Proofs of two conjectures on generalized Fibonacci cubes

0.22 MB·English

by  Jianxin Wei

Proofs of two conjectures on generalized
∗
Fibonacci cubes
Jianxin Weia,b, Heping Zhang a,†
5 a School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000,P. R. China
1
0 b
School of Mathematics and Statistics Science, Ludong University, Yantai, Shandong 264025,P. R. China
2

n
E-mail addresses: [e-mail], [e-mail]
a
J

2

]
O
Abstract
C
.
h A binarystring f is a factor of string u if f appears as a sequence of |f|consecu-
t
a tive bits of u, where |f| denotes the length of f. Generalized Fibonacci cube Q (f)
m d
is the graph obtained from the d-cube Q by removing all vertices that contain a
[ d

1 given binary string f as a factor. A binary string f is called good if Qd(f) is an
v
8 isometric subgraph of Qd for all d ≥ 1, it is called bad otherwise. The index of a
7
3 binary string f, denoted by B(f), is the smallest integer d such that Qd(f) is not
0
an isometric subgraph of Q . Ili´c, Klavˇzar and Rho conjectured that B(f) < 2|f|
0 d
.
1 for any bad string f. They also conjectured that if Q (f) is an isometric subgraph
d
0
5 of Qd, then Qd(ff) is an isometric subgraph of Qd. We confirm the two conjectures
1
: by obtaining a basic result: if there exist p-critical words for QB(f)(f), then p=2 or
v
i p = 3.
X
r
a
Key words: Fibonacci cube, Generalized Fibonacci cube, Isometric subgraph, Isometric
embedding, Good string
1 Introduction
Fibonacci cube Γ was introduced by Hsu [3] as a model for interconnection networks,
d
which has beneficial properties similar to those of the hypercube. It can be seen as
∗This work is supported by NSFC (Grant no. 61073046).
†Corresponding author.
1
the subgraph obtained from hypercube Q by removing all the vertices that contain two
d
consecutive 1s. As Klavˇzar and Zˇigert [10] showed that Fibonaccenes have resonance
graphs that are exactly the Fibonacci cubes. More generally, Zhang et al. [15] described
the class of planar bipartite graphs that have Fibonacci cubes as their resonance graphs.
Taranenko and Vesel [14] showed that Fibonacci cubes can be recognized in O(mlogn)
time (where m is the size and n the order of a given graph). There are more results on
the structural properties of Fibonacci cubes, see [1, 2, 7, 12], and [8] for a recent survey.
A binary string f is a factor of string u if f appears as a sequence of |f| consecutive
bits of u, where |f| denotes the length of f. Ili´c et al. [4] introduced generalized Fibonacci
cube, Q (f), as the graph obtained from Q by removing all strings that contain a given
d d
binary string f as a factor. In this notation the classical Fibonacci cube Γ is the graph
d
Q (11). The subclass Q (1s) of generalized Fibonacci cube has been studied in [3, 11, 13].
d d
A binary string f is called good if Q (f) is an isometric subgraph of Q for all d ≥ 1,
d d
bad otherwise. It was estimated that about eight percent of all strings are good [9]. The
index of a binary string f, denoted B(f), is the smallest integer d such that Q (f) is not
d
an isometric subgraph of Q . Ili´c et al. [5] showed B(f) < 2|f| for almost all bad string
d
f and thus posed the following conjecture:
Conjecture 1.1.([5]) For any bad string f, B(f) < 2|f|.
They also posed the following conjecture.
Conjecture 1.2.([4] ) If Q (f) is an isometric subgraph of Q , then Q (ff) is an iso-
d d d
metric subgraph of Q .
d
In this paper, our main aim is to confirm such conjectures. The organization of the
remainder of this paper is as follows.
In Section 2, we introduce some preliminary concepts and results of generalized Fi-
bonacci cubes. In Section 3, we obtain a basic result that if there exist p-critical words
for Q (f), then p=2 or p=3 by revealing the structures of p-critical words for Q (f).
B(f) B(f)
In section 4, we prove that both Conjectures 1.1 and 1.2 to be true.
2 Preliminaries
In this section we introduce some concepts and results of generalized Fibonacci cube need
in this paper.
