Encasement: A method to compute geometric arrangements Joseph - - PowerPoint PPT Presentation

Encasement: A method to compute geometric arrangements

Joseph Masterjohn

University of Miami Department of Computer Science

Sept 30, 2015

Overview

Overview

What is an Arrangement?

Start with a sheet of material.

What is an Arrangement?

Cut it with some curves

What is an Arrangement?

The arrangement is the resulting shapes

What is an Arrangement?

An arrangement in 2D is a decomposition of the plane into disjoint sets, which is induced by a set of curves in the plane.

What is an Arrangement?

An arrangement in 2D is a decomposition of the plane into disjoint sets, which is induced by a set of curves in the plane. The idea generalizes to 3D and higher, but for the purposes of visualization we will stick to 2D.

Overview

Applications

An arrangement is a fundamental operation that is the basis for many other geometric algorithms.

Applications

An arrangement is a fundamental operation that is the basis for many other geometric algorithms. Useful for range searching

Applications

An arrangement is a fundamental operation that is the basis for many other geometric algorithms. Useful for range searching Various geometric optimization problems (Distance Selection, Diameter in 3D, Biggest line segment in a simple polygon, etc.)

Applications

An arrangement is a fundamental operation that is the basis for many other geometric algorithms. Useful for range searching Various geometric optimization problems (Distance Selection, Diameter in 3D, Biggest line segment in a simple polygon, etc.) Motion planning in robotics, arrangement of ”contact surfaces” defines the configuration space of a robotic body.

Applications

An arrangement is a fundamental operation that is the basis for many other geometric algorithms. Useful for range searching Various geometric optimization problems (Distance Selection, Diameter in 3D, Biggest line segment in a simple polygon, etc.) Motion planning in robotics, arrangement of ”contact surfaces” defines the configuration space of a robotic body. Assembly planning, given two sets of objects placed rigidly and non-overlapping, can they be merged by some motion (i.e. assembled) while still remaining rigid and non-overlapping?

Overview

Overview

Resultants

Exact computation of intersections using resultants

Resultants

Exact computation of intersections using resultants For P(x), Q(x) ∈ k[x] : res(P, Q) =

• (x,y):P(x)=Q(y)=0

(x − y)

Resultants

Exact computation of intersections using resultants For P(x), Q(x) ∈ k[x] : res(P, Q) =

• (x,y):P(x)=Q(y)=0

(x − y) If the two polynomials share a root, then this product is zero.

Resultants

Exact computation of intersections using resultants For P(x), Q(x) ∈ k[x] : res(P, Q) =

• (x,y):P(x)=Q(y)=0

(x − y) If the two polynomials share a root, then this product is zero. The resultant can be computed by the determinant of the Sylvester Matrix.

Resultants

Exact computation of intersections using resultants For P(x), Q(x) ∈ k[x] : res(P, Q) =

• (x,y):P(x)=Q(y)=0

(x − y) If the two polynomials share a root, then this product is zero. The resultant can be computed by the determinant of the Sylvester Matrix. For polynomials in two variables, express as polynomials in x whose coefficients are polynomials in y

Example

Let P(x, y) = {(x, y) ∈ R : 1x3 + 0x2 + 1x + (1 − y3) = 0}

Example

Let P(x, y) = {(x, y) ∈ R : 1x3 + 0x2 + 1x + (1 − y3) = 0} Let Q(x, y) = {(x, y) ∈ R : −1x3 + 0x2 − 1x + (1 − y3) = 0}

Example

Example

res(P, Q) =

• 1

1 (1 − y3) 1 1 (1 − y3) 1 1 (1 − y3) −1 −1 (1 − y3) −1 −1 (1 − y3) −1 −1 (1 − y3)

Example

res(P, Q) = 8 − 24y3 + 24y6 − 8y9

Resultants

The degree of the resulting polynomial is proportional to the product of the degrees of the input polynomials

Resultants

The degree of the resulting polynomial is proportional to the product of the degrees of the input polynomials This can easily blow up for curves of reasonable degree

Resultants

The degree of the resulting polynomial is proportional to the product of the degrees of the input polynomials This can easily blow up for curves of reasonable degree For 3D NURBS, a common surface represntation, it blows up to degree > 1000

Resultants

The degree of the resulting polynomial is proportional to the product of the degrees of the input polynomials This can easily blow up for curves of reasonable degree For 3D NURBS, a common surface represntation, it blows up to degree > 1000 What’s the problem?

