The sum numbers and
the integral sum numbers of Pn and Fn
Haiying Wang1∗
Jingzhen Gao2 †
1. The School of Information Engineering
China University of Geosciences(Beijing)
Beijing 100083, P.R.China
2. Department of Mathematics and Science
Shandong Normal University
Jinan, Shandong, 250014,P.R.China
Abstract Let G = (V, E) be a simple graph with the vertex set V
and the edge set E. G is a sum graph if there exists a labelling f of
the vertices of G into distinct positive integers such that uv ∈ E if and
only if f (w) = f (u) + f (v) for some vertex w ∈ V . Such a labelling f is
called a sum labelling of G. The sum number σ(G) of G is the smallest
number of isolated vertices which result in a sum graph when added to G.
Similarly, the integral sum graph and the integral sum number ζ(G) are
also defined. The difference is that the labels may be any distinct integers.
In this paper, we will determine that
⎧
0 = ζ(P4 ) < σ(P4 ) = 1;
⎪
⎪
⎪
⎪
⎨ 1 = ζ(P5 ) < σ(P5 ) = 2;
3 = ζ(P6 ) < σ(P6 ) = 4;
⎪
⎪
⎪
ζ(Pn ) = σ(Pn ) = 0, n = 1, 2, 3;
⎪
⎩
ζ(Pn ) = σ(Pn ) = 2n − 7, n ≥ 7.
and
⎧
0 = ζ(F5 ) < σ(F5 ) = 1;
⎪
⎪
⎨
2 = ζ(F6 ) < σ(F6 ) = 3;
ζ(Fn ) = σ(Fn ) = 0, n = 3, 4;
⎪
⎪
⎩
ζ(Fn ) = σ(Fn ) = 2n − 8, n ≥ 7.
Keywords
The sum graph; The integral sum graph; The sum
number; The integral sum number; Path; Fan.
∗ E-mail:
[email protected].
research is supported by 2008 Foundation of China University of Geosciences (Beijing)(NO.51900961144) and National Nature Science Foundation of China (NO.10671014).
† This
1
1. Introduction
Let G = (V, E) be a simple graph with the vertex set V and the edge set E.
The complement G of G with order n is the graph with the vertex set V and the
edge set E(Kn ) − E. A path Pn is a graph with the vertex set {a1 ,a2 ,...,an } and
the edge set {a1 a2 ,a2 a3 ,...,an−1 an }, and a1 and an are called the end vertices
of Pn . A fan Fn is a graph with the vertex set {c, a1 , a2 , ..., an } and the edge
set {ca1 ,ca2 , ..., can } ∪ {a1 a2 , a2 a3 ,...,an−1 an }. It is obvious that Fn = Pn ∪ K1 .
A sum graph and an integral sum graph were introduced by Frank Harary
in [2] and [3]. G is a sum graph if there exists a labelling f of the vertices of G
into distinct positive integers such that uv ∈ E if and only if f (w) = f (u)+ f (v)
for some vertex w ∈ V . Such a labelling f is called a sum labelling of G. A sum
graph cannot be connected. There must always be at least one isolated vertex.
The sum number σ(G) of G is the smallest number of isolated vertices which
result in a sum graph when added to G. Similarly, an integral sum graph and
an integral sum number ζ(G) are also defined. The difference is that the labels
may be any distinct integers. Obviously ζ(G) ≤ σ(G).
A vertex w of G is working if its label corresponds to an edge uv of G. G is
exclusive if none of the vertices in V is working. For example, Kn and W2n−1
are exclusive in [9].
To simplify the notations, we may assume that the vertices of G are identified
with their labels throughout this paper. And let Vi and Ei denote the set of
the vertices independent of ai and the set of the edges adjacent to ai in Pn
respectively. Besides, some results have been obtained as follow.
Lemma 1([2]) σ(Pn ) = 1 and ζ(Pn ) = 0 for n ≥ 2.
Lemma 2([1][8]) ζ(Kn ) = σ(Kn ) = 2n − 3 for n ≥ 4.
Lemma 3([8]) ζ(Cn ) = ζ(Wn ) = 0 for n= 5.
2, n = 4,
Lemma 4([2]) For n ≥ 3,
σ(Cn ) =
3, n = 4.
n
2 + 2, n even,
Lemma 5([10][7]) For n ≥ 3, σ(Wn ) =
n,
n odd.
In this paper, we will determine the sum numbers and the integral sum
numbers of Pn and Fn for n ≥ 1.
2. Main results
Let Pn = (V, E) and S = V ∪ C, where V = {a1 , a2 , ..., an } and C is
the isolated vertex set. It is clear that P2 = 2K1 and P3 = P2 ∪ K1 and
P4 = P4 . By Lemma 1 and Lemma 2, ζ(Pi ) = σ(Pi ) = 0 for i = 1, 2, 3 and
0 = ζ(P4 ) < σ(P4 ) = 1. In this section, we only consider n ≥ 5.
2
Lemma 2.1 Pn is not an integral sum graph for n ≥ 5.
Proof: Let |ax | = max{|a| : a ∈ V } and ax ∈ V . Assume that ax > 0 (A
similar argument work for ax < 0). By contradiction. If Pn is an integral sum
graph for n ≥ 5 then 0 ∈ V and ax + ai ∈ V for all ax ai ∈ E. Then ax + ai > 0
and ai < 0 according to the choice of ax . So we get at least n − 3 distinct
positive vertices ax + ai in V . Meanwhile, we also get at least n − 3 distinct
negative vertices ai . So 2(n − 3) + 1 ≤ n, that is, n ≤ 5. Since n ≥ 5, only
n = 5.
Assume that V (P5 ) = {ax , a1 , a2 , ax + a1 , ax + a2 } with ai < 0 (i = 1, 2).
By the choice of ax , (ax + ai )ax ∈ E with i = 1, 2 and Vx = {ax + a1 , ax + a2 }
(see Figure 1). So (ax + a1 )(ax + a2 ) ∈ E, that is, (ax + a1 ) + (ax + a2 ) ∈
{ax , ax + a1 , ax + a2 }. Thus, (ax + a1 ) + (ax + a2 ) = ax , that is, ax + a1 = −a2
and ax + a2 = −a1 . So (ax + a1 )a2 ∈ E and (ax + a2 )a1 ∈ E and a1 a2 ∈ E (see
Figure 1), contracting the structure of P5 .
Thus, Lemma 2.1 holds.
2
ax
0<−a1 =a x +a2
ax +a1=−a >0
2
a2 <0
a1 <0
Figure 1
Lemma 2.2 ζ(P5 ) = 1.
Proof: By Lemma 2.1, ζ(P5 ) ≥ 1. Below we will give an integral sum
labelling of P5 ∪ K1 (see Figure 2). So ζ(P5 ) ≤ 1. Thus, ζ(P5 ) = 1. 2
_1
1
2
4
3
Figure 2
3
6
Lemma 2.3 σ(P5 ) = 2.
Proof: It is clear that the sum number of a graph must be at least as large
as the minimum degree of the graph, so σ(P5 ) ≥ 2. Figure 3 below shows that
σ(P5 ) ≤ 2. Thus, σ(P5 ) = 2. 2
1
5
6
12
16
10
11
Figure 3
Lemma 2.4 If ax ∈ V with |ax | = max{|a| : a ∈ V }, then there exists one
edge ax aj0 ∈ E such that ax + aj0 ∈ C for n ≥ 6.
