1. No category

The 27th Workshop on Combinatorial Mathematics and Computation Theory
Displacements of Weighted Graphs
∗
Chiang Lin
Department of Mathematics
National Central University, Chung-Li, Taiwan, R.O.C.
[email protected]
Jen-Ling Shang
Department of Banking and Finance
Kainan University, Tao-Yuan, Taiwan, R.O.C.
[email protected]
Abstract
x in G, the status sG (x) of x is defined by
X
dG (x, y).
sG (x) =
In this paper we investigate the relationship between the status and the displacement of a connected weighted graph. We also obtain a product
result for the displacements.
y∈V (G)
The status s(G) of the graph G is defined by
s(G) = min sG (x).
x∈V (G)
1
The median of G is the set of vertices in G with
status equal to s(G) (i.e, the set of vertices with
minimum status).
For the results about statuses one may refer to
[1]. The following characterization of the median
of a tree follows from the results of A. Kang and
D. Ault [2], and was mentioned by O. Kariv and
S.L. Hakimi in [4] for trees with weights on the
vertices and edges.
Introduction
All the graphs considered in this paper are finite and simple. We introduce two parameters of
graphs: displacement and status.
We first give the definition of the displacement
of a graph. Let G be a connected graph and let
φ be a permutation of V (G) (i.e., φ is a bijection
from V (G) to V (G)). The displacement DG (φ) of
φ is defined by
X
DG (φ) =
dG (x, φ(x)).
Proposition 1.1 [3] Let T be a tree and v be a
vertex in T . Then v is in the median of T if and
only if |V (T 0 )| ≤ 21 |V (T )| for every component T 0
of T − v.
2
x∈V (G)
The displacement D(G) of G is defined by
In the following the notions of displacement and
status are extended to the graphs with weights on
the edges.
If G is a graph and there exists a weight function w : E(G) → R+ , then (G, w) is called a
weighted graph.
Let (G, w) be a connected weighted graph. For
a path P in (G, w) the weight wG (P ) of P is defined by
X
wG (P ) =
w(e).
D(G) = max DG (φ),
where the maximum is taken over all permutations
φ of V (G).
E.T.H. Wang proposed a problem [5] which is
equivalent to determining the displacements of all
permutations of the path Pn (a path on n vertices).
He and other solvers obtained that all the possible
2
values are 0, 2, 4, · · · , b n2 c.
Next we give the definition of the status of a
graph. Let G be a connected graph. For a vertex
e∈E(P )
For two vertices x, y in (G, w), the weight distance dG,w (x, y) between x and y is defined by
∗ This research was supported by NSC of R.O.C. under
grant NSC 98-2115-M-008-004
dG,w (x, y) = min wG (P ),
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The 27th Workshop on Combinatorial Mathematics and Computation Theory
where the minimum is taken over all paths P
which join x and y. Note that if w(e) = 1 for every
edge e in (G, w), then the weight distance is the
distance in usual sense, i.e. dG,w (x, y) = dG (x, y).
It is easy to see that for a connected weighted
graph (G, w), we have dG,w (x, z) + dG,w (z, y) ≥
dG,w (x, y) where x, y, z ∈ V (G), and if v is a cut
vertex of G and x, y are vertices in different components of G − v, then dG,w (x, v) + dG,w (v, y) =
dG,w (x, y).
Let (G, w) be a connected weighted graph and
let φ be a permutation of V (G). The displacement
DG,w (φ) of φ is defined by
X
DG,w (φ) =
dG,w (x, φ(x)).
X
+
(dG,w (y, x) − dG,w (y, v))
y∈V (G)−V (G0 )
≥
X
−dG,w (x, v) +
y∈V (G0 )
X
dG,w (v, x)
y∈V (G)−V (G0 )
= (|V (G)| − 2|V (G0 )|) dG,w (v, x)
≥ 0.
Thus sG,w (x) ≥ sG,w (v) for all vertices x 6= v.
Hence v is in the median of (G, w).
2
In this paper, we investigate the relationship
between the status and the displacement of a connected weighted graph and also the displacement
about the product of two weighted graphs.
x∈V (G)
The displacement D(G, w) of (G, w) is defined
by
2
D(G, w) = max DG,w (φ),
where the maximum is taken over all permutations
φ of V (G).
