dsyrfs function
void
dsyrfs()
Implementation
void dsyrfs(
final String UPLO,
final int N,
final int NRHS,
final Matrix<double> A_,
final int LDA,
final Matrix<double> AF_,
final int LDAF,
final Array<int> IPIV_,
final Matrix<double> B_,
final int LDB,
final Matrix<double> X_,
final int LDX,
final Array<double> FERR_,
final Array<double> BERR_,
final Array<double> WORK_,
final Array<int> IWORK_,
final Box<int> INFO,
) {
final A = A_.having(ld: LDA);
final AF = AF_.having(ld: LDAF);
final IPIV = IPIV_.having();
final B = B_.having(ld: LDB);
final X = X_.having(ld: LDX);
final FERR = FERR_.having();
final BERR = BERR_.having();
final WORK = WORK_.having();
final IWORK = IWORK_.having();
const ITMAX = 5;
const ZERO = 0.0;
const ONE = 1.0;
const TWO = 2.0;
const THREE = 3.0;
bool UPPER;
int COUNT, I, J, K, NZ;
double EPS, LSTRES, S, SAFE1, SAFE2, SAFMIN, XK;
final ISAVE = Array<int>(3);
final KASE = Box(0);
// Test the input parameters.
INFO.value = 0;
UPPER = lsame(UPLO, 'U');
if (!UPPER && !lsame(UPLO, 'L')) {
INFO.value = -1;
} else if (N < 0) {
INFO.value = -2;
} else if (NRHS < 0) {
INFO.value = -3;
} else if (LDA < max(1, N)) {
INFO.value = -5;
} else if (LDAF < max(1, N)) {
INFO.value = -7;
} else if (LDB < max(1, N)) {
INFO.value = -10;
} else if (LDX < max(1, N)) {
INFO.value = -12;
}
if (INFO.value != 0) {
xerbla('DSYRFS', -INFO.value);
return;
}
// Quick return if possible
if (N == 0 || NRHS == 0) {
for (J = 1; J <= NRHS; J++) {
FERR[J] = ZERO;
BERR[J] = ZERO;
}
return;
}
// NZ = maximum number of nonzero elements in each row of A, plus 1
NZ = N + 1;
EPS = dlamch('Epsilon');
SAFMIN = dlamch('Safe minimum');
SAFE1 = NZ * SAFMIN;
SAFE2 = SAFE1 / EPS;
// Do for each right hand side
for (J = 1; J <= NRHS; J++) {
COUNT = 1;
LSTRES = THREE;
while (true) {
// Loop until stopping criterion is satisfied.
// Compute residual R = B - A * X
dcopy(N, B(1, J).asArray(), 1, WORK(N + 1), 1);
dsymv(UPLO, N, -ONE, A, LDA, X(1, J).asArray(), 1, ONE, WORK(N + 1), 1);
// Compute componentwise relative backward error from formula
// max(i) ( abs(R(i)) / ( abs(A)*abs(X) + abs(B) )(i) )
// where abs(Z) is the componentwise absolute value of the matrix
// or vector Z. If the i-th component of the denominator is less
// than SAFE2, then SAFE1 is added to the i-th components of the
// numerator and denominator before dividing.
for (I = 1; I <= N; I++) {
WORK[I] = B[I][J].abs();
}
// Compute abs(A)*abs(X) + abs(B).
if (UPPER) {
for (K = 1; K <= N; K++) {
S = ZERO;
XK = X[K][J].abs();
for (I = 1; I <= K - 1; I++) {
WORK[I] += A[I][K].abs() * XK;
S += A[I][K].abs() * X[I][J].abs();
}
WORK[K] += A[K][K].abs() * XK + S;
}
} else {
for (K = 1; K <= N; K++) {
S = ZERO;
XK = X[K][J].abs();
WORK[K] += A[K][K].abs() * XK;
for (I = K + 1; I <= N; I++) {
WORK[I] += A[I][K].abs() * XK;
S += A[I][K].abs() * X[I][J].abs();
}
WORK[K] += S;
}
}
S = ZERO;
for (I = 1; I <= N; I++) {
if (WORK[I] > SAFE2) {
S = max(S, WORK[N + I].abs() / WORK[I]);
} else {
S = max(S, (WORK[N + I].abs() + SAFE1) / (WORK[I] + SAFE1));
}
}
BERR[J] = S;
// Test stopping criterion. Continue iterating if
// 1) The residual BERR[J] is larger than machine epsilon, and
// 2) BERR[J] decreased by at least a factor of 2 during the
// last iteration, and
// 3) At most ITMAX iterations tried.
if (BERR[J] > EPS && TWO * BERR[J] <= LSTRES && COUNT <= ITMAX) {
// Update solution and try again.
dsytrs(UPLO, N, 1, AF, LDAF, IPIV, WORK(N + 1).asMatrix(N), N, INFO);
daxpy(N, ONE, WORK(N + 1), 1, X(1, J).asArray(), 1);
LSTRES = BERR[J];
COUNT++;
continue;
}
break;
}
// Bound error from formula
// norm(X - XTRUE) / norm(X) <= FERR =
// norm( abs(inv(A))*
// ( abs(R) + NZ*EPS*( abs(A)*abs(X)+abs(B) ))) / norm(X)
// where
// norm(Z) is the magnitude of the largest component of Z
// inv(A) is the inverse of A
// abs(Z) is the componentwise absolute value of the matrix or
// vector Z
// NZ is the maximum number of nonzeros in any row of A, plus 1
// EPS is machine epsilon
// The i-th component of abs(R)+NZ*EPS*(abs(A)*abs(X)+abs(B))
// is incremented by SAFE1 if the i-th component of
// abs(A)*abs(X) + abs(B) is less than SAFE2.
// Use DLACN2 to estimate the infinity-norm of the matrix
// inv(A) * diag(W),
// where W = abs(R) + NZ*EPS*( abs(A)*abs(X)+abs(B) )))
for (I = 1; I <= N; I++) {
if (WORK[I] > SAFE2) {
WORK[I] = WORK[N + I].abs() + NZ * EPS * WORK[I];
} else {
WORK[I] = WORK[N + I].abs() + NZ * EPS * WORK[I] + SAFE1;
}
}
KASE.value = 0;
while (true) {
dlacn2(N, WORK(2 * N + 1), WORK(N + 1), IWORK, FERR.box(J), KASE, ISAVE);
if (KASE.value == 0) break;
if (KASE.value == 1) {
// Multiply by diag(W)*inv(A**T).
dsytrs(UPLO, N, 1, AF, LDAF, IPIV, WORK(N + 1).asMatrix(N), N, INFO);
for (I = 1; I <= N; I++) {
WORK[N + I] = WORK[I] * WORK[N + I];
}
} else if (KASE.value == 2) {
// Multiply by inv(A)*diag(W).
for (I = 1; I <= N; I++) {
WORK[N + I] = WORK[I] * WORK[N + I];
}
dsytrs(UPLO, N, 1, AF, LDAF, IPIV, WORK(N + 1).asMatrix(N), N, INFO);
}
}
// Normalize error.
LSTRES = ZERO;
for (I = 1; I <= N; I++) {
LSTRES = max(LSTRES, X[I][J].abs());
}
if (LSTRES != ZERO) FERR[J] /= LSTRES;
}
}