Constructing Patterns of (Many) ESSs Under Support Size Control

As is well known, equilibrium analysis of evolutionary partnership games can be done by studying a so-called standard quadratic optimization problem, where a possibly indefinite quadratic form is maximized over the standard (probability) simplex. Despite the mathematical simplicity of this model, the nonconvex instances in this problem class allow for remarkably rich patterns of coexisting (strict) local solutions, which correspond to evolutionarily stable states (ESSs) in the game; seen from a dynamic perspective, ESSs form the asymptotically stable fixed points under the continuous-time replicator dynamics. In this study, we develop perturbation methods to enrich existing ESS patterns by a new technique, continuing the research strategy started by Chris Cannings and coworkers in the last quarter of the past century.


Motivation
We consider the Standard Quadratic Optimization Problem (StQP) given by where A ∈ S n (S n denoting symmetric n × n-matrices) and n is the standard-simplex where N = {1, . . . , n}, also denoted as [1 : n].
B Immanuel M. Bomze Despite its mathematical simplicity, above model serves in different domains of applied mathematics: apart from straightforward applications like in portfolio optimization [15], it can be used in mathematical biology [12,13] as well as in Evolutionary Game Theory and Game Dynamics [1,11,17,20]. For a more detailed account on the interrelation between these domains, we refer to [3,4] and references therein. Here, we want to focus on perturbation approaches to generate richer ESSs patterns from existing ones, building upon the seminal work of Chris Cannings and coworkers [6][7][8][9]18,19], as well as upon the more recent study [5].

Notation and Preliminaries
Given any set S ⊆ R n , we denote by conv S the convex hull of S, and the linear hull is denoted by span S, while closure and interior of S are denoted by cl S and S • , respectively. For a (convex) set C, the set of its extreme points is denoted by Ext C. Later, we will use the relation Ext conv S ⊆ S holding for any set S ⊆ R n . For matrices {A, B} ⊂ S n we write Clearly we have rank A ≥ rank B(A). Furthermore, for x ∈ n , let be the support of x, and I := {x ∈ n : I (x) ⊆ I }. The extended support with respect to A is given by We will use some more convenient notation (again, we refer to [3] for background on the interplay of optimization and game equilibrium notions): given A ∈ S n , we denote by ESS(A) := x ∈ n : x is an ESS for A the set of all strict local maximizers of (1), which is always finite but may be empty; by NSS(A) := x ∈ n : x is an NSS for A the set of all local maximizers of (1), and by NES(A) := x ∈ n : x is an NES for A, i.e., Ax ≤ (x Ax) e the set of all KKT points of (1). The latter two sets are never empty; however, they may be infinite. Given A ∈ S n with finite NES(A), the procedures FINDEQ and CHECKSTAB discussed, e.g., in [2,5], The significance of the latter two sets is the following: We try to find a perturbation A of a matrix A such that ESS( A) contains perturbations of some members of E(A). These will be quasistrict ESSs of A, with supports not contained in pattern(A), and the enrichment of the pattern can be expressed as qpattern(A) being a strict subset of qpattern( A).
For the readers' convenience, we provide a short glossary of the sets used in our analysis, all refer to a fixed instance of (1) given by a symmetric matrix A ∈ We collect some basic properties of the set E(A) in the following

Proposition 1 (a) E(A) is nonempty and finite;
to be more precise, we have and and all set inclusions above are strict in general.
Ax excludes the equilibrium condition (Ax) ≤ (x Ax)e at x. Let us turn towards the set relations in (6); we deal with them from left to right; (i) SOL(A) is nonempty and closed, so conv SOL(A) = ∅ is compact, thus Ext conv SOL(A) = ∅. (ii) Let p ∈ Ext conv SOL(A). Then p ∈ SOL(A) ⊆ NSS(A), and extremality in this set is obviously settled in case |I (p)| = 1. Turning to the case |I (p)| > 1, we note extremality of p in conv SOL(A) implies Hence we deduce I (p) ∩ SOL(A) = {p}. But this in turn even implies Obviously then also I (p) ∩ Ext conv NSS(A) = {p} holds. If now {x, y} ⊆ conv NSS(A) \ {p} and 0 < λ < 1 such that p = λx + (1 − λ)y, then {x, y} ⊆ I (p) , implying x = y = p, so p is extremal in conv NSS(A). For A = Diag (1, 2) we have SOL(A) = {e 2 } but also e 1 ∈ Ext conv NSS(A).
Otherwise, since f is quadratic, there would be a whole segment (the intersection of a straight line with I (p) ) of points along which f is constantly equalling f (p), because ∂ x−p f (p) = 0 for all x ∈ I (p) . Due to extremality of p, this segment cannot contain two members of NSS(A) on either side of p. So there must be points p k on this segment arbitrarily close to p, say p − p k < 1 k , that are not in NSS(A), thus there are points q k ∈ n with p k − q k < 1 k and f (q k ) > f (p k ) = f (p). These points q k get arbitrarily close to p, meaning that p is not in NSS(A), a contradiction. we have NES(A) = {q} ∪ conv {e 1 , p}, but p / ∈ E(A) and p is not quasistrict.

