/* glpnpp04.c */ /*********************************************************************** * This code is part of GLPK (GNU Linear Programming Kit). * * Copyright (C) 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, * 2009, 2010, 2011, 2013 Andrew Makhorin, Department for Applied * Informatics, Moscow Aviation Institute, Moscow, Russia. All rights * reserved. E-mail: . * * 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 "glpenv.h" #include "glpnpp.h" /*********************************************************************** * NAME * * npp_binarize_prob - binarize MIP problem * * SYNOPSIS * * #include "glpnpp.h" * int npp_binarize_prob(NPP *npp); * * DESCRIPTION * * The routine npp_binarize_prob replaces in the original MIP problem * every integer variable: * * l[q] <= x[q] <= u[q], (1) * * where l[q] < u[q], by an equivalent sum of binary variables. * * RETURNS * * The routine returns the number of integer variables for which the * transformation failed, because u[q] - l[q] > d_max. * * PROBLEM TRANSFORMATION * * If variable x[q] has non-zero lower bound, it is first processed * with the routine npp_lbnd_col. Thus, we can assume that: * * 0 <= x[q] <= u[q]. (2) * * If u[q] = 1, variable x[q] is already binary, so further processing * is not needed. Let, therefore, that 2 <= u[q] <= d_max, and n be a * smallest integer such that u[q] <= 2^n - 1 (n >= 2, since u[q] >= 2). * Then variable x[q] can be replaced by the following sum: * * n-1 * x[q] = sum 2^k x[k], (3) * k=0 * * where x[k] are binary columns (variables). If u[q] < 2^n - 1, the * following additional inequality constraint must be also included in * the transformed problem: * * n-1 * sum 2^k x[k] <= u[q]. (4) * k=0 * * Note: Assuming that in the transformed problem x[q] becomes binary * variable x[0], this transformation causes new n-1 binary variables * to appear. * * Substituting x[q] from (3) to the objective row gives: * * z = sum c[j] x[j] + c[0] = * j * * = sum c[j] x[j] + c[q] x[q] + c[0] = * j!=q * n-1 * = sum c[j] x[j] + c[q] sum 2^k x[k] + c[0] = * j!=q k=0 * n-1 * = sum c[j] x[j] + sum c[k] x[k] + c[0], * j!=q k=0 * * where: * * c[k] = 2^k c[q], k = 0, ..., n-1. (5) * * And substituting x[q] from (3) to i-th constraint row i gives: * * L[i] <= sum a[i,j] x[j] <= U[i] ==> * j * * L[i] <= sum a[i,j] x[j] + a[i,q] x[q] <= U[i] ==> * j!=q * n-1 * L[i] <= sum a[i,j] x[j] + a[i,q] sum 2^k x[k] <= U[i] ==> * j!=q k=0 * n-1 * L[i] <= sum a[i,j] x[j] + sum a[i,k] x[k] <= U[i], * j!=q k=0 * * where: * * a[i,k] = 2^k a[i,q], k = 0, ..., n-1. (6) * * RECOVERING SOLUTION * * Value of variable x[q] is computed with formula (3). */ struct binarize { int q; /* column reference number for x[q] = x[0] */ int j; /* column reference number for x[1]; x[2] has reference number j+1, x[3] - j+2, etc. */ int n; /* total number of binary variables, n >= 2 */ }; static int rcv_binarize_prob(NPP *npp, void *info); int npp_binarize_prob(NPP *npp) { /* binarize MIP problem */ struct binarize *info; NPPROW *row; NPPCOL *col, *bin; NPPAIJ *aij; int u, n, k, temp, nfails, nvars, nbins, nrows; /* new variables will be added to the end of the column list, so we go from the end to beginning of the column list */ nfails = nvars = nbins = nrows = 0; for (col = npp->c_tail; col != NULL; col = col->prev) { /* skip continuous variable */ if (!col->is_int) continue; /* skip fixed variable */ if (col->lb == col->ub) continue; /* skip binary variable */ if (col->lb == 0.0 && col->ub == 1.0) continue; /* check if the transformation is applicable */ if (col->lb < -1e6 || col->ub > +1e6 || col->ub - col->lb > 4095.0) { /* unfortunately, not */ nfails++; continue; } /* process integer non-binary variable x[q] */ nvars++; /* make x[q] non-negative, if its lower bound is non-zero */ if (col->lb != 0.0) npp_lbnd_col(npp, col); /* now 0 <= x[q] <= u[q] */ xassert(col->lb == 0.0); u = (int)col->ub; xassert(col->ub == (double)u); /* if x[q] is binary, further processing is not needed */ if (u == 1) continue; /* determine smallest n such that u <= 2^n - 1 (thus, n is the number of binary variables needed) */ n = 2, temp = 4; while (u >= temp) n++, temp += temp; nbins += n; /* create transformation stack entry */ info = npp_push_tse(npp, rcv_binarize_prob, sizeof(struct binarize)); info->q = col->j; info->j = 0; /* will be set below */ info->n = n; /* if u < 2^n - 1, we need one additional row for (4) */ if (u < temp - 1) { row = npp_add_row(npp), nrows++; row->lb = -DBL_MAX, row->ub = u; } else row = NULL; /* in the transformed problem variable x[q] becomes binary variable x[0], so its objective and constraint coefficients are not changed */ col->ub = 1.0; /* include x[0] into constraint (4) */ if (row != NULL) npp_add_aij(npp, row, col, 1.0); /* add other binary variables x[1], ..., x[n-1] */ for (k = 1, temp = 2; k < n; k++, temp += temp) { /* add new binary variable x[k] */ bin = npp_add_col(npp); bin->is_int = 1; bin->lb = 0.0, bin->ub = 1.0; bin->coef = (double)temp * col->coef; /* store column reference number for x[1] */ if (info->j == 0) info->j = bin->j; else xassert(info->j + (k-1) == bin->j); /* duplicate constraint coefficients for x[k]; this also automatically includes x[k] into constraint (4) */ for (aij = col->ptr; aij != NULL; aij = aij->c_next) npp_add_aij(npp, aij->row, bin, (double)temp * aij->val); } } if (nvars > 0) xprintf("%d integer variable(s) were replaced by %d binary one" "s\n", nvars, nbins); if (nrows > 0) xprintf("%d row(s) were added due to binarization\n", nrows); if (nfails > 0) xprintf("Binarization failed for %d integer variable(s)\n", nfails); return nfails; } static int rcv_binarize_prob(NPP *npp, void *_info) { /* recovery binarized variable */ struct binarize *info = _info; int k, temp; double sum; /* compute value of x[q]; see formula (3) */ sum = npp->c_value[info->q]; for (k = 1, temp = 2; k < info->n; k++, temp += temp) sum += (double)temp * npp->c_value[info->j + (k-1)]; npp->c_value[info->q] = sum; return 0; } /**********************************************************************/ struct elem { /* linear form element a[j] x[j] */ double aj; /* non-zero coefficient value */ NPPCOL *xj; /* pointer to variable (column) */ struct elem *next; /* pointer to another term */ }; static struct elem *copy_form(NPP *npp, NPPROW *row, double s) { /* copy linear form */ NPPAIJ *aij; struct elem *ptr, *e; ptr = NULL; for (aij = row->ptr; aij != NULL; aij = aij->r_next) { e = dmp_get_atom(npp->pool, sizeof(struct elem)); e->aj = s * aij->val; e->xj = aij->col; e->next = ptr; ptr = e; } return ptr; } static void drop_form(NPP *npp, struct elem *ptr) { /* drop linear form */ struct elem *e; while (ptr != NULL) { e = ptr; ptr = e->next; dmp_free_atom(npp->pool, e, sizeof(struct elem)); } return; } /*********************************************************************** * NAME * * npp_is_packing - test if constraint is packing inequality * * SYNOPSIS * * #include "glpnpp.h" * int npp_is_packing(NPP *npp, NPPROW *row); * * RETURNS * * If the specified row (constraint) is packing inequality (see below), * the routine npp_is_packing returns non-zero. Otherwise, it returns * zero. * * PACKING INEQUALITIES * * In canonical format the packing inequality is the following: * * sum x[j] <= 1, (1) * j in J * * where all variables x[j] are binary. This inequality expresses the * condition that in any integer feasible solution at most one variable * from set J can take non-zero (unity) value while other variables * must be equal to zero. W.l.o.g. it is assumed that |J| >= 2, because * if J is empty or |J| = 1, the inequality (1) is redundant. * * In general case the packing inequality may include original variables * x[j] as well as their complements x~[j]: * * sum x[j] + sum x~[j] <= 1, (2) * j in Jp j in Jn * * where Jp and Jn are not intersected. Therefore, using substitution * x~[j] = 1 - x[j] gives the packing inequality in generalized format: * * sum x[j] - sum x[j] <= 1 - |Jn|. (3) * j in Jp j in Jn */ int npp_is_packing(NPP *npp, NPPROW *row) { /* test if constraint is packing inequality */ NPPCOL *col; NPPAIJ *aij; int b; xassert(npp == npp); if (!(row->lb == -DBL_MAX && row->ub != +DBL_MAX)) return 0; b = 1; for (aij = row->ptr; aij != NULL; aij = aij->r_next) { col = aij->col; if (!(col->is_int && col->lb == 0.0 && col->ub == 1.0)) return 0; if (aij->val == +1.0) ; else if (aij->val == -1.0) b--; else return 0; } if (row->ub != (double)b) return 0; return 1; } /*********************************************************************** * NAME * * npp_hidden_packing - identify hidden packing inequality * * SYNOPSIS * * #include "glpnpp.h" * int npp_hidden_packing(NPP *npp, NPPROW *row); * * DESCRIPTION * * The routine npp_hidden_packing processes specified inequality * constraint, which includes only binary variables, and the number of * the variables is not less than two. If the original inequality is * equivalent to a packing inequality, the routine replaces it by this * equivalent inequality. If the original constraint is double-sided * inequality, it is replaced by a pair of single-sided inequalities, * if necessary. * * RETURNS * * If the original inequality constraint was replaced by equivalent * packing inequality, the routine npp_hidden_packing returns non-zero. * Otherwise, it returns zero. * * PROBLEM TRANSFORMATION * * Consider an inequality constraint: * * sum a[j] x[j] <= b, (1) * j in J * * where all variables x[j] are binary, and |J| >= 2. (In case of '>=' * inequality it can be transformed to '<=' format by multiplying both * its sides by -1.) * * Let Jp = {j: a[j] > 0}, Jn = {j: a[j] < 0}. Performing substitution * x[j] = 1 - x~[j] for all j in Jn, we have: * * sum a[j] x[j] <= b ==> * j in J * * sum a[j] x[j] + sum a[j] x[j] <= b ==> * j in Jp j in Jn * * sum a[j] x[j] + sum a[j] (1 - x~[j]) <= b ==> * j in Jp j in Jn * * sum a[j] x[j] - sum a[j] x~[j] <= b - sum a[j]. * j in Jp j in Jn j in Jn * * Thus, meaning the transformation above, we can assume that in * inequality (1) all coefficients a[j] are positive. Moreover, we can * assume that a[j] <= b. In