For a binary string f = f f ···f let fR = f ···f f be the reverse of f and f =
1 2 d d 2 1
2
f f ···f be the binary complement of f, where f = 1−f , i = 1,...,d. Let |f| denote
1 2 d i i
the length of f. Denote the string with 1 in coordinate i and 0 elsewhere with e . For
i
strings α and γ of equal length, let α+γ denote their sum computed bitwise modulo 2.
In particular, α+e is the string obtained from α by complementing its ith bit.
i
For a connected graph G, the distance d (µ,ν) between vertices µ and ν is the length
G
of a shortest µ,ν-path. Given two binary strings α and β with the same length, their
Hamming distance H(α,β) is the number of bits in which they differ. It is known that
[6] for any vertices α and β of Q , H(α,β) = d (α,β).
d Q
d
Obviously, for any subgraph H of G, d (µ,ν) ≥ d (µ,ν). If d (µ,ν) = d (µ,ν) for
H G H G
all µ,ν ∈ V(H), then H is called an isometric subgraph of G, and simply write H ֒→ G,
and H 6֒→ G otherwise. More generally, let H and G be any arbitrary graphs. A mapping
h : V(H) → V(G) is an isometric embedding of H into G if d (µ,ν) = d (h(µ),h(ν)) for
H G
any µ and ν ∈ V(H).
For two vertices µ and ν of graph G, the set of vertices lying on shortest µ,ν-paths
is called the interval between µ and ν, denoted by I (µ,ν). Let α and β ∈ Q (f) and
G d
p ≥ 2. Then α and β are called p-critical words [4] for Q (f) if d (α,β) = H(α,β) = p,
d Q
d
but none of the neighbors of α in I (α,β) belongs to Q (f) or none of the neighbors of
Q d
d
β in I (α,β) belongs to Q (f).
Q d
d
The sufficiency of the following lemma was given by Lemma 2.4 of [4].
Lemma 2.1. Suppose that f is a nonempty binary string and d ≥ 1. Then Q (f) 6֒→ Q
d d
if and only if there exist p-critical words for Q (f) for some p ≥ 2.
d
Proof. We only show the necessity. Assume that Q (f) 6֒→ Q , then there exist vertices α
d d
and β ∈ V(Q (f)) such that d (α,β) > H(α,β) = p ≥ 2. Let α and β be such vertices
d Qd(f)
with the minimum p. Set α = a a ···a , β = b b ···b , a 6= b , and α =α+e , where
1 2 d 1 2 d ij ij j ij
j = 1,2,...,p. We claim that α 6∈ V(Q (f)) for all k ∈ {1,2,...,p}. On the contrary
k d
suppose that α ∈ V(Q (f)) for some k ∈ {1,2,...,p}, then H(α ,β) = p − 1. If
k d k
p−1 = 1 or p−1 = d (α ,β), then p = H(α,β) = d (α,β), a contradiction. So
Qd(f) k Qd(f)
2 ≤ p−1 = H(α ,β) < d (α ,β). It is a contradiction to the minimality of p. Hence
k Qd(f) k
the claim holds. Thus α and β are p-critical words for Q (f). (cid:4)
d
The following lemma holds for a bad string.
Lemma 2.2 ([9]) Suppose that Qd(f) 6֒→ Qd. Then Qd′(f) 6֒→ Qd′ for all d′ > d.
Letting f be any bad string, Q (f) ֒→ Q for only a finite number of dimensions
d d
d < B(f) by Lemma 2.2.
3
Lemma 2.3 holds obviously. It will be used repeatedly in proving the main results of
next sections.
