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PROCEEDINGS of theAMERICAN MATHEMATICAL SOCIETYVolume 87. Number 2. February 1983
RATIONAL TILINGS BY «DIMENSIONAL CROSSES
SÁNDOR SZABÓ
Abstract. Consider the set of closed unit cubes whose edges are parallel to the
coordinate unit vectors e,,...,e„ and whose centers are re., 0<|;|*i, in n-
dimensional Euclidean space. The union of these cubes is called a cross. This cross
consists of 2kn + 1 cubes; a central cube together with In arms of length k. A
family of translates of a cross whose union is «-dimensional Euclidean space and
whose interiors are disjoint is a tiling. Denote the set of translation vectors by L If
the vector set L is a vector lattice, then we say that the tiling is a lattice tiling. If
every vector of L has rational coordinates, then we say that the tiling is a rational
tiling, and, similarly, if every vector of L has integer coordinates, then we say that
the tiling is an integer tiling. Is there a noninteger tiling by crosses? In this paper we
shall prove that if there is an integer lattice tiling by crosses, if 2kn + 1 is not a
prime, and if p > k for every prime divisor p of 2 kn + I, then there is a rational
noninteger lattice tiling by crosses and there is an integer nonlattice tiling by crosses.
We will illustrate this in the case of a cross with arms of length 2 in 55-dimensional
Euclidean space. Throughout, the techniques are algebraic.
1. Definitions. Let S", R, and Z be /i-dimensional Euclidean space, the real
number field, and the integer number ring, respectively. Translations of the space
belong to the «-dimensional vector space E" over the number field R. Let O be a
point in &" and let e,,... ,e„ be an orthonormal basis for E".
We call the point set
eo:= {P: 0?=c,e, + • • •+c„en; 0 ^ c, <1.0 < cn < 1; c,,... ,c„ G R)
an «-dimensional unit cube with preferential vertex O.
Let Tbe a finite subset of Z containing (0,...,0), (1,0,... ,0),... ,(0,... ,0,1) and
let T and 5"0 be the sets
T:= {f,e, + -+iA:((./,)Ei)
and
%•■ = {£p- wet}.
If L is a subset of E", then we use the following notation (5"0, L) : = {^P: OP G L}.
We say that the system (5"0, L) is a tiling of the space S" if the following conditions
hold:
(1) S" = U [ßP: ep G S"0 G (%, L)}, and
(2) if int ep n int 6Q^ 0 and Qp G ^R G (%, L), Ce G ?F5 G (%, L), then GP =
eQ.-Received by the editors December 3, 1981 and, in revised form. May 19, 1982.
1980 Mathematics Subject Classification. Primary 20K01; Secondary 52A45, 05B40, 05B45, 10E30.
Key words and phrases. Exact sequence, factorization of abelian groups, lattices, tiling, star body.
©1983 American Mathematical Society
0O02-9947/82/O0OO-0555/$03.25
213
214 SÁNDOR SZABÓ
Let qx,... ,qn be fixed positive integers. The vector set
x = {(*i/?i)ei + /•• + (xn/qn)en:xx,...,xrlG Z)
we call a lattice. Clearly the lattice X is a free abelian group with n generators. We
also call every subgroup of the group X a lattice. Thus L is a lattice if L < X. The
standard integer lattice is the lattice
M := {m,e, + ••• +m„en: m.mn G Z).
The system (?T0,L) is called a lattice system if L o X and an integer system if
L Ç M.
Let G be an abelian group written additively and let H, Ax,... ,Ar be subsets of G.
If each element h of H is uniquely expressible in the form
h = ax + ■ ■ ■ +ar; o, G Ax,... ,ar G Ar
(that is, if h = a\ + • ■ • +a'r; a\ G Ax_,a'r G Ar, then a, = a\.ar = a'r), then
we write H = Ax + ■ ■ ■ +Ar and speak of a factorization of the subset H by subsets
A.Ar.Let T C Z" and H Ç G. If there is a set of n elements, g,,... ,gn in G such that
each element of H is uniquely expressible in the form txgx + ■ ■ • +t„g„, (t\.tn)
G T, then we write H - T(gx,.. .,g„).
