All Good (Bad) Words Consisting of 5 Blocks

Acta Mathematica Sinica, English Series Jun, 2017, Vol 33, No 6, pp 851 860 Published online: January 25, 2017 DOI: 101007/s10114-017-6134-2 Http://wwwActaMathcom Acta Mathematica Sinica, English Series Springer-Verlag Berlin Heidelberg & The Editorial Office of AMS 2017 All Good (Bad) Words Consisting of 5 Blocks Jian Xin WEI School of Mathematics Statistics Science, Ludong University, Yantai 264025, P R China E-mail : wjx0426@163com Abstract Generalized Fibonacci cube Q d (f), introduced by Ilić, Klavžar Rho, is the graph obtained from the hypercube Q d by removing all vertices that contain f as factor A word f is good if Q d (f) is an isometric subgraph of Q d for all d 1, bad otherwise A non-extendable sequence of contiguous equal digits in a word µ is called a block of µ Ilić, Klavžar Rho shown that all the words consisting of one block are good, all the words consisting of three blocks are bad So a natural problem is to study the words consisting of other odd number of blocks In the present paper, a necessary condition for a word consisting of odd number of blocks being good is given, all the good (bad) words consisting of 5 blocks is determined Keywords Generalized Fibonacci cube, isometric subgraph, good word, bad word MR(2010) Subject Classification 05C12, 05C60 1 Introduction Fibonacci cubes were introduced by Hsu [5] as a model for interconnection networks, which have similar properties as hypercube The vertex set of a Fibonacci cube Γ d is the set of all words of length d containing no two consecutive 1s, two vertices are adjacent if they differ in precisely one bit Fibonacci cubes have been studied from numerous points of view [3, 4, 11, 15, 17], see [12] for a survey Awordf is called a factor of a word μ if f appears as a sequence of f consecutive bits of μ, where f denotes the length of f A Fibonacci cube Γ d can be seen as the graph obtained from Q d by removing all words that contain the factor 11 Inspired by this, Ilić, Klavžar Rho [7] introduced the generalized Fibonacci cube, Q d (f), as the graph obtained from Q d by removing all words that contain the fixed factor f In this notation Fibonacci cube Γ d is the graph Q d (11) Note that the subclass Q d (1 s ) of generalized Fibonacci cube has been studied in [6, 16, 18] under the same name of generalized cubes More research on generalized Fibonacci cubes can be found in [1, 2, 8, 9, 14, 21] Let d G (μ, ν) denote the distance of vertices μ ν in G It is known [10] that for any vertices α β of Q d, d Qd (α, β) is the number of bits in which they differ Obviously, for any subgraph H of G, d H (μ, ν) d G (μ, ν) If d H (μ, ν) =d G (μ, ν) for all μ, ν V (H), then H is called an isometric subgraph of G, simply write H G, H G otherwise More Received March 4, 2016, accepted October 17, 2016 Supported by National Natural Science Foundation of China (Grant Nos 11671347, 11605083), Shong Provincial Natural Science Foundation (Grant No ZR2015PA006) Research Fund for the Doctoral Program of Ludong University (Grant No LY2015006)

