/*************************************************************************
*                                                                       *
* Open Dynamics Engine, Copyright (C) 2001,2002 Russell L. Smith.       *
* All rights reserved.  Email: russ@q12.org   Web: www.q12.org          *
*                                                                       *
* This library is free software; you can redistribute it and/or         *
* modify it under the terms of EITHER:                                  *
*   (1) The GNU Lesser General Public License as published by the Free  *
*       Software Foundation; either version 2.1 of the License, or (at  *
*       your option) any later version. The text of the GNU Lesser      *
*       General Public License is included with this library in the     *
*       file LICENSE.TXT.                                               *
*   (2) The BSD-style license that is included with this library in     *
*       the file LICENSE-BSD.TXT.                                       *
*                                                                       *
* This library 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 files    *
* LICENSE.TXT and LICENSE-BSD.TXT for more details.                     *
*                                                                       *
*************************************************************************/

/*


THE ALGORITHM
-------------

solve A*x = b+w, with x and w subject to certain LCP conditions.
each x(i),w(i) must lie on one of the three line segments in the following
diagram. each line segment corresponds to one index set :

     w(i)
     /|\      |           :
      |       |           :
      |       |i in N     :
  w>0 |       |state[i]=0 :
      |       |           :
      |       |           :  i in C
  w=0 +       +-----------------------+
      |                   :           |
      |                   :           |
  w<0 |                   :           |i in N
      |                   :           |state[i]=1
      |                   :           |
      |                   :           |
      +-------|-----------|-----------|----------> x(i)
             lo           0           hi

the Dantzig algorithm proceeds as follows:
  for i=1:n
    * if (x(i),w(i)) is not on the line, push x(i) and w(i) positive or
      negative towards the line. as this is done, the other (x(j),w(j))
      for j<i are constrained to be on the line. if any (x,w) reaches the
      end of a line segment then it is switched between index sets.
    * i is added to the appropriate index set depending on what line segment
      it hits.

we restrict lo(i) <= 0 and hi(i) >= 0. this makes the algorithm a bit
simpler, because the starting point for x(i),w(i) is always on the dotted
line x=0 and x will only ever increase in one direction, so it can only hit
two out of the three line segments.


NOTES
-----

this is an implementation of "lcp_dantzig2_ldlt.m" and "lcp_dantzig_lohi.m".
the implementation is split into an LCP problem object (dLCP) and an LCP
driver function. most optimization occurs in the dLCP object.

a naive implementation of the algorithm requires either a lot of data motion
or a lot of permutation-array lookup, because we are constantly re-ordering
rows and columns. to avoid this and make a more optimized algorithm, a
non-trivial data structure is used to represent the matrix A (this is
implemented in the fast version of the dLCP object).

during execution of this algorithm, some indexes in A are clamped (set C),
some are non-clamped (set N), and some are "don't care" (where x=0).
A,x,b,w (and other problem vectors) are permuted such that the clamped
indexes are first, the unclamped indexes are next, and the don't-care
indexes are last. this permutation is recorded in the array `p'.
initially p = 0..n-1, and as the rows and columns of A,x,b,w are swapped,
the corresponding elements of p are swapped.

because the C and N elements are grouped together in the rows of A, we can do
lots of work with a fast dot product function. if A,x,etc were not permuted
and we only had a permutation array, then those dot products would be much
slower as we would have a permutation array lookup in some inner loops.

A is accessed through an array of row pointers, so that element (i,j) of the
permuted matrix is A[i][j]. this makes row swapping fast. for column swapping
we still have to actually move the data.

during execution of this algorithm we maintain an L*D*L' factorization of
the clamped submatrix of A (call it `AC') which is the top left nC*nC
submatrix of A. there are two ways we could arrange the rows/columns in AC.

(1) AC is always permuted such that L*D*L' = AC. this causes a problem
when a row/column is removed from C, because then all the rows/columns of A
between the deleted index and the end of C need to be rotated downward.
this results in a lot of data motion and slows things down.
(2) L*D*L' is actually a factorization of a *permutation* of AC (which is
itself a permutation of the underlying A). this is what we do - the
permutation is recorded in the vector C. call this permutation A[C,C].
when a row/column is removed from C, all we have to do is swap two
rows/columns and manipulate C.

*/

#include "dart/external/odelcpsolver/odeconfig.h"
#include "dart/external/odelcpsolver/lcp.h"
#include "dart/external/odelcpsolver/matrix.h"
#include "dart/external/odelcpsolver/misc.h"

//***************************************************************************
// code generation parameters

// LCP debugging (mostly for fast dLCP) - this slows things down a lot
//#define DEBUG_LCP

#define dLCP_FAST		// use fast dLCP object

// option 1 : matrix row pointers (less data copying)
#define ROWPTRS
#define ATYPE dReal **
#define AROW(i) (m_A[i])

// option 2 : no matrix row pointers (slightly faster inner loops)
//#define NOROWPTRS
//#define ATYPE dReal *
//#define AROW(i) (m_A+(i)*m_nskip)

#define NUB_OPTIMIZATIONS

//***************************************************************************

// swap row/column i1 with i2 in the n*n matrix A. the leading dimension of
// A is nskip. this only references and swaps the lower triangle.
// if `do_fast_row_swaps' is nonzero and row pointers are being used, then
// rows will be swapped by exchanging row pointers. otherwise the data will
// be copied.

static void swapRowsAndCols (ATYPE A, int n, int i1, int i2, int nskip, 
  int do_fast_row_swaps)
{
  dAASSERT (A && n > 0 && i1 >= 0 && i2 >= 0 && i1 < n && i2 < n &&
    nskip >= n && i1 < i2);

