本文Google Docs地址
Guy Rosman，Alexander M. Bronstein，Michael M. Bronstein，Avram Sidi，Ron Kimmel
Fast Multidimensional Scaling using Vector Extrapolation
Technion - Computer Science Department - Technical Report CIS-2008-01 --2008
摘要：
MDS是一类低维表示方法，对象是给定距离矩阵的点集。在很多MDS应用中，算法的速
度和效率是关键问题，向量外推方法可以用来提高固定点迭代算法的收敛速度。本文
将提出用向量外推加速MDS数值解速度的方法。
2 多维标度
2.1 最小二乘MDS
在n维欧式空间中，坐标表达：$latex {X=\left( {x_{ij} } \right)}&fg=000000$
，是N$latex {\times }&fg=000000$m度量值；$latex {d_{ij} \left( X \right)}
&fg=000000$表示i、j之间的距离；$latex {\delta _{ij} }&fg=000000$表示输入距
离，本文将使用一个新的表示形式。
通常MDS问题可以用下述优化模型表示：
$latex \displaystyle
\mathop {\min }\limits_X \sum\limits_{i<j} {w_{ij} f_{ERR} \left( {d_
{ij} \left( X \right),\delta _{ij} } \right)} &fg=000000$
$latex {w_{ij} }&fg=000000$是表征每一对距离重要性的权值；$latex {f_{ERR}
}&fg=000000$是计算输入距离和近似距离的误差代价函数，选择距离误差函数的二次
项形式，即「stress 」函数：
$latex \displaystyle
f_{STRESS} \left( {d,\delta } \right)=\left( {d-\delta } \right)^2
&fg=000000$
，
定义距离平方之差的形式：
$latex \displaystyle f_{SSTRESS}
\left( {d,\delta } \right)=\left( {d^2-\delta ^2} \right)^2 &fg=000000$
得到一个函数，通常称之为「sstress」函数。
2.2 SMACOF 算法
为了最小化stress函数，$latex {s\left( {\rm
{\bf X}} \right)=\sum\limits_{i<j}^ {w_{ij} \left( {d_{ij} \left( {\rm
{\bf X}} \right)-\delta _{ij} } \right)^2} }&fg=000000$，我们需要求出$latex
{s\left( {\rm {\bf X}} \right)}&fg=000000$关于$latex {{\rm {\bf X}}}
&fg=000000$梯度：
$latex \displaystyle \nabla s\left( {\rm
{\bf X}} \right)=2{\rm {\bf VX}}-2{\rm {\bf B}}\left( {\rm {\bf X}}
\right){\rm {\bf X}} &fg=000000$
，
这里$latex {{\rm {\bf V}}}&fg=000000$和$latex {{\rm {\bf B}}}&fg=000000$通
过下式给出：
$latex \displaystyle \left(
{\rm {\bf V}} \right)_{ij} =\left\{ {\begin{array}{c} -w_{ij} \mbox{ } if
i\ne j \\ \sum\limits_{k\ne i}^ {w_{ik} } if i=j \\ \end{array}} \right. \
\ \ \ \ (1)&fg=000000$
$latex \displaystyle \left( {\rm {\bf
B}} \right)_{ij} =\left\{ {\begin{array}{c} \mbox{-w}_{ij} \delta _{ij} d_
{ij}^{-1} \left( {\rm {\bf X}} \right) \mbox{if }i\ne j \mbox{and} d_{ij}
\left( {\rm {\bf X}} \right)\ne 0 \\ 0 \mbox{if }i\ne j \mbox{and} d_{ij}
\left( {\rm {\bf X}} \right)=0 \\ -\sum\nolimits_{k\ne i} {b_{ik} } \mbox
{if }i=j \\ \end{array}} \right. \ \ \ \ \ (2)&fg=000000$
使用一阶优化，则通过下式得到：
$latex {2{\rm {\bf VX}}=2{\rm {\bf B}}\left( {\rm {\bf X}} \right){\rm {\bf
X}}}&fg=000000$，或者：
$latex \displaystyle
{\rm {\bf X}}={\rm {\bf V}}^\dag {\rm {\bf B}}\left( {\rm {\bf X}}
\right){\rm {\bf X}} \ \ \ \ \ (3)&fg=000000$
这里$latex {\dag }&fg=000000$表示矩阵伪逆。
由（3）可以得到迭代形式（方法见文献[25]）：
$latex \displaystyle {\rm {\bf X}}^{\left( {k+1} \right)}={\rm {\bf V}}^
\dag {\rm {\bf B}}\left( {{\rm {\bf X}}^{\left( k \right)}} \right){\rm
{\bf X}}^{\left( k \right)} \ \ \ \ \ (4)&fg=000000$
（3）式可视为一个固定点，将（3）迭代使之收敛到stress 代价函数的局部极