Resultants

The degree of the resulting polynomial is proportional to the product of the degrees of the input polynomials This can easily blow up for curves of reasonable degree For 3D NURBS, a common surface represntation, it blows up to degree > 1000 What’s the problem? We’re paying for both the complex and the real roots, when we are only interested in the real roots.

Resultants

The degree of the resulting polynomial is proportional to the product of the degrees of the input polynomials This can easily blow up for curves of reasonable degree For 3D NURBS, a common surface represntation, it blows up to degree > 1000 What’s the problem? We’re paying for both the complex and the real roots, when we are only interested in the real roots. In this case the polynomial of degree 9 has 1 real root (of multiplicity 3).

Overview

Curve Tracing

More practical and efficient ”engineering” solution

Curve Tracing

More practical and efficient ”engineering” solution

1

Start with a point on the curve

Curve Tracing

More practical and efficient ”engineering” solution

1

Start with a point on the curve

2

Compute the tangent line to the curve at that point

Curve Tracing

More practical and efficient ”engineering” solution

1

Start with a point on the curve

2

Compute the tangent line to the curve at that point

3

Move some amount along that line

Curve Tracing

More practical and efficient ”engineering” solution

1

Start with a point on the curve

2

Compute the tangent line to the curve at that point

3

Move some amount along that line

4

Use a Newton-like method to get back to the curve

Curve Tracing

Curve Tracing

Curve Tracing: Problem

Curve Tracing: Problem

Curve Tracing: Problem

Each step is very fast, but

Curve Tracing: Problem

Each step is very fast, but In general we can’t guarantee correctness.

Curve Tracing: Problem

Each step is very fast, but In general we can’t guarantee correctness. We seek a method with running time comparable to the curve tracing method, with the correctness of the mathematical approach.

Overview

Idea

What do we already know how to do reliably?

Arrangements of Lines

L1 : ax + by = e

Arrangements of Lines

L1 : ax + by = e L2 : cx + dy = f

Arrangements of Lines

L1 : ax + by = e L2 : cx + dy = f Solve the linear system... easy

Roots of univariate polynomials

f (x) = a0 + a1x + a2x2 + . . . + anxn = 0

Roots of univariate polynomials

f (x) = a0 + a1x + a2x2 + . . . + anxn = 0 Root finding is a well studied area of research

Roots of univariate polynomials

f (x) = a0 + a1x + a2x2 + . . . + anxn = 0 Root finding is a well studied area of research Many methods exists, pick your poison

Roots of univariate polynomials

f (x) = a0 + a1x + a2x2 + . . . + anxn = 0 Root finding is a well studied area of research Many methods exists, pick your poison Interval Newton’s method will suffice

Intersection of Lines and Curves

σ(t) = (x(t), y(t)) = (At + B, Ct + D)

Intersection of Lines and Curves

σ(t) = (x(t), y(t)) = (At + B, Ct + D) f (x, y) = (a00) + (a10)x + (a01)y + (a11)xy + ... + (anm)xnym

Intersection of Lines and Curves

σ(t) = (x(t), y(t)) = (At + B, Ct + D) f (x, y) = (a00) + (a10)x + (a01)y + (a11)xy + ... + (anm)xnym f (x, y) ◦ σ(t) = f (x(t), y(t)) = 0

Intersection of Lines and Curves

σ(t) = (x(t), y(t)) = (At + B, Ct + D) f (x, y) = (a00) + (a10)x + (a01)y + (a11)xy + ... + (anm)xnym f (x, y) ◦ σ(t) = f (x(t), y(t)) = 0 That’s just a univariate polynomial in t, we know how to solve those.

