A characterization of linearizable instances of the quadratic minimum spanning tree problem
AA characterization of linearizable instances of thequadratic minimum spanning tree problem
Ante ´Custi´c ∗ Abraham P. Punnen † October 11, 2018
Abstract
We investigate special cases of the quadratic minimum spanning treeproblem (QMSTP) on a graph G = ( V, E ) that can be solved as a linearminimum spanning tree problem. Characterization of such problems ongraphs with special properties are given. This include complete graphs,complete bipartite graphs, cactuses among others. Our characterizationcan be verified in O ( | E | ) time. In the case of complete graphs and whenthe cost matrix is given in factored form, we show that our characterizationcan be verified in O ( | E | ) time. Related open problems are also indicated. Keywords:
Minimum spanning tree, quadratic 0-1 problems, quadraticminimum spanning tree, polynomially solvable cases, linearization.
The minimum spanning tree problem (MSTP) is well studied in the combi-natorial optimization literature. A generalization of this problem, called the quadratic minimum spanning tree problem (QMSTP), recently received consid-erable attention from the research community. Some of these papers focus onexact algorithms [2, 23] while the majority of published works deal with heuristicalgorithms [5, 9, 16, 20, 21, 27, 31]. Isolated results on some theoretical prop-erties of the problem are also available. Special cases of QMSTP studied in theliterature include multiplicative objective functions [11, 14, 19], spanning treeswith conflict constraints [7, 30], and spanning tree problems with one quadraticterm [3, 8]. Some polynomially solvable special cases of QMSTP are discussedin [29] along with various complexity results.Let G = ( V, E ) be a graph such that | V | = n and E = { , , . . . , m } . Foreach ( e, f ) ∈ E × E a cost q ( e, f ) is prescribed. Let F be the family of all ∗ [email protected] . Department of Mathematics, Simon Fraser University Surrey, 250-13450102nd AV, Surrey, British Columbia, V3T 0A3, Canada † [email protected] . Department of Mathematics, Simon Fraser University Surrey, 250-13450102nd AV, Surrey, British Columbia, V3T 0A3, Canada a r X i v : . [ m a t h . O C ] O c t panning trees of G and Q be the m × m matrix with its ( i, j )-th element as q ( i, j ). For each T ∈ F its cost Q ( T ) is given by Q ( T ) = (cid:88) e ∈ T (cid:88) f ∈ T q ( e, f ) . Note that the notation e ∈ T means e belongs to the edge set of T . Thenthe QMSTP is to find a spanning tree T in F such that Q ( T ) is as small aspossible. Similarly, for each T ∈ F , let C ( T ) = (cid:80) e ∈ T c ( e ), where c ( e ) is aprescribed cost of edge e ∈ E . Given a cost matrix Q , the quadratic spanningtree linearization problem (QST-LP) is to determine if there exists a linear costvector C = ( c (1) , c (2) , . . . , c ( m )) such that Q ( T ) = C ( T ) for all T ∈ F . If theanswer to this decision problem is ‘yes’, the quadratic cost matrix Q is said tobe linearizable and C is called a linearization of Q . Note that |F| could be aslarge as n n − and hence QST-LP is a non-trivial problem. In fact, there is noimmediate direct way to test if QST-LP belongs to NP.The linearization problem for the quadratic assignment problem (QAP) wasconsidered by Kabadi and Punnen [13], Adams and Wadell [1], and C¸ ela etal. [4]. The special case of Koopmans-Beckman QAP linearization problemwas studied by Punnen and Kabadi [24] and C¸ ela et al. [4]. In this paper, weprovide a characterization of linearizable instances of QMSTP on a wide classof graphs, including the complete graph, complete bipartite graph, cactus etc.Our characterization can be tested in O ( m ) time. Also an O ( n ) algorithm forrecognizing an n × n sum matrix represented in factored form is given. In thecase of complete (bipartite) graphs, this leads to an O ( m ) algorithm to test ifsymmetric matrix Q is linearizable when represented in factored form. As abyproduct of these results, we have new polynomially solvable special cases ofthe QMSTP. QMSTP is well known to be strongly NP-hard. In fact, it is NP-hard even if Q is of rank one [25]. A special case of the rank one QMSTP is called the multiplicative minimum spanning tree problem (MMSTP) considered by variousauthors [11, 14, 15, 19, 20, 25, 26, 28]. The MMSTP on the graph G can bestated as Minimize (cid:32)(cid:88) e ∈ T d e + δ (cid:33) (cid:32)(cid:88) e ∈ T d e + δ (cid:33) Subject to T ∈ F , where d e , d e are two prescribed weights of the edge e ∈ E and δ , δ are con-stants. If (cid:0)(cid:80) e ∈ T d e + δ (cid:1) > (cid:0)(cid:80) e ∈ T d e + δ (cid:1) > T ∈ F , MMSTPcan be solved in polynomial time using the parametric minimum spanning treeproblem [15, 20, 25, 26, 28]. If d e , d e are allowed to take any real values, then2MSTP is known to be NP-hard [25]. We now observe that MMSTP is NP-hardeven if d e , d e ≥ δ and δ are arbitrary. Theorem 1.
