图论不变量是一些特殊的拓扑指标,常被用来描述某些化学分子结构图的基础性质。基于顶点间距离的不变量如Wiener指标[1-3]、度距离指标[4-6]等都得到学者们的广泛关注,并得出大量相关结论。Gupta等[7]于2002年首次提出了新的图论不变量,即离心距离指标(EDS)。此后,学者们对树、单圈图和仙人掌图的EDS做了大量研究。Yu等[8]刻画出具有最小、第二小ξd(G)值的单圈图,并计算出树的最小离心距离指标;Ilić等[9]确定了具有最大ξd(G)值的树,并对EDS与其他图论不变量之间的关系作了一系列的探究;Hua等[10]得出了n个顶点仙人掌图的ξd(G)值的下确界。
三圈图是一类具有特殊结构的图,当且仅当图G的边数等于顶点数加2时,图G就是一个三圈图。在对三圈图的EDS的研究过程中,Li等[9]通过确定EDS与边数m(m=|E(G)|)之间的关系,最终得到了三圈图的最小ξd(G)值。但是文献[11]的方法不具一般性和适用性,不能被用来研究更一般图的离心距离指标。本文构思了一种完全不同于文献[11]的新方法,首先研究了文献[12]中3种移边变换对离心距离指标的数值大小的影响,然后基于这3种变换,利用数学归纳法得到了三圈图的最小ξd(G)值以及最小值所对应的特殊结构的三圈图。
1 特定移边变换对ξd(G)的影响对于任意连通图G=(V, E),定义图G的EDS[7]为
$ {\xi ^d}\left( G \right) = \sum\limits_{\left\{ {u,v} \right\} \subset V\left( G \right)} {\left( {{\varepsilon _G}\left( u \right) + {\varepsilon _G}\left( v \right)} \right){d_G}\left( {u,v} \right)} $ |
其中距离dG(u, v)表示顶点u和v之间的最短路长度,离心率εG(·)表示对应顶点到图G所有其他顶点的距离最大值。
Li等[12]在研究离心阻尼距离指标(ERDS)的数学性质时刻画了3种移边变换,具体内容如下。
第一变换 给定一个简单连通图H和一个星图(中心为v,叶子为v1, v2, …, vp+1)如图 1(a)所示。对于图H中的某一顶点u,将顶点vp+1和u重合,记新得到的图为G。则第一变换G→G′为
$ G' = G - \left\{ {v{v_1},v{v_2}, \cdots ,v{v_p}} \right\} + \left\{ {u{v_1},u{v_2}, \cdots ,u{v_p}} \right\} $ |
第二变换 给定一个简单连通图G如图 1(b),设顶点u在图G中具有p个悬挂点(u1, u2, …, up),顶点v(不同于顶点u)在图G中具有q个悬挂点(v1, v2, …, vq)。则第二变换G→G′和G→G″分别为
$ G' = G - \left\{ {v{v_1},v{v_2}, \cdots ,v{v_q}} \right\} + \left\{ {u{v_1},u{v_2}, \cdots ,u{v_q}} \right\} $ |
$ G'' = G - \left\{ {u{u_1},u{u_2}, \cdots ,u{u_p}} \right\} + \left\{ {v{u_1},v{u_2}, \cdots ,v{u_p}} \right\} $ |
第三变换给定简单连通图G如图 1(c),n>3,e=uv是图G中的一条非悬挂割边。设
$ G - e = {G_1} \cup {G_2},u \in {V_{{G_1}}},v \in {V_{{G_2}}} $ |
将图 1(c)中的G→G′变换记为第三变换。
本文通过计算验证了这3种移边变换对ξd(G)的数值大小的影响,同时给出下列3条相应的定理。
定理1 设图G和图G′分别为图 1(a)中的两个图,则ξd(G′)≤ξd(G),当且仅当图G是以顶点v为中心的星图时等号成立。
定理2 设图G、图G′和图G″分别为图 1(b)中的图,则ξd(G′) < ξd(G)或ξd(G″) < ξd(G)。
定理3 给定简单连通图G(n>3),e=uv是图G中的一条非悬挂割边,对图G中的边e做第三变换得到图G′,则ξd(G′) < ξd(G)。
由于定理1~3的证明思路、证明方法与文献[10]非常相似,所以在此只给出定理1的详细证明。
为了方便计算,在后续证明中,对于图G或图G′中任意两点x和y,假设ε(x):=εG(x)或ε′(x):=εG′(x),d(x, y):=dG(x, y)或d′(x, y):=dG′(x, y)
证明(定理1)
对于图 1中的图G和G′,假设|V(H)|>1,W={v1, v2, …, vp},可以得到
$ \left\{ \begin{array}{l} V\left( G \right) = V\left( {G'} \right),\varepsilon \left( v \right) = \varepsilon '\left( v \right)\\ \forall x \in V\left( H \right),\varepsilon \left( x \right) \ge \varepsilon '\left( x \right)\\ \forall y \in W,\varepsilon \left( y \right) = \varepsilon '\left( y \right) + 1 \end{array} \right. $ | (1) |
$ \left\{ \begin{array}{l} \forall x,y \in V\left( H \right),d\left( {x,y} \right) = d'\left( {x,y} \right)\\ \forall x,y \in W,d\left( {x,y} \right) = d'\left( {x,y} \right) \end{array} \right. $ | (2) |
