J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
R
EINHARD
W
INKLER
Sets of block structure and discrepancy estimates
Journal de Théorie des Nombres de Bordeaux, tome
9, n
o2 (1997),
p. 337-349
<http://www.numdam.org/item?id=JTNB_1997__9_2_337_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Sets of
block
structure anddiscrepancy
estimates
par REINHARD WINKLER
RÉSUMÉ. Soient x =
(xn)n~N
une suite d’éléments d’un ensemblefini M et f =
(fn)n~N
une suited’applications
fn :
M ~ M.Quelle
information sur x et f permet d’obtenir des estimations dela
discrépance
de la suitef(x) = (fn(xn))n~N?
Nous donnons dans cet article desréponses
à cettequestion,
en utilisant unrésultat
qualitatif
récent.ABSTRACT. Given a sequence x =
(xn)n~N
on the finite set Mand a sequence f =
(fn)n~N
of mapsfn :
M ~ M. Whichinformation about x and f is suitable for
getting
estimates for thediscrepancy
of the sequencef(x) = (fn (xn))n~N?
Thepaper’s
object is, using
a recentqualitative result,
togive
answers to thisquestion.
1. Introduction
Let M be a finite
set,
w.l.o.g.
M ={1, ... , m~.
For a sequence x =on M and a sequence f = of
transformations In :
M -~ Mwe are
going
tostudy
the distribution behaviour of the sequencef(x) =
y = =
fn(xn).
Thequalitative question
is: Which f have theproperty
that y isuniformly
distributed(u.d.)
whenever x is u.d.?(As
general
references for thetheory
of u.d. sequences cf.(Kui-N), [H]
or(D-T~.)
An answer is contained in
[W]
by
means of a characterization of all thoseso-called
u.d.p.
(uniform
distributionpreserving)
f.Proposition
2 recalls the resultadopted
for our purposes. In this paper we show in which way one can derivequantitative
versions which are necessary forapplications
similar to Monte Carlo and
Quasi-Monte
Carlo methods. Moreprecise:
Weare
looking
for functionsFN
such thatholds.
DN(x)
denotes thediscrepancy
which,
for anarbitrary
sequence x on a finite set M ofcardinality #M
= m, is defined in thefollowing
way.
For
arbitrary
T C Nput
and
(for
N >
minT,
T 0
0)
If T = N we
shortly
writeA(N, x, a)
resp.
DN(x)
for thediscrepancy.
Drr(x)
tends to 0 for N -~ oo if andonly
if x is u.d.Similarly
s-blockdiscrepancy
is definedby
= whereis considered as a sequence on M~
consisting
of the membersA sequence x is called s-block u.d. if
D~r,~(x) ->
0,
x is calledcompletely
u.d. if this holds for all s
Section 2
presents
Theorem 1 which showswhy
FN
has todepend
on x in a morecomplicated
way thanjust
viaDN(x).
Interesting
and muchdeeper
results on similarcomplexity questions
on uniform distribution canbe found in
[G]
and[Ki-Li].
Section 3 is devoted to the notion of sets of blockstructure which has been introduced in
[W]
and is useful for our purposes.This notion is
closely
related to theconcept
of almost constant sequenceswhich has been studied
by
Rindler andLosert,
cf. for instance[Ri],
[Ri-Lo]
and[Lo].
They
continuedinvestigations
due toRauzy,
cf.[Rau],
whoseems to be the first one who
investigated
questions
onu.d.p.
sequences ina similar way. For more references cf.
~W~.
In this paper(section 3)
severalsimple properties
of sets of block structure areproved
which are relevantfor us. Section 4
analyses
how thequalitative
characterization ofu.d.p.
f inProposition
2 can be modified in such a way that it can be used for concreteand
quantitative
discrepancy
estimates. The discussion isessentially
leadby
the results of Section 2 and 3. Adiscrepancy
estimate of the desiredtype
is carried out in the final section 5. Theproblem
is discussed also for s-blockdiscrepancy.
Oneeasily
gets
a characterization of thoseu.d.p.
fwhich also preserve s-block or
complete
uniform distribution.2.
Discrepancy
alone is notenough
informationOur main
question
is: In which way shouldFnr
depend
on itsarguments?
If
FN
is allowed to be anarbitrary
function thesetting
of thequestion
is toogeneral,
sinceFN(x, f)
=DN(f(x)),
in a trivial way, is theoptimal
solutionbut not
interesting
forcomputations.
The first restriction one thinks of is torequire
thatFN
depends
in the firstargument
not on all informationabout x but
just
on thediscrepancy:
FN
=FN(DN(x), f).