Lemma 2.3. Let f = f f ···f be a binary string, r,s and t be positive integers such
1 2 |f|
that |f| ≥ t+ r +s. If any two equations of the three equations (1) hold, then the third
must hold,
f = f ,f = f ,f = f . (1)
t t+r t+r t+r+s t t+r+s
Lemma 2.4. Let f = f f ···f be a binary string and r,s,t be positive integers such
1 2 |f|
that |f| > r +s−g and t ≤ |f|+g −r −s, where g is the greatest common divisor of r
and s. If s+r −1 ones of the s+r equations (2) and (3) hold, then the remaining one must
g g
hold,
r
f = f ,i = 1,..., , (2)
t+(i−1)g t+(i−1)g+s
g
s
f = f ,j = 1,..., . (3)
t+(j−1)g t+(j−1)g+r
g
Proof. Letting k = r and k = s, k and k are relatively prime. We first construct the
1 g 2 g 1 2
following 2-regular bipartite graph Ghr,si with the bipartition (X,Y), where
X = {v1 | i = 1,...,k }∪{v3 | i = 1,...,k } and
i 1 i 2
Y = {v2 | i = 1,2,...,k +k }.
i 1 2
The edge set of Ghr,si is
E(G(thr,si)) = {v1v2,v1v2 | i = 1,...,k }∪{v3v2,v3v2 | k = 1,...,k }.
i i i i+k2 1 k k k k+k1 2
An example of Ghr,si that k = 10 and k = 3 is shown in Fig . 1.
1 2
v1 v1 v1 v1 v1 v1 v1 v1 v1 v1
1 2 3 4 5 6 7 8 9 10
v2 v2 v2 v2 v2 v2 v2 v2 v2 v2 v2 v2 v2
1 2 3 4 5 6 7 8 9 10 11 12 13
v3 v3 v3
1 2 3
Fig. 1. An example of the graph Ghr,si such that k = 10 and k = 3
1 2
Letting v1 denote f for i = 1,...,k , v2 denote f for j = 1,2,...,k +k
i t+(i−1)g 1 j t+(j−1)g 1 2
and v3 denote f for k = 1,...,k , a vertex in X is joined to a vertex in Y if they
k t+(k−1)g+r 2
denote some one equation of (2), (3) and (4), where (4) are the following equations,
4
r +s
f = f ,i = 1,..., . (4)
t+(i−1)g t+(i−1)g
g
Now we show that Ghr,si is connected. Obviously if this holds, then Ghr,si is a cycle,
and so this lemma is proved. Since v1 is joined to v2 for i = 1,...,k , and v3 is joined
i i+k2 1 k
to v2 for k = 1,...,k , we only need to show that any two vertices of Y are connected,
k 2
and this can be done by proving that v2 and v2 are connected for j = 1,...,k +k . Two
1 j 1 2
claims are needed.
Claim 1. Let Z = {h | h := (j − 1)k mod (k + k ), 0 ≤ h ≤ k + k and j =
j j 1 1 2 j 1 2
1,2,...,k +k }. Then Z = {0,1,...,k +k −1}, and h = k if and only if j = k +k .
1 2 1 2 j 2 1 2
Proof. First we show that Z = {0,1,...,k + k − 1}. We only need to prove that
1 2
if 1 ≤ q < q ≤ k + k , then h 6= h . Otherwise there exist m ≥ 1 such that
1 2 1 2 q2 q1
(q −q )k = m(k + k ), that is, (q −q −m)k = mk . If k = k , then k = k = 1
2 1 1 1 2 2 1 1 2 1 2 1 2
and q − q = 2m, but this is a contradiction to 1 ≤ q < q ≤ 2; if k 6= k , then
2 1 1 2 1 2
q −q = k +m and k = m, but this is a contradiction to 1 ≤ q < q ≤ k +k . Hence
2 1 2 1 1 2 1 2
Z = {0,1,...,k +k −1}.
1 2
Now we show that h = k if and only if j = k +k . Since Z = {0,1,...,k +k −1},
j 2 1 2 1 2
we only need to show the sufficiency. Suppose that (k + k − 1)k = m(k + k ) + h ,
1 2 1 1 2 j
where m ≥ 0. Then (k +k )(k −m) = k +h . Obviously, k −m ≥ 1. If k −m ≥ 2,
1 2 1 1 j 1 1
then k +h = (k +k )(k −m) ≥ 2(k +k ), it is a contradiction to h < k +k . Hence
1 j 1 2 1 1 2 j 1 2
k −m = 1 and so h = k . ✷
1 j 2
Recall that v2 is joined to v3, and v3 is joined to v2 in graph Ghr,si. Hence we get a
1 1 1 1+k1
path of Ghr,si that start from the first vertex v2 of Y and reach the second vertex v2 of
1 1+k1
Y, if we continue to walk along this path as far as possible, we have the following claim:
Claim 2. The kth vertex of Y that walking along this path is v2 , k = 1,2,...,k +k .