# ifdef ROWPTRS
  dReal *A_i1 = A[i1];
  dReal *A_i2 = A[i2];
  for (int i=i1+1; i<i2; ++i) {
    dReal *A_i_i1 = A[i] + i1;
    A_i1[i] = *A_i_i1;
    *A_i_i1 = A_i2[i];
  }
  A_i1[i2] = A_i1[i1];
  A_i1[i1] = A_i2[i1];
  A_i2[i1] = A_i2[i2];
  // swap rows, by swapping row pointers
  if (do_fast_row_swaps) {
    A[i1] = A_i2;
    A[i2] = A_i1;
  }
  else {
    // Only swap till i2 column to match A plain storage variant.
    for (int k = 0; k <= i2; ++k) {
      dReal tmp = A_i1[k];
      A_i1[k] = A_i2[k];
      A_i2[k] = tmp;
    }
  }
  // swap columns the hard way
  for (int j=i2+1; j<n; ++j) {
    dReal *A_j = A[j];
    dReal tmp = A_j[i1];
    A_j[i1] = A_j[i2];
    A_j[i2] = tmp;
  }
# else
  dReal *A_i1 = A+i1*nskip;
  dReal *A_i2 = A+i2*nskip;
  for (int k = 0; k < i1; ++k) {
    dReal tmp = A_i1[k];
    A_i1[k] = A_i2[k];
    A_i2[k] = tmp;
  }
  dReal *A_i = A_i1 + nskip;
  for (int i=i1+1; i<i2; A_i+=nskip, ++i) {
    dReal tmp = A_i2[i];
    A_i2[i] = A_i[i1];
    A_i[i1] = tmp;
  }
  {
    dReal tmp = A_i1[i1];
    A_i1[i1] = A_i2[i2];
    A_i2[i2] = tmp;
  }
  dReal *A_j = A_i2 + nskip;
  for (int j=i2+1; j<n; A_j+=nskip, ++j) {
    dReal tmp = A_j[i1];
    A_j[i1] = A_j[i2];
    A_j[i2] = tmp;
  }
# endif
}


// swap two indexes in the n*n LCP problem. i1 must be <= i2.

static void swapProblem (ATYPE A, dReal *x, dReal *b, dReal *w, dReal *lo,
                         dReal *hi, int *p, bool *state, int *findex,
                         int n, int i1, int i2, int nskip,
                         int do_fast_row_swaps)
{
  dReal tmpr;
  int tmpi;
  bool tmpb;
  dIASSERT (n>0 && i1 >=0 && i2 >= 0 && i1 < n && i2 < n && nskip >= n && i1 <= i2);
  if (i1==i2) return;
  
  swapRowsAndCols (A,n,i1,i2,nskip,do_fast_row_swaps);
  
  tmpr = x[i1];
  x[i1] = x[i2];
  x[i2] = tmpr;
  
  tmpr = b[i1];
  b[i1] = b[i2];
  b[i2] = tmpr;
  
  tmpr = w[i1];
  w[i1] = w[i2];
  w[i2] = tmpr;
  
  tmpr = lo[i1];
  lo[i1] = lo[i2];
  lo[i2] = tmpr;

  tmpr = hi[i1];
  hi[i1] = hi[i2];
  hi[i2] = tmpr;

  tmpi = p[i1];
  p[i1] = p[i2];
  p[i2] = tmpi;

  tmpb = state[i1];
  state[i1] = state[i2];
  state[i2] = tmpb;

  if (findex) {
    tmpi = findex[i1];
    findex[i1] = findex[i2];
    findex[i2] = tmpi;
  }
}


// for debugging - check that L,d is the factorization of A[C,C].
// A[C,C] has size nC*nC and leading dimension nskip.
// L has size nC*nC and leading dimension nskip.
// d has size nC.

#ifdef DEBUG_LCP

static void checkFactorization (ATYPE A, dReal *_L, dReal *_d,
                                int nC, int *C, int nskip)
{
  int i,j;
  if (nC==0) return;

  // get A1=A, copy the lower triangle to the upper triangle, get A2=A[C,C]
  dMatrix A1 (nC,nC);
  for (i=0; i<nC; i++) {
    for (j=0; j<=i; j++) A1(i,j) = A1(j,i) = AROW(i)[j];
  }
  dMatrix A2 = A1.select (nC,C,nC,C);

  // printf ("A1=\n"); A1.print(); printf ("\n");
  // printf ("A2=\n"); A2.print(); printf ("\n");

  // compute A3 = L*D*L'
  dMatrix L (nC,nC,_L,nskip,1);
  dMatrix D (nC,nC);
  for (i=0; i<nC; i++) D(i,i) = 1/_d[i];
  L.clearUpperTriangle();
  for (i=0; i<nC; i++) L(i,i) = 1;
  dMatrix A3 = L * D * L.transpose();

  // printf ("L=\n"); L.print(); printf ("\n");
  // printf ("D=\n"); D.print(); printf ("\n");
  // printf ("A3=\n"); A2.print(); printf ("\n");

  // compare A2 and A3
  dReal diff = A2.maxDifference (A3);
  if (diff > 1e-8)
    dDebug (0,"L*D*L' check, maximum difference = %.6e\n",diff);
}

#endif


// for debugging

#ifdef DEBUG_LCP

static void checkPermutations (int i, int n, int nC, int nN, int *p, int *C)
{
  int j,k;
  dIASSERT (nC>=0 && nN>=0 && (nC+nN)==i && i < n);
  for (k=0; k<i; k++) dIASSERT (p[k] >= 0 && p[k] < i);
  for (k=i; k<n; k++) dIASSERT (p[k] == k);
  for (j=0; j<nC; j++) {
    int C_is_bad = 1;
    for (k=0; k<nC; k++) if (C[k]==j) C_is_bad = 0;
    dIASSERT (C_is_bad==0);
  }
}

#endif

//***************************************************************************
// dLCP manipulator object. this represents an n*n LCP problem.
//
// two index sets C and N are kept. each set holds a subset of
// the variable indexes 0..n-1. an index can only be in one set.
// initially both sets are empty.
//
// the index set C is special: solutions to A(C,C)\A(C,i) can be generated.

