We construct (k+-1)-regular graphs which provide sequences of expanders by
adding or substracting appropriate 1-factors from given sequences of k-regular
graphs. We compute numerical examples in a few cases for which the given
sequences are from the work of Lubotzky, Phillips, and Sarnak (with k-1 the
order of a finite field). If k+1 = 7, our construction results in a sequence of
7-regular expanders with all spectral gaps at least about 1.52

de la Harpe, Pierre

Musitelli, Antoine

English

arXiv

We construct (k+-1)-regular graphs which provide sequences of expanders by adding or substracting appropriate 1-factors from given sequences of k-regular graphs. We compute numerical examples in a few cases for which the given sequences are from the work of Lubotzky, Phillips, and Sarnak (with k-1 the order of a finite field). If k+1 = 7, our construction results in a sequence of 7-regular expanders with all spectral gaps at least about 1.52

De La Harpe, Pierre

Archive ouverte UNIGE

Expanding graphs, Ramanujan graphs, and 1-factor perturbations

de la Harpe, P.

Musitelli, A.

Infoscience - École polytechnique fédérale de Lausanne

We construct (k±1)-regular graphs which provide sequences of expanders by adding or substracting appropriate 1- factors from given sequences of k- regular graphs. We compute numerical examples in a few cases for which the given sequences are from the work of Lubotzky, Phillips, and Sarnak (with k − 1 the order of a finite field). If k + 1 = 7, our construction results in a sequence of 7- regular expanders with all spectral gaps at least 6−2√5 ≈ 1.52; the corresponding minoration for a sequence of Ramanujan 7- regular graphs (which is not known to exist) would be 7 − 2√6 ≈ 2.10

