The crossing number of folded hypercubes

0.29 MB·English

by  Haoli Wang

The crossing number of folded hypercubes
∗
Haoli Wang, Yuansheng Yang, Yan Zhou
†
Department of Computer Science
Dalian University of Technology, Dalian, 116024, P.R. China
Wenping Zheng
1
1 Key Laboratory of Computational Intelligence and Chinese Information
0
Processing of Ministry of Education,
2
n Shanxi University, Taiyuan, 030006, P.R. China
a
J

8 Guoqing Wang
1
Center for Combinatorics, LPMC-TJKLC

]
Nankai University, Tianjin, 300071, P.R. China
O
C
.
h
t
a
m Abstract
[

ThecrossingnumberofagraphGistheminimumnumberofpairwiseintersections
1
v of edges in a drawing of G. The n-dimensional folded hypercube FQn is a graph
6 obtained from n-dimensional hypercube by adding all complementary edges. In this
8
paper, we obtain upper and lower bounds of the crossing number of FQ .
3 n
3
1. Keywords: Drawing; Crossing number; Folded hypercube
0
1
1
: 1 Introduction
v
i
X
r Let G be a simple connected graph with vertex set V(G) and edge set E(G). The
a
crossing number cr(G) of a graph G is the minimum number of pairwise intersections of
edges in a drawing of G in the plane. In the past thirty years, it turned out that crossing
number played an important role not only in various fields of discrete and computational
geometry (see [2, 12, 14, 16]), but also in VLSI theory and wiring layout problems (see [1,
9, 10, 13]). For this reason, the study of crossing numberof some popularparallel network
topologies such as hypercube and its variants which have good topological properties and
applications in VLSI theory, would be of theoretical importance and practical value. An
n-dimensional hypercube Q is a graph in which the nodes can be one-to-one labeled
n
∗The research is supported byNSFC (60973014, 60803034, 11001035) and SRFDP(200801081017)
†corresponding author’s email : [email protected]
1
with 0-1 binary sequences of length n, so that the labels of any two adjacent nodes differ
in exactly one bit. Determining the crossing number of an arbitrary graph is proved to
be NP-complete [7]. Even for hypercube, for a long time the only known result on the
exact value of crossing number of Q has been cr(Q ) = 0, cr(Q ) = 8 [3], cr(Q ) 56
n 3 4 5
≤
[11]. Hence, it is more practical to find upper and lower bounds of crossing numbers of
hypercubeanditsvariants. Concernedwithupperboundofcrossingnumberofhypercube,
Erd˝os and Guy [5] in 1973 conjectured the following:
5 n2+1
cr(Q ) 4n 2n 2.
n −
≤ 32 −⌊ 2 ⌋
In 2008, Sykora and Vrt’o [6] constructed a drawing of Q in the plane which has the
n
conjectured number of crossings mentioned above. Early in 1993 they [15] also proved a
lower bound of cr(Q ):
n
1
cr(Q ) > 4n (n2+1)2n 1.
n −
20 −
Since the hypercube does not have the smallest possible diameter for its resources, to
achieve smaller diameter with the same number of nodes and links as an n-dimensional
hypercube, a variety of hypercube variants were proposed. Folded hypercube is one of
these variants. The n-dimensional folded hypercube FQ was proposed by El-Amawy
n
and Latifi [4] in 1991. The folded hypercube has many appealing features of the n-
dimensional hypercube such as node and edge symmetry. In addition, it has been shown
to be superior over the n-dimensional hypercube in many communication aspects such
as halved diameter, better average distance, shorter delay in communication links, less
message traffic density and lower cost. Therefore, it would be more attractive to study
the crossing number of the folded hypercube.
The n-dimensional folded hypercube FQ is a graph obtained from Q by adding all
n n
complementaryedges,whichjoinavertexx = x x ...x toanothervertexx = x x ...x
1 2 n 1 2 n
for every x V(Q ), where x = 0,1 x . The graphs shown in Figure 1.1 are FQ ,
n i i 1
∈ { }\{ }
FQ , FQ and FQ , respectively, where thick lines represent the complementary edges.
