On Friendly Index Sets of Barycentric
Subdivision of Wheels
Sin-Min Lee, San Jose State University
Hsin-hao Su*, Stonehill College
45th Southeastern International Conference on
Combinatorics, Graph Theory, and Computing
at
Florida Atlantic University
March 6, 2014
Vertex Labeling
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
Let G be a graph with vertex set V(G) and
edge set E(G) Let A be an abelian group.
A labeling f : V(G)  A induces an edge
labeling f* : E(G)  A defined by
f*(xy) = f(x) + f(y).
Here, we consider the abelian group Z2 =
{0,1}.
[1] G. Chartrand, S-M. Lee and P. Zhang, Uniformly cordial graphs,
Discrete Math 206 (2006), 726-737.
Example : A tree L4
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FI(L4) is {0,2,4,6}.
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In general, FI(Ln) is {0,2,4,…,2(n-1)}.
Definition of Friendly Index
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For convenience, we use the following
notations: for i  Z2, let vf(i) = |{v  V(G) :
f(v) = i}| and ef(i) = |{e  E(G) : f*(e) = i}|.
Definition: A labeling f of a graph G is
said to be friendly if | vf(0)  vf(1) |  1.
Definition: The friendly index set of the
graph G, FI(G), is defined as {|ef(0) –
ef(1)| : the vertex labeling f is friendly.}
Examples

There are two non-isomorphic cubic graphs
of order 6, K3,3 and the prism C3P2.

They have
different friendly
index sets:
FI(K3,3) = {1, 9}
and FI(C3P2) =
{1, 3, 5}.
Cordial


A graph G is called cordial if there exists a
labeling f : V(G)  Z2 which induces an edge
labeling f* : E(G)  Z2 defined by f*(xy) = f(x)
+ f(y) such that |ef(0) – ef(1)| ≤ 1.
If the friendly index set of G contains 0 or 1,
then G is cordial.
[2] I. Cahit, Cordial graphs: a weaker version of graceful and harmonious graphs,
Ars Combinatoria 23 (1987), 201-207.
[3] M. Hovey, A-Cordial graphs,
Discrete Math 93 (1991), 183-194.
Wheels

The wheel graph Wn = N1 +Cn where
V(Wn) = {c0}{c1,…,cn} and E(Wn) =
{(c0,ci): i=1,…,n} E(Cn).
W5
W6
Barycentric Subdivision

The subdivision of some edge e of a
graph G with endpoints (u,v) yields a
graph containing one new vertex w, and
with an edge set replacing e by two new
edges, (u,w) and (w,v). The barycentric
subdivision subdivides each edge of the
graph. We denote the barycentric
subdivision of G by S(G).
Example: S(W3)
|e(0)-e(1)|= 0
|e(0)-e(1)|= 2
|e(0)-e(1)|= 4
FI(S(W3))={0,2,4,6,8}
|e(0)-e(1)|= 6
|e(0)-e(1)|= 8
Idea

Because of the nature of a barycentric
subdivided wheel, we separate a wheel
into three pieces.
Separation

Because each edge label is determined
only by the two adjacent vertices, we can
calculate the difference between the
number of 0-edges and 1-edges for each
piece and sum them up to get the total
friendly index (TFI), i.e., ef(0) – ef(1).
Notation

Let f be a friendly labeling of S(Wn). We
denote the number of edges of C2n
which are labeled by 0 and 1 by f* by
eC(0) and eC(1), the number of edges in
bridges which are labeled by 0 and 1 by
f* by eB(0) and eB(1), and the number of
edges in Stn which are labeled by 0 and
1 by f* by eS(0) and eS(1).
Useful Facts

For any labeling, if we swap the labels of 0 and
1, the friendly index remains the same.

