Alain Hertz 1 and Sacha Varone 2. Introduction A NOTE ON TREE REALIZATIONS OF MATRICES. RAIRO Operations Research Will be set by the publisher

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1 RAIRO Operations Research Will be set by the publisher A NOTE ON TREE REALIZATIONS OF MATRICES Alain Hertz and Sacha Varone 2 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 of M We extend this result to the class of symmetric matrices M with zero diagonal, positive entries, and such that m ij + m kl max{m ik + m, m il + m } for all distinct i, j, k, l Keywords: matrices, tree metrics, -point condition Mathematics Subject Classification 05C50,05B20,68R0,68U99 Introduction An n n matrix M = (m ij ) with zero diagonal is a tree metric if it satisfies the followng -point condition: m ij + m kl max{m ik + m,m il + m } i,j,k,l in {,,n} The -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 [,6 9] It is well known that a tree metric M = (m ij ) can be represented by an unrooted tree T such that {,,n} is a subset of the vertex set of T, and the length of the unique chain connecting two vertices i and j in T ( i < j n) is equal to m ij March 2005 This wo has been partially funded by grant PA / from the Swiss National Science Foundation, received by the second author Département de mathématiques et de génie industriel, École Polytechnique, Montréal, Canada, alainhertz@geradca 2 LARIM, Département de génie informatique, École Polytechnique, Montréal, Canada, sachavarone@polymtlca c EDP Sciences 200

2 2 TITLE WILL BE SET BY THE PUBLISHER Let G = (V,E,d) be the graph with vertex set V, edge set E, and where d is a function assigning a positive length d ij to each edge (i,j) of G The length of the shortest chain between two vertices i and j in G is denoted d G ij Definition 0 Let M be a symmetric n n matrix with zero diagonal and such that 0 m ij m ik + m kj for all i,j,k in {,,n} A graph G = (V,E,d) is a realization of M = (m ij ) if and only if {,,n} is a subset of V, and d G ij = m ij for all i,j in {,,n} As mentioned above, tree metrics have a realization as a tree A realization G of a matrix M is said optimal if the total length of the edges in G is minimal among all realizations of M Hakimi and Yau [6] have proved that tree metrics have a unique realization as a tree, and this realization is optimal We propose to extend the above definition to matrices whose entries do not necessarily satisfy the triangle inequality Given a symmetric n n matrix M = (m ij ) with zero diagonal and positive entries, let K M denote the complete graph on n vertices in which each edge (i,j) has length m ij Definition 02 Let M be a symmetric n n matrix with zero diagonal and positive entries A graph G = (V,E,d) is a realization of M = (m ij ) if and only if {,,n} is a subset of V, and d G ij = dkm ij for all i,j in {,,n} Obviously, if M satisfies the triangle inequality, then d KM ij = m ij, and Definition 02 is then equivalent to Definition 0 Figure illustrates this new definition Notice that the matrix in Figure is not a tree metric, while it has a realization as a tree A matrix M Its associated complete graph K M A realization of M as a tree 5 Figure a tree realization of a tree metric Let M n denote the set of symmetric n n matrices M = (m ij ) with zero diagonal, positive entries, and such that m ij + m kl max{m ik + m,m il + m } for all distinct points i,j,k,l in {,,n} Since we only impose the -point condition on distinct points i,j,k,l, the entries of a matrix in M 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 to M n However, we prove in this

3 TITLE WILL BE SET BY THE PUBLISHER paper that all matrices in M n have a unique realization as a tree, and that this realization is optimal A matrix M that does not belong to M 2 Its associated complete graph K M 2 A realization of M as a tree Figure 2 a tree realization of a matrix that does not belong to M n The main result Let M = (m ij ) be any matrix in M n, and consider the matrix M = (m ij ) obtained from M by setting m ij equal to the length dkm ij of the shortest chain between i and j in K M 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 algorithm [5] for the computation of M Floyd s algorithm [5] Set M 0 equal to M; For r := to n do For all i and j in {,,n} do Set m r ij equal to min{mr ij,m r ir + m r }; Set M equal to M n ; We shall prove that each matrix M r ( r n) is in M n Since the entries of M = M n satisfy the triangle inequality, we will be able to conclude that M is a tree metric Theorem Let M = (m ij ) be a matrix in M n, and let M = (m ij ) be the n n matrix obtained from M by setting m ij = dkm ij for all i and j in {,,n} Then M is a tree metric Proof Following Floyd s algorithm, define M 0 = M and let M r be the matrix obtained from M r by setting m r ij = min{mr ij,m r ir + m r } for all i and j in {,,n} Given four distinct points i,j,k,l in {,,n}, we denote s r il = m r ij + mr kl We prove by induction that each Mr (r =,,n) is in M n By hypothesis, M 0 = M is in M n, so assume M r M n It is sufficient to show that s r il max{sr ik,sr il } for all distinct i,j,k,l in {,,n}, or equivalently,