Proof: Let |ax | = max{|a| : a ∈ V } with ax ∈ V . Assume that ax > 0 (A
similar argument works for ax < 0.). By contradiction. Suppose to the contrary
that ax + aj ∈ V for all ax aj ∈ E. According to the choice of ax , ax + aj > 0
and aj < 0. Then there are at least n − 3 distinct positive vertices ax + aj and
n − 3 distinct negative vertices adjacent to ax . So (n − 3) + (n − 3) + 1 ≤ n,
i.e., n ≤ 5, contradicting the fact n ≥ 6.
Thus, Lemma 2.4 holds. 2
Lemma 2.5
ζ(P6 ) = 3.
Proof: Below we give an integral sum labelling of P6 (see Figure 4). So
ζ(P6 ) ≤ 3. What we need to do is only to prove ζ(P6 ) ≥ 3.
6
8
11
7
13
5
15
1
_
1
Figure 4
4
Let |ax | = max{|a| : a ∈ V } with ax ∈ V . Assume that ax > 0 (A similar
argument works for ax < 0). By Lemma 2.2, there exists one edge ax aj0 ∈ Ex
such that ax + aj0 ∈ C. Firstly, we will show Claim 1 and Claim 2 and Claim 3.
Claim 1: There exists another edge ax al ∈ Ex −ax aj0 such that ax +al ∈ C.
By contradiction. Suppose to the contrary that ax + al ∈ V for all ax al ∈
Ex − ax aj0 . By the choice of ax , ax + al > 0 and al < 0. Since ax + aj0 ∈ C,
(ax + aj0 ) + al = (ax + al ) + aj0 ∈ S. Then ax + al ∈ {aj0 } ∪ Vj0 , denoted (1).
If ax is an end vertex of P6 then |Ex | = 4. So there are four distinct
positive vertices and at least three distinct negative vertices in V . So n > 6, a
contraction. So ax is not an end vertex of P6 .
Let Vx = {ax+1 , ax−1 } and V = {ax , ax+1 , ax−1 , aj0 , al1 , al2 } (see Figure 5).
Assume ai , aj are two end vertices of P6 . Then ai aj ∈ E and ai + aj ∈ S.
ax >0
a x+1
ax−1
al1 <0
aj <0
0
ax +a j
0
al <0
2
Figure 5
If Vj0 ⊆ {al1 , al2 } and ai aj ∈ {ax+1 al1 , aj0 al2 , al2 ax−1 } then we may assume
that al1 aj0 ∈ E (see Figure 6). By (1), ax + al1 = aj0 and ax + al2 = aj0 , a
contraction. So Vj0 ⊆ {al1 , ax+1 }.
ax >0
ax+1
ax−1
0> al1
al <0
2
aj0
Figure 6
5
ax +a j
0
By (1) and al1 < 0, {ax + al1 , ax + al2 } = {aj0 , ax+1 } and ax+1 aj0 ∈ E (see
Figure 7).
ax >0
a x+1
ax−1
aj0
al2 <0
ax +a j0
al1 <0
Figure 7
If ax + al1 = ax+1 and ax + al2 = aj0 then al1 + aj0 = ax+1 + al2 ∈ S. So
al1 aj0 ∈ E. Since ax−1 + aj0 = ax−1 + (ax + al2 ) = ax + (ax−1 + al2 ) ∈ S,
ax−1 + al2 = al1 , contracting ax−1 al2 ∈ E.
If ax + al1 = aj0 and ax + al2 = ax+1 then al1 + ax+1 = aj0 + al2 ∈ S. Uniting
al1 + ax+1 = aj0 + al2 ∈ S and ax−1 + al1 = al2 , we have aj0 + ax−1 = ax+1 .
If al1 + ax+1 = aj0 + al2 ∈ V then al1 + ax+1 = aj0 + al2 = ax−1 . Since
ax−1 = aj0 + al2 = aj0 + (ax−1 + al1 ), aj0 + al1 = 0. Thus, (ax + aj0 ) + al1 =
ax + (aj0 + al1 ) = ax ∈ S, contracting ax + aj0 ∈ C. If al1 + ax+1 = aj0 + al2 ∈ C
then ax+1 + al2 = (aj0 + ax−1 ) + al2 = (aj0 + al2 ) + ax−1 ∈ S, contracting
aj0 + al2 ∈ C.
Thus, Claim 1 holds.
Up to now, we may assume that ax + aj0 ∈ C and ax + al1 ∈ C with
{ax aj0 , ax al1 } ⊆ Ex .
Claim 2: If ax is one end vertex of P6 then ζ(P6 ) ≥ 3.
In fact, if ax is an end vertex of P6 then |Vx | = 1. Let Vx = {ax+1 }. Assume
that aj0 ax−1 ∈ E (see Figure 8) (if not, we can consider al1 ax−1 ∈ E).
ax
ax+1
ax−1
ax +a j
0
ax +a
al1
aj0
al2
Figure 8
6
l1
By contradiction. Suppose to the contrary that ζ(P6 ) ≤ 2. By Claim 1,
ζ(P6 ) ≥ 2. So ζ(P6 ) = 2. Let C = {ax + aj0 , ax + al1 }. Then {ax + ax−1 , ax +
al2 } ⊂ ({aj0 } ∪ Vx ) ∩ ({al1 } ∪ Vl1 ).
According to the choice of ax , ax−1 < 0 and al2 < 0. So there is only one case
of {ax +ax−1 , ax +al2 } = {aj0 , al1 } and P6 = ax ax+1 aj0 al2 al1 ax−1 (see Figure 9).
ax >0
a x+1
ax−1<0
aj0
ax +a j0
ax +a l1
al
1
al2 <0
Figure 9
If ax + ax−1 = aj0 and ax + al2 = al1 then aj0 + al1 = (ax + ax−1 ) + al1 =
(ax + al1 ) + ax−1 ∈ S, contracting ax + al1 ∈ C.
If ax + ax−1 = al1 and ax + al2 = aj0 then aj0 + al1 = (ax + al2 ) + aj0 =
(ax + aj0 ) + al2 ∈ S, contracting ax + aj0 ∈ C.
Thus, Claim 2 holds.
Claim 3: If ax is not an end vertex of P6 then ζ(P6 ) ≥ 3.
In fact, if ax is not an end vertex of P6 then |Vx | = 2. Let Vx = {ax+1 , ax−1 }.
Then ax ax+1 ∈ E and ax ax−1 ∈ E. Let ax al2 ∈ Ex − {ax aj0 , ax al1 }. Since
{ax + aj0 , ax + al1 } ⊆ C, ax + al2 ∈ [({aj0 } ∪ Vj0 ) ∩ ({al1 } ∪ Vl1 )] ∪ C.
By contradiction. Suppose to the contrary that ζ(P6 ) ≤ 2. By Claim 1,
ζ(P6 ) ≥ 2. So ζ(P6 ) = 2. Let C = {ax + aj0 , ax + al1 }. Then ax + al2 ∈ V . So
al2 < 0 and it is impossible that both of aj0 and al1 are adjacent to al2 . Assume ai aj ∈ {ax+1 aj0 , aj0 al1 , al1 al2 , al2 ax−1 }. Then ax + al2 ∈ {aj0 , al1 } with
aj0 al1 ∈ E ( if aj0 al1 ∈ E then ax + al2 ∈ ({aj0 } ∪ Vj0 ) ∩ ({al1 } ∪ Vl1 ) = ∅. It is
impossible.)(see Figure 10).
ax
ax−1
ax+1
ax +a j
0
ax +a l
al2
aj0
al1
7
1
Figure 10
Similarly, aj0 + ax−1 ∈ {ax , ax+1 } ∪ C; ax+1 + al1 ∈ {ax , ax−1 } ∪ C; ax−1 +
al1 ∈ {ax , ax+1 } ∪ C; ax+1 + aj0 ∈ {ax , ax−1 } ∪ C; aj0 + al2 ∈ {ax−1 , ax+1 } ∪ C.
