Bounding Tree Cover Number and Positive Semidefinite Zero Forcing Number

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1 Bounding Tree Cover Number and Positive Semidefinite Zero Forcing Number Sofia Burille Mentor: Micael Natanson September 15, 2014 Abstract Given a grap, G, wit a set of vertices, v, and edges, various properties can be calculated. Two of tese properties, or grap parameters, are te tree cover number and te positive semidefinite zero forcing numbers, T (G and Z + (G respectively. Metods of easily bounding T (G and Z + (G are presented. Te general procedure is to break a grap G, into multiple smaller subgraps G i at specific vertices, separating vertices. Tree cover number of smaller graps are easier to calculate, so T (G i for eac subgrap is calculated and combined to estimate te tree cover number of T (G. Tese estimations and bounds can be more or less accurate depending of te structure of te grap. Different cases are examined to conclude results specific to certain kind of graps. Te same process is applied to positive semidefinite zero forcing, Z + (G. Activity Report To begin tis Matematics researc project, I ad to first understand wat grap teory is and te more recent discoveries relating to te field. So te for te first part of te summer I was reading textbooks and academic papers relating to te topic. Never aving read Matematics researc papers I ad to learn te various vocabulary, style, and formatting of it. Ten I started collecting data. Collecting data consisted of some-wat systematically generating graps, ten calculating various parameters of eac grap. Essentially I was looking for patterns and insigts on ow te graps were beaving in relation to one anoter. Tis eventually led me to some observations, wic I would develop into propositions and some eventually teorems. I also worked on understanding some proofs in academic papers and even sowing te proof eld true for an example. Trougout te summer I also tried to develop more efficient ways of collecting data by using computer programs suc as Sage or Matematica (wic required also learning te programs temselves. Trougout te summer I would discuss my progress and questions wit my mentor, Professor Natanson. In our discussions, we elped eac oter wit te understanding of te material, a proof, or ideas from te data. 1

2 1 Introduction Grap Teory is te study of graps. Graps are a way to model many tings from everyday life. Grap parameters are measurable properties of graps. I investigated metods to bound 2 parameters, tree cover number and te positive semidefinite zero forcing number, wic elp bound a tird more important parameter, minimum semidefinite rank, T (G, Z + (G, and mr + respectively. In [4], cut vertex reduction is formulated. A cut vertex is a vertex suc tat wen removed disconnects te grap. In graps tat contain a cut vertex, te grap can be divided into smaller graps, called subgraps, G i, to more easily calculate T (G and Z + (G. We expand tis idea to multiple separating vertices, in wic removing tose vertices disconnects te grap. We make te following definitions X = {v 1,v 2,...,v n } is te set of separating vertices, v i is a vertex in X, T 1 (G is te size of te minimum tree cover of G wic as all v i in te same tree, T 2 (G is te size of te minimum tree cover of G in wic te separating vertices are comprised of 2 trees, and T k (G is te size of te minimum tree cover of G in wic te separating vertices are comprised of k trees. 2 Preliminaries A grap is made up of vertices and edges between tem. Two vertices are neigbors, or connected, of tere is an edge between tem. Te main portion of my researc was focused on te grap parameter tree cover number, T (G. In [6], te tree cover property is created and defined. Te tree cover number of a multigrap G, denoted T (G, is te minimum number of vertex disjoint simple trees occurring as induced subgraps of G tat cover all of te vertices of G. Essentially wat tat means is te minimum number of trees, wic is a type of grap, needed to cover all of a grap G. An induced subgrap means tat one take a portion a grap, but tis part of te grap, if it contains any 2 vertices tat contain an edge in G, ten te edge between tem must also be present in te subgrap. So a tree cover of any grap will ave a collection of trees, all of wic will be induced subgraps of G. So tree covers are collections of induced subgraps. Anoter parameter i investigated was positive semidefinite zero forcing number. Te positive semidefinite zero forcing number is given a set of initial black vertices tat using te forcing rules can force te oter vertices to be black, wat is te minimum lengt of te initial set needed to be able to turn all of te vertices black? 3 Induced Subgraps Te simplest expansion of te cut vertex reduction is two separating vertices reduction. Even tis extra step complicates te problem significantly. Al- 2