因为式(1)和式(2)成立,所以有
$ \begin{array}{l} {\xi _1}: = \left( {\sum\limits_{\left\{ {x,y} \right\} \subseteq {V_H}} + \sum\limits_{\left\{ {x,y} \right\} \subseteq W} {} } \right)\left[ {\left( {\varepsilon \left( x \right) + \varepsilon \left( y \right)} \right)} \right.\\ \left. {d\left( {x,y} \right) - \left( {\varepsilon '\left( x \right) + \varepsilon '\left( y \right)} \right)d'\left( {x,y} \right)} \right] + \left[ {\sum\limits_{x \in {V_H}\backslash \left\{ u \right\},y \in W} + } \right.\\ \left. {\sum\limits_{x \in {V_H},y = v} {} } \right)\left[ {\left( {\varepsilon \left( x \right) + \varepsilon \left( y \right)} \right)d\left( {x,y} \right) - \left( {\varepsilon '\left( x \right) + } \right.} \right.\\ \left. {\left. {\varepsilon '\left( y \right)} \right)d'\left( {x,y} \right)} \right] > 0 \end{array} $ |
$ \begin{array}{l} {\xi _2}: = \sum\limits_{x \in W,y = u} {\left( {\varepsilon \left( x \right) + \varepsilon \left( y \right)} \right)d\left( {x,y} \right)} - \\ \sum\limits_{x \in W,y = u} {\left( {\varepsilon '\left( x \right) + \varepsilon '\left( y \right)} \right)d'\left( {x,y} \right)} + \sum\limits_{x \in W,y = v} {\left( {\varepsilon \left( x \right) + } \right.} \\ \left. {\varepsilon \left( y \right)} \right)d\left( {x,y} \right) - \sum\limits_{x \in W,y = v} {\left( {\varepsilon '\left( x \right) + \varepsilon '\left( y \right)} \right)d'\left( {x,y} \right)} = \\ \left| W \right|\left[ {2\left( {\varepsilon \left( u \right) + \varepsilon \left( {{v_1}} \right)} \right) - \left( {\varepsilon '\left( u \right) + \varepsilon '\left( {{v_1}} \right)} \right)} \right] + \left| W \right|\\ \left[ {\left( {\varepsilon \left( v \right) + \varepsilon \left( {{v_1}} \right)} \right) - 2\left( {\varepsilon '\left( v \right) + \varepsilon '\left( {{v_1}} \right)} \right)} \right] \ge 3\left| W \right|\left( \varepsilon \right.\\ \left. {\left( {{v_1}} \right) + \varepsilon '\left( {{v_1}} \right)} \right) + \left| W \right|\left( {2\varepsilon \left( u \right) - 2\varepsilon '\left( v \right) + \varepsilon \left( v \right) - } \right.\\ \left. {\varepsilon '\left( u \right)} \right) \ge 3\left| W \right| + \left| W \right|\left( {\varepsilon \left( u \right) - \varepsilon '\left( v \right)} \right) = \left| W \right|\left( {\varepsilon \left( u \right) - } \right.\\ \left. {\varepsilon \left( v \right) + 3} \right) \end{array} $ |
如果εH(u)=1,那么ε(u)=ε(v)=2,则
$ \varepsilon \left( u \right) - \varepsilon \left( v \right) + 3 > 0 $ |