For elements gx,_g„ of group G denote by (gx,-g„> the group generated by
g,.g„, that is, the smallest subgroup of G containing g,.g„. In other words,
(g.gn) '■= Ol#l + ••• +x„g„'-x\.Xn G Z)-
If G is the internal direct product of the cyclic groups C,.Cr of orders
m,,...,mr and generators g,,...,gr, respectively, then we will say that G is defined
by the equations mxgx = 0.wrgr = 0.
2. Tiling space and factoring groups. The next lemma, due to S. K. Stein (see [8, p.
545]), enables us to lift a factoring from a group G' to any group G of which G' is a
homomorphic image.
Lemma 2.1. Let G and G' be abelian groups and let G' = A' + B' be a factorization
of G' and finally let <p: G — G' be a homomorphism from G onto G'. Assume that A is a
subset of G such that the restriction of y to A is a bijection between A and A'. Then
G — A + B<p~ ' is a factorization of G.
Theorem 2.1. (1) There is a tiling (íí0, L), L G X, if and only if there is a group G,
elements gt,... ,g„ that generate G, and positive integers qx_,qn such that
(2.1) G=T(qxgx.i„gj+{0,g,,2g,.(qx - l)g,}
+ •••+ {0,g„,2g„.(qn- l)g„} +H,
where H G G.
Furthermore,
(2) L < X if and only if H o G, and
(3)LGMifandonlyif(T(qxgx.q„gn) + H)<3G.
RATIONAL TILINGS BY «-DIMENSIONAL CROSSES 215
Proof. (1) Assume that the system (S"0,L), L G X, is a tiling. So the abelian
group X is expressible in the form
(2.2) X=T(e.,en)+{0,(\/qx)ex,...,((qx-\)/qx)ex}
+ --- + {0,(\/q„)en,...,((qn-\)/qn)en}+L.
Assume that the abelian group G is expressible in the form (2.1). Let <p: X -» G be
the mapping defined as
((*i/$i)«i + ••• + (xn/qn)e„)<p = xxgx + ■■■ +x„gn; xx,...,x„GZ.
Clearly <p is a homomorphism from X onto G and the restriction of <p to set
r(e,,...,ej + {0,(l/<7,)e,,..., ((<?, - l)/?,)e,}
+ •••+ {0,0/<7>„.((qn- \)/q„)en}
is a bijection between this set and the set
T(^g.,qngn)+{0,gx,..., (qx - l)g,} + ••■ + {0,gn,...,(qn- l)g„}.
So by Lemma 2.1, X is expressible in the form (2.2), where L = //<?"'. Thus the
system (5"0, L) is a tiling of &".
(2) If L o X, then the equations L<p = H and Xqp = G imply that H < G.
Conversely, in case H <¡ G, then, as is well known, H<p~ ' < C7<p~ '. Thus L<X.
(3) Assume that (T(qxg.,qngn) + H)<¡ G. We shall show that if 1 G L, then
1 GM.
Since I G L Ç X, we may write
-l = (lx/qx)ex + ••• + (/„/?>„; /.,/„GZ.
Let u¡, /", be integers with the property that /, = u,q, + /-,-, 0 < r¡ < q,. Thus
(2.3) - 1 = W,e, + • • • +K„e„ + (rx/qt)ex + ■■■ + (r„/qn)en.
Now we apply the homomorphism <p. Since Im = : h G H,
(T(qxgx,...,qngn) + H)<3G; e,,... ,e„ G T,
and
T<p= T(qxg],...,qng„),
therefore(utqxgx + ■•• +u„qng„ + h) G(T(qxgx.qngn) + H); thus
0 = txqxgx + ■■■ +tnqng„ + rxgx + ■■■ +rngn + h',
where h' G H and (tx,...,t„) G T. According to the factorization (2.1),
(2.4) (r,./J = (rx,...,rn) = (0,...,0); W = 0.
By virtue of (2.3) and (2.4) we have 1 G M, which was to be proved.
Assume that L G M. We shall show that (Tep + H) < G. Let t, t' G T and 1,
I' G L. Since (t + I - (f + 1')) G M and the system (%, L) is a tiling so t + 1 - (f
+ I') = t" + 1", where t" G T, I" G L. We have
t<p + h - (ftp + «') = t"m + h"; h, h', h" G H,
with the application of homomorphism <p, which was to be proved.
216 SÂNDOR SZABÓ
(2.5) {*«"
(2.6)
Remarks. (1) Theorem 2.1 is known for certain sets T in the following cases:
H = {0} in [2, 15]; qx = ■ ■ ■ = qn = 1 in [8, 10]; H = {0} and qx = ■ ■ ■ = q„ = 1 in
[6, 7, 10 and 14].
