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godot/thirdparty/bullet/BulletDynamics/MLCPSolvers/btDantzigLCP.cpp
Oussama 22b7c9dfa8 Update Bullet to the latest commit 126b676
2019-01-07 12:30:35 +01:00

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/*************************************************************************
* *
* 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 (btLCP) and an LCP
driver function. most optimization occurs in the btLCP 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 btLCP 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 "btDantzigLCP.h"
#include <string.h> //memcpy
bool s_error = false;
//***************************************************************************
// code generation parameters
#define btLCP_FAST // use fast btLCP object
// option 1 : matrix row pointers (less data copying)
#define BTROWPTRS
#define BTATYPE btScalar **
#define BTAROW(i) (m_A[i])
// option 2 : no matrix row pointers (slightly faster inner loops)
//#define NOROWPTRS
//#define BTATYPE btScalar *
//#define BTAROW(i) (m_A+(i)*m_nskip)
#define BTNUB_OPTIMIZATIONS
/* solve L*X=B, with B containing 1 right hand sides.
* L is an n*n lower triangular matrix with ones on the diagonal.
* L is stored by rows and its leading dimension is lskip.
* B is an n*1 matrix that contains the right hand sides.
* B is stored by columns and its leading dimension is also lskip.
* B is overwritten with X.
* this processes blocks of 2*2.
* if this is in the factorizer source file, n must be a multiple of 2.
*/
static void btSolveL1_1(const btScalar *L, btScalar *B, int n, int lskip1)
{
/* declare variables - Z matrix, p and q vectors, etc */
btScalar Z11, m11, Z21, m21, p1, q1, p2, *ex;
const btScalar *ell;
int i, j;
/* compute all 2 x 1 blocks of X */
for (i = 0; i < n; i += 2)
{
/* compute all 2 x 1 block of X, from rows i..i+2-1 */
/* set the Z matrix to 0 */
Z11 = 0;
Z21 = 0;
ell = L + i * lskip1;
ex = B;
/* the inner loop that computes outer products and adds them to Z */
for (j = i - 2; j >= 0; j -= 2)
{
/* compute outer product and add it to the Z matrix */
p1 = ell[0];
q1 = ex[0];
m11 = p1 * q1;
p2 = ell[lskip1];
m21 = p2 * q1;
Z11 += m11;
Z21 += m21;
/* compute outer product and add it to the Z matrix */
p1 = ell[1];
q1 = ex[1];
m11 = p1 * q1;
p2 = ell[1 + lskip1];
m21 = p2 * q1;
/* advance pointers */
ell += 2;
ex += 2;
Z11 += m11;
Z21 += m21;
/* end of inner loop */
}
/* compute left-over iterations */
j += 2;
for (; j > 0; j--)
{
/* compute outer product and add it to the Z matrix */
p1 = ell[0];
q1 = ex[0];
m11 = p1 * q1;
p2 = ell[lskip1];
m21 = p2 * q1;
/* advance pointers */
ell += 1;
ex += 1;
Z11 += m11;
Z21 += m21;
}
/* finish computing the X(i) block */
Z11 = ex[0] - Z11;
ex[0] = Z11;
p1 = ell[lskip1];
Z21 = ex[1] - Z21 - p1 * Z11;
ex[1] = Z21;
/* end of outer loop */
}
}
/* solve L*X=B, with B containing 2 right hand sides.
* L is an n*n lower triangular matrix with ones on the diagonal.
* L is stored by rows and its leading dimension is lskip.
* B is an n*2 matrix that contains the right hand sides.
* B is stored by columns and its leading dimension is also lskip.
* B is overwritten with X.
* this processes blocks of 2*2.
* if this is in the factorizer source file, n must be a multiple of 2.
*/
static void btSolveL1_2(const btScalar *L, btScalar *B, int n, int lskip1)
{
/* declare variables - Z matrix, p and q vectors, etc */
btScalar Z11, m11, Z12, m12, Z21, m21, Z22, m22, p1, q1, p2, q2, *ex;
const btScalar *ell;
int i, j;
/* compute all 2 x 2 blocks of X */
for (i = 0; i < n; i += 2)
{
/* compute all 2 x 2 block of X, from rows i..i+2-1 */
/* set the Z matrix to 0 */
Z11 = 0;
Z12 = 0;
Z21 = 0;
Z22 = 0;
ell = L + i * lskip1;
ex = B;
/* the inner loop that computes outer products and adds them to Z */
for (j = i - 2; j >= 0; j -= 2)
{
/* compute outer product and add it to the Z matrix */
p1 = ell[0];
q1 = ex[0];
m11 = p1 * q1;
q2 = ex[lskip1];
m12 = p1 * q2;
p2 = ell[lskip1];
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z12 += m12;
Z21 += m21;
Z22 += m22;
/* compute outer product and add it to the Z matrix */
p1 = ell[1];
q1 = ex[1];
m11 = p1 * q1;
q2 = ex[1 + lskip1];
m12 = p1 * q2;
p2 = ell[1 + lskip1];
m21 = p2 * q1;
m22 = p2 * q2;
/* advance pointers */
ell += 2;
ex += 2;
Z11 += m11;
Z12 += m12;
Z21 += m21;
Z22 += m22;
