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A study of a covering dimension of finite lattices
Applied Mathematics and Computation 333 (2018) 276–285
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A study of a covering dimension of finite lattices D. Boyadzhiev a, D.N. Georgiou b,∗, A.C. Megaritis c, F. Sereti b a b c
Faculty of Mathematics and Informatics, Department of Applied Mathematics and Modeling, University of Plovdiv, Bulgaria Department of Mathematics, University of Patras, Patras 265 00, Greece Department of Computer Engineering, Technological Educational Institute of Peloponnese, Sparta 23100, Greece
a r t i c l e
i n f o
keywords: Covering dimension Finite lattice Minimal cover Matrix theory
a b s t r a c t Indubitably, the notion of covering dimension of frames was, extensively, studied. Many searches such as Charalambous, Banashewski and Gilmour (see, for example (Charalambous, 1974; Charalambous, 1974 [11]; Banaschewski and Gilmour, 1989 [12]) studied this dimension. Also, in the study [5], the covering dimension of finite lattices has been characterized by using the so called minimal covers. This approach gave the motive to other searches such as Zhang et al., to study properties of this dimension (see Zhang et al. (2017) [9]). In this paper, we study the covering dimension of finite lattices in combination with matrix theory. Essentially, we characterize the minimal covers of finite lattices and the order of those covers using matrices and we compute the covering dimension of the corresponding finite lattices. © 2018 Elsevier Inc. All rights reserved.
1. Introduction In the study [5], adapting the usual definition of the topological covering dimension (see, for example [4]), the meaning of the covering dimension, dim(X), of a bounded lattice X and the meaning of the order, ord(C), of a subset C of X were studied in the class of finite lattices based on the study [10], where the covering dimension in the class of frames was studied. Many properties of this dimension were studied, mentioning the characterization of the covering dimension of finite lattices by their minimal covers: Theorem 1.1. Let X be a finite lattice and k ∈ {0, 1, 2, . . .}. Then dim(X) ; k if and only if every minimal cover C of X satisfies ord(C) k. Corollary 1.2. Let X be a finite bounded lattice and
k = max{ord (C ) : C is a minimal cover of X }. Then dim(X ) = k. Also, in the study [5], it was observed and proved that each minimal cover of a finite lattice is an antichain and open problems were posted, some of which were answered in the study [9].
∗
Corresponding author. E-mail addresses: [email protected] (D. Boyadzhiev), [email protected] [email protected] (F. Sereti). https://doi.org/10.1016/j.amc.2018.03.041 0 096-30 03/© 2018 Elsevier Inc. All rights reserved.
(D.N. Georgiou), [email protected] (A.C. Megaritis),
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In studies [6,7], the classical topological covering dimension was characterized through the matrix theory, and in study [8], the order dimension of a finite poset was studied by using matrices and the notion of the order-matrix. In this study we are going to characterize the covering dimension of an arbitrary finite lattice with matrices. In particular, the rest of the paper is organized as follows: In Section 2, we give basic definitions and notations which will be used in this study. In Section 3, we characterize the minimal covers of finite lattices using order-matrices. In Section 4, we characterize the meaning of the order of those covers and, consequently, the covering dimension of finite lattices through the matrix theory. Finally, in Sections 5 and 6, we present algorithms for computing the covering dimension of an arbitrary finite lattice. 2. Preliminaries Our references are [1–3,5]. In what follows, we use the notation of a finite lattice (X, ) writing X = {x1 , x2 , . . . , xn }, where x1 = 0X and xn = 1X are the bottom and the top elements of the lattice X, respectively. Also, if C⊆X, then by C and C, we denote the supremum and the infimum of C, respectively and by UB (C ), we denote the set of all upper bounds of C. Two elements xi and xj of X are said to be comparable if xi xj or xj xi . Otherwise they are said to be incomparable (we write xi xj ). A subset of X in which every pair of elements is comparable is called a chain. A subset of X in which every two distinct elements are incomparable is called an antichain. If A is an antichain, then
P l (A ) = {x ∈ X : x a for every a ∈ A}. For every xi ∈ X we set ↓∗ xi = {x ∈ X : x xi } \ {0X }.
Definition 2.1. We say that a cover of the lattice X is a subset C of X such that 0X ∈ C and C = 1X . A cover D of X is called refinement of a cover C of X, written D C, if for every d ∈ D, there exists c ∈ C such that d c. We recall that the relation is quasiorder (or preorder) (reflexive and transitive) but not in general antisymmetric. We also use the following notation:
N0 = {0, 1, 2, . . .}
and
N = N0 ∪ {∞}.
Definition 2.2. Let X be a bounded lattice and k ∈ N0 . The order of a subset C of X is defined as follows: (a) ord (C ) = k, if the infimum of any k + 2 distinct elements of C is 0X and there exist k + 1 distinct elements of C whose infimum is not 0X . (b) ord (C ) = ∞, if for every k ∈ N0 , there exist k distinct elements of C whose infimum is not 0X . Definition 2.3. Let BLat be the class of bounded lattices. We define the covering dimension function
dim : BLat −→ N0 as follows: (1) dim(X) k, if every finite cover C of X has a finite refinement R with ord(R) k. (2) dim(X ) = k, where k ≥ 1, if dim(X) k and dim(X ) k − 1. (3) dim(X ) = ∞, if the inequality dim(X) k does not hold for any k ∈ N0 . Definition 2.4. A cover C of X is called minimal if C⊆R for every refinement R of C. We mention that a minimal cover is the bottom element of the poset (Cov(X ), ⊆ ), where Cov(X ) is the set of all covers of X. Clearly, the cover {xn } is minimal if and only if every cover of X contains the top element xn . Definition 2.5. Let (X, ) be a finite poset, where X = {x1 , x2 , . . . , xn }. The n × n matrix TX = (ti j ), where
ti j =
1, 0,
if xi x j otherwise
is called the incidence matrix of X. We denote by c1 (TX ), . . . , cn (TX ) the n columns of the incidence matrix TX . Definition 2.6. Let (X, ) be a finite poset, where X = {x1 , x2 , . . . , xn }. The n × n matrix A = (ai j ), where X
⎧ ⎪ ⎨1,
2, ai j = ⎪ ⎩−2, 0,
if i = j i f xi < x j if x j < xi if xi x j
is called the order-matrix of X.
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x6 x5
x4 x2
x3 x1
Fig. 1. The poset (X, ).
For example, we consider the finite poset (X = {x1 , x2 , x3 , x4 , x5 , x6 }, ), represented by the diagram of Fig. 1. The incidence matrix TX and the order-matrix A of X are respectively: X
⎛
TX
1 ⎜0 ⎜0 =⎜ ⎜0 ⎝ 0 0
1 1 0 0 0 0
1 0 1 0 0 0
1 1 1 1 0 0
1 1 0 0 1 0
⎞
⎛
1 1 1⎟ ⎜−2 ⎜ 1⎟ ⎟, A = ⎜−2 X ⎟ ⎜−2 1 ⎠ ⎝ 1 −2 1 −2
2 1 0 −2 −2 −2
2 0 1 −2 0 −2
2 2 2 1 0 −2
2 2 0 0 1 −2
⎞
2 2⎟ 2⎟ ⎟. 2⎟ ⎠ 2 1
We denote by r1 (A ), . . . , rn (A ) the n rows of the order-matrix A and by E (A ) the set of all elements of the orderX X X X
matrix A . In what follows, for the simplicity of the writing, if α is an element of a matrix A (respectively, if α is not an X element of a matrix A), then we shall write α ∈ A (respectively, α ∈ A). 3. A characterization of minimal covers using matrices
In this section, we give a characterization of the minimal covers of an arbitrary finite lattice (X = {x1 , . . . , xn }, ) with order-matrix A = ( ai j ). X The following propositions follows immediately from the definition of the matrix A . X
Proposition 3.1. Let xk1 , xk2 ∈ X. The following statements are true: (1) xk1 xk2 if and only if 1 ∈ rk1 (A ) + rk2 (A ), X X (2) xk1 < xk2 if and only if 3 ∈ rk1 (A ) − rk2 (A ) X X
Proposition 3.2. The subset {x j1 , . . . , x jm } of the lattice X is an antichain if and only if 1 ∈ r ji (A ) + r ji (A ) for every i1 , i2 ∈ X X
{1, . . . , m} with i1 = i2 .
