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Journal of Discrete Algorithms
www.elsevier.com/locate/jda
Lattices, whose incomparability graphs have horns
Meenakshi Wasadikar ∗, Pradnya Survase 1
Department of Mathematics, Dr. Babasaheb Ambedkar Marathwada University, Aurangabad 431004, India
a r t i c l e i n f o a b s t r a c t
Article history:Available online 24 July 2013
Keywords:Incomparability graphWidth of a latticeBipartite graphHornDouble star graphZero-divisor graphCozero-divisor graph
In this paper, we study some graphs which are realizable and some which are notrealizable as the incomparability graph (denoted by Γ ′(L)) of a lattice L with at least twoatoms. We prove that the complete graph Kn with two horns is realizable as Γ ′(L). Weshow that the complete graph K3 with three horns is not realizable as Γ ′(L), however it isrealizable as the zero-divisor graph of L. Also we give a necessary and sufficient conditionfor a complete bipartite graph with one horn and with two horns to be realizable as Γ ′(L)
Filipov [5] discuses the comparability graphs of partially ordered sets by defining the adjacency between two elementsof a poset by using the comparability relation, that is a, b are adjacent if either a � b or b � a. Duffus and Rival [4] discussthe covering graph of a poset. The papers of Gedenova [6], Bollobas and Rival [2] discuss the properties of covering graphsderived from lattices. Nimbhorkar, Wasadikar and DeMeyer [9] and Nimbhorkar, Wasadikar and Pawar [10] defined thezero-divisor graph of a meet-semilattice and a lattice L with 0, by defining the adjacency of two elements x, y ∈ L byx ∧ y = 0. Wasadikar and Survase [11,14] defined the zero-divisor graph of a lattice and a meet-semilattice L with 0, bydefining the adjacency of two nonzero elements in the same way. This graph is denoted by Γ (L).
Also, the concept of the cozero-divisor graph of a commutative ring was introduced by Afkhami and Khashyarmanesh in[1]. Let R be a commutative ring with identity and let W (R)∗ be the set of all nonzero and nonunit elements of R . Thecozero-divisor graph of R , which is denoted by Γ ′(R), is an undirected graph with vertex set W (R)∗ , and for two distinctvertices a and b, a is adjacent to b if and only if a /∈ bR and b /∈ aR .
Recently, Bresar et al. [3] introduced the cover incomparability graphs of posets and called these graphs C − I graphsof P . They defined the graph in which the edge set is the union of the edge sets of the corresponding covering graph andthe corresponding incomparability graph.
Wasadikar and Survase [12,13] introduced the incomparability graph of a lattice. In a lattice L, if a,b are incomparablethen we write a ‖ b. Let L be a finite lattice and let W (L) = {x | there exists y ∈ L such that x ‖ y}. The incomparability graphof L, denoted by Γ ′(L), is a graph with the vertex set W (L) and two distinct vertices a,b ∈ W (L) are adjacent if and only ifthey are incomparable. In other words, a is adjacent to b if and only if a /∈ {b}l and b /∈ {a}l , where {z}l = {x ∈ L | x � z} is aprincipal ideal of L. Therefore, in fact Γ ′(L) is the cozero-divisor graph of a lattice L. Note that Γ ′(L) does not contain anyisolated vertex.
1 The second author gratefully acknowledges the financial assistance in the form of Rajiv Gandhi National Senior Research Fellowship from UGC, NewDelhi, India.
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Fig. A. Fig. B.
In this paper, we study some more properties of Γ ′(L). In Section 2 we show that, if G is a graph on five vertices withoutany isolated vertex then G is realizable as Γ ′(L) for some lattice L if and only if G is not isomorphic to a member of a setof four graphs. Also we show when the zero-divisor graph and the incomparability graph of a lattice L are isomorphic.
Yu and Wu in [16] have studied graphs of po-semirings with horns. Let G be a graph. All pendant vertices which areadjacent to the same vertex of G together with edges is called a horn. In Section 3 we show that, the complete graph K3with three horns is not realizable as the incomparability graph of a lattice. However it is realizable as the zero-divisor graphof a lattice L.
Definition 1. In a lattice L with 0, a nonzero element a ∈ L is called an atom if there is no x ∈ L such that 0 < x < a.
Throughout this paper, L is a finite lattice.
Definition 2. Let L1 and L2 be two lattices. The linear sum of L1 and L2, denoted by L1 ⊕ L2, is obtained by placingthe diagram of L1 directly below the diagram of L2 and adding a line segment from the maximum element of L1 to theminimum element of L2.
For example see Figs. A and B. Fig. A is a linear sum of two lattices and Fig. B is a linear sum of a lattice and a chain.Fig. B shows that if the lattice L has only one atom and if W (L) is nonempty then L has the form L = L1 ⊕ L2 where L1
is a chain and its elements do not appear in W (L). Hence throughout this paper, we assume that L is a finite lattice withat least two atoms.
