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