Some three and four elements codes in Fibonacci languages · Computer Communication & Collaboration...
Transcript of Some three and four elements codes in Fibonacci languages · Computer Communication & Collaboration...
Computer Communication & Collaboration (Vol. 7, Issue 1, 2019)
ISSN 2292-1028(Print) 2292-1036(Online)
DOIC:2292-1036-2019-02-001-97
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Some three and four elements codes in Fibonacci languages1
Chunhua Cao (Correspondence author)
School of Mathematics and Statistics, Yunnan University
Kunming, Yunnan, 650091, China
E-mail: [email protected]
Qin Liu, Ling Li, Yanling Li , Lijuan Zhu
School of Mathematics and Statistics, Yunnan University
Kunming, Yunnan, 650091, China
Abstract: In the theory of codes, it is known that the language },{ yx is a code if and only if
yxxy . In 2004, Zheng-Zhu Li, Y.S. Tsai and Gian-Chi Yih provided a characterization for codes
consisting of three words in Fibonacci languages. In this paper, we present some results of codes
with three elements in the Fibonacci languages. Then we give a characterization on codes with four
elements in Fibonacci languages.
Keywords: Three Element, Four Element, Code, Fibonacci Languages
2010 Mathematics Subject Classification: 68Q45; 68R15
1 Introduction and Preliminaries
Let X be a finite alphabet consisting of at least two letters and X be the number of letters in
X . Every finite string over X is called a word. The word that contains no letter is called the
empty word, denoted by 1. The set of all words is denoted by *X , which is a free monoid with
concatenation. Let }1{\*XX . For any word
*Xv , the length of v , denoted by v , is the
number of letters occurring in v . In particular, 01 . A word Xx is said to be primitive if it
is not a proper power of another word. A word *Xu is a prefix (or suffix) of a word
*Xv if
there is a word *Xx such that vux (or vxu ). We write vu p
(or vu s ) when
1 This work is supported by a National Natural Science Foundation of China \# 11861071, an
Applied Basic Research Program of Science and Technology Department of Yunnan Province of
China \# 2014FB101, and an Educational Committee Major Natural Science Foundation of Yunnan
Province of China \# ZD2015013. Corresponding author: Chunhua Cao, School of Mathematics and
Statistics, Yunnan University, [email protected].
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u is a prefix (or suffix) of v . By vu p , we mean vu p but vu and 1v . A nonempty
language XC is a code if nm yyyxxx 2121 , where Cyx ji , and ,,,2,1 mi
,,,2,1 nj implies that nm and ii yx for all mi ,,1 . (see [1])
We assume that },{ baX and define some special codes in the Fibonacci languages. Let
aw 1 , bw 2 where Xba , . The two types of Fibonacci sequences are defined recursively
as follows: (see [2])
;,,,,,, 1112321 nnnnnn wwwwwwabwbwaw
.,,,,,, 1121321 nnnnnn zzzzzzbazbzaz
Let 1
,baF be the set consisting of those words in the first sequence and 0
,baF be the set consisting
of those words in the second sequence respectively.
That is
},,,,,{},,,,,{ 54321
1
, a b b a bb a babbawwwwwF ba
and
},,,,,{},,,,,{ 54321
0
, babbababbabazzzzzF ba
And let
;,,,,
},,,,{},7,5,3,1{ 7531
1
,
1
1
b a ba b b a b b a b a ba b b a baba
wwwwnFwF ban
};,,,,{
},,,,{},8,6,4,2{ 8642
1
,
1
2
bbabbababbabababbababbababbabbabb
wwwwnFwF ban
};,,,,{
},,,,{},7,5,3,1{ 7531
0
,
0
1
bbababbababbababbabaa
zzzznFzF ban
}.,,,,{
},,,,{},8,6,4,2{ 8642
0
,
0
2
bbbababbabababbababbababbababbabb
zzzznFzF ban
Remarks. The Fibonacci words have following properties:
(1) No words in 1
1F is a prefix of any words in 1
2F , vice versa.
(2) For all 1
1Fwi , 1
2Fw j , we have ip wa , jp wb .
(3) The Fibonacci languages 1
1F and 1
2F are codes.
(4) If 1
,ban Fw for 3n , then 1531 nkkkkn wwwwww for any nk , kn |2 .
(5) For Fibonacci words 32 ,, nnn www in 1
,baF , we have 322 nnnn wwww .
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(6) Let 1
,1 },,{ batrr Fwww , where tr 1 . Then
},{ 1rrt www .
The 0
,baF has same properties as above.
Definitions which are used in the paper but not stated here can be found in [1]. If 1X , we
know any three-element languages is not a code. So in this paper, we always let 2X .