(b)
The leftmost inclusion follows similarly as in the proof of (5), while the rightmost is a consequence of (6) and of the generally valid relation Ext conv S ⊆ S which we used already in our proofs of (ai) and (aii). Examples of p ∈ E(A) \ NSS(A) or q ∈ NSS(A)\E(A) are p = e 1 for A = Diag (0, 1) -see the proof of (aiii) -and q = [ 1 2 , 1 2 ] for A = O, respectively. (c) If J A (p) = I (p) and p ∈ E(A), then any x ∈ n with x Ap = p Ap must lie in I (p) , therefore as in the proof of (aiv) we have p Ax = x Ap > x Ax unless x = p, which shows p ∈ ESS(A). The example p given in the proof of (aiv) proves the last assertion.

Main Results
Our aim is to increase the number of ESSs of a matrix A by perturbing it in a way such that members of E(A) \ ESS(A) get perturbed into (quasistrict) ESSs of the perturbed matrix. To this end, we consider perturbations of A of the form A = A + εB, with ε small. We seek simple sufficient conditions on B that lead to successful perturbations in the sense that |qpattern( A)| > |qpattern(A)|. We further aim at results telling us that we have found a matrix with the largest number of quasistrict ESSs of prescribed support size.
The plan of this section is as follows: In Sect. 2.1 (Lemma 2), we keep both A and x ∈ E(A) fixed and derive a sufficient condition on B guaranteeing I (x) ∈ qpattern( A) for ε > 0 small enough. This is accomplished by studying the first-order expansion in ε of the first-order optimality condition for the StQP with matrix A. In Sect. 2.2 (Lemma 3), we prove, loosely speaking, that if close to A there are sufficiently many matrices A satisfying I (x) ∈ qpattern(A), then there will be B satisfying the sufficient condition of Lemma 2, such that also I (x) ∈ qpattern( A) for ε > 0 small enough. Further results (Theorems 8 and 9) explore ways to restrict A and B to smaller sets of matrices, varying either A or B while not losing any generality in qpattern( A). Negative-semidefinite matrices A are particularly interesting starting points for perturbation, as they have large sets E(A). Proposition 6 and Corollary 10 deal with those. In Sect. 2.3, we introduce n × n cyclically symmetric matrices, forming a vector space C n of dimension n−1 2 . With those, a further simplification allows to restrict our perturbation procedure to a finite set of candidate matrices A, with B ranging in a certain linear subspace of C n depending on A. In Sect. 2.4, the preceding results are applied to the setting of negative-semidefinite candidate matrices A ∈ C n , where the task is to find a maximal subset P ⊆ E(A), such that there is B ∈ C n satisfying the sufficient condition of Lemma 2 for any x ∈ P. Examples 12 and 13 for n ∈ {6, 7} explicate the procedure. In Sect. 2.5, we report on our results for orders n ∈ [4 : 23]. The subsections are interspersed with several examples illustrating our results, some also show limitations.

First-Order Conditions
The following lemma deals with perturbations of the form A + εB and fathoms what can be deduced from a first-order expansion of the resulting FINDEQ inequalities.

Lemma 2 Let
then for ε > 0 small enough, A + εB will have an ESS x ε such that I (x ε ) = J A+εB (x ε ) = I , and hence we have For ε > 0 small enough,Ā 1 +εB 1 will therefore be invertible as well, and with (x ε ) N \I := o gives rise to a candidate x ε for an ESS of A + εB, satisfying I (x ε ) = I . By continuity, for ε > 0 small enough we will have is the only remaining task in the proof of a). The latter inequality can be restated as , so that indeed we have, using for ε > 0 small enough, by our assumption (7). Turning to b), by the assumption made there, at least one component of the L.H.S. of (8) will become positive for ε > 0 small enough, showing x ε / ∈ NES(A + εB).