如果εH(u)=r≥2,那么ε(u)=r,ε(v)=r+1,则
$ \varepsilon \left( u \right) - \varepsilon \left( v \right) + 3 > 0 $ |
由此可知ξ2>0。根据ξd(G)的定义可以得到
$ {\xi ^d}\left( G \right) - {\xi ^d}\left( {G'} \right) = {\xi _1} + {\xi _2} > 0 $ |
显然,当且仅当|V(H)|=1,即图G是以顶点v为中心的星图时,ξd(G)=ξd(G′)。
定理1由此得证。证毕。
2 三圈图的最小离心距离指标对任意三圈图G,其基图
设图G是具有最小ξd值的三圈图,
1) T1(n, r, s, t)={G},其中r、s、t分别表示图
2) T2(n, x, y, z, w)={G},其中x, y, z分别表示
3) T3(n, x, y, z, w)={G},其中x, y, z分别表示
引理4 若三圈图G∈T1(7, r, s, t),则ξd(G)≥126;若三圈图G∈T2(5, x, y, z, w),则ξd(G)≥48;若三圈图G∈T3(6, x, y, z, w),则ξd(G)≥83。
证明 显然,如图 2所示,若三圈图G∈T1(7, r, s, t),则
$ G \in \left\{ {{T^{1\left( 1 \right)}},{T^{1\left( 2 \right)}}} \right\} $ |
若三圈图G∈T2(5, x, y, z, w),则
$ G \in \left\{ {{T^{2\left( 1 \right)}},{T^{2\left( 2 \right)}},{T^{2\left( 3 \right)}}} \right\} $ |
若三圈图G∈T3(6, x, y, z, w),则
$ G \in \left\{ {{T^{3\left( 1 \right)}},{T^{3\left( 2 \right)}}} \right\} $ |
计算可得
$ \left\{ \begin{array}{l} {\xi ^d}\left( {{T^{1\left( 1 \right)}}} \right) = 126,{\xi ^d}\left( {{T^{1\left( 2 \right)}}} \right) = 211\\ {\xi ^d}\left( {{T^{2\left( 1 \right)}}} \right) = 48,{\xi ^d}\left( {{T^{2\left( 2 \right)}}} \right) = 48,{\xi ^d}\left( {{T^{2\left( 3 \right)}}} \right) = 48\\ {\xi ^d}\left( {{T^{3\left( 1 \right)}}} \right) = 83,{\xi ^d}\left( {{T^{3\left( 2 \right)}}} \right) = 124 \end{array} \right. $ |
由计算结果可知引理4成立。证毕。
引理5 若三圈图G∈T1(n, r, s, t),n≥7,则ξd(G)≥4n2-9n-7,当且仅当
证明 对n进行归纳。由引理4可知,当n=7时,ξd(G)≥126,当且仅当
情况1 若图G中存在悬挂点,不妨取G中任意悬挂点w,顶点w′为w在图G中的唯一邻点。显然有
$ \left\{ \begin{array}{l} \varepsilon \left( w \right) \ge 2,\varepsilon \left( {w'} \right) \ge 1,d\left( {w,w'} \right) = 1\\ \varepsilon \left( a \right) \ge 2,d\left( {a,w} \right) \ge 2,\forall a \in V\left( G \right)\backslash \left\{ {w,w'} \right\} \end{array} \right. $ | (3) |
记G′=G-w,G′仍是三圈图,由分析可知
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge \varepsilon '\left( a \right),\varepsilon \left( b \right) \ge \varepsilon '\left( b \right),\forall a,b \in V\left( {G'} \right)\\ d\left( {a,b} \right) \ge d'\left( {a,b} \right),\forall a,b \in V\left( {G'} \right)\\ {\xi ^d}\left( {G'} \right) \ge 4{\left( {n - 1} \right)^2} - 9\left( {n - 1} \right) - 7 = \\ \;\;\;\;\;\;\;\;\;\;4{n^2} - 17n + 6 \end{array} \right. $ | (4) |
因为式(3)和式(4)成立,所以有