/* end of inner loop */
}
/* compute left-over iterations */
j += 2;
for (; j > 0; j--)
{
/* compute outer product and add it to the Z matrix */
p1 = ell[0];
q1 = ex[0];
m11 = p1 * q1;
q2 = ex[lskip1];
m12 = p1 * q2;
p2 = ell[lskip1];
m21 = p2 * q1;
m22 = p2 * q2;
/* advance pointers */
ell += 1;
ex += 1;
Z11 += m11;
Z12 += m12;
Z21 += m21;
Z22 += m22;
}
/* finish computing the X(i) block */
Z11 = ex[0] - Z11;
ex[0] = Z11;
Z12 = ex[lskip1] - Z12;
ex[lskip1] = Z12;
p1 = ell[lskip1];
Z21 = ex[1] - Z21 - p1 * Z11;
ex[1] = Z21;
Z22 = ex[1 + lskip1] - Z22 - p1 * Z12;
ex[1 + lskip1] = Z22;
/* end of outer loop */
}
}
void btFactorLDLT(btScalar *A, btScalar *d, int n, int nskip1)
{
int i, j;
btScalar sum, *ell, *dee, dd, p1, p2, q1, q2, Z11, m11, Z21, m21, Z22, m22;
if (n < 1) return;
for (i = 0; i <= n - 2; i += 2)
{
/* solve L*(D*l)=a, l is scaled elements in 2 x i block at A(i,0) */
btSolveL1_2(A, A + i * nskip1, i, nskip1);
/* scale the elements in a 2 x i block at A(i,0), and also */
/* compute Z = the outer product matrix that we'll need. */
Z11 = 0;
Z21 = 0;
Z22 = 0;
ell = A + i * nskip1;
dee = d;
for (j = i - 6; j >= 0; j -= 6)
{
p1 = ell[0];
p2 = ell[nskip1];
dd = dee[0];
q1 = p1 * dd;
q2 = p2 * dd;
ell[0] = q1;
ell[nskip1] = q2;
m11 = p1 * q1;
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z21 += m21;
Z22 += m22;
p1 = ell[1];
p2 = ell[1 + nskip1];
dd = dee[1];
q1 = p1 * dd;
q2 = p2 * dd;
ell[1] = q1;
ell[1 + nskip1] = q2;
m11 = p1 * q1;
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z21 += m21;
Z22 += m22;
p1 = ell[2];
p2 = ell[2 + nskip1];
dd = dee[2];
q1 = p1 * dd;
q2 = p2 * dd;
ell[2] = q1;
ell[2 + nskip1] = q2;
m11 = p1 * q1;
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z21 += m21;
Z22 += m22;
p1 = ell[3];
p2 = ell[3 + nskip1];
dd = dee[3];
q1 = p1 * dd;
q2 = p2 * dd;
ell[3] = q1;
ell[3 + nskip1] = q2;
m11 = p1 * q1;
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z21 += m21;
Z22 += m22;
p1 = ell[4];
p2 = ell[4 + nskip1];
dd = dee[4];
q1 = p1 * dd;
q2 = p2 * dd;
ell[4] = q1;
ell[4 + nskip1] = q2;
m11 = p1 * q1;
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z21 += m21;
Z22 += m22;
p1 = ell[5];
p2 = ell[5 + nskip1];
dd = dee[5];
q1 = p1 * dd;
q2 = p2 * dd;
ell[5] = q1;
ell[5 + nskip1] = q2;
m11 = p1 * q1;
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z21 += m21;
Z22 += m22;
ell += 6;
dee += 6;
}
/* compute left-over iterations */
j += 6;
for (; j > 0; j--)
{
p1 = ell[0];
p2 = ell[nskip1];
dd = dee[0];
q1 = p1 * dd;
q2 = p2 * dd;
ell[0] = q1;
ell[nskip1] = q2;
m11 = p1 * q1;
m21 = p2 * q1;
m22 = p2 * q2;
Z11 += m11;
Z21 += m21;
Z22 += m22;
ell++;
dee++;
}
/* solve for diagonal 2 x 2 block at A(i,i) */
Z11 = ell[0] - Z11;
Z21 = ell[nskip1] - Z21;
Z22 = ell[1 + nskip1] - Z22;
dee = d + i;
/* factorize 2 x 2 block Z,dee */
/* factorize row 1 */
dee[0] = btRecip(Z11);
/* factorize row 2 */
sum = 0;
q1 = Z21;
q2 = q1 * dee[0];
Z21 = q2;
sum += q1 * q2;
dee[1] = btRecip(Z22 - sum);
/* done factorizing 2 x 2 block */
ell[nskip1] = Z21;
}
/* compute the (less than 2) rows at the bottom */
switch (n - i)
{
case 0:
break;
case 1:
btSolveL1_1(A, A + i * nskip1, i, nskip1);
/* scale the elements in a 1 x i block at A(i,0), and also */
/* compute Z = the outer product matrix that we'll need. */
Z11 = 0;
ell = A + i * nskip1;
dee = d;
for (j = i - 6; j >= 0; j -= 6)
{
p1 = ell[0];
dd = dee[0];
q1 = p1 * dd;
ell[0] = q1;
m11 = p1 * q1;
Z11 += m11;
p1 = ell[1];
dd = dee[1];
q1 = p1 * dd;
ell[1] = q1;
m11 = p1 * q1;
Z11 += m11;
p1 = ell[2];
dd = dee[2];
q1 = p1 * dd;
ell[2] = q1;
m11 = p1 * q1;
Z11 += m11;
p1 = ell[3];
dd = dee[3];
q1 = p1 * dd;
ell[3] = q1;
m11 = p1 * q1;
Z11 += m11;
p1 = ell[4];
dd = dee[4];
q1 = p1 * dd;
ell[4] = q1;
m11 = p1 * q1;
Z11 += m11;
p1 = ell[5];
dd = dee[5];
q1 = p1 * dd;
ell[5] = q1;
m11 = p1 * q1;
Z11 += m11;
ell += 6;
dee += 6;
}
/* compute left-over iterations */
j += 6;
for (; j > 0; j--)
{
p1 = ell[0];
dd = dee[0];
q1 = p1 * dd;
ell[0] = q1;
m11 = p1 * q1;
Z11 += m11;
ell++;
dee++;
}
/* solve for diagonal 1 x 1 block at A(i,i) */
Z11 = ell[0] - Z11;
dee = d + i;
/* factorize 1 x 1 block Z,dee */
/* factorize row 1 */
dee[0] = btRecip(Z11);
/* done factorizing 1 x 1 block */
break;
//default: *((char*)0)=0; /* this should never happen! */
}
}
/* solve L*X=B, with B containing 1 right hand sides.
