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Proper distance in edge-colored hypercubes
Applied Mathematics and Computation 313 (2017) 384–391
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Proper distance in edge-colored hypercubes Eddie Cheng a, Colton Magnant b,∗, Dhruv Medarametla c a b c
Department of Mathematics and Statistics, Oakland Universityi, Rochester, MI 48309, USA Department of Mathematical Sciences, Georgia Southern University, Statesboro, GA 30460, USA Stanford University, Stanford, CA 94305, USA
a r t i c l e
i n f o
Keywords: Edge-coloring Proper connection Hypercube
a b s t r a c t An edge-colored path is called properly colored if no two consecutive edges have the same color. An edge-colored graph is called properly connected if, between every pair of vertices, there is a properly colored path. Moreover, the proper distance between vertices u and v is the length of the shortest properly colored path from u to v. Given a particular class of properly connected colorings of the hypercube, we consider the proper distance between pairs of vertices in the hypercube. © 2017 Elsevier Inc. All rights reserved.
1. Introduction All graph colorings in this work are colorings of the edges. A coloring of a graph (or subgraph) is called proper if no two adjacent edges in the graph (respectively subgraph) share a color. A colored graph is called properly connected if between every pair of vertices, there is a properly colored path. In recent years, there have been numerous works on proper connectivity and the proper connection number. See the dynamic survey [] for more details. The study of proper connection in graphs is motivated in part by applications in networking and network security. In addition to security, of course it is critical that message transmission in a network be fast. For this reason, we consider shortest properly colored paths in edge-colored graphs. Because of its abundant symmetry and numerous applications to network design and reliability (see [1,3] for two of many examples) we consider colorings of the hypercube. For n ≥ 2, let Hn be the n-dimensional hypercube with vertex set being the set of binary strings of length n and edges between pairs of vertices whose strings differ in exactly one position. Given 1 ≤ j < n, define the (j)-coloring of Hn to be the 2-coloring of the edges of Hn in which all edges in dimension i where 1 ≤ i ≤ j have color 1 and all other edges have color 2. For example, the (1)-coloring of Hn has a perfect matching in color 1 and all other edges in color 2. For the sake of notation, given two vertices u and v in a hypercube, let o(u, v ) denote the number of positions i for 1 ≤ i ≤ j in which u and v differ and let t (u, v ) denote the number of positions i for j < i ≤ n in which u and v differ. Let γ (u, v ) be the indicator variable which takes the value 1 if o(u, v ) + t (u, v ) is odd and 0 if o(u, v ) + t (u, v ) is even. Given a properly connected coloring of a graph, the work of Coll et al. [2] initiated the study of the proper distance between pairs of vertices. The proper distance between two vertices u and v, denoted by pd (u, v ), is the minimum length (number of edges) of a properly colored path between u and v. Although the proper connection number has been studied and applications to computer science and networking have been described, precious little attention has been paid to the proper distance which is, in some sense, a measure of the efficiency of the properly connected network. ∗
Corresponding author. E-mail addresses: [email protected], [email protected] (C. Magnant).
http://dx.doi.org/10.1016/j.amc.2017.05.065 0 096-30 03/© 2017 Elsevier Inc. All rights reserved.
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
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In Section 2, we consider the proper distance between two arbitrary vertices of a (j)-colored hypercube and in Section 3, we count the number of different shortest proper paths between two vertices in the (1)-colored hypercube. In Section 4, we count the number of vertices at each proper distance from a given fixed point in the (j)-colored hypercube. Finally in Section 5, we find the expected proper distance between two randomly selected vertices in a (j)-colored hypercube. 2. Proper distance We prove the following result which establishes the proper distance between any pair of vertices based on the number of positions in which they differ and which color those dimensions are assigned. Theorem 1. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then H is properly connected. Furthermore, let u and v be any pair of vertices in H. Then
pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ) Proof. We first constructively show that H is properly connected, and then show the desired result, that pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ). Note that in the hypercube, any path from u to v consists of flipping the bits of u until we arrive at v. A properly colored path from u to v then consists of flipping the bits of u until we arrive at v with the restriction that the bits we flip must alternate between the first j bits and the last n − j bits. To construct such a properly colored path, flip the bits of u that differ with the respective bits of v, making sure to flip the bits in an order that alternates between the first j bits and the last n − j bits, until you arrive at a vertex u such that either o(u , v ) = 0 or t (u , v ) = 0. If o(u , v ) = 0, we alternate flipping the first bit with flipping the remaining bits in the last n − j positions that differ with the respective bits of v. Otherwise if t (u , v ) = 0, then we alternate flipping the last bit with flipping the remaining bits in the first j positions that differ with the respective bits of v. Since there are a finite number of positions in which u and v differ and the process will not repeat vertices, this process will eventually terminate, resulting in a properly colored path from u to v. See Example 1 for an example of this process. Example 1. If n = 7 and j = 4 and we consider u = (0, 1, 0, 1, 0, 0, 0 ), and v = (0, 0, 1, 1, 1, 1, 1 ), then some examples of proper colored paths from u to v created by the algorithm above would be as follows:
u
=
u
=
v
=
( 0, 1, 0, 1, 0, 0, 0 ) → ( 0, 0, 0, 1, 0, 0, 0 ) → ( 0, 0, 0, 1, 1, 0, 0 ) → ( 0, 0, 1, 1, 1, 0, 0 ) → ( 0, 0, 1, 1, 1, 1, 0 ) → ( 1, 0, 1, 1, 1, 1, 0 ) → ( 1, 0, 1, 1, 1, 1, 1 ) → ( 0, 0, 1, 1, 1, 1, 1 )
u
=
u
= =
v
( 0, 1, 0, 1, 0, 0, 0 ) → ( 0, 1, 0, 1, 1, 0, 0 ) → ( 0, 0, 0, 1, 1, 0, 0 ) → ( 0, 0, 0, 1, 1, 1, 0 ) → ( 0, 0, 1, 1, 1, 1, 0 ) → ( 0, 0, 1, 1, 1, 1, 1 )
