω and (λ1 (τ ), 0) = (λ1 (τ ), 0) + 2k (λ1 (τ ), 0) + ... + 2k(−1) (λ1 (τ ), 0). Furthermore, any (β1 , β2 ) ∈ Tτ with β2 ≥ 1 can be written as (β1 , β2 ) = 2k (0, 1) + (λ1 (τ ), 0) −
k−1
2j γj .
(3.5)
j=0
To continue our discussion let us note that Ω is convex, E ⊂ Ω, and (0, 0) is an inner point of Ω. Thus, if (0, −p) with some p ≥ 1 belongs to Ω, then (0, −1) is in Ω. In this case we choose τ such that k−1+τ = 0. With this in mind we obtain (0, 1) = 2(0, 0) + (k−1+τ , 0) − (0, −1). Clearly, (3.5) is still valid if we replace k by k − 1, i.e. for some γj ∈ Ω and large enough (β1 , β2 ) = 2k−1 (0, 1) + (λ1 (τ ), 0) −
k−2
2j γj .
j=0 It follows from the last two displays that with γj = γj for j ≤ k − 2 and γk−1 = (0, −1)
(β1 , β2 ) = 2k−1 (0, 1) + (λ1 (τ ), 0) −
k−2
2j γj
j=0
= 2k (0, 0) + (λ1 (τ ), 0) −
k−1
2j γj .
j=0
Consequently, we obtain from (3.4) and the last display, that (3.4) is true for all (β1 , β2 ) ∈ Tτ . As k−1 j k we conclude from (3.4) that j=0 2 γj ∈ Ω (0, 0) ∈
Tτ − (λ1 (τ ), 0) + Ωk ∩ Z 2 = Tτ . 2k
The same holds also for Lτ . Hence, (0, 0) ∈ Tτ ∩ Lτ = ∅, which implies T ∩ L = ∅ as mentioned above. To finish the proof for Case 3 we have to address the situation where any integer of the form (0, −p), p ≥ 1, does not belong to Ω. Since Ω is convex, E ⊂ Ω, and (0, 0) is an inner integer of Ω, there is (−p, −q) or (p, −q) with p, q ≥ 1 in Ω. The both cases can be treated in the same way. We observe below the case (−p, −q). We may express (−p, −q) in binary form. Thus, for some δi ∈ E we have (−p, −q) = −
ν
2i δi , δν = 0.
i=0
ν However, as E ⊂ Ω and Ω is convex, we obtain {(−p, −q) + δ0 }/2 ∈ Ω. Hence, − i=1 2i−1 δi ∈ Ω. We ν conclude recursively αη := − i=η 2i−η δi ∈ Ω for η = 0, 1, ..., ν. Because (0, −p) ∈ / Ω for all p ≥ 1, the first
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coordinate of αη cannot be zero. On the other hand, there exists an η such that the second coordinate of αη equals −1. So αη = (−α, −1) for some α ≥ 1. If α is even, we have (0, 1) = 2(−t, 0) + (0, 0) − (−α, −1) with (−t, 0) ∈ Ω and t ≥ 1. Choosing τ such that τ = 0 we obtain (see (3.4)) for some γj ∈ Ω (−t, 0) = 2k−1 (0, 0) + (λ1 (τ )/2, 0) −
k−2
2j γj .
j=0
Therefore, for some other γj ∈ Ω (0, 1) = 2k (0, 0) + (λ1 (τ ), 0) −
k−1
2j γj .
j=0
Consequently, by (3.5) any (β1 , β2 ) ∈ Tτ with β2 ≥ 1 can be expressed as (β1 , β2 ) = 22k (0, 0) + (λ1 (τ ), 0) + 2k (λ1 (τ ), 0) −
2k−1
2j γj .
j=0
Combining this with (3.4) we get that the above holds for all (β1 , β2 ) ∈ T . In other words, (0, 0) ∈
Tτ − (λ1 (τ ), 0) − 2k (λ1 (τ ), 0) + Ω2k = Tτ . 22k
This relation is also true for Lτ . Thus, (0, 0) ∈ Tτ ∩ Lτ = ∅ or T ∩ L = ∅. If α is odd, we have (0, 1) = 2(−t, 0) + (1, 0) − (−α, −1) with (−t, 0) ∈ Ω and t ≥ 1. We choose τ such that τ = 1. Repeating the above process we obtain again T ∩ L = ∅. The proof for Case 3 is complete. There are other situations of (λ1 , λ2 ) that we have not treated, namely, (λ1 , λ2 ) where at least one coordinate is 2k − 1. However, the proof for these situations can be carried out from Cases 1-3. For example, if (λ1 , λ2 ) = (2k − 1, λ2 ) with 0 < λ2 < 2k − 1, then with T = T + (1, 0) we have T − (λ1 , λ2 ) + Ωk ∩ Z2 ⊆ T 2k Therefore, the proof follows from Case 3.
⇔
T − (0, λ1 ) + Ωk ∩ Z2 ⊆ T . 2k
2
4. Example and algorithm Let X = {x1 , ..., xκ } ⊂ Z2 \{(0, 0)} and κ ≥ 2. The zonotope generated by the vectors in X is defined by Z(X) := {
κ
vi xi : 0 ≤ vi ≤ 1}.
i=1
Corollary 4.1. Let X = {x1 , ..., xκ } ⊂ Z2 \{(0, 0)} with κ ≥ 2. Let the support Ω of a nonnegative mask {a(i, j)} be Z(X) ∩ Z2 and satisfy the sum rule (1.2). The corresponding bivariate scheme converges if and only if the matrix A is connected. The following algorithm tells us that the complexity to check the connectivity of matrix A from Theorem 1.5 is O(|Ω|2 ).
