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Convertible and m-convertible matricesAdam H. Berliner aa Department of Mathematics, Statistics, and Computer Science ,St. Olaf College , 1520 St. Olaf Ave., Northfield , MN 55057 , USAPublished online: 25 Aug 2011.
To cite this article: Adam H. Berliner (2012) Convertible and m-convertible matrices, Linear andMultilinear Algebra, 60:3, 267-283, DOI: 10.1080/03081087.2011.591394
To link to this article: http://dx.doi.org/10.1080/03081087.2011.591394
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Linear and Multilinear AlgebraVol. 60, No. 3, March 2012, 267–283
Convertible and m-convertible matrices
Adam H. Berliner*
Department of Mathematics, Statistics, and Computer Science, St. Olaf College,1520 St. Olaf Ave., Northfield, MN 55057, USA
Communicated by B. Shader
(Received 25 August 2010; final version received 22 May 2011)
In this article, we consider the calculation of the permanent of a(0, 1)-matrix using determinants. We investigate when determinants ofmore than one signing of the original are used to calculate the permanentand we show that non-convertible matrices require at least four differentsignings. Then, we loosen the restriction that the matrices used to convertthe permanent are signings of the original and use this to reduce thenumber of determinants necessary to convert the permanent of the all 1’smatrix by considering a particular partition of the set Sn of permutationsof {1, . . . , n}. Finally, we construct a sequence of maximal convertiblematrices with a small number of nonzero entries, thus lowering the possibleupper bound for the number of nonzero entries of such a matrix, relative tothe order.
Keywords: permanent; convertible matrix; m-convertible matrix;sign-nonsingular matrix
AMS Subject Classifications: 15A15; 05C50; 15B35
1. Introduction
Suppose that A¼ [aij] is a (0, 1)-matrix of order n. Where appropriate, we will alsouse A[i, j] to denote the entry in the i-th row and j-th column of A. The permanent ofA, denoted per(A), is defined to be
perðAÞ ¼X�2Sn
Yni¼1
ai,�ðiÞ
!
where the sum extends over the set Sn of all permutations of {1, 2, . . . , n}. Much likethe determinant, the permanent of a matrix has one term for each of the nonzerodiagonals of the matrix. However, each term counts positively in the expansion of thepermanent. For example, if Jn is the n� n matrix for which every entry is equal to 1,we note that (Jn)¼ n!.
A matrix A is fully indecomposable if the rows and columns of A cannot bepermuted into a matrix that has a p� q block of 0’s where pþ q¼ n. For the entiretyof this article, we generally consider only fully indecomposable matrices. Let G be the
*Email: [email protected]
ISSN 0308–1087 print/ISSN 1563–5139 online
� 2012 Taylor & Francis
http://dx.doi.org/10.1080/03081087.2011.591394
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bipartite graph whose vertices are {x1, . . . , xn, y1, . . . , yn} where xi and yj are joined byan edge if and only if aij¼ 1. Then, G is called the bipartite graph associated with Aand A is the bipartite adjacency matrix of G. Calculating per(A) is thus equivalent tocounting the number of perfect matchings of G.
Polya [16] asked if it was possible to characterize the matrices A whose permanentis equal to their determinant. This question carries great significance, as the
calculation of the permanent was shown by Valiant [19] to be a computationallydifficult problem, specifically a #P-complete problem. Calculating the determinant,on the other hand, is computationally simple and can be calculated in polynomialtime. Polya’s problem leads naturally to sign-nonsingular (SNS) matrices – matriceswhich are determined to be nonsingular based solely upon the positions of the zero,positive and negative entries (c.f. [3]). If A is SNS, then per(jAj)¼�det(A), and so wehave a way of using a determinant to calculate a permanent.
If A0 is a (0,�1)-matrix such that jA0j ¼A (entry-wise), then we say that A0 is asigning of A. We say that A is convertible if A has a nonzero permanent and there is asigning A0 of A such that
perðAÞ ¼ detðA0Þ
In this case, we say that A0 is a conversion of A.For example, the matrix
A ¼
1 1 0
1 1 1
1 1 1
264
375
is convertible, and one possible conversion is given by
A0 ¼
1 �1 0
1 1 �1
1 1 1
264
375
Suppose A is an n� n (0, 1)-matrix with the following properties:
(I) There is a row p containing exactly two nonzero entries apr and aps.(II) Column r contains exactly two nonzero entries apr and aqr.(III) aqs¼ 0.
Given A, we construct a matrix B of order n� 1 by making aqs¼ 1 and deletingrow p and column r. We call B the matrix obtained from A by elementary contractionof the columns r, s on the row p. We could similarly perform an elementarycontraction of two rows r, s on a column p of a matrix. The following theoremcharacterizes matrices which are not convertible [3,14].
THEOREM 1.1 Suppose A is a (0, 1)-matrix of order n whose determinant is not
identically zero. Then A is not convertible if and only if there exists a (0, 1)-matrix A0 oforder k and permutation matrices P and Q such that P(A0 � In�k)Q�A and A0 can becontracted to J3 by a series of elementary contractions.
