ThomasFernique
LIRMMCNRS-UMR5506andUniversitéMontpellierII,
161rueAda34392MontpellierCedex5-Frane,
PONCELETLab.CNRS-UMI2615andIndependentUniversityofMosow,
Bol'shojVlas'evskijper.11.119002Mosow-Russia,
thomas.ferniqueens-lyon.org
Abstrat. We provideinthispaperamultidimensionalgeneralization
ofsubstitutionsonwords,whihisdenedastheationonmultidimen-
sionalsequenesofanon-pointed substitution endowedwithloalrules.
Thenon-pointed substitutionsand the loalrules havein themultidi-
mensionalaserespetivelytherolesplayedbythesubstitutionsdened
onlettersand bythe onatenation onwords. This denition thenal-
lowsusto providea(yetpartial)multidimensionalgeneralization ofan
algebrai haraterization of Sturmian words whih are xed-point or
morphiimageofaxed-pointofanon-trivialsubstitutiononwords.
Introdution
Asubstitutionatsonawordinthisway:theimageofeahletterisaword,and
theimageof thewholewordis thenjust theonatenationof theimages ofits
letters.Substitutionsarepowerfulombinatorialtools,andhavenaturalinter-
ations with languagetheory,geometry of tilings,automata theory, and many
others(seee.g. [14℄andthereferenesinside).It thus wouldbeusefulto dene
asimilartoolinthemoregeneralframeworkofmultidimensonalsequenes,that
are sequenesof letters indexedby Z n
(whereaswordsare sequenesof letters
indexed by N). It is however adiult problem, mainly for lak of a natural
multidimensionalonatenation.
Suhageneralizationhasalreadybeenintroduedin[15℄:forp
1
;:::;p
n xed
inN,aletteruindexedby(i
1
;:::;i
n
)ismappedtoaset(u)oflettersindexed
byf(j
1
;:::;j
n
)j8k; p
k i
k j
k
<p
k (i
k
+1)g(thatis,ap
1
:::p
n
-retangle).
But it generalizes in fat only onstant-length substitutions on words (whih
map lettersto words allof thesame length). An algebraiharaterization of
all the multidimensionalsequenes whih are xed point of suh substitutions
isalsoproved(seeagain[15℄),whatgeneralizesasimilarresultforwordswhih
arexed-pointofaonstant-length substitution(seee.g.[1℄).
A rstaim of this paperis to introdue anotion of multidimensionalsub-
stitutionwhihgeneralizesanytypeofsubstitutionsonwords,andnotonlythe
givean algebraiharaterizationof themultidimensionalsequenes whih are
xed-pointofsuhamultidimensionalsubstitution.More preisely,Theorem2
generalizesthefollowingresult(seee.g.[6,9℄):
Let be an irrational number in [0;1℄. One denes the Sturmian sequene
u
=(u
n
)overthe alphabetf1;2gby:
8n1; u
n
=1 , (n)mod12I
;
where I
= (0;1 ℄ or I
= [0;1 ). Then u
is a xed point (resp. the
morphiimageofaxedpoint)ofasubstitution onwordsifandonlyifhasa
purely periodi (resp. eventuallyperiodi)ontinuedfrationexpansion.
Notie that this haraterization onerns only Sturmian sequenes, that is, a
subsetofthesetofallthesequenes.Thus,generalizingthisresultalsorequires
todene anotionofmultidimensionalSturmiansequene.
Thepaperisorganizedasfollows.Intherstsetion,wedenenon-pointed
substitutions and loal rules, that are our multidimensional equivalents of the
lassi substitutionsdenedonletters,andoftheonatenationprodutused
to makesuh substitutions aton sequenes.It allowsus, under onditionson
theloalrules,todeneournotionofmultidimensionalsubstitution.InSetion
2,wedesribeatypeof loalruleswhih satisfytheonditionsrequiredto de-
ne amultidimensionalsubstitution: the loal rulesderived from aglobal rule.