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Algorithm CHECK(B): (1) build the directed graph G(Γ, K) from B; (2) calculate all strongly connected components of G(Γ, K), say Γ1 , Γ2 , ..., Γm ; (3) build a new directed graph G (V, E), where V = {1, ..., m} and E = {(i, j) | i, j ∈ V, if there is an edge of K from Γi to Γj }; (4) if there are at least two vertexes of V whose in-degree is zero, return false, otherwise return true. A strongly connected component of a directed graph G(Γ, K) is the maximal set of vertexes Γ ⊆ Γ such that every pair of vertexes u and v in Γ are reachable from each other. The in-degree of a vertex is the number of edges which is incident to this vertex. Let the size of B be N × N , so to build G(Γ, K) in form of adjacency list one needs the complexity O(N 2 ). To find all strongly connected components of G(Γ, K) and to construct the new graph G (V, E) one needs O(|Γ| + |K|) = O(N 2 ) as shown in [4] (see Section 23 of [4]). Finally, finding the vertex in G (V, E) with zero in-degree costs O(|Γ|). Thus, the complexity of CHECK(B) is O(N 2 ). The matrix B is connected if and only if the output of CHECK(B) is true. Indeed, if G (V, E) has more than one vertex with zero in-degree then G(Γ, K) cannot be connected. To see this let a and b two vertexes of G (V, E). So there is no path in G (V, E) for a and b, which in turn implies that there are no paths among the corresponding strongly connected components of G(Γ, K). On the other hand, as G (V, E) is acyclic so there is at least one vertex with in-degree zero. If there is only one vertex v ∈ V with zero in-degree then for any u ∈ V \ {v} one can always find a path from v to u. Assume the strongly component corresponding to v is Γ1 , hence, any vertex from Γj , j = 1 can be reached from the vertexes of Γ1 , in particular from a vertex of Γ1 . In other words, B is connected. Acknowledgments The author is supported partly by National Natural Science Foundation of China No. 11701246 and Scientific Research Foundation of the First-Class Discipline of Zhejiang Province (B) No. 201601. References [1] A.S. Cavaretta, W. Dahmen, C.A. Micchelli, Stationary subdivision, Mem. Amer. Math. Soc. 453 (1991) 1–186. [2] L. Cheng, X. Zhou, Necessary conditions for the convergence of subdivision schemes with finite masks, Appl. Math. Comput. 303 (2017) 34–41. [3] L. Cheng, X. Zhou, A new characterization of convergent multivariate subdivision schemes with nonnegative masks, Arch. Math. 108 (2017) 197–207. [4] T.H. Cormen, C.E. Leiserson, R.L. Rivest, Introduction to Algorithms, MIT Press, 1990. [5] I. Daubechies, J.C. Lagarias, Two-scale difference equations I. Existence and global regularity of solutions, SIAM J. Math. Anal. 22 (1991) 1388–1410. [6] T.N.T. Goodman, C.M. Micchelli, J. Ward, Spectral radius formulas for subdivision operators, in: L.L. Schumaker, G. Webb (Eds.), Recent Advances in Wavelet Analysis, Academic Press Inc., 1994, pp. 335–360. [7] B. Han, R.Q. Jia, Multivariate refinement equations and convergence of subdivision schemes, SIAM J. Math. Anal. 29 (1998) 1177–1199. [8] K. Jetter, X. Li, SIA matrices and non-negative subdivision, Results Math. 62 (2012) 355–375. [9] R.Q. Jia, D.X. Zhou, Convergence of subdivision schemes associated with nonnegative masks, SIAM J. Matrix Anal. Appl. 21 (1999) 418–430. [10] A.A. Melkman, Subdivision schemes with non-negative masks always converge - unless they obviously cannot?, Ann. Numer. Math. 4 (1997) 451–460. [11] J.N. Tsitsiklis, V.D. Blondel, The Lyapunov exponent and joint spectral radius of pairs of matrices are hard, when not impossible, to compute and to approximate, Math. Control Signals Systems 10 (1997) 31–41. [12] X. Zhou, Characterization of convergent subdivision schemes, Approx. Theory Appl. 14 (1998) 11–24. [13] X. Zhou, Subdivision schemes with nonnegative masks, Math. Comp. 74 (2005) 819–839. [14] X. Zhou, Positivity of refinable functions defined by nonnegative finite masks, Appl. Comput. Harmon. Anal. 27 (2009) 133–156.
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Contents lists available at ScienceDirect
Applied and Computational Harmonic Analysis www.elsevier.com/locate/acha
Letter to the Editor
Characterization of some convergent bivariate subdivision schemes with nonnegative masks Li Cheng Vocational and Technical College, Institute of Nonlinear Analysis and Department of Mathematics, Lishui University, 323000 Lishui, China
a r t i c l e
i n f o
Article history: Received 2 October 2018 Received in revised form 29 August 2019 Accepted 25 September 2019 Available online xxxx Communicated by Ming-Jun Lai MSC: 65D17 26A18 39B12 Keywords: Convergence Nonnegative mask Bivariate subdivision scheme
a b s t r a c t Knowing that the convergence of a multivariate subdivision scheme with a nonnegative mask can be characterized by whether or not some finite products of row-stochastic matrices induced by this mask have a positive column. However, the number of those products is exponential with respect to the size of matrices. For nonnegative univariate subdivision, this problem is completely solved. Thus, the convergence in this case can be checked in linear time with respect to the size of a square matrix. This paper will demonstrate the necessary and sufficient conditions for the convergence of some nonnegative bivariate subdivision schemes by means of the so-called connectivity of a square matrix, which is derived by a given mask. Moreover, the connectivity can be examined in linear time with respect to the size of this matrix. © 2019 Elsevier Inc. All rights reserved.
1. Introduction Bivariate subdivision schemes are efficient iterative procedures for generating recursively discrete functions that are defined on finer and finer grids of points in R2 . Additionally, they are useful for surface modelling in computer aided geometric design (CAGD) as well as the animation industry. Let Z2 be the integer lattice. A subdivision scheme is defined by a fixed, finitely supported, real sequence (mask) {a(i, j)} := {a(i, j) : (i, j) ∈ Z2 }. Given an initial finite sequence of data values, v 0 = {v 0 (i, j)}, a subdivision scheme with a mask {a(i, j)} defines a sequence of values v k (i, j) recursively by the rule
v k (i, j) =
v k−1 (m, n)a(i − 2m, j − 2n).
