On The Edge-Balance Index Sets of Fans and Wheels

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On The Edge-Balance Index Sets of Fans and Wheels
Dharam Chopra, Wichita State University Sin-Min
Lee, San Jose State University Hsin-hao Su,
Stonehill College MCCCC 22 At University of
Nevada, Las Vegas October 23, 2008
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Example nK2

• EBI(nK2 ) is 0 if n is even and 2if n is odd.

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Edge Labeling

• A labeling f E(G) ? Z2 induces a vertex partial
  labeling f V(G) ? A defined by f(x) 0 if
  the edge labeling of f(x,y) is 0 more than 1 and
  f(x) 1 if the edge labeling of f(x,y) is 1
  more than 0. f(x) is not defined if the number
  of edge labeled by 0 is equal to the number of
  edge labeled by 1.

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Definition of Edge-balance

• Definition A labeling f of a graph G is said to
  be edge-friendly if ef(0) ? ef(1) ? 1.
• Definition The edge-balance index set of the
  graph G, EBI(G), is defined as vf(0) vf(1)
  the edge labeling f is edge-friendly.

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Example Pn

• Lee, Tao and Lo showed that

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Example Pn
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Wheels

• The wheel graph Wn N1 Cn-1 where V(Wn) c0?
  c1,,cn-1 and E(Wn) (c0,ci) i1,,n-1?
  E(Cn-1).

W6
W5
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Notation

• Let f be an edge-friendly labeling of Wn. We
  denote the number of edges of Cn-1 which are
  labeled by 0 and 1 by f by eC(0) and eC(1) and
  the number of edges in E(Wn) which are labeled by
  0 and 1 by f by eN(0) and eN(1).

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Idea

• Because of the nature of a wheel, we separate a
  wheel into a cycle and a star.

W6
W5
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Separation

• Because of the nature of a wheel, we separate a
  wheel into a cycle and a star.

W5C5St5
W6C6St6
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Vertex of Order 2

• For a vertex of order 2 with an edge-labeling
  (not necessary friendly), it can only be labeled
  as one of the following three cases.

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Add an Edge to an Vertex of Order 2

• If the vertex was labeled, then the label of the
  vertex is not changed.
• If the vertex was not labeled then the label of
  the vertex is labeled by the same as the adding
  edge.

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Edge-balance Index Sets of a Cycle

• We focus on the cycle part and get several
  algebraic equations that we can use later.
• We denote the number of vertices on Cn-1 which
  are labeled by 0, 1, and not labeled by the
  restricted f by vC(0), vC(1), and vC(x),
  respectively.

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Lemma

• In a cycle Cn with a labeling f (not necessary
  edge friendly), we have two equations
• 2 vC(0) vC(x)2eC(0), and
• 2 vC(1) vC(x)2eC(1).

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Lemmas

• In a cycle Cn with a labeling f (not necessary
  edge friendly), eC(0) or eC(1) is zero if and
  only if vC(x) 0.
• In a cycle Cn with a labeling f (not necessary
  edge friendly), if vC(0) and vC(1) are both
  positive then vC(x) gt 0.

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Lemma

• In a cycle Cn with a labeling f (not necessary
  edge friendly), assume that eC(0) gt eC(1) gt 1. We
  have
• 0 lt vC(0) lt eC(0)-1
• and
• 0 lt vC(1) lt eC(1)-1.

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Do we have all possible labelings of a cycle?

• We know that v(x) must be even. For an even
  number 2k gt 0, We can construct a labeling of Cn
  by putting k pairs of 0- and 1-edges
  alternatively in the middle of Pn1 and all the
  rest 0s on the left side edges and all the rest
  1s on the right side edges and then glue the two
  sided vertices together.

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Do we have all possible labelings of a cycle?
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Do we have all possible labelings of a cycle?

• Since we put k pairs of 0 and 1 alternatively,
  these 2k edges produce exactly 2k-1 unlabeled
  vertices. Additionally, the glued two sided
  vertices become an unlabeled vertex. All the
  other vertices are either in the middle of
  0-edges or 1-edges. Thus, we have exactly 2k
  unlabeled vertices.

