On the tree cover number of a graph

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1 On the tree cover nmber of a graph Chassidy Bozeman Minerva Catral Brendan Cook Oscar E. González Carolyn Reinhart Abstract Given a graph G, the tree cover nmber of the graph, denoted T (G), is the minimm nmber of vertex disjoint simple trees occrring as indced sbgraphs that cover all the vertices of G. This graph parameter was introdced in 2011 as a tool for stdying the maximm positive semidefinite nllity of a graph, and little is known abot it. It is conjectred that the tree cover nmber of a graph is at most the maximm positive semidefinite nllity of the graph. In this paper, we establish bonds on the tree cover nmber of a graph, characterize when an edge is reqired to be in some tree of a minimm tree cover, and show that the tree cover nmber of the d-dimensional hypercbe is 2 for all d 2. Keywords: graphs, matrices, tree cover nmber, maximm positive semidefinte nllity MSC2010: 05C50, 05C05, 05C76 1 Introdction A simple graph is a pair G = (V, E), where V = {1, 2,..., n} is the vertex set, and E, the edge set, is a set of 2-element sbsets (edges) of the vertices. A mltigraph is a pair G = (V, E), where V = {1, 2,..., n}, and E is a mltiset of 2-element sbsets of the vertices. That is, a mltigraph allows mltiple edges between a pair of vertices. (Note that all simple graphs are mltigraphs.) Two Department of Mathematics, Iowa State University, Ames, IA 50011, USA (cbozeman@iastate.ed) Department of Mathematics and Compter Science, Xavier University, Cincinnati, OH 5207, USA (catralm@xavier.ed) Department of Mathematics, Carleton College, Northfield, MN 55057, USA (cookb@carleton.ed Department of Mathematics, University of Perto Rico, San Jan, PR,0091 (oscar.gonzalez@pr.ed) Department of Mathematics, Minneapolis, MN 5555, USA (reinh196@mn.ed) Research spported by NSF DSM-157 1

2 vertices, v V (G) are said to be adjacent if {, v} E(G). All graphs in this paper are considered to be mltigraphs nless otherwise stated. For a mltigraph G, S(G) denotes the set of real valed symmetric n n matrices (a i,j) satisfying: 1. a i,j = 0 if i j and i, j are nonadjacent, 2. a i,j 0 if i j and i, j are adjacent via one edge, and. a i,j R if i = j or i, j are adjacent via mltiple edges. The maximm nllity of a mltigraph G is defined to be M(G) = max{nll(a) : A S(G)}. The maximm nllity of a simple graph G is eqivalent to the maximm mltiplicity of an eigenvale among all matrices in S(G). This graph parameter has connections to many other concepts in linear algebra (as can be seen in [5] and []), and has been given a significant amont of consideration as it is very difficlt to compte. A related and eqally important parameter is the maximm positive semidefinite nllity of a graph. A symmetric n n real matrix A is said to be positive semidefinte if x T Ax 0 for all x R n. The maximm positive semidefinite nllity of a mltigraph G is defined to be M +(G) = max{nll(a) : A S +(G)}, where S +(G) = {A S(G) : A is positive semidefinite}. It follows that for a mltigraph G, M +(G) M(G). In some cases, one can se tools sch as orthogonal representations (see [5]) to compte M +(G), obtaining a lower bond for M(G). The tree cover nmber of a graph was introdced in 2011 in [1] as another tool for stdying the maximm positive semidefinite nllity of a mltigraph. The (simple) path on n vertices, denoted P n, is the graph with vertex set V (P n) = {1,..., n} and edge set E(P n) = {{i, i + 1} i 1,..., n 1}. A simple graph G = (V, E) is said to be a tree if for every, v V (G), there is exactly one path from to v. Given a graph G = (V, E), a sbgraph, G = (V, E ), is a graph sch that V (G ) V (G) and E(G ) E(G), i.e, a sbgraph of a graph G can be obtained be deleting edges and/or vertices (and edges incident to the deleted vertices) of G. A sbgraph G = (V, E ) of G is said to be an indced sbgraph of G if for each edge v E(G) with, v V (G ), it follows that v E(G ), i.e, an indced sbgraph of G can be obtained by only deleting vertices (and any edges incident to the deleted vertices). For a sbset S V (G), the graph indced by 2

