羅 宇, 王力工
(西北工業(yè)大學(xué)理學(xué)院應(yīng)用數(shù)學(xué)系, 中國(guó) 西安 710072)
一類圖及其線圖的Wiener指數(shù)
羅 宇, 王力工
(西北工業(yè)大學(xué)理學(xué)院應(yīng)用數(shù)學(xué)系, 中國(guó) 西安 710072)
Wiener指數(shù)W(G)是指一個(gè)連通圖G中所有頂點(diǎn)對(duì)之間的距離之和.本文定義了一類具有圈數(shù)為r,圍長(zhǎng)為n的平面圖Gr,s,t,n,證明了對(duì)于滿足特定條件的正整數(shù)r,s,t,n,存在無窮個(gè)這樣的圖Gr,s,t,n,滿足性質(zhì)W(Gr,s,t,n)=W(L(Gr,s,t,n)),這里L(fēng)(Gr,s,t,n)表示圖Gr,s,t,n的線圖, 推廣了蘇曉海等人的結(jié)果.
Wiener指數(shù);線圖;圍長(zhǎng)
設(shè)圖G是一個(gè)具有頂點(diǎn)集V(G)={v1,v2,…,vn}和邊集E(G)={e1,e2,…,em}的簡(jiǎn)單無向連通圖.圖G的頂點(diǎn)數(shù)n稱為圖G的階數(shù), 也記為n(G). 圖G的線圖L(G)是指它的頂點(diǎn)集V(L(G))=E(G),線圖L(G)中兩個(gè)互異的頂點(diǎn)相鄰當(dāng)且僅當(dāng)在圖G中相對(duì)應(yīng)的兩條邊有一個(gè)公共頂點(diǎn). 一個(gè)圖G的圈數(shù)λ定義為λ(G)=m-n+1.
在本文中,作者的主要目標(biāo)是找出滿足下列等式的具有特定結(jié)構(gòu)的圖類,
W(L(G))=W(G) .
(1)
已知樹的Wiener指數(shù)和其線圖的Wiener指數(shù)總是不相等的[5],對(duì)于單圈圖,除了圈圖Cn之外,均滿足W(L(G))
為了便于計(jì)算圖的Wiener指數(shù),先介紹兩個(gè)引理:
引理1[3]設(shè)G1和G2是任意兩個(gè)圖,v1∈V(G1),v2∈V(G2),且圖G是將圖G1的頂點(diǎn)v1和圖G2的頂點(diǎn)v2重疊后得到的圖,則圖G的Wiener指數(shù)為:
W(G)=W(G1)+W(G2)+(n(G1)-1)dG2(v2)+(n(G2)-1)dG1(v1).
引理2[15]設(shè)a為正整數(shù),則不定方程x2-y2=a有解的充分必要條件是a為奇數(shù)或4 的倍數(shù).
考慮如圖1所示的圖Gr,s,t,n,它的圈數(shù)為r,圍長(zhǎng)是n(n為正整數(shù)且n≥ 3),圖中的每個(gè)圈都是Cn,階為nr+s+t+2,對(duì)于每一組正整數(shù)r,s,t和n,Gr,s,t,n都是簡(jiǎn)單平面圖,其線圖L(Gr,s,t,n)的具體結(jié)構(gòu)見圖1,其中它的完全子圖中部分邊沒有畫出.
定理1 圖1中圖Gr,s,t,n的Wiener指數(shù)為:
A.n=2k-1時(shí),L(Gr,s,t,n) B.n=2k時(shí),L(Gr,s,t,n)
(2)當(dāng)n=2k時(shí),W(Gr,s,t,n)=(2k3+4k2)r2+s2+t2+3st+(k2+4k)sr+(k2+6k)rt-(k3+2k2-6k)r+2s+2t+1.
(2)當(dāng)n=2k時(shí),由Wiener指數(shù)的定義可得:W(G1)=(2k3+4k2)r2-(k3+3k2-2k)r,dG1(v)=(k2+2k)r,n(G1)=2kr+1,W(G2)=s2+t2+3st+2s+2t+1,dG2(v)=s+2t+1,n(G2)=s+t+2. 由引理1可得(2)成立.
定理2 圖1中圖Gr,s,t,n和L(Gr,s,t,n),設(shè)ΔW(Gr,s,t,n)=W(L(Gr,s,t,n))-W(Gr,s,t,n),則有:
定理3 對(duì)于圖1中圖Gr,s,t,n,當(dāng)r,s,t,k,l均為正整數(shù),且滿足下列條件之一,則有ΔW(Gr,s,t,n)=0,即W(L(Gr,s,t,n))=W(Gr,s,t,n).
[2s+ 2t+ 2(k-3)r+ 3]2- 4k2r2=(8k2+28)r2-8sr-(8k2+28k+4)r+1.
引入任意正整數(shù)l,上式兩邊分別減去(8kl+4l2)r2,左邊配方得
[2s+2t+2(k-3)r+3]2-[(2k+2l)r]2=(8k2-8kl-4l2+28)r2-8sr-(8k2+28k+4)r+1.
令x=2s+2t+2(k-3)r+3,y=(2k+2l)r,有
x2-y2=(8k2-8kl-4l2+28)r2-8sr-(8k2+28k+4)r+1.
顯然上式等號(hào)右邊是一個(gè)奇數(shù),故x+y和x-y也是奇數(shù),不妨令
其中s+t+1=(l+3)r.于是定理3(1)的結(jié)果成立.
[2s+2t+2(k-2)r+3]2- 4k2r2=(8k2+8k+20)r2-8sr-(8k2+28k+4)r+1.
引入任意正整數(shù)l,上式兩邊分別減去(8kl+4l2)r2左邊配方得到:
[2s+2t+2(k-2)r+3]2-[(2k+2l)r]2=[8k2-(8l-8)k-4l2+20]r2-8sr-(8k2+28k+20)r+1.
其中s+t+=(l+2)r.于是定理3(2)的結(jié)果成立.
推論1 當(dāng)正整數(shù)n,k,r,s,t,l滿足下列條件之一,W(L(Gr,s,t,n))=W(Gr,s,t,n)成立.
(4)n=6,k=3,l=2,s=r-25,t=4r+24,r≥25是整數(shù);
(5)n=7,k=4,l=3,s=3r-34,t=3r+33,r≥12是整數(shù);
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