小值。以上处理思路称为「SMACOF」------standing from Scaling by Majorizing a
Complicated Function。
stress函数有一个重要特性就是可以保证一个stress值的单调下降序列。这是其他优
化问题少见的。另一方面，SMACOF算法的收敛速度是比较慢的。
2.3 经典标度
另一类使用广泛的代价函数是「strain」
（定义见下文），其引出了一类代数MDS算法，称为「经典标度」（classical
scaling）。
令$latex {\Delta ^{\left( 2 \right)}}&fg=000000$表示输入距离矩阵的平方；
$latex {{\rm {\bf D}}^{\left( \mbox{2} \right)}\left( {\rm {\bf X}}
\right)}&fg=000000$表示目标欧式距离的平方：
$latex \displaystyle {\rm {\bf D}}^\mbox{2}\left( {\rm {\bf X}}\right)\mbox{=}{\rm {\bf c1}}^T+{\rm {\bf 1c}}^T-2{\rm {\bf XX}}^T \ \ \ \ \ (5)&fg=000000$
这里$latex {c_i =\left\langle {x_i ,x_i } \right\rangle }&fg=000000$
令$latex {{\rm {\bf J}}\mbox{=}{\rm {\bf I}}\mbox{-}\frac{1}{N}{\rm {\bf
11}}^T}&fg=000000$，表示中心矩阵。
假设给定距离均为欧式距离，得到这些点的内积矩阵（Gram 矩阵）：
$latex \displaystyle {\rm {\bf B}}_\Delta =\frac{1}{2}{\rm
{\bf J}}\Delta ^{\left( 2 \right)}{\rm {\bf J}} &fg=000000$
对于欧式距离的情形，$latex {\left( {{\rm {\bf B}}_\Delta } \right)_{ij} =
\left\langle {x_i ,x_j } \right\rangle }&fg=000000$
但实际上，输入矩阵往往不是欧式的，二者的差异度可以通过一个测量值近似表示，
称之为「strain」：
$latex \displaystyle \begin
{array}{l} \left\| {-\frac{1}{2}{\rm {\bf J}}\left( {{\rm {\bf D}}^{\left(
2 \right)}\left( {\rm {\bf X}} \right)-\Delta ^{\left( 2 \right)}} \right)
{\rm {\bf J}}} \right\|_F^2 \\ {\rm {\bf =}}\left\| {{\rm {\bf XX}}^T+
\frac{1}{2}{\rm {\bf J}}\left( {\Delta ^{\left( 2 \right)}} \right){\rm
{\bf J}}} \right\|_F^2 \\ =\left\| {{\rm {\bf XX}}^T-{\rm {\bf B}}_\Delta }
\right\|_F^2 \\ \end{array} &fg=000000$
通过对$latex {{\rm {\bf B}}_\Delta }&fg=000000$进行特征值分解，可以找到
strain函数的全局最优解。
经典标度方法的缺陷在于它的自适应性不强，而且计算开销也较大。因此在后文的论
述中，将只关注stress函数以及SMACOF算法。然而，全局优化可保证经典标度方法在
SMACOF的初始阶段具有良好性能。
3 向量外推方法
考虑使用历次迭代方法来加速SMACOF算法的收敛速度，如使用向量外推方法来预测收
敛极限。
本文考察了两个向量外推方法：
1、 「MPE」------最小多项式外推（minimal polynomial extrapolation，文献[11]
）；
2、「RRE」------减秩外推（reduced rank extrapolation，文献[31,19]）。
两个方法在加速非线性/线性/大型稀疏系统方程中固定点迭代的向量序列收敛速度方
面都具有高效的性能。
二者均需要考察一个经过线性固定点迭代过程得出的序列（$latex {{\rm {\bf x}}_0
,{\rm {\bf x}}_1 ,...}&fg=000000$）：
$latex
\displaystyle x_{n+1} ={\rm {\bf A}}x_n +b, n=0,1,... \ \ \ \ \ (6)
&fg=000000$
此处$latex {{\rm {\bf A}}}&fg=000000$是给定$latex {N\times N}
&fg=000000$矩阵，$latex {{\rm {\bf b}}}&fg=000000$是给定N维向量，$latex
{x_0 }&fg=000000$是用户选取的初始向量。这个序列有一个极限s，它是下述方程的
唯一解：
$latex \displaystyle {\rm {\bf
x}}={\rm {\bf Ax}}+{\rm {\bf b}} \ \ \ \ \ (7)&fg=000000$
假设$latex {\rho \left( {\rm {\bf A}} \right)}&fg=000000$是谱半
径，$latex {\rho \left( {\rm {\bf A}} \right)<{\rm x}}&fg=000000$，另一
方面（7）式也可以写为$latex {\left( {{\rm {\bf I}}-{\rm {\bf A}}} \right)
x={\rm {\bf b}}}&fg=000000$，由于1不是$latex {{\rm {\bf A}}}&fg=000000$的奇
异值，所以矩阵$latex {{\rm {\bf I}}-{\rm {\bf A}}}&fg=000000$是非奇异的。
给定(6)式的序列，令
$latex \displaystyle