2 3 4
It is easy to see that FQ1 ∼= K2, FQ2 ∼= K4 and FQ3 ∼= K4,4 (see Figure 1.2). So
cr(FQ ) = cr(FQ )= 0 and cr(FQ ) = 4.
1 2 3
011 111 0011 0111 1011 1111
00 10 010 110 0010 1010
0110 1110
0001 1001
001 101 1101
0 1 0101
01 11 000 100 0000 0100 1000 1100
FQ1 FQ2 FQ3 FQ4
Figure 1.1: Folded hypercube FQ1, FQ2, FQ3 and FQ4
2
000
011
10
001 010 100 111
00
101
01 11 110
FQ2 FQ3
Figure 1.2: Drawings of FQ2 ∼=K4 and FQ3 ∼=K4,4
In this paper, we prove the following bounds of cr(FQ ):
n
4n 11
(n2+2n+4)2n 1 < cr(FQ ) 4n (n2+3n)2n 3.
− n −
20 (1 2 1 )2 − ≤ 32 −
× −qπ√2⌈n2⌉+1
2 Upper bound
A drawing of G is said to be a good drawing, provided that no edge crosses itself, no
adjacent edges cross each other, no two edges cross more than once, and no three edges
cross in a point. It is well known that the crossing number of a graph is attained only
in good drawings of the graph. So, we always assume that all drawings throughout this
paper are good drawings. For a good drawing D of a graph G, let ν (G) be the number
D
of crossings in D.
For a binary string x x x , let
1 2 n
···
D(x x x )= 2n 1x +2n 2x + +20x
1 2 n − 1 − 2 n
··· ···
be the corresponding decimal number of x x x . For any integers n > m 1 and
1 2 n
··· ≥
binary string x x x , let
1 2 m
···
n = y y y : y ,y ,...,y 0,1 ,y =x for all i 1,2,...,m .
Fx1···xm { 1 2··· n 1 2 n ∈ { } i i ∈ { }}
For a vertex subset S of a graph G, let S be the subgraph of G induced by S.
h i
We need the following observations which will be useful for the proofs of the upper
bound.
Observation 2.1. Let n be a vertex subset of FQ , where n > m 1 and
Fx1···xm n ≥
x ,...,x 0,1 . Then there exists an isomorphism µ of n onto Q such
1 m ∈ { } hFx1···xmi n−m
that µ(x x x x ) = x x for all x ,...,x 0,1 .
1 m m+1 n m+1 n m+1 n
··· ··· ··· ∈ { }
Observation 2.2. For any m 1, let R and S be two non-horizontal bunches of m
≥
parallel lines starting from points (0,0),(1,0),...,(m 1,0) respectively (see Figure 2.1),
−
which are above the real axis x. Then the number of crossings between R and S is m(m−1).
2
3
R S
(0,0) (1,0) (2,0) (m−2,0)(m−1,0) x
Figure 2.1: The crossings between two bunches of m parallel lines R and S
Now we shall introduce a recursive drawing Γ of Q . Consider the real axis x in the
n n
2-dimensional Euclidean plane. Let Γ be a drawing of Q in the plane such that
n 1 n 1
− −
all vertices of Q are drawn in the points (0,0),(1,0),...,(2n 1 1,0). Produce an
n 1 −
− −
identical drawing to Γ in the points (2n 1,0),(2n 1 +1,0),...,(2n 1,0). Then join
n 1 − −
− −
point (i,0) and point (2n 1 +i,0) for all i 0,1,...,2n 1 1 by “parallel” curves. In
− −
∈ { − }
Figure 2.2, we show as an example drawings of Γ , Γ and Γ , where the “parallel” curves
1 2 3
joining (i,0) and (2n 1+i,0) are drawn in red.
−
x x
(1)Γ1 (2)Γ2
x
(3)Γ3
Figure 2.2: Drawings of Γ1, Γ2 and Γ3
By the construction of Γ , it is easy to observe
n
Observation 2.3. In the drawing Γ , any vertex x x x V(Q ) and its comple-
n 1 2 n n
··· ∈
mentary vertex x¯ x¯ x¯ are drawn in point (D(x x x ),0) and point ((2n 1)
1 2 n 1 2 n
··· ··· − −
D(x x x ),0), respectively.
1 2 n
···
We still need to introduce a necessary definition.