InSetion3,weresumethenotionofgeneralizedsubstitutions,deneSturmian
hyperplanesequenesandthenweshowthatthesegeneralizedsubstitutionspro-
videglobalrulesfromwhihweanderiveloalrulesasdesribedinSetion2.
ItyieldsmultidimensionalsubstitutionsonSturmianhyperplanesequenes,and
allows usto give(Theorem 2) apartial generalization ofthe algebraihara-
terizationofxed-pointsstatedabove.
1 Non-pointed substitutions and loal rules
LetAbeanite alphabet.Apointedletter isanelementL=(x;l)ofZ n
A,
wherexistheloation oftheletterl.WedenotebyLthesetofpointedletters.
Apointedpattern isasetofpointedletterswithdistintloations.Thesup-
port ofapointedpatternisdenedasthesetoftheloationsofitsletters.Two
pointed patterns are said onsistent if two letters with the same loation are
idential. The notionsof union,intersetion and inlusion are then dened for
onsistentpatternsasforusualsets.WedenotebyPthesetofpointedpatterns.
The lattieZ n
ats on pointed letters (resp.pointed patterns) by transla-
tion onthe loations (resp.supports): the lassesof equivalene of this ation
areallednon-pointedletters anddenotedbyL(resp.non-pointedpatterns,de-
Thus, toeah pointedpatternP orrespondsauniquenon-pointedpattern,
alled its underlying non-pointed pattern and denoted P. Conversely, to eah
non-pointed pattern P orresponds all the ongruent pointed patterns, alled
realizations of P, that haveP asunderlying non-pointedpattern. If P and P 0
areongruentpointedpatterns,onedenotesv(P;P 0
)2Z n
thevetorthatmaps
P ontoP 0
bytranslation.
Wearenowin apositionto giveourmultidimensionalgeneralizationofthe
denition onlettersofasubstitution onwords:
Denition1. A non-pointedsubstitutionisamap fromL toP.
Inwhat follows, denote a non-pointed substitution. We now dene loal
rules,whiharethemainingredientofourmultidimensionalonatenation.
Denition2. Wedene twotypesof loalrulesfor :
an initialrule
isdenedon aset I(
)=fLgofone pointedletter,and
maps Ltoarealizationof (L);
an extension rule is dened on a set E()=fL;L 0
gof two pointed let-
ters with distint loations, and maps L and L 0
to disjoint realizations of
respetively (L)and(L 0
).
Roughlyspeaking,aninitialruletellsushowtoposition(L)forapartiular
pointedletterL,whileanextensionrulesuhthatE()=fL;L 0
gisused,for
apointedpattern fA;A 0
gongruentto fL;L 0
g,to position (A 0
) relatively to
(A )inthesameway(L 0
)ispositionedrelativelyto (L).Werstdenethe
ationof on-paths:
Denition3. LetU beapointedpatternand beasetof loal rulesfor .A
-path ofU isasequeneR=(R
1
;:::;R
k
) ofpointedlettersofU suhthat:
thereexistsaninitial rule
2 suhthat I(
)=fR
1 g;
for i = 1:::k 1, there exist an extension rule
i
2 and x
i 2 Z
n
suh
thatE(
i )=fL
i
;L 0
i
gwith R
i
=L
i +x
i andR
i+1
=L 0
i +x
i .
One then denes by indution a map denoted by (;;R ) on the letters of R
(see Fig.1):
(;;R )(R
1 )=
(R
1 );
fori=1:::k 1,(;;R )(R
i+1 )=
i (L
0
i )+v(
i (L
i
);(;;R )(R
i )).
Notiethat, when omputingthe ation of asubstitution on aword, we
proeed in the same way: the image by of the rstletter of theword (here
seenas apath)hasaspeiedposition(heregivenbyaninitialrule),whilethe
position oftheimage ofaletterfollows,byindution,from theposition ofthe
onatenationoftheimagesofthepreviousletters(here,weuseextensionrules
Fig.1.Top:fromlefttoright,aninitialruleandtwoextensionrules;bottom:ompu-
tationoftheimageofapathusingsuessivelythethreepreviousloalrules.
Denition4. Let be a set of loal rules for and U be a pointed pattern.
