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Discrete Applied Mathematics
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Characterization of common-edge sigraph
Deepa Sinha ∗, Somya Upadhyaya, Priya KatariaCentre for Mathematical Sciences, Banasthali University, Banasthali-304022, Rajasthan, India
a r t i c l e i n f o
Article history:Received 17 May 2009Received in revised form 12 September2012Accepted 7 October 2012Available online 8 November 2012
Keywords:SigraphCommon-edge graphIterated line sigraphCommon-edge sigraph
a b s t r a c t
A sigraph is a graph G in which each edge x carries a value s(x) ∈ −1,+1 called its sign,denoted specially as S = (G, s). Given a sigraph S, a new sigraph CE(S), called the common-edge sigraph of S is that sigraph whose vertex-set is the set of pairs of adjacent edges inS and two vertices of CE(S) are adjacent if the corresponding pairs of adjacent edges ofS have exactly one edge in common, and the sign of the edge is the sign of the commonedge. If all the edges of the sigraph S carry + sign then S is actually a graph and thecorresponding common-edge sigraph is termed as the common-edge graph. In this paper,we characterize common-edge graph and common-edge sigraph and write an algorithmto obtain a corresponding common-edge root graph and common-edge root sigraph from agiven common-edge graph and common-edge sigraph respectively.
For all terminology and notations in graph theory, except for those that are specifically defined here, the reader is referredto West [12]. The graphs considered here are finite, undirected, without self-loops or multiple edges. By a (p, q)-graph Gwemean a graph having p vertices and q edges; p is called the order and q is called the size of G.
In the spirit of a study of graph-valued functions, obtaining the line graph L(G) of a given graph G = (V , E)may be treatedas amapping L that operates on G to give rise to L(G) as the graphwhose vertices are the edges of Gwith two of these verticesjoined to each other (or, adjacent) whenever the edges of G they represent have a common vertex in G or equivalently thetwo edges form a P3 in G.
Broersma and Hoede [4] defined in general path graphs Pk(G) of G for any positive integer k as follows: Pk(G) has for itsvertex-set the set P G
k of all distinct paths in G having k vertices, and two vertices in Pk(G) are adjacent if they represent twopaths P,Q ∈ P G
k whose union forms either a path Pk+1 or a cycle Ck inG. Some improvement of their paperwas subsequentlygiven by [10,2,11].
Much earlier, making independently the same observation as above on the formation of a line graph L(G) of a given graphG, Kulli [8] had defined the common-edge graph CE(G) of G as the intersection graph of the family P3(G) of 2-paths (i.e., pathsof length two) each member of which is treated as a set of edges of the corresponding 2-path; as shown by him, it is notdifficult to see that
CE(G) ∼= L2(G) (1)
for any isolate-free graph G, where L(G) := L1(G) and Lt(G) denotes the t-th iterated line graph of G for any integer t ≥ 2.
Research is supported by the Department of Science and Technology (Govt. of India), New Delhi, India under the project SR/S4/MS:409/06.∗ Correspondence to: South Asian University, Akbar Bhawan, Chanakyapuri, New Delhi-110021, India.
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Fig. 1. Common-edge graph and Path graph.
Further, it is easy to verify that for any graph G, P3(G) can be considered as a spanning subgraph of CE(G). Fig. 1 illustratesCE(G) and P3(G) for a given graph G. Furthermore, it is not hard to establish the following two results.
Lemma 1. For any graph G of order at least three, P3(G) is triangle-free if and only if G is triangle-free.
Proposition 2. For any connected graph G of order at least three, CE(G) ∼= P3(G) if and only if G is isomorphic to either a pathor a cycle.
The degree of a vertex v ∈ V (G) in G is denoted dG(v) or just as d(v) if G is clear from the context, and for any edge e = uvits edge-degree is defined as the number d1(e) = d(u)+ d(v)− 2. The following result has been proved by Kulli.
Theorem 3 ([8]). If G is a (p, q)-graph, then CE(G) has,
12
pi=1
d(vi)2 − q
vertices and
12
qj=1
d1(ej)2 −12
pi=1
d(vi)2 + q
edges.
A graph G is called a common-edge graph whenever there exists a graph H such that G ∼= CE(H); H is then called acommon-edge root graph of G.