1+hk 1 2
Proof. Recall that k and k are relatively prime. If k = k , then Ghr,si is 4-cycle, and so
1 2 1 2
the claim holds.
Now we prove this claim by induction on k for the case k 6= k . First we show it holds
1 2
k = 1,2. In fact, the first vertex is v2 = v2 and the second one is v2 = v2 .
1 1+h1 1+k1 1+h2
We suppose that it holds for 2 ≤ k < k +k and h = (k−1)k −m(k +k ) for some
1 2 k 1 1 2
m ≥ 0. As shown in Claim 1, h 6= k , and so h < k or h > k . If h < k , then v2 is
k 2 k 2 k 2 k 2 1+hk
joined to v3 and v3 is joined to v2 . Now we show that h +k = h . In fact,
1+hk 1+hk 1+hk+k1 k 1 1+k
kk = h +k +m(k +k ). So h = kk mod (k +k ) = h +k . If h > k , then v2
1 k 1 1 2 k+1 1 1 2 k 1 k 2 1+hk
5
is joined to v1 and v1 is joined to v2 . Now we show that h −k = h .
1+hk−k2 1+hk−k2 1+hk−k2 k 2 1+k
In fact, kk = h −k +(m+1)(k +k ) and so h ≡ kk (mod(k +k )) = h −k . So
1 k 2 1 2 k+1 1 1 2 k 2
the claim holds for k +1. Hence by inductive assumption the claim holds. ✷
So by Claims 1 and 2, we show that v2 and v2 are connected for j = 1,...,k + k .
1 j 1 2
(cid:4)
This completes the proof.
Now we consider a special case of Lemma 2.4 such that r +s = |f|.
Corollary 2.5. Let f = f f ···f be a binary string and r,s be positive integers such
1 2 |f|
that |f| = r +s. Suppose f = f for i = 1,...,r and f = f for j = 1,...,s, then
i i+s j j+r
equations (2) and (3) holds for t = 1,...,g.
By the structure of the graph Ghr,si of Lemma 2.4, we have the following lemma.
Lemma 2.6. Let r,s,t, and g be the same as in Lemma 2.4.
(i) If equations (2) hold for all i ∈ {1,..., r}\{i ,i } and equations (3) hold for all
g 1 2
j ∈ {1,..., s}, then f = f and f = f .
g t+(i1−1)g t+(i2−1)g+s t+(i2−1)g t+(i1−1)g+s
(ii) If equations (2) hold for all i ∈ {1,..., r} and equations (3) hold for all j ∈
g
{1,..., s}\{j ,j }, then f = f and f = f .
g 1 2 t+(j1−1)g t+(j2−1)g+r t+(j2−1)g t+(j1−1)g+r
(iii) If equations (2) hold for all i ∈ {1,..., r} \ {i } and equations (3) hold for all
g 1
j ∈ {1,..., s}\{j }, then f = f and f = f .
g 1 t+(i1−1)g t+(j1−1)g t+(i1−1)g+s t+(j1−1)g+r
3 A basic result on p-critical words
In this section we prove that for every bad string f, the p-critical words of Q (f) satisfy
B(f)
that p = 2 or p = 3. More generally, there exist 2-critical words or 3-critical words for
Q (f), where d ≥ B(f).
d
Suppose that f is a binary string, α ∈ V(Q (f)) and α′ = α+e 6∈ V(Q (f)) for some
d j d
j ∈ {1,2,...,d}, in other words, α′ contains the factor f. We denote f(uj) as the factor f
appears in α′ starting from coordinate u . Note that if α′′ = α+e 6∈ V(Q (f)) for some
j k d
k ∈ {1,2,...,d}\{j}, then u 6= u .
j k
In the lemmas and theorems of this section, we suppose that f = f f ···f is a bad
1 2 |f|
string and d = B(f); α = a a ···a and β = b b ···b are p-critical words for Q (f),
0 1 2 d0 1 2 d0 d0
a 6= b and α = α+e 6∈ V(Q (f)) for j = 1,2,...,p.
ij ij j ij d0
Lemma 3.1. For any k ∈ {1,2,...,p}, there exists some one j ∈ {1,2,...,p}\{k} such
that i ∈ [u ,u +|f|−1].
k ij ij
Proof. We distinguish three cases: p = 2, p = 3 and p ≥ 4.