The MMSTP is NP-hard even if d e , d e ≥ for all e ∈ E .Proof. We reduce the subset sum problem to the MMSTP. The subset sumproblem can be stated as follows. Given non-negative numbers a , a , . . . , a n and a constant K , determine if there exists a subset S of { , , . . . , n } suchthat (cid:80) i ∈ S a i = K . From an instance of subset sum problem, we construct aninstance of MMSTP as follows. For each i = 1 , , . . . , n create a 3-cycle on nodes v i , v i , v i . Link these 3-cycles using the path v − v − · · · − v n . For the edge e i = ( v i , v i ) assign cost d e i = a i , i = 1 , , . . . , n . For any other edge e , weset d e = 0. Choose the vector d = d and δ = δ = − K . It can be verifiedthat the resulting MMSTP has an optimal spanning tree with objective functionvalue zero if and only if the subset sum problem has a solution. The result nowfollows from the NP-completeness of the subset sum problem.We continue this section by presenting some useful basic facts about theQMSTP.Let M m × m be the vector space of all real valued m × m matrices. The set oflinearizable quadratic cost matrices for QMSTP on a given graph with m edgesforms a subspace of M m × m . As a consequence we have the following. Observation 2.
Let Q and Q be two cost matrices for the QMSTP on agraph G . If Q and Q are linearizable, then αQ + βQ is also linearizable forany scalars α and β . Furthermore, if C is a linearization of Q and C is alinearization of Q , then αC + βC is a linearization of αQ + βQ . A square matrix A is said to be a skew-symmetric matrix if A T = − A . Observation 3. If Q is a cost matrix for the QMSTP on a graph G , A is anyskew-symmetric matrix and D is a diagonal matrix, all of the same size, then Q is linearizable if and only if Q + A + D is linearizable. It may be noted that if Q is skew-symmetric then Q ( T ) = 0 for any spanningtree T . Thus a skew-symmetric matrix is linearizable for any graph G . Observation 4. If Q is a cost matrix for the QMSTP on a graph G . Then Q is linearizable if and only if ( Q + Q T ) is linearizable. Furthermore, C is alinearization of Q if and only if C is a linearization of ( Q + Q T ) . Proof.
Note that Q = ( Q − Q T )+ ( Q + Q T ). As ( Q − Q T ) is skew-symmetric,the result follows from Observation 2, Observation 3 and the fact that the null-vector is a linearization of a skew-symmetric matrix.As noted earlier, any skew-symmetric matrix is linearizable regardless thestructure of the underlaying graph. We now observe that if the underlying graphis a cycle, then the resulting QMSTP is linearizable regardless the structure ofthe cost matrix Q . 3 heorem 5. QMSTP on a cycle is linearizable for any cost matrix Q . Further,the linearization C = ( c (1) , c (2) , . . . , c ( n )) is given by c ( e ) = q ( e, e ) + (cid:88) i ∈ E \{ e } ( q ( i, e ) + q ( e, i )) − (cid:80) i ∈ E (cid:80) j ∈ E,j (cid:54) = i q ( i, j ) n − . (1) Proof.
Let G be a cycle with edges e , e , . . . , e n . Then the spanning trees of G are precisely T , T , . . . , T n where T i = G \ { e i } . We need to find a vector C = ( c (1) , c (2) , . . . , c ( n )) such that (cid:88) e ∈ T i c ( e ) = Q ( T i )for all i = 1 , . . . , n . Equivalently, we want to find a solution to the linear systemabove, where the variables being c (1) , c (2) , . . . , c ( n ). It can be verified that thecoefficient matrix is invertible and hence the system has a unique solution. Theformula for the linearization can be verified by simple algebra.The following is an immediate corollary of Theorem 5. Corollary 6.
The QMSTP is linearizable for any cost matrix Q on the graph T ∪ { e } where T is a tree and e is an edge (not necessarily in T ) joining twovertices of T . Note that the result of Theorem 5 can be extended to any real valued objec-tive function for a spanning tree, not simply the quadratic objective function.We conclude this section with few more definitions and observations that wemake use later in this paper.
Definition 7. An n × n matrix H = ( h ( i, j )) is called a sum matrix if thereexist vectors a = ( a (1) , a (2) , . . . , a ( n )) and b = ( b (1) , b (2) , . . . , b ( n )) such that h ( i, j ) = a ( i )+ b ( j ) for all i = 1 , . . . , n , j = 1 , . . . , n . A square matrix is calleda weak sum matrix if the relation above is not mandatory for the elements onthe diagonal. Note that if an n × n square sum matrix H = ( h ( i, j )) is symmetric, then h ( i, j ) = a ( i )+ a ( j ) for all i, j = 1 , , . . . , n , for some vector a = ( a (1) , . . . , a ( n )).Similarly, if an n × n square weak sum matrix H = ( h ( i, j )) is symmetric,then h ( i, j ) = a ( i ) + a ( j ) for all i, j = 1 , , . . . , n, i (cid:54) = j , for some vector a =( a (1) , . . . , a ( n )). Definition 8.
A maximal biconnected subgraph of a simple graph G is called a biconnected component of G . The following fact is straightforward to prove, for example see [6, Ch. 5, p.101].
Proposition 9.
An instance of the MSTP on a graph G has the property thatevery spanning tree has the same cost, if and only if all edges from the samebiconnected component of G have the same cost. ( Q + Q T ) is a symmetric matrix. Thus in view of Observa-tion 4 hereafter we assume without loss of generality that the cost matrix Q issymmetric. In this section we investigate characterizations of linearizable QMSTP instanceson various graph classes. Our findings are summarized in Theorem 17. We beginwith a sufficient condition.
Lemma 10.
