A note on tree realizations of matrices

RAIRO Oper. Res. 41(2007) 361–366 DOI: 10.1051/ro:2007028

A NOTE ON TREE REALIZATIONS OF MATRICES

Alain Hertz

and Sacha Varone

Abstract. It is well known that each tree metric M has a unique realization as a tree, and that this realization minimizes the total length of the edges among all other realizations ofM. We extend this result to the class of symmetric matricesM with zero diagonal, positive entries, and such thatmij+mkl≤max{mik+mjl, mil+mjk}for all distinct i, j, k, l.

Keywords.Matrices, tree metrics, 4-point condition.

Mathematics Subject Classification. 05C50, 05B20, 68R10, 68U99.

Introduction

Ann×nmatrixM = (mij) with zero diagonal is atree metricif it satisfies the following4-point condition:

mij+mkl≤max{m_ik+mjl, mil+mjk} ∀i, j, k, lin{1, . . . , n}.

By denotingsijkl=mij+mkl, the 4-point condition is equivalent to imposing that two of the three sumssijkl, sikjl and siljk are equal and not less than the third.

The 4-point condition entails the triangle inequality (for k = l) and symmetry (for i = k and j = l). There is an extensive literature on tree metrics; see for example [1–3, 7–10].

Received March 07, 2005. Accepted January 19, 2007.

∗ This work has been partially funded by grant PA002-104974/1 from the Swiss National Science Foundation, received by the second author.

1 D´epartement de math´ematiques et de g´enie industriel, ´Ecole Polytechnique, Montr´eal, Canada;alain.hertz@gerad.ca

2Haute ´ecole de gestion de Gen`eve, ´Economie d’Entreprise, Gen`eve, Switzerland;

sacha.varone@hesge.ch

c EDP Sciences, ROADEF, SMAI 2007

Article published by EDP Sciences and available at http://www.rairo-ro.org or http://dx.doi.org/10.1051/ro:2007028

⎛

⎜⎜

⎜⎜

⎝

0 12 20 22 4

12 0 6 8 6

20 6 0 4 14

22 8 4 0 16

4 6 14 16 0

⎞

⎟⎟

⎟⎟

⎠

A matrixM.

12 8

22

4 14

6 6

1

4

4 2

3

5 16 20

Its associated complete graphKM.

4

1 3 5 6

4 1

2

3

5

A realization ofM as a tree.

Figure 1. A tree realization of a tree metric.

It is well known that a tree metricM = (mij) can be represented by an unrooted treeT such that{1, . . . , n}is a subset of the vertex set ofT, and the length of the unique chain connecting two verticesiandj inT (1≤i < j≤n) is equal tomij. LetG= (V, E, d) be the graph with vertex setV, edge setE, and wheredis a function assigning a positive lengthdij to each edge (i, j) of G. The length of the shortest chain between two verticesiandj inGis denoted d^G_ij.

Definition 0.1. LetM be a symmetricn×nmatrix with zero diagonal and such that 0≤mij ≤mik+mkj for all i, j, k in {1, . . . , n}. A graphG= (V, E, d) is a realization ofM = (mij) if and only if{1, . . . , n}is a subset ofV, andd^G_ij=mij

for alli, j in{1, . . . , n}.

As mentioned above, tree metrics have a realization as a tree. A realization Gof a matrix M is saidoptimal if the total length of the edges inGis minimal among all realizations ofM. Hakimi and Yau [7] have proved that tree metrics have a unique realization as a tree, and this realization is optimal. Culberson and Rudnicki [4] have designed anO(n²) time algorithm for constructing a realization as a tree of tree metrics.

We propose to extend the above definition to matrices whose entries do not necessarily satisfy the triangle inequality. Given a symmetricn×n matrixM = (mij) with zero diagonal and positive entries, letKM denote the complete graph onnvertices in which each edge (i, j) has lengthmij.

Definition 0.2. Let M be a symmetric n×n matrix with zero diagonal and positive entries. A graphG= (V, E, d) is a realization of M = (mij) if and only if{1, . . . , n}is a subset ofV, and d^G_ij =d^K_ij^M for alli, j in{1, . . . , n}.