A graph G is connected if there exists a path between any two distinct vertices. A graph G is complete if each pair ofdistinct vertices is joined by an edge. For a positive integer n, we use Kn to denote the complete graph with n vertices.A complete bipartite graph is a simple bipartite graph such that two vertices are adjacent if and only if they belong todifferent partite sets. The complete bipartite graph is denoted by Km,n . A graph in which one vertex is adjacent to everyother vertex and no other adjacencies is called a star graph. A vertex of a graph G is called a pendant vertex if its degree is 1.A graph which is the union of two star graphs whose centers a and b are connected by a single edge is called a double stargraph.
The undefined terms are from West [15], Harary [8] and Grätzer [7].
2. Some realizable and nonrealizable graphs
Wasadikar and Survase [12] have shown that all connected graphs with at most four vertices can be realized as Γ ′(L).In this section we discuss graphs with five vertices. There are 34 graphs with five vertices (see [8, Appendix 1]). We
show that out of these only 19 graphs are realizable as Γ ′(L).
Definition 3. The width of a lattice L is n, where n is a natural number, if and only if there is an antichain in L with nelements and no antichain with n + 1 elements exists in L.
Lemma 1. Kn is a maximal complete subgraph of Γ ′(L) if and only if the width of the lattice L is n.
Proof. If the width of L is n then there is an antichain A = {a1,a2, . . . ,an} in L with n elements and there does not existan antichain with n + 1 elements. For all i �= j, ai ‖ a j , and hence are adjacent. Thus A forms a Kn . �
In [11] we define a graph of a lattice L with 0. We say that an element x ∈ L is a zero-divisor if there exists a nonzeroy ∈ L such that x ∧ y = 0. We denote by Z(L) the set of zero-divisors of L. We associate a graph to L with the vertex set
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Fig. a. Fig. b. Fig. c.
Fig. 1. Fig. 2. Fig. 3. Fig. 4.
Z∗(L) = Z(L) − {0}, the set of all nonzero zero-divisors of L and distinct a,b ∈ Z∗(L) are adjacent if and only if a ∧ b = 0.We call this graph as the zero-divisor graph of L and denote it by Γ (L).
Theorem 2. If Γ (L) has a cycle, then Γ ′(L) has a cycle. However in general the converse is not true.
Proof. Suppose that Γ (L) contains a 3-cycle so x ∧ y = 0, y ∧ z = 0 and z ∧ x = 0 for some x, y, z ∈ Z∗(L). Then we havex ‖ y, y ‖ z and z ‖ x for these x, y, z ∈ W (L). Hence we get a 3-cycle in Γ ′(L). Similar proof can be given if Γ (L) containsan n-cycle.
However, the converse need not hold. Consider the lattice shown in Fig. a. Its zero-divisor graph Γ (L) is a star graphshown in Fig. b and its incomparability graph Γ ′(L) is shown in Fig. c which contains a 3-cycle. �
For a nonempty subset A of a lattice L, we denote the set of all lower bounds of A by Al . Thus Al = {x ∈ L |x � y for all y ∈ A}.
The next theorem characterizes which graphs on five vertices are realizable as the incomparability graph of a lattice.
Theorem 3. Let G be a graph on five vertices without any isolated vertex. Then G is realizable as Γ ′(L) for some L if and only if G isnot isomorphic to any of the four graphs shown in Figs. 1 to 4.
Proof. We know that, in a lattice the greatest lower bound of any nonempty finite subset of L exists. Here we show that,the greatest lower bound of some nonempty finite subset of L does not exist.
Consider Fig. 1. Suppose that G = Γ ′(L) for some lattice L. Since Γ ′(L) contains a 3-cycle, L can contain two or threeatoms and any two atoms are adjacent in Γ ′(L). we have the following cases.
Case (i): Suppose, without loss of generality, L has two atoms d, b. We show that a ∧ c does not exist. From Fig. 1, d anda are comparable and d is an atom hence d � a. Similarly d � c. Also a, e are comparable. If a � e, then d � a implies d � e,a contradiction since d and e are adjacent. Hence e � a. Similarly e � c. Thus {a, c}l = {0,d, e} but d ‖ e hence a ∧ c does notexist.
Now suppose that d, e are the two atoms in L then in a similar manner {a, c}l = {0,d, e} but d ‖ e. Thus a ∧ c does notexist.
Case (ii): Suppose that L has three atoms a, b and c. We show that d ∧ e does not exist. We note that, {d, e}l = {0,a, c}but a ‖ c hence d ∧ e does not exist. So Fig. 1 cannot be realizable as Γ ′(L).
Now consider Fig. 2. Suppose that G = Γ ′(L) for some lattice L. We have the following cases.Case (i): Suppose that L has two atoms a, b. We show that d ∧ e does not exist. Since a is an atom we have a � d and
a � e. From Fig. 2, c and d are comparable. If d � c, then a � d implies a � c, a contradiction since a and c are adjacent.Hence c � d. Similarly c � e. Thus {d, e}l = {0,a, c} but a ‖ c. Hence d ∧ e does not exist.
Suppose that e,d are the two atoms in L then e � a, e � c and d � a, d � c that is {a, c}l = {0,d, e} but e ‖ d. Hence a ∧ cdoes not exist.
Case (ii): Suppose that L has three atoms a, b and c. Then by similar arguments as in case (ii) of Fig. 1, d ∧ e does notexist. So Fig. 2 is not realizable as Γ ′(L).