But this cannot be done in a reasonable way as Theorem 1explains.
THEOREM 1. There is no
family of functions
FN
=FN(DN(X), f),
N EN,
eachFN
depending
on xonly
viaDN(x)
and on f =(In)nEN,
fn :
M -M,
in anarbitrary
way with thefollowing
properties:
for
all N andwhenever x is
uniforml y
distributed and f preservesuniform
distribution.Proof:
Suppose, by contradiction,
that such a sequence ofFN
exists. Fortechnical convenience suppose M =
(0, 1)
and define g : M - ~VIby
g(i)~i,i=0,1.
Letfn = IdM if aj
j =
0,1,
... , where
the sequenceis defined in such a way that aj =
3aj-l
and2aj
for j
> 2. Thenf = is
u.d.p.
This follows from the main result in[W],
here restatedas
Proposition
2 in section 4. Consider the sequence x = definedby xn
= i E M with n - i mod 2. x is u.d.and, by assumption,
Now define y =
(Y,,),,Elq
in such a waythat yn
= 1 ifIn
=IdM
and yn
= 0 iffn
= g. HenceIn(Yn)
= 1 for all n E N andDN(f(y))
= 2
for all N E N. NoteDa;(Y)
=Da;(x)
= 0 forall j
> 1. Forarbitrary -
> 0 andsufficiently
large j
thisyields
contradiction.
q.e.d.
Thus one has to make finer
investigations
of u.d. sequences x and useproperties
which fit tou.d.p.
sequences f in the sense ofProposition
2(cf.
section4).
For that we have to turn to theconcept
of sets with blockstructure.
3. Sets of block structure and related
concepts
For a better
understanding
of ourproblem
we considerpartitions
of the set N ofpositive
integers
into blocks. For this reason we use thefollowing
notations. If
0
is a
given
sequence ofintegers
letIk
=(ak-1,
ak]
n N denote the induced blocks(intervals)
and call thefamily
I =(Ik )kEN
thepartition
of N inducedby
a = If 0(or,
equivalently,
ak+llak
-1)
for h --~ o0 we call I = shortpartition.
We introduce the abbreviationwhere
k(N)
is thegreatest
index such thatak(N)
N. If thisquantity
is small this means that almost each blockIk
is almost contained either in T or in itscomplement
Hence it is obviouswhy
we say that T has blockstructure if
L(T,
I,
N)
= 0 for all shortpartitions
I. It is easy tosee that this
is,
forinstance,
the case if T iscompatible
with a sequence a = with > 1(example:
a (q) =
q >
1),
inthe
following
sense: A set T C N is calledcompatible
with a sequence a(or
with the inducedpartition
I =I(a) = (Ik)ken)
if andonly
if eachIk
is contained either in T or in itscomplement
~1 ~
T . With ,~ we denote thesystem of all sets T C N with block structure. Sets of block structure have been characterized in
[W]
in several ways.A related
topic
which isimportant
for us is the distribution of restrictedsequences. For TON define
and for xT =
1,
call thepositive
numberthe restricted
discrepancy.
We say that XT is u.d. w.r.t. T ifDN(xT)
-0 for N -- oo. Note that
DN(x, T )
where,
ingeneral,
theinequality
is strict.The sequence of numbers
densN(T)
E~0,1~,
N EN,
has an upper limitdens(T),
called upperdensity
of T.Similarly
the lowerdensity
dens(T)
is defined as the lower limit. In case of coincidence of both the common valuedens(T)
is calleddensity
of T. It is easy to checkthat,
forarbitrary
setsS, T
C N of naturalnumbers,
d(S, T)
=W(S4lT)
definesa pseudometric
dinducing
atopology
on the power set of N. Here =(S ~ T)
U(T B
S)
denotes thesymmetric
difference of the involved sets. Note that thetopological
closure of thesingleton
{0},
forinstance,
containsexactly
thesets T with
dens(T)
= 0.Here are
descriptions
from[W]
of sets with block structure:PROPOSITION 1. For T C N the
following
three conditions areequivalent:
(1)
T E B.
(2)
For all - > 0 there is a q > 1 and an Scompatible
witha (q)
suchthat
d(S, T)
E.(3)
For all u.d. sequences x on afinite
set M with#M >
2 therestric-tion XT is u, d. zv. r. t. T.
Proof: Follows from
[W],
Theorem 1.q.e.d.
