4-Regular Leaves of Partial 4-cycle System

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Transcript 4-Regular Leaves of Partial 4-cycle System

Packing Graphs with 4-Cycles
學生: 徐育鋒
指導教授: 高金美教授
2013組合新苗研討會 (2013.08.10 ~ 2013.08.11)
國立高雄師範大學
1. Definition
2. Known Results
3. 4-Regular Graphs
4. Main Results
5. Future Works
Definition
Definitions
1. A graph G is an order pair (V, E), where V is a non-empty set
called a vertex set and E is a set of two-element subsets of V
called an edge set.
2. degG(v) = the number of edges incident with a vertex v in G.
3. If all the vertices of a graph have the same degree r, then the
graph is called r-regular.
Definitions
V = {v1, v2, v3, v4, v5, v6}.
E = {v1v2, v1v3, v1v5, v1v6, v2v3, v2v4,
v2v6, v3v4, v3v5, v4v5, v4v6, v5v6}.
v6
v1
G: v5
v2
v4
v3
The graph G is 4-regular.
Definitions
5. Cn = (v1,v2, ..., vn) : n-cycle
v1
v5
v2
v4
v3
C5 = (v1, v2, v3, v4, v5)
Definitions
6. Kn : the complete graph of order n.
v1
v5
v2
v4
v3
K5
Definitions
7. KU,V : the complete bipartite graph with partite set U, V.
If |U| = m, |V| = n, then KU,V can be denoted by Km,n.
v1
v2
v3
v4
v5
v6
U = {v1, v2, v3}, V = {v4, v5, v6}
KU,V = K3,3
Definitions
Let = {H1, H2, , Hs} be a set of subgraphs of G.
If E(H1)  E(H2)    E(Hs) = E(G) and
E(Hi)  E(Hj) =  for i  j, then we call  is a
decomposition (packing) of G.
If Hi is isomorphic to a subgraph H of G for each i = 1, 2, , s,
then we say that G has an H decomposition (H system) or
 is a H packing of G.
If Hi is isomorphic to a subgraph H of G for each
i = 1, 2, , s–1, then we say that G can be packed with H
and leave Hs. That is, 
a Hanpacking
of G – E(Hs).
G –– {H
E(H
H decomposition.
s}s)ishas
v1
v2
v3
HG:
1:
v4
v5
v6
v7
v8
H2:
G =can
decomposed
intoofH1G.
, H2.

