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J. Fractal Geom. 11 (2024), 205–217
DOI 10.4171/JFG/149
© 2024 European Mathematical Society
Published by EMS Press
This work is licensed under a CC BY 4.0 license
Intersecting the Twin Dragon with rational lines
Shigeki Akiyama, Paul Großkopf, Benoît Loridant, and Wolfgang Steiner
Abstract. The Knuth Twin Dragon is a compact subset of the plane with fractal boundary of
Hausdorff dimension sD.log /=.log p2/,3D2C2. Although the intersection with a
generic line has Hausdorff dimension s1, we prove that this does not occur for lines with
rational parameters. We further describe the intersection of the Twin Dragon with the two diag-
onals as well as with various axis parallel lines.
Dedicated to Professor Jörg Thuswaldner on the occasion of his 50th birthday
1. Introduction
We investigate the intersections of the Knuth Twin Dragon with rational lines. Let
˛D 1Ci, then
KD²1
X
kD1
dk
˛kWdk2 ¹0; 1º³
is the Knuth Twin Dragon. The Hausdorff dimension of its boundary @K is sD
log
log p21:5236, where is the real number satisfying 3D2C2. For lines
p;q;r D ¹xCiy 2CWpx Cqy Drº(1.1)
with p; q; r 2Z, we show that the ˛-expansions of K\p;q;r are recognized by a
finite automaton.
By a result of John Marstrand [5], the intersection of @Kwith Lebesgue almost all
lines going through Khas Hausdorff dimension s1, meaning that in the set of all
parameter triples .p; q; r / 2R3for which p;q;r \K¤ ;, the exceptional cases form
a Lebesgue null set. We obtain here that the Hausdorff dimension of the intersection
of the boundary of the Twin Dragon with rational lines is never equal to s1.
Further, we revisit results by Akiyama and Scheicher [1] and add uncountably
many examples of horizontal, vertical, and diagonal lines.
Mathematics Subject Classification 2020: 52C20 (primary); 28A80 (secondary).
Keywords: number system, Hausdorff dimension.
S. Akiyama, P. Großkopf, B. Loridant, and W. Steiner 206
We mention that similar results were obtained in [4] for lines intersecting the
Sierpinski carpet F. The set Fhas Hausdorff dimension log 8
log 3. Manning and Simon
showed that, given a slope ˛2Q, the intersection of Fwith the line yD˛x Cˇis
strictly less than log8
log 31for Lebesgue almost every ˇ.
2. Main statement and proof
We first recall the notions of a canonical number system and its fundamental domain.
Let ˇbe an algebraic integer and ND ¹0; 1; : : : ; jN.ˇ/j 1º, where N .x/ denotes
the norm of xover Q.ˇ/=Q. The pair .ˇ; N/is called a canonical number system
(CNS) if each 2ZŒˇ admits a representation of the form
D
n
X
kD0
dkˇk; dk2N:(2.1)
We call ˇthe radix or base and Nthe set of digits. The representation (2.1) is unique
up to leading zeros.
The Knuth Twin Dragon Kappears as the fundamental domain of the CNS .˛; N/,
where ˛D 1Ciis the root of the polynomial x22x 2and ND ¹0; 1º. The
fundamental domain of a CNS is the set of all numbers that can be expressed with
purely negative exponents. Since ˛4D 4, it is often useful to consider groups of
four digits:
1
X
kD1
dk
˛kD1
X
kD1P3
jD0d4kj˛j
˛4k D1
X
kD1
bk
.4/k;
with the possibilities for bkDP3
jD0d4kj˛jbeing
Œ0000˛D0; Œ0001˛D1; Œ0010˛D 1Ci; Œ0011˛Di;
Œ0100˛D 2i; Œ0101˛D12i; Œ0110˛D 1i; Œ0111˛D i;
Œ1000˛D2C2i; Œ1001˛D3C2i; Œ1010˛D1C3i; Œ1011˛D2C3i;
Œ1100˛D2; Œ1101˛D3; Œ1110˛D1Ci; Œ1111˛D2Ci:
In other words, we have
KD²1
X
kD1
bk
.4/kWbk2D³;
with
DD ¹1i; 1Ci ; 2i; i; 0; i ; 12i; 1; 1Ci ; 1C3i; 2; 2Ci; 2C2i ; 2C3i; 3; 3C2i º:
Points in the intersection of Kwith lines p;q;r D ¹xCiy Wpx Cqy Drºcan now
be characterized by their digit expansion in the following way.
