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<H2><A NAME="SECTION00073000000000000000">
Scale Free网络的偏好依附模型</A>
</H2>
对偏好依附模型的一般处理方法包含连续性方法[<A
 HREF="node23.html#modelBA">18</A>,<A
 HREF="node23.html#modelBAJ">21</A>]，速率方程[<A
 HREF="node23.html#modelRed">20</A>,<A
 HREF="node23.html#age">23</A>]，Master方程[<A
 HREF="node23.html#modelDM">19</A>]与
生成函数方法[<A
 HREF="node23.html#arbitrary">9</A>]，这也是统计物理学经常使用的方法。这一节，我们围绕偏好依附模型的幂率分布形式介绍连续性方法，
速率方程与Master方程在复杂网络研究中的应用。

<P>
考虑Scale Free网络的偏好依附模型，从顶点的度值变化的角度，假设其度值连续，我们有如下方程，

<P></P>
<DIV ALIGN="CENTER">
<A NAME="continuum"></A><!-- MATH
 \begin{equation}
\frac{\partial k_{i}}{\partial t}=m\Pi\left(k_{i}\right)=m\frac{k_{i}}{\sum k_{j}}
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<IMG
 WIDTH="33" HEIGHT="56" ALIGN="MIDDLE" BORDER="0"
 SRC="img190.gif"
 ALT="$\displaystyle {\frac{{\partial k_{i}}}{{\partial t}}}$"> = <I>m</I><IMG
 WIDTH="17" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img118.gif"
 ALT="$\displaystyle \Pi$"><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img191.gif"
 ALT="$\displaystyle \left(\vphantom{k_{i}}\right.$"><I>k</I><SUB>i</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img192.gif"
 ALT="$\displaystyle \left.\vphantom{k_{i}}\right)$"> = <I>m</I><IMG
 WIDTH="44" HEIGHT="56" ALIGN="MIDDLE" BORDER="0"
 SRC="img193.gif"
 ALT="$\displaystyle {\frac{{k_{i}}}{{\sum k_{j}}}}$">
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(28)</TD></TR>
</TABLE>
</DIV>
每一步我们加入<I>m</I>条边，即增加2<I>m</I>个度值，于是<!-- MATH
 $\sum k_{j}=2mt+m_{0}$
 -->
<IMG
 WIDTH="21" HEIGHT="33" ALIGN="MIDDLE" BORDER="0"
 SRC="img194.gif"
 ALT="$ \sum$"><I>k</I><SUB>j</SUB> = 2<I>mt</I> + <I>m</I><SUB>0</SUB>，则

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
\frac{\partial k_{i}}{\partial t}=\frac{k_{i}}{2t}
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<IMG
 WIDTH="33" HEIGHT="56" ALIGN="MIDDLE" BORDER="0"
 SRC="img190.gif"
 ALT="$\displaystyle {\frac{{\partial k_{i}}}{{\partial t}}}$"> = <IMG
 WIDTH="23" HEIGHT="56" ALIGN="MIDDLE" BORDER="0"
 SRC="img195.gif"
 ALT="$\displaystyle {\frac{{k_{i}}}{{2t}}}$">
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(29)</TD></TR>
</TABLE>
</DIV>
顶点<I>k</I><SUB>i</SUB>是在<I>t</I><SUB>i</SUB>时刻进入系统的，且满足初始条件<!-- MATH
 $k_{i}\left(t_{i}\right)=m$
 -->
<I>k</I><SUB>i</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img196.gif"
 ALT="$ \left(\vphantom{t_{i}}\right.$"><I>t</I><SUB>i</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img197.gif"
 ALT="$ \left.\vphantom{t_{i}}\right)$"> = <I>m</I>，于是