The notion of CE(G) has been extended to the realm of sigraphs [1]. As in [7] (also see, [5]) by a sigraph S wemean a graphG = (V , E), called the underlying graph of S and denoted by Su, in which each edge x carries a value s(x) ∈ −1,+1 calledits sign; an edge x is positive or negative according to whether s(x) = +1 or s(x) = −1. The set of positive edges of S isdenoted by E+(S) and E−(S) = E(G)− E+(S) is the set of negative edges of S. We regard graphs themselves as sigraphs inwhich every edge is positive. Given a graph G, letψ(G) denote the set of all sigraphs whose underlying graph is G. In general,a subgraph S ′ of a sigraph S is said to be all-positive (all-negative) if all the edges of S ′ are positive (negative). A sigraph is saidto be homogeneous if it is either all-positive or all-negative and heterogeneous otherwise. Cliques are defined as the completesubgraphs of the graph.
Given a sigraph S, its common-edge sigraph CE(S) is that sigraph whose vertex-set is the set of pairs of adjacent edges inS and two vertices of CE(S) are adjacent if the corresponding pairs of adjacent edges of S have exactly one edge in common,with the same sign as that of their common edge.
A sigraph S is called a common-edge sigraph if the set ζ EΓS = H ∈ S : S ∼= CE(H) is nonempty, where S denotes the
set of all sigraphs and each member ζ EΓS is then called a common-edge root sigraph (or CE-root in short) of S. We shall call
SCE-primitive if ζ EΓS = ∅.
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Fig. 2. Beineke’s nine forbidden subgraphs.
2. Characterization of common-edge graph and common-edge sigraph
In order to give the characterization of a common-edge graph, we would refer the following theorem which is wellknown characterization of a line graph given in most of the standard text-books on graph theory (e.g., see [6, Chapter 8,p. 74]), originally due to Beineke [3], where a triangle of a graph G is called odd if there is a vertex of G adjacent to an oddnumber of its vertices and is even otherwise.
Theorem 4. The following statements are equivalent:
(a) G = (V , E) is a line graph.(b) The edges of G can be partitioned into some of its complete subgraphs in such a way that no vertex lies in more than two of
the subgraphs.(c) G does not have K1,3 as an induced subgraph, and if two odd triangles have a common edge then the subgraph induced by
their vertices is K4.(d) None of the nine subgraphs shown in Fig. 2 is an induced subgraph of G.
We shall now embark on obtaining characterization of common-edge graph:
Theorem 5. The following statements are equivalent for any simple connected graph G:
(1) G is a common-edge graph,(2) The edges of G can be partitioned into complete subgraphs Si having the following properties:
(a) Every vertex of G lies in at most two such subgraphs,(b) Each such subgraph contains at most one vertex that lies in no other of the subgraphs,(c) The graph Q whose vertices are the complete subgraphs along with the vertices of G that appear in exactly one such
subgraph, with edges between any two that have a vertex of G in common, does not contain any of the nine Beinekegraphs (Fig. 2) as induced subgraphs,
(d) Every edge of G lies in exactly one such subgraph.
Proof. (1) implies (2): Let G be a connected common-edge graph CE(H) of some graph H , having at least three vertices. Thenthe pairs of paths of length two in H that have a particular edge in common form a complete subgraph of G, and every edgeof G lies in exactly one such subgraph. Since each vertex of G consists of two edges of H it can consist of at most the twosubgraphs of G corresponding to those two edges. A vertex of G will lie in only one such subgraph only if it corresponds toa pair of edges of H one of which has degree one at one of its vertices and degree two at the other (so that no two two-edgepaths of H both contain that edge of G). The graph Q described above will then be the line graph of H , its vertices eachcorresponding to a unique edge of H with edges between any two that have a vertex of G in common; this means that thetwo edges meet at a vertex of H . This line graph must not have any of the nine Beineke graphs as induced subgraphs. Itfollows that Gmust obey all the conditions of (2).
(2) implies (1): If G obeys all these conditions, then Q will be the line graph of some graph H and we will haveG = CE(H).
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Next, we go for the characterization of common-edge sigraphs:
Theorem 6. The following statements are equivalent for any signed graph S:
(1) S is a common-edge sigraph.(2) Su is a common-edge graph and the edges of S can be decomposed into complete sub-sigraphs, each of which is homogeneous.
