Research Article  Open Access
A Note on the Adversary Degree Associated Reconstruction Number of Graphs
Abstract
A vertexdeleted subgraph of a graph is called a card of . A card of with which the degree of the deleted vertex is also given is called a degree associated card (or dacard) of . The degree associated reconstruction number drn () of a graph is the size of the smallest collection of dacards of that uniquely determines . The adversary degree associated reconstruction number of a graph , adrn(), is the minimum number such that every collection of dacards of that uniquely determines . In this paper, we show that adrn of wheels and complete bipartite graphs on at least 4 vertices is 2 or 3.
1. Introduction
All graphs considered are simple, finite, and undirected. We will mostly follow the standard graph theoretic terminology of [1]. A vertexdeleted subgraph or card of a graph is the unlabeled graph obtained from by deleting the vertex and all edges incident with . The ordered pair is called a degree associated card or dacard of the graph where is the degree of in . The deck (dadeck) of a graph is its collection of cards (dacards). Ulam's Conjecture [2], also called Reconstruction Conjecture (RC), asserts that every graph on at least three vertices is determined uniquely (up to isomorphism) by its deck. Graphs that obey RC are called reconstructible.
For a reconstructible graph , Harary and Plantholt [3] have defined the reconstruction number to be the size of the smallest subcollection of the deck of which is not contained in the deck of any other graph ; . Myrvold [4] referred to this number as allyreconstruction number of . Myrvold [5] also studied adversary reconstruction number of which is the smallest such that no subcollection of the deck of of size is contained in the deck of any other graph ; .
An extension of RC to digraphs, the Digraph Reconstruction Conjecture, was disproved when Stockmeyer exhibited [6] several infinite families of counterexamples. In view of this, Ramachandran [7] studied the degree (degree triple) associated reconstruction of graphs (digraphs). For a vertex of a digraph, the ordered triple is called the degree triple of where , , and are, respectively, the number of unpaired outarcs, unpaired inarcs, and symmetric pairs of arcs incident with . A graph (digraph) is called degree associated reconstructible if it can be determined uniquely from its dadeck. For a degree associated reconstructible graph (digraph) , the degree (degree triple) associated reconstruction number, , of is the size of the smallest subcollection of the dadeck of which is not contained in the dadeck of any other graph (digraph) ; . Barrus and West [8] have shown that for all caterpillars except stars and one 6vertex example, and that for all vertextransitive graphs (not complete or edgeless).
The following weakening of the reconstruction problem has also been considered by Harary and Plantholt [3]. A graph , in a given class of graphs , is called classreconstructible if whenever has the same deck as , then . If a graph is degree associated reconstructible then it is classreconstructible, and viceversa, where the class is the class of graphs with a given number of edges.
In this paper, we study the parameter adversary degree associated reconstruction number adrn(G) of a graph . For a reconstructible graph from its dadeck, is the minimum number such that every collection of dacards of is not contained in the dadeck of any other graph ; . From their definitions, it is clear that for any graph ; the equality holds for vertextransitive graphs (where all the dacards are necessarily identical). In this paper, we show that is 2 or 3 for wheels and complete bipartite graphs on at least 4 vertices.
2. of Standard Graphs
Since and are exactly equal for any vertextransitive graph , the next theorem follows from [7].
Theorem 1. (i) .
(ii) for .
(iii) for , where is a cycle on vertices.
It is clear, from the definition of , that . In fact, it is true that the of the complement of a graph is equal to the of the graph.
Lemma 2. For any graph , .
Proof. The latter inequality follows immediately from the definitions of and . To prove the first equality, let be a graph of order ; let . Then, there exists a graph such that and have dacards in common. If ( is a dacard of , then is a dacard of and viceversa. The graph has therefore dacards in common with those of . Consequently, we have .
On the other hand, if , then there exists a graph such that has dacards of . It follows that the graph has dacards of . Therefore, , giving a contradiction. Hence, .
An extension of a dacard of is a graph obtained from the dacard by adding a new vertex and joining it with vertices of the dacard, and it is denoted by (or simply by ). Throughout this paper, and are used in the sense of this definition.