The
following
theorempresents
further useful observations on B.THEOREM 2. B is a Boolean set
algebra
(not
aa-algebra)
onN,
and,
as asubset
of the
power setof N, it is topologically
closed w.r.t. d.Proof: ,Ci is not a
Q-algebra
since,
forinstance,
An
={2n}
El3,
but A =UnEF1
An
B.Obviously
,CB contains0,
Nand, immediately by definition,
isclosed under
complements,
i.e. Timplies
N B
T E Ci. To show that Bis a Boolean set
algebra
on N it therefore suffices to prove that S U T E ,Ciunder the
assumption
thatS,
T E B: For anygiven
shortpartition
I definesk + tk . This
implies
since ,S‘ and T are
supposed
to have block structure. Thus ~5’ U T E B.Finally Proposition
1.2 says ~3 = C forhence Ci is indeed a closed set.
q.e.d.
From the results up to now it follows
that,
for T E ? and E >0,
there isa set S E Ci
compatible
with some a = withak+l~ak > q
for someq > 1 such that
d(S, T)
E. Since d is apseudometric
and not a metricthe
question
arises whether one can even take 6’ = 0. That this is not thecase follows from Theorem 3.
THEOREM 3. There are sets T with block structure such that
d(S,T)
> 0for
all Scompatible
with some I =I(a),
a =(ak)kEN
with q > 1or,
equivalently,
Proof: Construction of T: For
j,
I > 1 define the numbersr( j, l)
=2 2j+l
+2’,
=22j+l
+ 2 ~ 21 =
22j +1
+2’+’
and the setsWe claim that the set T =
Too =
Aj
has the desiredproperties.
(Note
that n EAi
if andonly
if thebinary representation
of n - 1 starts with ablock 100 ...001 where the number of
0-digits
between the first two1-digits
is
2j - 1.)
T E B: The set
Aj
is,
by definition, compatible
with the sequenceCheck that the
quotients
of succeding
members of this sequence are boundedbelow
by
1 +(2 2j
+1)-1
= qj > 1.Hence Aj
forall j
andthus,
by
Theorem for all i E N. Now observeJ=Z
for
sufficiently large i.
Thus T EN}
C .C~ = .f3by
Theorem 2. It is obvious that the bothremaining
assertions of the theorem are indeedequivalent.
Weprefer
to prove the second one. For that reason suppose now that I =I(a),
a = is apartition
with q for some q > 1. If thequotients
are not bounded it is easy to see thatHence we may suppose
Q
for some fixedQ
to show thatL(T,
I,
N)
has apositive upper
limit.For each n let
kn
be chosen in such a way that Forq = such that
2-2~
~/3.
2j
+ 2 and I = n -2j
wedistinguish
two cases:First case:
ak"
With the notationand,
using
From the definition of T follows
Second case:
r( j, l)
akn. Similar to the first case,just replacing j by
j
+1,
1by
1 - 2and kn
by
onegets
Combining
both cases andchoosing E
=22Ji2Q)
onegets
Theorem 3 and the
following
fact will also be relevant on the search forpossible
estimates,
cf. the discussion in section 4.THEOREM 4. Let
(I(’))’EN,
1(1)
=I(a(l)) =
a(l) =
be anarbitrary
countablefamily of
shortpartitions.
Then there is a set Tç
N withL(T, I~i~, ~ ) -~
0for
atI l E l~ butT ~
B.Proof: Since all are short
partitions
there is a sequencesuch that
#
1
whenever I d and and withN1+1/N1 --1-
oo.la _ k _ t
1+1’
ik
Furthermore there exists a short
partition
J = =J(b), b
= such thatsufficintly large
Ifulfill t2
ifJk n
0.
Define
..-- i 1 .
To estimate
note that
only
those k contribute anotvanishing
term for whichbk, E
Iki)
for some k’.Using
the bounds for and~Jk/bk
we concludeOn the other hand let I =
I(a)
be a shortpartition
with )
ifIk
n(Ni,
Ni+i ) # @
forsufficiently large
1. Since#-~I~’-
- 0 forand
k, k’
2013~ 00 a further use of the bounds for and those foryields
implying
limL(T,
I,
~V) >
3.
Thus T does not have block structure.q.e.d.
4.
Reconsidering
aqualitative
resultFor
given
f = ~VI --~M,
we consider the setsT~ _ ~ n
EI
fn
=7r}
for everypermutation
7r ESM
of the set M(SM
denotes thesymmetric group
acting
on the set C =~n
EN I In
=c,~~
of all nfor
which fn
is a constant functiontaking
some value cn and theremaining
set D =N B
C B
T where T =T1I".