{\rm {\bf u}}_{\rm {\bf n}} {\rm {\bf =\Delta x}}_{\rm {\bf n}} {\rm {\bf
=x}}_{{\rm {\bf n+1}}} {\rm {\bf -x}}_{\rm {\bf n}} {\rm {\bf , n=0,1,...}}
&fg=000000$
再定义误差向量：
$latex \displaystyle ?_n
=x_n -s\mbox{, }n=0,1,... \ \ \ \ \ (8)&fg=000000$
考虑$latex {{\rm {\bf s}}={\rm {\bf As}}+{\rm {\bf b}}}&fg=000000$，我
们可以通过n步迭代从初始误差得到当前误差：
$latex \displaystyle ?_n =\left( {{\rm {\bf Ax}}_{n-1} +b} \right)-\left(
{{\rm {\bf As}}+{\rm {\bf b}}} \right)={\rm {\bf A}}\left( {{\rm {\bf x}}_
{n-1} -{\rm {\bf s}}} \right)={\rm {\bf A}}?_{n-1} \ \ \ \ \ (9)
&fg=000000$
由此可以得到：
$latex \displaystyle
?_n ={\rm {\bf A}}^n?_0 \mbox{, }n=0,1,... \ \ \ \ \ (10)&fg=000000$
通过对k+1连续$latex {x_i }&fg=000000$（k +1 consecutive）采用「加权平
均」来近似s：
$latex \displaystyle
{\rm {\bf s}}_{n,k} =\sum\limits_{i=0}^k {\gamma _i {\rm {\bf x}}_{n+i} }
\mbox{; }\sum\limits_{i=0}^k {\gamma _i } =1 \ \ \ \ \ (11)&fg=000000$
将(8)带入（11），并考虑$latex {\sum\nolimits_{i=0}
^k {\gamma _i } =1}&fg=000000$，可以得到：
$latex \displaystyle {\rm {\bf s}}_{n,k} =\sum\limits_{i=0}^k {\gamma _i
\left( {{\rm {\bf s}}+?_{n+i} } \right)} ={\rm {\bf s}}+\sum\limits_{i=0}^k
{\gamma _i } ?_{n+i} \ \ \ \ \ (12)&fg=000000$
再带入(10)，得到
$latex \displaystyle {\rm {\bf s}}_{n,k} ={\rm {\bf s}}+\sum\limits_
{i=0}^k {\gamma _i {\rm {\bf A}}^{n+i}} ?_0 \ \ \ \ \ (13)&fg=000000$
为了使$latex {?_{n+i\mbox{, }} i=0,1,...}&fg=000000$的加权和$latex
{\sum\nolimits_{i=0}^k {\gamma _i {\rm {\bf A}}^{n+i}?_0 } }&fg=000000$尽可
能小，我们必须选择合适的$latex {\gamma _i }&fg=000000$。
现在，给定一个N$latex {\times }&fg=000000$N矩阵B，以及任意一个N维向
量u，则存在具有最小阶数（最多N）的唯一莫尼多项式（monic polynomial）
$latex {P\left( z \right)}&fg=000000$，是u的零化多项式：$latex {P
\left( {\rm {\bf B}} \right){\rm {\bf u}}=0}&fg=000000$，这个多项式称为
B关于u的「极小多项式」（minimal polynomial）。$latex {P\left(
z \right)}&fg=000000$的零解部分或者全部是B的特征值。
因此，若A关于$latex {?_n }&fg=000000$的极小多项式是：
$latex \displaystyle P\left( z \right)=\sum\limits_{i=0}^k
{c_i z^i} ;\mbox{ }c_k =1 &fg=000000$
即：$latex {P\left( {\rm {\bf A}} \right)?_n =0}&fg=000000$，联立(10)式，得：
$latex
\displaystyle \sum\limits_{i=0}^k {c_i {\rm {\bf A}}^i?_n } =\sum
\limits_{i=0}^k {c_i ?_{n+i} } =0 \ \ \ \ \ (14)&fg=000000$
(14)是N个线性方程的集，k个未知参数$latex {c_0 ,c_1
,...,}&fg=000000$同时$latex {c_k =1}&fg=000000$。这N个方程的解是唯一的，因
为$latex {P\left( z \right)}&fg=000000$的解是唯一的。我们可以由x$latex {_
{i}}&fg=000000$的信息单独得到c$latex {_{i}}&fg=000000$，联立(14)和(9)，有：
$latex
\displaystyle 0=\sum\limits_{i=0}^k {c_i {\rm {\bf A}}?_{n+i} } =\sum
\limits_{i=0}^k {c_i ?_{n+i+1} } &fg=000000$
减去(14)，由此得到线性系统：
$latex \displaystyle \sum\limits_{i=0}^k {c_i {\rm {\bf u}}
_{n+i} } ={\rm {\bf 0}} \ \ \ \ \ (15)&fg=000000$
一旦$latex {c_0 ,c_1 ,...,c_{k-1} }&fg=000000$确定，我们令$latex {c_k
}&fg=000000$=1、$latex {\gamma _i ={c_i } \mathord{\left/ {\vphantom {{c_i