For a vertex x in a connected weighted graph
(G, w), the status sG,w (x) of x is
X
sG,w (x) =
dG,w (y, x) .
Relationship between Displacement and Status
In this section, we investigate the relationship
between the displacement and the status of a connected weighted graph. The displacement of a
graph being the maximum of the displacements of
all permutations of the vertices in the graph and
the status being the minimum of the statuses of
all vertices, the following is an min-max inequality
for these two variants.
y∈V (G)
The status s(G, w) of the graph (G, w) is
s(G, w) = minx∈V (G) sG,w (x).
Theorem 2.1 Suppose that (G, w) is a connected
weighted graph. Then
The median of (G, w) is the set of vertices in (G, w)
with status equal to s(G, w). If w(e) = 1 for every
edge e in (G, w), then sG,w (x) = sG (x) for all
x ∈ V (G), s(G, w) = s(G), and the median of
(G, w) is the same as the median of G.
D(G, w) ≤ 2s(G, w).
Proof. Let φ be an arbitrary permutation of
V (G), and y be an arbitrary vertex of G. Then
X
DG,w (φ) =
dG,w (x, φ(x))
Theorem 1.2 Suppose that (G, w) is a connected
weighted graph which has a cut vertex v such that
|V (G0 )| ≤ 21 |V (G)| for every component G0 of G −
v. Then v is in the median of (G, w).
x∈V (G)
≤
X
(dG,w (x, y) + dG,w (φ(x), y))
x∈V (G)
Proof. Suppose that x is an arbitrary vertex of G
other than v, and x ∈ V (G0 ) for some component
G0 of G − v. Then
=
=
X
X
=
y∈V (G)
2sG,w (y).
Since φ is an arbitrary permutation, and y is an
arbitrary vertex, we have
(dG,w (y, x) − dG,w (y, v))
X
dG,w (φ(x), y)
x∈V (G)
y∈V (G)
=
dG,w (x, y) +
x∈V (G)
sG,w (x) − sG,w (v)
X
X
=
dG,w (y, x) −
dG,w (y, v)
y∈V (G)
X
max DG,w (φ) ≤ min 2sG,w (y).
(dG,w (y, x) − dG,w (y, v))
Thus D(G, w) ≤ 2s(G, w).
y∈V (G0 )
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The 27th Workshop on Combinatorial Mathematics and Computation Theory
Let (G, w) be a weighted graph. A permutation
π of V (G) is optimal for the displacement of (G, w)
if DG,w (π) = D(G, w). In Theorem 2.3 the minmax inequality about displacement and status will
be shown to be an equality for some class of connected graphs, and the optimal permutations for
the displacement are also characterized for these
graphs. Let us begin with the following observation.
dG,w (x, v) + dG,w (v, π(x)). Hence
X
DG,w (π) =
dG,w (x, π(x))
x∈V (G)
=
2
X
dG,w (x, v)
x∈V (G)
= 2sG,w (v)
≥ 2s(G, w).
On the other hand, by Theorem 2.1,
2s(G, w) ≥ D(G, w) ≥ DG,w (π).
Proof. Let X = {x1 , x2 , · · · , x|X| } such that
Xm
(dG,w (x, v) + dG,w (π(x), v))
x∈V (G)
Lemma 2.2 Let X1 , X2 , · · · , Xm be disjoint
nonempty subsets of a set X. If |Xi | ≤ 12 |X| for
i = 1, 2, · · · , m. Then there exists a permutation φ
of X such that φ(Xi ) ∩ Xi = ∅ for i = 1, 2, · · · , m.
X1
X2
X
=
Thus DG,w (π) = D(G, w) = 2s(G, w).
Hence we have proved (1) and the sufficiency
of (2). Now we prove the necessity of (2). Suppose, on the contrary, there exists a vertex z with
dG,w (z, π(z)) < dG,w (z, v) + dG,w (v, π(z)). Then
X
DG,w (π) =
dG,w (x, π(x))
= {x1 , x2 , · · · , x|X1 | },
= {x|X1 |+1 , x|X1 |+2 , · · · , x|X1 |+|X2 | },
..
.
= {x|X1 |+|X2 |+···+|Xm−1 |+1 ,
x|X1 |+|X2 |+···+|Xm−1 |+2 , · · · ,
x|X1 |+|X2 |+···+|Xm | }.
x∈V (G)
<
X
(dG,w (x, v) + dG,w (π(x), v))
x∈V (G)
Let A be an integer such that |Xi | ≤ A ≤ 12 |X|
for i = 1, 2, · · · , m.