1
2
Definition 3.3. (1) A subset {xj } of X is said to satisfy the cover condition if the following condition is satisfied:
3∈ / r j ( A ) + (−1 ) · rn (A ). X X (2) An antichain subset {x j1 , . . . , x jm } of X, where m ≥ 2, is said to satisfy the cover condition if the following condition is satisfied:
2m + 2 ∈ / r j 1 ( A ) + . . . + r j m ( A ) + (−1 ) · rn (A ). X X X Proposition 3.4. (1) A subset C = {x j } of X is a cover of X if and only if C satisfies the cover condition. (2) An antichain subset C = {x j1 , . . . , x jm } of X, where m ≥ 2, is a cover of X if and only if C satisfies the cover condition. Proof. (1) Let us C = {x j } be a cover of X. We suppose that C does not satisfy the cover condition. Then
3 ∈ r j ( A ) + (−1 ) · rn (A ). X X According to the definition of the order-matrix A , there exists k ∈ {1, . . . , n} such that a jk = 1 and ank = −2. That is, x j = xk X and xk < xn , which is a contradiction, since x j = xn . Conversely, we suppose that C satisfies the cover condition. We prove that C is a cover of X. It suffices to prove that x j = xn . Let suppose that x j = xk , where k = n. Then a jk = 1 and since, always, xk < xn , we have that akn = 2 or equivalently, ank = −2. Thus 3 ∈ r j (A ) + (−1 ) · rn (A ), which is a contradiction. Thus xn = x j . X X (2) Let us C = {x j1 , . . . , x jm }, where m ≥ 2, be a cover of X. We suppose that C does not satisfy the cover condition. Then
2 m + 2 ∈ r j 1 ( A ) + . . . + r j m ( A ) + (−1 ) · rn (A ). X X X
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According to the definition of the order-matrix A and considering that C is an antichain, there exists k ∈ {1, . . . , n} \ X { j1 , . . . , jm , n} such that a j1 k = . . . = a jm k = 2 and ank = −2. Therefore xk ∈ UB ({x j1 , . . . , x jm } ) and xk < xn , which is a contradiction since x j1 ∨ . . . ∨ x jm = xn . Conversely, we suppose that the set C = {x j1 , . . . , x jm } satisfies the cover condition. Since {x j1 , . . . , x jm } is an antichain, x1 ∈ / {x j1 , . . . , x jm }. We suppose that xk = x j1 ∨ . . . ∨ x jm , where k = n. Then xk ∈ UB ({x j1 , . . . , x jm } ) and so x j1 < xk , . . . , x jm < xk or equivalently, a j1 k = . . . = x jm k = 2. Also, we have that xk < xn and thus akn = 2 (or equivalently, ank = −2). Therefore 2m + 2 ∈ r j1 (A ) + . . . + r jm (A ) + (−1 ) · rn (A ), which is a contradiction. Thus xn = x j1 ∨ . . . ∨ x jm . X X X
Proposition 3.5. Let C = {x j1 , . . . , x jm } be an antichain cover of the lattice X. The cover C is minimal if and only if for every i ∈ {1, . . . , m} the following conditions hold: (M1) C \ {x ji } is not a cover of X. (M2) For every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 and xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk1 , xk2 } is not a cover of X. (M3) For every k ∈ {1, . . . , n} \ { j1 , . . . , jm } with xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk } is not a cover of X. Proof. We suppose that the conditions (M1), (M2) and (M3) are satisfied and the set {x j1 , . . . , x jm } is not a minimal cover of the lattice X. Then there exists a refinement R of C such that CࣰR. Hence there exists i ∈ {1, . . . , m} such that x ji ∈ C \ R. By the condition (M1), C \ {x ji } is not a cover of X. Therefore there exists xk ∈ R such that xk < x ji and xk ∈ P l (C \ {x ji } ). We have two cases: (i) There exist k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 and xk1 , xk2 ∈ R ∩ ↓∗ x ji ∩ P l (C \ {x ji } ). In this case (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X. Indeed, we suppose that ∨((C \ {x ji } ) ∪ {xk1 , xk2 } ) = a = 1. Since X is a lattice, xk1 , xk2 < x ji , xk1 , xk2 < a, and xk1 xk2 , we have either x ji < a or a < x ji . If x ji < a, then ∨C a, a contradiction. If a < x ji , then C is not an antichain, a contradiction. Therefore (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X which contradicts the condition (M2). (ii) {xk ∈ R : xk < x ji , xk ∈ P l (C \ {x ji } )} is a chain. In this case we set x p = max{xk ∈ R : xk < x ji , xk ∈ P l (C \ {x ji } )} and we prove that (C \ {x ji } ) ∪ {x p } is a cover of X. We suppose that ∨((C \ {x ji } ) ∪ {x p } ) = a p = 1. Then ∨R ap , a contradiction. Therefore (C \ {x ji } ) ∪ {x p } is a cover of X which contradicts the condition (M3). Conversely, we suppose that C is a minimal cover and let i ∈ {1, . . . , m}. Firstly, we observe that the set C \ {x ji } is not a cover of X. Indeed, if we suppose that the set C \ {x ji } is a cover of X, since it refines C which is a minimal cover of X, then by the definition of the minimal cover, we must have that C ⊆ C \ {x ji }, which is a contradiction. Thus the set C \ {x ji } is not a cover of X. Now, if there exist k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 , xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), and (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X, then R = (C \ {x ji } ) ∪ {xk1 , xk2 } is a refinement of C such that CࣰR, a contradiction. Finally, if there exists k ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ) and (C \ {x ji } ) ∪ {xk } is cover of X, then R = (C \ {x ji } ) ∪ {xk } is a refinement of C such that CࣰR, a contradiction. Proposition 3.6. The cover C = {xn } of X is minimal if and only if every antichain subset A of X, A = {xn }, does not satisfy the cover condition. Proof. Let us C = {xn } be a minimal cover of X. We suppose that there exists an antichain subset A of X, A = {xn }, which satisfes the cover condition. Then by Proposition 3.4, A is a cover of X. Since C = {xn } is a minimal cover of X, we must have xn ∈ A, a contradiction. Thus every antichain subset of X does not satisfy the cover condition. Conversely, we suppose that the cover C = {xn } of X is not minimal. Then there exists a cover R of X which refines C and CࣰR, that is xn ∈ R. By [5, Lemma 3.7], there exists a minimal cover R of X which refines R and xn ∈ R . Since R is a minimal cover of X, R is an antichain cover of X and by Proposition 3.4, R satisfies the cover condition. The last contradicts our assumption that every antichain subset of X, different from {xn }, does not satisfy the cover condition. Thus the cover C = {xn } is minimal. Theorem 3.7. Let C = {x j1 , . . . , x jm } be an antichain cover of the lattice X. The cover C is minimal if and only if for every i ∈ {1, . . . , m} the following conditions hold: (MC1) C \ {x ji } does not satisfy the cover condition.
(MC2) If for every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm }, 1 ∈ rk1 (A ) + rk2 (A ), 3 ∈ r ji (A ) + rk1 (A ), 3 ∈ r ji (A ) + rk2 (A ), 1 ∈ X X X X X X rk1 (A ) + r jl (A ), 1 ∈ rk2 (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}, X X X X then (C \ {x ji } ) ∪ {xk1 , xk2 } does not satisfy the cover condition.
(MC3) If for every k ∈ {1, . . . , n} \ { j1 , . . . , jm }, 3 ∈ r ji (A ) + rk (A ) and X X 1 ∈ rk (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}, X X then (C \ {x ji } ) ∪ {xk } does not satisfy the cover condition.
Proof. Suppose that the conditions (MC1), (MC2) and (MC3) are satisfied and the set {x j1 , . . . , x jm } is not a minimal cover of the lattice X. By Proposition 3.5 there exists i ∈ {1, . . . , m} such that at least one of the following conditions hold: (1) C \ {x ji } is a cover of X.
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(2) There exist k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 , xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), and (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X. (3) There exists k ∈ {1, . . . , n} \ { j1 , . . . , jm } with xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ) and (C \ {x ji } ) ∪ {xk } is a cover of X. By Propositions 3.1, 3.2, and 3.4 in each of these cases a contradiction arises. Thus the cover C is minimal. Conversely, suppose that the cover C is minimal and let i ∈ {1, . . . , m}. Then by Proposition 3.5, the following conditions hold: (1) C \ {x ji } is not a cover of X. (2) For every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 and xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk1 , xk2 } is not a cover of X. (3) For every k ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk } is not a cover of X. By the above conditions, Propositions 3.1, 3.2, and 3.4, the conditions (MC1), (MC2) and (MC3) are satisfied. / Proposition 3.8. If 0 ∈
E ( A ), X
then the cover {xn } is minimal.
Proof. Since 0 ∈ / E ( A ), by Georgiou et al. [8, Remark 2.4], the poset (X, ) is linearly ordered set. Thus every cover of X X contains the top element xn . Therefore {xn } is minimal. Algorithm 3.9. Let (X = {x1 , . . . , xn }, ) be a finite lattice, where n ≥ 2. The algorithm which finds the set MCov(X ) of all minimal covers of X consists of the following steps: Step 1: Find the n rows r1 (A ), . . . , rn (A ) of the order-matrix A of X and go to Step 2. X X X Step Step Step Step Step Step Step
2: 3: 4: 5: 6: 7: 8:
If 0 ∈ / E ( A ), then create MCov(X ) = {{xn }}. Otherwise, go to Step 3. X Set m = 1 and go to Step 4. Find the set Am (X ) of all antichains {x j1 , . . . , x jm } of X which satisfy the cover condition and go to Step 5. If m < n − 2, then put m → m + 1 (that is replace the old m with m + 1) and go to Step 4. Otherwise, go to Step 6. If Am (X ) = ∅ for all m = 2, . . . , n − 2, then create MCov(X ) = {{xn }}. Otherwise, go to Step 7. Create the set A(X ) = A1 (X ) ∪ . . . ∪ An−2 (X ) and go to Step 8. Find the set M(X ) of all {x j1 , . . . , x jm } ∈ A(X ) such that for every i ∈ {1, . . . , m} the following conditions hold: (1) C \ {x ji } does not satisfy the cover condition.