We now prove that pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ). Consider a properly colored path from u to v and label the vertices of this path as u = u0 → u1 → u2 → · · · → ud = v. In this path, a total of d bits have been flipped, counting with repetition. Also, flipping a bit an even number of times brings it back to its original position. Because u and v initially differed in o(u, v ) + t (u, v ) bits, we get d ≡ o(u, v ) + t (u, v ) ≡ o(u, v ) − t (u, v ) ≡ |o(u, v ) − t (u, v )| ≡ γ (u, v ) mod 2. Thus, we get d ≡ γ (u, v ) mod 2. Certainly the path contains at least o(u, v ) bits flipped from the first j bits and at least t (u, v ) bits flipped from the last n − j bits. Since, in order to be a properly colored path, the flipping must alternate between the first j bits and the last n − j bits, there must be at least 2 max{o(u, v ), t (u, v )} − 1 bits flipped in total. Thus, it is clear that d ≥ 2 max{o(u, v ), t (u, v )} − 1. Now, because 2 max{o(u, v ), t (u, v )} − γ (u, v ) is the smallest positive integer that is both at least 2 max{o(u, v ), t (u, v )} − 1 and is also congruent to γ (u, v ) mod 2, it is clear that pd (u, v ) ≥ 2 max{o(u, v ), t (u, v )} − γ (u, v ). Thus, we must now show that a properly colored path between u and v of length 2 max{o(u, v ), t (u, v )} − γ (u, v ) exists. Without loss of generality, assume o(u, v ) ≥ t (u, v ), so we must show that pd (u, v ) = 2o(u, v ) − γ (u, v ). Let a1 , a2 , . . . , ao(u,v ) be the o(u, v ) bits in which u and v differ with 1 ≤ a1 < a2 < · · · < ao(u,v ) ≤ j, and let b1 , b2 , . . . , bt (u,v ) be the t (u, v ) bits in which u and v differ in the last n − j bits with j < b1 < b2 < · · · < bt (u,v ) ≤ n. We now formalize a method for creating a path from u to v of length 2o(u, v ) − γ (u, v ). First flip the bit of u in position a1 , then position b1 , then position a2 , and so on, until you have flipped the bit in position bt (u,v ) . This is a total of 2t (u, v ) edges in the path from u to v. If o(u, v ) = t (u, v ), then you have gone from u to v in 2o(u, v ) − γ (u, v ) = 2t (u, v ) edges, and you are done. Otherwise, call the vertex that you are on u . By definition, u will only differ from v in positions at (u,v )+1 , . . . , ao(u,v ) . Then, continuing the path from before, and keeping in mind that the last edge in the path was of color 2, flip bit at (u,v )+1 , then bit n, then bit at (u,v )+2 , then bit n, and so on, until you flip bit ao(u,v ) . This is a total of 2(o(u, v ) − t (u, v )) − 1 edges, meaning that you have used a total of 2o(u, v ) − 1 edges in total. Let u be the terminal vertex of this path. If u = v, then the only position where it can differ from v in is bit n. This only occurs if 2
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divides o(u, v ) − t (u, v ), in which case we flip the bit in position n to complete the desired path. This necessarily produces a properly colored path from u to v of length 2o(u, v ) − γ (u, v ), completing the proof. Example 2. If n = 7 and j = 3 and we consider u = (0, 0, 0, 1, 0, 1, 0 ), and v = (1, 1, 1, 1, 1, 0, 0 ), then a proper colored path from u to v created by the algorithm above would be as follows:
u = ( 0, 0, 0, 1, 0, 1, 0 ) →
( 1, 0, 0, 1, 0, 1, 0 ) → ( 1, 0, 0, 1, 1, 1, 0 ) → ( 1, 1, 0, 1, 1, 1, 0 ) → ( 1, 1, 0, 1, 1, 0, 0 ) → v = ( 1, 1, 1, 1, 1, 0, 0 ) 3. Number of proper paths between vertices Assuming a properly connected coloring of a graph H, recall that pd (u, v ) is the proper distance between vertices u and v and let pp(u, v ) denote the number of distinct shortest properly colored paths in H from u to v. Theorem 2. Let n ≥ 2 and let H be the (1)-coloring of Hn . Then H is properly connected. Furthermore, let u and v be any pair of distinct vertices whose strings differ in t = t (u, v ) positions other than the first. If t = 0, then u and v are adjacent so pd (u, v ) = 1 and pp(u, v ) = 1. Otherwise, for t ≥ 1, we have pd (u, v ) = 2t − γ (u, v ) and
ppH (u, v ) = (2 − γ (u, v ))(t! ). Proof. Without loss of generality, let v = (0, 0, . . . , 0 ) and let u be an arbitrary vertex of Hn \ {v}. If u and v are adjacent, the edge between them is the only shortest proper path from u to v so there is nothing to prove. So suppose u and v are not adjacent. In particular, this means that t ≥ 1. Then t is precisely the number ‘1’s in u in positions other than the first, and γ (u, v ) = 1 if the total number of ‘1’s in u is odd and γ (u, v ) = 0 otherwise. In the (1)-coloring H of Hn , every proper path from u to v must use alternating edges in color 1. This means that every shortest proper path from u to v must alternate between switching 1 to 0 in a position other than the first position and switching the first position. First suppose γ (u, v ) = 0 so there is an even total number of positions in which u and v differ. Let π be an arbitrary ordering of the t ‘1’s in u in positions other than the first. For any choice of π , the sequence of switches of positions 1, π (1 ), 1, π (2 ), . . . , 1, π (t ) and the sequence of switches of positions π (1 ), 1, π (2 ), 1, . . . , π (t ), 1 both constitute shortest proper paths from u to v and are all distinct. Clearly these paths have length 2t = 2t − γ (u, v ). Also there can be no other shortest proper path from u to v because every proper path from u to v must include all t of the ‘1’s and must also alternate switching position 1. Since there are t! such orderings, there are a total of 2(t! ) = (2 − γ (u, v ))(t! ) distinct shortest proper paths from u to v. Finally suppose γ (u, v ) = 1 so there is an odd total number of positions in which u and v differ. Then every shortest proper path will start on an edge of color 2, by switching 1 to 0 in some position other than the first. Again letting π be an arbitrary ordering of the t ‘1’s in u in positions other than the first, for any choice of π , the sequence of switches of positions π (1 ), 1, π (2 ), 1, . . . , 1, π (t ) constitutes a shortest proper path from u to v and all are distinct. These paths have length 2t − 1 = 2t − γ (u, v ) and there is no other shortest proper path from u to v. Since there are t! such orderings, there are a total of t! = (2 − γ (u, v ))(t! ) distinct shortest proper paths from u to v. 4. Counting vertices at each proper distance Now, given H the (j)-coloring of Hn , we define the surface area SAn, j (k ) = #{v| pd (u, v ) = k}, where u is any arbitrary vertex of H. Note that the value of SAn, j (k) does not depend on the choice of u due to the symmetry of the hypercube. Theorem 3. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then
SAn, j (k ) =
where
a b
x+y≡k mod 2 max{x,y}= k+1 2
j x
n− j y
is defined to be 0 for b < 0 or b > a.
