# Degree-associated reconstruction parameters of complete multipartite graphs and their complements

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1 Degree-associated reconstruction parameters of complete multipartite graphs and their complements Meijie Ma, Huangping Shi, Hannah Spinoza, Douglas B. West January 23, 2014 Abstract Avertex-deleted subgraphofagraphgisacard. Adacard consistsofacardandthe degree of the missing vertex. The degree-associated reconstruction number of a graph G, denoted drn(g), is the minimum number of dacards that suffice to reconstruct G. The adversary degree-associated reconstruction number adrn(g) is the least k such that every set of k dacards determines G. The analogous parameters for degree-associated edge reconstruction are dern(g) and adern(g). We determine these four parameters for all complete multipartite graphs and their complements. The answer is usually 2 for all four parameters, but there are exceptions in each case. Key words: degree-associated reconstruction number, complete multipartite graph, adversary reconstruction. 1 Introduction The Reconstruction Conjecture of Kelly [5, 6] and Ulam [13] has been open for more than 50 years. A subgraph of a graph G obtained by deleting one vertex is a card. Cards are unlabeled; that is, only the isomorphism class of a card is known. The multiset of cards is the deck of G. The Reconstruction Conjecture asserts that every graph with at least three vertices is uniquely determined by its deck. Such a graph is reconstructible. For a reconstructible graph G, Harary and Plantholt [4] introduced the reconstruction number, denoted by rn(g); it is the least k such that some multiset of k cards from the deck of G determines G (meaning that every graph not isomorphic to G shares at most rn(g) 1 Department of Mathematics, Zhejiang Normal University, Jinhua , China; Research supported by National Natural Science Foundation of China grant Department of Mathematics, Zhejiang Normal University, Jinhua , China; Department of Mathematics, University of Illinois, Urbana IL 61801, U.S.A; Departments of Mathematics, Zhejiang Normal University, Jinhua , China, and University of Illinois, Urbana IL 61801, U.S.A.; Research supported by State Administration of Foreign Experts Affairs, China. 1

4 Proof. The graph H arises from C by adding a vertex v having d neighbors in C. In all cases other than those listed above, every card other than H v has a component that is not a complete graph. Hence the listed cases are the only graphs other than G n that share at least two dacards with G n. The other cases to consider are (1) v has neighbors in at least three components of C, (2) v has at least two neighbors in each of two components of C, (3) v has d 1 neighbors in a j-component of C with j > d 1 and one neighbor in another component, and (4) v has all neighbors in a j-component of C with j > d+1. Lemma 2.2. Any two distinct dacards of G n determine G n. Proof. Given n = (n 1,...,n r ), let ˆn t denote the r-tuple obtained from n by decreasing the t- th entry by 1. Since the given dacards are distinct, we may assume that they are (Gˆni,n i 1) and (Gˆnj,n j 1), where n i < n j. If n i = 1, then G n has an isolated vertex and is determined by the dacard (Gˆni,0). Henceforth assume n i 2, so n j 3. Let H be a reconstruction from (Gˆnj,n j 1) that also has (Gˆni,n i 1) as a dacard. By Lemma 2.1, when H G n there are only two cases for the neighborhood of v that permit Gˆni to also be a card of H. Case 1: All neighbors of v lie in one n j -component of Gˆnj. In this case, the component C of H containing v is not a complete graph and remains in any card obtained by deleting a vertex of degree n i 1, since δ(c ) = n j 1. Hence (Gˆni,n i 1) is not a dacard of H. Case 2: v has n j 2 neighbors