(1.1) If ax +al2 = aj0 and aj0 +ax−1 = ax then ax +al1 = (aj0 +ax−1 )+al1 =
aj0 + (ax−1 + al1 ) ∈ S. So ax−1 + al1 = ax+1 , which implies ax + al1 = aj0 + ax+1
and aj0 ax+1 ∈ E. Since (ax + aj0 ) + ax+1 = ax + (aj0 + ax+1 ) ∈ S, aj0 + ax+1 ∈
{ax−1 } ∪ C.
(1.1.1) If aj0 + ax+1 = ax−1 then al1 + aj0 = 0 (since ax−1 + al1 = ax+1 ).
So ax + al1 = (aj0 + ax−1 ) − aj0 = ax−1 ∈ V , contracting ax + al1 ∈ C.
(1.1.2) If ax+1 + aj0 ∈ C, (ax+1 + aj0 ) + al1 = (ax+1 + al1 ) + aj0 ∈ S.
Then ax+1 + al1 ∈ C. So {ax+1 + aj0 , ax+1 + al1 } ⊆ C = {ax + aj0 , ax + al1 }, a
contraction.
(1.2) If ax + al2 = aj0 and aj0 + ax−1 = ax+1 then ax+1 + al1 = (aj0 +
ax−1 ) + al1 = aj0 + (ax−1 + al1 ) ∈ S. So ax−1 + al1 = ax . Since ax+1 + al2 =
(aj0 +ax−1 )+al2 = (aj0 +al2 )+ax−1 ∈ S, aj0 +al2 = ax−1 . Uniting ax +al2 = aj0
and aj0 + al2 = ax−1 , we have ax + 2al2 = ax−1 , contracting the choice of ax .
(1.3) If ax + al2 = aj0 and aj0 + ax−1 ∈ C then al2 + (aj0 + ax−1 ) =
(al2 + aj0 ) + ax−1 ∈ S. So al2 + aj0 = ax−1 ( If not, then al2 + aj0 ∈ C, but
al2 + aj0 ∈ C = {ax + aj0 , ax + al1 }, a contraction.). Then (ax+1 + al2 ) + aj0 =
ax+1 + (al2 + aj0 ) = ax+1 + ax−1 ∈ S. So ax+1 + al2 = ax , contracting the choice
of ax .
(2) If ax +al2 = al1 then al1 +ax+1 = (ax +al2 )+ax+1 = ax +(al2 +ax+1) ∈ S.
So al2 + ax+1 = aj0 and ax + aj0 = ax + (al2 + ax+1 ) = (ax + al2 ) + ax+1 =
al1 + ax+1 ∈ C. Since aj0 + ax−1 = (al2 + ax+1 )+ ax−1 = al2 + (ax+1 + ax−1 ) ∈ S
and (al1 + ax+1 ) + ax−1 = al1 + (ax+1 + ax−1 ) ∈ S, ax+1 + ax−1 ∈ {al1 , al2 }.
(2.1) If ax+1 + ax−1 = al1 then aj0 + ax−1 = al1 + al2 ∈ S. So al1 al2 ∈ E
and aj0 + ax−1 = al2 + al1 = ax+1 (Since aj0 + ax−1 = al2 + al1 = ax+1 ∈ C =
{ax + aj0 , ax + al1 }).
Since aj0 + ax−1 = (al2 + ax+1 ) + ax−1 = ax+1 , al2 + ax−1 = 0. Uniting
al1 = ax + al2 , we have al1 + ax−1 = ax .
Since C = {ax + aj0 , ax + al1 } and the choice of ax , we have aj0 + al2 ∈ C.
So aj0 + al2 = ax−1 (note that aj0 + ax−1 = ax+1 ).
Assume that ax−1 = x. By the above, aj0 = 2x and ax+1 = 3x and ax = 5x.
Since ax+1 + aj0 = 5x = ax , contracting ax+1 aj0 ∈ E.
(2.2) If ax+1 + ax−1 = al2 then 2al2 = aj0 + ax−1 = ax+1 (If aj0 + ax−1 ∈ C
then (aj0 + ax−1 ) + ax+1 = aj0 + (ax−1 + ax+1 ) = aj0 + al2 ∈ S, a contraction).
Since aj0 + al2 = ax+1 = aj0 + ax−1 , aj0 + al2 = ax−1 or aj0 + al2 ∈ C.
(2.2.1) If aj0 + al2 = ax−1 then −al2 = ax−1 = aj0 + al2 = (al2 + ax+1 ) + al2 .
So ax+1 = −3al2 , contracting ax+1 = 2al2 .
(2.2.2) If aj0 + al2 ∈ C = {ax + aj0 , ax + al1 } then aj0 + al2 = ax + al1 =
ax + (ax + al2 ) = 2ax + al2 . So aj0 = 2ax , contracting the choice of ax .
Therefore, Claim 3 holds.
Thus, Lemma 2.5 holds. 2
8
Lemma 2.6 σ(P6 ) = 4.
Proof: Let V = {ax , ax+1 , ax−1 , al1 , al2 , aj0 } and ax = max{a : a ∈ V }.
Then ax +ai ∈ C for all ax ai ∈ E. Firstly, Figure 11 below shows that σ(P6 ) ≤ 4.
What we need is to prove σ(P6 ) ≥ 4.
19
21
24
26
7
31
2
17
36
12
Figure 11
If ax is an end vertex of Pn then σ(P6 ) ≥ 4. Otherwise, it is clear that
σ(P6 ) ≥ 3. Below we will prove that σ(P6 ) = 3.
By contradiction. Suppose to the contrary that σ(P6 ) = 3. Then C =
{ax + al1 , ax + al2 , ax + aj0 }. Assume that Vx = {ax+1 , ax−1 } and ai aj ∈
{al1 ax+1 , al2 al1 , aj0 al2 , ax−1 aj0 }, where ai and aj are two end vertices of P6
(see Figure 12).
ax
a x+1
ax−1
ax +a l1
ax+a l
2
al1
aj0
ax +a j
0
al2
Figure 12
Since (ax + aj0 ) + al1 = ax + (aj0 + al1 ) ∈ S, aj0 + al1 ∈ {ax , ax+1 , ax−1 } ∪ C.
Similarly, ax−1 + al1 ∈ {ax , ax+1 } ∪ C; ax−1 + al2 ∈ {ax , ax+1 } ∪ C; ax+1 + al2 ∈
{ax , ax−1 } ∪ C; ax+1 + aj0 ∈ {ax , ax−1 } ∪ C.
(I) If aj0 +al1 = ax then ax +ax−1 = (aj0 +al1 )+ax−1 = (ax−1 +al1 )+aj0 ∈
S, which implies ax−1 + al1 ∈ {aj0 } ∪ Vj0 ∪ C. By the above, ax−1 + al1 ∈ C. So
9
(ax−1 +al1 )+al2 = al1 +(ax−1 +al2 ) ∈ S, which implies ax−1 +al2 ∈ {ax+1 }∪C.
Similarly, ax−1 + ax+1 ∈ {al1 } ∪ Vl1 ∪ C with Vl1 ⊆ {ax+1 , al2 }.
