Minimal weight expansions in Pisot bases

0.3 MB·English

by  Christiane Frougny

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n Minimal weight expansions in Pisot bases
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9 ChristianeFrougny and WolfgangSteiner

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D Abstract.Forapplicationstocryptography,itisimportanttorepresentnumberswithasmallnumber
ofnon-zerodigits(Hammingweight)orwithsmallabsolutesumofdigits. Theproblemoffinding
.
s representations with minimal weight has been solved for integer bases, e.g. by the non-adjacent
c forminbase2.Inthispaper,weconsidernumerationsystemswithrespecttorealbasesβwhichare
[ Pisotnumbersandprovethattheexpansionswithminimalabsolutesumofdigitsarerecognizable

byfiniteautomata.WhenβistheGoldenRatio,theTribonaccinumberorthesmallestPisotnumber,
3
wedetermineexpansionswithminimalnumberofdigits±1andgiveexplicitelythefiniteautomata
v
recognizingalltheseexpansions.Theaverageweightislowerthanforthenon-adjacentform.
4
7
Keywords.Minimalweight,beta-expansions,Pisotnumbers,Fibonaccinumbers,automata.
8
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AMSclassification.11A63,11B39,68Q45,94A60.
.
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8 1 Introduction
0
: Let A be a set of (integer) digits and x = x x x be a word with letters x in A.
v Theweightofxistheabsolutesumofdigits1 x2··=· n n x . TheHammingjweight
Xi of x is the number of non-zero digits in x. Okf ckoursPe, jw=h1e|njA| 1,0,1 , the two
⊆ {− }
r definitionscoincide.
a
Expansions of minimal weight in integer bases β have been studied extensively.
When β = 2, it is known since Booth [4] and Reitwiesner [23] how to obtain such
anexpansionwiththedigitset 1,0,1 . Thewell-knownnon-adjacent form(NAF)
{− }
isaparticular expansion ofminimal weight withthepropertythatthe non-zerodigits
are isolated. It has many applications to cryptography, see in particular [20, 17, 21].
Other expansions of minimal weight in integer base are studied in [14, 16]. Ergodic
propertiesofsignedbinaryexpansionsareestablishedin[6].
Non-standard number systems — where the base is not an integer — have been
studied from various points of view. Expansions in a real non-integral base β > 1
have been introduced by Rényi [24] and studied initially by Parry [22]. Number the-
oretic transforms where numbers are represented in base the Golden Ratio have been
introduced in [7] for application to signalprocessing and fast convolution. Fibonacci
representations have been used in [19] to design exponentiation algorithms based on
addition chains. Recently, the investigation of minimal weight expansions has been
extended totheFibonacci numeration systembyHeuberger [15], whogavean equiv-
alenttotheNAF. Solinas [26]hasshownhowtorepresentascalar inacomplex base
τ relatedtoKoblitzcurves,andhasgivenaτ-NAFform,andtheHammingweightof
theserepresentationshasbeenstudiedin[9].
ThisresearchwassupportedbytheFrenchAgenceNationaledelaRecherche, grantANR-JCJC06-134288
“DyCoNum”.
2 ChristianeFrougny and WolfgangSteiner
In this paper, we study expansions in a real base β > 1 which is not an integer.
Any number z in the interval [0,1) has a so-called greedy β-expansion given by the
β-transformation τ , which relies on a greedy algorithm: let τ (z) = βz βz and
β β
−⌊ ⌋
define, for j ≥ 1, xj = ⌊βτβj−1(z)⌋. Then z = ∞j=1xjβ−j, where the xj’s are
integer digits in the alphabet 0,1,..., β . We wrPite z = .x1x2 . If there exists
{ ⌊ ⌋} ···
a n such that x = 0 for all j > n, the expansion is said to be finite and we write
j
z =.x x x . Byshifting,anynon-negativerealnumberhasagreedyβ-expansion:
1 2 n
···
Ifz [0,βk),k 0,andβ kz =.x x ,thenz =x x .x x .
− 1 2 1 k k+1 k+2
∈ ≥ ··· ··· ···
Weconsiderthesequences ofdigitsx x aswords. Sincewewanttominimize
1 2
···
the weight, we are only interested in finite words x = x x x , but we allow a
1 2 n
···
priori arbitrary digits x in Z. The corresponding set of numbers z = .x x x is
j 1 2 n
therefore Z[β 1]. Note that we do not require that the greedy β-expansion o··f·every
−
z Z[β 1] [0,1) is finite, although this property (F) holds for the three numbers β
−
∈ ∩
studiedinSections4to6,see[12,1].