852 Wei J X generally, if H G are arbitrary graphs, then a mapping I : V (H) V (G) isanisometric embedding if d H (α, β) =d G (I(α),I(β)) holds for any α, β V (H) Klavžar [11] showed that Fibonacci cube Q n (11) is an isometric subgraph of Q n, he [12] sensed that to characterize words f integers n, forwhichq n (f) is an isometric subgraph of Q n seems a difficult question This question was first studied by Ilić et al [7], it leads to the concept of the so called good (bad) word Awordf is good if Q d (f) Q d for all d 1, bad otherwise Klavžar Shpectorov [13] showed that about eight percent of all words are good Wei Zhang [22] proved that if Q n (f) Q n,thenq n (f) cannot be isometrically embedded into any hypercube Wei [19] characterized the structures of good (bad) words A non-extendable sequence of contiguous equal digits in a word α is called a block of α Ilić et al [7] shown that all the words consisting of one block are good, all the words consisting of three blocks are bad So a natural problem is to study the isometricity of the words consisting of other odd number of blocks Obviously, the next case is the words consisting of 5 blocks, this is what is studied in the present paper Let F = { 0 y1 1 x n x 1,x i,y j 1, 1 i n, 1 j n 1n 3} Several subfamilies of bad words of F are founded as shown in Lemmas 21 22, a necessary condition for a word among F being good is given in the following theorem: Theorem 11 Let f F be a good word Then either y 1 2 x n = x 1 +1,ory 1 =1 x n = x 1 + x 2 +1 In particular, we study the words consisting of 5 blocks Let F = {1 r 0 s 1 t 0 k 1 m r, s, t, k 1m r} All the good words among F are determined in Table 1 in the following theorem Theorem 12 Let f F Thenf is good if only if it is one of words in Table 1 (i) r s t k m (1) r 1 s =1 t r k =1 m = t + r +1 (2) r 1 s =1 t 1 k 2 m = t + r +1 (3) r 2 s k +3 t 1 k 1 m = r +1 (4) r t +1 s = k +2 t 1 k 1 m = r +1 (5) r 1 s = k +1 t 1 k 1 m = r +1 (6) r t +1 s 2 t 1 k = s m = r +1 (7) r t +1 s 2 t 1 k = s +1 m = r +1 (8) r 2 s 2 t r +2 k = s +1 m = r +1 (9) r 2 s 2 r t 1 k = s +2 m = r +1 (10) r =2 s 2 t =1 k s +3 m = r +1 (11) r =2 s 2 t =2 k s +3 m = r +1 (12) r 3 s 2 t 1 k s +3 m = r +1 Table 1 All the good words among F For a word f = f 1 f 2 f d, f R = f d f 2 f 1 denotes the reverse of f f = f 1 f 2 f d the complement of f, wheref i =1 f i, i =1,,dIlić et al showed the following result:

All Good (Bad) Words Consisting of 5 Blocks 853 Proposition 13 ([7]) Let f be any word d 1 ThenQ d (f) = Q d (f) = Q d (f R ) The isometricity of all the good words consisting of 5 blocks can be covered by the words among Table 1 in view of Proposition 13 For instance, by Table 1 (5) the words 1 r 0 k+1 1 t 0 k 1 r+1, 0 r 1 k+1 0 t 1 k 0 r+1,1 r+1 0 k 1 t 0 k+1 1 r 0 r+1 1 k 0 t 1 k+1 0 r all are good Ilić et al [9] studied good (bad) words from another point of view A word μ is called f-free if it does not contain f as a factor Let s be a positive integer Then f is called s-isometric if for any f-free words μ ν of the same length that differ in s bits, the following holds: μ can be transformed into ν by complementing one by one all the s bits on which