3 toug te tree cover number is te minimal amount of disjoint trees, tese trees can often be created in several ways. Tere are often multiple optimal tree covers. Tese tree covers can ave te separating vertices in many combinations of same or different trees. Since more tan a single vertex is included in every G i we need to verify tat tey still create a tree covering for G. It is obvious tat wen starting wit a tree cover for G, te resulting tree covers for G i still cover eac vertex in G i and eac vertex of te tree cover. However edges could be missing and result in disconnected trees, causing a general induced forest instead of a specific induced tree cover. We prove te following Lemma to eliminate tis problem. Lemma 3.1. Given a tree cover T of G, let H be an induced subgrap of G. Ten, H T is a collection of subgraps wic serve as a tree cover of H. Start wit G. T is a collection of subgraps tat contains all of te vertices in G. Ten we take te subgrap of T, H T, wic is te restriction of T. T contains all te vertices H could possibly ave, terefore te vertices of H T are exactly tose of H. Since te edges of H T are no more tan T, wic is comprised only of trees, ten H T can only be eiter tose trees, or tose trees minus some edges, wic simply creates more trees. Terefore H T is still an induced tree cover. 4 Relating T k s In relating different T k s to eac oter some more interesting results arose; owever, we ave not been able to find an appropriate application for tem. Teorem 4.1. Limits For any grap G, wit 1 < k te number of separating vertices, T k 1 (G T k (G 1 (1 If k = n ten wen looking at T k 1, since tere are n 1 trees but n vertices, ten 1 tree as to ave 2 of te v i togeter. If k < n ten at least 1 tree as to ave 2 or more of te v i togeter. Let v 1 and v 2 be in te same tree togeter. Tere exists a distinct pat between v 1 and v 2. If we remove one of te edges in tat pat ten v 1 and v 2 will be disconnected. Wat was once 1 tree, will now be 2 trees, wit v 1 and v 2 being in 2 different trees, wic adds exactly 1 to T. So for a particular G, if T k 1 (G < T k (G, we can make a T k cover from a T k 1 cover by adding exactly one tree. 3

4 Tis teorem was in part created to be able to relate tese terms we created to eac oter and to T (G in general. Since T k 1 is often greater tan or equal to T k, an example being T 1 is usually greater tan or equal to T 2, I investigated wen is T 1 smaller tan T 2? Tis was in opes to better understand T (G in general. However, tis question was difficult to answer and is open to furter researc. 5 Initial Point Te simplest expansion of te cut vertex reduction is two separating vertices reduction. Te goal of expanding te cut vertex reduction is tat many graps do not ave a single vertex, in wic case te teorem cannot be applied. Expanding to two separating vertices, more graps can be included, wit te end goal of expanding to any amount of separating vertices. 5.1 Graps wit Two Separating Vertices Teorem 5.1. Let G be a grap wit vertex set V. Let X be fixed {v 1,v 2 } V, suc tat X is a separating set for G. Let {W 1,...,W } be te vertices of te connected components of G\{v 1,v 2 }; and let G i = G[W i X]. Let = te number of subgraps. T (G T (G i 2( 1 (2 Start wit an optimal tree cover, T, for G. Break apart G at v i into subgraps, G i, and consequently te optimal tree cover into tree covers for eac G i, wic using Lemma 3.1 Trees not containing v i will be in only one subgrap. Since tere are 2 separating vertices in v i and terefore only 2 vertices tat are in multiple subgraps, te igest amount of trees te v i can be broken up into is 2 trees. We are calculating a lower bound, we want to subtract off te greatest possible amount of extra trees to get te smallest value. Since 2 is te igest value, we will focus on v i being in 2 different trees in te tree cover for G to prove our bound. So eac G i will ave at least 2 trees. Te 2 trees eac wit a different v i, will be in eac subgrap, meaning tey will eac be counted times. We want tem counted only once so tey are counted 1 extra times. In our final equation we want to subtract tis extra. Since we ave 2 many different trees ten we are subtracting 2( 1, wic is te igest we can subtract. Te bound is as small as possible. Putting all tese trees togeter and subtracting te extra will give us a size of a tree cover tat is a lower bound because tis tree covering cannot be made any less optimally. Terefore T (G T (G i 2( 1. 4

