Note on decipherability of three word codes

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PII. S0161171202011729 http://ijmms.hindawi.com © Hindawi Publishing Corp.

NOTE ON DECIPHERABILITY OF THREE-WORD CODES

F. BLANCHET-SADRI and T. HOWELL

Received 29 January 2001

The theory of uniquely decipherable (UD) codes has been widely developed in connection with automata theory, combinatorics on words, formal languages, and monoid theory. Recently, the concepts of multiset decipherable (MSD) and set decipherable (SD) codes were developed to handle some special problems in the transmission of information. Unique decipherability is a vital requirement in a wide range of coding applications where distinct sequences of code words carry diﬀerent information. However, in several applications, it is necessary or desirable to communicate a description of a sequence of events where the information of interest is the set of possible events, including multiplicity, but where the order of occurrences is irrelevant. Suitable codes for these communication purposes need not possess the UD property, but the weaker MSD property. In other applications, the information of interest may be the presence or absence of possible events. The SD property is adequate for such codes. Lempel (1986) showed that the UD and MSD properties coincide for two-word codes and conjectured that every three-word MSD code is a UD code. Guzmán (1995) showed that the UD, MSD, and SD properties coincide for two-word codes and conjectured that these properties coincide for three-word codes. In an earlier paper (2001), Blanchet-Sadri answered both conjectures positively for all three-word codes

{c1,c2,c3}satisfying|c1| = |c2| ≤ |c3|. In this note, we answer both conjectures positively for other special three-word codes. Our procedures are based on techniques related to dominoes.

2000 Mathematics Subject Classiﬁcation: 94A15, 05A99, 94B35.

1. Introduction. LetAbe a nonempty finite set or an alphabet;A∗denotes the set of all sequences of finite length (greater than or equal to 0) of elements ofA(such sequences are called words onA). The unique sequence of length 0, denoted by, is called the empty word.AcodeConAis a nonempty finite subset ofA+=A∗\ {}. The words inC are called code words. Amessage onC is a word inA∗ that is a concatenation of code words. The sequence of these code words is a decoding or factorizationof the message. The codeCis called

• uniquely decipherable, if every message onC has a unique factorization into code words;

• multiset decipherable, if any two factorizations of the same message onCyield the same multiset of code words;

• set decipherable, if any two factorizations of the same message onCyield the same set of code words.

Every UD code is MSD, and every MSD code is SD. It has been shown that these rela-tionships are proper. The codeC= {0,0111110,10101,1111}on{0,1}is an example of aproperMSD code (i.e., an MSD code that is not UD). In fact, the message

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onChas two distinct factorizations into code words [3,5]. The code

C= {0,010,11011,101101} (1.2)

on{0,1}is an example of a proper SD code (i.e., an SD code that is not MSD). The message

(0)(101101)(11011)(0)(11011)(010)=(010)(11011)(101101)(101101)(0) (1.3)

on C has two distinct factorizations with distinct multisets of code words [3, 5]. Guzmán presents a complete list of proper MSD and proper SD four-word codes on

{0,1}with code words of length less than or equal to 7 [3]. It is decidable whether or not a codeCis UD [1,5,6,9] (resp., MSD [5]).

For two-word codes, the UD, MSD, and SD properties coincide [3,7]. For three-word codes, it is an open question whether or not they coincide. Lempel [7] conjectured that every three-word MSD code is a UD code, and Guzmán [3] conjectured that the UD, MSD, and SD properties coincide for three-word codes. We answered both conjectures positively for all three-word codes{c1,c2,c3}satisfying|c1| = |c2| ≤ |c3|[2]. In this

note, we give (inSection 1.1) a brief overview of Head’s and Weber’s domino technique [5], and give (inSection 2) an application of this approach by proving that Lempel’s and Guzmán’s conjectures are true for some special three-word codes.

1.1. A domino technique. LetAbe an alphabet andC a code onA. The set of all preﬁxes of words inCwill be denoted by Preﬁx(C). Thedomino graphassociated with

Cis the directed graphG=(V ,E)where

V=

open, close,

u

,

u

u∈Preﬁx(C)\{}

, (1.4)

andE=E1∪E2∪E3∪E4where

E1=

open,

u

,

open,

u

u∈C

,

E2=

u

,close

,

u

,close

u∈C

,

E3=

u

,

uv

,

u

,

uv

v∈C

,

E4=

u

,

v

,

u

,

v

uv∈C

.

(1.5)

Thedomino functionassociated withCis the mappingdfromEtou,u

|u∈C

deﬁned on

• E1by

open,u u

andopen,_uu

,

• E2by

u

, closeu

and_u, close u

,

• E3by

u

,uv

v

and_u,uv

v

,

• E4by

_u

,v

uv

and_u,v

uv

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Thedomino associated with an edgeeofE is the dominod(e)=d1(e) d2(e)

. The func-tiondinduces mappingsd1andd2fromE toC∪ {}also called domino functions.

If p= e1···em is a path in G, the word d(e1)···d(em) (resp., d1(e1)···d1(em), d2(e1)···d2(em)) will be denoted byd(p)(resp.,d1(p),d2(p)).