3 S, denoted G[S], is the indced sbgraph of G with vertex set S. A tree cover is a set of vertex disjoint simple trees occrring as indced sbgraphs that cover all the vertices of the graph. The tree cover nmber of a graph G, denoted T (G), is defined as Conjectre 1. [1] T (G) M +(G). T (G) = min{ T : T is a tree cover of G}. This bond has been proven to be tre for several families of graphs, inclding oterplanar graphs [1] and chordal graphs [1]. In fact, eqality holds for oterplanar graphs (and in fact for all graphs of tree-width at most 2, as observed in []). In section 2 we give bonds on the tree cover nmber, provide an example in which the tree cover nmber behaves like the maximm positive semidefinite nllity, and provide an example in which the tree cover nmber does not behave like the maximm positive semidefinite nllity. See [1] and [] for definitions of oterplanar and tree-width. In section, we characterize when an edge is reqired to be in some tree of a minimm tree cover. In section, we prove that the tree cover nmber of the d dimensional hypercbe is 2 for all d More Notation and Terminology The cycle on n vertices, denoted C n, is the graph with vertex set V (C n) = {1,..., n} and edge set E(C n) = {{i, i + 1} i 1,..., n 1} {1, n}. The star K 1,n is the graph with vertex set {1,..., n} and edge set {{1, j} j {2,..., n}}. The complete graph, denoted K n, is the graph on n vertices sch that there is an edge between any two vertices. A graph is said to be connected if there is a path from any vertex to any other vertex. If G is not connected, then it is said to be disconnected. Given a graph G = (V, E), a connected component of G is a sbgraph, C, where C is connected and no vertex in C is adjacent to any vertex of V (G) \ V (C). A graph is said to be a forest if each of its connected components is a tree. If vertices and v are adjacent, we say that they are neighbors. The neighborhood of a vertex v, denoted N(v), is the set of neighbors of v. The degree of v is given by deg(v) = N(v). For a graph G = (V, E), a cover of G is a partition of V (G). An independent set, S, is a sbset of V (G) sch that no two vertices in S are adjacent. The independence nmber of G, denoted α(g), is defined by α(g) = max{ S : S is an independent set in G}.

4 Given two simple graphs G and H, the cartesian prodct of G and H, denoted G H is the graph whose vertex set is the cartesian prodct V (G) V (H), and any two vertices (, ) and (v, v ) are adjacent in G H if and only if either = v and is adjacent to v in H, or = v and is adjacent to v in G. The nion of G and H, denoted G H is the graph with vertex set V (G) V (H) and edge set E(G) E(H). Throghot this paper, we often denote an edge {, v} by v. For an edge e = v E(G), we se G e to denote the graph obtained from G by adding a new vertex w into V (G), inserting edges w and wv into E(G), and deleting v from E(G). In this case, we say that the edge e = v has been sbdivided, and we call G e an edge sbdivision of G. An edge v is called a bridge of G if C v is disconnected, where C is the component of G with v E(C) and C v denotes the sbgraph obtained from C by deleting the edge v. 2 Some Bonds for the Tree Cover Nmber In this section, we give an pper bond on the tree cover nmber of a graph sing the size of an independent set in the graph. We also provide pper and lower bonds on the tree cover nmber of a sbgraph of G obtained by deleting an edge from G. In addition, we observe that sbdividing an edge of a graph does not change the tree cover nmber. The following proposition shows that, for a connected graph, we are able to bond the tree cover nmber by the difference between the order of the graph and the size of an independent set of vertices of the graph. Proposition 2. Let G = (V, E) be a connected graph, and let S V (G) be an independent set. Then, T (G) G S. In particlar, T (G) G α(g), where α(g) is the independence nmber of G. Frthermore, this bond is tight. Proof. We constrct a tree cover of G of size G S. For v V (G) \ S, let T v denote the tree indced by (N(v) S) {v}. Since G is connected, for each s S, there mst exist a v V (G) \ S sch that s V (T v). If there exists more than one sch vertex, we simply pick one v and assign s to V (T v). This gives a tree cover of size G S. Ths, T (G) G S. The star K 1,n shows that the bond T (G) G α(g) is tight. In connection with the conjectre that T(G) M +(G), we show that for some bonds on M +(G), analogos bonds hold for T(G). For a graph G = (V, E) and e E(G), let G e denoted the graph obtained from G be deleting the edge e. In [2], it was shown that M +(G) 1 M +(G e) M +(G) + 1. We show that an analogos bond holds for the tree cover nmber.

