MATHEMATICA APPLICANDA
Vol. 42(2) 2014, p. 259–272
doi: 10.14708/ma.v42i2.524
Andrzej Tomski (Cracow)
A toy model of a weighted voting game including
two strong players
Abstract The aim of this article is to compare a specific toy model of voting with
the results derived from normal approximation techniques by Słomczyński et al. in
previous papers. For this purpose, we construct a model in which the optimal quota
defining a qualified majority is estimated. The optimal quota is set in such a way that
the voting power of each member of the voting body, measured by the (normalised)
Penrose-Banzhaf index, is proportional to its voting weight. We present the ‘FranceGermany’ model of two strong players, both of which are c > 1 times stronger than
each of the others and we estimate the quota in the case of c = 2. We check that
these results are consistent with a formula derived from the normal approximation,
where the quota we are looking for is the inflection point of the density function for
this distribution.
2000 Mathematics Subject Classification:
Primary 91B14, Secondary 91B12.
Key words and phrases: toy model of voting with two strong players, PenroseBanzhaf index, optimal quota, normal approximation.
1. Introduction The theory of weighted voting games has become a
significant area in game theory. According to this theory, each representative
of a population, being a member of the voting body, has a vote of a certain
(and usually normalised) weight. Obviously, different members may have
different voting weights. It is assumed that all the voters can only accept
(vote ‘yes’) or reject (vote ‘no’) a proposition. A decision is passed if the sum
of the weights reaches a certain number, called the quota. Thus, a winning
coalition is a group of members which reaches the required quota (in other
words: has a qualified majority). After World War II the concept of a priori
voting power was introduced by a British mathematician, Lionel Penrose, as
the probability that a voter is decisive, i.e. a change in his vote will simultaneously create a change in the result to opposite; equivalently a winning
coalition loses a qualified majority (see [5], [6]). In general, a voting power
index describes a “reasonable expectation” of the share of decisional power
among members of the voting body, given by their impact on the formation
of a winning coalition ( [10]). Although a number of competing definitions of
voting power indexes are known, Penrose’s idea led to the Penrose-Banzhaf
index (the other most widely used index is called the Shapley-Shubik index ),
260
A toy model of a weighted voting game including two strong players
also named after John F. Banzhaf III, who investigated the theory of public
choice independently. In this paper we will use the Penrose-Banzhaf index
as a voting power index for all of the voters.
Since 2014, according to the Treaty of Lisbon, the Council of the European Union will take decisions by a double majority rule. This system
requires at least 55 percent of voters saying ‘yes’, while representing at least
65 percent of the whole European Community for a decision to be made.
It is believed that this system favours the biggest and smallest countries in
the European Union, whereas it to disfavours medium-sized ones. In addition, the system does not seem to be representative enough, since individual
inhabitants have unequal influence on the results of voting.
However, the following central question arises: what type of relationship
should exist between these two different terms: the voting weight and the
voting power? How to set a quota needed to a reach qualified majority? The
rule proposed in [8] states that the voting weight should be proportional to
the voting power. Under this assumption, the optimal quota, that is the
number for which the distance (in the ‘mean-quadratic’ sense) between the
voting weight and the voting power is minimised, is to be investigated and
found. Under this system, each individual resident has almost the same
influence.
This paper is organised as follows. In Section 2 we recall some basic
definitions, namely the Penrose-Banzhaf index, mean discrepancy and the
optimal quota. Next, in Section 3 we construct and analyse a toy model
with two strong players (the ‘France-Germany’ model) where both are c > 1
times stronger than the rest of the group. This is an extension of the model
with one strong player analysed in [9]. We present a general method for
analysing such models using Gosper’s algorithm (introduced in [3]) and we
estimate the optimal quota succesfully in the case of c = 2. Section 4 is
devoted to a comparison of the formula taken from [8] and the estimate from
our toy model. Section 5 presents an alternative idea for estimating the
optimal quota, which can be applied to more complicated types of such toy
models. The final section summarizes the results of the study and contains
an overwiew of some open problems to investigate in the future.
2. The theory of weighted voting games Let N = {1, ..., M } be
a non-empty finite set. This set is called a voting body, every element of N
is a player and every A ⊂ N is called a coalition. It is obvious that the
set of all the possible coalitions is equal to the power set of N. Moreover,
we assume that each player casts a ‘yes’ or ‘no’ vote (there is no abstention)
independently with equal probability 21 and thus all 2M coalitions are equally
likely (see [1], [9]).