A pathpinGfromopento some vertexu, respectively,_u, is trying to ﬁnd two factorizations of the same message onCinto code words beginning with distinct code words. The decodings obtained so far ared1(p)andd2(p). The worduinA∗denotes

the backlog of the first (resp., second) decoding as against the second (resp., first) one. Guzmán [4] suggested to look at thesimplified domino graphand thedomino func-tionofC. The simplified domino graph ofCis a subgraph of the domino graph ofC. ReplaceE1byE1=

open,_u|u∈C ,E2byE2=

u

,close|u∈C ,V byV

which consists of open, close, and those verticesvinVsuch that there is a path from open to close that goes throughv, andE byEwhich consists of those edgeseinE

such that there is a path from open to close going throughe. The simpliﬁed domino graph ofCis denoted byG(C).

The UD, MSD, and SD properties of a codeCcan be characterized in terms of its simpliﬁed domino graphG(C)and functionsd1andd2,

•Cis not UD if and only if there is a path inG(C)from open to close [6].

•Cis not MSD if and only if there is a pathpinG(C)from open to close such that

d1(p)andd2(p)do not have the same multiset of code words [5].

•C is not SD if and only if there is a pathpinG(C)from open to close such that

d1(p)andd2(p)do not have the same set of code words [3].

As an example, we consider the codec= {c1,c2,c3,c4}on{0,1}discussed in [5,4]

(c1=0,c2=0111110,c3=10101,andc4=1111). The simpliﬁed domino graph and

function associated with this code are shown inFigure 1.1.

111

( c4)

open (

c1

)

0

₍c4

) 01111

( c2) 10

(c3

)

₁

( c1)

(c4

)

close

_0111110

(c2

) 011111

(c2

) (

c1) 01

(

c4)

101

₍c1

) (

c3)

1010

( c3)

Figure_1.1

Each edgeeis labeled byd(e). The path

p=open,

0

,

01111

,

10

,

101

,

01

,

011111

,

0111110

,close (1.6)

is from open to close showing thatCis not UD. However, every pathpfrom open to close is such thatd1(p)and d2(p)have the same multiset of code words showing

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1.2. Preliminary lemmas. We give some preliminary lemmas that are used in

Section 2to prove our main results.

Definition_1.1_{(see Lothaire [}₈_]). _{A nonempty word}_u_{on an alphabet}_A_{is called} primitive, ifu=vn_{for some nonempty word}_v_on_A_{implies that}_n₌_1.

Lemma1.2(see Lothaire [8]). Letube a nonempty word on an alphabetA. There exist a unique primitive wordvand a unique positive integernsuch thatu=vn₍_v_,

denoted by√u, is called the root ofuandn, denoted byexp(u), is called the exponent ofu). Moreover, all positive powers ofuhave the same root.

Lemma1.3(see Lothaire [8]). LetC= {u,v}be a two-word on an alphabet A. The

UD,MSD, andSDproperties are equivalent to the following properties: (1)uv≠vu.

(2)√u≠√v.

Lemma_1.4_{(see Blanchet-Sadri [}₂_]). _Let_u_,_x_{be nonempty words on an alphabet}_A and letkbe a positive integer. If{ux,(ux)k_u}_is_SD_{, then}_ux_≠_xu_.

Lemma_1.5. _Let_u_,_v_,_x_,_y_{be nonempty words on an alphabet}_A_satisfying_u≠_v_.

(1)The equalitiesvyx=yxu=xuycannot hold simultaneously. (2)The equalitiesxvy=vyx=yxucannot hold simultaneously.

Proof. _{We prove assertion (1) (assertion (2) is proved similarly). Since}_yxu=_xuy_,

by Lemmas1.2 and1.3 there exist a unique primitive wordw and unique positive integerskandksuch thaty=wk_and_xu₌_wk_{. Put}_x₌_w_z_and_u₌_z_w _where wis not a preﬁx ofzandwis not a suﬃx ofz. Ifz,z=, thenu=v, a contradiction. Otherwise,z,z≠and the equalityvyx=xuyimplies thatw=zz=zz. Lemmas

1.2and 1.3imply the existence of a unique primitive wordwand unique positive integersm andm satisfyingz=(w)m _and _z₌_(w₎m_{. We get}_w₌_(w₎m+m_{, a}

contradiction with the fact thatwis primitive.

Lemma1.6. Letu,v,x,ybe nonempty words on an alphabetAsatisfyingu≠v. If the equalitiesvyx=yxv=yuxhold simultaneously, thenu=(zz)k_,_v₌_(zz₎k_, x=(zz)_z_{, and}_y₌_(zz₎m_z_{for some}_{k >}₀_,_≥₀_,_m_≥₀_{, and words}_z,z_{on A.}

Proof. Sincevyx=yxv, by Lemmas1.2and1.3there exist a unique primitive

wordwand unique positive integerskandksuch thatv=wk_and_yx₌_wk_{. Put} y=w_z_and _x₌_z_w _where_w _{is not a preﬁx of}_z_and _w_{is not a suﬃx of}_z_{. If}

z,z=, thenu=v, a contradiction. Otherwise,z,z≠andw=zz, and the equality

yxv=yuximplies thatu=(zz)k_{and the result follows.}

Lemma_1.7_{(see Blanchet-Sadri [}₂_]). _Let_u_,_v_,_x_{be nonempty words on an alphabet} A satisfyingu≠v. The equalitiesux=xvandvx=xucannot hold simultaneously.