} {\sum\nolimits_{j=0}^k {c_j } }}} \right. \kern-\nulldelimiterspace}
{\sum\nolimits_{j=0}^k {c_j } },\mbox{ }i=0,1,...,k.}&fg=000000$
综上所述，若k是A关于$latex {?_n }&fg=000000$的极小多项式的阶数，则存在序列
满足$latex {\sum\nolimits_{i=0}^k {\gamma _i =1} }&fg=000000$，则有：$latex
{\sum\nolimits_{i=0}^k {\gamma _i {\rm {\bf x}}_{n+i} ={\rm {\bf s}}} }
&fg=000000$（Why not $latex {{\rm {\bf s}}_{n,k} }&fg=000000$？）
值得注意的是，不论是否$latex {\rho \left( {\rm {\bf A}} \right)<1}
&fg=000000$，s都是$latex {\left( {{\rm {\bf I}}-{\rm {\bf A}}}
\right){\rm {\bf x}}={\rm {\bf b}}}&fg=000000$的解，因此，无论$latex
{\mathop {\lim }\limits_{n\rightarrow \infty } {\rm {\bf x}}_n }&fg=000000$
是否存在，$latex {{\rm {\bf s}}=\sum\nolimits_{i=0}^k {\gamma _i {\rm {\bf
x}}_{n+i} } }&fg=000000$都成立。
为简化表述，使用如下标记：
$latex
\displaystyle {\rm {\bf U}}_s^{\left( j \right)} =\left[ {{\rm {\bf u}}_j
\vert {\rm {\bf u}}_{j+1} \vert ...\vert {\rm {\bf u}}_{j+s} } \right] \ \
\ \ \ (16)&fg=000000$
因此，$latex {{\rm {\bf U}}_s^{\left( j \right)} }&fg=000000$是一个N
$latex {\times }&fg=000000$（j+1）矩阵，则(14)可以表述
为：
$latex \displaystyle {\rm {\bf U}}
_k^{\left( n \right)} {\rm {\bf c}}={\rm {\bf 0}}\mbox{; }{\rm {\bf c}}=
\left[ {c_0 ,c_1 ,...c_k } \right]^T \ \ \ \ \ (17)&fg=000000$
当然，用$latex {\sum\nolimits_{i=0}^k {c_i } }&fg=000000$分解（17），
可以得到：
$latex \displaystyle {\rm {\bf
U}}_k^{\left( n \right)} \gamma ={\rm {\bf 0}}\mbox{; }\gamma =\left[
{\gamma _0 ,\gamma _1 ,...\gamma _k } \right]^T \ \ \ \ \ (18)
&fg=000000$
3.1 MPE派生
前文所述，极小多项式的阶数可以达到N，若N很大
，这就可能会导致比较大的存储开销；此外，我们还没有方法可以准确得知这个阶数
，鉴于此，可以采取一种解决途径：先任意找一个远远小于阶数的正指数k，带入(15)，则(15)则不再是一致的，因此也不再
有对于$latex {c_0 ,c_1 ,...,c_{k-1} }&fg=000000$（其中$latex {c_\mbox{k}
=1}&fg=000000$）的唯一解。通常对于这样的问题，我们可以采取最小二乘方法求解
，沿着这个思路，计算(15)中描述的$latex {\gamma _0 ,
\gamma _1 ,...\gamma _k }&fg=000000$，然后计算向量$latex {{\rm {\bf s}}_
{n,k} =\sum\limits_{i=0}^k {\gamma _i {\rm {\bf x}}_{n+i} } }&fg=000000$，
计算结果即为s的近似。以上方法称为「极小多项式外推算法」（MPE------
minimal polynomial extrapolation），其算法流程见表-1：
\centerline{\includegraphics[width=5.83in,height=1.41in]{mytex31.eps}}
\caption{MPE算法流程}
3.2 RRE派生
同3.1，在(18)中带入k，则不再对$latex {\gamma _0 ,
\gamma _1 ,...\gamma _k }&fg=000000$具有唯一解，同样应用最小二乘算法（约束
为$latex {\sum\nolimits_{i=0}^k {\gamma _i =1} }&fg=000000$），然后计算
s的近似$latex {{\rm {\bf s}}_{n,k} =\sum\limits_{i=0}^k {\gamma _i
{\rm {\bf x}}_{n+i} } }&fg=000000$。这个方法称为「减秩外推」（RRE------
reduced rank extrapolation），算法流程见表-2：
\centerline{\includegraphics[width=5.83in,height=1.22in]{mytex32.eps}}
\caption{RRE算法}
3.3 对非线性方程的处理
前文所述，在SMACOF算法中存在非线性方程，现在用向量外推方法处理这个问题。假
设非线性方程记作：
$latex \displaystyle
{\rm {\bf x}}={\rm {\bf F}}\left( {\rm {\bf x}} \right) \ \ \ \ \ (19)
&fg=000000$
这里$latex {{\rm {\bf F}}\left( {\rm {\bf x}} \right)}&fg=000000$是N维
向量值函数，x是N维未知向量，通过下式得到近似值x$latex {_{n}}