Discrete Mathematics 306 (2006) 3302–3306www.elsevier.com/locate/discNoteNew polynomial-time algorithms for Camion basesKomei Fukuda1, Antoine MusitelliROSO/IMA, EPFL, Lausanne, Ch-1015, SwitzerlandReceived 26 July 2005; received in revised form 1 June 2006; accepted 25 June 2006Available online 7 September 2006AbstractLet M be a ﬁnite set of vectors in Rn of cardinality m andH(M) = {{x ∈ Rn : cTx = 0} : c ∈ M} the central hyperplanearrangement represented by M. An independent subset of M of cardinality n is called a Camion basis, if it determines a simplexregion in the arrangementH(M). In this paper, we ﬁrst present a new characterization of Camion bases, in the case where M is thecolumn set of the node-edge incidence matrix (without one row) of a given connected digraph. Then, a general characterization ofCamion bases as well as a recognition procedure which runs in O(n2m) are given. Finally, an algorithm which ﬁnds a Camion basisis presented. For certain classes of matrices, including totally unimodular matrices, it is proven to run in polynomial time and fasterthan the algorithm due to Fonlupt and Raco.© 2006 Elsevier B.V. All rights reserved.Keywords: Camion basis; Graph theory; Hyperplane arrangement; Pivot algorithm1. IntroductionIn what follows, we will consider a matrix as a set of column vectors. Let M ∈ Rn×m be a matrix of rank n,H(M) = {{x ∈ Rn : cTx = 0} : c ∈ M} and B a basis of M.H(M) splits up Rn into a set S of full dimensionalcones (regions). A simplex region is one which has exactly n facets. B is called a Camion basis if the correspondinghyperplanes determine the facets of a simplex region in S. It is known that there always exists a Camion basis. Aftersome column permutations, we may write M = (BN). Deﬁne a signing of a vector as multiplying it by −1. If A is amatrix, we write A0 if each entry of A is nonnegative. It is possible to show that B is a Camion basis if and onlyif there exists a signing of some columns of M so that B−1N0. Geometrically, B−1N0 means that the columnvectors of N are contained in the cone generated by B. We deﬁne A=B−1N and denote by m′ =m− n the number ofcolumns of A.Camion bases were ﬁrst investigated by Camion [1], who proved that one always exists. The geometric counterpartwas studied by Shannon [6], who gave the best lower bound of the number of simplex regions, namely twice the numberof distinct hyperplanes inH(M). Note that this lower bound does not translate immediately to a lower bound for theCamion bases, because one Camion basis might be associated with many (and at least two) simplex regions. Supported by the Swiss National Science Foundation Project 200021–105202, “Polytopes, Matroids and Polynomial Systems”.1 Also afﬁliated with Institute for Operations Research and Institute of Theoretical Computer Science, ETHZ, Zurich, Switzerland.E-mail address: antoine.musitelli@epﬂ.ch (A. Musitelli).0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved.doi:10.1016/j.disc.2006.06.015K. Fukuda, A. Musitelli / Discrete Mathematics 306 (2006) 3302–3306 3303There is no known polynomial-time algorithm to ﬁnd a Camion basis in general. Fonlupt and Raco [2] describeda ﬁnite procedure to ﬁnd one based on the results of Camion. They also gave an algorithm which runs in time O(n3m2)for totally unimodular matrices.In this paper, we present a new characterization of Camion bases, in the case where M is the column set of the node-edge incidence matrix (without one row) of a given connected digraph. Then, a general characterization of Camionbases as well as a recognition procedure which runs in O(n2m) are given. Finally, an algorithm which ﬁnds a Camionbasis is presented. For totally unimodular matrices, it is proven to run in time O((nm)2).2. Camion bases of digraphsLet G = (V ,E) be a connected digraph and M the V × E-incidence matrix of G. Let M˜ be any submatrix of Mobtained by deleting one row. It is well-known that a basis of M˜ corresponds to a spanning tree of G. Let B be a basisof M˜ , T = (V ,E0) the corresponding spanning tree and suppose M˜ = (BN). We call T a Camion tree if B is a Camionbasis. Let A = B−1N . Such a matrix is called a network matrix in the literature. It is known that for e ∈ E0 and anedge f = (u, v) ∈ E:ae,f :={1 if the unique u-v-path in T passes through e forwardly,−1 if the unique u-v-path in T passes through e backwardly,0 if the unique u-v-path in T does not pass through e.The unique closed path going from u to v through f and