The setissaid tooverU if anypointedletterofU belongstoa-path ofU
and issaid tobe onsistent on U if for any two-paths R and R 0
of U whih
both ontainapointedletter L,(;;R )(L)=(;;R 0
)(L).
If oversU andisonsistenton U,one thendenesthe ation of endowed
with theset ofloal rules, denotedby (;),asfollows:
(;)(U)= [
f(;;R )(L)j Ris a-path ofU andL2R g:
Thus, (;)is ournotionof multidimensionalsubstitutionon pointedpat-
terns.It anbeshownthat itgeneralizesthesubstitutions onwordsaswell as
themultidimensionalsubstitutionsdesribedin[15℄.Thepossibilitiesaremuh
larger,butitisingeneralnoteasytoobtainsetsofloalrulesthatareonsistent
onasetofpointedpatternsandoverthisset: thenextsetionpresentsaway
toobtainsuh setsofloalrules.
2 Loal rules derived from a global rule
Letbeanon-pointedsubstitutionandHbeasetofpointedpatterns.Weare
hereinterestedinageneriwaytoobtainsetsof loalrulesfor thatoverH
and are onsistenton it(that is, that overany pointed pattern ofH and are
onsistentonanyofthem).Wederivesuhsetsofloalrulesfromglobal rules:
Denition5. AglobalruleonHfor isamap denedonthesetofpointed
lettersfL2U j U 2H g suhthat:
apointedletterL ismappedtoarealizationof (L);
pointedletterswithdistintloationsaremappedtodisjointpointedpatterns.
Letusdenotebyd(L;L 0
)thedistane P
jx
i x
0
i
jbetweentheloations(x
i )
and(x 0
)ofLand L 0
.Weintrodueanotionofweakonnexity:
Denition6. The spanbetweentwopointedlettersLandL 0
ofU 2H ,denoted
by sp(L;L 0
), is the smallest integer D suh that there exists asequene (L
1
=
L;L
2
;:::;L
k
=L 0
) ofpointedletters of U whih veries: 8j,d(L
j
;L
j+1 )D.
The spansof U andHarethen denedby:
sp(U)= sup
L;L 0
2U sp (L;L
0
) and sp (H )= sup
U2H sp(U):
Forexample,sp(U)=1ifandonlyifU is4-onneted.Letusnowderivea
set ofloalrulesfromaglobalrule:
Denition7. Let H
0
be apointed pattern and a global rule on H for . A
set of loal rules for issaidtobe derived from(H ;H
0
; )if itveries:
1. if
isaninitialruleofwithI(
)=fLg,thenL2H
0 and
(L)= (L);
2. if is an extension ruleof with E()=fL;L 0
g, then d(L;L 0
)sp(H ),
(L)= (L)and(L 0
)= (L 0
);
3. ifand 0
areextensionrulesof,thenE()andE(
0
)arenotongruent.
Suhderivedsetsofloalsruleshaveinterestingproperties:
Proposition1. IfH
0
isniteandsp (H )isbounded,thenanysetofloal rules
derivedfrom(H ;H
0
; )isnite.
Proof. Letbederivedfrom(H ;H
0
; ).ThereisnomorethanjH
0
jinitialrules
in .There arejAj
j(sp(H)+1) n
=Z n
j
non-ongruentpointedpatterns fL;L 0
gthat
verifyd(L;L 0
)sp(H ):itfollowsthatthereisanitenumberofextensionrules
in .Thus,isnite. ut
Denition8. Aglobalrule onHissaid ontext-freeif,forU 2H ,L;L 0
2U
andx2Z n
suhthat L+x;L 0
+x2U,onehas:
v( (L); (L+x))=v( (L 0
); (L 0
+x)):
WepresentexamplesofsuhglobalrulesinSetion3.
Proposition2. If is a ontext-free global rule on H , then any set of loal
rules derivedfrom(H ;H
0
; )isonsistenton H .