&fg=000000$：
$latex \displaystyle {\rm
{\bf x}}_{n+1} ={\rm {\bf F}}\left( {{\rm {\bf x}}_n } \right)\mbox{, }
n=0,1,... \ \ \ \ \ (20)&fg=000000$
并且假设这个序列收敛于解s，$latex {{\rm {\bf F}}}&fg=000000$是
SMACOF迭代的右边函数，由于x接近s，$latex {{\rm {\bf F}}\left(
{\rm {\bf x}} \right)}&fg=000000$可通过泰勒级数展开：
$latex \displaystyle {\rm {\bf F}}\left( {\rm {\bf x}} \right)={\rm {\bf
F}}\left( {\rm {\bf s}} \right)+{\rm {\bf {F}'}}\left( {\rm {\bf s}}
\right)\left( {{\rm {\bf x}}-{\rm {\bf s}}} \right)+o\left( {\left\| {{\rm
{\bf x}}-{\rm {\bf s}}} \right\|^2} \right)\mbox{ as }{\rm {\bf
x}}\rightarrow {\rm {\bf s}} &fg=000000$
此处$latex {{\rm {\bf {F}'}}\left( {\rm {\bf x}} \right)}&fg=000000$是
$latex {{\rm {\bf F}}\left( {\rm {\bf x}} \right)}&fg=000000$的雅各比矩阵，
又因为$latex {{\rm {\bf F}}\left( {\rm {\bf s}} \right)={\rm {\bf s}}}
&fg=000000$，则上式可写为：
$latex \displaystyle {\rm {\bf
F}}\left( {\rm {\bf x}} \right)={\rm {\bf s}}+{\rm {\bf {F}'}}\left( {\rm
{\bf s}} \right)\left( {{\rm {\bf x}}-{\rm {\bf s}}} \right)+o\left(
{\left\| {{\rm {\bf x}}-{\rm {\bf s}}} \right\|^2} \right)\mbox{ as }{\rm
{\bf x}}\rightarrow {\rm {\bf s}} &fg=000000$
假设序列$latex {{\rm {\bf x}}_0 ,{\rm {\bf x}}_1 ,...,}&fg=000000$收敛于
s，则当n足够大时，$latex {{\rm {\bf x}}_n }&fg=000000$逼近s，
因此有：
$latex \displaystyle {\rm {\bf x}}_{n+1} ={\rm
{\bf s}}+{\rm {\bf {F}'}}\left( {\rm {\bf s}} \right)\left( {{\rm {\bf x}}
_n -{\rm {\bf s}}} \right)+o\left( {\left\| {{\rm {\bf x}}_n -{\rm {\bf
s}}} \right\|^2} \right)\mbox{ as n}\rightarrow \infty &fg=000000$
即：$latex {{\rm {\bf x}}_{n+1} -{\rm {\bf s}}={\rm {\bf {F}'}}\left( {\rm
{\bf s}} \right)\left( {{\rm {\bf x}}_n -{\rm {\bf s}}} \right)+o\left(
{\left\| {{\rm {\bf x}}_n -{\rm {\bf s}}} \right\|^2} \right)\mbox{ as
n}\rightarrow \infty }&fg=000000$。
对于所有的大n，向量$latex {{\rm {\bf x}}_n }&fg=000000$可视为从形如$latex
{\left( {{\rm {\bf I}}-{\rm {\bf A}}} \right){\rm {\bf x}}={\rm {\bf b}}}
&fg=000000$的线性系统产生：
$latex
\displaystyle {\rm {\bf x}}_{n+1} ={\rm {\bf Ax}}_n +{\rm {\bf b}}\mbox{,
}n=0,1,..., \ \ \ \ \ (21)&fg=000000$
此处$latex {{\rm {\bf A}}={\rm {\bf {F}'}}\left( {\rm {\bf s}}
\right)}&fg=000000$，$latex {{\rm {\bf b}}=\left[ {{\rm {\bf I}}-{\rm {\bf
{F}'}}\left( {\rm {\bf s}} \right)} \right]{\rm {\bf s}}}&fg=000000$。
MPE和RRE已经应用在大型稀疏数值系统中，如计算流体动力学、半导体研究和X-射线
断层扫描系统。
3.4 MPE和RRE的有效应用
MPE和RRE的关键问题是关于最小二乘问题的精确解
和尽可能减少计算时间和存储空间开销。
关于最小二乘问题的解决，可以采用对基$latex {{\rm {\bf U}}_k^{\left( n
\right)} }&fg=000000$进行QR分解：
$latex \displaystyle
{\rm {\bf U}}_k^{\left( n \right)} ={\rm {\bf Q}}_k {\rm {\bf R}}_k
&fg=000000$
此处$latex {{\rm {\bf Q}}_k }&fg=000000$是N$latex {\times }&fg=000000$（k
+1）阶酉矩阵，可记为
$latex \displaystyle
{\rm {\bf Q}}_k =\left[ {{\rm {\bf q}}_0 \vert {\rm {\bf q}}_1 \vert ...