5 Tere is difficulty in building an optimal tree cover, terefore attaining T (G, G from optimal tree covers of eac G i. Tis is wy we ave instead of = in our formula. However in certain cases an optimal tree cover of G can be made from optimal tree covers of eac G i. Terefore wit some certain assumptions made, we can ave = in our formula. Tis means under certain cases, combining for all i,t (G i can be used to calculate T (G. Teorem 5.2. Let G be a grap wit vertex set V. Let X be fixed {v 1,v 2 } V, suc tat X is a separating set for G. Let {W 1,...,W } be te vertices of te connected components of G\{v 1,v 2 }; and let G i = G[W i X]. For any G suc tat v1 v 2, T 1 (G = T 1 (G i ( 1 (3 and T 2 (G = T 2 (G i 2( 1 (4 Start wit a minimal tree cover, T i, for eac G i. For eac T i tere exists some T i1 suc tat v 1 V (T i1, and some different T i2 suc tat v 2 V (T i2. For T i {T i1,t i2 },T i is contained in only one G i, terefore appears only once in T (G i, and translates simply to be only 1 tree in exactly 1 T i terefore exactly 1 T i in T. Let T v1 be (T i1. Because all T i s are induced trees and we are just glueing all te trees togeter at teir one common vertex, v 1, T v1 is still a connected tree. Since we are glueing all T i 1 s togeter at v 1 tere is no way for tere to be an edge between 1 vertex in one G i to anoter except troug v 1, terefore tere is a unique pat from any vertex to anoter. So T v1 is te induced tree in G tat contains v 1. Let T v2 be (T i,2, meaning T v2 is te induced tree in G tat contains v 2. Ten T = ( (T i \ {T i,1,t i,2 } {T v1 Tv2 } is a tree cover for G because all T i are induced trees and V (T = V (G (All te vertices in G are in T. Since we found a tree cover te minimum tree cover can be no bigger tan te tree cover we just created. Terefore, T 2 (G T 2 (G i 2( 1 Combining wit 5.1 we get T 2 (G = T 2 (G i 2( 1 5

6 Start wit a minimal tree cover, T i, for eac G i. Since v 1 v 2, and v 1 and v 2 are in te same tree ten te tree containing tem, because tree covers are collections of induced subgraps, must go troug te edge between tem. All te trees in eac G i containing v 1 and v 2 are all glued togeter at v 1 and v 2. Terefore wen glueing all tese trees to make one tree, we know it is still a connected tree because tere is no way for tere to be an edge between 1 vertex in one G i to anoter except troug v 1 or v 2. Adding tis tree to all te oter trees gives us a tree cover for G. Since we found a tree cover te minimum tree cover can be no bigger tan te tree cover we just created. Terefore, T 1 (G T 1 (G i ( 1 Combining wit 5.1 we get T 1 (G = T 1 (G i ( 1 From tese we see certain patterns arising and build to multiple vertices. 6 Multiple separating vertices Te 2 vertice general bound conjecture can be expanded to multiple vertices. Eac increasing vertex simply adds one more vertex, and terefore tree tat can be counted extra times. Teorem 6.1. General Bound Wit assumptions and definitions above, for any grap G, T (G T (G i n( 1 (5 Start wit an optimal tree cover, T, for G. Break apart G at v i into subgraps, G i, and consequently te optimal tree cover into tree covers for eac G i, wic using Lemma 3.1 Trees not containing v i will be in only one subgrap. Since tere are n separating vertices in v i and terefore only n vertices tat are in multiple subgraps, te igest amount of trees te v i can be broken up into is 6

