A Note on the Adversary Degree Associated Reconstruction Number of Graphs

A vertex-deleted subgraph of a graph G is called a card of G. A card of G with which the degree of the deleted vertex is also given is called a degree associated card (or dacard) of G. The degree associated reconstruction number drn (G) of a graph G is the size of the smallest collection of dacards of G that uniquely determines G. The adversary degree associated reconstruction number of a graph G, adrn(G), is the minimum number k such that every collection of k dacards of G that uniquely determines G. In this paper, we show that adrn of wheels and complete bipartite graphs on at least 4 vertices is 2 or 3.


Introduction
All graphs considered are simple, finite, and undirected. We will mostly follow the standard graph theoretic terminology of [1]. A vertex-deleted subgraph or card − V of a graph is the unlabeled graph obtained from by deleting the vertex V and all edges incident with V. The ordered pair ( (V), − V) is called a degree associated card or dacard of the graph where (V) is the degree of V 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 ally-reconstruction 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 counter-examples. In view of this, Ramachandran [7] studied the degree (degree triple) associated reconstruction of graphs (digraphs). For a vertex V of a digraph, the ordered triple ( , , ) is called the degree triple of V where , , and are, respectively, the number of unpaired outarcs, unpaired inarcs, and symmetric pairs of arcs incident with V. 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 ( ) = 2 for all caterpillars except stars and one 6-vertex example, and that ( ) ≥ 3 for all vertex-transitive 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 C, is called class-reconstructible if whenever ∈ C has the same deck as , then ≅ . If a graph is degree associated reconstructible then it is 2 Journal of Discrete Mathematics class-reconstructible, and vice-versa, 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 vertex-transitive 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.

of Standard Graphs
Since ( ) and ( ) are exactly equal for any vertextransitive graph , the next theorem follows from [7].
where is a cycle on vertices.
It is clear, from the definition of , that ( ) = ( ) = 1. 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 − 1 dacards in common. If ( , ) is a dacard of , then ( − 1 − , ) is a dacard of and vice-versa. The graph (≇ ) has therefore − 1 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 ( (V), − V) of is a graph obtained from the dacard by adding a new vertex and joining it with (V) vertices of the dacard, and it is denoted by ( (V), − V) (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 − 1 dacards in common with those of , and that at least one extension has precisely − 1 dacards in common with those of .
A vertex of degree is called an -vertex. We call a neighbour of degree of a vertex V by an -neighbour of V. 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 ( − 1)vertex, the dacard (obtained by deleting the ( − 1)-vertex of ) has a unique extension, and so ( ) = 1 by Remark 3. We now show that ( ) = 3 for > 4.

Theorem 4.
If is a wheel on (≥4) vertices, then (1) Proof. The dadeck of = 1 + −1 consists of one copy of the dacard ( − 1, −1 ) and − 1 copies of the dacard . We consider four cases depending on the value of . Case 1. = 4. Now, all the four dacards are isomorphic and they are (3, 3 ). Since it has a unique extension, it follows that ( 4 ) = 1.
Case 2. = 5. In the dacard (3, 1 + 3 ), there are two 2vertices and two 3-vertices. If we join the newly added vertex to the two 2-vertices and a 3-vertex, then the extension is isomorphic to . Therefore, we join the newly added vertex to a 2-vertex and the two 3-vertices. The extension so has one 2-vertex, two 3-vertices, and two 4-vertices ( Figure 1). The dacard of the extension corresponding to each of the two 3-vertices is (3, 1 + 3 ). The extension has thus only two dacards in common to those of , and, hence, Case 3. = 6. In the dacard (3, 1 + 4 ), there are exactly two 2-vertices, one 4-vertex, and two 3-vertices.
Case 3.1. Join to none of the two 2-vertices. 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 3-vertex of the extension is clearly in common with that of . The dacard of obtained by deleting the 5-vertex has endvertices, and, hence, it is not a dacard of . The extension has thus only one dacard in common with that of .  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 3-vertices is a dacard of . The extension has thus only two dacards in common with those of .  In , is a 3-vertex and is a 4-vertex. The dacards of corresponding to and the 3-vertex other than are dacards of . The dacard corresponding to of is not a dacard of , since the subgraph of obtained by deleting has three 2-vertices. The dacard corresponding to the 5-vertex of is not a dacard of . The extension has thus only two dacards in common with those of . The dacards corresponding to the 3-vertices and in are dacards of . The dacard corresponding to the other 3-vertex is not a dacard of , since the 3-vertex is adjacent to a 2-vertex in . Also in , the 5-vertex is adjacent to a 2vertex, and so the dacard corresponding to the 5-vertex is not a dacard of . The extension has thus only two dacards in common with those of . If is the 4-vertex, then is isomorphic to . Otherwise, has four 3-vertices and two 4-vertices. The dacards of corresponding to and the 3-vertex not adjacent to are dacards of . But the dacard obtained from by deleting a 3-vertex other than and has no 4-vertex, and so it is not a dacard of . Thus, has only two dacards in common with those of .
Case 4. ≥ 7. The dacard (3, 1 + −2 ) has two vertices of degree 2, − 4 (≥3) vertices of degree 3, and one vertex of degree − 2 (≥5). If the newly added vertex is not joined to the ( − 2)-vertex, then the extension cannot have more than one dacard in common with that of . If is joined to the ( −2)-vertex and the two 2-vertices, then is isomorphic to . Therefore, it is enough to consider the case in which is joined to the ( − 2)-vertex and not joined to at least one 2-vertex. If the 4-vertex is adjacent to the 2-vertex in , then there are two 3-vertices adjacent to the 4-vertex. 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 4-vertex, is a dacard of . Otherwise, the 3-neighbour closest to the 2-vertex yields a dacard with a cut vertex, while no dacard of has a cut vertex. The extension has two vertices of degree 2, − 5 (≥2) vertices of degree 3, two vertices of degree 4, and one vertex of degree − 1. The dacard (3, 1 + −2 ) cannot have a vertex of degree 4. Therefore, only dacards corresponding to the 3-vertices which are common neighbours of the two 4-vertices can be isomorphic to the dacard (3, 1 + −2 ). In the extension , there are at most two 3-vertices which are the common neighbours of the two 4-vertices. Thus, the extension has at most two dacards in common with those of , which completes the proof.