(m,n)∈Z2
E-mail addresses: [email protected], [email protected]. https://doi.org/10.1016/j.acha.2019.09.004 1063-5203/© 2019 Elsevier Inc. All rights reserved.
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This scheme is said to be convergent if, for each v 0 , there exists a continuous function f such that sup |f (
lim
k→∞ (i,j)∈Z2
i j , ) − v k (i, j)| = 0 2k 2k
and f ≡ 0 for at least one v 0 . Let H be the hat function defined by H(y) = 1 − |y|, if |y| ≤ 1 and 0 for all other y. For (x1 , x2 ) ∈ R2 we set ψ(x1 , x2 ) = H(x1 ) · H(x2 ). Using v k (i, j) we get a “polygon” f k (x1 , x2 ) = (m,n)∈Z2 v k (m, n)ψ(2k x1 − m, 2k x2 − n). Clearly, f k (m/2k , n/2k ) = v k (m, n) and therefore the convergence of the subdivision scheme is equivalent to the uniform convergence of f k . On the other hand, write a1 (i, j) = a(i, j) and ak (i, j) = k−1 (m, n)a(i − 2m, j − 2n). In particular, taking v 0 (i, j) = δ0 (i, j), where δ0 (0, 0) = 1 and m,n∈Z2 a δ0 (i, j) = 0 if (i, j) = (0, 0), one has f k (x1 , x2 ) = (m,n) ak (m, n)ψ(2k x1 − m, 2k x2 − n). Therefore, the convergence of the subdivision scheme is equivalent to the uniform convergence of
ak (m, n)ψ(2k x1 − m, 2k x2 − n).
(1.1)
(m,n)∈Z2
In this paper, when we say the subdivision scheme converges to ϕ, we mean (1.1) converges to ϕ, which is also equivalent to the scheme with δ0 (i, j) converging to ϕ. Cavaretta, Dahmen and Micchelli present a comprehensive discussion of this subject in [1]. The characterization of convergent subdivision schemes with the finitely mask can be found in [6,7,12]. Let E = {(0, 0), (1, 0), (0, 1), (1, 1)}. Theorem 1.1. A subdivision scheme associated with a fixed finitely supported real sequence (mask) {a(i, j) : (i, j) ∈ Z2 } converges if and only if
a(i + 2m, j + 2n) = 1, ∀ (i, j) ∈ Z2
(1.2)
(m,n)∈Z2
and lim
sup
k→∞ (i,j)∈Z2 , (e1 ,e2 )∈E
|ak (i, j) − ak (i − e1 , j − e2 )| = 0.
(1.3)
The first condition (1.2) is called sum rule and is easy to check. However, the second one is rather difficult to verify and can be determined by whether the so-called joint spectral radius of some square matrices is less than 1 (see [1,5,7]). However, by a result in [11] the decision problem of joint spectral radius is generally NP-hard. In 2005 the convergence of nonnegative univariate subdivision had been completely characterized (see [10,13]). The result is as follows. Theorem 1.2. Let {a(i) : i = 0, ..., N } be a nonnegative mask, which satisfies a(0), a(N ) = 0. Then, the univariate subdivision scheme associated with this mask converges if and only if 1) i a(2i) = i a(2i + 1) = 1 and 0 < a(0), a(N ) < 1, and 2) the greatest common divisor of {i : a(i) = 0} is 1, i.e. gcd{i : a(i) = 0} = 1. In this paper, we focus on the sufficient and necessary conditions for the convergence of the bivariate subdivision schemes with nonnegative masks. We begin with the following observation. For a square matrix B = {B(i, j)}1≤i,j≤N , we should define a directed graph G(Γ, K), where Γ = {1, ..., N } is the set of vertexes and the set of edges is given by K = {(i, j) : i, j ∈ Γ, B(i, j) = 0}.
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Definition 1.3. Let B = {B(i, j)}1≤i,j≤N be a square matrix and G(Γ, K) be the associated directed graph. The matrix B is connected if the directed graph G(Γ, K) is connected, i.e., for some vertex j ∈ Γ and any i ∈ Γ \ {j }, there exists a directed path from j to i. As pointed out in [14], roughly speaking, in the univariate case the convergence of subdivision schemes with nonnegative masks is equivalent to the connectivity of a square matrix. Remark 1.4. Let B 2 (m, n) = 0 for some 1 ≤ m, n ≤ N . Hence, N
B(m, j) · B(j, n) = 0.
