The maximum multiplicity and the two largest multiplicities of eigenvalues in a Hermitian matrix whose graph is a tree

© 2015 Rosário Fernandes, licensee De Gruyter Open.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License.

Rosário Fernandes*

The maximum multiplicity and the two largest

multiplicities of eigenvalues in a Hermitian

matrix whose graph is a tree

Abstract:The maximum multiplicity of an eigenvalue in a matrix whose graph is a tree,M1, was understood 5 fully (from a combinatorial perspective) by C.R. Johnson, A. Leal-Duarte (Linear Algebra and Multilinear Alge-bra 46(1999) 139-144). Among the possible multiplicity lists for the eigenvalues of Hermitian matrices whose graph is a tree, we focus uponM2, the maximum value of the sum of the two largest multiplicities when the

largest multiplicity isM1. Upper and lower bounds are given forM2. Using a combinatorial algorithm, cases

of equality are computed forM2. 10

Keywords:Eigenvalue multiplicities; Symmetric matrices; Trees; Two largest multiplicities

MSC:15A18, 05C38, 05C50

DOI 10.1515/spma-2015-0001

Received September 10, 2013; accepted December 4, 2014

1 Introduction

LetTbe a tree onn≥2vertices. We denote byS_{(T)the collection of alln-by-ncomplex Hermitian matrices} whose graph isT. No restriction is placed upon the diagonal entries of matrices inS_(T).

For convenience, when A ∈ S_{(T), we place in non-increasing order the multiplicities of the} eigen-values ofA. We refer to such a list of multiplicities as theunordered multiplicity listand we denote it by (m1(A),m2(A),. . .,mk(A)(A)), wherek(A)is the number of distinct eigenvalues ofA. So,mj(A)is the jth 20

largest multiplicity of an eigenvalue in the multiplicity list ofA.

Definition 1.1. LetL_{(T)be the set of all positive integer lists (unordered multiplicity lists)(p1,p2,. . .,ps)} satisfying:

(1) p1≥p2≥. . .≥ps≥1;

(2) ∑︀s

i=1pi=n; 25

(3) There is anA∈S_{(T)with(m1(A),m2(A),. . .,m}

k(A)(A)) = (p1,p2,. . .,ps).

Forj≥1, we denote by

Mj(T) = max

(p1,p2,...,ps)∈L(T)(p1+. . .+pj).

It is well known thatM1(T)is equal to the path cover numberP(T), the smallest number of

nonintersect-ing induced paths ofTthat cover all the vertices ofT; this is the same asmax(p−q), wherepis the number of paths remaining whenqvertices have been removed fromTin such a way as to leave only induced paths

[3]. 30

Remark 1.2.In [7] a combinatorial algorithm was given to computeM2(T). It is easy to see that if(p1,p2,. . .,ps)∈

L_(T)then

(1) p1≤M1(T).

(2) p1+p2≤M2(T).

(3) p1+p2≥2,p2≠0(because ifTis a tree andA∈S(T)then the largest and the smallest eigenvalues of

Ahave multiplicities one. So, each list inL_{(T)has at least two1’s, [4]).}

(4) Using the definition ofM1(T), there exists(p1,p2,. . .,ps)∈L(T)such thatp1=M1(T). 5

GivenM1(T)andM2(T), we cannot say there exists a list(p1,p2,. . .,ps) ∈ L(T)such thatp1 = M1(T)

and p2 = M2(T)− M1(T). For example, [7], the double star D3,3has M1(D3,3) = 4, M2(D3,3) = 6 but (4,2,1,1) ∉ L_{(D3,3)(we can prove this using the Parter-Wiener theorem [5]). M1(D3,3) = 4 because}

(4,1,1,1,1)∈L_{(D3,3), for example, consider the matrix} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

0 0 0 1 0 0 0 0

0 0 0 1 0 0 0 0

0 0 0 1 0 0 0 0

1 1 1 1 1 0 0 0

0 0 0 1 1 1 1 1

0 0 0 0 1 0 0 0

0 0 0 0 1 0 0 0

0 0 0 0 1 0 0 0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ .

M2(D3,3) = 6because(3,3,1,1)∈L(D3,3), for example, consider the matrix

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

0 0 0 1 0 0 0 0

0 0 0 1 0 0 0 0

0 0 0 1 0 0 0 0

1 1 1 −2 1 0 0 0

0 0 0 1 3 1 1 1

0 0 0 0 1 1 0 0

0 0 0 0 1 0 1 0

0 0 0 0 1 0 0 1

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ .

So, it is important to know when givenM2(T), we can say that there is a list(p1,p2,. . .,ps)∈L(T)such

thatp1=M1(T)andp2=M2(T)−M1(T).

LetM2(T)(or simplyM2) denote the maximum value of the sum of the two largest integers among the

lists(p1,p2,. . .,ps)∈L(T), whenp1=M1(T), i.e.,

M2(T) = max

(M1(T),p2,...,ps)∈L(T)(M1(T) +p2).

Using the definition ofM2(T), we haveM2(T)≤M2(T). In this paper we give upper and lower bounds forM2

and in some cases, a method for calculatingM2.

2 Assignments

LetT be a tree onn ≥ 2vertices. IfA ∈ S_{(T)andvis a vertex ofT thenA(v)denotes the principal} submatrix ofAresulting from deleting row and column associated withv, andmA(λ)denotes the multiplicity

of eigenvalueλof matrixA. The Parter theorem, [8], indicates that ifA ∈S_(T)andmA(λ)≥2, then there is at least one vertexvofT, of degree at least3, such thatmA(v)(λ) = mA(λ) + 1. Moreover,vmay be chosen

so thatλis an eigenvalue of at least three principal submatrices ofAassociated with branches ofTatv. So, 15

for any multiple eigenvalue. If a principal submatrix ofAassociated with some branch atvagain hasλas a multiple eigenvalue, then this theorem may again be applied to that branch. Parter vertices forλmay be removed in this fashion until (fully) fragmentingTinto many subtrees whenλoccurs as an eigenvalue in such a submatrix associated with the subtree at most once. Such a set of Parter vertices is called afully fragmented Parterset forλ, and it is known that each successive Parter vertex is also a Parter vertex forAandλin the 5 original tree.