Let Q be a symmetric cost matrix of the QMSTP on a graph G = ( V, E ) such that for every pair I , J of biconnected components of G , thesubmatrix of Q defined by rows I and columns J is a sum matrix if I (cid:54) = J , or asymmetric weak sum matrix if I = J . Then Q is linearizable and a linearizationof Q can be computed in O ( | E | ) time.Proof. Let Q be a symmetric matrix that satisfies the hypothesis of the lemma.Then Q can be expressed as Q = A + A T + D, (2)where D = ( d ( i, j )) is a diagonal matrix, and matrix A = ( a ( i, j )) has theproperty that a ( i, j ) = a ( i, k ) if j and k are edges from the same biconnectedcomponent. Note that matrices A and D can be found in O ( | E | ) time. FromProposition 9 it follows that an MSTP instance defined by any row of A (i.e.for some fixed row i we define the length of an edge j to be a ( i, j )) has theproperty that every spanning tree has the same cost. Let r ( i ) denotes theconstant objective function value of the MSTP corresponding to the i -th row of A . Then the objective value of the QMSTP for some spanning tree T is Q ( T ) = (cid:88) e ∈ T (cid:88) f ∈ T ( a ( e, f ) + a ( f, e ) + d ( e, f ))= (cid:88) e ∈ T (cid:88) f ∈ T ( a ( e, f ) + a ( f, e )) + (cid:88) e ∈ T d ( e, e )= (cid:88) e ∈ T ( r ( e ) + r ( e )) + (cid:88) e ∈ T d ( e, e )= (cid:88) e ∈ T (2 r ( e ) + d ( e, e )) . Hence, by setting c ( i ) := 2 r ( i ) + d ( i, i ) (3)we obtain a linearization of Q . Note that the choice of matrix A is not alwaysuniquely determined, hence a linearization is not necessarily unique.Next, we present characterizations of linearizable instances of QMSTP forsome special types of graphs. We start with the complete graph. In this case,5he linearization characterization shows that Q must be a weak sum matrix,which, according to Lemma 10, is the most restrictive possible characterization. Theorem 11.
A symmetric cost matrix Q of the QMSTP on a complete graph K n is linearizable if and only if it is a symmetric weak sum matrix. Further, alinearization of a linearizable symmetric matrix Q is given by (3) .Proof. If Q is a weak sum matrix, then from Lemma 10 it follows that Q islinearizable and a linearization is given by (3). Note that in the case of acomplete graph K n , entries of the matrix A in the expression (2) are the samefor every fixed row. Hence, r ( i ) = ( n − a ( i, j ) for any column j .Next we assume that Q is linearizable. For n ≤ n ≥ (cid:0) n (cid:1) × (cid:0) n (cid:1) sum matrix M = ( m ( i, j )) of the form m ( i, j ) = a ( i ) + a ( j ), where a (1) = 0 and a ( i ) = q ( i,
1) for i = 2 , . . . , (cid:0) n (cid:1) . By subtracting M and an appropriate diagonal matrix from Q , we could obtain zeros on thefirst row, the first column and the diagonal. By Lemma 10 M is linearizable, andfurthermore, any diagonal matrix is linearizable. Hence, from Observation 2 itfollows that without loss of generality we can assume that elements of the firstrow, the first columns and the diagonal of Q are equal to zero. In that case, Q is a weak sum matrix if and only if all elements of Q that are not in the firstrow, the first column or on the diagonal, have the same value.Now we assume the contrary, i.e. there are two elements of Q (not in thefirst row/column or on the diagonal) that have different values. Moreover, dueto the symmetry of Q , there is a row b that contains such two distinct valueelements q ( b, x ) and q ( b, y ). As any element of row b (except q ( b,
1) and q ( b, b ))can be a member of such pair, without loss of generality we can assume thatedges 1 and x are nonadjacent.Next show that there exists a cycle C that contains edges 1 and b , and a cycle C that contains edges x and y with the following property: C ∪ C \ { e, f } does not contains a cycle for all e ∈ { , b } , f ∈ { x, y } . Namely, in the casewhen there are no two pairs of edges from { , b } × { x, y } that are adjacent,it is straightforward to construct C and C that are edge disjoint and satisfythe above property, see Figure 1(a). In the case when there are at least two b y x (a) xy b (b) xy b (c) y xb (d) Figure 1: Configurations of { , b, x, y } and corresponding C ∪ C pairs of edges from { , b } × { x, y } that are adjacent, every possible 1 , b, x, y { , b } and { x, y } , and by exchanging elements inside of those two sets), andedge contractions (that can induce more incidences and only make the case morecomplicated). These configurations in Figure 1(b) and Figure 1(c) are extendedwith a (dashed) edge that constitutes feasible C and C . Note that we usedthe fact that 1 and x are nonadjacent, otherwise there are instance for which C and C with the property above do not exist, see Figure 1(d).Let T be a minimum cardinality set of edges of a tree connected to both C and C that spans the remaining vertices. Then we define B to be T ∪ C ∪ C \{ , b, x, y } . It is easy to see that B extended by any two edges e ∈ { , b } and f ∈ { x, y } forms a spanning tree.Let C = ( c ( i )) be a cost vector that linearizes Q . Since both B ∪ { b, x } and B ∪ { , x } form a spanning tree, we have that C ( B ∪ { b, x } ) − C ( B ∪ { , x } ) = (cid:88) e ∈ B ∪{ b,x } c ( e ) − (cid:88) e ∈ B ∪{ ,x } c ( e ) = c ( b ) − c (1) . Analogously, C ( B ∪ { b, y } ) − C ( B ∪ { , y } ) = c ( b ) − c (1), hence Q ( B ∪ { b, x } ) − Q ( B ∪ { , x } ) = Q ( B ∪ { b, y } ) − Q ( B ∪ { , y } ) . (4)Now let us express the cost of the spanning tree B ∪ { b, x } in terms of thequadratic cost matrix Q . Since q ( e, e ) = 0 ∀ e, we have Q ( B ∪ { b, x } ) = (cid:88) e ∈ B ∪{ b,x } (cid:88) f ∈ B ∪{ b,x } q ( e, f )= (cid:88) e ∈ B (cid:88) f ∈ B q ( e, f ) + (cid:88) e ∈ B q ( b, e ) + (cid:88) e ∈ B q ( x, e ) + 2 q ( b, x ) . Since q (1 , e ) = 0 ∀ e we analogously have Q ( B ∪ { , x } ) = (cid:88) e ∈ B (cid:88) f ∈ B q ( e, f ) + (cid:88) e ∈ B q ( x, e ) . Therefore Q ( B ∪ { b, x } ) − Q ( B ∪ { , x } ) = (cid:88) e ∈ B q ( b, e ) + 2 q ( b, x ) . (5)Analogously Q ( B ∪ { b, y } ) − Q ( B ∪ { , y } ) = (cid:88) e ∈ B q ( b, e ) + 2 q ( b, y ) . (6)Then from (4), (5) and (6) follows that q ( b, x ) = q ( b, y ) which is a contradictionto our choice of b, x and y . 7ext we will generalize the approach above to obtain a characterization oflinearizable cost matrices for QMSTP for a more general class of graphs (seeTheorem 17). First we present a tool which is used to prove such characteriza-tions. Definition 12.