Proof. (1) implies (2) Suppose S is a common-edge sigraph. Then there exists a sigraph T , such that, Su ∼= CE(T u), so that Suis a common-edge graph.
To prove the other part, first of all, if S is homogeneous, then there is nothing to prove. Hence, suppose that S is aheterogeneous sigraph. Then, by the characterization of a common-edge graph we can say that the edges of S can bedecomposed into complete subgraphs such that no vertex lies in more than two of the subgraphs and each edge lies inexactly one such subgraph. Then take T u to be the edge intersection graph.
Let, on the contrary, the complete subgraphs so obtained be heterogeneous. This implies that the sign of an edge whichcorresponds to the complete subgraphs will not be unique and will contradict the definition of common-edge graph.
(2) implies (1) Suppose S is a sigraph satisfying (2). We shall show that S is the common-edge sigraph, that is, there existsa sigraph T such that S ∼= CE(T ).
Now, Su being a common-edge graph, by Theorem 5 its lines can be partitioned into a family ℑ of complete subgraphs ofSu such that no vertex of Su lies in more than two of these subgraphs. Then let Su to be the edge intersection graphΩ(F) ofthe family F = ℑ ∪ V ′(S)where V ′(S) denotes the family of all one-vertex subsets of V (S) each of which belongs to exactlyone of the complete subgraphs in ℑ. Then by the construction described in the characterization of common-edge graph,Su ∼= CE(Ω(F)) where if F = (C1, C2, C3, . . . ., Cq) then there exists an isomorphism ψ : V (Su) → V (CE(Ω(F))) such thatψ(ν) = CiCj, if and only if, Ci ∩ Cj = ν. We shall show that S ∼= CE(T ), where T u ∼= Ω(F) and s1 : E(Ω(F)) → +,− issuch that s1(Ci) = − ⇔ Ci is a homogeneous all-negative clique, and s1(Ci) = + ⇔ Ci is a homogeneous all-positive clique.
Let (CiCj, CjCk) ∈ E(CEΩ(F)), then there exists (u, ν) ∈ V (S), such that, Ci ∩ Cj = u and Cj ∩ Ck = ν. Hence for thesign of (CiCj, CjCk) in CE(T ), we have, se(CiCj, CjCk) = +1, if Cj is positive and se(CiCj, CjCk) = −1, if Cj is negative.
Therefore, in CE(T ), if CiCj, CjCk ∈ V (CE(T )) then the edge (CiCj, CjCk) has the same sign as that of Cj ∈ E(T ), whichcorresponds to the clique in S that is homogeneous and has same sign as that of Cj.
Note: Since the sign of the pendant edge does not contribute to the sign of the edge in the common-edge root sigraph, the signof the edge in the common-edge root graph is immaterial. It can be positive or negative. Therefore, for a given common-edgesigraph having ‘n’ pendant edges 2n common-edge root sigraphs are possible. Hence, the following problems are open.
Problem 7. Characterize common-edge sigraph having exactly one, up to isomorphism, common-edge root sigraph.
Problem 8. Characterize common-edge sigraph having all distinct, up to isomorphism, common-edge root sigraphs;particularly, solve this problem for 2-connected (2-edge connected) common-edge sigraphs.
3. Construction: finding H from CE(H)
In light of Theorem 5, one can find the line graph ofH from CE(H) if one can find the graphQ of that theorem, from CE(H).This can be accomplished by finding all the maximal cliques containing each edge of CE(H) since some of these representthe vertices of Q .
In fact, for all but certain special graphs, finding thosemaximal cliques actually determinesH itself, as we shall see below.The general problem of finding maximal cliques in a graph is NP complete, but for common-edge graphs the problem is
actually easy, because each edge is in at most two of these.We first note some properties of common-edge graphs that justify these statements. Each vertex of CE(H) is a two-edge
path ofH and has some vertex of H as its middle vertex. H can be recovered by identifying the set of vertices of CE(H)whosemiddle vertex is v for each vertex v of H , along with finding which edges of H contain each v.