6
Obviously if p = 2, then i ∈ [u ,u +|f|−1] and i ∈ [u ,u +|f|−1], otherwise
1 i2 i2 2 i1 i1
β contains the factor f.
Let p = 3. On the contrary we distinguish the following three cases:
i 6∈ [u ,u +|f|−1] and i 6∈ [u ,u +|f|−1],
1 i2 i2 1 i3 i3
i 6∈ [u ,u +|f|−1] and i 6∈ [u ,u +|f|−1] and
2 i1 i1 2 i3 i3
i 6∈ [u ,u +|f|−1] and i 6∈ [u ,u +|f|−1].
3 i1 i1 3 i2 i2
We only show that the first and second cases do not hold since the third one can be
proved similarly as the first one. In the first case, we claim that i ∈ [u ,u +|f|−1] and
2 i3 i3
i ∈ [u ,u +|f|−1], otherwise β contains f as a factor, a contradiction. Let α′ and β′ be
3 i2 i2
the strings obtained by deleting the first to ith bit of α and β, respectively. Then α′ and
1
β′ are 2-critical words for Qd′(f), where d′ = d0 −i1, it is a contradiction to d0 = B(f).
In the second case, obviously the factor that starts from the uth to (u +|f|−1)th bit of
i1 i1
β is f, it is a contradiction.
Now we consider p ≥ 4. Assume for some one k ∈ {1,2,...,p}, i 6∈ [u ,u +|f|−1]
k ij ij
holds for all j ∈ {1,2,...,p}\{k}. Then u +|f|−1 < i for all j ∈ X = {1,...,k−1}
ij k
and u > i for all j ∈ Y = {k + 1,...,p}. Note that X = ∅ if k = 1, and Y = ∅ if
ij k
k = p. Since p ≥ 4, at least one of |X| and |Y| not less than 2. Without loss of generality,
suppose |Y| ≥ 2. Letting α′ and β′ be the strings obtained by deleting the first bit to
the itkh bit of α and β respectively, then α′ and β′ are p′-critical words for Qd′(f), where
p′ = p−k and d′ = d −i , it is a contradiction to that d = B(f). (cid:4)
0 k 0
Theorem 3.2. If there exist p-critical words for Q (f), then p = 2 or p = 3.
d0
Proof. Letting k = 1, there must exist j ∈ {2,...,p} such that i ∈ [u ,u +|f|−1] by
1 ij ij
Lemma 3.1. Suppose that j be the smallest one among all such coordinates. According
to i belongs to [u ,u +|f|−1] or not, we distinguish two cases.
j i1 i1
Case 1. i ∈ [u ,u +|f|−1].
j i1 i1
There are two subcases: u < u or u > u .
i1 ij i1 ij
Subcase 1.1. u < u .
i1 ij
Let r = u −u . Obviously, |f|−2 ≥ r ≥ 1. Let l = i −u +1 and m = i −u +1.
ij i1 1 i1 j i1
Set α′ = a′ ···a′ = a ···a . Then a′ = a and a′ = a . It follows that
1 r+|f| ui1 uij+|f|−1 l i1 m ij
α′+e and α′+e both contain the factor f . Obviously the first coordinate of f(ul) and
l m
f(um) are u = 1 and u = r+1, respectively. Since α′ is a factor of α, α′ does not contain
l m
the factor f. Let β′ = α′ + e + e . We claim that β′ contains no factor f. Obviously
l m
it holds for r = 1. On the contrary suppose that β′ contains factor f for r ≥ 2, and
denote this factor as f′. Then the first coordinate v of f′ must satisfy with 1 ≤ v ≤ r by
7
|β′| = r+|f| ≤ 2|f|−2. So this factor must contain a′ and a′ of β′. Set s′ = r+1−v
l m
and r′ = v −1. Hence there are three copies of factor f: f(r+1), f′ and f(1). They have
the relative positions as shown in Fig. 2 and satisfy that:
f(r+1) = f′ = f(1) for all t ∈ {1,...,|f|};
t t t
f(r+1) = f′ = a′ for all t ∈ {1,...,|f|−s′}\{l−r},
t t+s′ t+r
specially, f(r+1) = f′ = a′ , but f(r+1) = a′, f′ = a′ ;
m−r m−r′ m l−r l l−r′ l
f′ = f(1) , t ∈ {1,...,|f|−r′}\{m−r′},
t t+r′
specially, f′ = f(1) = a′ , but f′ = a′ , f(1) = a′ .