Idea

The only thing we don’t know how to do explicitly is intersecting two implicit curves. We won’t ever try to do that. We will just use our existing reliable techniques to compute an arrangement of lines, from which the arrangement of curves can be extracted.

Overview

Overview

What is an Encasement?

A collection of connected convex cells in the plane that isolate the features of the arrangement of curves is an Encasement if:

What is an Encasement?

A collection of connected convex cells in the plane that isolate the features of the arrangement of curves is an Encasement if:

1

No vertex of a cell lies on a curve.

What is an Encasement?

A collection of connected convex cells in the plane that isolate the features of the arrangement of curves is an Encasement if:

1

No vertex of a cell lies on a curve.

2

A cell edge intersects up to one curve in a single point.

What is an Encasement?

A collection of connected convex cells in the plane that isolate the features of the arrangement of curves is an Encasement if:

1

No vertex of a cell lies on a curve.

2

A cell edge intersects up to one curve in a single point.

3

A cell intersects up to two curves as simple segments.

What is an Encasement?

A collection of connected convex cells in the plane that isolate the features of the arrangement of curves is an Encasement if:

1

No vertex of a cell lies on a curve.

2

A cell edge intersects up to one curve in a single point.

3

A cell intersects up to two curves as simple segments.

4

If a cell intersects two curves, those curves have one intersection point.

Valid Encasement

Valid Encasement

Invalid Encasement

Invalid Encasement

Invalid Encasement

Full Encasement

How does an encasement solve the original problem of finding an arrangement?

How does an encasement solve the original problem of finding an arrangement? How can we compute some feature of the arrangement using the Encasement?

Full Encasement

Full Encasement

OK we’ll start out with a single cell, the boundary cell.

OK we’ll start out with a single cell, the boundary cell. If no rules are violated, we’re done, we have a valid Encasement.

OK we’ll start out with a single cell, the boundary cell. If no rules are violated, we’re done, we have a valid Encasement. If not, how do we fix those violating situations we encountered before?

OK we’ll start out with a single cell, the boundary cell. If no rules are violated, we’re done, we have a valid Encasement. If not, how do we fix those violating situations we encountered before? One simple operation: split.

Fixing Invalid Encasement

Fixing Invalid Encasement

Fixing Invalid Encasement

Overview

Split Edges

Split Edges

Overview

Isolate segments of one curve

Isolate segments of one curve

Isolate segments of one curve

Isolate segments of one curve

Overview

Isolate segments of two curves

Isolate segments of two curves

Isolate segments of two curves

Isolate segments of two curves

Overview

Isolate intersections

We know two curves intersect at least once in a cell by looking at the order of the intersections with the boundary.

Isolate intersections

But that doesn’t sufficiently tell us that there is only one. There could be an odd number of intersections inside the cell.

Isolate intersections

Notice the tangents at the intersections though. If the two curves intersect more than once, then their tangent cones intersect.

Isolate intersections

Thus if we can show that their tangent cones do not intersect, then we have shown that the curves intersect only once.

Isolate intersections

For implicit curves, the tangents of f (x, y) = 0 are all rotations of ▽f = fx(x, y), fy(x, y). So we can formulate the argument for tangent cones equivalently with gradient cones.

Isolate intersections

For implicit curves, the tangents of f (x, y) = 0 are all rotations of ▽f = fx(x, y), fy(x, y). So we can formulate the argument for tangent cones equivalently with gradient cones. We choose a direction v such that every gradient vector of some curve f (x, y) lie on the same side of v within the cell. And on the opposite side for g(x, y). (We hope...)