Consider Fig. 3. Suppose that P5 = Γ ′(L) for some lattice L. Then by Lemma 1 L has exactly two atoms.Let b and c be the two atoms. Then we have b � d, b � e and c � a, c � e.Also we have a � d or d � a.
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Fig. 5. Fig. 6. Fig. 7. Fig. 8.
Fig. 9. Fig. 10. Fig. 11. Fig. 12.
Fig. 13. Fig. 14. Fig. 15. Fig. 16.
Fig. 17. Fig. 18. Fig. 19. Fig. 20.
Fig. 21. Fig. 22. Fig. 23.
If a � d, then c � a implies c � d, a contradiction since c and d are adjacent.If d � a, then b � d implies b � a, a contradiction since a and b are adjacent.Hence neither a � d nor d � a, a contradiction since a and d are not adjacent.Now let d and e be the two atoms in L. We show that a ∧b does not exist. We note that {a,b}l = {0,d, e} but d ‖ e hence
a ∧ b does not exist. So the path P5 cannot be realized as Γ ′(L).Consider Fig. 4. Suppose that G = Γ ′(L) for some lattice L. By Lemma 1, L has exactly two atoms. Let, without any loss
of generality, a and b be the two atoms. Then a � c, a � d and b � e, b � d.Also we have c � e or e � c. If c � e, then a � c implies a � e, a contradiction since a and e are adjacent.If e � c, then b � e implies b � c, a contradiction since b and c are adjacent. Thus neither c � e nor e � c, a contradiction
since c and e are nonadjacent. Hence Γ ′(L) cannot be a 5-gon.To show the converse, as mentioned earlier, Γ ′(L) cannot have any isolated vertex. There are 23 graphs on five vertices
without isolated vertices. Hence there are 19 graphs other than the graphs shown in Figs. 1 to 4. These graphs are shownin Figs. 5 to 23. Each of these 19 graphs is realizable as the incomparability graph of a lattice.
Figs. 5a to 23a are examples of lattices whose incomparability graphs are shown in Figs. 5 to 23 respectively. �However each graph shown in Figs. 1 to 4 can be realized as a subgraph of Γ ′(L) for some lattice L. We have the
following two theorems.
Theorem 4. The graphs shown in Figs. 1 to 3 can be realized as induced subgraphs of Γ ′(L) for some lattice L.
Proof. The graphs shown in Figs. 1 to 3 are the induced subgraphs of Γ ′(L) shown in Figs. 24 to 26 respectively.Figs. 24a to 26a respectively, are the examples of lattices corresponding to the above graphs. �
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Fig. 5a. Fig. 6a. Fig. 7a. Fig. 8a.
Fig. 9a. Fig. 10a. Fig. 11a. Fig. 12a.
Fig. 13a. Fig. 14a. Fig. 15a. Fig. 16a.
Fig. 17a. Fig. 18a. Fig. 19a. Fig. 20a.
Fig. 21a. Fig. 22a. Fig. 23a.
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Fig. 24. Fig. 25. Fig. 26.
Fig. 24a. Fig. 25a. Fig. 26a.
Theorem 5. The graph shown in Fig. 4 can be realized as a subgraph of Γ ′(L) for some lattice L but it cannot be realized as an inducedsubgraph of Γ ′(L).
Proof. The graph shown in Fig. 4 is realizable as a subgraph of Γ ′(L) for some lattice L, see Fig. 17 and the correspondinglattice is shown in Fig. 17a.
Suppose that the graph C5 shown in Fig. 4 is an induced subgraph of Γ ′(L) for some lattice L. Then Γ ′(L) contains acycle C5 that is a − b − c − d − e.
The pairs a, c; a,d; b,d; c, e are comparable.We have (1) a � c or c � a (2) a � d or d � a (3) b � d or d � b (4) c � e or e � c.If a � c then d � a implies d � c, a contradiction since c and d are adjacent. Hence if a � c then a � d.If d � b then a � d implies a � b, a contradiction since a and b are adjacent. Hence if a � c and a � d then b � d.If c � e then a � c implies a � e, a contradiction since a and e are adjacent. Hence if a � c, a � d and b � d then e � c.So from (1), (2), (3) and (4) we have a � c, a � d, b � d and e � c.Also the pair b, e is comparable. Hence we have e � b or b � e.If e � b then b � d implies e � d, a contradiction since e and d are adjacent.If b � e then e � c implies b � c, a contradiction since b and c are adjacent.Hence neither e � b nor b � e. This is a contradiction since b and e are not adjacent.Similarly if we start from c � a then we can get a contradiction.Hence C5 cannot be an induced subgraph of Γ ′(L) for any lattice L. �
Definition 4. Let L be a lattice. A nonzero element a ∈ L is called meet-irreducible if a = b ∧ c implies a = b or a = c.Otherwise it is called meet-reducible.
For example, in Fig. 17a, the elements a, c, d and e are meet-irreducible whereas the element b is meet-reducible.
Theorem 6. The zero-divisor graph and the incomparability graph of a lattice L are isomorphic if and only if L does not contain anymeet-reducible element.
Proof. Suppose that Γ (L) and Γ ′(L) are isomorphic for some lattice L. We want to show that, L does not contain anymeet-reducible element.