in an (n j 2)-component of Gˆnj and one neighbor x in another component. In this case the only card of H distinct from Gˆnj that is a clique-union is H x. Since (Gˆni,n i 1) is a dacard of H, we have d H (x) = n i 1. Now H x is obtained from Gˆnj by adding v to turn an (n j 2)-component into an (n j 1)-component and deleting x from an (n i 1)-component; hence H x has the same number of n j -components as Gˆnj, which is fewer than G n has. Since n i < n j, in Gˆni the number of n j -components is the same as in G n. However, the number of such components in H x is smaller. Hence H x cannot be Gˆni, which eliminates this case. Lemma 2.3. Any three identical dacards of G n determine G n. Proof. We may assume that the three dacards are copies of (Gˆni,n i 1). Let H be a graph having three copies of (Gˆni,n i 1) in its dadeck, obtained from Gˆni by adding a vertex v with degree n i 1 (in H). If n i = 1, then H is formed by adding an isolated vertex to Gˆni. Hence H = G n. If n i = 2, then v has degree 1; let x be its neighbor in Gˆni. Let x be in an n j -component in Gˆni. If n j 3, then (Gˆni,1) occurs only once in the dadeck of H. If n j = 2, then the component of H containing v is K 1,2, and H has two copies of (Gˆni,1) in its dadeck. Having three copies of (Gˆni,1) requires n j = 1, which yields H = G n. 4

5 Hence we may assume n i 3. ByLemma 2.1, v has neighborsinat most two components of Gˆni. If two components, then one is an (n i 2)-component and the other has just one neighbor of v. If one component, then it is an n i -component or H = G n. Suppose H G n. Case 1: All neighbors of v lie in an n i -component of Gˆni. Here the component C of H containing v is not complete. It remains in all but two cards where vertices of degree n i 1 are deleted. Since H has three copies of (Gˆni,n i 1) in its dadeck, this case does not arise. Case 2: v has n i 2 neighbors in an (n i 2)-component of Gˆni and one neighbor x in another component. In this case, H u is a clique-union only for u {v,x}. Since H has at least three such cards, this case does not arise. With these lemmas, the results for drn and adrn on G n and K n do not require much more work. We will use two lemmas proved by Barrus and West [1] (the proofs are not difficult). Lemma 2.4 ([1]). A graph G satisfies drn(g) = 1 if and only if G or its complement has an isolated vertex or has a leaf whose deletion leaves a vertex-transitive graph. Lemma 2.5 ([1]). If G is vertex-transitive and is not complete or edgeless, then drn(g) 3. Since complete graphs and their complements are determined by any one dacard, we consider only r 2 when studying K n. Theorem 2.6. If r 2 and n 1 n r, then 1, if n 1 = 1; drn(k n ) = drn(g n ) = 3, if n 1 = n r 2. 2, otherwise. Proof. Since G n is the complement of K n, it suffices to determine drn(g n ), where n = (n 1,...,n r ). By Lemma 2.4, drn(g n ) = 1 when n 1 = 1, and otherwise drn(g n ) 2. If n 1 < n r, then the dadeck of G n has two distinct dacards. By Lemma 2.2, drn(g n ) 2. The only remaining case is n 1 = n r 2, where G n is vertex-transitive and not complete or edgeless. By Lemma 2.5, drn(g n ) 3. Since r 2, in G n there are at least four vertices, so G n has at least three dacards, and Lemma 2.3 yields drn(g n ) 3. Theorem 2.7. For the complete r-partite graph K n with n 1 n r, 1, if n 1 = n r 1 = 1 and n r {1,2}; adrn(k n ) = adrn(g n ) = 3, if n j n i {0,2} for some i and j with n j > 1; 2, otherwise. Proof. Again it suffices to determine adrn(g n ). By Lemmas 2.2 and 2.3, adrn(g n ) 3. When n 1 = n r 1 = 1 and n r {1,2}, any dacard of G n determines G n. 5