2. Some special three-word codes. We show that the UD, MSD, and SD properties are equivalent to the special three-word codes described in Theorems 2.1,2.2, and

2.3.

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of u, c1 is not a preﬁx of v,u is not a preﬁx of v, k >0, andu,v ≠. For each

of our theorems, we need only to show thatC is UD. It is easy to see that{c1,c2}, {c1,c3},and{c2,c3}are two-word SD codes, and are therefore MSD and UD codes by Lemma 1.3. When constructingG(C), we see thatE1consists of the edges

open,_c

i

andE2 consists of the edges

ci

,close. NoE3- norE4-edge leaves

c2

and noE3

-norE4-edge leaves

c3

.

Letui(resp.,vi,wi) be the preﬁx of lengthiofc1(resp.,c2,c3). We writec1=uixi.

When trying to build a path from open to close, a list of vertices gets generated. The vertices are amongui

,vi

,wi

,_u

i

,_v

i

, and_w

i

where 0< i≤ |c3|. TheE3- or E4-edges leaving

_u

i

,vi

,wi

,_u

i

,_v

i

, and_w

i

are easily described.

We need to check that none of the generated vertices is of the formc1

,c2

, orc3

(otherwise there would be a path from open to close). It is obvious thatc1

is not the

end vertex of anyE3-edge and

c3

is not the end vertex of anyE4-edge. If

c1

is the

end vertex of anE4-edge of the form

u_i

,c1

(resp.,_v

i

,c1

,_w

i

,c1

), then

c2=uic1orc3=uic1(resp.,c2=vic1orc3=vic1,c2=wic1orc3=wic1). Similarly, if

c2

is the end vertex of anE4-edge of the form

ui

,c2

, respectively,_v

i

,c2

, w_i

,c2

, thenc3=uic2 (resp.,c3=vic2,c3=wic2). If

c2

is the end vertex of

anE3-edge of the form

ui

,uic1

, respectively,vi

,vic1

,wi

,wic1

, then

c2=uic1(resp.,c2=vic1,c2=wic1). If

c3

is the end vertex of anE3-edge of the form

ui

,uic1

(resp.,vi

,vic1

, wi

,wic1

,ui

,uic2

, vi

,vic2

,wi

,

wic2

), thenc3=uic1(resp.,c3=vic1, c3=wic1, c3=uic2, c3=vic2, c3=wic2).

We need to consider the casesc2=vic1,c3=wic1, andc3=wic2.

For the rest of the discussion, we can assume thatuorvis a preﬁx ofc1(otherwise,

it is not difficult to see that there is no path inG(C)fromopentoclose). Sinceuis not a prefix of v and |u|<|v|, u and v cannot be both prefixes ofc1. In either

case,|u|<|c1|. To simplify the notation, putx|u|=x, and whenever|v|<|c1|, put x|v|=y.

First, assume thatv is a preﬁx ofc1. Here,y≠andvy≠yvbyLemma 1.4. To

simplify the notation, putu|u|=w. in the case wherec3=wic1, sincei= |c1k−1v|,

we havewi=c1k−1v and c3=c1k−1vc1=ck1v yieldingvy=yv, a contradiction. In

the case where c3=wic2, we have i= |v| − |u|, wi= ui, c3 =uic1ku=ck1v, and c1=uixi=vy. Putv =uitwheret≠. The equalityuick1u=ck1uitimplies that t=uanduixi=xiui. But then,c1=uixi=uiuy=uyui. We then conclude that u is a preﬁx ofc1, which is a contradiction. In the case where c2=vic1, we have i= |ck−1

1 u|. Ify∈Preﬁx(C)\{}, we getFigure 2.1.

Otherwise, putvy=yv wherev≠v (v∈Preﬁx(C)\ {}). Ifuis not a preﬁx ofv, then add the edgev,y

toFigure 2.1. Otherwise, putv=uswheres≠. We havevi=c1k−1w,c2=c1k−1wc1=c1ku, and thereforec1=wx=xu=vy. We have v=wtwheret≠, and thereforewty=tyu=yuswithu≠w. ByLemma 1.5(1), we gets≠t. Since|s| = |t|, we conclude thats∈Preﬁx(C)\{}. Add the edgesv,_y

andck1−1v

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c1k−1v

ck1−2v

c1v

v

open

c1

c21

···

ck1−1

c1k

Figure_2.1

In the rest of the note, we assume thatuis a preﬁx ofc1. Here,x≠andux≠xuby Lemma 1.4. In the case wherec2=vic1, we havevi=c1k−1uandc2=c1k−1uc1=ck1u.

We conclude thatux=xu, a contradiction. In the cases wherec3=wic1orc3=wic2,

ifx∈Preﬁx(C)\{}, we getFigure 2.2.

ck1−1u

ck1−2u

c1u

u

open

c1

c21

···

ck1−1

ck1

Figure_2.2

Otherwise, putux=xuwhereu≠u (u∈Preﬁx(C)\ {}). Ifu is not a preﬁx ofv, then add the edgeu,_xtoFigure 2.2. Otherwise, putv=ucq1zwherec1

is not a preﬁx ofzandq≥0. Note that ifz=, thenc3=c1kuc q

1=c1k−1uc q+1 1 and

soc2cq+ 1

1 =c1c3, a contradiction with the fact thatC is SD. Note also that if q≥k

and z=u, thenc1c3=c2cq1−kc2c1, a contradiction with the fact thatC is SD. It is

not difficult to see thatc1iz∈Prefix(C)\{}(0≤i < k)unlesszis a prefix ofc1, and c1kz∈Prefix(C)\ {}only ifzis a prefix ofuorv. Wheneverzis a prefix ofc1, put x|z|=x1and wheneverzis a prefix ofu, putc2=ck1zc

p

1 ywherep≥0 andc1is

not a preﬁx ofy. In the case wherezis a preﬁx ofv, putc3=ck1zc q

1zwhereq≥0

andc1is not a preﬁx ofz.