(2) If for every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm }, 1 ∈ rk1 (A ) + rk2 (A ), 3 ∈ r ji (A ) + rk1 (A ), 3 ∈ r ji (A ) + rk2 (A ), X X X X X X
1 ∈ rk1 (A ) + r jl (A ), 1 ∈ rk2 (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}. then (C \ {x ji } ) ∪ {xk1 , xk2 } does not satX X X X isfy the cover condition. (3) If for every k ∈ {1, . . . , n} \ { j1 , . . . , jm }, 3 ∈ r ji (A ) + rk (A ) and 1 ∈ rk (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}, X X X X then (C \ {x ji } ) ∪ {xk } does not satisfy the cover condition. Subsequently go to Step 9. Step 9: Create MCov(X ) = M(X ). Example 3.10. We consider the finite lattice (X, ) which is represented by the diagram of Fig. 2. The order-matrix of this lattice is
⎛1 ⎜−2 ⎜−2 ⎜ ⎜−2 A = ⎜−2 X ⎜ ⎜−2 ⎝ −2 −2
2 1 0 0 −2 −2 0 −2
2 0 1 0 0 −2 −2 −2
2 0 0 1 −2 0 −2 −2
2 2 0 2 1 0 0 −2
2 2 2 0 0 1 0 −2
2 0 2 2 0 0 1 −2
⎞
2 2⎟ 2⎟ ⎟ 2⎟ ⎟. 2⎟ 2⎟ ⎠ 2 1
x8 x5 x2
x6 x3
x7 x4
x1 Fig. 2. The lattice (X, ).
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Using Algorithm 3.9 we find A1 (X ) = {{x8 }}, A2 (X ) = {{x2 , x7 }, {x3 , x5 }, {x4 , x6 }, {x5 , x6 }, {x5 , x7 }, {x6 , x7 }}, A3 (X ) = {{x2 , x3 , x4 }, {x5 , x6 , x7 }}, A4 ( X ) = A 5 ( X ) = A 6 ( X ) = ∅. Therefore MCov(X ) = {{x2 , x3 , x4 }}. For example, we observe that the set C = {x2 , x3 , x4 } is a minimal cover of X: (i) C satisfies the cover condition. 2·3+2=8∈ / r2 (A ) + r3 (A ) + r4 (A ) + (−1 ) · r8 (A ) X X X X (ii) C \ {x2 } = {x3 , x4 } does not satisfy the cover condition. 2 · 2 + 2 = 6 ∈ r3 (A ) + r4 (A ) + (−1 ) · r8 (A ) X X X C \ {x3 } = {x2 , x4 } does not satisfy the cover condition. 2 · 2 + 2 = 6 ∈ r2 (A ) + r4 (A ) + (−1 ) · r8 (A ) X X X C \ {x4 } = {x2 , x3 } does not satisfy the cover condition. 2 · 2 + 2 = 6 ∈ r2 (A ) + r2 (A ) + (−1 ) · r8 (A ) X X X Also, the others conditions in Step 8 of the above Algorithm are satisfied. 4. The meaning of the order with matrices In this section we are going to characterize the covering dimension of an arbitrary finite lattice (X = {x1 , . . . , xn }, ) with incidence matrix TX = (ti j ) and order-matrix A = (ai j ) providing, firstly, basic results for the meaning of the order. X Notation 4.1. For every m ∈ {1, . . . , n} we consider the n × 1 matrix Bm = (bim ), where
bim
=
m,
if i = 1
0,
otherwise.
Theorem 4.2. Let {x j1 , . . . , x jm } be an antichain subset of X and
k = max(c j1 (TX ) + . . . + c jm (TX ) − Bm ), that is k is the maximum element of the n × 1 matrix c j1 (TX ) + . . . + c jm (TX ) − Bm . Then ord ({x j1 , . . . , x jm } ) = k − 1. Proof. Since max(c j1 (TX ) + . . . + c jm (TX ) − Bm ) = k, by the definition of the matrices TX and Bm , there exist i0 ∈ {2, . . . , n} and jqλ ∈ { j1 , . . . , jm }, λ = 1, . . . , k, such that ti0 jq = 1 for every λ = 1, . . . , k. This means that xi0 x jq for every λ = 1, . . . , k λ
λ
and, hence, x jq ∧ . . . ∧ x jq = x1 . Also, by the definition of k, the infimum of any k + 1 distinct elements of {x j1 , . . . , x jm } is 1
k
equal to x1 . Thus ord({x j1 , . . . , x jm } ) = k − 1.
Example 4.3. We consider the finite lattice (X, ) which is represented by the diagram of Fig. 3. The incidence matrix of this lattice is
⎛1
⎜0 TX = ⎜0 ⎝ 0 0
1 1 0 0 0
1 1 1 0 0
1 1 0 1 0
⎞
1 1⎟ 1⎟. ⎠ 1 1
Based on Theorem 4.2, we shall compute the order of the antichain {x3 , x4 }. We have
⎛1⎞
⎛1⎞
⎛2⎞
⎛ 0⎞
⎜1⎟ ⎜1⎟ ⎜0⎟ ⎜2⎟ c3 (TX ) + c4 (TX ) − B2 = ⎜1⎟ + ⎜0⎟ − ⎜0⎟ = ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 0 1 0 1 0 0 0 0 x5 x3
x4 x2 x1
Fig. 3. The lattice (X, ).
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x9 x6
x7
x8
x3
x4
x5
x2 x1 Fig. 4. The lattice (X, ).
and max(c3 (TX ) + c4 (TX ) − B2 ) = 2. Hence ord({x3 , x4 } ) = 2 − 1 = 1. Example 4.4. We consider the finite lattice (X, ), represented by the diagram of Fig. 4. Based on Theorem 4.2, we show that ord({x3 , x4 , x5 } ) = 2. The incidence matrix TX of X is
⎛
TX
1 ⎜0 ⎜0 ⎜ ⎜0 ⎜ = ⎜0 ⎜0 ⎜ ⎜0 ⎝ 0 0
1 1 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0
1 1 0 1 0 0 0 0 0
1 1 0 0 1 0 0 0 0
1 1 0 1 1 1 0 0 0
Then
1 1 1 0 1 0 1 0 0
1 1 1 1 0 0 0 1 0
⎛ ⎞
⎞
1 1⎟ 1⎟ ⎟ 1⎟ ⎟ 1⎟. 1⎟ ⎟ 1⎟ ⎠ 1 1
⎛ ⎞
⎛ ⎞
3 3 0 ⎜3⎟ ⎜0⎟ ⎜3⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ c3 (TX ) + c4 (TX ) + c5 (TX ) − B3 = ⎜1⎟ − ⎜0⎟ = ⎜1⎟. ⎜ 0 ⎟ ⎜ 0⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ 0⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 0 0 0 0 0 0 Hence
ord({x3 , x4 , x5 } ) = max(c3 (TX ) + c4 (TX ) + c5 (TX ) − B3 ) − 1 = 3 − 1 = 2. Thus according to Corollary 1.2, Algorithm 3.9 and Theorem 4.2, we can succeed a computation of the covering dimension of an arbitrary finite lattice using order-matrices and incidence matrices. 5. An algorithm for computing the covering dimension of finite lattices In this section, we present an algorithm which is based on Corollary 1.2 and the results of Sections 3 and 4. This algorithm computes the covering dimension of finite lattices. Notation 5.1. In what follows we denote by M(X ) the set of all antichain subsets {x j1 , . . . , x jm } of X which satisfy the cover condition and the conditions (MC1), (MC2) and (MC3) of Theorem 3.7. Theorem 5.2. Let X be a finite lattice and
L = max{max(c j1 (TX ) + . . . + c jm (TX ) − Bm ) − 1 : {x j1 , . . . , x jm } ∈ M(X )}. Then dim(X ) = L. Proof. It follows by Corollary 1.2, Theorems 3.7 and 4.2. Proposition 5.3. If 0 ∈ /
E ( A ), X
then dim(X ) = 0.
Proof. According to Proposition 3.8, {xn } is a minimal cover of X. Thus every cover of X contains the element xn and so the cover {xn } is refinement of every cover of X. Since ord ({xn } ) = 0, we have dim(X ) = 0. Algorithm 5.4. Let (X = {x1 , . . . , xn }, ) be a finite lattice.
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x6 x3
x4
x5
x2 x1 Fig. 5. The lattice (X, ).