//***************************************************************************
// fast implementation of dLCP. see the above definition of dLCP for
// interface comments.
//
// `p' records the permutation of A,x,b,w,etc. p is initially 1:n and is
// permuted as the other vectors/matrices are permuted.
//
// A,x,b,w,lo,hi,state,findex,p,c are permuted such that sets C,N have
// contiguous indexes. the don't-care indexes follow N.
//
// an L*D*L' factorization is maintained of A(C,C), and whenever indexes are
// added or removed from the set C the factorization is updated.
// thus L*D*L'=A[C,C], i.e. a permuted top left nC*nC submatrix of A.
// the leading dimension of the matrix L is always `nskip'.
//
// at the start there may be other indexes that are unbounded but are not
// included in `nub'. dLCP will permute the matrix so that absolutely all
// unbounded vectors are at the start. thus there may be some initial
// permutation.
//
// the algorithms here assume certain patterns, particularly with respect to
// index transfer.

#ifdef dLCP_FAST

struct dLCP {
  const int m_n;
  const int m_nskip;
  int m_nub;
  int m_nC, m_nN;				// size of each index set
  ATYPE const m_A;				// A rows
  dReal *const m_x, * const m_b, *const m_w, *const m_lo,* const m_hi;	// permuted LCP problem data
  dReal *const m_L, *const m_d;				// L*D*L' factorization of set C
  dReal *const m_Dell, *const m_ell, *const m_tmp;
  bool *const m_state;
  int *const m_findex, *const m_p, *const m_C;

  dLCP (int _n, int _nskip, int _nub, dReal *_Adata, dReal *_x, dReal *_b, dReal *_w,
    dReal *_lo, dReal *_hi, dReal *_L, dReal *_d,
    dReal *_Dell, dReal *_ell, dReal *_tmp,
    bool *_state, int *_findex, int *_p, int *_C, dReal **Arows);
  int getNub() const { return m_nub; }
  void transfer_i_to_C2 (int i);
  void transfer_i_to_N2 (int /*i*/) { m_nN++; }			// because we can assume C and N span 1:i-1
  void transfer_i_from_N_to_C2 (int i);
  void transfer_i_from_C_to_N2 (int i, void *tmpbuf);
  static size_t estimate_transfer_i_from_C_to_N_mem_req(int nC, int nskip) { return dEstimateLDLTRemoveTmpbufSize(nC, nskip); }
  int numC() const { return m_nC; }
  int numN() const { return m_nN; }
  int indexC (int i) const { return i; }
  int indexN (int i) const { return i+m_nC; }
  dReal Aii (int i) const  { return AROW(i)[i]; }
  dReal AiC_times_qC (int i, dReal *q) const { return dDot (AROW(i), q, m_nC); }
  dReal AiN_times_qN (int i, dReal *q) const { return dDot (AROW(i)+m_nC, q+m_nC, m_nN); }
  void pN_equals_ANC_times_qC (dReal *p, dReal *q);
  void pN_plusequals_ANi (dReal *p, int i, int sign=1);
  void pC_plusequals_s_times_qC (dReal *p, dReal s, dReal *q);
  void pN_plusequals_s_times_qN (dReal *p, dReal s, dReal *q);
  void solve1 (dReal *a, int i, int dir=1, int only_transfer=0);
  void unpermute();
};


dLCP::dLCP (int _n, int _nskip, int _nub, dReal *_Adata, dReal *_x, dReal *_b, dReal *_w,
            dReal *_lo, dReal *_hi, dReal *_L, dReal *_d,
            dReal *_Dell, dReal *_ell, dReal *_tmp,
            bool *_state, int *_findex, int *_p, int *_C, dReal **Arows):
  m_n(_n), m_nskip(_nskip), m_nub(_nub), m_nC(0), m_nN(0),
# ifdef ROWPTRS
  m_A(Arows),
#else
  m_A(_Adata),
#endif
  m_x(_x), m_b(_b), m_w(_w), m_lo(_lo), m_hi(_hi),
  m_L(_L), m_d(_d), m_Dell(_Dell), m_ell(_ell), m_tmp(_tmp),
  m_state(_state), m_findex(_findex), m_p(_p), m_C(_C)
{
  {
    dSetZero (m_x,m_n);
  }

  {
# ifdef ROWPTRS
    // make matrix row pointers
    dReal *aptr = _Adata;
    ATYPE A = m_A;
    const int n = m_n, nskip = m_nskip;
    for (int k=0; k<n; aptr+=nskip, ++k) A[k] = aptr;
# endif
  }

  {
    int *p = m_p;
    const int n = m_n;
    for (int k=0; k<n; ++k) p[k]=k;		// initially unpermuted
  }

  /*
  // for testing, we can do some random swaps in the area i > nub
  {
    const int n = m_n;
    const int nub = m_nub;
    if (nub < n) {
    for (int k=0; k<100; k++) {
      int i1,i2;
      do {
        i1 = dRandInt(n-nub)+nub;
        i2 = dRandInt(n-nub)+nub;
      }
      while (i1 > i2); 
      //printf ("--> %d %d\n",i1,i2);
      swapProblem (m_A,m_x,m_b,m_w,m_lo,m_hi,m_p,m_state,m_findex,n,i1,i2,m_nskip,0);
    }
  }
  */

  // permute the problem so that *all* the unbounded variables are at the
  // start, i.e. look for unbounded variables not included in `nub'. we can
  // potentially push up `nub' this way and get a bigger initial factorization.
  // note that when we swap rows/cols here we must not just swap row pointers,
  // as the initial factorization relies on the data being all in one chunk.
  // variables that have findex >= 0 are *not* considered to be unbounded even
  // if lo=-inf and hi=inf - this is because these limits may change during the
  // solution process.