The following points enable us to assume that|z|<|u| and z is a preﬁx of u

wheneverzis a preﬁx ofc1andc3=wic1:

• if|z| = |u|, thenz=u=u, a contradiction;

• if|z|>|u|, then putz=ut=tufore somet,t≠. Here,c1=utx1=tx1u= tux1implies thatt=t. We also havex=tx1=x1tandc1=ux1t=x1tu= tux1. These equalities cannot hold simultaneously byLemma 1.5(1).

Ifc3=wic2and|z| = |u|, we havez=uandzis not a preﬁx ofv. In this case, if q < k−1,u=u, a contradiction, and ifq≥k−1,Cis not SD(c1c3=c2cq1−k+1c2). The

above and the following points enable us to assume that|z|<|u|,zis a preﬁx ofu, andq > k−1 wheneverzis a preﬁx ofc1andc3=wic2:

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• if|z|<|u|, then putu=zt=tzfor somet,t≠, and putu=zt. In this case, ifq < k, thenck1uc

q 1z=c

q

1zck1uandc1=ztx=txz=xztwitht=t.

These equalities cannot hold simultaneously byLemma 1.5(1).

Trying to build a path fromopentoclose, addu,_xandFigure 2.3toFigure 2.2.

ck1−1u

ck1u

ck1uc1

··· c1kucq− 1 1

c1kuc1q

Figure_2.3

The edgeck1−1u

,c1k−1uc2

is added in caseq > k−1, orq=k−1,uis a proper

preﬁx ofz, andzis not a preﬁx ofv. The edgeck1uci1

,ck1uci1c2

is added in case

0≤i < q−k. The edge c1kuc1q−k

,c1kuc1q−kc2

is added in case q > k−1, u is a

proper preﬁx ofz, andzis not a preﬁx ofv. The edgec1k−1u

,_cq+1 1 z

is added in caseq < k−1 andzis a preﬁx ofu, orq=k−1 andzis a preﬁx ofuorv. The edge

c₁kucq₁−k

,_ck

1z

is added in caseq > k−1 andzis a preﬁx ofuorv. Ifi≥0 and

q−k < i≤q, the edgeck1uci1

,_cq−i

1 z

is added in casezis a preﬁx ofu. Note that

noE3- norE4-edge leaves

x

, and noE3- norE4-edge leaves the vertices inFigure 2.3

other than the edges discussed above.

In the rest of the discussion, wheneveruis a proper preﬁx ofz, we putz=urwhere

r≠andx is not a prefix ofr, and wheneveruis a prefix ofz, we putz=ur. It is not difficult to see thatr∈Prefix(C)\{}(otherwise,ur=zis a prefix ofux=c1

and thereforezis a preﬁx ofu, a contradiction). Note the following points:

•NoE3-edge leaves

ck1−1uc2

unlessq=kanduis a preﬁx ofz. NoE4-edge leaves

c1k−1uc2

other thanc1k−1uc2

,_xzin caseq=kandxz∈Preﬁx(C)\{}.

•NoE3- norE4-edges leave

ck1uc1ic2

(0≤i < q−k−1).

•Forq > kandi=q−k−1, noE3-edge leaves

_ck

1uc1ic2

unlessuis a preﬁx ofz, and

noE4-edge leaves

c1kuc1ic2

other thanck1uci1c2

,xz

in casexz∈Preﬁx(C)\{}.

•NoE3- norE4-edges leave

_ck

1uc1q−kc2

. We now state and prove our main results.

Theorem_2.1. _Let_C_,_u_,_v_{, and}_z_{be deﬁned as described above. If}_z_{is a preﬁx of} bothuandv, thenCisUD.

Proof. _Here|z|_<|u|_{. Put} _u=_zt_and _u=_zt_where_t,t≠ _and_t≠_t_{. First,}

assume that c3=wic1. Hereu=tz and c1=ztx= xzt=xtz=zxt =txz.

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Theorem2.2. LetC,u,v, andzbe defined as described above. Ifzis a prefix ofu but not a prefix ofv, thenCisUD.

Proof. _Here,|z|_<|u|_{. Put}_u=_zt_and_u=_z_t_{, where}_z≠_z _and_t,t≠_{. We}

havexz∈Preﬁx(C)\{}sincec1=xztandz≠z.