μ differs from ν, such that all of the new words obtained in this process are f-free Such a transformation is called an f-free transformation of μ to ν It is easy to see that a word f is good if only if f is s-isometric for all s 1, a word f is bad if only if f is not s-isometric for some s Note that Wei [19] showed that a word is bad if only if it is not 2-isometric or not 3-isometric Ilić et al [9] showed that the words among the family {1 2r 1 01 2r 1 01 2r+1 r 0} are bad 2-isometric Wei Zhang [20] showed that the words (or their reverse) of the family {1 r 01 r 01 2r+2 r 0} are all the words that are bad 2-isometric among those with exactly two 0s Obviously, if r 1, then the words among F = {1 r 01 r 01 2r+2 } are bad 2-isometric with 5 blocks In the proof of Theorem 12 we find that all the bad words among F are not 2-isometric except that these among F By applying Proposition 13, the following corollary holds: Corollary 14 Let f be a bad word consisting of 5 blocks Then f is 2-isometric if only if f =1 r 01 r 01 2r+2,or1 2r+2 01 r 01 r,or0 r 10 r 10 2r+2,or0 2r+2 10 r 10 r for some r 1 In the rest of this section, some necessary definitions results are introduced With e i we denote the binary word with 1 in the i-th bit 0 elsewhere For words α β of the same length, let α + β denote their sum computed bitwise modulo 2 In particular, α + e i is the word obtained from α by reversing its i-th bit The empty word, denoted by λ, isaword of length zero For convenience of use, let μ s,t be the factor of μ that starts from the s-th bit to the t-th bit if t s λ if t<s,wheres μ t μ For two vertices μ ν of a graph G, the set of vertices lying on shortest μ, ν-paths is called the interval between μ ν, denoted by I G (μ, ν) Let α β V (Q d (f)) p 2 Then α β are called p-critical words for Q d (f) ifd Qd (α, β) =p, but none of the neighbors of α in I Qd (α, β) belongs to Q d (f) or none of the neighbors of β in I Qd (α, β) belongs to Q d (f) The following proposition gives a tool to prove Q d (f) Q d Proposition 15 ([7]) If there exist p-critical words for Q d (f), thenq d (f) Q d We proceed as follows In the next section, several classes of bad words with odd number of blocks are given Theorem 11 is proved In the last section, Theorem 12 is proved so all good (bad) words consisting of 5 blocks are determined 2 Results on Words with Odd Number of Blocks In this section, several classes of bad words are given in Lemmas 21 22, Theorem 11 is proved Note that all the words considered in the following lemmas comes from F Lemma 21 Let f F Iff has one of the following properties, then it is bad

854 Wei J X (i) x 1 = x n ; (ii) x n x 1 +2 y 1 3; (iii) y 1 =2 x 1 + x 2 +1 x n x 1 +2; (iv) y 1 =2 x n x 1 + x 2 +2; (v) y 1 =1, y 2 2 x n x 1 + x 2 +2; (vi) y 1 = + x 2 x n x 1 +1, (vii) y 1 = y 2 =1 x n x 1 + x 2 +2 Proof We first show that all words from (i) are bad Let f be any word among (i) Set d =2 n i=2 x i +2 n 1 i=1 y i + x 1 1 Let 0 y2 1 0 yn 1 1 11 x1 1 10 y1 1 0 y2 1 0 y2 1 0 yn 1 1 01 x1 1 00 y1 1 0 y2 1 Note that α = β = d, α β differ in two bits The only vertices on the two shortest