1 subdivision termination criteria in subdivision multivariate solvers iddo hanniel, gershon elber...
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Subdivision Termination Subdivision Termination Criteria in Subdivision Criteria in Subdivision
Multivariate SolversMultivariate Solvers
Iddo Hanniel, Gershon ElberIddo Hanniel, Gershon ElberCGGC, CS, TechnionCGGC, CS, Technion
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The ProblemThe ProblemConsider the following set of d polynomial equations:
In Rd.
We seek the simultaneous solution, ,
such that for all i =1,…,d.
,0),...,,(
,0),...,,(
21
211
dd
d
uuuF
uuuF
0)( siF u
ds Ru).,...,,( 21 duuuu
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Motivation for the SolverMotivation for the Solver Many geometric problems are reduced to the
simultaneous solution of a set of constraints. Ray-surface, curve-curve and surface-surface
intersections. Voronoi diagram of curves. Curve-curve bi-tangents, convex hull of curves. Minimum enclosing circle/sphere of a set of
curves/surfaces.
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Here, one can find four
solutions that satisfy the
equations. We are interested in finding all solutions.
Example – Intersection of 3 Example – Intersection of 3 Explicit Surfaces in Explicit Surfaces in RR33
,0),(),(
,0),(),(
213211
212211
uuFuuF
uuFuuF
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Previous Work on Multivariate Previous Work on Multivariate Polynomial SolversPolynomial Solvers
Algebraic Solvers: Grobner bases [Cox et al., 1992]. Multivariate Sturm sequences [Milne, 1992].
Geometric Solvers: Based on the Bernstein/Bezier/Bspline properties. Bezier clipping methods [Sherbrooke and Patrikalakis,
1993], [Mourrain and Pavone, 2005]. Subdivision step and numeric improvement [Elber and
Kim, 2001].
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The Bezier/B-Spline SolverThe Bezier/B-Spline Solver A subdivision stage in a multidimensional space, using B-spline/Bezier subdivision. A multivariate Newton-Raphson (NR) numeric step.
Elber and Kim, 2001: Geometric Constraint Solver Elber and Kim, 2001: Geometric Constraint Solver using Multivariate Rational Spline Functionsusing Multivariate Rational Spline Functions
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Subdivision Illustration – Subdivision Illustration – Univariate CaseUnivariate Case
Subdivision. CH containment.
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The Termination QuestionThe Termination Question
We can identify domain cells with no root by the CH property. Can we do any better?
Question: When do we stop the subdivision?
If we know there is at most a single root in some domain cell, we can move to the NR step. How can we know that?
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The Hodograph – Univariate CaseThe Hodograph – Univariate Case
)(' tC
)(tC
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Single Solution GuaranteeSingle Solution Guarantee
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Extension to Higher DimensionsExtension to Higher Dimensions
Theorem: Given d implicit hyper-surfaces Fi(u) = 0,
i = 1,….,d, in Rd, there is at most one common
solution if where 0 is the origin. ,01
d
i
ciC
ciC
niC
Elber and Kim, 2001: Geometric Constraint Solver using Elber and Kim, 2001: Geometric Constraint Solver using Multivariate Rational Spline Functions Multivariate Rational Spline Functions
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Implementing the Single Solution TestImplementing the Single Solution Test
How can this condition be tested
efficiently in Rd? Clearly the gradient field of Fi(u)
provides a bound on the normal
space of Fi(u).
Optimal bounding cones over
vectors in Rd could be found
in average linear time.
niC
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Implementing the SST ( Cont.)Implementing the SST ( Cont.)
The complementary cone can easily be derived, given the normal cone. and share the same axis but with complementary angles.
ciC
niC
)90(cos,| 222
0 uvuu
222
0 cos,| uvuu
niC c
iC
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Implementing the SST (Cont.)Implementing the SST (Cont.)
But now we have to test for the intersection of d complementary cones in Rd. A difficult task, even in R3 !
cC1
cC2cC3
d = 3
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Implementing the SST (Cont.)Implementing the SST (Cont.)
Reexamine the univariate case (two curves):
…intersectiff the parallellines domain’s Intersection…
Two cones of two planar curves…
….is not confinedto the
unit circle
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Implementing the SST (Cont.)Implementing the SST (Cont.)
The efficient solution is found by using a different representation for the complementary cones.
Two infinite parallel hyper-planes in Rd.