Proof. Suppose that is ontext-free, and let be a set of loal rules de-
rived from (H ;H
0
; ). Let R = (R
1
;:::;R
k
) be a -path of U 2 H . Let us
prove by indution that for all i, (;;R )(R
i
) = (R
i
). Sine R is a -path,
there exists an initial rule
2 suh that I(
)=fR
1
g,and sine is de-
rived from (H ;H
0
; ), (;
k
;R )(R
1 ) =
(R
1
) = (R
1
). Suppose now that
(;
k
;R )(R
i
) = (R
i
). Aording to Denition 3, there exists an extension
rule
i
2 and x
i 2 Z
n
suh that E(
i
) = fL
i
;L 0
i
g with R
i
= L
i +x
i
and R
i+1
= L 0
i +x
i
, and (;
k
;R )(R
i+1
) = (L 0
i
)+v((L
i );(;
k
;R )(R
i )).
But is derived from (H ;H
0
; ), hene (L
i
) = (L
i
) and (L 0
i
) = (L 0
i ).
Moreover, (;
k
;R )(R
i
) = (R
i
) = (L
i +x
i
). Thus, (;
k
;R )(R
i+1 ) =
(L 0
i
)+v( (L
i ); (L
i +x
i
).Finally,sine isontext-free,(;
k
;R )(R
i+1 )=
(L 0
i
)+v( (L 0
i ); (L
0
i +x
i
))= (L 0
i +x
i
)= (R
i+1
).It yieldsthat is on-
Proposition3. IfH
0
intersetsanypointedpattern ofH ,thenthereexistsets
of loalrules derivedfrom(H ;H
0
; )thatoverH .
Proof. Let us dene E = ffL;L 0
gjL;L 0
2U; U 2Handd(L;L 0
)sp(H )g,
and let E 0
be amaximal subsetof E that does not ontain ongruent pointed
patterns.Letbethesetofthefollowingloalrules:
foreahL2H
0
,theinitialrule
denedonI(
)=fLgby
(L)= (L);
for eah fL;L 0
g 2 E 0
, the extension rule dened on E() = fL;L 0
g by
(L)= (L)and(L 0
)= (L 0
).
Oneeasilyheksthatisderivedfrom (H ;H
0
; ).Letusprovethatovers
H .LetU 2HandL 0
2U.SineH
0
intersets anypointedpattern ofH ,there
exists L2U[H
0
. Bydenition,therealso existsasequeneofpointedletters
(L
1
= L;L
2
;:::;L
k
= L 0
) suh that 8i, d(L
i
;L
i+1
) sp(H ). Then, for all i
there exists x
i 2 Z
n
suh that fL
i
;L
i+1 g+x
i 2E
0
, and thereexists an initial
ruleofdenedonfL
1
g.Ityieldsthat(L
1
;:::;L
k
)isa-pathwhihontains
L 0
.Thus, oversH . ut
Weanresumethepreviouspropositionsinthefollowingtheorem:
Theorem1. Let beaontext-freeglobal ruleonHfor.Ifsp (H )isbounded
andif H
0
2P is anite pointedpattern intersetingany pointedpattern of H ,
then onean derive from(H ;H
0
; )anite setof loal rules thatis onsistent
on Handoversit.
We thus have a way to derive, from a ontext-free global rule, loal rules
onsistentonagivensetofpointedpatternandoveringthisset.Thisresultis
applied inthenextsetiontoapartiulartypeofontext-freeglobalrule.
3 Sturmian hyperplane sequenes and algebraiity
Werstbrieyresumethenotionofgeneralizedsubstitution (seee.g.[4,5,14℄).
Lete
1
;:::;e
n
denotetheanonialbasisofR n
andleth:;:idenotetheanonial
salarprodutonR n
.
Afae (x;i
),forx2Z n
andi2f1;:::;ngisdened by:
(x;i
)=fx+ X
j6=i r
j e
j j0r
j 1g:
Suh faes generate the Z-mo dule of the formal sums of weighted faes G =
f P
m
x;i (x;i
)jm
x;i
2Zg, on whih the lattie Z n
ats by translation: y +
(x;i
)=(y+x;i
).Faesareusedtoapproximatehyperplanes ofR n
:
Denition9. Let2R n
+
,6=0. The hyperplane P
ofR
n
isdenedby:
P =fx2R n
j hx; i=0g:
The steppedhyperplaneS
assoiatedtoP
isdenedby:
S
=f(x;i
)j hx; i>0andhx e
i
;i0g;
anda path ofS
is anite subsetof the setof faesof S
.