α, β-paths in Q d are μ = 0 y2 1 0 yn 1 1 01 x1 1 10 y1 1 0 y2 1 ν = 0 y2 1 0 yn 1 1 11 x1 1 00 y1 1 0 y2 1 Yet none of μ ν is a vertex of Q d (f) To complete the proof, we would like to show that α β are 2-critical words for Q d (f) by Proposition 15 Obviously, it suffices to show that both α β V (Q d (f)) To do this, we can check every factor of consecutive 2n 1blocks contained in α or β whether it is f by a system of equations inequalities has a solution in positive integers As f has 2n 1blocks,thereare2n 1 equations inequalities in every system The left side of an equation (inequality) is the length of a block of f, theright side is the length of the corresponding block of the compared factor from α Obviously, we only need to consider the factors that begin from the block consisting of 1s For the word α, therearen systems of equations inequalities since it has 4n 3blocks The first last are (1) x 1 x 1, x t = x t, t =2,,n 1, x n x t +1, t = n, y t = y t, t =1,,n 2, y n 1 = y n 1 1 (n) x 1 x n +1, x t = x t, t =2,,n 1, x n x n, t = n, y 1 = y 1 1, y t = y t, t =2,,n 1 The i-th (i =2,,n 1,n 3) system of equations inequalities is: x t x i, t =1, x t = x t+i 1, t =2,,n i, x t = x n +1, t = n i +1, x t = x t+i n, t = n i +2,,n 1, (i) x n x i, t = n, y t = y t+i 1, t =1,,n i 1, y t = y n 1 1, t = n i, y t = y 1 1, t = n i +1, y t = y t+i n, t = n i +2,,n 1 In the first system, the last equality is y n 1 = y n 1 1, in the last system, the (n+1)-th equality is y 1 = y 1 1, obviously those are impossible For i =2,,n 1, if we add up the

All Good (Bad) Words Consisting of 5 Blocks 855 left the right respectively of the (n + 1)-th to (2n 1)-th equations in the i-th system, then 0 = 2 It is impossible Hence all those systems of equations inequalities have no solution in positive integers, so α is f-free Similarly it can be shown that β is f-free Thus α β are 2-critical words for Q d (f) This proves that the words with x 1 = x n are bad The other cases (ii) (vii) can be treated analogously The detailed proofs are omitted to save space, but the p-critical words for Q d (f) forsomed are presented for those words For the case (ii), let 1 x2 1 xn 2 100 y1 2 1 x2 1 x2 1 xn 2 010 y1 2 1 x2 Then α β are 2-critical words for Q d (f) ford =2 n i=2 x i +2 n 1 i=1 y i + x 1 2 For the case (iii), let 0 2 1 x2 101 x2 0 2 1 x2 011 x2 Then α β are 2-critical words for Q d (f) ford =2 n 1 i=1 x i +2 n 1 i=2 y i + x n +4 For the case (iv), let 0 2 1 x2 x 2 2 101 x2 0 2 1 x2 x 2 2 011 x2 Then α β are 2-critical words for Q d (f) ford =2 n i=3 x i +2 n 1 i=2 y i + x 1 + x 2 +2 For the case (v), if n =3,let 0 0 y 2 1 x 3 x 2 2 1 00 y2 1 1 x 3 0 0 y 2 1 x 3 x 2 2 0 10 y2 1 1 x 3 ; if n 4, let 01 x2 x 2 2 1 00 y2 1 1 x3 01 x2 x 2 2 0 10 y2 1 1 x3 Then α β are 2-critical words for Q d (f) ford =2 n i=3 x i +2 n 1 i=2 y i + x 1 + x 2 For the case (vi), let 01 x2 0 yn 1 1 11 x1 1 11 x2 01 x2 0 yn 1 1 01 x1 1 01 x2 Then α β are 2-critical words for Q d (f) ford =2 n 1 i=1 x i +2 n 1 i=2 y i + x n +1 For the case (vii), we distinguish n 4n =3 If n 4, then 0 01 x3 x 2 2 1 01 x3 0 01 x3 x 2 2 0 11 x3 are 2-critical words for Q d (f) ford =2 n i=3 x i +2 n 1 i=3 y i + x 1 + x 2 +2 If n =3x 1 x 2,then 0 1 01 x 3 0 0 11 x 3 are 2-critical words for Q d (f) ford =2x 1 +2x 2 + x 3 +4 If n =3,x 1 = x 2 x 3 = x 1 + x 2 +2,then 1 11 2x1+2 1 01 2x 1+2