Remark 1 For
A and x fixed, also I , J and v are fixed, so the matrices B satisfying (7) form the interior of a polyhedral cone, and thus a convex cone.

Perturbation
If there are (in a certain sense) enough arbitrarily small perturbationsĀ of A that transform , then there will be a matrix B that satisfies (7), as the following lemma claims.
. Fix k ≥ 1 and let V be a linear subspace of S n of dimension k + 1, and for a > 0 define Proof We have to show that the polyhedral cone C : of points on a k-sphere S k of radius 1 has an accumulation point denotes the Gram matrix of a set of vectors (or vectorized matrices), and δ 1 > 0 is chosen such that σ k (N 1 ) ≤ s 2 , and Again, the sequence has an accumulation point B (2) ∈ V , that satisfies (1) and B (2) are linearly independent. Assume now that for r ≤ k we have found linearly independent elements B (1) , . . . , B (r ) ∈ C. Then define has a limit point B (r +1) ∈ C. Summarizing, we have found spanning a subcone of C of full dimension k + 1. Therefore C has nonempty interior C • , and any B ∈ C • satisfies (9) for all (I , K ) ∈ I P .
Does Lemma 3 also hold for k = 0, in which case we always have s ∈ {0, 1} ? That is, if there is B ∈ S n and a sequence (ε m ) of positive reals such that ε m → 0 and {I (p) : p ∈ P} ⊆ qpattern(A + ε m B) holds for all m ≥ 1, does that imply that (9) is satisfied for all (I , K ) ∈ I P ? The answer is no, as the following example shows.
s a = 0 (this can happen, as the next example shows), then for each For A ∈ S n and P ⊆ E(A) we now define the following convex cone Further members of C A (P), in case that A is negative-semidefinite, will be identified in the next result.
it is sufficient to have that p ∈ P m and q ∈ I (p) \ {p} together imply q / ∈ ker H . This would follow if the map from I (p) to R m−1 taking x to B F x is one-to-one, which again is satisfied by all B in an open dense subset of R k×(m−1) . As P m is a finite set, we find B as needed in a finite intersection of open dense subsets of R k×(m−1) .
So when searching for negative-semidefinite matrices A of order n with many points of support size m in their sets E(A), we may restrict our search to matrices of the minimal possible rank m − 1. However, not every element of C A ({p}) need have a representation as in Proposition 6(a): Suppose now that (9) is satisfied for certain A, B, I , K . In which ways can we change either A or B while keeping (9) intact? We start considering variations B of B.
Theorem 8 Let A ∈ S n and p ∈ E(A) such that |I (p)| < |J A (p)|. Suppose that (9) holds for some B ∈ S n , where I = I (p) and K = J A (p) \ I . Then for anyB ∈ span C A ({p}) and any λ ∈ R, the matrix B := B + λB will also satisfy (9).

Proof From the observations
we conclude that h I ,K , defined in (9), satisfies Since h I ,K is linear, we conclude h I ,K (B ) = h I ,K (B), and this completes the proof.
Now we turn to variations A of A. Any matrix A ∈ C A (P) is a candidate for some perturbation A + εB to yield {I (p) : p ∈ P} ⊆ qpattern(A + εB); however, in general, not every A , if any, will achieve this goal. Anyway, the search for a suitable A can be restricted to the set C A (P) Fixing now p ∈ E(A) with |J A (p)| > |I (p)|, and fixing B satisfying (9) with I = I (p), K = J A (p) \ I , we ask whether (9) will stay satisfied if we replace A with some A ∈ C A ({p}). The answer will clearly be yes, if J A (p) = J A (p) =: J and A J ×J = γ A J ×J + λee for some γ > 0, λ ∈ R, but there are other cases as well, as the following theorem tells.  O is equivalent to x A I ×I x < 0 for all x ∈ e ⊥ \ {o}, and then, because of A I ×I p I = o, we have (λp I + x) A I ×I (λp I + x) ≤ 0, for λ ∈ R and x ⊥ e, which means A I ×I O. If for some I ⊂ I with |I | = |I | − 1 the matrix A I ×I had not full rank, there would be q ∈ |I | satisfying I (q) = I , such that A I ×I q = o. Thus the dimension of the kernel of A I ×I would be at least 2, and the rank of A I ×I would have to be less than |I | − 1, which is a contradiction.  g(A, B) = g(A , B).