The simple voting game is a collection W of subsets of the voting body,
which satisfies the following conditions:
261
Andrzej Tomski
1. N ∈ W,
2. ∅ ∈
/ W,
3. if X ⊆ Y ⊆ N and X ∈ W, then Y ∈ W.
A simple voting game is proper if X ∈ W implies N \X ∈
/ W. We say that a
coalition A is winning, if A ∈ W.
A pair (N, W ) is called a weighted voting game if there exists a nonnegative function w : N → [0, ∞) and a number q ∈ [0, ∞) such that
X
A ∈ W if and only if
w(k) ≥ q.
k∈A
We denote such a game by [q; w1 , ..., wk ]. If
P
k∈N
wk = 1 and 1 > q > 12 ,
then the weighted voting game is proper. The function w is called the voting
weight and q is called the quota.
In this paper we consider a voting body with M participants.
Let wk (k =
PM
1, ..., M ) be the voting weight of the k-th player and i=1 wi = 1. We say that
the voting body takes a decision if the sum of the weights reaches or exceeds
q ∈ ( 12 , 1] and we analyse a weighted voting game V := [q; w1 , ..., wM ].
We call k a decisive player in a coalition A ∈ W if k ∈ A and A\{k} 6∈ W .
In other words, k is decisive if the coalition fails without his support.
The Banzhaf score of the k-th participant, nk (V ), is the cardinality of
the set
{C ∈ W : k ∈ C, C\{k} ∈
/ W },
that is the number of the coalitions in which k is decisive.
The absolute Penrose-Banzhaf index (PBI) , ψk (V ), of the k-th participant is the probability that k is decisive, in other words k is a critical voter
in the coalition. If k is not critical in any coalition, we call him a null player.
Moreover, the absolute Penrose-Banzhaf index of the k-th participant is equal
to (see [9])
nk (V )
ψk (V ) := M −1 .
2
The normalised Penrose-Banzhaf index is
ψk (V )
nk (V )
βk (V ) := PM
= PM
.
i=1 ψi (V )
i=1 ni (V )
(1)
For the sake of simplicity, we denote βk := βk,q := βk (V ) for the rest
of the article. We define a function which (in the ‘mean-quadratic’ sense)
measures the difference between the voting power and the voting weight.
This function
v
u
M
u1 X
t
∆(q) :=
(βi − wi )2
M
i=1
262
A toy model of a weighted voting game including two strong players
is called the mean discrepancy and an optimal quota is a quota which minimizes ∆. In general, this number is not unique and often an interval of
optimal quotas exists. In such a case, the length of this interval decreases
as M increases. The analysis done in [8] leads to the conclusion that in a
generic situation the ratios βk /wk , k = 1, ..., M are approximately close to 1
for some q∗ . In other words,
βk,q∗ ≈ wk ,
for k = 1, ..., M and this point q∗ is the optimal quota.
In the case of only two types of player exist: small ones and large ones,
where the weights of all players of the same type are equal, we have just two
functions: βk /wk , (k = S, L), so it is relatively easy to find q∗ , a point at
which
βL,q∗
βS,q∗
≈
.
wL
wS
This remark is a starting point to analysing toy models in the subsequent
chapters. The first of the models was introduced in [9]. It describes a voting
body where one strong player is two times stronger than each of m := M − 1
weak players. It has been shown that the optimal quota q∗ satisfies the
following inequalities:
√
√
M+ M
2+M + M
6 q∗ 6
.
2(M + 1)
2(M + 1)
This means that for any large number M, we have q∗ '
1
2
1+
√1
M
.
3. The ‘France-Germany’ model Let us consider a model with two
strong players. Both of them (we can call them ‘France’ and ‘Germany’)
are stronger than each of the other participants of the voting body and have
voting weights equal to w1 = wG , w2 = wF , respectively and m := M − 2
small players with weights equal to w3 = ... = wR = 1−(wFm+wG ) . We assume
that wF + wG 6 12 in order to prevent the two strong players from deciding
on their own. We also assume that the number of small players is greater
than four.
We introduce the weight ratios:
c1 =
wG
wF
, and c2 =
.
wR
wR
We can compute the weights:
wR =
wG
wF
=
.
c1
c2
Andrzej Tomski
263
It leads to the formulas:
wF =
c2
c1
, and wG =
.
c1 + c2 + m
c1 + c2 + m
and thus we obtain the game [q, c1 +cc12 +m , c1 +cc22 +m , c1 +c12 +m , ..., c1 +c12 +m ].
