【发布时间】：2016-08-15 17:48:20
【问题描述】：
我有一个对方阵执行奇异值分解 (SVD) 的代码。然而，代码完成了这项工作，它很慢，当矩阵大小增加时，它变得难以忍受。由于我不熟悉并行编程，因此在我开始深入挖掘并最终意识到我想要实现的操作是不可能的之前，我会征求专家的建议。
提前谢谢你。
void SVD::decompose() {
bool flag;
int i, its, j, jj, k, l, nm;
double anorm, c, f, g, h, s, scale, x, y, z;
Row rv1(n);
g = scale = anorm = 0.0; //Householder reduction to bidiagonal form.
for (i = 0; i < n; i++) {
l = i + 2;
rv1[i] = scale*g;
g = s = scale = 0.0;
if (i < m) {
for (k = i; k < m; k++) scale += abs(u[k][i]);
if (scale != 0.0) {
for (k = i; k < m; k++) {
u[k][i] /= scale;
s += u[k][i] * u[k][i];
}
f = u[i][i];
g = -SIGN(sqrt(s), f);
h = f*g - s;
u[i][i] = f - g;
for (j = l - 1; j < n; j++) {
for (s = 0.0, k = i; k < m; k++) s += u[k][i] * u[k][j];
f = s / h;
for (k = i; k < m; k++) u[k][j] += f*u[k][i];
}
for (k = i; k < m; k++) u[k][i] *= scale;
}
}
w[i] = scale *g;
g = s = scale = 0.0;
if (i + 1 <= m && i + 1 != n) {
for (k = l - 1; k < n; k++) scale += abs(u[i][k]);
if (scale != 0.0) {
for (k = l - 1; k < n; k++) {
u[i][k] /= scale;
s += u[i][k] * u[i][k];
}
f = u[i][l - 1];
g = -SIGN(sqrt(s), f);
h = f*g - s;
u[i][l - 1] = f - g;
for (k = l - 1; k < n; k++) rv1[k] = u[i][k] / h;
for (j = l - 1; j < m; j++) {
for (s = 0.0, k = l - 1; k < n; k++) s += u[j][k] * u[i][k];
for (k = l - 1; k < n; k++) u[j][k] += s*rv1[k];
}
for (k = l - 1; k < n; k++) u[i][k] *= scale;
}
}
anorm = MAX(anorm, (abs(w[i]) + abs(rv1[i])));
}
for (i = n - 1; i >= 0; i--) { //Accumulation of right-hand tranformations.
if (i < n - 1) {
if (g != 0.0) {
for (j = l; j < n; j++) // Double division to avoid possible underflow.
v[j][i] = (u[i][j] / u[i][l]) / g;
for (j = l; j < n; j++) {
for (s = 0.0, k = l; k < n; k++) s += u[i][k] * v[k][j];
for (k = l; k < n; k++) v[k][j] += s*v[k][i];
}
}
for (j = l; j < n; j++) v[i][j] = v[j][i] = 0.0;
}
v[i][i] = 1.0;
g = rv1[i];
l = i;
}
for (i = MIN(m, n) - 1; i >= 0; i--) { //Accumulation of left-hand transformations.
l = i + 1;
g = w[i];
for (j = l; j < n; j++) u[i][j] = 0.0;
if (g != 0.0) {
g = 1.0 / g;
for (j = l; j < n; j++) {
for (s = 0.0, k = l; k < m; k++) s += u[k][i] * u[k][j];
f = (s / u[i][i])*g;
for (k = i; k < m; k++) u[k][j] += f*u[k][i];
}
for (j = i; j < m; j++) u[j][i] *= g;
}
else for (j = i; j < m; j++) u[j][i] = 0.0;
++u[i][i];
}
for (k = n - 1; k >= 0; k--) { //Diagonalization of the bidiagonal form: Loop over
for (its = 0; its < 30; its++) { //singular values, and over allowed iterations.
flag = true;
for (l = k; l >= 0; l--) { //Test ofr splitting.
nm = l - 1;
if (l == 0 || abs(rv1[l]) <= eps*anorm) {
flag = false;
break;
}
if (abs(w[nm]) <= eps*anorm) break;
}
if (flag) {
c = 0.0; //Cancellatin of rv[l], if l>0.
s = 1.0;
for (i = l; i < k + 1; i++) {
f = s*rv1[i];
rv1[i] = c*rv1[i];
if (abs(f) <= eps*anorm) break;
g = w[i];
h = pythag(f, g);
w[i] = h;
h = 1.0 / h;
c = g*h;
s = -f*h;
for (j = 0; j < m; j++) {
y = u[j][nm];
z = u[j][i];
u[j][nm] = y*c + z*s;
u[j][i] = z*c - y*s;
}
}
}
z = w[k];
if (l == k) { //Convergence.
if (z < 0.0) { //Singular value is made nonnegative.
w[k] = -z;
for (j = 0; j < n; j++) v[j][k] = -v[j][k];
}
break;
}
x = w[l]; //Shift from bottom 2-by-2 minor.
nm = k - 1;
y = w[nm];
g = rv1[nm];
h = rv1[k];
f = ((y - z)*(y + z) + (g - h)*(g + h)) / (2.0*h*y);
g = pythag(f, 1.0);
f = ((x - z)*(x + z) + h*((y / (f + SIGN(g, f))) - h)) / x;
c = s = 1.0; //Next QR transformation:
for (j = l; j <= nm; j++) {
i = j + 1;
g = rv1[i];
y = w[i];
h = s*g;
g = c*g;
z = pythag(f, h);
rv1[j] = z;
c = f / z;
s = h / z;
f = x*c + g*s;
g = g*c - x*s;
h = y*s;
y *= c;
for (jj = 0; jj < n; jj++) {
x = v[jj][j];
z = v[jj][i];
v[jj][j] = x*c + z*s;
v[jj][i] = z*c - x*s;
}
z = pythag(f, h);
w[j] = z; //Rotation can be arbitrary if z = 0.
if (z) {
z = 1.0 / z;
c = f*z;
s = h*z;
}
f = c*g + s*y;
x = c*y - s*g;
for (jj = 0; jj < m; jj++) {
y = u[jj][j];
z = u[jj][i];
u[jj][j] = y*c + z*s;
u[jj][i] = z*c - y*s;
}
}
rv1[l] = 0.0;
rv1[k] = f;
w[k] = x;
}
}
}
标签： c++ for-loop visual-studio-2015 parallel-processing openmp