* L is an n*n lower triangular matrix with ones on the diagonal.
* L is stored by rows and its leading dimension is lskip.
* B is an n*1 matrix that contains the right hand sides.
* B is stored by columns and its leading dimension is also lskip.
* B is overwritten with X.
* this processes blocks of 4*4.
* if this is in the factorizer source file, n must be a multiple of 4.
*/
void btSolveL1(const btScalar *L, btScalar *B, int n, int lskip1)
{
/* declare variables - Z matrix, p and q vectors, etc */
btScalar Z11, Z21, Z31, Z41, p1, q1, p2, p3, p4, *ex;
const btScalar *ell;
int lskip2, lskip3, i, j;
/* compute lskip values */
lskip2 = 2 * lskip1;
lskip3 = 3 * lskip1;
/* compute all 4 x 1 blocks of X */
for (i = 0; i <= n - 4; i += 4)
{
/* compute all 4 x 1 block of X, from rows i..i+4-1 */
/* set the Z matrix to 0 */
Z11 = 0;
Z21 = 0;
Z31 = 0;
Z41 = 0;
ell = L + i * lskip1;
ex = B;
/* the inner loop that computes outer products and adds them to Z */
for (j = i - 12; j >= 0; j -= 12)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
p2 = ell[lskip1];
p3 = ell[lskip2];
p4 = ell[lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[1];
q1 = ex[1];
p2 = ell[1 + lskip1];
p3 = ell[1 + lskip2];
p4 = ell[1 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[2];
q1 = ex[2];
p2 = ell[2 + lskip1];
p3 = ell[2 + lskip2];
p4 = ell[2 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[3];
q1 = ex[3];
p2 = ell[3 + lskip1];
p3 = ell[3 + lskip2];
p4 = ell[3 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[4];
q1 = ex[4];
p2 = ell[4 + lskip1];
p3 = ell[4 + lskip2];
p4 = ell[4 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[5];
q1 = ex[5];
p2 = ell[5 + lskip1];
p3 = ell[5 + lskip2];
p4 = ell[5 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[6];
q1 = ex[6];
p2 = ell[6 + lskip1];
p3 = ell[6 + lskip2];
p4 = ell[6 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[7];
q1 = ex[7];
p2 = ell[7 + lskip1];
p3 = ell[7 + lskip2];
p4 = ell[7 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[8];
q1 = ex[8];
p2 = ell[8 + lskip1];
p3 = ell[8 + lskip2];
p4 = ell[8 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[9];
q1 = ex[9];
p2 = ell[9 + lskip1];
p3 = ell[9 + lskip2];
p4 = ell[9 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[10];
q1 = ex[10];
p2 = ell[10 + lskip1];
p3 = ell[10 + lskip2];
p4 = ell[10 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* load p and q values */
p1 = ell[11];
q1 = ex[11];
p2 = ell[11 + lskip1];
p3 = ell[11 + lskip2];
p4 = ell[11 + lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* advance pointers */
ell += 12;
ex += 12;
/* end of inner loop */
}
/* compute left-over iterations */
j += 12;
for (; j > 0; j--)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
p2 = ell[lskip1];
p3 = ell[lskip2];
p4 = ell[lskip3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
Z21 += p2 * q1;
Z31 += p3 * q1;
Z41 += p4 * q1;
/* advance pointers */
ell += 1;
ex += 1;
}
/* finish computing the X(i) block */
Z11 = ex[0] - Z11;
ex[0] = Z11;
p1 = ell[lskip1];
Z21 = ex[1] - Z21 - p1 * Z11;
ex[1] = Z21;
p1 = ell[lskip2];
p2 = ell[1 + lskip2];
Z31 = ex[2] - Z31 - p1 * Z11 - p2 * Z21;
ex[2] = Z31;
p1 = ell[lskip3];
p2 = ell[1 + lskip3];
p3 = ell[2 + lskip3];
Z41 = ex[3] - Z41 - p1 * Z11 - p2 * Z21 - p3 * Z31;
ex[3] = Z41;
/* end of outer loop */
}
/* compute rows at end that are not a multiple of block size */
for (; i < n; i++)
{
/* compute all 1 x 1 block of X, from rows i..i+1-1 */
/* set the Z matrix to 0 */
Z11 = 0;
ell = L + i * lskip1;
ex = B;
/* the inner loop that computes outer products and adds them to Z */
for (j = i - 12; j >= 0; j -= 12)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[1];
q1 = ex[1];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[2];
q1 = ex[2];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[3];
q1 = ex[3];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[4];
q1 = ex[4];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[5];
q1 = ex[5];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[6];
q1 = ex[6];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[7];
q1 = ex[7];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[8];
q1 = ex[8];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[9];
q1 = ex[9];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[10];
q1 = ex[10];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* load p and q values */
p1 = ell[11];
q1 = ex[11];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* advance pointers */
ell += 12;
ex += 12;
/* end of inner loop */
}
/* compute left-over iterations */
j += 12;
for (; j > 0; j--)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
/* compute outer product and add it to the Z matrix */
Z11 += p1 * q1;
/* advance pointers */
ell += 1;
ex += 1;
}
/* finish computing the X(i) block */
Z11 = ex[0] - Z11;
ex[0] = Z11;
}
}
/* solve L^T * x=b, with b containing 1 right hand side.