{Hbe
1, H
2} is a packing
v9
v1
v2
v3
H1:
v4
v5
v6
v1
v7
v2
v8
v9
v8
v9
v3
H2:
v4
v5
v6
v7
‘= {(v1, v5, v3, v6G
), (v
has
a2,4-cycle
v5, v4), system.
(v1, v7, v2, v9), (v2, v3, v7, v6),
1, v
(v2, v4, v3, v8), (v1, v3, v9, v8)} is a 4-cycle packing of G.
Cycle Decomposition
Alspach Conjecture :
Let 3  m1, m2, ..., mt  n such that m1 + m2 + ... + mt = n(n–1)/2
for odd n (m1 + m2 + ... + mt = n(n–2)/2 for even n).
Then Kn (Kn – F) can be decomposed into cycles C1, C2, ..., Ct
such that Ci is a mi-cycle for i = 1, 2, ..., t.
D. Bryant, D. Horsley and W. Pettersson,
Cycle decompositions V: Complete graphs into cycles of
arbitrary lengths, arXiv:1204.3709v2 [math.CO], 2013.
Cycle Decomposition
D. Sotteau, Decomposition of Km,n (Km,n*) into cycles (circuits)
of length 2k, J. Combin. Theory B, 30 (1981) 75.81.
Theorem 1:
There exists a 2k-cycle decomposition of Km,n if and only if
each vertex has even degree, mn is divisible by 2k, and m, n  k.
Does there exist a 4-cycle system of Kn – E(G) for
any 4-regular subgraph G of Kn?
Known Results
Known Results
A. Kotzig, On decomposition of the complete graph into
4k-gons, Mat.-Fyz. Cas., 15 (1965), 227-233.
Theorem 2:
There exists a 4-cycle system of Kn if and only if n ≡ 1 (mod 8).
Known Results
B. Alspach and S. Marshall, Even cycle decompositions of
complete graphs minus a 1-factor, J. Combin. Des., 2 (1994),
441-458.
Theorem 3:
There exists a 4-cycle system on Kn – F, where F is a
1-factor of Kn, if and only if n ≡ 0 (mod 2).
Known Results
H.-L. Fu and C. A. Rodger, Four-Cycle Systems with TwoRegular Leaves, Graphs and Comb., 17 (2001), 457-461.
Theorem 4:
Let F be a 2-regular subgraph of Kn. There exists a 4-cycle
system of Kn – F if and only if n is odd and 4 divides the
number of edges of Kn – F.
Known Results
C.-M. Fu, H.-L. Fu, C. A. Rodger and T. Smith, All graphs with
Maximum degree three whose complements have 4-cycle
Decompositions, Discrete Math., 308 (2008), 2901-2909.
Theorem 5:
Let G be a graph on n vertices, where n is even and (G)  3.
Then there exists a 4-cycle system of Kn – E(G) if and only if
(1) All vertices in G have odd degree,
(2) 4 divides n(n–1)/2 – |E(G)|, and
(3) G is not one of the two graphs of order 8 as follows.
Let G be a 4-regular subgraph of Kn.
Does there exist a 4-cycle system of Kn – E(G)?
4-Regular Graphs
Some 4-regular graphs
Question:
Does there exist a 4-cycle system of Kn – E(K5) ?
1. n = 5, Yes!
2. n = 6, No!
3. n = 7, No!
4. n = ?, Yes!
Question:
Does there exist a 4-cycle system of Kn – E(K5) ?
n  5 is odd and 4 | n(n – 1) / 2 – 10
⇒ 4 | (n2 – n – 20) / 2
⇒ 8 | (n – 4)(n – 5)
⇒ n  5 (mod 8).
Question:
Does there exist a 4-cycle system of Kn – E(K5) ?
Answer: n  5 (mod 8), Yes!
Let n = 8k + 5.
K8k+5 – E(K5) = K8k+1  K4, 8k.
...
K8k+1
K4, 8k
Lemma 6:
There exists a 4-cycle system of Kn – E(K5)
if and only if n  5 (mod 8).
Question:
Does there exist a 4-cycle system of Kn – E(G) ?
n  6 is odd and 4 | n(n – 1) / 2 – 12
⇒ 4 | n(n – 1) / 2
⇒ 8 | n(n – 1)
⇒ n  1 (mod 8).
G:
Question:
Does there exist a 4-cycle system of K9 – E(G) ?
G:
K9 – E(G) :
Question:
Does there exist a 4-cycle system of K9 – E(G) ?
Answer: Yes !
Question:
Does there exist a 4-cycle system of Kn – E(G) ?
Answer: n  1 (mod 8), Yes !
Let n = 8k + 1.
G:
Kn – E(G)
= (K9 – E(G))  K8k–8  K8k–8,9
= (K9 – E(G))  K8k–7  K8k–8,8
Kn – E(G)
K9 – E(G)
G
K8k–8
Lemma 7:
There exists a 4-cycle system of Kn – E(G)
if and only if n  1 (mod 8).
G:
Question:
Does there exist a 4-cycle system of Kn – E(G)?
n  6 is odd and 4 | n(n – 1) / 2 – 12
⇒ 4 | n(n – 1) / 2
⇒ 8 | n(n – 1)
⇒ n  1 (mod 8).
G:
Question:
Does there exist a 4-cycle system of K9 – E(G)?
Answer: No!
G:
K9 – E(G) :
K9 – E(G) :
Lemma 8:
There exists a 4-cycle system of Kn – E(G)
if and only if n  1 (mod 8) and n  17.
G:
Q(t) = {G | G is any connected 4-regular graph with t vertices}.
t
5
6
7
8
9
10
11
12
13
|Q(t)|
1
1
2
6
16
59
265
1544
10778
s-reducible
Definition 9:
Let G be a 4-regular graph of order t.
If there exists S  V(G), |S| = s and
a graph H where V(H) = N1(S) and E(H)  E(G) = ∅
such that (G – S)  H is 4-regular,
then we call the graph G is s-reducible.
S = {∞1}
V(H) = N1(S) = {v1, v2, v3, v4}
E(H) = {v1v4, v2v3}
∞1
v1
v2 v1
G – S:
G: (G – S)  H:
v3
vv34
v5
v6
v2v1
v2
v3v4
v5
v6
G is 1-reducible.
v5
v4
v6
3-reducible
Theorem 10:
Let t  8 and G be a 4-regular graph of order t.
If G contains a component with at least 6 vertices,
then G is 3-reducible.
Sufficient Condition
Theorem 11:
Let G be a 4-regular of order t.
If there exists a 4-cycle system of Kn – E(G), then
(1) n ≣ 1 (mod 8), for t is even and
(2) n ≣ 5 (mod 8), for t is odd.
Construction
G is 3-reducible 4-regular graph
of order t.
Kn – E(G)
= [Kn–4 – E((G – S)  H)]  R.
n ≣ 1 (mod 8), t is even.
n ≣ 5 (mod 8), t is odd.
Kn – E(G)
(G – S)  H
H
Kn–4 – E((G – S)  H)
S
Main Results
Main Theorem:
Let G be any 4-regular graph with t vertices.
There exists a 4-cycle system of Kn – E(G), if
n is odd, 4 | n(n – 1)/2 – 2t, and
(1) G is a vertex-disjoint union of t/5 copies of K5.
(2) n  (4t – 5)/3.
(3) n > 9 for the following two graphs.
Future Work
Question 1.
Let G be a 4-regular graph of order t.
Does there exist a 4-cycle system of Kn – E(G) for
t  n < (4t – 5)/3?
Question 2.
Let G be a 4-regular graph of order t and t ≣ 5 (mod 8).
Is G 5-reducible?
Question 3.
Let G be a 4-regular spanning subgraph of Kn.
Does there exist a 4-cycle system of Kn – E(G) for
n ≡ 5 (mod 8)?
Thanks for your patient.