l−r′ l l m−r′ m m m
r'+1 s'+1
a, a,
l m f(r+1)
a, a,
l m f '
a, a,
l m f(1)
Fig. 2. Illustration of the claim that β′ contains no factor f in Subcase 1.1.
Obviously f′ = f1 ,f(r+1) = f′, a′ = f(r+1) andf(1) = a′ , inother words, equations
l−r l−s′ l−s′ l l l−r l l
(1) holds for t = l −r. By Lemma 2.3, a′ = a′ . It is impossible. Thus β′ contains no f.
l l
Hence α′ and β′ are 2-critical words for Qd′(f), where d′ = r+|f| and 1 ≤ r ≤ |f|−2.
So α = α′, β = β′ and B(f) = d′ = r + |f| by the fact that α′ is a factor of α and
d = B(f). Thus p = 2 in this subcase.
0
Subcase 1.2. u < u .
ij i1
Letting r = u −u , |f|−2 ≥ r ≥ 1. Set l = i +1−u and m = i +1−u . Denote
i1 ij 1 ij j ij
α′ = a′ ···a′ = a ···a . So a′ = a and a′ = a . It follows that α′ + e′
1 r+|f| uij ui1+|f|−1 l i1 m ij l
and α′ +e′ both contain factor f. Obviously, the first coordinate of f(ul) and f(um) are
m
u = r + 1 and u = 1, respectively. Since α′ is a factor of α, α′ contains no factor f.
l m
Let β′ = α′ + e +e . If we show that f is not a factor of β′, then α = α′, β = β′ and
l m
d′ = r +|f| = B(f). So α and β are 2-critical words for Q (f). Thus p = 2.
d
Now we assume that β′ contains factor f and denote it as f′, then r ≥ 2. Suppose
the first coordinate of f′ is v. Then 1 < v ≤ r and f′ must contain the lth and mth bits.
Of course α′ 6= α, β 6= β′ and p ≥ 3, in other words, there exist k ∈ {1,2,...,p}\{1,j}
such that i ∈ (i ,u +v +|f|−2]. In order to determine the coordinate of i , we need
k 1 ij k
the following claims.
Claim 1. r is even, m = l+ r and a′ = a′ .
2 l m
Proof. Letting r = v −1 and r = r +1−v, there are three copies of f: f(r+1), f′ and
1 2
f(1). They have the relative positions as shown in Fig. 3 and satisfy that:
8
f(r+1) = f′ = f(1) = f , t ∈ {1,...,|f|};
t t t t
f(r+1) = f′ = a′ , t ∈ {1,...,|f|−r }\{m−r},
t t+r2 t+r 2
specially, f(r+1) = f(1) = a′ , but f(r+1) = a′ , f′ = a′ ;
l−r l−r1 l m−r m m−r1 m
f′ = f(1) = a′ , t ∈ {1,...,|f|−r }\{l −r };
t t+r1 t+r1 1 1
specially, f′ = f(1) = a′ , but f′ = a′ , f(1) = a′.
m−r1 m m l−r1 l l l
r +1 r +1
1 2 a, a，
l m f(r+1)
a, a,
l m f '
a, a,
l m f(1)
Fig. 3. Illustration of the Claim 1 of Subcase 1.2.