* L is an n*n lower triangular matrix with ones on the diagonal.
* L is stored by rows and its leading dimension is lskip.
* b is an n*1 matrix that contains the right hand side.
* b is overwritten with x.
* this processes blocks of 4.
*/
void btSolveL1T(const btScalar *L, btScalar *B, int n, int lskip1)
{
/* declare variables - Z matrix, p and q vectors, etc */
btScalar Z11, m11, Z21, m21, Z31, m31, Z41, m41, p1, q1, p2, p3, p4, *ex;
const btScalar *ell;
int lskip2, i, j;
// int lskip3;
/* special handling for L and B because we're solving L1 *transpose* */
L = L + (n - 1) * (lskip1 + 1);
B = B + n - 1;
lskip1 = -lskip1;
/* compute lskip values */
lskip2 = 2 * lskip1;
//lskip3 = 3*lskip1;
/* compute all 4 x 1 blocks of X */
for (i = 0; i <= n - 4; i += 4)
{
/* compute all 4 x 1 block of X, from rows i..i+4-1 */
/* set the Z matrix to 0 */
Z11 = 0;
Z21 = 0;
Z31 = 0;
Z41 = 0;
ell = L - i;
ex = B;
/* the inner loop that computes outer products and adds them to Z */
for (j = i - 4; j >= 0; j -= 4)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
p2 = ell[-1];
p3 = ell[-2];
p4 = ell[-3];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
m21 = p2 * q1;
m31 = p3 * q1;
m41 = p4 * q1;
ell += lskip1;
Z11 += m11;
Z21 += m21;
Z31 += m31;
Z41 += m41;
/* load p and q values */
p1 = ell[0];
q1 = ex[-1];
p2 = ell[-1];
p3 = ell[-2];
p4 = ell[-3];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
m21 = p2 * q1;
m31 = p3 * q1;
m41 = p4 * q1;
ell += lskip1;
Z11 += m11;
Z21 += m21;
Z31 += m31;
Z41 += m41;
/* load p and q values */
p1 = ell[0];
q1 = ex[-2];
p2 = ell[-1];
p3 = ell[-2];
p4 = ell[-3];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
m21 = p2 * q1;
m31 = p3 * q1;
m41 = p4 * q1;
ell += lskip1;
Z11 += m11;
Z21 += m21;
Z31 += m31;
Z41 += m41;
/* load p and q values */
p1 = ell[0];
q1 = ex[-3];
p2 = ell[-1];
p3 = ell[-2];
p4 = ell[-3];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
m21 = p2 * q1;
m31 = p3 * q1;
m41 = p4 * q1;
ell += lskip1;
ex -= 4;
Z11 += m11;
Z21 += m21;
Z31 += m31;
Z41 += m41;
/* end of inner loop */
}
/* compute left-over iterations */
j += 4;
for (; j > 0; j--)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
p2 = ell[-1];
p3 = ell[-2];
p4 = ell[-3];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
m21 = p2 * q1;
m31 = p3 * q1;
m41 = p4 * q1;
ell += lskip1;
ex -= 1;
Z11 += m11;
Z21 += m21;
Z31 += m31;
Z41 += m41;
}
/* finish computing the X(i) block */
Z11 = ex[0] - Z11;
ex[0] = Z11;
p1 = ell[-1];
Z21 = ex[-1] - Z21 - p1 * Z11;
ex[-1] = Z21;
p1 = ell[-2];
p2 = ell[-2 + lskip1];
Z31 = ex[-2] - Z31 - p1 * Z11 - p2 * Z21;
ex[-2] = Z31;
p1 = ell[-3];
p2 = ell[-3 + lskip1];
p3 = ell[-3 + lskip2];
Z41 = ex[-3] - Z41 - p1 * Z11 - p2 * Z21 - p3 * Z31;
ex[-3] = Z41;
/* end of outer loop */
}
/* compute rows at end that are not a multiple of block size */
for (; i < n; i++)
{
/* compute all 1 x 1 block of X, from rows i..i+1-1 */
/* set the Z matrix to 0 */
Z11 = 0;
ell = L - i;
ex = B;
/* the inner loop that computes outer products and adds them to Z */
for (j = i - 4; j >= 0; j -= 4)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
ell += lskip1;
Z11 += m11;
/* load p and q values */
p1 = ell[0];
q1 = ex[-1];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
ell += lskip1;
Z11 += m11;
/* load p and q values */
p1 = ell[0];
q1 = ex[-2];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
ell += lskip1;
Z11 += m11;
/* load p and q values */
p1 = ell[0];
q1 = ex[-3];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
ell += lskip1;
ex -= 4;
Z11 += m11;
/* end of inner loop */
}
/* compute left-over iterations */
j += 4;
for (; j > 0; j--)
{
/* load p and q values */
p1 = ell[0];
q1 = ex[0];
/* compute outer product and add it to the Z matrix */
m11 = p1 * q1;
ell += lskip1;
ex -= 1;
Z11 += m11;
}
/* finish computing the X(i) block */
Z11 = ex[0] - Z11;
ex[0] = Z11;
}
}
void btVectorScale(btScalar *a, const btScalar *d, int n)
{
btAssert(a && d && n >= 0);
for (int i = 0; i < n; i++)
{
a[i] *= d[i];
}
}
void btSolveLDLT(const btScalar *L, const btScalar *d, btScalar *b, int n, int nskip)
{
btAssert(L && d && b && n > 0 && nskip >= n);
btSolveL1(L, b, n, nskip);
btVectorScale(b, d, n);
btSolveL1T(L, b, n, nskip);
}
//***************************************************************************
// 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 btSwapRowsAndCols(BTATYPE A, int n, int i1, int i2, int nskip,
int do_fast_row_swaps)
{
btAssert(A && n > 0 && i1 >= 0 && i2 >= 0 && i1 < n && i2 < n &&
nskip >= n && i1 < i2);
#ifdef BTROWPTRS
btScalar *A_i1 = A[i1];
btScalar *A_i2 = A[i2];
for (int i = i1 + 1; i < i2; ++i)
{
btScalar *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)
{
btScalar 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)
{
btScalar *A_j = A[j];
btScalar tmp = A_j[i1];
A_j[i1] = A_j[i2];
A_j[i2] = tmp;
}
#else
btScalar *A_i1 = A + i1 * nskip;
btScalar *A_i2 = A + i2 * nskip;