2 Main result
In this section we characterize three-element and four-element codes which are Fibonacci words. In
these two types of Fibonacci languages 1
,baF and
0
,baF , the corresponding terms of Fibonacci
words have the same length. We know that they are in fact a reverse pair. (see [1])
Lemma 2.1.[3]
Let 1
,},,{ batsr Fwww , where tsr . Then },,{ tsr wwwA is a code if and
only if the following two conditions hold:
(1) 1 rs ;
(2) )3,2(),( rrts .
Lemma 2.2. [3] )(1
, )( n
baF is a prefix code for 2n .
Lemma 2.3. [3]
)(1
, )( m
baF is a prefix code for 3m .
The Fibonacci languages )( 0
1
1
1 FF and )( 0
2
1
2 FF are codes, in addition, )(1
, )( n
baF is a biffix code
for 3n . A language containing three elements only in )( 0
1
1
1 FF or )( 0
2
1
2 FF or )(1
, )( n
baF is a
code. Then we are thinking about a language which both contains words in )(1
, )( n
baF and 1
,baF is
a code or not. And we also want to know the language which contains four Fibonacci words is a
code or not.
Proposition 2.4. Let },,{ ts
n
r wwwA and 1
,},,{ batsr Fwww , where tsr and 2n .
Then A is a code.
Proof. Suppose A is not a code , then there exists a word Xz with minimal length, which
has two representations over A . That is, qp yyyxxxz 2121 , Ayx ji , with 11 yx ,
pi ,,2,1 and qj ,2,1 . Since 11 yx p or 11 xy p , we have 1
111 },{ Fyx or 1
2F .
Without loss of generality, we may suppose 1
111 },{ Fyx .
(1) If s
n
r wywx 11 , , then qsp
n
r yywxxw 22 . By remark(4), we have 1 rrs www ,
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so qrrp
n
rr yywwxxww 212
1
, then qrp
n
r yywxxw 212
1
. Since rp wa but
1 rp wb , which is a contradiction.
(2) If t
n
r wywx 11 , , then qtp
n
r yywxxw 22 . Since 2rt , then 1 rrt www , we
have a contradiction like (1).
(3) If ts wywx 11 , , then qtps yywxxw 22 . Since 1 sst www , this implies that
qtsp yywwxx 2112 . If 1
12 Fx , then 2xa p , which contradicts that
1 sp wb . If
1
22 Fx , then n
rwx 2 and 2xb p . Since 11 rrs www , so prrp
n
r ywwxxw 13 , we
have a contradiction like (1).
Proposition 2.5. Let },,{ t
n
sr wwwA and 1
,},,{ batsr Fwww , where tsr , and 2n .
Then A is a code if and only if )3,2(),( rrts .
Proof. )(
Immediately.
)( Suppose the condition holds and A is not a code , then qp yyyxxx 2121 for some
Ayyyxxx qp ,,,,,, 2121 with 11 yx . It’s similar to proposition 2.4, we suppose that
1
111 },{ Fyx .
(1) If n
sr wywx 11 , , then q
n
spr yywxxw 22 . Since 1 rrs www , so pxx 2
q
n
sr yyww 2
1
1
. If 1
12 Fx , then 2xa p , which contradicts that
1 rp wb . If 1
22 Fx ,
then twx 2 and tp wb , this implies that q
n
srpt yywwxxw 2
1
13
.
(1-1) If 4rs , then 31 rrrs wwww and 1421 trrrt wwwww . Therefore we have
q
n
srrprr yywwwxxww 2
1
31321
. That is, qrpr yywxxw 2332 , since
2 rp wa but 3 rp wb , which is a contradiction.
(1-2) If 2rs , then 1 rrs www and q
n
srpt yywwxxw 2
1
13
.
(1-2-1) When 2n . If 3 rt , then },,{ 3
2
2 rrr wwwA , this contradicts the condition. If
5 rt , then 421 rrrt wwww , and it must have qrrprrr yywwxwww 221421 . So
rwy 2 or 2
2 swy . If rwy 2 , then qprr yyxxww 3331
. So twy 3 and we will
have qrrprr yywwxxww 421331 , since 2 rp wa but 3 rp wb , which is a
contradiction. If22
2
2 rrs wwwy , then qrrptrr yywwxxwww 3223132 , since
2 rp wa but 3 rp wb , which is a contradiction.
(1-2-2) When 3n . If 3 rt , 21 rrt www , then },,{ 3
3
2 rrr wwwA , this contradicts the
condition. If 5 rt , then 421 rrrt wwww , so qrrprrr yywwxxwww 2
2
213421 .
