Mixtures of star trees and deficiency graphs

European Journal of Combinatorics 44 (2015) 140–143

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Mixtures of star trees and deficiency graphs Yaroslav Shitov National Research University Higher School of Economics, 20 Myasnitskaya Ulitsa, Moscow 101000, Russia

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Article history: Received 23 June 2014 Accepted 29 September 2014

The star tree rank of a matrix A is the smallest k for which A is an entrywise minimum of k distance matrices of weighted trees without internal edges. The deficiency graph of A has an edge connecting (i, j) and (k, l) if and only if Aij + Akl < Ail + Akj . We answer the question asked by Cartwright and Chan: There is a pair of matrices which have different star tree ranks and share the same deficiency graph. The same result is established for another rank function known as symmetric Barvinok rank. © 2014 Elsevier Ltd. All rights reserved.

1. Introduction The tropical arithmetic operations on R are defined by a ⊕ b = min{a, b} and a ⊙ b = a + b. The operations with tropical matrices are defined in the same way as with matrices over a field but with classical addition and multiplication replaced by the tropical operations ⊕ and ⊙. A symmetric tropical matrix A has symmetric Barvinok rank k if A = B ⊙ B⊤ , for some matrix B ∈ Rn×k , and k is a minimal integer satisfying this property. Now consider a weighted star tree S on {1, . . . , n}, that is, a weighted tree which has leaves 1, . . . , n and no internal edges. Assume that, for every i and j, the (i, j)th entry of a matrix B equals the weight of a path connecting i and j in S; in this case, B is called a matrix of star tree rank one. The star tree rank of an arbitrary symmetric matrix A with zeros on diagonal is the smallest k such that A is a mixture (that is, a tropical sum) of k matrices with star tree rank one. As we can see from definitions, there are tropical matrices on which these rank functions are not defined. For instance, the second of these definitions requires a matrix to have a zero diagonal, so that the star tree rank actually depends only on off-diagonal entries. Also, we can see that a matrix of symmetric Barvinok rank one turns to a matrix of star tree rank one if we replace its diagonal entries

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.ejc.2014.09.010 0195-6698/© 2014 Elsevier Ltd. All rights reserved.

Y. Shitov / European Journal of Combinatorics 44 (2015) 140–143

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by zeros. In other words, we can say that the difference between symmetric Barvinok rank and star tree rank is only in the treatment of the diagonal. 2. Geometric characterization and deficiency graphs Cartwright and Chan [1] give a more general pattern of constructing tropical rank functions. The tropical hypersurface defined by a tropical polynomial t

c

p(x1 , . . . , xm ) = min ci ⊙ x1i1 ⊙ · · · ⊙ xcmim



i=1



(2.1)

is the set of all x ∈ Rm such that the minimum in (2.1) is achieved at least twice. Let F be a set of tropical polynomials; the tropical prevariety V (F ) is the intersection of hypersurfaces defined by polynomials from F . The rank of a tuple a ∈ Rm with respect to F is the smallest k such that a is a tropical sum of k vectors from V (F ). Now let us follow [1] to define the deficiency hypergraph of a with respect to F . Assume that the exponents cij in every tropical polynomial p ∈ F are nonnegative. If the minimum is achieved uniquely when p is evaluated at a ∈ Rm , then we add a hyperedge E whose elements correspond to the coordinates that appear with non-zero exponent in the unique minimal term. The deficiency hypergraph of a consists of hyperedges coming from all polynomials in F with a unique minimum at a. Theorem 2.1 ([1]). The rank of a tuple a ∈ Rm with respect to a set F of tropical polynomials is greater than or equal to the chromatic number of the corresponding deficiency hypergraph. Now consider the set of variables xij with i, j ∈ {1, . . . , n} and xuv , xv u identified. It is shown in [1] that the symmetric Barvinok rank arises as the rank with respect to polynomials xpq ⊙ xrs ⊕ xps ⊙ xrq . Restricting the set of variables xij to those with i ̸= j, one can think of star tree rank as the rank with respect to xpq ⊙ xrs ⊕ xps ⊙ xrq with pairwise different p, q, r, s. Note that these polynomials have degree two, so the deficiency graphs of symmetric Barvinok and star tree ranks are indeed graphs. The following question arises naturally from Theorem 2.1. Question 2.2 ([1, Section 9]). Is it true that the rank of a matrix, according to any of our notions, is always equal to the chromatic number of the corresponding deficiency graph? A further investigation on the symmetric Barvinok rank has been undertaken in [3]. It has been shown that matrices allowing the presentation B ⊙ B⊤ arise as the tropicalization of a cone of positive semidefinite matrices over an ordered field. The paper [2] is devoted to studying the mixtures of star tree metrics. A full characterization of matrices with star tree metric rank two has been given. However, Question 2.2 remained open for both the symmetric Barvinok rank and star tree rank. Our paper aims to give a negative answer to the above question. It turns out that, in fact, not only the chromatic number but the deficiency graph itself is not sufficient to determine the rank of a matrix. We construct an explicit example of a pair of 16 × 16 matrices which, with respect to any of our rank concepts, have different ranks and share the same deficiency graph. 3. The result The key example is the matrix 0 0 2  4 A(α) =  2  4  0 0