<P></P>
<DIV ALIGN="CENTER">
<A NAME="ki"></A><!-- MATH
 \begin{equation}
k_{i}\left(t\right)=m\left(\frac{t}{t_{i}}\right)^{\frac{1}{2}}
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>k</I><SUB>i</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img106.gif"
 ALT="$\displaystyle \left(\vphantom{t}\right.$"><I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img107.gif"
 ALT="$\displaystyle \left.\vphantom{t}\right)$"> = <I>m</I><IMG
 WIDTH="18" HEIGHT="55" ALIGN="MIDDLE" BORDER="0"
 SRC="img198.gif"
 ALT="$\displaystyle \left(\vphantom{\frac{t}{t_{i}}}\right.$"><IMG
 WIDTH="20" HEIGHT="54" ALIGN="MIDDLE" BORDER="0"
 SRC="img199.gif"
 ALT="$\displaystyle {\frac{{t}}{{t_{i}}}}$"><IMG
 WIDTH="28" HEIGHT="68" ALIGN="MIDDLE" BORDER="0"
 SRC="img200.gif"
 ALT="$\displaystyle \left.\vphantom{\frac{t}{t_{i}}}\right)^{{\frac{1}{2}}}_{}$">
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(30)</TD></TR>
</TABLE>
</DIV>
利用上式，结合<I>t</I><SUB>i</SUB>的分布，就可以知道任意<I>t</I>时刻<I>k</I><SUB>i</SUB>的分布，即

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
P\left(k_{i}\left(t\right)<k\right)=P\left(t_{i}>\frac{m^{2}t}{k^{2}}\right).
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>P</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img201.gif"
 ALT="$\displaystyle \left(\vphantom{k_{i}\left(t\right)&lt;k}\right.$"><I>k</I><SUB>i</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img106.gif"
 ALT="$\displaystyle \left(\vphantom{t}\right.$"><I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img107.gif"
 ALT="$\displaystyle \left.\vphantom{t}\right)$"> &lt; <I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img202.gif"
 ALT="$\displaystyle \left.\vphantom{k_{i}\left(t\right)&lt;k}\right)$"> = <I>P</I><IMG
 WIDTH="19" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
 SRC="img203.gif"
 ALT="$\displaystyle \left(\vphantom{t_{i}&gt;\frac{m^{2}t}{k^{2}}}\right.$"><I>t</I><SUB>i</SUB> &gt; <IMG
 WIDTH="37" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"
 SRC="img204.gif"
 ALT="$\displaystyle {\frac{{m^{2}t}}{{k^{2}}}}$"><IMG
 WIDTH="19" HEIGHT="65" ALIGN="MIDDLE" BORDER="0"
 SRC="img205.gif"
 ALT="$\displaystyle \left.\vphantom{t_{i}&gt;\frac{m^{2}t}{k^{2}}}\right)$">.
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(31)</TD></TR>
</TABLE>
</DIV>
在<I>t</I>时刻，<!-- MATH
 $t_{i}\in \left[0,t\right)$
 -->
<I>t</I><SUB>i</SUB> <IMG
 WIDTH="16" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
 SRC="img29.gif"
 ALT="$ \in$"> <IMG
 WIDTH="11" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img206.gif"
 ALT="$ \left[\vphantom{0,t}\right.$">0, <I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img207.gif"
 ALT="$ \left.\vphantom{0,t}\right)$">的分布函数为

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
P(t_{i})=\frac{1}{m_{0}+t}.
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>P</I>(<I>t</I><SUB>i</SUB>) = <IMG
 WIDTH="59" HEIGHT="55" ALIGN="MIDDLE" BORDER="0"
 SRC="img208.gif"
 ALT="$\displaystyle {\frac{{1}}{{m_{0}+t}}}$">.
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(32)</TD></TR>
</TABLE>
</DIV>
结合以上两式就可以得到<I>k</I><SUB>i</SUB>的分布，

