📄 r1updt_.c
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/* r1updt.f -- translated by f2c (version 20020621). You must link the resulting object file with the libraries: -lf2c -lm (in that order)*/#include <math.h>#include "minpack.h"#define abs(x) ((x) >= 0 ? (x) : -(x))#define TRUE_ (1)#define FALSE_ (0)/* Subroutine */ void r1updt_(const int *m, const int *n, double *s, const int * ls, const double *u, double *v, double *w, int *sing){ /* Initialized data */#define p5 .5#define p25 .25 const int c__3 = 3; /* System generated locals */ int i__1, i__2; double d__1, d__2; /* Local variables */ int i__, j, l, jj, nm1; double tan__; int nmj; double cos__, sin__, tau, temp, giant, cotan;/* ********** *//* subroutine r1updt *//* given an m by n lower trapezoidal matrix s, an m-vector u, *//* and an n-vector v, the problem is to determine an *//* orthogonal matrix q such that *//* t *//* (s + u*v )*q *//* is again lower trapezoidal. *//* this subroutine determines q as the product of 2*(n - 1) *//* transformations *//* gv(n-1)*...*gv(1)*gw(1)*...*gw(n-1) *//* where gv(i), gw(i) are givens rotations in the (i,n) plane *//* which eliminate elements in the i-th and n-th planes, *//* respectively. q itself is not accumulated, rather the *//* information to recover the gv, gw rotations is returned. *//* the subroutine statement is *//* subroutine r1updt(m,n,s,ls,u,v,w,sing) *//* where *//* m is a positive integer input variable set to the number *//* of rows of s. *//* n is a positive integer input variable set to the number *//* of columns of s. n must not exceed m. *//* s is an array of length ls. on input s must contain the lower *//* trapezoidal matrix s stored by columns. on output s contains *//* the lower trapezoidal matrix produced as described above. *//* ls is a positive integer input variable not less than *//* (n*(2*m-n+1))/2. *//* u is an input array of length m which must contain the *//* vector u. *//* v is an array of length n. on input v must contain the vector *//* v. on output v(i) contains the information necessary to *//* recover the givens rotation gv(i) described above. *//* w is an output array of length m. w(i) contains information *//* necessary to recover the givens rotation gw(i) described *//* above. *//* sing is a logical output variable. sing is set true if any *//* of the diagonal elements of the output s are zero. otherwise *//* sing is set false. *//* subprograms called *//* minpack-supplied ... dpmpar *//* fortran-supplied ... dabs,dsqrt *//* argonne national laboratory. minpack project. march 1980. *//* burton s. garbow, kenneth e. hillstrom, jorge j. more, *//* john l. nazareth *//* ********** */ /* Parameter adjustments */ --w; --u; --v; --s; /* Function Body *//* giant is the largest magnitude. */ giant = dpmpar_(&c__3);/* initialize the diagonal element pointer. */ jj = *n * ((*m << 1) - *n + 1) / 2 - (*m - *n);/* move the nontrivial part of the last column of s into w. */ l = jj; i__1 = *m; for (i__ = *n; i__ <= i__1; ++i__) { w[i__] = s[l]; ++l;/* L10: */ }/* rotate the vector v into a multiple of the n-th unit vector *//* in such a way that a spike is introduced into w. */ nm1 = *n - 1; if (nm1 < 1) { goto L70; } i__1 = nm1; for (nmj = 1; nmj <= i__1; ++nmj) { j = *n - nmj; jj -= *m - j + 1; w[j] = 0.; if (v[j] == 0.) { goto L50; }/* determine a givens rotation which eliminates the *//* j-th element of v. */ if ((d__1 = v[*n], abs(d__1)) >= (d__2 = v[j], abs(d__2))) { goto L20; } cotan = v[*n] / v[j];/* Computing 2nd power */ d__1 = cotan; sin__ = p5 / sqrt(p25 + p25 * (d__1 * d__1)); cos__ = sin__ * cotan; tau = 1.; if (abs(cos__) * giant > 1.) { tau = 1. / cos__; } goto L30;L20: tan__ = v[j] / v[*n];/* Computing 2nd power */ d__1 = tan__; cos__ = p5 / sqrt(p25 + p25 * (d__1 * d__1)); sin__ = cos__ * tan__; tau = sin__;L30:/* apply the transformation to v and store the information *//* necessary to recover the givens rotation. */ v[*n] = sin__ * v[j] + cos__ * v[*n]; v[j] = tau;/* apply the transformation to s and extend the spike in w. */ l = jj; i__2 = *m; for (i__ = j; i__ <= i__2; ++i__) { temp = cos__ * s[l] - sin__ * w[i__]; w[i__] = sin__ * s[l] + cos__ * w[i__]; s[l] = temp; ++l;/* L40: */ }L50:/* L60: */ ; }L70:/* add the spike from the rank 1 update to w. */ i__1 = *m; for (i__ = 1; i__ <= i__1; ++i__) { w[i__] += v[*n] * u[i__];/* L80: */ }/* eliminate the spike. */ *sing = FALSE_; if (nm1 < 1) { goto L140; } i__1 = nm1; for (j = 1; j <= i__1; ++j) { if (w[j] == 0.) { goto L120; }/* determine a givens rotation which eliminates the *//* j-th element of the spike. */ if ((d__1 = s[jj], abs(d__1)) >= (d__2 = w[j], abs(d__2))) { goto L90; } cotan = s[jj] / w[j];/* Computing 2nd power */ d__1 = cotan; sin__ = p5 / sqrt(p25 + p25 * (d__1 * d__1)); cos__ = sin__ * cotan; tau = 1.; if (abs(cos__) * giant > 1.) { tau = 1. / cos__; } goto L100;L90: tan__ = w[j] / s[jj];/* Computing 2nd power */ d__1 = tan__; cos__ = p5 / sqrt(p25 + p25 * (d__1 * d__1)); sin__ = cos__ * tan__; tau = sin__;L100:/* apply the transformation to s and reduce the spike in w. */ l = jj; i__2 = *m; for (i__ = j; i__ <= i__2; ++i__) { temp = cos__ * s[l] + sin__ * w[i__]; w[i__] = -sin__ * s[l] + cos__ * w[i__]; s[l] = temp; ++l;/* L110: */ }/* store the information necessary to recover the *//* givens rotation. */ w[j] = tau;L120:/* test for zero diagonal elements in the output s. */ if (s[jj] == 0.) { *sing = TRUE_; } jj += *m - j + 1;/* L130: */ }L140:/* move w back into the last column of the output s. */ l = jj; i__1 = *m; for (i__ = *n; i__ <= i__1; ++i__) { s[l] = w[i__]; ++l;/* L150: */ } if (s[jj] == 0.) { *sing = TRUE_; } return;/* last card of subroutine r1updt. */} /* r1updt_ */
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