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Diffstat (limited to 'thirdparty/bullet/src/BulletDynamics/MLCPSolvers/btDantzigLCP.cpp')
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diff --git a/thirdparty/bullet/src/BulletDynamics/MLCPSolvers/btDantzigLCP.cpp b/thirdparty/bullet/src/BulletDynamics/MLCPSolvers/btDantzigLCP.cpp new file mode 100644 index 0000000000..986f214870 --- /dev/null +++ b/thirdparty/bullet/src/BulletDynamics/MLCPSolvers/btDantzigLCP.cpp @@ -0,0 +1,2080 @@ +/************************************************************************* +* * +* 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; +} + |