Kasteleyn [8–10] worked with SNS matrices while studying the dimer problem instatistical mechanics. Building on this, Galluccio and Loebl [5] and Tesler [18] found
268 A.H. Berliner
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that if A is the bipartite adjacency matrix of a genus g bipartite graph, then 2g � per(A)can be written as a sum of the determinants of 4g signings of A. Codenotti andResta [4] then used this result to calculate the permanents for several families ofsparse circulant matrices using four signings. In Section 2, we investigate thecalculation of the permanent of a matrix using determinants of several othermatrices.
In Section 3, we consider the structure of (0, 1)-matrices that are maximal withrespect to having a SNS signing. In particular, we construct a sequence of matriceswhich is maximal with respect to convertibility and yet have very few nonzeroentries. These results are based on those found in [1]. In Section 4, we conclude with abrief discussion of the notion of maximality as applied to the matrices studied inSection 2.
2. m-Convertible matrices
In this section, we consider the case in which A is non-convertible. A naturalquestion would be to ask how many matrices Ai are needed so that (a multiple of)per(A) can be written as the sum of determinants of the Ai. Thus, if A is a(0, 1)-matrix of order n, we want matrices A1, . . . ,Am such that
C perðAÞ ¼Xmi¼1
detðAiÞ ð1Þ
for some positive integer C. We investigate two ways of accomplishing this goal.
2.1. (Generic) m-convertibility
First, as in the case with convertibility, we require the Ai to be signings of A.Furthermore, we desire (1) to be generic. That is, replacing 1’s in A by algebraicallyindependent indeterminates yields an algebraic identity. In other words, for eachpermutation � 2Sn for which
Qni¼1 ai,�ðiÞ ¼ 1, the coefficient of
Qni¼1 ai,�ðiÞ on the
right-hand side of (1) equals C. If this happens for a particular permutation � 2Sn,we call � generic.
In general, suppose A1, . . . ,Am are signings of A and C is a positive integer suchthat (1) is generic. In this case, A is m-convertible and the collection A1, . . . ,Am iscalled an m-conversion of A. It is usually convenient, although not necessary, toconsider m to be as small as possible.
Ringel [17] proved that the complete bipartite graph Ki, j has genus
ði� 2Þð j� 2Þ þ 3
4
� �
Thus, using the results of Tesler [18] and Galluccio and Loebl [5], we know that Ji, j ism-convertible for
m ¼ 4 ðði�2Þð j�2Þþ3Þ=4b c ð2Þ
THEOREM 2.1 Suppose A is a non-convertible (0, 1)-matrix of order n� 3 whosedeterminant is not identically zero. If A is m-convertible, then m� 4.
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Proof Since A is not convertible, we cannot have m¼ 1. If m¼ 2, then we must have
2 perðAÞ ¼ detðA1Þ þ detðA2Þ
since an m-conversion gives an algebraic identity. Thus, per(A)¼ det(A1)¼ det(A2),
which implies A is convertible. In the case m¼ 3, C¼ 1 or C¼ 3 in order for
C perðAÞ ¼ detðA1Þ þ detðA2Þ þ detðA3Þ
to be generic. If C¼ 3, then per(A)¼ det(A1)¼ det(A2)¼det(A3) and A would be
convertible. Therefore, we have C¼ 1 and
perðAÞ ¼ detðA1Þ þ detðA2Þ þ detðA3Þ
Suppose that A is such a matrix of smallest possible order n. If n¼ 3, A¼ J3 as there
is only one 3� 3 matrix which is not convertible. However, it is not possible to
obtain
perðJ3Þ ¼ detðA1Þ þ detðA2Þ þ detðA3Þ
where A1, A2, A3 are signings of J3, as the only possible determinants for signings of
J3 are 0,�4. Now suppose n43. We want each � 2Sn to be generic, so we assume
without loss of generality that A can be contracted to J3 by elementary contractions
and has the form
A ¼
0
B ... ..
.
0
� � � 0 1
0 � � � 0 1 1
26666664
37777775
where the first elementary contraction is of columns n� 1, n on row n. Now, suppose
we can write
perðAÞ ¼ detðA1Þ þ detðA2Þ þ detðA3Þ
where jA1j ¼ jA2j ¼ jA3j ¼A and each permutation is generic. Without loss of
generality, we can assume that each Ai has the form
Ai ¼
0
Bi... ..
.
0
� � � 0 1
0 � � � 0 1 1
26666664
37777775
If we let A0 be the (n� 1)� (n� 1) matrix obtained from A by the elementary
contraction of columns n� 1, n on row n, we have (with each permutation generic)
perðA0Þ ¼ detðA01Þ þ detðA02Þ þ detðA03Þ
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where
A0i ¼
Bi...
� � � �1
26664
37775
This results in a 3-conversion of the (n� 1)� (n� 1) matrix A0, a contradiction.Therefore, m� 4. g
If m¼ 4, then we must have
2 perðAÞ ¼ detðA1Þ þ detðA2Þ þ detðA3Þ þ detðA4Þ ð3Þ
where each permutation � 2Sn for whichQn
i¼1 ai,�ðiÞ 6¼ 0 contributes þ1 three timesand �1 one time on the right-hand side of (3).