$ \begin{array}{l} {\xi ^d}\left( G \right) - {\xi ^d}\left( {G'} \right) = \sum\limits_{a,b \in V\left( {G'} \right)} {\left[ {\left( {\varepsilon \left( a \right) + \varepsilon \left( b \right)} \right)} \right.} \\ \left. {d\left( {a,b} \right) - \left( {\varepsilon '\left( a \right) + \varepsilon '\left( b \right)} \right)d'\left( {a,b} \right)} \right] + \left( {\varepsilon \left( {w'} \right) + } \right.\\ \left. {\varepsilon \left( w \right)} \right)d\left( {w',w} \right) + \sum\limits_{a \in V\left( {G'} \right)\backslash \left\{ {w'} \right\}} {\left( {\varepsilon \left( a \right) + \varepsilon \left( w \right)} \right)d\left( {a,} \right.} \\ \left. w \right) \ge 0 + \left( {1 + 2} \right) \times 1 + \left( {n - 2} \right) \times \left( {2 + 2} \right) \times 2 = \\ 8n - 13 \end{array} $ |
即对于情况1下的三圈图G,有ξd(G)≥ξd(G′)+8n-13≥4n2-9n-7,当且仅当
情况2 若图G中不存在悬挂点,由于n≥8,r+s+t-2=n,且r≥s≥t,则有r≥4。不妨取G中顶点w∈V(Cr)且w∉V(Cs)∪V(Ct),w′和w″为w在图G中的两个邻点。显然有
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge 2,\forall a \in V\left( G \right)\\ d\left( {w,w'} \right) = 1,d\left( {w,w''} \right) = 1,d\left( {w',w''} \right) = 2\\ d\left( {a,w} \right) \ge 2,\forall a \in V\left( G \right)\backslash \left\{ {w,w',w''} \right\} \end{array} \right. $ | (5) |
记G′=G-w+w′w″,G′仍是三圈图,由分析可得
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge \varepsilon '\left( a \right),\varepsilon \left( b \right) \ge \varepsilon '\left( b \right),\forall a,b \in V\left( {G'} \right)\\ d\left( {a,b} \right) \ge d'\left( {a,b} \right),\forall a,b \in V\left( {G'} \right)\\ {\xi ^d}\left( {G'} \right) \ge 4{\left( {n - 1} \right)^2} - 9\left( {n - 1} \right) - 7 = \\ \;\;\;\;\;\;\;\;\;\;4{n^2} - 17n + 6 \end{array} \right. $ | (6) |
因为式(5)和式(6)成立,所以有
$ \begin{array}{l} {\xi ^d}\left( G \right) - {\xi ^d}\left( {G'} \right) \ge 0 + \left( {\varepsilon \left( {w'} \right) + \varepsilon \left( {w''} \right)} \right)d\left( {w',} \right.\\ \left. {w''} \right)\left( {\varepsilon' \left( {w'} \right) + \varepsilon' \left( {w''} \right)} \right)d'\left( {w',w''} \right) + \sum\limits_{a \in \left\{ {w',w''} \right\}} {\left( \varepsilon \right.} \\ \left. {\left( a \right) + \varepsilon \left( w \right)} \right)d\left( {a,w} \right) + \sum\limits_{a \in V\left( {G'} \right)\backslash \left\{ {w',w''} \right\}} {\left( {\varepsilon \left( a \right) + \varepsilon \left( w \right)} \right)} \\ d\left( {a,w} \right) \ge \left( {2 + 2} \right) + 2 \times \left( {2 + 2} \right) + 2 \times \left( {2 + 2} \right) \times \left( {n - } \right.\\ \left. 3 \right) = 8n - 12 \end{array} $ |
即对于情况2下的三圈图G,有
$ {\xi ^d}\left( G \right) \ge {\xi ^d}\left( {G'} \right) + 8n - 12 > 4{n^2} - 9n - 7 $ |
引理5由此得证。证毕。
引理6 若G∈T2(n, x, y, z, w),n≥5,则ξd(G)≥4n2-9n-7当且仅当G∈{G1, G4, G5}时等号成立(其中{G1, G4, G5}见图 3)。
证明 显然,n=4时
情况1 若图G中存在悬挂点,不妨取G中任意悬挂点u,顶点u′为u在图G中的唯一邻点。显然有
