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].

© 2019 Academic Research Centre of Canada

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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 .

Computer Communication & Collaboration (2019) Vol. 7, Issue 1: 1-13

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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


Then A is a code if and only if 1 rs .

Proposition 2.7. Let },,{ t

n

s

n

r wwwA and 1


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 .


ts

n

r wwwA and 1


Then A is a code .


t

n

sr wwwA and 1


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 ,



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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 ,


(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 ,


(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 ,


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


~ 11 ~

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 ,


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 ,


(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


~ 12 ~

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 , ,


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 ,


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 ,


3 ip wb . If jwy 3, then we have

qpii yyxxww 4363 , so nwy 4 and

qiiipii yywwwxxww 5643363 ,


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 ,


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.

Some three and four elements codes in Fibonacci languages · Computer Communication & Collaboration...

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