2.2. General m-convertibility
Here, we still want to be able to write, in a generic way,
C perðAÞ ¼Xmi¼1
detðAiÞ ð4Þ
for some positive integer C. However, we no longer require the matrices Ai to besignings of the original matrix A. All we require is that Ai is a fully indecomposable(0, 1,�1)-matrix where jAij �A for 1� i�m. In this case, we call the collectionA1, . . . ,Am in (4) a general m-conversion of A. In particular, we consider A¼ Jn as acase where a general m-conversion uses fewer determinants than a genericm-conversion.
Setting i¼ j¼ n in (2), we know that Jn is m-convertible for
m ¼ 4 ððn�2Þ2þ3Þ=4b c ð5Þ
Our goal is to find a general m-conversion of Jn which uses fewer matrices than (5).For n41, let Pn be the (0, 1)-matrix of order n where Pn[i, j]¼ 1 if and only if
j¼ iþ 1 or i¼ n and j¼ 1. If the rows and columns of A can be permuted to result inthe matrix A0 ¼ InþPn, we will say that A is a full cycle matrix. If A is a full cyclematrix, then we calculate that per(A)¼ 2.
Suppose A is a (0, 1)-matrix of order n41. For 2� k� n, we will say that A isa simple k-permanent matrix if A has a full cycle matrix of order n as a submatrix,a row (column) with exactly k nonzero entries, and �(A)¼ 2nþ k� 2. As aconsequence, we note that if A is a simple k-permanent matrix, then per(A)¼ k. Inaddition, a full cycle matrix is a simple 2-permanent matrix. The main result here isthe following theorem.
THEOREM 2.2 For all n� 4 and 2� k� n, there is a collection fAi : 1 � i � n!kg of
simplek-permanent matrices of order n such that for each � 2Sn,Ynj¼1
Ai j, �ð j Þ½ ¼ 1 for exactly one value of i:
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Thus, each of the n! terms in per(Jn) occurs as a nonzero term in the permanent of
exactly one Ai. Furthermore, simple k-matrices are convertible. Since InþPn is
convertible, we can change some of the extra 1’s into �1’s where appropriate to make
sure each nonzero permutation is positive in the determinant expansion. Therefore,
we get the following corollary.
COROLLARY 2.3 For n� 4, Jn has a general n!k-conversion for each k, 2� k� n.
For large enough values of n, this is a smaller number of matrices than (5). Before
we prove the theorem, we give some definitions and lemmas.Suppose that s and t are positive integers where s� t. We define P(t, s) to be the
set of permutations of length s on a set of size t. Given two elements �1, �2 of P(t, s),we say that they form a full cycle if, up to simultaneous reordering of the positions,
they have the form
�1 ¼ a1 a2 � � � as�1 as�2 ¼ a2 a3 � � � as a1
and we say that they form a full path if, up to simultaneous reordering of the
positions, they have the form
�1 ¼ a1 a2 � � � as�1 as
�2 ¼ a2 a3 � � � as asþ1
where a1, a2, . . . , asþ1 are distinct.We can interpret Theorem 2.2 to mean the following: for each k, 2� k� n, the
elements of Sn can be partitioned into sets of size k such that each set contains two
permutations �1, �2 which form a full cycle. Each of the other k� 2 permutations
agree with either �1 or �2 in every position except in one fixed position (dependent on
the particular set) in which none of the permutations agree. We will call such a set of
k elements of Sn a simple k-permanent set.
LEMMA 2.4 For t� 4, the set P(t, 2) can be partitioned into pairs such that each pair
forms a full path.
Proof Let t¼ 4. Then we take the following pairs: {12, 31}, {21, 13}, {14, 43},
{41, 34}, {23, 42} and {32, 24}.Now suppose t44. By induction, we assume we already have a partition for the
elements of P(t� 1, 2). Thus, we are left to partition the elements of P(t, 2)
containing t as an element. To do this we take the pairs {it, t(iþ 1)} along with the
pair {(t� 1)t, t1} for 1� i� t� 2. g
LEMMA 2.5 Theorem 2.2 holds for k¼ 2.
Proof We must partition the elements of Sn into pairs so that each pair forms a full
cycle. By Lemma 2.4, we can find a pairing of the elements of P(n, 2) such that each
pair has the form {ab, bc} where a, b, c are distinct. For a fixed pair {ab, bc}, let
d1, . . . , dn�3 denote the elements of {1, . . . , n} n {a, b, c}. We start by pairing the
following two elements of Sn:
�1 ¼ a b c d1 � � � dn�4 dn�3�2 ¼ b c d1 d2 � � � dn�3 a
ð6Þ
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We take all pairs resulting from the (n� 2)! simultaneous permutations ofpositions 3, . . . , n of (6). This results in a pairing for each element of Sn thatbegins with either ab or bc. We do this for each pairing of P(n, 2) given byLemma 2.4, and the result follows. g
Suppose that s, t are positive integers and that s5t. The Johnson graph J(t, s) isthe graph whose vertices are the s-subsets of [t]¼ {1, . . . , t} (we could use any set ofsize t), and two vertices v1 and v2 are connected by an edge if and only ifjv1\ v2j ¼ s� 1. For example, the Johnson graph J(n, 1) is the complete graph on nvertices and the graph J(5, 2) is the complement of the Petersen graph (Figure 1).
