显然我可以做y = x’* A * x,但我需要表现,似乎应该有办法利用
> A是对称的
>左右乘数是相同的向量
如果没有单一的内置函数,是否有比x’* A * x更快的方法?或者,Matlab解析器是否足够智能以优化x’* A * x?如果是这样,你能指点我在文件中的一个地方验证这个事实吗?
我找不到这样的内置函数,我知道为什么.y = x’* A * x可以写成n ^ 2项A(i,j)* x(i)* x(j)之和,其中i和j从1到n(其中A是a) nxn矩阵). A是对称的:对于所有i和j,A(i,j)= A(j,i).由于对称性,每个项在总和中出现两次,除了那些i等于j的项.所以我们有n *(n 1)/ 2个不同的术语.每个都有两个浮点乘法,所以一个朴素的方法总共需要n *(n 1)次乘法.很容易看出,x’* A * x的朴素计算,即计算z = A * x,然后计算y = x’* z,也需要n *(n 1)次乘法.然而,有一种更快的方法来求和我们的n *(n 1)/ 2个不同的项:对于每个i,我们可以分解x(i),这意味着只有n *(n-1)/ 2 3 * n乘法就足够了.但这并没有真正帮助:y = x’* A * x的计算运行时间仍为O(n ^ 2).
因此,我认为二次形式的计算不能比O(n ^ 2)更快地完成,并且因为这也可以通过公式y = x’* A * x来实现,所以没有特殊的优势. “二次型”功能.
===更新===
我在C中写了函数“quadraticform”,作为Matlab扩展:
// y = quadraticform(A, x)
#include "mex.h"
/* Input Arguments */
#define A_in prhs[0]
#define x_in prhs[1]
/* Output Arguments */
#define y_out plhs[0]
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[])
{
mwSize mA, nA, n, mx, nx;
double *A, *x;
double z, y;
int i, j, k;
if (nrhs != 2) {
mexErrMsgTxt("Two input arguments required.");
} else if (nlhs > 1) {
mexErrMsgTxt("Too many output arguments.");
}
mA = mxGetM(A_in);
nA = mxGetN(A_in);
if (mA != nA)
mexErrMsgTxt("The first input argument must be a quadratic matrix.");
n = mA;
mx = mxGetM(x_in);
nx = mxGetN(x_in);
if (mx != n || nx != 1)
mexErrMsgTxt("The second input argument must be a column vector of proper size.");
A = mxGetPr(A_in);
x = mxGetPr(x_in);
y = 0.0;
k = 0;
for (i = 0; i < n; ++i)
{
z = 0.0;
for (j = 0; j < i; ++j)
z += A[k + j] * x[j];
z *= x[i];
y += A[k + i] * x[i] * x[i] + z + z;
k += n;
}
y_out = mxCreateDoubleScalar(y);
}
我将此代码保存为“quadraticform.c”,并使用Matlab编译:
mex -O quadraticform.c
我写了一个简单的性能测试来比较这个函数和x’Ax:
clear all; close all; clc;
sizes = int32(logspace(2, 3, 25));
nsizes = length(sizes);
etimes = zeros(nsizes, 2); % Matlab vs. C
nrepeats = 100;
h = waitbar(0, 'Please wait...');
for i = 1 : nrepeats
for j = 1 : nsizes
n = sizes(j);
A = randn(n);
A = (A + A') / 2;
x = randn(n, 1);
if randn > 0
start = tic;
y1 = x' * A * x;
etimes(j, 1) = etimes(j, 1) + toc(start);
start = tic;
y2 = quadraticform(A, x);
etimes(j, 2) = etimes(j, 2) + toc(start);
else
start = tic;
y2 = quadraticform(A, x);
etimes(j, 2) = etimes(j, 2) + toc(start);
start = tic;
y1 = x' * A * x;
etimes(j, 1) = etimes(j, 1) + toc(start);
end;
if abs((y1 - y2) / y2) > 1e-10
error('"x'' * A * x" is not equal to "quadraticform(A, x)"');
end;
waitbar(((i - 1) * nsizes + j) / (nrepeats * nsizes), h);
end;
end;
close(h);
clear A x y;
etimes = etimes / nrepeats;
n = double(sizes);
n2 = n .^ 2.0;
i = nsizes - 2 : nsizes;
n2_1 = mean(etimes(i, 1)) * n2 / mean(n2(i));
n2_2 = mean(etimes(i, 2)) * n2 / mean(n2(i));
figure;
loglog(n, etimes(:, 1), 'r.-', 'LineSmoothing', 'on');
hold on;
loglog(n, etimes(:, 2), 'g.-', 'LineSmoothing', 'on');
loglog(n, n2_1, 'k-', 'LineSmoothing', 'on');
loglog(n, n2_2, 'k-', 'LineSmoothing', 'on');
axis([n(1) n(end) 1e-4 1e-2]);
xlabel('Matrix size, n');
ylabel('Running time (a.u.)');
legend('x'' * A * x', 'quadraticform(A, x)', 'O(n^2)', 'Location', 'NorthWest');
W = 16 / 2.54; H = 12 / 2.54; dpi = 100;
set(gcf, 'PaperPosition', [0, 0, W, H]);
set(gcf, 'PaperSize', [W, H]);
print(gcf, sprintf('-r%d',dpi), '-dpng', 'quadraticformtest.png');
结果非常有趣. x’* A * x和二次型(A,x)的运行时间收敛到O(n ^ 2),但前者的因子较小:
