i
Chapter V
DIVISOR GRAPHS
J
'
Divisor graphs
173
CHAPTER V
DIVISOR GRAPHS
5.1 Introd uction
In this chapter we introduce the concept of a divisor graph.
Adivisor graph 0(8) of a finite subset S of Zis the graph (V; E) where
V = Sand uv
E
E if and only if either u divides v or v divides u. A graph
Gis a divisor· graph if it is isomorphic to the divisor graph 0(8) of
some subset S of Z.
We also give an equivalent definition of a divisor graph as follows.
Agraph G is called a divisor graph if there exists 'a labeling / of its
vertices with distinct integers such that for any two distinct vertices u
and v, uv is an edge of G if and only if either / (u) divides / (v) or / (v)
divides/(u). In this case we say that/is a divisor labeling of G.
First, we give an example of a divisor graph. Consider the
star KI,n' Take the set S = { 1, PI' P2' ... ,Pn} where Pi stands for the
lU1
prime. Clearly
KI,n
==
O(S).
5.2 On Divisor Graphs
In this section we give some results on cycle related graphs,
complete graphs and trees and also study some general results on
divisor graphs..
Divisor graphs
174
Theorem 5.2.1[154] The cycles C2n are divisor graphs for all n
Proof. Let { v l1 v 2 ' v 3 '
...
,v2n
},
n= 2, the resulting graph is C 4
n
•
~
~
2.
2 be the vertex set of C2n . When
Let SI = {V l ,V2 ,V3 ,V 4
v2 = 12, v 3 = 3, v 4 = 18. It is easy to see that C4
}
==
where
VI
= 2,
O(SJ. Next,
consider all even cycles of length greater than 4.
Set
V 2i
where
Pi
Pi' i
V 2i _ l =
is the
Let S = {
=
1, 2, ... , n
=
i = 1, 2, ... , (n - 1)
PiPi+l'
Z'th
VI'
prime.
V2 ,
above. Then, obviously
V 3 , ... ,
C2n
==
v 2n } where the
O(S) when
n>
Vi'S
are as defined
o
2.
Theorem 5.2.2 [154] Odd cycles C2n +Jor all n > '1 are not divisor
graphs.
Proof. On the contrary, let us assume that C2n + l (n> 1) is a divisor
graph. Let S
aiE
=
V (C2n + l
)
= { a p a2 , a3 ,
...
,
a 2n +l } where a i
':f.
0,
Z, 1 ~ i ~ 2n + 1. Since C2n + l is a divisor graph, note that a
vertex a i in S can divide only the vertices immediately preceding it
and succeeding it along the rim of the cycle. Without loss of
generali ty, let us assume ~
given by a l a 2 a 3
•••
Ia2
(al divides a 2 ). The cycle being
a 2n + 1 ~, by definition, a 2 1 a 3 or a 3 1 a 2 • If a 2 1 a 3
Divisor graphs
175
all a 3
implying the given cycle is a triangle. Hence a 3 a 2 . Thus
then
all a2
1
and a 3 a 2 • Similarly a 3 and as must divide a 4 • Proceeding like
1
this we get, a 2n - l and a 2n + l must divide a 2n . But then, either a 1 must
divide a 2n + l in which case a l I a 2n (or) a 2n + l must divide a l in which
case a 2n + l l a 2 , both of which are contradictions. Thus C2n + l for n> 1
o
is not a divisor graph.
Theorem 5.2.3 [154] An induced subgraph of a divisor graph is a
divisor graph.
Proof. Let G be the divisor graph of the set 5 c Z. Let 51
~
5. Then
the induced subgraph (81 ) is clearly the divisor graph of 51'
D
Theorem 5.2.4 [154] Any graph with an induced subgraph which
is an odd cycle of length greater than or equ8.l to 5 is not a
divisor graph.
Proof. By Theorem 5.2.2, it is clear that an odd cycle of length
~
5
is not a divisor graph. Then, if we use Theorem 5.2.3, the result
follows.
D
Corollary 5.2.1 The Petersen graph is not a divisor graph.
Proof. Since the Petersen graph has an induced subgraph which is a
cycle of length 5, by Theorem 5.2.4, Petersen graph is not a divisor
graph.
D
Divisor graphs
176
Theorem 5.2.5 [154] Let G be a divisor graph containing both
v and -v such that
v
E
Z - { 0 }. Then G contains at least one
triangle if <5 (G ) ~ 2.
Proof. Since <5 (G )
one vertex u
-:;e
~
2, we have d (v )
~
2. Thus there exists at least
-v which is adjacent to v. Then u must be adjacent
o
to -v also. i.e., v, -v, u form a triangle in G.
Theorem 5.2.6 [154] Every graph is a subgraph of a divisor graph.
Proof. Let a be a nonzero integer. Consider the set
s = {a,
a2 , a3 ,
•.• ,
an} where n
E
N. It follows that
Kn
==
0(8). Since
any graph on n vertices is a subgraph of K n , the proof is complete.
Let Hm,n denote the complete n-partite graph on n
m ~ 2 independent vertices. Also let [a 1 , a 2
,
a3 ,
•••
~
2 sets of
,an 1 denote the
least common multiple of n positive integers a 1 , a 2 , a 3 ,
•••
,an'
Theorem 5.2.7 [154] H mn
, is a divisor graph.
Proof. Let Pi denote the ith prime. Let £1 =
2 s; k ~ n. Consider the sets
[
0
PI' P2 ,P3' ... , Pm 1and
Dilfisor graphs
177
..................................................................
Vn
=
Let S = V;
U
{PI L 1 L 2 ... L n - 1 , P2 L1 L2 '" Ln _ 1 ,
V2
U
... U
... ,
Vn . We claim that Hm,n
==
Pm L1 L2
...
Ln _ 1
}
O(S).
To prove this claim we observe that Pi does not divide
Pj for I :;; j, 1 :s; I, j :s; m. Consider u = P r L1 L2
v = P s L1 L 2
•••
L j -1 E V j for 1 :s; r, s :s; m, r:;;
5,
...
Lj _ 1
E
Vj and
2 ::;; j :s; n. If u I v then
we get Prl Ps' a contradiction. If v I u then also we get a similar
contradiction. Hence u does not divide v and v does not divide u.
Also it is easy to find the vertices
x I y or y
XJ
yES
such that either
I x. This completes the proof of the theorem.
Corollary 5.2;2 The cocktail party graph H 2,n is a divisor graph.
0
0
Theorem 5.2.8 [154] mKn , the union of m copies of K n is a divisor
graph for all m > 1.
Proof. Recall that the complete graphs K n are divisor graphs. Let Vj
denote the vertex set of the j
th
copy of K n , 1 ::;; j::;; m.
Let Vj = { P j ' P~, P~, ... ,
pi } for
1 ::;; j::;; m where Pi denotes
the ith prime.. Consider the set S = V;
U
mK n == 0(8). Let u
j, 1 ::;; i, j ::;; m.
Then u =
E
p7,
V; and v E Vj , i
:j;
1 :s; s:S; n and v =
V2
U
... U
Vm • We show that
P;, 1 :s; t:s; n.. If u I v then we get
Pi I Pj' a contradiction. Similarly if vlu then pjl Pi' a contradiction.
DilJisor graphs
178
Hence u does not divide v and v does not divide u.
It is easy to find the vertices x) yES such that either x I y or
Y I x. This completes the proof of the theorem.
CJ
Theorem 5.2.9 [154] If F be the union of m complete graphs of
different orders with a vertex in common then F is a divisor graph.
Proof. Let K n. denote the complete graph of order n i , 1
I
~
~
i
m,
having vertex set Vi' Let Vi = { 1, Pi' P~, ... , p'(-l } for 1 ~ i.~ m
where Pi is the i th prime. Consider the set S = ~
show that F == O(S). For i :;t.j, 1 ~ i,) ~ m, let U
that u:;t. 1 and v:;t. 1. Then
1~ t ~ n j
-
1, i
:;t.
U
=
P:,
E
U
V2
U ... U
v: and v
for 1 ~ s ~ n i
-
E
Vm • We
Vj' Suppose
1 and v =p}, for
j, 1 ~ i,) ~ m. If u I v then Pi I P j ' and if v I u then
Pjl Pi' both are contradictions. Hence u does not divide v and v does
not divide u. It is easy to find the vertices x) yES such that either
x I y or y
I x. This completes the proof of the theorem.
CJ
Notations
Let G = (V, E ) be a divisor graph on n vertices. Let d j denote
the number of divisors of i in V and m i , the number of multiples
of i in V. Also let d (v ) denotes the degree of a vertex v in V.
Let ~ = {
Va
== {
°}.
U E
V
I u:;t. 0,
- u ~ V }, V2
= { U E
V
I u:;t. 0, -
u
E
V },
Clearly ~, V2 , V3 are mutually disjoint subsets of V and
Divisor graphs
~
U
V2
179
V3 = V. Let I ~
U
I=~
and I V2
I = n2 . Any vertex in
G must
be adjacent to all its divisors and multiples in V, except itself.
Thus, d (u )
=
d u + m u - 2 for all
E
U
V1 , For any
W
E
V2 ,
-
w is a
divisor of w and also a multiple of w. Hence d (w ) = d w + m w - 3 for
all
W E
°)= n - 1.
V2 . Also d (
Now using these symbols we prove the following theorem.
Theorem 5.2.10 [154] Let G = (V, E) be a graph on n vertices and
8 edges.
If G is a divisor graph, then
2&
Ld
=
vE
Lm
+
u
Vi U V~
2n1
u -
-
3n 2 + (n -1)
V E Vi U V~
where ~ = I ~ I and n 2 = I V2 I .
Proof. Suppose G = (V, E ) is a divisor graph on n vertices and
8 edges.
Let V; ={U
E
V I U"* 0, -
I
Then, 2& =
~ V }, V2
U
d(v)
=
UEV
=
I
= {U E V IU:f. 0, -
d(v) +
UEV1
L (d
u
+ mu
-
Ld
UEViUV~
+
I
Ld(v) + (n - 1)
2) + I (dv + mv - 3) + (n - 1)
VE
v
}, V3 = {O }.
UEV2
UE Vi
=
U EV
mv
-
V2
2n1
-
3n2 + (n - 1) .
o
VEViUV~
Theorem 5.2.11 Let G be a divisor graph with no triangles. Then for
every vertex v of G, the label I (v) is a common multiple or common
Divisor graphs
180
Ix
divisor of { 1 (x)
E
No (v)} where No (v) denotes the neighbourhood
of v and 1 (v) denotes the number labeled at v.
Proof. Let u be a vertex of G such that 1 (u) is a multiple of 1 (a) and
a divisor of 1 (b) where a, b
E
No (u). Then 1 (a) divides 1(b) and the
o
three vertices u, a, b form a triangle in G.
Theorem 5.2.12 [158] Every tree is a divisor graph.
Proof. We proceed by induction on the order n. The result is true if
n = 1. Assume that all trees of order k are divisor graphs. Let T be a
tree of order k + 1. Let v be an end-vertex of T and let u be adjacent
to v in T. Then T1 = T - v is a tree of order k and is therefore a
divisor graph.
Let f be a divisor labeling of
~.
Let No (v) denote the neighbourhood
of v. By Theorem 5.2.11, we need only consider two cases.
Case 1 f(u) is a common divisor of {f(x): x
E
No (u) }
Define a labeling 9 of T as follows:
9 (x)
= f(x) for all x
9 (v) = p
f
E
V(T1 ) and
(u) where p is a prime not occurring as a factor of
any labelf(x), x
E
V(T1 ).
Case 2 f(u) is a common multiple of {f(x): x
E
No (u) }.
Define a labeling 9 of T as follows:
9 (x)
=
pf (x) for all x in T1 where p is a prime not occurring as a
factor of any labelf(x), x
E
V(T1 ) and 9 (v)
= f(u).
Divisor graphs
181
In both cases it is easy to see that 9 gives a divisor labeling of
T and this completes the proof.
0
Theorem 5.2.13 [158] If G1 and G2 are two divisor graphs then
we can label the vertices of G1 and G2 so that for all x
E
V (G 1 ) and
y E V (G2 ), I (x) and I (y) are relatively prime where I (x) denotes the
number labeled at x.
Proof. Consider labelings of the divisor graphs G1 and G2 in which
the label of every vertex is either a prime or a product of distinct
primes. Also choose the labeling of G2 such that, for all v
E
V (G 2 ),
every prime factor of I (v) is greater than M where
M = max { Pi I Pi is a prime factor of I (x),
for all x
E
V (G 1 ) and y
E
X E
V (G1 )
}.
It follows that
V (G2 ), I (x) and I (y) are relatively prime.
0
Corollary 5.2.3 If G1 and G2 are two divisor graphs with disjoint
vertex sets then G1 u G2 is also a divisor graph.
0
Corollary 5.2.4 If G is a divisor graph so is mG for m ~ 2.