For further calculation, we assume that q 6 m+cm1 +c2 , bearing in mind
that without this restriction some of the formulas below are not true or they
can be simplified significantly.
We want to know when ‘Germany’s’ vote is decisive (the case of ‘France’
is almost identical). This happens in two cases. First, ‘Germany’ can form a
coalition with k + 1 countries when one of those k + 1 countries is ‘France’,
thus we get
k(1 − (wF + wG ))
+ wF < q
m
and
k(1 − (wF + wG ))
+ wF + wG > q.
m
Using the ceiling function dxe, which is the smallest positive integer
greater than or equal to x, we consider
j1 + 1 ≤ k ≤ j2 ,
where
j1 :=
(q − wF − wG )m
−1
1 − (wF + wG )
and
(q − wF )m
j2 :=
− 1.
1 − (wF + wG )
The second option for ‘Germany’ could be to form a coalition omitting
‘France’, choosing l countries such that
l(1 − (wF + wG ))
<q
m
and
264
A toy model of a weighted voting game including two strong players
l(1 − (wF + wG ))
+ wG > q,
m
i1 + 1 ≤ l ≤ i2 ,
where
(q − wG )m
i1 :=
− 1,
1 − (wF + wG )
qm
i2 :=
− 1.
1 − (wF + wG )
Thus the value of the Penrose-Banzhaf index for ‘Germany’ is
Pj2
k=j1 +1
ψG =
m
k
+
Pi2
k=i1 +1
m
k
.
2m+1
Arguing similarly, the value of the Penrose-Banzhaf index for ‘France’ is
Pi1
m
k
k=j1 +1
+
Pi2
k=j2 +1
m
k
.
2m+1
Now we will analyse the options of a weak player. Firstly, he can form
a coalition with i2 other weak players. Secondly, he can cooperate with one
of the strong players and i1 or j2 of the small ones, respectively. Finally, he
can vote in the same way as the two strong players and j1 small ones.
The Penrose-Banzhaf index of a weak player is equal to
m−1
+ m−1
+ m−1
+ m−1
j1
j2
i1
i2
ψS =
.
2m+1
We assume now that ‘France’ and ‘Germany’ have equal voting weights.
Thus, the indicators will be as follows:
ψF =
j1 :=
j2 :=
i1 :=
i2 :=
(q − 2wG )m
− 1,
1 − 2wG
(q − wG )m
− 1,
1 − 2wG
(q − wG )m
− 1 and
1 − 2wG
qm
− 1.
1 − 2wG
265
Andrzej Tomski
As i1 = j2 , we deduce that
Pi2
m
k=j1 +1 k
2m+1
ψG =
= ψF
(2)
and
ψS =
m−1
j1
m−1
j2
m+1
2
+2
+
m−1
i2
.
(3)
Using (1), (2) and (3) we get
r
∆(q) =
=
=
=
=
1p
2(βG − wG )2 + m(βS − wS )2
M
r s
2βG − 2wG 2
1
2
2(βG − wG ) + m
M
m
√
2
√ |βG − wG |
m
Pi2
√ m
2 c k=j1 +1 k
√ P
− m + 2c m
m−1
m−1
m−1
m 2 i2
+
m
+
2
+
k=j1 +1 k
j1
j2
i2
√
2
√
m
Pdde−1
m)
(
c k=dd−2ce k
− m + 2c ,
Pdde−1
m−1
m−1
m−1
2 k=dd−2ce m
+
+
m
+
2
dde−1
dd−ce−1
k
dd−2ce−1
where d = (m + 2c)q.
The case c = 2 can be treated analytically. We introduce the parameter
r := dde and now we consider ∆ as a function of r. The whole formula,
consequently, can be expressed as follows:
r
2 ˆ
∆(r)
m
r
2 1 a(m, r) =
,
m m + 4 b(m, r) ∆(r) =
where
ˆ
∆(r)
m
m
m
( m )+(r−3
)+(r−2
)+(r−1
)
−
:= 2 m + m + r−4
m
m
m−1
+ m−1
((r−4) (r−3) (r−2)+(r−1))+m(( r−5 )+2(m−1
r−3 ) ( r−1 ))
2 m+4 ,
266
A toy model of a weighted voting game including two strong players
a(m, r) := −192 − 142m − 57m2 − 12m3 − m4 + 400r + 230mr + 60m2 r +
6m3 r − 280r2 − 110mr2 − 14m2 r2 + 80r3 + 16mr3 − 8r4
and
b(m, r) := 124 + 120m + 55m2 + 12m3 + m4 − 220r − 144mr − 40m2 r − 4m3 r +
144r2 + 60mr2 + 8m2 r2 − 40r3 − 8mr3 + 4r4 .