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Do we have all possible labelings of a cycle?

• For an edge-labeling f of Cn with v(x)2kgt0, pick
  an unlabeled vertex and split this vertex into
  two vertices.
• Then, we have a Pn1 with the very left edge is
  labeled by 0 and the very right edge is labeled
  by 1.

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Do we have all possible labelings of a cycle?
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Do we have all possible labelings of a cycle?

• Let a maximal chain of 0-edges is a path with
  only 0-edges where its length is greater than 1
  and it is not a sub chain of 0-edges with longer
  length.
• If a maximal chain of 0-edges does not contain
  the very left 0-edge, we can cut it off and
  insert into the left side of the first from the
  left unlabeled vertex without altering the number
  of 0-vertices, 1-vertices and unlabeled vertices.
  

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Lemmas

• In a cycle Cn with a labeling f (not necessary
  edge friendly), assume that eC(0) gt eC(1) gt 1 and
  vC(x)2kgt0. Then vC(1) eC(1)-k.
• In a cycle Cn with a labeling f (not necessary
  edge friendly), assume that eC(0) gt eC(1) gt 1 and
  vC(x)2kgt0. Then vC(0) n- eC(1)-k.

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Proof

• By the above construction and rearrangement, we
  can put k pairs of 0- and 1-edges in the middle
  of Pn1.
• Since these k pairs of edges occupy k edges
  labeled by 1, there are only eC(1)-k 1-edges left
  on the right part of Pn1.

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Proof (continues)

• Since the very right edge of the k pairs of 0-
  and 1-edges is an edge labeled by 1, we have a
  chain of 1-edges with eC(1)-k1 edges and eC(1)-k
  vertices in between edges. Thus, vC(1)
  eC(1)-k.
• In Cn, n vC(0) vC(1) vC(x). Thus, vC(0) n-
  vC(1)- vC(x) n- (eC(1)-k) -2k n- eC(1)-k.

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Assumption

• By the symmetry of the role of 0 and 1 in the
  labeling to calculate the edge balance index,
  without loss of generosity, we may assume that
  eC(0) gt eC(1) gt 0. Since eC(0) gt eC(1) and
  eC(0)eC(1)n-1. We have
• 0 lt eC(1) lt (n-1)/2 if n is odd, or,
• 0 lt eC(1) lt n/2 if n is even.

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Three Cases on Cn-1

• Its easy to see that there are three possible
  cases
• eC(0) gt eC (1) gt 1, or,
• eC(0) eC(1) gt 1 only when n is odd (which
  implies eC(0) eC(1) (n-1)/2,) or,
• eC(1) 0 (which implies that eC(0)n-1.)

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Examples

• v(0)-v(1) 4 v(0)-v(1) 2
  v(0)-v(1) 9
• W8 W7
  W11

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h

• Note that when we put all E(Wn) edges and c0 into
  Cn-1 to get our wheel back, all the vertices of
  Cn-1 become of order 3.
• Thus, all the unlabeled vertices in Cn-1 become
  labeled by either 0 or 1.
• All other labeled vertices remain labeled by the
  same value.
• Assume that there are h 0-edges connected to an
  unlabeled vertex of Cn-1. Obviously, we have 0 lt
  h lt eN(0)eC(1).

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Edge Balance Index of a Wheel

• Theorem Let Wn be a wheel.
• If n is even, there are two types of edge-balance
  indexes
• v(0)-v(1) n-2 2(eC(1)k-h) where 0 lt h lt
  eC(1), 0 lt k lt eC(1) and 0 lt eC(1) lt n/2
• v(0)-v(1) n-2.