7 n trees. We are calculating a lower bound, we want to subtract off te greatest possible amount of extra trees to get te smallest value. Since n is te igest value, we will focus on v i being in n different trees in te tree cover for G to prove our bound. So eac G i will ave at least n trees. Te n many trees eac wit a different v i, will be in eac subgrap, meaning tey will eac be counted times. We want tem counted only once so tey are counted 1 extra times. In our final equation we want to subtract tis extra. Since we ave n many different trees ten we are subtracting n( 1, wic is te igest we can subtract. Te bound is as small os possible. Putting all tese trees togeter and subtracting te extra will give us a size of a tree cover tat is a lower bound because ( tis tree covering cannot be made any less optimally. Terefore T (G (T (G i n( 1. In te 2 connected vertices equality proof, wen bot vertices were in te different trees, te problem was simplified somewat. Tis is in part due to ow tis specific format some some way mirrors te original cut vertex reduction, but instead of one vertex be in all G i tere are more vertices in all G i but tey do not interact wit eac oter. Tis brings us to a a teorem wit a substantial assumption, but very useful result. Recall tat T k (G is te size of te minimum tree cover of G in wic te separating vertices are comprised of k trees. Also, we define n as te number of separating vertices in te grap, G. An Example No assumptions: T (G ( T (G i n( 1, were n = te number of separating vertices and = te number of subgraps G i 7

8 4 v3 v2 5 v G = G 1 = 5 4 v3 v2 v1 G 2 = v3 v2 6 v Figure 1: An example of a grap, G, wit 3 separating vertices, v 1, v 2, and v 3. and its two subgraps G 1 and G 2 It was found tat = 2, n = 3 because tere are 2 subgraps and 3 separating vertices. It was also found tat T (G = 2 and for bot i, T (G i = 2. Using our formula of an estimate of te tree cover number: T (G (2 1 = 1, terefore T (G 1, wic is tree because T (G = 2. Conjecture 6.2. T k EQUALITY For any grap G, were n is te number of separating vertices and eac of te separating vertices are in different trees. T n (G = T n (G i n( 1 (6 Start wit a minimal tree cover, T i, for eac G i. For eac T i tere exists some T i1 suc tat v 1 V (T i1, some different T i2 suc tat v 2 V (T i2,... and some oter different T in suc tat v n V (T in. For T i {T i1,t i2,...,t in },T i is contained in only one G i, terefore appears only once in T (G i, and translates simply 8

9 to be only 1 tree in exactly 1 T i terefore exactly 1 T i in T. Let T v1 be (T i1. Because all T i s are induced trees and we are just glueing all te trees togeter at teir one common vertex, v 1, T v1 is still a connected tree. Since we are glueing all T i 1 s togeter at v 1 tere is no way for tere to be an edge between 1 vertex in one G i to anoter except troug v 1, terefore tere is a unique pat from any vertex to anoter. So T v1 is te induced tree in G tat contains v 1. Let T v2 be (T i,2, meaning T v2 is te induced tree in G tat contains v 2. Let T vn be (T i,n, meaning T vn is te induced tree in G tat contains v n. Ten T = ( (T i \ {T i,1,t i,2,...,t i,n } {T v1 Tv2... T vn } is a tree cover for G because all T i are induced trees and V (T = V (G (All te vertices in G are in T. Since we found a tree cover te minimum tree cover can be no bigger tan te tree cover we just created. Terefore, T n (G T n (G i Combining wit te general bound result, T n (G T n (G i n( 1 n( 1 we get: T n (G i n( 1 T n (G T n (G i n( 1. Terefore, T n = ( T n(g i n( 1, wen all v i are in different trees. Te problem wit trying to ave an exact calculation, instead of just an estimation, of T (G using T (G i is wen we are building a tree cover of G using te tree covers of G i te graps ave to be able to glue togeter at a distinct part of a tree. Tis is a problem wen two of te separating vertices are in te same tree, but not troug an edge between tem, or if tere is not consistency among te trees wit separating vertices. Wit more assumptions we can get around some of tese problems. Conjecture 6.3. For any grap G wit v i tree connected, wit tree connected meaning tat tere is a distinct pat between all v i, T 1 (G = (T 1 (G i ( 1 (7 Start wit a minimal tree cover, T i, for eac G i. For eac T i tere exists some T iv suc tat v i V (T iv. For T i {T iv },T i is contained in only one G i, 9