Step 1: Find the n rows r1 (A ), . . . , rn (A ) of the order-matrix A of X. If 0 ∈ / E ( A ), then print dim(X ) = 0. Otherwise, X X X X continue with Step 2. Step 2: Apply Algorithm 3.9 to find the set A(X ) = nm−2 =1 Am (X ) and go to Step 3. Step 3: If Am (X ) = ∅ for all m = 2, . . . , n − 2, then print dim(X ) = 0. Otherwise, go to Step 4. Step 4: Apply Algorithm 3.9 to find the set M(X ) and go to Step 5. Step 5: Find the n columns c1 (TX ), . . . , cn (TX ) of the incidence matrix TX of X and go to Step 6. Step 6: For each m ∈ {1, . . . , n − 2}, find
dm = max{max(c j1 (TX ) + . . . + c jm (TX ) − Bm ) − 1 : {x j1 , . . . , x jm } ∈ M(X )}. Moreover, if Am (X ) ∩ M(X ) = ∅ for some m, then put dm = 0. Subsequently go to Step 7. Step 7: Print dim(X ) = max{d1 , d2 , . . . , dn−2 }. Example 5.5. Following the Steps of Algorithms 3.9 and 5.4 for the finite lattice which is represented in Example 4.3 we have that A1 (X ) = {{x5 }} and d1 = 0, A2 (X ) = {{x3 , x4 }} and d2 = 1, A3 (X ) = ∅ and d3 = 0. Therefore dim(X ) = max{d1 , d2 , d3 } = 1. Example 5.6. (1) We consider the finite lattice (X, ), represented by the diagram of Fig. 5. Following Algorithm 5.4, we find that MCov(X ) = A2 (X ), where
A2 (X ) = {{x3 , x4 }, {x3 , x5 }, {x4 , x5 }}. Moreover,
ord({x3 , x4 } ) = ord({x3 , x5 } ) = ord({x4 , x5 } ) = 1. Therefore dim(X ) = 1. (2) We consider the finite lattice (X, ) which is represented in Example 4.4. Its order-matrix is the following 9 × 9 matrix
⎛
A X
1 ⎜−2 ⎜−2 ⎜ ⎜−2 ⎜ = ⎜−2 ⎜−2 ⎜ ⎜−2 ⎝ −2 −2
2 1 −2 −2 −2 −2 −2 −2 −2
2 2 1 0 0 0 −2 −2 −2
2 2 0 1 0 −2 0 −2 −2
2 2 0 0 1 −2 −2 0 −2
2 2 0 2 2 1 0 0 −2
2 2 2 0 2 0 1 0 −2
2 2 2 2 0 0 0 1 −2
⎞
2 2⎟ 2⎟ ⎟ 2⎟ ⎟ 2⎟. 2⎟ ⎟ 2⎟ ⎠ 2 1
Following Algorithms 3.9 and 5.4, we can observe that {x3 , x4 , x5 } is the only minimal cover with order equal to 2, that is M(X ) = {{x3 , x4 , x5 }} and d3 = 2 (see Example 4.4). Thus we conclude that dim(X ) = d3 = 2. (3) We consider the finite lattice (X, ), represented by the diagram of Fig. 6. The incidence matrix TX and the order-matrix A of X are respectively X
⎛
TX
1 ⎜0 ⎜0 =⎜ ⎜0 ⎝ 0 0
1 1 0 0 0 0
1 1 1 0 0 0
1 1 0 1 0 0
1 1 1 1 1 0
⎞
1 1⎟ 1⎟ ⎟, 1⎟ ⎠ 1 1
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x6 x5 x3
x4 x2 x1
Fig. 6. The lattice (X, ).
⎛
A X
1 ⎜−2 ⎜−2 =⎜ ⎜−2 ⎝ −2 −2
2 1 −2 −2 −2 −2
2 2 1 0 −2 −2
2 2 0 1 −2 −2
2 2 2 2 1 −2
⎞
2 2⎟ 2⎟ ⎟. 2⎟ ⎠ 2 1
We have that A2 (X ) = ∅ since the antichain set C = {x3 , x4 } does not satisfy the cover condition: 2 · 2 + 2 ∈ r3 (A )+ X r4 (A ) + (−1 ) · r6 (A ). Moreover, A3 (X ) = A4 (X ) = ∅. Thus MCov(X ) = {{x6 }} for which d1 = 0 since X X
⎛ ⎞
⎛ ⎞
⎛ ⎞
1 1 0 ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎟ ⎜ ⎟ ⎜ ⎟ c6 (TX ) + B1 = ⎜ ⎜1⎟ − ⎜0⎟ = ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 0 1 1 0 1 and max(c6 (TX ) − B1 ) = 1. 6. Computing the covering dimension of certain finite lattices In this section, we present an another algorithm in order to compute the covering dimension of the linear sum and of the lexicographic product of two finite lattices. Definition 6.1. The lexicographic product X◦Y of two lattices (X, 1 ) and (Y, 2 ) is the lattice (X × Y, ), where the relation is defined as follows:
( x1 , y1 ) ( x2 , y2 ) ⇔
x1 <1 x2 or x1 = x2 and y1 2 y2 .
Definition 6.2. The linear sum XY of two lattices (X, 1 ) and (Y, 2 ), where X ∩ Y = ∅, is the lattice (X ∪ Y, ) where the relation is defined as follows:
xy⇔
x, y ∈ X and x 1 y x, y ∈ Y and x 2 y x ∈ X and y ∈ Y.
Remark 6.3. In study [5], the following two propositions were proved: (1) Let X and Y be finite lattices and
= max{|C | : C is a minimal cover of Y }. Then dim(X ◦ Y ) = − 1. (2) Let X and Y be finite lattices and
k = max{m : there exists a minimal cover {a1 , . . . , am } of Y }. Then dim(X Y ) = k − 1. Using this result, we have the following: Theorem 6.4. Let (X = {x1 , . . . , xn1 }, 1 ) and (Y = {y1 , . . . , yn2 }, 2 ) be two finite lattices and L = max{m : {x j1 , . . . , x jm } ∈ M(Y )}. Then dim(X ◦ Y ) = dim(X Y ) = L − 1.
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285
y4 x2
y2
x1
y3 y1
Fig. 7. The lattices (X, 1 ) and (Y, 2 ).
(x2 , y4 ) y4
(x2 , y2 )
y2
(x2 , y3 )
y3
(x2 , y1 )
y1
(x1 , y4 )
x2
(x1 , y2 )
x1
(x1 , y3 ) (x1 , y1 )
Fig. 8. The lattices (XY, ) and (X◦Y, ∗ ).
Proof. It follows by Theorem 3.7 and Remark 6.3.
Thus using Algorithms 3.9 and 5.4, we have the following: Algorithm 6.5. Let (X = {x1 , . . . , xn1 }, 1 ) and (Y = {y1 , . . . , yn2 }, 2 ) be two finite lattices.
Step 1: Find the n2 rows r1 (AY 2 ), . . . , rn2 (AY 2 ) of the order-matrix AY 2 of Y and go to Step 2.
then print dim(X ◦ Y ) = dim(X Y ) = 0. Otherwise, go to Step 3. n −2 Apply Algorithm 3.9 to find the set A(Y ) = m2=1 Am (Y ) and go to Step 4. If Am (Y ) = ∅ for all m = 2, . . . , n2 − 2, then print dim(X ◦ Y ) = dim(X Y ) = 0. Otherwise, go to Step 5. Apply Algorithm 3.9 to find the set M(Y ) and go to Step 6. Print
Step 2: If 0 ∈ / Step Step Step Step
3: 4: 5: 6:
E (AY ),
dim(X ◦ Y ) = dim(X Y ) = max{m : {x j1 , . . . , x jm } ∈ M(Y )} − 1. Example 6.6. We consider the finite lattices (X, 1 ) and (Y, 2 ), represented by the diagrams of Fig. 7, respectively. The finite lattice (XY, ) and (X◦Y, ∗ ) are represented, respectively, by the diagrams of Fig. 8. Following Algorithm 6.5 we have that MCov(Y ) = {{y2 , y3 }} and thus, dim(X Y ) = dim(X ◦ Y ) = 2 − 1 = 1. Acknowledgments The authors would like to thank the referee for the careful reading of the paper and the useful comments. The fourth author of this paper F. Sereti (with application code 2547) would like to thank the General Secretariat for Research and Technology (GSRT) and the Hellenic Foundation for Research and Innovation (HFRI) for the scholarship during her studies. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
B. Dushnik, E.W. Miller, Partially ordered sets, Am. J. Math. 63 (1941) 600–610. B.A. Davery, H.A. Priestley, Introduction to Lattices and Order, Cambridge University Press, 2002. S. Roman, Lattices and Ordered Sets I, Springer, New York, 2008. R. Engelking, Theory of Dimensions, Finite and Infinite, Heldermann Verlag, Berlin, 1995. T. Dube, D.N. Georgiou, A.C. Megaritis, S.P. Moshoka, A study of covering dimension for the class of finite lattices, Discret. Math. 338 (2015) 1096–1110. D.N. Georgiou, A.C. Megaritis, Covering dimension and finite spaces, Appl. Math. Comput. 218 (2011) 3122–3130. D.N. Georgiou, A.C. Megaritis, An algorithm of polynomial order for computing the covering dimension of a finite space, Appl. Math. Comput. 231 (2014) 276–283. D.N. Georgiou, A.C. Megaritis, F. Sereti, A study of the order dimension of a poset through the matrix theory, Quaest. Math. 39 (6) (2016) 797–814. H.-f. Zhang, M. Zhou, G.-j. Zhang, Answer to some open problems on covering dimension for finite lattices, Discret. Math. 340 (5) (2017) 1086–1091. M.G. Charalambous, Dimension theory for σ -frames, J. Lond. Math. Soc. s2-8 (1) (1974) 149–160. M.G. Charalambous, The dimension of paracompact normal κ -frames, Topol. Proc. 20 (1995) 49–66. ˇ B. Banaschewski, C. Gilmour, Stone-Cech compactifiaction and dimension theory for regular σ -frames, J. Lond. Mat. Soc. 2 (39) (1989) 1–8.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
A study of a covering dimension of finite lattices D. Boyadzhiev a, D.N. Georgiou b,∗, A.C. Megaritis c, F. Sereti b a b c
Faculty of Mathematics and Informatics, Department of Applied Mathematics and Modeling, University of Plovdiv, Bulgaria Department of Mathematics, University of Patras, Patras 265 00, Greece Department of Computer Engineering, Technological Educational Institute of Peloponnese, Sparta 23100, Greece
a r t i c l e
i n f o
keywords: Covering dimension Finite lattice Minimal cover Matrix theory
a b s t r a c t Indubitably, the notion of covering dimension of frames was, extensively, studied. Many searches such as Charalambous, Banashewski and Gilmour (see, for example (Charalambous, 1974; Charalambous, 1974 [11]; Banaschewski and Gilmour, 1989 [12]) studied this dimension. Also, in the study [5], the covering dimension of finite lattices has been characterized by using the so called minimal covers. This approach gave the motive to other searches such as Zhang et al., to study properties of this dimension (see Zhang et al. (2017) [9]). In this paper, we study the covering dimension of finite lattices in combination with matrix theory. Essentially, we characterize the minimal covers of finite lattices and the order of those covers using matrices and we compute the covering dimension of the corresponding finite lattices. © 2018 Elsevier Inc. All rights reserved.