7 The condition in Lemma 3.1 is sufficient but not necessary. For example, if G consists of one edge uv joining two disjoint complete graphs, then the condition fails, but dern(g) = 1. In general, dern(g) = 1 if and only if G has an edge e such that all nonadjacent pairs in G e with degree-sum d(e) are in the same edge-orbit in the complement of G e. Lemma 3.2. For clique-unions, dern(g n ) = 1 if and only if (1) G n has a 2-component, or (2) there exists k with k 3 such that G n has a k-component, has no (k 1)-component, and does not have both a (k + j)-component and a (k j 2)-component for any j with 0 j k 3. Proof. Sufficiency. If (1) holds, then Lemma 3.1 applies. Now suppose that G n has a k- component and that G n has neither a (k 1)-component nor both a (k + j)-component and a (k j 2)-component with 0 j k 3. For an edge uv in the k-component, d(uv) = 2k 4. In G n uv, the only pair of non-adjacent vertices with degree-sum 2k 4 is {u,v}, so Lemma 3.1 applies. Necessity. We prove the contrapositive; assume that (1) and (2) fail. Since (1) fails, G n has no 2-component. Consider any edge e in G n, it belongs to some k-component with k 3. Since (2) fails, G n also either has a (k 1)-component or has a (k + j)-component and a (k j 2)-component for some j with 0 j k 3. In both cases, we can add an edge with degree 2k 4 to G n e that creates a graph not isomorphic to G n. Lemma 3.3. Any two decards of G n determine G n, except for two identical decards having degree 2 for the deleted edge. Proof. Let e 1 and e 2 be the edges deleted to form these decards, with d(e 1 ) d(e 2 ). By Lemma 3.1, we may assume d(e 1 ) > 0. Since all edges in G n have even degree, we may assume d(e 1 ) 2. We exclude the case d(e 2 ) = 2, so d(e 2 ) 4. Let H be a reconstruction from (G n e 2,d(e 2 )) that also has (G n e 1,d(e 1 )) as a decard; H is obtained from G n e 2 by adding an edge e of degree d(e 2 ). If H G n, then e joins two vertices in distinct components of G n e 2. Let H be the component of H containing e. Since d(e 2 ) 4, there are at least six vertices in H. For any edge e of degree d(e 1 ) in H, the graph H e has a component of order at least 4 in which e is a cut-edge. Since every component with at least four vertices in any decard of G n has no cut-edge, (G n e 1,d(e 1 )) is not a decard of H. This contradiction yields H = G n. For r 2, the disjoint union of K 1,3 with r 2 isolated vertices has the same dedecks as G (1,...,1,3), so they are not reconstructible from their dedecks. Theorem 3.4. For the graph G n with n 1 n r and n (1,...,1,3), 1, if n satisfies the condition of Lemma 3.2; dern(g n ) = 4, if n = (1,...,1,3,...,3) with n 1 = 1 and n r 1 = 3; 2, otherwise. 7

8 Proof. We have dern(g n ) = 1 when Lemma 3.2 applies (including all cases with n 1 > 1), and dern(g n ) 2 otherwise. If n r 4, then Lemma 3.3 yields dern(g n ) 2. In the remaining case, G n has at least two 3-components, at least one 1-component, and no components of other sizes. All decards of G n are copies of (G,2), where G is obtained from G n by deleting one edge. The graph obtained from G n by replacing a 1-component and a 3-component with K 1,3 also has three copies of (G,2) in its dedeck, so dern(g n ) 4. Let H be a graph having four copies of (G,2) in its dedeck; H is obtained from G by adding an edge e joining two vertices with degree-sum 2. If e joins the two vertices of degree 1 in G, then H = G n. Otherwise, e joins an isolated vertex with a vertex v of degree 2 in G. The component of G containing v may be a triangle or a path of length 2. Each resulting graph shares at most three decards with G n. Hence H = G n, yielding dern(g n ) 4. Let F +F denote the disjoint union of graphs F and F. Theorem 3.5. For the graph G n with n 1 n r and n (1,...,1,3), 1, if every edge of G n satisfies the condition of Lemma 3.1; 3, if n 1 = 2 and G n has a 3-component; adern(g n ) = 4, if n 1 = 1 and G n has a 3-component; 2, otherwise. Proof. Case 1: Every edge of G n satisfies