fact, let a[j] > b; then the following * three cases are possible: * * 1) b < 0. In this case inequality (1) is infeasible, so the problem * has no feasible solution (see the routine npp_analyze_row); * * 2) b = 0. In this case inequality (1) is a forcing inequality on its * upper bound (see the routine npp_forcing row), from which it * follows that all variables x[j] should be fixed at zero; * * 3) b > 0. In this case inequality (1) defines an implied zero upper * bound for variable x[j] (see the routine npp_implied_bounds), from * which it follows that x[j] should be fixed at zero. * * It is assumed that all three cases listed above have been recognized * by the routine npp_process_prob, which performs basic MIP processing * prior to a call the routine npp_hidden_packing. So, if one of these * cases occurs, we should just skip processing such constraint. * * Thus, let 0 < a[j] <= b. Then it is obvious that constraint (1) is * equivalent to packing inquality only if: * * a[j] + a[k] > b + eps (2) * * for all j, k in J, j != k, where eps is an absolute tolerance for * row (linear form) value. Checking the condition (2) for all j and k, * j != k, requires time O(|J|^2). However, this time can be reduced to * O(|J|), if use minimal a[j] and a[k], in which case it is sufficient * to check the condition (2) only once. * * Once the original inequality (1) is replaced by equivalent packing * inequality, we need to perform back substitution x~[j] = 1 - x[j] for * all j in Jn (see above). * * RECOVERING SOLUTION * * None needed. */ static int hidden_packing(NPP *npp, struct elem *ptr, double *_b) { /* process inequality constraint: sum a[j] x[j] <= b; 0 - specified row is NOT hidden packing inequality; 1 - specified row is packing inequality; 2 - specified row is hidden packing inequality. */ struct elem *e, *ej, *ek; int neg; double b = *_b, eps; xassert(npp == npp); /* a[j] must be non-zero, x[j] must be binary, for all j in J */ for (e = ptr; e != NULL; e = e->next) { xassert(e->aj != 0.0); xassert(e->xj->is_int); xassert(e->xj->lb == 0.0 && e->xj->ub == 1.0); } /* check if the specified inequality constraint already has the form of packing inequality */ neg = 0; /* neg is |Jn| */ for (e = ptr; e != NULL; e = e->next) { if (e->aj == +1.0) ; else if (e->aj == -1.0) neg++; else break; } if (e == NULL) { /* all coefficients a[j] are +1 or -1; check rhs b */ if (b == (double)(1 - neg)) { /* it is packing inequality; no processing is needed */ return 1; } } /* substitute x[j] = 1 - x~[j] for all j in Jn to make all a[j] positive; the result is a~[j] = |a[j]| and new rhs b */ for (e = ptr; e != NULL; e = e->next) if (e->aj < 0) b -= e->aj; /* now a[j] > 0 for all j in J (actually |a[j]| are used) */ /* if a[j] > b, skip processing--this case must not appear */ for (e = ptr; e != NULL; e = e->next) if (fabs(e->aj) > b) return 0; /* now 0 < a[j] <= b for all j in J */ /* find two minimal coefficients a[j] and a[k], j != k */ ej = NULL; for (e = ptr; e != NULL; e = e->next) if (ej == NULL || fabs(ej->aj) > fabs(e->aj)) ej = e; xassert(ej != NULL); ek = NULL; for (e = ptr; e != NULL; e = e->next) if (e != ej) if (ek == NULL || fabs(ek->aj) > fabs(e->aj)) ek = e; xassert(ek != NULL); /* the specified constraint is equivalent to packing inequality iff a[j] + a[k] > b + eps */ eps = 1e-3 + 1e-6 * fabs(b); if (fabs(ej->aj) + fabs(ek->aj) <= b + eps) return 0; /* perform back substitution x~[j] = 1 - x[j] and construct the final equivalent packing inequality in generalized format */ b = 1.0; for (e = ptr; e != NULL; e = e->next) { if (e->aj > 0.0) e->aj = +1.0; else /* e->aj < 0.0 */ e->aj = -1.0, b -= 1.0; } *_b = b; return 2; } int npp_hidden_packing(NPP *npp, NPPROW *row) { /* identify hidden packing inequality */ NPPROW *copy; NPPAIJ *aij; struct elem *ptr, *e; int kase, ret, count = 0; double b; /* the row must be inequality constraint */ xassert(row->lb < row->ub); for (kase = 0; kase <= 1; kase++) { if (kase == 0) { /* process row upper bound */ if (row->ub == +DBL_MAX) continue; ptr = copy_form(npp, row, +1.0); b = + row->ub; } else { /* process row lower bound */ if (row->lb == -DBL_MAX) continue; ptr = copy_form(npp, row, -1.0); b = - row->lb; } /* now the inequality has the form "sum a[j] x[j] <= b" */ ret = hidden_packing(npp, ptr, &b); xassert(0 <= ret && ret <= 2); if (kase == 1 && ret == 1 || ret == 2) { /* the original inequality has been identified as hidden packing inequality */ count++; #ifdef GLP_DEBUG xprintf("Original constraint:\n"); for (aij = row->ptr; aij != NULL; aij = aij->r_next) xprintf(" %+g x%d", aij->val, aij->col->j); if (row->lb != -DBL_MAX) xprintf(", >= %g", row->lb); if (row->ub != +DBL_MAX) xprintf(", <= %g", row->ub); xprintf("\n"); xprintf("Equivalent packing inequality:\n"); for (e = ptr; e != NULL; e = e->next) xprintf(" %sx%d", e->aj > 0.0 ? "+" : "-", e->xj->j); xprintf(", <= %g\n", b); #endif if (row->lb == -DBL_MAX || row->ub == +DBL_MAX) { /* the original row is single-sided inequality; no copy is needed */ copy = NULL; } else { /* the original row is double-sided inequality; we need to create its copy for other bound before replacing it with the equivalent inequality */ copy = npp_add_row(npp); if (kase == 0) { /* the copy is for lower bound */ copy->lb = row->lb, copy->ub = +DBL_MAX; } else { /* the copy is for upper bound */ copy->lb = -DBL_MAX, copy->ub = row->ub; } /* copy original row coefficients */ for (aij = row->ptr; aij != NULL; aij = aij->r_next) npp_add_aij(npp, copy, aij->col, aij->val); } /* replace the original inequality by equivalent one */ npp_erase_row(npp, row); row->lb = -DBL_MAX, row->ub = b; for (e = ptr; e != NULL; e = e->next) npp_add_aij(npp, row, e->xj, e->aj); /* continue processing lower bound for the copy */ if (copy != NULL) row = copy; } drop_form(npp, ptr); } return count; } /*********************************************************************** * NAME * * npp_implied_packing - identify implied packing inequality * * SYNOPSIS * * #include "glpnpp.h" * int npp_implied_packing(NPP *npp, NPPROW *row, int which, * NPPCOL *var[], char set[]); * * DESCRIPTION * * The routine npp_implied_packing processes specified row (constraint) * of general format: * * L <= sum a[j] x[j] <= U. (1) * j * * If which = 0, only lower bound L, which must exist, is considered, * while upper bound U is ignored. Similarly, if which = 1, only upper * bound U, which must exist, is considered, while lower bound L is * ignored. Thus, if the specified row is a double-sided inequality or * equality constraint, this routine should be called twice for both * lower and upper bounds. * * The routine npp_implied_packing attempts to find a non-trivial (i.e. * having not less than two binary variables) packing inequality: * * sum x[j] - sum x[j] <= 1 - |Jn|, (2) * j in Jp j in Jn * * which is relaxation of the constraint (1) in the sense that any * solution satisfying to that constraint also satisfies to the packing * inequality (2). If such relaxation exists, the routine stores * pointers to descriptors of corresponding binary variables and their * flags, resp., to locations var[1], var[2], ..., var[len] and set[1], * set[2], ..., set[len], where set[j] = 0 means that j in Jp and * set[j] = 1 means that j in Jn. * * RETURNS * * The routine npp_implied_packing returns len, which is the total * number of binary variables in the packing inequality found, len >= 2. * However, if the relaxation does not exist, the routine returns zero. * * ALGORITHM * * If which = 0, the constraint coefficients (1) are multiplied by -1 * and b is assigned -L; if which = 1, the constraint coefficients (1) * are not changed and b is assigned +U. In both cases the specified * constraint gets the following format: * * sum a[j] x[j] <= b. (3) * j * * (Note that (3) is a relaxation of (1), because one of bounds L or U * is ignored.) * * Let J be set of binary variables, Kp be set of non-binary (integer * or continuous) variables with a[j] > 0, and Kn be set of non-binary * variables with a[j] < 0. Then the inequality (3) can be written as * follows: * * sum a[j] x[j] <= b - sum a[j] x[j] - sum a[j] x[j]. (4) * j in J j in Kp j in Kn * * To get rid of non-binary variables we can replace the inequality (4) * by the following relaxed inequality: * * sum a[j] x[j] <= b~, (5) * j in J * * where: * * b~ = sup(b - sum a[j] x[j] - sum a[j] x[j]) = * j in Kp j in Kn * * = b - inf sum a[j] x[j] - inf sum a[j] x[j] = (6) * j in Kp j in Kn * * = b - sum a[j] l[j] - sum a[j] u[j]. * j in Kp j in Kn * * Note that if lower bound l[j] (if j in Kp) or upper bound u[j] * (if j in Kn) of some non-binary variable x[j] does not exist, then * formally b = +oo, in which case further analysis is not performed. * * Let Bp = {j in J: a[j] > 0}, Bn = {j in J: a[j] < 0}. To make all * the inequality coefficients in (5) positive, we replace all x[j] in * Bn by their complementaries, substituting x[j] = 1 - x~[j] for all * j in Bn, that gives: * * sum a[j] x[j] - sum a[j] x~[j] <= b~ - sum a[j]. (7) * j in Bp j in Bn j in Bn * * This inequality is a relaxation of the original constraint (1), and * it is a binary knapsack inequality. Writing it in the standard format * we have: * * sum alfa[j] z[j] <= beta, (8) * j in J * * where: * ( + a[j], if j in Bp, * alfa[j] = < (9) * ( - a[j], if j in Bn, * * ( x[j], if j in Bp, * z[j] = < (10) * ( 1 - x[j], if j in Bn, * * beta = b~ - sum a[j]. (11) * j in Bn * * In the inequality (8) all coefficients are positive, therefore, the * packing relaxation to be found for this inequality is the following: * * sum z[j] <= 1. (12) * j in P * * It is obvious that set P within J, which we would like to find, must * satisfy to the following condition: * * alfa[j] + alfa[k] > beta + eps for all j, k in P, j != k, (13) * * where eps is an absolute tolerance for value of the linear form. * Thus, it is natural to take P = {j: alpha[j] > (beta + eps) / 2}. * Moreover, if in the equality (8) there exist coefficients alfa[k], * for which alfa[k] <= (beta + eps) / 2, but which, nevertheless, * satisfies to the condition (13) for all j in P, *one* corresponding * variable z[k] (having, for example, maximal coefficient alfa[k]) can * be included in set P, that allows increasing the number of binary * variables in (12) by one. * * Once the set P has been built, for the inequality (12) we need to * perform back substitution according to (10) in order to express it * through the original binary variables. As the result of such back * substitution the relaxed packing inequality get its final format (2), * where Jp = J intersect Bp, and Jn = J intersect Bn. */ int npp_implied_packing(NPP *npp, NPPROW *row, int which, NPPCOL *var[], char set[]) { struct elem *ptr, *e, *i, *k; int len = 0; double b, eps; /* build inequality (3) */ if (which == 0) { ptr = copy_form(npp, row, -1.0); xassert(row->lb != -DBL_MAX); b = - row->lb; } else if (which == 1) { ptr = copy_form(npp, row, +1.0); xassert(row->ub != +DBL_MAX); b = + row->ub; } /* remove non-binary variables to build relaxed inequality (5); compute its right-hand side b~ with formula (6) */ for (e = ptr; e != NULL; e = e->next) { if (!(e->xj->is_int && e->xj->lb == 0.0 && e->xj->ub == 1.0)) { /* x[j] is non-binary variable */ if (e->aj > 0.0) { if (e->xj->lb == -DBL_MAX) goto done; b -= e->aj * e->xj->lb; } else /* e->aj < 0.0 */ { if (e->xj->ub == +DBL_MAX) goto done; b -= e->aj * e->xj->ub; } /* a[j] = 0 means that variable x[j] is removed */ e->aj = 0.0; } } /* substitute x[j] = 1 - x~[j] to build knapsack inequality (8); compute its right-hand side beta with formula (11) */ for (e = ptr; e != NULL; e = e->next) if (e->aj < 0.0) b -= e->aj; /* if beta is close to zero, the knapsack inequality is either infeasible or forcing inequality; this must never happen, so we skip further analysis */ if (b < 1e-3) goto done; /* build set P as well as sets Jp and Jn, and determine x[k] as explained above in comments to the routine */ eps = 1e-3 + 1e-6 * b; i = k = NULL; for (e = ptr; e != NULL; e = e->next) { /* note that alfa[j] = |a[j]| */ if (fabs(e->aj) > 0.5 * (b + eps)) { /* alfa[j] > (b + eps) / 2; include x[j] in set P, i.e. in set Jp or Jn */ var[++len] = e->xj; set[len] = (char)(e->aj > 0.0 ? 0 : 1); /* alfa[i] = min alfa[j] over all j included in set P */ if (i == NULL || fabs(i->aj) > fabs(e->aj)) i = e; } else if (fabs(e->aj) >= 1e-3) { /* alfa[k] = max alfa[j] over all j not included in set P; we skip coefficient a[j] if it is close to zero to avoid numerically unreliable results */ if (k == NULL || fabs(k->aj) < fabs(e->aj)) k = e; } } /* if alfa[k] satisfies to condition (13) for all j in P, include x[k] in P */ if (i != NULL && k != NULL && fabs(i->aj) + fabs(k->aj) > b + eps) { var[++len] = k->xj; set[len] = (char)(k->aj > 0.0 ? 0 : 1); } /* trivial packing inequality being redundant must never appear, so we just ignore it */ if (len < 2) len = 0; done: drop_form(npp, ptr); return len; } /*********************************************************************** * NAME * * npp_is_covering - test if constraint is covering inequality * * SYNOPSIS * * #include "glpnpp.h" * int npp_is_covering(NPP *npp, NPPROW *row); * * RETURNS * * If the specified row (constraint) is covering inequality (see below), * the routine npp_is_covering returns non-zero. Otherwise, it returns * zero. * * COVERING INEQUALITIES * * In canonical format the covering inequality is the following: * * sum x[j] >= 1, (1) * j in J * * where all variables x[j] are binary. This inequality expresses the * condition that in any integer feasible solution variables in set J * cannot be all equal to zero at the same time, i.e. at least one * variable must take non-zero (unity) value. W.l.o.g. it is assumed * that |J| >= 2, because if J is empty, the inequality (1) is * infeasible, and if |J| = 1, the inequality (1) is a forcing row. * * In general case the covering inequality may include original * variables x[j] as well as their complements x~[j]: * * sum x[j] + sum x~[j] >= 1, (2) * j in Jp j in Jn * * where Jp and Jn are not intersected. Therefore, using substitution * x~[j] = 1 - x[j] gives the packing inequality in generalized format: * * sum x[j] - sum x[j] >= 1 - |Jn|. (3) * j in Jp j in Jn * * (May note that the inequality (3) cuts off infeasible solutions, * where x[j] = 0 for all j in Jp and x[j] = 1 for all j in Jn.) * * NOTE: If |J| = 2, the inequality (3) is equivalent to packing * inequality (see the routine npp_is_packing). */ int npp_is_covering(NPP *npp, NPPROW *row) { /* test if constraint is covering inequality */ NPPCOL *col; NPPAIJ *aij; int b; xassert(npp == npp); if (!(row->lb != -DBL_MAX && row->ub == +DBL_MAX)) return 0; b = 1; for (aij = row->ptr; aij != NULL; aij = aij->r_next) { col = aij->col; if (!