\vert {\rm {\bf q}}_k } \right] \ \ \ \ \ (22)&fg=000000$
其中，$latex {{\rm {\bf q}}_i^\ast {\rm {\bf q}}_j =\delta _{ij} }
&fg=000000$；
$latex {{\rm {\bf R}}_k }&fg=000000$是（k+1）$latex {\times }
&fg=000000$（k+1）阶上三角矩阵，对角线元素均为正数：
$latex \displaystyle {\rm {\bf R}}_k =\left[ {{\begin
{array}{*{20}c} {r_{00} } \hfill } \hfill & \cdots \hfill } \hfill \\ \hfill } \hfill & \cdots \hfill }
\hfill \\ \hfill & \hfill & \ddots \hfill & \vdots \hfill \\ \hfill &
\hfill & \hfill } \hfill \\ \end{array} }} \right]\mbox{; }r_{ii}
\mbox{>0, }i=0,1,...,k \ \ \ \ \ (23)&fg=000000$
QR分解可采用改良Gram-Schmidt正交方法（MGS），表-3描述了对$latex {{\rm
{\bf U}}_k^{\left( n \right)} }&fg=000000$应用MGS的流程：
\centerline{\includegraphics[width=5.93in,height=1.48in]{mytex33.eps}}
\caption{MGS算法}
此处$latex {\left\| {\rm {\bf x}} \right\|\mbox{-}\sqrt {{\rm {\bf x}}^\ast
{\rm {\bf x}}} }&fg=000000$，$latex {{\rm {\bf u}}_i^{\left( {j+1} \right)}
}&fg=000000$重写为$latex {{\rm {\bf u}}_i^{\left( j \right)} }&fg=000000$，
如此则$latex {{\rm {\bf u}}_{n+i} ,{\rm {\bf u}}_i^{\left( j \right)} ,{\rm
{\bf q}}_i }&fg=000000$占有相同的存储空间。
很重要的一点是，当构建矩阵$latex {{\rm {\bf U}}_k^{\left( n \right)} }
&fg=000000$时，我们将$latex {{\rm {\bf x}}_{n+i} }&fg=000000$重写为$latex
{{\rm {\bf u}}_{n+i} ={\rm {\bf x}}_{n+i} }&fg=000000$，而只保存$latex
{{\rm {\bf x}}_n }&fg=000000$；下一步，当计算矩阵$latex {{\rm {\bf Q}}_k }
&fg=000000$时，我们把$latex {{\rm {\bf q}}_i \mbox{,i=0,1,...,k}}
&fg=000000$重写为$latex {{\rm {\bf u}}_{n+i} }&fg=000000$。这就意味着，在计
算$latex {{\rm {\bf Q}}_k }&fg=000000$和$latex {{\rm {\bf R}}_x }
&fg=000000$的每一阶段，我们都只保留k+2个向量，$latex {{\rm {\bf x}}_{n+1}
,...,{\rm {\bf x}}_{n+k+1} }&fg=000000$无需保存。
表-4展示了采用QR分解的MPE和RRE算法，它们具有一致的框架：
\centerline{\includegraphics[width=6.00in,height=5.63in]{mytex34.eps}}
\caption{采用QR分解的MPR/RRE算法}
3.5 误差估计
1、 对于线性序列：当(6)中的迭代向量$latex {{\rm {\bf
x}}_i }&fg=000000$是线性时，有：
$latex \displaystyle {\rm
{\bf r}}\left( {\rm {\bf x}} \right)={\rm {\bf b}}-\left( {{\rm {\bf I}}-
{\rm {\bf A}}} \right){\rm {\bf x}}=\left( {{\rm {\bf Ax}}+{\rm {\bf b}}}
\right)-{\rm {\bf x}} &fg=000000$
即：$latex {{\rm {\bf r}}\left( {{\rm {\bf x}}_n } \right)={\rm {\bf x}}_
{n+1} -{\rm {\bf x}}_n ={\rm {\bf u}}_n }&fg=000000$
在应用MPE和RRE算法时，考虑$latex {\sum\nolimits_{i=0}^k {\gamma _i =1} }
&fg=000000$，可得
$latex \displaystyle
{\rm {\bf r}}\left( {{\rm {\bf s}}_{n,k} } \right)=\sum\limits_{i=0}^k
{\gamma _i {\rm {\bf u}}_{n+i} } ={\rm {\bf U}}_k^{\left( n \right)} \gamma
\ \ \ \ \ (24)&fg=000000$
我们考察其$latex {l_2 }&fg=000000$范数，即：
$latex
\displaystyle \left\| {{\rm {\bf r}}\left( {{\rm {\bf s}}_{n,k} } \right)}
\right\|=\left\| {{\rm {\bf U}}_k^{\left( n \right)} \gamma } \right\|
&fg=000000$
2、 对于非线性序列：当(20)中的$latex {{\rm {\bf x}}_i
}&fg=000000$是线性时，有：
$latex \displaystyle {\rm {\bf
r}}\left( {{\rm {\bf s}}_{n,k} } \right)={\rm {\bf F}}\left( {{\rm {\bf
s}}_{n,k} } \right)-{\rm {\bf s}}_{n,k} \approx {\rm {\bf U}}_k^{\left( n