Each edge of CE(H) contains three edges ofH one of which is the common edge of its two vertices. The edges of CE(H) thatall contain the edge e ofH form a clique in CE(H)whichwill be amaximal clique in all but some special cases; preciselywhichcases wewill consider later. Finding the cliques containing all edges of CE(H) that overlap in e for each e inH determines theline graph of H . An edge of CE(H) can have three possible forms, as edges of H . These are: a triangle, a three edge claw, anda path of length three. In the first two of these cases, there are three vertices of CE(H) that do not all contain any single edgeof H that form a triangle in CE(H). This triangle will always be a maximal clique in CE(H), since no other vertex of CE(H) cancontain more than one of the three edges of H in it, and must therefore fail to have a common edge with at least one of thethree. For example the edges in G of the vertex 12, 23 of CE(H)meet the edges in G of the vertex 42, 23 of CE(H) in a claw,and these two vertices and 12, 24 form a triangle in CE(H). Similarly, 12, 23 and 23, 31 form a triangle with 31, 12.
These and the cliques that contain a single edge of H are the only possible cliques among edges in CE(H), so that at mosttwomaximal cliques of edges of CE(H) contain any one such edge, and if there are twomaximal cliques that contain an edgeof CE(H), at least one has cardinality 3. An edge of CE(H) that forms a three-edge path in H lies in only the cliques of suchedges that contain the middle edge of the path.
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We summarize these statements into:
Remark 9. Each edge of CE(H) lies in at most twomaximal cliques of such edges, and if there are two suchmaximal cliques,at least one of them is a K3.
Remark 10. An open path of length 3 in H defines an edge of CE(H) that lies in only one maximal clique of edges of CE(H),namely the clique consisting of edges of CE(H) containing the middle edge of the path.
Wewill start by describing an algorithm for findingH from CE(H) in the special case inwhich all vertices inH have degreeat least 3. In that case, each edge of H will have at least four edges of CE(H) for which it is the common edge. This means, thecliques consisting of all edges of CE(H)whose common edge is any edge of H will each contain at least four edges, while themaximal cliques corresponding to triangle or 3 edge-claw edges all consist of exactly three edges. It follows that the cliquesof the two kinds cannot be confused with one another, and all the maximal cliques containing a single edge of CE(H) thatcontain at least four such edges must consist of all the edges of CE(H) that contain a particular edge of H .
(In general, we can always distinguish cliques corresponding to triangles or claws from those containing a single edge.Any vertex which meets one of the former triangles meets at least two of its vertices, while except for some very smallgraphs, there are vertices of CE(H) that meet only one vertex of a single edge type clique.)
The following procedure allows us to find all maximal cliques of edges of G if G is CE(H) for some graph H , and hence tofind the line graph of H .Step 0: Set k to be 1Step 1: Choose an edge of G, that is, a pair of vertices vka and vkb of G that are adjacent in G. Test all other vertices vj ofG to find those such that (vka, vj) and (vkb, vj) are both edges of G. Examine G restricted to those vertices that obey thiscondition.
If this graph is neither a complete graph nor the union of a complete graph and an isolated vertex, thenG is not a common-edge graph.
If this graph is a complete graph, give each edge of the complete graphwhose vertices are those vj in that complete graphand also vka and vkb, the label k. Give each vertex of this complete subgraph of G the label k. If this graph is a complete graphand an isolated vertex, then give each edge of the complete graph on the vertices vj of this one and vka and vkb (but not theisolated vertex) the label k. Give each vertex of this complete subgraph of G the label k.Step 2: Raise k by 1, choose any edge of G that is not already labeled and repeat step 1. Continue until all edges of G arelabeled.
If any edge of G receives two or more labels then G is not a common-edge graph.If any vertex of G receives three or more labels, or one or no label then G is not a common-edge graph.The graph Q is defined as follows: its vertices are the sets of edges of Gwith the same label k. Each vertex of Gwith labels
k and j defines an edge between the sets with labels k and labels j. If two or more vertices of G have the same pair of labels,then G is not a common-edge graph.
The graph Q will be the line graph of some H if it obeys the conditions that it contains none of Beineke’s nine excludedgraphs as an induced subgraph.