5 Theorem. For a graph G = (V, E) and e E(G), T(G) 1 T(G e) T (G) + 1. Proof. Let, v V (G) sch that e = v. Consider the graph G e obtained from G by deleting e. Let T be a minimm tree cover of G e. If and v are in the same tree in T, denoted by T v, then the graph indced by the vertices of T v contains a cycle in G, so T is not a tree cover of G. However, we may partition the vertices of T v into two sets A and B, sch that the tree indced by the vertices in A contains and the tree indced by the vertices in B contains v. Denote these trees by T A and T B. Then, (T \ T v) T A T B is a tree cover of G of size T (G e) + 1, so T(G) 1 T(G e). If and v are in disjoint trees in T, denoted by T and T v, then T is a tree cover of G. So, T (G) T (G e). Frthermore, the edge e connects T to T v. If a cycle is created, then T (G) = T (G e). If no cycle is created, then the graph indced in G by the vertices of T and T v is a single tree, and T (G) + 1 = T (G e), which completes the proof. It was also shown in [6] that sbdividing an edge in a graph does not change the positive semidefinite maximm nllity of the graph. Observation. Let G = (V, E) be a graph, e E(G), and let G e denote the graph obtained from G by sbdividing the edge e. Then T(G e) = T(G). This observation can be immediately dedced from Proposition. of [1]. The next theorem gives a bond that holds for the positive semidefinite maximm nllity of a graph, bt the example that follows demonstrates that the analogos bond for the tree cover nmber fails. A 2-separation of a graph G = (V, E) is a pair of sbgraphs (G 1, G 2) sch that V (G 1) V (G 2) = V, V (G 1) V (G 2) = 2, E(G 1) E(G 2) = E, and E(G 1) E(G 2) =. Theorem 5. ([6], Theorem 2.8) Let (G 1, G 2) be a 2-separation of a graph G = (V, E), and let H 1 and H 2 be obtained from G 1 = (V 1, E 1) and G 2 = (V 2, E 2), respectively, by adding an edge between the vertices of R = {r 1, r 2} = V 1 V 2. Then M +(G) = max{m +(G 1) + M +(G 2) 2, M +(H 1) + M +(H 2) 2} The analogos bond does not hold for the tree cover nmber. The next example provides a conterexample. Example 6. For the graphs G, G 1, G 2, H 1, H 2 given in Figre 1 below, we have that M +(G i) = 2, M +(H i) =, and T (G i) = T (H i) = 2 for i {1, 2}. So by Theorem 5, M +(G) =. However, = T(G) > max {T(G 1) + T(G 2) 2, T(H 1) + T(H 2) 2} = 2. 5

6 v v 6 v v v 10 Figre 1: G (top left), G 1 (top middle), G 2 (top right), H 1 (bottom left), H 2 (bottom right) Characterizing Edges Reqired in a Minimm Tree Cover Proposition 7. Let G = (V, E) be a graph sch that v E(G) is a bridge. Then v is in a tree in every minimm tree cover of G. Proof. Note that there is no path from to v that does not inclde v. Therefore, for any tree cover that does not inclde v, it mst be the case that and v are in separate trees. These two trees can be consolidated into one tree by adding the edge v. We then ask the qestion: If an edge is reqired in every minimm tree cover, mst it be a bridge? Figre 2 below shows that sch an edge is not necessarily a bridge. Example 8. Figre 2 below gives a graph whose tree cover nmber is 2. However, althogh v is not a bridge, any tree cover that does not inclde v is of size at least. The next lemma gives s a way to determine if an edge is reqired in every minimm tree cover, given that we are able to compte the necessary tree cover nmbers. Lemma 9. Let G be a graph,, v V (G), and v E(G). Let H be the graph obtained from G by adding a vertex sch that V (H) = V (G) {w} and E(H) = E(G) {w, vw}. If v is reqired in every minimm tree cover of 6

