AbstractThe Fibonacci (p, r)-cube is an interconnection topology, which unifies a wide range of connection topologies, such as hypercube, Fibonacci cube, postal network, etc. It is known that the Fibonacci cubes are median graphs [S. Klavžar, On median nature and enumerative properties of Fibonacci-like cubes, Discrete Math. 299 (2005) 145–153]. The question for determining which Fibonacci (p, r)-cubes are median graphs is solved completely in this paper. We show that Fibonacci (p, r)-cubes are median graphs if and only if either r≤p and r≤2, or p=1 and r=n

Ou, Lifeng

Zhang, Heping

Elsevier - Publisher Connector 

Discrete Applied Mathematics 161 (2013) 441–444
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Discrete Applied Mathematics
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Note
Fibonacci (p, r)-cubes which are median graphs✩
Lifeng Ou a, Heping Zhang b,∗
a School of Mathematics and Computer Science Institute, Northwest University for Nationalities, Lanzhou, Gansu 730030, People’s Republic of China
b School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, People’s Republic of China
a r t i c l e i n f o
Article history:
Received 24 April 2012
Received in revised form 14 September
2012
Accepted 18 September 2012
Available online 9 October 2012
Keywords:
Hypercube
Fibonacci (p, r)-cube
Median graph
a b s t r a c t
The Fibonacci (p, r)-cube is an interconnection topology, which unifies a wide range of
connection topologies, such as hypercube, Fibonacci cube, postal network, etc. It is known
that the Fibonacci cubes aremedian graphs [S. Klavžar, Onmedian nature and enumerative
properties of Fibonacci-like cubes, Discrete Math. 299 (2005) 145–153]. The question for
determining which Fibonacci (p, r)-cubes are median graphs is solved completely in this
paper. We show that Fibonacci (p, r)-cubes are median graphs if and only if either r ≤ p
and r ≤ 2, or p = 1 and r = n.
© 2012 Elsevier B.V. All rights reserved.
1. Introduction
Hypercube is an interconnection network for large multicomputer systems. However, the number of nodes 2n in an
n-dimensional hypercube Qn grows rapidly as n increases. Based on Fibonacci numbers [3], a more flexible interconnection
topology called the Fibonacci cube was proposed by Hsu [4]. A Fibonacci cube can be viewed as resulting from a hypercube
after some nodes become faulty and the system is reconfigured. It has been shown that the Fibonacci cube can efficiently
emulatemany hypercube algorithms [4,17]. Structural and enumerative properties of the Fibonacci cubeswere given in [15].
There are also some results on the Fibonacci cubes; see [9] for survey.When the Fibonacci cube became increasingly popular,
manymodifications appeared [2,5,6,14,18,19], one being the Fibonacci (p, r)-cube. The Fibonacci (p, r)-cube presented in [2]
unifies a wide range of connection topologies such as the hypercube, the Fibonacci cube, etc. Its topological properties have
also been studied in [2].
A Fibonacci (p, r)-string α(p,r)n is a binary string of length n in which there are at most r consecutive ones, and at least p
zeros between two substrings composed of at most r consecutive ones, where p, r ≤ n are positive integers. For example,
01110001, 10000110, 00000010 and 00000000 are Fibonacci (3, 3)-strings of length 8. Thus, α(1,n)n is a usual binary string of
length n, and α(1,1)n is a Fibonacci string of length n. We use the notation F
(p,r)
n to denote the set of Fibonacci (p, r)-strings of
length n. Then the Fibonacci (p, r)-cube Γ (p,r)n of order n is the graphwhose vertex set is F
(p,r)
n and two vertices are adjacent if
they differ in exactly one position. Thus, Γ (1,n)n is the classical n-dimensional hypercube Qn, and Γ
(1,1)
n is the usual Fibonacci
cube Γn. In Fig. 1 we give two specific examples of Fibonacci (p, r)-cubes.