$ \left\{ \begin{array}{l} \varepsilon \left( u \right) \ge 2,\varepsilon \left( {u'} \right) \ge 1,d\left( {u,u'} \right) = 1\\ \varepsilon \left( a \right) \ge 2,d\left( {a,u} \right) \ge 2,\forall a \in V\left( G \right)\backslash \left\{ {u,u'} \right\} \end{array} \right. $ | (7) |
记G′=G-u,G′仍是三圈图。由分析可知
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge \varepsilon '\left( a \right),\varepsilon \left( b \right) \ge \varepsilon '\left( b \right),\forall a,b \in V\left( {G'} \right)\\ d\left( {a,b} \right) \ge d'\left( {a,b} \right),\forall a,b \in V\left( {G'} \right)\\ {\xi ^d}\left( {G'} \right) \ge 4{\left( {n - 1} \right)^2} - 9\left( {n - 1} \right) - 7 = \\ \;\;\;\;\;\;\;\;\;\;4{n^2} - 17n + 6 \end{array} \right. $ | (8) |
因为式(7)和式(8)成立,所以有
$ \begin{array}{l} {\xi ^d}\left( G \right) - {\xi ^d}\left( {G'} \right) = \sum\limits_{a,b \in V\left( {G'} \right)} {\left[ {\left( {\varepsilon \left( a \right) + \varepsilon \left( b \right)} \right)} \right.} \\ \left. {d\left( {a,b} \right) - \left( {\varepsilon '\left( a \right) + \varepsilon '\left( b \right)} \right)d'\left( {a,b} \right)} \right] + \left( {\varepsilon \left( {u'} \right) + } \right.\\ \left. {\varepsilon \left( u \right)} \right)d\left( {u,u'} \right) + \sum\limits_{a \in V\left( {G'} \right)\backslash \left\{ {u'} \right\}} {\left( {\varepsilon \left( a \right) + \varepsilon \left( u \right)} \right)d\left( {a,u} \right)} \ge \\ 0 + \left( {1 + 2} \right) \times 1 + \left( {n - 2} \right) \times \left( {2 + 2} \right) \times 2 = 8n - 13 \end{array} $ |
即对于情况1下的三圈图G,有ξd(G)≥ξd(G′)+8n-13≥4n2-9n-7,当且仅当G∈{G1, G4, G5}时等号成立。
情况2 若图G中不存在悬挂点,由于n≥6,x+y+z+w-6=n且w≥x≥y≥z,则w≥4。不妨取G中顶点u∈V(Pw)且u∉V(Px)∪V(Py)∪V(Pz),顶点u′和u″为u在图G中的两个邻点。显然有
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge 2,\forall a \in V\left( G \right)\\ d\left( {u,u'} \right) = 1,d\left( {u,u''} \right) = 1,d\left( {u',u''} \right) = 2\\ d\left( {a,u} \right) \ge 2,\forall a \in V\left( G \right)\backslash \left\{ {u,u',u''} \right\} \end{array} \right. $ | (9) |
记G′=G-u+u′u″,G′仍是三圈图,由分析可知
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge \varepsilon '\left( a \right),\varepsilon \left( b \right) \ge \varepsilon '\left( b \right),\forall a,b \in V\left( {G'} \right)\\ d\left( {a,b} \right) \ge d'\left( {a,b} \right),\forall a,b \in V\left( {G'} \right)\\ {\xi ^d}\left( {G'} \right) \ge 4{n^2} - 17n + 6 \end{array} \right. $ | (10) |
因为式(9)和式(10)成立,所以有