  {
    int *findex = m_findex;
    dReal *lo = m_lo, *hi = m_hi;
    const int n = m_n;
    for (int k = m_nub; k<n; ++k) {
      if (findex && findex[k] >= 0) continue;
      if (lo[k]==-dInfinity && hi[k]==dInfinity) {
        swapProblem (m_A,m_x,m_b,m_w,lo,hi,m_p,m_state,findex,n,m_nub,k,m_nskip,0);
        m_nub++;
      }
    }
  }

  // if there are unbounded variables at the start, factorize A up to that
  // point and solve for x. this puts all indexes 0..nub-1 into C.
  if (m_nub > 0) {
    const int nub = m_nub;
    {
      dReal *Lrow = m_L;
      const int nskip = m_nskip;
      for (int j=0; j<nub; Lrow+=nskip, ++j) memcpy(Lrow,AROW(j),(j+1)*sizeof(dReal));
    }
    dFactorLDLT (m_L,m_d,nub,m_nskip);
    memcpy (m_x,m_b,nub*sizeof(dReal));
    dSolveLDLT (m_L,m_d,m_x,nub,m_nskip);
    dSetZero (m_w,nub);
    {
      int *C = m_C;
      for (int k=0; k<nub; ++k) C[k] = k;
    }
    m_nC = nub;
  }

  // permute the indexes > nub such that all findex variables are at the end
  if (m_findex) {
    const int nub = m_nub;
    int *findex = m_findex;
    int num_at_end = 0;
    for (int k=m_n-1; k >= nub; k--) {
      if (findex[k] >= 0) {
        swapProblem (m_A,m_x,m_b,m_w,m_lo,m_hi,m_p,m_state,findex,m_n,k,m_n-1-num_at_end,m_nskip,1);
        num_at_end++;
      }
    }
  }

  // print info about indexes
  /*
  {
    const int n = m_n;
    const int nub = m_nub;
    for (int k=0; k<n; k++) {
      if (k<nub) printf ("C");
      else if (m_lo[k]==-dInfinity && m_hi[k]==dInfinity) printf ("c");
      else printf (".");
    }
    printf ("\n");
  }
  */
}


void dLCP::transfer_i_to_C2 (int i)
{
  {
    if (m_nC > 0) {
      // ell,Dell were computed by solve1(). note, ell = D \ L1solve (L,A(i,C))
      {
        const int nC = m_nC;
        dReal *const Ltgt = m_L + nC*m_nskip, *ell = m_ell;
        for (int j=0; j<nC; ++j) Ltgt[j] = ell[j];
      }
      const int nC = m_nC;
      m_d[nC] = dRecip (AROW(i)[i] - dDot(m_ell,m_Dell,nC));
    }
    else {
      m_d[0] = dRecip (AROW(i)[i]);
    }

    swapProblem (m_A,m_x,m_b,m_w,m_lo,m_hi,m_p,m_state,m_findex,m_n,m_nC,i,m_nskip,1);

    const int nC = m_nC;
    m_C[nC] = nC;
    m_nC = nC + 1; // nC value is outdated after this line
  }

# ifdef DEBUG_LCP
  checkFactorization (m_A,m_L,m_d,m_nC,m_C,m_nskip);
  if (i < (m_n-1)) checkPermutations (i+1,m_n,m_nC,m_nN,m_p,m_C);
# endif
}


void dLCP::transfer_i_from_N_to_C2 (int i)
{
  {
    if (m_nC > 0) {
      {
        dReal *const aptr = AROW(i);
        dReal *Dell = m_Dell;
        const int *C = m_C;
#   ifdef NUB_OPTIMIZATIONS
        // if nub>0, initial part of aptr unpermuted
        const int nub = m_nub;
        int j=0;
        for ( ; j<nub; ++j) Dell[j] = aptr[j];
        const int nC = m_nC;
        for ( ; j<nC; ++j) Dell[j] = aptr[C[j]];
#   else
        const int nC = m_nC;
        for (int j=0; j<nC; ++j) Dell[j] = aptr[C[j]];
#   endif
      }
      dSolveL1 (m_L,m_Dell,m_nC,m_nskip);
      {
        const int nC = m_nC;
        dReal *const Ltgt = m_L + nC*m_nskip;
        dReal *ell = m_ell, *Dell = m_Dell, *d = m_d;
        for (int j=0; j<nC; ++j) Ltgt[j] = ell[j] = Dell[j] * d[j];
      }
      const int nC = m_nC;
      m_d[nC] = dRecip (AROW(i)[i] - dDot(m_ell,m_Dell,nC));
    }
    else {
      m_d[0] = dRecip (AROW(i)[i]);
    }

    swapProblem (m_A,m_x,m_b,m_w,m_lo,m_hi,m_p,m_state,m_findex,m_n,m_nC,i,m_nskip,1);

    const int nC = m_nC;
    m_C[nC] = nC;
    m_nN--;
    m_nC = nC + 1; // nC value is outdated after this line
  }

  // @@@ TO DO LATER
  // if we just finish here then we'll go back and re-solve for
  // delta_x. but actually we can be more efficient and incrementally
  // update delta_x here. but if we do this, we wont have ell and Dell
  // to use in updating the factorization later.