(col->is_int && col->lb == 0.0 && col->ub == 1.0)) return 0; if (aij->val == +1.0) ; else if (aij->val == -1.0) b--; else return 0; } if (row->lb != (double)b) return 0; return 1; } /*********************************************************************** * NAME * * npp_hidden_covering - identify hidden covering inequality * * SYNOPSIS * * #include "glpnpp.h" * int npp_hidden_covering(NPP *npp, NPPROW *row); * * DESCRIPTION * * The routine npp_hidden_covering processes specified inequality * constraint, which includes only binary variables, and the number of * the variables is not less than three. If the original inequality is * equivalent to a covering inequality (see below), the routine * replaces it by the equivalent inequality. If the original constraint * is double-sided inequality, it is replaced by a pair of single-sided * inequalities, if necessary. * * RETURNS * * If the original inequality constraint was replaced by equivalent * covering inequality, the routine npp_hidden_covering returns * non-zero. Otherwise, it returns zero. * * PROBLEM TRANSFORMATION * * Consider an inequality constraint: * * sum a[j] x[j] >= b, (1) * j in J * * where all variables x[j] are binary, and |J| >= 3. (In case of '<=' * inequality it can be transformed to '>=' format by multiplying both * its sides by -1.) * * Let Jp = {j: a[j] > 0}, Jn = {j: a[j] < 0}. Performing substitution * x[j] = 1 - x~[j] for all j in Jn, we have: * * sum a[j] x[j] >= b ==> * j in J * * sum a[j] x[j] + sum a[j] x[j] >= b ==> * j in Jp j in Jn * * sum a[j] x[j] + sum a[j] (1 - x~[j]) >= b ==> * j in Jp j in Jn * * sum m a[j] x[j] - sum a[j] x~[j] >= b - sum a[j]. * j in Jp j in Jn j in Jn * * Thus, meaning the transformation above, we can assume that in * inequality (1) all coefficients a[j] are positive. Moreover, we can * assume that b > 0, because otherwise the inequality (1) would be * redundant (see the routine npp_analyze_row). It is then obvious that * constraint (1) is equivalent to covering inequality only if: * * a[j] >= b, (2) * * for all j in J. * * Once the original inequality (1) is replaced by equivalent covering * inequality, we need to perform back substitution x~[j] = 1 - x[j] for * all j in Jn (see above). * * RECOVERING SOLUTION * * None needed. */ static int hidden_covering(NPP *npp, struct elem *ptr, double *_b) { /* process inequality constraint: sum a[j] x[j] >= b; 0 - specified row is NOT hidden covering inequality; 1 - specified row is covering inequality; 2 - specified row is hidden covering inequality. */ struct elem *e; int neg; double b = *_b, eps; xassert(npp == npp); /* a[j] must be non-zero, x[j] must be binary, for all j in J */ for (e = ptr; e != NULL; e = e->next) { xassert(e->aj != 0.0); xassert(e->xj->is_int); xassert(e->xj->lb == 0.0 && e->xj->ub == 1.0); } /* check if the specified inequality constraint already has the form of covering inequality */ neg = 0; /* neg is |Jn| */ for (e = ptr; e != NULL; e = e->next) { if (e->aj == +1.0) ; else if (e->aj == -1.0) neg++; else break; } if (e == NULL) { /* all coefficients a[j] are +1 or -1; check rhs b */ if (b == (double)(1 - neg)) { /* it is covering inequality; no processing is needed */ return 1; } } /* substitute x[j] = 1 - x~[j] for all j in Jn to make all a[j] positive; the result is a~[j] = |a[j]| and new rhs b */ for (e = ptr; e != NULL; e = e->next) if (e->aj < 0) b -= e->aj; /* now a[j] > 0 for all j in J (actually |a[j]| are used) */ /* if b <= 0, skip processing--this case must not appear */ if (b < 1e-3) return 0; /* now a[j] > 0 for all j in J, and b > 0 */ /* the specified constraint is equivalent to covering inequality iff a[j] >= b for all j in J */ eps = 1e-9 + 1e-12 * fabs(b); for (e = ptr; e != NULL; e = e->next) if (fabs(e->aj) < b - eps) return 0; /* perform back substitution x~[j] = 1 - x[j] and construct the final equivalent covering inequality in generalized format */ b = 1.0; for (e = ptr; e != NULL; e = e->next) { if (e->aj > 0.0) e->aj = +1.0; else /* e->aj < 0.0 */ e->aj = -1.0, b -= 1.0; } *_b = b; return 2; } int npp_hidden_covering(NPP *npp, NPPROW *row) { /* identify hidden covering inequality */ NPPROW *copy; NPPAIJ *aij; struct elem *ptr, *e; int kase, ret, count = 0; double b; /* the row must be inequality constraint */ xassert(row->lb < row->ub); for (kase = 0; kase <= 1; kase++) { if (kase == 0) { /* process row lower bound */ if (row->lb == -DBL_MAX) continue; ptr = copy_form(npp, row, +1.0); b = + row->lb; } else { /* process row upper bound */ if (row->ub == +DBL_MAX) continue; ptr = copy_form(npp, row, -1.0); b = - row->ub; } /* now the inequality has the form "sum a[j] x[j] >= b" */ ret = hidden_covering(npp, ptr, &b); xassert(0 <= ret && ret <= 2); if (kase == 1 && ret == 1 || ret == 2) { /* the original inequality has been identified as hidden covering inequality */ count++; #ifdef GLP_DEBUG xprintf("Original constraint:\n"); for (aij = row->ptr; aij != NULL; aij = aij->r_next) xprintf(" %+g x%d", aij->val, aij->col->j); if (row->lb != -DBL_MAX) xprintf(", >= %g", row->lb); if (row->ub != +DBL_MAX) xprintf(", <= %g", row->ub); xprintf("\n"); xprintf("Equivalent covering inequality:\n"); for (e = ptr; e != NULL; e = e->next) xprintf(" %sx%d", e->aj > 0.0 ? "+" : "-", e->xj->j); xprintf(", >= %g\n", b); #endif if (row->lb == -DBL_MAX || row->ub == +DBL_MAX) { /* the original row is single-sided inequality; no