7 1 v 6 Figre 2: for Example 8 G, then T(H) = T(G) + 1 Frthermore, if T (H) = T (G), then there exists a minimm tree cover of G not inclding v. Proof. First observe that T (H) T (G) + 1 since any tree cover of G together with {w} is a tree cover for H. Let T = {T 1, T 2,... T k } be a minimm tree cover of H sch that w T i for some i. Since w,, and v cannot all be in the same tree, then either w is a leaf in T i or T i = {w}. If w is a leaf in T i, then T 1, T 2,..., T i w, T i+1,..., T k is a tree cover of G, so T (G) T (H). If T i = w, then T \ T i is a tree cover for G, so T (G) T (H) 1. This shows that T (H) = T (G) or T (H) = T (G) + 1. Sppose that v is reqired in every minimm tree cover of G. If w is a leaf in T i, then T 1, T 2,..., T i w, T i+1,..., T k is a tree cover of G with and v in separate trees, so it follows that k = T (G) + 1. If T i = w, then we also have that T (H) = T (G) + 1. Sppose that there exists a minimm tree cover T = {T 1, T 2,... T k } of G sch that and v are in different trees. If T i, we can create a tree cover of H of size k by adding the edge v to E(T i). In this case, T (G) = T (H). One might think that if H is a graph obtained from G by adding the edge v, and v is reqired in every minimm tree cover of H, then T (G) = T (H) + 1. However, this is not tre. Example 10 provides a conterexample. Example 10. It is easy to see that T (G) = T (H) = 2. (For H, take the set {1,, v, 5} and {2,, } for example.) It can also be verified that T (Ĥ) =. By Lemma 9, it follows that the edge v is reqired in every minimm tree cover of H w 5 v 5 v 5 v Figre : G, H and Ĥ for Example 10. 7

8 Tree Cover Nmber of the Hypercbe The d-dimensional hypercbe, denoted Q d, is the graph with vertex set {0, 1} d where two vertices are adjacent if and only if they differ in exactly one position. For example, the 2 dimensional hypercbe is a sqare, shown in Figre below, and the dimensional hypercbe is a cbe, shown in Figre 5. Eqivalently, hypercbes can be indctively defined as the cartesian prodct of d copies of the complete graph K 2. Hypercbes are a particlar case of a larger family of graphs called Hamming graphs. The d dimensional Hamming graph, denoted H(d, q) is the graph with vertex set {0,..., n 1} d where two vertices are adjacent if and only if they differ in exactly one position. Hamming graphs are of se in many areas inclding error-correcting codes, modeling heat diffsion, and association schemes in statistics. In this section, we show that the tree cover nmber of the d-dimensional hypercbe is 2 for all d Figre : Q 2. Figre 5: Q. Theorem 11. Let Q d be the d dimensional hypercbe graph. For all d 2, T (Q d ) = 2. Proof. For d {2,,, 5}, the sets which indce a tree cover of size for Q d are explicitly listed in the appendix. Throghot this proof, the sets ˆT 1j and ˆT 2j are covers that will be sed as preliminary steps to obtain the sets T 1j and T 2j that will indce a tree cover of size two for Q j. The proof proceeds as follows: First we constrct a cover and a tree cover of size two for Q 6. Using this tree cover, we constrct a cover and a tree cover of size two for Q 7. We then indctively show that for d 8 we can systematically constrct a tree cover of size two sing the covers and tree covers constrcted for Q d 1 and Q d 2. Consider the sets ˆT 16 and ˆT 26 given in the appendix. Note that { ˆT 16, ˆT 26 } is a cover for Q 6, and that Q 6[ ˆT 16 ] and Q 6[ ˆT 17 ] are both forests, each consisting of two disjoint trees. Let x 16 = (001101), x 26 = (110010), y 16 = (001001), y 26 = (110100). Then x 16 and x 26 are in ˆT 16, and they are not in the same tree in Q 6[ ˆT 16 ]. Similarly, y 16 and y 26 are in ˆT 26, and they are not in the same tree in Q 6[ ˆT 26 ]. By swapping x 16 and y 16, the reslting sets T 16 and T 26 (listed in the appendix) indce a tree cover for Q 6 of size two. To obtain a tree cover of size two for Q 7, we begin by adding a 0 to the beginning of each element in T 16, and a 1 to the beginning of each element in T 26. 8