Now we areready
to formulate the characterization of
u.d.p.
sequences f = of maps which is the motivation for ourquantitative
results.PROPOSITION 2. The sequence f =
f,~ : M ->
M,
isu.d.p. if
andonly if
thefollowing
three conditions hold:1.
dens(D)
= 0.2. All sets
C, D
andT1r,
7r ESM,
have block structure.3. The sequence cc =
(Cn)nEC
is u.d. w.r.t. C.Proof:
Follows, by specialization, immediately
from[W],
Theorem 3.q.e.d.
If we say that f is
compatible
modulo constant maps with apartition P
of N if on each T either all
In
coincide or allIn
are constant maps then we can reformulate condition 2 ofProposition
2 as follows: f iscompatible
modulo constant maps with a finite
partition
of N into sets with blockstructure.
In order to use the conditions from
Proposition
2 fordiscrepancy
esti-mates we have toanalyse
howthey
can be made concreteby
means ofsuit-able
quantities.
For condition 1 and condition 3 this is clear:dens(D)
= 0 if andonly
if thequantity
densN(D)
tends to 0 and c = is u.d.w.r.t. C if and
only
if thequantity
DN(CC)
tends to 0.Which
quantity
describes howgood
the block structure of some T 6 ~is? This
question
is not so trivial. The reason is that in the definition forT the limit relation for
L(T,
I,
N)
must hold forall,
evenuncountably
many, shortpartitions
I.Unfortunately
the behaviour of one fixed I oreven of
countably
many of them does notgive
enough
information. This has been illustratedby
Theorem 4.Looking
at certainspecial examples
of sequences with block structure(as
the sequencesa (q)
for q >1),
one observes thatL(T,
I,
N) -
0 for somesuitably
chosen"long"
partition
I with > 1. Thehope
that thismight
happen
for all T has beendestroyed by
Theorem 3. Butwhat we may
expect
is thatL(T,
I,
N)
gets
small independence
on N andb if 1 + 6-
(Let
as callpartitions
I =I(a)
with thisproperty
6-partitions.)
Thisapproach
indeed ispossible
as Theorem 5.1 will show.Theorem 5.2 is an
application
of Theorem 5.1 tou.d.p.
f.But how to match a u.d. x with a
u.d.p.
f? Condition 2 inProposition
2 tells us that in a
u.d.p.
f thepermutations
7r : M --~ M occur(with
fewexceptions)
inlong
blocks. On such blocks the distribution behaviour of the xn is the same as the behaviour of the7r(xn) (up
to a fixedpermutation
7r of the elements ofM).
Hence we have toguarantee,
that the blocks arelong enough
such that the uniform distribution of x can beobserved. This will be the
object
of Theorem 5.3.THEOREM 5.
( 1 )
Suppose
that T has block structure and > 0 isgiven.
Then there exist a
positive
b =6T(E)
> 0 and a natural numbersuch that
1 11 -, 1 /
holds
for
all N >NT(£,6)
and allb-partitions
I =(1k)kEIq-(2)
Suppose
that E > 0 and fu.d.p.
aregiven.
Then there is apositive
b =
~f(~)
> 0 and a natural numberN1=
6)
such thatfor
allN >
N1
and all6-partitions
I there is an f’ =compatible
with I modulo constant maps such that
Furthermore we may assume
DN(f(x)c)
-for
such N.(3)
Suppose
that x is u.d. and E, b > 0 aregiven.
Then there is apositive integer No
=b)
E ~I such thatfor
all I =(N’,
Ni]
n N with N’ >No
and(1
+6)N’
and all N >N1
Proof:
(1)
Suppose,
by contradiction,
that there is some E > 0 such that for ah 6 > 0 and all N there is a6-partition
1(6, N) = (Ik(6, N))keN
By
a standarddiagonal
argument
(similar
as in theproof
of The-orem4)
thepartitions
1(/)
=I( t, N1)
withsufhciently rapidly
in-creasing
Ni
> andcorresponding Ni
can be combined to ashort
partition
I withimplying
contradiction.
(2)
1
x E?M}
whereC, T7r
~ 8 andD ~
S are asabove. All S
E S, by Proposition
2,
have block structure. Thuspart
1 of the theorem
guarantees
that there are, for each S E S and for2~+2) ?
numbers and with theproperties
stated there. Let b =6f(s)
> 0 be the minimum of thes E S,
w.l.o.g.
E/2.
Furthermore, by Proposition
2,
thereis a natural number N’ =
ND(£1,8D(£1)),
such thatdensN(D)
£1 for all N > lV’, DefineNow choose any
6-partition
I =I(a).
We have to find an f’ = with theproperties
stated in the theorem. For all n E11
let
f ~
be the samearbitrary
but fixedmapping.
