##### https://github.com/VincentYu68/SymmetryCurriculumLocomotion
Tip revision: b50478f
lcp.cpp
``````/*************************************************************************
*                                                                       *
* Open Dynamics Engine, Copyright (C) 2001,2002 Russell L. Smith.       *
*                                                                       *
* This library is free software; you can redistribute it and/or         *
* modify it under the terms of EITHER:                                  *
*       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     *
*   (2) The BSD-style license that is included with this library in     *
*                                                                       *
* 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    *
*                                                                       *
*************************************************************************/

/*

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++) {
for (k=0; k<nC; k++) if (C[k]==j) 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
//
// `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
#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
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++;
}
}
}

/*
{
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))
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);

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

return;
}

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);

// 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)
{

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;
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;
}
``````