9 Denote these sets by T 16,0 and T 26,1, respectively, and let ˆT 17 := T 16,0 T 26,1. Similarly, we constrct the sets T 16,1 and T 26,0, and let ˆT 27 := T 16,1 T 26,0 (see appendix). Then, both Q 7[ ˆT 17 ] and Q 7[ ˆT 27 ] are forests consisting of two disjoint trees. By swapping 0x 26 and 0y 26, the reslting sets T 17 and T 27 (given in appendix) indce a tree cover of size two for Q 7. We proceed by indction to prove the claim for Q d with d 8. Sppose that we have constrcted the sets ˆT 1d 2 = {x 1, x 2,..., x n} and ˆT 2d 2 = {y 1, y 2,..., y n} sch that { ˆT 1d 2, ˆT 2d 2 } gives a cover for Q d 2 satisfying the following conditions: (1.) Q d 2 [ ˆT 1d 2 ] and Q d 2 [ ˆT 2d 2 ] are forest composed of two disjoint trees. (2.) Swapping x 1 and y 1 reslts in sets T 1d 2 = {y 1, x 2,..., x n} T 2d 2 = {x 1, y 2,..., y n} that indce a tree cover of Q d 2 of size two. (.) For the cover ˆT 1d 1 = T 1d 2,0 T 2d 2,1 = {0y 1, 0x 2, 0x..., 0x n, 1x 1, 1y 2,..., 1y n} ˆT 2d 1 = T 2d 2,0 T 1d 2,1 = {0x 1, 0y 2, 0y..., 0y n, 1y 1, 1x 2,..., 1x n} of Q d 1, swapping 0x 2 ˆT 1d 1 and 0y 2 ˆT 2d 1 reslts in sets T 1d 1 = {0y 1, 0y 2, 0x..., 0x n, 1x 1, 1y 2,..., 1y n} T 2d 1 = {0x 1, 0x 2, 0y..., 0y n, 1y 1, 1x 2,..., 1x n}. for Q d 1 that indced a tree cover of Q d 1 of size two. (.) x 1 and x 2 are not in the same indced tree in ˆT 1d 2 (5.) y 1 and y 2 are not in the same indced tree in ˆT 2d 2 (We are also assming that x i x j, y i y j for i j, and x i y j for all i, j). Then we can constrct a cover for Q d sch that swapping two of the elements in the cover will reslt in a tree cover of size two for Q d. Frthermore, we show that the constrcted cover and tree cover for Q d, together with the constrcted cover and tree cover for Q d 1, still satisfy the above hypotheses, which proves the claim for all d 8. We first constrct a cover { ˆT 1d, ˆT 2d } for Q d in the following way: ˆT 1d = T 1d 1,0 T 2d 1,1 = {00y 1, 00y 2, 00x..., 00x n, 01x 1, 01y 2, 01y,..., 01y n, 10x 1, 10x 2, 10y,..., 10y n, 11y 1, 11x 2, 11x,..., 11x n} ˆT 2d = T 2d 1,0 T 1d 1,1 = {00x 1, 00x 2, 00y..., 00y n, 01y 1, 01x 2, 01x,..., 01x n, 10y 1, 10y 2, 10x,..., 10x n, 11x 1, 11y 2, 11y,..., 11y n}. 9