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Edge Balance Index of a Wheel

• If n is odd,
• v(0)-v(1) n-2 2(eC(1)k-h) where 0 lt h lt eC
  (1), 0 lt k lt eC (1) and 0 lt eC (1) lt (n-1)/2
• v(0)-v(1) 2(h-k)
• where 0 lt k lt (n-1)/2
• and 0 lt h lt (n-1)/2 if 0 lt k lt (n-1)/4 and
  k-(n-1)/4 lt h lt (n-1)/2 if (n-1)/4 lt k lt (n-1)/2.
• v(0)-v(1) n-2.

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Proof (Case 1)

• By lemma, vC(x) is not zero. Since vC(x) must be
  even, we assume that vC(x)2kgt0 where k is an
  integer. By lemmas, we have vC(1) eC(1)-k and
  vC(0) n- eC(1)-k.
• Since the number of edges of our wheel Wn is
  even, (actually, it is 2n-2,) and f is an
  edge-friendly labeling of Wn, we have eN(0)
  eC(1) and eN(1) eC(0). Therefore, by the
  assumption that eC(0) gt eC(1), we have eN(1) gt
  eN(0). This implies that c0 must be labeled by 1.

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Case 1 Examples

• v(0)-v(1) 0 v(0)-v(1) 2
  v(0)-v(1) 1
• W8 W8
  W7

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Proof (Case 1 Continues)

• If eC(1) lt vC(x), then we have h 0-vertices
  converted from unlabeled vertices and the rest
  vC(x)-h unlabeled vertices are converted into
  vertices labeled by 1. Therefore, the number of
  vertices labeled by 0 v(0) vC(0) h and the
  number of vertices labeled by 1 v(1) vC(1)
  vC(x)-h 1.

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Proof (Case 1 Continues)

• So, the edge-balance index v(0)-v(1) (vC(0)
  h) (vC(1) vC(x)-h 1) vC(0) 2h vC(1)
  2k 1 (n-1-eC(1)-k) 2h (eC(1)-k) 2k
  1 n-2 2(eC(1)k-h).
• Note here that we have 0 lt h lt eC(1) and 0 lt k lt
  eC(1). Also, since eC(0) gt eC(1) gt 1, we have 0 lt
  eC(1) lt (n-1)/2 if n is odd, or, 0 lt eC(1) lt n/2
  if n is even.

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Proof (Case 1 Continues)

• If vC(x) lt eN(0)eC(1), then h must be less than
  vC(x) since those are all unlabeled vertices
  available to be connected by an 0-edge. We have a
  similar inequality 0 lt h lt vC(x) lt eC(1). In the
  calculation of edge-balance index set, this case
  will be override by the previous one since we
  have to calculate the balance index for all
  possible edge friendly labelings. Thus, it will
  not generate any new edge-balance index.

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Case 2 Examples

• v(0)-v(1) 0 v(0)-v(1) 2
  v(0)-v(1) 4
• W7 W11
  W11

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Proof (Case 2)

• The argument is very similar except c0 is not
  labeled because eC(0) eC(1) (n-1)/2.
• Therefore, we get similar equations v(0) vC(0)
  h and v(1) vC(1) vC(x)-h.
• Thus, v(0)-v(1) (vC(0) h) (vC(1) vC(x)-h)
  2(h-k).

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Proof (Case 2 continues)

• Since eN(0) gt 2k, it provides enough 0-edges to
  be used.
• And k-(n-1)/4 lt h lt (n-1)/2 if (n-1)/4 lt k lt
  (n-1)/2 since the number of unlabeled vertices is
  greater or equal to eN(1).
• Therefore, in this case, 0 lt k lt (n-1)/2 and 0 lt
  h lt (n-1)/2 if 0 lt k lt (n-1)/4

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Case 3 Examples

• v(0)-v(1) 6 v(0)-v(1) 5
  v(0)-v(1) 11
• W8 W7
  W13

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Proof (Case 3)

• Since the number of edges of our wheel Wn is even
  and f is an edge-friendly labeling of Wn, we have
  eN(0) eC(1) and eN(1) eC(0).
• Therefore, by the assumption that eC(1)0, we
  have eN(0)0 and eN(1)n-1. This implies that c0
  must be labeled by 1.
• Since all the edges in Cn-1 are labeled by 0, we
  have n-1 vertices labeled by 0. Thus, the
  edge-balance index v(0)-v(1)n-2.