To obtain formulas for a and b, we use Gosper’s algorithm (see [3] for
details). This algorithm accelerates the calculation of Newton sums remarkably. We define f (m, r) := |a(m, r)| . To find the minimum of ∆, we would
like to find a positive number r which solves the equation
f (m, r) = 0.
(4)
It is easy to see that the following lemma holds:
Lemma 3.1 The graph of the function f (m, ·) is symmetric with respect to
r = m+5
2 .
This means that if the function f has a zero (and it is not the center
of symmetry), then it must have at least two real zeroes. Because of the
definition of r and the assumption that the quota should not be less than 12 ,
we look for the largest one.
ˆ
It should be emphasised that ∆(r) (and so ∆(r))
is not a continuous
function of the quota and the problem does not have a unique solution (this
case was discussed in [9]). However, let
m 5
t := r −
+
,
(5)
2
2
then (4) takes the form:
1
9
9
1
−8t4 + 10mt2 + 20t2 − 2t2 m2 + m3 + m2 − m − = 0.
2
2
2
2
The equation above has four solutions, two of them are complex and the
remaining two are real. We choose the largest solution:
1
t0 =
4
q
p
−2m2 + 10m + 20 + 2 m4 − 6m3 + 9m2 + 64m + 64.
Substituting M = m + 2 and using (5) we get
q
p
1
3 1
r0 = M + +
−2M 2 + 18M − 8 + 2 M 4 − 14M 3 + 69M 2 − 76M + 36.
2
2 4
267
Andrzej Tomski
It remains to verify that the numerator and the denominator of ∆ do not
have common zeroes. To do this, we can compute the resultant of both of
the polynomials a(m, ·) and b(m, ·), and use the following lemma:
Lemma 3.2 ( [2], Ch. XII) Two polynomials have a common zero if and
only if their resultant is equal to 0.
We calculate the resultant of P and Q:
R(P, Q) = 9437184+81788928m+321519616m2 +758841344m3 +1199771648m4 +
1342738432m5 +1096000512m6 +662433792m7 +297848832m8 +99151872m9 +
24066048m10 + 4132864m11 + 475136m12 + 32768m13 + 1024m14 .
We conclude that the polynomials a(m, ·) and b(m, ·) do not have common
zeroes.
From the definition of r0 , it can be shown that there exists a point q∗ for
which the function ∆ reaches its minimum and which satisfies the following
inequalities:
(M + 2)q∗ 6 r0 6 (M + 2)q∗ + 1.
We get
M +1
+
2M + 4
p
√
−2M 2 + 18M − 8 + 2 M 4 − 14M 3 + 69M 2 − 76M + 36
6 q∗
4(M + 2)
and
M +3
q∗ 6
+
2M + 4
p
√
−2M 2 + 18M − 8 + 2 M 4 − 14M 3 + 69M 2 − 76M + 36
.
4(M + 2)
We examine the asymptotic behaviour of q∗ . As in the model from [9], it
can be shown that
√
lim
M →∞
√
−2M 2 +18M −8+2 M 4 −14M 3 +69M 2 −76M +36
4(M +2)
√1
2 M
so
1
q∗ ' qs :=
2
1
1+ √
M
.
= 1,
268
A toy model of a weighted voting game including two strong players
4. A comparison of the results obtained from the toy model
with the conclusions from the normal approximation Let us assume
that the voting body contains M participants and wk , where k = 1, ..., M,
denote their respective normalised voting weights. We consider a weighted
voting game [q; w1 , ..., wM ] with q ∈ ( 12 , 1]. The histogram of the sum of the
weights, divided by the total number of all coalitions:
P
card{I ⊂ {1, ..., M } : i∈I wi = z}
n(z) =
2M
was introduced in [1] and proved
distribuP to have1 an approximately
PM normal
2 , that is:
2 := 1
w
=
w
tion with parameters µ := 12 M
and
σ
i=1 i
i=1 i
2
4
N (q) :=
X
z≤q
Zq
n(z) ≈
−∞
−(x−µ)2
1
√ e 2σ2 dx = Φ
σ 2π
q−µ
σ
.