10 Note that since Q d 1 [T 1d 1 ] and Q d 1 [T 2d 1 ] are two disjoint trees, it follows that Q d [ ˆT 1d ] is a forest consisting of two disjoint trees. Similarly, Q d [ ˆT 2d ] is a forest consisting of two disjoint trees. By swapping 01x 1 and 01y 1, we obtain the sets T 1d = {00y 1, 00y 2, 00x..., 00x n, 01y 1, 01y 2, 01y,..., 01y n, 10x 1, 10x 2, 10y,..., 10y n, 11y 1, 11x 2, 11x,..., 11x n} T 2d = {00x 1, 00x 2, 00y..., 00y n, 01x 1, 01x 2, 01x..., 01x n, 10y 1, 10y 2, 10x,..., 10x n, 11x 1, 11y 2, 11y,..., 11y n}. We now show that {Q d [T 1d ], Q d [T 2d ]} is a tree cover for Q d of size two by showing: 1. Q d [T 1d ] and Q d [T 2d ] are forests (i.e., there are no cycles in each of Q d [T 1d ] and Q d [T 2d ]). 2. Both Q d [T 1d ] and Q d [T 2d ] are connected graphs. We show that Q d [T 1d ] is a forest (a similar argment shows that Q d [T 2d ] is a forest). From or constrction Q d [ ˆT 1d ] is a forest composed of 2 trees, Q d [Â] and Q d [B] where  := {00y 1, 00y 2, 00x..., 00x n, 01x 1, 01y 2, 01y,..., 01y n} B := {10x 1, 10x 2, 10y,..., 10y n, 11y 1, 11x 2, 11x,..., 11x n}. By definition T 1d = ( ˆT 1d \ {01x 1}) {01y1}. By removing 01x 1 from ˆT 1d, B is not affected, and Q d [Â\{01x1}] is now the nion of deg(01x1) disjoint trees. We now show that by adding 01y 1 to ˆT 1d \ {01x 1}, no cycles are created in Q d [ T 1d ]. Define A = {00y 1, 00y 2, 00x..., 00x n, 01y 1, 01y 2, 01y,..., 01y n} (note that T 1d = A B). Between A and B, the only vertices that are adjacent are 01y 1 and 11y 1 (everything else differs in more than one position). Hence, if there is a cycle in Q d [T 1d ], it mst be in Q d [A]. Since Q d [A \ {01y 1}](= Q d [Â\{01x1}]) is a forest composed of deg(01x 1) trees, if there is a cycle in Q d [A] it mst involve 01y 1. We will now show that it is not possible to have a cycle involving 01y 1, hence no cycle is possible in Q d [T 1d ]. Note that there is an edge between 00y 1 and 01y 1, and that there are no edges between 01y 1 and any of 00y 2, 00x..., 00x n. Ths, the neighbors of 01y 1 in Q d [A] are 00y 1 and a sbset of {01y, 01y,..., 01y n} (since y 1 is not adjacent to y 2 by condition (5) above, then 01y 1 is not adjacent to 01y 2). Let 01y i and 01y j, i j, be arbitrary neighbors of 01y 1. We show that: (a) There is no path from 01y i to 01y j in Q d [A] for i, j {,,..., n} that does not inclde 01y 1. (b) There is no path from 00y 1 to 01y i in Q d [A] that does not inclde 01y 1. 10

11 To see (a), note that from condition (1), Q d 2 [{y 1,... y n}] is a forest of two disjoint trees. This implies that Q d [{01y 1,... 01y n}] is a forest of two disjoint trees. Then, within Q d [{01y 1,... 01y n}] there is no path from 01y i to 01y j that does not inclde 01y 1. Note that vertices of {01y, 01y,..., 01y n} are not adjacent to any vertices in A except for possibly each other and 01y 1 and 01y 2. Ths, any path from 01y i to 01y j not inclding 01y 1 mst inclde 01y 2. By condition (1), y 1 and y 2 are not in the same indced tree of Q d 2 [{y 1,..., y n}], so 01y 1 and 01y 2 are not in the same indced tree of Q d [{01y 1,..., 01y n}]. Since 01y i and 01y j are neighbors of 01y 1, and 01y 1 is not in the same indced tree as 01y 2 in Q d [{01y 1,..., 01y n}], then 01y i and 01y j are not in the same indced tree as 01y 2. Ths, the only path from 01y i to 01y j is (01y i, 01y 1, 01y j). For (b), we have that the vertices in the set {01y, 01y,..., 01y n} are not connected in Q d [A] to any vertices in A except for possibly each other and 01y 1. We also have that 01y i is not adjacent to 00y 1 in Q d [A]. So any path from 01y i to 00y 1 mst inclde 01y 1. Next we show that Q d [T 1d ] is connected (a similar argment shows that Q d [T 2d ] is connected). Recall from the hypotheses that Q d 2 [ ˆT 2d 2 ] = Q d 2 [{y 1, y 2,..., y n}] is a forest consisting of two disjoint trees, and Q d 2 [T 2d 2 ] = Q d 2 [{x 1, y 2,..., y n}] is a tree. This implies that y 1 has exactly one fewer neighbor among y 2,..., y n than x 1. To see this, note that Q d 2 [ ˆT 2d 2 \ {y 1}] is composed of 1 + deg(y 1) trees. Since Q d 2 [T 2d 2 ] = Q d 2 [( ˆT 2d 2 \ {y 1}) {x 1}] is a tree, we mst have deg(x 1) = 1 + deg(y 1). Therefore, 01y 1 mst have one less neighbor than 01x 1 among 01y 2,..., 01y n. Hence, 01y 1 and 01x 1 have the same nmber of neighbors in A, and ths 01y 1 has one more neighbor than 01x 1 in T 1d.We will now show that this last statement implies that Q d [T 1d ] is connected. Since the graphs indced by T 1d 1 and T 2d 1 are trees, then Q d [T 1d ] = Q d [T 1d 1,0 T 2d 1,1 ] is a forest consisting of two disjoint trees. Hence, Q d [ ˆT 1d \ {01x 1}] is a forest consisting of 1 + deg(01x 1) trees. Since deg(01y 1) = 1 + deg(01x 1), and since Q d [T 1d ] has no cycles, we have that each of the edges of 01y 1 mst be connected to a different component of the forest. Therefore, Q d [T 1d ] is a tree. An analogos argment shows that Q d [T 2d ] is a tree. Ths, {Q d [T 1d ], Q d [T 2d ]} is a tree cover of size two of Q d. We now show that the covers and tree covers constrcted for Q d 1 and Q d satisfy the indction hypotheses. Note that since Q d 2 [T 1d 2 ] and Q d 2 [T 2d 2 ] are two disjoint trees, it follows from constrction that Q d [ ˆT 1d 1 ] is a forest consisting of two disjoint trees. Similarly, Q d [ ˆT 2d 1 ] is a forest consisting of two disjoint trees, satisfying condition (1). For clarity, we relabel the vertices of ˆT 1d 1 and ˆT 2d 1 sch that ˆT 1d 1 = {w 1,..., w m} and ˆT 2d 1 = {z 1,..., z m} where w 1 = 0x 2, w 2 = 1x 1, z 1 = 0y 2, z 2 = 1y 1. Then by condition (), swapping w 1 and z 1 reslts in sets T 1d 1 = {z 1, w 2,..., w m} and T 2d 1 = {w 1, z 2,..., z m} 11