j=1
So, there is at least one 1 ≤ τ ≤ N such that B(m, τ ) · B(τ, n) = 0. It follows from the above definition that we have edges (n, τ ) and (τ, m). In other words, in G(Γ, K) there exists a path from n to m via τ : n −→ τ −→ m. Consequently, if B p (m, n) = 0 for some p, there is a path from n to m in G(Γ, K). In particular, if B p (m, n) = 0 for all 1 ≤ m ≤ N , then B is connected. Denote the support of a mask by Ω = {(i, j) : a(i, j) = 0} and [Ω] the convex cover of Ω. Our main result of this paper is as follows: Theorem 1.5. Let the support Ω of a nonnegative mask {a(i, j)} in R2 be convex, i.e. Ω = [Ω] ∩Z2 , and satisfy the sum rule (1.2). The corresponding bivariate scheme converges if and only if the matrix A is connected, where A is a square matrix given by A((i, j), (m, n)) = a(−i + 2m, −j + 2n), (i, j), (m, n) ∈ [Ω] ∩ Z2 . The contents of the paper are as follows: Section 2 gives a brief view of some commonly used notations such as unimodular matrix, and a number of lemmas which lead to our main result, being stated as Theorem 1.5. Section 3 devotes to the proof of Theorem 1.5. The proof splits into three cases according to the difference k−1 values of (λ1 , λ2 ), where (λ1 , λ2 ) = l=0 2l (δ1,l , δ2,l ) for any k ∈ N, (δ1,0 , δ2,0 ), ..., (δ1,k−1 , δ2,k−1 ) ∈ E. In Section 4 we present a bivariate example for our main result; An algorithm to check the connectivity of matrix A is also given there. 2. Preliminaries and lemmas We begin with some notations. For a given mask {a(i, j)} let Q ⊂ Z2 be a finite quadrat satisfying [Ω] ∩ Z2 ⊆ Q. Let N = |Q| be the cardinality of the set Q. We define that for every = (1 , 2 ) ∈ E, the N × N matrix A , whose entries are given by A ((i, j), (m, n)) = a(−i + 1 + 2m, −j + 2 + 2n), (i, j), (m, n) ∈ Q. Clearly, if the mask {a(i, j)} is nonnegative and satisfies the sum rule (1.2), then A is row-stochastic matrix for all ∈ E. It is known (see e.g. [9]) that, among others, if the subdivision scheme with nonnegative mask {a(i, j)} is convergent, then for sufficient large p each product Ae1 · · · Aep has a positive column, e1 , ..., ep ∈ E. Later we will use this fact. A unimodular matrix M is a square matrix with integer entries having determinant 1 or −1. Let M2 be the set of 2 × 2 unimodular matrices, namely,
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M2 = {M : M is a 2 × 2 matrix with integer entries and | det M | = 1}. According to the properties of unimodular matrices, we note that the transformation of masks under a unimodular matrix does not affect the convergence and the divergence of the corresponding subdivision scheme. As a result, there holds Lemma 2.1. Let {a(i, j)} be a finite mask in R2 that satisfies the sum rule (1.2). Further, let b(i, j) = a((i, j)M ) for any given M ∈ M2 . Then, {b(i, j)} satisfies the sum rule. Moreover, the convergence behavior of the subdivision schemes associated with {a(i, j)} and {b(i, j)}, respectively, are the same. Proof. The mask {b(i, j)} satisfies the sum rule, because for any given M ∈ M2 and (i, j) ∈ Z2 there holds
b(i + 2m, j + 2n) =
(m,n)∈Z2
a((i + 2m, j + 2n)M ) = 1.
(m,n)∈Z 2
To show the second assertion we use induction to verify bl (i, j) = al ((i, j)M ), l = 1, 2, ... Therefore, if (1.3) in Theorem 1.1 holds for {a(i, j)}, so does {b(i, j)}, and vice versa. 2 For any given (σ, τ ) ∈ Z2 the translational mask is denoted as {b(i, j) : b(i, j) = a(i +σ, j +τ )}. Hence, the support of {b(i, j)} is Ω − (σ, τ ). We also find that the translation of a mask does not affect the convergence of the corresponding subdivision scheme, and can be described as follows. Lemma 2.2. Let {a(i, j)} be a finite mask in R2 and satisfy the sum rule (1.2). Further, let b(i, j) = a(i + σ, j + τ ) for any given (σ, τ ) ∈ Z2 . Then, {b(i, j)} satisfies the sum rule. Moreover, the convergence behavior of the subdivision schemes associated with {a(i, j)} and {b(i, j)} respectively are the same. Proof. Clearly, {b(i, j)} satisfies the sum rule. To show the second assertion we use induction to obtain l−1 l−1 bl (i, j) = al (i + i=0 2i σ, j + i=0 2i τ, ), l = 1, 2, .... Therefore, if (1.3) in Theorem 1.1 holds for {a(i, j) : α ∈ Z2 }, so does {b(i, j)}, and vice versa. 2 Next we focus on the connectivity of the matrix A (see [2]). Lemma 2.3. Let the nonnegative finite mask {a(i, j)} satisfy the sum rule (1.2). If the matrix A is connected, then for any M ∈ M2 and any finite convex set Γ ⊂ Z2 , i.e., [Γ] ∩ Z2 = Γ, such that [ΩM ] ∩ Z2 ⊆ Γ, the matrix B defined by B((i, j), (m, n)) = a(−(i, j)M −1 + 2(m, n)M −1 ), ∀ (i, j) (m, n) ∈ Γ is row-stochastic and connected with some (m, n) ∈ ([ΩM ] ∩ Z2 ) \ ∂[ΩM ]. About the connectivity of products of matrices can also be found in [8]. Recently, we established a new characterization of convergent multivariate subdivision schemes with nonnegative masks [3]. Let Γ ⊂ Z2 be finite and the direct sum Γk , k ∈ N, be defined by Γk = Γ + 2Γ + · · · + 2k−1 Γ. In the bivariate case this result is as follows: Lemma 2.4. The bivariate subdivision scheme with a nonnegative mask {a(i, j)}, whose support is Ω and that satisfies the sum rule (1.2), converges if and only if for any k ∈ N, (δ1,0 , δ2,0 ), ..., (δ1,k−1 , δ2,k−1 ) ∈ E k−1 and (λ1 , λ2 ) = l=0 2l (δ1,l , δ2,l ), the inclusion relations for any nonempty sets T and T of Q