# ifdef DEBUG_LCP
  checkFactorization (m_A,m_L,m_d,m_nC,m_C,m_nskip);
# endif
}


void dLCP::transfer_i_from_C_to_N2 (int i, void *tmpbuf)
{
  {
    int *C = m_C;
    // remove a row/column from the factorization, and adjust the
    // indexes (black magic!)
    int last_idx = -1;
    const int nC = m_nC;
    int j = 0;
    for ( ; j<nC; ++j) {
      if (C[j]==nC-1) {
        last_idx = j;
      }
      if (C[j]==i) {
        //printf("HERERERERE\n");
        for (int id = 0; id < j; id++) {
          if (m_d[id] == dReal(0.0))
            //printf("%d, is bad\n", id);
            m_d[id]=1e-10;
        }
        dLDLTRemove (m_A,C,m_L,m_d,m_n,nC,j,m_nskip,tmpbuf);
        int k;
        if (last_idx == -1) {
          for (k=j+1 ; k<nC; ++k) {
            if (C[k]==nC-1) {
              break;
            }
          }
          dIASSERT (k < nC);
        }
        else {
          k = last_idx;
        }
        C[k] = C[j];
        if (j < (nC-1)) memmove (C+j,C+j+1,(nC-j-1)*sizeof(int));
        break;
      }
    }
    dIASSERT (j < nC);

    swapProblem (m_A,m_x,m_b,m_w,m_lo,m_hi,m_p,m_state,m_findex,m_n,i,nC-1,m_nskip,1);

    m_nN++;
    m_nC = nC - 1; // nC value is outdated after this line
  }

# ifdef DEBUG_LCP
  checkFactorization (m_A,m_L,m_d,m_nC,m_C,m_nskip);
# endif
}


void dLCP::pN_equals_ANC_times_qC (dReal *p, dReal *q)
{
  // we could try to make this matrix-vector multiplication faster using
  // outer product matrix tricks, e.g. with the dMultidotX() functions.
  // but i tried it and it actually made things slower on random 100x100
  // problems because of the overhead involved. so we'll stick with the
  // simple method for now.
  const int nC = m_nC;
  dReal *ptgt = p + nC;
  const int nN = m_nN;
  for (int i=0; i<nN; ++i) {
    ptgt[i] = dDot (AROW(i+nC),q,nC);
  }
}


void dLCP::pN_plusequals_ANi (dReal *p, int i, int sign)
{
  const int nC = m_nC;
  dReal *aptr = AROW(i) + nC;
  dReal *ptgt = p + nC;
  if (sign > 0) {
    const int nN = m_nN;
    for (int j=0; j<nN; ++j) ptgt[j] += aptr[j];
  }
  else {
    const int nN = m_nN;
    for (int j=0; j<nN; ++j) ptgt[j] -= aptr[j];
  }
}

void dLCP::pC_plusequals_s_times_qC (dReal *p, dReal s, dReal *q)
{
  const int nC = m_nC;
  for (int i=0; i<nC; ++i) {
    p[i] += s*q[i];
  }
}

void dLCP::pN_plusequals_s_times_qN (dReal *p, dReal s, dReal *q)
{
  const int nC = m_nC;
  dReal *ptgt = p + nC, *qsrc = q + nC;
  const int nN = m_nN;
  for (int i=0; i<nN; ++i) {
    ptgt[i] += s*qsrc[i];
  }
}

void dLCP::solve1 (dReal *a, int i, int dir, int only_transfer)
{
  // the `Dell' and `ell' that are computed here are saved. if index i is
  // later added to the factorization then they can be reused.
  //
  // @@@ question: do we need to solve for entire delta_x??? yes, but
  //     only if an x goes below 0 during the step.

  if (m_nC > 0) {
    {
      dReal *Dell = m_Dell;
      int *C = m_C;
      dReal *aptr = AROW(i);
#   ifdef NUB_OPTIMIZATIONS
      // if nub>0, initial part of aptr[] is guaranteed unpermuted
      const int nub = m_nub;
      int j=0;
      for ( ; j<nub; ++j) Dell[j] = aptr[j];
      const int nC = m_nC;
      for ( ; j<nC; ++j) Dell[j] = aptr[C[j]];
#   else
      const int nC = m_nC;
      for (int j=0; j<nC; ++j) Dell[j] = aptr[C[j]];
#   endif
    }
    dSolveL1 (m_L,m_Dell,m_nC,m_nskip);
    {
      dReal *ell = m_ell, *Dell = m_Dell, *d = m_d;
      const int nC = m_nC;
      for (int j=0; j<nC; ++j) ell[j] = Dell[j] * d[j];
    }

    if (!only_transfer) {
      dReal *tmp = m_tmp, *ell = m_ell;
      {
        const int nC = m_nC;
        for (int j=0; j<nC; ++j) tmp[j] = ell[j];
      }
      dSolveL1T (m_L,tmp,m_nC,m_nskip);
      if (dir > 0) {
        int *C = m_C;
        dReal *tmp = m_tmp;
        const int nC = m_nC;
        for (int j=0; j<nC; ++j) a[C[j]] = -tmp[j];
      } else {
        int *C = m_C;
        dReal *tmp = m_tmp;
        const int nC = m_nC;
        for (int j=0; j<nC; ++j) a[C[j]] = tmp[j];
      }
    }
  }
}


void dLCP::unpermute()
{
  // now we have to un-permute x and w
  {
    memcpy (m_tmp,m_x,m_n*sizeof(dReal));
    dReal *x = m_x, *tmp = m_tmp;
    const int *p = m_p;
    const int n = m_n;
    for (int j=0; j<n; ++j) x[p[j]] = tmp[j];
  }
  {
    memcpy (m_tmp,m_w,m_n*sizeof(dReal));
    dReal *w = m_w, *tmp = m_tmp;
    const int *p = m_p;
    const int n = m_n;
    for (int j=0; j<n; ++j) w[p[j]] = tmp[j];
  }
}

#endif // dLCP_FAST


//***************************************************************************
// an optimized Dantzig LCP driver routine for the lo-hi LCP problem.