for (int k = 0; k < i1; ++k)
{
btScalar tmp = A_i1[k];
A_i1[k] = A_i2[k];
A_i2[k] = tmp;
}
btScalar *A_i = A_i1 + nskip;
for (int i = i1 + 1; i < i2; A_i += nskip, ++i)
{
btScalar tmp = A_i2[i];
A_i2[i] = A_i[i1];
A_i[i1] = tmp;
}
{
btScalar tmp = A_i1[i1];
A_i1[i1] = A_i2[i2];
A_i2[i2] = tmp;
}
btScalar *A_j = A_i2 + nskip;
for (int j = i2 + 1; j < n; A_j += nskip, ++j)
{
btScalar 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 btSwapProblem(BTATYPE A, btScalar *x, btScalar *b, btScalar *w, btScalar *lo,
btScalar *hi, int *p, bool *state, int *findex,
int n, int i1, int i2, int nskip,
int do_fast_row_swaps)
{
btScalar tmpr;
int tmpi;
bool tmpb;
btAssert(n > 0 && i1 >= 0 && i2 >= 0 && i1 < n && i2 < n && nskip >= n && i1 <= i2);
if (i1 == i2) return;
btSwapRowsAndCols(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;
}
}
//***************************************************************************
// btLCP 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 btLCP. see the above definition of btLCP 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'. btLCP 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 btLCP_FAST
struct btLCP
{
const int m_n;
const int m_nskip;
int m_nub;
int m_nC, m_nN; // size of each index set
BTATYPE const m_A; // A rows
btScalar *const m_x, *const m_b, *const m_w, *const m_lo, *const m_hi; // permuted LCP problem data
btScalar *const m_L, *const m_d; // L*D*L' factorization of set C
btScalar *const m_Dell, *const m_ell, *const m_tmp;
bool *const m_state;
int *const m_findex, *const m_p, *const m_C;
btLCP(int _n, int _nskip, int _nub, btScalar *_Adata, btScalar *_x, btScalar *_b, btScalar *_w,
btScalar *_lo, btScalar *_hi, btScalar *l, btScalar *_d,
btScalar *_Dell, btScalar *_ell, btScalar *_tmp,
bool *_state, int *_findex, int *p, int *c, btScalar **Arows);
int getNub() const { return m_nub; }
void transfer_i_to_C(int i);
void transfer_i_to_N(int i) { m_nN++; } // because we can assume C and N span 1:i-1
void transfer_i_from_N_to_C(int i);
void transfer_i_from_C_to_N(int i, btAlignedObjectArray<btScalar> &scratch);
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; }
btScalar Aii(int i) const { return BTAROW(i)[i]; }
btScalar AiC_times_qC(int i, btScalar *q) const { return btLargeDot(BTAROW(i), q, m_nC); }
btScalar AiN_times_qN(int i, btScalar *q) const { return btLargeDot(BTAROW(i) + m_nC, q + m_nC, m_nN); }
void pN_equals_ANC_times_qC(btScalar *p, btScalar *q);
void pN_plusequals_ANi(btScalar *p, int i, int sign = 1);
void pC_plusequals_s_times_qC(btScalar *p, btScalar s, btScalar *q);
void pN_plusequals_s_times_qN(btScalar *p, btScalar s, btScalar *q);
void solve1(btScalar *a, int i, int dir = 1, int only_transfer = 0);
void unpermute();
};
btLCP::btLCP(int _n, int _nskip, int _nub, btScalar *_Adata, btScalar *_x, btScalar *_b, btScalar *_w,
btScalar *_lo, btScalar *_hi, btScalar *l, btScalar *_d,
btScalar *_Dell, btScalar *_ell, btScalar *_tmp,
bool *_state, int *_findex, int *p, int *c, btScalar **Arows) : m_n(_n), m_nskip(_nskip), m_nub(_nub), m_nC(0), m_nN(0),
#ifdef BTROWPTRS
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)
{
{
btSetZero(m_x, m_n);
}
{
#ifdef BTROWPTRS
// make matrix row pointers
btScalar *aptr = _Adata;
BTATYPE 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);
btSwapProblem (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;
btScalar *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] == -BT_INFINITY && hi[k] == BT_INFINITY)
{
btSwapProblem(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;
{
btScalar *Lrow = m_L;
const int nskip = m_nskip;
for (int j = 0; j < nub; Lrow += nskip, ++j) memcpy(Lrow, BTAROW(j), (j + 1) * sizeof(btScalar));
}
btFactorLDLT(m_L, m_d, nub, m_nskip);
memcpy(m_x, m_b, nub * sizeof(btScalar));
btSolveLDLT(m_L, m_d, m_x, nub, m_nskip);
btSetZero(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)
{
btSwapProblem(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]==-BT_INFINITY && m_hi[k]==BT_INFINITY) printf ("c");
else printf (".");
}
printf ("\n");
}
*/
}
void btLCP::transfer_i_to_C(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;
btScalar *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] = btRecip(BTAROW(i)[i] - btLargeDot(m_ell, m_Dell, nC));
}
else
{
m_d[0] = btRecip(BTAROW(i)[i]);
}
btSwapProblem(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
}
}
void btLCP::transfer_i_from_N_to_C(int i)
{
{
if (m_nC > 0)
{
{
btScalar *const aptr = BTAROW(i);
btScalar *Dell = m_Dell;
const int *C = m_C;
#ifdef BTNUB_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
}
btSolveL1(m_L, m_Dell, m_nC, m_nskip);
{
const int nC = m_nC;
btScalar *const Ltgt = m_L + nC * m_nskip;
btScalar *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] = btRecip(BTAROW(i)[i] - btLargeDot(m_ell, m_Dell, nC));
}
else
{
m_d[0] = btRecip(BTAROW(i)[i]);
}
btSwapProblem(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.