First, assume thatc3=wic1. Hereu=tzandc1=ztx=zxt=txz=xzt= xtz. If t=t, thenzxt=xtz=xzt. ByLemma 1.6,z=(ss)_, _z₌_(ss₎_,_t₌ (ss)m_s_,_x₌_(ss₎n_s_{for some}_>_0,_m_≥_0,_n_≥_{0, and words}_s,s_on_A_{. The equality}

txz=xtzimplies that ss=ss, and therefore z=z, a contradiction. Otherwise, since|t|<|u|andc1=txz, we getu=tzfor somez. We havezx=xz and zx=xz. Certainlyz≠z, and byLemma 1.7,z≠z. ByLemma 1.6, sinceztx= txz=tzx, putz=(ss)_,_z₌_(ss₎_,_x₌_(s_s)m_s_{, and}_t₌_(ss₎n_s_{for some}_>_0, m≥0,n≥0, and wordss,sonA. Wheneverm >0, the equalityztx=xztimplies thatss=ssandz=z, a contradiction. Otherwise,xis a preﬁx ofzand we put

z=xy andz=yx, wherey=(ss)−1_s_{. Here}_y_{cannot be a preﬁx of}_z _{or the} equalityxzt=zxtimplies thatz=z. Trying to build a path fromopentoclose, the edgeu,x

together withFigure 2.3are added toFigure 2.2. We now discuss the other edges added.

First, assume thatq≤k. Wheneverq < k, addck1−1u

,_cq+1 1 z

to the graph and

wheneverq=k, add c1k−1u

,ck1−1uc2

. In either case, we then also addck1uci1

,

c1q−iz

(0≤i≤q),_ci

1z

,_ci+1 1 z

(0≤i < k),_z,tx,tx,_z,_ci

1z

,ck1−it

(0≤i≤k),c1itx

,ci1+1tx

(0≤i < k), andci1tx

,_ck−i

1 y

(0≤i≤k).

Second, assume thatq > k. Addc1k−1u

,c1k−1uc2

, c1kuc1i

,c1kuc1ic2

(0≤i <

q−k), andck1uci1

,_cq−i

1 z

(q−k≤i≤q). Also add _ci

1z

,_ci+1 1 z

(0≤i < k),

z

,tx

, tx,z

,_ci

1z

,c1k−it

(0≤i≤k),ci1tx

,c1i+1tx

(0≤i < k), and

c1itx

,_ck−i

1 y

(0≤i≤k).

Now, assume thatc3=wic2. Hereu=tzandc1=ztx=tzx=zxt=xzt= txz. Ift=t, thenztx=txz =xztand these equalities cannot hold simulta-neously byLemma 1.5(1). Otherwise,t≠tand t∈Preﬁx(C)\ {}. trying to build a path fromopen toclose, the edgeu,x

together withFigure 2.3 are added to

Figure 2.2. We now discuss the other edges added. First, assume thatq=k. Addck1−1u

,c1k−1uc2

andc1kuc1i

,_cq−i

1 z

(0≤i≤q).

Second, assume thatq > k. Addc1k−1u

,c1k−1uc2

, c1kuc1i

,c1kuc1ic2

(0≤i <

q−k), andck1uc1i

,_cq−i

1 z

(q−k≤i≤q).

Theorem2.3. LetC,u,v, andzbe defined as described above. Ifzis a prefix ofv but not a prefix ofu, thenCisUD.

Proof. Herec₃=ck₁uc₁qz=c₁kzcq

1zand|cq

1z| = |cq1u|. The latter and the fact

that|u|<|c1|imply thatq≥q. Note thatuis not a preﬁx ofz,uis a suﬃx ofzin

casec3=wic1, anduis a suﬃx ofzin casec3=wic2. Trying to build a path from

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First, assume thatq < k−1. In this case, no more edges are added. Second, assume thatq=k−1. We have|c1qz| = |ck1−1u|and add

_ck−1 1 u

,_ck

1z

along with

ck₁z

c₁kzc1

···

ck1zcq− 1 1

c1kzcq

1

Figure2.4

In case c3=wic2, we have cq

1 z=c1k−1u (q=k−1 andz=u) and also add

c1kzc1q−i

,ci1u

(0≤i < k).

Third, assume thatq=kandc3=wic2. We havecq

1z=ck1u(q=kandz=u) and |z|≠|u|here. If|z|<|u|, thenu=ztandu=tzfor somet≠. The latter together with the equalityc1kuc1kz=c1kzc1kuyield thatc1=tzx=xzt=zxtand thereforez

is a preﬁx ofu, a contradiction. If|z|>|u|, putz=uc1pt=tuwheret≠andc1

is not a preﬁx oftnort. To simplify the notation, whenever|t|<|c1|, putx|t_|=_x. From the equalityck

1uc1kz=ck1zc1ku, it follows thatc1kuc p 1t=c

p 1tc1ku. •Ifp > k, we contradict the fact thatc1is not a preﬁx ofz.

•Ifp < k, then|t|<|c1|or we contradict the fact thatc1is not a preﬁx oft(t

is a preﬁx ofc1here). If|t|<|u|, putu=tr1=r1tand u=tr1. We then have c1=tr1x=r1tx=xtr1=r1xt andr1=r1. Thent≠t and we havetr1x= r1xt=xtr1, which cannot hold simultaneously byLemma 1.5(1). If|t| = |u|, then t=uand, whenk−p=1,C is not SD(c1c3=c23). Whenk−p >1, we getu=u,

a contradiction. So we have|t_|_>_|u|_{and put}_t₌_ur₁₌_r

1u. Then c1=ur1x= r1ux=r1xu=xur1andr1=r1. Consequently,ur1x=r1xu=xur1, which

cannot hold simultaneously byLemma 1.5(1).