There is an algorithmwhich requires somewhat more work, but which gives H itself as output rather than the line graphof H . The information contained in a line graph consists of which pairs of adjacent edges contain each edge. The middlevertex of each such pair must be one of the two vertices ofH that define that edge. If we also knowwhich of the two verticeseach middle vertex is, we can read off H itself.
In the algorithmabove, step 1was applied only once for all pairs of vertices ofG that both contain one edge, e or (va, vb), ofH , and there were two different outcomes that were consistent with G being a common-edge graph: the graph on commonneighbors of va and vb in G could be a complete graph or the union of a complete graph and an isolated vertex. In the formercase the edges of H contained in va and vb of G consist of a path of length three in H , and that means, the middle vertices ofva and vb lie at opposite ends of the edge in which these two vertices overlap. If step 1 is applied to all pairs of vertices ofG among va, vb and their common neighbors, we can deduce which of these pairs have middle vertices that lie on oppositeends of that edge, and do not form a triangle.
The following remarks are easily proven:
Remark 11. If the graph R, of pairs of vertices of G that we know contain e and have middle vertices in H on opposite endsof e, is connected, then we know which vertices of G have middle vertices on the same side of e.
Remark 12. When all vertices of H have degrees at least 3, then the graph R determinable by applying step 1 to all pairs ofvertices of G that contain e in H , which contains all pairs that form a 3-edge simple path, is connected, unless both verticesof e have degree three and e is the common edge of two triangles in H .
In the exceptional case of Remark 10 either H is a complete graph on four vertices, or the information from other edgesof the two triangles allows determination of which vertices of G that contain e have the same common vertex.
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The exceptional case here, which occurs when H is a complete graph on four vertices appears because there are in facttwo possible answers to this question (namely, which middle vertices are on the same side of e) for that graph.
These two are isomorphic for ordinary graphs but not generally so for sigraphs.To understand this strange phenomenon, realize that K4 is a planar graph that is isomorphic to its dual graph. Suppose
we give all of its edges a positive sign except those which share a single vertex. This sigraph will have a certain common-edge sigraph. The dual to this sign assignment gives a positive sign to all edges except those on a single triangle. Since thesegraphs and assignments are dual to one another, their common-edge sigraphs will be so as well, and these two differentsign assignments will have common-edge sigraphs that are isomorphic to one another.
Thus far, we have restricted ourselves to finding algorithms for finding sigraphs from their common-edge sigraphs whenall degrees are at least 3. We now extend these results to the case in which there are also vertices of degree 2 in H . If v hasdegree 2 in H then only one vertex of G has center in H at v. If v does not lie in a triangle of H , then all edges of G containingv form three-edge paths in H , the vertex of G with center at v stands alone on either of its two edges, and all other verticeson either edge are together on the opposite sides of that edge from v. Moreover, v lies in exactly two maximal cliques in Gand not three or more (those containing each of its two edges in H and at least one coming from the claw that it forms).
Thismeans that upon determining allmaximal cliques inG using our first algorithmwe can identify each vertex of degree2 and also determine which vertices of its edges its neighbors in G belong to.
When a vertex in H has degree 2 and is part of a triangle in H , again it is sometimes possible that H is not uniquelydeterminable from CE(H). For example, if H is a triangle, CE(H) is also a triangle, and is also the common-edge graph of athree edge claw. A similar phenomenon occurswhenH consists of two triangles that share an edge. For this graph the sigraphin which two adjacent edges of its four-edge cycle are negative and the other edges are positive, has the same common-edge sigraph independent of which two edges are chosen. Thus there are two distinct sigraphs with the same common-edgesigraph in this case.
When two adjacent vertices of H have degree 2, and they lie in a triangle of H , the clique containing the edge betweenthem is a proper subclique of the triangle clique, so it is not a maximal clique but this causes no significant complication.
The examples above appear to be the only sigraphs in which CE(H) does not determine H uniquely when the minimaldegree is 2.
When the minimum degree of vertices in H is 1, and a vertex v of degree 1 is adjacent to one,w, of degree 2, in H , thereis no edge of CE(H) both of whose vertices in that graph contain the edge (v,w). The sigraph CE(H) will therefore be thesame whatever sign the edge (v,w) possesses. Furthermore, that the vertex of CE(H)with middle vertexw lies in only onecomplete graph vertex of Q .