Observe that if k = 0, we get SAn, j (0 ) = 1 as one might expect. Although this is not a closed form, it is an explicit formula, which is computable in a polynomial number of steps as a function of n and k. Proof. Let u be any arbitrary vertex. Then, we must find the number of vertices v ∈ V (H ) such that pd (u, v ) = k. By Theorem 1, this is the number of vertices v such that 2 max{o(u, v ), t (u, v )} − γ (u, v ) = k. In addition, note that
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
387 k+γ (u,v )
2 max{o(u, v ), t (u, v )} − γ (u, v ) = k implies that γ (u, v ) ≡ k mod 2. Therefore, max{o(u, v ), t (u, v )} = = k+1 2 2 , as 0 ≤ γ (u, v ) ≤ 1. As γ (u, v ) ≡ |o(u, v ) − t (u, v )| ≡ o(u, v ) − t (u, v ) ≡ o(u, v ) + t (u, v ) mod 2, we have that o(u, v ) + t (u, v ) ≡ k mod 2. Therefore, we now know that for pd (u, v ) = k, we must have
max{o(u, v ), t (u, v )} =
k+1 2
and o(u, v ) + t (u, v ) ≡ k mod 2. In addition, note that these are sufficient conditions; if both of those conditions are true, then pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ) = 2 k+1 2 − γ (u, v ). If k is odd, then o(u, v ) + t (u, v ) ≡ k mod 2 implies that γ (u, v ) = 1, so pd (u, v ) = 2 k+1
− γ ( u, v ) = k + 1 − 1 = k, and if k is even, o(u, v ) + t (u, v ) ≡ k mod 2 implies 2
that γ (u, v ) = 0, so pd (u, v ) = 2 k+1 2 − γ (u, v ) = k + 0 − 0 = k. Therefore, pd (u, v ) = k if and only if max{o(u, v ), t (u, v )} = k+1
and o ( u, v ) + t ( u, v ) ≡ k mod 2. 2 Given two integers x and y, we need to count the number of vertices v satisfying o(u, v ) = x and t (u, v) = y. Since exactly j x of the first j bits must differ from the first j bits of u, the number of choices for the first j bits of v is . Similarly since x
exactly y of the last n − j bits must differ from the last n − j bits of u, the number of choices for the last j bits is
These imply that the total number of such vertices v is
j x
n− j . y
n− j . y
Because pd (u, v ) = k if and only if max{o(u, v ), t (u, v )} = k+1 2 and o(u, v ) + t (u, v ) ≡ k mod 2, we can write that
SAn, j (k ) =
#{v|o(u, v ) = x, t (u, v ) = y}
x+y≡k mod 2 max{x,y}= k+1 2
which then implies that
SAn, j (k ) =
n− j y
j x
x+y≡k mod 2 max{x,y}= k+1 2
by the previous paragraph, completing the proof.
Corollary 4. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Without loss of generality, assume j ≤ n − j. Then if k > 2j,
SAn, j (k ) = 2
j−1
n− j
Proof. By Theorem 3,
SAn, j (k ) =
.
k+1
2
x+y≡k mod 2 max{x,y}= k+1 2
n− j . y
j x
2 j+2 The assumption that k > 2j implies that k ≥ 2 j + 1. If x = k+1 2 , then we have x ≥ 2 = j + 1 > j, so
if x = k+1 2 , then
jn− j x
y
j x
x+y≡k mod 2 max{x,y}= k+1 2
x+y≡k mod 2 y= k+1 2
=
Since
n− j
x≡0 mod 2
x≡0 mod 2
j x
or
j x
+
j x
k+1
2
n− j y
=
x
= 0. Therefore,
= 0, which simplifies the calculation for surface area to the following:
SAn, j (k ) =
j
n− j y
x+
k+1 2
≡k mod 2
x≡1 mod 2
x≡1 mod 2
j x
=
j . x
x∈Z
j x
=
2 j,
it is clear that
x+ k+1
≡k mod 2 2
j x
is equivalent to either
j . We observe that these two sums are equal, each being equal to 2 j−1 , which holds x
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E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
because j
1
( j−x )
(−1 )
j x
x
x=0
Therefore,
SAn, j (k ) =
n− j
= ( 1 − 1 ) j = 0.
k+1
2
j x
x+ k+1 2 ≡k mod 2
= 2 j−1
n− j
k+1
2
,
completing the proof.
Theorem 5. Let j ≥ 1, and H be the (j)-coloring of H2j . Then, for integers a and b with a + b = 2 j − 1 + (−1 )a , we have
SA2 j, j (a ) + SA2 j, j (b) = 2 j
j
a+1
2
Proof. Note that a + b = 2 j − 1 + (−1 )a ≡ 0 mod 2 (since if a is even, a + b = 2 j; otherwise a is odd and hence a + b = a 2 j+(−1 )a −a 2 j − 2). Also, b+1
= j + (−12) −a = j − a+1 2 = 2 2 . Utilizing Theorem 3, we have that
SA2 j, j (a ) =
j x
x+y≡a mod 2 max{x,y}= a+1 2
and
SA2 j, j (b) =
j x
x+y≡b mod 2 max{x,y}= b+1 2
x+y≡a mod 2 max{x,y}= j− a+1 2
=
j y
=
j y
j x
x+y≡a mod 2 max{x,y}= j− a+1 2
( j−x )+( j−y )≡a mod 2 min{ j−x, j−y}= a+1 2
j j−y
j j−x
j j−y
=
j j−x
=
j y
j x
(x )+(y )≡a mod 2 min{x ,y }= a+1 2
j . y
Therefore,
SA2 j, j (a ) + SA2 j, j (b)
=
x+y≡a mod 2 max{x,y}= a+1 2
= x+
=
=
a+1 2
≡a mod 2
j
a+1 2
j
j x
j y
j x
a+1
2
(x )+(y )≡a mod 2 min{x ,y }= a+1
j
x+
≡a mod 2
j x
(2 j−1 + 2 j−1 ) = 2 j
j x
2
+
a+1
2
a+1 2
+
a+1 2
+y≡a mod 2
+
j
j y
a+1 2
+y≡a mod 2
j
a+1
2
j y
j y
a+1
2
due to the properties of the sum of alternating binomial coefficients, completing the proof.
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
389
5. Expected proper distance In addition to surface area, we also consider the expected proper distance between two randomly selected vertices. First two technical lemmas. Lemma 1.
x+y≡1 mod 2
Proof.
j x
x+y≡1 mod 2
=
= 2n−1 .
n− j y
j x
x≥0
n− j y
j x
n− j y
y≡1+x mod 2
=
j n− j−1 2 x
x≥0
= 2 j 2n− j−1 = 2n−1 . Lemma 2.
x,y≥0
Proof.