Suppose on the contrary, L has a meet-reducible element say b. Then there exist a, c ∈ L and a, c �= b such that b = a ∧ c.Hence a and c are incomparable. So there is an edge a − c in Γ ′(L) but a ∧ c �= 0. So a and c are not adjacent in Γ (L),a contradiction to the assumption that Γ (L) and Γ ′(L) are isomorphic.
Conversely suppose that L does not contain any meet-reducible element. We want to show that, Γ (L) and Γ ′(L) areisomorphic. Let a − b be an edge in Γ ′(L). Then a ‖ b in L. Since L does not contain any meet-reducible element, a ∧ b �= cfor any nonzero c ∈ L. Hence c = 0, that is a − b is an edge in Γ (L). On the other hand if a − b is an edge in Γ (L) thena ∧ b = 0, hence a ‖ b in L. Thus a − b is an edge in Γ ′(L). Hence Γ (L) and Γ ′(L) are isomorphic. �
The lattice shown in Fig. 27 is denoted by Mn .
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Fig. 27. The lattice Mn . Fig. 28. A lattice whose Γ ′(L) is Km,n . Fig. 29. A graph with a horn. Fig. 30. A lattice whose Γ ′(L) is Kn(1).
Corollary 7. The zero-divisor graph and the incomparability graph of the lattice Mn are isomorphic.
Proof. Suppose that L = Mn . Since Mn does not contain any meet-reducible element, Γ (L) and Γ ′(L) are isomorphic byTheorem 6. �Theorem 8. The complete graph Kn is realizable as the incomparability graph of a lattice.
Proof. Consider the complete graph Kn . Let ai , i = 1,2, . . . ,n be the vertices of Kn . The corresponding lattice is as shownin Fig. 27. �Theorem 9. Any complete bipartite graph Km,n is realizable as the incomparability graph of a lattice.
Proof. Consider the complete bipartite graph Km,n . Let V 1 = {a1,a2, . . . ,an} and V 2 = {b1,b2, . . . ,bm} be the two partitions.The corresponding lattice is shown in Fig. 28. �3. Graphs with horns
Let G be a graph. All pendant vertices which are adjacent to the same vertex of G together with edges is called a horn.For example, in Fig. 29, X = {x1, x2, x3, x4} together with the edges a − x1, a − x2, a − x3, a − x4 is a horn at a, and is
denoted as a − X .We denote the complete graph Kn together with m horns X1, X2, . . . , Xm by Kn(m) where a1 − X1,a2 − X2, . . . ,am − Xm ,
ai ∈ V (Kn) and 0 � m � n.We note that K1(1), K2(1) and K2(0) are star graphs, K2(2) is a double star graph.We prove some results for complete graphs with horns.
Theorem 10. The double star graph K2(2) is realizable as the incomparability graph of a lattice.
Proof. Let G = Γ ′(L) be a double star graph with centers a1, b1 and end vertices b j , j = 2,3, . . . ,m and ai , i = 2, . . . ,n. Thecorresponding lattice is shown in Fig. 36. �Theorem 11. The complete graph Kn with 1 horn, Kn(1), n � 3 is realizable as the incomparability graph of a lattice.
Proof. Consider the complete graph Kn . Let X be a horn in Kn at the vertex an where X = {x1, x2, . . . , xm} and let ai ,i = 1,2, . . . ,n be the vertices of Kn . The corresponding lattice is shown in Fig. 30. �Corollary 12. The complete graph K3 with 1 horn, K3(1) is realizable as the incomparability graph of a lattice.
Proof. Consider the complete graph K3. Let a,b and c be the three vertices of K3 and let X be a horn at c. Let X ={x1, x2, . . . , xn}. The corresponding lattice is shown in Fig. 31. �Lemma 13. The complete graph K3 with 2 horns, K3(2), is realizable as the incomparability graph of a lattice.
Proof. Consider the complete graph K3. Let a,b and c be the three vertices of K3 and let X and Y be horns at a and brespectively. Let X = {x1, x2, . . . , xn} and Y = {y1, y2, . . . , ym}. The corresponding lattice is as shown in Fig. 32. �Theorem 14. The complete graph K3 with 3 horns, K3(3), is not realizable as the incomparability graph of a lattice.
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Fig. 31. A lattice whose Γ ′(L) is K3(1). Fig. 32. A lattice whose Γ ′(L) is K3(2). Fig. 33. The graph K3(3). Fig. 34. A lattice whose Γ (L) is K3(3).
Proof. Consider the complete graph K3. Let a,b and c be the three vertices of K3 and let X , Y and Z be horns at a, b andc respectively. Let X = {x1, x2, . . . , xn}, Y = {y1, y2, . . . , ym} and Z = {z1, z2, . . . , zp}.
Case (i): Suppose that L has two atoms a and b. Then a � z j for j = 1,2, . . . , p, a � yk for k = 1,2, . . . ,m and b � xi fori = 1,2, . . . ,n, b � z j for j = 1,2, . . . , p.