(I.1) If ax−1 + al2 = ax+1 then ax−1 + ax+1 ∈ {al1 } ∪ C. Furthermore,
ax−1 + ax+1 ∈ C. (If not, then ax−1 + ax+1 = al1 . Then ax = aj0 + al1 =
aj0 +(ax−1 +ax+1 ) = (aj0 +ax+1 )+ax−1 ∈ S. So aj0 +ax+1 = al1 (= ax−1 +ax+1 ),
which implies aj0 + al2 = ax+1 , a contradiction with ax−1 + al2 = ax+1 .). So
(ax−1 +ax+1 )+al2 = ax−1 +(ax+1 +al2 ) ∈ S. By the above, ax+1 +al2 ∈ {ax }∪C.
(I.1.1) If ax+1 + al2 = ax then ax + aj0 = (ax+1 + al2 ) + aj0 = (ax+1 +
aj0 ) + al2 ∈ S, which implies ax+1 + aj0 ∈ V . By the above, ax+1 + aj0 = ax−1 .
Uniting ax+1 = ax−1 + al2 , we have al2 + aj0 = 0, a contradiction.
(I.1.2) If ax+1 + al2 ∈ C then (ax+1 + al2 ) + aj0 = (ax+1 + aj0 ) + al2 ∈ S,
which implies ax+1 + aj0 ∈ {al2 } ∪ Vl2 ∪ C. By the above, ax+1 + aj0 ∈ C.
(I.1.2.1) If ax+1 al1 ∈ E then ax+1 + al1 ∈ S. So ax+1 + al1 = (ax−1 + al2 ) +
al1 = al2 + (ax−1 + al1 ) ∈ S, contradicting ax−1 + al1 ∈ C.
(I.1.2.2) If al1 al2 ∈ E then al1 + al2 ∈ S. Uniting ax+1 + al1 = (ax−1 + al2 )+
al1 = ax−1 +(al1 +al2 ) ∈ S and ax +al2 = (aj0 +al1 )+al2 = aj0 +(al1 +al2 ) ∈ S,
we have al1 + al2 = aj0 . So (ax + al2 ) + al1 = ax + (al1 + al2 ) = ax + aj0 ∈ S,
contradicting ax + al2 ∈ C.
(I.1.2.3) If al2 aj0 ∈ E then al2 + aj0 ∈ S. Since ax + al2 = (aj0 + al1 ) + al2 =
(aj0 + al2 ) + al1 ∈ S, aj0 + al2 ∈ {al1 , ax−1 }. Uniting (aj0 + al2 ) + ax = (ax +
al2 ) + aj0 ∈ S, we have aj0 + al2 = ax−1 . So ax+1 + ax−1 = ax+1 + (aj0 + al2 ) =
(ax+1 + al2 ) + aj0 ∈ S, a contradiction.
(I.1.2.4) If ax−1 aj0 ∈ E then ax−1 +aj0 ∈ S. Since ax+1 +aj0 = (ax−1 +al2 )+
aj0 = (ax−1 + aj0 ) + al2 ∈ S, ax−1 + aj0 = al2 . So ax + al2 = ax + (ax−1 + aj0 ) =
(ax + aj0 ) + ax−1 ∈ S, contradicting ax + aj0 ∈ C.
(I.2) If ax−1 + al2 ∈ C then ax−1 + ax+1 ∈ {al2 } ∪ Vl2 ∪ C. Since ax−1 + al1 ∈
C, ax−1 + ax+1 ∈ {al1 } ∪ Vl1 ∪ C. So ax−1 + ax+1 ∈ {al1 , al2 } ∪ C. Since
ax+1 + aj0 = ax = aj0 + al1 , ax+1 + aj0 ∈ {ax−1 } ∪ C. Since {ax−1 + al1 , ax−1 +
al2 } ⊂ C = {ax + al1 , ax + al2 , ax + aj0 }, ax−1 + al1 ∈ {ax + al2 , ax + aj0 } and
ax−1 + al2 ∈ {ax + al1 , ax + aj0 }.
(I.2.1) If ax−1 + al1 = ax + al2 (= (aj0 + al1 ) + al2 ) then ax−1 = aj0 + al2
and ax−1 + al2 = ax + aj0 . So 2al2 = al1 + aj0 = ax . Since ax+1 + aj0 = ax−1 =
aj0 +al2 , ax+1 +aj0 ∈ C. Uniting ax +al2 = ax−1 +al1 and ax +aj0 = ax−1 +al2 ,
we have ax+1 + aj0 = ax + al1 . So ax+1 = 2al1 and ax−1 + ax+1 ∈ C, but
ax−1 + ax+1 ∈ {ax + al1 , ax + al2 , ax + aj0 }, a contradiction.
(I.2.2) If ax−1 + al1 = ax + aj0 (= (aj0 + al1 ) + aj0 ) then ax−1 + al2 = ax + al1
and ax−1 = 2aj0 . So ax+1 + aj0 ∈ {ax + al1 , ax + al2 } ⊂ C.
(I.2.2.1) If ax+1 + aj0 = ax + al1 (= (aj0 + al1 )+ al1 = 2al1 + aj0 = al2 + ax−1 )
then 2al1 = ax+1 . If ax+1 + aj0 = ax−1 then 2al1 + aj0 = ax−1 = 2aj0 . So
2al1 = aj0 , contradicting ax+1 = 2al1 . So ax+1 +aj0 = ax +al1 = al2 +ax−1 ∈ C.
By the choice of ax , ax+1 + ax−1 ∈ C. Since ax+1 + ax−1 = 2al1 + 2aj0 =
2(al1 + aj0 ) = 2ax ∈ S, contradicting the choice of ax .
(I.2.2.2) If ax+1 + aj0 = ax + al2 (= (aj0 + al1 ) + al2 ) then al1 + al2 = ax+1 .
So ax−1 + ax+1 ∈ C, but, ax−1 + ax+1 ∈ {ax + al1 , ax + al2 , ax + aj0 } =
{ax−1 + al2 , ax+1 + aj0 , ax−1 + al1 }, a contradiction.
10
(II) If aj0 + al1 = ax+1 then ax+1 + ax−1 = (aj0 + al1 ) + ax−1 = (aj0 +
ax−1 ) + al1 ∈ S. Then ax−1 + aj0 = ax . Since ax + al2 = (ax−1 + aj0 ) + al2 =
(ax−1 + al2 ) + aj0 ∈ S, ax−1 + al2 = ax+1 (= aj0 + al1 ).
Uniting ax−1 + aj0 = ax and ax−1 + al2 = ax+1 , we have ax + al2 = ax+1 +
aj0 ∈ C.
Uniting ax−1 + aj0 = ax and aj0 + al1 = ax+1 , we have ax + al1 = ax+1 +
ax−1 ∈ C.
Uniting (ax+1 + aj0 ) + al2 = (ax+1 + al2 ) + aj0 ∈ S and ax−1 + al2 = ax+1 ,
we have ax+1 + al2 ∈ C. Then ax+1 + al2 = ax + aj0 , that is, (ax−1 + aj0 ) + aj0 =
(ax−1 + al2 ) + al2 . So 2aj0 = 2al2 , a contradiction.
A similar argument works for aj0 + al1 = ax+1 .
(III) If aj0 + al1 ∈ C then aj0 + al1 = ax + al2 .
Since (aj0 + al1 ) + ax−1 = (al1 + ax−1 ) + aj0 ∈ S, al1 + ax−1 ∈ C. So
al1 + ax−1 = ax + aj0 .
Since aj0 + al1 ∈ C, (aj0 + al1 ) + ax+1 = (aj0 + ax+1 ) + al1 ∈ S, aj0 + ax+1 =
ax + al1 ∈ C.
Since aj0 + ax+1 ∈ C, (aj0 + ax+1 ) + ax−1 = aj0 + (ax+1 + ax−1 ) ∈ S. So
ax+1 + ax−1 = ax + al2 = aj0 + al1 ∈ C.