Isolate intersections

For implicit curves, the tangents of f (x, y) = 0 are all rotations of ▽f = fx(x, y), fy(x, y). So we can formulate the argument for tangent cones equivalently with gradient cones. We choose a direction v such that every gradient vector of some curve f (x, y) lie on the same side of v within the cell. And on the opposite side for g(x, y). (We hope...) Verify by checking ▽f × v = 0 inside the cell

Isolate intersections

For implicit curves, the tangents of f (x, y) = 0 are all rotations of ▽f = fx(x, y), fy(x, y). So we can formulate the argument for tangent cones equivalently with gradient cones. We choose a direction v such that every gradient vector of some curve f (x, y) lie on the same side of v within the cell. And on the opposite side for g(x, y). (We hope...) Verify by checking ▽f × v = 0 inside the cell How do we do that?

Isolate intersections

▽f × v = fx(x, y) vy − fy(x, y) vx

Isolate intersections

▽f × v = fx(x, y) vy − fy(x, y) vx fx(x, y) vy − fy(x, y) vx = 0 is just an implicit curve.

Isolate intersections

▽f × v = fx(x, y) vy − fy(x, y) vx fx(x, y) vy − fy(x, y) vx = 0 is just an implicit curve. So to show fx(x, y) vy − fy(x, y) vx = 0, just show that fx(x, y) vy − fy(x, y) vx = 0 doesn’t exist in the cell.

Isolate intersections

▽f × v = fx(x, y) vy − fy(x, y) vx fx(x, y) vy − fy(x, y) vx = 0 is just an implicit curve. So to show fx(x, y) vy − fy(x, y) vx = 0, just show that fx(x, y) vy − fy(x, y) vx = 0 doesn’t exist in the cell. Check for intersection with the cell’s edges, and also check for loops.

Isolate intersections

▽f × v = fx(x, y) vy − fy(x, y) vx fx(x, y) vy − fy(x, y) vx = 0 is just an implicit curve. So to show fx(x, y) vy − fy(x, y) vx = 0, just show that fx(x, y) vy − fy(x, y) vx = 0 doesn’t exist in the cell. Check for intersection with the cell’s edges, and also check for loops. Wait...

Overview

Eliminate Loops

Let red be f (x, y) > 0 and blue be f (x, y) < 0

Eliminate Loops

Let red be f (x, y) > 0 and blue be f (x, y) < 0 Because we’re dealing with smooth polynomial curves, either the region inside the loop is strictly positive or strictly negative.

Eliminate Loops

Let red be f (x, y) > 0 and blue be f (x, y) < 0 Because we’re dealing with smooth polynomial curves, either the region inside the loop is strictly positive or strictly negative. Either way, that means that there is a critical point in that region (i.e. there exists some (x, y) such that fx(x, y) = 0 and fy(x, y) = 0)

Eliminate Loops

If there is a loop inside this cell, then there must be a critical point of f (x, y) inside this cell. Cutting through the critical point will cut the loop.

Eliminate Loops

If there is a loop inside this cell, then there must be a critical point of f (x, y) inside this cell. Cutting through the critical point will cut the loop. How do we find the critical points?

Eliminate Loops

If there is a loop inside this cell, then there must be a critical point of f (x, y) inside this cell. Cutting through the critical point will cut the loop. How do we find the critical points? We need to find where fx(x, y) = 0 and fy(x, y) = 0 simultaneously... in other words, detect the intersection points

• f those two curves inside this cell.

Eliminate Loops

If there is a loop inside this cell, then there must be a critical point of f (x, y) inside this cell. Cutting through the critical point will cut the loop. How do we find the critical points? We need to find where fx(x, y) = 0 and fy(x, y) = 0 simultaneously... in other words, detect the intersection points

• f those two curves inside this cell.

First answer: Recursion, we are writing an algorithm to compute the intersection of implicit curves, aren’t we?

Eliminate Loops

If there is a loop inside this cell, then there must be a critical point of f (x, y) inside this cell. Cutting through the critical point will cut the loop. How do we find the critical points? We need to find where fx(x, y) = 0 and fy(x, y) = 0 simultaneously... in other words, detect the intersection points

• f those two curves inside this cell.