IfXis a set or collection (or graph), then|X|denotes the cardinality of (number of vertices in)X. IfVis a set of vertices andXis a graph thenV∩Xdenotes the set of vertices in bothVandX. IfXis a tree thenP_(X) denotes the collection of all subtrees ofX, includingX.

Definition 2.1. [7] (Assignment)Let T be a tree on n≥ 2vertices and let

(︁

p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

be a non-increasing list of positive integers, with∑︀k

i=1pi≤n. The notation1ldenotes that the last l entries of 10

the list are1. Note that some of the pi‘s may be1. AnassignmentAof

(︁

p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

to T is a collectionA_{= ((}A_{1,V1),. . ., (}A_{k,Vk))of k collections}A_{iof subtrees of T and k collections Viof vertices of T,} with the following properties.

(1) (Specification of Parter vertices)For each integer i between1and k,

(1a) Each subtree inA_{iis a connected component of T−Vi.} 15

(1b) |A_i|=pi+|Vi|.

(1c) For each vertex v∈Vi, there exists a vertex x adjacent to v such that x is in one of the subtrees in

A_i.

(2) (No overloading)We require that no subtree S of T is assigned more than|S|integers; define

ci(S) =|Ai∩P(S)|−|Vi∩S|,

the difference between the number of subtrees contained in S and the number of Parter vertices in S for the ith integer. So, we require that

k

∑︁

i=1

max(0,ci(S)) ≤|S|, for each S∈P(T).

If this condition is violated at any subtree, then thatsubtreeis said to beoverloaded.

Definition 2.2. [7]A collectionA_{= ((}A_{1,V1),. . ., (}A_{k,Vk))of k collections}A_{iof subtrees of T and k collec-} 20

tions Viof vertices of T is:

(1) anassignment candidateof(︁p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

toTwhenA_{satisfies condition1, but not} nec-essarily2ofDefinition 2.1.

(2) anear-assignmentof(︁p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

toTwhenA_{satisfies conditions1a, 1b, 2, but not}

necessarily1cofDefinition 2.1. 25

(3) anear-assignment candidateof(︁p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

toTwhenAsatisfies conditions1a, 1b, but not necessarily1cor2ofDefinition 2.1.

In [7] a simplification of assignments of the list(p1,p2, 1l)is considered.

Lemma 2.3. (Overloading Lemma)IfT is a tree andA_{is an assignment candidate(or a near-assignment} candidate)of the list(p1,p2, 1l)toT, butAis not an assignment(or a near-assignment, respectively), then 30

there must exist a single vertex inTthat is overloaded byA_.

✉ ✉ ✉ ✉ ✉

✉ ✉

✉

1

2

4 5

7 8

6 3

and let(3,2,13_{)be a list.}

If we considerA_{= ((}A_1,V1),(A_2,V2))where

A_1=T−{4,5}, A_{2=T−{5}, V1={4,5}andV2={5},}

thenA_{1has5connected components and}A_{2has3connected components. So,|}A_{1|= 5and|}A_{2|= 3.}

A_{is anassignment candidateof(3,2,1}3_{) toTbut not an assignment because the subtree{6}ofTsatisfies}

max(0,c1({6})) + max(0,c2({6})) = 1 + 1 = 2>1 =|{6}|.

If we considerA′_{= ((}A′

1,V1′),(A′2,V2′)), where

A′₁₌T−{4}, A′₂₌T−{5}V′₁={4}andV₂′ ={5}

thenA′

1has4connected components andA′2has3connected components. So,|A′1|= 4and|A′2|= 3. A′_{satisfies condition1ofDefinition2.1.}

5

IfS={1}orS={2}orS={3}, then

max(0,c1(S)) + max(0,c2(S)) = 1 + 0 =|S|.

IfS={4}orS={5}orS={7}orS={8}, then

max(0,c1(S)) + max(0,c2(S)) = 0 + 0<|S|= 1.

IfS={6}then

max(0,c1(S)) + max(0,c2(S)) = 0 + 1 =|S|. UsingLemma 2.3,A′_{is anassignmentof(3,2,1}3_{) toT.}

Example 2.5. LetTbe the following tree

✉ ✉ ✉

✉ ✉

✉ ✉

✉

❅ ❅ ❅

1

3 2

5 6

8

7 4

and let(2,2,14_{)be a list.}

If we considerA_{= ((}A_1,V1),(A_{2,V2)), where}

A₁₌T−{5,6,7,8}, A₂₌T−{6}, V1={5,6}andV2={6}

thenA_{1has4connected components and}A_{2has3connected components. So,|}A_{1|= 4and|}A_{2|= 3.}

10

Using the Overloading Lemma (Lemma2.3), another important result appears.

Lemma 2.6. LetTbe a tree. Then

there exists a near-assignment of the list(p1,p2, 1l)toTif and only if there exists an assignment of the list (p1,p2, 1l)toT.

ProofSuppose there exists a near-assignmentA_{= ((}A_{1,V1), (}A_{2,V2))of the list(p1,p2, 1}l_{)toT. If}A_{satisfies 5}

1cof Definition 2.1, thenA_{is an assignment of(p1,p2, 1}l_)toT.

Suppose thatA_{does not satisfy1c. ThenV1orV2does not satisfy1c. Suppose, without loss of} general-ization thatV1does not satisfy1c. So, there exists a vertexv1∈V1such that there is not a vertexxadjacent

tov1in a subtree ofA1.

Since|A_{1| = p1+|V1|, removev1fromV1and remove a subtreeR1from}A_{1. We obtain}A′

1 = A1\R1 10

andV′₁= V1\{v1}. Since|A′1|=p1+|V′1|, we conclude thatA′= ((A′1,V′1), (A2,V2))is a near-assignment

candidate of the list(p1,p2, 1l)toT.