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Edge Balance Index Set of Wheels

• Theorem If n is even, then EBI(Wn)
  0,2,,2i,,n-2.
• Theorem If n is odd, then EBI(Wn) 1,3,
  ,2i1,,n-2?0,1,2,,(n-1)/2.

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EBI(W6) 0,2,4

• v(0)-v(1) 0 v(0)-v(1) 2
  v(0)-v(1) 4

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Proof (Even Case)

• Since n is even, we have v(0)-v(1) n-2 and
  v(0)-v(1) n-2 2(eC(1)k-h) where 1 lt
  eC(1)k-h lt 2eC(1) and 0 lt eC(1) lt n/2.
• This implies that 1 lt eC(1)k-h lt n.
• Let t eC(1)k-h.
• Since k can be any integer between 1 and eC(1)
  including 1 and eC(1) and h can be any integer
  between 0 and eC(1) including 0 and eC(1), we
  have 1 lt t lt n.

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Proof (Even Case Continues)

• Thus, v(0)-v(1) n-2 -2t where 1lttltn-1 where t
  is an integer.
• Hence, -(n-2) lt v(0)-v(1) lt n-4 and v(0)-v(1)
  must be even.
• Therefore, EBI(Wn) 0,2,,2i,,n-2

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EBI(W5) 0,1,2,3

• v(0)-v(1) 0 v(0)-v(1) 1
  v(0)-v(1) 2 v(0)-v(1) 3

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Proof (Odd Case)

• Since n is odd, we have
• v(0)-v(1) n-2 2(eC(1)k-h) where 0 lt h lt
  eC(1), 0 lt k lt eC(1) and 0 lt eC(1) lt (n-1)/2
• v(0)-v(1) 2(h-k) where 0 lt k lt (n-1)/2 and 0 lt
  h lt (n-1)/2 if 0 lt k lt (n-1)/4 and k-(n-1)/4 lt h
  lt (n-1)/2 if (n-1)/4 lt k lt (n-1)/2.
• v(0)-v(1) n-2.

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Proof (Odd Case Continues)

• For the first case, a similar argument applies.
  Thus, we have v(0)-v(1) n-2 and v(0)-v(1) n-2
  2(eC(1)k-h) where 1 lt eC(1)k-h lt 2eC(1) and 0
  lt eC(1) lt (n-1)/2.
• This implies that 1 lt eC(1)k-h lt n-1. Let t
  eC(1)k-h.
• Since k can be any integer between 1 and eC(1)
  including 1 and eC(1) and h can be any integer
  between 0 and eC(1) including 0 and eC(1), we
  have 1 lt t lt n-1.

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Proof (Odd Case Continues)

• Thus, v(0)-v(1) n-2 -2t where 1lttltn-2 where t
  is an integer.
• Thus, -(n-2) lt v(0)-v(1) lt n-4 and v(0)-v(1) must
  be odd since n is odd.
• Therefore, EBI(Wn) contains1,3,,2i1,,n-2.

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Proof (Odd Case Continues)

• For the second case, v(0)-v(1) 2(h-k) where 0 lt
  k lt (n-1)/2.
• For h, 0 lt h lt (n-1)/2 if 0 lt k lt (n-1)/4 and
  k-(n-1)/4 lt h lt (n-1)/2 if (n-1)/4 lt k lt (n-1)/2
• We have - (n-1)/2 lt v(0)-v(1) 2(h-k) lt (n-1)/2.
  Therefore, EBI(Wn) contains0,1,2,,(n-1)/2.

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Fans

• A Fan graph is F1,n N1 Pn where V(F1,n) c?
  v1,,vn and E(F1,n) (c,vi) i1,,n? E(Pn).

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Turn a Fan into a Wheel

• If we add an extra edge (v1,vn) into F1,n, then
  it becomes W1n.

