Alain Hertz 1 and Sacha Varone 2

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1 RAIRO Operations Research RAIRO Oper Res (2007) 6 66 DOI: 005/ro: 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, andsuchthatm ij + m kl max{m ik + m,m + 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 following -point condition: m ij + m kl max{m ik + m,m + m } i, j, k, l in {,,n} By denoting s = m ij +m kl, the -point condition is equivalent to imposing that two of the three sums s,s ik and s are equal and not less than the third The -point condition entas 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 [, 7 0] Received March 07, 2005 Accepted January 9, 2007 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 Haute école de gestion de Genève, Économie d Entreprise, Genève, Switzeand; sachavarone@hesgech c EDP Sciences, ROADEF, SMAI 2007 Article published by EDP Sciences and avaable at or

2 62 A HERTZ AND S VARONE AmatrixM Its associated complete graph K M A realization of M as a tree Figure A tree realization of a tree metric 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) isequaltom ij Let G =(V,E,d) be the graph with vertex set V,edgesetE, andwhered 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) isa realization of M =(m ij ) if and only if {,,n} is a subset of V,andd 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 [7] have proved that tree metrics have a unique realization as a tree, and this realization is optimal Culberson and Rudnicki [] have designed an O(n 2 ) 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 necessary 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) haslengthm 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 )ifandonly if {,,n} is a subset of V,andd 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 lustrates this new definition Notice that the matrix in Figure is not a tree metric, whe it has a realization as a tree 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 + m } for all distinct points i, j, k, l in {,,n}

3 ANOTEONTREEREALIZATIONSOFMATRICES A matrixm 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 Since we only impose the -point condition on distinct points i, j, k, l, theentries of a matrix in M n do not necessary satisfy the triangle inequality Whe all tree metrics belong to M n, the example in Figure 2 shows that a matrix having a realization as a tree does not necessary belong to M n However, we prove in this paper that all matrices in M n have a unique realization as a tree, and that this realization is optimal 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 O(n )time algorithm [6] for the computation of M Floyd s algorithm [6] 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 Set M equal to M n,m r ir + m r rj }; We shall prove that each matrix M r ( r n) isinm n Since the entries of M = M n satisfy the triangle inequality, we wl be able to conclude that M is a tree metric Theorem Let M =(m ij ) be a matrix in M n,andletm =(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 rj } for all i and j

4 6 A HERTZ AND S VARONE in {,,n} Given four distinct points i, j, k, l in {,,n}, wedenotes r = m r ij + mr kl We prove by induction that each M r (r =,,n)isinm n By hypothesis, M 0 = M is in M n, so assume M r M n It is sufficient to show that s r max{sr ik,sr } for all distinct i, j, k, l in {,,n}, orequivalently, that two of the three sums s r,sr ik and sr 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}, sayr = l Since M r M n, we may assume, without loss of generality (wlog) that s r ri sr rjik = sr ij If mr ik = mr ik and m r ij = mr ij,thens r ri s r rjik = sr ij and we are done So, we can assume wlog mr ik <mr ik It then follows that m r ri + s r rjik = mr ri + s r ij <mr ik + m r ij,which means that m r ij = mr ri + m r rj <m r ij We therefore have s r ri m r ri + m r rj + m r = s r rjik = sr ij Case B: r/ {i, j, k, l} If s r = sr,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, whe sr rjik = sr rjik and sr ri = sr case A that s r rjik = sr ri Hence, s r rjik + sr ri = sr ik and sr = sr, there is nothing to prove So Notice that if m r ik = mr ik, mr = mr,, then we are done Indeed, since M r ri,weknowfrom In a simar way, we also have sr rj = s r ri, which means that sr ri + sr rj M r M n,s r ik = sr ik,sr = sr and sr <sr ik = sr Since we conclude that s r <sr ik = sr 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 rj + m r and m r >m r rj + m r Since s r = mr + m r rj + m r >s r,weknowfromcaseathat s r = s r, which means that mr kl = m r + m r Hence, s r < s r = sr ik Case B2: m r <m r rj + m r and m r m r rj + m r We can assume m r kl = mr kl points j and k are exchanged We can also assume m r Indeed, if m r m r ri + m r, else we are in case B, where the roles of <m r ri + m r then s r = mr ri ri + + s r, sr ik = mr s r,andsr = mr ri + s r and we are done since M r M n But now, s r ik >sr, and we know from case A that sr rikl = s r ik, which means that m r kl = mr + mr Hence, s r >sr,andweknow + m r We from case A that s r = sr, which means that mr = mr rj therefore have s r <sr = sr ik

5 ANOTEONTREEREALIZATIONSOFMATRICES 65 Case B: m r m r rj + m r and m r >m r rj + m r It follows from cases B and B2 that i, j, k and l satisfy the -point condition in M r if m r ij <mr ij, m r ik <mr ik,andmr <m r rj + m r 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 -point condition in M r if m r ij <mr ij, m r <mr,andm r <m r ri + m r Sincem r ij <mr ij and m r <mr in case B, we can assume m r m r ri + m r Hence, s r sr ik = sr Case B: m r m r rj + m r and m r m r rj + m r Since M r M n,ands r ri <s r ij we know that sr rj = s r ij,which means that m r <mr Ifm r = m r rj +m r then s r sr ik = sr Else, m r <m r rj + m r, which implies s r <sr We then know from case A that s r = sr, which means that mr kl = mr + m r We therefore have s r ik <sr = sr 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,andletm =(m ij )bethen 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 metrics L =(l ij )andu =(u ij ), the matrix sandwich problem [5] is to find (if possible) a tree metric M =(m ij ) such that l ij m ij u ij for all i, j {,,n} Typically, the information concerning the distance matrix associated with a netwo may be inaccurate, and 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, andthen constructing the tree metric M =(m ij )withm 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,andletm =(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, j {,,n}, thenm is a solution to the matrix sandwich problem

6 66 A HERTZ AND S VARONE Proof Let M =(m ij ) be a matrix in M n, such that l ij m ij u ij for all i, j {,,n}, andletm =(m ij )bethen 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 metric, we have l ij m ij m ij for all i, j {,,n} References [] H-J Bandelt, Recognition of tree metrics SIAM J Algebr Discrete Methods (990) 6 [2] J-P Barthélémy and A Guénoche, Trees and proximity representations John Wey & Sons Ltd, Chichester (99) [] P Buneman, A note on metric properties of trees J Combin Theory Ser B 7 (97) 8 50 [] JC Culberson and P Rudnicki, A fast algorithm for constructing trees from distance matrices In Inf Process Lett 0 (989) [5] M Farach, S Kannan and T Warnow, A robust model for finding optimal evolutionary trees Algorithmica (995) [6] RW Floyd, Algorithm 97 Shortest path Comm ACM 5 (962) 5 [7] SL Hakimi and SS Yau, Distance matrix of a graph and its realizabity Q Appl Math 22 (96) 05 7 [8] AN Patrinos and SL Hakimi, The distance matrix of a graph and its tree realization Q Appl Math 0 (972) [9] JMS Simões-Pereira, A note on the tree realizabity of a distance matrix J Combin Theory 6 (969) 0 0 [0] SC Varone, Trees related to realizations of distance matrices Discrete Math 92 (998) 7 6 To access this journal online: wwwedpsciencesorg