ThesetoffinitewordswithlettersinanalphabetAisdenotedbyA ,asusual. We
∗
definearelationonwordsx=x x x Z ,y =y y y Z by
1 2 n ∗ 1 2 m ∗
··· ∈ ··· ∈
x y ifandonlyif .x x x =βk .y y y forsomek Z.
β 1 2 n 1 2 m
∼ ··· × ··· ∈
A word x Z is said to be β-heavy if there exists y Z such that x y and
∗ ∗ β
∈ ∈ ∼
y < x . Wesaythatyisβ-lighterthanx. Thismeansthatanappropriateshiftofy
k k k k
providesaβ-expansionofthenumber.x x x withsmallerabsolutesumofdigits
1 2 n
···
than x . If x is not β-heavy, then we call x a β-expansion of minimal weight. It is
k k
easytoseethateverywordcontaining aβ-heavyfactor isβ-heavy. Thereforewecan
restrictourattentiontostrictlyβ-heavywordsx=x x Z ,whichmeansthatx
1 n ∗
··· ∈
isβ-heavy,andx x andx x arenotβ-heavy.
1 n 1 2 n
··· − ···
Inthefollowing,weconsiderrealbasesβ satisfyingthecondition
(D ): thereexistsawordb 1 B,...,B 1 suchthatB band b B
B ∗ β
∈{ − − } ∼ k k≤
for some positive integer B. Corollary 3.2 and Remark 3.4 show that every class of
words (with respect to ) contains a β-expansion of minimal weight with digits in
β
∼
1 B,...,B 1 ifandonlyifβ satisfies(D ).
B
{ − − }
Ifβ isaPisotnumber,i.e.,analgebraicintegergreaterthan1suchthatalltheother
rootsofitsminimalpolynomialareinmoduluslessthanone,thenitsatisfies(D )for
B
someB >0byProposition3.5. Thecontraryisnottrue: Thereexistalgebraicintegers
β > 1 satsfying (D ) which are not Pisot, e.g. the positive solution of β4 = 2β +1
B
isnotaPisotnumber butsatisfies(D )since2 = 1000.( 1). Thefollowingexample
2
−
providesalargeclassofnumbersβ satisfying(D ).
B
Example1.1. If1 = .t t t (t )ω withintegerst t t > t 0,
1 2 d d+1 1 2 d d+1
··· ≥ ≥ ··· ≥ ≥
thenβ satisfies(D )withB =t +1= β +1,since
B 1
⌊ ⌋
t
βd+1−t1βd−···−tdβ−td+1 = βd+11 =βd−t1βd−1−···−td
−
andthus
βd+1 (1+t )βd+(t t )βd 1+ +(t t )β+(t t )=0.
1 1 2 − d 1 d d d+1
− − ··· − − −
MinimalweightexpansionsinPisotbases 3
Recall that the set of greedy β-expansions is recognizable by a finite automaton
whenβisaPisotnumber[3]. Inthiswork,weprovethatthesetofallβ-expansionsof
minimalweightisrecognizedbyafiniteautomatonwhenβ isaPisotnumber.
We then consider particular Pisot numbers satisfying (D ) which have been exten-
2
sivelystudiedfromvariouspointsofview. Whenβ istheGoldenRatio,weconstruct
afinite transducer which gives, for a strictlyβ-heavy word as input, aβ-lighter word
asoutput. SimilarlytotheNon-AdjacentForminbase2,wedefineaparticularunique
expansion of minimal weight avoiding a certain given set of factors. We show that
there is a finite transducer which converts all words of minimal weight into these ex-
pansions avoiding these factors. From these transducers, we derive the minimal au-
tomaton recognizing the set of β-expansions of minimal weight in 1,0,1 . We
∗
{− }
giveabranchingtransformationwhichprovidesallβ-expansionsofminimalweightin
1,0,1 of a given z Z[β 1]. Similar results areobtained for the representation
∗ −
{− } ∈
ofintegersintheFibonacci numerationsystem. Theaverage weightofexpansions of
thenumbers M,...,M is 1log M,whichmeansthattypicallyonlyeveryfifthdigit
− 5 β
isnon-zero. Notethatthecorrespondingvaluefor2-expansions ofminimalweightis
1log M,see[2,5],andthat 1log M 0.288log M.
3 2 5 β ≈ 2
We obtain similar results for the case where β is the so-called Tribonacci number,
which satisfies β3 = β2 + β + 1 (β 1.839), and the corresponding representa-
≈
tions forintegers. Inthis case, theaverage weight is β3 log M 0.282log M
β5+1 β ≈ β ≈
0.321log M.