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T − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk 2 ∩ Z ⊆ T and ∩ Z2 ⊆ T 2k 2k imply T ∩ T = ∅. In the remainder of this section, we continue to study the properties for the connectivity of the matrix A and the convexity of the support Ω. Lemma 2.5. Let the nonnegative finite mask a(i, j) satisfy the conditions of Theorem 1.5 and Ω be convex. If the matrix A is connected, then, for some suitable unimodular matrix M ∈ M2 and some integer (m, n), there holds (m, n) + E ⊆ ΩM . Proof. Denote Ω(1 , 2 ) = Ω ∩ {2Z2 + (1 , 2 )}, for (1 , 2 ) ∈ E and |Ω(1 , 2 )| to be the number of the elements in the set Ω(1 , 2 ). By Theorem 1.3 of [2], the connectivity of the matrix A implies that there are at least one inner integer in [Ω] and at most one Ω(1 , 2 ), with |Ω(1 , 2 )| = 1. Moreover, if |Ω(1 , 2 )| = 1, then the only one element of Ω(1 , 2 ) belongs to [Ω]o . In other words, this element is an inner integer in [Ω]. Hence, there are at least 7 integers in the convex cover [Ω]. With this in mind, let us prove that for some suitable unimodular matrix M ∈ M2 and some (m, n) ∈ Z2 there holds (m, n) + E ⊂ ΩM . We observe first the following simple fact: let a convex quadrilateral, whose four extreme points are integer, be given. If except those four integers there are no any integers on this quadrilateral, then this quadrilateral is parallelogram, that under some matrix M ∈ M2 is (m, n) + E for some (m, n) ∈ Z2 . Indeed, we may suppose that the four extreme points are (0, 0), a, b and c. The four edges are [(0, 0), a], [a, c], [c, b] and [(0, 0), b]. Thus the determinant of ab is ±1. Consequently, M =: ab is a matrix in M2 . So aM −1 = (1, 0) and bM −1 = (0, 1). Since the convex quadrilateral contains no any other integers we must have cM −1 = (1, 1). Hence, [(0, 0), a, b, c] = [E]M . We begin with the construction of such a convex quadrilateral. Let a, b, c, d be four elements of Ω, that belong to different sets Ω(1 , 2 ). Hence, each pair of these four elements is not congruent modulo 2. We observe the convex cover [a, b, c, d]. Clearly, [a, b, c, d] ∩ Z2 ⊂ Ω. This set is either a convex quadrilateral (whose extreme points are a, b, c, d) or a triangle, whose extreme points are three of these four elements (say a, b and c). The element d lies in [a, b, c]. We may assume that any edge of [a, b, c, d] contains only two integers, namely, the extreme points. Indeed, if, for example, the edge [a, b] contains an integer x, that is different from a and b, then x ≡ a (mod 2) or x ≡ b (mod 2). Let x ≡ a (mod 2). We replace [a, b, c, d] by [x, b, c, d]. The set [x, b, c, d] has less integers than [a, b, c, d] and [x, b, c, d] ⊂ [a, b, c, d]. Moreover, these four elements belong to different sets Ω(1 , 2 ). Under this assumption let integer y ∈ [a, b, c, d] and be different from a, b, c, d. Hence, y is congruent exact one of these four integers modulo 2 (say b). In case [a, b, c, d] is a convex quadrilateral, we replace [a, b, c, d] by [a, y, c, d]. If [a, b, c, d] is a triangle and y ≡ d (mod 2), then in the segment [y, d] there is an integer y that is not congruent d modulo 2. However, y must be congruent exact one of a, b or c modulo 2 (say c). We replace [a, b, c, d] by [a, b, y , d]. Thus, beginning with [a, b, c, d] we obtain a convex cover deduced by four elements of Ω, that belong to different sets Ω(1 , 2 ). Moreover, this new convex cover is a subset of [a, b, c, d] and contains less integers than [a, b, c, d]. Repeat the above steps we get finally a convex cover [a , b , c , d ], that is a subset of [a, b, c, d] and contains exact four integers a , b , c and d . Those four integers belong to different sets Ω(1 , 2 ). Clearly, [a , b , c , d ] ∩ Z2 ⊂ Ω. If this set is a quadrilateral, we have nothing more to do as above shows. Otherwise, [a , b , c , d ] is a triangle, whose extreme points are three of these four elements (say a , b and c ). The element d lies in [a , b , c ]. We define three half-lines L1 , L2 , L3 beginning at d through a , b , c , respectively. We remember that Ω is
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convex and contains at least 7 integers. Let x ∈ Ω be different from a , b , c , d . Moreover, x does not lie on L1 , L2 and L3 . Without loss of generality, we may assume that x lies between L1 and L2 . On the other hand, x must be congruent to one of a , b , c and d modulo 2. Put x1 :=
x + {a , b , c , d } ∩ Z2 . 2
It is easy to see that, if x1 = d , the set [a , b , x, d ] is the desired convex quadrilateral. If x1 = d , the convexity of Ω implies x1 ∈ Ω. Obviously, x1 lies also between L1 and L2 and x1 ∈ / [a , b , c ]. We replace x by x1 and repeat the above process until x1 = d . It remains to show the existence of such an x. Indeed, there are at least three integers in Ω \ {a, b , c , d }. If all integers from Ω \ {a , b , c , d } lie on L1 ∪ L2 ∪ L3 , then at most one of these three half-lines contains no any those integers because there is maximal one set Ω(1 , 2 ) with |Ω(1 , 2 )| = 1. Let L1 and L2 contain y1 , y2 ∈ Ω \ {a , b , c , d }, respectively. The convexity of Ω implies that we must have integers z1 ∈ [a , y1 ] ∩ Ω and z2 ∈ [b , y2 ] ∩ Ω such that z1 ≡ d (mod 2) and z2 ≡ d (mod 2). Clearly, (z1 + z2 )/2 is again an integer and belongs to Ω. However, (z1 + z2 )/2 ∈ / L1 ∪ L2 ∪ L3 . Therefore, we have always such an x. 2 We also have the following fact: Lemma 2.6. Let the nonnegative finite mask a(i, j) satisfy the conditions of Theorem 1.5 and Ω be convex. If the matrix A is connected, then one of the extreme points of [(m, n) + E] is an inner point of [Ω]M . Proof. From Lemma 2.5 we know [(m, n) +E] ⊆ ΩM for some M ∈ M2 . On the other hand, the connectivity of A implies [Ω]o ∩ Z2 = ∅ (see Theorem 1.3 of [2]). Hence, [Ω]o M ∩ Z2 = ∅. We may assume (m, n) = (0, 0). Let (ε1 , ε2 ) ∈ [Ω]o M ∩ Z2 . If (ε1 , ε2 ) is an extreme point of [E], we have nothing more to do. Otherwise, for all (δ1 , δ2 ) ∈ [E] and 0 ≤ t < 1 the vertex (e1 , e2 ) = (ε1 , ε2 ) + (δ1 − ε1 , δ2 − ε2 )t is an inner point of [Ω]M because [E] ⊆ [Ω]M . In case of εi = 0 or 1, i = 1, 2, we choose ei = δi = 0 or 1, respectively. Denote η = maxi,εi >1 (εi − 1)/εi if there is εi > 1, otherwise η = 0; and ζ = maxi,εi <0 |εi |/(1 + |εi |) if there is εi < 0, otherwise ζ = 0. Finally, t = max{η, ζ}. Since (ε1 , ε2 ) is not an extreme point of [E], we must have 0 < t < 1. Our (e1 , e2 ) is given by ei =
1,
εi ≥ 1,
0,
εi ≤ 0
and our (δ1 , δ2 ) by δi =
εi , εi +
εi = 0 or 1, ei −εi t ,
otherwise.