}
void btRemoveRowCol(btScalar *A, int n, int nskip, int r)
{
btAssert(A && n > 0 && nskip >= n && r >= 0 && r < n);
if (r >= n - 1) return;
if (r > 0)
{
{
const size_t move_size = (n - r - 1) * sizeof(btScalar);
btScalar *Adst = A + r;
for (int i = 0; i < r; Adst += nskip, ++i)
{
btScalar *Asrc = Adst + 1;
memmove(Adst, Asrc, move_size);
}
}
{
const size_t cpy_size = r * sizeof(btScalar);
btScalar *Adst = A + r * nskip;
for (int i = r; i < (n - 1); ++i)
{
btScalar *Asrc = Adst + nskip;
memcpy(Adst, Asrc, cpy_size);
Adst = Asrc;
}
}
}
{
const size_t cpy_size = (n - r - 1) * sizeof(btScalar);
btScalar *Adst = A + r * (nskip + 1);
for (int i = r; i < (n - 1); ++i)
{
btScalar *Asrc = Adst + (nskip + 1);
memcpy(Adst, Asrc, cpy_size);
Adst = Asrc - 1;
}
}
}
void btLDLTAddTL(btScalar *L, btScalar *d, const btScalar *a, int n, int nskip, btAlignedObjectArray<btScalar> &scratch)
{
btAssert(L && d && a && n > 0 && nskip >= n);
if (n < 2) return;
scratch.resize(2 * nskip);
btScalar *W1 = &scratch[0];
btScalar *W2 = W1 + nskip;
W1[0] = btScalar(0.0);
W2[0] = btScalar(0.0);
for (int j = 1; j < n; ++j)
{
W1[j] = W2[j] = (btScalar)(a[j] * SIMDSQRT12);
}
btScalar W11 = (btScalar)((btScalar(0.5) * a[0] + 1) * SIMDSQRT12);
btScalar W21 = (btScalar)((btScalar(0.5) * a[0] - 1) * SIMDSQRT12);
btScalar alpha1 = btScalar(1.0);
btScalar alpha2 = btScalar(1.0);
{
btScalar dee = d[0];
btScalar alphanew = alpha1 + (W11 * W11) * dee;
btAssert(alphanew != btScalar(0.0));
dee /= alphanew;
btScalar gamma1 = W11 * dee;
dee *= alpha1;
alpha1 = alphanew;
alphanew = alpha2 - (W21 * W21) * dee;
dee /= alphanew;
//btScalar gamma2 = W21 * dee;
alpha2 = alphanew;
btScalar k1 = btScalar(1.0) - W21 * gamma1;
btScalar k2 = W21 * gamma1 * W11 - W21;
btScalar *ll = L + nskip;
for (int p = 1; p < n; ll += nskip, ++p)
{
btScalar Wp = W1[p];
btScalar ell = *ll;
W1[p] = Wp - W11 * ell;
W2[p] = k1 * Wp + k2 * ell;
}
}
btScalar *ll = L + (nskip + 1);
for (int j = 1; j < n; ll += nskip + 1, ++j)
{
btScalar k1 = W1[j];
btScalar k2 = W2[j];
btScalar dee = d[j];
btScalar alphanew = alpha1 + (k1 * k1) * dee;
btAssert(alphanew != btScalar(0.0));
dee /= alphanew;
btScalar gamma1 = k1 * dee;
dee *= alpha1;
alpha1 = alphanew;
alphanew = alpha2 - (k2 * k2) * dee;
dee /= alphanew;
btScalar gamma2 = k2 * dee;
dee *= alpha2;
d[j] = dee;
alpha2 = alphanew;
btScalar *l = ll + nskip;
for (int p = j + 1; p < n; l += nskip, ++p)
{
btScalar ell = *l;
btScalar Wp = W1[p] - k1 * ell;
ell += gamma1 * Wp;
W1[p] = Wp;
Wp = W2[p] - k2 * ell;
ell -= gamma2 * Wp;
W2[p] = Wp;
*l = ell;
}
}
}
#define _BTGETA(i, j) (A[i][j])
//#define _GETA(i,j) (A[(i)*nskip+(j)])
#define BTGETA(i, j) ((i > j) ? _BTGETA(i, j) : _BTGETA(j, i))
inline size_t btEstimateLDLTAddTLTmpbufSize(int nskip)
{
return nskip * 2 * sizeof(btScalar);
}
void btLDLTRemove(btScalar **A, const int *p, btScalar *L, btScalar *d,
int n1, int n2, int r, int nskip, btAlignedObjectArray<btScalar> &scratch)
{
btAssert(A && p && L && d && n1 > 0 && n2 > 0 && r >= 0 && r < n2 &&
n1 >= n2 && nskip >= n1);
#ifdef BT_DEBUG
for (int i = 0; i < n2; ++i)
btAssert(p[i] >= 0 && p[i] < n1);
#endif
if (r == n2 - 1)
{
return; // deleting last row/col is easy
}
else
{
size_t LDLTAddTL_size = btEstimateLDLTAddTLTmpbufSize(nskip);
btAssert(LDLTAddTL_size % sizeof(btScalar) == 0);