1. Introduction In the study [5], adapting the usual definition of the topological covering dimension (see, for example [4]), the meaning of the covering dimension, dim(X), of a bounded lattice X and the meaning of the order, ord(C), of a subset C of X were studied in the class of finite lattices based on the study [10], where the covering dimension in the class of frames was studied. Many properties of this dimension were studied, mentioning the characterization of the covering dimension of finite lattices by their minimal covers: Theorem 1.1. Let X be a finite lattice and k ∈ {0, 1, 2, . . .}. Then dim(X) ; k if and only if every minimal cover C of X satisfies ord(C) k. Corollary 1.2. Let X be a finite bounded lattice and
k = max{ord (C ) : C is a minimal cover of X }. Then dim(X ) = k. Also, in the study [5], it was observed and proved that each minimal cover of a finite lattice is an antichain and open problems were posted, some of which were answered in the study [9].
∗
Corresponding author. E-mail addresses: [email protected] (D. Boyadzhiev), [email protected] [email protected] (F. Sereti). https://doi.org/10.1016/j.amc.2018.03.041 0 096-30 03/© 2018 Elsevier Inc. All rights reserved.
(D.N. Georgiou), [email protected] (A.C. Megaritis),
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In studies [6,7], the classical topological covering dimension was characterized through the matrix theory, and in study [8], the order dimension of a finite poset was studied by using matrices and the notion of the order-matrix. In this study we are going to characterize the covering dimension of an arbitrary finite lattice with matrices. In particular, the rest of the paper is organized as follows: In Section 2, we give basic definitions and notations which will be used in this study. In Section 3, we characterize the minimal covers of finite lattices using order-matrices. In Section 4, we characterize the meaning of the order of those covers and, consequently, the covering dimension of finite lattices through the matrix theory. Finally, in Sections 5 and 6, we present algorithms for computing the covering dimension of an arbitrary finite lattice. 2. Preliminaries Our references are [1–3,5]. In what follows, we use the notation of a finite lattice (X, ) writing X = {x1 , x2 , . . . , xn }, where x1 = 0X and xn = 1X are the bottom and the top elements of the lattice X, respectively. Also, if C⊆X, then by C and C, we denote the supremum and the infimum of C, respectively and by UB (C ), we denote the set of all upper bounds of C. Two elements xi and xj of X are said to be comparable if xi xj or xj xi . Otherwise they are said to be incomparable (we write xi xj ). A subset of X in which every pair of elements is comparable is called a chain. A subset of X in which every two distinct elements are incomparable is called an antichain. If A is an antichain, then
P l (A ) = {x ∈ X : x a for every a ∈ A}. For every xi ∈ X we set ↓∗ xi = {x ∈ X : x xi } \ {0X }.
Definition 2.1. We say that a cover of the lattice X is a subset C of X such that 0X ∈ C and C = 1X . A cover D of X is called refinement of a cover C of X, written D C, if for every d ∈ D, there exists c ∈ C such that d c. We recall that the relation is quasiorder (or preorder) (reflexive and transitive) but not in general antisymmetric. We also use the following notation:
N0 = {0, 1, 2, . . .}
and
N = N0 ∪ {∞}.
Definition 2.2. Let X be a bounded lattice and k ∈ N0 . The order of a subset C of X is defined as follows: (a) ord (C ) = k, if the infimum of any k + 2 distinct elements of C is 0X and there exist k + 1 distinct elements of C whose infimum is not 0X . (b) ord (C ) = ∞, if for every k ∈ N0 , there exist k distinct elements of C whose infimum is not 0X . Definition 2.3. Let BLat be the class of bounded lattices. We define the covering dimension function
dim : BLat −→ N0 as follows: (1) dim(X) k, if every finite cover C of X has a finite refinement R with ord(R) k. (2) dim(X ) = k, where k ≥ 1, if dim(X) k and dim(X ) k − 1. (3) dim(X ) = ∞, if the inequality dim(X) k does not hold for any k ∈ N0 . Definition 2.4. A cover C of X is called minimal if C⊆R for every refinement R of C. We mention that a minimal cover is the bottom element of the poset (Cov(X ), ⊆ ), where Cov(X ) is the set of all covers of X. Clearly, the cover {xn } is minimal if and only if every cover of X contains the top element xn . Definition 2.5. Let (X, ) be a finite poset, where X = {x1 , x2 , . . . , xn }. The n × n matrix TX = (ti j ), where
ti j =
1, 0,
if xi x j otherwise
is called the incidence matrix of X. We denote by c1 (TX ), . . . , cn (TX ) the n columns of the incidence matrix TX . Definition 2.6. Let (X, ) be a finite poset, where X = {x1 , x2 , . . . , xn }. The n × n matrix A = (ai j ), where X
⎧ ⎪ ⎨1,
2, ai j = ⎪ ⎩−2, 0,
if i = j i f xi < x j if x j < xi if xi x j
is called the order-matrix of X.
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x6 x5
x4 x2
x3 x1
Fig. 1. The poset (X, ).
For example, we consider the finite poset (X = {x1 , x2 , x3 , x4 , x5 , x6 }, ), represented by the diagram of Fig. 1. The incidence matrix TX and the order-matrix A of X are respectively: X
⎛
TX
1 ⎜0 ⎜0 =⎜ ⎜0 ⎝ 0 0
1 1 0 0 0 0
1 0 1 0 0 0
1 1 1 1 0 0
1 1 0 0 1 0
⎞
⎛
1 1 1⎟ ⎜−2 ⎜ 1⎟ ⎟, A = ⎜−2 X ⎟ ⎜−2 1 ⎠ ⎝ 1 −2 1 −2
2 1 0 −2 −2 −2
2 0 1 −2 0 −2
2 2 2 1 0 −2
2 2 0 0 1 −2
⎞
2 2⎟ 2⎟ ⎟. 2⎟ ⎠ 2 1
We denote by r1 (A ), . . . , rn (A ) the n rows of the order-matrix A and by E (A ) the set of all elements of the orderX X X X
matrix A . In what follows, for the simplicity of the writing, if α is an element of a matrix A (respectively, if α is not an X element of a matrix A), then we shall write α ∈ A (respectively, α ∈ A). 3. A characterization of minimal covers using matrices
In this section, we give a characterization of the minimal covers of an arbitrary finite lattice (X = {x1 , . . . , xn }, ) with order-matrix A = ( ai j ). X The following propositions follows immediately from the definition of the matrix A . X
Proposition 3.1. Let xk1 , xk2 ∈ X. The following statements are true: (1) xk1 xk2 if and only if 1 ∈ rk1 (A ) + rk2 (A ), X X (2) xk1 < xk2 if and only if 3 ∈ rk1 (A ) − rk2 (A ) X X
Proposition 3.2. The subset {x j1 , . . . , x jm } of the lattice X is an antichain if and only if 1 ∈ r ji (A ) + r ji (A ) for every i1 , i2 ∈ X X
{1, . . . , m} with i1 = i2 .