From this we obtain
That is, for every P ⊆ E(A) and for every O, if {I (p) : p ∈ P} ⊆ qpattern(A + εB) for some B ∈ S n and for ε > 0 small enough, then also {I (p) : p ∈ P} ⊆ qpattern(A + εB) for that same B ∈ S n and for ε > 0 small enough.

Cyclic Symmetry
We call a subset P ⊆ R n closed under cyclic permutations, if p ∈ P ⇒ Pp ∈ P, where P = ( p i j ) satisfies p i j = 1 for j ≡ i + 1( mod n), and otherwise p i j = 0. We call A ∈ S n cyclically symmetric (or symmetric circulant), if it satisfies A = P AP, and employ the notation C(a) for a cyclically symmetric matrix whose first column is a. Note that for an n × n cyclically symmetric matrix A the set E(A) is closed under cyclic permutations. In case that A is cyclically symmetric, there are well-behaved subsets of E(A), where the "transfer" can be achieved by cyclically symmetric perturbations, whenever it can be achieved by certain symmetric perturbations, as the following theorem tells. Thus, if (9) is satisfied, it stays satisfied if we replace B by (P i ) BP i . Summing over i then finishes the proof of (10). Finally, I (p) ∈ qpattern(A+εC), for ε > 0 small enough, is true for p ∈ P by Lemma 2, and true for p ∈ P\P by a continuity argument (quasistrictness of NESs is preserved, as is negative-definiteness).
The vector space C n := {A ∈ S n : A is cyclically symmetric}, on which we now concentrate, has dimension n + 1, where n := n−1 2 , a basis being {P i + P −i : i ∈ [0 : n ]}. As C n is a subspace of the space of all circulant matrices (see [14]) of order n, all members of C n are simultaneously diagonalizable, and because of their symmetry, a common orthogonal basis of real eigenvectors can be used for that purpose.  (Note that, if n is odd then k := rank (A) = 2|L| is even.) Also 1 n e ∈ n ∩ ker F , therefore Corollary 10 applies. So when perturbing negative-semidefinite matrices A ∈ C n \ {O} satisfying A • V 0 = 0, we may restrict our attention to the finite set of candidate matrices A n := {A L : ∅ = L ⊆ [1 : n ]}, where A L := − i∈L V i . Moreover, by Theorem 8, we may restrict the matrix B, used to perturb A L via A L + εB, to span {V j : j ∈ [1 : n ] \ L}.

Now consider negative-semidefinite matrices
As we can restrict ourselves to matrices with zero diagonal, yet another basis will even better serve our needs: The last two sentences of the preceding paragraph remain true with A L := i∈L W i and B ∈ span {W j : j ∈ [1 : n ] \ L}.
We are only interested in the size of qpattern(A), so we would not distinguish between A ∈ C n and A :=PAP , whereP is a permutation matrix. If A =PAP ∈ C n , but A = A, this allows for a further reduction of the set of candidate matrices. In particular, if for some permutation τ of the set [1 : n ], we havePW iP = W τ (i) for i ∈ [1 : n ], then for sets L ⊆ [1 : n ] and τ L : where B =PBP . That is, we can further restrict our set of candidate matrices to a sub-setĀ n ⊆ A n by ensuring that for every L ⊆ [1 : n ] exactly one matrix from the set The only permutation τ that can be introduced via a permutation matrixP as above is the identity, so the set of candidate matrices will beĀ 6 For n = 7, we obtain W 0 = ee , W 1 = C([0, sin 2 π 7 , sin 2 2π 7 , sin 2 3π 7 , sin 2 3π 7 , sin 2 2π 7 , sin 2 π 7 ] ), n e}, and the latter set will not change, if A is slightly perturbed. Therefore we exclude that candidate matrix from further considerations. For the remaining candidate matrices A L we have L := [1 : n ] \ L = ∅, and for α ∈ R L we define B α := i∈L α i W i . For every p ∈ E(A L ), we denote by L L,p the solution set of inequality (9), where I = I (p) and K = J A L (p) \ I , and clearly that inequality is linear in α. Next, we denote by R L a maximal subset of E(A L ) consisting only of p of maximal support size, i.e., |I (p)| = rank (A L ) + 1, and satisfying Z p ∩ R L = {p} for every p ∈ R L , where Z p is introduced in Remark 3. Our aim is then to find a subsetR L ⊆ R L of largest size(R L ) := p∈R L |Z p |, such that Any α in that nonempty intersection will then lead to a matrix B α , such that for ε > 0 small enough A natural further reduction of the set of candidate matrices is via the definitionĀ n := {A L ∈ A n : R L = ∅}. 2 (e 1 + e 2 ), p 5 := 1 6 (e 1 + 2e 2 + 2e 3 + e 4 ), resulting in the following sets R L for L ⊆ [1 : 3], 0 < |L| < 3:

Example 13 (Continuation of Example 12) For
with maximizing setsR L = R L for all L and respective sizes Corresponding matrices with 9 ESSs of support size 2 (resp. 8 ESSs of support size 3, resp. 6 ESSs of support size 4) are given in Table 2. Also note that we have just derived   Figure 1 shows part of the search space for n = 6. Note that we may get rid of two of the four parameters, that matrices from C 6 depend on, by assuming a zero diagonal and a fixed sum of entries. In the figure we have fixed the sum of entries of the first row to 2. The interior of the white triangle corresponds to matrices with one single ESS, namely 1 6 e. The boundary of the white triangle consists of negative-semidefinite matrices (up to an additive multiple of ee ), the three vertices corresponding to (positive multiples of) matrices W 1 , W 2 and W 3 . The matrix M from above then belongs to the dark gray region. There is a circle attached to each shaded region containing information regarding the pattern of matrices in the interior of that region, e.g., 8 3 indicates that there are 8 ESSs, each of support size 3. Note that the figure clearly misses matrices where entries sum to 0 or a negative value, and indeed the pattern 6 1 attained at the matrix C([0, −1, −1, −1, −1, −1] ) does not show up in the figure; see, however, [5, Figure 1] for the whole search space.
In terms of our perturbation approach, matrix A {2,3} is our entry to region 6 4 , matrix A {3} to regions 3 2 , 6 2 and 9 2 , matrix A {2} to regions 2 3 , 6 3 and 8 3 , and matrix A {1} is our entry to the interior of cl(2 3 ∪ 2 3 3 2 ). Note that, in the latter case, p 2 ∈ E(A {1} ) \ Z p 1 has insufficient support size and thus does not enter the inequalities restricting α. Matrices found near A {1} may lie on either side of, or straight on, the boundary 2 3 -2 3 3 2 .
We conclude this example by some further remarks concerning the regions in Fig. 1, and their boundaries. For matrices A from one of the regions, we always have ESS(A) = E(A).  Table 1. We observe two types of behavior. Either for a matrix A on a boundary, |ESS(A)| is the minimum of the numbers of ESSs found in the two regions separated by that boundary, while |E(A)| is the maximum of those, and NSS(A) = ESS(A). Or ESS(A) = ∅, while is infinite. The relative interiors of the sides of the white triangle, that we added to the table, are also of the latter type, with the simplification, resulting from J A (p) = [1 : 6] in the second to last column, that NSS(A) = conv E(A). There may be further types of n − 2-dimensional boundaries for higher orders n.
In general, we would desire a more detailed picture, such as given in Fig. 2, where pattern(A) enters the definition of a "region" directly and not just via the sizes of the elements of pattern(A). This would, however, not change Fig. 1. The question arises, if those newly defined "regions", which in general have to be thought of as being confined to a hyperplane not containing the origin, are connected or even convex (the latter is not always the case, as can be seen in Fig. 1).     ([0, a, b, c, c, b, a] )