Intersecting the Twin Dragon with rational lines 207
g1g2
g3
g4
g5g6
1
0 0,1
01
0,1
0
1
Figure 1. An automaton characterizing @K(in base ˛), where all states are initial and terminal.
Lemma 2.1. We have z2K\p;q;r if and only if there is a digit sequence b1b2 2
DNwith
zD1
X
kD1
bk
.4/kand rD1
X
kD1
pR.bk/CqI.bk/
.4/k:
Here, R.b/ denotes the real part and I.b/ denotes the imaginary part of b2C.
We show that we can characterize the digit expansion of the points in the intersec-
tion p;q;r \Kvia a Büchi automaton, that is a finite automaton that accepts infinite
paths. Using this representation, we are able to calculate the Hausdorff dimension of
the intersection K\p;q;r as well as the Hausdorff dimension of @K \p;q;r .
Definition 2.2. ABüchi automaton is a 5-tuple .Q; A; E; I ; T /, where the set QD
¹q1; : : : ; qNºis a finite set of states,Ais a finite alphabet,EQAQis a set
of edges and I; T Qthe set of initial and terminal states. Let Adenote the set of
all (finite) words and A!denote the set of all (right) infinite words. A word w2A,
wDw1wn, is accepted by the automaton if and only if there are states qi0; : : : ; qin
such that qi02I,qin2Tand .qik1; wk; qik/2Efor all k. We call such a finite
path successful, and we call an infinite path successful if and only if infinitely many
subpaths are successful. An infinite word w2A!is accepted by the automaton if
there exists an infinite successful path with label w. The set of all w2A!that are
accepted by the automaton is called its !-language.
Büchi automata are really helpful to describe self-similar sets. The automaton in
Figure 1characterizes all infinite sequences of digits 0; 1 in base ˛that give rise to
boundary points in @K; see [3,7].
S. Akiyama, P. Großkopf, B. Loridant, and W. Steiner 208
Let L1; L2be two !-languages on the same alphabet that are accepted by A;B,
respectively. It can be necessary to create automata accepting the union of the lan-
guages or their intersection. The union is not difficult: one just uses the union of states
and edges, as well as the union of terminal and initial states. The intersection gener-
ally requires heavy computations, especially in the non-deterministic case, where a
larger framework than Büchi automata needs to be used. But it becomes easy in some
cases. We prove one particular case that is useful to prove our main statements.
Lemma 2.3. Let L1; L2be two !-languages on the same alphabet Aaccepted by
Büchi automata. If one of the automata has only terminal states, then there is a Büchi
automaton accepting L1\L2.
Proof. Define ABD.QAQB; A; E; IAIB; TATB/, where Econsists of
the edges .a; b/ d
!.a0; b0/with ad
!a0and bd
!b0. Let w2A!be a word that is
accepted by AB. Then there exists an infinite path in the automaton. Projecting to
the first coordinate gives an infinite path through A. Therefore, we have w2L1and
with the same reasoning w2L2. Now let w2L1\L2. There exists a path a0a1
through Aand a path b0b1 through B. Then .a0; b0/.a1; b1/ is a path in the
product automaton. Assume w.l.o.g. that all states of Aare terminal. Then, for every
finite subpath b0b1 bkaccepted by B, the corresponding path a0a1 akin Ais
also accepted, hence .a0; b0/.a1; b1/ is successful.
In general, if p;q;r \Kis described by a Büchi automaton Aand the boundary
@Kby a Büchi automaton G, then @K\p;q;r is described by the product automaton
AG. Interpreting this Büchi automaton as a graph directed construction for the set
@K\p;q;r , we have a way to compute the Hausdorff dimension of this set via
results of Mauldin and Williams [6]. Let us state and prove our main statements.
Theorem 2.4. Let p; q; r 2Z,p;q;r as in (1.1)and Kthe Knuth Twin Dragon. Then
the intersection K\p;q;r can be described by a Büchi automaton.