Hence 432 rrt wwwy , then qrp yywxx 343 . So rwx 3 or 3
3 swx . If rwx 3 ,
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then qrrrpr yywwwxxw 3314 , we have twx 4 . It follows that prrr xxwww 5421
qrr yyww 331 , since
2 rp wa but 3 rp wb , which is a contradiction. If
3
3 swx , then
qrrrprrr yywwwxxwww 3314
2
21 , a contradiction similar to rwx 3 . If 7 rt , then
6421 rrrrt wwwww and qrrprrrr yywwxxwwww 2
2
2136421 . So twy 2 . We
have prrrqrrrprr xxwwwyywwwxxww 35433643363 , since
6 rp wa
but 5 rp wb , which is a contradiction.
(1-2-3) When 4n . If 3 rt , then 21 rrt www , },,{ 32 r
n
rr wwwA , this contradicts the
condition. If 5 rt , then 421 rrrt wwww , and q
n
rrprrr ywwxwww 1
21421
, that
is q
n
rpr ywxw 3
23
, which contradicts the fact that 2 rp wa and
3 rp wb .
(2) If tr wywx 11 , , then qtpr yywxxw 22 . Since 2rt , so 1 rrt www , we
must have qtrrpr yywwwxxw 21-12 . If
1
12 Fx , then 2xa p , which contradicts
1 rp wb . If 1
22 Fx , then 2xb p and
n
swx 2, so qtrp
n
s yywwxxw 2113 .
(2-1) If 2rt , then 1 rs , and qrp yywxx 212 . So
n
swx 2, then we obtain
n
swy 2, continuing this process, we know that
n
sii wyx for all 2i , which indicates the
length of the word z is infinite, which is a contradiction.
(2-2) If 4rt , then 31 rrrt wwww .
(2-2-1) When 2n . If 1 rs , then qrrrpr yywwwxxw 22113
2
1 . So rwx 3 or
twx 3 . If rwx 3 , then qrp yywxx 214 . We must have 2
4 swx . Therefore
qrpr yywxxw 215
2
1 , so 2
2 swy . Continuing this process, we obtain that 4,2 iwx si
and 2
sj wy , 2j , which contradicts the finite length of z . If twx 3 , then we have
qrprrr yywxxwww 22431 . So n
swy 2, then qrprr yywxxww 3
2
1421 , which
contradicts the fact that 2 rp wa and 1 rp wb . If 3 rs , then 21 rrs www , then
qrrpsrr yywwxxwww 231321 . Since 2 rp wa but
3 rp wb , which is a
contradiction.
(2-2-2) When 3n . If 1 rs or 3 rs , then qrrrp
n
r yywwwxxw 221131 or
qrrp
n
sr yywwxxww 2213
1
2
, which contradicts the fact that 2 rp wa and
1 rp wb .
(2-3) If 6rt , then 1531 trrrrt wwwwww .
(2-3-1) If 2n , then qtrrrrpr yywwwwwxxw 215312 , so 2
2 swx . If 1 rs ,
then qrrrrprr yywwwwxxww 25211311 , so rwx 3 or twx 3 . If twx 3 , then
qrrrrqrrrprrrr ywwwwyywwwxxwwww 4312514531 , since 4 rp wa but
5 rp wb , which is a contradiction. If rwx 3 , then qrrp yywwxx 2514 , so 2
4 swx
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and 42115
2
rrrrps wwwwxxw qyy 2
. So rwx 5 or twx 5 . If rwx 5 , then we
obtain that qrrrpr yywwwxxw 2416 , then
2 2
6 1s rx w w , it implies that pr xxw 7
2
1
qrr yyww 221 , which contradicts the fact that
2 rp wa and 1 rp wb . If twx 5 , then
qrrprrrr yywwxxwwww 2426531 , which contradicts the fact that 4 rp wa
and
3 rp wb . If 3 rs , then 21 rrs www and psrr xxwww 321
qrr yww 31 ,
which contradicts the fact that 2 rp wa and
3 rp wb .
(2-3-2) If 3n and 1 rs , then qrrrqrrrp
n
r ywwwywwwxxw 21153131 ,
which contradicts the fact that 2 rp wa and
1 rp wb . If 3 rs , then 21 rrs www ,
then qrrrp
n
srr yywwwxxwww 25313
1
21
, since 2 rp wa but
3 rp wb , which
is a contradiction.
(3) If t
n
s wywx 11 , , then we have qtp
n
s yywxxw 22 . For 11 tsst wwww , then
qtssp
n
s yywwwxxw 21-12 , which contradicts the fact that sp wa and
1 sp wb .