void dSolveLCP (int n, dReal *A, dReal *x, dReal *b,
                dReal *outer_w/*=nullptr*/, int nub, dReal *lo, dReal *hi, int *findex)
{
  dAASSERT (n>0 && A && x && b && lo && hi && nub >= 0 && nub <= n);
# ifndef dNODEBUG
  {
    // check restrictions on lo and hi
    for (int k=0; k<n; ++k) dIASSERT (lo[k] <= 0 && hi[k] >= 0);
  }
# endif


  // if all the variables are unbounded then we can just factor, solve,
  // and return
  if (nub >= n) {
    dReal *d = new dReal[n];
    dSetZero (d, n);

    int nskip = dPAD(n);
    dFactorLDLT (A, d, n, nskip);
    dSolveLDLT (A, d, b, n, nskip);
    memcpy (x, b, n*sizeof(dReal));

    return;
  }

  const int nskip = dPAD(n);
  dReal *L = new dReal[ (n*nskip)];
  dReal *d = new dReal[ (n)];
  dReal *w = outer_w ? outer_w : (new dReal[n]);
  dReal *delta_w = new dReal[ (n)];
  dReal *delta_x = new dReal[ (n)];
  dReal *Dell = new dReal[ (n)];
  dReal *ell = new dReal[ (n)];
#ifdef ROWPTRS
  dReal **Arows = new dReal* [n];
#else
  dReal **Arows = nullptr;
#endif
  int *p = new int[n];
  int *C = new int[n];

  // for i in N, state[i] is 0 if x(i)==lo(i) or 1 if x(i)==hi(i)
  bool *state = new bool[n];

  // create LCP object. note that tmp is set to delta_w to save space, this
  // optimization relies on knowledge of how tmp is used, so be careful!
  dLCP lcp(n,nskip,nub,A,x,b,w,lo,hi,L,d,Dell,ell,delta_w,state,findex,p,C,Arows);
  int adj_nub = lcp.getNub();

  // loop over all indexes adj_nub..n-1. for index i, if x(i),w(i) satisfy the
  // LCP conditions then i is added to the appropriate index set. otherwise
  // x(i),w(i) is driven either +ve or -ve to force it to the valid region.
  // as we drive x(i), x(C) is also adjusted to keep w(C) at zero.
  // while driving x(i) we maintain the LCP conditions on the other variables
  // 0..i-1. we do this by watching out for other x(i),w(i) values going
  // outside the valid region, and then switching them between index sets
  // when that happens.

  bool hit_first_friction_index = false;
  for (int i=adj_nub; i<n; ++i) {
    bool s_error = false;
    // the index i is the driving index and indexes i+1..n-1 are "dont care",
    // i.e. when we make changes to the system those x's will be zero and we
    // don't care what happens to those w's. in other words, we only consider
    // an (i+1)*(i+1) sub-problem of A*x=b+w.

    // if we've hit the first friction index, we have to compute the lo and
    // hi values based on the values of x already computed. we have been
    // permuting the indexes, so the values stored in the findex vector are
    // no longer valid. thus we have to temporarily unpermute the x vector. 
    // for the purposes of this computation, 0*infinity = 0 ... so if the
    // contact constraint's normal force is 0, there should be no tangential
    // force applied.

    if (!hit_first_friction_index && findex && findex[i] >= 0) {
      // un-permute x into delta_w, which is not being used at the moment
      for (int j=0; j<n; ++j) delta_w[p[j]] = x[j];

      // set lo and hi values
      for (int k=i; k<n; ++k) {
        dReal wfk = delta_w[findex[k]];
        if (wfk == 0) {
          hi[k] = 0;
          lo[k] = 0;
        }
        else {
          hi[k] = dFabs (hi[k] * wfk);
          lo[k] = -hi[k];
        }
      }
      hit_first_friction_index = true;
    }

    // thus far we have not even been computing the w values for indexes
    // greater than i, so compute w[i] now.
    w[i] = lcp.AiC_times_qC (i,x) + lcp.AiN_times_qN (i,x) - b[i];

    // if lo=hi=0 (which can happen for tangential friction when normals are
    // 0) then the index will be assigned to set N with some state. however,
    // set C's line has zero size, so the index will always remain in set N.
    // with the "normal" switching logic, if w changed sign then the index
    // would have to switch to set C and then back to set N with an inverted
    // state. this is pointless, and also computationally expensive. to
    // prevent this from happening, we use the rule that indexes with lo=hi=0
    // will never be checked for set changes. this means that the state for
    // these indexes may be incorrect, but that doesn't matter.

    // see if x(i),w(i) is in a valid region
    if (lo[i]==0 && w[i] >= 0) {
      lcp.transfer_i_to_N2 (i);
      state[i] = false;
    }
    else if (hi[i]==0 && w[i] <= 0) {
      lcp.transfer_i_to_N2 (i);
      state[i] = true;
    }
    else if (w[i]==0) {
      // this is a degenerate case. by the time we get to this test we know
      // that lo != 0, which means that lo < 0 as lo is not allowed to be +ve,
      // and similarly that hi > 0. this means that the line segment
      // corresponding to set C is at least finite in extent, and we are on it.
      // NOTE: we must call lcp.solve1() before lcp.transfer_i_to_C()
      lcp.solve1 (delta_x,i,0,1);

      lcp.transfer_i_to_C2 (i);
    }
    else {
      // we must push x(i) and w(i)
      for (;;) {
        int dir;
        dReal dirf;
        // find direction to push on x(i)
        if (w[i] <= 0) {
          dir = 1;
          dirf = REAL(1.0);
        }
        else {
          dir = -1;
          dirf = REAL(-1.0);
        }

        // compute: delta_x(C) = -dir*A(C,C)\A(C,i)
        lcp.solve1 (delta_x,i,dir);