•Ifp=kthen from the equalityc1kuc1pt=c1ptc1kuit follows thatuc1kt=tck1u.

Here, clearly|t_|_≠_|u|_{. If}_|t_|_<_|u|_{, put}_u₌_t_r

1andu=r1t=tr1where|t| = |t|.

This then yields thatc1=r1tx=tr1x=xtr1=txr1 and t=t. So we have r1≠r1andr1tx=txr1=tr1x. UsingLemma 1.6and the fact thattxr1=xtr1,

we conclude thatu=u, a contradiction. Thus|t_|_>_|u|_{and put}_t₌_u_cp1

1 t1=t1ufor

somet1andt1wheret1≠andc1is not a preﬁx oft1nort1. From the equalityuc1kt= tck1u, it follows thatc1kuc1p1t1=c1p1t1ck1uand we continue as above, comparingp1

tok. Sincec3is ﬁnite, we must reacht_i=ucp1i+1ti+1=ti+1uwheret_i₊1≠andc1

is not a preﬁx ofti+1nort_i₊1, and consequently we get the equalityc1kuc pi+1 1 ti+1= cpi+1

1 ti+1c1ku. Then we havepi+1> k,pi+1< k, orpi+1=k, and|t_i₊1| ≤ |u|. In all cases

we reach the same types of contradictions as previously stated. Fourth, assume thatq > k and c3=wic2. Since |c1qu| =c

q

1z, it follows thatc2

is a suﬃx ofcq1zand we putc3=c1kuc q

1z=ck1zc q

1z=c1kzsck1uwhere|s| = |c q−k 1 |.

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putz=uc1pt=tuwheret≠andc1is not a preﬁx oftnort. To simplify the

nota-tion, whenever|t_|_<_|c

1|, putx|t|=x. From the equalityc1kuc q

1z=c1kzsc1ku, it

fol-lows thatc1quc p 1t=c

p

1tsc1ku. Add

c₁k−1u

,c1k−1uc2

,c1kucq

−k

1

,_ck

1z

,Figure 2.4,

and c1kuc1i

,ck1uci1c2

(0 ≤i < q−k) along with c1kucq

−k−1 1 c2

,c1kucq1u

and

Figure 2.5:

ck1ucq1u

c1kuc1quc1

··· ck₁uc₁quc₁p−1

c1kuc1quc1p

Figure_2.5

•Ifp > q, then we contradict the fact thatc1is not a preﬁx ofz.

•Ifp < q, then|t_|_<_|c₁_|_{or we contradict the fact that}_c₁_{is not a preﬁx of}_t₍_t_{is a} preﬁx ofc1here). If|t|>|u|, putt=ur1=r1u. Thenc1=ur1x=r1ux=r1xu= xur1 andr1=r1. Sour1x=r1xu=xur1, which cannot hold simultaneously

byLemma 1.5(1). If|t_{| = |u|}_{, then}_t₌_u_{and it follows from the equality}_cq 1uc

p 1u= c1pusc1kuthatc

q−p 1 uc

p

1 =usc1k. Here ifk−p >1, we getu=u, a contradiction. If k−p≤1 thenC is not SD(c1c3=c2c1q−kc2c1p−k+1c2). If |t|<|u|, put u=tr1= r1t and u=tr1. We then havec1=tr1x =r1tx=xtr1=txr1 =r1xt. In

casek > p, from the equalityc1quc p 1t=c

p

1tsc1ku, it follows thatc q−p

1 t=tsc k−p 1 .

Consequently, we also havec1=r1xtandr1=r1. Thent≠tandtr1x=r1xt= xtr1, which cannot hold simultaneously byLemma 1.5(1). So we havep≥k,r1≠r1,

and xz=c1p+1t∈Preﬁx(C)\ {}. Note that q=0 here or we haver1=r1. Also, xt∈Preﬁx(C)\ {}or we gett=t, which leads to a contradiction byLemma 1.6.

Add

c1kuc q 1uc

p−i 1

,

ci

1t

(0≤i≤k)along with

c1kuc q 1uc1i

,

c1kuc q 1uc1ic2

(0≤i < p−k)

(2.1)

andck1ucq−k− 1 1 c2

,c1kuc1quck1−1u

.

•Ifp=q, from the equalityc1quc p 1t=c

p

1tsck1u, it follows thatuc q

1t=tsck1u.

Clearly,|t_|_≠_|u|_{, and if}_|t_|_<_|u|_{,then put}_u₌_t_r₁_and_u₌_r₁_t₌_t_r

1where|t| = |t_|_{. We then have}_r₁_t_x₌_t_r

1x=xtr1=txr1andt=t. Consequently,r1≠r1

and we getr1tx=txr1=tr1x. UsingLemma 1.6and the fact thatxtr1=txr1,

we reach a contradiction with the fact thatu≠u. So we have|t_|_>_|u| _{and put}

t=ucp1

1 t1=t1u, wheret1≠andc1is not a preﬁx oft1nort1. From the equality uc1qt =tsc1ku, it follows thatc

q 1uc

p1

1 t1=c p1

1 t1sc1ku and we continue as above,

comparingp1withq. Sincec3is ﬁnite, we must reach the equalityt_i=uc1pi+1ti+1= ti+1uwhereti+1≠andc1is not a preﬁx ofti+1norti+1, and consequently we get

the equalityc1quc p_i₊1

1 ti+1=c p_i₊1

1 ti+1sc1ku. Then one of the following must occur:

(1) pi+1> q, thus yielding the same type of contradiction as above;

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(3) pi+1< q,|t_i₊1|<|u|, andpi+1< k, thus yielding the same type of contradiction

as above;

(4) pi+1< q,|t_i₊1|<|u|, andpi+1≥k, in which case we add

c1ku

c1qu

i+2 cpi+1−j

1

,

c1jti+1

(0≤j≤k),

c1ku

c1qu

i+2 cj1

,

c1ku

c1qu

i+2 cj1c2

0≤j < pi+1−k

,

ck1u

cq1u

i+1

c1q−k−1c2

,

ck1u

c1qu

i+2 c1k−1u

;

(2.2)

(5) pi+1=qand|t_i₊1| ≤ |u|, thus yielding the same types of contradiction as above.

Last, assume that q ≥ k and c3 = wic1. In this case, add

c1k−1u

,ck1−1uc2

, _ck

1ucq1−k

,_ck

1z

, andFigure 2.4. Also, add the edgesc1kuc1i

,c1kuci1c2

(0≤i < q−k).

If|z|<|u|putu=zt=tzandu=ztwhere|t| = |t|,z≠z, andt,t≠. Then we havec1=xzt=xtz=ztx=txz. Here, ift=twe have the equalities ztx =txz=xzt, which cannot hold by Lemma 1.5(1). So t≠t and we add _ck−1

1 uc2

,_xzwheneverq=kandck1ucq−k− 1 1 c2

,_xzwheneverq > k. Note that hereq=0 nor we contradict the fact thatt≠t.

If|z| = |u|thenz=uandCis not SD(c1c3=c2cq1−kc2c1).

If |z|>|u| putz=ucp1t=tu wherec1 is not a preﬁx of t or t, andt≠

(otherwise,c1c3=c1ck1uc q 1uc

p

1 =c2c1q−kc2cp+ 1

1 ). To simplify the notation, whenever |t_|_<_|c

1|, putx|t|=x. Then we havec3=ck1uc q

1z=c1kzc q

1 z=c1kzsc1, where|s| = |cq−1

1 u|and it follows thatc q 1uc

p 1t=c

p

1tsc1. Here, we add

ck1−1uc2

,ck1ucq1u

if

q=k, andck1ucq

−k−1 1 c2

,c1kuc1qu

ifq > k. In either case, also addFigure 2.5.

•Ifp > q, then we contradict the fact thatc1is not a preﬁx ofz.

• If p < q, then |t_|_< _|c₁_| _{or we contradict the fact that} _c₁ _{is not a preﬁx of}

t(tis a preﬁx ofc1here). Clearly,|t|≠|u|or we contradict the fact that u≠u.

If|t_|_>_|u|_{, put}_t₌_ur₁₌_r

1uandc1=ur1x=r1ux=r1xu=xr1u=uxr1.

Thenuxr1=xr1u=r1ux, which cannot hold simultaneously byLemma 1.5(1).

If |t_|_<_|u|_{, put}_u₌_t_r

1=r1t and u=r1t=tr1 where|t| = |t|. Then we

have c1= tr1x =xr1t =txr1 =r1xt= xtr1 =r1tx and r1=r1. Here, if t=t, thenr1≠r1and we havetr1x=r1xt=xtr1, which cannot hold

simultane-ously byLemma 1.5(2). So we assume thatt≠t; and ifr1=r1, then the equalities txr1=xr1t=xtr1together withLemma 1.6and the fact thatxr1t=r1xtyield

a contradiction with the fact thatu≠u. Also, ift=t thenxt=tx=xt and

t=t, a contradiction. So we havet≠t,t≠t,r1≠r1, andxz=c p+1

1 t. Here,

note thatq≤q−p−1 or we contradict the fact thatt≠t. Thus|z_{| = |c}q−q

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In the case whereq=k, we add

ck−1 1 uc2

, xz , ck 1uc

q 1uci1

,

cp1−it

(0≤i≤p). (2.3)

In the case whereq > kandp < k, addck1ucq−k− 1 1 c2

,xz

with (2.3). The edge

ck

1zc1i

,

ck

1zc1ic2

(2.4)

is added in case 0≤i < q−k, ori=q−kanduis a preﬁx ofz.

In the case where q > k and p ≥ k, in addition to (2.1) and ck1ucq−k− 1 1 c2

,

c1kuc1quck1−1u

, add c1kucq1ucp1−i

,ci

1t

(0≤i≤k). Also, (2.4) is added in case 0≤i < q−k, ori=q−kanduis a preﬁx ofz.

In all cases, also add_ci

1t

,_ci+1 1 t

(0≤i < k)and_ci

1t

,c1k−ir1

(0≤i≤k)

as well as _t

,r1x

, r1x

,t

, and c1ir1x

,ci1+1r1x

(0≤ i < k−1). In the case where|t_|_>_|x|_{, put}_t₌_xs_and _t₌_sx₌_xs_{. It follows from the equality}

txr1=xtr1 thatt=sx and consequently,s≠s. Add

_ck−1 1 r1x

,c1kr1x

along

withci1r1x

,_ck−i

1 s

(0≤i≤k).