✩ This work was supported by NSFC (Grant Nos. 61073046, 11126322) and by the Fundamental Research Funds for the Central Universities (Grant No.
zyz2011076).∗ Corresponding author.
E-mail addresses: olfaff@tom.com (L. Ou), zhanghp@lzu.edu.cn (H. Zhang).
0166-218X/$ – see front matter© 2012 Elsevier B.V. All rights reserved.
doi:10.1016/j.dam.2012.09.008
442 L. Ou, H. Zhang / Discrete Applied Mathematics 161 (2013) 441–444
Fig. 1. Fibonacci (p, r)-cubes Γ (3,2)5 and Γ
(2,4)
4 .
The distance dG(u, v) between two vertices u and v in a graph G is the length of a shortest path connecting u and v. Given
two binary strings a = a1a2 · · · an and b = b1b2 · · · bn, the Hamming distance H(a, b) between a and b is the number of
positions where the two strings differ. A path connecting two vertices is called Hamming distance path if its length is equal
to the Hamming distance between these two vertices. For any two vertices a and b of Qn,H(a, b) = dQn(a, b); see [7]. Thus,
for the subgraph Γ (p,r)n of Qn, we have the inequality
H(a, b) ≤ d
Γ
(p,r)
n
(a, b). (1)
We should point out that inequality (1) may be strict. See Fig. 1 for an example. For two vertices 1001 and 1111 of Γ (2,4)4 ,
we see that H(1001, 1111) = 2 and d
Γ
(2,4)
4
(1001, 1111) = 4.
A median of vertices u, v, w of a graph G is a vertex z that simultaneously lies on a shortest u, v-path, on a shortest
u, w-path and on a shortest v,w-path. Here z can be one of the vertices u, v, w. For example, in Fig. 1, the median of 10001,
01100, 00110 in Γ (3,2)5 is 00100 and the median of 1001, 1110, 0111 in Γ
(2,4)
4 does not exist. A graph G is a median graph
if every triple of its vertices has a unique median. Median graphs are bipartite and can also be characterized as retracts of
hypercubes [1]. For more on median graphs, see [1,7,10,12,13,20]. It is well known that the hypercubes and the Fibonacci
cubes possess the median nature [7,8,11]. But Γ (2,4)4 is not a median graph. So it is natural to raise a question of determining
which Fibonacci (p, r)-cubes are median graphs. In the next section we solve the problem completely.
2. The main result
We present first two lemmas.
Lemma 1. Let a = a1a2 · · · an and b = b1b2 · · · bn be two Fibonacci (p, r)-strings of length n. Let H(a, b) = s and
aik = bik , k = 1, 2, . . . , n − s. If dΓ (p,r)n (a, b) = H(a, b) = s, then for every vertex x = x1x2 · · · xn on a shortest path
joining a and b in Γ (p,r)n , we have xik = aik = bik .
Proof. We proceed by induction on s. It is obvious for the base case s = 1. For s ≥ 2 suppose it is true for distance s− 1 and
let a(=x(0))x(1) · · · x(s−1)(x(s) =)b be any shortest path of length s inΓ (p,r)n between a and b. We use x(i)j to denote the element
of the jth position of x(i), j = 1, 2, . . . , n. Clearly,H(x(0), x(1)) = 1. Suppose that x(0)t ≠ x(1)t . Then t ≠ ik, k = 1, 2, . . . , n−s.
Otherwise, if t is some ik, then H(x(1), x(s)) = s + 1. But, by (1), H(x(1), x(s)) ≤ dΓ (p,r)n (x
(1), x(s)) = s − 1, a contradiction.
So aik = x(0)ik = x(1)ik = x(s)ik = bik and x(1)t = x(s)t = bt . Thus, H(x(1), x(s)) = dΓ (p,r)n (x
(1), x(s)) = s − 1. From the induction
hypothesis, we have x(l)ik = x(1)ik = x(s)ik , l = 2, 3, . . . , s− 1. The result follows. 