IfA′_{is not a near-assignment, by Lemma 2.3, there must exist a single vertexyinTthat is overloaded by} A′_{. Using the fact that}A_{is a near-assignment,y=v1. Butv1does not belong to}A′

1. Consequently,S={v1}

satisfies condition2of Definition 2.1. Contradiction. Therefore,A′_{is a near-assignment. 15} IfA′_{satisfies1cof Definition 2.1, then}A′_{is an assignment of(p1,p2, 1}l_{)toT. If}A′_{does not satisfy1cof} Definition 2.1, repeat the process.

Repeating this process we obtain an assignment becausep1,p2 ≥1and in each process we have a

col-lection of subtrees ofTsatisfying condition1aof Definition 2.1.

Conversely, the proof is trivial. 20

Definition 2.7. IfA∈S_{(T)andSis a subgraph ofTthen}

(1) A[S]denotes the principal submatrix ofAlying on rows and columns associated with the vertices ofS. (2) A(S)denotes the principal submatrix ofAresulting from deleting rows and columns associated with the

vertices ofS.

Using the interlacing theorem for Hermitian matrices [2], ifxis a vertex ofT(tree) andλis an eigenvalue of A∈S_{(T), then there is a simple relation betweenmA(x)(λ)andmA(λ):}

m_A(x)(λ) =mA(λ)−1 or mA(x)(λ) =mA(λ) or mA(x)(λ) =mA(λ) + 1.

Definition 2.8. [7]LetTbe a tree onn ≥ 2vertices. We call an assignmentA _{= ((}A_1,V1),. . ., (Ak,Vk))of 25

(︁

p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

to T realizable if there exists a matrix B ∈ S_{(T)with unordered multiplicity list} (︁

p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

, such that, for each i between1and k, if siis the eigenvalue of B associated with

pi, i.e, mB(si) =pi, then:

(1) For each subtreeRofTinA_i,m

B[R](si) = 1.

(2) For each connected componentQofT−Vithat is not inAi,mB[Q](si) = 0. 30 (3) For eachx∈Vi,xis a Parter vertex forBandsi.

Remark 2.9. Note that ifC∈S_{(T)is a matrix that satisfies conditions1and2ofDefinition 2.8, then for eachi}

between1andk,mC(si) =pi≥1.

Using the interlacing theorem for Hermitian matrices, ifx∈Vi, thenmC(x)(si)is equal to

mC(si)−1 or mC(si) or mC(si) + 1.

By conditions1and2ofDefinition 2.8,mC(Vi)(si) =|Ai|. ButAis an assignment, so,|Ai|=pi+|Vi|. Thus,

m_C(x)(si) =mC(si) + 1.

Using the last remark, we can rewrite Definition2.8.

Definition 2.8 LetT be a tree on n ≥ 2 vertices. We call an assignmentA _{= ((}A_{1,V1),. . ., (}A_{k,Vk)) of} (︁

p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

toT realizable if there exists a matrixB ∈ S_{(T)with unordered multiplicity list} (︁

p1,p2,. . .,pk, 1n−

∑︀k i=1pi

)︁

, such that, for eachibetween1andk, ifsiis the eigenvalue ofBassociated with

pi, i.e,mB(si) =pi, then: 5

(1) For each subtreeRofTinA_i,m

B[R](si) = 1.

(2) For each connected componentQofT−Vithat is not inAi,mB[Q](si) = 0.

Definition 2.10. IfTis a tree onn≥ 2vertices,A_{is a realizable assignment of}(︁_{p1,p2,. . .,p}

k, 1n−

∑︀k i=1pi

)︁

to

TandB∈S_{(T)is a matrix that satisfies Definition 2.8, then we say thatBrealizes the assignment}A_.

There are assignments that are not realizable. For instance see Example 2.3 in [7]. However when we study 10

the list(p1,p2, 1l)we have the following result.

Theorem 2.11. [7] Given a treeTonn = p1+p2+lvertices, a near-assignment of the list(p1,p2,ll)toT,

A _{= ((}A_{1,V1), (}A_{2,V2)), and any distinct real numbers αand β, then there existsA ∈} S_{(T)satisfying the}

following conditions:

IfRis a connected component ofT−V1, then

αis an eigenvalue ofA[R]if and only ifR∈A_1.

Similarly, ifSis a connected component ofT−V2, then

βis an eigenvalue ofA[S]if and only ifS∈A_2.

Using Lemma 2.6, Theorem 2.11 and the new version of Definition 2.8 we obtain the following result. 15

Theorem 2.12. Given a treeTonn=p1+p2+lvertices, a near-assignmentAof the list(p1,p2,ll)toT, and

any distinct real numbersαandβ, then

(1) there exists a realizable assignmentB_of(︁_{p1,p2, 1}l)︁_toT.

(2) there existsA∈S_{(T)that realizes the assignment}B_{withmA(α) =p1andmA(β) =p2.}

Therefore, we immediately have as a consequence: 20

Corollary 2.13. For any treeT, if there exists a near-assignment of the list(M1(T),p2, 1l)toT, then

M2(T) ≥M1(T) +p2.

3

Upper and lower bounds for

_M

In this section, using the reduction theorem forM2, [7], we directly computeM2for particular trees. For other

kind of trees, we give bounds onM2.

In [7], the authors directly computedM2for generalized stars (for the notion of generalized star see [6]).

Definition 3.1. [6]LetTbe a tree andx0be a vertex ofT. Ageneralized starTwith central vertexx0is a tree 25

such thatT−{x0}is a union of paths (arms), each one of them is adjacent tox0by an endpoint.

Proposition 3.2. [7]LetTbe a generalized star onn≥ 2vertices, withfarms of length1andgarms of length at least2. Then:

(B) Ifg≤ 1andTis not a path, thenM2(T) =f+g.

(C) IfTis a path, thenM2(T) = 2.

Definition 3.3. [7](Peripheral HDV, peripheral arm)Given a treeTand a high-degree vertexv,vis aperipheral HDVofTif and only if there is a branch ofTatvthat contains all the other high-degree vertices inT. Aperipheral armof a treeTis a branch ofTat a peripheral HDV such that the branch does not itself contain any HDV. 5

Definition 3.4. Throughout this section, we will consider aperipheral HDVvin a treeT.