856 Wei J X are 3-critical words for Q d (f) ford =7x 1 +7 If n =3,x 1 = x 2 x 3 x 1 + x 2 +3,then 01 x1+1 1 01 x 3 01 x1+1 11 x 3 are 2-critical words for Q d (f) ford =4x 1 + x 3 +5 Now we give the proof of Theorem 11 Proof of Theorem 11 We first divide F into the following 13 subfamilies: x n = x 1 (1 ) { y1 2 (2 x n = x 1 +1 ) y 1 =1 (3 ) y 1 3 (4 ) x n x 1 + x 2 y 1 =2 (5 ) y 1 =1 (6 ) x n x 1 y x n x 1 +1 1 3 (7 ) x x n x 1 +2 n = x 1 + x 2 +1 y 1 =2 (8 ) y 1 =1 (9 ) y 1 3 (10 ) y x n x 1 + x 2 +2 1 =2 { (11 ) y2 2 (12 y 1 =1 ) y 2 =1 (13 ) We would like to show that the good words only exist in (2 )(9 ) to complete the proof, that is, all the words among other subfamilies are bad This process depend on Lemma 21 By (i), the words from (1 ) are bad By (vi), the words from (3 )(6 ) are bad By (ii), the words from (4 ), (7 )(10 ) are bad By (iii), the words among (5 )(8 ) are bad By (iv), the words from (11 ) are bad By (v), the words from (12 ) are bad By (vii), the words from (13 ) are bad The next lemma gives other classes of bad words from F It will be used in the next section to classify the good words consisting of 5 blocks Lemma 22 Let f F Iffhas one of the following properties, then it is bad (i ) y 1 = y n 1 +2, x n 1 x 1 x 2 +2 x n 2; (ii ) x n = x 2 +1, x n 1 +1 x 1 y i = y 1 +1, i =2,,n 1; (iii ) y 1 = y n, x n 1 +1 x 1 x 2 +1 x n ; (iv ) x n =2, y 1 y n 1 +3 x n 1 x 1 ; (v ) x 1 =1, y n 1 y 1 2 x 2 +1 x n ; (vi ) x 1 x n, x 2 = = x n, y 1 = = y n 2 y n 1 = y 1 +1, (vii ) x 1 >x 2 +1, x n >x 2 +1, x 2 = = x n 1 y 1 = = y n 1 =1 Proof By Proposition 15, for every word f above it need to give p-critical words for Q d (f) for some d The proving process is similar to Lemma 21 Here we only give the p-critical words for Q d (f) forsomed for every word, but omit the detailed proof For the case (i ), let 0 yn 1+2 1 x2 101 x2 0 yn 1+2 1 x2 011 x2 Then α β are 2-critical words for Q d (f) ford =2 n 1 i=2 (x i + y i )+x 1 + x n + y n 1 +4

All Good (Bad) Words Consisting of 5 Blocks 857 For the case (ii ), we distinguish cases n =3n =4 If n =3,let 1 1 1 x2+1 0 0 1 x2+1 ;ifn 4, let +1 1 1 10 y1 1 0 y1+1 1 x2+1 +1 1 0 00 y1 1 0 y1+1 1 x2+1 Then α β are 2-critical words for Q d (f) ford =2 n 1 i=2 x i +(2n 3)(y 1 +1)+x 1 + x 2 For the case (iii ), let 0 y2 1 10 y1 1 1 0 y2 1 0 y2 1 00 y1 1 0 0 y2 1 Then α β are 2-critical words for Q d (f) ford =2 n 1 i=2 x i +2 n 2 i=1 y i + x 1 + y 1 + x n +1 For the case (iv ), let 0 y1 1 10 y n 1 2 1 x2 1 2 0 y1 1 01 y n 1 2 1 x2 1 2 Then α β are 2-critical words for Q d (f) ford =2 n 1 i=2 (x i + y i )+x n + y 1 +2 For the case (v ), let α =10 y1 1 y 1 10 y1 1 11 x2 β =10 y1 1 y 1 00 y1 1 01 x2 Then α β are 2-critical words for Q d (f) ford =2 n 1 i=2 (x i + y i )+x n + y 1 +2 For the case (vi ), let x 1 = x, x n = z, y 1 = = y n 2 = y Then f =1 x ( z ) n 2 0 y+1 1 z Let α =1 x ( z ) n 2 1 z 1 1 z