Let a, b, x and y be distinct edges of a simple graph G with n vertices. We say that a set B of n − edges is an { a, b } - { x, y } -backbone of G ifadding any two edges e ∈ { a, b } and f ∈ { x, y } to B generates a spanning treeof G . Lemma 13.
Let Q be a linearizable symmetric cost matrix of the QMSTP ona simple graph G , and let a and x be two fixed distinct edges of G . If for alladditional edges b and y there exist a sequence of k ≥ edges z , z , . . . , z k suchthat x = z , y = z k and there exists an { a, b } - { z i , z i +1 } -backbone B i for every i = 1 , . . . , k − , then Q is a symmetric weak sum matrix.Proof. Let Q be linearizable and let a, b, x, y be four distinct edges such thatthere exists { a, b } - { x, y } -backbone B . Since Q is linearizable it follows that Q ( B ∪ { b, x } ) − Q ( B ∪ { a, x } ) = Q ( B ∪ { b, y } ) − Q ( B ∪ { a, y } ) . (7)Namely, by expressing spanning tree objective values from (7) with a lineariza-tion costs C = ( c ( i )), one gets c ( b ) − c ( a ) = c ( b ) − c ( a ). However, by expressingspanning tree objective vales from (7) with quadratic costs Q = ( q ( i, j )), onegets q ( b, x ) − q ( a, x ) = q ( b, y ) − q ( a, y ) . Now assume that a and x are fixed and there exist edges b, y and z , . . . , z k with z = x, z k = y , such that there exists { a, b } - { z i , z i +1 } -backbone B i forevery i = 1 , . . . , k −
1. Then by the same reasoning as above for all i = 1 , . . . , k −
1, we obtain the following system of equations: q ( b, x ) − q ( a, x ) = q ( b, z ) − q ( a, z ) ,q ( b, z ) − q ( a, z ) = q ( b, z ) − q ( a, z ) , ... q ( b, z k − ) − q ( a, z k − ) = q ( b, y ) − q ( a, y ) . As the right-hand side of every i -th equation is identical to the left-hand sideof ( i + 1)-th equation, it follows that q ( b, x ) − q ( a, x ) = q ( b, y ) − q ( a, y ), whichcan be rearranged to q ( b, y ) = q ( b, x ) + q ( a, y ) − q ( a, x ) . (8)Note that (8) is satisfied also for b = a or y = x .By the assumption of the lemma, we can obtain (8) for all b and y , thereforeit follows that q ( b, y ) is a sum of a function of b and a function of y (as a and x are fixed), i.e. q ( i, j ) = s ( i ) + t ( j ) ∀ i (cid:54) = j, s = ( s ( i )) and t = ( t ( i )). As Q is symmetric it follows that q ( i, j ) = w ( i ) + w ( j ) ∀ i (cid:54) = j, for some vector w = ( w ( i )), which proves the lemma.Note that Theorem 11 can be proven using Lemma 13 and the fact that if a and x are two nonadjacent edges of a complete graph, then for any other pairof edges b and y there exists an { a, b } - { x, y } -backbone. Corollary 14.
Let Q be a linearizable symmetric cost matrix of the QMSTPon a simple graph G . Let I and J be two disjoint sets of edges of G , and let a ∈ I and x ∈ J be two fixed edges. Let Q IJ be the submatrix of Q defined byrows I and columns J . If for all additional edges b ∈ I and y ∈ J there exist asequence of k ≥ edges z , z , . . . , z k such that x = z , y = z k and there existsan { a, b } - { z i , z i +1 } -backbone B i for every i = 1 , . . . , k − , then Q IJ is a summatrix.Proof. The proof is similar as that of Lemma 13.In the majority of cases we use Lemma 13 and Corollary 14, k will be equalto 2, i.e. we will not need additional edges z i .Given the edges a , b , x and y , usually we try to build an { a, b } - { x, y } -backbone in the following way. We aim to find a cycle C that contains a and b and a cycle C that contains x and y , such that if intersection of C and C is nonempty, then it is connected and does not contains a pair of edges from { a, b } × { x, y } . We call such C and C as feasible backbone cycles for a, b, x ,and y . For feasible backbone cycles C and C , ( C \ { a, b } ) ∪ ( C \ { x, y } )extended by a tree which is connected to C and C and spans the remainingset of vertices, forms an { a, b } - { x, y } -backbone. Lemma 15.