The one vertex of CE(H)which contains (v,w)meets all other vertices of CE(H) in 3-edge paths, which means that theyall have centers at the same vertex of H . This fact makes finding H from CE(H) easier.
When a vertex v of degree 1 in H is adjacent to a vertex w of degree three, only one edge of CE(H) has common edge(v,w), and this clique that defines a vertex ofQ contains only one edge. The three-edge clique in CE(H) corresponding to thethree-edge clawat vertexw inH contains this edge, so that this edge,which determines a vertex ofQ , is not amaximal clique.This causes no problem in determining H except in one small case, in which this three-edge clique cannot be distinguishedfrom the similar clique that defines a vertex of Q .
This happens for the graph of a trianglewith one additional edge: (1, 2), (2, 3), (3, 1), (1, 4). In this case, having all edgespositive except (1,4) gives rise to the same common-edge sigraph as one gets if only the edge (2,3) is negative.
The details of finding H from CE(H)when it exists in this case are straightforward and will be omitted here.Non-uniqueness of H given CE(H) means that no specific algorithm can find H; it can only find the set of H ′s that have
the same CE(H). In the case of the sigraphs whose graph is K4, and the graph of two triangles with an edge in common,this indeterminism arises from the confusion of cliques corresponding to triangles with cliques corresponding to 3-edgeclaws. In the other strange case, the indeterminism arises because it is impossible to distinguish which 3-cliques come fromvertices of Q and which come from triangles or three-edge claws. This latter confusion is impossible in larger connectedgraphs in which one can always distinguish these. Only the case in which there are adjacent vertices of H with degree sum3 can occur in larger graphs.
Figs. 3 and 4 show the construction of a CE-root graph and CE-root sigraph from a common-edge graph and from acommon-edge sigraph, respectively.
4. Algorithm to output CE-root sigraph of a given common-edge sigraph
An algorithm to detect a line graph and output its L-root graphwas given by [9] in which for a given graphH with E edgesand N nodes, a graph G is sought such that H is a line graph of G, if G exists. The algorithm does this within the order of Esteps, in fact in E + O(N) steps. This algorithm is optimal in its complexity.
Following is the algorithm, code-named ‘COMMON-EDGE SIGRAPH ’, to output common-edge root sigraph for a givencommon-edge sigraph where the vertices of the graph are referred as nodes from the algorithmic point of view:
Step 1: Input a sigraph H whose common-edge root sigraph S is required.
Step 2: Pick up any two adjacent nodes in H . Name them 1 − 2 and 2 − 3. They are the two basic nodes to start with.
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Fig. 3. Common-edge graph G and its common-edge root graph H .
Fig. 4. Common-edge sigraph S and its common-edge root sigraph H .
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Fig. 5.
Step 3: Find all the nodes adjacent to both basic nodes.3.1 If there is only one such node, go to step 4.3.2 If there are two such nodes, go to step 5.3.3 If there are three or more nodes, go to step 6.3.4 If there is none, go to step 7.Step 4: There is only one such node. Call this unique node x. We have got a triangle (x, 1 − 2, 2 − 3). Check whether thetriangle so obtained is odd or even.4.1 If the triangle is odd, that is, there exists a node in H which is adjacent to only one node of the triangle without beingadjacent to any other node of the triangle then x = 2 − 4. Go to step 7.4.2 If the triangle is even then x = 1 − 3 (here x is the cross-node). Go to step 8.Step 5: There are two nodes adjacent to both first step basic nodes. Call them x and y.5.1 If x and y are adjacent, then there is no cross-node. Go to step 7.5.2.1 If both the triangles are odd then H is not a common-edge sigraph.5.2.2 If one triangle out of two is odd then corresponding summit, say x, is 2 − 4 and y will be named 1 − 3 and will be thecross-node. Go to step 8.5.2.3 If both the triangles are even, then there are only three remaining possibilities as shown below in 5.2.3. If both thetriangles are even, then there are only three remaining possibilities as shown below in Figs. 5–7.Step 6: There is a group of three or more nodes adjacent to both basic nodes. Check if there is a cross-node in this group. Ifthere is a node m of this group which is not adjacent to a certain node n(chosen at random but once for all) then go to step6.1 else go to step 6.2.6.1 In this case there are two possibilities,6.1.1 If m is the first node of the group under investigation then either m is a cross-node or n is a cross-node. This tie isbroken by examining the adjacency ofm to other nodes of the group. Again, there are three possibilities,6.1.1.a Ifm is adjacent to other nodes then n is declared a cross-node. Go to step 8.6.1.1.b Ifm is not adjacent to other nodes of the group, but n is adjacent to them, thenm is declared a cross-node. Go to step8.6.1.1.c If there are three or more nodes in the group which are not adjacent to each other then H is not a common-edgesigraph.6.1.2 Ifm is not the first node under investigation, then in this casem is the cross-node if n is adjacent to all other nodes.6.2 If there is no node in the group that is non-adjacent to n, then there is no cross-node. If there is no cross-node go to step7 else go to step 8.Step 7: All the nodes of the clique are named. Check if the clique is homogeneous or not.7.1 If the clique is homogeneous, then ‘half-name’ successively all the ‘‘not-yet-fully-named’’ nodes which are adjacent tothe nodes of the clique. Go to step 9.
