A note on assignment games with the same nucleolus

We show that the family of assignment matrices which give rise to the same nucleolus forms a compact join-semilattice with one maximal element. The above family is, in general, not a convex set, but path-connected.


Introduction
The assignment game (Shapley and Shubik 1972) is the cooperative viewpoint of a twosided market. There are two sides of the market, i.e., two disjoint sets of agents, buyers, and sellers, who can trade. The profits are collected in a matrix, the assignment matrix. The allocation of the optimal profit should be such that no coalition has incentives to depart from the grand coalition and act on its own. In doing so, a first game-theoretical analysis of cooperation focuses on the core of the game. Shapley and Shubik show that the core of any assignment game is always nonempty. It coincides with the set of solutions of the linear program, dual to the classical optimal assignment problem. A recent survey on assignment games is Núñez and Rafels (2015).
Among other solutions, the nucleolus (Schmeidler 1969) is a "fair" solution in the general context of cooperative games. It is a unique core selection that lexico- graphically minimizes the excesses 1 arranged in a nondecreasing way. The standard procedure for computing the nucleolus proceeds by solving a finite (but large) number of related linear programs. As a solution concept, the nucleolus has been analyzed and computed in many cooperative games. Solymosi and Raghavan (1994) gives an algorithm for the computation of the nucleolus of the assignment game, computed in polynomial time. Recently, Martínez-de-Albéniz et al. (2013b) provides a new procedure to compute the nucleolus of the assignment game. An interesting survey on the nucleolus and its computational complexity is given in Greco et al. (2015). The description of the 2 × 2 case is discussed in Martínez- de-Albéniz et al. (2013a).
In this paper, we focus on the structure of the family of assignment matrices that give rise to the same nucleolus. The main contributions of the paper are the following: • The family of matrices with the same nucleolus forms a join-semilattice, i.e., closed by entry-wise maximum. The family has a unique maximum element which is always a valuation matrix. 2 • We show that the above family is a path-connected set and give a precise path, connecting any matrix of the family with its maximum element.

Preliminaries on the assignment game
An assignment market M, M , A is defined to be two disjoint finite sets: M the set of buyers and M the set of sellers, and a nonnegative matrix A = a i j i∈M, j∈M which represents the profit obtained by each mixed-pair (i, j) ∈ M × M . To distinguish the jth seller from the jth buyer, we will write the former as j when needed. The assignment market is called square whenever |M| = M . Usually, we denote by m = |M| and m = M . M + m denotes the set of nonnegative square matrices with m rows and columns, and M + m×m the set of nonnegative matrices with m rows and m columns.
Recall that M + m×m forms a lattice with the usual ordering ≤ between matrices. The maximum C = A ∨ B of two matrices A, B ∈ M + m×m is defined entry-wise, i.e., as c i j = max{a i j , b i j } for all (i, j) ∈ M × M . Given an ordered subset of matrices (i, j)∈μ a i j , and any coalition formed only by buyers or sellers has a worth of zero.
The main goal is to allocate the total worth among the agents, and a prominent solution for cooperative games is the core. Shapley and Shubik (1972) prove that the core of the assignment game is always nonempty. Given an optimal matching μ ∈ M * A M, M , the core of the assignment game, C(w A ), can be easily described as the set of nonnegative payoff vectors (x, y) ∈ R M + × R M + satisfying the following: and all agents unmatched by μ get a null payoff. Now, we define the nucleolus (Schmeidler 1969) of an assignment game, taking into account that its core is always nonempty. The excess of a coalition ∅ = R ⊆ M ∪ M with respect to an allocation in the core, (x, y) ∈ C(w A ), is defined as e (R, (x, y)) := w A (R) − i∈R∩M x i − j∈R∩M y j . By the bilateral nature of the market, it is known that the only coalitions that matter are the individual and mixedpair ones (Núñez 2004). Given an allocation (x, y) ∈ C(w A ), define the excess vector θ (x, y) = (θ k ) k=1,...,r as the vector of individual and mixed-pair coalitions excesses arranged in a non-increasing order, i.e., θ 1 ≥ θ 2 ≥ · · · ≥ θ r . Then, the nucleolus of the game M ∪ M , w A is the unique core allocation ν (w A ) ∈ C(w A ) which minimizes θ (x, y) with respect to the lexicographic order 3 over the whole set of core allocations. For ease of notation, we will use, for We use the characterization of the nucleolus of a square assignment game of Llerena and Núñez (2011) (see also Llerena et al. (2015)). To introduce this characterization, we define the maximum transfer from a coalition to another coalition.
Given any square assignment game M ∪ M , w A , and two arbitrary coalitions ∅ = S ⊆ M and ∅ = T ⊆ M , we define the following: for any core allocation (x, y) ∈ C (w A ). Llerena and Núñez (2011) gives a geometric characterization of the nucleolus of a square assignment game. They prove that the nucleolus of a square assignment game is characterized as the unique core allocation (x, y) ∈ C(w A ), such that for any ∅ = S ⊆ M and ∅ = T ⊆ M with |S| = |T |. In certain cases, the number of equalities can be reduced. Indeed, note that if T = μ(S) for some μ ∈ M * A M, M , then it holds δ A S,T (x, y) = δ A T ,S (x, y) = 0. Therefore, for this characterization, we only have to check (3) for the cases T = μ(S) for some optimal matching μ ∈ M * A M, M and any ∅ = S ⊆ M, that is: To analyze the non-square case, we can use two different approaches and we will apply both of them.
The first and classical approach consists in adding null rows or columns to make the initial matrix square. The added rows or columns correspond to dummy agents and they receive a null payoff at any core allocation and hence also in the nucleolus. At this extended square assignment matrix, we apply the previous geometric characterization. Notice that the number of coalitions to be checked grows quickly for each added agent.
To fix our first approach, we introduce some notation. Given any arbitrary assign- We know that the matching μ 0 = {(1, 1), (2, 2), . . . , (m , m )} is optimal for matrix A 0 . The second approach is an adaptation of the consistency property of the nucleolus (see Llerena et al. (2015)), and it can be found in Martínez-de-Albéniz et al. (2015). It keeps the dimension of the problem as low as we can and it has an interest on its own. Basically, it consists in reducing the assignment problem to an appropriate square matrix, dropping out those agents unassigned by an optimal matching, and reassessing the matrix entries. Apart from the dimension issue, the main feature of this approach is that we must not care about the added zero rows or columns when we deal with the matrix.
To introduce the second approach, we need some notations. Let M, M , A , A ∈ M + m×m be a non-square assignment market with m < m , and let μ ∈ M * A M, M be an optimal matching. Define the vector a μ = a μ i i∈M ∈ R M + by the following: and define the square matrix A μ ∈ M + m by Then, the relationship between their nucleolus is the following one: Moreover, the fixed matching μ is also optimal for matrix A μ . An example of the application of this second approach is the following. Consider matrix: where the optimal matching is denoted in boldface, μ = {(1, 1), (2, 2)}. Now, vector a μ = (7, 5) and matrix A μ ∈ M + 2 is given by the following: The nucleolus of the game w A μ is ν(A μ ) = (0.5, 6.25; 0.5, 0.75), and then, the nucleolus of the game w A is ν(A) = (7.5, 11.25; 0.5, 0.75, 0, 0, 0).