$ \begin{array}{l} {\xi ^d}\left( G \right) - {\xi ^d}\left( {G'} \right) \ge 0 + \left( {\varepsilon \left( {u'} \right),\varepsilon \left( {u''} \right)} \right)d\left( {u',} \right.\\ \left. {u''} \right) - \left( {\varepsilon '\left( {u'} \right),\varepsilon '\left( {u''} \right)} \right)d'\left( {u',u''} \right) + \sum\limits_{a \in \left\{ {u',u''} \right\}} {\left( {\varepsilon \left( a \right) + } \right.} \\ \left. {\varepsilon \left( u \right)} \right)d\left( {a,u} \right) + \sum\limits_{a \in V\left( {G'} \right)\backslash \left\{ {u',u''} \right\}} {\left( {\varepsilon \left( a \right) + \varepsilon \left( u \right)} \right)d\left( {a,} \right.} \\ \left. u \right) \ge \left( {2 + 2} \right) + 2 \times \left( {2 + 2} \right) + 2 \times \left( {2 + 2} \right) \times \left( {n - 3} \right) = \\ 8n - 12 \end{array} $ |
对于情况2下的三圈图G,有
$ {\xi ^d}\left( G \right) \ge {\xi ^d}\left( {G'} \right) + 8n - 12 > 4{n^2} - 9n - 7 $ |
引理6由此得证。证毕。
引理7 若G∈T3(n, x, y, z, w),n≥6,则ξd(G)≥4n2-9n-7当且仅当
证明 对n进行归纳。由引理4可知,当n=6时,ξd(G)≥83当且仅当
情况1 若图G中存在悬挂点,不妨取G中任意悬挂点u,顶点u′为u在图G中的唯一邻点。显然有
$ \left\{ \begin{array}{l} \varepsilon \left( u \right) \ge 2,\varepsilon \left( {u'} \right) \ge 1,d\left( {u,u'} \right) = 1\\ \varepsilon \left( a \right) \ge 2,d\left( {a,u} \right) \ge 2,\forall a \in V\left( G \right)\backslash \left\{ {u,u'} \right\} \end{array} \right. $ | (11) |
记G′=G-u,G′仍是三圈图。由分析可知
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge \varepsilon '\left( a \right),\varepsilon \left( b \right) \ge \varepsilon '\left( b \right),\forall a,b \in V\left( {G'} \right)\\ d\left( {a,b} \right) \ge d'\left( {a,b} \right),\forall a,b \in V\left( {G'} \right)\\ {\xi ^d}\left( {G'} \right) \ge 4{\left( {n - 1} \right)^2} - 9\left( {n - 1} \right) - 7 = \\ \;\;\;\;\;\;\;\;\;\;4{n^2} - 17n + 6 \end{array} \right. $ | (12) |
因为式(11)和式(12)成立,所以有
$ \begin{array}{l} {\xi ^d}\left( G \right) - {\xi ^d}\left( {G'} \right) = \sum\limits_{a,b \in V\left( {G'} \right)} {\left[ {\left( {\varepsilon \left( a \right) + \varepsilon \left( b \right)} \right)} \right.} \\ \left. {d\left( {a,b} \right) - \left( {\varepsilon '\left( a \right) + \varepsilon '\left( b \right)} \right)d'\left( {a,b} \right)} \right] + \left( {\varepsilon \left( {u'} \right) + } \right.\\ \left. {\varepsilon \left( u \right)} \right)d\left( {u,u'} \right) + \sum\limits_{a \in V\left( {G'} \right)\backslash \left\{ {u'} \right\}} {\left( {\varepsilon \left( a \right) + \varepsilon \left( u \right)} \right)d\left( {a,u} \right)} \ge \\ 0 + \left( {1 + 2} \right) \times 1 + \left( {n - 2} \right) \times \left( {2 + 2} \right) \times 2 = 8n - 13 \end{array} $ |
