update the EliminateHybrid note to be more specific

release/4.3a0
Varun Agrawal 2022-05-27 18:46:52 -04:00
parent 6cd20fba4d
commit c3a59cfd62
1 changed files with 28 additions and 25 deletions

View File

@ -236,43 +236,46 @@ hybridElimination(const HybridFactorGraph &factors, const Ordering &frontalKeys,
/* ************************************************************************ */
std::pair<HybridConditional::shared_ptr, HybridFactor::shared_ptr> //
EliminateHybrid(const HybridFactorGraph &factors, const Ordering &frontalKeys) {
// NOTE(fan): Because we are in the Conditional Gaussian regime there are only
// NOTE: Because we are in the Conditional Gaussian regime there are only
// a few cases:
// continuous variable, we make a GM if there are hybrid factors;
// continuous variable, we make a GF if there are no hybrid factors;
// discrete variable, no continuous factor is allowed (escapes CG regime), so
// we panic, if discrete only we do the discrete elimination.
// 1. continuous variable, make a Gaussian Mixture if there are hybrid
// factors;
// 2. continuous variable, we make a Gaussian Factor if there are no hybrid
// factors;
// 3. discrete variable, no continuous factor is allowed
// (escapes Conditional Gaussian regime), if discrete only we do the discrete
// elimination.
// However it is not that simple. During elimination it is possible that the
// multifrontal needs to eliminate an ordering that contains both Gaussian and
// hybrid variables, for example x1, c1. In this scenario, we will have a
// density P(x1, c1) that is a CLG P(x1|c1)P(c1) (see Murphy02)
// hybrid variables, for example x1, c1.
// In this scenario, we will have a density P(x1, c1) that is a Conditional
// Linear Gaussian P(x1|c1)P(c1) (see Murphy02).
// The issue here is that, how can we know which variable is discrete if we
// unify Values? Obviously we can tell using the factors, but is that fast?
// In the case of multifrontal, we will need to use a constrained ordering
// so that the discrete parts will be guaranteed to be eliminated last!
// Because of all these reasons, we need to think very carefully about how to
// Because of all these reasons, we carefully consider how to
// implement the hybrid factors so that we do not get poor performance.
//
// The first thing is how to represent the GaussianMixtureConditional. A very
// possible scenario is that the incoming factors will have different levels
// of discrete keys. For example, imagine we are going to eliminate the
// fragment:
// $\phi(x1,c1,c2)$, $\phi(x1,c2,c3)$, which is perfectly valid. Now we will
// need to know how to retrieve the corresponding continuous densities for the
// assi- -gnment (c1,c2,c3) (OR (c2,c3,c1)! note there is NO defined order!).
// And we also need to consider when there is pruning. Two mixture factors
// could have different pruning patterns-one could have (c1=0,c2=1) pruned,
// and another could have (c2=0,c3=1) pruned, and this creates a big problem
// in how to identify the intersection of non-pruned branches.
// One possible approach is first building the collection of all discrete
// keys. After that we enumerate the space of all key combinations *lazily* so
// that the exploration branch terminates whenever an assignment yields NULL
// in any of the hybrid factors.
// The first thing is how to represent the GaussianMixtureConditional.
// A very possible scenario is that the incoming factors will have different
// levels of discrete keys. For example, imagine we are going to eliminate the
// fragment: $\phi(x1,c1,c2)$, $\phi(x1,c2,c3)$, which is perfectly valid.
// Now we will need to know how to retrieve the corresponding continuous
// densities for the assignment (c1,c2,c3) (OR (c2,c3,c1), note there is NO
// defined order!). We also need to consider when there is pruning. Two
// mixture factors could have different pruning patterns - one could have
// (c1=0,c2=1) pruned, and another could have (c2=0,c3=1) pruned, and this
// creates a big problem in how to identify the intersection of non-pruned
// branches.
// Our approach is first building the collection of all discrete keys. After
// that we enumerate the space of all key combinations *lazily* so that the
// exploration branch terminates whenever an assignment yields NULL in any of
// the hybrid factors.
// When the number of assignments is large we may encounter stack overflows.
// However this is also the case with iSAM2, so no pressure :)