For a graph , to prove , we proceed as follows. (i)First, find the dadeck of . (ii)Determine next all possible extensions of every dacard of .(iii)Finally, show that every extension other than has at most dacards in common with those of , and that at least one extension has precisely dacards in common with those of .
A vertex of degree is called an vertex. We call a neighbour of degree of a vertex by an neighbour of . The union of graphs and is the graph with vertex set and edge set . The join of disjoint graphs and is the graph obtained from by joining each vertex of to each vertex of .
Remark 3. Note that if a dacard of a graph is vertextransitive and the degree associated with the dacard is one or else the degree associated with the dacard equals the number of vertices in the dacard, then the dacard has a unique extension (up to isomorphism), and, hence, .
Since the wheel on (≥4) vertices has an vertex, the dacard (obtained by deleting the vertex of ) has a unique extension, and so by Remark 3. We now show that for .
Theorem 4. If is a wheel on vertices, then
Proof. The dadeck of consists of one copy of the dacard and copies of the dacard , where is a vertex of . Let be an extension of the dacard . We consider four cases depending on the value of .
Case 1. . Now, all the four dacards are isomorphic and they are . Since it has a unique extension, it follows that .
Case 2. . In the dacard , there are two 2vertices and two 3vertices. If we join the newly added vertex to the two 2vertices and a 3vertex, then the extension is isomorphic to . Therefore, we join the newly added vertex to a 2vertex and the two 3vertices. The extension so has one 2vertex, two 3vertices, and two 4vertices (Figure 1). The dacard of the extension corresponding to each of the two 3vertices is . The extension has thus only two dacards in common to those of , and, hence, .
Case 3. . In the dacard , there are exactly two 2vertices, one 4vertex, and two 3vertices.
Case 3.1. Join to none of the two 2vertices.
The extension has one vertex of degree 3, two vertices of degree 2, two vertices of degree 4, and one vertex of degree 5 (Figure 2). The dacard corresponding to the 3vertex of the extension is clearly in common with that of . The dacard of obtained by deleting the 5vertex has endvertices, and, hence, it is not a dacard of . The extension has thus only one dacard in common with that of .
Case 3.2. Join to one 2vertex and two 3vertices.
The extension has two vertices of degree 3, one vertex of degree 2, and three vertices of degree 4. The dacard corresponding to each of the two 3vertices is a dacard of . The extension has thus only two dacards in common with those of .
Case 3.3. Join to one 2vertex (say ), one 3vertex (say ), and the unique 4vertex.
Case 3.3.1. Vertices and are nonadjacent.
In , is a 3vertex and is a 4vertex. The dacards of corresponding to and the 3vertex other than are dacards of . The dacard corresponding to of is not a dacard of , since the subgraph of obtained by deleting has three 2vertices. The dacard corresponding to the 5vertex of is not a dacard of . The extension has thus only two dacards in common with those of .
Case 3.3.2. Vertices and are adjacent.
The dacards corresponding to the 3vertices and in are dacards of . The dacard corresponding to the other 3vertex is not a dacard of , since the 3vertex is adjacent to a 2vertex in . Also in , the 5vertex is adjacent to a 2vertex, and so the dacard corresponding to the 5vertex is not a dacard of . The extension has thus only two dacards in common with those of .
Case 3.4. Join to two 2vertices and a 3vertex or 4vertex (say ).
If is the 4vertex, then is isomorphic to . Otherwise, has four 3vertices and two 4vertices. The dacards of corresponding to and the 3vertex not adjacent to are dacards of . But the dacard obtained from by deleting a 3vertex other than and has no 4vertex, and so it is not a dacard of . Thus, has only two dacards in common with those of .
Case 4. . The dacard has two vertices of degree 2, (≥3) vertices of degree 3, and one vertex of degree (≥5). If the newly added vertex is not joined to the vertex, then the extension cannot have more than one dacard in common with that of . If is joined to the vertex and the two 2vertices, then is isomorphic to . Therefore, it is enough to consider the case in which is joined to the vertex and not joined to at least one 2vertex.
Case 4.1. Join to the vertex, one 2vertex, and one 3vertex.