Results
We used the construction outlined in the previous subsection to find cyclically symmetric matrices of orders n ∈ [4 : 23] with many ESSs of support size k ∈ [2 : n − 2], with k either 2 or odd, when n is odd. We wanted those matrices to have integer entries to facilitate the computation of the set of ESSs. While our main goal was to maximize the number of ESSs, we also made efforts to keep the integer entries small. Our results can be found in Tables 2, 3  ESSs of support size 5, but also 2 "spurious" ESSs of support size 4. It was not our aim to get a "pure" result 8 5 by introducing further inequalities on α. Indeed, cases of "spurious" ESSs of slightly smaller support sizes than aimed at appear also in other places of the tables, mainly for even orders. In some cases that information was suppressed due to lack of space, but we give it here; the corresponding vector a can be found in row [x|a ] in Tables 2, 3   Table 2 Cyclically symmetric matrices of orders from 4 up to 19 with many ESSs of prescribed support size The bold number in the lower left of each box indicates the order of matrices in that box, and the row [14 3 | 1 5 8] in the box corresponding to order 7 tells us that the matrix C([0, 1,5,8,8,5,1] ) has 14 ESSs of support size 3 Table 3 Cyclically symmetric matrices of orders 20 and 21 with many ESSs of prescribed support size 1  8004  21  14  30  5  15  5  30  14  21  −1  11845254  39  151  298  435  523  544  505 and 4, where x is the first part of the regular-type faced entries in the five row vectors of the following four lines: For instance, for 24 7 2 6 3 4 with order n = 12 we have a = [2,7,7,5,6,7] from Table 2.
All calculations were done with Maple TM and the matrices listed in the tables were doublechecked with Reinhard Ullrich's program ref (rational equilibrium finder). For fixed n, the first task is to find the setĀ n of candidate matrices. Containing at most one matrix indexed by any of the subsets L of [1 : n ], its size is bounded by 2 n , but a further reduction in size to roughly 2 ϕ(n) 2 n , where ϕ denotes the Euler totient function, is possible, by keeping just one member of each orbit {A L , A τ L , A τ 2 L , . . .}, where τ is a permutation on [1 : n ] generating a cyclic group of order φ(n) 2 . That resulted in, e.g., |Ā 22 | = 394, which is much smaller then 2 11 .
Next, we fixed A = A L ∈Ā n of rank k − 1 and computed R L . Potential supports of members of R L are all subsets of [1 : n] of size k. Among a certain subset and all its shifts and reflected shifts only one set has to be taken into account, which reduces the number of potential supports to roughly 1 2n n k . For any such support I we then checked if there is p ∈ E(A) with I (p) = I , i.e., we numerically solved A I ×I p I = o, e p I = 1. If the system appeared to lack a unique solution, we discarded that support. We also did that, if the solution was outside the simplex k or apparently on its boundary. Only if the solution p was clearly in the interior of k , and L L,p = ∅ was fulfilled, did we include I in R L . E.g., for n = 22 and k = 10 we obtain 14938 (shift-and reflection-reduced) potential supports, of which not more than 280 belong to any set R L : There are 42 candidate matrices of rank 9 inĀ 22 . Corresponding sets R L satisfy 280 ≥ |R L | ≥ 1 and 11011 ≥ size(R L ) ≥ 22 for L in a suitable 42-element set. For each such L we have to maximize size(R L ) under the constraintsR L ⊆ R L , p∈R L L L,p = ∅. We would work our way down starting with the most promising L corresponding to size(R L ) = 11011, thus finding the largest value 4422 of size(R L ) for L = {1, 2, 3, 4, 11}, with size(R L ) = 6138 being the 28th-largest among the corresponding 42 values.
We replaced the "< 0"-inequalities originating in (9), that are present in the constraints of the maximization problem, by "≤ −δ"-inequalities for some fixed δ > 0. The problem can be regarded as a Maximum Feasible Subsystem Problem, see [16]. For small n, we solved the problem by exhaustive search, for larger n we used a heuristic approach to generate good lower bounds for the optimal value (basically, we repeatedly started with a list built from a random permutation of the points in R L and, starting at the beginning of the list, greedily included points p while maintaining nonempty intersection of the corresponding sets L L,p ). For n up to 19 we also computed upper bounds for the optimal values, derived from formulations as Min IIS (irreducible inconsistent subsystem) Cover problems [16], with those upper bounds matching the lower bounds. So we can be sure to have solved the maximization problems for n ∈ [4 : 19] to optimality, but we do not know whether some of our lower bounds for n ∈ [20 : 23] leading to entries in Tables 3 and 4 are actually the optimal objective values for the corresponding problems.
Having solved (or having found a good lower bound m for) the maximization problem means that we now know a matrix B α such that A + εB α will have many quasistrict ESSs.
As we want integer entries for our matrices, we would rather consider matrices r 1 A + r 2 B α with r 1 large and r 2 of moderate size, and we would round the entries of that matrix to the nearest integers. Clearly r 2 may not be too small, otherwise B α 's influence will get wiped out by the rounding. Also r 1 r 2 = 1 ε may not be too small, as we need ε small. So suitable r 1 , r 2 will lead to a matrix with m quasistrict ESSs of support size k, having typically very large integer entries. Computing a series of ever smaller multiples of that matrix and rounding again-which is what we did-can reduce the entries of the matrix to some extent. Trying to have integer entries close to the minimal values possible would require considering all α in the nonempty intersection of the sets L L,p and all matrices A as defined in Corollary 10, for all L for which size(R L ) = m. We did not pursue that approach.
Summarizing, for orders n ∈ [4 : 19] and prescribed support sizes k ∈ [2 : n − 2], with k odd when n is odd, we have constructed matrices in C n with a maximum number of quasistrict ESSs of support size k. Here maximality is meant among matrices in C n occurring, in positive proportion, arbitrarily close to some negative-semidefinite matrix of rank k − 1. We do, however, not claim that the matrices we found are close to negative-semidefinite ones in the sense that there is a continuous path in C n from one to the other along which the pattern does not change, as we do not know if there may be disconnected regions within C n that share the same pattern, so that our rounding procedure could have led to a jump from one region to another. This holds in particular in some cases where n = n 1 n 2 is not a prime number. If the best result we found for some fixed k was not better than what we got by constructing a matrix from suitable matrices of orders n 1 and n 2 using construction techniques from [5], we would list the latter in our table, when it had entries of smaller size. Note that, in view of Lemma 3 and Example 4, our approach need not have worked for k = n − 2, but it did: It is known from [6,Thm. 5], that max A∈C n |{p ∈ ESS(A) : |I (p)| = n − 2}| = n, and our construction method produced matrices with integer entries achieving this maximal value for all n ∈ [4 : 23].