(2) Theorem 2.1 was proved for general Tin the special case H = {0}, qx — ■ • • —
q„ = 1 in [13].
We shall construct a vector set L for a system (50,L),LC X, in the special case
when a basis for G is a subset of (g,,... ,g„}.
Let G he an abelian group expressible in the form (2.1), where H = {/i,,...,«,}.
We shall assume that gx,...,gs, s < n, is a basis for G, that is, G is given by the
relations w,g, = 0,...,m,$, = 0. Thus
'gv = avlg\ + ■■■ +ausgs'> (l+l <€><»),
Define n + t vectors in X as follows:
'l. * (««/ftXi (Ku^s),
K - (a„i/?i)ei + • ■ • + (avl/qs)es - (\/qv)ev; (s + 1< V < n),
}.+* = (*wi/9i)e, + • ■ • + (*„,/?>,; (1 «5 w< r),
and the following vector set
L:= <I,,...,ln)+{ln+l,...,l„+,}.
Theorem 2.2. H<p~] = L.
Proof. We shall show that if 1 G L, then I G H<p~ ', that is, Iqp G H. Assume that
I G L so I = xx\x + • - • +xnln + lK+w, where xx,...,xnG Z; 1 < w =s t. Thus l<p =
x,l,<p + • • ■ +xnln<p + ln+M,<p. It is obvious that lu<p = 0, 1 *£ u *s n. Indeed
K<P = (mjqu)*jp = (mu/qu)qugu = 0; (1 < u < s)
and
. av\ , avs 11»9 = — e,<P + • • • + — e5m - — e„m
q\ qs qv
av\ , °vs i= — iigi + • • • + —q,g, - —?«,«„
= o„ic?i + • • • +avsgs - gv = °; (s + 1 <»< »).
Finally l<p = ln+K,<p. It is obvious that \n+w<p G H. Indeed
\n + w<P = -fe.rp + • • • + -f es<p = bwXgx + ■■■ + b„sgs = hwGH.
Thus L G Hq>\ which was to be proved.
Now we shall show that L D Hq>~]. In other words we shall show that if x G X
and x<p G H, then x G L.
Assume that x<p = hw. Since 1„ + M,<p = hw, we have (x — l„+w)<p — 0. Since (x —
K+w) e x,
(2.7) x - L+w = (xx/qx)ex + ••• + (x„/qn)e„; xx,...,x„GZ.
RATIONAL TILINGS BY «-DIMENSIONAL CROSSES 217
By virtue of (2.6) and (2.7) it is clear that
x - \n + w + xs+x\s+x + ■■■ +xn\n = (x\/qx)ex + ■ ■ ■ + (x's/qs)es
for some integers x\,...,x's. Let w, and r, be integers with the property that
x'¡ = ujm¡ + /-,, 0 < r, < m¡, 1 < i < s. Thus
(2.8)
x - \B+W + x,+ xl,+, + • • • +xnln = «,l, + • • • +us\s + £-e, + • • • + £-e,.
Applying the homomorphism<p, we find that 0 = rxgx + ■ ■ ■ +rsgs. Since {g,,... ,gs)
is a basis for G and 0 < rt < mt, 1 * j «C s, it follows that r, = • • • = rs — 0.
Returning to (2.8), we conclude that (x - l,+„) E L, which was to be proved.
The following theorem serves as a basis for our construction of nonlattice tiling
(%,L).
Theorem 2.3. Assume that the abelian group G is expressible in the form (2.1), where
H - {°}> (9i>-••>?■') ^ 0,...,1) and (qxgx,... ,q„g„)= G. Then G is expressible in
the form
(2-9) G=T(g\,...,g'n) + H',
where (g'x,... ,g'„)— G and H' <¡b G.
Proof. Assume that qx > \,...,qs> \, qs+x = • ■ • = q„ = 1 (s > 1). Let g[ -
H'={0,gx,...,(qx - l)g,} + ••• + {0,gs,...,(qs-\)gs}.
So G is expressible in the form (2.9) and ( g\,... ,g'„ ) = G.