        // note that delta_x[i] = dirf, but we wont bother to set it

        // compute: delta_w = A*delta_x ... note we only care about
        // delta_w(N) and delta_w(i), the rest is ignored
        lcp.pN_equals_ANC_times_qC (delta_w,delta_x);
        lcp.pN_plusequals_ANi (delta_w,i,dir);
        delta_w[i] = lcp.AiC_times_qC (i,delta_x) + lcp.Aii(i)*dirf;

        // find largest step we can take (size=s), either to drive x(i),w(i)
        // to the valid LCP region or to drive an already-valid variable
        // outside the valid region.

        int cmd = 1;		// index switching command
        int si = 0;		// si = index to switch if cmd>3
        dReal s = -w[i]/delta_w[i];
        if (dir > 0) {
          if (hi[i] < dInfinity) {
            dReal s2 = (hi[i]-x[i])*dirf;	// was (hi[i]-x[i])/dirf	// step to x(i)=hi(i)
            if (s2 < s) {
              s = s2;
              cmd = 3;
            }
          }
        }
        else {
          if (lo[i] > -dInfinity) {
            dReal s2 = (lo[i]-x[i])*dirf;	// was (lo[i]-x[i])/dirf	// step to x(i)=lo(i)
            if (s2 < s) {
              s = s2;
              cmd = 2;
            }
          }
        }

        {
          const int numN = lcp.numN();
          for (int k=0; k < numN; ++k) {
            const int indexN_k = lcp.indexN(k);
            if (!state[indexN_k] ? delta_w[indexN_k] < 0 : delta_w[indexN_k] > 0) {
                // don't bother checking if lo=hi=0
                if (lo[indexN_k] == 0 && hi[indexN_k] == 0) continue;
                dReal s2 = -w[indexN_k] / delta_w[indexN_k];
                if (s2 < s) {
                  s = s2;
                  cmd = 4;
                  si = indexN_k;
                }
            }
          }
        }

        {
          const int numC = lcp.numC();
          for (int k=adj_nub; k < numC; ++k) {
            const int indexC_k = lcp.indexC(k);
            if (delta_x[indexC_k] < 0 && lo[indexC_k] > -dInfinity) {
              dReal s2 = (lo[indexC_k]-x[indexC_k]) / delta_x[indexC_k];
              if (s2 < s) {
                s = s2;
                cmd = 5;
                si = indexC_k;
              }
            }
            if (delta_x[indexC_k] > 0 && hi[indexC_k] < dInfinity) {
              dReal s2 = (hi[indexC_k]-x[indexC_k]) / delta_x[indexC_k];
              if (s2 < s) {
                s = s2;
                cmd = 6;
                si = indexC_k;
              }
            }
          }
        }

        //static char* cmdstring[8] = {0,"->C","->NL","->NH","N->C",
        //			     "C->NL","C->NH"};
        //printf ("cmd=%d (%s), si=%d\n",cmd,cmdstring[cmd],(cmd>3) ? si : i);

        // if s <= 0 then we've got a problem. if we just keep going then
        // we're going to get stuck in an infinite loop. instead, just cross
        // our fingers and exit with the current solution.
        if (s <= REAL(0.0)) {
          //dMessage (d_ERR_LCP, "LCP internal error, s <= 0 (s=%.4e)",(double)s);
          if (i < n) {
            dSetZero (x+i,n-i);
            dSetZero (w+i,n-i);
          }
          s_error = true;
          break;
        }

        // apply x = x + s * delta_x
        lcp.pC_plusequals_s_times_qC (x, s, delta_x);
        x[i] += s * dirf;

        // apply w = w + s * delta_w
        lcp.pN_plusequals_s_times_qN (w, s, delta_w);
        w[i] += s * delta_w[i];

        // void *tmpbuf;
        // switch indexes between sets if necessary
        switch (cmd) {
        case 1:		// done
          w[i] = 0;
          lcp.transfer_i_to_C2 (i);
          break;
        case 2:		// done
          x[i] = lo[i];
          state[i] = false;
          lcp.transfer_i_to_N2 (i);
          break;
        case 3:		// done
          x[i] = hi[i];
          state[i] = true;
          lcp.transfer_i_to_N2 (i);
          break;
        case 4:		// keep going
          w[si] = 0;
          lcp.transfer_i_from_N_to_C2 (si);
          break;
        case 5:		// keep going
          x[si] = lo[si];
          state[si] = false;
          lcp.transfer_i_from_C_to_N2 (si, nullptr);
          break;
        case 6:		// keep going
          x[si] = hi[si];
          state[si] = true;
          lcp.transfer_i_from_C_to_N2 (si, nullptr);
          break;
        }

        if (cmd <= 3) break;
      } // for (;;)
    } // else

    if (s_error) {
      break;
    }
  } // for (int i=adj_nub; i<n; ++i)

  lcp.unpermute();

  if (!outer_w)
	  delete[] w;
  delete[] L;
  delete[] d;
  delete[] delta_w;
  delete[] delta_x;
  delete[] Dell;
  delete[] ell;
#ifdef ROWPTRS
  delete[] Arows;
#endif
  delete[] p;
  delete[] C;

  delete[] state;
}

size_t dEstimateSolveLCPMemoryReq(int n, bool outer_w_avail)
{
  const int nskip = dPAD(n);

  size_t res = 0;

  res += (sizeof(dReal) * (n * nskip)); // for L
  res += 5 * (sizeof(dReal) * n); // for d, delta_w, delta_x, Dell, ell
  if (!outer_w_avail) {
    res += (sizeof(dReal) * n); // for w
  }
#ifdef ROWPTRS
  res += (sizeof(dReal *) * n); // for Arows
#endif
  res += 2 * (sizeof(int) * n); // for p, C
  res += (sizeof(bool) * n); // for state