Thus, w.l.o.g., we can only consider half of the
friendly labelings by assigning a specific vertex
labeled 0. Here, we will assume that the center,
c, is labeled 0.
Star Piece
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Let vS(0) and vS(1) be the number of 0- or
1-vertices on the star piece, respectively.
Since we assume that the center is
labeled 0, we can easily find TFI to be
vS(0) - vS(1).
In S(Wn), there are n vertices on the star
piece and n = vS(0) + vS(1).
Therefore, TFIStar= 2vS(0) - n.
Star Piece on S(W3)
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In S(W3), there are 3 vertices on the star
piece and n = vS(0) + vS(1).
Therefore, TFIStar= 2vS(0) - 3, where 0
≤ vS(0) ≤ 3.
Thus, total friendly index set on the star
piece is {-3, -1, 1, 3}.
Bridge Piece
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To distinguish, we call the vertices on
both bridge piece and cycle piece major
vertices.
Let t be the number of major vertices
labeled 0.
Let m be the number of 0-vertex pair
connecting a major vertex labeled 0 and a
star vertex labeled 0.
Bridge Piece
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Therefore, we have t-m 0-vertices in the
cycle piece connecting to a 1-vertex in
the star piece and vS(0)-m 0-vertices in
the star piece connecting to a 1-vertex in
the cycle piece.
Thus, we have (t-m)+(vS(0)-m) 1-edges in
the star piece.
So, TFIBridge= n - 2((t - m) + (vS(0) - m)) =
n - 2(t + vS(0) - 2m).
Bridge Piece on S(W3)
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In S(W3), there are 3 major vertices.
Note that, in S(W3), there are 10 vertices.
Thus, for a friendly labeling, we have 5
0-vertices and 5 1-vertices.
So, there are (4 - vS(0)) 0-vertices on the
cycle piece, where 0 ≤ vS(0) ≤ 3.
At the same time, we have (2 + vS(0)) 1vertices on the cycle piece.
Bridge Piece on S(W3)
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For vS(0) = 0, t could be 0, 1, 2, 3 and m
must be 0.
Thus, for vS(0) = 0, TFIBridge= 3-2(t+02m) = {3, 1, -1, -3}.
For vS(0) = 3, t could be 0 or 1 and m
must be 0 or 1, respectively.
Thus, for vS(0) = 3, TFIBridge= 3-2(t+32m) = {-3, -1}.
Bridge Piece on S(W3)
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For vS(0) = 1, t could be 0, 1, 2, 3.
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When t = 0, m must be 0.
When t = 1, m could be 0 or 1.
When t = 2, m could be 0 or 1.
When t = 3, m must be 1.
Thus, for vS(0) = 1, TFIBridge= 3-2(t+1-2m)
= {1, -1, 3, -3, 1, -1}.
Bridge Piece on S(W3)
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For vS(0) = 2, t could be 0, 1, 2.
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When t = 0, m must be 0.
When t = 1, m could be 0 or 1.
When t = 2, m could be 1 or 2.
Thus, for vS(0) = 2, TFIBridge= 3-2(t+22m) = {-1, -3, 1, -1, 3}.
Total Friendly Index of Cycles
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A chain of 0-vertices or a chain of 1vertices yields only 0-edges.
Let q be the number of a chain starting
from a 0-vertex and then a segment of 1vertex and ends at another 0-vertex.
Each chain above yields 2 1-edges.
Each pair of 1-edges takes away spaces
for two 0-edges. Thus, it reduces the TFI
by 4.
Total Friendly Index of Cycles
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The number q could run from 0 to vC(1).
Total Friendly Index of a cycle C2n is 2n4q, where 0 ≤ q ≤ min{vC(0) , vC(1)}
Therefore, it is {2n4min{vC(0),vC(1)} , …, 2n-4, 2n}.
Note here that the maximum of q only
happens when vC(0) = vC(1) and you label
the cycle by 0- and 1-vertices
alternatively. This, in this case, t can only
be 0 or 3.
Cycle Piece
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In S(Wn), we have assume that the center
is labeled 0 and there are vS(0) 0-vertices
in the star piece. Thus, we have vC(0) =
(3k+2) - vS(0) – 1 = 3k+1 - vS(0) and vC(1)
= 2n - vC(0) = k+1+ vS(0).
Note here that vC(1) is greater than 0.
Therefore, q must be greater or equal to 1,
i.e., 1 ≤ q ≤ min{3k+1 - vS(0) , k+1+
vS(0)}.
Cycle Piece on S(W3)
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In S(W3), we have n=3 and k=1. Thus, TFICycle=
2n-4q = 6-4q = {2, -2, …, -2-4q}, where 1 ≤ q ≤
min{3k+1-vS(0), k+1+ vS(0)}.
For vS(0) = 0, 1 ≤ q ≤ 2. Thus, TFICycle= {2, -2}.
For vS(0) = 1, 1 ≤ q ≤ 3. Thus, TFICycle= {2, -2, 6}.
For vS(0) = 2, 1 ≤ q ≤ 2. Thus, TFICycle= {2, -2}.
For vS(0) = 3, 1 ≤ q ≤ 1. Thus, TFICycle= {2}.
Note here that TFICycle= -6 only happens when q =
3 and t = 0 or 3.
Total Friendly Index Set of S(W3)
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In S(W3), we can now add all three TFI together to
get the TFI of S(W3).
For vS(0) = 0, TFI = {-3} + {-3, -1, 1, 3} + {2, -2}
= {-8, -6, -4, -2, 0, 2}.
For vS(0) = 1, TFI = {-1} + {-3, -1, 1, 3} + {2, -2,
-6} = {-8, -6, -4, -2, 0, 2}. Note here that we don’t
get TFI=-10 because that’s when q=3 and t=3.
For vS(0) = 2, TFI = {1} + {-3, -1, 1, 3} + {2, -2}
= {-4, -2, 0, 2, 4, 6}.
For vS(0) = 3, TFI = {3} + {-3, -1} + {2} = {2, 4}.