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Fig. 6.
Fig. 7.
7.2 If the clique is heterogeneous then check whether new basic nodes can be selected or not (other than those selected instep 2). If new basic nodes can be selected, go to step 2, else H is not a common-edge sigraph.Step 8: All the nodes of the clique are named and the cross-node is fully named. Check if the clique is homogeneous or not.8.1 If the clique is homogeneous, then only half-name successively all the ‘‘not-yet-fully-named’’ nodes which are adjacentto nodes of clique (cross-node is not used for this naming). Go to step 9.8.2 If the clique is heterogeneous, then go to step 7.2.Step 9: 9.1 If all the nodes of H are fully named, go to step 10.9.2.1 Else choose any half-named node and find a fully named nodewhich is adjacent to this half-named node and belongs toan already discovered clique. These are two new basic nodes. Complete the names of all those nodes which are half-named.9.2.2 Discover the nodes (among the list of not-yet-fully-named nodes) which are adjacent to both the basic nodes andhalf-name them and then complete their names. All the nodes of the clique are now fully named.9.2.3 Find any ‘‘not-yet-fully-named’’ node which is adjacent to nodes of the clique(s) found in step 9.2.2 and half-namethem with the number which is not the clique number and which is in the ‘name’ of the clique node.9.2.4 If ∃ a ‘‘not-yet-fully-named’’ node such that it is not possible to half-name it then H is not a common-edge sigraph,else go to step 9.1.
1284 D. Sinha et al. / Discrete Applied Mathematics 161 (2013) 1275–1285
Step 10: Take cliques as edges of the common-edge root sigraph S with the sign same as that of the clique. These edges of Sare adjacent if the cliques have a node in common.10.1 Find any two cliques (P and Q ) which have any node in common. Draw them as two adjacent edges of S.10.2.1 Now, take a new clique T (which is not in S yet).10.2.2 If T has a node common with only one clique in H go to step 10.3, else go to step 10.4.10.3 If T has a point in common with only one clique, say P , in H then join T directly with the edge corresponding to P in S,go to step 10.7.10.4 If T has a node in common with two cliques, say P and Q , in H then there are two possibilities for T in S. ‘‘It may makea triangle with P and Q ’’ or ‘‘It may make a K1,3 with P and Q ’’. Draw dotted edges (temporary) corresponding to both thesepossibilities in S and name them T1 and T2.10.5 Find any clique R which is adjacent to P but not to Q and T1.10.5.1 : If ∃ any such clique delete T1 and make T2 as permanent edge (bold). Go to step 10.6.10.5.2 If there does not exist any such clique, find any clique R adjacent to Q but not to P and T1.10.5.2.a If there exists any such clique go to step 10.5.1.10.5.2.b If there is no such clique delete T2 and make T1 as permanent edge.10.6 To join clique R in S rename this clique as T and go to step 10.2.2.10.7 If ∃ any clique in H which cannot be connected to S in the above manner then H is not a common-edge sigraph,
else if there is any clique in H which can be connected to S but have not been connected yet, go to step 10.2,else if all cliques have been connected to S as edges then go to step 10.8.