  // Use n instead of nC as nC varies at runtime while n is greater or equal to nC
  size_t lcp_transfer_req = dLCP::estimate_transfer_i_from_C_to_N_mem_req(n, nskip);
  res += (lcp_transfer_req); // for dLCP::transfer_i_from_C_to_N

  return res;
}


//***************************************************************************
// accuracy and timing test

//static size_t EstimateTestSolveLCPMemoryReq(int n)
//{
//  const int nskip = dPAD(n);

//  size_t res = 0;

//  res += 2 * (sizeof(dReal) * (n * nskip)); // for A, A2
//  res += 10 * (sizeof(dReal) * n); // for x, b, w, lo, hi, b2, lo2, hi2, tmp1, tmp2

//  res += dEstimateSolveLCPMemoryReq(n, true);

//  return res;
//}

extern "C" ODE_API int dTestSolveLCP()
{
  const int n = 100;

  //size_t memreq = EstimateTestSolveLCPMemoryReq(n);
  //dxWorldProcessMemArena *arena = dxAllocateTemporaryWorldProcessMemArena(memreq, nullptr, nullptr);
  //if (arena == nullptr) {
  //  return 0;
  //}

  //int i,nskip = dPAD(n);
  int i;
  int nskip = n;
#ifdef dDOUBLE
  const dReal tol = REAL(1e-9);
#endif
#ifdef dSINGLE
  const dReal tol = REAL(1e-4);
#endif
  printf ("dTestSolveLCP()\n");

  dReal *A = new dReal[n*nskip];
  dReal *x = new dReal[n];
  dReal *b = new dReal[n];
  dReal *w = new dReal[n];
  dReal *lo = new dReal[n];
  dReal *hi = new dReal[n];
  
  dReal *A2 = new dReal[n*nskip];
  dReal *b2 = new dReal[n];
  dReal *lo2 = new dReal[n];
  dReal *hi2 = new dReal[n];
  
  dReal *tmp1 = new dReal[n];
  dReal *tmp2 = new dReal[n];

  // double total_time = 0;
  for (int count=0; count < 1000; count++) {

      // form (A,b) = a random positive definite LCP problem
      dMakeRandomMatrix (A2,n,n,1.0);
      dMultiply2 (A,A2,A2,n,n,n);
      dMakeRandomMatrix (x,n,1,1.0);
      dMultiply0 (b,A,x,n,n,1);
      for (i=0; i<n; i++) b[i] += (dRandReal()*REAL(0.2))-REAL(0.1);

      // choose `nub' in the range 0..n-1
      int nub = 50; //dRandInt (n);

      // make limits
      for (i=0; i<nub; i++) lo[i] = -dInfinity;
      for (i=0; i<nub; i++) hi[i] = dInfinity;
      //for (i=nub; i<n; i++) lo[i] = 0;
      //for (i=nub; i<n; i++) hi[i] = dInfinity;
      //for (i=nub; i<n; i++) lo[i] = -dInfinity;
      //for (i=nub; i<n; i++) hi[i] = 0;
      for (i=nub; i<n; i++) lo[i] = -(dRandReal()*REAL(1.0))-REAL(0.01);
      for (i=nub; i<n; i++) hi[i] =  (dRandReal()*REAL(1.0))+REAL(0.01);

      // set a few limits to lo=hi=0
      /*
      for (i=0; i<10; i++) {
      int j = dRandInt (n-nub) + nub;
      lo[j] = 0;
      hi[j] = 0;
      }
      */

      // solve the LCP. we must make copy of A,b,lo,hi (A2,b2,lo2,hi2) for
      // SolveLCP() to permute. also, we'll clear the upper triangle of A2 to
      // ensure that it doesn't get referenced (if it does, the answer will be
      // wrong).

      memcpy (A2,A,n*nskip*sizeof(dReal));
      dClearUpperTriangle (A2,n);
      memcpy (b2,b,n*sizeof(dReal));
      memcpy (lo2,lo,n*sizeof(dReal));
      memcpy (hi2,hi,n*sizeof(dReal));
      dSetZero (x,n);
      dSetZero (w,n);

 
      dSolveLCP (n,A2,x,b2,w,nub,lo2,hi2,0);

      // check the solution

      dMultiply0 (tmp1,A,x,n,n,1);
      for (i=0; i<n; i++) tmp2[i] = b[i] + w[i];
      dReal diff = dMaxDifference (tmp1,tmp2,n,1);
      // printf ("\tA*x = b+w, maximum difference = %.6e - %s (1)\n",diff,
      //	    diff > tol ? "FAILED" : "passed");
      if (diff > tol) dDebug (0,"A*x = b+w, maximum difference = %.6e",diff);
      int n1=0,n2=0,n3=0;
      for (i=0; i<n; i++) {
        if (x[i]==lo[i] && w[i] >= 0) {
          n1++;	// ok
        }
        else if (x[i]==hi[i] && w[i] <= 0) {
          n2++;	// ok
        }
        else if (x[i] >= lo[i] && x[i] <= hi[i] && w[i] == 0) {
          n3++;	// ok
        }
        else {
          dDebug (0,"FAILED: i=%d x=%.4e w=%.4e lo=%.4e hi=%.4e",i,
            x[i],w[i],lo[i],hi[i]);
        }
      }

      // pacifier
      printf ("passed: NL=%3d NH=%3d C=%3d   ",n1,n2,n3);
  }
 
  delete[] A;
  delete[] x;
  delete[] b;
  delete[] w;
  delete[] lo ;
  delete[] hi ;

  delete[] A2 ;
  delete[] b2 ;
  delete[] lo2;
  delete[] hi2;
  
  delete[] tmp1;
  delete[] tmp2;

  return 1;
}