= =
n− j y
k
k≥0
n k
= n2n−1 +
+
= n2
+
j x
= n2
+
|y + ( j − x ) − j | |y + z − j |
y,z≥0
= n2n−1 +
|i − j |
i≥0
= n2n−1 +
i≥0
n− j y
y+z=i
|i − j |
n− j y
j j−x
j z
n− j y
j z
n− j y
j x
|x − y|
n− j y
x,y≥0 n−1
j x
n . i
x,y≥0
|x − y|
x,y≥0 n−1
+
|x − y|
x,y≥0
|i − j |
i=0
j (x + y ) x
n
= n2n−1 +
n− j y
2 max{x, y}
x,y≥0
j x
x,y≥0
n− j y
j x
2 max{x, y}
n− j y
n i
by Vandermonde’s Identity and other well-known facts about properties of sums of binomial coefficients. We are now able to prove our main result of this section. Theorem 6. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then
n n−1 1 n E[ pd (u, v )] = + n |i − j | 2 2 i i=0
where u and v are vertices of H that are chosen independently and randomly.
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E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
Proof. Note that we can fix u without affecting the value of E[ pd (u, v )] due to the symmetry of the hypercube. Let u be any arbitrary vertex of H; then, because there are 2n vertices in H, we have that
E[ pd (u, v )] =
∞ 1 1 pd ( u, v ) = kSAn, j (k ). 2n 2n
v∈V (H )
k=0
By Theorem 3, we have that
E[ pd (u, v )] =
∞ ∞ 1 1 kSAn, j (k ) = n k n 2 2 k=0
k=0
x+y≡k mod 2 max{x,y}= k+1 2
n− j . y
j x
Through algebraic manipulation, we can see that
⎛
∞ 1⎜ ⎜ k 2n ⎝ k=0
1 = n 2
x+y≡k mod 2 max{x,y}= k+1 2
j x
n− j y
2 max{x, y}
⎟ ⎟ ⎠
n− j y
j x
x,y≥0
⎞
1 − n 2
Utilizing Lemmas 1 and 3, we see that
1 E[ pd (u, v )] = n 2
=
1 2n
2 max{x, y}
x,y≥0
n2n−1 +
n
n− j y
n i
|i − j |
i=0
n−1 1 = + n 2 2 completing the proof.
n
j x
−
x+y≡1 mod 2
1 − n 2
j x
n− j y
.
x+y≡1 mod 2
j x
n− j y
1 n−1 (2 ) 2n
n i
|i − j |
i=0
Corollary 7. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then
1 2
E[ pd (u, v )] ≥ max{n − j, j} −
where u and v are vertices of H that are chosen independently and randomly. Proof. We first show that E[ pd (u, v )] ≥ n − j − By Theorem 6,
n n−1 1 n E[ pd (u, v )] = + n |i − j | 2 2 i i=0
n n−1 1 n ≥ + n (i − j ) 2 2 i i=0
n n−1 1 n = − j+ n i 2 2 i
then show that E[ pd (u, v )] ≥ j − 12 .
1 2
=
i=0
n−1 1 − j + n (n2n−1 ) 2 2
1 =n− j− . 2 Also by Theorem 6,
E[ pd (u, v )] =
n n−1 1 n + n |i − j | 2 2 i i=0
n 1 n n−1 = + j− n i 2 2 i i=0
≥
n n−1 1 n + n ( j − i) 2 2 i i=0
1 1 n−1 = + j − n (n2n−1 ) = j − . 2 2 2
Therefore, E[ pd (u, v )] ≥ max{n − j − 12 , j − 12 }, completing the proof. Corollary 8. Let j ≥ 1, and H be the (j)-coloring of H2j . Then
1 j E[ pd (u, v )] = j − + 2 j 2 2
2j j
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
391
where u and v are vertices of H that are chosen independently and randomly. Proof. By Theorem 6, E[ pd (u, v )] =
n−1 2
1 2n
+
2j i i
i=0
|i − j |
i=0
Note that j
n
=
j
2j i
i
i=1
=
j
n , where n = 2 j. i
( 2 j )! ( 2 j )! = i ! ( 2 j − i )! ( i − 1 )! ( 2 j − i )! i=1 i=1 j−1
j 2j − 1 = 2j i−1
j
i
= 2j
i=1
2j − 1 i
= 2 j22 j−2 = j22 j−1
i=0
due to the fact that the sum of the first half of the binomial coefficients of the form In addition,
j
j
i=0
2j i
j 2j i
= j
i=0
Therefore, j
= j
( j − i)
i=0
2j i
=
j
j
i=0
2j i
22 j 1 2j + 2 2 j
−
j i=0
j 2j = . 2 j
Because
2 j
|i − j | i=0
2j i
=2
j
2j i
= j22 j−1 +
i=0
Corollary 8 can be rephrased as
− j22 j−1
1 j = j − + 2j 2 2
By well established bounds on
j 2j 2 j
2 j 2j 2j , we have that i=0 |i − j| i i
n n−1 1 n E[ pd (u, v )] = + n |i − j | 2 2 i
is half the total sum by symmetry.
j 2j . 2 j
( j − i) i=0
i
= j22 j−1 +
i
2 j−1
= j
2j , implying that j
2j j
2j , namely, the fact that j
2j j
2j
2 = √
πj
(1 −
cn n
), where
1 9
< cn <
1 8
for all n ≥ 1,
j j 1 1 1 1 j− + √ − < E[ pd (u, v )] < j − + √ − . 2 2 π 8 πj π 9 πj 6. Conclusion Within this new and exciting area of properly colored paths, the results in this work consider the proper distance between vertices and the number of different shortest proper paths between vertices. Although we confine our investigation to (j)-colorings of a hypercube, the questions already demonstrate their complexity and difficulty. Much work is yet to come, particularly considering other classes of colorings. Acknowledgment The authors would like to thank the referees very much for their careful reading and helpful comments. References [1] M. Adler, E. Halperin, R.M. Karp, V.V. Vazirani, A stochastic process on the hypercube with applications to peer-to-peer networks, in: Proceedings of the Thirty-Fifth Annual ACM Symposium on Theory of Computing, ACM, New York, 2003, pp. 575–584 (electronic), doi:10.1145/780542.780626. [2] V. Coll, J. Hook, C. Magnant, K. McCready, K. Ryan, Proper diameter of graphs, Manuscript. [3] F. Harary, J.P. Hayes, H.-J. Wu, A survey of the theory of hypercube graphs, Comput. Math. Appl. 15 (4) (1988) 277–289, doi:10.1016/0898-1221(88) 90213-1. [4] X. Li, C. Magnant, Properly colored notions of connectivity - a dynamic survey, Theory Appl. Graphs 0 (1) (2015) 1–28. Article 2.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Proper distance in edge-colored hypercubes Eddie Cheng a, Colton Magnant b,∗, Dhruv Medarametla c a b c
Department of Mathematics and Statistics, Oakland Universityi, Rochester, MI 48309, USA Department of Mathematical Sciences, Georgia Southern University, Statesboro, GA 30460, USA Stanford University, Stanford, CA 94305, USA
a r t i c l e
i n f o
Keywords: Edge-coloring Proper connection Hypercube
a b s t r a c t An edge-colored path is called properly colored if no two consecutive edges have the same color. An edge-colored graph is called properly connected if, between every pair of vertices, there is a properly colored path. Moreover, the proper distance between vertices u and v is the length of the shortest properly colored path from u to v. Given a particular class of properly connected colorings of the hypercube, we consider the proper distance between pairs of vertices in the hypercube. © 2017 Elsevier Inc. All rights reserved.