Also we have yk � xi or xi � yk .If yk � xi , then a � yk implies a � xi , a contradiction since a and xi are adjacent.If xi � yk , then b � xi implies b � yk , a contradiction since b and yk are adjacent.Hence neither yk � xi nor xi � yk , a contradiction since xi and yk are not adjacent.Suppose that a and x1 are the two atoms in L. We have a � z j for j = 1,2, . . . , p, a � yk for k = 1,2, . . . ,m and x1 � xi
for i = 2, . . . ,n, x1 � z j for j = 1,2, . . . , p, x1 � b, x1 � c, x1 � yk for k = 1,2, . . . ,m.Also we have (i) yk � c or c � yk , (ii) yk � z j or z j � yk , (iii) b � z j or z j � b.If yk � c, then a � yk implies a � c, a contradiction since a and c are adjacent. Hence c � yk .If yk � z j , then c � yk implies c � z j , a contradiction since c and z j are adjacent. Hence z j � yk .We have b � z j or z j � b.If b � z j , then z j � yk implies b � yk , a contradiction since b and yk are adjacent.If z j � b, then a � z j implies a � b, a contradiction since a and b are adjacent.Hence neither b � z j nor z j � b, a contradiction since b and z j are not adjacent.Case (ii): Suppose that L has three atoms a, b and c. Then we have a � z j , a � yk , b � xi , b � z j , c � xi and c � yk .Also we have yk � xi or xi � yk .If yk � xi , then a � yk implies a � xi , a contradiction since a and xi are adjacent.If xi � yk , then b � xi implies b � yk , a contradiction since b and yk are adjacent.Hence neither yk � xi nor xi � yk since xi and yk are not adjacent. Hence K3(3) cannot be realized as Γ ′(L). �
Remark 1. However K3(3) realizable as the zero-divisor graph of a lattice L. Fig. 33 is the zero-divisor graph of the latticeshown in Fig. 34.
Theorem 15. The complete graph Kn with 2 horns, Kn(2), n � 4 is realizable as the incomparability graph of a lattice.
Proof. Consider the complete graph Kn . Let ai , i = 1,2, . . . ,n be the vertices of Kn and let X and Y be horns at a1 and anrespectively. Let X = {x1, x2, . . . , xm} and Y = {y1, y2, . . . , yp}. The corresponding lattice is shown in Fig. 35. �Theorem 16. The complete graph Kn with m horns, Kn(m), n � m � 3 is not realizable as the incomparability graph of a lattice.
Proof. Consider the complete graph Kn . Let ai , i = 1,2, . . . ,n be the vertices of Kn . Let X , Y and Z be the horns at verticesa1, a2 and a3 respectively with X = {x1, x2, . . . , xm}, Y = {y1, y2, . . . , yp} and Z = {z1, z2, . . . , zq}.
Case (i): Suppose, without any loss of generality, L has two atoms a1 and a2. Then a1 � yk for each k, a2 � xt for each t ,a1 � zr and a2 � zr for each r.
Also we have xt � yk or yk � xt .If xt � yk then a2 � xt implies a2 � yk , a contradiction since a2 and yk are adjacent.If yk � xt then a1 � yk implies a1 � xt , a contradiction since a1 and xt are adjacent.Hence neither xt � yk nor yk � xt , a contradiction since yk and xt are not adjacent.Case (ii): Suppose that a1 and x1 are the two atoms. Then a1 � yk , x1 � yk for each k and a1 � zr and x1 � zr for each r.Also we have (1) a3 � y1 or y1 � a3 and (2) a2 � zr or zr � a2 and (3) y1 � z1 or z1 � y1.If y1 � a3 then a1 � y1 implies a1 � a3, a contradiction since a1 and a3 are adjacent. Hence a3 � y1.If zr � a2 then a1 � zr implies a1 � a2, a contradiction since a1 and a2 are adjacent. Hence a2 � zr .
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Fig. 35. A lattice whose Γ ′(L) is Kn(2). Fig. 36. A lattice whose Γ ′(L) is K2(2). Fig. 37. A lattice whose Γ ′(L) is K2,2(2).
If y1 � z1 then a3 � y1 implies a3 � z1, a contradiction since a3 and z1 are adjacent.If z1 � y1 then a2 � z1 implies a2 � y1, a contradiction since a2 and y1 are adjacent.Hence neither y1 � z1 nor z1 � y1, a contradiction since y1 and z1 are not adjacent.Case (iii): Suppose that a1 and ai (i �= 2,3) are the two atoms. Then a1 � yk , ai � yk for each k and a1 � zr and ai � zr
for each r, ai � xt for each t .We have xt � zr or zr � xt .If zr � xt then a1 � zr implies a1 � xt , a contradiction since a1 and xt are adjacent. Hence xt � zr .Also we have a3 � xt or xt � a3.If a3 � xt then xt � zr implies a3 � zr , a contradiction since a3 and zr are adjacent.If xt � a3 then ai � xt implies ai � a3, a contradiction since a3 and ai are adjacent.Hence neither a3 � xt nor xt � a3, a contradiction since a3 and xt are not adjacent.Case (iv): Suppose that ai , a j ; i, j �= 1,2,3 are the two atoms. Then a j � yk , ai � yk for each k and a j � zr and ai � zr
for each r, ai � xt , a j � xt for each t .We have (I) a2 � xt or xt � a2 and (II) a1 � yk or yk � a1.If xt � a2 then ai � xt implies ai � a2, a contradiction since ai and a2 are adjacent. Hence a2 � xt .If yk � a1 then ai � yk implies ai � a1, a contradiction since ai and a1 are adjacent. Hence a1 � yk .Also we have xt � yk or yk � xt .If xt � yk then a2 � xt implies a2 � yk , a contradiction since a2 and yk are adjacent.If yk � xt then a1 � yk implies a1 � xt , a contradiction since a1 and xt are adjacent.Hence neither xt � yk nor yk � xt , a contradiction since yk and xt are not adjacent. �
Remark 2. Let Km,n be the complete bipartite graph with partitions V 1 = {a1,a2, . . . ,an} and V 2 = {b1,b2, . . . ,bm}. Then byTheorem 9, Km,n is realizable as Γ ′(L). Since the ai are nonadjacent in Γ ′(L), they are comparable in L. So we can arrangethem as a1 < a2 < a3 < · · · < an . Similarly, we can arrange b j ’s as b1 < b2 < · · · < bm .
