/* btf.c (sparse block triangular LU-factorization) */ /*********************************************************************** * This code is part of GLPK (GNU Linear Programming Kit). * Copyright (C) 2013-2014 Free Software Foundation, Inc. * Written by Andrew Makhorin . * * GLPK is free software: you can redistribute it and/or modify it * under the terms of the GNU General Public License as published by * the Free Software Foundation, either version 3 of the License, or * (at your option) any later version. * * GLPK is distributed in the hope that it will be useful, but WITHOUT * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY * or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public * License for more details. * * You should have received a copy of the GNU General Public License * along with GLPK. If not, see . ***********************************************************************/ #include "btf.h" #include "env.h" #include "luf.h" #include "mc13d.h" #include "mc21a.h" /*********************************************************************** * btf_store_a_cols - store pattern of matrix A in column-wise format * * This routine stores the pattern (that is, only indices of non-zero * elements) of the original matrix A in column-wise format. * * On exit the routine returns the number of non-zeros in matrix A. */ int btf_store_a_cols(BTF *btf, int (*col)(void *info, int j, int ind[], double val[]), void *info, int ind[], double val[]) { int n = btf->n; SVA *sva = btf->sva; int *sv_ind = sva->ind; int ac_ref = btf->ac_ref; int *ac_ptr = &sva->ptr[ac_ref-1]; int *ac_len = &sva->len[ac_ref-1]; int j, len, ptr, nnz; nnz = 0; for (j = 1; j <= n; j++) { /* get j-th column */ len = col(info, j, ind, val); xassert(0 <= len && len <= n); /* reserve locations for j-th column */ if (len > 0) { if (sva->r_ptr - sva->m_ptr < len) { sva_more_space(sva, len); sv_ind = sva->ind; } sva_reserve_cap(sva, ac_ref+(j-1), len); } /* store pattern of j-th column */ ptr = ac_ptr[j]; memcpy(&sv_ind[ptr], &ind[1], len * sizeof(int)); ac_len[j] = len; nnz += len; } return nnz; } /*********************************************************************** * btf_make_blocks - permutations to block triangular form * * This routine analyzes the pattern of the original matrix A and * determines permutation matrices P and Q such that A = P * A~* Q, * where A~ is an upper block triangular matrix. * * On exit the routine returns symbolic rank of matrix A. */ int btf_make_blocks(BTF *btf) { int n = btf->n; SVA *sva = btf->sva; int *sv_ind = sva->ind; int *pp_ind = btf->pp_ind; int *pp_inv = btf->pp_inv; int *qq_ind = btf->qq_ind; int *qq_inv = btf->qq_inv; int *beg = btf->beg; int ac_ref = btf->ac_ref; int *ac_ptr = &sva->ptr[ac_ref-1]; int *ac_len = &sva->len[ac_ref-1]; int i, j, rank, *iperm, *pr, *arp, *cv, *out, *ip, *lenr, *lowl, *numb, *prev; /* determine column permutation matrix M such that matrix A * M * has zero-free diagonal */ iperm = qq_inv; /* matrix M */ pr = btf->p1_ind; /* working array */ arp = btf->p1_inv; /* working array */ cv = btf->q1_ind; /* working array */ out = btf->q1_inv; /* working array */ rank = mc21a(n, sv_ind, ac_ptr, ac_len, iperm, pr, arp, cv, out); xassert(0 <= rank && rank <= n); if (rank < n) { /* A is structurally singular (rank is its symbolic rank) */ goto done; } /* build pattern of matrix A * M */ ip = pp_ind; /* working array */ lenr = qq_ind; /* working array */ for (j = 1; j <= n; j++) { ip[j] = ac_ptr[iperm[j]]; lenr[j] = ac_len[iperm[j]]; } /* determine symmetric permutation matrix S such that matrix * S * (A * M) * S' = A~ is upper block triangular */ lowl = btf->p1_ind; /* working array */ numb = btf->p1_inv; /* working array */ prev = btf->q1_ind; /* working array */ btf->num = mc13d(n, sv_ind, ip, lenr, pp_inv, beg, lowl, numb, prev); xassert(beg[1] == 1); beg[btf->num+1] = n+1; /* A * M = S' * A~ * S ==> A = S' * A~ * (S * M') */ /* determine permutation matrix P = S' */ for (j = 1; j <= n; j++) pp_ind[pp_inv[j]] = j; /* determine permutation matrix Q = S * M' = P' * M' */ for (i = 1; i <= n; i++) qq_ind[i] = iperm[pp_inv[i]]; for (i = 1; i <= n; i++) qq_inv[qq_ind[i]] = i; done: return rank; } /*********************************************************************** * btf_check_blocks - check structure of matrix A~ * * This routine checks that structure of upper block triangular matrix * A~ is correct. * * NOTE: For testing/debugging only. */ void btf_check_blocks(BTF *btf) { int n = btf->n; SVA *sva = btf->sva; int *sv_ind = sva->ind; int *pp_ind = btf->pp_ind; int *pp_inv = btf->pp_inv; int *qq_ind = btf->qq_ind; int *qq_inv = btf->qq_inv; int num = btf->num; int *beg = btf->beg; int ac_ref = btf->ac_ref; int *ac_ptr = &sva->ptr[ac_ref-1]; int *ac_len = &sva->len[ac_ref-1]; int i, ii, j, jj, k, size, ptr, end, diag; xassert(n > 0); /* check permutation matrices P and Q */ for (k = 1; k <= n; k++) { xassert(1 <= pp_ind[k] && pp_ind[k] <= n); xassert(pp_inv[pp_ind[k]] == k); xassert(1 <= qq_ind[k] && qq_ind[k] <= n); xassert(qq_inv[qq_ind[k]] == k); } /* check that matrix A~ is upper block triangular with non-zero * diagonal */ xassert(1 <= num && num <= n); xassert(beg[1] == 1); xassert(beg[num+1] == n+1); /* walk thru blocks of A~ */ for (k = 1; k <= num; k++) { /* determine size of k-th block */ size = beg[k+1] - beg[k]; xassert(size >= 1); /* walk thru columns of k-th block */ for (jj = beg[k]; jj < beg[k+1]; jj++) { diag = 0; /* jj-th column of A~ = j-th column of A */ j = qq_ind[jj]; /* walk thru elements of j-th column of A */ ptr = ac_ptr[j]; end = ptr + ac_len[j]; for (; ptr < end; ptr++) { /* determine row index of a[i,j] */ i = sv_ind[ptr]; /* i-th row of A = ii-th row of A~ */ ii = pp_ind[i]; /* a~[ii,jj] should not be below k-th block */ xassert(ii < beg[k+1]); if (ii == jj) { /* non-zero diagonal element of A~ encountered */ diag = 1; } } xassert(diag); } } return; } /*********************************************************************** * btf_build_a_rows - build matrix A in row-wise format * * This routine builds the row-wise representation of matrix A in the * right part of SVA using its column-wise representation. * * The working array len should have at least 1+n elements (len[0] is * not used). */ void btf_build_a_rows(BTF *btf, int len[/*1+n*/]) { int n = btf->n; SVA *sva = btf->sva; int *sv_ind = sva->ind; double *sv_val = sva->val; int ar_ref = btf->ar_ref; int *ar_ptr = &sva->ptr[ar_ref-1]; int *ar_len = &sva->len[ar_ref-1]; int ac_ref = btf->ac_ref; int *ac_ptr = &sva->ptr[ac_ref-1]; int *ac_len = &sva->len[ac_ref-1]; int i, j, end, nnz, ptr, ptr1; /* calculate the number of non-zeros in each row of matrix A and * the total number of non-zeros */ nnz = 0; for (i = 1; i <= n; i++) len[i] = 0; for (j = 1; j <= n; j++) { nnz += ac_len[j]; for (end = (ptr = ac_ptr[j]) + ac_len[j]; ptr < end; ptr++) len[sv_ind[ptr]]++; } /* we need at least nnz free locations in SVA */ if (sva->r_ptr - sva->m_ptr < nnz) { sva_more_space(sva, nnz); sv_ind = sva->ind; sv_val = sva->val; } /* reserve locations for rows of matrix A */ for (i = 1; i <= n; i++) { if (len[i] > 0) sva_reserve_cap(sva, ar_ref-1+i, len[i]); ar_len[i] = len[i]; } /* walk thru columns of matrix A and build its rows */ for (j = 1; j <= n; j++) { for (end = (ptr = ac_ptr[j]) + ac_len[j]; ptr < end; ptr++) { i = sv_ind[ptr]; sv_ind[ptr1 = ar_ptr[i] + (--len[i])] = j; sv_val[ptr1] = sv_val[ptr]; } } return; } /*********************************************************************** * btf_a_solve - solve system A * x = b * * This routine solves the system A * x = b, where A is the original * matrix. * * On entry the array b should contain elements of the right-hand size * vector b in locations b[1], ..., b[n], where n is the order of the * matrix A. On exit the array x will contain elements of the solution * vector in locations x[1], ..., x[n]. Note that the array b will be * clobbered on exit. * * The routine also uses locations [1], ..., [max_size] of two working * arrays w1 and w2, where max_size is the maximal size of diagonal * blocks in BT-factorization (max_size <= n). */ void btf_a_solve(BTF *btf, double b[/*1+n*/], double x[/*1+n*/], double w1[/*1+n*/], double w2[/*1+n*/]) { SVA *sva = btf->sva; int *sv_ind = sva->ind; double *sv_val = sva->val; int *pp_inv = btf->pp_inv; int *qq_ind = btf->qq_ind; int num = btf->num; int *beg = btf->beg; int ac_ref = btf->ac_ref; int *ac_ptr = &sva->ptr[ac_ref-1]; int *ac_len = &sva->len[ac_ref-1]; double *bb = w1; double *xx = w2; LUF luf; int i, j, jj, k, beg_k, flag; double t; for (k = num; k >= 1; k--) { /* determine order of diagonal block A~[k,k] */ luf.n = beg[k+1] - (beg_k = beg[k]); if (luf.n == 1) { /* trivial case */ /* solve system A~[k,k] * X[k] = B[k] */ t = x[qq_ind[beg_k]] = b[pp_inv[beg_k]] / btf->vr_piv[beg_k]; /* substitute X[k] into other equations */ if (t != 0.0) { int ptr = ac_ptr[qq_ind[beg_k]]; int end = ptr + ac_len[qq_ind[beg_k]]; for (; ptr < end; ptr++) b[sv_ind[ptr]] -= sv_val[ptr] * t; } } else { /* general case */ /* construct B[k] */ flag = 0; for (i = 1; i <= luf.n; i++) { if ((bb[i] = b[pp_inv[i + (beg_k-1)]]) != 0.0) flag = 1; } /* solve system A~[k,k] * X[k] = B[k] */ if (!flag) { /* B[k] = 0, so X[k] = 0 */ for (j = 1; j <= luf.n; j++) x[qq_ind[j + (beg_k-1)]] = 0.0; continue; } luf.sva = sva; luf.fr_ref = btf->fr_ref + (beg_k-1); luf.fc_ref = btf->fc_ref + (beg_k-1); luf.vr_ref = btf->vr_ref + (beg_k-1); luf.vr_piv = btf->vr_piv + (beg_k-1); luf.vc_ref = btf->vc_ref + (beg_k-1); luf.pp_ind = btf->p1_ind + (beg_k-1); luf.pp_inv = btf->p1_inv + (beg_k-1); luf.qq_ind = btf->q1_ind + (beg_k-1); luf.qq_inv = btf->q1_inv + (beg_k-1); luf_f_solve(&luf, bb); luf_v_solve(&luf, bb, xx); /* store X[k] and substitute it into other equations */ for (j = 1; j <= luf.n; j++) { jj = j + (beg_k-1); t = x[qq_ind[jj]] = xx[j]; if (t != 0.0) { int ptr = ac_ptr[qq_ind[jj]]; int end = ptr + ac_len[qq_ind[jj]]; for (; ptr < end; ptr++) b[sv_ind[ptr]] -= sv_val[ptr] * t; } } } } return; } /*********************************************************************** * btf_at_solve - solve system A'* x = b * * This routine solves the system A'* x = b, where A' is a matrix * transposed to the original matrix A. * * On entry the array b should contain elements of the right-hand size * vector b in locations b[1], ..., b[n], where n is the order of the * matrix A. On exit the array x will contain elements of the solution * vector in locations x[1], ..., x[n]. Note that the array b will be * clobbered on exit. * * The routine also uses locations [1], ..., [max_size] of two working * arrays w1 and w2, where max_size is the maximal size of diagonal * blocks in BT-factorization (max_size <= n). */ void btf_at_solve(BTF *btf, double b[/*1+n*/], double x[/*1+n*/], double w1[/*1+n*/], double w2[/*1+n*/]) { SVA *sva = btf->sva; int *sv_ind = sva->ind; double *sv_val = sva->val; int *pp_inv = btf->pp_inv; int *qq_ind = btf->qq_ind; int num = btf->num; int *beg = btf->beg; int ar_ref = btf->ar_ref; int *ar_ptr = &sva->ptr[ar_ref-1]; int *ar_len = &sva->len[ar_ref-1]; double *bb = w1; double *xx = w2; LUF luf; int i, j, jj, k, beg_k, flag; double t; for (k = 1; k <= num; k++) { /* determine order of diagonal block A~[k,k] */ luf.n = beg[k+1] - (beg_k = beg[k]); if (luf.n == 1) { /* trivial case */ /* solve system A~'[k,k] * X[k] = B[k] */ t = x[pp_inv[beg_k]] = b[qq_ind[beg_k]] / btf->vr_piv[beg_k]; /* substitute X[k] into other equations */ if (t != 0.0) { int ptr = ar_ptr[pp_inv[beg_k]]; int end = ptr + ar_len[pp_inv[beg_k]]; for (; ptr < end; ptr++) b[sv_ind[ptr]] -= sv_val[ptr] * t; } } else { /* general case */ /* construct B[k] */ flag = 0; for (i = 1; i <= luf.n; i++) { if ((bb[i] = b[qq_ind[i + (beg_k-1)]]) != 0.0) flag = 1; } /* solve system A~'[k,k] * X[k] = B[k] */ if (!flag) { /* B[k] = 0, so X[k] = 0 */ for (j = 1; j <= luf.n; j++) x[pp_inv[j + (beg_k-1)]] = 0.0; continue; } luf.sva = sva; luf.fr_ref = btf->fr_ref + (beg_k-1); luf.fc_ref = btf->fc_ref + (beg_k-1); luf.vr_ref = btf->vr_ref + (beg_k-1); luf.vr_piv = btf->vr_piv + (beg_k-1); luf.vc_ref = btf->vc_ref + (beg_k-1); luf.pp_ind = btf->p1_ind + (beg_k-1); luf.pp_inv = btf->p1_inv + (beg_k-1); luf.qq_ind = btf->q1_ind + (beg_k-1); luf.qq_inv = btf->q1_inv + (beg_k-1); luf_vt_solve(&luf, bb, xx); luf_ft_solve(&luf, xx); /* store X[k] and substitute it into other equations */ for (j = 1; j <= luf.n; j++) { jj = j + (beg_k-1); t = x[pp_inv[jj]] = xx[j]; if (t != 0.0) { int ptr = ar_ptr[pp_inv[jj]]; int end = ptr + ar_len[pp_inv[jj]]; for (; ptr < end; ptr++) b[sv_ind[ptr]] -= sv_val[ptr] * t; } } } } return; } /*********************************************************************** * btf_at_solve1 - solve system A'* y = e' to cause growth in y * * This routine is a special version of btf_at_solve. It solves the * system A'* y = e' = e + delta e, where A' is a matrix transposed to * the original matrix A, e is the specified right-hand side vector, * and delta e is a vector of +1 and -1 chosen to cause growth in the * solution vector y. * * On entry the array e should contain elements of the right-hand size * vector e in locations e[1], ..., e[n], where n is the order of the * matrix A. On exit the array y will contain elements of the solution * vector in locations y[1], ..., y[n]. Note that the array e will be * clobbered on exit. * * The routine also uses locations [1], ..., [max_size] of two working * arrays w1 and