1. Introduction All graph colorings in this work are colorings of the edges. A coloring of a graph (or subgraph) is called proper if no two adjacent edges in the graph (respectively subgraph) share a color. A colored graph is called properly connected if between every pair of vertices, there is a properly colored path. In recent years, there have been numerous works on proper connectivity and the proper connection number. See the dynamic survey [] for more details. The study of proper connection in graphs is motivated in part by applications in networking and network security. In addition to security, of course it is critical that message transmission in a network be fast. For this reason, we consider shortest properly colored paths in edge-colored graphs. Because of its abundant symmetry and numerous applications to network design and reliability (see [1,3] for two of many examples) we consider colorings of the hypercube. For n ≥ 2, let Hn be the n-dimensional hypercube with vertex set being the set of binary strings of length n and edges between pairs of vertices whose strings differ in exactly one position. Given 1 ≤ j < n, define the (j)-coloring of Hn to be the 2-coloring of the edges of Hn in which all edges in dimension i where 1 ≤ i ≤ j have color 1 and all other edges have color 2. For example, the (1)-coloring of Hn has a perfect matching in color 1 and all other edges in color 2. For the sake of notation, given two vertices u and v in a hypercube, let o(u, v ) denote the number of positions i for 1 ≤ i ≤ j in which u and v differ and let t (u, v ) denote the number of positions i for j < i ≤ n in which u and v differ. Let γ (u, v ) be the indicator variable which takes the value 1 if o(u, v ) + t (u, v ) is odd and 0 if o(u, v ) + t (u, v ) is even. Given a properly connected coloring of a graph, the work of Coll et al. [2] initiated the study of the proper distance between pairs of vertices. The proper distance between two vertices u and v, denoted by pd (u, v ), is the minimum length (number of edges) of a properly colored path between u and v. Although the proper connection number has been studied and applications to computer science and networking have been described, precious little attention has been paid to the proper distance which is, in some sense, a measure of the efficiency of the properly connected network. ∗
Corresponding author. E-mail addresses: [email protected], [email protected] (C. Magnant).
http://dx.doi.org/10.1016/j.amc.2017.05.065 0 096-30 03/© 2017 Elsevier Inc. All rights reserved.
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
385
In Section 2, we consider the proper distance between two arbitrary vertices of a (j)-colored hypercube and in Section 3, we count the number of different shortest proper paths between two vertices in the (1)-colored hypercube. In Section 4, we count the number of vertices at each proper distance from a given fixed point in the (j)-colored hypercube. Finally in Section 5, we find the expected proper distance between two randomly selected vertices in a (j)-colored hypercube. 2. Proper distance We prove the following result which establishes the proper distance between any pair of vertices based on the number of positions in which they differ and which color those dimensions are assigned. Theorem 1. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then H is properly connected. Furthermore, let u and v be any pair of vertices in H. Then
pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ) Proof. We first constructively show that H is properly connected, and then show the desired result, that pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ). Note that in the hypercube, any path from u to v consists of flipping the bits of u until we arrive at v. A properly colored path from u to v then consists of flipping the bits of u until we arrive at v with the restriction that the bits we flip must alternate between the first j bits and the last n − j bits. To construct such a properly colored path, flip the bits of u that differ with the respective bits of v, making sure to flip the bits in an order that alternates between the first j bits and the last n − j bits, until you arrive at a vertex u such that either o(u , v ) = 0 or t (u , v ) = 0. If o(u , v ) = 0, we alternate flipping the first bit with flipping the remaining bits in the last n − j positions that differ with the respective bits of v. Otherwise if t (u , v ) = 0, then we alternate flipping the last bit with flipping the remaining bits in the first j positions that differ with the respective bits of v. Since there are a finite number of positions in which u and v differ and the process will not repeat vertices, this process will eventually terminate, resulting in a properly colored path from u to v. See Example 1 for an example of this process. Example 1. If n = 7 and j = 4 and we consider u = (0, 1, 0, 1, 0, 0, 0 ), and v = (0, 0, 1, 1, 1, 1, 1 ), then some examples of proper colored paths from u to v created by the algorithm above would be as follows:
u
=
u
=
v
=
( 0, 1, 0, 1, 0, 0, 0 ) → ( 0, 0, 0, 1, 0, 0, 0 ) → ( 0, 0, 0, 1, 1, 0, 0 ) → ( 0, 0, 1, 1, 1, 0, 0 ) → ( 0, 0, 1, 1, 1, 1, 0 ) → ( 1, 0, 1, 1, 1, 1, 0 ) → ( 1, 0, 1, 1, 1, 1, 1 ) → ( 0, 0, 1, 1, 1, 1, 1 )
u
=
u
= =
v
( 0, 1, 0, 1, 0, 0, 0 ) → ( 0, 1, 0, 1, 1, 0, 0 ) → ( 0, 0, 0, 1, 1, 0, 0 ) → ( 0, 0, 0, 1, 1, 1, 0 ) → ( 0, 0, 1, 1, 1, 1, 0 ) → ( 0, 0, 1, 1, 1, 1, 1 )