10 terefore appears only once in T (G i, and translates simply to be only 1 tree in exactly 1 T i terefore exactly 1 T i in T. Let T v be (T iv. Since all te v i are in te same tree togeter ten all te edges between tem will be in all T i. Eac vertex tat is not v i, but is in a tree wit v i, will ave a unique pat to v i. Because all T iv s are induced trees and we are just glueing all te trees togeter at teir one common root, te v i tree, and glueing tem togeter nowere else, T v is still a connected tree. Since we are glueing all T i v s togeter at v i tere is no way for tere to be an edge between 1 vertex in one G i to anoter except troug v i, terefore tere is a unique pat from any vertex to anoter. So T vi is te induced tree in G tat contains all v i. Ten T = ( (T i \ {T i,v } {T v } is a tree cover for G because all T i are induced trees and V (T = V (G (All te vertices in G are in T. Since we found a tree cover te minimum tree cover can be no bigger tan te tree cover we just created. Terefore, T 1 (G T 1 (G i ( 1 Combining wit te general bound result, T 1 (G T 1 (G i ( 1 (8 we get: ( T 1 (G i ( 1 T 1 (G ( T 1 (G i ( 1. Terefore, T 1 = ( T 1(G i ( 1, wen all v i are tree connected. Te various results and specific cases are summed up in a few tables. 10

11 v i E T 1 T = = = Table 1: Summary of results wen tere are 2 separating vertices, wen n = 2 v i E T 1 T 2 T DNE = = = = = Table 2: Summary of results wen tere are 2 separating vertices, wen n = 3 ( T k (G T k(g i n( 1, n = number of separating vertices v i : Example configuration of ow v i could be arranged in G. E : Te number of edges between v i : Case wit only inequality proven in formula = : Case wit equality proven in formula Tere can be a variety of assumptions made so similar results are attained. One way to enforce consistency witout restricting te tree cover to suc specific types of configuration, we simply enforce consistency by starting wit a tree cover of v i. Conjecture 6.4. MN CONJECTURE 4 Let G be a grap wit vertex set V suc tat X V is a separating set for G, as usual, and let S be a fixed tree cover of X. For any grap H wic contains X, define T S (H to be te size of te minimum 11

12 tree cover of H wic reduces to S on X. Ten T S (G = T S (G i S ( 1 (9 Start wit a tree cover, T for G, break it into tree covers, T i for eac G i. Trees not containing v i will only be in 1 G i and remain as induced trees, as known from Lemma 3.1. Since we started wit S, we extend tose trees to te rest of G and consequently to eac G i. Eac G i will ave S as an induced subgrap. For any two vertices tat are in te same tree, if te pat between tem passes troug anoter G i, it does so troug S, wic is a tree cover of X tat is in G i. Terefore T i is still an induced tree. Tere are at least S trees in eac G i. So te trees in S will be counted for eac G i, we want tem counted only once, so tey are counted extra 1 times. Tis appens for eac tree in S, so we ave S ( 1 total extra. Terefore, T S (G T S (G i S ( 1 (10 Starting wit a tree cover for eac G i we build tem up to get a tree cover for G. Trees not containing v i will only be in 1 G i and are just one tree in G. Eac G i will ave te trees tat are in S. Tese trees can all be glued at eac tree in S, wic are in every G i. Since we are glueing all tese trees at a single common base vertex, tey form an induced tree covering, wic covers all te vertices not in v i trees. Adding tese trees to te trees not containing v i, gives a tree cover for G. Te optimal tree cover cannot be any smaller tan tis tree cover. Terefore, T S (G Combining (10 and (11 we get, Terefore, T S (G i S ( 1 T S (G T S (G = 7 Relating T (G to Z + (G T S (G i S ( 1 (11 T S (G i S ( 1 T S (G i S ( 1 Zero forcing, Z(G, and positive semidefinite zero forcing, Z + (G, are te minimum amount of initial vertices, wic make up te Forcing Set, wic 12