Corollary 2.6. Let },,{ n
tsr wwwA and 1
,},,{ batsr Fwww , where tsr and 2n .
Then A is a code if and only if 1 rs .
Proposition 2.7. Let },,{ t
n
s
n
r wwwA and 1
,},,{ batsr Fwww , where tsr and 2n .
Then A is a code .
Proof. The process is the same as proof of proposition 2.4, we suppose that 1
111 },{ Fyx .
(1) If n
rwx 1 ,n
swy 1, then q
n
sp
n
r yywxxw 22 . Since 11 srrs wwww , we have
q
n
srrp
n
r yywwwxxw 2
1
12
, which contradicts the fact that rp wa and
1 rp wb .
(2) If n
rwx 1 , twy 1 , then qtp
n
r yywxxw 22 . Since 2rt , so 11 trrt wwww ,
hence qrrp
n
r yywwxxw 212 , which contradicts the fact that rp wa and
1 rp wb .
(3) If n
swx 1, twy 1 , then qtp
n
s yywxxw 22 . Since 11 tsst wwww , so we have
qssp
n
s yywwxxw 212 , which contradicts the fact that sp wa and
1 sp wb .
Corollary 2.8. Let },,{ n
ts
n
r wwwA and 1
,},,{ batsr Fwww , where tsr and 2n .
Then A is a code .
Corollary 2.9. Let },,{ n
t
n
sr wwwA and 1
,},,{ batsr Fwww , where tsr and 2n .
Then A is a code .
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Proposition 2.10. Let 1
,},,,{ baktsr FwwwwA , where ktsr . Then A is a code if
and only if the following four conditions are true:
(1) 1 rs ;
(2) 1 st ;
(3) )3,2(),( rrts ;
(4) )3,2(),( sskt .
Proof. )( If 1 rs , then tw ,
},{ 1rrk www . If 1 st , then
},{ 1ssk www . If
)3,2(),( rrts , then 3
2
2 rrr www . If )3,2(),( sskt , then 3
2
2 sss www . All
contradicts that A is a code.
)( If 1
1FA or 1
2FA , then A is a code. Let 1
1FA and
1
2FA , we show that A is a
code. Suppose the conditions hold and A is not a code, then there exists a word w with
minimal length such that qp yyyxxxw 2121 , where Ayyyxxx qp ,,,,,,, 2121 ,
11 yx . Since 11 yx p or
11 xy p , we have 1
111 },{ Fyx or 1
211 },{ Fyx . Without
loss of generality , we assume that 1
111 },{ Fyx and },{\, 1111 yxAvu . There are two
cases: (1) 1
211, Fvu ; (2) 1
11 Fu and 1
21 Fv . Let iwx 1 , jwy 1, mwu 1 , nwv 1
and ji , ij 2 .
(1) If 1
111 },{ Fyx and 1
211, Fvu , then mn 2 . Let nm .
(1-1) If 4j i , by remark (4), we have 31 iiij wwww , then qiip ywwxx 312 .
So mwx 2 or nwx 2 .
(1-1-1) If mwx 2 , then qiipm yywwxxw 2313 .
(1-1-1-1) If 1 im or 1 im , then },,,{ 1 njii wwwwA or },,,{ 1 njii wwwwA ,
which contradicts condition (1).
(1-1-1-2) If 3 im , then 311 mmmi wwww , we have 131 ipmmm wwww . Therefore
qimmmpm yywwwwxxw 23313 , then iwx 3 or jwx 3 . Since 1
1, Fww ji , we
have ipmm www 21 and
jpmm www 21, then
qmmpmm ywwxxww 31421 . Since
2 mp wb but 3 mp wa , which is a contradiction.
(1-1-1-3) If 3 im , then 212 iim wwwx . So qiipii ywwxxww 31321 ,
which contradicts the fact that 2 ip wa and 3 ip wb .
(1-1-2) If nwx 2 , then qiipn yywwxxw 2313 .
(1-1-2-1) If 1 in or 1 in , then },,,{ 1 jiim wwwwA or },,,{ 1 jiim wwwwA ,
which contradicts condition (2).
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(1-1-2-2) If 3 in , then 311 nnni wwww . So qinnp yywwwxx 23313 . It
must have iwx 3 or jwx 3
. Since 1
1, Fww ji , we have ipnn www 21
and
jpnn www 21, then
qinnpnn yywwwxxww 2331421 , which contradicts the
fact that 3 np wa and
2 np wb .
(1-1-2-3) If 3 in , then 21 iin www . Hence qiipii ywwxww 3121 , which
contradicts 2 ip wa and
3 ip wb .
