12345qiupeng
/
gtsam
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

[REFACTOR] Move Comment

release/4.3a0
= 2016-06-16 10:40:49 -04:00
parent f3a4788193
commit dcd2b81bb1
2 changed files with 16 additions and 16 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

16
gtsam_unstable/linear/ActiveSetSolver.cpp
View File

@ -97,6 +97,22 @@ GaussianFactorGraph::shared_ptr ActiveSetSolver::buildDualGraph(
* This constraint will be added to the working set and become active
* in the next iteration.
*/
/* We have to make sure the new solution with alpha satisfies all INACTIVE inequality constraints
* If some inactive inequality constraints complain about the full step (alpha = 1),
* we have to adjust alpha to stay within the inequality constraints' feasible regions.
*
* For each inactive inequality j:
* - We already have: aj'*xk - bj <= 0, since xk satisfies all inequality constraints
* - We want: aj'*(xk + alpha*p) - bj <= 0
* - If aj'*p <= 0, we have: aj'*(xk + alpha*p) <= aj'*xk <= bj, for all alpha>0
* it's good!
* - We only care when aj'*p > 0. In this case, we need to choose alpha so that
* aj'*xk + alpha*aj'*p - bj <= 0 --> alpha <= (bj - aj'*xk) / (aj'*p)
* We want to step as far as possible, so we should choose alpha = (bj - aj'*xk) / (aj'*p)
*
* We want the minimum of all those alphas among all inactive inequality.
*/
boost::tuple<double, int> ActiveSetSolver::computeStepSize(
const InequalityFactorGraph& workingSet, const VectorValues& xk,
const VectorValues& p) const {

16
gtsam_unstable/linear/QPSolver.h
View File

@ -53,22 +53,6 @@ public:
JacobianFactor::shared_ptr createDualFactor(Key key,
const InequalityFactorGraph& workingSet, const VectorValues& delta) const;
/* We have to make sure the new solution with alpha satisfies all INACTIVE inequality constraints
* If some inactive inequality constraints complain about the full step (alpha = 1),
* we have to adjust alpha to stay within the inequality constraints' feasible regions.
*
* For each inactive inequality j:
* - We already have: aj'*xk - bj <= 0, since xk satisfies all inequality constraints
* - We want: aj'*(xk + alpha*p) - bj <= 0
* - If aj'*p <= 0, we have: aj'*(xk + alpha*p) <= aj'*xk <= bj, for all alpha>0
* it's good!
* - We only care when aj'*p > 0. In this case, we need to choose alpha so that
* aj'*xk + alpha*aj'*p - bj <= 0 --> alpha <= (bj - aj'*xk) / (aj'*p)
* We want to step as far as possible, so we should choose alpha = (bj - aj'*xk) / (aj'*p)
*
* We want the minimum of all those alphas among all inactive inequality.
*/
/// Iterate 1 step, return a new state with a new workingSet and values
QPState iterate(const QPState& state) const;