β =1 x ( z ) n 2 0 y 01 z 1 0 z Then α β are 2-critical words for Q d (f) ford = n(z + y)+x +1 For the case (vii ), let x 1 = x, x n = z x 2 = = x n 1 = y Thenf =1 x (01 y ) n 2 01 z Let 1 y (01 y ) n 3 01 y 11 z β =1 x 01 y (01 y ) n 3 01 y 01 z Then α β are 2-critical words for Q d (f) ford = x +(n 1)y + z + n 3 Good Words Consisting of 5 Blocks Theorem 12 is proven in this section, hence all good (bad) words consisting of 5 blocks are determined by Proposition 13 Recall that the family of words considered in Theorem 12 is F = {1 r 0 s 1 t 0 k 1 m r, s, t, k 1m r} Proof of Theorem 12 To prove this theorem, we would like to show that if f F is good, then f must be one of the words in Table 1, then to show that every word f in Table 1 is indeed good By Theorem 11, the good words among F satisfy s =1m = r + t +1,orm = r +1 s 2 But there still exist bad words among those cases Now we further choose the bad words among them by Lemma 22

858 Wei J X Firstwediscussthecases =1m = r +t+1, divide it into the following 4 subcases: { k 2 (1 r t ) k =1 s =1,m= t + r +1 { (2 ) k 2 (3 r t +1 ) k =1 (4 ) By (vii ), the words among (4 ) are bad Hence the words among (1 ) (3 ) may be good, ie, the words (1) (2) in Table 1 may be good Now we turn to the case m = r+1 s 2 We divide this case into 7 subcases: s k+3, s = k +2,s = k +1,s = k, k = s +1,k = s +2k s +3 For the case s k +3, if r = 1, then those words are bad by (iv ) So the words satisfy s k +3r 2, ie, (3) in Table 1 may be good For the case s = k +2, if t r, then those words are bad by (i ) So the words satisfy s = k +2r t + 1, ie, (4) in Table 1 may be good For s = k + 1, ie, (5) in Table 1 may be good For the case s = k, ift r 1, then those words are bad by (iii ) So the words satisfy s = k r t + 1, ie, (6) in Table 1 may be good For the case k = s + 1, we divide it into the following 5 subcases: r t +1 (a) t = r (b) k = s +1 t = r +1 { (c) r =1 (d) t r +2 r 2 (e) By (ii ), the words among (b) are bad By (vi ), the words among (c) are bad By (v ), the words among (d) are bad Hence the words among (a) (e), ie, (7) (8) in Table 1 may be good For k = s + 2, we divide it into the following 4 subcases: { r =1 (a ) r t r 2 (b k = s +2 ) t r +1 { r =1 (c ) r 2 (d ) By (i ), all words among (d ) are bad By (v ), all words among (a )(c ) are bad So the words among (b ), ie, (9) in Table 1 may be good For k s +3,ifr = 1, then the words are bad by (v ); if r =2t 3, then the words are bad by (iv ) So the words (10), (11) (12) in Table 1 may be good Next we show that every word in Table 1 is indeed good We only give the proof of (1) to save space since the others can be proved similarly Let f be a word from (1) We need to show that for any vertices α β of Q d (f), d Qd (f)(α, β) =d Qd (α, β), where d 1 Suppose that α = a 1 a 2 a d, β = b 1 b 2 b d, a ij b ij α j = α + e j,wherej =1,,p Without loss of generality, assume that a i1 =1 The following claim holds Claim If p 2, then there exists k {1, 2,,p} such that α k V (Q d (f))