(1-2) If 2 ij , then 12 iiij wwww and we have
qip yywxx 212 . Therefore
mwx 2 or nwx 2 .
(1-2-1) If mwx 2 , then qipm yywxxw 213 .
(1-2-1-1) If 1 im or 1 im , then },,,{ 21 niii wwwwA or },,,{ 21 niii wwwwA ,
which contradicts condition (1).
(1-2-1-2) If 3 im , then 311 mmmi wwww , we have 131 ipmmm wwww . Therefore
qimmmpm yywwwwxxw 2313 , then iwx 3 or jwx 3
. Since 1
1, Fww ji , we
have ipmm www 21 and
jpmm www 21, then
qmmpmm yywwxxww 231421 ,
which contradicts the fact that 3 mp wa and
2 mp wb .
(1-2-1-3) If 3 im , then },,,{ 32 niii wwwwA , which contradicts condition (3).
(1-2-1-4) If 5 im , then 421 iiim wwww . We have qipiii yywxxwww 213421 .
Hence iwy 2 or jwy 2. If iwy 2 , then
qpii yyxxww 3341 . Therefore mwy 3
or nwy 3 . Since 1
2, Fww nm , we have mpiii wwww 421
and npiii wwww 421
, then
piiiqiiipii xxwwwyywwwxxww 33214421341 , since 4 ip wa but
3 ip wb ,
which is a contradiction. If jwy 2 , then 334 yxxw pi qy . So iwy 3 or jwy 3 . If
iwy 3 , then qpii yyxxww 4331
. So mwy 4 or nwy 4 . Since mpii www 21
and
npii www 21 , we have qiipii ywwxxww 21331 , since 2 ip wa but 3 ip wb ,
which is a contradiction. If jwy 3, then
qpi yyxxw 433 . So mwy 4 or nwy 4 . If
mwy 4 , then qip yywxx 543 . We obtain that iwx 3 or
jwx 3. If iwx 3 , then we
have qiip yywwxx 5314 . It follows that mwx 4 or nwx 4 , then we have same
contradictions as above. If jwx 3, similar to the above, we have mwx 4 or nwx 4 and then
iwy 5 or jwy 5
. Continuing this process, we have jss wyx , mss wyx 11 for
3s and s is an odd number. This implies that the length of the word w is infinite, which is a
contradiction. And nwy 4 has the same contradiction with mwy 4 .
(1-2-1-5) If 7 im , then 6421 iiiim wwwww . So qpiii yyxxwww 23642 .
We have iwy 2 or jwy 2 . If iwy 2 , then qpiii yyxxwww 33641 , so we have
mwy 3 or nwy 3 . Since mpiii wwww 421 and npiii wwww 421 . Therefore we have
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qiiipiii yywwwxwww 4421641 , which is a contradiction. If jwy 2, then we have
qpii yyxxww 3364 , so iwy 3 or
jwy 3. If iwy 3 , then piii xwww 631
qyy 4, so we have mwy 4 or nwy 4 . Since
mpii www 21 and
npii www 21, we
have qiipiii yywwxwww 521631 , a contradiction. If
jwy 3, then pii xww 63
qyy 4, so mwy 4 or nwy 4 . We have
qiiipii yywwwxww 564363 , which is a
contradiction.
(1-2-2) If nwx 2 , then qipn yywxxw 213 .
(1-2-2-1) If 1 in or 1 in , then },,,{ 21 iiim wwwwA or },,,{ 21 iiim wwwwA ,
which contradicts condition (2).
(1-2-2-2) If 3 in , the proof is the same as (1-1-2-2).
(1-2-2-3) If 3 in , then },,,{ 32 iimi wwwwA . If 1 im , then },,,{ 3211 iii wwwwA .
If 1 im , then },,,{ 32 iimi wwwwA , both contradicts condition (1). If 3 im , then
},,,{ 32 iiim wwwwA , which contradicts condition (4).
(1-2-2-4) If 5 in , we have 421 iiin wwww and },,,{ 52 iimi wwwwA . If 3 im ,
then },,,{ 532 iiii wwwwA , a contradiction with condition (3). If 1 im or 1 im , then
},,,{ 521 iiii wwwwA or },,,{ 521 iiii wwwwA , which is a contradiction with condition (1).
