大 家 好 !

完全图的 Cartesian 积的 L(j,k)- 标定报告人：吕大梅导师：宋增民 林文松专业：运筹学与控制论方向：图论及其应用学校 : 东南大学

L(j,k)-labelings of Cartesian Products of Complete GraphsSpeaker: Damei Lü

Supervisor: Zengmin SongAdvisor: Wensong Lin

Speciality: Operational research and cyberneticsField: Graph theory and its application

school:southeast university

ContentsDefinitions and BackgroundCartesian products

L(j,k)-labeling Number λj,k of Kn□Km

□ Kl (n≥m≥l )

Definitions and Background L(j,k)-labelings and L(j,k)-labeling number λj,k

≥ j

0,1, … , t

Channel Assignment Problem

≥k

Cartesian productsCartesian products of complete graphs

Example: Kn□Km

V(Kn□Km )=V(Kn) ×V(Km ) E(Kn□Km )={{(a1 , b1) , (a2 , b2)} |

a1=a2 and (b1 , b2) E(∈ Km ) or b1=b2 and (a1 , a2) E(∈ Kn ) }

Preliminary results on Cartesian products of complete graphs

L(j,k)-labeling Number λj,k of Kn□Km □ Kl (n≥m≥l )

n=m

n ＞ 2m

n＝2m

m ＜ n ＜ 2m 3m+2 ＜ 2n ＜ 4m

2m ＜ 2n≤3m+2

n ＞ 2m If n ＞ 2m ＞ 4 and j/k≤m then

λj,k( Kn□Km □ Kl )=(nm-1)k

If n ＞ 2m ＞ 4 and j/k≥m then λj,k ( Kn□Km □ Kl )=(n-1)j + (m-1)k

jk

jk jk jk jk

n＝2mIf n=2m ＞ 4 and j/k≤m-1 then λj,k ( Kn□Km □ Kl )=(nm-1)k

If n=2m ＞ 4 and j/k≥m-1 then λj,k ( Kn□Km □ Kl )=(n-1)j+(m-1)k

3m+2＜2n＜4mIf 3m+2 ＜ 2n ＜ 4m and j/k≤d=n-m-1 then λj,k ( Kn□Km □ Kl )=(nm-1)k

If 3m+2 ＜ 2n ＜ 4m and j/k≥d=n-m-1 then λj,k ( Kn□Km □ Kl ) ≤(n-1)[j+(m-d)k]+(m-1)k

2m ＜ 2n≤3m+2If 2m ＜ 2n ≤ 3m+2 and j/k≤d=[(m+1)/2] then λj,k ( Kn□Km □ Kl )=(nm-1)k

Suppose m is odd. If 2m ＜ 2n ≤ 3m+2 and j/k≥d=(m+1)/2 then λj,k ( Kn□Km □ Kl ) ≤(n-1)[j+(m-d)k]+(m-1)k

Thank you you !