2
FinallyweconsiderthesmallestPisotnumber,β3 =β+1(β 1.325),whichpro-
≈
videsrepresentationsofintegerswithevenlowerweightthantheFibonaccinumeration
system: 1 log M 0.095log M 0.234log M.
7+2β2 β ≈ β ≈ 2
SincetheprooftechniquesfortheTribonaccinumberandthesmallestPisotnumber
are quite similar to the Golden Ratio case (but more complicated), some parts of the
proofs are not contained in the final version of this paper. The interested reader can
findthemin[13].
2 Preliminaries
A finite sequence of elements of a set A is called a word, and the set of words on A
is the free monoid A . The set A is called alphabet. The set of infinite sequences
∗
or infinite words on A is denoted by AN. Let v be a word of A , denote by vn the
∗
concatenationofvtoitselfntimes,andbyvω theinfiniteconcatenationvvv .
···
Afinitewordvisafactorofa(finiteorinfinite)wordxifthereexistsuandwsuch
that x = uvw. When u is the empty word, v is a prefix of x. The prefix v is strict if
v =x. Whenwisempty,vissaidtobeasuffixofx.
6
Werecallsomedefinitionsonautomata,see[10]and[25]forinstance. Anautoma-
tonoverA, =(Q,A,E,I,T),isadirectedgraphlabelledbyelementsofA. Theset
A
ofvertices,traditionallycalledstates,isdenotedbyQ,I Qisthesetofinitialstates,
⊂
T Q is the set of terminal states and E Q A Q is the set of labelled edges.
If(⊂p,a,q) E,wewritep a q. Theautom⊂aton×isfin×iteifQisfinite. AsubsetH of
∈ →
A issaidtoberecognizablebyafiniteautomatonifthereexistsafiniteautomaton
∗
A
4 ChristianeFrougny and WolfgangSteiner
suchthatH is equaltothesetof labels of paths startinginaninitial stateandending
inaterminalstate.
Atransducerisanautomaton =(Q,A A ,E,I,T)wheretheedgesofE are
∗ ′∗
T ×
labelled by couples of words in A A . It is said to be finite if the set Q of states
∗ ′∗
×
uv
andthesetE ofedgesarefinite. If(p,(u,v),q) E, wewritep | q. Inthispaper
∈ −→
we consider letter-to-letter transducers, where the edges are labelled by elements of
A A. Theinputautomatonofsuchatransducerisobtainedbytakingtheprojection
′
×
ofedgesonthefirstcomponent.
3 General case
Inthissection,ouraimistoprovethatonecanconstructafiniteautomatonrecognizing
thesetofβ-expansionsofminimalweightwhenβ isaPisotnumber.
Weneedfirstsomecombinatorialresultsforbasesβ satisfying(D ). Notethatβ is
B
notassumedtobeaPisotnumberhere.
Proposition 3.1. If β satisfies (D ) with some integer B 2, then for every word
B
≥
x Z thereexistssomey 1 B,...,B 1 withx y and y x .
∗ ∗ β
∈ ∈{ − − } ∼ k k≤k k
Corollary3.2. Ifβsatisfies(D )withsomeintegerB 2,thenforeverywordx Z
B ∗
≥ ∈
thereexistsaβ-expansionofminimalweighty 1 B,...,B 1 withx y.
∗ β
∈{ − − } ∼
Remark3.3. Ifβ satisfies(D )forsomepositiveintegerB,thenitiseasytoseethat
B
β satisfies(D )foreveryintegerC >B.
C
Remark3.4. Ifβ doesnotsatisfy(D ),thenallwordsx 1 B,...,B 1 with
B ∗
∈{ − − }
x Bareβ-heavierthanB. Itfollowsthatthesetofβ-expansionsofminimalweight
β
∼
x Bis0 B0 .