Lemma 2. Let Γ (p,r)n be the Fibonacci (p, r)-cube with r ≤ p. Then for any two vertices a = a1a2 · · · an and b = b1b2 · · · bn of
Γ
(p,r)
n ,H(a, b) = dΓ (p,r)n (a, b).
Proof. Let H(a, b) = s and i1 < i2 < · · · < is be the positions in which a and b differ. By induction on swe prove that there
is a Hamming distance path between a and b in Γ (p,r)n . For the case s = 1 the result is obvious. Since ai1 ≠ bi1 , without loss
of generality we may assume ai1 = 1 and bi1 = 0.
L. Ou, H. Zhang / Discrete Applied Mathematics 161 (2013) 441–444 443
If i1 = 1 or ai1−1 = bi1−1 = 0, let a′ be a binary string obtained from a by changing ai1 from one to zero. Then a′ is
a Fibonacci (p, r)-string adjacent to a in Γ (p,r)n , and H(a′, b) = s − 1. By the induction hypothesis, s − 1 = H(a′, b) =
d
Γ
(p,r)
n
(a′, b). Hence there is a path of length s connecting a and b in Γ (p,r)n , and H(a, b) ≥ dΓ (p,r)n (a, b). So inequality (1)
implies the lemma.
Assume that ai1−1 = bi1−1 = 1. Then 2 ≤ r ≤ p since ai1−1 = ai1 = 1. Suppose that ai1−1 = ai1 = ai1+1 = · · · = ai1+t =
1, and ai1+t+1 = 0 or i1+t = n, t ≥ 0. Then t+2 ≤ r ≤ p. Since b is a Fibonacci (p, r)-string, bi1 = bi1+1 = · · · = bi1+t = 0,
and bi1+t+1 = 0 or i1 + t = n. As H(a, b) = s, it follows that t + 1 ≤ s. Replacing position i1 + t of a with 0 yields a
Fibonacci (p, r)-string a′ = a1 · · · ai1−211 · · · 100ai1+t+2 · · · an. Then a and a′ are adjacent, and H(a′, b) = s − 1. By the
induction hypothesis, s − 1 = H(a′, b) = d
Γ
(p,r)
n
(a′, b). Hence there is a path of length s connecting a and b in Γ (p,r)n , and
H(a, b) ≥ d
Γ
(p,r)
n
(a, b). So inequality (1) implies the lemma. 
Next, we introduce the majority rule [7,8]. Let a = a1a2 · · · an, b = b1b2 · · · bn and c = c1c2 · · · cn be arbitrary binary
strings. Noting that ai, bi, ci ∈ {0, 1}, two of ai, bi, ci must be equal. Thus, we can obtain a unique binary string z = z1z2 · · · zn
by using the majority rule: zi is equal to the element that appears at least twice among ai, bi and ci.
We now turn to our main result.
Theorem 3. Γ (p,r)n is a median graph if and only if either r ≤ p and r ≤ 2, or p = 1 and r = n.
Proof. To prove that Γ (p,r)n is a median graph for r ≤ p and r ≤ 2, we show that every three vertices of Γ (p,r)n have exactly
one median. Let a = a1a2 · · · an, b = b1b2 · · · bn and c = c1c2 · · · cn be any three vertices of Γ (p,r)n . By the majority rule, we
can uniquely obtain a binary string z = z1z2 · · · zn, which is a median vertex in hypercube Qn. As we shall see, z is exactly
the only median of a, b, c also in Γ (p,r)n for r ≤ p, r ≤ 2.
We first claim that z is a vertex of Γ (p,r)n . Let xn denote the concatenation of string x with itself done recursively
n-times. Suppose that z = z1 · · · zi−11r+1zi+r+1 · · · zn = z1 · · · zi−111r−11zi+r+1 · · · zn. Then the majorities of ai, bi, ci and
ai+r , bi+r , ci+r are equal to 1. Thus, there is at least one of the vertices a, b, c in which both position i and position i+ r are
equal to 1, say ai = ai+r = 1. Since r ≤ p, we have ai+1 = ai+2 = · · · = ai+r−1 = 1. Hence a = a1 · · · ai−11r+1ai+r+1 · · · an,
which is impossible. So z does not contain more than r consecutive 1’s.
