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

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}.

The graph G is 4-regular.

v1

v5

v4 v3

v2

v6

G:

Definitions

5. Cn = (v1,v2, ..., vn) : n-cycle

Definitions

v1

C5 = (v1, v2, v3, v4, v5)

v2

v3v4

v5

6. Kn : the complete graph of order n.

Definitions

v1

K5

v2

v3v4

v5

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.

Definitions

v2

U = {v1, v2, v3}, V = {v4, v5, v6}KU,V = K3,3

v1 v3

v5v4 v6

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, G – E(Hs) has an H decomposition.

Definitions

and leave Hs. That is, – {Hs} is a H packing of G – E(Hs).

v4 v5 v6 v7 v8 v9

v1 v2 v3

v4 v5 v6 v7 v8 v9

v1 v2 v3

G:

G can be decomposed into H1, H2.

H1:

H2:

= {H1, H2} is a packing of G.

v4 v5 v6 v7 v8 v9

v1 v2 v3

v4 v5 v6 v7 v8 v9

v1 v2 v3

G‘= {(v1, v5, v3, v6), (v1, v2, v5, v4), (v1, v7, v2, v9), (v2, v3, v7, v6), (v2, v4, v3, v8), (v1, v3, v9, v8)} is a 4-cycle packing of G.

H1:

H2:

G has a 4-cycle system.

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.

Cycle Decomposition

Does there exist a 4-cycle system of Kn – E(G) for

any 4-regular subgraph G of Kn?

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 Two-

Regular 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.

Known Results

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.

Kn – E(G)

= (K9 – E(G)) K8k–8 K8k–8,9

= (K9 – E(G)) K8k–7 K8k–8,8

G:

K8k–8

Kn – E(G)

K9 – E(G)

G

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:

t |Q(t)|

5 1

6 1

7 2

8 6

9 16

10 59

11 265

12 1544

13 10778

Q(t) = {G | G is any connected 4-regular graph with t vertices}.

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-reducible

S = {∞1}

V(H) = N1(S) = {v1, v2, v3, v4}

E(H) = {v1v4, v2v3}

∞1

v3

v1 v2

v4

v5 v6

G: G – S:v3

v1 v2

v4

v5 v6

v3

v1 v2

v4

v5 v6

(G – S) H:

G is 1-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.

3-reducible

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.

Sufficient Condition

G is 3-reducible 4-regular graph

of order t.

Kn – E(G)

= [Kn–4 – E((G – S) H)] R.

(G – S) H

Kn–4 – E((G – S) H)S

Kn – E(G)

H

Construction

n ≣ 1 (mod 8), t is even.

n ≣ 5 (mod 8), t is odd.

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.