scratch.resize(nskip * 2 + n2);
btScalar *tmp = &scratch[0];
if (r == 0)
{
btScalar *a = (btScalar *)((char *)tmp + LDLTAddTL_size);
const int p_0 = p[0];
for (int i = 0; i < n2; ++i)
{
a[i] = -BTGETA(p[i], p_0);
}
a[0] += btScalar(1.0);
btLDLTAddTL(L, d, a, n2, nskip, scratch);
}
else
{
btScalar *t = (btScalar *)((char *)tmp + LDLTAddTL_size);
{
btScalar *Lcurr = L + r * nskip;
for (int i = 0; i < r; ++Lcurr, ++i)
{
btAssert(d[i] != btScalar(0.0));
t[i] = *Lcurr / d[i];
}
}
btScalar *a = t + r;
{
btScalar *Lcurr = L + r * nskip;
const int *pp_r = p + r, p_r = *pp_r;
const int n2_minus_r = n2 - r;
for (int i = 0; i < n2_minus_r; Lcurr += nskip, ++i)
{
a[i] = btLargeDot(Lcurr, t, r) - BTGETA(pp_r[i], p_r);
}
}
a[0] += btScalar(1.0);
btLDLTAddTL(L + r * nskip + r, d + r, a, n2 - r, nskip, scratch);
}
}
// snip out row/column r from L and d
btRemoveRowCol(L, n2, nskip, r);
if (r < (n2 - 1)) memmove(d + r, d + r + 1, (n2 - r - 1) * sizeof(btScalar));
}
void btLCP::transfer_i_from_C_to_N(int i, btAlignedObjectArray<btScalar> &scratch)
{
{
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)
{
btLDLTRemove(m_A, C, m_L, m_d, m_n, nC, j, m_nskip, scratch);
int k;
if (last_idx == -1)
{
for (k = j + 1; k < nC; ++k)
{
if (C[k] == nC - 1)
{
break;
}
}
btAssert(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;
}
}
btAssert(j < nC);
btSwapProblem(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
}
}
void btLCP::pN_equals_ANC_times_qC(btScalar *p, btScalar *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;
btScalar *ptgt = p + nC;
const int nN = m_nN;
for (int i = 0; i < nN; ++i)
{
ptgt[i] = btLargeDot(BTAROW(i + nC), q, nC);
}
}
void btLCP::pN_plusequals_ANi(btScalar *p, int i, int sign)
{
const int nC = m_nC;
btScalar *aptr = BTAROW(i) + nC;
btScalar *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 btLCP::pC_plusequals_s_times_qC(btScalar *p, btScalar s, btScalar *q)
{
const int nC = m_nC;
for (int i = 0; i < nC; ++i)
{
p[i] += s * q[i];
}
}
void btLCP::pN_plusequals_s_times_qN(btScalar *p, btScalar s, btScalar *q)
{
const int nC = m_nC;
btScalar *ptgt = p + nC, *qsrc = q + nC;
const int nN = m_nN;
for (int i = 0; i < nN; ++i)
{
ptgt[i] += s * qsrc[i];
}
}
void btLCP::solve1(btScalar *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)
{
{
btScalar *Dell = m_Dell;
int *C = m_C;
btScalar *aptr = BTAROW(i);
#ifdef BTNUB_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
}
btSolveL1(m_L, m_Dell, m_nC, m_nskip);
{
btScalar *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)
{
btScalar *tmp = m_tmp, *ell = m_ell;
{
const int nC = m_nC;
for (int j = 0; j < nC; ++j) tmp[j] = ell[j];
}
btSolveL1T(m_L, tmp, m_nC, m_nskip);
if (dir > 0)
{
int *C = m_C;
btScalar *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;
btScalar *tmp = m_tmp;
const int nC = m_nC;
for (int j = 0; j < nC; ++j) a[C[j]] = tmp[j];
}
}
}
}
void btLCP::unpermute()
{
// now we have to un-permute x and w
{
memcpy(m_tmp, m_x, m_n * sizeof(btScalar));
btScalar *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(btScalar));
btScalar *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 // btLCP_FAST
//***************************************************************************
// an optimized Dantzig LCP driver routine for the lo-hi LCP problem.