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
P(k)=\frac{\partial P\left(k_{i}\left(t\right)<k\right)}{\partial k}=\frac{2m^2t}{m_{0}+t}\frac{1}{k^3}\sim 2m^2k^{-3}(t\rightarrow\infty).
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>P</I>(<I>k</I>) = <IMG
 WIDTH="119" HEIGHT="58" ALIGN="MIDDLE" BORDER="0"
 SRC="img209.gif"
 ALT="$\displaystyle {\frac{{\partial P\left(k_{i}\left(t\right)&lt;k\right)}}{{\partial k}}}$"> = <IMG
 WIDTH="59" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"
 SRC="img210.gif"
 ALT="$\displaystyle {\frac{{2m^2t}}{{m_{0}+t}}}$"><IMG
 WIDTH="25" HEIGHT="55" ALIGN="MIDDLE" BORDER="0"
 SRC="img211.gif"
 ALT="$\displaystyle {\frac{{1}}{{k^3}}}$"> <IMG
 WIDTH="18" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
 SRC="img121.gif"
 ALT="$\displaystyle \sim$"> 2<I>m</I><SUP>2</SUP><I>k</I><SUP>-3</SUP>(<I>t</I> <IMG
 WIDTH="22" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
 SRC="img212.gif"
 ALT="$\displaystyle \rightarrow$"> <IMG
 WIDTH="22" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
 SRC="img213.gif"
 ALT="$\displaystyle \infty$">).
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(33)</TD></TR>
</TABLE>
</DIV>

<P>
方程(<A HREF="#continuum">28</A>)是一种平均意义上的描述，实际随机过程中所加入的边只是按照<!-- MATH
 $\Pi\left(k_{i}\right)$
 -->
<IMG
 WIDTH="17" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
 SRC="img111.gif"
 ALT="$ \Pi$"><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img214.gif"
 ALT="$ \left(\vphantom{k_{i}}\right.$"><I>k</I><SUB>i</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img215.gif"
 ALT="$ \left.\vphantom{k_{i}}\right)$">的分布来选择，并不能保
证确实让每一个顶点的度值增加那么大，也正是因此称为连续性方法，所以对实际随机过程分布函数演化的描述需要用到其他方法。引入分布
函数<!-- MATH
 $p\left(k;t_{i},t\right)$
 -->
<I>p</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img216.gif"
 ALT="$ \left(\vphantom{k;t_{i},t}\right.$"><I>k</I>;<I>t</I><SUB>i</SUB>, <I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img217.gif"
 ALT="$ \left.\vphantom{k;t_{i},t}\right)$">，表示<I>t</I><SUB>i</SUB>时刻引入系统的顶点在<I>t</I>时刻度值为<I>k</I>的几率，其初始条件为
<!-- MATH
 $p\left(k;t_{i},t_{i}\right)=\delta\left(k-m\right)$
 -->
<I>p</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img218.gif"
 ALT="$ \left(\vphantom{k;t_{i},t_{i}}\right.$"><I>k</I>;<I>t</I><SUB>i</SUB>, <I>t</I><SUB>i</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img219.gif"
 ALT="$ \left.\vphantom{k;t_{i},t_{i}}\right)$"> = <IMG
 WIDTH="13" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
 SRC="img20.gif"
 ALT="$ \delta$"><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img220.gif"
 ALT="$ \left(\vphantom{k-m}\right.$"><I>k</I> - <I>m</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img221.gif"
 ALT="$ \left.\vphantom{k-m}\right)$">，，满足如下Master方程，