LEMMA 2.6 The vertices of J(t, s) can be partitioned into sets V1, . . . ,Vp such that
(i) jVij41 for 1� i� p,(ii) The graph induced on each Vi is complete.
Proof Without loss of generality, we may assume s � t2
� �, as J(t, t� s) is isomorphic
to J(t, s). For the case s¼ 1, J(t, 1) is a complete graph on t41 vertices.
Let s41. By induction, we assume that the result holds for J(t0 s0) where s0 � s andt05t (and 05s05t0). Thus, we can partition the s-subsets of [t� 1] to form completesubgraphs of J(t� 1, s) and we are left with partitioning the s-subsets of [t] whichcontain t as an element. In addition, we can partition the (s� 1)-subsets of [t� 1] toform complete subgraphs of J(t� 1, s� 1). Leaving that partition intact, we add theelement t to each of the (s� 1)-subsets of [t� 1], resulting in a partition of thes-subsets of [t] which contain t as an element into complete subgraphs of J(t, s). g
LEMMA 2.7 Suppose that s, t are positive integers and s5t. The elements of P(t, s)can be partitioned into L ordered sets ð�1,�2, . . . ,�pl Þ such that for 1� l�L, pl� 2 andthe pairs �1,�pl and �i, �iþ1 (1� i� pl� 1) form full paths. If t� 4 and if s¼ 2 ors¼ t� 1, then P(t, s) can be partitioned so that pl¼ 2 for all l.
{1, 2} {1,3}
{1,4}
{2,4}
{4,5}
{1,5}
{3,5}
{2,5}
{2,3}
{3,4}
Figure 1. A representation of J(5, 2).
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Proof If s¼ 1, then the ordered set (1,2, . . . , t) works.Suppose s¼ 2. If t¼ 3, then we can partition P(3, 2) into the ordered sets
(12, 23, 31) and (13, 21, 32). If t� 4, the result follows by Lemma 2.4.Suppose s¼ t� 1 and t� 4. For each permutation � in St we remove the element
in the last position, resulting in a permutation �0 in P(t, t� 1). All elements ofP(t, t� 1) are uniquely obtained in this manner. By Lemma 2.5, we can partition theelements of St into pairs {�1,�2} so that each pair forms a full cycle. If we pair theelements of P(t, t� 1) in the same manner, then the resulting pairs f�01,�
02g will each
form a full path.Finally, suppose 3� s5t� 1. By Lemma 2.6, we can partition the vertices of the
Johnson graph J(t, s) into subsets of size at least 2 where the graph induced on eachof the subsets is complete. Consider one of the subsets V¼ {v1, . . . , vq} and make itinto an ordered set (v1, . . . , vq). For 1� i5q, let ri be the element of vi not in viþ1 andlet rq be the element of vq not in v1. Similarly, for 15i� q, let li be the element of vinot in vi�1 and let l1¼ lqþ1 be the element of v1 not in vq.
Since we are considering permutations in P(t, s), we let Wi be the set of allpermutations in P(t, s) whose elements come from vi. For 1� i� p and for any v2Wi,we define a mapping R by defining R(v) to be the element of Wiþ1 (whereWqþ1¼W1) obtained by the following procedure:
(i) Take the elements of v and shift them one position to the right (where thelast element becomes the first element).
(ii) Then, replace ri by liþ1 (if i¼ q, we replace rq with l1).
For example,
ri a1 a2 � � � as�1
becomes
as�1 liþ1 a1 � � � as�2
By construction, we see that v and R(v) form a full path. Thus, pick any �12W1 andcontinue to apply R. Since R is injective, the process of repeatedly applying R musteventually return to the original element �12W1. Thus, we obtain an ordered set(�1,�2, . . . ,�p) of size at least 2 such that �1, �p and �i, �iþ1 (1� i� p� 1) each formfull paths. If all of the elements of W1 have not been used, then we pick an unusedelement of W1 and repeat the process. Since W1, . . . ,Wq all have the same size, eachof the elements of W1, . . . ,Wq will eventually be in one of the ordered sets. Since(v1, . . . , vq) was arbitrary, the result follows. g
We are now ready to prove Theorem 2.2.
Proof of Theorem 2.2 By Lemma 2.5, we know the theorem holds for k¼ 2 and thuswe may assume k� 3. Suppose n� t (mod k) where 0� t� k� 1. We will split theproof up into cases based on the value of t. If t� 2, we will break the process into tsteps. At step j, we find k-permanent sets for some of elements of Sn which start withcertain elements of P(t, j). The unused elements of Sn will start with certain elementsof P(t, jþ 1) and are grouped together in such a way as to make it possible for severalof them to become part of a k-permanent set at the next step.
First, we partition the elements of {tþ 1, tþ 2, . . . , n} into sets Z1,Z2, . . . ,Zl ofsize k where l ¼ n�t
k and we write Zi ¼ fzi1, z
i2, . . . , zikg.