bool btSolveDantzigLCP(int n, btScalar *A, btScalar *x, btScalar *b,
btScalar *outer_w, int nub, btScalar *lo, btScalar *hi, int *findex, btDantzigScratchMemory &scratchMem)
{
s_error = false;
// printf("btSolveDantzigLCP n=%d\n",n);
btAssert(n > 0 && A && x && b && lo && hi && nub >= 0 && nub <= n);
btAssert(outer_w);
#ifdef BT_DEBUG
{
// check restrictions on lo and hi
for (int k = 0; k < n; ++k)
btAssert(lo[k] <= 0 && hi[k] >= 0);
}
#endif
// if all the variables are unbounded then we can just factor, solve,
// and return
if (nub >= n)
{
int nskip = (n);
btFactorLDLT(A, outer_w, n, nskip);
btSolveLDLT(A, outer_w, b, n, nskip);
memcpy(x, b, n * sizeof(btScalar));
return !s_error;
}
const int nskip = (n);
scratchMem.L.resize(n * nskip);
scratchMem.d.resize(n);
btScalar *w = outer_w;
scratchMem.delta_w.resize(n);
scratchMem.delta_x.resize(n);
scratchMem.Dell.resize(n);
scratchMem.ell.resize(n);
scratchMem.Arows.resize(n);
scratchMem.p.resize(n);
scratchMem.C.resize(n);
// for i in N, state[i] is 0 if x(i)==lo(i) or 1 if x(i)==hi(i)
scratchMem.state.resize(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!
btLCP lcp(n, nskip, nub, A, x, b, w, lo, hi, &scratchMem.L[0], &scratchMem.d[0], &scratchMem.Dell[0], &scratchMem.ell[0], &scratchMem.delta_w[0], &scratchMem.state[0], findex, &scratchMem.p[0], &scratchMem.C[0], &scratchMem.Arows[0]);
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)
{
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) scratchMem.delta_w[scratchMem.p[j]] = x[j];
// set lo and hi values
for (int k = i; k < n; ++k)
{
btScalar wfk = scratchMem.delta_w[findex[k]];
if (wfk == 0)
{
hi[k] = 0;
lo[k] = 0;
}
else
{
hi[k] = btFabs(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_N(i);
scratchMem.state[i] = false;
}
else if (hi[i] == 0 && w[i] <= 0)
{
lcp.transfer_i_to_N(i);
scratchMem.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(&scratchMem.delta_x[0], i, 0, 1);
lcp.transfer_i_to_C(i);
}
else
{
// we must push x(i) and w(i)
for (;;)
{
int dir;
btScalar dirf;
// find direction to push on x(i)
if (w[i] <= 0)
{
dir = 1;
dirf = btScalar(1.0);
}
else
{
dir = -1;
dirf = btScalar(-1.0);
}
// compute: delta_x(C) = -dir*A(C,C)\A(C,i)
lcp.solve1(&scratchMem.delta_x[0], 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(&scratchMem.delta_w[0], &scratchMem.delta_x[0]);
lcp.pN_plusequals_ANi(&scratchMem.delta_w[0], i, dir);
scratchMem.delta_w[i] = lcp.AiC_times_qC(i, &scratchMem.delta_x[0]) + 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
btScalar s = -w[i] / scratchMem.delta_w[i];
if (dir > 0)
{
if (hi[i] < BT_INFINITY)
{
btScalar 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] > -BT_INFINITY)
{
btScalar 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 (!scratchMem.state[indexN_k] ? scratchMem.delta_w[indexN_k] < 0 : scratchMem.delta_w[indexN_k] > 0)
{
// don't bother checking if lo=hi=0
if (lo[indexN_k] == 0 && hi[indexN_k] == 0) continue;
btScalar s2 = -w[indexN_k] / scratchMem.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 (scratchMem.delta_x[indexC_k] < 0 && lo[indexC_k] > -BT_INFINITY)
{
btScalar s2 = (lo[indexC_k] - x[indexC_k]) / scratchMem.delta_x[indexC_k];
if (s2 < s)
{
s = s2;
cmd = 5;
si = indexC_k;
}
}
if (scratchMem.delta_x[indexC_k] > 0 && hi[indexC_k] < BT_INFINITY)
{
btScalar s2 = (hi[indexC_k] - x[indexC_k]) / scratchMem.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 <= btScalar(0.0))
{
// printf("LCP internal error, s <= 0 (s=%.4e)",(double)s);
if (i < n)
{
btSetZero(x + i, n - i);
btSetZero(w + i, n - i);
}
s_error = true;
break;
}
// apply x = x + s * delta_x
lcp.pC_plusequals_s_times_qC(x, s, &scratchMem.delta_x[0]);
x[i] += s * dirf;
// apply w = w + s * delta_w
lcp.pN_plusequals_s_times_qN(w, s, &scratchMem.delta_w[0]);
w[i] += s * scratchMem.delta_w[i];
// void *tmpbuf;
// switch indexes between sets if necessary
switch (cmd)
{
case 1: // done
w[i] = 0;
lcp.transfer_i_to_C(i);
break;
case 2: // done
x[i] = lo[i];
scratchMem.state[i] = false;
lcp.transfer_i_to_N(i);
break;
case 3: // done
x[i] = hi[i];
scratchMem.state[i] = true;
lcp.transfer_i_to_N(i);
break;
case 4: // keep going
w[si] = 0;
lcp.transfer_i_from_N_to_C(si);
break;
case 5: // keep going
x[si] = lo[si];
scratchMem.state[si] = false;
lcp.transfer_i_from_C_to_N(si, scratchMem.m_scratch);
break;
case 6: // keep going
x[si] = hi[si];
scratchMem.state[si] = true;
lcp.transfer_i_from_C_to_N(si, scratchMem.m_scratch);
break;
}
if (cmd <= 3) break;
} // for (;;)
} // else
if (s_error)
{
break;
}
} // for (int i=adj_nub; i<n; ++i)
lcp.unpermute();
return !s_error;
}
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