It’s easy to see that 0 ≤ δi ≤ 1, so (δ1 , δ2 ) belongs to [E]. As 0 < t < 1 and (e1 , e2 ) = (ε1 , ε2 ) + (δ1 − ε1 , δ2 − ε2 )t , the vertex (e1 , e2 ) is an inner point of [Ω]M and belongs to E. 2 3. Proof of the main result We are now in the position to prove Theorem 1.5. Proof of Theorem 1.5. At the beginning of Section 2 we have defined the matrices A for ∈ E and mentioned that for convergent subdivision scheme with a finite nonnegative mask there is p > 1 such that A1 · · · Ap has a positive column with any 1 , ..., p ∈ E (see [9] for the detail). As matrix A is a submatrix
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of A0 and row-stochastic, the product Ap must also have a positive column if the scheme is convergent. Consequently, A is connected (see Remark 1.4), which proves the necessity. To prove the sufficiency we will take advantage of Lemma 2.4. Let integer k ≥ 1 and (λ1 , λ2 ) = (1,0 , 2,0 ) + 2(1,1 , 2,1 ) + ... + 2k−1 (1,k−1 , 2,k−1 ). We have to show that (see Lemma 2.4), for subsets T and L of a finite set, L − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk 2 ∩ Z ⊆ T and ∩ Z2 ⊆ L 2k 2k
(3.1)
implies T ∩ L = ∅. Let us make some simplifications. We observe the first inclusion. Denote T1 − (λ1 , λ2 ) + Ωk Tj − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk ∩ Z2 = T1 , ∩ Z2 = T2 , ..., ∩ Z2 = Tj+1 , ... k k 2 2 2k Thus, Tj ⊆ T for all j ∈ N. Consequently, since T is a finite set there are ≥ 1 and τ ≥ 1 such that T = T+τ . In other words, with (λτ1 , λτ2 ) = (λ1 , λ2 ) + 2k (λ1 , λ2 ) + ... + 2k(τ −1) (λ1 , λ2 ) T − (λτ1 , λτ2 ) + Ωkτ ∩ Z2 = T . 2kτ Clearly, the last identity is again true if we replace τ by τ p for all p ≥ 1. Obviously, this consideration is also valid for the second inclusion of (3.1). Therefore, to finish the proof we need only to show: for subsets T, L of a finite set L − (λ1 , λ2 ) + Ωk T − (λ1 , λ2 ) + Ωk 2 ∩ Z = T and ∩ Z2 = L 2k 2k
(3.2)
implies T ∩ L = ∅. We divide the proof into three cases according to different values of (λ1 , λ2 ). Case 1. (λ1 , λ2 ) = (0, 0). We may write the first inclusion of (3.2) as T1 + Ω Tk−1 + Ω T +Ω ∩ Z2 = T1 , ∩ Z2 = T2 , ..., ∩ Z2 = T. 2 2 2 We remember (see [3]) that T ⊆ [Ω] ∩ Z2 . As Ω is convex, T ⊆ Ω. Now let (x1 , x2 ) ∈ T ⊆ Ω. Then ((x1 , x2 ) + (x1 , x2 ))/2 = (x1 , x2 ) ∈ T1 . Consequently, T ⊆ T1 ⊆ T2 ⊆ ... ⊆ T , i.e. T = T1 = ... = Tk−1 . We conclude that the first inclusion of (3.2) implies T +Ω ∩ Z2 = T. 2
(3.3)
On the other hand, the matrix A is connected. Hence, we have (i, j) ∈ [Ω] ∩ Z2 = Ω such that for all fixed (m, n) ∈ Ω \ {(i, j)}, there is a path from (i, j) to (m, n). In particular, for a given (m, n) ∈ T , let the path from (i, j) to (m, n) be (i, j) −→ (m1 , n1 ) −→ (m2 , n2 ) −→ ... −→ (mp−1 , np−1 ) −→ (mp , np ) −→ (m, n). By the definition of a directed graph deduced from the matrix A, we then have (μ, ν), (μ1 , ν1 ), ..., (μp , νp ) in Ω satisfying
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(m + μ, n + ν) (mp + μp , np + νp ) (m1 + μ1 , n1 + ν1 ) = (mp , np ), = (mp−1 , np−1 ), ..., = (i, j). 2 2 2 We conclude from (3.3) that (mp , np ), ..., (m1 , n1 ), (i, j) ∈ T . The same is also true for L from the second inclusion of (3.2). We obtain that (i, j) ∈ L. Thus (i, j) ∈ T ∩ L = ∅. This concludes the first case. Case 2. (λ1 , λ2 ), 0 < λ1 , λ2 < 2k − 1. Remember that Ω is convex. Lemmas 2.1-2.3 and 2.5 allow us to suppose that E ⊂ Ω. On the other hand, the inclusions of (3.2) are still valid if we replace k by k for any ≥ 1. In the following proof we may choose an that is large enough. Let (β1 , β2 ) ∈ T and (λ1 , λ2 ) = (λ1 , λ2 ) + 2k (λ1 , λ2 ) + ... + 2k(−1) (λ1 , λ2 ) for a large enough . Because 0 < λ1 , λ2 < 2k − 1, the numbers λ1 and λ2 are between (2k − 1)/(2k − 1) and (2k − 2)(2k − 1)/(2k − 1). Noticing that T is finite, we have for a large enough and some γj ∈ E ⊂ Ω (binary expression) (β1 , β2 ) = 2k (0, 0) + (λ1 , λ2 ) −
k−1
2j γj .