The subtree ofTconsisting ofvand its peripheral arms will be calledS- however, ifvis the only HDV in T, we will letSbevand all but one of its peripheral arms (chosen arbitrarily). The point is thatSshould be a generalized star containing everything except a single branch ofTatv.

Letwbe the one vertex adjacent tovthat is not inS. We denote by(T−S) +wK1the tree obtained fromT−S 10

by putting a vertex adjacent tow.

Theorem 3.5. [7] (M2Reduction Theorem)LetTbe a tree andva peripheral HDV, withSas defined earlier in

this section. Suppose thatShasfarms of length1andgarms of length at least2. Then:

(A) Ifg≥ 2, thenM2(T−S) =M2(T) −f− 2g+ 2.

(B) Ifg≤ 1, thenM2((T−S) +wK1) =M2(T) −f −g+ 1. 15

In [1] a class of trees was introduced that contains the generalized stars, the superstars.

Definition 3.6. [1]LetTbe a tree andx0be a vertex ofT. AsuperstarTwith central vertexx0is a tree such

thatT−{x0}is a union of paths.

The focus of this section is to directly computeM2for a subclass of superstars.

Definition 3.7. LetTbe a superstar with central vertexx0. Asmall pincerofTis a path,P, ofT−{x0}such 20

that:

(1) Pis adjacent tox0by a vertexuof degree two inP.

(2) At least one path ofP−uis a vertex.

Definition 3.8. LetTbe a superstar with central vertexx0.Tis a small superstar if all paths ofT−{x0}are

small pincers or are adjacent tox0by an endpoint (arms). 25

Example 3.9. The superstarTofExample 2.4is a small superstar with cental vertex4. The superstarTof

Example 2.5is a small superstar with central vertex5. All stars and generalized stars are small superstars. The following superstar is not a small superstar

✉ ✉ ✉ ✉

✉ ✉

✉

✉ ✉

✉

❅ ❅ ❅

1

3 2

5 6

9 10

7

8 4

Definition 3.10. LetTbe a tree andA_{an assignment of(M1(T),p2, 1}l_)toT. 30

(1) We refer toA_{as anM2assignment toT.}

Remark 3.11. LetTbe a tree andA_{= ((}A_1,V1),(A_{2,V2))anM2assignment of(M1(T),p2,1}l_{)toT. Because} M1(T) =|A1|−|V1|,

(1) All components ofT−V1are inA1.

(2) We can assume that ifv∈V1thenvis a HDV.

(3) Since all components ofT−V1are paths, ifvif a peripheral HDV of degree greater or equal to4inTthen

5

v∈V1.

(4) Ifvis a peripheral HDV,v∈V1and all peripheral arms have length1then they are inA1and no one is in A_{2(see Lemma 2.3).}

Remark 3.12. LetTbe a tree andA_{= ((}A_1,V1),(A_{2,V2))anM2-maximal assignment of(M1(T),p2,1}l_)toT.

BecauseM2(T) =|A1|−|V1|+|A2|−|V2|,

10

(1) All components ofT−V2with more than one vertex are inA2. (2) We can assume that ifv∈V2thenvis a HDV.

(3) All components ofT−V2with one vertex that are not components ofT−V1are inA2.

(4) Ifvis a peripheral HDV,v ∈ V1and all peripheral arms have length1then using Remark 3.11, 4, we conclude thatv∉V2.

15

(5) Ifvif a peripheral HDV,v∈ V1and all peripheral arms have length1, except one, then there is anM2 -maximal assignment of(M1(T),p2,1l)toTsuch thatv∉V2.

Remark 3.13. In some proofs we construct anM2-maximal (or simply anM2) assignment of(M1(T),p2,1l)

to T. for some integer p2. In these cases, first we construct anM2assignment of (M1(T),p2,1l_{)toT,A =}

((A_1,V1),(A_{2,V2))by putting the elements in}A_{1and inV1, next we put the elements in}A_{2and inV2,}

us-20

ing Remarks 3.11 and 3.12. This construction is in such a way that M1(T) = |A_1|−|V1|and M1(T) +p2 =

|A_1|−|V1|+|A2|−|V2|. After using Lemma 2.3 we conclude condition2of Definition 2.1 and by Corollary 2.13,

we say thatM2(T)≥M1(T) +p2.

Proposition 3.14. LetTbe a small superstar onn≥2vertices, withfarms of length1,garms of length at least

2andhsmall pincers, withf+g≥2orh≥2. Then:

25

(A) Ifg≥2, thenM2(T) = 2h+f+ 2g−2.

(B) Ifg≤1andTis not a path, thenM2(T) = 2h+f+g. (C) IfTis a path, thenM2(T) = 2.

ProofLetxbe the central vertex ofT. IfSis a small pincer ofT, by Theorem 3.5,

M2((T−S) +xK1) =M2(T)−1. SinceThashsmall pincers,

M2(T′) =M2(T)−h,

whereT′is obtained fromTby removing all small pincers and by puttinghvertices adjacent tox. Conse-quently,T′is a generalized star withf+harms of length1andgarms of length at least2. Using Proposition 3.2

M2(T′) =

⎧ ⎪ ⎨

⎪ ⎩

f +h+ 2g−2 if g≥2

f +h+g if g≤1 andT′is not a path 2 ifT′is a path.

Therefore,

M2(T) = ⎧ ⎪ ⎨

⎪ ⎩

f+ 2h+ 2g−2 ifg≥2

f+ 2h+g ifg≤1 andT′is not a path 2 +h ifT′is a path.

Note that ifT′is a path withh = 2andf = g = 0thenTis not a path andM2(T) = M2(T′) +h = 2 + 2 = 4 =f + 2h+g. By hypothesis, ifh< 2thenf +g≥ 2. In this case, ifT′is a path thenh= 0andf +g = 2. 30

So, we conclude that

M2(T) =

⎧ ⎪ ⎨

⎪ ⎩

f + 2h+ 2g− 2 ifg≥ 2

f + 2h+g ifg≤ 1 andTis not a path

2 ifTis a path.