β ∗ ∗
∼
ProofofProposition3.1. Let A = 1 B,...,B 1 . If x = x x x A ,
1 2 n ∗
{ − − } ··· ∈
then there is nothing to do. Otherwise, we use (D ): there exists some word b =
B
b b A such that b b (b B)b b 0 and b B. We use
k d ∗ k 1 0 1 d β
th−is·r·e·latio∈n to decrease the−ab·so··lut−e valu−e of a d·i·g·it x∼ A withkoukt≤increasing the
h
6∈
weightofx,andweshowthatweeventuallyobtainawordinA ifwealwayschoose
∗
the rightmost such digit. More precisely, set x(0) = x for 1 j n, x(0) = 0 for
j j ≤ ≤ j
j 0 and j > n, b = 0 for j < k and j > d. Define, recursively for i 0,
j
≤ − ≥
h =max j Z: x(i) B ,
i { ∈ | j |≥ }
x(i+1) =x(i)+sgn(x(i))(b B), x(i+1) =x(i) +sgn(x(i))b forj =0,
hi hi hi 0− hi+j hi+j hi j 6
aslongash exists. Thenwehave x(0) = x , x(i+1)β j = x(i)β j and
i jP∈Z| j | k k jP∈Z j − jP∈Z j −
x(i+1) = x(i+1) + x(i+1) x(i) + b B+ (x(i) + b ) x(i) .
| j | | hi | | hi+j|≤| hi| | 0|− | hi+j| | j| ≤ | j |
Xj Z Xj=0 Xj=0 Xj Z
∈ 6 6 ∈
MinimalweightexpansionsinPisotbases 5
wIfihthioduotetshenoletaedxiinsgt,atnhdentrwaielinhgavzeer|xos(ji)i|s<awBorfdoryallAj ∈wZit,hanthdetdheessireeqduepnrocpee(rxti(jei)s).j∈Z
∗
∈
Since x isfinite,wehave x(i+1) < x(i) onlyforfinitelymanyi 0.
k k j Z| j | j Z| j | ≥
Inparticular,thealgorithmterPmin∈atesafteratmPos∈t x B+1stepsif b <B.1
k k− k k
If b =B and x(i+1) = x(i) ,thenwehave
k k j Z| j | j Z| j |
∈ ∈
P P
hi−1 x(i+1) = hi−1 x(i) + k b and ∞ x(i+1) = ∞ x(i) + d b .
j=X−∞| j | j=X−∞| j | Xj=1| −j| j=Xhi+1| j | j=Xhi+1| j | Xj=1| j|
Assume that h exists for all i 0. If (h ) has a minimum, then there exists an
i i i 0
increasingsequenceofindices(i≥) sucht≥hath h forallℓ>i ,m 0,thus
m m≥0 im ≤ ℓ m ≥
x him−1 x(im+1) him−1−1 x(im−1+1) + k b (m+1) k b .
k k≥ j=X | j |≥ j=X | j | Xj=1| −j|≥···≥ Xj=1| −j|
−∞ −∞
If k b > 0, this is not possible since x is finite. Similarly, (h ) has no
j=1| −j| k k i i≥0
maPximum if d b > 0. Since x(i+1) can differ from x(i) only for h k j
j=1| j| j j i − ≤ ≤
hi+d,wehaPvehi+1 hi+dforalli 0. Ifhi < hi′,i < i′,thenthereistherefore
≤ ≥
a sequence (i ) , i i < i < < i = i, with M (h h )/d
m 0 m M 0 1 M ′ i′ i
such that h h≤ f≤or all ℓ≤ i ,i +1·,·.·..,i , m 0,...,M ≥. As ab−ove, we
im ≤ ℓ ∈ { m m ′} ∈ { }
obtain x (M+1) k b ,butM canbearbitrarilylargeif(h ) hasneither
minimukmkn≥or maximumP.j=H1e|n−cej|we have shown that hi cannot existifio≥r0all i 0 if
k b >0and d b >0. ≥
PjI=t1re|m−aji|ns to consPidje=r1th|ej|case b = B with k = 0 or d = 0. Assume, w.l.o.g.,
k k
d= 0. Thenwehaveh h . Ifh existsforalli 0,thenboth k x(i) and
∞j=1|x(hii)+j|areeventuia+l1ly≤conistant.iThereforewem≥usthavesomePi,ij′=w0i|thhhi−i′j|<hi
Pasuncdhxt(hhii)a−tjx=(hii′x′)−(hiki′′)·−·j·=x(hii0′′)fo=raxll(hiij)−>k·k·.·xT(hhii)i,sixm(hiip′′)l+i1exs(hxii′(h′)i+i)−2k······=x(hi0i)h∼i−βh0i′xor(hii)β+h1ix−(hhii)i+′2=··1·.,
In the first case, x(i) x(i) without the trailing zeros is a word y A with the
hi+1 hi+2··· ∈ ∗
desiredproperties. Inthelattercase,eachx Z canbeeasilytransformedintosome
∗
∈
y 1,0,1 withy xand y = x ,andthepropositionisproved. 2
∗ β
∈{− } ∼ k k k k
Thefollowingpropositionshowsslightlymorethantheexistenceofapositiveinte-
gerB suchthatβ satisfies(D )whenβ isaPisotnumber.