We now prove that pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ). Consider a properly colored path from u to v and label the vertices of this path as u = u0 → u1 → u2 → · · · → ud = v. In this path, a total of d bits have been flipped, counting with repetition. Also, flipping a bit an even number of times brings it back to its original position. Because u and v initially differed in o(u, v ) + t (u, v ) bits, we get d ≡ o(u, v ) + t (u, v ) ≡ o(u, v ) − t (u, v ) ≡ |o(u, v ) − t (u, v )| ≡ γ (u, v ) mod 2. Thus, we get d ≡ γ (u, v ) mod 2. Certainly the path contains at least o(u, v ) bits flipped from the first j bits and at least t (u, v ) bits flipped from the last n − j bits. Since, in order to be a properly colored path, the flipping must alternate between the first j bits and the last n − j bits, there must be at least 2 max{o(u, v ), t (u, v )} − 1 bits flipped in total. Thus, it is clear that d ≥ 2 max{o(u, v ), t (u, v )} − 1. Now, because 2 max{o(u, v ), t (u, v )} − γ (u, v ) is the smallest positive integer that is both at least 2 max{o(u, v ), t (u, v )} − 1 and is also congruent to γ (u, v ) mod 2, it is clear that pd (u, v ) ≥ 2 max{o(u, v ), t (u, v )} − γ (u, v ). Thus, we must now show that a properly colored path between u and v of length 2 max{o(u, v ), t (u, v )} − γ (u, v ) exists. Without loss of generality, assume o(u, v ) ≥ t (u, v ), so we must show that pd (u, v ) = 2o(u, v ) − γ (u, v ). Let a1 , a2 , . . . , ao(u,v ) be the o(u, v ) bits in which u and v differ with 1 ≤ a1 < a2 < · · · < ao(u,v ) ≤ j, and let b1 , b2 , . . . , bt (u,v ) be the t (u, v ) bits in which u and v differ in the last n − j bits with j < b1 < b2 < · · · < bt (u,v ) ≤ n. We now formalize a method for creating a path from u to v of length 2o(u, v ) − γ (u, v ). First flip the bit of u in position a1 , then position b1 , then position a2 , and so on, until you have flipped the bit in position bt (u,v ) . This is a total of 2t (u, v ) edges in the path from u to v. If o(u, v ) = t (u, v ), then you have gone from u to v in 2o(u, v ) − γ (u, v ) = 2t (u, v ) edges, and you are done. Otherwise, call the vertex that you are on u . By definition, u will only differ from v in positions at (u,v )+1 , . . . , ao(u,v ) . Then, continuing the path from before, and keeping in mind that the last edge in the path was of color 2, flip bit at (u,v )+1 , then bit n, then bit at (u,v )+2 , then bit n, and so on, until you flip bit ao(u,v ) . This is a total of 2(o(u, v ) − t (u, v )) − 1 edges, meaning that you have used a total of 2o(u, v ) − 1 edges in total. Let u be the terminal vertex of this path. If u = v, then the only position where it can differ from v in is bit n. This only occurs if 2
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divides o(u, v ) − t (u, v ), in which case we flip the bit in position n to complete the desired path. This necessarily produces a properly colored path from u to v of length 2o(u, v ) − γ (u, v ), completing the proof. Example 2. If n = 7 and j = 3 and we consider u = (0, 0, 0, 1, 0, 1, 0 ), and v = (1, 1, 1, 1, 1, 0, 0 ), then a proper colored path from u to v created by the algorithm above would be as follows:
u = ( 0, 0, 0, 1, 0, 1, 0 ) →
( 1, 0, 0, 1, 0, 1, 0 ) → ( 1, 0, 0, 1, 1, 1, 0 ) → ( 1, 1, 0, 1, 1, 1, 0 ) → ( 1, 1, 0, 1, 1, 0, 0 ) → v = ( 1, 1, 1, 1, 1, 0, 0 ) 3. Number of proper paths between vertices Assuming a properly connected coloring of a graph H, recall that pd (u, v ) is the proper distance between vertices u and v and let pp(u, v ) denote the number of distinct shortest properly colored paths in H from u to v. Theorem 2. Let n ≥ 2 and let H be the (1)-coloring of Hn . Then H is properly connected. Furthermore, let u and v be any pair of distinct vertices whose strings differ in t = t (u, v ) positions other than the first. If t = 0, then u and v are adjacent so pd (u, v ) = 1 and pp(u, v ) = 1. Otherwise, for t ≥ 1, we have pd (u, v ) = 2t − γ (u, v ) and
ppH (u, v ) = (2 − γ (u, v ))(t! ). Proof. Without loss of generality, let v = (0, 0, . . . , 0 ) and let u be an arbitrary vertex of Hn \ {v}. If u and v are adjacent, the edge between them is the only shortest proper path from u to v so there is nothing to prove. So suppose u and v are not adjacent. In particular, this means that t ≥ 1. Then t is precisely the number ‘1’s in u in positions other than the first, and γ (u, v ) = 1 if the total number of ‘1’s in u is odd and γ (u, v ) = 0 otherwise. In the (1)-coloring H of Hn , every proper path from u to v must use alternating edges in color 1. This means that every shortest proper path from u to v must alternate between switching 1 to 0 in a position other than the first position and switching the first position. First suppose γ (u, v ) = 0 so there is an even total number of positions in which u and v differ. Let π be an arbitrary ordering of the t ‘1’s in u in positions other than the first. For any choice of π , the sequence of switches of positions 1, π (1 ), 1, π (2 ), . . . , 1, π (t ) and the sequence of switches of positions π (1 ), 1, π (2 ), 1, . . . , π (t ), 1 both constitute shortest proper paths from u to v and are all distinct. Clearly these paths have length 2t = 2t − γ (u, v ). Also there can be no other shortest proper path from u to v because every proper path from u to v must include all t of the ‘1’s and must also alternate switching position 1. Since there are t! such orderings, there are a total of 2(t! ) = (2 − γ (u, v ))(t! ) distinct shortest proper paths from u to v. Finally suppose γ (u, v ) = 1 so there is an odd total number of positions in which u and v differ. Then every shortest proper path will start on an edge of color 2, by switching 1 to 0 in some position other than the first. Again letting π be an arbitrary ordering of the t ‘1’s in u in positions other than the first, for any choice of π , the sequence of switches of positions π (1 ), 1, π (2 ), 1, . . . , 1, π (t ) constitutes a shortest proper path from u to v and all are distinct. These paths have length 2t − 1 = 2t − γ (u, v ) and there is no other shortest proper path from u to v. Since there are t! such orderings, there are a total of t! = (2 − γ (u, v ))(t! ) distinct shortest proper paths from u to v. 4. Counting vertices at each proper distance Now, given H the (j)-coloring of Hn , we define the surface area SAn, j (k ) = #{v| pd (u, v ) = k}, where u is any arbitrary vertex of H. Note that the value of SAn, j (k) does not depend on the choice of u due to the symmetry of the hypercube. Theorem 3. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then
SAn, j (k ) =
where
a b
x+y≡k mod 2 max{x,y}= k+1 2
j x
n− j y
is defined to be 0 for b < 0 or b > a.
