1 Introduction
Simple games are cooperative games that are commonly used to describe realworld voting systems. Considering a fixed, finite set of voting members, a simple game is given by a collection of subsets of satisfying the monotonicity property: and implies . The sets in are called winning coalitions, and each subset of that is not in is called a losing coalition. A fundamental class of simple games are weighted games whose winning coalitions can be written as
for some and . It is a basic fact that every simple game is the intersection of finitely many weighted games, and hence we may define the dimension of a simple game to be the smallest number of weighted games whose intersection is .
Determining the dimension of (simple games associated to) realworld voting systems has been of particular interest in social choice theory, see, e.g., Taylor and Pacelli [8]. While many voting rules are actually weighted and hence have dimension one, examples of dimension two are given by the US federal legislative system [9] and the amendment of the Canadian constitution [6]. A voting rule of dimension three has been adopted by the Legislative Council of Hong Kong [2].
A new record was set with the change of the EU (European Union) council’s voting system by the Treaty of Lisbon in 2014. Based on the population data of 2014, Kurz and Napel [7] showed that its dimension is at least and at most , and they posed the exact determination as a challenge to the community. In response, Chen, Cheung, and Ng [1] were able to reduce the upper bound to .
We provide the first improved lower bound and show that the dimension is at least . Although we will not rely on this interpretation in what follows, the idea behind our lower bound is based on the observation that the dimension of a simple game can be seen as the chromatic number of a particular hypergraph : the nodes of are the losing coalitions, and a set of losing coalitions forms a hyperedge iff for every weighted game . The proof of Kurz and Napel [7] establishes that contains a clique of cardinality , which directly implies that the chromatic number of is at least . This idea has been used previously in the context of lower bounds on sizes of integer programming formulations [4, 5, 3]. While we have not found any simple subgraph of larger chromatic number, we will show that contains a hypergraph on nodes whose chromatic number is .
Outline.
In Section 2 we introduce the concept of nonseparable subsets of the losing coalitions of a simple game . A family of such subsets can be thought of as a subgraph of the above hypergraph. Moreover, we consider the notion of a cover for such a set , which can be seen as a nodecoloring of the respective subgraph with colors. Accordingly, we will see that if the dimension of is at most , then there exists a cover for each . In Section 3 we consider the simple game associated to the EU council and give a construction of a set , for which no cover exists. A proof of the latter fact will be given in Section 4.
2 Strategy
In what follows, we consider simple games on a common fixed ground set .
Definition 1.
Let be a simple game and be any set of losing coalitions of . We say that is nonseparable with respect to if every weighted game satisfies .
From the definition it is immediate that a simple game is weighted if and only if no set of losing coalitions is nonseparable. So, the existence of a single nonseparable set yields that the dimension of a simple game is at least two. To obtain a larger lower bound, the following notion will be useful.
Definition 2.
Let be a simple game with losing coalitions , and let be nonseparable with respect to . A cover of is a collection of sets such that

and

for all , .
In order to obtain a lower bound on the dimension, we will exploit the following observation.
Lemma 1.
Let be a simple game with nonseparable sets . If has dimension at most , then there exists a cover for .
Proof.
If has dimension at most , then there exist weighted games such that . For define as the intersection of the losing coalitions in and .
We claim that is a cover of . In order to show Property 1, first observe that holds. Now, for any we have and hence there is an with , which implies .
For Property 2, assume that holds for some and . This means that each coalition in is losing for , meaning that and are disjoint. This contradicts the fact that is nonseparable with respect to since is weighted. ∎
In what follows, we will consider the simple game associated with the EU council and construct a collection of nonseparable losing coalitions that does not permit a covering. By Lemma 1 this implies that the dimension must be at least .
3 Our Construction
Let us give a formal definition of the simple game associated to the EU council based on the population data of 2014, as considered by Kurz and Napel [7]. In 2014, the European Union consisted of members and hence we may fix . In the voting system of the EU council, a coalition is winning if

it contains at least of all members states and

it unites at least of the total EU population,
or

it consists of at least of the member states.
Denoting the weighted game associated with rule by and the simple game that represents the voting system of the EU council by , we thus have
Note that depends on the population of each member state. As in [7], we will work with the data depicted in Table 1. From these numbers, it can be seen that the following coalitions are losing with respect to .
Next, we construct nonseparable subsets with respect to that consist of the above losing coalitions. In order to verify that these subsets are indeed nonseparable, the following lemma is helpful.
#  Member state  Population  #  Member state  Population 

1  Germany  80 780 000  15  Austria  8 507 786 
2  France  65 856 609  16  Bulgaria  7 245 677 
3  United Kingdom  64 308 261  17  Denmark  5 627 235 
4  Italy  60 782 668  18  Finland  5 451 270 
5  Spain  46 507 760  19  Slovakia  5 415 949 
6  Poland  38 495 659  20  Ireland  4 604 029 
7  Romania  19 942 642  21  Croatia  4 246 700 
8  Netherlands  16 829 289  22  Lithuania  2 943 472 
9  Belgium  11 203 992  23  Slovenia  2 061 085 
10  Greece  10 992 589  24  Latvia  2 001 468 
11  Czech Republic  10 512 419  25  Estonia  1 315 819 
12  Portugal  10 427 301  26  Cyprus  858 000 
13  Hungary  9 879 000  27  Luxembourg  549 680 
14  Sweden  9 644 864  28  Malta  425 384 
Lemma 2.
Let be a simple game and let and be sets of winning and losing coalitions for , respectively, such that . If
holds for all , then is nonseparable with respect to .
Proof.
Consider any weighted game with and . Then we have
The last inequality holds because all elements of are contained in . Thus, there must exist some , such that
Therefore, we have and hence . Since this holds for any weighted game , is nonseparable with respect to . ∎
We claim that the following element subsets of the above losing coalitions are nonseparable.
(1) 
To see that each above set is nonseparable, we make use of Lemma 2 as follows. If , we have that and are contained in . Pick a set of states of minimum total population such that is contained in . For all above pairs it can be checked that is contained in . By construction, and satisfy the assumptions of Lemma 2 and hence is indeed nonseparable.
Otherwise, we may assume that . For all above pairs, exchanging the two members with the least population in with the member of largest population in , results in two winning sets . Again, and satisfy the assumptions of Lemma 2, implying that is nonseparable.
4 Proof that no 7cover can exist
For the sake of contradiction, let us assume that the nonseparable sets in (1) and (2) admit a cover. This implies that there exist sets such that

.
It can be easily verified that the only sets satisfying (i) are the following.
(3) 
In what follows, for a weightvector
, let us define the weight of a set as .Suppose first that none of the sets is equal to . In this case, consider the weightvector
and observe that the weight of each set in (3) that is distinct from is at most . Thus, the weight of each set is at most , and we obtain
a contradiction.
It remains to consider the case that one of the sets is equal to , say . Consider the weightvector
and observe that the weight of each set in (3) is at most , and that . Thus, we have
another contradiction. This completes our proof.
Acknowledgements
The second author would like to thank Gerhard Woeginger for bringing this topic to his attention.
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