•Ifp=q, then it follows fromcq1ucp1t=cp1tsc1thatuc1qt=tsc1. Here, if|t| = |u|, thent=uandCis not SD(c1c3=c2c1q−kc2cq1−kc2c1). If|t|<|u|then putu= tr1=r1tandu=tr1, where|t| = |t|. So we havec1=xtr1=xr1t=tr1x= r1xt. Ift=t, thenr1≠r1and we gettr1x=r1xt=xtr1, which cannot hold

si-multaneously byLemma 1.5(2). So we assume thatt≠t, and ifr1=r1thentr1x= r1xt=xtr1, which cannot hold byLemma 1.5(1). Thust≠t,r1≠r1, andxz= xuc1qt=c

q+1

1 t∈preﬁx(C)\{}. Note that hereq=0 or we contradict the fact that r1≠r1. Whenq=k, we add

ck₁−1uc2

,ck1uc1kuck1−1u

as well asc1kuc1kuck1−1u

,xt

andck1uc1ku

,_ck

1t

. Whenq > k, add (2.1) andc1kucq−k− 1 1 c2

,ck1uc1quc1k−1u

along

with ck1ucq1ucq−k− 1 1 c2

,_xt

and ck1ucq1uc1q−k

,_ck

1t

. If |t_| _> _|u|_{, put} _t ₌

ucp1

1 t1=t1u, wherec1is not a preﬁx oft1 nort1, and t1 ≠(otherwise,c1c3= c1c1ku(c

q 1u)2c

p1

1 =(c2c1q−k)2c2cp1+ 1

1 ). From the equalityuc q

1t=tsc1, it follows that c1quc

p1

1 t1=c p1

1 t1sc1and we continue as above, comparingp1toq. Sincec3is ﬁnite,

we must reachti=uc p_i+1

1 ti+1=ti+1u wherec1is not a preﬁx ofti+1ort_i₊1, and ti+1≠ (otherwise,c1c3=c1ck1u(c

q 1u)i+2c

p_i+1

1 =(c2cq1−k)i+2c2cpi+1+ 1

1 ), and

conse-quently we get the equality cq1uc pi+1

1 ti+1=c pi+1

1 ti+1sc1. Then one of the following

occurs:

(1) pi+1> q, in which case we reach the same type of contradiction as above;

(2) pi+1< qand|t_i₊1| ≥ |u|, in which case we reach the same type of contradictions

as above;

(3) pi+1 < q and |t_i₊1|<|u|, in which case we put u=ti+tri+2= r_i₊2ti+1 and u=ri+2ti+1=ti+1ri+2where|ti+1| = |ti+1|. As above, we haveti+1≠ti+1,ti+1≠ ti+1,ri+2≠r_i₊2, andxz=c

pi+1+1

1 ti+1∈preﬁx(C)\ {}. Also,q≤q−pi+1−1, |z_{| = |c}q−q

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Addc1ku(c1qu)i+2c1j

,_cPi+1−j 1 ti+1

(0≤j≤pi+1)wheneverq=k, orq > k, and pi+1< k. Wheneverq > kandpi+1≥k, add

ck 1u

cq1u

i+1

c1q−k−1c2 , ck 1u

cq1u

i+2 ck−1

1 u , ck 1u

c1qu

i+2 cj1 , ck 1u

c1qu

i+2 cj1c2

, 0≤j < pi+1−k,

(2.5)

and c1ku(c1qu)i+2c

p i+1−j 1

,_cj

1ti+1

(0≤ j ≤ k). In all cases, we also include

c1jti+1

,_cj+1 1 ti+1

(0≤j < k),

c1jti+1

,c1k−jri+2

(0≤j≤k),_t i+1

,ri+2x

,

ri+2x

,_t i+1

, andc1jri+2x

,c1j+1ri+2x

(0≤j < k−1), In the case where

q > k, add

ck1zc1j

,_ck

1zc1jc2

whenever 0≤j < q−k, orj=q−kand uis

a preﬁx ofz. In the case where|t_i₊1|>|x|, putti+1=xsi+1, t_i₊1=si+1x= xsi+1, andti+1=si+1xwheresi+1≠s_i₊1. Add

c1k−1ri+2x

,ck1ri+2x

along with _cj

1ri+2x

,_ck−j

1 si+1

(0≤j≤k);

(4) pi+1=qand|t_i₊1| = |u|, in which case we reach the same type of contradiction

as above;

(5) pi+1=qand|ti+1|<|u|, in which case, whenq=k, we add

c1ku

i+1 ck1−1uc2

,

c1ku

i+3 ck1−1u

,

c1ku

i+3

, ck

1ti+1

, (2.6)

andck1u

i+3

c1k−1u

,_xt i+1

. Wheneverq > k, we add

ck 1u

c1qu

i+1

cq1−k−1c2 , ck 1u

c1qu

i+2 ck−1

1 u

,

ck1u

cq1u

i+2 c1j

,

ck1u

c1qu

i+2 c1jc2

, 0≤j < q−k,

c1ku

c1qu

i+2 cq1−k

,

c1kti+1

,

c1ku

cq1u

i+2

c1q−k−1c2

,

xti+1

.

(2.7)

Acknowledgments. _{This material is based upon work supported by the National}

Science Foundation (NSF) under Grant No. CCR-9700228. We thank the referees of a preliminary version of this note for their very valuable comments and suggestions.

References

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