If 3 im , we have qpii yyxxww 2342
. So iwy 2 or jwy 2. If iwy 2 , then
qpii yyxxww 3341 , so mwy 3 or nwy 3 . If 4213 iiin wwwwy , then
piiiqiiipii xxwwwyywwwxxww 33214421341 which contradicts the fact that
4 ip wa and 3 ip wb . If mwy 3 , then 311 mmmi wwww . We have pmm xww 31
qyy 4. Hence iwy 4 or
jwy 4. Since
ipmm www 21 and
jpmm www 21, we have
qmmpmm yywwxww 52131 which contradicts that 2 mp wb and 3 mp wa . If
jwy 2, then
qpi yyxxw 334 . So iwy 3 or
jwy 3. If iwy 3 , then qyy 4
pii xxww 331 , so we have mwy 4 or nwy 4 . If 4214 iiin wwwwy , we must have
piiqiii xxwwyywww 3315421 , since 3 ip wb but
2 ip wa , which is a contradiction.
If mwy 4 , then 311 mmmi wwww , we have qpimm yyxxwww 53331
. Therefore
iwy 5 or jwy 5. Since
ipmm www 21 and
jpmm www 21, we have contradictions like
above. If jwy 3 , then qpi yyxxw 433 . So mwy 4 or nwy 4 . If mwy 4 , then
we have 313 mmmi wwww and qpmm yyxxww 5331 . Therefore iwy 5 or
jwy 5 . Since ipmm www 21 and jpmm www 21 , we have contradictions like above. If
434 iin wwwy , then qip yywxx 543 . So iwx 3 or jwx 3 . If iwx 3 , then
pqii xxyyww 4531 . So mwx 4 or nwx 4 . If 4214 iiin wwwwx , then we have
piiiqii xxwwwyyww 5421531 , which contradicts the fact that 2 ip wa and 3 ip wb .
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If mwx 4 , then 311 mmmi wwww and qimmp yywwwxx 53315 . So iwx 5
or jwx 5. Since
ipmm www 21 and
jpmm www 21, we have contradictions like above. If
jwx 3, then
pxx 4 qi yyw 53 . So mwx 4 or nwx 4 . If mwx 4 , then
313 mmmi wwww . so we have iwx 5 or jwx 5
. Since ipmm www 21
and
jpmm www 21, we have contradictions as above. If nwx 4 , then
qpi yyxxw 554 .
Therefore iwy 5 or jwy 5
. Continuing this process, we know nss wyx 11 and
jss wyx for 3s and s is an odd number. This implies the length of w is infinite,
which is a contradiction.
(1-2-2-5) If 7 in , then 421 iiin wwww . We have qpii yyxxww 2342
. So
iwy 2 or jwy 2. If iwy 2 , then
qpii yyxxww 3341 , so mwy 3 or nwy 3 .
If 4213 iiin wwwwy , then piiiqiiipii xwwwyywwwxxww 3214421341 ,
which contradicts the fact that 4 ip wa and
3 ip wb . If mwy 3 , we compare the size of
m and 1i . If 5 im , then 421 iiim wwww and we have a same contradiction as
nwy 3 . If 3 im , then },,,{ 32 niii wwwwA , which is a contradiction with condition (3).
If 1 im or 1 im , then },,,{ 21 niii wwwwA or },,,{ 21 niii wwwwA , which are
contradictions with condition (3) and (1). If 3 im , then 311 mmmi wwww . So we have
qpimm yyxxwww 43431 . Hence iwy 4 or
jwy 4. Since 21 mm ww is a prefix
of iw and jw , then we have
qmmpimm yywwxxwww 5213431 , which
contradicts the fact that 3 mp wa and
2 mp wb . If jwy 2, then
qpi yyxxw 334 .
So iwy 3 or jwy 3. If iwy 3 , then qyy 4 pii xxww 331
, so we have mwy 4
or nwy 4 . If 4214 iiin wwwwy , then piiqiii xxwwyywww 3315421 ,
which contradicts the fact that 2 ip wa and
3 ip wb . If mwy 4 , then we have
311 mmmi wwww , so qpimm yyxxwww 53331
. Hence iwy 5 or jwy 5.
Since ipmm www 21 and jpmm www 21 , we have a contradiction like iwy 4 . If jwy 3 ,
then qpii yyxxww 4363
. So mwy 4 or nwy 4 . If 6434 iiin wwwwy ,
then qiiipiii yywwwxxwww 56433543 , which contradicts the fact that
6 ip wa
and 5 ip wb . If mwy 4 , then
qpmm yyxxww 5331 . Hence iwy 5 or
jwy 5.
Since ipmm www 21 and
jpmm www 21, we have a contradiction like iwy 4 .
(2) If 1
1111 ,, Fuyx and 1
21 Fv , then ij 2 and mj 2 .
(2-1) If 4 ij , then 1 3 1 j i i i jw w w w w L , then qiip ywwxx 312 . So nwx 2 .
(2-1-1) If 1 in or 1 in , then },,,{ 1 mjii wwwwA or },,,{ 1 mjii wwwwA , which
contradicts condition (1).