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Suppose t¼ 0. We denote the elements of [n]nZi by fdi1, . . . , difg. We then form the
k-permanent set given by the following k elements of Sn (where the first two
permutations form the required full cycle):
�1 ¼ zi1 zi2 � � � zik�1 zik di1 � � � dif�1 dif�2 ¼ zi2 zi3 � � � zik di1 di2 � � � dif zi1
�3 ¼ zi3 zi2 � � � zik di1 di2 � � � dif zi1
..
. ... ..
. ... ..
.
�k ¼ zik zi2 � � � zik�1 di1 di2 � � � dif zi1
ð7Þ
The (n� 1)! simultaneous permutations of columns 2, . . . , n of (7) yield a simple
k-permanent set for each element of Sn beginning with one of zi1, . . . , zik. Since Zi was
arbitrary, we obtain one simple k-permanent set for each element of Sn.Suppose t¼ 1. We now denote the elements of [n]n({1}[Zi) by fd
i1, . . . , difg. We
then form the k-permanent set given by the following k elements of Sn:
�1 ¼ zi1 zik zi2 � � � zik�2 zik�1 di1 � � � dif�1 dif 1
�2 ¼ zi2 1 zi3 � � � zik�1 di1 di2 � � � dif zik zi1
�3 ¼ zi3 1 zi2 � � � zik�1 di1 di2 � � � dif zik zi1
..
. ... ..
. ... ..
.
�k ¼ zik 1 zi2 � � � zik�2 zik�1 di1 � � � dif�1 dif zi1
ð8Þ
The (n� 1)! simultaneous permutations of columns 2, . . . , n of (8) yield a simple
k-permanent set for each element of Sn beginning with one of zi1, . . . , zik. Since Zi was
arbitrary, we obtain a simple k-permanent set for each element of Sn not beginning
with 1.Each element of Sn that begins with 1 must, more specifically, begin with 1 zij for
some i2 {1, . . . , l} and j2 {1, . . . , k}. Consider each of the k-permanent sets resulting
from one of the (n� 2)! possible simultaneous permutations of columns 3, . . . , n of
(8). The final permutation in the list begins with zik 1. By our construction, we can
interchange the first two columns of this last row and not change the fact that the k
permutations form a k-permanent set. If we do this, then instead of being left to find
a simple k-permanent set for each element of Sn beginning with 1, we are left to find
a simple k-permanent set for each element of Sn beginning with 1 zij for each
i2 {1, . . . , l} and j2 {1, . . . , k� 1} or beginning with zik 1 for each i2 {1, . . . , l}.For each i2 {1, . . . , l}, we form the following k-permanent set given by the
following k elements of Sn:
�01 ¼ 1 zi1 zi2 � � � zik�2 zik�1 di1 � � � dif�1 dif zik�02 ¼ zik 1 zi3 � � � zik�1 di1 di2 � � � dif zi1 zi2
�03 ¼ 1 zi2 zi3 � � � zik�1 di1 di2 � � � dif zi1 zik
..
. ... ..
. ... ..
. ... ..
. ...
�0k ¼ 1 zik�1 zi2 � � � zik�2 di1 di2 � � � dif zi1 zik
ð9Þ
The (n� 2)! simultaneous permutations of columns 3, . . . , n of (9) yield a simple
k-permanent set for each element of Sn beginning with one of 1 zi1, . . . , 1 zik�1, zik 1.
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Since i was arbitrary, we obtain a simple k-permanent set for each element of Sn
beginning with 1 zij for each i2 {1, . . . , l} and j2 {1, . . . , k� 1} or beginning with zik 1
for each i2 {1, . . . , l}. As these simple k-permanent sets are pairwise disjoint, we are
finished.Now, suppose t41. As we did in the case t¼ 0, we can find simple k-permanent
sets for each of the permutations in Sn beginning with one of tþ 1, . . . , n. Thus, we
are left with the permutations beginning with one of 1, . . . , t (the elements of P(t, 1)).
Writing this out, we are left with the permutations in Sn beginning with i j where i 6¼ j,
i2 [t], and j2 [n]. We now break the process into t steps:
Step 1 (t42) If t¼ 2, see step (t� 1) below. For each value of m2 [t], we denote the
elements of [n]n({m, m1}[Zi) by fdi1, . . . , difg (where m1¼ 1 if m¼ t and m1¼mþ 1
otherwise). We form the k-permanent set given by the following k elements of Sn:
�1 ¼ m zi1 zik m1 di1 � � � dif�1 dif zi2 � � � zik�2 zik�1�2 ¼ m1 zik m di1 di2 � � � dif zi2 zi3 � � � zik�1 zi1
�3 ¼ m zi2 zik m1 di1 � � � dif�1 dif zi3 � � � zik�1 zi1
..
. ... ..
. ... ..
. ... ..
. ... ..
.
�k ¼ m zik�1 zik m1 di1 � � � dif�1 dif zi2 � � � zik�2 zi1
ð10Þ
The (n� 2)! simultaneous permutations of columns 3, . . . , n of (10) yield a simple
k-permanent set for elements of Sn beginning with one of mzi1, . . . ,mzik�1,m1 zik.
Since Zi was arbitrary and since we do this for each value of m, we obtain a simple
k-permanent set for each element of Sn beginning with c d where c2 [t] and
d2 {tþ 1, . . . , n}. Thus, we are left to deal with permutations of Sn beginning with
one of the elements of P(t, 2).