Notie that a path of S
belongs to the Z-module G, but is geometri,
that is,withoutmultiplefaes. Letusreallthattheinidenematrix M
ofa
substitutiononwords givesatposition (i;j)thenumberof ourenesofthe
letteriin theword(j).IfdetM
=1,then issaidunimodular.
Denition10. The generalizedsubstitution assoiatedtothe unimodular sub-
stitution isthe endomorphism
of G denedby:
8
>
>
>
>
<
>
>
>
>
:
8i2A;
(0;i
)= P
3
j=1 P
s:(j)=pis M
1
(f(s));j
;
8x2Z 3
; 8i2A;
(x;i
)=M 1
x+
(0;i
);
8 P
m
x;i (x;i
)2G;
(
P
m
x;i (x;i
))= P
m
x;i
(x;i
);
wheref(w)=(jwj
1
;jwj
2
;jwj
3
)andjwj
i
isthenumberof ourenesof theletter
i inw.
Thefollowingtypeofsubstitution ispartiularlyinteresting:
Denition11. A substitution is of Pisot type if its inidene matrix M
has eigenvalues ;
1
;:::;
n 1
satisfying 0 < j
i
j < 1 < . The generalized
substitution
isthen alsosaid ofPisot type.
Indeed,thefollowingresultisprovedin[4,5℄:
Proposition4 ([4,5℄). If is of Pisot type andif is a left eigenvetor of
M
for the dominant eigenvalue , then
(S
) S
and
maps distint
faes ofthe steppedhyperplane S
todisjoint pathes ofS
.
Thestepped hyperplane S
is alled the invariant hyperplane of
. It is
alsoprovedin[11℄:
Proposition5 ([11℄). If the modied Jaobi-Perron algorithm ([8℄) yields a
purely periodi (resp.eventuallyperiodi) ontinuedfration expansionfor 2
R n
,thenthesteppedhyperplane S
isaxedpoint(resp.the imagebyagener-
alizedsubstitution ofaxedpoint)ofageneralizedsubstitution ofPisot type.
Wethen dene hyperplanesequenes, mapping stepped hyperplanes of R n
to (n 1)-dimensional sequenes over the alphabet f1;:::;ng. The following
proposition(provedin Appendix)resumesaresultgivenin[2,3℄:
Proposition6. LetV
Z
n
bethe set of the verties that belong to the faes
of S
.Letv
and
bethemaps denedrespetively onS
andV
by:
v
(x;i
)=x+e
1
+:::+e
i 1
and
(x
1
;:::;x
n )=(x
1 x
n
;:::;x
n 1 x
n ):
Then,v (resp. )isabijetionfromS ontoV (resp.fromV ontoZ n 1
).
Let
bedenedonS
by
(x;i
)=(
(v
(x;i
));i):itmapsbijetively
thefaesofS
tothelettersofa(n 1)-dimensionalsequeneoverf1;:::;ng.
Notie that not all these (n 1)-dimensionalsequenesoverf1;:::;ngorre-
spondtoasteppedhyperplane.Wethusintroduethefollowingdenition:
Denition12. An hyperplane sequene is an (n 1)-dimensional sequene
over f1;:::;ng dened, for 2 R n
, by
(S
). One denotes by H
suh an
hyperplanesequene.Moreover,if=(
1
;:::;
n
)issuhthat1;
1
;:::;
n are
linearlyindependentoverQ,thenH
isalleda Sturmianhyperplanesequene.