the condition of Lemma 3.1. By Lemma 3.1, every decard (G n e,d(e)) determines G n. Cases 2 and 3: n 1 2 and G n has a 3-component. Let c = 5 n 1 ; we prove adern(g n ) = c. The graph K 2 + K 3 shares two decards with a 5-vertex path, and the graph K 1 + K 3 shares three decards with K 1,3. Hence adern(g n ) c. For the upper bound, we show that any c decards determine G n. By Lemma 3.3, we may assume that all are copies of (G,2), where G arises by deleting an edge of a 3-component in G n. Let H be a graph having c copies of (G,2) in its dedeck; H arises from G by adding an edge e joining vertices with degree-sum 2. When e joins vertices of degree 1 in G, there are three possible graphs; one is G n. The others have a 5-vertex path or 3- and 4-vertex paths as components. The dedecks have at most two copies of (G,2), since G has only one path component with at least three vertices. When n 1 = 1 and c = 4, also e may join a vertex of degree 2 to an isolated vertex, creating a component that is K 1,3 or K 1,3 plus an edge. Since G has no such component, decards shared with G n must arise by deleting an edge of this component. Three edges yield G when the component is K 1,3, but only one in the other case. Hence adern(g n ) c. Case 4: Some edge fails the condition of Lemma 3.1, and G n does not have both a 3- component and a smaller component. Since some edge fails the condition of Lemma 3.1, adern(g n ) > 1 and n r 4. Also, n 1 = 3 or G n has no 3-component. By Lemmas 3.1 and 3.3, any two decards determine G n. 8

9 4 dern(k n ) and adern(k n ) Since Monikandan and Sundar Raj [9] determined these parameters for complete bipartite graphs, we henceforth assume r 3. In the complete multipartite graph K n, let a k-part be a partite set of size k (analogous to k-component in G n ). Let m = n i, so any vertex in an n j -part has degree m n j. Lemma 4.1. A decard (K n e,d(e)) determines K n if and only if (1) e joins a 1-part and a 2-part, or (2) e joins a k-part and an l-part such that k l / {1,2} and K n has no ( k+l 2 +1)-part. Proof. Sufficiency. Under(1),d(e) = 2m 5. InK n e, theonlynonadjacentpairsofvertices with degree-sum 2m 5 are the endpoints of e and the vertices of the 2-part containing an endpoint of e. Adding either edge yields K n. Under (2), d(e) = 2m (k + l) 2. Since k l / {1,2}, no two vertices within one of these parts has degree-sum d(e) in K n e. Since also K n has no ( k+l+1)-part, the only nonadjacent vertices in K 2 n e with degree-sum d(e) are the endpoints of e. Necessity. An edge e failing the condition joins a k-part and an l-part such that (1) {k,l} {1,2} and (2) k l {1,2} or K n has a ( k+l +1)-part. In either case, we can add 2 an edge with degree 2m (k+l) 2 to K n e to create a graph not isomorphic to K n. Lemma 4.2. Any two decards of K n determine K n, except when the two decards are the same and are generated by deleting an edge joining a 3-part to a smaller part. Proof. Let (K n e 1,d(e 1 )) and (K n e 2,d(e 2 )) be two decards. Suppose that e 1 joins an i- partandaj-part, withi j, ande 2 joinsak-partandanl-part, withk l. Wemayassume that neither edge satisfies the condition of Lemma 4.1. Thus (i,j),(k,l) / {(1,1),(1,2)}. Hence, i+j 4 and k +l 4. The complement of K n is G n, with r components, all complete. The complement of K n e 1 is G n +e 1, where e 1 joins an i-component and a j-component of G n. Hence G n +e 1 has r 1 components. Let H be a reconstruction from (K n e 1,d(e 1 )) that also has (K n e 2,d(e 2 )) as a decard. Let e be the added edge, so H = K n e 1 +e. Let e be the edge deleted from H to obtain K n e 2 ; note that e has degree d(e 2 ) in H. We may assume e e, even if K n e 1 = Kn e 2. Let H = K n e 1 +e e; the complement of H is G n +e 1 e +e. If e = e 1, then H = K n, so we may assume e e 1. Our task is to show that in this case, H cannot be isomorphic to K n e 2, and hence no graph other than K n shares these two decards with K n. Since e e 1, the edge e is not a cut-edge of G n + e 1. Hence G n + e 1 e has r 1 components. If the endpoints of e lie in different components of G n + e 1 e, then the complement of H has r 2 components. Since the complement of K n e 2 has r 1 9