12 that indce a tree cover of Q d 1 of size two, which shows that condition (2) is satisfied. Note that with this relabeling, the sets ˆT 1d and ˆT 2d become ˆT 1d = T 1d 1,0 T 2d 1,1 = {0z 1, 0w 2, 0w..., 0w m, 1w 1, 1z,..., 1z m} ˆT 2d = T 2d 1,0 T 1d 1,1 = {0w 1, 0z 2, 0y..., 0z m, 1z 1, 1w 2,..., 1w m}, and we have shown above that swapping 0w 2 = 01x 1 and 0z 2 = 01y 1 reslts in the sets T 1d and T 2d which indce a tree cover of size two for Q d, satisfying condition (). Frthermore, since w 1 = 0x 2 T 1d 2,0 and w 2 = 1x 1 T 2d 2,1, we have that w 1 and w 2 are not in the same indced tree in Q d 1 [ ˆT 1d 1 ]. Similarly, z 1 = 0y 2 T 2d 2,0 and z 2 = 1y 1 T 1d 2,1, so z 1 and z 2 are not in the same indced tree in Q d 1 [ ˆT 2d 1 ], showing that conditions () and (5) are satisfied. Since the hypotheses still hold with the constrcted covers and tree covers of Q d 1 and Q d, then it follows, by indctively applying the above argment, that T(Q d ) = 2 for all d. One may wonder why the base case of the proof starts with Q 6 and Q 7. We wold like to note that starting as early as d = 2, we were able to se a tree cover of Q d to prodce a cover for Q d+1 sch that there exists two vertices that cold be swapped in order to prodce a tree cover for Q d+1. In fact, this is how we constrcted the tree covers for Q, Q, Q 5 given in the appendix. However, there is a choice to be made when switching vertices, and the point at which the above constrctive pattern holds is dependent pon the initial choice of vertices that are swapped. For example, we experimented with sing a different initial swap and fond that the pattern did not hold ntil d = 11 or later. It may also be the case that there is an initial swap that allows the pattern to begin sooner than d = 8. This is a very interesting phenomenon that is worth frther exploration. We also investigated the idea of generalizing the above proof to all Hamming graphs. For H(2, ), we fond that T (H(2, )) =, and evidence sggests that T (H(d, q)) = q. Conjectre 12. Let H(d, q) be the Hamming graph of dimension d. T (H(d, q)) = q. Then, Acknowledgements: This research was condcted at Iowa State University s 2015 Mathematics REU Program. We wold like to thank the Department of Mathematics at Iowa State University, as well as the NSF for spporting the research. In addition we wold like to thank Dr. Leslie Hogben for her contribted insight to this project and for her sggestions that improved the presentation of this paper. 12