For n = 2, Sturmian hyperplane sequenes are nothing but Sturmian se-
quenes over f1;2g (see [12℄), and for n = 3, one retrieves the notionof two-
dimensional Sturmiansequene of[7℄. Notiethat an hyperplanesequeneH
is dened on the whole Z n 1
: it yields sp(H
) = 1. Let us now derive,from
generalizedsubstitution, ontext-freeglobalrulesonhyperplanesequenes:
Proposition7. LetbeaPisotunimodularsubstitutiononwordsoverf1;:::;ng.
Let
be the assoiated generalized substitution,and S
its invariant stepped
hyperplane.LetH
=
(S
). We setL=Z n 1
f1;:::;nganddene:
=
Æ
Æ
1
and
: (0;i)2L7!
(0;i)2P:
Then,
isaontext-freeglobal ruleonH
for thenon-pointedsubstitution
.
Proof. For(x;i)2H
andy2Z n 1
,oneomputes:
((x;i)+y)=
(x;i)+
(M
1
1
(y)):
Itfollowsthat
(x;i)=
(0;i)=
(0;i)
.Moreover,sine
mapsdistint
faes of S
to disjoint pathes of S
(see Proposition 4) and sine
maps
bijetivelythefaesofS
tothelettersofH
,
=
Æ
Æ
1
mapsletters
withdistintloationsto disjointpointedpatterns. Thus,
isaglobalruleon
H
for
.
Then,if(x;i)2H
,(x
0
;i)2H
andy2Z n 1
,one has:
v(
(x;i);
((x;i)+y))=
(M
1
1
(y))=v(
(x
0
;i);
((x
0
;i)+y)):
Hene
isontext-free,aordingtoDenition 8. ut
Finally,ombiningTheorem1andProposition5and7,weobtain:
Theorem2. If themodiedJaobi-Perronalgorithm ([8℄)yields apurelyperi-
odi (resp. eventually periodi) ontinued fration expansion for 2R n
, then
theSturmianhyperplane sequeneH
isaxedpoint(resp.theimagebyamul-
tidimensional substitutionof axedpoint)of amultidimensional substitution.
This result an thus be seen as a multidimensional generalization of the
algebraiharaterizationresumedin theintrodution,thoughitprovidesonly
multidimensionalsubstitutionortheimagebyamultidimensionalsubstitutionof
suhaxedpoint.Infat,theproofofthealgebraiharaterizationresumedin
theintrodutionusesthenotionofreturnwords of[10℄.Thisnotionhasalready
beengeneralized,in termsof tilings, in [13℄:it thus givesus a possible way to
ahievetheharaterizationofTheorem2.
Example 1. Let bethelassi substitutiondened on f1;2;3gby(1)=13,
(2)=1and(3)=2.Oneomputes:
M 1
= 0
00 1
10 1
01 0 1
A
; and
:
(0;1
)7!((1; 1;0);1
)+(0;2
)
(0;2
)7!(0;1
)
(0;3
)7!(0;2
)
;
whihyieldsthenon-pointedsubstitution:
: 1
0;0 7!f1
0;0
;2
0;1
g; 2
0;0 7!f3
0;0
g; 3
0;0 7!f1
0;0 g;
whihoneanalsorepresentasfollows:
: 17!
2
1
; 27!3; 37!1:
Letus dene H=f n
((0;0);1);n1g. Oneanprovein this partiularase
that sp(H )=1.Thus,oneanompute(Theorem 1) anite set ofloal rules
that oversHand is onsistenton it. Oneobtainsfor examplethe initial rule
dened by:
: ((0;0);1)7!f((0;0);1);((0;1);2)g;
andveextensionrules,representedasfollows(theboldedlettersaremappedto
theboldedletters,sotheinformationaboutrelative loationsisstillonserved):
1 :
2
1 7!
2
31
;
2
: 31 7!
2
1
1
;
3 :
1
1 7!
2
21
1
;
4
: 21 7!
2
1
3
;
5 1
3 7!
11
2 :
Forexample, omputing the sequene (
;f
;
1
;:::;
5 g)
n
((0;0);1) for n =
1;:::;7gives(theletterwithloation(0;0)isbolded):
1 7!
2
1 7!
2
31 7!
2
31
1 7!
2
21
31
1 7!
2 2
3121
31
1 7!