Assignment games with the same nucleolus
We introduce the family of matrices with a given nucleolus. To this end, for an arbitrary assignment matrix A ∈ M + m×m , we denote by the following: the family of matrices that share the same nucleolus as A.
It is clear that matrices with the same nucleolus must have the same worth for the grand coalition even if they do not have any optimal matching in common, consider, e.g., matrices: 1 0 0 1 and 0 1 1 0 .
For any assignment game, there exists a unique matrix, its buyer-seller exact representative, among those leading to the same core, such that no matrix entry can be raised without modifying its core (Núñez and Rafels 2002). From Núñez (2004), assignment games with the same core have the same nucleolus. In particular for each matrix in [A] ν , its corresponding buyer-seller exact representative always belongs to the family. Nevertheless, as we will see, assignment matrices with different cores may also share the same nucleolus.
We focus now on the structure of this family [A] ν : it is a nonempty compact joinsemilattice 4 with a unique maximal element. Second, we characterize this maximum and show that it is a specific type of assignment matrix, a valuation matrix.
Theorem 1 Let A ∈ M + m×m be an assignment matrix. The family [A] ν forms a compact join-semilattice with a unique maximal element.
Proof First, we prove that this family is a join-semilattice. Let B, B ∈ [A] ν . If m = m , we add zero rows or columns to make the matrices square, recall (5). It is known that these rows or columns correspond to dummy players which obtain zero payoff at any core allocation, and also in the nucleolus. Therefore, we can assume from now on that matrices are square. We have B, B ≤ B ∨ B , and also C(w B ) ∩ C(w B ) = ∅, since both games share the nucleolus. We claim the following: Then, for any optimal matching μ of matrix B ∨ B , we have the following: As a consequence, since (x, y) is the nucleolus of w B and w B , we obtain the equality δ B∨B S,T (x, y) = δ B∨B T ,S (x, y) , proving that B ∨ B ∈ [A] ν . Now, we show that this family is a compact set, and, therefore, with a unique maximal element. We show that it is bounded and closed. It is bounded, In contrast with the previous result, the minimum defined entry-wise of two matrices with the same nucleolus may not have the same nucleolus, see matrices in (10). Now, we introduce a kind of assignment matrices, useful for our purposes. A matrix A ∈ M + m×m is a valuation matrix 5 if for any i 1 , i 2 ∈ {1, . . . , m} and j 1 , j 2 ∈ {1, . . . , m }, we have a i 1 j 1 + a i 2 j 2 = a i 1 j 2 + a i 2 j 1 . Clearly, this definition is equivalent to see that any 2 × 2 submatrix has two optimal matchings. Obviously, any fully optimal 6 square matrix is a valuation matrix, and for square matrices, the converse also holds. This characterization fails for non-square matrices as the following matrix shows: D = ⎛ ⎝ 3 6 8 1 0 4 7 9 2 1 6 9 11 4 3 ⎞ ⎠ .
This is a valuation matrix, but clearly not all matchings are optimal. Finally, we want to point out two general properties for non-square valuation matrices. Let A ∈ M + m×m be a non-square valuation matrix with m < m and μ ∈ M * A M, M any optimal matching, Then: (ii) The entries of matrix A satisfy a i j 1 ≥ a i j 2 for all i ∈ M, j 1 ∈ μ(M) and j 2 ∈ M \μ(M).