then from v to u in T is called a fundamental cycle. Deﬁnean auxiliary graph H associated with T as follows. The set of vertices is E0 and for e, f ∈ E0, e and f are adjacent ifand only if they share a common end-point in G and are in a same fundamental cycle. Then we have:Proposition 1. A spanning tree T is a Camion tree if and only if the auxiliary graph H is bipartite.Proof. A spanning tree T is a camion tree if and only if there exists an orientation of the edges of G so that for eachnon-basic edge g = uv of G, all basic edges of the unique u-v-path in T are forward edges. Such an orientation will becalled a proper orientation.Suppose that there is a minimal odd cycleC in H. By minimality ofC, the subgraph of H induced byC has no chord.Using the deﬁnition of H, we deduce that the vertices of C determine a star in T. Denote the central vertex of the starby v0. Since C is an odd cycle, for each orientation of G, there are two adjacent vertices in C corresponding to edgesin G that are both either entering v0 or leaving it. Thus, there is no proper orientation of the edges of T.Now suppose that H is bipartite. We may suppose H is connected (otherwise what follows can be applied to eachconnected component of H). Here is a little procedure that ﬁnds a proper orientation of G.Procedure Orientation_Propagation. Choose any element e0 ∈ E0 and orient it in an arbitrary way in G. LetF = {e0}. Then successively choose an element e ∈ E0\F adjacent in H to some f ∈ F ; put e in F and orient itso that {e, f } determines a directed path in G.Denote by T (F ) the subgraph of G whose edge set is F. Let us prove by induction on the cardinality of F that duringthe above procedure, F always satisﬁes properties (a) and (b) below:(a) T (F ) is a tree.(b) For all f1, f2∈F such that (f1, f2) is an edge in H, the reoriented edges f1 and f2 determine a directed path in G.Let k = |F |. If k = 1, then clearly (a) and (b) are true. Now suppose k1 and let e ∈ E0\F , f ∈ F such that (e, f )is an edge of H. Using the deﬁnition of H and the induction hypothesis, we have that T (F ∪ {e}) is a tree in G. LetNF (e) = {g ∈ F : (e, g) ∈ H }. Since e is a hanging edge of the tree T (F ∪ {e}), from the deﬁnition of H we deducethat NF (e) ∪ {e} determines a star in G with a central node, say v0.Let f1, f2 ∈ NF (e). As H is bipartite and the subgraph of H induced by F is connected by construction, there is aminimal path of even length in H between f1 and f2. Since the path linking f1 to f2 is of minimal length, all its verticescorrespond to edges of the star. So f1 and f2 are both entering into v0 or leaving this node. It follows that the elementsof NF (e) are all either entering into v0 or leaving it. Thus we can orient e in such a way that F ∪ {e} satisﬁes (b). Theorientation of F given by the above procedure induces a proper orientation of G. 3304 K. Fukuda, A. Musitelli / Discrete Mathematics 306 (2006) 3302–3306Hoffman andKruskal [4] andHeller andHoffman [3] gave a characterization of positive networkmatrices.A directedgraph is called alternating if in each circuit the edges are oriented alternately forwards and backwards. In Schrijver[5, pp. 278–279], it is shown that a {0, 1}-matrix is a network matrix if and only if its columns are the incidencevectors of some directed paths in an alternating digraph. To prove the necessary part of the condition, the alter-nating digraph G′ = (E0, F ′) is deﬁned, where for edges e, e′ ∈ E0, (e, e′) is in F ′ if and only if the head ofe is the same as the tail of e′. In Proposition 1, H is simply a subgraph of G′ considered as an unorientedgraph.3. A polynomial-time recognition algorithmWe are going to see a characterization of Camion bases and a procedure that recognizes them in polynomial time.Let us remark that when one of m or n is ﬁxed, there is a trivial polynomial algorithm. We will present an algorithm tocheck for a given basis B whether B is a Camion basis in time O(n2m).Now suppose that B (a basis), N and A = B−1N are given. Consider the bipartite digraph G(A), whose vertex setis the index set of the rows and columns of A and e = (i, j) is an edge of G(A) if and only if aij = 0. Set a weightfunction w on the set of edges:w((i, j)) ={2 if sign(aij ) = +,0 if sign(aij ) = −.If C is a cycle of G(A), set w(C) =∑e∈C w(e). If A′ is obtained from A by multiplying a row or a column of A by−1 and keeping the rest of A unchanged, we will say that A′ is obtained from A by a basic signing operation. Theweight