The extension has one 2vertex, vertices of degree 3, one vertex of degree 4, and one vertex of degree . The dacard cannot have a vertex of degree 4. Therefore, only dacards corresponding to the 3neighbours of the 4vertex in can be isomorphic to the dacard .
If the 4vertex is adjacent to the 2vertex in , then there are two 3vertices adjacent to the 4vertex. The extension therefore can have at most two dacards in common with those of , and the dacard, corresponding to each of the two 3vertices adjacent to the 4vertex, is a dacard of . Otherwise, the 3neighbour closest to the 2vertex yields a dacard with a cut vertex, while no dacard of has a cut vertex.
Case 4.2. Join to the vertex and two 3vertices.
The extension has two vertices of degree 2, (≥2) vertices of degree 3, two vertices of degree 4, and one vertex of degree . The dacard cannot have a vertex of degree 4. Therefore, only dacards corresponding to the 3vertices which are common neighbours of the two 4vertices can be isomorphic to the dacard . In the extension , there are at most two 3vertices which are the common neighbours of the two 4vertices. Thus, the extension has at most two dacards in common with those of , which completes the proof.
From Remark 3, it follows that . But can be greater than one for by the next theorem.
Theorem 5. For
Proof. The dadeck of consists of one copy of and copies of . The extension is clearly isomorphic to . Consider the extension . If we join the newly added vertex to the vertex of the dacard , then the extension is isomorphic to . We join therefore the vertex to any one of the endvertices of the dacard. For , the extension is isomorphic to , and, hence, . For , the extension is isomorphic to . The dadeck of consists of 2 copies of and 2 copies of . The extension therefore has exactly two dacards in common with those of and . Now, let us assume that . Then, in the extension , there is exactly one 2vertex which is adjacent to the newly added endvertex and a unique vertex. Since there is no 2vertex in for , only the dacard corresponding to the 1neighbour of the 2vertex in the extension can be a dacard of . Hence, .
Ramachandran [7] proved that for . We shall show that of is not always two.
Theorem 6. For ,
Proof. The dadeck of consists of copies of and copies of . To get an extension having at least one dacard in common with that of , augment the dacard or by adding a new vertex and joining it to precisely or vertices, respectively, in the dacard.
Case 1. Augmenting the dacard .
Let be the bipartition of the dacard , where . Then, if we join to every vertex of , then . We join therefore to at least one vertex of . If we join to all the vertices of , then no vertex other than can have degree in . The extension therefore has only one dacard isomorphic to . In this extension, the vertices have degrees , and only. The degrees of the vertices in the dacard corresponding to an vertex of the extension are , and where . The dacard is therefore not a dacard of the extension . Thus, has only one dacard in common with that of . If at least one vertex of is not joined to , then in , there are vertices of degrees , and . This extension clearly has a dacard (dacard corresponding to the vertex ). The degrees of the vertices in the dacard corresponding to an vertex other than of are , and . Therefore, has only one dacard isomorphic to . The degrees of the vertices in the dacard corresponding to an vertex of are , and where . Hence, this extension has only one dacard in common with that of .
Case 2. Augmenting the dacard .
Let be the bipartition of the dacard , where is the set of vertices. If we join to every vertex of , then the extension is isomorphic to . We join therefore to at least one vertex of .
Case 2.1. Join to no vertex of .
If , then the extension is isomorphic to . If , then the extension has two dacards isomorphic to , and, hence, it has only two dacards in common with those of . So, we take that . Now, the extension has vertices of degrees and only. Clearly, this extension has a dacard corresponding to the vertex . The removal of any other vertex from this extension would give a dacard with vertices of degrees , and only, and, hence, this dacard would not be isomorphic to . Thus, has only one dacard isomorphic to , and this is the only dacard in common with that of .
Case 2.2. Join to at least one vertex of .
Case 2.2.1. Join to exactly one vertex (say ) of and exactly one vertex (say ) of .