Definition 2.1. In the drawing Γ , for a vertex v of Q , let Cn(v) and Cn(v) be the
n n a b
number of curves which cover v from the upper semi-plane of real axis x and from the
lower semi-plane of real axis x, respectively (see Figure 2.3).
In Figure 2.3, the curve joining u and u covers v ,v from the upper semi-plane of
1 2 1 2
real axis x, and covers v from the lower semi-plane of real axis x.
3
4
u1 v1 v2 v3 u2 x
Figure 2.3: Vertices covered by a curve
Lemma 2.1. For n 1,
≥
Cn(v) = Cn(v) = 4n 1 (n+1)2n 2 .
a b − −
−
v XV(Qn) v XV(Qn)
∈ ∈
Proof. We argue by induction on n. Lemma 2.1 is true for n= 1. Now assume n> 1. By
the construction of Γ , we see that the curves joining point (i,0) and point (2n 1 +i,0)
n −
would cover points (i+1,0),(i+2,0),...,(2n 1 1,0) from the upper semi-plane of axis
−
−
x and cover points (2n 1,0),(2n 1 +1,0),...,(2n 1+i 1,0) from the lower semi-plane
− − −
−
of axis x for all i 0,1,...,2n 1 1 . Therefore, it follows that
−
∈ { − }
Cn(v) = (1+2+ +(2n 1 1))+2 Cn 1(v)
a − a−
··· − ×
X X
v V(Qn) v V(Qn−1)
∈ ∈
and
Cn(v) = (1+2+ +(2n 1 1))+2 Cn 1(v).
b ··· − − × b−
X X
v V(Qn) v V(Qn−1)
∈ ∈
Then the lemma follows from immediate verification.
Lemma 2.2. For n 1,
≥
ν (Q )= 4n 1 (n2+n+2)2n 3 .
Γn n − − −
Proof. Theproofwillbebyinductiononn. Lemma2.2istrueforn = 1sinceν (Q ) =0.
Γ1 1
Now assume n > 1. We see that the curves joining point (i,0) and point (2n 1 + i,0)
−
would cross all curves which cover point (i,0) from the upper semi-plane of axis x and
belong to the induced subgraph n of Q , similarly would cross all curves which cover
hF0i n
point (2n 1+i,0) from the lower semi-plane of axis x and belong to the induced subgraph
−
n for all i 0,1,...,2n 1 1 . Therefore, it follows from Lemma 2.1 that
hF1i ∈ { − − }
ν (Q ) = 2 ν (Q )+ Cn 1(v)+ Cn 1(v)
Γn n × Γn−1 n−1 v VP(Qn−1) a− v VP(Qn−1) b−
∈ ∈
= 2 ν (Q )+(4n 2 n 2n 3) 2
× Γn−1 n−1 − − × − ×
= 2 ν (Q )+22n 3 n 2n 2
× Γn−1 n−1 − − × −
= 22 ν (Q )+22n 4 (n 1) 2n 2+22n 3 n 2n 2
. × Γn−2 n−2 − − − × − − − × −
.
.
= 2n 1 ν (Q )+(2n 1+2n + +22n 3) (2+3+ +n) 2n 2
− × Γ1 1 − ··· − − ··· × −
= 2n 1 (2n 1 1)+(n2+n 2) 2n 3
− − −
× − − ×
= 4n 1 (n2+n+2)2n 3.
− −
−
5
Theorem 2.1. For n 3,
≥
11
cr(FQ ) 4n (n2+3n)2n 3.
n −
≤ 32 −
Proof. Toprovethetheorem,itsufficestoconstructadrawingD ofFQ withν (FQ )=
n n Dn n
114n (n2 +3n)2n 3 for all n 3. If n = 3, let D be the drawing shown in Figure 2.4,
32 − − ≥ 3
in which the number of crossings is 4 = 11 43 (32 +3 3) 23 3.