1
2
Definition 3.3. (1) A subset {xj } of X is said to satisfy the cover condition if the following condition is satisfied:
3∈ / r j ( A ) + (−1 ) · rn (A ). X X (2) An antichain subset {x j1 , . . . , x jm } of X, where m ≥ 2, is said to satisfy the cover condition if the following condition is satisfied:
2m + 2 ∈ / r j 1 ( A ) + . . . + r j m ( A ) + (−1 ) · rn (A ). X X X Proposition 3.4. (1) A subset C = {x j } of X is a cover of X if and only if C satisfies the cover condition. (2) An antichain subset C = {x j1 , . . . , x jm } of X, where m ≥ 2, is a cover of X if and only if C satisfies the cover condition. Proof. (1) Let us C = {x j } be a cover of X. We suppose that C does not satisfy the cover condition. Then
3 ∈ r j ( A ) + (−1 ) · rn (A ). X X According to the definition of the order-matrix A , there exists k ∈ {1, . . . , n} such that a jk = 1 and ank = −2. That is, x j = xk X and xk < xn , which is a contradiction, since x j = xn . Conversely, we suppose that C satisfies the cover condition. We prove that C is a cover of X. It suffices to prove that x j = xn . Let suppose that x j = xk , where k = n. Then a jk = 1 and since, always, xk < xn , we have that akn = 2 or equivalently, ank = −2. Thus 3 ∈ r j (A ) + (−1 ) · rn (A ), which is a contradiction. Thus xn = x j . X X (2) Let us C = {x j1 , . . . , x jm }, where m ≥ 2, be a cover of X. We suppose that C does not satisfy the cover condition. Then
2 m + 2 ∈ r j 1 ( A ) + . . . + r j m ( A ) + (−1 ) · rn (A ). X X X
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According to the definition of the order-matrix A and considering that C is an antichain, there exists k ∈ {1, . . . , n} \ X { j1 , . . . , jm , n} such that a j1 k = . . . = a jm k = 2 and ank = −2. Therefore xk ∈ UB ({x j1 , . . . , x jm } ) and xk < xn , which is a contradiction since x j1 ∨ . . . ∨ x jm = xn . Conversely, we suppose that the set C = {x j1 , . . . , x jm } satisfies the cover condition. Since {x j1 , . . . , x jm } is an antichain, x1 ∈ / {x j1 , . . . , x jm }. We suppose that xk = x j1 ∨ . . . ∨ x jm , where k = n. Then xk ∈ UB ({x j1 , . . . , x jm } ) and so x j1 < xk , . . . , x jm < xk or equivalently, a j1 k = . . . = x jm k = 2. Also, we have that xk < xn and thus akn = 2 (or equivalently, ank = −2). Therefore 2m + 2 ∈ r j1 (A ) + . . . + r jm (A ) + (−1 ) · rn (A ), which is a contradiction. Thus xn = x j1 ∨ . . . ∨ x jm . X X X
Proposition 3.5. Let C = {x j1 , . . . , x jm } be an antichain cover of the lattice X. The cover C is minimal if and only if for every i ∈ {1, . . . , m} the following conditions hold: (M1) C \ {x ji } is not a cover of X. (M2) For every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 and xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk1 , xk2 } is not a cover of X. (M3) For every k ∈ {1, . . . , n} \ { j1 , . . . , jm } with xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk } is not a cover of X. Proof. We suppose that the conditions (M1), (M2) and (M3) are satisfied and the set {x j1 , . . . , x jm } is not a minimal cover of the lattice X. Then there exists a refinement R of C such that CࣰR. Hence there exists i ∈ {1, . . . , m} such that x ji ∈ C \ R. By the condition (M1), C \ {x ji } is not a cover of X. Therefore there exists xk ∈ R such that xk < x ji and xk ∈ P l (C \ {x ji } ). We have two cases: (i) There exist k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 and xk1 , xk2 ∈ R ∩ ↓∗ x ji ∩ P l (C \ {x ji } ). In this case (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X. Indeed, we suppose that ∨((C \ {x ji } ) ∪ {xk1 , xk2 } ) = a = 1. Since X is a lattice, xk1 , xk2 < x ji , xk1 , xk2 < a, and xk1 xk2 , we have either x ji < a or a < x ji . If x ji < a, then ∨C a, a contradiction. If a < x ji , then C is not an antichain, a contradiction. Therefore (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X which contradicts the condition (M2). (ii) {xk ∈ R : xk < x ji , xk ∈ P l (C \ {x ji } )} is a chain. In this case we set x p = max{xk ∈ R : xk < x ji , xk ∈ P l (C \ {x ji } )} and we prove that (C \ {x ji } ) ∪ {x p } is a cover of X. We suppose that ∨((C \ {x ji } ) ∪ {x p } ) = a p = 1. Then ∨R ap , a contradiction. Therefore (C \ {x ji } ) ∪ {x p } is a cover of X which contradicts the condition (M3). Conversely, we suppose that C is a minimal cover and let i ∈ {1, . . . , m}. Firstly, we observe that the set C \ {x ji } is not a cover of X. Indeed, if we suppose that the set C \ {x ji } is a cover of X, since it refines C which is a minimal cover of X, then by the definition of the minimal cover, we must have that C ⊆ C \ {x ji }, which is a contradiction. Thus the set C \ {x ji } is not a cover of X. Now, if there exist k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 , xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), and (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X, then R = (C \ {x ji } ) ∪ {xk1 , xk2 } is a refinement of C such that CࣰR, a contradiction. Finally, if there exists k ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ) and (C \ {x ji } ) ∪ {xk } is cover of X, then R = (C \ {x ji } ) ∪ {xk } is a refinement of C such that CࣰR, a contradiction. Proposition 3.6. The cover C = {xn } of X is minimal if and only if every antichain subset A of X, A = {xn }, does not satisfy the cover condition. Proof. Let us C = {xn } be a minimal cover of X. We suppose that there exists an antichain subset A of X, A = {xn }, which satisfes the cover condition. Then by Proposition 3.4, A is a cover of X. Since C = {xn } is a minimal cover of X, we must have xn ∈ A, a contradiction. Thus every antichain subset of X does not satisfy the cover condition. Conversely, we suppose that the cover C = {xn } of X is not minimal. Then there exists a cover R of X which refines C and CࣰR, that is xn ∈ R. By [5, Lemma 3.7], there exists a minimal cover R of X which refines R and xn ∈ R . Since R is a minimal cover of X, R is an antichain cover of X and by Proposition 3.4, R satisfies the cover condition. The last contradicts our assumption that every antichain subset of X, different from {xn }, does not satisfy the cover condition. Thus the cover C = {xn } is minimal. Theorem 3.7. Let C = {x j1 , . . . , x jm } be an antichain cover of the lattice X. The cover C is minimal if and only if for every i ∈ {1, . . . , m} the following conditions hold: (MC1) C \ {x ji } does not satisfy the cover condition.
(MC2) If for every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm }, 1 ∈ rk1 (A ) + rk2 (A ), 3 ∈ r ji (A ) + rk1 (A ), 3 ∈ r ji (A ) + rk2 (A ), 1 ∈ X X X X X X rk1 (A ) + r jl (A ), 1 ∈ rk2 (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}, X X X X then (C \ {x ji } ) ∪ {xk1 , xk2 } does not satisfy the cover condition.
(MC3) If for every k ∈ {1, . . . , n} \ { j1 , . . . , jm }, 3 ∈ r ji (A ) + rk (A ) and X X 1 ∈ rk (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}, X X then (C \ {x ji } ) ∪ {xk } does not satisfy the cover condition.
Proof. Suppose that the conditions (MC1), (MC2) and (MC3) are satisfied and the set {x j1 , . . . , x jm } is not a minimal cover of the lattice X. By Proposition 3.5 there exists i ∈ {1, . . . , m} such that at least one of the following conditions hold: (1) C \ {x ji } is a cover of X.
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(2) There exist k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 , xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), and (C \ {x ji } ) ∪ {xk1 , xk2 } is a cover of X. (3) There exists k ∈ {1, . . . , n} \ { j1 , . . . , jm } with xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ) and (C \ {x ji } ) ∪ {xk } is a cover of X. By Propositions 3.1, 3.2, and 3.4 in each of these cases a contradiction arises. Thus the cover C is minimal. Conversely, suppose that the cover C is minimal and let i ∈ {1, . . . , m}. Then by Proposition 3.5, the following conditions hold: (1) C \ {x ji } is not a cover of X. (2) For every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk1 xk2 and xk1 , xk2 ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk1 , xk2 } is not a cover of X. (3) For every k ∈ {1, . . . , n} \ { j1 , . . . , jm } such that xk ∈ ↓∗ x ji ∩ P l (C \ {x ji } ), (C \ {x ji } ) ∪ {xk } is not a cover of X. By the above conditions, Propositions 3.1, 3.2, and 3.4, the conditions (MC1), (MC2) and (MC3) are satisfied. / Proposition 3.8. If 0 ∈
E ( A ), X
then the cover {xn } is minimal.