j=0
Since
k−1 j=0
2j γj ∈ Ωk we get (0, 0) ∈
T − (λ1 , λ2 ) + Ωk ∩ Z2 = T. 2k
Clearly, the same holds also for L. Consequently, (0, 0) ∈ T ∩ L = ∅. Case 3. (λ1 , λ2 ) = (λ1 , 0) with 0 < λ1 < 2k − 1 or (λ1 , λ2 ) = (0, λ2 ) with 0 < λ2 < 2k − 1. We prove the assertion only for (λ1 , λ2 ) = (λ1 , 0) with 0 < λ1 < 2k − 1. k−1 Write (λ1 , 0) = j=0 2j (j , 0) where j = 0 or 1. Since 0 < λ1 < 2k − 1 there exists at least one j with j = 1 and one j with j = 0. Denote for 0 ≤ τ ≤ k − 1
(λ1 (τ ), 0) =
k−1
2j (j+τ , 0),
j=0
where j+τ is cyclic defined, i.e. j+τ = j+τ if 0 ≤ j + τ ≤ k − 1 and j+τ = j+τ −k if j + τ ≥ k. Let us observe the first identity of (3.2). This identity means T1 − (1 , 0) + Ω Tk−1 − (k−1 , 0) + Ω T − (0 , 0) + Ω ∩ Z2 = T1 , ∩ Z2 = T2 , ..., ∩ Z2 = T. 2 2 2 Thus, for any 0 ≤ τ ≤ k − 1 Tτ − (λ1 (τ ), 0) + Ωk ∩ Z 2 = Tτ , 2k where T0 = T . Similarly, the second identity of (3.2) implies Lτ − (λ1 (τ ), 0) + Ωk ∩ Z2 = Lτ . 2k It is easy to see that, if for some 0 ≤ τ ≤ k − 1 there holds Tτ ∩ Lτ = ∅ then T ∩ L = ∅. We know that Ω is convex and E ⊂ Ω by Lemma 2.5. Moreover, by using Lemma 2.6 we may assume that (0, 0) is an inner integer of Ω. At first, just like in Case 2, for a fixed k and large enough ∈ N, any (β1 , β2 ) ∈ Tτ with β2 ≤ 0 can be written as
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(β1 , β2 ) = 2k (0, 0) + (λ1 (τ ), 0) −
k−1
2j γj ,
9
(3.4)
j=0
for some γj ∈ Ω and (λ1 (τ ), 0) = (λ1 (τ ), 0) + 2k (λ1 (τ ), 0) + ... + 2k(−1) (λ1 (τ ), 0). Furthermore, any (β1 , β2 ) ∈ Tτ with β2 ≥ 1 can be written as (β1 , β2 ) = 2k (0, 1) + (λ1 (τ ), 0) −
k−1
2j γj .
(3.5)
j=0
To continue our discussion let us note that Ω is convex, E ⊂ Ω, and (0, 0) is an inner point of Ω. Thus, if (0, −p) with some p ≥ 1 belongs to Ω, then (0, −1) is in Ω. In this case we choose τ such that k−1+τ = 0. With this in mind we obtain (0, 1) = 2(0, 0) + (k−1+τ , 0) − (0, −1). Clearly, (3.5) is still valid if we replace k by k − 1, i.e. for some γj ∈ Ω and large enough (β1 , β2 ) = 2k−1 (0, 1) + (λ1 (τ ), 0) −
k−2
2j γj .
j=0 It follows from the last two displays that with γj = γj for j ≤ k − 2 and γk−1 = (0, −1)
(β1 , β2 ) = 2k−1 (0, 1) + (λ1 (τ ), 0) −
k−2
2j γj
j=0
= 2k (0, 0) + (λ1 (τ ), 0) −
k−1
2j γj .
j=0
Consequently, we obtain from (3.4) and the last display, that (3.4) is true for all (β1 , β2 ) ∈ Tτ . As k−1 j k we conclude from (3.4) that j=0 2 γj ∈ Ω (0, 0) ∈
Tτ − (λ1 (τ ), 0) + Ωk ∩ Z 2 = Tτ . 2k
The same holds also for Lτ . Hence, (0, 0) ∈ Tτ ∩ Lτ = ∅, which implies T ∩ L = ∅ as mentioned above. To finish the proof for Case 3 we have to address the situation where any integer of the form (0, −p), p ≥ 1, does not belong to Ω. Since Ω is convex, E ⊂ Ω, and (0, 0) is an inner integer of Ω, there is (−p, −q) or (p, −q) with p, q ≥ 1 in Ω. The both cases can be treated in the same way. We observe below the case (−p, −q). We may express (−p, −q) in binary form. Thus, for some δi ∈ E we have (−p, −q) = −
ν
2i δi , δν = 0.
i=0
ν However, as E ⊂ Ω and Ω is convex, we obtain {(−p, −q) + δ0 }/2 ∈ Ω. Hence, − i=1 2i−1 δi ∈ Ω. We ν conclude recursively αη := − i=η 2i−η δi ∈ Ω for η = 0, 1, ..., ν. Because (0, −p) ∈ / Ω for all p ≥ 1, the first
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coordinate of αη cannot be zero. On the other hand, there exists an η such that the second coordinate of αη equals −1. So αη = (−α, −1) for some α ≥ 1. If α is even, we have (0, 1) = 2(−t, 0) + (0, 0) − (−α, −1) with (−t, 0) ∈ Ω and t ≥ 1. Choosing τ such that τ = 0 we obtain (see (3.4)) for some γj ∈ Ω (−t, 0) = 2k−1 (0, 0) + (λ1 (τ )/2, 0) −
k−2
2j γj .
j=0
Therefore, for some other γj ∈ Ω (0, 1) = 2k (0, 0) + (λ1 (τ ), 0) −
k−1
2j γj .
j=0
Consequently, by (3.5) any (β1 , β2 ) ∈ Tτ with β2 ≥ 1 can be expressed as (β1 , β2 ) = 22k (0, 0) + (λ1 (τ ), 0) + 2k (λ1 (τ ), 0) −
2k−1
2j γj .
j=0
Combining this with (3.4) we get that the above holds for all (β1 , β2 ) ∈ T . In other words, (0, 0) ∈
Tτ − (λ1 (τ ), 0) − 2k (λ1 (τ ), 0) + Ω2k = Tτ . 22k
This relation is also true for Lτ . Thus, (0, 0) ∈ Tτ ∩ Lτ = ∅ or T ∩ L = ∅. If α is odd, we have (0, 1) = 2(−t, 0) + (1, 0) − (−α, −1) with (−t, 0) ∈ Ω and t ≥ 1. We choose τ such that τ = 1. Repeating the above process we obtain again T ∩ L = ∅. The proof for Case 3 is complete. There are other situations of (λ1 , λ2 ) that we have not treated, namely, (λ1 , λ2 ) where at least one coordinate is 2k − 1. However, the proof for these situations can be carried out from Cases 1-3. For example, if (λ1 , λ2 ) = (2k − 1, λ2 ) with 0 < λ2 < 2k − 1, then with T = T + (1, 0) we have T − (λ1 , λ2 ) + Ωk ∩ Z2 ⊆ T 2k Therefore, the proof follows from Case 3.