B
Proposition 3.5. For every Pisotnumber β, there exists some positive integer B and
somewordb Z suchthatB band b <B.
∗ β
∈ ∼ k k
Proof. If β is an integer, then we can choose B = β and b = 1. So let β be a Pisot
number of degree d 2, i.e., β has d 1 Galois conjugates β(j), 2 j d, with
≥ − ≤ ≤
β(j) <1. Foreveryz Q(β)setz(j) =P(β(j))ifz =P(β),P Q[X].
| | ∈ ∈
1FortheproofofTheorem3.11,itissufficienttoconsiderthecasekbk < B. However, Corollary3.2is
particularlyinterestinginthecasekbk=B,andweuseitinthefollowingsectionsforB=2.
6 ChristianeFrougny and WolfgangSteiner
LetBbeapositiveinteger,L= logB/logβ ,andx x thegreedyβ-expansion
1 2
⌈ ⌉ ···
ofz =β LB [0,1). Since
−
∈
k
τβk(z)=βτk−1(z)−xk =···=βkz− xℓβk−ℓ,
Xℓ=1
wehave
k
(cid:12)(cid:12)(cid:12)(τβk(z))(j)(cid:12)(cid:12)(cid:12)=(cid:12)(cid:12)(cid:12)(cid:12)(β(j))kz(j)−Xℓ=1xℓ(β(j))k−ℓ(cid:12)(cid:12)(cid:12)(cid:12)<(cid:12)(cid:12)β(j)(cid:12)(cid:12)k(cid:12)(cid:12)z(j)(cid:12)(cid:12)+ 1−⌊β|β⌋(j)|
forallk 0and2 j d. Setk =max log z(j) /log β(j) . Thenτk(z)is
aneleme≥ntofthefin≤ites≤et 2≤j≤d⌈− | | | |⌉ β
β
Y = y Z[β−1] [0,1): y(j) <1+ ⌊ ⌋ for 2 j d .
(cid:26) ∈ ∩ | | 1 β(j) ≤ ≤ (cid:27)
−| |
Foreveryy Y,wecanchooseaβ-expansiony =.a a . LetW bethemaximal
1 m
∈ ···
weight of all these expansions and τk(z) = .a a . Since z = .x ...x +τk(z),
β ′1··· ′m 1 k β
thedigitwiseadditionofx x anda a providesawordbwithb B and
1··· k ′1··· ′m ∼β
logB logB
b k β +W = max β +W = (logB).
k k≤ ⌊ ⌋ 2 j d(cid:24)(cid:24)logβ(cid:25)− log β(j) (cid:25)⌊ ⌋ O
≤ ≤ | |
IfBissufficientlylarge,wehavetherefore b <B. 2
k k
Inordertounderstandtherelation onA ,A= 1 B,...,B 1 ,wehaveto
β ∗
∼ { − − }
considerthewordsz (A A) withz 0. Thereforeweset
∗ β
∈ − ∼
n
Z = z z 2(1 B),...,2(B 1) n 0, z β j =0
β 1 n ∗ j −
n ··· ∈{ − − } (cid:12)(cid:12) ≥ Xj=1 o
(cid:12)
andrecallaresultfrom[11]. Alltheautomataconsideredinthispaperprocesswords
fromlefttoright,thatistosay,mostsignificantdigitfirst.
Lemma3.6([11]). IfβisaPisotnumber,thenZ isrecognizedbyafiniteautomaton.
β
Forconvenience,wequicklyexplaintheconstructionoftheautomaton recogniz-
β
A
ingZβ. ThestatesofAβ are0andalls∈Z[β]∩(2(β1−B1),2(βB−11))whichareaccessible
from0bypathsconsistingoftransitionss e s with−e A −Asuchthats =βs+e.
′ ′
→ ∈ −
The state 0 is both initial and terminal. When β is a Pisot number, then the set of
statesisfinite. Notethat issymmetric, meaningthatifs e s isatransition,then
β ′
A →
s−e s′isalsoatransition. Theautomaton β isaccessibleandco-accessible.
− →− A
The redundancy automaton (or transducer) is similar to . Each transition
β β
R A
s e s′ of β is replaced in β by a set of transitions s a|b s′, with a,b A and
→ A R −→ ∈
a b=e. FromLemma3.6,oneobtainsthefollowinglemma.
−
MinimalweightexpansionsinPisotbases 7
Lemma3.7. Theredundancytransducer recognizestheset
β
R
(x x ,y y ) A A n 0, .x x =.y y .