2 2
3121
31
21
31
7! :::
Weaninthiswaygeneratearbitrarelylargepathesofthehyperplanesequene
H
,where isaleft eigenvetorofM
.Moreover,H
is axed-pointof this
multidimensionalsubstitution.
Aknowledgements.WewouldliketothankValérieBerthéandPierreArnoux
formanyusefulsuggestions.
Referenes
1. J.-P.Allouhe,J.O.Shallit,Automatisequenes:TheoryandAppliations,Cam-
bridgeUniversityPress, 2002.
2. P.Arnoux,V.Berthé,S.Ito,Disreteplanes,Z 2
-ations,Jaobi-Perronalgorithm
andsubstitutions. Ann.Inst.Fourier(Grenoble)52(2002),10011045.
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SimonStevin8no.2(2001),181207.
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Arithmetia,toappear.
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CentreTrats145, MatematishCentrum,Amsterdam,1981.
9. D. Crisp,W. Moran, A. Pollington,P.Shiue, Substitution invariant utting se-
quenes.J.Théor.NombresBordeaux5(1993),123137.
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port05024(2005),LIRMM.
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Voronoïtessellations.Geom.Dediata79(2000),239265.
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Sér.IMath305(1987),501504.
Appendix
Proof ofProposition6:
Let(x;i
)and(y;j
)betwofaesofS
suhthatv
(x;i
)=v
(y;j
).Ifi<j,
then x=y+e
i
+:::+e
j 1
,and hx e
i
; i=h(y+e
i+1
+:::+e
j 1
;i=
hy;i +he
i+1
+::: +e
j 1
;i. Sine (y;j
) 2 S
, hy;i > 0. Moreover,
he
i+1
+:::+e
j 1
;i 0. Thus, i < j would yield hx e
i
;i > 0, what
wouldontradit(x;i
)2 S
.Similarly, i>j is impossible.Hene i =j, and
x=y follows.Itprovesthatv
isone-to-onefrom S
toV
.
Ify2V
,then thereexist (x;i
)2S
andI f1;:::;ng,i2=I, suh that
y=x+ P
j2I e
j
. Letusdenotef :k7!hx+ P
j2I e
j e
1
::: e
k
;i.One
has:
f(0)=hx; i+ X
j2I he
j
;i>0; f(n)=hx e
i
;i
X
j=2 I;j6=i he
j
;i0;
and f is dereasing. Letk
0
suh that f(k
0
1) >0 and f(k
0
) 0. Lety
0
=
y e
1
::: e
k0 1
.Then,hy
0
;i=f(k
0
1)>0,andhy
0 e
k0
;i=f(k
0 )0.
Thus,(y
0
;k
0 )2S
.Sinev
(y
0
;k
0
)=y,itprovesthatv
isonto fromS
on
V
.
Letus denote by(
1
;:::;
n
).Reallthat the
i
are positiveandnotall
equal to zero. Letthen x =(x
1
;:::;x
n ) 2V
and (x
0
;i
)= v 1
(x). Onehas
0<hx 0
;ihe
i
;i=
i .Thus:
0<
n
X
j=1 x
j
j i 1
X
j=1
j
i :
Supposenow
(x)=(y
1
;:::;y
n 1
).Thepreviousformulayields:
0<
n 1
X
j=1 y
j
j +x
n n
X
j=1
j
i 1
X
j=1
j +
i
n
X
j=1
j
;
andperformingthedivision by P
n
j=1
j
>0,itthengives:
0<
P
n 1
j=1 y
j
j
P
n
j=1
j +x
n 1;
that is,sinex
n 2Z:
x
n
=1
&
P
n 1
j=1 y
j
j
P
n
j=1
j '
:
Conversely,given(y
1
;:::;y
n 1 )2Z
n 1
,setting x
n
2Zasaboveandthen, for
i=1:::n 1,x
i
=y
i +x
n
yields
(x
1
;:::;x
n )=(y
1
;:::;y
n 1
).Thus,
isa
bijetionfromV
toZ
n 1
(andtheproofprovidesanexpliitformulafor 1
).