copy is needed */ copy = NULL; } else { /* the original row is double-sided inequality; we need to create its copy for other bound before replacing it with the equivalent inequality */ copy = npp_add_row(npp); if (kase == 0) { /* the copy is for upper bound */ copy->lb = -DBL_MAX, copy->ub = row->ub; } else { /* the copy is for lower bound */ copy->lb = row->lb, copy->ub = +DBL_MAX; } /* copy original row coefficients */ for (aij = row->ptr; aij != NULL; aij = aij->r_next) npp_add_aij(npp, copy, aij->col, aij->val); } /* replace the original inequality by equivalent one */ npp_erase_row(npp, row); row->lb = b, row->ub = +DBL_MAX; for (e = ptr; e != NULL; e = e->next) npp_add_aij(npp, row, e->xj, e->aj); /* continue processing upper bound for the copy */ if (copy != NULL) row = copy; } drop_form(npp, ptr); } return count; } /*********************************************************************** * NAME * * npp_is_partitioning - test if constraint is partitioning equality * * SYNOPSIS * * #include "glpnpp.h" * int npp_is_partitioning(NPP *npp, NPPROW *row); * * RETURNS * * If the specified row (constraint) is partitioning equality (see * below), the routine npp_is_partitioning returns non-zero. Otherwise, * it returns zero. * * PARTITIONING EQUALITIES * * In canonical format the partitioning equality is the following: * * sum x[j] = 1, (1) * j in J * * where all variables x[j] are binary. This equality expresses the * condition that in any integer feasible solution exactly one variable * in set J must take non-zero (unity) value while other variables must * be equal to zero. W.l.o.g. it is assumed that |J| >= 2, because if * J is empty, the inequality (1) is infeasible, and if |J| = 1, the * inequality (1) is a fixing row. * * In general case the partitioning equality may include original * variables x[j] as well as their complements x~[j]: * * sum x[j] + sum x~[j] = 1, (2) * j in Jp j in Jn * * where Jp and Jn are not intersected. Therefore, using substitution * x~[j] = 1 - x[j] leads to the partitioning equality in generalized * format: * * sum x[j] - sum x[j] = 1 - |Jn|. (3) * j in Jp j in Jn */ int npp_is_partitioning(NPP *npp, NPPROW *row) { /* test if constraint is partitioning equality */ NPPCOL *col; NPPAIJ *aij; int b; xassert(npp == npp); if (row->lb != row->ub) return 0; b = 1; for (aij = row->ptr; aij != NULL; aij = aij->r_next) { col = aij->col; if (!(col->is_int && col->lb == 0.0 && col->ub == 1.0)) return 0; if (aij->val == +1.0) ; else if (aij->val == -1.0) b--; else return 0; } if (row->lb != (double)b) return 0; return 1; } /*********************************************************************** * NAME * * npp_reduce_ineq_coef - reduce inequality constraint coefficients * * SYNOPSIS * * #include "glpnpp.h" * int npp_reduce_ineq_coef(NPP *npp, NPPROW *row); * * DESCRIPTION * * The routine npp_reduce_ineq_coef processes specified inequality * constraint attempting to replace it by an equivalent constraint, * where magnitude of coefficients at binary variables is smaller than * in the original constraint. If the inequality is double-sided, it is * replaced by a pair of single-sided inequalities, if necessary. * * RETURNS * * The routine npp_reduce_ineq_coef returns the number of coefficients * reduced. * * BACKGROUND * * Consider an inequality constraint: * * sum a[j] x[j] >= b. (1) * j in J * * (In case of '<=' inequality it can be transformed to '>=' format by * multiplying both its sides by -1.) Let x[k] be a binary variable; * other variables can be integer as well as continuous. We can write * constraint (1) as follows: * * a[k] x[k] + t[k] >= b, (2) * * where: * * t[k] = sum a[j] x[j]. (3) * j in J\{k} * * Since x[k] is binary, constraint (2) is equivalent to disjunction of * the following two constraints: * * x[k] = 0, t[k] >= b (4) * * OR * * x[k] = 1, t[k] >= b - a[k]. (5) * * Let also that for the partial sum t[k] be known some its implied * lower bound inf t[k]. * * Case a[k] > 0. Let inf t[k] < b, since otherwise both constraints * (4) and (5) and therefore constraint (2) are redundant. * If inf t[k] > b - a[k], only constraint (5) is redundant, in which * case it can be replaced with the following redundant and therefore * equivalent constraint: * * t[k] >= b - a'[k] = inf t[k], (6) * * where: * * a'[k] = b - inf t[k]. (7) * * Thus, the original constraint (2) is equivalent to the following * constraint with coefficient at variable x[k] changed: * * a'[k] x[k] + t[k] >= b. (8) * * From inf t[k] < b it follows that a'[k] > 0, i.e. the coefficient * at x[k] keeps its sign. And from inf t[k] > b - a[k] it follows that * a'[k] < a[k], i.e. the coefficient reduces in magnitude. * * Case a[k] < 0. Let inf t[k] < b - a[k], since otherwise both * constraints (4) and (5) and therefore constraint (2) are redundant. * If inf t[k] > b, only constraint (4) is redundant, in which case it * can be replaced with the following redundant and therefore equivalent * constraint: * * t[k] >= b' = inf t[k]. (9) * * Rewriting constraint (5) as follows: * * t[k] >= b - a[k] = b' - a'[k], (10) * * where: * * a'[k] = a[k] + b' - b = a[k] + inf t[k] - b, (11) * * we can see that disjunction of constraint (9) and (10) is equivalent * to disjunction of constraint (4) and (5), from which it follows that * the original constraint (2) is equivalent to the following constraint * with both coefficient at variable x[k] and right-hand side changed: * * a'[k] x[k] + t[k] >= b'. (12) * * From inf t[k] < b - a[k] it follows that a'[k] < 0, i.e. the * coefficient at x[k] keeps its sign. And from inf t[k] > b it follows * that a'[k] > a[k], i.e. the coefficient reduces in magnitude. * * PROBLEM TRANSFORMATION * * In the routine npp_reduce_ineq_coef the following implied lower * bound of the partial sum (3) is used: * * inf t[k] = sum a[j] l[j] + sum a[j] u[j], (13) * j in Jp\{k} k in Jn\{k} * * where Jp = {j : a[j] > 0}, Jn = {j : a[j] < 0}, l[j] and u[j] are * lower and upper bounds, resp., of variable x[j]. * * In order to compute inf t[k] more efficiently, the following formula, * which is equivalent to (13), is actually used: * * ( h - a[k] l[k] = h, if a[k] > 0, * inf t[k] = < (14) * ( h - a[k] u[k] = h - a[k], if a[k] < 0, * * where: * * h = sum a[j] l[j] + sum a[j] u[j] (15) * j in Jp j in Jn * * is the implied lower bound of row (1). * * Reduction of positive coefficient (a[k] > 0) does not change value * of h, since l[k] = 0. In case of reduction of negative coefficient * (a[k] < 0) from (11) it follows that: * * delta a[k] = a'[k] - a[k] = inf t[k] - b (> 0), (16) * * so new value of h (accounting that u[k] = 1) can be computed as * follows: * * h := h + delta a[k] = h + (inf t[k] - b). (17) * * RECOVERING SOLUTION * * None needed. */ static int reduce_ineq_coef(NPP *npp, struct elem *ptr, double *_b) { /* process inequality constraint: sum a[j] x[j] >= b */ /* returns: the number of coefficients reduced */ struct elem *e; int count = 0; double h, inf_t, new_a, b = *_b; xassert(npp == npp); /* compute h; see (15) */ h = 0.0; for (e = ptr; e != NULL; e = e->next) { if (e->aj > 0.0) { if (e->xj->lb == -DBL_MAX) goto done; h += e->aj * e->xj->lb; } else /* e->aj < 0.0 */ { if (e->xj->ub == +DBL_MAX) goto done; h += e->aj * e->xj->ub; } } /* perform reduction of coefficients at binary variables */ for (e = ptr; e != NULL; e = e->next) { /* skip non-binary variable */ if (!(e->xj->is_int && e->xj->lb == 0.0 && e->xj->ub == 1.0)) continue; if (e->aj > 0.0) { /* compute inf t[k]; see (14) */ inf_t = h; if (b - e->aj < inf_t && inf_t < b) { /* compute reduced coefficient a'[k]; see (7) */ new_a = b - inf_t; if (new_a >= +1e-3 && e->aj - new_a >= 0.01 * (1.0 + e->aj)) { /* accept a'[k] */ #ifdef GLP_DEBUG xprintf("+"); #endif e->aj = new_a; count++; } } } else /* e->aj < 0.0 */ { /* compute inf t[k]; see (14) */ inf_t = h - e->aj; if (b < inf_t && inf_t < b - e->aj) { /* compute reduced coefficient a'[k]; see (11) */ new_a = e->aj + (inf_t - b); if (new_a <= -1e-3 && new_a - e->aj >= 0.01 * (1.0 - e->aj)) { /* accept a'[k] */ #ifdef GLP_DEBUG xprintf("-"); #endif e->aj = new_a; /* update h; see (17) */ h += (inf_t - b); /* compute b'; see (9) */ b = inf_t; count++; } } } } *_b = b; done: return count; } int npp_reduce_ineq_coef(NPP *npp, NPPROW *row) { /* reduce inequality constraint coefficients */ NPPROW *copy; NPPAIJ *aij; struct elem *ptr, *e; int kase, count[2]; double b; /* the row must be inequality constraint */ xassert(row->lb < row->ub); count[0] = count[1] = 0; for (kase = 0; kase <= 1; kase++) { if (kase == 0) { /* process row lower bound */ if (row->lb == -DBL_MAX) continue; #ifdef GLP_DEBUG xprintf("L"); #endif ptr = copy_form(npp, row, +1.0); b = + row->lb; } else { /* process row upper bound */ if (row->ub == +DBL_MAX) continue; #ifdef GLP_DEBUG xprintf("U"); #endif ptr = copy_form(npp, row, -1.0); b = - row->ub; } /* now the inequality has the form "sum a[j] x[j] >= b" */ count[kase] = reduce_ineq_coef(npp, ptr, &b); if (count[kase] > 0) { /* the original inequality has been replaced by equivalent one with coefficients reduced */ if (row->lb == -DBL_MAX || row->ub == +DBL_MAX) { /* the original row is single-sided inequality; no copy is needed */ copy = NULL; } else { /* the original row is double-sided inequality; we need to create its copy for other bound before replacing it with the equivalent inequality */ #ifdef GLP_DEBUG xprintf("*"); #endif copy = npp_add_row(npp); if (kase == 0) { /* the copy is for upper bound */ copy->lb = -DBL_MAX, copy->ub = row->ub; } else { /* the copy is for lower bound */ copy->lb = row->lb, copy->ub = +DBL_MAX; } /* copy original row coefficients */ for (aij = row->ptr; aij != NULL; aij = aij->r_next) npp_add_aij(npp, copy, aij->col, aij->val); } /* replace the original inequality by equivalent one */ npp_erase_row(npp, row); row->lb = b, row->ub = +DBL_MAX; for (e = ptr; e != NULL; e = e->next) npp_add_aij(npp, row, e->xj, e->aj); /* continue processing upper bound for the copy */ if (copy != NULL) row = copy; } drop_form(npp, ptr); } return count[0] + count[1]; } /* eof */