1 n 1 n ∗ ∗ 1 n 1 n
··· ··· ∈ × ≥ ··· ···
(cid:8) (cid:12) (cid:9)
Ifβ isaPisotnumber,then isfinite. (cid:12)
β
R
From the redundancy transducer , one constructs another transducer with
β β
R T
states of the form (s,δ), where s is a state of and δ Z. The transitions are of
β
R ∈
ab ab
theform(s,δ) | (s′,δ′)ifs | s′ isatransitionin β andδ′ =δ+ b a. The
−→ −→ R | |−| |
initialstateis(0,0),andterminalstatesareoftheform(0,δ)withδ <0.
Lemma3.8. Thetransducer recognizestheset
β
T
(x x ,y y ) A A .x x =.y y , y y < x x .
1 n 1 n ∗ ∗ 1 n 1 n 1 n 1 n
··· ··· ∈ × ··· ··· k ··· k k ··· k
(cid:8) (cid:12) (cid:9)
Ofcourse,thetransducer isn(cid:12)otfinite,andthecoreoftheproofofthemainresult
β
T
consistsinshowingthatweneedonlyafinitepartof .
β
T
Wealsoneedthefollowingwell-knownlemma, andgiveaproofforitbecausethe
constructionintheproofwillbeusedinthefollowingsections.
Lemma 3.9. Let H A and M = A A HA . If H is recognized by a finite
∗ ∗ ∗ ∗
⊂ \
automaton,thensoisM.
Proof. Suppose that H is recognized by a finite automaton . Let P be the set of
H
strict prefixes of H. We construct the minimal automaton of M as follows. The
M
set of states of is the quotient P/ where p q if the paths labelled by p end in
thesamesetofMstates in as thepath≡s labelled b≡yq. Since is finite, P/ isfinite.
TransitionsaredefinedasHfollows. LetabeinA. IfpaisinPH,thenthereisa≡transition
[p] a [pa] . IfpaisnotinH P,thenthereisatransition[p] a [v] withvinP
ma≡xi→malin≡lengthsuchthatpa=∪uv. Everystateisterminal. ≡ → ≡ 2
Now, wecanprovethefollowingtheorem. Themainresult,Theorem3.11,willbe
aspecialcaseofit.
Theorem 3.10. Let β be a Pisot number and B a positive integer such that (D )
B
holds. Thenonecanconstructafiniteautomaton recognizingthesetofβ-expansions
ofminimalweightin 1 B,...,B 1 .
∗
{ − − }
Proof. Let A = 1 B,...,B 1 , x A be a strictly β-heavy word and y
∗
{ − − } ∈ ∈
A be a β-expansion of minimal weight with x y. Such a y exists because of
∗ β
∼
Proposition3.1. Extendx,y towords x,y byadding leading and trailingzeros such
′ ′
that x = x x , y = y y and .x x = .y y . Then there is a path in
′ 1 n ′ 1 n 1 n 1 n
··· ··· ··· ···
thetransducer β composedof transitions(sj 1,δj 1) xj|yj (sj,δj),1 j n,with
s =0,δ =0,Ts =0,δ <0. − − −→ ≤ ≤
0 0 n n
We determine bounds for δ , 1 j n, which depend only on the state s = s .
j j
≤ ≤
Choose a β-expansion of s, s = a a .a a , and set w = a a . If
1 i i+1 m s 1 m
··· ··· k ··· k
δ >w ,thenwehave y y > x x +w . Sinces =(x y ) (x y ).,
j s 1 j 1 j s j 1 1 j j
k ··· k k ··· k − ··· −
8 ChristianeFrougny and WolfgangSteiner
the digitwise subtraction of 0max(i j,0)x x 0m i and 0max(j i,0)a a provides
− 1 j − − 1 m
··· ···
a word which is β-lighter than y y , which contradicts the assumption that y is a
1 j
···
β-expansionofminimalweight.