SinceM2(T) ≤M2(T), we have

(A) Ifg≥ 2, thenM2(T) ≤ 2h+f+ 2g− 2.

(B) Ifg≤ 1andTis not a path, thenM2(T) ≤ 2h+f+g.

(C) IfTis a path, thenM2(T) ≤ 2.

Conversely, sinceTis a tree,

M1(T) =

{︃

f +h+g− 1 iff +g≥ 2

h+ 1 iff +g≤ 1.

We are going to construct anM2assignment of(M1(T),p2, 1l), for some integerp2, toT(see Remark3.13). 5

Case 1Iff+g≥ 2, we put the central vertex ofTinV1and we put thef+h+gpaths obtained by removing

the central vertex ofTinA_1.

Ifg ≥ 2, we put the central vertex ofTinV2and we put theh+gpaths of length at least2, obtained

by removing the central vertex ofTin A_{2. So,|}A_{1|−|V1| = f + g+h− 1 = M1(T). Using Remark 3.13,}

M2(T) ≥f +h+g− 1 +h+g− 1 =f+ 2h+ 2g− 2. 10

Ifg≤ 1, we put the central vertex of each small pincer ofTinV2, we put the2h+ 1subtrees obtained by

removing the central vertex of all small pincers ofTinA_{2. Since|}A_{1|−|V1|=f +g+h− 1 =M1(T), using} Remark 3.13,M2(T) ≥f+h+g− 1 + 2h+ 1 −h=f+ 2h+g.

Note that ifTis a path andf+g≥ 2thenf+g= 2andh= 0. Thus, ifg= 2, thenM2(T) ≥f+2h+2g−2 = 2

and ifg≤ 1, thenM2(T) ≥f+ 2h+g= 2. 15

Case 2Iff +g≤ 1thenh≥ 2andTis not a path. We put the central vertex of each small pincer ofTin

V1and we put the2h+ 1subtrees obtained by removing the central vertex of all small pincers ofTinA1. We

put the central vertex ofTinV2and we put thef+h+gpaths obtained by removing the central vertex ofT

inA_{2. Since|}A_{1|−|V1|=h+ 1 =M1(T), by Remark 3.13,M2(T) ≥h+ 1 +f +g+h− 1 =f +g+ 2h.}

Consequently, 20

(A) Ifg≥ 2, thenM2(T) ≥ 2h+f+ 2g− 2.

(B) Ifg≤ 1andTis not a path, thenM2(T) ≥ 2h+f+g.

(C) IfTis a path, thenM2(T) ≥ 2.

Therefore,

(A) Ifg≥ 2, thenM2(T) = 2h+f+ 2g− 2. 25

(B) Ifg≤ 1andTis not a path, thenM2(T) = 2h+f+g.

(C) IfTis a path, thenM2(T) = 2.

Proposition 3.15. LetTbe a tree andva peripheral HDV, withSas defined earlier in this section. Suppose that

Shas3arms of length1and0arms of length at least2andT≠S. Then

M2(T−S) + 2≤M2(T)≤M2(T−S) + 3.

ProofLetA _{= ((}A_{1,V1), (}A_{2,V2))be anM2-maximal assignment toT. We are going to construct anM2} assignment toT−S,A′_{= ((}A′

1,V1′), (A′2,V2′))(see Remark 3.13). Note thatM1(T−S) =M1(T)−2. Becausev

has degree4, by Remark 3.11, 3,v∈V1, the peripheral arms ofSare inA1and no one is inA2. Using Remark 30

3.12, 4,v∉V2. So, letFbe the component ofT−V2containingS. By Remark 3.12, 1,Fis inA2.

LetA′

1=A1\{the peripheral arms ofS},V1′ =V1\{v},V2′ =V2and

A′₂₌ {︃

A₂\{F} ifA₂₌̸{F}

By Remark3.13,A′_{= ((}A′

1,V′1), (A′2,V2′))is anM2assignment toT−SandM2(T−S)≥M2(T)−2−1 =

M2(T)−3.

LetA _{= ((}A_{1,V1), (}A_{2,V2))be anM2-maximal assignment toT}−S. We are going to construct anM2

assignment toT,A′_{= ((}A′

1,V1′), (A′2,V2′))(see Remark 3.13). Note thatM1(T) = M1(T−S) + 2. Letwbe the

vertex ofT−Sadjacent tovinT. Ifw∉V2then letRbe the component of(T−S)−V2containingwand let

5

Pbe the component ofT−V2containingS.

LetA′

1=A1∪ {the peripheral arms ofS},V1′ =V1∪ {v},V2′ =V2and

A′₂₌ {︃

(A₂\{R})∪ {P} ifR∈A_2andw∉V2

A_{2 otherwise} .

By Remark 3.13,A′_{= ((}A′

1,V′1), (A2′,V2′))is anM2assignment toTandM2(T)≥M2(T−S) + 2.

Proposition 3.16. LetTbe a tree andva peripheral HDV, withSas defined earlier in this section. Suppose that

Shas1arm of length1and1arm of length at least2 (orThas2arms of length1and0arms of length at least

2)andT≠S. Then

M2(T−S) + 1≤M2(T)≤M2(T−S) + 2.

ProofLetA _{= ((}A_{1,V1), (}A_{2,V2))be anM2-maximal assignment toT. We are going to construct anM2} assignment toT−S,A′_{= ((}A′

1,V1′), (A′2,V′2))(see Remark 3.13). Note thatM1(T−S) =M1(T)−1.

Using Remark 3.11, 1, ifvis inV1, then the peripheral arms ofS−vare inA1. Using Remark 3.12, 5, without

10

loss of generality, we can assume thatv∉V2. LetFbe the component ofT−V2containingS. By Remark 3.12,

1 and 3,Fis inA_2. LetA′

1=A1\{the peripheral arms ofS},V1′ =V1\{v},V2′ =V2and

A′₂₌ {︃

A₂\{F} ifA₂₌̸{F}

T−S ifA_2={F} . By Remark 3.13,A′_{= ((}A′

1,V′1), (A′2,V2′))is anM2assignment toT−SandM2(T−S)≥M2(T)−1−1 =

M2(T)−2.