(2-1-2) If 3 in , then 1 1 3 i n n n iw w w w w L
, we have qnnp ywwxx 313 . It
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follows that iwx 3 or jwx 3 or mwx 3 . Since
1
1, Fww ji , then ipnn www 21
and jpnn www 21, so
qnnpnn ywwxxww 31421 , which contradicts the fact that
3 np wa and 2 np wb . If mwx 3 , then },,,{ jimn wwwwA . If 1 nm , then
},,,{ 1 jinn wwwwA . If 1 nm , then },,,{ 1 jinn wwwwA . Both contradicts condition (1).
If 3 nm , then 311 mmmn wwww , we have qmmp yywwxx 2314 . So
nwx 4 . this implies that qmmpmm yywwxxww 231521 , which contradicts the
fact that 3 mp wb and
2 mp wa . If 3 nm , then 21 nnm www , this implies that
pnn xxww 421 qnn yww 31 . Since
3 np wa but 2 np wb , which is a contradiction.
(2-1-3) If 3 in , then 212 iin wwwx , we have qiipii ywwxxww 31321 ,
which contradicts the fact that 2 ip wa and
3 ip wb .
(2-2) If 2 ij , then 1 iij www , we have
qip yywxx 212 , so nwx 2 .
(2-2-1) If 1 in or 1 in , then },,,{ 21 miii wwwwA or },,,{ 21 miii wwwwA ,
which contradicts condition (1).
(2-2-2) If 3 in , then 11 nni www and it must have qnnp ywwxx 313 . We
have a same contradiction like (2-1-2).
(2-2-3) If 3 in , then },,,{ 32 miii wwwwA . If 1 im or 1 im or 4 im ,
then },,,{ 321 iiii wwwwA or },,,{ 32 iiim wwwwA or },,,{ 32 miii wwwwA , we all
have contradictions with condition(1), (4), (3).
(2-2-4) If 5 in , then 421 iiin wwww . We have qpii yyxxww 2342
. So
iwy 2 or jwy 2 or mwy 2 .
(2-2-4-1) If iwy 2 , then qpii yyxxww 3341 . This implies that nwy 3 , we have
piiiqiiipii xxwwwyywwwxxww 33214421341 , which contradicts the fact that
4 ip wa but 3 ip wb .
(2-2-4-2) If mwy 2 , then qmpii yywxxww 3342
. If 4 im , then we have
32 iim www , so qiipii yywwxxww 332342 . Since 4 ip wa but 3 ip wb ,
which is a contradiction. If 2 im , then we have 312 mmmi wwww and
qpmm yyxww 431 . So nwy 4 . Hence
qmmpmm ywwxww 2131 , which
contradicts the fact that 3 mp wa and 2 mp wb .
(2-2-4-3) If jwy 2 , then qpi yyxxw 334 , we obtain that iwy 3 or jwy 3 or
mwy 3 . If iwy 3 , then qpii yyxxww 4331 . This implies that nwy 4 . Hence we
have qiiipii yywwwxxww 5421331 , which contradicts the face that 2 ip wa and
3 ip wb . If jwy 3 , then we have qpi yyxxw 433 . We obtain that nwy 4 . So
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qip yywxx 543 . It follows that iwx 3 or jwx 3 or mwx 3 . If iwx 3 , then
qiip yywwxx 5314 . So nwx 4 . Since qiipiii yywwxxwww 5315421 , ,
which contradicts the fact that 2 ip wa and
3 ip wb . If mwx 3 , then we have
qipm yywxxw 544 . If 2 im , then 314 mmmi wwww . It follows that pxx 4
qmm yyww 531 . So 42214 iimmn wwwwwx . That is, qmm yyww 531
piimm xxwwww 54221 , which contradicts the fact that
2 mp wa and 3 mp wb . If
4 im , then qp yyxx 54 , which contradicts the minimal length of the word w .
If 6 im , then qpi yyxxw 545
. So nwy 5 and qp yyxx 64 , which
contradicts the minimal length of the word w . If 8 im , then 754 iiim wwww . So
qpii yyxxww 5475 . It follows that nwy 5 . Hence
qpi yyxxw 647 . So
nwy 6 , and qpi yyxxw 746 . This implies that iwy 7 or
jwy 7 or mwy 7 . If
iwy 7 , then qpiii yyxxwww 84531 , which implies that 4218 iiin wwwwy . So
qiiipiii yywwwxxwww 94214531 , which contradicts the fact that 2 ip wa and
3 ip wb . If jwy 7, then
qpii yyxxww 8453 . So 438 iin wwwy . Hence we
have qiipii yywwxxww 943453 , which contradicts the fact that
4 ip wa and
3 ip wb . If mwy 7 , then qip yywxx 874 . So nwx 4 . Similar to above, we
know that mwx 5 and nwy 8 . Continuing this process, we get that nkk wyx 222 ,
mkk wyx 3212 for any 2k . This implies the length of w is infinite, which is a
contradiction. If 10 im , then 9754 iiiim wwwww . So qpiii yyxwww 5975
.