13 5 Appendix 5.1 Sets which indce tree covers for Q d for d {2,,, 5} For d = 2: T 12 = {(00), (01)}, T 22 = {(11), (10)}. For d = : T 1 = {(010), (000), (001), (110)} T 2 = {(111), (011), (100), (101)}. For d = : T 1 = {(0011), (0010), (0000), (0110), (1111), (1011), (1100), (1101)} T 2 = {(0001), (0111), (0100), (0101), (1010), (1000), (1001), (1110)}. For d = 5 T 15 = {(00100), (00011), (00000), (00110), (01111), (01011), (01100), (01101), (10001), (10111), (10100), (10101), (11010), (11000), (11001), (11110)} T 25 = {(00010), (00001), (00111), (00101), (01010), (01000), (01001), (01110), (10011), (10010), (10000), (10110), (11111), (11011), (11100), (11101)}. 5.2 Sets Used in Proof of Theorem 11 ˆT 16 = {(000000), (000011), (000100), (000110), (001011), (001100), (001101), (001111), (010001), (010100), (010101), (010111), (011000), (011001), (011010), (011110), (100001), (100010), (100101), (100111), (101000), (101001), (101010), (101110), (110000), (110010), (110011), (110110), (111011), (111100), (111101), (111111)}, ˆT 26 = {(000001), (000010), (000101), (000111), (001000), (001001), (001010), (001110), (010000), (010010), (010011), (010110), (011011), (011100), (011101), (011111), (100000), (100011), (100100), (100110), (101011), (101100), (101101), (101111), (110001), (110100), (110101), (110111), (111000), (111001), (111010), (111110)}. T 16 = {(001001), (000000), (000011), (000100), (000110), (001011), (001100), (001111), (010001), (010100), (010101), (010111), (011000), (011001), (011010), (011110), (100001), (100010), (100101), (100111), (101000), (101001), (101010), (101110), (110000), (110010), (110011), (110110), (111011), (111100), (111101), (111111)}, 1

14 T 26 = {(001101), (000001), (000010), (000101), (000111), (001000), (001010), (001110), (010000), (010010), (010011), (010110), (011011), (011100), (011101), (011111), (100000), (100011), (100100), (100110), (101011), (101100), (101101), (101111), (110001), (110100), (110101), (110111), (111000), (111001), (111010), (111110)} ˆT 17 = {( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), (010100), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), (101111), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( )} ˆT 27 = {( ), ( ), ( ), ( ), ( )( ), ( ), ( ), ( , ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( )}. T 17 = {( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), (010100), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), (101111), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( )} 1

15 T 27 = {( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( ), ( )}. References [1] F. Barioli, S. Fallat, L. Mitchell, and S. Narayan. Minimm semidefinite rank of oterplanar graphs and the tree cover nmber. Elec. J. Lin. Alg., 22: 10 21, [2] J. Ekstrand, C. Erickson, H.T. Hall, D. Hay, L. Hogben, R. Johnson, N. Kingsley, S. Osborne, T. Peters, J. Roat, A. Ross, D.D. Row, N. Warnberg, and M. Yong. Positive semidefinite zero forcing. Lin. Alg. Appl., 9: , 201. [] J. Ekstrand, C. Erickson, D. Hay, L. Hogben, and J. Roat. Note on positive semidefinite maximm nllity and positive semidefinite zero forcing nmber of partial 2-trees. Elec. J. Lin. Alg., 2: 79 97, [] S. Fallat and L. Hogben. The minimm rank of symmetric matrices described by a graph: A srvey. Lin. Alg. Appl., 26: , [5] S. Fallat and L. Hogben. Minimm Rank, Maximm Nllity, and Zero Forcing Nmber of Graphs. In Handbook of Linear Algebra, 2nd ed., L. Hogben editor, CRC Press, Boca Raton, 201. [6] H. van der Holst. On the maximm positive semi-definite nllity and the cycle matroid of graphs. Elec. J. Lin. Alg., 18: ,