0
Theorem 5.2.14 [158] If G and H are vertex disjoint divisor graphs
then G + H is also a divisor graph.
Proof. Let
f and
9 be the divisor graph labelings of G and H respectively
such that f (x) and 9 (y) are relatively prime for all x
y
E
E
V (H ). We define a labeling function h of G + H as follows.
V (G) and
Divisor graphs
182
f(v)
h(v) =
for all v
E
V(G)
L1g(v) for all v E V(H) where L1 is the least common
multiple of {f(x): x E V(G)}
It is easy to see that h is a divisor labeling of G + H.
0
Corollary 5.2.5 If G is a divisor graph so is G + K n for any n.
0
Corollary 5.2.6 The complete bipartite graphs K m•n are divisor
o
graphs.
Corollary 5.2.7 Wheels Wn are divisor graphs for all even n (n > 2).
Theorem 5.2.15 [158] The subdivision graph of a divisor graph is a
divisor graph.
Proof. Let G be a divisor graph and f denote its divisor labeling.
Let w ll W 2 ,
ell e2 ,
...
•••
,wm denote the newly added vertices on the edges
,em of G. We define a labeling function 9 of the subdivision
graph 5 (G ) as follows.
9 (Wi) = Pi for 1
f
~
i ~ m where Pi is a prime such that Pi and
(v j) are relatively prime for all 1 ~ i ~ m, 1 ~ j ~ n.
9 (v j ) = k j
f
(v j
)
for 1 ~ j
~
n where k j is the product of primes
adjacent to v j in 5(G).
It follows that 9 is a divisor labeling of 5 (G ).
It is easy to prove the following lemma.
Divisor graphs
183
Lemma 5.2.1 Suppose u i is a product of midistinct primes for i = 1,2.
If u1 and u 2 have k common prime factors where 0 ~ k < min { m1 , m2
then u1 does not divide u 2 and u 2 does not divide
~
},
o
.
Theorem 5.2.16 [158] The ladder graph L n is a divisor graph.
ladder Ln'
V 2i + 1
U 2i
= P 2 i + 2 for 1 < i
=
P2i + 1
for 1 < i
~
~
-lj
I
n
-2,
l;J'
where P j denotes the j
th
prime.
defined above. The labeling is illustrated in Figure 5.2.1.
We claim that Ln == O(S). To prove this claim we observe that
V2i + 1 'S
and u 2j 's are primes. Hence we get the following results.
""'I does not divide
V 2j ,1 for
1 "j, 0 " i, j"
ln; J. u
1
2;
does not
Divisor graphs
184
divide u 2j for i "j, 1
does not divide
~
l;j.
"2id
does not divide
O~i5ln;1j, 15
for
V 2i + 1
i,j ~
i, (i - 1). Hence
for J.
~ I,.
(.I
and v 2j 's
U 2i + 1 'S
are products of distinct primes. By definition, the prime
~
and u 2j
j5l;j.
Next we consider primes and product of primes.
factor of u 2j + 1 if j
~j
U 2i
is not a
U 2i
does not divide u 2j + 1 and
1) , 1
5. 5lnj2' 0 5J. In-2-,-lj .
~
u2j +1 does not divide
U 2i
Similarly we get
does not divide v2j and v2j does not divide
for j
V 2i + 1
" i, ( i + 1), 0
~i~
is not a factor of v 2j
divide
~,
for i "j, 1
does not divide
.
ln; j,
1
Hence
~ i, j ~
U 2i
l;j.
1
-
~ j~
I
l;j.
V 2i + 1
For i "j, the prime
u"
does not divide v2j and v2j does not
A similar argument shows that
u 2j + 1 and u 2j + 1 does not divide
V 2i + 1
if i
"2,.1
~
j,
In
.. 5 -2-lj .
O~ I,}
Finally we consider products of primes. By definition,
U 2i + 1 'S
and v2j 's are products of three distinct prime factors and both have at
most two common factors. Hence by Lemma 5.2.1,
divide
"OJ
and
"2j
Also if i ~ j, both
does not divide u 2i • , for 0
U 2i + 1
~i~
U 2i + 1
ln; j.
1
does not
1~ j
~
l;J
and u 2j + 1 have at most one common factor.
Divisor graphs
185
By Lemma 5.2.1, U 2i + 1 does not divide
Bya similar argument,
V 2i
U 2j + 1
for i '" j,
-lj .