For n EIk,
k > 2distinguish
two cases:°
Case 1: If
E Ik ~
fn
=7r}
>#Ik/2
for some 7r ESM
put
for all n ~Ik.
Case 2: Otherwise define
f,~
=cn
for some constantcn
wherecn
= Cn for the casethat f n =
c,~ is constant.By
definition f’ iscompatible
with I modulo constant maps. We have to estimatedensN(X)
for X =~n
EN
I
fn}
and N >Nr(e,6).
LetAk =
and,
for S ES,
AS,k = min(#Ik n
s, ~Ik ~ S).
In case 1 we haveAk -
ATr,k
for some ~r EsM.
Incase 2 we have
Ak
=#(I~ 1 C)
and wedistinguish
two subcases: IfC) > ~(Ik 1
C)
thenAk
=Ac,k .
In the other subcase weget
lY ’ I
which is the assertion of the second
part
of the theorem.(3)
Follows with the same method as Lemma 1 in[W].
q.e.d.
5.Discrepancy
estimates and s-blocksUsing
Theorem 5 we are now in theposition
to deduce adiscrepancy
estimate for
f(x)
of thefollowing
type:
THEOREM 6. Given a u.d. sequence x and a
u.d.p.
sequence fof
maps.For
arbirarily given
e > 0 let b =bf(s)
1~
andNl
=Nf(E,
6)
(w.l.o.g.
N1
2::
a)
be as in Theorem 5.2. Furthermore let80
E(0, b -
2
and
No
N’N1
whereNo
= as in Theorem 5.3. Thenfor
all N >Nl
thediscrepancy of
the sequencef(x)
can be estimatedby
Proof: Fix any
partition
I =I(a)
with 1 + 1-E- b
and~Nl/2
EN,.
(Note
that this ispossible by
theassumption
onbo
since
1.)
Let f’ be as in Theorem 5.2compatible
with I moduloconstant maps and let the sets resp.
C~D~T~T~
be associatedto f resp. f’ as in section 4. Define for
arbitrary
Nand the numbers
I~o
and1~1
by
N’ ako resp. 0~1 N ak1 +1.Combination of these estimates
gives
Since E and 6 can be chosen
arbitrarily small,
this resultgives
a method to determine howlarge
N must be such thatDN(f(x))
gets
smaller thanany
given
positive
value. Forgiven
u.d. x andu.d.p.
f one has to know theinteger
functionsNx(£,6)
andNf(ê,6)
which existby
Theorem 5.It is clear that the
explicit
bounds forDN(f(x))
stemming
from theproofs
of Theorem 5 and Theorem 6 are notoptimal.
But a more carefulestimation would have taken so much technical effort that the view to the main
object
would have been obscuredby
boring
calculations. What wewant to
emphasize
is that functionsNx(s)
andNf(s)
depending only
on oneargument E
do notgive enough
information toguarantee
DN(f (x))
6~ but the functionsN x( ê, b)
and6)
guaranteed by
Theorem 5 and used inTheorem 6 do.
Finally
we have toinvestigate
how our methodsapply
to s-blockdiscrep-ancy. If one assumes that f is
u.d.p.
then thegeneralizations
arestraight
forward. The
only
modifications are as follows:In
estimating
one has to look at the blocksIf
only
one ofthe Xi
or fz, i = n, ... , n + s -1,
does not behaveregularly,
this may affect not
only
one but s members in the sequenceyes) =
Hence
discrepancy
estimates have to take this into accountby
a factor s.It is clear
that,
for every fixed s EN,
theasymptotic
definition of blockstructure causes that this
problem
does not affect the nature of the results. Of course we have tomodify
the definition of restricteddiscrepancy
in thefollowing
way: For T C N and a E MS define-- ----
I ... -
I
x is called s-block u.d. w.r.t. T if
DN(x) -
0 for N - oo.Similarly
onecalls x
completely
u.d. w.r.t. T if this is the case for all s E N.In the natural way we call f s-block resp.
complete u.d.p.
if f (x)
is s-blockresp.
completely
u.d. whenever x is s-block resp.completely
u.d. With the same notations as inProposition
2 thequalitative
result is thefollowing:
THEOREM 7. Let f be
u.d.p.
Then f is s-block resp.completely u.d.p. if
Proof:
Carry
out the program described above. We omit details.q.e.d.
Note that in Theorem 7 we have assumed that f is
u.d.p.
It is aconjecture
but
yet
unproved
whether s-block orcomplete u.d.p. always implies u.d.p.
REFERENCES
[D-T]
M. Drmota and R. F.Tichy. Sequences,
discrepancy
andapplications.
Springer,
Lecture Notes in Mathematics 1651(1997).
[G]
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