Uniting aj0 +ax+1 = ax +al1 and ax+1 +ax−1 = ax +al2 , we have ax−1 +al1 =
aj0 + al2 . Then aj0 al2 ∈ E.
Uniting ax+1 + ax−1 = ax + al2 and al1 + ax−1 = ax + aj0 , we have al1 + al2 =
ax+1 + aj0 ∈ S. Then al1 al2 ∈ E, a contradiction.
So σ(P6 ) = 3.
Thus, Lemma 2.6 holds. 2
Lemma 2.7 Let |ax | = max{|a| : a ∈ V } with ax ∈ V . Then ax + ap ∈ C
for all ax ap ∈ E with n = 7.
Proof: Let |ax | = max{|a| : a ∈ V } with ax ∈ V . Assume that V = {ax ,
ax+1 , ax−1 , aj0 , al1 , al2 , al3 } and ax > 0 (A similar argument works for ax < 0).
By Lemma 2.4, there exists one edge ax aj0 ∈ Ex such that ax + aj0 ∈ C.
Since (ax + aj0 ) + al = (ax + al ) + aj0 ∈ S, ax + al ∈ {aj0 } ∪ Vj0 ∪ C for all
ax al ∈ Ex − ax aj0 .
Claim There exists at least one edge ax al ∈ Ex −ax aj0 such that ax al ∈ C.
In fact, if |Vx | = 2 then |Ex | = 5. Since |{aj0 } ∪ Vj0 | = 3, Claim 1 holds. If
|Vx | = 1 then suppose to the contrary that ax + al ∈ V for all ax al ∈ Ex − ax aj0 .
Then al < 0 and ax + al > 0. So there are at least eight distinct vertices,
contradicting n = 7.
Thus, Claim holds.
Assume that ax ax+1 ∈ E and ax + al1 ∈ C with ax al1 ∈ Ex − ax aj0 . Since
ax al1 ∈ C, ax + ali ∈ {aj0 } ∪ Vj0 ∪ C for all ax ali ∈ Ex − {ax aj0 , ax al1 } with
i = 2, 3. So ax + ali ∈ [({aj0 } ∪ Vj0 ) ∩ ({al1 } ∪ Vl1 )] ∪ C.
By contradiction. Suppose to the contrary that ax al2 ∈ V . By the above,
ax + al2 ∈ ({aj0 } ∪ Vj0 ) ∩ ({al1 } ∪ Vl1 ). There are at most two cases (I) (II) in
11
all (see Figure 13,15). Let ai and aj be two end vertices of P7 .
(I) If ai aj ∈ {ax+1 aj0 , aj0 al1 , al1 al2 , al2 al3 , al3 ax−1 , ax−1 ax } then ax + al2 ∈
{aj0 , al1 } (see Figure 13).
ax
a x+1
ax−1
ax +a j
aj0
al3
ax+a l
al
0
1
al2
1
I: Figure 13
(I.1) If ax + al2 = aj0 then aj0 + al3 = (ax + al3 ) + al2 ∈ S. So ax + al3 ∈ V .
Uniting ax + aj0 ∈ C and ax + al1 ∈ C, we have ax + al3 = al1 , which implies
al1 + al2 = al3 + aj0 and al3 < 0. Then al1 al2 ∈ E (see Figure 14).
ax
a x+1
ax +a l =a j
2
0
al1
ax−1
ax +a j0
al3
ax +a l
1
al2
I.1: Figure 14
Since aj0 + ax−1 = (ax + al2 ) + ax−1 = ax + (al2 + ax−1 ) ∈ S, al2 + ax−1 ∈
{al1 , al3 }.
Since aj0 + ax+1 = (ax + al2 ) + ax+1 = ax + (al2 + ax+1 ) ∈ S, al2 + ax+1 ∈
{ax−1 } ∪ C.
If al2 + ax+1 ∈ C then (al2 + ax+1 ) + ax−1 = (al2 + ax−1 ) + ax+1 ∈ S,
contradicting al2 + ax−1 ∈ {al1 , al3 }. So al2 + ax+1 = ax−1 .
Since al3 +ax−1 = al3 +(ax+1 +al2 ) = al2 +(al3 +ax+1 ) ∈ S and al3 +ax+1 ∈
{aj0 , al2 }, al3 + ax+1 = al2 . Note that al1 + al2 = al3 + aj0 ∈ {ax+1 , ax−1 } ∪ C,
then (al1 + al2 =)al3 + aj0 ∈ C. So (al3 + aj0 ) + ax+1 = (al3 + ax+1 ) + aj0 =
al2 + aj0 ∈ S, contradicting al3 + aj0 ∈ C.
(I.2) If ax + al2 = al1 then ax + al3 = aj0 , which implies al1 + al3 = al2 + aj0 .
So ax +al3 = aj0 (If not, then ax +al3 ∈ C. So (ax +al3 )+al2 = (ax +al2 )+al3 =
al1 + al3 ∈ S, contradicting ax + al3 ∈ C).
12
Since al1 + ax−1 = (ax + al2 ) + ax−1 = ax + (ax−1 + al2 ) ∈ S, ax−1 + al2 ∈
{aj0 , al3 }.
(I.2.1) If ax−1 + al2 = aj0 then ax + aj0 = al1 + ax−1 ∈ C. So al1 + aj0 =
al1 + (ax−1 + al2 ) = (al1 + ax−1 ) + al2 ∈ S, contradicting al1 + ax−1 ∈ C.
(I.2.2) If ax−1 + al2 = al3 then ax−1 + al1 = aj0 (since al1 + al3 = al2 + aj0 ).
So ax + aj0 = ax + (ax−1 + al1 ) = (ax + al1 ) + ax−1 ∈ S, contradicting
ax + al1 ∈ C.
(II) If ai aj ∈ {ax+1 aj0 , aj0 al3 , al3 al1 , al1 al2 , al2 ax−1 , ax−1 ax } then ax +al2 =
al3 and al1 al3 ∈ E and aj0 al3 ∈ E. According to the choice of ax , ax + al3 ∈ C
and al3 > 0 (see Figure 15).
ax
a x+1
ax−1
al
aj0
2
ax +a =a
l2
l
3
ax+a j0
ax+a l
1
ax+a l
3
al
1
II: Figure 15
Since al3 + ax+1 = (ax + al2 ) + ax+1 = ax + (al2 + ax+1 ) ∈ S, al2 + ax+1 ∈
{ax−1 , aj0 , al1 }.
Since (ax + al3 ) + ax−1 = ax + (al3 + ax−1 ) ∈ S, al3 + ax−1 ∈ {ax , ax+1 } ∪ C.
Similarly, ax+1 + al3 ∈ {ax−1 } ∪ C; al1 + ax−1 ∈ {ax , ax+1 } ∪ C; aj0 + ax−1 ∈
{ax , ax+1 } ∪ C; ax+1 + al1 ∈ {ax , ax−1 } ∪ C; al2 + al3 ∈ {ax+1 , ax−1 } ∪ C.
(II.1) If ax+1 + al3 = ax−1 then ax+1 + al3 = ax+1 + (ax + al2 ) = (ax+1 +
al2 ) + ax (= ax−1 ), contradicting ax+1 + al2 ∈ {ax−1 , aj0 , al1 }.
(II.2) If ax+1 + al3 ∈ C then ax+1 + al1 ∈ C (since (ax+1 + al3 ) + al1 =
(ax+1 + al1 ) + al3 ∈ S). Then (ax+1 + al1 ) + al2 = (ax+1 + al2 ) + al1 ∈ S.