function on the edges of the graph G(A′) will be denoted by w′. Then, we have the following characterizationof Camion bases.Proposition 2. A basis B is a Camion basis if and only if each non-oriented cycle of G(A) has a total weight equal to0mod (4).Lemma 1. Assume that A′ is obtained from A by successive applications of basic signing operations. If C is a cycle ofG(A), w′(C) ≡ w(C)mod (4).Proof. We can assume thatA′ is obtained from A by application of a basic signing operation. Note thatG(A)=G(A′).We can assume that C is an elementary cycle of G(A) = G(A′). But clearly w′(C) = w(C) ± 4 or w′(C) = w(C) andthe result follows. Proof of Proposition 2. First we prove the necessity. Let B be a Camion basis. There exists a matrix A′0 which canbe obtained fromA by successive applications of basic signing operations. IfC is a cycle ofG(A′)=G(A),w′(C)=2|C|and w′(C) ≡ 0mod (4) since C is even. The result follows from the previous lemma.To prove the sufﬁciency part, we can assume that G(A) is connected and we use the procedure below. In particular,the procedure returns a cycle with a total weight equal to 2mod (4) if B is not a Camion basis. For a given spanningtree T of G(A) and an edge e ∈ G(A) − T , let Ce denote the unique cycle in T ∪ {e}.Procedure IS_CAMIONInput: A matrix A ∈ Rn×m′ , where A = B−1N such that G(A) is connected.Output:Yes (if B is a Camion basis) with a matrix A′0 or No (otherwise) with a certiﬁcate, that is a cycle in G(A′)of total weight equal to 2mod (4), where A′ is A after some basic signing operations.(1) Let T be a subset of edges of G(A) which induces a spanning tree of G(A).Assign to an initial node v of G(A) the label l(v) = +1, the other nodes are unlabelled.(2) while ∃(i, j) ∈ T with i (resp. j) labelled and j (resp. i) unlabelled dolabel j (resp. i) with the label +1 or −1 in such a way that aij · l(i) · l(j)> 0.(3) Let A′ be the matrix obtained by multiplying each row i of A by its label l(i) and each column j of A by its labell(j). If A′0, outputYes and A′. Otherwise, there exists e = (i, j) /∈ T such that a′ij < 0. Output No and Ce.K. Fukuda, A. Musitelli / Discrete Mathematics 306 (2006) 3302–3306 3305All the nodes of G(A) are labelled by +1 or −1 at the end of this procedure. Let A′ be the matrix constructed(at step 3). Note that for all (i, j) ∈ T , a′ij = aij · l(i) · l(j)> 0.If B is not a Camion basis, there exists e = (i, j) /∈ T such that a′ij < 0 and w′(e) = 0. We have w′(Ce) ≡ 2mod (4)since w′(Ce) = 2(|Ce| − 1) and Ce is even. But by the previous lemma we have also w(Ce) ≡ 2mod (4).If B is Camion, then for each edge e = (i, j) ∈ G(A) − T , w′(Ce) ≡ 0mod (4) as proved above and sincew′(Ce) = 2(|Ce| − 1) + w′(e), it follows that w′(e) = 2 and a′ij > 0. Corollary 1. There is a polynomial-time algorithm to check whether a basis B is Camion.Proof. To check whether a given basis B is a Camion basis, we simply apply the procedure IS_CAMION to the matrixA=B−1N . Since the complexity of calculatingB−1N is O(n2m), computingA dominates the procedure IS_CAMION.Thus, the complexity of checking a given basis B is O(n2m). 4. A new algorithm to ﬁnd a Camion basisSuppose that B, N (and A=B−1N ) are given. If A=[ cb D], where  is a non-zero entry, b is a column vector, cis a row vector, and D is a submatrix of A, then the transforsmationA −→ A′ =⎡⎢⎣1c−bD − bc⎤⎥⎦is called a pivoting operation on the element . Such an operation corresponds to the replacement of a basic vectorby a nonbasic one. Remark that a pivoting operation is described in general on the standard matrix M = [I, A].For our purpose, it is convenient to use the “condensed” matrix by ignoring the identity matrix part. Finding aCamion basis is equivalent to transforming the matrix A into a nonnegative one by application of the following twooperations:• pivoting operations,• basic signing operations.Signing a row ofA is equivalent to signing an associated basic vector.AmatrixA′ obtained fromA after some signingand pivoting operations will be called equivalent to A. Denote by (A) the set of matrices that are equivalent to A. Sincethe number of bases of A and possibilities of signing some columns of A is ﬁnite, the cardinality of (A) is ﬁnite.Let us remark that for an implementation of the algorithm that is described below, it is important to maintain thecorrespondence between the rows of A and the basic vectors and between the columns of A and the nonbasic ones.We will see an algorithm, called Simp, that runs in