0 0 4 4 4 4 0 0

2 4 0 0 2 4 0 0

4 4 0 0 4 4 0 0

2 4 2 4 0 0

α 4

4 4 4 4 0 0 4 4

0 0 0 0

α 4 0 0

0 0 0  0 , 4  4  0 0



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Y. Shitov / European Journal of Combinatorics 44 (2015) 140–143

which, as we will see from further considerations, gives a negative answer for Question 2.2 in the case of symmetric Barvinok  rank. To treat  the case of star tree rank, it will be convenient to consider the block matrix B (α) =

A(α) A(α)

A(α) A(α)

. The first eight rows and columns of B (α) will be indexed by

1 , . . . , 8 , and the last eight rows and columns by 1− , . . . , 8− . Since [B (α)]iσ jτ = [A(α)]ij for any σ , τ ∈ {+, −}, we can make the following observation. +

+

Observation 3.1. Let G(α) and G′ (α) be deficiency graphs of A(α) and B (α) with respect to polynomials xpq ⊙ xrs ⊕ xps ⊙ xrq . Then, G′ (α) has an edge between (iσ , jτ ) and (kπ , lυ ) if and only if G(α) contains an edge connecting (i, j) with (k, l). Proposition 3.2. The star tree rank and symmetric Barvinok rank of B (1) do not exceed 4. Proof. To prove the result, it suffices to express B (1) as a tropical sum of four matrices from prevariety generated by polynomials xpq ⊙ xrs ⊕ xps ⊙ xrq . (Here, p, q, r , s are arbitrary indices, not necessarily distinct.) Namely, one can check that B (1) = v1⊤ ⊙v1 ⊕v2⊤ ⊙v2 ⊕v3⊤ ⊙v3 ⊕v4⊤ ⊙v4 , where vi = (ui ui ) and u1 = (0 0 4 4 4 4 0 0), u2 = (4 4 0 0 4 4 0 0), u3 = (4 4 4 4 0 0 4 4), u4 = (1 4 1 4 1 4 0 4).  The computer experiment, performed by the reviewer, shows that the deficiency graphs have chromatic number 4, for both A and B and both the star tree rank and symmetric Barvinok rank. Therefore, we can actually change ‘do not exceed’ to ‘equal’ in the above proposition. Proposition 3.3. Assume α < 1. Then, the star tree rank and symmetric Barvinok rank of B (α) are greater than or equal to 5. Proof. Assume the converse. Then, there are four 8 × 8 matrices C (1) , C (2) , C (3) , C (4) which satisfy, for any t and arbitrary (not necessarily different) i, j, k, l, the conditions (t )

(t )

(t )

(t )

Cij + Ckl = Cil + Ckj

(1)

and (2)

(3.1) (3)

(4)

[B (α)]i+,j− = min{Cij , Cij , Cij , Cij }.