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Implementing the SST - Main LemmaImplementing the SST - Main Lemma
Lemma: contains a vector other than {0}, if and only if the intersection of the unit hyper-sphere Sd-1 with the regions between the bounding hyper-planes is non-empty.
d
i
ciC
1
Implication: In order to test for a single solution, we do not have to intersect d-dimensional cones.
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Implementing the SST (Cont.)Implementing the SST (Cont.) Two options:
The entire intersection is inside the unit sphere. Then SST holds. Otherwise, SST failed.
Hence, only test if all the vertices of the intersection are inside the unit sphere.
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Implementing the SST (Cont.)Implementing the SST (Cont.)
SST holds SST failed
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Implementing the SST (Cont.)Implementing the SST (Cont.)The set of 2d hyper-plane intersection (HPI) equations to solve is:
222
0 cos, uuv |cos|,0 uv
|cos|...
...
|cos|...
11
111
11
dd
dd
d
dd
uvuv
uvuv
ibuV
Or in matrix (Or in matrix (d d dd) form:) form:
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Implementing the SST (Cont.)Implementing the SST (Cont.)
Due to symmetry, only 2d-1 hyper-plane intersection (HPI) tests are required. 4 HPI tests for d = 3 (x, y, z) constraints in R3.
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Implementing the SST (Cont.)Implementing the SST (Cont.)An efficient way to solve the 2d-1 systems of equations is due to the inherent symmetry of the problem:
d
iiiebuV
1
d
iii eVbu
1
1
Therefore, given one solution (e.g., {-,-,…,-,-}), computing a second solution that differs by a single sign (e.g., {-,-,…,-,+}), takes only O(d) operations and not O(d2 ).
ibuV
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Implementing the SST – Implementing the SST – ExampleExample
Without SST With SST
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Implementing the SST – Implementing the SST – Example (Zoom)Example (Zoom) With SSTWithout SST
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Purging Away Zero-solution Purging Away Zero-solution DomainsDomains
Problem: SST does not guarantee that a solution exists in the sub-domain. We might start the numerical iterations only to find that no root exists in this sub-domain.
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Purging Away Zero-solution Purging Away Zero-solution Domains (Cont.)Domains (Cont.)
Therefore, we seek to purge away, as much as possible, sub-domains that contain no solution.
We present a second criterion, in addition to the CH criterion, for identifying no-solution sub-domains.
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Purging Away Zero-solution Purging Away Zero-solution Domains (Cont.)Domains (Cont.)
The basic idea: Bound the function Fi=0 by a pair of parallel hyper-planes in Rd. If the intersection of the hyper-planes is entirely outside the sub-domain, then the solution is outside the sub-domain.
FF11
FF22
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Purging Away Zero-solution Purging Away Zero-solution Domains (Cont.)Domains (Cont.)
Problem: In order to bound Fi=0 in Rd we need to have at least a sample point u on Fi=0.
Solution: Bound Fi in Rd+1 and intersect the resulting (d+1)-dimensional hyper-planes with ud+1=0, resulting bounding hyper-planes in Rd.
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Purging Away Zero-solution Purging Away Zero-solution Domains (Cont.)Domains (Cont.)
Promote Fi from a scalar function to a hyper-surface in Rd+1, . Evaluate the gradient (i.e., the normal) at the sub-domain mid-point and project all control points of the hyper-surface onto it. The two hyper-planes orthogonal to the gradient and passing through the extreme projection points, bound Fi.
),,...,,(ˆ21 idi FuuuF
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Center for Graphics and Geometric Computing, TechnionCenter for Graphics and Geometric Computing, Technion 30
Purging Away Zero-solution Domains Illustration (d=1)
iF̂
ud+1 = 0
)( midi uF
][
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Purging Away Zero-solution Purging Away Zero-solution Domains (Cont.)Domains (Cont.)
How to determine whether the intersection is entirely outside the sub-domain?
Possible solutions:1. Linear programming techniques.2. Similarly to the SST for finding all
vertices of the intersection.
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Purging Away Zero-solution Purging Away Zero-solution sub-domains – Examplesub-domains – Example
Without hyper-plane test With hyper-plane test
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Future WorkFuture Work Compare the SST with algebraic/numeric termination criteria (Kantorovich). Use the vertices, which were computed for purging away zero-solutions, for clipping the sub-domain. Extend the presented ideas for under- and over-constrained systems. Search for tighter bounding volumes on the solutions (not just cones).
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EndEnd