A symmetric cost matrix Q of the QMSTP on a complete bipartitegraph K n ,n with min { n , n } ≥ , is linearizable if and only if Q is a sym-metric weak sum matrix. A linearization of a lineariable symmetric matrix Q is given by (3) .Proof. If Q is a weak sum matrix, then from Lemma 10 it follows that Q islinearizable and a linearization is given by (3). Note that in the case of thecomplete bipartite graph K n ,n , entries of the matrix A in the expression (2)are the same for every fixed row. Hence, r ( i ) = ( n + n − a ( i, j ) for anycolumn j .Let min { n , n } ≥ Q is linearizable. We fix two arbitrarynonadjacent edges a and x . We will show that for any two additional edges b and y , ( b (cid:54) = y ) conditions of Lemma 13 are satisfied, which completes the proof.In Figure 2 all possible essentially different configurations of incidences be-tween a , b , x and y , up to symmetries, are presented. (The symmetries aredefined by exchanging sets { a, b } and { x, y } , and by exchanging elements in-side of those two sets.) Configurations in Figure 2 are of two types. In the9 b yx (a) ab yx (b) a byx (c) ay bx (d) ay bxz (e) ayx b (f) axb y (g) aybxz (h) a byx (i) a by xz (j) a byx (k) a ybx (l) ax by (m) Figure 2: The a, b, x, y configurationscases where we apply Lemma 13 with k = 2, the configurations are extended by(dashed) edge(s) that form feasible backbone cycles. In other cases we use oneauxiliary edge of Lemma 13 ( k = 3), therefore configurations are extended bythe edge z which plays the role of z in Lemma 13.In the previous lemma, linearization characterization only for min { n , n } ≥ { n , n } ≥
3. If min { n , n } < Q is not necessary a weak sum matrix. Namely, if n or n equals to 1, K n ,n is a tree, and if n = n = 2, K n ,n is a cycle. Inboth cases arbitrary Q is linearizable. For the remaining case of n = 2 and n ≥
3, we present the following counterexample of a symmetric matrix Q thatis linearizable but not a weak sum matrix. For i (cid:54) = j , cost element q ( i, j ) is equalto 1 if edges i and j are adjacent through an n -set vertex, and 0 otherwise.Then the linearization costs are given by c ( i ) = q ( i, i ) + 2 / ( n + 1). Lemma 16.
Let Q be a linearizable symmetric cost matrix of the QMSTP ona graph G . Then for every two distinct biconnected components I , J of G ,submatrix of costs q ( i, j ) , i ∈ I , j ∈ J is a sum matrix.Proof. If I or J is just one edge, i.e. a bridge, then there is nothing to prove,as every 1 × n matrix is a sum matrix. In the rest of the proof we assume thatmin {| I | , | J |} ≥ a ∈ I and x ∈ J . It is easy to see that for every pair of additional edges b ∈ I and y ∈ J there exist an { a, b } - { x, y } -backbone. Namely, in every biconnectedcomponent, there exist a cycle that contains any pair of edges. Hence, thereexist a cycle in I that contains a and b , and a cycle in J that contains x and y . As their intersection contains at most one vertex, they are feasible backbonecycles. Hence, by Corollary 14, the lemma follows.10emma 10, 15, 16 and Theorem 5 and 11 can be combined to produce thelinearization characterization for more general class of graphs. Theorem 17.
Let G be a graph such that its every biconnected component iseither a clique, a cycle or a biclique (with vertex partition sets of sizes at leastthree). Then a symmetric cost matrix Q of the QMSTP on G is linearizable ifand only if the submatrices of Q that correspond to different biconnected compo-nents are sum matrices, and submatrices that correspond to single biconnectedcomponents that are either a clique or a biclique are symmetric weak sum ma-trices. Furthermore, if Q is linearizable, a linearization can be computed in O ( | E | ) time.Proof. Let Q be of the form described in the theorem. We denote by k the num-ber of biconnected components of G that are cycles. Then Q can be expressedas Q = M + B + · · · + B k , where M satisfies the hypothesis of Lemma 10,and B i is a matrix in which all entries that are not in the submatrix defined bythe i -th cycle, are equal 0. Note that matrices M and B i , i = 1 , . . . , k , can befound in O ( | E | ) time. From Lemma 10 it follows that M is linearizable and itslinearization vector C M can be computed by (3). From Theorem 5, it followsthat for every i = 1 , . . . , k , B i is linearizable and its linearization C B i is givenby (1). Therefore by Observation 2, Q is also linearizable and its linearizationvector is given by C = C M + C B + · · · + C B k .Conversely, if Q is linearizable then it has to be of the form described in thetheorem. This follows directly from Lemma 16 and the proofs of Theorem 11and Lemma 15. Namely, backbones of biconnected components can be extendedinto backbones of G by adding edges that span remaining vertices.We present an example that illustrates Theorem 17. Let G = ( V, E ) be thegraph presented by Figure 3(a). Graph G has four biconnected componentswith its corresponding edge sets being E = { e , e , e } , E = { e } , E = { e , e , e , e , e , e } and E = { e } . Let the symmetric matrix Q = ( q ( i, j )),presented in Figure 3(b), be a QMSTP cost matrix associated to G , such that q ( i, j ) is the QMSTP cost associated to the edge pair ( e i , e j ). We denote by Q E i E j the submatrix of Q consisting of elements q ( k, (cid:96) ) for e k ∈ E i and e (cid:96) ∈ E j .In Figure 3(b), Q is divided into submatrices Q E i E j , i, j ∈ { , . . . , } , usingdashed lines.Biconnected components E , . . . , E are cycles and cliques, so according toTheorem 17, matrix Q is linearizable if and only if submatrices Q E i E j havesome specific properties. In particular, submatrices that correspond to a pairof different biconnected components, i.e. Q E i E j with i (cid:54) = j , have to be subma-trices. There are 12 such submatrices, and 10 of them have one row and/orone column in which case sum matrix property is trivially satisfied. The re-maining 2 submatrices are Q E E and Q E E . Since Q is symmetric, they aretranspose of each other, hence it is enough to check sum matrix property onlyfor one of them. Indeed they are sum matrices, since they are a sum of vectors(3 , ,