Since ax+1 + al3 ∈ C, (ax+1 + al3 ) + al2 = (ax+1 + al2 ) + al3 ∈ S. Uniting
al2 + ax+1 ∈ {ax−1 , aj0 , al1 }, we have ax+1 + al2 = al1 . So al1 + al3 = (ax+1 +
al2 ) + al3 = ax+1 + (al2 + al3 ) ∈ S, which implies al2 + al3 ∈ {ax+1 } ∪ C.
So al1 + aj0 = (ax+1 + al2 ) + aj0 = ax+1 + (al2 + aj0 ) ∈ S, which implies
al2 + aj0 ∈ {ax } ∪ Vx . Then ax+1 + (al1 + aj0 ) = (ax+1 + al1 ) + aj0 ∈ S, which
implies that al1 + aj0 ∈ {ax } ∪ C.
(II.2.1) If al1 + aj0 ∈ C then (al1 + aj0 ) + al2 = al1 + (al2 + aj0 ) ∈ S, a
contradiction al2 + aj0 ∈ {ax } ∪ Vx .
(II.2.2) If al1 +aj0 = ax then ax +al2 = (al1 +aj0 )+al2 = al1 +(al2 +aj0 ) ∈ S,
which implies al2 + aj0 ∈ {ax−1 , ax+1 }.
(II.2.2.1) If al2 + al3 ∈ C then (al2 + al3 ) + aj0 = (al2 + aj0 ) + al3 ∈ S,
contradicting al2 + aj0 ∈ {ax−1 , ax+1 }.
(II.2.2.2) If al2 + al3 = ax+1 then al2 + aj0 = ax−1 . Since ax−1 + al1 =
(al2 + aj0 ) + al1 = al2 + (al1 + aj0 ) = al2 + ax = al3 , ax−1 + al1 = al3 , a
contradiction.
13
Thus, Lemma 2.7 holds.
2
Lemma 2.8 If ax ∈ V with |ax | = max{|a| : a ∈ V }, then ax + ap ∈ C for
any ax ap ∈ E with n ≥ 8.
Proof: Let |ax | = max{|a| : a ∈ V } with ax ∈ V . Assume that ax >
0 (A similar argument works for ax < 0). By contradiction. Suppose to
the contrary that there exist ap0 ∈ V and ak0 ∈ V − {ap0 , ax } such that
ax + ap0 = ak0 . According to the choice of ax , ax + ap0 > 0 and ap0 < 0. Let
V0 = {ak0 , ax } ∪ Vk0 ∪ Vx . Then ax al ∈ E and ak0 al ∈ E for all al ∈ V − V0 . So
ak0 +al = (ax +ap0 )+al = (ax +al )+ap0 ∈ S. Thus, ax +al ∈ V −{ax, ak0 , ap0 }
with ax + al > 0 and al < 0. Since n ≥ 8, there exists at least one such vertex
al above (see Figure 16).
ax
ax+1
ax−1
ax +a j
0
ap
0
al
ak
0
II. Figure 16
On the other hand, by Lemma 2.2, there exists one edge ax aj0 ∈ E such
that ax + aj0 ∈ C for n ≥ 8. Then ax aj ∈ E for all aj ∈ V − ({ax } ∪ Vx ). So
(ax + aj0 ) + aj = (ax + aj ) + aj0 ∈ S. Thus, ax + aj ∈ {aj0 } ∪ Vj0 ∪ C for all
aj ∈ V − ({ax } ∪ Vx ).
For all al ∈ V − V0 , ax + al ∈ {aj0 } ∪ Vj0 . Since |Vi | ∈ {1, 2} for all ai ∈ V ,
n − 6 ≤ |V − V0 | ≤ |{aj0 } ∪ Vj0 | ≤ 3, that is, n ≤ 9. So we only consider n = 9
and n = 8. If |Vi | = 2 then let Vi = {ai−1 , ai+1 } for any ai ∈ V . If |Vi | = 1 then
let Vi = {ai+1 } for any ai ∈ V .
Case 1 n = 9
(I) If aj0 is an end vertex of Pn then |{aj0 } ∪ Vj0 | = 2, contradicting
|V − V0 | ≥ 3.
(II) If aj0 is not an end vertex of Pn then |{aj0 } ∪ Vj0 | = 3. So ax is not
an end vertex of Pn (If not, |{ax } ∪ Vx | = 2. So |V − V0 | ≥ 4, contradicting
|{aj0 } ∪ Vj0 | ≤ 3). Thus, only |{ax } ∪ Vx | = |{ak0 } ∪ Vk0 | = |{aj0 } ∪ Vj0 | = 3.
Note: n = 9 and none of the vertices in {ax , aj0 , ak0 } is an end vertex of Pn .
If ak0 aj0 ∈ E then ak0 + aj0 = (ax + ap0 ) + aj0 = (ax + aj0 ) + ap0 ∈ S,
contradicting ax + aj0 ∈ C.
14
If ak0 aj0 ∈ E then there exists one vertex ay ∈ V − V0 such that ax + ay =
aj0 +1 with aj0 aj0 +1 ∈ E and aj0 +1 ∈ V . So ak0 + aj0 +1 = ak0 + (ax + ay ) =
(ax + ak0 ) + ay ∈ S, contradicting ax + ak0 ∈ C.
Case 2 n = 8
(I) If aj0 is an end vertex of Pn then |{aj0 } ∪ Vj0 | = 2. Let Vj0 = {aj0 +1 }.
(I.1) If ax is the other end vertex of Pn then we consider the below.
If aj0 = ak0 = ax + ap0 then |V0 | = 4, contradicting |{aj0 } ∪ Vj0 | = 2.
If ax + ap0 = ak0 ∈ {aj0 , aj0 +1 } then aj0 ak0 ∈ E. So aj0 + ak0 = aj0 + (ax +
ap0 ) = (ax + aj0 ) + ap0 ∈ S, contradicting ax + aj0 ∈ C (See Figure 17).
aj0
ax+a j
0
ax
aj +1
0
ak−1
ax−1
0
ax +ap =a
k
0
ap
0
0
a k +1
0
Figure 17
If aj0 = ak0 and aj0 +1 = ak0 = ax + ap0 then |V − V0 | = 3, contradicting
|{aj0 } ∪ Vj0 | = 2 (See Figure 18).
aj0 =a
k−1
0
ax +ap =ak =aj +1
0
0
ax +aj0
ax
0
ak +1
ax−1
0
a i+1
ai−1
ai
Figure 18
(I.2) If ax is not the other end vertex of Pn then |{ax }∪Vx | = 3, so |V −V0 | ≥
n−6 = 2. If |V −V0 | > 2 then it is a contradiction with |{aj0 }∪Vj0 | = 2, so only
|V0 | = 6 and |{ak0 } ∪ Vk0 | = 3. There are only two subcases in the following.
(I.2.1) If aj0 = ak0 −1 with aj0 aj0 +1 ∈ E, then there exist two distinct
vertices ay , aj0 −1 ∈ V − V0 , then {ax + ay , ax + aj0 −1 } = {aj0 , aj0 +1 }. Since
ax + aj0 ∈ C, (ax + ak0 +1 ) + aj0 = (ax + aj0 ) + ak0 +1 ∈ S, then ax + ak0 +1 ∈ C.
Select any vertex az ∈ {ay , aj0 −1 } and then aj0 + ak0 +1 = (ax + az ) + ak0 +1 =
(ax + ak0 +1 ) + az ∈ S, contradicting ax + ak0 +1 ∈ C (See Figure 19).