(3.2)

(In the rest of the proof, we use the notation Aij instead of [B (α)]i+,j− , for the sake of simplicity.) (τ )

Denote by τ (i, j) an arbitrary index τ satisfying Aij = Cij . Step 1. Assume that the condition τ (2p, 2p) = τ (2q, 2q) = t holds for distinct p, q ∈ {1, 2, 3}. (t ) (t ) (t ) (t ) Then, we have C2p,2p = A2p,2p and C2q,2q = A2q,2q . The definition of A implies C2p,2p = C2q,2q = 0. (t )

(t )

From (3.1) we deduce C2p,2q + C2q,2p = 0; from (3.2) it follows that A2p,2q + A2q,2p ≤ 0, which is a contradiction as A2p,2q = A2q,2p = 4. Therefore, indices τ (2, 2), τ (4, 4), τ (6, 6) are pairwise different; we assume τ (2p, 2p) = p up to renumbering the matrices C (τ ) . Step 2. Let (u, v) = (r , s), where r and s either are distinct elements from {1, 3, 5} or satisfy {r , s} = {5, 7}. Assume that the condition τ (u, v) = p holds for some p ∈ {1, 2, 3}. Then, we have (p) (p) (p) (p) C2p,2p = A2p,2p and Cuv = Auv . The definition of A implies C2p,2p = 0 and Cuv < 4. From (3.1) we (p)

(p)

deduce C2p,v + Cu,2p < 4; from (3.2) it follows that A2p,v + Au,2p < 4. This is a contradiction: by the construction of A, the numbers A2p,v and Au,2p are nonnegative and one of them equals 4. Now we see that τ (u, v) = 4. (4)

Step 3. From Step 2 it follows that Cuv = Auv . In particular, for any distinct r , s ∈ {1, 3, 5}, we have = 2. Equality (3.1) implies that Cr(,4r) = Cr(,4s) + Cs(,4r) − Cs(,4s) = 4 − Cs(,4s) . So we conclude that Cr(,4r) = 2.

(4) Cr ,s

(4)

(4)

Step 4. By Step 2, we have τ (5, 7) = τ (7, 5) = 4, so that C5,7 = C7,5 = α . By Step 3, we have = 2. Equality (3.1) implies C7(4,7) = C5(4,7) + C7(4,5) − C5(4,5) , so that C7(4,7) = 2α − 2 < 0. From (3.2) it follows that A7,7 < 0, which is a contradiction.  (4) C5,5

The following proposition completes the proof of the main result.

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Proposition 3.4. Let F (respectively, F ′ ) be the set of polynomials xpq ⊙ xrs ⊕ xps ⊙ xrq , where p, q, r, s are arbitrary (respectively, pairwise different). Then, for any ε ∈ (0, 1), the matrices B (ε) and B (1) have the same deficiency graph with respect to F and the same deficiency graph with respect to F ′ . Proof. By definition, the deficiency graph with respect to F ′ can be obtained from the deficiency graph with respect to F by removing the edges containing repeating indices. So it suffices to consider the set F only. By Observation 3.1, it suffices to prove that A(1) and A(ε) share the same deficiency graph. We need to check that the numbers H0 = [A(1)]ij + [A(1)]kl − [A(1)]il − [A(1)]jk and Hε = [A(ε)]ij + [A(ε)]kl − [A(ε)]il − [A(ε)]jk have the same sign, for any indices i, j, k, l. If i = k or j = l, then H0 = Hε = 0; moreover, we have H0 = Hε if {5, 7} is not a subset of {i, j, k, l} because [A(1)]rs = [A(ε)]rs unless {r , s} = {5, 7}. So we can assume that k = 5, l = 7, i ̸= 5, j ̸= 7; in this case, H0 is an odd (and, therefore, non-zero) integer. It remains to note that |Hε − H0 | ≤ 1 − ε to complete the proof.  From Propositions 3.2 and 3.3 it follows that B (ε) and B (1) have different symmetric Barvinok ranks and different star tree ranks. Proposition 3.4 shows that B (ε) and B (1) share the same deficiency graphs with respect to both of these rank functions. In particular, we get a negative answer for Question 2.2. Acknowledgments I would like to thank the anonymous reviewers for their detailed comments. References [1] D. Cartwright, M. Chan, Three notions of tropical rank for symmetric matrices, Combinatorica 32 (1) (2012) 55–84. [2] M.A. Cueto, Tropical mixtures of star tree metrics, Ann. Comb. 16 (2) (2012) 233–251. [3] J. Yu, Tropicalizing the positive semidefinite cone, Proc. Amer. Math. Soc. (2013) in press. arXiv:1309.6011.