Observe that if k = 0, we get SAn, j (0 ) = 1 as one might expect. Although this is not a closed form, it is an explicit formula, which is computable in a polynomial number of steps as a function of n and k. Proof. Let u be any arbitrary vertex. Then, we must find the number of vertices v ∈ V (H ) such that pd (u, v ) = k. By Theorem 1, this is the number of vertices v such that 2 max{o(u, v ), t (u, v )} − γ (u, v ) = k. In addition, note that
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
387 k+γ (u,v )
2 max{o(u, v ), t (u, v )} − γ (u, v ) = k implies that γ (u, v ) ≡ k mod 2. Therefore, max{o(u, v ), t (u, v )} = = k+1 2 2 , as 0 ≤ γ (u, v ) ≤ 1. As γ (u, v ) ≡ |o(u, v ) − t (u, v )| ≡ o(u, v ) − t (u, v ) ≡ o(u, v ) + t (u, v ) mod 2, we have that o(u, v ) + t (u, v ) ≡ k mod 2. Therefore, we now know that for pd (u, v ) = k, we must have
max{o(u, v ), t (u, v )} =
k+1 2
and o(u, v ) + t (u, v ) ≡ k mod 2. In addition, note that these are sufficient conditions; if both of those conditions are true, then pd (u, v ) = 2 max{o(u, v ), t (u, v )} − γ (u, v ) = 2 k+1 2 − γ (u, v ). If k is odd, then o(u, v ) + t (u, v ) ≡ k mod 2 implies that γ (u, v ) = 1, so pd (u, v ) = 2 k+1
− γ ( u, v ) = k + 1 − 1 = k, and if k is even, o(u, v ) + t (u, v ) ≡ k mod 2 implies 2
that γ (u, v ) = 0, so pd (u, v ) = 2 k+1 2 − γ (u, v ) = k + 0 − 0 = k. Therefore, pd (u, v ) = k if and only if max{o(u, v ), t (u, v )} = k+1
and o ( u, v ) + t ( u, v ) ≡ k mod 2. 2 Given two integers x and y, we need to count the number of vertices v satisfying o(u, v ) = x and t (u, v) = y. Since exactly j x of the first j bits must differ from the first j bits of u, the number of choices for the first j bits of v is . Similarly since x
exactly y of the last n − j bits must differ from the last n − j bits of u, the number of choices for the last j bits is
These imply that the total number of such vertices v is
j x
n− j . y
n− j . y
Because pd (u, v ) = k if and only if max{o(u, v ), t (u, v )} = k+1 2 and o(u, v ) + t (u, v ) ≡ k mod 2, we can write that
SAn, j (k ) =
#{v|o(u, v ) = x, t (u, v ) = y}
x+y≡k mod 2 max{x,y}= k+1 2
which then implies that
SAn, j (k ) =
n− j y
j x
x+y≡k mod 2 max{x,y}= k+1 2
by the previous paragraph, completing the proof.
Corollary 4. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Without loss of generality, assume j ≤ n − j. Then if k > 2j,
SAn, j (k ) = 2
j−1
n− j
Proof. By Theorem 3,
SAn, j (k ) =
.
k+1
2
x+y≡k mod 2 max{x,y}= k+1 2
n− j . y
j x
2 j+2 The assumption that k > 2j implies that k ≥ 2 j + 1. If x = k+1 2 , then we have x ≥ 2 = j + 1 > j, so
if x = k+1 2 , then
jn− j x
y
j x
x+y≡k mod 2 max{x,y}= k+1 2
x+y≡k mod 2 y= k+1 2
=
Since
n− j
x≡0 mod 2
x≡0 mod 2
j x
or
j x
+
j x
k+1
2
n− j y
=
x
= 0. Therefore,
= 0, which simplifies the calculation for surface area to the following:
SAn, j (k ) =
j
n− j y
x+
k+1 2
≡k mod 2
x≡1 mod 2
x≡1 mod 2
j x
=
j . x
x∈Z
j x
=
2 j,
it is clear that
x+ k+1
≡k mod 2 2
j x
is equivalent to either
j . We observe that these two sums are equal, each being equal to 2 j−1 , which holds x
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E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
because j
1
( j−x )
(−1 )
j x
x
x=0
Therefore,
SAn, j (k ) =
n− j
= ( 1 − 1 ) j = 0.
k+1
2
j x
x+ k+1 2 ≡k mod 2
= 2 j−1
n− j
k+1
2
,
completing the proof.
Theorem 5. Let j ≥ 1, and H be the (j)-coloring of H2j . Then, for integers a and b with a + b = 2 j − 1 + (−1 )a , we have
SA2 j, j (a ) + SA2 j, j (b) = 2 j
j
a+1
2
Proof. Note that a + b = 2 j − 1 + (−1 )a ≡ 0 mod 2 (since if a is even, a + b = 2 j; otherwise a is odd and hence a + b = a 2 j+(−1 )a −a 2 j − 2). Also, b+1
= j + (−12) −a = j − a+1 2 = 2 2 . Utilizing Theorem 3, we have that
SA2 j, j (a ) =
j x
x+y≡a mod 2 max{x,y}= a+1 2
and
SA2 j, j (b) =
j x
x+y≡b mod 2 max{x,y}= b+1 2
x+y≡a mod 2 max{x,y}= j− a+1 2
=
j y
=
j y
j x
x+y≡a mod 2 max{x,y}= j− a+1 2
( j−x )+( j−y )≡a mod 2 min{ j−x, j−y}= a+1 2
j j−y
j j−x
j j−y
=
j j−x
=
j y
j x
(x )+(y )≡a mod 2 min{x ,y }= a+1 2
j . y
Therefore,
SA2 j, j (a ) + SA2 j, j (b)
=
x+y≡a mod 2 max{x,y}= a+1 2
= x+
=
=
a+1 2
≡a mod 2
j
a+1 2
j
j x
j y
j x
a+1
2
(x )+(y )≡a mod 2 min{x ,y }= a+1
j
x+
≡a mod 2
j x
(2 j−1 + 2 j−1 ) = 2 j
j x
2
+
a+1
2
a+1 2
+
a+1 2
+y≡a mod 2
+
j
j y
a+1 2
+y≡a mod 2
j
a+1
2
j y
j y
a+1
2
due to the properties of the sum of alternating binomial coefficients, completing the proof.
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
389
5. Expected proper distance In addition to surface area, we also consider the expected proper distance between two randomly selected vertices. First two technical lemmas. Lemma 1.
x+y≡1 mod 2
Proof.
j x
x+y≡1 mod 2
=
= 2n−1 .
n− j y
j x
x≥0
n− j y
j x
n− j y
y≡1+x mod 2
=
j n− j−1 2 x
x≥0
= 2 j 2n− j−1 = 2n−1 . Lemma 2.
x,y≥0
Proof.