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
p\left(k;t_{i},t+1\right)=p\left(k;t_{i},t\right)+w^{get}\left(k-1\right)p\left(k-1;t_{i},t\right)-w^{get}\left(k-1\right)p\left(k;t_{i},t\right)
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>p</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img222.gif"
 ALT="$\displaystyle \left(\vphantom{k;t_{i},t+1}\right.$"><I>k</I>;<I>t</I><SUB>i</SUB>, <I>t</I> + 1<IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img223.gif"
 ALT="$\displaystyle \left.\vphantom{k;t_{i},t+1}\right)$"> = <I>p</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img224.gif"
 ALT="$\displaystyle \left(\vphantom{k;t_{i},t}\right.$"><I>k</I>;<I>t</I><SUB>i</SUB>, <I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img225.gif"
 ALT="$\displaystyle \left.\vphantom{k;t_{i},t}\right)$"> + <I>w</I><SUP>get</SUP><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img226.gif"
 ALT="$\displaystyle \left(\vphantom{k-1}\right.$"><I>k</I> - 1<IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img227.gif"
 ALT="$\displaystyle \left.\vphantom{k-1}\right)$"><I>p</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img228.gif"
 ALT="$\displaystyle \left(\vphantom{k-1;t_{i},t}\right.$"><I>k</I> - 1;<I>t</I><SUB>i</SUB>, <I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img229.gif"
 ALT="$\displaystyle \left.\vphantom{k-1;t_{i},t}\right)$"> - <I>w</I><SUP>get</SUP><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img226.gif"
 ALT="$\displaystyle \left(\vphantom{k-1}\right.$"><I>k</I> - 1<IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img227.gif"
 ALT="$\displaystyle \left.\vphantom{k-1}\right)$"><I>p</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img224.gif"
 ALT="$\displaystyle \left(\vphantom{k;t_{i},t}\right.$"><I>k</I>;<I>t</I><SUB>i</SUB>, <I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img225.gif"
 ALT="$\displaystyle \left.\vphantom{k;t_{i},t}\right)$">
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(34)</TD></TR>
</TABLE>
</DIV>
其中<!-- MATH
 $w^{get}\left(k\right)=\frac{k}{2t}$
 -->
<I>w</I><SUP>get</SUP><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img78.gif"
 ALT="$ \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img79.gif"
 ALT="$ \left.\vphantom{k}\right)$"> = <IMG
 WIDTH="20" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"
 SRC="img230.gif"
 ALT="$ {\frac{{k}}{{2t}}}$">。所有顶点的度值分布定义所有不同时刻加入系统的顶点的平均行为，即

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
P\left(k\right)=\lim_{t\rightarrow\infty}\frac{\sum_{t_{i}}p\left(k;t_{i},t\right)}{t}.
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>P</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img231.gif"
 ALT="$\displaystyle \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img232.gif"
 ALT="$\displaystyle \left.\vphantom{k}\right)$"> = <IMG
 WIDTH="36" HEIGHT="42" ALIGN="MIDDLE" BORDER="0"
 SRC="img233.gif"
 ALT="$\displaystyle \lim_{{t\rightarrow\infty}}^{}$"><IMG
 WIDTH="107" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"
 SRC="img234.gif"
 ALT="$\displaystyle {\frac{{\sum_{t_{i}}p\left(k;t_{i},t\right)}}{{t}}}$">.
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(35)</TD></TR>
</TABLE>
</DIV>
有以上两式可以求得度分布

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
P\left(k\right)=\frac{2m\left(m+1\right)}{k\left(k+1\right)\left(k+2\right)}\sim k^{-3}.
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>P</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img231.gif"
 ALT="$\displaystyle \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img232.gif"
 ALT="$\displaystyle \left.\vphantom{k}\right)$"> = <IMG
 WIDTH="130" HEIGHT="58" ALIGN="MIDDLE" BORDER="0"
 SRC="img235.gif"
 ALT="$\displaystyle {\frac{{2m\left(m+1\right)}}{{k\left(k+1\right)\left(k+2\right)}}}$"> <IMG
 WIDTH="18" HEIGHT="31" ALIGN="MIDDLE" BORDER="0"
 SRC="img121.gif"
 ALT="$\displaystyle \sim$"> <I>k</I><SUP>-3</SUP>.
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(36)</TD></TR>
</TABLE>
</DIV>

<P>
速率方程从另一个分布空间描述。<I>t</I>时刻度值为<I>k</I>的顶点组成一个集团，集团大小为<!-- MATH
 $N_{k}\left(t\right)$
 -->
<I>N</I><SUB>k</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img74.gif"
 ALT="$ \left(\vphantom{t}\right.$"><I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img75.gif"
 ALT="$ \left.\vphantom{t}\right)$">。当新顶点进入系统时，
<I>N</I><SUB>k</SUB>可能发生改变，其变化率满足如下方程，