Let W = max w sisastatein . If δ W B, then let h j be such
s β j
{ | A } ≤ − − ≤
that x = 0, x = 0 for h < i j. Since x < B, we have δ < δ +B
h i h h 1 j
W 6 w , hence x x ≤ > y | y| +w . Let−a a be th≤e
w−ord≤wh−ichshw−a1s used fokr1th·e··dehfi−n1iktion kof1w··· h,−i1.ke., s sh−=1 a 1a··.a· m a ,
w = a a . Thenthedigitwiseadditsiho−n1of0max(hi−h1+1,0)y1···yi i+01m··i·anmd
0msahx−(1h 1 ki,01)a··· mak providesawordwhichisβ-lighterth−anx 1x··· h.−S1ince−x =
− − 1 m 1 h 1 h
0,thiscontradic·t·s·theassumptionthatxisstrictlyβ-heavy. ··· − 6
Let betherestrictionof tothestates(s,δ)with W B <δ w withsome
β β s
S T − − ≤
additionalinitialandterminalstates: Everystatewhichcanbereachedfrom(0,0)by
a path with input in 0 is initial, and every state with a path to (0,δ), δ < 0, with an
∗
input in 0 is terminal. Then the set H which is recognized by the input automaton
∗
of consists only of β-heavy words and contains all strictly β-heavy words in A .
β ∗
S
ThereforeM =A A HA isthesetofβ-expansionsofminimalweightinA ,and
∗ ∗ ∗ ∗
\
M isrecognizablebyafiniteautomatonbyLemma3.9. 2
Theorem 3.11. Let β be a Pisot number. Then one can construct a finite automaton
recognizingthesetofβ-expansionsofminimalweight.
Proof. Proposition 3.5 shows that β satisfies (D ) for some positive integer B, and
B
that no β-expansion of minimal weight y Z can contain a digit y with y B,
∗ j j
∈ | | ≥
sinceweobtainaβ-lighterwordifwereplaceBbybasintheproofofProposition3.1.
ThereforeTheorem3.10impliesTheorem3.11. 2
4 GoldenRatiocase
Inthissectionwegiveexplicitconstructionsforthecasewhereβ istheGoldenRatio
1+√5. Wehave1 = .11, hence 2 = 10.01andβ satisfies(D ), seealsoExample1.1.
2 2
Corollary 3.2 shows that every z Z[β 1] can be represented by a β-expansion of
−
∈
minimal weight in 1,0,1 . For most applications, only these expansions are in-
∗
{− }
teresting. Remark that the digits of arbitrary β-expansions of minimal weight are in
2, 1,0,1,2 bytheproofofTheorem3.11,since3=100.01.
{−For−typograp}hicalreasons,wewritethedigit 1as1¯ inwordsandtransitions.
−
4.1 β-expansionsofminimalweightforβ = 1+√5
2
Our aim in this section is to construct explicitly the finite automaton recognizing the
β-expansionsofminimalweightinA ,A= 1,0,1 .
∗
{− }
Theorem4.1. Ifβ = 1+√5,thenthesetofβ-expansionsofminimalweightin 1,0,1
2 {− }∗
isrecognizedbythefiniteautomaton ofFigure1whereallstatesareterminal.
β
M
MinimalweightexpansionsinPisotbases 9
1 100
1 0 1 0 0¯10 ¯10 10 010
0 ¯1 1 0
0 0
0 ¯1 0 0 ¯1 0100 00 0 00 0¯100
¯1 ¯100
Figure1. Automaton recognizingβ-expansionsofminimalweightforβ = 1+√5
Mβ 2
(left)andacompactrepresentationof (right).
β
M
ItisofcoursepossibletofollowtheproofofTheorem3.10,butthestatesof are
β
A
1 1 1 1 1 1 1
0, , , , 1, β, β2, β , β , β2 , β2 ,
±β3 ±β2 ±β ± ± ± ± ± β2 ± ± β3 ± ± β2 ± ± β3
thus W = 2 and the transducer has 160 states. For other bases β, the number of
β
S
statescanbemuchlarger. Thereforewehavetorefinethetechniquesifwedonotwant
computer-assisted proofs. Itis possibleto show that a large partof is not needed,
β
S
e.g. by excluding some β-heavy factors such as 11 from the output, and to obtain
finally the transducer in Figure 2. However, it is easier to prove Theorem 4.1 by an
indirectstrategy,whichincludessomeresultswhichareinterestingbythemselves.
Lemma4.2. Allwordsin 1,0,1 whicharenotrecognizedbytheautomaton
∗ β
{− } M
inFigure1areβ-heavy.