If X = {x1, x2, . . . , xp} is a horn then all xi, x j are nonadjacent in Γ ′(L) and so are comparable in L. Hence we can arrangethem as x1 < x2 < x3 < · · · < xk .
We now prove some theorems for complete bipartite graphs with horns. We shall arrange the vertices in ascending orderusing Remark 2. We denote the complete bipartite graph Km,n together with p horns by Km,n(p).
Theorem 17. K2,2(2) is realizable as Γ ′(L) if and only if the two horns are at vertices a1 and b2 , as shown in Fig. i below.
Proof. Consider the complete bipartite graph K2,2. Let V 1 = {a1,a2} and V 2 = {b1,b2} be the two partitions. Let X ={x1, x2, . . . , xn} and Y = {y1, y2, . . . , ym} be the two horns. If the two horns are at a1 and b2 respectively as shown inFig. i, then the corresponding lattice is as shown in Fig. 37.
To prove the converse, we consider the two cases.Case (i): Suppose that the horns X and Y are at a1 and a2 respectively, see Fig. ii. Let this graph be realizable as Γ ′(L)
for some lattice L. Clearly L does not contain three atoms as K2,2 does not contain K3 as a subgraph.Subcase (i): Suppose that a1 and b1 are the two atoms.Then a1 � a2, a1 � yk , b1 � yk for each k and b1 � xi for each i.Also we have a2 � x1 or x1 � a2.If a2 � x1, then a1 � a2 implies a1 � x1, a contradiction since a1 and x1 are adjacent.If x1 � a2, then b1 � x1 implies b1 � a2, a contradiction since a2 and b1 are adjacent.
72 M. Wasadikar, P. Survase / Journal of Discrete Algorithms 23 (2013) 63–75
Fig. i. K2,2 having horns ata1 and b2.
Fig. ii. K2,2 having horns ata1 and a2.
Fig. iii. K2,2 having horns at a1
and b1.
Fig. iv. K2,2 having three horns. Fig. v. K2,2 having four horns.
Hence neither a2 � x1 nor x1 � a2, a contradiction since x1 and a2 are not adjacent.Subcase (ii): Suppose that a1 and x1 are the two atoms. Then a1 � a2, x1 � b1, x1 � b2, x1 � a2, a1 � yk and x1 � yk for
each k.We know that, in a lattice the greatest lower bound of any nonempty finite subset of L exists. We now show that the
greatest lower bound of A = {a2, y1, y2, . . . , ym} does not exist. The possible set of lower bounds of A is {0,a1, x1, . . . , xn}.If a1 is the greatest lower bound, then xi � a1, a contradiction since a1 is an atom.
If any xi is the greatest lower bound then a1 � xi , a contradiction since a1 ‖ xi . Hence the greatest lower bound of Adoes not exist. So K2,2 is not realizable as Γ ′(L) if both the horns are at vertices a1 and a2 respectively.
Case (ii): Suppose that both the horns are at vertices a1 and b1 respectively see Fig. iii.Subcase (i): Suppose that a1 and b1 are the two atoms. Then by similar manner as in case (i) we get a contradiction.Subcase (ii): If a1 and x1 are the two atoms then a1 � a2, x1 � b1, x1 � b2, x1 � a2, a1 � yk and x1 � yk for each k.By Remark 2 we have b1 � b2.Also we have b2 � yk or yk � b2.If b2 � yk , then b1 � b2 implies b1 � yk , a contradiction since b1 and yk are adjacent.If yk � b2, then a1 � yk implies a1 � b2, a contradiction since a1 and b2 are adjacent.Hence neither b2 � yk nor yk � b2, a contradiction since b2 and yk are not adjacent.Hence K2,2(2) is not realizable as Γ ′(L) if both the horns are at vertices a1 and b1 respectively. �
Theorem 18. K2,2(p), p = 3,4 is not realizable as Γ ′(L) for any lattice L.