Define φ : X → X by π(xi ) = xi+A (the
subscripts being taken modulo |X|) for i =
1, 2, · · · , |X|. It is easy to see that φ satisfies the
required property.
2
=
=
2sG,w (v)
2s(G, w).
The last equality follows from Theorem 1.2. By
(1) of this theorem, we have DG,w (π) < D(G, w),
a contradiction. This complete the proof.
2
Theorem 2.3 Suppose that (G, w) is a connected
weighted graph which has a cut vertex v such that
|V (G0 )| ≤ 12 |V (G)| for every component G0 of G −
v. Then we have
(1) D(G, w) = 2s(G, w),
(2) a permutation π of V (G) is optimal for the
displacement of (G, w) if and only if
dG,w (x, π(x)) = dG,w (x, v) + dG,w (v, π(x)) whenever the vertices x and π(x) are in the same component of G − v.
Corollary 2.4 If (T, w) is a weighted tree, then
(1) D(T, w) = 2s(T, w),
(2) a permutation π of V (T ) is optimal for the
displacement of (T, w) if and only if π(V (T 0 )) ∩
V (T 0 ) = ∅ for every component T 0 of T − v where
v is in the median of T .
Proof.
This follows from Proposition 1.1,
Lemma 2.2 and Theorem 2.3.
2
An easy consequence follows from Corollary 2.4
(1).
Proof. We first show (1) and the sufficiency of
(2). Since |V (G0 )| ≤ 12 |V (G)| for every component
G0 of G − v, by Lemma 2.2 there exists a permutation φ of V (G) such that φ(V (G0 ))∩V (G0 ) = ∅ for
every component G0 of G−v. Let π be any permutation of V (G) with this property dG,w (x, π(x)) =
dG,w (x, v) + dG,w (v, π(x)) whenever the vertices x
and π(x) are in the same component of G − v.
(The permutation φ satisfies this property trivially.) Then for every x ∈ V (G), dG,w (x, π(x)) =
Corollary 2.5 If (T, w) is a tree with integral
weights (i.e., w(e) is an integer for each e ∈
E(T )), then D(T, w) is an even integer.
2
Corollary 2.5 can also follow from the following
result.
Theorem 2.6 Let (T, w) be a weighted tree with
integral weights. Suppose that π is a permutation
of V (T ). Then DT,w (π) is an even integer.
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The 27th Workshop on Combinatorial Mathematics and Computation Theory
Proof. For y, z ∈ V (T ), we use [y, z] to denote
the set of edges in the path which joins y and z.
We have
X
DT,w (π) =
dT,w (x, π(x))





x∈V (T )
=
X
X
w(e)
=
X
w(e)
e∈E(T ) x:e∈[x,π(x)]
=
X
a12
a22
..
.
···
···
..
.
a1i
a2i
..
.
···
···
..
.
a1n
a2n
..
.
am1
am2
···
ami
···
amn





such that each element of X appears n times in A.
Then for each i (i = 1, 2, · · · , m), there exists a
permutation ai1 ai2 · · · ain of ai1 ai2 · · · ain such
that for each j (j = 1, 2, · · · , n), a1j , a2j , · · · , am
j are
distinct elements (i.e., {a1j , a2j , · · · , am
j } = {x1 , x2 ,
· · · , xm }).
x∈V (T ) e∈[x,π(x)]
X
a11
a21
..
.
w(e)|{x : e ∈ [x, π(x)]}|.
e∈E(T )
Proof. First we choose a11 , a21 , · · · , am
1 as follows:
For i = 1, 2, · · · , m, let Bi = {aij : 1 ≤ j ≤ n}.
Let 1 ≤ i1 < i2 < · · · < ik ≤ m. Since each
element in X appears n times in A, it appears at
most n times in the cells of i1 -th, i2 -th, · · ·, ik -th
rows of A. The i1 -th, i2 -th, · · ·, ik -th rows has kn
cells. Thus |Bi1 ∪ Bi2 ∪ · · · ∪ Bik | ≥ k. By Hall’s
Theorem, there exist
Let e be an arbitrary edge in T , we show that
|{x : e ∈ [x, π(x)]}| is an even integer. Let B1 and
B2 be the components of T − e. Then
{x : e ∈ [x, π(x)]} =
S
{x : x ∈ V (B1 ), π(x) ∈ V (B2 )}
{x : x ∈ V (B2 ), π(x) ∈ V (B1 )}.
