subdivision termination criteria in subdivision multivariate solvers

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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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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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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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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|...

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111

11

dd

dd

d

dd

uvuv

uvuv

ibuV

Or in matrix (Or in matrix (d d dd) form:) form:

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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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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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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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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

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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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

subdivision termination criteria in subdivision multivariate solvers

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