Step j (2� j� t� 2) At the end of the previous step, we are left to find simple
k-permanent sets for the elements of Sn that begin with an element of P(t, j). We
partition P(t, j) into ordered sets according to Lemma 2.7. Fix one of the ordered sets
(�1, �2, . . . ,�p) and then consider �m and �mþ1 (where �pþ1¼�1) for any fixed value
of m. Without loss of generality, we can assume �m and �mþ1 to have the form
�m ¼ x1 x2 � � � xj
�mþ1¼ x2 x3 � � � xjþ1
for some distinct x1, . . . ,xjþ12 [t].We now denote the elements of [n]n({x1, . . . , xjþ1}[Zi) by fd
i1, . . . , difg. We form
the k-permanent set given by the following k elements of Sn:
�1 ¼ x1 � � � xj zi1 zik xjþ1 zi2 � � � zik�2 zik�1 di1 � � � dif�1 dif�2 ¼ x2 � � � xjþ1 zik x1 zi2 zi3 � � � zik�1 di1 di2 � � � dif zi1
�3 ¼ x1 � � � xj zi2 zik xjþ1 zi3 � � � zik�1 di1 di2 � � � dif zi1
..
. ... ..
. ... ..
. ... ..
. ... ..
. ...
�k ¼ x1 � � � xj zik�1 zik xjþ1 zi2 � � � zik�2 di1 di2 � � � dif zi1
ð11Þ
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The (n� j� 1)! simultaneous permutations of columns jþ 2, . . . , n of (11) yield a
simple k-permanent set for each element of Sn beginning with one of
�m zi1, . . . ,�m zik�1,�mþ1 zik
Since Zi was arbitrary and since we do this for each value of m, we obtain a simple
k-permanent set for each element of Sn beginning with �d where �2 {�1, . . . ,�p} andd2 {tþ 1, . . . , n}. Also, since we do this for each ordered set in the partition of P(t, j),
this process results in a simple k-permanent set for each element of Sn beginning with
�d where �2P(t, j) and d2 {tþ 1, . . . , n}. Thus, we are left with permutations of Sn
beginning with one of the elements of P(t, jþ 1).
Step (t� 1) At the end of the previous step, we are left with grouping the elements
of Sn that begin with one of the elements of P(t, t� 1). By Lemma 2.7, we can
partition P(t, j) into pairs which form full paths. Fix one of the pairs {�1,�2}.Without loss of generality, we can assume �1 and �2 to have the form
�1 ¼ x1 x2 � � � xt�1
�2 ¼ x2 x3 � � � xt
where x1, . . . , xt are the elements of [t].We now denote the elements of [n]n({x1, . . . , xt}[Zi) by fd
i1, . . . , difg and form the
k-permanent set given by the following k elements of Sn:
�1 ¼ x1 � � � xt�1 zi1 zik xt zi2 � � � zik�2 zik�1 di1 � � � dif�1 dif�2 ¼ x2 � � � xt zik x1 zi2 zi3 � � � zik�1 di1 di2 � � � dif zi1
�3 ¼ x1 � � � xt�1 zi2 zik xt zi3 � � � zik�1 di1 di2 � � � dif zi1
..
. ... ..
. ... ..
. ... ..
. ... ..
. ...
�k ¼ x1 � � � xt�1 zik�1 zik xt zi2 � � � zik�2 di1 di2 � � � dif zi1
ð12Þ
The (n� t)! simultaneous permutations of columns tþ 1, . . . , n of (12) yield a simple
k-permanent set for elements of Sn beginning with one of �1zi1, . . . ,�1z
ik�1,�2z
ik.
Swapping �1 and �2, we get a simple k-permanent set for each element of Sn
beginning with one of �2zi1, . . . ,�2z
ik�1,�1z
ik. Since Zi was arbitrary, we do this for
Z1, . . . ,Zl�1, but not for Zl¼ {y1, . . . , yk}. Thus, we are left with elements of Sn
beginning with �d where �2P(t, t� 1) and d2Zl.As above, we get a simple k-permanent set for each element of Sn beginning with
�1y1, . . . ,�1yk�1, or �2 yk. We then form the k-permanent set given by the following k
elements of Sn:
�1 ¼ x2 � � � xt y1 x1 y2 � � � yk�2 yk�1 yk d1 � � � df�1 df�2 ¼ x1 � � � xt�1 xt y2 y3 � � � yk�1 yk d1 d2 � � � df y1
�3 ¼ x2 � � � xt y2 x1 y3 � � � yk�1 yk d1 d2 � � � df y1
..
. ... ..
. ... ..
. ... ..
. ... ..
. ...
�k ¼ x2 � � � xt yk�1 x1 y2 � � � yk�2 yk d1 d2 � � � df y1
ð13Þ
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The (n� t)! simultaneous permutations of columns tþ 1, . . . , n of (13) yields a simple
k-permanent set for elements of Sn beginning with one of �2y1, . . . ,�2yk�1, �1xt.Thus, for each of the pairs {�1, �2} we are left with finding k-permanent sets for the
elements of Sn beginning with either �1yk or �2x1.