13 ten force oter vertices so tat te wole grap is covered. Z(G and Z + (G are comparable to P(G and T (G respectively. From a forcing set, one can produce a forcing tree cover, wic is similar to trees in tree covers. We try to establis connections between T (G and Z + (G. Since Z + (G is more restrictive ten T (G, not every tree cover can be made into a forcing tree cover. I created a algoritm to find out if a specific tree cover could be turned into a forcing tree cover, and if yes, ow to do so. Te algoritm was never completely finised and is open to furter investigation. However, in trying to understand te previously stated question of wen is T 1 (G < T 2 (G, in a different way, I was able to relate T (G and Z + (G in a very specific way in certain cases. Tere is known relationsip between T (G and Z + (G tat Z + (G T (G. From tis, it can be seen tat forcing tree covers are similar in some ways to tree covers. So saying two vertices are in te same tree is similar to saying to vertices are in te same forcing tree, wic means tey are not in te same forcing set for tat particular forcing set. If two vertices are in different forcing trees, owever, we cannot say weter or not tey can be in te same forcing set togeter. Proposition 7.1. Fix {v 1,v 2 } and let v 1 v 2. Let G be a grap suc tat T (G = Z + (G. If T 1 (G < T 2 (G ten no optimal zero forcing set contains bot v 1 and v 2. Let G be a grap suc tat T 1 (G < T 2 (G. Te edge v 1 v 2 will be included in every single optimal tree cover of G. Every forcing tree cover can be created from a tree cover. Because T (G = Z + (G, every optimal forcing tree is also an optimal tree cover. Since v 1 v 2, one v always as to force te oter v. Terefore v 1 and v 2 can never be in te same optimal forcing set. 8 Positive Semidefinite Zero Forcing Since positive semidefinite zero forcing as similarities to tree cover number, I was able to make similar conclusions wit positive semidefinite zero forcing. Lemma 8.1. Let T be a forcing tree cover of G, let H be an induced subgrap of G. H T is te restriction of T onto H, making it a subgrap of T. Ten, H T is a collection of subgraps wic serve as forcing tree cover of H. Because T is a forcing tree cover and H T is a induced subgrap of T, ten H T must be a forest cover or tree cover. Wen restricting T, if trees become disconnected, wat was once 1 tree simply becomes multiple trees tat cover 13

14 te same vertices. Since H T is an induced subgrap, te degree of a vertex in a component can only decrease, meaning u cannot be prevented from forcing because of too many neigbors. So if u v and bot u and v are still present, ten u v. If u is no longer present ten v is made it be part of te initial forcing set and forcing will continue as before. Tis appens for every uv pair in H T. Terefore H T is still a collection of trees serving as an induced forcing tree cover. Tis allows furter results, similar to te T (G results. Teorem 8.2. Fix v 1,v 2,...,v n to be our separating set Ten, Z + (G Z + (G i n( 1 (12 Start wit an optimal forcing set F for G. Te optimal forcing set F creates an optimal forcing tree cover, T, of G. Tere are at most n trees in T intersecting our separating set, one containing v 1, one containing v 2,..., anoter containing v n. Because F is an optimal forcing set we know tat T = Z + (G. We break down tis optimal forcing tree cover into covers for eac G i. Using our result from Lemma 8.1, we know tese covers, T i, are comprised of forcing trees for eac G i. Tese forcing trees give us forcing sets for eac G i. So we ave a forcing tree cover, we just do not know if it is te optimal one, so Z + (G cannot be any bigger tan tat. For any T i, T i Z + (G i. Ten, sum up all te forcing trees. Of all te forcing trees in all G i, only n at most, one for eac v i, can be counted for eac G i.tey are counted times, meaning tey are counted extra 1 times. Since tere are at most n, we get at most n( 1 extra counted trees. So from our past proofs we know tat Since T i Z + (G i ten T T i T and since T = Z + (G, terefore, Z + (G T i n( 1 Z + (G i n( 1 (13 Z + (G i n( 1 (14 n( 1 (15 14