w2, where max_size is the maximal size of diagonal * blocks in BT-factorization (max_size <= n). */ void btf_at_solve1(BTF *btf, double e[/*1+n*/], double y[/*1+n*/], double w1[/*1+n*/], double w2[/*1+n*/]) { SVA *sva = btf->sva; int *sv_ind = sva->ind; double *sv_val = sva->val; int *pp_inv = btf->pp_inv; int *qq_ind = btf->qq_ind; int num = btf->num; int *beg = btf->beg; int ar_ref = btf->ar_ref; int *ar_ptr = &sva->ptr[ar_ref-1]; int *ar_len = &sva->len[ar_ref-1]; double *ee = w1; double *yy = w2; LUF luf; int i, j, jj, k, beg_k, ptr, end; double e_k, y_k; for (k = 1; k <= num; k++) { /* determine order of diagonal block A~[k,k] */ luf.n = beg[k+1] - (beg_k = beg[k]); if (luf.n == 1) { /* trivial case */ /* determine E'[k] = E[k] + delta E[k] */ e_k = e[qq_ind[beg_k]]; e_k = (e_k >= 0.0 ? e_k + 1.0 : e_k - 1.0); /* solve system A~'[k,k] * Y[k] = E[k] */ y_k = y[pp_inv[beg_k]] = e_k / btf->vr_piv[beg_k]; /* substitute Y[k] into other equations */ ptr = ar_ptr[pp_inv[beg_k]]; end = ptr + ar_len[pp_inv[beg_k]]; for (; ptr < end; ptr++) e[sv_ind[ptr]] -= sv_val[ptr] * y_k; } else { /* general case */ /* construct E[k] */ for (i = 1; i <= luf.n; i++) ee[i] = e[qq_ind[i + (beg_k-1)]]; /* solve system A~'[k,k] * Y[k] = E[k] + delta E[k] */ luf.sva = sva; luf.fr_ref = btf->fr_ref + (beg_k-1); luf.fc_ref = btf->fc_ref + (beg_k-1); luf.vr_ref = btf->vr_ref + (beg_k-1); luf.vr_piv = btf->vr_piv + (beg_k-1); luf.vc_ref = btf->vc_ref + (beg_k-1); luf.pp_ind = btf->p1_ind + (beg_k-1); luf.pp_inv = btf->p1_inv + (beg_k-1); luf.qq_ind = btf->q1_ind + (beg_k-1); luf.qq_inv = btf->q1_inv + (beg_k-1); luf_vt_solve1(&luf, ee, yy); luf_ft_solve(&luf, yy); /* store Y[k] and substitute it into other equations */ for (j = 1; j <= luf.n; j++) { jj = j + (beg_k-1); y_k = y[pp_inv[jj]] = yy[j]; ptr = ar_ptr[pp_inv[jj]]; end = ptr + ar_len[pp_inv[jj]]; for (; ptr < end; ptr++) e[sv_ind[ptr]] -= sv_val[ptr] * y_k; } } } return; } /*********************************************************************** * btf_estimate_norm - estimate 1-norm of inv(A) * * This routine estimates 1-norm of inv(A) by one step of inverse * iteration for the small singular vector as described in [1]. This * involves solving two systems of equations: * * A'* y = e, * * A * z = y, * * where A' is a matrix transposed to A, and e is a vector of +1 and -1 * chosen to cause growth in y. Then * * estimate 1-norm of inv(A) = (1-norm of z) / (1-norm of y) * * REFERENCES * * 1. G.E.Forsythe, M.A.Malcolm, C.B.Moler. Computer Methods for * Mathematical Computations. Prentice-Hall, Englewood Cliffs, N.J., * pp. 30-62 (subroutines DECOMP and SOLVE). */ double btf_estimate_norm(BTF *btf, double w1[/*1+n*/], double w2[/*1+n*/], double w3[/*1+n*/], double w4[/*1+n*/]) { int n = btf->n; double *e = w1; double *y = w2; double *z = w1; int i; double y_norm, z_norm; /* compute y = inv(A') * e to cause growth in y */ for (i = 1; i <= n; i++) e[i] = 0.0; btf_at_solve1(btf, e, y, w3, w4); /* compute 1-norm of y = sum |y[i]| */ y_norm = 0.0; for (i = 1; i <= n; i++) y_norm += (y[i] >= 0.0 ? +y[i] : -y[i]); /* compute z = inv(A) * y */ btf_a_solve(btf, y, z, w3, w4); /* compute 1-norm of z = sum |z[i]| */ z_norm = 0.0; for (i = 1; i <= n; i++) z_norm += (z[i] >= 0.0 ? +z[i] : -z[i]); /* estimate 1-norm of inv(A) = (1-norm of z) / (1-norm of y) */ return z_norm / y_norm; } /* eof */