Ifvis not inV1, sincevhas degree3inT, thenw∈V1. By Remark 3.11, 1,Sis inA1. By Remark 3.12, 3,

15

we can assume, without loss of generality, thatv∈V2and the peripheral arms ofSare inA2.

LetA′

1=A1\{S},V′1=V1,V′2=V2\{v},A2′ =A2\{the peripheral arms ofS}.

By Remark 3.13,A′_{= ((}A′

1,V′1), (A′2,V2′))is anM2assignment toT−SandM2(T−S)≥M2(T)−1−1 =

M2(T)−2.

LetA _{= ((}A_{1,V1), (}A_{2,V2))be anM2-maximal assignment toT−S. We are going to construct anM2} 20

assignment toT,A′_{= ((}A′

1,V1′), (A′2,V2′))(see Remark 3.13). Note thatM1(T) = M1(T−S) + 1. Letwbe the

vertex ofT−Sadjacent tovinT. Ifw∉V2then letFbe the component of(T−S)−V2containingwand let

Pbe the component ofT−V2containingS.

LetA′

1=A1∪ {the peripheral arms ofS},V1′ =V1∪ {v},V2′ =V2and

A′₂₌ {︃

(A₂\{F})∪ {P} ifF∈A_2andw∉V2

A_{2 otherwise} .

By Remark 3.13,A′_{= ((}A′

1,V′1), (A2′,V2′))is anM2assignment toTandM2(T)≥M2(T−S) + 1.

Example 3.17. LetTbe the following tree

✉ ✉ ✉ ✉ ✉ ✉

✉

✉

✉ ✉ ✉

✉ ✉

1 2 4 7 8 12 13

5

3 9 11

LetHbe the subtree obtained fromTby removing vertices11,12,13. ByProposition3.16,

M2(H) + 1≤M2(T)≤M2(H) + 2.

SinceHis a small superstar with central vertex4, byProposition 3.14,M2(H) = 4 + 0 + 4−2 = 2. So,

5≤M2(T)≤6.

4

An algorithm for

_M

The purpose of this section is to find simple reductions of the initial tree in such a way that we know the effect of each reduction onM2. The process may be continued until a small superstar, for whichM2is known (Proposition 3.14), or until a subtree for whichM2has bounds (Section 3). 5

Definition 4.1. (Peripheral SHDV, peripheral super path) LetTbe a tree that is not a small superstar. A periph-eral superstar high degree vertex(SHDV)vofTis an HDV vertex ofTsuch that

[(1) there is a unique subtree ofT−v,R, that contains high-degree vertices;

[(2) T−Ris a small superstar;

[(3) ifw∈Randwis adjacent tov, thenwdoes not satisfy1,2. 10

Aperipheral super pathofTatv(vis a SHDV)is a path of(T−R)−v. There are two kinds of peripheral super paths ofTatv(SHDV): peripheral arms and small pincers.

Example 4.2. Consider the treeTofExample 3.17.

The vertices4and8are peripheral superstar high degree vertices.

The vertex2is not a peripheral superstar high degree vertex because it is adjacent to vertex4and this vertex 15

satisfies conditions1and2ofDefinition 4.1.

The subtree ofTgenerated by vertices1, 2, 3is a peripheral super path ofTat4, but it is not a peripheral arm ofTat4(it is a small pincer).

Definition 4.3. Throughout this section, we will consider aperipheral SHDVvin a treeTthat is not a small superstar. ThesubtreeofTconsisting ofvand its peripheral super pathswill be calledQ. Letwbe the one 20

vertex adjacent tovthat is not inQ.

Remark 4.4. LetTbe a tree andA_{= ((}A_1,V1),(A_{2,V2))anM2assignment of(M1(T),p2,1}l_{)toT. Because} M1(T) =|A_1|−|V1|,

(1) All components ofT−V1are inA1.

(2) We can assume that ifv∈V1thenvhas degree greater than two inT. 25 (3) Since all components ofT−V1are paths, ifvis a peripheral SHDV of degree greater or equal to4inTthen

v∈ V1or there is at most one peripheral arm adjacent tovand the central vertex of each small pincer

adjacent tovis inV1.

(4) Ifvis a peripheral SHDV,v∈V1and all peripheral super paths adjacent tovhave length1then they are

Remark 4.5. LetTbe a tree andA_{= ((}A_1,V1),(A_{2,V2))anM2-maximal assignment of(M1(T),p2,1}l_)toT.

BecauseM2(T) =|A1|−|V1|+|A2|−|V2|,

(1) All components ofT−V2with more than one vertex are inA2. (2) We can assume that ifv∈V2thenvhas degree greater than two inT.

(3) All components ofT−V2with one vertex that are not components ofT−V1are inA2.

5

(4) Ifvis a peripheral SHDV,v∈V1and all peripheral super paths adjacent tovhave length1, then using Remark 4.4, 4, we conclude thatv∉V2.

(5) Ifvis a peripheral SHDV,v∈V1and all peripheral super paths adjacent tovhave length1, except one, then there is anM2-maximal assignment of(M1(T),p2,1l)toTsuch thatv∉V2.

Remark 4.6. In some proofs we construct anM2-maximal (or simply anM2) assignment of(M1(T),p2,1l)

10

to T. for some integer p2. In these cases, first we construct anM2assignment of (M1(T),p2,1l_{)toT,A =}

((A_1,V1),(A_{2,V2))by putting the elements in}A_{1and inV1, next we put the elements in}A_{2and inV2,}

us-ing Remarks 4.4 and 4.5. This construction is in such a way that M1(T) = |A_1|−|V1|and M1(T) +p2 =

|A_1|−|V1|+|A2|−|V2|. After using Lemma 2.3 we conclude condition2of Definition 2.1 and by Corollary 2.13,

we say thatM2(T)≥M1(T) +p2.

15

Proposition 4.7. LetTbe a tree that is not a small superstar andva peripheral SHDV, withQas defined earlier in this section. Suppose thatQhash≥1small pincers and the degree ofvinTis greater than4. LetHbe the graph obtained fromTby removing one small pincer ofQ. Then

M2(H) =M2(T)−2.