Step t At the end of step t� 1, we are left with grouping the elements of Sn that
begin with �1yk or �2x1 for each pair {�1, �2}.We denote the elements of [n]n({x1, . . . , xt, yk}[Zi) by fd
i1, . . . , difg. We then form
the k-permanent sets:
�1 ¼ x1 � � � xt�1 yk zi1 zi2 � � � zik�2 zik�1 zik xt di1 � � � dif�1 dif
�2 ¼ x2 � � � xt x1 zik zi3 � � � zik�1 di1 yk zi2 di2 � � � dif zi1
�3 ¼ x1 � � � xt�1 yk zi2 zi3 � � � zik�1 di1 zik xt di2 � � � dif zi1
..
. ... ..
. ... ..
. ... ..
. ... ..
. ... ..
.
�k ¼ x1 � � � xt�1 yk zik�1 zi2 � � � zik�2 di1 zik xt di2 � � � dif zi1
ð14Þ
and
�01 ¼ x2 � � � xt x1 zi1 zi2 � � � zik�2 zik�1 zik yk di1 � � � dif�1 dif
�02 ¼ x1 � � � xt�1 yk zik zi3 � � � zik�1 di1 xt zi2 di2 � � � dif zi1
�03 ¼ x2 � � � xt x1 zi2 zi3 � � � zik�1 di1 zik yk di2 � � � dif zi1
..
. ... ..
. ... ..
. ... ..
. ... ..
. ... ..
.
�0k ¼ x2 � � � xt x1 zik�1 zi2 � � � zik�2 di1 zik yk di2 � � � dif zi1
ð15Þ
The (n� t� 1)! simultaneous permutations of columns tþ 2, . . . , n of (14) and (15)
yield a simple k-permanent set for each element of Sn beginning with �1yk z or �2x1zfor all z2Zi. Since Zi is arbitrary, we do this for Z1, . . . ,Zl�1, but not for
Zl¼ {y1, . . . , yk}. Thus, for each pair {�1,�2} we are left with elements of Sn
beginning with �1yk j1 or �2x1 j2, where j12 {xt, y1, . . . , yk�1} and j22 {y1, . . . , yk}.We now denote the elements of [n]n({x1, . . . , xt}[Zl) by {d1, . . . , df} and form the
k-permanent sets:
�1 ¼ x1 x2 � � � xt�1 yk y1 y2 � � � yk�2 yk�1 xt d1 � � � df�1 df
�2 ¼ x2 x3 � � � xt x1 yk y3 � � � yk�1 d1 y2 d2 � � � df y1
�3 ¼ x1 x2 � � � xt�1 yk y2 y3 � � � yk�1 d1 xt d2 � � � df y1
..
. ... ..
. ... ..
. ... ..
. ... ..
. ... ..
.
�k ¼ x1 x2 � � � xt�1 yk yk�1 y2 � � � yk�2 d1 xt d2 � � � df y1
ð16Þ
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and
�01 ¼ x2 x3 � � � xt x1 y1 y2 � � � yk�2 yk�1 yk d1 � � � df�1 df�02 ¼ x1 x2 � � � xt�1 yk xt y3 � � � yk�1 d1 y2 d2 � � � df y1
�03 ¼ x2 x3 � � � xt x1 y2 y3 � � � yk�1 d1 yk d2 � � � df y1
..
. ... ..
. ... ..
. ... ..
. ... ..
. ... ..
.
�0k ¼ x2 x3 � � � xt x1 yk�1 y2 � � � yk�2 d1 yk d2 � � � df y1
ð17Þ
The (n� t� 1)! simultaneous permutations of columns tþ 2, . . . , n of (16) and (17)yield a simple k-permanent set for each element of Sn beginning with �1 yk j1 or�2x1j2, where j12 {xt, y1, . . . , yk�1} and j22 {y1, . . . , yk}. If we do this for each pair{�1, �2}, then we have successfully found a simple k-permanent set for each elementof Sn. g
3. Maximal convertible matrices
We now turn our attention back to a problem involving 1-convertible matrices.Several questions regarding the permanents, ranks and number of nonzero entries ofconvertible matrices have been addressed in [6,11–13]. We say that a matrix A ismaximal convertible if A is convertible and if it not possible to change a 0 to a 1 andremain convertible.
Given an n� n matrix A, we construct an (nþ 1)� (nþ 1) matrix B in thefollowing way:
B ¼
1 a11 � � � a1n
1
0
..
.
0
A
2666666664
3777777775
The matrix B is called the ( first) row copy of the matrix A. Other rows (or columns)may be copied similarly. The following theorem can be found in [2].
THEOREM 3.1 Suppose A and B are (0, 1)-matrices such that B is a row (column) copyof A. Then A is maximal convertible if and only if B is maximal convertible.
For a matrix A, denote the number of nonzero entries of A by �(A). We define
H3 ¼
1 1 1
1 1 1
0 1 1
264
375
and for n43 we define Hn by taking the first row copy of Hn�1. We call Hn theHessenberg matrix of order n, and �(Hn)¼ (n2þ 3n� 2)/2. It is shown in [3,6] that upto column and row permutations, the matrix Hn is the unique convertible matrix oforder n with the maximum number of nonzero entries.
