xref: /petsc/src/mat/impls/baij/seq/dgefa5.c (revision 93c39befdc2c545b92559b48e3c3f7c5a930faa6)
1 #ifndef lint
2 static char vcid[] = "$Id: dgefa4.c,v 1.1 1997/06/10 03:18:18 bsmith Exp bsmith $";
3 #endif
4 /*
5     Inverts 5 by 5 matrix using partial pivoting.
6 */
7 #include "petsc.h"
8 
9 #undef __FUNC__
10 #define __FUNC__ "Kernel_A_gets_inverse_A_5"
11 int Kernel_A_gets_inverse_A_5(Scalar *a)
12 {
13     int     i__2, i__3, kp1, j, k, l,ll,i,ipvt_l[5],*ipvt = ipvt_l-1,kb,k3;
14     int     k4,j3;
15     Scalar  *aa,*ax,*ay,work_l[25],*work = work_l-1,stmp;
16     double  tmp,max;
17 
18 /*     gaussian elimination with partial pivoting */
19 
20     /* Parameter adjustments */
21     a       -= 6;
22 
23     for (k = 1; k <= 4; ++k) {
24 	kp1 = k + 1;
25         k3  = 5*k;
26         k4  = k3 + k;
27 /*        find l = pivot index */
28 
29 	i__2 = 5 - k;
30         aa = &a[k4];
31         max = PetscAbsScalar(aa[0]);
32         l = 1;
33         for ( ll=1; ll<i__2; ll++ ) {
34           tmp = PetscAbsScalar(aa[ll]);
35           if (tmp > max) { max = tmp; l = ll+1;}
36         }
37         l       += k - 1;
38 	ipvt[k] = l;
39 
40 	if (a[l + k3] == 0.) {
41 	  SETERRQ(k,0,"Zero pivot");
42 	}
43 
44 /*           interchange if necessary */
45 
46 	if (l != k) {
47 	  stmp      = a[l + k3];
48 	  a[l + k3] = a[k4];
49 	  a[k4]     = stmp;
50         }
51 
52 /*           compute multipliers */
53 
54 	stmp = -1. / a[k4];
55 	i__2 = 5 - k;
56         aa = &a[1 + k4];
57         for ( ll=0; ll<i__2; ll++ ) {
58           aa[ll] *= stmp;
59         }
60 
61 /*           row elimination with column indexing */
62 
63 	ax = &a[k4+1];
64         for (j = kp1; j <= 5; ++j) {
65             j3   = 5*j;
66 	    stmp = a[l + j3];
67 	    if (l != k) {
68 	      a[l + j3] = a[k + j3];
69 	      a[k + j3] = stmp;
70             }
71 
72 	    i__3 = 5 - k;
73             ay = &a[1+k+j3];
74             for ( ll=0; ll<i__3; ll++ ) {
75               ay[ll] += stmp*ax[ll];
76             }
77 	}
78     }
79     ipvt[5] = 5;
80     if (a[30] == 0.) {
81 	SETERRQ(3,0,"Zero pivot,final row");
82     }
83 
84     /*
85          Now form the inverse
86     */
87 
88    /*     compute inverse(u) */
89 
90     for (k = 1; k <= 5; ++k) {
91         k3    = 5*k;
92         k4    = k3 + k;
93 	a[k4] = 1.0 / a[k4];
94 	stmp  = -a[k4];
95 	i__2  = k - 1;
96         aa    = &a[k3 + 1];
97         for ( ll=0; ll<i__2; ll++ ) aa[ll] *= stmp;
98 	kp1 = k + 1;
99 	if (5 < kp1) continue;
100         ax = aa;
101         for (j = kp1; j <= 5; ++j) {
102             j3        = 5*j;
103 	    stmp      = a[k + j3];
104 	    a[k + j3] = 0.0;
105             ay        = &a[j3 + 1];
106             for ( ll=0; ll<k; ll++ ) {
107               ay[ll] += stmp*ax[ll];
108             }
109 	}
110     }
111 
112    /*    form inverse(u)*inverse(l) */
113 
114     for (kb = 1; kb <= 4; ++kb) {
115 	k   = 5 - kb;
116         k3  = 5*k;
117 	kp1 = k + 1;
118         aa  = a + k3;
119 	for (i = kp1; i <= 5; ++i) {
120 	    work[i] = aa[i];
121 	    aa[i]   = 0.0;
122 	}
123 	for (j = kp1; j <= 5; ++j) {
124 	    stmp  = work[j];
125             ax    = &a[5*j + 1];
126             ay    = &a[k3 + 1];
127             ay[0] += stmp*ax[0];
128             ay[1] += stmp*ax[1];
129             ay[2] += stmp*ax[2];
130             ay[3] += stmp*ax[3];
131             ay[4] += stmp*ax[4];
132 	}
133 	l = ipvt[k];
134 	if (l != k) {
135             ax = &a[k3 + 1];
136             ay = &a[5*l + 1];
137             stmp = ax[0]; ax[0] = ay[0]; ay[0] = stmp;
138             stmp = ax[1]; ax[1] = ay[1]; ay[1] = stmp;
139             stmp = ax[2]; ax[2] = ay[2]; ay[2] = stmp;
140             stmp = ax[3]; ax[3] = ay[3]; ay[3] = stmp;
141             stmp = ax[4]; ax[4] = ay[4]; ay[4] = stmp;
142 	}
143     }
144     return 0;
145 }
146 
147