Remark 5
Special attention to cyclically symmetric matrices has been payed by [6], from which we can deduce that and also if n is even and k = 2. The L.H.S. is, however, strictly smaller than the R.H.S. if n ≥ 3 and k = n −1, and also if n ≥ 3 is odd and k = 2. (See also the next section for the case k = 2.) Furthermore, the best lower bound for the constant γ introduced in [6, Thm. 2], that is currently known, is 15120 1 24 ≈ 1.4933, and originates from a matrix A ∈ C 24 satisfying |ESS(A)| = 15120, see [5]. The main reason we concentrated on cyclically symmetric matrices is however that they are particularly well suited for our perturbation method. There are only finitely many column spaces associated with matrices in C n , which, by invoking Corollary 10, led to finite setsĀ n of candidate matrices, something we cannot hope for if we replace C n by (other interesting subsets of) S n .

Even Support Sizes of ESSs When the Order is Odd
The construction method described in Sect. 2 is not suited for finding cyclically symmetric matrices of odd order that have many ESSs of even support size, simply becauseĀ n does not contain matrices of odd rank when n is odd. Remark 4 above addresses a related issue. Figure 2 nevertheless shows that for order n = 7 matrices with 7 ESSs of support size 4 can be found by perturbing the matrix A {1} ∈Ā 7 . Furthermore, it is seen in Fig. 2 that for some matrices of order 7, there are 7 ESSs of support size 2, but those matrices do not result from small perturbations of matrices in the setĀ 7 . In Fig. 2 regions of matrices sharing the same pattern are visualized by their shading, a particular support attached to some region always indicating just one member of a whole class of 7 cyclic shifts of that support. The (middle gray) 7 3 regions actually overlap, and their intersections, the dark gray triangle shaped regions, correspond to matrices with 14 ESSs of support size 3. The white triangle corresponds to matrices with 1 ESS of support size 7. The 7 4 regions are bounded by segments of the three ellipses (b − c) 2 = ab, (c − a) 2 = bc, and (a − b) 2 = ca. Two sides of the dark gray triangle shaped regions are actually curved, and are segments of the three hyperbolae ac + 3bc = ab + 2c 2 , ab + 3ac = bc + 2a 2 , and bc + 3ab = ac + 2b 2 , the third sides are the straight lines a = 0, b = 0, c = 0. The 7 2 regions are bounded by the lines a = 0, 2b = c, resp. b = 0, 2c = a, resp. c = 0, 2a = b. The union of the three lines forming the boundary of the 1 7 region satisfies the equation a 3 + b 3 + c 3 + 3(a 2 b + b 2 c + c 2 a) − 4(ab 2 + bc 2 + ca 2 ) − abc = 0. Figure 2 suggests that for given odd n, matrices in C n with many ESSs of even support size k could be found by perturbing a negative-semidefinite cyclically symmetric matrix A of rank k − 2 in such a way that points p ∈ conv {p 1 , p 2 }, with p 1 , p 2 ∈ E(A), |I (p 1 )| = |I (p 2 )| = |I (p 1 )∩ I (p 2 )|+1 = k −1, get perturbed into quasistrict ESSs of the perturbed matrix. (This should also be considered in the case of even n and k of any parity, as suggested by Fig. 1, where we can enter the 6 3 region by perturbing A {3} .) Since there is no straightforward way to adapt our construction method to that setting, we leave this approach to future research. We do however have results for k = 2, which are also given in Tables 2, 3 and 4.