Proof. (I) Suppose that p = 3.Consider the complete bipartite graph K2,2. Let V 1 = {a1,a2} and V 2 = {b1,b2} be the two partitions. Let X =
{x1, x2, . . . , xn}, Y = {y1, y2, . . . , ym} and Z = {z1, z2, . . . , zr} be the three horns.Suppose the three horns are at a1, a2 and b2 respectively as shown in Fig. iv.Clearly L does not contain three atoms as K2,2 does not contain K3 as a subgraph.Case (i): Suppose that a1 and b1 are the two atoms. Then a1 � a2, a1 � z j , b1 � z j for each j, a1 � yk , b1 � yk for each k
and b1 � xi for each i.Also we have a2 � xi or xi � a2 for each i.If a2 � xi then a1 � a2 implies a1 � xi , a contradiction since a1 and xi are adjacent.If xi � a2 then b1 � xi implies b1 � a2, a contradiction since a2 and b1 are adjacent.Hence neither a2 � xi nor xi � a2, a contradiction since a2 and xi are not adjacent.Case (ii): Suppose that a1 and x1 are the two atoms. Then a1 � a2, a1 � z j , x1 � z j for each j, a1 � yk , x1 � yk for each k,
x1 � b1, x1 � b2 and x1 � a2.By Remark 2 we have b1 � b2.Also we have (1) b2 � yk or yk � b2 and (2) a2 � z j or z j � a2 and (3) b1 � z j or z j � b1.If z j � b1 in (3) then b1 � b2 implies z j � b2, a contradiction since z j and b2 are adjacent. Hence b1 � z j .If z j � a2 in (2) then b1 � z j implies b1 � a2, a contradiction since b1 and a2 are adjacent. Hence a2 � z j .
M. Wasadikar, P. Survase / Journal of Discrete Algorithms 23 (2013) 63–75 73
Fig. 38. Fig. 39. Fig. 40. Fig. 41.
If yk � b2 in (1) then a1 � yk implies a1 � b2, a contradiction since a1 and b2 are adjacent. Hence b2 � yk .Thus from (1), (2) and (3) we have the only possibilities b1 � z j , a2 � z j and b2 � yk .Also we have z j � yk or yk � z j since yk, z j are not adjacent.If z j � yk then a2 � z j implies a2 � yk , a contradiction since a2 and yk are adjacent.If yk � z j then b2 � yk implies b2 � z j , a contradiction since b2 and z j are adjacent.Hence neither z j � yk nor yk � z j , a contradiction since z j and yk are not adjacent.In a similar manner we can prove that if the three horns are at vertices a1, a2 and b1 then the graph cannot be realized
as Γ ′(L). Hence K2,2(3) is not realizable as Γ ′(L) if the three horns are at vertices a1, a2 and b1 or at vertices a1, a2 andb2 respectively.
(II) Suppose that p = 4.Let X , Y , Z and W be the four horns are at vertices a1, a2, b2 and b1 respectively as shown in Fig. v.Case (1): Suppose that a1 and b1 are the two atoms. Then in a similar manner as in case (i) we get a contradiction for
the pair a2 and xi .Case (2): Suppose that a1 and x1 are the two atoms. Then in a similar manner as in case (ii) we get a contradiction for
the pair yk and z j .Hence K2,2(4) cannot be realized as Γ ′(L). �
Theorem 19. A complete bipartite graph with a horn, that is Km,n(1) is realizable as Γ ′(L) for some lattice L if and only if the horn isat vertex a1 or bm or an or b1 .
Proof. Consider the complete bipartite graph Km,n . Suppose that V 1 = {a1,a2, . . . ,an} and V 2 = {b1,b2, . . . ,bm} are the twopartitions. Let X = {x1, x2, . . . , xr} be a horn.
If the horn is at vertex a1 then the corresponding lattice is shown in Fig. 38.If the horn is at vertex an then the corresponding lattice is shown in Fig. 39.If the horn is at vertex b1 then the corresponding lattice is shown in Fig. 40.If the horn is at vertex bm then the corresponding lattice is shown in Fig. 41.Conversely, consider the complete bipartite graph Km,n and without any loss of generality let the horn be at some vertex
from the partite set V 1.Let the horn be at ai , i �= 1,n.Let this graph be realizable as Γ ′(L) for some lattice L. Clearly L does not contain three atoms as Km,n does not contain
K3 as a subgraph.Suppose that a1 and b1 are the two atoms. We have a1 � xi , b1 � xi for each i.Also we have ai+1 � x1 or x1 � ai+1.If ai+1 � x1 then ai � ai+1 implies ai � x1, a contradiction since ai and x1 are adjacent.If x1 � ai+1 then b1 � x1 implies b1 � ai+1, a contradiction since b1 and ai+1 are adjacent.Hence neither ai+1 � x1 nor x1 � ai+1, a contradiction since ai+1 and x1 are not adjacent. Thus Km,n(1) is not realizable
if the horn is at some ai (i �= 1,n) or is at some b j ( j �= 1,m). �Theorem 20. A complete bipartite graph with two horns, that is Km,n(2), m > 2 or n > 2 is realizable as Γ ′(L) for some lattice L ifand only if the two horns are at vertices a1 , an or at vertices a1 , bm.
74 M. Wasadikar, P. Survase / Journal of Discrete Algorithms 23 (2013) 63–75
Fig. 42. Lattice whose Γ ′(L) is Km,n
with horns at a1, an .Fig. 43. Lattice whose Γ ′(L) is Km,n
with horns at a1, bm .