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
\frac{dN_{k}\left(t\right)}{dt}=mW^{get}\left(k-1\right)-mW^{get}\left(k\right)+\delta_{k,m}.
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<IMG
 WIDTH="62" HEIGHT="58" ALIGN="MIDDLE" BORDER="0"
 SRC="img236.gif"
 ALT="$\displaystyle {\frac{{dN_{k}\left(t\right)}}{{dt}}}$"> = <I>mW</I><SUP>get</SUP><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img226.gif"
 ALT="$\displaystyle \left(\vphantom{k-1}\right.$"><I>k</I> - 1<IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img227.gif"
 ALT="$\displaystyle \left.\vphantom{k-1}\right)$"> - <I>mW</I><SUP>get</SUP><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img231.gif"
 ALT="$\displaystyle \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img232.gif"
 ALT="$\displaystyle \left.\vphantom{k}\right)$"> + <IMG
 WIDTH="36" HEIGHT="33" ALIGN="MIDDLE" BORDER="0"
 SRC="img237.gif"
 ALT="$\displaystyle \delta_{{k,m}}^{}$">.
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(37)</TD></TR>
</TABLE>
</DIV>
其中<!-- MATH
 $W^{get}\left(k\right)$
 -->
<I>W</I><SUP>get</SUP><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img78.gif"
 ALT="$ \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img79.gif"
 ALT="$ \left.\vphantom{k}\right)$">为所有度值为<I>k</I>的顶点获得新的联接的几率，

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
W^{get}\left(k\right)=\frac{kN_{k}\left(t\right)}{\sum_{d}dN_{d}\left(t\right)}.
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>W</I><SUP>get</SUP><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img231.gif"
 ALT="$\displaystyle \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img232.gif"
 ALT="$\displaystyle \left.\vphantom{k}\right)$"> = <IMG
 WIDTH="89" HEIGHT="58" ALIGN="MIDDLE" BORDER="0"
 SRC="img238.gif"
 ALT="$\displaystyle {\frac{{kN_{k}\left(t\right)}}{{\sum_{d}dN_{d}\left(t\right)}}}$">.
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(38)</TD></TR>
</TABLE>
</DIV>
<I>N</I><SUB>k</SUB>同样满足方程<!-- MATH
 $\sum_{k}kN_{k}\left(t\right)=2mt+m_{0}$
 -->
<IMG
 WIDTH="29" HEIGHT="33" ALIGN="MIDDLE" BORDER="0"
 SRC="img239.gif"
 ALT="$ \sum_{{k}}^{}$"><I>kN</I><SUB>k</SUB><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img74.gif"
 ALT="$ \left(\vphantom{t}\right.$"><I>t</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img75.gif"
 ALT="$ \left.\vphantom{t}\right)$"> = 2<I>mt</I> + <I>m</I><SUB>0</SUB>，且

<P></P>
<DIV ALIGN="CENTER">
<!-- MATH
 \begin{equation}
P\left(k\right)=\lim_{t\rightarrow\infty}\frac{N_{k}\left(t\right)}{\sum_{d}N_{d}\left(t\right)}.
\end{equation}
 -->
<TABLE WIDTH="100%" ALIGN="CENTER">
<TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP>
<I>P</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img231.gif"
 ALT="$\displaystyle \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img232.gif"
 ALT="$\displaystyle \left.\vphantom{k}\right)$"> = <IMG
 WIDTH="36" HEIGHT="42" ALIGN="MIDDLE" BORDER="0"
 SRC="img233.gif"
 ALT="$\displaystyle \lim_{{t\rightarrow\infty}}^{}$"><IMG
 WIDTH="80" HEIGHT="58" ALIGN="MIDDLE" BORDER="0"
 SRC="img240.gif"
 ALT="$\displaystyle {\frac{{N_{k}\left(t\right)}}{{\sum_{d}N_{d}\left(t\right)}}}$">.
</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(39)</TD></TR>
</TABLE>
</DIV>
结合一上方程可以解得度分布<!-- MATH
 $P\left(k\right)$
 -->
<I>P</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img78.gif"
 ALT="$ \left(\vphantom{k}\right.$"><I>k</I><IMG
 WIDTH="13" HEIGHT="35" ALIGN="MIDDLE" BORDER="0"
 SRC="img79.gif"
 ALT="$ \left.\vphantom{k}\right)$">，具体计算见文章[<A
 HREF="node23.html#modelRed">20</A>,<A
 HREF="node23.html#age">23</A>]。

<P>
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2004-01-04
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