4) and (1 , , , , , Q E i E i , i ∈ { , . . . , } . If E i is a clique or (big enough) biclique then we need11 e e e e e e e e e e (a) (b) Figure 3: A linearizable QMSTP instanceto check whether Q E i E i is a weak sum matrix. If E i is a cycle then there areno necessary conditions on Q E i E i . Edges E form the complete graph on threevertices, but in the same time E forms a cycle. This is not in contradiction, asevery symmetric 3 × E and E are trivial cliques, hence the weak sum property of Q E E and Q E E is trivially satisfied. E is a complete graph, hence it remains to check whether Q E E is a weak sum matrix. It is easy to see that Q E E is a symmetric weaksum matrix generated by the vector a = (3 , , , , , i (cid:54) = j , i -th rowand j -th column of the submatrix Q E E contains the value a i + a j . We seethat all submatrices satisfy necessary properties, therefore Q is linearizable.At this point, it is straightforward to obtain a linearization of Q . We canexpress Q as Q = M + B , where M is the matrix obtained from Q by re-placing elements in the submatrix Q E E by 0. Matrix M satisfies Lemma 10,and its linearization vector C M can be calculated as described in the proofof Lemma 10 using vectors obtained in the analysis above. Furthermore, B is linearizable and its linearization vector C B can be calculated as describedin Theorem 5. Then vector C = C M + C B is a linearization of Q . And C = (54 , , , , , , , , , ,
2) is one such vector in the case of matrix Q . Theorem 17 gives us a solution for the quadratic spanning tree linearizationproblem (QST-LP) for the class of graphs in which every biconnected componentis either a clique, a biclique or a cycle. Given such graph G = ( V, E ), one canfind in linear time its biconnected components (see [12]), and determine whichtype they are. Now for a given (not necessary symmetric) cost matrix Q , fromObservation 4 if follows that Q is linearizable if and only if the symmetric matrix12 ( Q + Q T ) is linearizable. According to Theorem 17, to determine whether Q is linearizable we need to check whether appropriate submatrices of ( Q + Q T )are sum matrices or symmetric weak sum matrices. In worst case this takesΘ( | E | ) time, since potentially every element of Q which is not on the maindiagonal has to be examined. Next we examine whether the recognition can bedone faster if the cost matrix is given in the factored form.Let M = ( m ( i, j )) be an n × n matrix of rank p . Then the elements of M are of the form M ( i, j ) = p (cid:88) k =1 a ki b kj , (9)for some vectors a k and b k , k = 1 , . . . , p . Hence, an n × n matrix of a rank p canbe represented with 2 pn values. We say that (9) is a factored form representation of matrix M .Note that every sum matrix can be written as the sum of a constant rowmatrix and a constant column matrix. Since rank( M + M ) ≤ rank( M ) +rank( M ) for every matrices M and M , it follows that every sum matrixhas the rank at most 2. Therefore, the problem of recognizing sum matricesrepresented in a factored form (9) is reduced to the following question. Given a i , b i , c i , d i , i = 1 , . . . , n , is it possible to decide in O ( n ) time whether the matrix M = ( m ( i, j )) with m ( i, j ) = a i b j + c i d j is a sum matrix? An affirmative answerto this question follows from the following theorem.
Theorem 18.
Let an n × n matrix M = ( m ( i, j )) be of the form m ( i, j ) = a i b j + c i d j , i, j = 1 , . . . , n . • If at least one of the vectors a, b, c, d is a constant vector, then M is a summatrix if and only if a or b is a constant vector, and c or d is a constantvector. • If none of the vectors a, b, c, d is a constant vector, then M is a summatrix if and only if there exist three constants K (cid:54) = 0 , K and K suchthat a i = Kc i + K and d i = − Kb i + K , i = 1 , . . . , n .Proof. Let a matrix M = ( m ( i, j )) be of the form m ( i, j ) = a i b j + c i d j . Letus assume M is a sum matrix, i.e. there exist two vectors e and f such that m ( i, j ) = e i + f j , i, j = 1 , . . . , n . Then for arbitrary i, j, k, (cid:96) ∈ { , . . . , n } m ( i, k ) − m ( i, (cid:96) ) = f k − f (cid:96) and m ( j, k ) − m ( j, (cid:96) ) = f k − f (cid:96) . Hence m ( i, k ) − m ( i, (cid:96) ) = m ( j, k ) − m ( j, (cid:96) ). Now from m ( i, j ) = a i b j + c i d j itfollows that a i b k + c i d k − a i b (cid:96) − c i d (cid:96) = a j b k + c j d k − a j b (cid:96) − c j d (cid:96) , which can be rearranged to a i ( b k − b (cid:96) ) + c i ( d k − d (cid:96) ) = a j ( b k − b (cid:96) ) + c j ( d k − d (cid:96) ) . M is a sum matrix then( a i − a j )( b k − b (cid:96) ) = − ( c i − c j )( d k − d (cid:96) ) , (10)for every i, j, k, (cid:96) ∈ { , . . . , n } . Now we divide our investigation into two cases. Case 1 : At least one of the vectors a, b, c, d is a constant vector.