ProofBy Proposition 3.16,M2(H)≥M2(T)−2.

LetA_{= ((}A_1,V1),(A_{2,V2))be anM2-maximal assignment toH. We are going to construct anM2} assign-ment toT,A′_{= ((}A′_1,V′₁),(A′_2,V′₂))(see Remark 4.6). Note thatM1(T) =M1(H) + 1. Since the degree ofvinT is greater than4, we conclude that the degree ofvinHis greater than3. By Remark 4.4, 3 and Remark 4.5, 1, 3, we havev∈V1∪V2.

20

Suppose thatv∈V1∩V2. LetPbe the small pincerT−H. LetA′₁₌A_{1∪ {}P},V₁′ =V1,V2′ =V2andA′2=A2∪ {P}. By Remark 4.6,A′_{= ((}A′

1,V1′),(A′2,V2′))is anM2assignment toTandM2(T)≥M2(H) + 2. Suppose thatv∈V1\V2. Letxbe the central vertex of the small pincer,P, ofT−H. LetA′

1=A1∪ {P},V1′ =V1,A′2=A2∪ {the peripheral arms ofPatx}andV′2=V2∪ {x}. 25

By Remark 4.6,A′_{= ((}A′_1,V₁′),(A′_2,V₂′))is anM2assignment toTandM2(T)≥M2(H) + 2. Suppose thatv∈V2\V1. Letxbe the central vertex of the small pincer,P, ofT−H. LetV₁′ =V1∪ {x},A′1=A1∪ {the peripheral arms ofPatx}.,A′2=A2∪ {P}andV2′ =V2. By Remark 4.6,A′_{= ((}A′

1,V1′),(A′2,V2′))is anM2assignment toTandM2(T)≥M2(H) + 2.

Consequently,M2(T) =M2(H) + 2.

30

Lemma 4.8. LetTbe a tree that is not a small superstar. Suppose thatvis a peripheral SHDV inTwithQ, w

as defined earlier in this section. Then, there exists anM2-maximal assignment toT,A_{= ((}A_1,V1),(A_2,V2)),

in whichv∈V1∪V2.

Moreover,

(1) Ifvhas at least two peripheral arms of length at least2, then there exists anM2-maximal assignment,

35

A′_{= ((}A′_1,V₁′),(A′_2,V′₂))in whichv∈V₁′ ∩V₂′.

(2) Ifvhas at most one peripheral arm of length at least2andwhas degree two inT, then there exists an

M2-maximal assignment,A′′= ((A1′′,V′′1),(A′′2,V′′2))such thatvis in exactly oneV1′′orV2′′.

(3) IfQhasf peripheral arms of length1andg ≤ 1peripheral arms of length at least2,f +g > 2and

A′′′_{= ((}A′′′_1,V′′′₁),(A′′′_2,V₂′′′))is anM2-maximal assignment toT, thenv∈V1′′′.

ProofLetA _{= ((}A_{1,V1), (}A_{2,V2))be anM2-maximal assignment toTin whichv∈}̸V1∪V2. Suppose that

Qhasf peripheral arms of length1andgperipheral arms of length at least2. We are going to construct an

M2-maximal assignment toT,B= ((B1,U1), (B2,U2))(see Remark4.6).

Iff +g ≥ 2, then by Remark 4.4, 1, the component,R, ofT −V1containingvis inA1. Note that the

peripheral arms ofQmight be inR. 5

LetB_{1= (}A_{1\{R})∪ {two peripheral arms ofQ},U1=V1∪ {v},}B₂₌A_2andU2=V2.

By Remark 4.6 and the cardinality ofB_,B_{= ((}B_{1,U1), (}A_{2,V2))is anM2-maximal assignment toTin} whichv∈U1.

Iff +g ≤1, by Remark 4.4, 3 and Remark 4.5, 1 and 3, the central vertex of each small pincer ofQis in

V1\V2. By Remark 4.5, 1, the component,R, ofT−V2containingv, is inA2. 10

LetB₁₌A_1,U1=V1,B_{2= (}A₁\{R})∪ {two peripheral super paths ofQ}andU2=V2∪ {v}.

By Remark 4.6 and the cardinality ofB_,B_{= ((}B_{1,U1), (}B_{2,U2))is anM2-maximal assignment toTin} whichv∈U2. So, there exists anM2-maximal assignment toTA= ((A1,V1), (A2,V2))in whichv∈V1∪V2.

(1) By what we just proved, there exists anM2-maximal assignment toT,

A_{= ((}A_{1,V1), (}A_2,V2)),

in whichv∈V1∪V2. Suppose without loss of generality thatv∈V1\V2. We are going to construct

anM2-maximal assignment toT,A′= ((A′1,V1′), (A′2,V′2)), in whichv∈V1′∩V′2. (see Remark 4.6). By 15

Remark 4.5, 1 and 3, the component,R, ofT−V2containingv, is inA2. Note that the peripheral arms

ofQmight be inR. LetA′

1=A1,nV1′ =V1,V2′ =V2∪ {v}and

A′_{2= (}A_{2\{R})∪ {two peripheral arms of length at least two ofQ}.}

Since|A_2|−|V2|=|A′_2|−|V2′|, by Remark 4.6 and the cardinality ofA′,A′= ((A′1,V1′), (A′2,V2′))is an

M2-maximal assignment toT, in whichv∈V1′ ∩V2′.

(2) By what we just proved, there exists anM2-maximal assignment,

A_{= ((}A_{1,V1), (}A_2,V2)),

in whichv∈ V1∪V2. Supposev ∈ V1∩V2. We are going to construct anM2-maximal assignment 20

toT,A′′ _{= ((}A′′_1,V1′′_{), (}A₂′′_,V2′′_{)), in whichv ∈ V1}′′\V2′′. (see Remark 4.6) Using Remark 4.4, 1, each

peripheral super path ofQis inA_{1. By Remark 4.5, 1, the longer arm ofQand the small pincers of}

Qare inA_{2 and there is not a peripheral arm of length1of Qin}A_{2. By Remark 4.5, 2,w ∈}̸ V2.