10 components, H cannot be a decard of K n and hence cannot be K n e 2. Hence we may assume that the endpoints of e lie in one component of G n +e 1 e. Let G denote the non-complete component of G n +e 1, in which e 1 is a cut-edge. Since e 1 does not satisfy the condition of Lemma 4.1, we have that j i {1,2} or that K n has an ( i+j 2 +1)-part, or both. We distinguish cases based on the location of e. Case 1: j i {1,2} and e joins two vertices in the j-part having an endpoint of e 1. Since (i,j) (1,2), we have j 3. Since e E(G ) in this case, G e is the only component of G n +e 1 e that is not complete. Hence G e is the component of G n +e 1 e containing both endpoints of e (recall that K n e 2 = H e). If j 4, then G e +e has no cut-edge. Thus no component with at least four vertices in the complement of H has a cut-edge. However, since k+l 4, the complement of K n e 2 has a component with at least four vertices in which e 2 is a cut-edge. Hence H K n e 2. If j = 3 and i = 1, then G e = K1,3. We have G e +e = G and G n +e 1 e +e = G n +e 1. Hence, H = Kn e 1. In this case we have assumed that the two decards are not the same. That is, K n e 2 K n e 1, which yields H K n e 2, as desired. If j = 3 and i = 2, then G e is a 5-vertex path. Up to isomorphism, there are four ways to add e to form G e +e. One of them creates G ; as in the previous subcase, this is forbidden by assuming K n e 2 K n e 1. In the other three cases (5-cycle, 4-cycle plus pendant edge, 3-cycle plus two pendant edges), G e + e cannot be produced by adding an edge joining two complete graphs. That is, H e is not a decard of K n. Case 2: e joins two vertices of K n in an ( i+j 2 +1)-part. In this case, i+j is even. By Case 1, we may assume that the part containing the endpoints of e does not contain an endpoint of e 1. Hence the complement G n +e 1 e of H has two non-complete components, one of which contains both endpoints of e. Since we have assumed e e, and e is the only edge missing from the component of G n +e 1 e containing its endpoints, again e must join two vertices of G, as in Case 1. Since G has at least four vertices, G +e is not a complete graph. Hence the complement of H has two non-complete components. This implies that H cannot be a decard of a complete multipartite graph and hence cannot be K n e 2. The problem is simpler for r = 2, because K m,n is edge-transitive. Monikandan and Sundar Raj [9] solved it; we include the statement for completeness. Theorem 4.3 ([9]). For 1 m n and (m,n) (1,3), 3, if (m,n) = (2,3); adern(k m,n ) = dern(k m,n ) = 2, if n 4 and m {n 1,n 2}; 1, otherwise ; 10

11 Theorem 4.4. If r 3 and n 1 n r, then { 1, if (n 1,...,n r ) satisfies the condition of Lemma 4.1; dern(k n ) = 2, otherwise ; Proof. We have dern(k n ) = 1 when (n 1,...,n r ) satisfies the condition of Lemma 4.1. Otherwise, dern(k n ) > 1 and n 1 n r. Since also r 3, there are distinct decards of K n, and Lemma 4.2 yields dern(k n ) 2. Theorem 4.5. If r 3 and n 1 n r, then 1, if every edge of K n satisfies the condition of Lemma 4.1; 3, if n 1 = 2 and K n has a 3-part; adern(k n ) = 4, if n 1 = 1 and K n has a 3-part; 2, otherwise ; Proof. Case 1: Every edge of K n satisfies the condition of Lemma 4.1. By Lemma 4.1, every decard (G n e,d(e)) determines K n. Case 2: K n has a 3-part and a 2-part but no 1-part. Let S be the union of a 3-part and a 2-part. Let F be a 4-cycle plus a pendant