If //' G G were a subgroup of G, then it would follow that
({0,g,,...,(<?, -l)g,} +•■•+ {0,gs,...,(qs-\)gs})<iG
and therefore, according to Hajós' theorem on the factorization of finite abelian
groups (see [2]), at least one set (0, g,,... ,(q, — l)g,} would be a subgroup of G. In
other words, at least one qtgt = 0. Finally, since (2.1) is a factorization of G and
therefore 0 ¥= q,g,, it follows that H' is not a subgroup of G.
3. The algebra of crosses. Let G, G', G" be finite abelian groups, A' G G',
A" G G" and let
{0} -G-A'-*A"-{0}
be an exact sequence. (That is, a is an isomorphism, ß is a homomorphism,
Ker/5 = Im a, and Im ß = A".) In this case if A is a subset of A' with the property
that the restriction of ß to A is a bijection between A and A", then A' = A + Get is a
factorization of A'.
Conversely, let A' = A + B he a factorization of A' such that B is a subgroup of
G'. If a is the inclusion mapping and ß is the restriction to A' of the natural
218 SÁNDOR SZABÓ
homomorphism from G' to G'/B, then
{0} -+B^A'-*A -{0}
is an exact sequence.
Let 5,.sk be distinct nonzero integers and let Sn k he the following subset of
Z":
S„,k--= {(0.0),(;,0.0).(0.0,i):,G {j,.sk)}.
Theorem 3.1. Assume that s x.s k are relatively prime to \ G\ and that
a ß
{0} -G-A'-A"-{0}
M an exact sequence. Furthermore, assume that G and A" are expressible in the form
G = Srk(gx.gr). A" = Slk(bx.5,"). For each j, 1 =£;r «S t, select <S, G A' such
that <5;/? = 8". Then A' is expressible in the form A' = Sv k(S'x.5t'), where
(5;.S;.)= {g,«,r5, + ga: \<i<r,l</<t,gGG}.
Proof. This statement was proved by W. Hamaker and S. K. Stein in [5, p. 322]
with the assumption that A' is a group. The reader can readily verify that this
assumption can be removed.
Theorem 3.2. Assume that sx,...,sk are relatively prime to \G\. Then G is
expressible in the form G = Sr k(gx,... ,gr) if and only if for each prime divisor p of
| G | the cyclic group C of order p is expressible in the form C' — S, k( c,,..., c, ).
For a proof see [5, p. 324].
Theorem 3.3. Assume that sx,... ,sk are relatively prime to \G\ . Assume that G is
given by «i,g, = 0,... ,wJgJ = 0 and that G is expressible in the form G —
Srk(g'x.g'r). Then G is expressible in the form G = 5rA(g¡',_g") such that
{g^.gs) £ UÍ'.-•-.«"}•
Proof. If a is the inclusion mapping and ß is the natural homomorphism between
<*M---.«, + i>and<g.g,+ l>/(g,+ 1),then
{0} -<&+,')■<*,.ft+l>-(g,.g,)-{0}
is an exact sequence. On the basis of the assumption that the statement of the
theorem is true in the case s — 1 it follows that the statement of the theorem is true
for every s. (Use Theorem 3.1 and induction on /.)
We shall restrict our consideration to the case 5=1. Let g be a generator of the
cyclic group G. By assumption, there is an integer s, and an element g't such that
Sjg¡ = g. Since Sj is relatively prime to G, the homomorphism \p: G -» G defined by
h\p — Sjh, h G G, is an automorphism. Thus G = SrJt(g!>//,■ • -^g'^) and, since
g¡\¡/ = g, the theorem is proved.
RATIONAL TILINGS BY «-DIMENSIONAL CROSSES 219
Theorem 3.4. Assume that sx,...,sk are relatively prime to \G\ and that G is
expressible in the form G = Sn k(gx,. ..,g„), where n > 2. If G is not a prime number,
then there is an abelian group T which is expressible in the form
(3.1) T = SnJi(qxyx,...,qny„)
+ {0,y„..., (qx - l)y,} + ••• + {0,y„,. ..,(<?„ - 1)y„},
where Snk(qxyx.qnyn) <£ T, (qxy.,q„yn)=T.