O " In-2~ 1,J~
does not divide v2j for i '" j, 1 ~ i, j
~
l; j.
..
..
~
Ls
Figure 5.2.1
.
S such that either x
Also it is easy to fmd the vertlces x, Y E
Y I x. This completes the proof of the theorem.
I y or
o
Divisor graphs
186
em
Theorem 5.2.17 [158] Prisms
x Pn are divisor graphs for all even
positive integers m and all positive integers n.
Proof. Let
1 :::; i :::; m, 1 :::; j :::; n denote the vertices of the
Vi)'
Pi j if i and j are both odd or both even
Set
vi)
= product of the labels of the neighbours of ViP if i and
j are of different parity, where Pij are distinct primes.
Let S
= {
vij I 1:::; i :::; m, 1 :::; j:::; n} where
Vi)
are as defined above.
The labeling is illustrated in Figure 5.2.2 .
•
!
PUP""P31
IL
•
/I
P31
I
P'J'J
Pl~'J'JP"J'33
I
P31P 'J" P 33 P4'J
I
JP""
/
I
I
P 33
P~33P3~44
P3~4'JPJs3
P44
..
II'
P31P4"P:51
PSI
\
Ill.
P4'J
P SIP S3 P 4"p6'J
\
\
P S3
P5~55PJ64
\
\
•
Figure 5.2.2
\
P
P~
Divisor graphs
187
We claim that C m x Pn == O(S).
To prove this claim we observe that if v pq and v rs are distinct
primes then v pq does not divide v rs and vrs does not divide v pq . If v ij
and v tw are products of primes, by Lemma 5.2.1, v ij does not divide
v1w and v t w does not divide vi}' From the definition it follows that if
vij is a prime, it is not a factor of v rs if vrs is not a neighbour of vi}'
It is easy to find the vertices x, yES such that either
x I y or y
I x. This completes the proof of the theorem.
Lemma 5.2.2 C 3
X
0
Pn is not a divisor graph for all integers n> 1.
Proof. We first show that C3 x P2 is not a divisor graph. On the
contrary, suppose that C3 x P2 is a divisor graph.
Let V (C3
X
P2
)
= { al' a 2 , a 3 , a 4 , as, a 6
}
where
aj
7;
0,
Then either a l divides a 2 or a 2 divides a l . (See Figure 5.2.3).
a 2 .------_a
Cax P2
Figure 5.2.3
a j E Z.
Divisor graphs
188
We consider these two cases.
a1 divides a 2
Case 1.
•
In this case if a 6 divides a 1 then a 6 divides a 2 , a contradiction.
Hence a 1 divides a 6
a4
divides a 3
,
•
Similarly we can show that
~
divides a 3 ,
a 2 divides a 3 , a 4 divides a 6 , as divides a 6 , and a 4
divides as' Finally, either a 2 divides as or as divides a 2 • If a 2 divides
as then since a 1 divides a 2 we get a 1 divides as, a contradiction.
If as divides a 2 then since a 4 divides as we get a 4 divides a 2 , which
is impossible.
Case 2.
a 2 divides a 1 •
In this case if a 1 divides a 6 then we get a2 divides a6 ,
a contradiction. Hence a 6 divides a 1 • Similarly we can show that a 3
divides
aI' a 3
divides
a 4, a3
divides
a 2 , a 6 divides a 4 , a 6
divides as,
and as divides a 4 • Finally, either a 2 divides as or as divides a 2 •
If a2 divides as then, since as divides a 4 we get a 2 divides a 4 ,
a contradiction. If as divides a 2 then, since a 2 divides
~
we get as
divides a 1 , a contradiction.
Since both these cases lead to contradictions C3 x P2 is not a
divisor graph. But C3 x P2 is an induced subgraph of C3 X Pn for all
integers n
~
3. By Theorem 5.2.3, any induced subgraph of a divisor
Dillisor graphs
189
graph is a divisor graph. Hence C3 X Pn is not a divisor graph for all
integers n
~
3. This completes the proof.
o
Theorem 5.2.18 [158] Prisms Cm x Pn are not divisor graphs for all
odd integers m ~ 3 and all integers n> 1.
Proof. In view of Lemma 5.2.2, we need only consider prisms Cm
x Pn
with odd integers m ~ 5 and all integers n > 1. Such a prism is a
graph with an induced subgraph which is a cycle of length m. Then
the theorem follows from Theorem 5.2.4.
0