\right)} \gamma &fg=000000$
因此：
$latex \displaystyle \left\| {{\rm {\bf r}}\left(
{{\rm {\bf s}}_{n,k} } \right)} \right\|\approx \left\| {{\rm {\bf U}}_k^
{\left( n \right)} \gamma } \right\| &fg=000000$
不论$latex {{\rm {\bf x}}_i }&fg=000000$是否是线性的，$latex {\left\|
{{\rm {\bf U}}_k^{\left( n \right)} \gamma } \right\|}&fg=000000$都可以无需
计算$latex {{\rm {\bf s}}_{n,k} }&fg=000000$而得出：
$latex
\displaystyle \left\| {{\rm {\bf U}}_k^{\left( n \right)} \gamma } \right
\|=\left\{ {{\begin{array}{*{20}c} {r_{kk} \left| {\gamma _k } \right|
\mbox{ for MPE}} \hfill \\ {\sqrt \lambda \mbox{ for RRE}} \hfill \\ \end
{array} }} \right. &fg=000000$
此处$latex {r_{kk} }&fg=000000$是矩阵$latex {{\rm {\bf R}}_k }&fg=000000$
对角线上的最后一个元素，$latex {\lambda }&fg=000000$是上一节算法的第三步中
得到的参数（文献[44]）。
3.6 MPE和RRE的误差分析
关于MPE和RRE的线性误差分析在文献[42、46、45、48、49]中已有论述。本文将重点
介绍非线性的情况下，向量外推方法可以达到怎样的性能。
定理1 假设向量$latex {{\rm {\bf x}}_n }&fg=000000$满足：
$latex \displaystyle {\rm {\bf x}}_n ={\rm {\bf s}}+\sum
\limits_{i=1}^p {{\rm {\bf v}}_i \lambda _i^n } &fg=000000$
此处，向量$latex {{\rm {\bf v}}_i }&fg=000000$是线性独立的，非零标量$latex
{\lambda _i \ne 1}&fg=000000$，取值不同，且满足：
$latex
\displaystyle \left| {\lambda _1 } \right|\ge \left| {\lambda _2 }
\right|\ge ... &fg=000000$
若有$latex {\left| {\lambda _k } \right|\ge \left| {\lambda _{k+1} }
\right|}&fg=000000$，则对于MPE和RRE，有：
$latex
\displaystyle {\rm {\bf s}}_{n,k} -{\rm {\bf s}}=o\left( {\lambda _{k+1}^n
} \right)\mbox{ as }n\rightarrow \infty &fg=000000$
以及：
$latex \displaystyle \mathop {\lim }\limits_{n
\rightarrow \infty } \sum\limits_{i=0}^k {\gamma _i^{\left( {n,k} \right)}
z^i} =\prod\limits_{i=1}^k {\frac{\lambda -\lambda _i }{1-\lambda _i }}
&fg=000000$
此处，$latex {\gamma _i^{\left( {n,k} \right)} }&fg=000000$代表$latex
{\gamma _i }&fg=000000$。
当$latex {{\rm {\bf x}}_n }&fg=000000$是(6)中迭代产生时
，$latex {\lambda _i }&fg=000000$部分或者全部是迭代矩阵$latex {{\rm {\bf
A}}}&fg=000000$（$latex {{\rm {\bf A}}}&fg=000000$是对角化的）的非零特征值
，$latex {{\rm {\bf v}}_i }&fg=000000$是对应的特征向量。
定理2 假设向量$latex {{\rm {\bf x}}_n }&fg=000000$由$latex {{\rm
{\bf x}}_{n+1} ={\rm {\bf Ax}}_n +{\rm {\bf b}}}&fg=000000$迭代产生，矩阵
$latex {{\rm {\bf I}}-{\rm {\bf A}}}&fg=000000$非奇异，$latex {\left( {{\rm
{\bf I}}-{\rm {\bf A}}} \right){\rm {\bf x}}={\rm {\bf b}}}&fg=000000$的解
为$latex {{\rm {\bf s}}}&fg=000000$，$latex {{\rm {\bf A}}}&fg=000000$的非
零特征值为：
$latex \displaystyle \left| {\lambda _1 }
\right|\ge \left| {\lambda _2 } \right|\ge ... &fg=000000$
不论是否$latex {\left| {\lambda _k } \right|\ge \left| {\lambda _{k+1} }
\right|}&fg=000000$，对于
（i）RRE无条件满足；
（ii）MPE在提供$latex {{\rm {\bf I}}-{\rm {\bf A}}}&fg=000000$的特征值（这
些特征值均位于一条复平面上通过原点的直线的同一侧，例如当$latex {{\rm {\bf
A}}+{\rm {\bf A}}^\ast }&fg=000000$是正定）时，
下式均成立：
$latex \displaystyle {\rm {\bf s}}_{n,k} -
{\rm {\bf s}}=o\left( {\lambda _{k+1}^n } \right)\mbox{ as }n\rightarrow
\infty &fg=000000$
综合定理1、2，注意到我们并未假设$latex {\lim _{n\rightarrow \infty } {\rm