Proof. | G| is not a prime number so\G\= u-v, u > \,v > I. The cyclic group T
is defined by 2uvy — 0. If a is the inclusion mapping from (vy) to (y) and ß is the
homomorphism (y)-» (2«y) given by yß = 2uy, then
{0}^(vy)t(y)i(2uy)^{0}
is an exact sequence. According to Theorem 3.2, (2uy) is expressible in the form
(2uy)= S, k(ax2uy,...,a,2uy), a],...,a1 G Z. According to Theorem 3.3 we may
assume that a, = 1. Define the set Ax G (y) by Ax = S, k(axy,...,a,y). It is clear
that (2uy) is isomorphic to (y)/(vy)= {iy + (vy): 0 < / < v}. Furthermore, it
is clear that the restriction of ß to Ax is a bijection between Ax and (2uy). Thus
(y) = Ax + (vy) is a factorization of (y).
There is an exact sequence
{0} -<2©Y>-<t>y>-<.at>y)-*{0}.
Define the setA2 C (vy) by A2 = {0, t>y). Then (vy)= A2 + (2t>y) is a factoriza-
tion of (t>y).
Finally, (y)= Ax + A2+ (2uy> is a factorization of (y). According to Theorem
3.2, <2«y) is expressible in the form (2uy) = S„ k(bx,...,bw). According to Theo-
rem 3.3 we may assume that bx — 2uy. Note that there is an exact sequence
(0) -{2cy>-^1'+(2»y>-»^I-{0}.
According to Theorem 3.1 we have/l, + (2uy)= Sn k(d.,dn), where dx - 2uy,
<Ch = Y- W qt = 2, q2 - • ■ • = qH = L Y, = vy, y2 = d2,..., yn = dn. then Y is
expressible in the form (3.1). It is clear from the definitions of qx.q„, y,,... ,y„
that<<7,Y,.qnyn)= (y)= r.
We shall show that Ax + (2uy)= Sn k(qxyx,. ..,qnyn) is not a subgroup of T.
Assume that (Ax + (2t)y>) < T. Since y G(AX + (2üy>). (y><3 (Ax + <2t>y»,
but 2uv =| y |>| Ax + (2üy)| = uv.
4. Tiling by semicrosses and full crosses. By considering a special type of set T, we
shall examine the problem of tiling Euclidean space by crosses and semicrosses.
Let K*(n, k) G Z" denote the set
K*(n,k) := {(0,...,0),(/,0,...,0).(0.0,/): 1 ̂ ¿<A:}
and let
K*(«, A:) := fie,; 0 < / < fc, 1 ̂ / < «) andi j i
%*(n,k) ■= {ep: OP GK*(«,/v)}.
220 SÁNDOR szabó
The set of cubes %£(n, k) consists of kn + 1 cubes formed of n arms of length k
attached at a central cube. In a similar manner we define the sets K(n, k), K(«, k),
%o(n, k):
K(n,k):= {(0,...,0),(.,0,...,0),...,(0,...,0,/): |<|/|<*},
K(n,k) := {&,: 0 <|» j< k, 1 </<«},
DC0(«,Ac) := {ep: 0? G K(«, *)}.
The set of cubes %0(n, k), which consists of 2kn + 1 cubes formed of 2« arms of
length k meeting at a central cube, is called a cross.
It is obvious that if 5, = 1, s2 — 2,... ,sk — k, then S„ k — K*(n, k) and if j, = 1,
s2 = 2-sA = fc,5A+1 = -\,sk + 2 = -2,...,*24 = -*, then S„M = K(n,k).
Theorem 4.1. If there is an integer lattice tiling (%*(n, k), L), L <3 M, ifkn 4- 1 ¿j
«oí a prime, and if p > k for every prime divisor p of kn + 1, f«e« there are
(1) a noninteger lattice tiling (%%(n, k), L'), L' (£ M, L' < X, and
(2) an integer nonlattice tiling (%£(n, /c),L"), L" G M, L" «jb M.
Proof. (1) follows immediately from Theorems 2.1 and 3.4. (2) follows from
Theorems 2.1, 2.3 and 3.4.
Theorem 4.2. // there is an integer lattice tiling (%n(n, k), L), L < M, such that
2kn + 1 is not a prime andp > k for every prime divisor p of2kn + 1, then there are
(\)a noninteger lattice tiling (%0(n, k), L'), L' <£ M, L' < X, and
(2) an integer nonlattice tiling (%0(n, k),L"), L" C M, L" <£ ML
The proof is similar to that of the preceding theorem.
The next theorem presents a contrast to the last two theorems.
Theorem 4.3. For any lattice tiling (5"0, L), where | ?F01 is a prime number, L G M.