Withoutloss of generality we can assume that a is a constant vector. From (10) it followsthat ( c i − c j )( d k − d (cid:96) ) = 0 ∀ i, j, k, (cid:96) ∈ { , . . . , n } . (11)Hence either c or d is a constant vector. Otherwise there would exist i, j forwhich c i − c j (cid:54) = 0, and k, (cid:96) for which d k − d (cid:96) (cid:54) = 0 which would contradict (11).Note that this is also a sufficient condition. Let us assume a i = α and d i = δ , i = 1 , . . . , n . Then, m ( i, j ) = αb j + c i δ = e i + f j ∀ i, j ∈ { , . . . , n } , where e i := δc i and f i := αb i , i = 1 , . . . , n . In the case when c (instead of d ) isa constant vector, in a similar way one gets that M is a sum matrix. Case 2 : None of the vectors a, b, c, d is a constant vector.
Assume thatthere are two elements of vector a that are the same, i.e. there exist i, j , i (cid:54) = j such that a i = a j . Then for the same i, j c i = c j holds. Assume the contrary,i.e. a i = a j and c i (cid:54) = c j . As d is not a constant vector, there exist k, (cid:96) suchthat d k (cid:54) = d (cid:96) . Now for such i, j, k, (cid:96) , equation (10) does not hold, which is acontradiction. Hence, a i = a j if and only if c i = c j . Using the same logic, forall i, j ∈ { , . . . , n } , b i = b j if and only if d i = d j .Let N ⊆ { , . . . , n } be a maximal set of indices i for which a i ’s (and c i ’s)are pairwise distinct. That is, for every i, j ∈ N , i (cid:54) = j , it follows that a i (cid:54) = a j (and hence c i (cid:54) = c j also). Let N be a set of indices with the same property forvectors b and d . Now from (10) it follows that a i − a j c i − c j = − d k − d (cid:96) b k − b (cid:96) , for every distinct i, j ∈ N and k, (cid:96) ∈ N . By fixing some distinct k, (cid:96) ∈ N , itfollows that ( a i − a j ) / ( c i − c j ) is a nonzero constant (which we denote by K ) forevery distinct i, j ∈ N . Analogously, it follows that ( d k − d (cid:96) ) / ( b k − b (cid:96) ) = − K for every distinct k, (cid:96) ∈ N . Hence a i − a j = K ( c i − c j ) = Kc i − Kc j ∀ i, j ∈ N , from which it follows that a i = Kc i + K for some constant K and for all i ∈ N . Note that from the way we defined N , this relation can be extendedto entire { , . . . , n } , i.e. we have that a i = Kc i + K i = 1 , . . . , n, (12)for some constants K (cid:54) = 0 and K . Analogously we obtain that d i = − Kb i + K i = 1 , . . . , n, (13)14or some additional constant K .Note that (12) and (13) are sufficient conditions also. Namely m ( i, j ) = a i b j + c i d j = ( Kc i + K ) b j + c i ( − Kb j + K )= K c i + K b j is a sum matrix relation. Corollary 19.
Given a i , b i , c i , d i , i = 1 , . . . , n , it is possible to decide in O ( n ) time whether the square matrix M = ( m ( i, j )) with m ( i, j ) = a i b j + c i d j is asum matrix.Proof. It follows directly form the statement and the proof of Theorem 18.Namely, the following is an O ( n ) time algorithm.First we check whether any of the vectors a, b, c, d are a constant vectors. Ifso, then M is a sum matrix if and only if a or b is a constant vector, and c or d is a constant vector. Else, find i and j such that a i − a j (cid:54) = 0 and define K tobe K = ( a i − a j ) / ( c i − c j ). Furthermore, define K , K to be K = a − Kc and K = d + Kb . Then M is a sum matrix if and only if (12) and (13) aresatisfied.In the case of complete graphs, the following result on the recognition oflinearizable cost matrices represented in factored form straightforwardly holds. Corollary 20.
Let G = ( V, E ) be a complete graph or a complete bipartitegraph. Let Q = ( q ( i, j )) be a symmetric cost matrix of a QMSTP on the graph G such that q ( i, j ) = a i b j + c i d j , i, j = 1 , . . . , | E | , i (cid:54) = j, for some given vectors a, b, c, d . Then in O ( | E | ) time it can be decided whether Q is linearizable, andif so, a linearization can be calculated in O ( | E | ) time.Proof. If follows directly form Corollary 19 and the fact that in the case ofcomplete (bipartite) graphs, r ( i ) from (3) can be calculated in O (1) time forevery i = 1 , . . . , | E | . We investigated the problem of characterizing linearizable QMSTP cost matri-ces, and we resolved the problem for a broad class of graphs. The main resultis presented as Theorem 17. In particular, given a graph G , Lemma 10 gives asufficient condition for a cost matrix to be linearizable, and in the case of com-plete and complete bipartite graphs, the condition is also necessary. A naturalquestion that imposes itself is: For which graphs the conditions of Lemma 10is necessary?
In the view of Lemma 16, this question can be rephrased as thefollowing open problem:
For which biconnected graphs a symmetric QMSTPcost matrix is linearizable only if it is a weak sum matrix.