One observation from [8] states that for quotas close to the inflection
point qn := µ + σ of the density function for the normal distribution, the
ratios wβkk (k = 1, ..., M ) are close to 1, so the quota qn approximates the
quota are looking for. This approximation leads to the formula ( [9]):
v


uM
uX
1
q∗ ' qn (w1 , ..., wM ) := µ + σ =
1+t
wi2  .
2
i=1
The results from the ‘France-Germany’ model and those from the normal
approximation are asymptotically equal. This means that if there are two
strong players and √each of them is twice as strong as each of the other players,
+6
and
then µ = 12 , σ = 12 MM+2
1
qn =
2
√
M +6
1
1
1+
'
1+ √
= qs for large M.
M +2
2
M
5. Estimation of the zeroes of ∆: an alternative solution In
Section 3 we found the minimum of the mean discrepancy analytically. Now
we will present one more method of finding the solution. If we can find the
zero of a(m, ·) and this function is increasing, then the minimum of ∆ is
attained at one of the two integers closest to the zero. This stems from the
fact that if a is increasing, then |a| decreases to the root, and then increases
again. Therefore, the function attains its minimum in this set. A similar
fact is true for decreasing functions. We want to minimize ∆ as a function of
r = d(m + 4)qe and 5 6 r 6 m. Later, we will use these formulas to estimate
the optimal quota.
Andrzej Tomski
269
Analysing the graph of this function we suppose that it is increasing in
(−∞, m+5
2 ), and by symmetry it decreases in the complement of this interval
to R.
We can use the fact that if a function is positive and increasing, then its
reciprocal is decreasing. We define
m
m
m
m
x :=
+
+
+
r−4
r−3
r−2
r−1
and
m−1
m−1
m−1
y :=
+2
+
,
r−5
r−3
r−1
˜ is equal to x −
then ∆
2x+my
2 m+4 .
We conclude that it is sufficient to show
˜
the monotonicity of the function ∆(r)
:= xy . After simplification, we get
˜
∆(r)
= ã(m,r)
, where:
b̃(m,r)
ã(m, r) := 96 + 78m + 39m2 + 10m3 + m4 − 200r − 120mr − 36m2 r − 4m3 r +
140r2 + 56mr2 + 8m2 r2 − 40r3 − 8mr3 + 4r4
and
b̃(m, r) := 14 + 7m + m2 − 10r − 2mr + 2r2 .
˜ in the
We will show that there exists only one zero of the derivative of ∆
interval (5, m), being the point of symmetry (due to its simplicity, we omit the
˜ is increasing in (5, m+5 ) and decreasing
case m = 5). This implies that ∆
2
in the complement. We use Sturm’s algorithm (see [7] for details). After
reduction, we obtain the following table showing changes in sign:
r=5
- + + - - +
r=m + + - - + +
Table 1: The table of changes in sign according to Sturm’s algorithm.
The number of changes in sign at r = 5 is greater than at r = m, so the
derivative of f˜ has exactly one zero in the interval (5, m). It can be easily
˜ is
checked that it is the point of symmetry, consequently concluding that ∆
m+5
m+5
decreasing in ( 2 , m), and increasing in (5, 2 ) (it is easy to show that
the reverse case is impossible).
˜ is monotonic
Example 5.1 We take m = 25 (see Figure 1). We know that ∆
m+5
for r > 2 , so the minimum should be obtained at one of the two integers
270
A toy model of a weighted voting game including two strong players
Figure 1: The graph of g :=
a(m,r)
b(m,r)
in the case m = 25.
closest to the zero. Moreover, the point of symmetry is r = 15, so the
minimum is reached at 18. We have r = d29qe , which implies the following
estimate:
17
18
6 q∗ 6 .
29
29
6. Conclusion In this paper we have constructed and discussed a particular model of voting: the ‘France-Germany’ model with two strong players each of which is c > 1 times stronger than the other players. We have
searched for the optimal quota for such a game, that is the quota which minimizes the difference between the voting power and the voting weight in the
‘mean-quadratic’ sense. We have estimated the optimal quota succesfully in
the case of c = 2. We compared these results with another formula for the
quota derived from normal approximation techniques. Approximation (for
large values of the total number M of players) of q∗ by the inflection point
of the normal curve is consistent with the one obtained from this model. For
further discussion on the advantages of this approach, we refer the reader
to [9]. However, it remains to find a method for analysing and estimating
the optimal quota under this model, in the case c 6= 2 and, in particular,
for much more complicated types of models. Especially, what happens if the
number of strong players increases and they have different voting weights ?