edge, and let F be a 5-cycle plus a chord. Both K 2,3 and F have two copies of (F,3) as decards. Hence substituting F for the subgraph of K n induced by S yields a graph sharing two decards with K n, so adern(k n ) 3. For the upper bound, since K n has no 1-part, Lemma 4.2 implies that we only need to prove that K n is determined by three copies of the decard (K n e,d(e)), where e joins two vertices of S. Note that d(e) = 2m 7, where m = V(K n ). Construct H from K n e by adding an edge e of degree 2m 7. If H K n, then e joins vertices of degrees m 4 and m 3 in the 3-part. The subgraph of H induced by S is F. Each vertex of S has m 5 neighbors outside S. Hence to obtain (K n e,2m 7) as a decard of H, we must delete an edge with degree 3 in F. There are four such edges, but only two yield F when deleted. Hence H has only two copies of (K n e,2m 7) in its dedeck, so adern(k n ) 3. Case 3: K n has a 3-part and a 1-part. Let S be the union of a 3-part and a 1-part. Substituting K 1 + K 3 for the subgraph of K n induced by S (which is K 1,3 ) yields a graph sharing three decards with K n, so adern(k n ) 4. For the upper bound, by Lemma 4.2 we only need to prove that four identical decards obtained by deleting an edge joining a 3-part to a smaller part determine K n. If the smaller part is a 2-part, then the argument for Case 2 applies. Hence we only need to prove that four copies of (K n e,2m 6) determine K n, where e joins a 3-part and a 1-part. Construct H from K n e by adding an edge of degree 2m 6. If H K n, then H is formed by adding an edge joining the two vertices of degree m 3 in a 3-part. The graph H has at most three copies of (K n e,2m 6) in its dedeck. Hence, adern(k n ) 4. 11

12 Case 4: Some edge fails the condition of Lemma 4.1, and K n does not have both a 3-part and a smaller part. Since some edge fails the condition of Lemma 4.1, adern(k n ) > 1. Every edge of K n joins i-part and j-part where (i,j) / {(1,2),(2,3)}. By Lemma 4.2, any two decards determine K n. References [1] M.D. Barrus and D.B. West, Degree-associated reconstruction number of graphs, Discrete Math. 310 (2010), [2] B. Bollobás, Almost every graph has reconstruction number three, J. Graph Theory 14 (1990), 1 4. [3] F. Harary, On the reconstruction of a graph from a collection of subgraphs, Theory of Graphs and its Applications (Proc. Sympos. Smolenice, 1963) (Publ. House Czechoslovak Acad. Sci., Prague, 1964), [4] F. Harary and M. Plantholt, The graph reconstruction number, J. Graph Theory 9 (1985), [5] P. J. Kelly, On isometric transformations, PhD Thesis, University of Wisconsin-Madison, [6] P. J. Kelly, A congruence theorem for trees, Pacific J. Math. 7 (1957), [7] M. Ma, H. Shi, and D. B. West, Adversary degree-associated reconstruction number of doublestars, submitted. [8] S. Monikandan and S. Sundar Raj, Adversary degree associated reconstruction number of graphs, Disc. Math. Alg. Appl., submitted. [9] S. Monikandan and S. Sundar Raj, Degree associated edge reconstruction number, in Proc. 23rd International Workshop on Combinatorial Algorithm (IWOCA 2012), S. Arumugam and B. Smyth, eds., Lect. Notes Comp. Sci (2012), [10] S. Monikandan, S. Sundar Raj, C. Jayasekaran, and A.P. Santhakumaran, A note on the adversary degree associated reconstruction number of graphs, J. Discrete Math. Volume 2013 (2013), Article ID , 5 pages, dx.doi.org/ /2013/ [11] W. J. Myrvold, The ally and adversary reconstruction problems, PhD thesis, University of Waterloo, [12] S. Ramachandran, Degree associated reconstruction number of graphs and digraphs, Mano. Int. J. Math. Sci. 1 (2000), [13] S. M. Ulam, A collection of mathematical problems, Interscience Tracts in Pure and Applied Mathematics 8 (Interscience Publishers, 1960). 12

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