Since |{x : x ∈ V (B1 ), π(x) ∈ V (B2 )}| =
|{x : x ∈ V (B2 ), π(x) ∈ V (B1 )}|, we have that
|{x : e ∈ [x, π(x)]}| is even. Thus DT,w (π) is even.
2
a1j1 ∈ B1 , a2j2 ∈ B2 , · · · , amjm ∈ Bm
such that a1j1 , a2j2 , · · · , amjm are distinct. Let
3
a11 = a1j1 , a21 = a2j2 , · · · , am
1 = amjm .
Product Result for Displacement
Next we choose a12 , a22 , · · · , am
2 as follows :
For i = 1, 2, · · · , m, let Bi 0 = {aij : 1 ≤ j ≤ n, j 6=
ji }. For 1 ≤ i1 < i2 < · · · < ik ≤ m, by the same
arguments as above, we have |Bi01 ∪ Bi02 ∪ · · · ∪
Bi0k | ≥ k. Again, by Hall’s Theorem, there exist
Let G and H be graphs. The Cartesian product
graph G × H of G and H is the graph with vertex
set V (G × H) = V (G) × V (H) and edge set E(G ×
H) = {((x, y), (x0 , y 0 )) : x = x0 ∈ V (G), yy 0 =
(y, y 0 ) ∈ E(H) or xx0 = (x, x0 ) ∈ E(G), y = y 0 ∈
V (H)}.
Let (G, w) and (H, u) be weighted graphs.
Then the Cartesian product graph of (G, w) and
(H, u) is the weighted graph (G × H, w × u) where
the weight function w × u is defined as follows: if
((x, y), (x0 , y 0 )) ∈ E(G × H) then
w × u((x, y), (x0 , y 0 )) =
0
u(yy ) if
w(xx0 ) if
0
0 ∈ B
a1j10 ∈ B10 , a2j20 ∈ B20 , · · · , amjm
m
0
such that a1j10 , a2j20 , · · · , amjm
are distinct. Let
0 .
a12 = a1j10 , a22 = a2j20 , · · · , am
2 = amjm
Repeating the above procedure, we eventually have aij 1 ≤ i ≤ m, 1 ≤ j ≤ n such
that a1j , a2j , · · · , am
j are distinct for every j (j =
1, 2, · · · , n). We can also see that ai1 ai2 · · · ain
is a permutation of ai1 ai2 · · · ain for each i(i =
1, 2, · · · , m).
2
0
x=x,
y = y0 .
It is easy to see that if (x, y) and (x0 , y 0 ) are
vertices in (G × H, w × u), then
dG×H,w×u ((x, y), (x0 , y 0 ))
= dG,w (x, x0 ) + dH,u (y, y 0 ).
Now we will establish a product result about
displacements of weighted graphs. We begin with
a lemma.
Now we prove the product result for displacements.
Theorem 3.2 Suppose that (G, w), (H, u) are
connected weighted graphs, and the orders of G
and H are m and n respectively. Then
Lemma 3.1 Let X be the set {x1 , x2 , · · · , xm },
and A be an m by n matrix
D(G × H, w × u) = nD(G, w) + mD(H, u).
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The 27th Workshop on Combinatorial Mathematics and Computation Theory
Proof. First we prove that D(G × H, w × u) ≥
nD(G, w) + mD(H, u). Let π1 be a permutation
of V (G) such that DG,w (π1 ) = D(G, w). Let π2
be a permutation of V (H) such that DH,u (π2 ) =
D(H, u). Define π : V (G × H) → V (G × H) by
π(x, y) = (π1 (x), π2 (y)). Then
Here we show that D1 ≤ n D(G, w). Let V (G) =
{x1 , x2 , · · · , xm }, V (H) = {y1 , y2 , · · · , yn }. Since π
is a permutation of V (G × H), we have {π(xi , yj ) :
1 ≤ i ≤ m, 1 ≤ j ≤ n} = {(xi , yj ) : 1 ≤ i ≤ m,
1 ≤ j ≤ n}. Thus each xi (1 ≤ i ≤ m) appears n
times in the array
(p1 (π(xi , yj )))1≤i≤m,1≤j≤n .