Suppose that z = z1 · · · zi−110t1zi+t+2 · · · zn for 1 ≤ t < p. Then the majorities of ai, bi, ci and ai+1, bi+1, ci+1 are 1 and 0,
respectively. Thus, there is at least one of the vertices a, b, c in which position i and position i+ 1 are 1 and 0, respectively,
say ai = 1 and ai+1 = 0. Since t < p, we have ai+2 = ai+3 = · · · = ai+t+1 = 0. Then bi+t+1 and ci+t+1 must be 1 because
zi+t+1 = 1, that is, the majority of ai+t+1, bi+t+1, ci+t+1 is 1. Again, since zi = 1, that is, the majority of ai, bi, ci is 1, at least
one of bi and ci must be 1. Without loss of generality, we may assume that bi = 1. As t < p, bi+1 = bi+2 = · · · = bi+t = 1.
Hence b = b1 · · · bi−11t+2bi+t+2 · · · bn, which is impossible since r ≤ 2 < 3 ≤ t + 2. This implies that in z there are at least
p zeros between two substrings composed of at most r consecutive ones. The claim holds.
It remains to prove that z simultaneously lies on a shortest a, b-path, a shortest a, c-path, and a shortest b, c-path.
Suppose that H(a, z) = p,H(z, b) = q, and H(a, b) = s. Since z is chosen by the majority rule, we deduce that if ai = bi,
then zi = ai = bi; if ai ≠ bi, then ai ≠ zi if and only if zi = bi. Therefore, s − p = q, that is p + q = s. From Lemma 2, we
know that z lies on a shortest a, b-path. Similarly, z lies on a shortest a, c-path, and a shortest b, c-path. This shows that z is
a median of a, b and c in Γ (p,r)n .
Suppose that z ′ is another median of a, b and c. By Lemma 2, for any two vertices of Γ (p,r)n , their Hamming distance is
equal to their distance in Γ (p,r)n . Then from Lemma 1, we know that z ′ satisfies the majority rule. Thus z ′ = z. The proof of
the uniqueness is complete.
For p = 1 and r = n,Γ (1,n)n = Qn. Proposition 1.29 of [7] tells us that every hypercube is a median graph.
To see that for all the remaining cases Fibonacci (p, r)-cubes are not median graphs, we will give three vertices whose
median does not exist. We distinguish three cases as follows:
Case 1: 3 ≤ r ≤ p. Choose a = 1000n−3, b = 0010n−3 and c = 1110n−3 in Γ (p,r)n . By Lemmas 1 and 2, if a, b and c have a
median, then it must satisfy the majority rule, and hence 1010n−3 is the only candidate for the median. But 1010n−3 ∉ F (p,r)n
since p ≥ 3. Thus a, b and c have no median.
Case 2: p < r < n. Choose a = 11r−100n−r−1, b = 01r−110n−r−1 and c = 10r−110n−r−1 in Γ (p,r)n . Since H(a, b) = 2 and
there is a Hamming distance path
P1 = a(01r−100n−r−1)b
in Γ (p,r)n , by (1), dΓ (p,r)n (a, b) = 2. Since H(a, c) = r and there is a Hamming distance path
P2 = a(11r−2020n−r−1)(11r−3030n−r−1) · · · (10r−100n−r−1)c
in Γ (p,r)n , by (1), dΓ (p,r)n (a, c) = r . Since H(b, c) = r and there is a Hamming distance path
P3 = b(021r−210n−r−1)(031r−310n−r−1) · · · (00r−110n−r−1)c
444 L. Ou, H. Zhang / Discrete Applied Mathematics 161 (2013) 441–444
inΓ (p,r)n , by (1), dΓ (p,r)n (b, c) = r . From Lemma 1, if a, b and c have amedian, then it must satisfy themajority rule, and hence
11r−110n−r−1 = 1r+10n−r−1 is the only candidate for the median. But 1r+10n−r−1 ∉ F (p,r)n . So a, b and c have no median.