Proof. Thetransducer inFigure2isapartofthetransducer intheproofof Theo-
β
S
rem3.10. Thismeansthateverywordwhichistheinputofapath(withfullordashed
transitions) going from (0,0) to (0, 1) is β-heavy, because the output has the same
−
value but less weight. Since a β-heavy word remains β-heavy if we omit the leading
andtrailingzeros,thedashedtransitionscanbeomitted. Thenthesetofinputsis
H =1(0100) 1 1(0100) 0101 1(001¯0) 1¯ 1(001¯0) 01¯
∗ ∗ ∗ ∗
∪ ∪ ∪
1¯(01¯00) 1¯ 1¯(01¯00) 01¯01¯ 1¯(0010) 1 1¯(0010) 01
∗ ∗ ∗ ∗
∪ ∪ ∪ ∪
and isconstructedasintheproofofLemma3.9. 2
β
M
SimilarlytotheNAFinbase2, where theexpansions ofminimal weightavoid the
set 11,11¯,1¯1¯,1¯1 ,weshowinthenextresultthat, forβ = 1+√5,everyrealnumber
{ } 2
admitsaβ-expansionwhichavoidsacertainfinitesetX.
Proposition 4.3. If β = 1+√5, then every z R has a β-expansion of the form
2 ∈
z = y y .y y with y 1,0,1 such that y y avoids the set
1 k k+1 k+2 j 1 2
X = 11··,·101,1001,11¯,·1··01¯, andthe∈iro{p−posites}. If z Z[β] =··Z·[β 1], then this
−
{ } ∈
expansionisuniqueuptoleadingzeros.
10 ChristianeFrougny and WolfgangSteiner
0|1 0|¯1
−1,1 0,0 1,1
1|0 ¯1|0
−1,0 −1/β,−1 1/β,−1 1,0
1|0 0|0 0|¯1 0|0 ¯1|0 1|0 0|0 0|1 0|0 ¯1|0
−1/β,0 −1,−1 1,−1 1/β,0
1|0 1|0 ¯1|0 ¯1|0
0|¯1 0|1
0,−1 −1/β,−2 1/β,−2 0,−1
Figure2. Transducerwithstrictlyβ-heavywordsasinputs,β = 1+√5.
2
Proof. We determine this β-expansion similarly to the greedy β-expansion in the In-
troduction. Notethatthemaximalvalueof.x x forasequencex x avoiding
1 2 1 2
theelementsofX is.(1000)ω =β2/(β2+1). Ifw··e·definethetransforma·t·io·n
β2 β2 β2 β2 β2+1
τ : − , − , , τ(z)=βz z+1/2 ,
(cid:20)β2+1 β2+1(cid:19)→(cid:20)β2+1 β2+1(cid:19) −(cid:22) 2β (cid:23)
aynd=se1tyfojr=so(cid:4)mβe22+βj1τj−11,(tzh)e+n w1/e2h(cid:5)afvoerτzj∈(zh)β−2β+β21,ββ2+21β(cid:17),j, ≥β21,then1z==.y11y2,·1·/·β. If,
j ≥ ∈ × β2+1 β2+1 − β−2+1 β2+1
hence yj+1 = 0, yj+2 = 0, and τj+2(z) ∈ β−2β+21,β(cid:2)2β+1 , hence(cid:1)yj+3 ∈(cid:2){1¯,0}. Th(cid:1)is
showsthatthegivenfactors areavoided. As(cid:2)imilarargu(cid:1)ment fory = 1showsthat
j
−
theoppositesareavoidedaswell,hencewehaveshowntheexistenceoftheexpansion
for z ∈ hβ−2β+21,ββ2+21(cid:17). For arbitrary z ∈ R, the expansion is given by shifting the
expansionofβ kz,k 0,totheleft.
−
Ifwechoosey =0≥incaseτj 1(z)>β/(β2+1)=.(0100)ω,thenitisimpossible
j −
to avoid the factors 11, 101 and 1001 in the following. If we choose y = 1 in case
j
τj 1(z) < β/(β2+1),thenβτj 1(z) 1 < 1/(β2+1) = .(001¯0)ω,andthusitis
− −
impossibletoavoid thefactors 11¯, 101¯−, 1¯1¯, 1¯0−1¯ and 1¯001¯. Since β/(β2+1) Z[β],
wehaveτj 1(z)= β/(β2+1)forz Z[β]. Similarrelationsholdfortheopp6∈osites,
−
6 ∈
thustheexpansionisunique. 2
Remark 4.4. Similarly, the transformation τ(z) = βz z +1/2 on [ β/2,β/2)
providesforeveryz Z[β]auniqueexpansionavoiding−th⌊efactors⌋11,10−1,11¯,101¯,
1001¯ andtheiroppos∈ites.
Proposition 4.5. If x is accepted by , then there exists y 1,0,1 avoiding
β ∗
X = 11,101,1001,11¯,101¯ andtheirMopposites with x y∈a{n−d x =} y . The
β
{ } ∼ k k k k
transducer inFigure3realizestheconversionfrom0x0toy.
β
N