Proof. For s; s02Z, we define an edge relation by
sb
!s0”s0DpR.b/ CqI.b/ 4s: (2.2)
Now consider a path rDs0
b1
!s1
b2
! bn
!sn. Then
snD.4/n.r/ C
n
X
kD1
.4/nkpR.bk/CqI.bk/;
i.e.,
sn
.4/nD rC
n
X
kD1
pR.bk/CqI.bk/
.4/k:
Intersecting the Twin Dragon with rational lines 209
Using Lemma 2.1, we immediately get that
.x; y / DŒ0:b1b2b3 42K\p;q;r if and only if lim
n!1
sn
.4/nD0:
We now show that the elements snlying on paths starting with s0D rand satisfying
limn!1 sn
.4/nD0are bounded by a constant c.p; q/. Indeed, we have
sn
.4/nD rC
n
X
kD1
pR.bk/CqI.bk/
.4/kD 1
X
kDnC1
pR.bk/CqI.bk/
.4/k;
and therefore
jsnjD4nˇ
ˇ
ˇ
ˇ
1
X
kDnC1
pR.bk/CqI.bk/
.4/kˇ
ˇ
ˇ
ˇmax¹jpR.b/ CqI.b/j W b2Dº
3Dc.p;q/:
Defining the set of states QD ¹s2ZW jsj c.p; q/º [ ¹rº,ID ¹rº,TDQand
edges as in 2.2, gives us the desired Büchi automaton.
Theorem 2.5. Let p; q; r 2Z,p;q;r as in (1.1)and Kthe Knuth Twin Dragon.
Then, the Hausdorff dimension of the intersection @K\p;q;r is never s1, where
sis the Hausdorff dimension of @K.
Proof. The Büchi automaton of Theorem 2.4 gives rise to a description of the inter-
section K\p;q;r as one of the attractors of a graph directed construction (GIFS)
with attractors .Ks/s2Q:
KrDK\p;q;r ;with KsD[
sb
!s02A
Ks0Cb
4.s 2Q/:
As mentioned above, @Kis also the attractor of a GIFS:
@KD[
g2Q0
Kg;with KgD[
gb
!g02G
Kg0Cb
4.g 2Q0/;
where Gis the automaton characterizing @Kin base 4. The automaton Gcan be
obtained from the automaton G0of Figure 1as follows.
• The set of states Q0is the same as for G0; all states are initial and terminal.
• There is an edge from gto g0in Gwhenever there is a path of length 4from gto
g0in G0. The label of this edge in Gis the digit vector Œd1d2d3d4˛corresponding
to the labels d1; d2; d3; d4in G0along the path of length 4.
S. Akiyama, P. Großkopf, B. Loridant, and W. Steiner 210
In that way, Aand Gare built on the same alphabet. By Lemma 2.3, the intersection
AGis a Büchi automaton describing the intersection p;q;r \@K. By Mauldin
and Williams [6], the Hausdorff dimension of a GIFS attractor can be computed from
the spectral radius ˇof the incidence matrix of a strongly connected component of
the associated automaton; see further details in Remark 2.6. In particular, in our case,
dimH.@K\p;q;r /Dlog ˇ
log 4;
where the involved number ˇis an algebraic integer.
Now, the dimension of the boundary of the Twin Dragon is sDlog
log p2, with 3D
2C2. To have log ˇ
log 4Ds1, we need ˇD4
4. However, the minimal polynomial
of 4
4is 4x39x 2C2x 1, thus 4
4is not an algebraic integer.
Remark 2.6. We shortly explain why the results of Mauldin and Williams [6] indeed
apply to our setting. All the similarities in our graphs are contractions of the form
T .x/ DxCb
4, with the same ratio 1
4. Therefore, if Gdenotes any of our graphs, we
only need to check the existence of nonoverlapping compact sets J1; : : : ; Jn(one for
each node 1; : : : ; n of G) with the property
8i2 ¹1; : : : ; nº; Ji[
i
T
!j2G
T .Jj/;
each union being nonoverlapping.
For the graph GDGof our paper (with states g2Q0), the intersections of K
with its six neighboring tiles in the plane tiling generated by Kare compact sets
playing the role of the Ji’s, that is, satisfying the above nonoverlapping conditions;
see for example [2]. These intersections are exactly the sets Kgdefined in the proof
of Theorem 2.5.