(x,y)∈V (G×H)
X
dG×H,w×u ((x, y), (π1 (x), π2 (y)))
(x,y)∈V (G×H)
X
=
(dG,w (x, π1 (x)) + dH,u (y, π2 (y)))
(x,y)∈V (G×H)
X
=
dG,w (x, π1 (x))
Let aij = p1 (π(xi , yj )), 1 ≤ i ≤ m, 1 ≤ j ≤ n. By
Lemma 3.1, for each i (i = 1, 2, · · · , m), there exists a permutation ai1 ai2 · · · ain of ai1 ai2 · · · ain such
that for each j (j = 1, 2, · · · , n), {a1j , a2j , · · · am
j } =
{x1 , x2 , · · · xm }. Thus
X X
D1 =
dG,w (xi , p1 (π(xi , yj )))
1≤i≤m 1≤j≤n
(x,y)∈V (G×H)
X
+
dH,u (y, p2 (π(x, y))).
(x,y)∈V (G×H)
DG×H,w×u (π)
X
=
dG×H,w×u ((x, y), π(x, y))
=
X
D2 =
=
dH,u (y, π2 (y))
X
X
dG,w (xi , aij )
1≤i≤m 1≤j≤n
(x,y)∈V (G×H)
=
= nDG,w (π1 ) + mDH,u (π2 )
= nD(G, w) + mD(H, u).
X
X
dG,w (xi , aij )
1≤i≤m 1≤j≤n
=
Since D(G × H, w × u) ≥ DG×H,w×u (π), we
have D(G × H, w × u) ≥ n D(G, w) + m D(H, u).
Next we prove the reverse inequality D(G ×
H, w × u) ≤ n D(G, w) + m D(H, u). Let π be an
arbitrary permutation of V (G × H). We will show
that DG×H,w×u (π) ≤ n D(G, w) + m D(H, u). Let
p1 be the function from V (G × H) to V (G) such
that p1 (x, y) = x for (x, y) ∈ V (G × H), and p2
be the function from V (G × H) to V (H) such that
p2 (x, y) = y for (x, y) ∈ V (G × H). Note
X
X
dG,w (xi , aij )
1≤j≤n 1≤i≤m
≤
X
D(G, w)
1≤j≤n
= n D(G, w).
Similarly, we can show that D2 ≤ m D(H, u).
Thus
DG×H,w×u (π)
dG×H,w×u ((x, y), π(x, y))
= dG×H,w×u ((x, y), (p1 (π(x, y)), p2 (π(x, y))))
= dG,w (x, p1 (π(x, y))) + dH,u (y, p2 (π(x, y))).
= D1 + D2
≤ n D(G, w) + m D(H, u).
Since π is an arbitrary permutation of V (G × H),
we obtain
D(G × H, w × u) ≤ n D(G, w) + m D(H, u).
This completes the proof.
2
Thus
DG×H,w×u (π)
X
=
dG×H,w×u ((x, y), π(x, y))
References
[1] F. Buckley and F. Harary, Distance in Graphs,
Addison-Wesley Publishing company (1990).
(x,y)∈V (G×H)
X
=
dG,w (x, p1 (π(x, y)))
[2] A. Kang and D. Ault, Some properties of a
centroid of a free tree, Inform. Process. Lett.
4, No. 1 (1975) 18-20.
(x,y)∈V (G×H)
+
X
dH,u (y, p2 (π(x, y))).
(x,y)∈V (G×H)
[3] C. Lin, Y.-J. Zhang and J.-L. Shang, Statuses
of Graphs, Proceedings of the 26th Workshop
on Combinatorial Mathematics and Computation Theory (2009) 282-287.
Let
D1 =
X
dG,w (x, p1 (π(x, y))),
(x,y)∈V (G×H)
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The 27th Workshop on Combinatorial Mathematics and Computation Theory
[4] O. Kariv and S.L. Hakimi, An algorithmic approach to network location problems. II: The
p-medians, Siam J. Appl. Math. 37 (1979) 539560.
[5] E.T.H. Wang, The symmetric group as a metric space, E2424, Amer. Math. Monthly (1974)
668-670.
101