Obviously, ifM satisfies the triangle inequality, thend^K_ij^M =mij, and Defini- tion 0.2 is then equivalent to Definition 0.1. Figure 1 illustrates this new definition.

Notice that the matrix in Figure 1 is not a tree metric, while it has a realization as a tree.

Let Mn denote the set of symmetric n×n matrices M = (mij) with zero diagonal, positive entries, and such thatmij+mkl≤max{mik+mjl, mil+mjk} for alldistinctpointsi, j, k, l in{1, . . . , n}.

⎛

⎜⎜

⎝

0 1 3 1

1 0 1 3

3 1 0 4

1 3 4 0

⎞

⎟⎟

⎠ A 4×4 matrixM

that does not belong toM4.

3

2

4 3

1

4 1

1 1

3

Its associated complete graph

KM.

3

1 2

1

4

1 1

A realization of M as a tree.

Figure 2. A tree realization of a matrix that does not belong toMn. Since we only impose the 4-point condition ondistinctpointsi, j, k, l, the entries of a matrix inM_n do not necessarily satisfy the triangle inequality. While all tree metrics belong to M_n, the example in Figure 2 shows that a matrix having a realization as a tree does not necessarily belong toM_n. However, we prove in this paper that all matrices inMn have a unique realization as a tree, and that this realization is optimal.

1. The main result

Let M = (mij) be any matrix in Mn, and consider the matrixM = (m_ij) obtained from M by setting m_ij equal to the length d^K_ij^M of the shortest chain between i and j in KM. Notice that the elements in M satisfy the triangle inequality. In order to prove that M has a realization as a tree, it is sufficient to prove that M is a tree metric. The proof is based on Floyd’s O(n³) time algorithm [6] for the computation ofM.

Floyd’s algorithm[6]

SetM⁰ equal toM; Forr:= 1 tondo

For alliandj in {1, . . . , n}do

Setm^r_ij equal tomin{m^r−1_ij , m^r−1_ir +m^r−1_rj }; SetM equal toMⁿ.

We shall prove that each matrixM^r(1≤r≤n) is inM_n. Since the entries of M=Mⁿ satisfy the triangle inequality, we will be able to conclude thatM is a tree metric.

Theorem 1.1. Let M = (mij) be a matrix in M_n, and let M = (m_ij) be the n×nmatrix obtained from M by settingm_ij =d^K_ij^M for all iandj in{1, . . . , n}. Then M is a tree metric.

Proof. Following Floyd’s algorithm, define M⁰ = M and let M^r be the matrix obtained fromM^r−1 by setting m^r_ij =min{m^r−1_ij , m^r−1_ir +m^r−1_rj } for all iand j

in {1, . . . , n}. Given four distinct points i, j, k, l in {1, . . . , n}, we denotes^r_ijkl = m^r_ij +m^r_kl. We prove by induction that eachM^r (r = 1, . . . , n) is in Mn. By hypothesis,M⁰= M is in M_n, so assume M^r−1 ∈ M_n. It is sufficient to show thats^r_ijkl ≤max{s^r_ikjl, s^r_iljk} for all distincti, j, k, lin {1, . . . , n}, or equivalently, that two of the three sums s^r_ijkl, s^r_ikjl and s^r_iljk are equal and not less than the third.

Notice thatm^r_ri=m^r−1_ri andm^r_ij ≤m^r−1_ij for all 1≤i≤j ≤n. Consider any four distinct points i, j, k and l. Sincer is possibly one of these four points, we divide the proof into two cases.

Case A: r∈ {i, j, k, l}, sayr=l.

SinceM^r−1∈ Mn, we may assume, without loss of generality (wlog) that s^r−1_rijk ≤ s^r−1_rjik = s^r−1_rkij. If m^r_ik = m^r−1_ik and m^r_ij = m^r−1_ij , then s^r_rijk ≤ s^r_rjik =s^r_rkij and we are done. So, we can assume wlogm^r_ik < m^r−1_ik . It then follows that m^r−1_ri +s^r−1_rjik = m^r−1_ri +s^r−1_rkij < m^r−1_ik +m^r−1_ij , which means that m^r_ij = m^r−1_ri +m^r−1_rj < m^r−1_ij . We therefore have s^r_rijk ≤ m^r−1_ri +m^r−1_rj +m^r−1_rk =s^r_rjik=s^r_rkij.