⇔
T − (0, λ1 ) + Ωk ∩ Z2 ⊆ T . 2k
2
4. Example and algorithm Let X = {x1 , ..., xκ } ⊂ Z2 \{(0, 0)} and κ ≥ 2. The zonotope generated by the vectors in X is defined by Z(X) := {
κ
vi xi : 0 ≤ vi ≤ 1}.
i=1
Corollary 4.1. Let X = {x1 , ..., xκ } ⊂ Z2 \{(0, 0)} with κ ≥ 2. Let the support Ω of a nonnegative mask {a(i, j)} be Z(X) ∩ Z2 and satisfy the sum rule (1.2). The corresponding bivariate scheme converges if and only if the matrix A is connected. The following algorithm tells us that the complexity to check the connectivity of matrix A from Theorem 1.5 is O(|Ω|2 ).
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Algorithm CHECK(B): (1) build the directed graph G(Γ, K) from B; (2) calculate all strongly connected components of G(Γ, K), say Γ1 , Γ2 , ..., Γm ; (3) build a new directed graph G (V, E), where V = {1, ..., m} and E = {(i, j) | i, j ∈ V, if there is an edge of K from Γi to Γj }; (4) if there are at least two vertexes of V whose in-degree is zero, return false, otherwise return true. A strongly connected component of a directed graph G(Γ, K) is the maximal set of vertexes Γ ⊆ Γ such that every pair of vertexes u and v in Γ are reachable from each other. The in-degree of a vertex is the number of edges which is incident to this vertex. Let the size of B be N × N , so to build G(Γ, K) in form of adjacency list one needs the complexity O(N 2 ). To find all strongly connected components of G(Γ, K) and to construct the new graph G (V, E) one needs O(|Γ| + |K|) = O(N 2 ) as shown in [4] (see Section 23 of [4]). Finally, finding the vertex in G (V, E) with zero in-degree costs O(|Γ|). Thus, the complexity of CHECK(B) is O(N 2 ). The matrix B is connected if and only if the output of CHECK(B) is true. Indeed, if G (V, E) has more than one vertex with zero in-degree then G(Γ, K) cannot be connected. To see this let a and b two vertexes of G (V, E). So there is no path in G (V, E) for a and b, which in turn implies that there are no paths among the corresponding strongly connected components of G(Γ, K). On the other hand, as G (V, E) is acyclic so there is at least one vertex with in-degree zero. If there is only one vertex v ∈ V with zero in-degree then for any u ∈ V \ {v} one can always find a path from v to u. Assume the strongly component corresponding to v is Γ1 , hence, any vertex from Γj , j = 1 can be reached from the vertexes of Γ1 , in particular from a vertex of Γ1 . In other words, B is connected. Acknowledgments The author is supported partly by National Natural Science Foundation of China No. 11701246 and Scientific Research Foundation of the First-Class Discipline of Zhejiang Province (B) No. 201601. References [1] A.S. Cavaretta, W. Dahmen, C.A. Micchelli, Stationary subdivision, Mem. Amer. Math. Soc. 453 (1991) 1–186. [2] L. Cheng, X. Zhou, Necessary conditions for the convergence of subdivision schemes with finite masks, Appl. Math. Comput. 303 (2017) 34–41. [3] L. Cheng, X. Zhou, A new characterization of convergent multivariate subdivision schemes with nonnegative masks, Arch. Math. 108 (2017) 197–207. [4] T.H. Cormen, C.E. Leiserson, R.L. Rivest, Introduction to Algorithms, MIT Press, 1990. [5] I. Daubechies, J.C. Lagarias, Two-scale difference equations I. Existence and global regularity of solutions, SIAM J. Math. Anal. 22 (1991) 1388–1410. [6] T.N.T. Goodman, C.M. Micchelli, J. Ward, Spectral radius formulas for subdivision operators, in: L.L. Schumaker, G. Webb (Eds.), Recent Advances in Wavelet Analysis, Academic Press Inc., 1994, pp. 335–360. [7] B. Han, R.Q. Jia, Multivariate refinement equations and convergence of subdivision schemes, SIAM J. Math. Anal. 29 (1998) 1177–1199. [8] K. Jetter, X. Li, SIA matrices and non-negative subdivision, Results Math. 62 (2012) 355–375. [9] R.Q. Jia, D.X. Zhou, Convergence of subdivision schemes associated with nonnegative masks, SIAM J. Matrix Anal. Appl. 21 (1999) 418–430. [10] A.A. Melkman, Subdivision schemes with non-negative masks always converge - unless they obviously cannot?, Ann. Numer. Math. 4 (1997) 451–460. [11] J.N. Tsitsiklis, V.D. Blondel, The Lyapunov exponent and joint spectral radius of pairs of matrices are hard, when not impossible, to compute and to approximate, Math. Control Signals Systems 10 (1997) 31–41. [12] X. Zhou, Characterization of convergent subdivision schemes, Approx. Theory Appl. 14 (1998) 11–24. [13] X. Zhou, Subdivision schemes with nonnegative masks, Math. Comp. 74 (2005) 819–839. [14] X. Zhou, Positivity of refinable functions defined by nonnegative finite masks, Appl. Comput. Harmon. Anal. 27 (2009) 133–156.