All Good (Bad) Words Consisting of 5 Blocks 859 Proof On the contrary we suppose that α k / V (Q d (f)) for all k {1, 2,,p} Thenwecan show that p 4 by the following possible cases of the factor f contained in α 1 Case 1 α i1 r,i 1 +r+2t+2 =1 r 11 t 01 t+r+1 Since β V (Q d (f)) α 2 V (Q d (f)), i 2 [i 1 +1,i 1 + r +2t +2] As t r, there exists some u 1 such that u 1 [r, r + t] i 2 = i 1 + t + u 1 +2 Thus α i1 r,i 2 +r+2t+2 = 1 r 11 t 01 u 1 11 t 01 t+r+1 Ifp =2,thenβ i2 r,i 2 +r+2t+2 =1 r 01 t 01 t+r+1 = f, a contradiction since β V (Q d (f)) So p 3 As β V (Q d (f)), α 3 V (Q d (f)) t r, thereexistssomeu 2 such that u 2 [r, r +t] i 3 = i 2 +t+u 2 +2 Thus α i1 r,i 3 +r+2t+2 =1 r 11 t 01 u 1 11 t 01 u 2 11 t 01 r+t+1 If p =3,thenβ i3 r,i 3 +r+2t+2 =1 r 01 t 01 t+r+1 = f, a contradiction since β V (Q d (f)) So p 4 Case 2 α i1 r t 1,i 1 +r+t+1 =1 r 01 t 11 t+r+1 Since β V (Q d (f)), α 2 V (Q d (f)) r t, thereexistssomev 1 [r, r + t] such that i 2 = i 1 + v 1 + 1 Hence α i1 r t 1,i 2 +r+2t+2 = 1 r 01 t 11 v 1 11 t 01 t+r+1 If p = 2, then β i2 r,i 2 +r+2t+2 = 1 r 01 t 01 t+r+1 = f, a contradiction since β V (Q d (f)) So p 3 As β V (Q d (f)), α 3 V (Q d (f)) t r, thereexistssomev 2 such that v 2 [r, r + t], i 3 = i 2 + t + v 2 +2 Thus α i1 r t 1,i 3 +r+2t+2 =1 r 01 t 11 v 1 11 t 01 v 2 11 t 01 t+r+1 If p =3,then β i3 r,i 3 +r+2t+2 =1 r 01 t 01 t+r+1 = f, a contradiction since β V (Q d (f)) So p 4 Let n be any integer that p>n 3 Now we show that there exists some k n 1 [r, r + t] such that i n = i n 1 + k n 1 +1α in k n 1,i n +r+2t+2 =1 k n 1 11 t 01 t+r+1 Weprovethisfact by induction on n Itholdsforn = 3 by the discussion of Cases 1 2 Now we assume that it holds for n Then α in k n 1,i n +r+2t+2 =1 r 11 t 01 t+r+1 As β V (Q d (f)), α n V (Q d (f)) t r, thereexistssomek n such that k n [r, r + t], i n+1 = i n + t + k n +2 So α in k n 1,i n+1 +r+2t+2 =1 k n 1 11 t 01 k n 11 t 01 t+r+1 Hence it holds for n +1 By the above discussion, α ip r,i p +r+2t+2 =1 r 11 t 01 r+t+1 Hence β ip r,i p +r+2t+2 =1 r 01 t 01 r+t+1 = f It is a contradiction since β V (Q d (f)) Thus the claim holds Now we show that d Qd (α, β) =d Qd (f)(α, β) =p by induction on p Obviously, it is trivial for p = 1 Suppose p 2 it holds for p 1 There exists some α j V (Q d (f)) by the above claim, so p 1=d Qd (f)(α j,β) by the induction hypothesis Hence d Qd (f)(α, β) =p Thusf is good This completes the proof Acknowledgements We are grateful to the anonymous referees for many valuable comments suggestions in improving the presentation of this paper References [1] Azarija, J, Klavžar, S, Lee, J, et al: On isomorphism classes of generalized Fibonacci cubes European J Combin, 51, 372 379 (2016) [2] Azarija, J, Klavžar, S, Lee, J, et al: Connectivity of Fibonacci cubes, Lucas cubes generalized cubes Discrete Math Theoret Comput Sci, 17(1), 79 88 (2015) [3] Dedó, E, Torri, D, Salvi, N Z: The observability of the Fibonacci the Lucas cubes Discrete Math, 255, 55 63 (2002) [4] Gregor P: Recursive fault-tolerance of Fibonacci cube in hypercubes Discrete Math, 306, 1327 1341 (2006) [5] Hsu, W J: Fibonacci cubes a new interconnection topology IEEE Trans Parallel Distrib Syst, 4(1), 3 12 (1993)

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