Support Size 2
Fix A ∈ S n and consider the graph G with vertex set V (G) = [1 : n] and an edge {i, j} ∈ E(G) iff {i, j} = I (p) for some p ∈ ESS(A). Then we know from [19,Thm. 2 and Corollaries] that G is triangle-free. For a given triangle-free graph G let now A(G) = (a i j ) ∈ S n be defined by a ii = 0 for i ∈ [1 : n], a i j = 1 ⇔ {i, j} ∈ E(G), 1 ≤ i < j ≤ n, and a i j = −1 otherwise. Then we know from [9, Thm. 1], that pattern(A(G)) = E(G). So, as noted in [9,Thm. 5 and Corollaries], the maximum number of ESSs of support size 2 a matrix A ∈ S n can have is n 2 −1 4 , corresponding to a complete bipartite graph with maximally balanced partition. For even n, the matrix A(G), obtained from the complete bipartite graph G with [1 : n] partitioned into odd and even numbers, turns out to belong to C n , therefore n 2 4 = max A∈C n |{p ∈ ESS(A) : |I (p)| = 2}| is attained. Not so for n odd. Note that A(G) ∈ C n means that G is a circulant graph on V (G) = [1 : n], being determined by a subset S ⊆ [1 : n − 1] such that {i, j} ∈ E(G) ⇔ i − j ≡ s mod n for some s ∈ S. In order that G be triangle-free, S has to be a symmetric sum-free subset of [1 : n − 1], i. e., s ∈ S ⇒ n − s ∈ S, and s 1 + s 2 ≡ s mod n has no solution with s 1 , s 2 , s ∈ S. For small odd n, the maximal sizes m n of such symmetric sum-free subsets of = (1, 1, 1, 2, 2, 3, 3, 3, 3,4,5,4,5,5,6,7,6,6,7,7,9,8,8,9,9,11,9,10,10,10,13), and it is observed that n − 3 6 ≤ m n 2 ≤ n 5 holds for that range. Note that we are interested in m n 2 , because |E(G)| = |S| 2 n. The inequality n−3 3 ≤ m n holds for all odd n, as is seen by observing that {2k − 1 : k ∈ L n } ∪ {n + 1 − 2k : k ∈ L n }, where L n = [1 : n−3 6 ], is a symmetric sum-free subset of [1 : n − 1] of at least n−3 3 elements. The inequality m n 2 ≤ n 5 , on the other hand, is not surprising in the light of [10, Prop. 1.3, Thm. 1.5], from which we can deduce that the maximal size possible for a sum-free subset of [1 : n − 1] (without the requirement of being symmetric) is upper bounded by n 5 . Interestingly, this upper bound is attained infinitely often also by symmetric sum-free sets, i. e., the equality m n 2 = n 5 holds for all n ∈ {10k −5 : k ≥ 1}. This is seen by observing that {1, 4, 6, 9, 11, 14, . . . , 10k −6} (with differences of consecutive elements alternating between 3 and 2) is a symmetric sum-free subset of [1 : 10k − 6] of size 4k − 2.
So we conclude, that M n := max A∈C n |{p ∈ ESS(A) : |I (p)| = 2}| satisfies M n ≥ n(n−3) 6 for any odd n, M n = n 2 5 for infinitely many odd n, and M n > n 2 5 for no odd n, in particular, the maximum n 2 −1 4 for matrices constructed from bipartite graphs is never attained by M n for odd n ≥ 3.