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Equivalence of Friendly Labelings

• For an edge friendly labeling, since F1,n has
  2n-1 edges, to calculate the edge balance index,
  we may assume that v(0)v(1)1. Thus, if we label
  the extra edge by 1, then we have an edge
  friendly labeling for W1n. Note that for an edge
  friendly labeling of W1n, if we remove an edge
  labeled by 1 from its cycle part, we get an F1,n
  with an edge friendly labeling. Thus, there is a
  1-1 correspondence between the edge friendly
  labelings of F1,n and the edge friendly labelings
  of W1n.

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V1 and Vn

• There are three different pairs of labeling of v1
  and vn
• v1 is labeled by 0 and vn is lebeled by 1 and
  vice versa.
• Both v1 and vn are labeled by 0.
• Both v1 and vn are labeled by 1.
• Since case 2 and 3 requires at least 4 edges
  labeled by the same value, it only happens when
  2n-1gt7, that is, ngt4.

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Changes of the Balance Index

• Because the extra edge is labeled by 1, if v1 or
  vn is labeled by 0, then it must have another two
  0-edges attached. Thus, it remains labeled by 0
  in F1,n after removing the extra edge.
• If v1 or vn is labeled by 1, then it could either
  remain labeled by 1 if there are another two
  1-edges attached or become unlabeled. The first
  case keeps the balance index the same. But, the
  second one reduces the amount of 1-vertices by 1.

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Edge Balance Index Set of Fans

• Theorem EBI(F1,n)
• 0,1,2 if n3
• 0,1,, n-2 for ngt 4.

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EBI(F1,3) 0,1,2
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EBI(F1,4) 0,1,2
59
EBI(F1,5) 0,1,2,3
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Proof

• By the 1-1 correspondence between the edge
  friendly labelings of W1n and F1,n, the edge
  balance calculation tells us that in W1n, -(n-1)
  lt v(0)-v(1) lt n-3 where v(0)-v(1) must be odd if
  n is even and v(0)-v(1) must be even if n is odd.
• Note that v(0)-v(1) n-3 only if k1 and
  heC(1)gt1 and v(0)-v(1) -(n-1) only if k0 and
  h0.

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Proof (continues)

• We first consider the extreme case v(0)-v(1)
  n-3 and v(1) is reduced by 2. In this case, we
  have k1. Thus, v1 and vn are the only two
  vertices which are unlabeled and both are
  converted to 1. But, since there are hgt1
  unlabeled vertices which are converted to 0, this
  cannot happen.

62
Proof (continues)

• The next extreme case is v(0)-v(1) -(n-1) and
  v(1) is reduced by 0. In this case, since h0, we
  know that all vC(x) unlabeled vertices are
  converted to 1. Also, since k0 and eC (1)n/2 or
  (n-1)/2 which depends on n is even or odd,
  respectively, we have vC(x)2eC (1)n if n is
  even and n-1 if n is odd. This cannot happen
  since we assume that eC (0)gteC (1) and the extra
  edge occupies the extra 1.

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Proof (continues)

• Therefore, we conclude that, in W1n, -(n-2) lt
  v(0)-v(1) lt n-2 where v(0)-v(1) must be odd if n
  is even and v(0)-v(1) must be even if n is odd.
• For n3, there is only case 1. So, v(1) can only
  be reduced by 0 or 1. Thus, the inequality
  becomes -(n-1) lt v(0)-v(1) lt n-1. Thus, the EBI
  0,1,2.

64
Proof (continues)

• For ngt4, since all the cases can only reduce v(1)
  by 0, 1 or 2, the inequality becomes -(n-2) lt
  v(0)-v(1) lt n-2.
• Also, it is easy to construct a friendly labeling
  that fits one of the above three cases. So, any
  integer between -(n-2) and n-2 can be the result
  of v(0)-v(1). Thus, the EBI 0,1,, n-2

65
Wheel-like Graphs

• Broken Wheels with Spikes

66
Wheel-like Graphs

• Halin Graph of Double Stars H(D(m,n)) N2Cmn.