4 TITLE WILL BE SET BY THE PUBLISHER that two of the three sums s r il,sr ik and sr il are equal and not less than the third Notice that m r ri = mr ri and m r ij mr ij for all i j n Consider any four distinct points i,j,k and l Since r is possibly one of these four points, we divide the proof into two cases Case A : r {i,j,k,l}, say r = l Since M r M n, we may assume, without loss of generality (wlog) that s r ri sr ik = sr ij If mr ik = mr ik and m r ij = m r ij, then s r ri sr ik = sr ij and we are done So, we can assume wlog mr ik < m r ik It then follows that mr ri + s r ik = mr ri + s r ij < mr ik + m r ij, which means that m r ij = m r ri + m r < m r ij We therefore have s r ri mr ri + m r + m r = s r ik = sr ij Case B : r / {i,j,k,l} If s r il = sr il,sr ik = sr assume wlog that m r ij < mr ij m r = mr and m r = mr M n and s r ij < sr ij, while sr ik = sr Case A that s r ik = sr ri Hence, s r ik + sr ri = sr ik and sr il = sr il, there is nothing to prove So Notice that if m r ik = mr ik, mr il = mr il,, then we are done Indeed, since M r ik and sr ri = sr ri, we know from In a similar way, we also have sr il = s r ri ri + sr il, which means that sr ik = sr il Since M r M n,s r ik = sr ik,sr il = sr il and sr il < sr il we conclude that s r il < sr ik = sr il Wlog, we can therefore assume mr ik < mr ik The rest of the proof is divided into four subcases Case B : m r < m r + m r and m r > m r + m r Since s r = mr + m r + m r > s r s r = s r, which means that mr kl = m r s r il = sr ik Case B2 : m r Case B : m r < m r + m r We can assume m r kl = mr kl and m r, we know from Case A that + m r Hence, s r il < m r + m r, else we are in Case B, where the roles of points j and k are exchanged We can also assume m r il Indeed, if m r il m r ri +m r then s r il = mr ri < m r ri + m r +s r, sr ik = mr ri + s r, and sr il = mr ri + s r and we are done since Mr M n But now, s r ik > sr il, and we know from Case A that sr rikl = sr ik, which means that m r kl = mr +mr Hence, s r > sr, and we know + m r from Case A that s r = sr, which means that mr = mr We therefore have s r il < sr il = sr ik m r + m r We may assume m r and m r > m r + m r il m r ri + m r, else the situation is equivalent either to Case B or B2 (where the roles of points i and j are exchanged, as well as those of k and l) Hence, s r il sr ik = sr il

5 Case B : m r TITLE WILL BE SET BY THE PUBLISHER 5 m r + m r and m r m r Since M r M n, and s r ri < s r ij means that m r il < mr il If m r + m r we know that sr il = s r ij, which = m r +m r then s r il sr ik = sr il, which implies s r < sr We then know + m r Else, m r < m r + m r from Case A that s r = sr, which means that mr kl = mr We therefore have s r ik < sr il = sr il Corollary 2 Each matrix in M n 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 setting m ij = dkm ij for all i < j n It follows from Definition 02 that a graph is a realization of M if and only if it is a realization of M We know from the above theorem that M is 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 two n n tree metrics L = (l ij ) and U = (u ij ), the matrix sandwich problem [] is to find (if possible) a tree metric M = (m ij ) such that l ij m ij u ij for all i {,,n} Typically, the information concerning the distance matrix associated with a netwo may be inaccurate, an we are only given lower and upper bound matrices L and U We prove here below that a solution to the matrix sandwich problem can be obtained by first finding a matrix M M n that lies between L and U, and then constructing the tree metric M = (m ij ) with m ij = dkm ij Finding a matrix M M n that lies between L and 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 in M n Proposition 2 Let M = (m ij ) be a matrix in M n, and let M = (m ij ) be the n n matrix obtained from M by setting m ij = dkm ij for all i and j in {,,n} If l ij m ij u ij for all i {,,n}, then M is a solution to the matrix sandwich problem Proof Let M = (m ij ) be a matrix in M n, such that l ij m ij u ij for all i {,,n} Since L and U are tree metrics, it follows that M has a zero diagonal and positive entries Let M = (m ij ) be the n n matrix obtained from M by setting m ij = dkm ij for all i < j n We know from Theorem that M is a tree metric Moreover, since L is a tree metric, we have l ij m ij m ij for all i,j {,,n}

6 6 TITLE WILL BE SET BY THE PUBLISHER References [] H-J Bandelt Recognition of tree metrics SIAM J on Algebraic Discrete Methods, (): 6, 990 [2] J-P Barthélémy and A Guénoche Trees and proximity representations John Wiley & Sons Ltd, Chichester, 99 [] P Buneman A note on metric properties of trees J Combin Theory Ser B, 7:8 50, 97 [] M Farach, S Kannan, and T Warnow A robust model for finding optimal evolutionary trees Algorithmica, :55 79, 995 [5] R W Floyd Algorithm 97 Shortest path Comm ACM, 5:5, 962 [6] S L Hakimi and S S Yau Distance matrix of a graph and its realizability Quart Appl Math, 22:05 7, 96 [7] A N Patrinos and S L Hakimi The distance matrix of a graph and its tree realization Quart Appl Math, 0: , 972 [8] J M S Simões-Pereira A note on the tree realizability of a distance matrix J Combin Theory, 6:0 0, 969 [9] S C Varone Trees related to realizations of distance matrices Discrete Mathematics, 92:7 6, 998