Proof. Since 0 ∈ / E ( A ), by Georgiou et al. [8, Remark 2.4], the poset (X, ) is linearly ordered set. Thus every cover of X X contains the top element xn . Therefore {xn } is minimal. Algorithm 3.9. Let (X = {x1 , . . . , xn }, ) be a finite lattice, where n ≥ 2. The algorithm which finds the set MCov(X ) of all minimal covers of X consists of the following steps: Step 1: Find the n rows r1 (A ), . . . , rn (A ) of the order-matrix A of X and go to Step 2. X X X Step Step Step Step Step Step Step
2: 3: 4: 5: 6: 7: 8:
If 0 ∈ / E ( A ), then create MCov(X ) = {{xn }}. Otherwise, go to Step 3. X Set m = 1 and go to Step 4. Find the set Am (X ) of all antichains {x j1 , . . . , x jm } of X which satisfy the cover condition and go to Step 5. If m < n − 2, then put m → m + 1 (that is replace the old m with m + 1) and go to Step 4. Otherwise, go to Step 6. If Am (X ) = ∅ for all m = 2, . . . , n − 2, then create MCov(X ) = {{xn }}. Otherwise, go to Step 7. Create the set A(X ) = A1 (X ) ∪ . . . ∪ An−2 (X ) and go to Step 8. Find the set M(X ) of all {x j1 , . . . , x jm } ∈ A(X ) such that for every i ∈ {1, . . . , m} the following conditions hold: (1) C \ {x ji } does not satisfy the cover condition.
(2) If for every k1 , k2 ∈ {1, . . . , n} \ { j1 , . . . , jm }, 1 ∈ rk1 (A ) + rk2 (A ), 3 ∈ r ji (A ) + rk1 (A ), 3 ∈ r ji (A ) + rk2 (A ), X X X X X X
1 ∈ rk1 (A ) + r jl (A ), 1 ∈ rk2 (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}. then (C \ {x ji } ) ∪ {xk1 , xk2 } does not satX X X X isfy the cover condition. (3) If for every k ∈ {1, . . . , n} \ { j1 , . . . , jm }, 3 ∈ r ji (A ) + rk (A ) and 1 ∈ rk (A ) + r jl (A ) for every l ∈ {1, . . . , m} \ {i}, X X X X then (C \ {x ji } ) ∪ {xk } does not satisfy the cover condition. Subsequently go to Step 9. Step 9: Create MCov(X ) = M(X ). Example 3.10. We consider the finite lattice (X, ) which is represented by the diagram of Fig. 2. The order-matrix of this lattice is
⎛1 ⎜−2 ⎜−2 ⎜ ⎜−2 A = ⎜−2 X ⎜ ⎜−2 ⎝ −2 −2
2 1 0 0 −2 −2 0 −2
2 0 1 0 0 −2 −2 −2
2 0 0 1 −2 0 −2 −2
2 2 0 2 1 0 0 −2
2 2 2 0 0 1 0 −2
2 0 2 2 0 0 1 −2
⎞
2 2⎟ 2⎟ ⎟ 2⎟ ⎟. 2⎟ 2⎟ ⎠ 2 1
x8 x5 x2
x6 x3
x7 x4
x1 Fig. 2. The lattice (X, ).
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Using Algorithm 3.9 we find A1 (X ) = {{x8 }}, A2 (X ) = {{x2 , x7 }, {x3 , x5 }, {x4 , x6 }, {x5 , x6 }, {x5 , x7 }, {x6 , x7 }}, A3 (X ) = {{x2 , x3 , x4 }, {x5 , x6 , x7 }}, A4 ( X ) = A 5 ( X ) = A 6 ( X ) = ∅. Therefore MCov(X ) = {{x2 , x3 , x4 }}. For example, we observe that the set C = {x2 , x3 , x4 } is a minimal cover of X: (i) C satisfies the cover condition. 2·3+2=8∈ / r2 (A ) + r3 (A ) + r4 (A ) + (−1 ) · r8 (A ) X X X X (ii) C \ {x2 } = {x3 , x4 } does not satisfy the cover condition. 2 · 2 + 2 = 6 ∈ r3 (A ) + r4 (A ) + (−1 ) · r8 (A ) X X X C \ {x3 } = {x2 , x4 } does not satisfy the cover condition. 2 · 2 + 2 = 6 ∈ r2 (A ) + r4 (A ) + (−1 ) · r8 (A ) X X X C \ {x4 } = {x2 , x3 } does not satisfy the cover condition. 2 · 2 + 2 = 6 ∈ r2 (A ) + r2 (A ) + (−1 ) · r8 (A ) X X X Also, the others conditions in Step 8 of the above Algorithm are satisfied. 4. The meaning of the order with matrices In this section we are going to characterize the covering dimension of an arbitrary finite lattice (X = {x1 , . . . , xn }, ) with incidence matrix TX = (ti j ) and order-matrix A = (ai j ) providing, firstly, basic results for the meaning of the order. X Notation 4.1. For every m ∈ {1, . . . , n} we consider the n × 1 matrix Bm = (bim ), where
bim
=
m,
if i = 1
0,
otherwise.
Theorem 4.2. Let {x j1 , . . . , x jm } be an antichain subset of X and
k = max(c j1 (TX ) + . . . + c jm (TX ) − Bm ), that is k is the maximum element of the n × 1 matrix c j1 (TX ) + . . . + c jm (TX ) − Bm . Then ord ({x j1 , . . . , x jm } ) = k − 1. Proof. Since max(c j1 (TX ) + . . . + c jm (TX ) − Bm ) = k, by the definition of the matrices TX and Bm , there exist i0 ∈ {2, . . . , n} and jqλ ∈ { j1 , . . . , jm }, λ = 1, . . . , k, such that ti0 jq = 1 for every λ = 1, . . . , k. This means that xi0 x jq for every λ = 1, . . . , k λ
λ
and, hence, x jq ∧ . . . ∧ x jq = x1 . Also, by the definition of k, the infimum of any k + 1 distinct elements of {x j1 , . . . , x jm } is 1
k
equal to x1 . Thus ord({x j1 , . . . , x jm } ) = k − 1.
Example 4.3. We consider the finite lattice (X, ) which is represented by the diagram of Fig. 3. The incidence matrix of this lattice is
⎛1
⎜0 TX = ⎜0 ⎝ 0 0
1 1 0 0 0
1 1 1 0 0
1 1 0 1 0
⎞
1 1⎟ 1⎟. ⎠ 1 1
Based on Theorem 4.2, we shall compute the order of the antichain {x3 , x4 }. We have
⎛1⎞
⎛1⎞
⎛2⎞
⎛ 0⎞
⎜1⎟ ⎜1⎟ ⎜0⎟ ⎜2⎟ c3 (TX ) + c4 (TX ) − B2 = ⎜1⎟ + ⎜0⎟ − ⎜0⎟ = ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 0 1 0 1 0 0 0 0 x5 x3
x4 x2 x1
Fig. 3. The lattice (X, ).
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x9 x6
x7
x8
x3
x4
x5
x2 x1 Fig. 4. The lattice (X, ).
and max(c3 (TX ) + c4 (TX ) − B2 ) = 2. Hence ord({x3 , x4 } ) = 2 − 1 = 1. Example 4.4. We consider the finite lattice (X, ), represented by the diagram of Fig. 4. Based on Theorem 4.2, we show that ord({x3 , x4 , x5 } ) = 2. The incidence matrix TX of X is
⎛
TX
1 ⎜0 ⎜0 ⎜ ⎜0 ⎜ = ⎜0 ⎜0 ⎜ ⎜0 ⎝ 0 0
1 1 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0
1 1 0 1 0 0 0 0 0
1 1 0 0 1 0 0 0 0
1 1 0 1 1 1 0 0 0
Then
1 1 1 0 1 0 1 0 0
1 1 1 1 0 0 0 1 0
⎛ ⎞
⎞
1 1⎟ 1⎟ ⎟ 1⎟ ⎟ 1⎟. 1⎟ ⎟ 1⎟ ⎠ 1 1
⎛ ⎞
⎛ ⎞
3 3 0 ⎜3⎟ ⎜0⎟ ⎜3⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ c3 (TX ) + c4 (TX ) + c5 (TX ) − B3 = ⎜1⎟ − ⎜0⎟ = ⎜1⎟. ⎜ 0 ⎟ ⎜ 0⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ 0⎟ ⎜ 0 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 0 0 0 0 0 0 Hence
ord({x3 , x4 , x5 } ) = max(c3 (TX ) + c4 (TX ) + c5 (TX ) − B3 ) − 1 = 3 − 1 = 2. Thus according to Corollary 1.2, Algorithm 3.9 and Theorem 4.2, we can succeed a computation of the covering dimension of an arbitrary finite lattice using order-matrices and incidence matrices. 5. An algorithm for computing the covering dimension of finite lattices In this section, we present an algorithm which is based on Corollary 1.2 and the results of Sections 3 and 4. This algorithm computes the covering dimension of finite lattices. Notation 5.1. In what follows we denote by M(X ) the set of all antichain subsets {x j1 , . . . , x jm } of X which satisfy the cover condition and the conditions (MC1), (MC2) and (MC3) of Theorem 3.7. Theorem 5.2. Let X be a finite lattice and
L = max{max(c j1 (TX ) + . . . + c jm (TX ) − Bm ) − 1 : {x j1 , . . . , x jm } ∈ M(X )}. Then dim(X ) = L. Proof. It follows by Corollary 1.2, Theorems 3.7 and 4.2. Proposition 5.3. If 0 ∈ /
E ( A ), X
then dim(X ) = 0.