LetR be the component ofT − V2 containingw and letF be the component ofT −((V2 \{v})∪ {the central vertex of each small pincer ofQ})containingvandw. By Remark 4.5, 1,F∈A_2. 25 LetA′′

1=A1,V1′′=V1,

A′′

2= (A2\{the peripheral super paths of length at least two ofQ,R})∪ ∪{the peripheral arms of each small pincer ofQ,F}

andV′′2= (V2\{v})∪ {the central vertex of each small pincer ofQ}.

IfQdoes not have a longer arm orR∉A_2then|A′′

2|−|V′′2|>|A2|−|V2|. This is impossible becauseA

is anM2-maximal assignment toT. So,v∉V1∩V2.

IfQhas a longer arm andR∈A_2then|A_2|−|V2|=|A′′

2|−|V′′2|. By Remark 4.6 and using the cardinality

ofA′′_,A′′_{= ((}A′′

1,V1′′), (A′′2,V2′′))is anM2-maximal assignment toT, in whichvinV1′′\V′′2. 30

(3) LetA′′′_{= ((}A′′′

1,V1′′′), (A′′′2,V2′′′))be anM2-maximal assignment toT. By Remark 4.4, 1, each peripheral

super path ofQbelongs toA′′′

1andv∈V1′′′.

Lemma 4.9. LetTbe a tree that is not a small superstar. Suppose thatvis a peripheral SHDV inTwithQ,

was defined earlier in this section. Suppose thatQhasf peripheral arms of length1andg ≤ 1peripheral arms of length at least2and the degree ofwinTis2. Then, there exists anM2-maximal assignment toT, 35

(1) Iff +g≥ 1, thenv∈V1and the central vertex of each small pincer ofQbelongs toV2.

(2) Iff +g= 0, thenv∈V2and the central vertex of each small pincer ofQbelongs toV1.

Proof

(1) By2of Lemma 4.8, letA_{= ((}A_{1,V1), (}A_{2,V2))be anM2-maximal assignment toTsuch thatvis exactly} oneV1orV2.

5

Iff +g > 1, sincevis a peripheral SHDV andw ∉ V1(the degree ofwinTis2), by Remark 4.4, 1,

each peripheral super path ofQbelongs toA_{1andv∈ V1. In this case, becausev ∈}̸ V2andAis an

M2-maximal assignment toT, we conclude that the central vertex of each small pincer ofQis inV2and

the peripheral arms of each small pincer ofQare inA_2..

Suppose thatf +g = 1andv ∈ V2. then by Remark 4.4, 1, the central vertex of each small pincer

10

ofQis inV1and the peripheral arms of each small pincer ofQare inA1. By Remark 4.5, 1 and 3,

the peripheral super paths ofQare inA_{2. Sincew has degree two inT, we can assume thatw ∈}̸

V1∪V2. We are going to construct an M2-maximal assignment toT, A′ = ((A′1,V1′), (A′2,V′2)), in

whichv ∈ V1and the central vertex of each small pincer ofQis in V2 (see Remark 4.6). LetRbe

the component of T− V1 containingv, w. By Remark 4.4, 1, R ∈ A1. Let Pbe the component of

15

T −V2, containingw. Since P ≠ R, by Remark 4.5, 1 and 3,P ∈ A2. Let Bbe the component of

T−((V2\{v})∪ {the central vertex of each small pincer ofQ}), containingvandw. LetCbe the

com-ponent ofT−(V1∪ {v}), containingw. Note thatB≠C.

Let

A′

1= (A1\{the peripheral arms of each small pincer ofQ,R})∪ ∪{C, the peripheral super paths ofQ},

V′₁= (V1\{the central vertex of each small pincer ofQ})∪ {v},

A′

2= (A2\{the peripheral super paths ofQ,P})∪ ∪{the peripheral arms of each small pincer ofQ,B}

andV₂′ = (V2\{v})∪ {the central vertex of each small pincer ofQ}.

Since|A′

1|−|V1′|=|A1|−|V1|and|A2′|−|V2′|=|A2|−|V2|and by Remark 4.6, we get anM2-maximal

20

assignment toT,A′_{= ((}A′

1,V1′), (A′2,V′2)), wherev∈V1′ and the central vertex of each small pincer of

Qbelongs toV′₂.

(2) By2of Lemma 4.8, letA_{= ((}A_{1,V1), (}A_{2,V2))be anM2-maximal assignment toTsuch thatvis exactly} oneV1orV2. Sincef +g= 0,vis a peripheral SHDV andw∉V1, ifv∈V1then by Remark 4.4, 1, the

peripheral super paths ofQare inA_{1. LetFbe the component ofT−V1containingw. By Remark 4.4,} 1,F ∈ A_{1. LetHbe the component ofT}−(V1\{v})∪ {the central vertex of each small pincer ofQ})

containingwandv. Let

A′

1= (A1\{the peripheral super paths ofQ,F})∪

∪{the peripheral arms of each small pincer ofQ,H},

V′₁= (V1\{v})∪ {the central vertex of each small pincer ofQ}. Since|A′1|−|V1′|=|A1|−|V1|+ 1|we

conclude thatA_{is not an}M2-maximal assignment toT. Impossible. Consequently,v∉V1andv∈V2.

Therefore, the central vertex of each small pincer ofQbelongs toV1.

25

Theorem 4.10. (M2Reduction Theorem)LetTbe a tree that is not a small superstar andva peripheral SHDV,

withQ,was defined earlier in this section. Suppose thatQhasfperipheral arms of length1,gperipheral arms of length at least2andhsmall pincers. Then:

(A) Ifg≥2, thenM2(T−Q) =M2(T)−f−2g−2h+ 2.

(B) Ifg≤1and the degree ofwinTis2, thenM2((T−Q) +wK1) = M2(T)−f −g−2h+ 1,where(+wK1) 30

means that we put a vertex adjacent tow.

(C) Ifg≤1, the degree ofwinTis greater than2andf +g> 2then