32 × − × × −
101 100
010 011
111 110
000 001
Figure 2.4: A drawing D3 of FQ3 with 4 crossings
Now assume n > 3. We set eight positive directions shown in Figure 2.5, where the
red arrows stand for the positive directions. By Observation 2.1, in the drawing D , we
3
replace every vertex x x x V(FQ ) in the “small” neighborhood of x x x by the
1 2 3 3 1 2 3
∈
induced subgraph n of FQ , in which the drawing of n is coincident with
hFx1x2x3i n hFx1x2x3i
Γ and the positive direction of Γ in Figure 2.2 is identical to that of Figure 2.5.
n 3 n 3
− −
101 100
010 011
111 110
000 001
Figure 2.5: D3 with positive direction(Thick lines represent the complementary edges)
Theneveryedgeeincidentwithvertex x x x andvertex y y y inD willbereplaced
1 2 3 1 2 3 3
by a bunchof 2n 3 edges which areincident with n and n anddrawn along the
− Fx1x2x3 Fy1y2y3
original edge e in D . Combined with Observation 2.3 and the arrangement of positive
3
directions shown in Figure 2.5, we conclude that all edges in an arbitrary bunch will be
parallel. Therefore, the total number of crossings in D will be the number of crossings in
n
6
the small neighborhoods of all induced subgraphs n plus 2n 3 2n 3 ν (FQ ),
hFx1x2x3i − · − · D3 3
where x x x V(FQ ).
1 2 3 3
∈
For the conveniences of the reader, we offer drawings for FQ , FQ , FQ and FQ
4 5 6 7
in Figures 2.6-2.9 obtained according to the rules of construction mentioned above, where
the vertices are represented by decimal numbers, the edges of n are drawn in red
hFx1x2x3i
and the rest edges are drawn in blue.
Claim A. For any vertex x x x of FQ , the number of crossings in the small neigh-
1 2 3 3
borhood of the induced subgraph n in the drawing D is
hFx1x2x3i n
9 4n 4 (n2+3n)2n 6.
− −
· −
Proof of Claim A. Let E = E( n ) and E be the set consisting of all edges of FQ
r hFx1x2x3i b n
which have exactly one end in V( n ). Then the number of crossings in the small
hFx1x2x3i
neighborhood of n is
hFx1x2x3i
ν (E )+ν (E )+ν (E ,E ).
Dn r Dn b Dn r b
By Observation 2.1 and Lemma 2.2, we have that
ν (E )= 4n 4 ((n 3)2 +(n 3)+2)2n 6.
Dn r − − − − −
By Observation 2.2, we have that
2n 3(2n 3 1)
− −
ν (E ) = 2 − .
Dn b · 2
By Lemma 2.1, we have that
ν (E ,E ) = 4 (4n 4 (n 2)2n 5).
Dn r b · − − − −
Then Claim A holds by immediate verification.
By Claim A, we have that ν (FQ ) = 8 (9 4n 4 (n2 +3n)2n 6)+2n 3 2n 3
Dn n · · − − − − · − ·
ν (FQ ) = 114n (n2+3n)2n 3. This completes the proof of the theorem.
D3 3 32 − −
7
10 8
11 9
4 6
5 7
15 13
14 12
1 3
0 2
Figure 2.6: A drawing D4 of FQ4
20 16
21 17
22 18
23 19
8 12
9 13
10 14
11 15
31 27
30 26
29 25
28 24
3 7
2 6
1 5
0 4
Figure 2.7: A drawing D5 of FQ5
8
40 32
41 33
42 34
43 35
44 36
45 37
46 38
47 39
16 24
17 25
18 26
19 27
20 28
21 29
22 30
23 31
63 55
62 54
61 53
60 52
59 51
58 50
57 49
56 48
7 15
6 14
5 13
4 12
3 11
2 10
1 9
0 8
Figure 2.8: A drawing D6 of FQ6
9
80 64
81 65
82 66
83 67
84 68
85 69
86 70
87 71
88 72
89 73
90 74
91 75
92 76
93 77
94 78
95 79
32 48
33 49
34 50
35 51
36 52
37 53
38 54
39 55
40 56
41 57
42 58
43 59
44 60
45 61
46 62
47 63
127 111
126 110
125 109
124 108
123 107
122 106
121 105
120 104
119 103
118 102
117 101
116 100
115 99
114 98
113 97
112 96
15 31
14 30
13 29
12 28
11 27
10 26
9 25
8 24
7 23
6 22
5 21
4 20
3 19
2 18
1 17
0 16
Figure 2.9: A drawing D7 of FQ7
10