7. Acknowledgements The author wishes to thank Professor Wojciech
Słomczyński for his support and Doctor Józef Piórek for help in preparing
the manuscript.
Andrzej Tomski
271
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[10] F. Turnovec, J. Mercik and M. Mazurkiewicz, Shapley-Shubik or Penrose-Banzhaf ? Badania operacyjne i systemowe. Podejmowanie decyzji. Podstawy metodyczne i zastosowania.
Pod red. Romana Kulikowskiego, Janusza Kacprzyka, Romana Słowińskiego, Warszawa:
”Exit”, 2004, 121–127.
1
2
Available at http://xxx.lanl.gov/ftp/cond-mat/papers/0405/0405396.pdf.
Available at http://www.homooeconomicus.org.
272
A toy model of a weighted voting game including two strong players
Model-zabawka z dwoma silnymi graczami w grze ważonej
większości
Streszczenie Celem niniejszego artykułu jest porównanie szczególnego typu wyborczego modelu-zabawki z wynikami uzyskanymi przy pomocy metod aproksymacji
normalnej przez Słomczyńskiego i innych w poprzednich pracach. Dla tego celu skonstruowaliśmy model, w którym zostaje oszacowany próg optymalny dla większości
kwalifikowanej. Próg optymalny jest to próg, który minimalizuje różnicę między siłą
głosu członka ciała decyzyjnego, mierzoną za pomocą (znormalizowanego) indeksu
Penrose‘a-Banzhafa, a jego wagą głosu. Wprowadzamy model ‘Francja-Niemcy’ z
dwoma silnymi graczami, w którym każdy z nich jest c > 1 razy silniejszy od reszty
i szacujemy próg optymalny w przypadku c = 2. Sprawdzamy, że te wyniki są
zgodne z oszacowaniem uzyskanym z aproksymacji normalnej, gdzie szukany próg
jest punktem przegięcia krzywej gęstości rozkładu normalnego.
2000 Klasyfikacja tematyczna AMS (2010):
Primary 91B14, Secondary 91B12.
Słowa kluczowe:
model-zabawka z dwoma silnymi graczami, indeks Penrose‘aBanzhafa, próg optymalny, aproksymacja normalna.
Andrzej Tomski obtained his MSc degree in applied
mathematics from the Jagiellonian University in 2011.
His research interests are: biomathematics, bioinformatics, weighted voting games and condensed matter
physics. He cooperates with the Institute of Pharmacology of the Polish Academy of Sciences in investigating
interactions between genes and with the Department of
Condensed Matter Theory and Nanophysics at the Jagiellonian University
in the field of unconventional superconductivity. Currently, he is preparing
his PhD thesis on biological models described by a piece-wise deterministic
Markov process under the supervision of Professor Ryszard Rudnicki. In his
free time, he enjoys hummus, good coffee and is inspired by Morrissey.
Andrzej Tomski ukończył zastosowania matematyki na Uniwersytecie Jagiellońskim w 2011 roku. Swoje zainteresowania naukowe koncentruje wokół
biomatematyki, bioinformatyki, ważonych gier wyborczych i fizyki materii
skondensowanej. Był współpracownikiem Instytutu Farmakologii Polskiej
Akademii Nauk w zakresie implementacji metod analizy danych z regulacji genów i jest współpracownikiem Zakładu Teorii Materii Skondensowanej
Instytutu Fizyki Uniwersytetu Jagiellońskiego w zakresie badania niekonwencjonalnego nadprzewodnictwa. Jest w trakcie przygotowywania pracy
doktorskiej z modeli biologicznych opartych o procesy Markowa kawałkami
deterministyczne pod opieką profesora Ryszarda Rudnickiego. Prywatnie;
lubi hummus, dobrą kawę i ceni twórczość Morrisseya.
Andrzej Tomski
Jagiellonian University
Institute of Mathematics
Łojasiewicza 6, 30-348 Kraków, Poland
E-mail: [email protected]
(Received:
15th of Fenruary 2014 ; revised: 3rd of October 2014)