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diff --git a/test/monniaux/glpk-4.65/src/npp/npp4.c b/test/monniaux/glpk-4.65/src/npp/npp4.c new file mode 100644 index 00000000..d7dd0e86 --- /dev/null +++ b/test/monniaux/glpk-4.65/src/npp/npp4.c @@ -0,0 +1,1414 @@ +/* npp4.c */ + +/*********************************************************************** +* This code is part of GLPK (GNU Linear Programming Kit). +* +* Copyright (C) 2009-2017 Andrew Makhorin, Department for Applied +* Informatics, Moscow Aviation Institute, Moscow, Russia. All rights +* reserved. E-mail: <mao@gnu.org>. +* +* 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 <http://www.gnu.org/licenses/>. +***********************************************************************/ + +#include "env.h" +#include "npp.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 */ |