即对于情况1下的三圈图G,有ξd(G)≥ξd(G′)+8n-13≥4n2-9n-7,当且仅当
情况2 若图G中不存在悬挂点,由于n=x+y+z+w-5≥7,x≥y≥z,w≥3,则x=y=z=w=3,或x≥4,或w≥4。
对于n=7,x=y=z=w=3的情况,经过简单计算可以直接得到
$ {\xi ^d}\left( G \right) > {\xi ^d}\left( {{G_2}} \right) = 126 $ |
对于n>7且x≥4的情况,选取G中顶点u∈V(Px)且u∉V(Cw)∪V(Py)∪V(Pz),顶点u′和u″为u在图G中的两个邻点。显然有
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge 2,\forall a \in V\left( G \right)\\ d\left( {u,u'} \right) = 1,d\left( {u,u''} \right) = 1,d\left( {u',u''} \right) = 2\\ d\left( {a,u} \right) \ge 2,\forall a \in V\left( G \right)\backslash \left\{ {u,u',u''} \right\} \end{array} \right. $ | (13) |
记G′=G-u+u′u″,G′仍是三圈图。由分析可知
$ \left\{ \begin{array}{l} \varepsilon \left( a \right) \ge \varepsilon '\left( a \right),\varepsilon \left( b \right) \ge \varepsilon '\left( b \right),\forall a,b \in V\left( {G'} \right)\\ d\left( {a,b} \right) \ge d'\left( {a,b} \right),\forall a,b \in V\left( {G'} \right)\\ {\xi ^d}\left( {G'} \right) \ge 4{n^2} - 17n + 6 \end{array} \right. $ | (14) |
因为式(13)和式(14)成立,所以有
$ \begin{array}{l} {\xi ^d}\left( G \right) - {\xi ^d}\left( {G'} \right) \ge 0 + \left( {\varepsilon \left( {u'} \right),\varepsilon \left( {u''} \right)} \right)d\left( {u',} \right.\\ \left. {u''} \right) - \left( {\varepsilon '\left( {u'} \right),\varepsilon '\left( {u''} \right)} \right)d'\left( {u',u''} \right) + \sum\limits_{a \in \left\{ {u',u''} \right\}} {\left( {\varepsilon \left( a \right) + } \right.} \\ \left. {\varepsilon \left( u \right)} \right)d\left( {a,u} \right) + \sum\limits_{a \in V\left( {G'} \right)\backslash \left\{ {u',u''} \right\}} {\left( {\varepsilon \left( a \right) + \varepsilon \left( u \right)} \right)d\left( {a,} \right.} \\ \left. u \right) \ge \left( {2 + 2} \right) + 2 \times \left( {2 + 2} \right) + 2 \times \left( {2 + 2} \right) \times \left( {n - 3} \right) = \\ 8n - 12 \end{array} $ |
即ξd(G)≥ξd(G′)+8n-12>4n2-9n-7成立。
对于n>7且w≥4的情况,选取G中顶点u∈V(Cw)且u∉V(Px)∪V(Py)∪V(Pz),顶点u′和u″为u在图G中的两个邻点。记G′=G-u+u′u″,显然G′仍是三圈图。类似n>7,x≥4的情况,通过证明可以得到ξd(G)>4n2-9n-7。
引理7由此得证。证毕。
结合定理1~3及引理5~7,可以得到关于三圈图ξd值的最终结论。
定理8 对任意n≥5个顶点的三圈图G,ξd(G)≥4n2-9n-7当且仅当G∈{G1, G2, G3, G4, G5}时等号成立(其中G1, G2, G3, G4, G5见图 3)。
证明 假设G*是具有最小ξd值的三圈图,
$ {G^ * } \notin \left\{ {{G_1},{G_2},{G_3},{G_4},{G_5}} \right\} $ |
由引理5~7的证明可知
$ {\xi ^d}\left( {{G^ * }} \right) > 4{n^2} - 9n - 7 $ |
因此当且仅当G*∈{G1, G2, G3, G4, G5}时ξd(G*)=4n2-9n-7达到最小值。
定理8由此得证。证毕。
3 结束语本文首先研究了移边变换对离心距离指标的数值影响,然后利用数学归纳计算出了三圈图的离心距离指标的最小值,同时确定了最小值所对应的图的结构。本文的研究方法具有较好的适应性,为研究更一般图的离心距离指标提供了一种新思路。
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