15n this paper so far we have encountered two types of biconnected graphsfor which a linearizable QMSTP cost matrix does not need to be a weak summatrix. These graphs were cycles and complete bipartite graphs K ,n . Notethat both of these graphs classes contain a vertex with degree 2. As a matterof fact, for every biconnected graph that contains a vertex with degree 2, theweak sum condition is not necessary. Namely, let G = ( V, E ) be a biconnectedgraph such that p ∈ V is of degree 2 and E p is the set of two edges adjacent to p . Then the following symmetric matrix Q = ( q ( i, j )) given by q ( e, f ) = / e, f ∈ E p , e (cid:54) = f, / (2( n − e, f ∈ E \ E p , e (cid:54) = f, c ( e ) = n − n − , e ∈ E . (Note that such cost matrices have even stronger propertythat the cost of every spanning tree is the same.) Therefore, an interestingquestion would be to identify how dense a graph needs to be in order that lin-earizable cost matrices are necessarily weak sum. Is it enough that the minimumvertex degree is at least 3? Acknowledgment
This research work was supported by an NSERC discovery grant and an NSERCdiscovery accelerator supplement awarded to Abraham P. Punnen.
References [1] W. Adams and L. Waddell, Linear programming insights into solvable casesof the quadratic assignment problem,
Discrete Optimization , 14 (2014) 46–60.[2] A. Assad and W. Xu, The quadratic minimum spanning tree problem,
Naval Research Logistics , 39 (1992) 399–417.[3] C. Buchheim and L. Klein, Combinatorial optimization with one quadraticterm: Spanning trees and forests,
Discrete Applied Mathematics
177 (2014)34–52.[4] E. C¸ ela, V. G. De˘ıneko and G. J. Woeginger, Linearizable special cases ofthe QAP,
Journal of Combinatorial optimization , doi:10.1007/s10878-014-9821-2[5] R. Cordone and G. Passeri, Solving the Quadratic Minimum Spanning TreeProblem,
Applied Mathematics and Computation
218 (2012) 11597–11612.[6] A. ´Custi´c, Efficiently solvable special cases of multidimensional assignmentproblems. Ph.D. thesis, TU Graz, 2014.167] A. Darmann, U. Pferschy, S. Schauer, and G. J. Woeginger, Paths, Treesand Matchings under Disjunctive Constraints,
Discrete Applied Mathemat-ics
159 (2011) 1726–1735.[8] A. Fischer and F. Fischer, Complete description for the spanning tree prob-lem with one linearised quadratic term,
Operations Research Letters
Ap-plied Mathematics and Computation , 164 (2005) 773–788.[11] V. Goyal, L. Genc-Kaya, R. Ravi, An FPTAS for minimizing the productof two non-negative linear cost functions,
Mathematical Programming
Communications of the ACM
16 (1973) 372–378.[13] S. N. Kabadi and A. P. Punnen, An O ( n ) algorithm for the QAP lineariza-tion problem, Mathematics of Operations Research
36 (2011) 754–761.[14] W. Kern and G. J. Woeginger, Quadratic programming and combinatorialminimum weight product problems,
Mathematical Programming
110 (2007)641–649.[15] H. Konno and T. Kuno, Linear multiplicative programming,
MathematicalProgramming
56 (1992) 51–64.[16] M. Lozano, F. Glover, C. Garcia-Martinez, F. Javier Rodriguez and R.Marti, Tabu search with strategic oscillation for the quadratic minimumspanning tree,
IIE Transactions
46 (2014) 414–428.[17] S. M. D. M. Maia, E. F. G. Goldbarg, M. C. Goldbarg, On the Biobjec-tive Adjacent Only Quadratic Spanning Tree Problem,
Electronic Notes inDiscrete Mathematics
41 (2013) 535–542.[18] S. M. D. M. Maia, E. F. G. Goldbarg, M. C. Goldbarg, Evolutionary algo-rithms for the bi-objective adjacent only quadratic spanning tree,
Interna-tional Journal of Innovative Computing and Applications
Mathematical Programming
141 (2013)103–120.[20] T. ¨Oncan and A. P. Punnen, The quadratic minimum spanning tree pro-belm: A lower bounding procedure and an efficient search algorithm,
Com-puters & Operations Research , 37 (2010) 1762–1773.1721] G. Palubeckis, D. Rubliauskas, A. Targamadz˙e, Metaheuristic approachesfor the quadratic minimum spanning tree problem,
Information technologyand control
29 (2010) 257–268.[22] D. L. Pereira, M. Gendreau, A. S. da Cunha, Stronger lower bounds for thequadratic minimum spanning tree problem with adjacency costs,
ElectronicNotes in Discrete Mathematics
41 (2013) 229–236.[23] D. L. Pereira, M. Gendreau, A. S. da Cunha, Branch-and-cut and Branch-and-cut-and-price algorithms for the adjacent only quadratic minimumspanning tree problem,
Networks
65 (2015) 367–379.[24] A. P. Punnen and S. N. Kabadi, A linear time algorithm for the Koopmans-Beckman QAP linearization and related problems,
Discrete Optimization
10 (2013) 200–209.[25] A. P. Punnen, Combinatorial optimization with multiplicative objectivefunction,
International Journal of Operations and Quantitative Manage-ment
Opsearch
34 (1997) 140–154.[27] S. Sundar and A. Singh, A swarm intelligence approach to the quadraticminimum spanning tree problem,
Information Science
180 (2010) 3182–3191.[28] H. Tuy and B. T. Tam, An efficient solution method for rank two quasi-concave minimization problems,
Optimization
24 (1992) 43–56.[29] R. Zhang, A. ´Custi´c, and A. P. Punnen, The quadratic minimum spanningtree problem and its variations. In preparation[30] R. Zhang, S. N. Kabadi, and A. P. Punnen, The minimum spanning treeproblem with conflict constraints and its variations.
Discrete Optimization