If l+r 6= m, then equations (1) hold for t = l −r and by Lemma 2.3, a′ = f(r+1) =
1 l l−r
f(1) = a′. Obviously it is impossible. Hence l + r = m. If m − r 6= l, then equations
l l 1 2
(1) hold for t = m − r and by Lemma 2.3, a′ = f(r+1) = f(1) = a′ . Obviously it is
m m−r m m
impossible. Hence l+r = m.
2
Thus r = r = r and so r = 2r′ is even. Hence a′ = f(r+1) = f′ = f(1) = f(r+1) =
1 2 2 l l−r l−r l−r′ l−r′
f(r+1) = a′ . ✷
m−r m
By Claim 1, i − i = r/2 = r′ and a = a . Note that the forbidden factor f′
j 1 i1 ij
contained in β′ means that f is a factor of α + e + e such that starts from the bit
i1 ij
u +r′, and denote this factor as f∗.
ij
By the above discussion, we know there must exist k ∈ {1,2,...,p}\{1,j} such that
i ∈ (i ,u +r′+|f|−1]. In the following claims we determine the position of i further.
k 1 ij k
Claim 2. There exists no k such that i < i < i . In other words, j = 2.
1 k j
On the contrary suppose that there exists i such that i < i < i . Since j is the
k 1 k j
minimal one such that i ∈ [u ,u +|f|−1], u > i .
1 ij ij ik 1
r+1 s+1
a a
ik ij f(uik)
a a a
i1 ik ij f(ui1)
a a a
i1 ik ij f(uij)
Fig. 4. Illustration of the proof of Claim 2 of Subcase 1.2.
Now we consider f(uik), f(ui1) and f(uij). Letting s = uik−ui1, those three copies have
the relative positions as shown in Fig. 4 and they satisfy:
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f(uik) = f(ui1) = f(uij), t ∈ {1,...,|f|};
t t t
f(uik) = f(ui1) = a , t ∈ {1,...,|f|−s}\{i −u +1},
t t+s t+uik−1 k ik
f(uik) = a , f(ui1) = a ;
ik−uik+1 ik ik−uik+1+s ik
f(ui1) = f(uij) = a , t ∈ {1,...,|f|−r}\{i −u +1, i −u +1},
t t+r t+ui1−1 1 i1 j ij
f(ui1) = a , f(uij) = a ,
i1−ui1+1 i1 i1−ui1+1+r i1
f(ui1) = a and f(uij) = a .
ij−uij+1 ij ij−uij+1+r ij
If i −i 6= s, then the equations (1) hold for t = i −u +1 and so by Lemma 2.3,
k 1 k ik
a = a , a contradiction. If i −i = s, then i −i 6= s, and so the equations (1) hold
ik ik k 1 j 1
for t = i −u +1 and so by Lemma 2.3, a = a , a contradiction.
j ik ij ij
Hence there exists no k such that i < i < i . In other words, j = 2. ✷
1 k j
By Claim 2, we get i ∈ [i +1,u +r′ +|f|−1].
k 2 i2
Claim 3. There exists no k such that u ≤ i .
ik 1
We first prove that there exist no k such that u < u . Otherwise there exist k such
ik i2
that ik < ui2 +|f|−1. Now we consider f(ui1), f(ui2) and f(uik). They have the positions
as shown in Fig. 5 (a), where s = u −u , and they satisfy:
i2 ik
f(ui1) = f(ui2) = a , i −u +1 < t ≤ |f|−r,
t t+r t+ui1−1 2 i1
f(ui2) = f(uik) = a , 1 ≤ t < i −u +1.
t t+s t+ui2−1 2 i2
s+1 r+1
a a a
i1 i2 ik f(ui1)
a a a
i1 i2 ik f(ui2)
a a a
i1 i2 ik f(uik)
(a)
r +1 r +1
1 2
a a a
i1 i2 ik f(ui1)
a a a
i1 i2 ik f(uik)
a a a
i1 i2 ik f(ui2)
(b)
r'+1 s+1
a a a
i1 i2 ik f(uik)
a a a
i1 i2 ik f(ui1)
a a a
i1 i2 ik f*
(c)
Fig. 5. Illustration of the Claim 3 of Subcase 1.2.
Note that i −u > r, u +|f|−i −1 > s. Hence the equations (2) and (3) hold for
2 i2 i2 2
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