So nww 5 and qpii yyxww 697 . So nwy 6 and
qpii yyxww 796 . This
implies that iwy 7 or jwy 7 or mwy 7 . If iwy 7 , then
qpii yyxww 831 .
So 4218 iii wwwy , then qiiipii yywwwxww 942131 , which contradicts the fact
that 2 ip wa and
3 ip wb . If 27 ij wwy , then
qpii yyxww 853 . So
438 ii wwy , then qiipii yywwxww 94353 , which contradicts the fact that
4 ip wa and 3 ip wb . If mwy 7 , then
qiiipii yywwwxww 997696 ,
which contradicts the fact that 8 ip wa and
9 ip wb . If mwy 3 , then pi xxw 34
qm yyw 4. So we compare the size of m and 4i . The process is the same as mwx 3 .
So we have a contradiction.
(2-2-5) If 7 in , then 6421 iiiin wwwww . So qpiii yyxxwww 23642 . It
follows that iwy 2 or jwy 2 or mwy 2 . If iwy 2 , then qpii yyxww 341 ,
so nwy 3 and it must have piiiqiiipii xwwwyywwwxww 321442141 ,
which contradicts the fact that 4 ip wa and 3 ip wb . If mwy 2 and 2 im , then
312 mmmi wwww . We have 331 xww mm qp yyx 3 . So 213 mmn wwwy
and qmmpmm yywwxxww 421331 , which contradicts the fact that 3 mp wa
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and 2 mp wb . If 4 im , then 32 iim www , so
qiipii ywwxxww 32342 , a
contradiction. If jwy 2, then
qpii yyxxww 3364 . Hence iwy 3 or
jwy 3
or mwy 3 . If iwy 3 , then qpiii yyxxwww 43631
, so 214 iin wwwy ,
which contradicts the fact that 4 ip wa and
3 ip wb . If jwy 3, then we have
qpii yyxxww 4363 , so nwy 4 and
qiiipii yywwwxxww 5643363 ,
which contradicts the fact that 6 ip wa and
5 ip wb . If mwy 3 and 2 im , then
214 mmmi wwww . So qpmm yyxxww 4331 . We obtain 214 mmn wwwy ,
then qmmpmm yywwxxww 521331 , which contradicts the fact that
3 mp wa
and 2 mp wb . If 6 im , then 54 iim www , we have
qiipii ywwxww 5464 ,
which contradicts the fact that 4 ip wa and
5 ip wb . If 4 im , then we have
qpi yyxxw 436 . So iwy 4 or
jwy 4 or mwy 4 . If iwy 4 or
jwy 4,
then qpiii yyxxwww 53531
or 53 ii wwqp yyxx 53 , so nwy 5 .
Hence qiiipiii ywwwxwww 421531 or
qiipii ywwxww 4353 , the first
contradicts the fact that 2 ip wa and
3 ip wb , the second contradicts the fact that
4 ip wa and 3 ip wb . If mwy 4 and 9 in , then
piiq xxwwyy 3855 , so
8655 iiin wwwwy , and we have piiqiii xxwwyywww 3856865 , since
6 ip wa but 7 ip wb , which is a contradiction. If 7 in , then
piq xxwyy 355 .
It follows that 655 iin wwwy , so we obtain that iwx 3 or jwx 3 or mwx 3 . If
iwx 3 or mwx 3 , we have same contradictions as above. So mwx 3 . Following this
process, we know mss wyx , nss wyx 11 for 3s and s is an odd number,
which indicates that the length of w is infinite. It is a contradiction.
Acknowledgments: The authors would like to thank the anonymous referees for their
careful reading of the manuscript and useful suggestions.
References
[1]. H.J. Shyr (2001), Free Monoids and Languages. 3rd ed. Hon Min Book Company,
Taichung, Taiwan.
[2].S.S Yu and Yu-Kuang Zhao (2000), “Properties of Fibonacci languages”, Discrete
Mathematics, 224 (1–3): 215-223.
[3]. Z.Z. Li, Y.S. Tsai and G.C. Yih (2004), “Characterizations on codes with three elements”,
Soochow Journal of Mathematics, 30(2): 177-196.