In this case, and the extension contains exactly one triangle, say . The dacard of the extension corresponding to the vertex is clearly . The extension may have two more dacards (corresponding to the vertices and ) in common with those of , since no dacard of contains a triangle. In the extension , there exists at least one mvertex in other than . Fix one such vertex and let it be . The vertex in the dacard obtained by deleting the vertex from is not adjacent to the vertex . The dacard of corresponding to the vertex is therefore not isomorphic to (this verification is needed only for the case when ). If we remove the vertex from the extension , then the vertex in the resulting dacard is not adjacent to the vertex , and, hence, the dacard of corresponding to the vertex is not isomorphic to . The extension has thus only one dacard in common with that of .
Case 2.2.2. Join to exactly one vertex (say ) of and at least two vertices of .
The extension has at least two triangles. Clearly, the dacard of corresponding to the vertex is . The extension may have one more dacard (corresponding to the vertex ) in common with that of . In , there exists at least one vertex in . Fix one such vertex and let it be . Then, the vertex in the dacard of obtained by deleting the vertex is not adjacent to the vertex , where . The dacard of (corresponding to the vertex ) is therefore not isomorphic to . The extension has thus only one dacard in common with that of .
Case 2.2.3. Join to at least two vertices of and exactly one vertex (say ) of .
The extension has at least two triangles. Clearly, the dacard of corresponding to the vertex is . The extension may have one more dacard (corresponding to the vertex ) in common with that of . In , there exists at least one vertex of . Fix one such vertex and let it be . The vertex in the dacard of , obtained by deleting the vertex , is then not adjacent to the vertex , where . The dacard of (corresponding to the vertex ) is therefore not isomorphic to . The extension has thus only one dacard in common with that of .
Case 2.2.4. Join to at least two vertices of and at least two vertices of .
Deleting from would give a dacard isomorphic to . The deletion of any vertex other than from will give a dacard containing a triangle. The extension has thus only one dacard isomorphic to , and this is the only dacard in common with that of , which completes the proof of Theorem 6.
3. Concluding Remarks
It is clear, from their definitions, that . But and are not comparable in general. For instance, it is proved [9] that and . Therefore, and . However, .
We summarize our results on the and the corresponding results on the in Table 1.

Acknowledgments
The second author is very grateful to Professor S. Ramachandran for drawing his attention to . The authors are thankful to anonymous referees for their many valuable comments which largely improved the style of the paper and the proof of Theorem 4. The work reported here is supported by the Project SR/S4/MS:628/09 awarded to the first author by the Department of Science and Technology, Government of India, New Delhi.
References
 F. Harary, Graph Theory, AddisonWesley, Reading, Mass, USA, 1969. View at: MathSciNet
 S. M. Ulam, A Collection of Mathematical Problems, Wiley, New York, NY, USA, 1960. View at: MathSciNet
 F. Harary and M. Plantholt, “The graph reconstruction number,” Journal of Graph Theory, vol. 9, no. 4, pp. 451–454, 1985. View at: Publisher Site  Google Scholar  Zentralblatt MATH  MathSciNet
 W. Myrvold, “The allyreconstruction number of a disconnected graph,” Ars Combinatoria, vol. 28, pp. 123–127, 1989. View at: Google Scholar  Zentralblatt MATH  MathSciNet
 W. J. Myrvold, The ally and adversary reconstruction problems [Ph.D. thesis], University of Waterloo, 1988.
 P. K. Stockmeyer, “The falsity of the reconstruction conjecture for tournaments,” Journal of Graph Theory, vol. 1, no. 1, pp. 19–25, 1977. View at: Publisher Site  Google Scholar  Zentralblatt MATH  MathSciNet
 S. Ramachandran, “Degree associated reconstruction number of graphs and digraphs,” Mano International Journal on Mathematical Sciences, vol. 1, pp. 41–53, 2000. View at: Google Scholar
 M. D. Barrus and D. B. West, “Degreeassociated reconstruction number of graphs,” Discrete Mathematics, vol. 310, no. 20, pp. 2600–2612, 2010. View at: Publisher Site  Google Scholar  Zentralblatt MATH  MathSciNet
 S. Sundar Raj and S. Monikandan, “Adversary degree associated reconstruction number of graphs obtained from complete graphs or cycles,” Journal of Combinatorics, Information & System Sciences, vol. 37, no. 1, pp. 75–94, 2012. View at: Google Scholar
Copyright
Copyright © 2013 S. Monikandan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.