Proof. According to Proposition 3.8, {xn } is a minimal cover of X. Thus every cover of X contains the element xn and so the cover {xn } is refinement of every cover of X. Since ord ({xn } ) = 0, we have dim(X ) = 0. Algorithm 5.4. Let (X = {x1 , . . . , xn }, ) be a finite lattice.
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x6 x3
x4
x5
x2 x1 Fig. 5. The lattice (X, ).
Step 1: Find the n rows r1 (A ), . . . , rn (A ) of the order-matrix A of X. If 0 ∈ / E ( A ), then print dim(X ) = 0. Otherwise, X X X X continue with Step 2. Step 2: Apply Algorithm 3.9 to find the set A(X ) = nm−2 =1 Am (X ) and go to Step 3. Step 3: If Am (X ) = ∅ for all m = 2, . . . , n − 2, then print dim(X ) = 0. Otherwise, go to Step 4. Step 4: Apply Algorithm 3.9 to find the set M(X ) and go to Step 5. Step 5: Find the n columns c1 (TX ), . . . , cn (TX ) of the incidence matrix TX of X and go to Step 6. Step 6: For each m ∈ {1, . . . , n − 2}, find
dm = max{max(c j1 (TX ) + . . . + c jm (TX ) − Bm ) − 1 : {x j1 , . . . , x jm } ∈ M(X )}. Moreover, if Am (X ) ∩ M(X ) = ∅ for some m, then put dm = 0. Subsequently go to Step 7. Step 7: Print dim(X ) = max{d1 , d2 , . . . , dn−2 }. Example 5.5. Following the Steps of Algorithms 3.9 and 5.4 for the finite lattice which is represented in Example 4.3 we have that A1 (X ) = {{x5 }} and d1 = 0, A2 (X ) = {{x3 , x4 }} and d2 = 1, A3 (X ) = ∅ and d3 = 0. Therefore dim(X ) = max{d1 , d2 , d3 } = 1. Example 5.6. (1) We consider the finite lattice (X, ), represented by the diagram of Fig. 5. Following Algorithm 5.4, we find that MCov(X ) = A2 (X ), where
A2 (X ) = {{x3 , x4 }, {x3 , x5 }, {x4 , x5 }}. Moreover,
ord({x3 , x4 } ) = ord({x3 , x5 } ) = ord({x4 , x5 } ) = 1. Therefore dim(X ) = 1. (2) We consider the finite lattice (X, ) which is represented in Example 4.4. Its order-matrix is the following 9 × 9 matrix
⎛
A X
1 ⎜−2 ⎜−2 ⎜ ⎜−2 ⎜ = ⎜−2 ⎜−2 ⎜ ⎜−2 ⎝ −2 −2
2 1 −2 −2 −2 −2 −2 −2 −2
2 2 1 0 0 0 −2 −2 −2
2 2 0 1 0 −2 0 −2 −2
2 2 0 0 1 −2 −2 0 −2
2 2 0 2 2 1 0 0 −2
2 2 2 0 2 0 1 0 −2
2 2 2 2 0 0 0 1 −2
⎞
2 2⎟ 2⎟ ⎟ 2⎟ ⎟ 2⎟. 2⎟ ⎟ 2⎟ ⎠ 2 1
Following Algorithms 3.9 and 5.4, we can observe that {x3 , x4 , x5 } is the only minimal cover with order equal to 2, that is M(X ) = {{x3 , x4 , x5 }} and d3 = 2 (see Example 4.4). Thus we conclude that dim(X ) = d3 = 2. (3) We consider the finite lattice (X, ), represented by the diagram of Fig. 6. The incidence matrix TX and the order-matrix A of X are respectively X
⎛
TX
1 ⎜0 ⎜0 =⎜ ⎜0 ⎝ 0 0
1 1 0 0 0 0
1 1 1 0 0 0
1 1 0 1 0 0
1 1 1 1 1 0
⎞
1 1⎟ 1⎟ ⎟, 1⎟ ⎠ 1 1
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x6 x5 x3
x4 x2 x1
Fig. 6. The lattice (X, ).
⎛
A X
1 ⎜−2 ⎜−2 =⎜ ⎜−2 ⎝ −2 −2
2 1 −2 −2 −2 −2
2 2 1 0 −2 −2
2 2 0 1 −2 −2
2 2 2 2 1 −2
⎞
2 2⎟ 2⎟ ⎟. 2⎟ ⎠ 2 1
We have that A2 (X ) = ∅ since the antichain set C = {x3 , x4 } does not satisfy the cover condition: 2 · 2 + 2 ∈ r3 (A )+ X r4 (A ) + (−1 ) · r6 (A ). Moreover, A3 (X ) = A4 (X ) = ∅. Thus MCov(X ) = {{x6 }} for which d1 = 0 since X X
⎛ ⎞
⎛ ⎞
⎛ ⎞
1 1 0 ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎟ ⎜ ⎟ ⎜ ⎟ c6 (TX ) + B1 = ⎜ ⎜1⎟ − ⎜0⎟ = ⎜1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 1 0 1 1 0 1 and max(c6 (TX ) − B1 ) = 1. 6. Computing the covering dimension of certain finite lattices In this section, we present an another algorithm in order to compute the covering dimension of the linear sum and of the lexicographic product of two finite lattices. Definition 6.1. The lexicographic product X◦Y of two lattices (X, 1 ) and (Y, 2 ) is the lattice (X × Y, ), where the relation is defined as follows:
( x1 , y1 ) ( x2 , y2 ) ⇔
x1 <1 x2 or x1 = x2 and y1 2 y2 .
Definition 6.2. The linear sum XY of two lattices (X, 1 ) and (Y, 2 ), where X ∩ Y = ∅, is the lattice (X ∪ Y, ) where the relation is defined as follows:
xy⇔
x, y ∈ X and x 1 y x, y ∈ Y and x 2 y x ∈ X and y ∈ Y.
Remark 6.3. In study [5], the following two propositions were proved: (1) Let X and Y be finite lattices and
= max{|C | : C is a minimal cover of Y }. Then dim(X ◦ Y ) = − 1. (2) Let X and Y be finite lattices and
k = max{m : there exists a minimal cover {a1 , . . . , am } of Y }. Then dim(X Y ) = k − 1. Using this result, we have the following: Theorem 6.4. Let (X = {x1 , . . . , xn1 }, 1 ) and (Y = {y1 , . . . , yn2 }, 2 ) be two finite lattices and L = max{m : {x j1 , . . . , x jm } ∈ M(Y )}. Then dim(X ◦ Y ) = dim(X Y ) = L − 1.
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285
y4 x2
y2
x1
y3 y1
Fig. 7. The lattices (X, 1 ) and (Y, 2 ).
(x2 , y4 ) y4
(x2 , y2 )
y2
(x2 , y3 )
y3
(x2 , y1 )
y1
(x1 , y4 )
x2
(x1 , y2 )
x1
(x1 , y3 ) (x1 , y1 )
Fig. 8. The lattices (XY, ) and (X◦Y, ∗ ).
Proof. It follows by Theorem 3.7 and Remark 6.3.
Thus using Algorithms 3.9 and 5.4, we have the following: Algorithm 6.5. Let (X = {x1 , . . . , xn1 }, 1 ) and (Y = {y1 , . . . , yn2 }, 2 ) be two finite lattices.
Step 1: Find the n2 rows r1 (AY 2 ), . . . , rn2 (AY 2 ) of the order-matrix AY 2 of Y and go to Step 2.
then print dim(X ◦ Y ) = dim(X Y ) = 0. Otherwise, go to Step 3. n −2 Apply Algorithm 3.9 to find the set A(Y ) = m2=1 Am (Y ) and go to Step 4. If Am (Y ) = ∅ for all m = 2, . . . , n2 − 2, then print dim(X ◦ Y ) = dim(X Y ) = 0. Otherwise, go to Step 5. Apply Algorithm 3.9 to find the set M(Y ) and go to Step 6. Print
Step 2: If 0 ∈ / Step Step Step Step
3: 4: 5: 6:
E (AY ),
dim(X ◦ Y ) = dim(X Y ) = max{m : {x j1 , . . . , x jm } ∈ M(Y )} − 1. Example 6.6. We consider the finite lattices (X, 1 ) and (Y, 2 ), represented by the diagrams of Fig. 7, respectively. The finite lattice (XY, ) and (X◦Y, ∗ ) are represented, respectively, by the diagrams of Fig. 8. Following Algorithm 6.5 we have that MCov(Y ) = {{y2 , y3 }} and thus, dim(X Y ) = dim(X ◦ Y ) = 2 − 1 = 1. Acknowledgments The authors would like to thank the referee for the careful reading of the paper and the useful comments. The fourth author of this paper F. Sereti (with application code 2547) would like to thank the General Secretariat for Research and Technology (GSRT) and the Hellenic Foundation for Research and Innovation (HFRI) for the scholarship during her studies. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]
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