From Remark 3, it follows that
( 1, ) = 1. But ( 1, ) can be greater than one for ≥ 3 by the next theorem. (2) Proof. The dadeck of 1, consists of one copy of ( , ) and copies of (1, 1, −1 ). The extension ( , ) is clearly isomorphic to 1, . Consider the extension (1, 1, −1 ). If we join the newly added vertex to the ( − 1)-vertex of the dacard (1, 1, −1 ), then the extension is isomorphic to 1, . We join therefore the vertex to any one of the endvertices of the dacard. For = 2, the extension is isomorphic to 1, , and, hence, ( 1,2 ) = 1. For = 3, the extension is isomorphic to 4 . The dadeck of 4 consists of 2 copies of 1,2 and 2 copies of 1 ∪ 2 . The extension therefore has exactly two dacards in common with those of 1, and ( 1,3 ) = 3. Now, let us assume that ≥ 4. Then, in the extension , there is exactly one 2-vertex which is adjacent to the newly added endvertex and a unique ( − 1)-vertex. Since there is no 2-vertex in 1, −1 for ≥ 4, only the dacard corresponding to the 1-neighbour of the 2vertex in the extension can be a dacard of 1, . Hence, Ramachandran [7] proved that ( , ) = 2 for 2 ≤ < . We shall show that of , is not always two.
Theorem 6. For 2 ≤ < , Proof. The dadeck of , consists of copies of ( , −1, ) and copies of ( , , −1 ). To get an extension having at least one dacard in common with that of , , augment the dacard ( , −1, ) or ( , , −1 ) by adding a new vertex and joining it to precisely or vertices, respectively, in the dacard. Let ( , ) be the bipartition of the dacard −1, , where | | = − 1. Then, if we join to every vertex of , then ≅ , . We join therefore to at least one vertex of . If we join to all the − 1 vertices of , then no vertex other than can have degree in . The extension therefore has only one dacard isomorphic to ( , −1, ). In this extension, the vertices have degrees − 1, , , and + 1 only. The degrees of the vertices in the dacard corresponding to an -vertex of the extension are − 1, , − 1 and where − 1 < ≤ − 1 < . The dacard ( , , −1 ) 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 − 1, , , and + 1. This extension clearly has a dacard ( , −1, ) (dacard corresponding to the vertex ). The degrees of the vertices in the dacard corresponding to an -vertex other than of are − 2, − 1, , and + 1. Therefore, has only one dacard isomorphic to ( , −1, ). The degrees of the vertices in the dacard corresponding to an -vertex of are − 1, , − 1, and where − 1 < ≤ − 1 < . Hence, this extension has only one dacard in common with that of , .
Let ( , ) be the bipartition of the dacard , −1 , 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 .

If
= + 1, then the extension is isomorphic to , . If = + 2, then the extension has two dacards isomorphic to ( , , −1 ), and, hence, it has only two dacards in common with those of , . So, we take that > + 2. Now, the extension has vertices of degrees , + 1 and − 1 only. Clearly, this extension has a dacard ( , , −1 ) corresponding to the -vertex . The removal of any othervertex from this extension would give a dacard with vertices of degrees , + 1, and − 2 only, and, hence, this dacard would not be isomorphic to , −1 . Thus, has only one dacard isomorphic to ( , , −1 ), and this is the only dacard in common with that of , . In this case, = 2 and the extension contains exactly one triangle, say . The dacard of the extension corresponding to the vertex is clearly ( , , −1 ). 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 m-vertex 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 ( − 1)vertex . The dacard of corresponding to the vertex is therefore not isomorphic to ( , −1, ) (this verification is needed only for the case when + 1 = ). If we remove the vertex from the extension , then the ( − 1)-vertex in the resulting dacard is not adjacent to the -vertex , and, hence, the dacard of corresponding to the vertex is not isomorphic to ( , , −1 ). 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 ( , , −1 ). The extension may have one more dacard (corresponding to the vertex ) in common with that of , . In , there exists at least one ( − 1)-vertex in . Fix one such vertex and let it be . Then, the ( − 1)-vertex in the dacard of obtained by deleting the vertex is not adjacent to the ( − 1)-vertex , where ̸ = . The dacard of (corresponding to the vertex ) is therefore not isomorphic to ( , −1, ). 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 ( , , −1 ). The extension may have one more dacard (corresponding to the vertex ) in common with that of , . In , there exists at least one ( − 1)-vertex of . Fix one such vertex and let it be . The ( − 1)-vertex in the dacard of , obtained by deleting the vertex , is then not adjacent to the ( −1)-vertex , where ̸ = . The dacard of (corresponding to the vertex ) is therefore not isomorphic to ( , −1, ). 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 ( , , −1 ). The deletion of any vertex other than from will give a dacard containing a triangle. The extension has thus only one dacard isomorphic to ( , , −1 ), and this is the only dacard in common with that of , , which completes the proof of Theorem 6.