Case B: r /∈ {i, j, k, l}.

Ifs^r_ijkl=s^r−1_ijkl, s^r_ikjl=s^r−1_ikjl ands^r_iljk =s^r−1_iljk, there is nothing to prove. So assume wlog thatm^r_ij < m^r−1_ij . Notice that ifm^r_ik =m^r−1_ik , m^r_il =m^r−1_il , m^r_jk =m^r−1_jk and m^r_jl =m^r−1_jl , then we are done. Indeed, sinceM^r−1 ∈ Mnands^r_rkij < s^r−1_rkij, whiles^r_rjik=s^r−1_rjik ands^r_rijk=s^r−1_rijk, we know from case A that s^r−1_rjik = s^r−1_rijk. In a similar way, we also have s^r−1_rjil = s^r−1_rijl. Hence,s^r−1_rjik+s^r−1_rijl =s^r−1_rijk+s^r−1_rjil, which means thats^r−1_ikjl =s^r−1_iljk. Since M^r−1∈ M_n, s^r_ikjl =s^r−1_ikjl, s^r_iljk =s^r−1_iljk ands^r_ijkl < s^r−1_ijkl we conclude that s^r_ijkl< s^r_ikjl=s^r_iljk. Wlog, we can therefore assume m^r_ik < m^r−1_ik .

The rest of the proof is divided into four subcases.

Case B1: m^r−1_jk < m^r−1_rj +m^r−1_rk andm^r−1_jl > m^r−1_rj +m^r−1_rl .

Since s^r_rkjl =m^r−1_rk +m^r−1_rj +m^r−1_rl > s^r_rljk, we know from case A that s^r_rjkl = s^r_rkjl, which means that m^r_kl = m^r−1_rk +m^r−1_rl . Hence, s^r_iljk <

s^r_ijkl=s^r_ikjl.

Case B2: m^r−1_jk < m^r−1_rj +m^r−1_rk andm^r−1_jl ≤m^r−1_rj +m^r−1_rl .

We can assume m^r_kl = m^r−1_kl , else we are in case B1, where the roles of points j and k are exchanged. We can also assumem^r−1_il < m^r−1_ri +m^r−1_rl . Indeed, ifm^r−1_il ≥m^r−1_ri +m^r−1_rl thens^r_ijkl=m^r−1_ri +s^r−1_rjkl,s^r_ikjl=m^r−1_ri + s^r−1_rkjl, and s^r_iljk=m^r−1_ri +s^r−1_rljk and we are done since M^r−1∈ M_n.

But now, s^r_rlik > s^r_rkil, and we know from case A that s^r_rikl = s^r_rlik, which means thatm^r_kl=m^r−1_rk +m^r−1_rl . Hence,s^r_rjkl> s^r_rljk, and we know from case A thats^r_rkjl=s^r_rjkl, which means thatm^r_jl =m^r−1_rj +m^r−1_rl . We therefore haves^r_iljk< s^r_ijkl=s^r_ikjl.

Case B3: m^r−1_jk ≥m^r−1_rj +m^r−1_rk andm^r−1_jl > m^r−1_rj +m^r−1_rl .

It follows from cases B1 and B2 thati, j, k and l satisfy the 4-point con- dition in M^r if m^r_ij < m^r−1_ij , m^r_ik < m^r−1_ik , and m^r−1_jk < m^r−1_rj +m^r−1_rk . By permuting the roles of points i and j as well as those of k and l, we also know that i, j, k and l satisfy the 4-point condition in M^r if m^r_ij < m^r−1_ij ,m^r_jl < m^r−1_jl , andm^r−1_il < m^r−1_ri +m^r−1_rl . Sincem^r_ij < m^r−1_ij andm^r_jl < m^r−1_jl in case B3, we can assumem^r−1_il ≥m^r−1_ri +m^r−1_rl . Hence, s^r_ijkl≤s^r_ikjl=s^r_iljk.

Case B4: m^r−1_jk ≥m^r−1_rj +m^r−1_rk andm^r−1_jl ≤m^r−1_rj +m^r−1_rl .