Now, the graph GDAGof our paper can be interpreted as a subgraph of G:
taking the product of Aand Gmeans to select paths of G. The states of AGare of
the form .r; g/, for some integers rand g2Q0. Defining
Kr;g WD p;q;r\Kg;
we obtain compact sets fulfilling the nonoverlapping requirements mentioned above.
3. Further results on intersections of the Twin Dragon with rational
lines
In this section, we want to extend the work of [1], where the intersections with the
x-and the y-axis are calculated. The intersections of these lines with @K are signi-
Intersecting the Twin Dragon with rational lines 211
1
1
Figure 2. The Knuth Twin Dragon Kand its intersection with 1;0;r for some ras in Theo-
rem 3.1 (red) and with 1;0;1=5 (blue).
ficatively different from the expected result for intersections of fractals and lines, as
they consist only of two points. First, we show that their result extends to uncountably
many axis-parallel lines (where we do not have finite automata), and using the self-
similar structure, to diagonal lines. Then, we give one example of a more complicated
intersection.
Theorem 3.1. Let a1a2 be a sequence in ¹0; 1º!not ending in .01/!, and
rD1
X
kD1
2ak
.4/k:
Then
@K\1;0;r D®rCr2
5i; r CrC3
5i¯;
and K\1;0;r is the closed line segment rCr2
5; r C3
5i.
Proof. We first use Lemma 2.1 to describe K\1;0;r , that is, we determine the
sequences b1b2 2 Dsuch that RP1
kD1bk.4/kDr, i.e.,
S. Akiyama, P. Großkopf, B. Loridant, and W. Steiner 212
1
X
kD1
2akR.bk/
.4/kD0:
Since R.bk/2 ¹1; 0; 1; 2; 3º, we have 2akR.bk/2 ¹3; 2; : : : ; 2; 3ºand thus
ˇ
ˇ
ˇ
ˇ
1
X
kDnC1
2akR.bk/
.4/kˇ
ˇ
ˇ
ˇ1
4nfor all n0:
Moreover, equality holds if and only if 2akR.bk/is alternately 3and 3, which
implies that akis alternately 1and 0, which we have excluded. This gives that
ˇ
ˇ
ˇ
ˇ
1
X
kDnC1
2akR.bk/
.4/kˇ
ˇ
ˇ
ˇ
<1
4nand 1
X
kDnC1
2akR.bk/
.4/kD
n
X
kD1
R.bk/2ak
.4/k2Z
4n
for all n1, hence R.bk/D2akfor all k1. For the corresponding sequences
d1d2 (with P3
jD0d4kj˛jDbk) this implies that
d4k3d4 k2d4k1d4k 2 ¹ak000; ak011; ak100; ak111ºfor all k1: (3.1)
Now consider sequences d1d2 of the form (3.1) in the boundary automaton G
given in Figure 1. The only paths labeled by abcc,a; b; c 2 ¹0; 1º, starting from g1,
g2,g5and g6, respectively, are
g1
0000
! g6; g1
0011
! g2; g2
1000
! g5; g2
1011
! g1;
g5
0100
! g6; g5
0111
! g2; g6
1100
! g5; g6
1111
! g1:
Therefore, for an infinite successful path of the form (3.1) starting from g1,g2,g5
or g6, the sequence a1a2 is alternately 0and 1, which we have excluded. Hence,
it suffices to consider paths that are in g3and g4after 4k steps for all k0. From
g3
a100
! g4and g4
a011
! g3.a 2 ¹0; 1º/;
we see that the only points in @K\1;0;r are
1
X
kD1
ak˛3
.4/kC1
X
kD1
˛6C˛C1
16kDr .1 Ci / C3i
5;
1
X
kD1
ak˛3
.4/kC1
X
kD1
˛5C˛4C˛2
16kDr .1 Ci / 2i
5:
Since r .1 Ci / 2K,K\1;0;r is the line segment between these points.
Intersecting the Twin Dragon with rational lines 213
1
1
Figure 3. The intersection of KD˛1K[.KC1/with lines 0;1;r=2 ,1;1;r, and
1;1;r=2 for some ras in Theorem 3.1.