15
a j =a k −1 =ax +az
0
0
ak0 =a j
ax +a j0
aj−1
0 +1
0
ak
ay
0 +1
ax−1
ax+1
ax
Figure 19
(I.2.2) If V = {ax−1 , ax , ax+1 , ak0 , ak0 +1 , ak0 −1 , aj0 , aj0 +1 }, then ax+1 is the
other end vertex of Pn . So ax ax+1 ∈ E. Since ax + ap0 = ak0 > 0, we have
ax +ak0 } ∈ C. Since {ax +ak0 , ax +aj0 } ⊆ C, we have {ax +aj0 +1 , ax +ak0 −1 } ⊆
C. So only ax + ak0 +1 = ak0 . Thus, ak0 + aj0 = (ax + ak0 +1 ) + aj0 =
(ax + aj0 ) + ak0 +1 ∈ S, contradicting ax + aj0 ∈ C (See Figure 20).
aj0
aj0+1
aj−1 =a x+1
0
ak−1
0
ax
0
ax +ak +1=a k
0
0
ax +a j
ak +1
ax−1
0
Figure 20
(II) If aj0 is not an end vertex of Pn then |{aj0 } ∪ Vj0 | = 3.
(II.1) If there exist two distinct vertices al1 , al2 ∈ V −V0 such that ax +al1 =
aj0 −1 > 0 and ax + al2 = aj0 +1 > 0 then ax + aj0 −1 ∈ C and ax + aj0 +1 ∈ C.
So aj0 −1 + aj0 +1 = (ax + al1 ) + aj0 +1 = (ax + aj0 +1 ) + al1 ∈ S, contradicting
ax + aj0 +1 ∈ C (See Figure 21).
al
1
ax +a l =a j −1
1
a x+1
0
aj0
ax
ax +al =a j +1
2
0
ax−1
al
2
Figure 21
16
ax+a j0
(II.2) Let {ay1 , ay2 } = {aj0 −1 , aj0 +1 }. If there exists at most one vertex
ay1 ∈ {aj0 −1 , aj0 +1 } such that ax + al1 = ay1 > 0, then we can consider ay1 as
ak0 in the following.
(II.2.1) If ax ay2 ∈ E then ax + ay2 ∈ V ∪ C.
If ax + ay2 ∈ C then ay1 + ay2 = (ax + al1 ) + ay2 = (ax + ay2 ) + al1 ∈ S,
contradicting ax + ay2 ∈ C.
If ax + ay2 ∈ V and al1 ay1 ∈ E then ax + ay2 ∈ {aj0 } ∪ Vj0 , a contradiction
(See Figure 22).
a l1=a p
0
ak =a x +a =a y
l1
0
1
ak +1 =a j
0
ax +a j0
a x+1
ax
0
ay
ax−1
2
ai
Figure 22
If ax + ay2 ∈ V and al1 ay1 ∈ E then exists one vertex az ∈ V − V0 such that
V − V0 such that ax + az ∈ {ay1 , ay2 , aj0 }. But it is impossible (See Figure 23).
ak −1
ax +a y =a
2
ax +a j
0
aj −1 =ak =a x +a l =ay
0
0
1
1
0
a x+1
ax
j0
ay
ax−1 =ap =a l
1
2
0
az
Figure 23
(II.2.2) If ax ay2 ∈ E then ax ay1 ∈ E and there exist two distinct vertices
ax1 , ax2 ∈ V − V0 such that ax ax1 ∈ E and ax ax2 ∈ E. Since ax + al1 = ay1 ,
we have {ax + ax1 , ax + ax2 } = {ay2 , aj0 }. Assume that ax + ax1 = ay2 . Then
ay1 + ay2 = ay1 + (ax + ax1 ) = ax1 + (ax + ay1 ) ∈ S, contradicting ax + ay1 ∈ C
(see Figure 24).
17
ak =ax +a l =a y
0
1
1
al =a p =a k −1
1
0
ax+a j
0
ak +1=a j
ax2
0
0
0
ax 1
ax +ax =ax−1=ay
1
2
ax+1
ax
Figure 24
Thus, Lemma 2.8 holds. 2
Lemma 2.9 Let |ax | = max{|a| : a ∈ V } and Ex = {ax ai |ax ai ∈ E} with
ax ∈ V . Then ak + al ∈ C for any ak al ∈ E − Ex for n ≥ 7.
Proof: Let |ax | = max{|a| : a ∈ V } and Ex = {ax ai |ax ai ∈ E} with ax ∈
V . If |Vi | = 2 then we may assume that Vi = {ai−1 , ai+1 } for ai ∈ V . Assume
ax > 0 (A similar argument works for ax < 0). By lemma 2.3, ax + ai ∈ C for
any ax ai ∈ Ex . For all ak al ∈ E − Ex , either there exists one vertex in {ak , al }
(we may assume ak ) such that ak ax ∈ E, or ak ax ∈ E and al ax ∈ E.
Claim 1 If ak and ay are the end vertices of Pn then all the sums of the
edges adjacent to ak or ay belong to C.
In fact, if ak and ay are the end vertices of Pn then ak + ay ∈ E and
dG (ak ) = dG (ay ) = n − 2. By lemma 2.3, ax + ak ∈ C. For all ak al ∈ E − Ex ,
(ax + ak ) + al = ax + (ak + al ) ∈ S. So ak + al ∈ {ax } ∪ Vx or ak + al ∈ C. Then
there are at most three edges ak ali ∈ E − Ex such that ak + ali ∈ {ax } ∪ Vx for
i = 1, 2, 3 (Since |{ax } ∪ Vx | ≤ 3). And others belong to C.
If there exist three edges ak ali ∈ E − Ex such that ak + ali ∈ V then we may
assume ak + ali = azi with azi ∈ {ax } ∪ Vx and i ∈ {1, 2, 3}. Since n ≥ 7 and
dG (ak ) = n − 2 ≥ 5, there exists one edge ak al4 ∈ E − Ex − {ak al1 , ak al2 , ak al3 }
such that azi0 al4 ∈ E with i0 ∈ {1, 2, 3}. Then ak + al4 ∈ C and azi0 + al4 ∈ S
(see Figure 25).
ak
ay =a
ax
l4
al3
al
1
al2=a zi
0
Figure 25
18
ax +a k
So azi0 + al4 = (ak + ali0 ) + al4 = (ak + al4 ) + ali0 ∈ S, contradicting the
fact ak + al4 ∈ C.
It is more easy to get contradictions when there exist two or one edge ak ali ∈
E − Ex such that ak + ali ∈ V for i ∈ {1, 2}. Thus, all the sums of the edges
adjacent to ak belong to C.
A similar argument works for ay .
Thus, Claim 1 holds.
Claim 2 If ah + al ∈ C and ah + al ∈ C with al al ∈ E then ah + al ∈ C
for any ah al ∈ E.
In fact, since ah + al ∈ C, (ah + al ) + al = (ah + al ) + al ∈ S for
all ah al ∈ E − {ah al , ah al }. Then ah + al ∈ {al } ∪ Vl ∪ C. Similarly,
ah + al ∈ {al } ∪ Vl ∪ C (see Figure 26).
ah
al’
al"
ah+a l’
ah+a l"
Figure 26
Since ah al ∈ E, ({al } ∪ Vl ) ∩ ({al } ∪ Vl ) = ∅. Thus, ah + al ∈ C.
Thus, Claim 2 holds.
