/* 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 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 "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;