Theorem 3.2. For 8
15 < r < 2
15 , we have
2i .K\0;1;r=2/D.K\1;0;r /C ¹0; i º;
.1Ci / .K\1;1;r/DK\1;0;r ;
.1Ci / .K\1;1;r=2/D.K\0;1;r=2/C ¹0; 1º;
2 .1 Ci / .K\1;1;r=2 /D.K\1;0;r /C ¹2i; i; 0; i º:
In particular, for ras in Theorem 3.1, the intersections K\0;1;r=2 ,K\1;1;r
and K\1;1;r=2 are closed line segments with endpoints
@K\0;1;r=2 D@.K\0;1;r =2/D®4
5r
2Cr
2i; 1
5r
2Cr
2i¯;
@K\1;1;rD@.K\1;1;r/D®1
5C1
5ri; 3
10 3
10 Cri¯;
@K\1;1;r=2 D@.K\1;1;r =2/D®3
5Cr
23
5i; 2
5Cr
2C2
5i¯:
Proof. Note that ˛KDK[.KC1/ and
˛ 1;1;rD1;0;r; ˛ 0;1;r=2 D1;1;r; ˛ 1;1;r=2 D0;1;r =2:
S. Akiyama, P. Großkopf, B. Loridant, and W. Steiner 214
Moreover, we have
.KC1/ \1;0;r D;D.K1/ \1;0;r D.KC˛/ \1;0;r
since 8
15 < r < 2
15 and
min¹xWxCiy 2Kº D 1
X
kD13
.4/2k1C1
.4/2k D 1
X
kD1
13
16kD 13
15;
max¹xWxCiy 2Kº D 1
X
kD11
.4/2k1C3
.4/2k D1
X
kD1
7
16kD7
15:
Using these geometric properties, we obtain that
˛ .K\1;1;r/DK[.KC1/\1;0;r DK\1;0;r ;
˛2.K\0;1;r=2 /DK[.KC1/ [.KC˛/ [.KC˛C1/\1;0;r
D.K\1;0;r /[.KCi/ \1;0;r
D.K\1;0;r /C ¹0; i º;
˛ .K\1;1;r=2 /DK[.KC1/\0;1;r=2 D.K\0;1;r =2 /C ¹0; 1º;
˛3.K\1;1;r=2 /D˛2.K\0;1;r=2 / ¹0; 2i º
D.K\1;0;r /C ¹2i; i; 0; i º:
For ras in Theorem 3.1, we have 8
15 < r < 2
15 since
min²1
X
kD1
2ak
.4/kWa1a2 2 ¹0; 1º!³D 1
X
kD1
8
16kD 8
15;
max²1
X
kD1
2ak
.4/kWa1a2 2 ¹0; 1º!³D1
X
kD1
2
16kD2
15;
and the minimum and maximum are attained only for the sequences .10/!and .01/!,
which we have excluded. Therefore, Theorem 3.1 and the formulae above give that
K\1;1;rD 1Ci
2r .1 Ci / C2
5;3
5iD r i C1
5;3
10 .1 i/;
K\0;1;r=2 Di
2r .1 Ci / C2
5;8
5iDr1Ci
2C4
5;1
5;
K\1;1;r=2 D1i
4r .1 Ci / C12
5;8
5iDr
2C3
5;2
5.1 Ci/;
which proves the statements for the intersection of Kwith lines. For the intersections
of @Kwith lines, it only remains to check that the points in
˛2.K\1;0;r /\..K\1;0;r /Ci/D°1
˛2r .1 Ci / C3i
5±
Intersecting the Twin Dragon with rational lines 215
and
˛1.K\0;1;r=2 /\..K\0;1;r=2 /C1/D®1
˛3r .1 Ci / 2
5i¯
are not in @K. By the proof of Theorem 3.1, the digit expansion
Œ:a1100a2011a3100a4011 ˛Dr .1 Ci/ C3
5i
is given by a path starting only from g3in the boundary automaton G. Dividing
by ˛2adds 00 in front of the expansion, but g3cannot be reached by 00, hence
1
˛2r .1 Ci / C3i
5is not on the boundary of K. Similarly, the digit expansion
Œ:a1011a2100a3011a4100 ˛Dr .1 Ci / 2
5i
is given by a path starting from g4in the boundary automaton G, and g4cannot be
reached by 000, thus 1
˛3r .1 Ci / 2
5iis not on the boundary of K. This proves
that all intersections of Kwith the given lines are line segments.
We can use this method to find a vertical line with a more interesting intersection.