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The problem of finding maximal convertible matrices with the fewest number of
nonzero entries is more complicated. Define �n to be the least number of nonzero
entries a maximal convertible (0, 1)-matrix of order n can have. It is shown in [2] that,
for n43,
3n � �n � 4n� 4 ð18Þ
In fact, if n45, the lower bound for �n can only hold if the matrix has three nonzero
entries in each row and column. Hwang et al. [7] note that finding an explicit formula
for �n seems quite difficult. Although it is not known if the limit exists, they raise the
question of finding
limn!1
�n
n
If the limit is defined then (18) implies that
3 � limn!1
�n
n� 4 ð19Þ
Our goal is to find a sequence of matrices that lowers the upper bound given in
(19). We will use the following lemma, which is proved in [12,15].
THEOREM 3.2 Suppose A and B are maximal convertible (0, 1)-matrices of orders j
and k (respectively) and are of the form
A ¼A1 a2
a1 1
" #and B ¼
1 b2
b1 B1
" #
Then the matrix
A ? B ¼A1 a2 0
a1 1 b2b1a1 b1 B1
24
35 ð20Þ
is a maximal convertible matrix of order jþ k� 1.
THEOREM 3.3 Suppose A is a maximal convertible (0, 1)-matrix of order k containing
at least two rows (columns) with exactly three nonzero entries. If �(A)¼ 3kþ l for some
non-negative integer l, then
lim infn!1
�n
n� 3þ
lþ 6
k
Proof Without loss of generality, the first and last rows of A¼ [aij] have exactly
three nonzero entries and akk¼ 1. Let B be the matrix of order kþ 1 obtained by
taking a first row copy of A. We now recursively define a sequence of matrices Mt of
order kt in the following way: let M1¼A, and for t41 we define Mt¼Mt�1 ?B.By Theorems 3.1 and 3.2, Mt is a maximal convertible matrix for all t.
By the construction in (20), we see that for any matrices A1, A2 we have
�ðA1 ? A2Þ ¼ �ðA1Þ þ �ðA2Þ � 1þ ð�ðlast row of A1Þ � 1Þð�ðfirst column of A2Þ � 1Þ
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We know that �(A)¼ 3kþ l and �(B)¼ 3kþ lþ 5. Since it is the case that the last row
of B has exactly three nonzero entries with bkþ1,1¼ 0 and bkþ1,kþ1¼ 1, we see by
induction that the last row of Mt has exactly three nonzero entries, one of which isthe last entry of the row. Therefore,
�ðMtþ1Þ ¼ �ðMtÞ þ ð3kþ lþ 5Þ � 1þ ð3� 1Þð2� 1Þ ¼ �ðMtÞ þ 3kþ lþ 6:
Thus, by our construction,
�ðMtÞ ¼ 3kþ lþ ðt� 1Þð3kþ lþ 6Þ ð21Þ
and
�ðMtÞ
kt¼
tð3kþ lþ 6Þ � 6
kt¼ 3þ
lþ 6
k�
6
kt
Therefore,
limt!1
�ðMtÞ
kt¼ lim
t!13þ
lþ 6
k�
6
kt
� �¼ 3þ
lþ 6
k
Thus, for n¼ kt, we have constructed matrices of order n with t(3kþ lþ 6)� 6
nonzero entries. Therefore,
lim infn!1
�n
n� 3þ
lþ 6
k
g
COROLLARY 3.4 lim infn!1�n
n �277 .
Proof Consider the maximal convertible matrix
A ¼
1 1 0 1 0 0 00 1 1 0 1 0 00 0 1 1 0 1 00 0 0 1 1 0 11 0 0 0 1 1 00 1 0 0 0 1 11 0 1 0 0 0 1
2666666664
3777777775
A satisfies the conditions of Theorem 3.3, where l¼ 0 and k¼ 7. g
4. Concluding remarks
The following result for convertible matrices easily extends to m-convertiblematrices.
PROPOSITION 4.1 Suppose A is a (0, 1)-matrix and B is a row (column) copy of A.Then A is m-convertible if and only if B is m-convertible.
Proof Without loss of generality, we assume that B is the first row copy of A. If B ism-convertible, it follows that A is m-convertible since I1 � A�B. Now suppose A is
m-convertible and consider an m-conversion A1, . . . ,Am of A. For each Ai, construct
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the (nþ 1)� (nþ 1) matrix Bi as follows:
Bi ¼
1 Ai 1, 1½ � � � Ai 1, n½
�1
0
..
.
0
Ai
26666664
37777775
Then B1, . . . ,Bm is an m-conversion of B. g
A natural question is to ask how the notion of maximality extends to
m-convertible matrices for any particular value of m. As noted in Theorem 3.1, if
A is a maximal convertible matrix, then any row or column copy of A is also a
maximal convertible matrix. This allowed us to study the structure of such matrices
more effectively. However, it appears much more difficult to check if this result
extends to m-convertible matrices.
CONJECTURE 4.2 If A is a maximal m-convertible matrix, then any row (or column)
copy of A is also a maximal m-convertible matrix.
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