Since M^r−1 ∈ Mn, ands^r−1_rijl < s^r−1_rlij we know that s^r−1_rjil =s^r−1_rlij, which means thatm^r_il< m^r−1_il . Ifm^r−1_jl =m^r−1_rj +m^r−1_rl thens^r_ijkl≤s^r_ikjl=s^r_iljk. Else, m^r−1_jl < m^r−1_rj +m^r−1_rl , which implies s^r_rkjl < s^r_rljk. We then know from case A that s^r_rjkl = s^r_rljk, which means thatm^r_kl =m^r−1_rk +m^r−1_rl . We therefore haves^r_ikjl< s^r_ijkl=s^r_iljk.

Corollary 1.2. Each matrix in Mn has a unique realization as a tree, and this realization is optimal.

Proof. Let M be any matrix in M_n, and let M = (m_ij) be the n×n matrix obtained from M by settingm_ij = d^K_ij^M for all 1 ≤ i < j ≤ n. It follows from Definition 0.2 that a graph is a realization ofM if and only if it is a realization of M. We know from the above theorem thatMis a tree metric. To conclude, it is sufficient to observe that each tree metric has a unique tree realization, and this

realization is optimal.

2. A related problem

Given twon×nmetricsL= (lij) andU = (uij), thematrix sandwich problem[5]

is to find (if possible) a tree metric M = (mij) such that lij ≤ mij ≤ uij for all i, j ∈ {1, . . . , n}. Typically, the information concerning the distance matrix associated with a network may be inaccurate, and we are only given lower and upper bound matricesLandU.

We prove here below that a solution to the matrix sandwich problem can be obtained by first finding a matrixM ∈ Mn that lies betweenL andU, and then constructing the tree metric M = (m_ij) with m_ij = d^K_ij^M. Finding a matrix M ∈ M_n that lies betweenLand U is possibly easier than finding a tree metric with the same lower and upper bound matrices, the reason being that the triangle inequality is not imposed on matrices inM_n.

Proposition 2.1. Let M = (mij)be a matrix in Mn, and let M= (m_ij)be the n×nmatrix obtained from M by settingm_ij =d^K_ij^M for all iandj in{1, . . . , n}.

If lij ≤ mij ≤ uij for all i, j ∈ {1, . . . , n}, then M is a solution to the matrix sandwich problem.

Proof. Let M = (mij) be a matrix in Mn, such that lij ≤ mij ≤ uij for all i, j ∈ {1, . . . , n}, and let M = (m_ij) be the n×n matrix obtained fromM by setting m_ij = d^K_ij^M for all 1 ≤i < j ≤ n. We know from Theorem 1 that M is a tree metric. Moreover, since L is a metric, we havelij ≤m_ij ≤mij for all

i, j∈ {1, . . . , n}.

References

[1] H.-J. Bandelt, Recognition of tree metrics.SIAM J. Algebr. Discrete Methods3(1990) 1–6.

[2] J.-P. Barth´el´emy and A. Gu´enoche, Trees and proximity representations. John Wiley &

Sons Ltd., Chichester (1991).

[3] P. Buneman, A note on metric properties of trees.J. Combin. Theory Ser. B 17(1974) 48–50.

[4] J.C. Culberson and P. Rudnicki, A fast algorithm for constructing trees from distance ma- trices. InInf. Process. Lett.30(1989) 215–220.

[5] M. Farach, S. Kannan and T. Warnow, A robust model for finding optimal evolutionary trees.Algorithmica13(1995) 155–179.

[6] R.W. Floyd, Algorithm 97.Shortest path. Comm. ACM5(1962) 345.

[7] S.L. Hakimi and S.S. Yau, Distance matrix of a graph and its realizability.Q. Appl. Math.

22(1964) 305–317.

[8] A.N. Patrinos and S.L. Hakimi, The distance matrix of a graph and its tree realization.Q.

Appl. Math.30(1972) 255–269.

[9] J.M.S. Sim˜oes-Pereira, A note on the tree realizability of a distance matrix. J. Combin.

Theory6(1969) 303–310.

[10] S.C. Varone, Trees related to realizations of distance matrices.Discrete Math.192(1998) 337–346.

To access this journal online:

www.edpsciences.org