= =
n− j y
k
k≥0
n k
= n2n−1 +
+
= n2
+
j x
= n2
+
|y + ( j − x ) − j | |y + z − j |
y,z≥0
= n2n−1 +
|i − j |
i≥0
= n2n−1 +
i≥0
n− j y
y+z=i
|i − j |
n− j y
j j−x
j z
n− j y
j z
n− j y
j x
|x − y|
n− j y
x,y≥0 n−1
j x
n . i
x,y≥0
|x − y|
x,y≥0 n−1
+
|x − y|
x,y≥0
|i − j |
i=0
j (x + y ) x
n
= n2n−1 +
n− j y
2 max{x, y}
x,y≥0
j x
x,y≥0
n− j y
j x
2 max{x, y}
n− j y
n i
by Vandermonde’s Identity and other well-known facts about properties of sums of binomial coefficients. We are now able to prove our main result of this section. Theorem 6. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then
n n−1 1 n E[ pd (u, v )] = + n |i − j | 2 2 i i=0
where u and v are vertices of H that are chosen independently and randomly.
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E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
Proof. Note that we can fix u without affecting the value of E[ pd (u, v )] due to the symmetry of the hypercube. Let u be any arbitrary vertex of H; then, because there are 2n vertices in H, we have that
E[ pd (u, v )] =
∞ 1 1 pd ( u, v ) = kSAn, j (k ). 2n 2n
v∈V (H )
k=0
By Theorem 3, we have that
E[ pd (u, v )] =
∞ ∞ 1 1 kSAn, j (k ) = n k n 2 2 k=0
k=0
x+y≡k mod 2 max{x,y}= k+1 2
n− j . y
j x
Through algebraic manipulation, we can see that
⎛
∞ 1⎜ ⎜ k 2n ⎝ k=0
1 = n 2
x+y≡k mod 2 max{x,y}= k+1 2
j x
n− j y
2 max{x, y}
⎟ ⎟ ⎠
n− j y
j x
x,y≥0
⎞
1 − n 2
Utilizing Lemmas 1 and 3, we see that
1 E[ pd (u, v )] = n 2
=
1 2n
2 max{x, y}
x,y≥0
n2n−1 +
n
n− j y
n i
|i − j |
i=0
n−1 1 = + n 2 2 completing the proof.
n
j x
−
x+y≡1 mod 2
1 − n 2
j x
n− j y
.
x+y≡1 mod 2
j x
n− j y
1 n−1 (2 ) 2n
n i
|i − j |
i=0
Corollary 7. Let n ≥ 2, 1 ≤ j < n, and H be the (j)-coloring of Hn . Then
1 2
E[ pd (u, v )] ≥ max{n − j, j} −
where u and v are vertices of H that are chosen independently and randomly. Proof. We first show that E[ pd (u, v )] ≥ n − j − By Theorem 6,
n n−1 1 n E[ pd (u, v )] = + n |i − j | 2 2 i i=0
n n−1 1 n ≥ + n (i − j ) 2 2 i i=0
n n−1 1 n = − j+ n i 2 2 i
then show that E[ pd (u, v )] ≥ j − 12 .
1 2
=
i=0
n−1 1 − j + n (n2n−1 ) 2 2
1 =n− j− . 2 Also by Theorem 6,
E[ pd (u, v )] =
n n−1 1 n + n |i − j | 2 2 i i=0
n 1 n n−1 = + j− n i 2 2 i i=0
≥
n n−1 1 n + n ( j − i) 2 2 i i=0
1 1 n−1 = + j − n (n2n−1 ) = j − . 2 2 2
Therefore, E[ pd (u, v )] ≥ max{n − j − 12 , j − 12 }, completing the proof. Corollary 8. Let j ≥ 1, and H be the (j)-coloring of H2j . Then
1 j E[ pd (u, v )] = j − + 2 j 2 2
2j j
E. Cheng et al. / Applied Mathematics and Computation 313 (2017) 384–391
391
where u and v are vertices of H that are chosen independently and randomly. Proof. By Theorem 6, E[ pd (u, v )] =
n−1 2
1 2n
+
2j i i
i=0
|i − j |
i=0
Note that j
n
=
j
2j i
i
i=1
=
j
n , where n = 2 j. i
( 2 j )! ( 2 j )! = i ! ( 2 j − i )! ( i − 1 )! ( 2 j − i )! i=1 i=1 j−1
j 2j − 1 = 2j i−1
j
i
= 2j
i=1
2j − 1 i
= 2 j22 j−2 = j22 j−1
i=0
due to the fact that the sum of the first half of the binomial coefficients of the form In addition,
j
j
i=0
2j i
j 2j i
= j
i=0
Therefore, j
= j
( j − i)
i=0
2j i
=
j
j
i=0
2j i
22 j 1 2j + 2 2 j
−
j i=0
j 2j = . 2 j
Because
2 j
|i − j | i=0
2j i
=2
j
2j i
= j22 j−1 +
i=0
Corollary 8 can be rephrased as
− j22 j−1
1 j = j − + 2j 2 2
By well established bounds on
j 2j 2 j
2 j 2j 2j , we have that i=0 |i − j| i i
n n−1 1 n E[ pd (u, v )] = + n |i − j | 2 2 i
is half the total sum by symmetry.
j 2j . 2 j
( j − i) i=0
i
= j22 j−1 +
i
2 j−1
= j
2j , implying that j
2j j
2j , namely, the fact that j
2j j
2j
2 = √
πj
(1 −
cn n
), where
1 9
< cn <
1 8
for all n ≥ 1,
j j 1 1 1 1 j− + √ − < E[ pd (u, v )] < j − + √ − . 2 2 π 8 πj π 9 πj 6. Conclusion Within this new and exciting area of properly colored paths, the results in this work consider the proper distance between vertices and the number of different shortest proper paths between vertices. Although we confine our investigation to (j)-colorings of a hypercube, the questions already demonstrate their complexity and difficulty. Much work is yet to come, particularly considering other classes of colorings. Acknowledgment The authors would like to thank the referees very much for their careful reading and helpful comments. References [1] M. Adler, E. Halperin, R.M. Karp, V.V. Vazirani, A stochastic process on the hypercube with applications to peer-to-peer networks, in: Proceedings of the Thirty-Fifth Annual ACM Symposium on Theory of Computing, ACM, New York, 2003, pp. 575–584 (electronic), doi:10.1145/780542.780626. [2] V. Coll, J. Hook, C. Magnant, K. McCready, K. Ryan, Proper diameter of graphs, Manuscript. [3] F. Harary, J.P. Hayes, H.-J. Wu, A survey of the theory of hypercube graphs, Comput. Math. Appl. 15 (4) (1988) 277–289, doi:10.1016/0898-1221(88) 90213-1. [4] X. Li, C. Magnant, Properly colored notions of connectivity - a dynamic survey, Theory Appl. Graphs 0 (1) (2015) 1–28. Article 2.