O(3(nm′)2) if M is an integral matrix, where  is the greatestdeterminant (in absolute value) of a basis. So, for the particular case of totally unimodular matrices (where  = 1),Simp is faster than the algorithm of Fonlupt and Raco [2]. Moreover, the procedure Simp applied to real matrices isalso ﬁnite.For i ∈ {1, . . . , n} and j ∈ {1, . . . , m′}, let us denote by Ai the ith row of A and A•j the jth column of A. Moreover,A(i,j) denotes the matrix obtained from A by pivoting on aij . For any vector or matrixV, the function sum(V ) evaluatesthe sum of all components of V.Let us present the algorithm SIMP.Procedure: SIMPInput: A real matrix A.Output: A nonnegative matrix A′ (obtained from A by a sequence of pivots and signings).(1) for i = 1, . . . , n doif sum(Ai)< 0, then multiply the ith row by −1.(2) for j = 1, . . . , m′ doif A•j 0, then multiply the jth column by −1.3306 K. Fukuda, A. Musitelli / Discrete Mathematics 306 (2006) 3302–3306(3) if A0, return A.otherwise, let k, k′ and l such that akl > 0 and ak′l < 0. doif sum(A•l ) < 1, then pivot on akl .otherwise pivot on ak′l .go to step 1.Theorem 1. For any real matrix A, the procedure Simp returns a nonnegative matrix after a ﬁnite number of steps.Furthermore if M is integral, then Simp runs in time O(3(nm′)2).Proof. Suppose that sum(Ai)< 0 at step 1 orA•j 0 at step 2 for some i or some j. LetA′ be the matrix obtained fromA by multiplying its ith row by −1 (at step 1) or its jth column by −1 (at step 2). Clearly, we have sum(A′)> sum(A).At the beginning of step 3, sum(Ai)0 for all i and there is no column A•j such that A•j 0. So, if there exists anentry ak′l < 0, then there exists k such that akl > 0. Denote by c the ith row of A without the lth column and b the lthcolumn of A without the ith row. We have thatA(i,l) − A =⎛⎜⎝1−  c− c−b− b −bc⎞⎟⎠= 1(1 − −b)· (1 +  c).(Note that in the middle of this equation, the matrix is written as if i = l = 1.) Thus,sum(A(i,l)) − sum(A) = 1 − sum(A•l )ail(sum(Ai) + 1) for 1 in.So, if sum(A•l ) = 1, sum(A) will increase by pivoting either on akl or ak′l . If sum(A•l ) = 1, then sum(A) nei-ther increases nor decreases. However, since∑j =l ak′j 0 and ak′l < 0, we deduce that sum(A(k′,l)k′ ) = 1/ak′l (1 +∑j =l ak′j )< 0. Therefore, sum(A) will increase at the next step 1.As |(A)|<∞, it follows that Simp generates a well-oriented matrix after a ﬁnite number of steps.Now, suppose thatM is integral.ByCramer’s rule, for eachmatrixA′ equivalent toA,wehavea′ij=± det(B ′′)/ det(B ′)∀i, j , where B ′, B ′′ are two bases of M. Since det(B ′′) and | det(B ′)|1, we deduce that a′ij  ∀i, j . Moreover,if A′ = (B ′)−1N ′ where B ′ is a basis and N ′ a submatrix of M after some signing operations of columns of M, thenwe may write each entry of A′ as a fraction of an integer over det(B ′). Thus sum(A′), respectively sum(A), is a ratioof some integer to det(B ′), respectively det(B). So, if A′ is obtained from A at some step and sum(A′) − sum(A)> 0,then sum(A′)− sum(A)1/2. Since sum(A) never decreases, but increases after at most two passages at step 1 andis between −nm and nm, the number of passages at step 1 is O(3nm).Since the number of elementary operations at steps 1,2 and 3 is O(nm), we conclude that the complexity of SIMPis O(3(nm)2). AcknowledgementsThe authors thank Thomas Liebling for many helpful comments on this paper. We are also grateful to Jean Fonluptwho considerably improved Section 2 by providing a much simpler and elegant proof of Proposition 2.References[1] P. Camion, Modules unimodulaires, J. Combin. Theory 4 (1968) 301–362.[2] J. Fonlupt, M. Raco, Orientation of matrices, Math. Program. Stud. 22 (1984) 86–98 (Mathematical programming at Oberwolfach, II,Oberwolfach, 1983).[3] I. Heller, A.J. Hoffman, On unimodular matrices, Paciﬁc J. Math. 12 (1962) 1321–1327.[4] A.J. Hoffman, J.B. Kruskal, Integral boundary points of convex polyhedra. In: Linear Inequalities and Related Systems, Annals of MathematicsStudies, vol. 38, Princeton University Press, Princeton, NJ, 1956, pp. 223–246.[5] A. Schrijver, Theory of Linear and Integer Programming, Wiley-Interscience Series in Discrete Mathematics, Wiley, Chichester, 1986(A Wiley-Interscience Publication).[6] R.W. Shannon, Simplicial cells in arrangements of hyperplanes, Geom. Dedic. 8 (1979) 179–187.

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Expanding graphs, Ramanujan graphs, and 1-factor perturbations

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