Case 3: 2 ≤ p < r = n. Notice that Γ (n−1,n)n = Γ (n,n)n . As we have seen in Case 1, Γ (n,n)n (n ≥ 3) is not a median graph. So
we may assume that p ≤ n− 2. Choose a = 11p00n−p−2, b = 01p10n−p−2 and c = 10p10n−p−2 in Γ (p,r)n . Since H(a, b) = 2
and there is a Hamming distance path
P1 = a(01p00n−p−2)b
in Γ (p,r)n , by (1), dΓ (p,r)n (a, b) = 2. Since H(a, c) = p+ 1 and there is a Hamming distance path
P2 = a(11p−10100n−p−2)(11p−20200n−p−2) · · · (10p00n−p−2)c
in Γ (p,r)n , by (1), dΓ (p,r)n (a, c) = p+ 1. Furthermore, we note that P2 is a unique shortest path joining a and c in Γ
(p,r)
n . Since
H(b, c) = p+ 1 and there is a Hamming distance path
P3 = b(001p−110n−p−2)(0021p−210n−p−2) · · · (00p10n−p−2)c
in Γ (p,r)n , by (1), dΓ (p,r)n (b, c) = p + 1. From Lemma 1, if a, b and c have a median, then it must satisfy the majority rule,
and hence 11p10n−p−2 is the only candidate for the median. But 11p10n−p−2 does not lie on P2. Thus a, b and c have no
median. 
3. Remarks
Theorem 3 implies that hypercube Qn, Fibonacci cube Γ
(1,1)
n , and the Fibonacci (2, 2)-cube Γ
(2,2)
n are median graphs.
Moreover, the other median graphs from a class of Fibonacci (p, r)-cubes are Γ (p,1)n (p ≥ 2) and Γ (p,2)n (p ≥ 3), which are
subgraphs of Fibonacci cube Γ (1,1)n and Γ
(2,2)
n , respectively.
Furthermore, Lemma 2 and Theorem 3 imply that if a Fibonacci (p, r)-cube Γ (p,r)n is a median graph, then the distance
between any two vertices is Hamming distance. The converse, however, is not true. For example, see Fibonacci (p, 3)-cube
Γ
(p,3)
n (p ≥ 4).
The Z-transformation graph of a planar (bipartite) graph G, denoted by Z(G), is such a graphwhose vertices are the perfect
matchings ofG, two perfectmatchings being adjacent if their symmetric difference is exactly a cycle ofG that is the boundary
of an interior face of G. Ref. [20] showed that every connected Z-transformation graph of planar bipartite graph is a median
graph. In [16, Theorem 5] it is shown that the Fibonacci (p, r)-cube Γ (p,r)n is the Z-transformation graph of a planar graph
if and only if Γ (p,r)n is one of the Qn,Γ
(1,1)
n ,Γ
(2,2)
n ,Γ
(2,r)
2 (r = 1, 2),Γ (p,r)3 (p ≥ 2, r = 1, 2),Γ (p,2)4 (p ≥ 3) or Γ (3,2)5 . From
Theorem 3 and the above fact we can see that if Γ (p,r)n is the Z-transformation graph of a planar graph, then it is a median
graph. However, not all Fibonacci (p, r)-cubes with the median property are the Z-transformation graphs of planar graphs.
For instance, see Γ (2,1)n (n ≥ 4).
Acknowledgments
The authors are grateful to the referees for their careful reading and many valuable suggestions.
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Fibonacci (p, r)-cubes which are median graphs 

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Fibonacci (p, r)-cubes which are median graphs

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