Theorem 2 Let A ∈ M + m×m be an assignment matrix. The maximal element of the family [A] ν is a valuation matrix.
Proof Let ν(A) = (x, y) ∈ R M + × R M + be the nucleolus of matrix A ∈ M + m×m , where we assume without loss of generality that m ≤ m and μ = {(1, 1), (2, 2), . . . , (m, m)} is an optimal matching for A.
Define now matrix A ∈ M + m×m as follows: We claim: A is the maximum of the family [A] ν , and clearly a valuation matrix.
To prove claim (i), let A 0 , M 0 , and μ 0 be the notation introduced in (5) to make square the initial matrix A. We denote by We know ν A 0 = x 0 , y 0 , and then, {y j }, and Here, we have used that y 0 j = y j = 0 for j ∈ M \μ(M) and From the above equality, we easily deduce x i ≥ min {y j }, and  (x, y). Clearly, a i j = x i + y j ≥ b i j for 1 ≤ i, j ≤ m.
If m = m , we are done, and B ≤ A. Otherwise, m < m . Consider matrix B 0 , see (5). We know ν(B 0 ) = (x 0 , y 0 ). Then {y j }, and We obtain for all i ∈ M and j ∈ M \μ(M), This ends our third claim, and proves the maximality of matrix A, since we have seen that, in the non-square case, B ≤ A. The fact that A is a valuation matrix is left to the reader.
From the proof of Theorem 2, we expect several valuation matrices if the initial assignment matrix is not square. In (11), we have introduced matrix D ∈ M + 3×5 which is an example of such a situation. By (4), (8), and (9), it is easy to check that the nucleolus of matrix D is ν(D) = (2, 3, 5; 1, 4, 6, 0, 0) and the maximum matrix of which is strictly greater than the valuation matrix D. Both valuation matrices share the same nucleolus.
In the proof of Theorem 2, we have found the expression of the maximum element of family [A] ν , with ν(A) = (x, y). It is matrix A ∈ M + m×m as follows: where μ ∈ M * A M, M is an optimal matching. A close look at (12) could raise expectations of different maximum matrices A depending on the chosen optimal matching μ, but this is not the case, as it is easy to check. The Let us define the set formed by the distances that appear in the geometric characterization of the nucleolus, see (3), except for the grand coalition: These elements are used for the characterization of the nucleolus and correspond to the minimum of some numbers. The elements of Δ(B) can be ordered increasingly: and then Δ(B) = {δ B 0 , δ B 1 , . . . , δ B r * }. From these parameters, we can define a new matrix B 0 with the same nucleolus. We set b 0 i j = b i j if x i + y j − b i j ∈ Δ(B), and we raise the worth of entry b i j to b 0 i j in such a way that x i + y j − b 0 i j equals the closest one-below element of Δ(B), that is, if δ B k < x i + y j − b i j < δ B k+1 for some k, then b 0 i j = x i + y j − δ B k . It is clear that matrix B 0 has the same nucleolus as matrix B since the equalities of the geometric characterization of the nucleolus have not changed, and therefore, B 0 ∈ [A] ν . Moreover Δ(B) = Δ(B 0 ). We may choose increasing linear paths from B to B 0 , one for each entry to raise. Notice that since we are moving up the entries that do not determine the distances of Δ(B), all matrices on these paths will preserve the original nucleolus. Now, we have a matrix B 0 ∈ [A] ν , such that x i + y j − b 0 i j ∈ Δ(B 0 ) for all (i, j) ∈ M × M . Moreover if δ B 0 S,T (x, y) = δ B r * , for some S ⊂ M and T ⊂ M with |S| = |T | = m, then we have, for all i ∈ S and j / ∈ T , x i + y j − b 0 i j = δ B r * . We finish the proof in the square case by raising the entries of matrix B 0 iteratively up to get A. 7 A path in X ⊆ M + m×m from A to B, A, B ∈ X , is a continuous function f from the unit interval I = [0, 1] to X , i.e., f : [0, 1] → X , with f (0) = A and f (1) = B. Moreover, a subset X ⊆ M + m×m is path-connected if, for any two elements A, B ∈ X , there exists a path from A to B entirely contained in X . j ∈ M \μ(M). For any i = i * , i ∈ M, such that x i > min i∈M x i or equivalently x i = x i − b μ i > min i∈M x i = min j∈μ(M) y j , that is b μ i < x i − min j∈μ(M) y j , we can raise at the same time entries b i j = b μ i to a i j = x i − min j∈μ(M) y j for all j ∈ M \μ(M) without changing the nucleolus, as the reader can check applying (8) and (9). This ends the proof.