For example, if we look at 1;0;1=4, we see that the only expansion P1
kD1
bk
.4/k
with bk2Dhaving real part 1=4 is b1b2 D 100 . In base ˛, we must therefore
have d1d2d3d42 ¹0001; 0101; 1010; 1110º, which correspond to the digits 1,12i,
1C3i,1Ci2D. The remaining digit sequences d5d6 give points in the set
1
˛4.K\1;0;0/, thus
K\1;0;1=4 D 1
4C9
10 ;13
20 [2
5;1
10 [7
20 ;3
5 i:
We go on with 1;0;1=4C1=16 and see that points in the intersection have imag-
inary part with an expansion in base 4starting with two digits in ¹2; 0; 1; 3ºand
ending with digits in ¹1; 0; 1; 2º. For the limit 1;0;1=5 of lines of this form, we
obtain the following intersection with K, see Figure 2.
Theorem 3.3. We have
K\1;0;1=5 D²1
5C1
X
kD1
dk
.4/kiWdk2 ¹2; 0; 1; 3ºfor all k1³;
and a point is in @K\1;0;1=5 if and only if it is of the form 1
5CP1
kD1dk.4/ki,
where d1d2 is a path in the automaton in Figure 4.
Proof. Since 1
5DP1
kD1.4/k, we obtain in the same way as in the proof of The-
orem 3.1 that RP1
kD1bk.4/kD 1
5with bk2Dif and only if R.bk/D1for
all k1, i.e., bk2 ¹12i ; 1; 1Ci; 1C3iº. The corresponding 4-digit blocks in base ˛
are 0101,0001,1110, and 1010. This proves the characterization of K\1;0;1=5.
S. Akiyama, P. Großkopf, B. Loridant, and W. Steiner 216
g3g4
-2
-2,0
3
1,3
Figure 4. Automaton recognizing the imaginary parts of points in @K\1;0;1=5 in base 4.
In the boundary automaton, the digit blocks 0101,0001,1110, and 1010 are
accepted only from g3and g4, and we have the transitions
g3
0101
! g3; g3
0001
! g4; g3
0101
! g4; g4
1010
! g4; g4
1010
! g3; g4
1110
! g3:
Taking imaginary parts of the corresponding numbers in Dgives the automaton in
Figure 4.
Theorem 3.4. The Hausdorff dimension of K\1;0;1=5 is 1and
dimH.@K\1;0;1=5/Dlog 3
log 40:7925 > s1:
Proof. We can interpret the intersection with 1;0;1=5 as the self-similar digit tile
in Rwith AD 4and DD ¹2; 0; 1; 3º. Since Dis a complete residue system
modulo 4, this tile has non-empty interior and therefore is of dimension 1.
For the boundary, we have @K\1;0;1=5 DK3[K4, with
4K3D.K32/ [.K42/ [K4;4K4D.K3C1/ [.K3C3/ [.K4C3/:
Therefore, by [6], the Hausdorff dimension of @K\1;0;1=5 is log ˇ= log 4, where
ˇis the Perron–Frobenius eigenvalue of the matrix 1 2
2 1, i.e., ˇD3.
Funding. The authors were supported by the project I3346 of the Japan Society for
the Promotion of Science (JSPS) and the Austrian Science Fund (FWF), the project
FR 07/2019 of the Austrian Agency for International Cooperation in Education and
Research (OeAD), the project PHC Amadeus 42314NC, the project ANR-18-CE40-
0007 CODYS and the project ANR-23-CE40-0024 SymDynAr / I 6750 of the Agence
Nationale de la Recherche (ANR) and the FWF.
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Received 14 February 2022.
Shigeki Akiyama
Institute of Mathematics, Tsukuba University, Tennodai-1-1-1, Tsukuba 350-8571, Japan;
akiyama@math.tsukuba.ac.jp
Paul Großkopf
Department of Mathematics, Université libre de Bruxelles, Boulevard du Triomphe,
1050 Bruxelles, Belgium; paul.grosskopf@gmx.at
Benoît Loridant
Lehrstuhl für Mathematik, Statistik und Geometrie, Montanuniversität Leoben, Franz
Josefstrasse 18, 8700 Leoben, Austria; benoit.loridant@unileoben.ac.at
Wolfgang Steiner
IRIF, CNRS, Université Paris Cité, 75013 Paris, France; steiner@irif.fr