Proof. This was proved for special ?f0 in [15] with the aid of Rèdei's theorem on
the factorization of abelian groups into sets with a prime number of elements. The
reader can readily verify that this proof holds for a general ?T0.
5. An example. We now illustrate our theorems by considering a cross with arms
of length two in 55-space. Now 4 • 55 + 1 = 221 = 13-17 and 13 > 2, 17 > 2. The
group C is defined by 13c = 0 and the group C is defined by 17c' = 0. It is clear
that
C = â:(3,2)(c,4c,3c), C' = A:(4,2)(c',4c',3c', 12c')-
So the abelian group G of order 221 is expressible in the form G =
K(55,2)(g,,...,g55). Thus there is an integer lattice tiling (DC0(55,2),L), L O M.
Now we shall construct a noninteger lattice tiling
(!Ko(55,2),L'), L' % M, L'< X, and an integer nonlattice tiling (Df0(55,2),L"),L" Ç M, L" <b. M.
Consider the group T = (y> defined by 2 • 22ly = 0. There is an exact sequence
{0}^(l7y>^(y>-.<26y)^{0}.
RATIONAL TILINGS BY «-DIMENSIONAL CROSSES 221
Thus (y)= Ax + (17y>, where Ax is A:(4,2)(y,4y,3y, 12y) = {0, y, 2y, -y, -2y,4y, 8y, — 4y, — 8y, 3y, 6y, — 3y, — 6y, 12y, 24y, — 12y, —24y}. There is an exact
sequence
{0} -(34y)^(l7y)^(221y)^{0}
so (17y)= A2 + <34y), where yl2 = {0,17y}.
Finally
(y) = At+A2+ (34y)=(Ax + (34y))+A2
= K(55,2)(34y, 4 • 34y, 3 • 34y, y, y + 34y, y + 2 ■ 34y,...,
y + 12-34y, 4y,4y + 34y, 4y + 2-34y,...,
4y + 12 • 34y, 3y, 3y + 34y, 3y + 2 • 34y_
3y + 12-34y, 12y, 12y + 34y, 12y + 2-34y,...,
12y + 12-34y) + {0, 17y}.
If qx = 2, q2 = • •• = q55 = 1; y, = 17y, y2 = 136y, y3 = 102y,
Y4 = Y, Y5 = 35y ,..., y16 = 409y,
Yn = 4Y, Yi8 = 38Y ,•••, Y29 = 412y,
Y3o = 3Y, Ï3i = 37y . y42 = 411y,
Y43 = 12Y, Y44 = 46y . Y55 = 420Y,
then T is expressible in the form
r = ii:(55)2)(i,yI,...,95Jy55)
+ {0,yx,...,(qx-l)yx} + ■■■+ {0,yS5.(qss - l)y5$}.
According to Theorem 2.1 there is a noninteger lattice tiling (%0(55,2), L'). Accord-
ing to Theorem 2.2 the vectors
l'i = -ie, + 17e4,1'2 = -e2 + 136e4, l'3 = -e3 + 102e4, l'4 = 442e4,
r5 = 35e4-e5,...,l',6 = 409e4-e,6,
I'n = 4e4 _ «n-l'is = 3&e4 - e18,...,l29 = 412e4 - e29,
'30 = 3e4 - e30,131 = 37e4 - e3,,...,l'42 = 411e4 - e42,
I« = 12e4 - e43, X'u = 46e4 - e^,... ,l'55 - 420e4 - e55
span the lattice L', that is, L' = (1',,... J55).
T is expressible in the form
T = K(55,2)(yx,...,yS5) + H,
where y, = 34y, y2 = y2,..., y55 = y55; H = {0,17y}; H is not a subgroup of T;
(y,,... ,y55)= T. Thus according to Theorem 2.1 there is an integer nonlattice tiling
(X0(55,2),L"). If 1',' = -e, + 34e4, Ç = l'2,..., I'5'5 = 155, l'5'6 = 0, l'5'7 = 17e4, then
according to Theorem 2.2, L" = (!',',...,r5'5) + {1'5'6,157}.
The author would like to thank the referee for his valuable suggestions and a short
proof of Theorem 3.3.
222 SÁNDOR SZABÓ
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Department of Civil Engineering Mathematics, Technical University, Budapest, 1111 Buda-
pest, Stoczek u.2., Hungary