By Claim 1, if ax is not an end vertex for n ≥ 7 then Claim 2 works for any
vertex in V − {ax , ak , ay } (see Figure 27,28,29).
ax
ax +a i
ay
ak
ai
Figure 27
19
ax
ax +a i
ai
ak
ay
Figure 28
ax
ax +a i
ai
ay
ak
Figure 29
If ax is an end vertex then assume that ax ax+1 ∈ E and ak ak−1 ∈ E (see
Figure 30).
ax
a x+1
ak
ak−1
ax +a k
a x +ai
ai
Figure 30
Firstly, Claim 2 works for every vertex in V − {ax , ax+1 , ak , ak−1 , ay }. Secondly, Claim 2 works for ax+1 and ak−1 . Thus, ak +al ∈ C for any ak al ∈ E −Ex
for n ≥ 7.
Thus, Lemma 2.9 holds. 2
20
Lemma 2.10 Pn is exclusive for n ≥ 7. 2
Lemma 2.11 ζ(Pn ) ≥ 2n − 7 for n ≥ 7.
2
Proof: Let V = {b1 , b2 , ..., bn }. Without loss of generality, we can assume
that b1 < b2 < ... < bn . So b1 + b2 < b1 + b3 < b1 + b4 < ... < b1 + bn < b2 + bn <
b3 +bn < ... < bn−1 +bn . Let C0 = {b1 +b2 , b1 +b3 , ..., b1 +bn , b2 +bn , ..., bn−1 +bn }.
Then there are at most four numbers which are not in S, but in C0 . On the
other hand, the others in C0 are the isolated vertices by Lemma 2.10. Thus,
ζ(Pn ) ≥ 2n − 7 for n ≥ 7. 2
Lemma 2.12 σ(Pn ) ≤ 2n − 7 for n ≥ 7.
Proof: Let V = {a1 , a2 , ..., an } and S = V ∪ C, where C is the isolated set.
case 1: n = 2k (k ≥ 4).
i = 1, 2, 3, ..., n,
ai = (i − 1) × 10 + 1,
cj = (j + 2) × 10 + 2,
j = 1, 2, 3, ..., n − 3, n − 1, n + 1, n + 2, ..., 2n − 5,
C = {c1 , c2 , ..., cn−3 , cn−1 , cn+1 , cn+2 , ..., c2n−5 }.
Let us verify this labelling is a sum labelling in detail.
(1) The vertices in S are distinct.
(2) For ∀i, j ∈ {1, 2, ..., n} and i = j, ai + aj = [(i + j − 4) + 2] × 10 + 2.
Since 1 ≤ i, j ≤ n and i = j, −1 ≤ i + j − 4 ≤ 2n − 5. So ai aj ∈ E ⇐⇒ ai + aj ∈
C ⇐⇒ ai +aj ∈ {12, 22, 10n+2, (n+2)×10+2} ⇐⇒ i+j −4 ∈ {−1, 0, n−2, n}.
That is, i + j − 4 = −1 ⇐⇒ i + j = 3 ⇐⇒ (i, j) ∈ {(1, 2), (2, 1)} ⇐⇒ a1 a2 ∈ E;
i + j − 4 = 0 ⇐⇒ i + j = 4 ⇐⇒ (i, j) ∈ {(1, 3), (3, 1)} ⇐⇒ a1 a3 ∈ E; i + j − 4 ∈
{n − 2, n} ⇐⇒ i + j ∈ {n + 2, n + 4} ⇐⇒ {(i, j), (j, i)} ⊆ {( n2 + 2, n2 ), ( n2 , n2 +
4), ( n2 + 4, n2 − 2), ( n2 − 2, n2 + 6), ( n2 + 6, n2 − 4), ..., (4, n), (n, 2), ..., (3, n − 1), (n −
1, 5), (5, n − 3), (n − 3, 7), (7, n − 5), ..., ( n2 − 1, n2 + 3), ( n2 + 3, n2 + 1)}. So Pn =
a n2 +2 a n2 a n2 +4 a n2 −2 a n2 +6 a n2 −4 ...a4 an a2 a1 a3 an−1 a5 an−3 a7 an−5 ...a n2 −1 a n2 +3 a n2 +1 .
Hence, for any ai aj ∈ E, ai + aj ∈ S; for any ai aj ∈ E, ai + aj ∈ S.
Therefore, the labelling is a sum labelling of Pn ∪ (2n − 7)K1 for n = 2k and
k ≥ 3.
case 2: n = 2k + 1 (k ≥ 3).
ai = (i − 1) × 10 + 1, i = 1, 2, 3, ..., n;
cj = (j + 2) × 10 + 2, j = 1, 2, 3, ..., n − 3, n − 1, n, n + 2, ..., 2n − 5;
C = {c1 , c2 , ..., cn−3 , cn−1 , cn , cn+2 , ..., c2n−5 }(For example Figure 31).
Let us verify this labelling is a sum labelling in detail.
(1) The vertices in S are distinct.
(2) For ∀i, j ∈ {1, 2, ..., n} and i = j, ai + aj = [(i + j − 4) + 2] × 10 + 2. Since
1 ≤ i, j ≤ n and i = j, −1 ≤ i+j −4 ≤ 2n−5. So ai aj ∈ E ⇐⇒ ai +aj ∈ C ⇐⇒
ai + aj ∈ {12, 22, 10n + 2, (n + 3) × 10 + 2} ⇐⇒ i + j − 4 ∈ {−1, 0, n − 2, n + 1}.
That is, i + j − 4 = −1 ⇐⇒ i + j = 3 ⇐⇒ (i, j) ∈ {(1, 2), (2, 1)} ⇐⇒
a1 a2 ∈ E; i + j − 4 = 0 ⇐⇒ i + j = 4 ⇐⇒ (i, j) ∈ {(1, 3), (3, 1)} ⇐⇒
a1 a3 ∈ E; i + j − 4 ∈ {n − 2, n + 1} ⇐⇒ i + j ∈ {n + 2, n + 5} ⇐⇒
n−3
n−3
n+3
n+3
n−3
n−3
{(i, j), (i, j)} ⊆ {( n+3
2 +1, 2 +1), ( 2 +1, 2 +4), ( 2 +4, 2 −2), ( 2 −
21
2, n+3
2 + 7), ..., (5, n), (n, 2), (2, 1), (1, 3), (3, n − 1), (n − 1, 6), (6, n − 4)}, ..., (n −
2, 4)}. So Pn = a n+3 +1 a n−3 +1 a n+3 +4 a n−3 −2 a n+3 +7 ...a8 an−3 a5 an a2 a1 a3 an−1
2
2
2
2
2
a6 an−4 ...an−2 a4 .
Hence, for any ai aj ∈ E, ai + aj ∈ S; for any ai aj ∈ E, ai + aj ∈ S. Therefore, the labelling is a sum labelling of Pn ∪ (2n − 7)K1 for n = 2k + 1 and k ≥ 3.
2
1
32
42
11
21
52
61
51
62
82
41
31
92
112
Figure 31
Theorem 2.1
⎧
0 = ζ(P4 ) < σ(P4 ) = 1;
⎪
⎪
⎪
⎪
⎨ 1 = ζ(P5 ) < σ(P5 ) = 2;
3 = ζ(P6 ) < σ(P6 ) = 4;
⎪
⎪
⎪
ζ(Pn ) = σ(Pn ) = 0, n = 1, 2, 3;
⎪
⎩
ζ(Pn ) = σ(Pn ) = 2n − 7, n ≥ 7.
Corollary 2.1
⎧
0 = ζ(F5 ) < σ(F5 ) = 1;
⎪
⎪
⎨
2 = ζ(F6 ) < σ(F6 ) = 3;
ζ(Fn ) = σ(Fn ) = 0, n = 3, 4;
⎪
⎪
⎩
ζ(Fn ) = σ(Fn ) = 2n − 8, n ≥ 7.
Acknowledgements
The authors are very grateful to the referee and Professor Liang Sun for
their helpful comments and suggestions.
22
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