10.8 S is the required common-edge root sigraph. Display S.
5. Complexity of COMMON-EDGE SIGRAPH
Let H be the given sigraph with N nodes and E edges. In the second statement of the algorithm, we have to choose twobasic nodes out of total N nodes. Thus, this statement runs at max N(N − 1)/2 times (worst in the case if the last selectedpair of nodes is the desired basic node).
Thus, complexity of step 2 is O(N2). After this, in the third statement, we check adjacency of the rest (N − 2) nodes withthe two basic nodes selected in step 2. Thus, third statement runs in at most 2(N − 2) time units.
Hence, complexity of step 3 is O(N).Now, depending on step 3, the algorithm executes either step 4 or step 5 or step 6 and each of these three steps have
complexity O(N).Then, algorithm proceeds to either step 7 or step 8. In both these stepswe have to checkwhether the clique so obtained is
homogeneous or not. For this wewill have to traverse all the edges of the clique. Theworst case would be if a clique containsall the edges of the sigraph.
Thus, maximum complexity for this step would be O(E).Now, if the above clique is found to be heterogeneous then we have to search for new basic nodes. Thus, in this case we
start again from step 2 and repeat the process. This process would continue until we get a homogeneous clique or a choice ofnew basic nodes is possible. Thus, it would be repeatedN(N−2)/2 times. Hence, the total complexity till this stepwould be,
O(N + E)((N2− 2N)/2) = O(N + E)(N2) = O(N3
+ N2E).
Now E can take at most value (N(N − 1))/2.Therefore, the the above complexity expression reduces to
O(N3+ N2(N(N − 1)/2)) = O(N3
+ N4) = O(N4).
After this, complexity of step 9 would be O(N) as each of the remaining nodes would be traversed twice, once for half-naming it and the second time for completing its name. Also, complexity of step 10 is less than O(N4) since, in step 10, wehave to compare cliques to each other and the number of cliques is less than or equal to the number of edges in sigraph.
Hence, complexity of step 10 of the algorithm is O(N4).
6. Conclusion
Since the complexity of the algorithm COMMON-EDGE SIGRAPH is bounded by a polynomial in N , we can say that thealgorithm is efficient.Scope: If, given a graph H , we are able to find whether H is a common-edge graph or not and detect a common-edge rootgraph of G then with the help of G and the two theorems given below we can say whether H is Hamiltonian or not.
D. Sinha et al. / Discrete Applied Mathematics 161 (2013) 1275–1285 1285
Theorem 13. If G is a nontrivial connected graphwith p vertices, which is not a path, then Ln(G) is Hamiltonian for all n ≥ (p−3).
Application: Since CE(G) ∼= L2(G), if H comes out to be a common-edge graph of G then H is Hamiltonian if G is a non-trivialconnected graph with p vertices, which is not a path, such that, p ≤ (2 + 3) = 5.
Theorem 14. If G is Eulerian, then L(G) is both Eulerian and Hamiltonian. If G is Hamiltonian then L(G) is Hamiltonian.
Application: Since CE(G) ∼= L2(G) = L(L(G)), H is found to be a common-edge graph of a graph G using algorithm,
(i) H is both Eulerian and Hamiltonian if G is Eulerian. Since if G is eulerian then L(G) is eulerian and hamiltonian both, andsince L(G) is Eulerian, L(L(G)) = L2(G)would also be Eulerian and hamiltonian.
(ii) H is Hamiltonian if G is Hamiltonian. Since, if G is Hamiltonian then L(G) is Hamiltonian and if L(G) is Hamiltonian thenL2(G) is Hamiltonian.
Lastly, it is expected that a solution of the following open problemwould lead to certain new insights into the nature theunary ‘sigraph operator’ CE .
Problem 15. Given a sigraph S, study the CE-orbit S, CE(S), C2E (S), C
3E (S), . . . of the iterated common-edge sigraphs of S.
Acknowledgments
Way back, during the first author’s earlier discussion with M. Acharya and B.D. Acharya, regarding the properties ofcommon-edge graphs and common-edge sigraphs, they posed the problem of characterizing common-edge graphs andcommon-edge sigraphs. The authors are deeply indebted to both of them even for having gone through the revised versionsof the paper, including posing three new open problems (Problems 7, 8 and 15), and to the referees who made extensiveand constructively critical comments on the first version of the paper.
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