15 Taking an optimal forcing set, making an optimal forcing tree cover, ten splitting tat into forcing tree covers for eac G i we find te minimum value of Z + (G. Using tis teorem, in certain cases even more can be said about T (G. Teorem 8.3. Fix v i to be our separating set and let G ave subgraps G i suc tat all v i can be made as part of an optimal forcing set togeter for eac G i. Ten, Z + (G = (Z + (G i n( 1 Start wit an optimal forcing set F i for eac G i in wic {v 1,v 2,...,v n } F i. Since eac G i ave v i as part of te optimal forcing set, ten wen we glue all te forcing sets togeter, defining F = F i, {v 1,v 2,...,v n } F. Also, since all {v i } are in a separating set ten tey will separate G into components, wic are te subgraps tat we started wit and we will still ave all te initial vertices tat we started wit. Terefore te set is a forcing set for G. We know te optimal forcing set can only be as big as te size of te entire sum (rigt and-side of te equation. Terefore, ( Z + (G (Z + (G i 2( 1 (16 Combining tis result wit Teorem 8.2 we get (Z + (G i 2( 1 Z + (G (Z + (G i 2( 1 Terefore, Z + (G = (Z + (G i 2( 1 wen given G i suc tat all v i are in an optimal forcing set togeter for eac G i. More may be able to be said about te bounds of Z + (G in general or in specific cases, but tese are unanswered questions/ 15

16 9 Open Questions As mentioned, one unanswered question tat arose was wen tere are two separating vertices, wen is T 1 smaller tan T 2? Tis would allow te T (G to be calculated muc more efficiently in many graps. Wen we know wic is smaller we know ow to construct te tree cover, and wen T 2 is smaller, ten T 2 (G = T (G, wic makes calculating T (G muc easier and efficient. Anoter question asks wat exactly are te connections between tree covers and zero forcing tree covers. Can tey be related in a more substantial way? Are are tere ways to turn tree covers into forcing tree covers? Lastlt, are tere more precise bounds tat can be made on T (G using T (G i? 10 Conclusion I worked on breaking a big grap, G, into smaller graps, G i, at te separating set, in order to bound T (G and Z + (G. Tis brougt up a variety of special cases in wic te bound became an exact calculation. I also related T (G to Z + (G in more specific ways. Since tere are many complicated graps, wenever a grap can be broken down into simpler parts and tose smaller parts are analyzed, it makes solving te problem muc easier. 16

17 References [1] X. Li, R. Pillips. Minimum Vector Rank and Complement Critical Graps. arxiv: [mat.co], [2] L Hogben. Ortogonal representations, minimum rank, and grap complements. Linear Algebra and its Applications, Vol. 428, [3] M. Boot, et al. On te Minimum Rank Among Positive Semidefinite matrices wit a Given Grap. Society for Industrial and Applied Matematics, Vol. 30, [4] Hogben et al. Positive Semidefinite Zero Forcing Number. Electronic Journal of Linear Algebra, Vol. 23, [5] F. Taklimi, S. Fallat, K. Meager. On te Relationsips between Zero Forcing Numbers and Certain Grap Coverings. Special Matrices, Vol. 2, [6] F. Barioli, et al. Minimum Semidefinite Rank Of Outerplanar Graps And Te Tree Cover Number. Electronic Journal of Linear Algebra. Vol. 22, [7] F. Barioli, et al. Parameters Related To Tree-Widt, Zero Forcing, And Maximum Nullity Of A Grap. Journal of Grap Teory, Vol. 72, [8] J. Ekstrand, et al. Note on positive semidefinite maximum nullity and positive semidefinite zero forcing number of partial 2-trees. Electronic Journal of Linear Algebra, Vol. 23,