Proof. Consider the complete bipartite graph Km,n . Suppose, without loss of generality, n > 2. Let V 1 = {a1,a2, . . . ,an} andV 2 = {b1,b2, . . . ,bm} be the two partitions. Let X = {x1, x2, . . . , xp} and Y = {y1, y2, . . . , yr} be the two horns. If the hornsare at a1 and an respectively then the corresponding lattice is shown in Fig. 42. If the horns are at a1 and bm respectivelythen the corresponding lattice is shown in Fig. 43.
Conversely, consider the complete bipartite graph Km,n with partitions V 1 = {a1,a2, . . . ,an} and V 2 = {b1,b2, . . . ,bm}.(I) Let both the horns be at vertices from the same partite set, say V 1.Suppose that X and Y are the two horns at a1 and ai , i �= n, respectively where X = {x1, x2, . . . , xp} and Y =
{y1, y2, . . . , yr}. Let this graph be realizable as Γ ′(L) for some lattice L. Clearly L does not contain three atoms as Km,n
does not contain K3 as a subgraph.Case (i): Suppose that a1 and b1 are the two atoms. We have a1 � y j and b1 � y j for j = 1,2, . . . , r and b1 � xl ,
l = 1,2, . . . , p.Also we have a2 � x1 or x1 � a2.If a2 � x1, then a1 � a2 implies a1 � x1, a contradiction since a1 and x1 are adjacent.If x1 � a2, then b1 � x1 implies b1 � a2, a contradiction since b1 and a2 are adjacent.Hence neither a2 � x1 nor x1 � a2, a contradiction since a2 and x1 are not adjacent.Case (ii): Suppose that a1 and x1 are the two atoms. Since x1, x2, . . . , xp are comparable we can arrange them as x1 <
x2 < · · · < xp . Similarly we have y1 < y2 < · · · < yr , a1 < a2 < · · · < an and b1 < b2 < · · · < bm . Now xk � y1 or y1 � xk foreach k. If y1 � xk then a1 � y1 implies a1 � xk , a contradiction. Hence xk � y1 for each k. Thus we have x1 < x2 < · · · <
xp < y1 < y2 < · · · < yr .Now yr � ai+1 or ai+1 � yr .If ai+1 � yr then ai � yr , a contradiction. Hence yr � ai+1. Thus we have the chain x1 < x2 < · · · < xp < y1 < y2 < · · · <
yr < ai+1 < · · · < an .Now for k � i − 1, either y j � ak or ak � y j for each j. If y j � ak then ak � ai implies y j � ai , a contradiction. Hence
ak � y j .Now since y1, b1 are not adjacent, we have y1 � b1 or b1 � y1. If y1 � b1 then a2 � y1 implies a2 � b1, a contradiction
since a2 and b1 are adjacent.If b1 � y1 then y1 � ai+1 implies b1 � ai+1, a contradiction since b1 and ai+1 are adjacent. Hence neither y1 � b1 nor
b1 � y1and y1 and b1 are not adjacent, a contradiction.(II) Let X and Y be the two horns at a1 and b j , j �= m, respectively. Let this graph be realizable as Γ ′(L) for some
lattice L.Case (i): Suppose that a1 and b1 are the two atoms. We have a1 � y j for j = 1,2, . . . , r, b1 � xk , k = 1,2, . . . , p.Also we have a2 � x1 or x1 � a2.If a2 � x1 then a1 � a2 implies a1 � x1, a contradiction since a1 and x1 are adjacent.If x1 � a2 then b1 � x1 implies b1 � a2, a contradiction since b1 and a2 are adjacent.Hence neither a2 � x1 nor x1 � a2, a contradiction since a2 and x1 are not adjacent.Case (ii): Suppose that a1 and x1 are the two atoms. We have a1 � y j , x1 � y j for each j, x1 � xk and x1 � b j for each j.Also we have b j+1 � y j or y j � b j+1.If y j � b j+1 then a1 � y j implies a1 � b j+1, a contradiction since a1 and b j+1 are adjacent.If b j+1 � y j then b j � b j+1 implies b j � y j , a contradiction since b j and y j are adjacent.Hence neither b j+1 � y j nor y j � b j+1, a contradiction since y j and b j+1 are not adjacent.(III) Let X and Y be the two horns at ai and b j , j �= 1,m and i �= 1,n respectively. Let this graph be realizable as Γ ′(L)
for some lattice L. Suppose that a1 and b1 are the two atoms. Then a1 � xi , b1 � xi for each i.
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Also we have ai+1 � x1 or x1 � ai+1.If ai+1 � x1 then ai � ai+1 implies ai � x1, a contradiction since ai and x1 are adjacent.If x1 � ai+1 then b1 � x1 implies b1 � ai+1, a contradiction since b1 and ai+1 are adjacent.Hence neither ai+1 � x1 nor x1 � ai+1, a contradiction since x1 and ai+1 are not adjacent. �From Theorems 19 and 20 we get the following.
Theorem 21. A complete bipartite graph Km,n, m > 2 or n > 2 with an additional third horn at any vertex is not realizable as Γ ′(L).
Acknowledgement
The authors are thankful to the referees for many fruitful suggestions for improvement of the paper.
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