{\bf x}}_n }&fg=000000$存在，对于$latex {\lim _{n\rightarrow \infty } {\rm
{\bf x}}_n }&fg=000000$不存在的情况，$latex {\left| {\lambda _1 } \right|
\ge 1}&fg=000000$的条件是必须的。
事实上，若用$latex {{\rm {\bf x}}_i }&fg=000000$逼近s，可得误差
$latex \displaystyle ?_n ={\rm {\bf x}}_n -{\rm {\bf s}}=o
\left( {\lambda _1^n } \right)\mbox{ as }n\rightarrow \infty
&fg=000000$
因此，无论$latex {\lim _{n\rightarrow \infty } {\rm {\bf x}}_n }
&fg=000000$存在与否，只要给定$latex {\left| {\lambda _{k+1} } \right|
<1}&fg=000000$，我们均可得到$latex {\lim _{n\rightarrow \infty } {\rm
{\bf s}}_{n,k} ={\rm {\bf s}}}&fg=000000$。此外，当给定$latex {\left|
{\lambda _{k+1} } \right|<\left| {\lambda _1 } \right|}&fg=000000$时，序
列$latex {{\rm {\bf x}}_n }&fg=000000$收敛于s、$latex {{\rm {\bf
s}}_{n,k} }&fg=000000$收敛于s的速度更快。即，MPE和RRE方法加速了序列
$latex {\left\{ {{\rm {\bf x}}_n } \right\}}&fg=000000$的收敛速度。
3.7 循环MPE/RRE
定理1、2需要固定k，然后令n趋于无限大，很显然
实际中无法满足这个条件；此外增加k也可以使得MPE和RRE的收敛速度加快，然而我们
同样无法无限增大k。
循环（或者再启动）方法可以用来解决上述问题，在循环模式中，n和k是固定的，
表-5的算法流程中1～3步称为一个「循环」，第i次循环的$latex {{\rm {\bf s}}_
{n,k} }&fg=000000$记作$latex {{\rm {\bf s}}_{n,k}^{\left( i \right)} }
&fg=000000$。
\centerline{\includegraphics[width=5.91in,height=1.30in]{mytex35.eps}}
\caption{循环模式下的向量外推算法}
循环模式带来的两个好处（文献[48,49]）：
1、产生一个紧致的误差上界；
2、防止了「广义最小残差停滞」（GMRES stagnates）
循环MPE/RRE在非线性系统下的性能分析（文献[50,51]）带来的启发意义是，当第i次
循环中k接近k$latex {_{i}}&fg=000000$时，矩阵$latex {{\rm {\bf {F}'}}\left(
{\rm {\bf s}} \right)}&fg=000000$关于$latex {?_\mbox{0} \mbox{=}{\rm {\bf
x}}_\mbox{0} -{\rm {\bf s}}}&fg=000000$的极小多项式的阶数------序列$latex
{\left\{ {{\rm {\bf s}}_{n,k}^{\left( i \right)} } \right\}_{i=0}^\infty }
&fg=000000$将二阶收敛于$latex {{\rm {\bf s}}}&fg=000000$。
但由于k$latex {_{i}}&fg=000000$可以等于N，且无法知道它的准确大小，对于大规
模问题的应用，其存储开销可能会很大；另一方面，尝试从循环MPE/RRE实现二阶收敛
可能是不现实的。
3.8 与Krylov子空间方法结合
对于线性序列应用，MPE和RRE方法与Krylov子空间
方法------如Arnoldi（文献[1]）、GMRES（文献[38]）------关系很大。文献[43]给
出了如下定理：
定理3 考察线性系统$latex {\left( {{\rm {\bf I}}-{\rm {\bf A}}}
\right){\rm {\bf x}}={\rm {\bf b}}}&fg=000000$，其中$latex {{\rm {\bf x}}_0
}&fg=000000$是初始向量，令向量序列$latex {\left\{ {{\rm {\bf x}}_n }
\right\}}&fg=000000$通过$latex {{\rm {\bf x}}_{n+1} ={\rm {\bf Ax}}_n +{\rm
{\bf b}}}&fg=000000$生成。分别应用MPE、RRE，从该序列中生成$latex {{\rm {\bf
s}}_{0,k}^{MPE} }&fg=000000$和$latex {{\rm {\bf s}}_{0,k}^{RRE} }
&fg=000000$；同样分别应用k步Arnoldi和GMRES，从$latex {\left( {{\rm {\bf
I}}-{\rm {\bf A}}} \right){\rm {\bf x}}={\rm {\bf b}}}&fg=000000$中生成
$latex {{\rm {\bf s}}_k^{\mbox{Arnoldi}} }&fg=000000$和$latex {{\rm {\bf
s}}_k^{\mbox{GMRES}} }&fg=000000$。则有如下关系：
$latex {{\rm {\bf s}}_{0,k}^{MPE} ={\rm {\bf s}}_k^{\mbox{Arnoldi}} }
&fg=000000$，$latex {{\rm {\bf s}}_{0,k}^{RRE} ={\rm {\bf s}}_k^{\mbox
{GMRES}} }&fg=000000$。
3.9 循环SMACOF算法
为了加速SMACOF的收敛速度，我们将循环模式应用进来，由于这是个非线性问题，故
外推法的近似极限向量并不一定会有低的stress值。因此我们必须采取某种保护措施
：
一种方法是检查外推极限的stress值，这个值若较高，则采用最后一次的迭代向量代
替，这个方法在表- 5 循环模式下的向量外推算法中的第五步做一个简单更改即可，
表-6做出了算法描述：
第二种方法是减小步长。
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