First answer: Recursion, we are writing an algorithm to compute the intersection of implicit curves, aren’t we? But then for each curve induced by a partial derivative, we have to intersect its partials. This ends up exponential in the degree of the original curves.

Eliminate Loops

Second answer: Well we don’t really need to precisely find the critical points. We just need to isolate a region where f (x, y) = 0 and has a point (x0, y0) where fx(x0, y0) = 0 and fy(x0, y0) = 0 We know such a region exists because the region surrounded by the loop is non-zero, and must obtain a maximum and a minimum.

Eliminate Loops

Second answer: Well we don’t really need to precisely find the critical points. We just need to isolate a region where f (x, y) = 0 and has a point (x0, y0) where fx(x0, y0) = 0 and fy(x0, y0) = 0 We know such a region exists because the region surrounded by the loop is non-zero, and must obtain a maximum and a minimum. How can we tell if a region contains a point that evaluates to zero, without knowing that point itself??

Eliminate Loops

Second answer: Well we don’t really need to precisely find the critical points. We just need to isolate a region where f (x, y) = 0 and has a point (x0, y0) where fx(x0, y0) = 0 and fy(x0, y0) = 0 We know such a region exists because the region surrounded by the loop is non-zero, and must obtain a maximum and a minimum. How can we tell if a region contains a point that evaluates to zero, without knowing that point itself?? Well we can’t, but we can tell that a region does not contain a point that evaluates to zero. How?

Eliminate Loops

Second answer: Well we don’t really need to precisely find the critical points. We just need to isolate a region where f (x, y) = 0 and has a point (x0, y0) where fx(x0, y0) = 0 and fy(x0, y0) = 0 We know such a region exists because the region surrounded by the loop is non-zero, and must obtain a maximum and a minimum. How can we tell if a region contains a point that evaluates to zero, without knowing that point itself?? Well we can’t, but we can tell that a region does not contain a point that evaluates to zero. How? Answer: Interval Arithmetic

Brief primer on interval arithmetic

We can use intervals to represent our coordinates.

Brief primer on interval arithmetic

We can use intervals to represent our coordinates. e.x. x = [−1, 1] y = [1, 2]

Brief primer on interval arithmetic

We can use intervals to represent our coordinates. e.x. x = [−1, 1] y = [1, 2] xy = [−2, 2], this interval contains the result for any combination of choices of points in the interval x and the interval y, guaranteed. Even in floating point if you round correctly.

Brief primer on interval arithmetic

We can use intervals to represent our coordinates. e.x. x = [−1, 1] y = [1, 2] xy = [−2, 2], this interval contains the result for any combination of choices of points in the interval x and the interval y, guaranteed. Even in floating point if you round correctly. Why is this useful? Consider f (x) = x2, does f (x) contain a zero on the interval [1, 2]?

Brief primer on interval arithmetic

We can use intervals to represent our coordinates. e.x. x = [−1, 1] y = [1, 2] xy = [−2, 2], this interval contains the result for any combination of choices of points in the interval x and the interval y, guaranteed. Even in floating point if you round correctly. Why is this useful? Consider f (x) = x2, does f (x) contain a zero on the interval [1, 2]? f ([1, 2]) = [1, 2]2 = [1, 2][1, 2] = [1, 4], and [1, 4] does not contain 0, so there cannot be any choice of x ∈ [1, 2] such that f (x) = 0.

Eliminate Loops

So we want to find a rectangular region (x, y) where x = [a, b] and y = [c, d] (The Cartesian product of two intervals), where:

Eliminate Loops

So we want to find a rectangular region (x, y) where x = [a, b] and y = [c, d] (The Cartesian product of two intervals), where: 0 ∈ f ([a, b], [c, d])

Eliminate Loops

So we want to find a rectangular region (x, y) where x = [a, b] and y = [c, d] (The Cartesian product of two intervals), where: 0 ∈ f ([a, b], [c, d]) 0 ∈ fx([a, b], [c, d])

Eliminate Loops

So we want to find a rectangular region (x, y) where x = [a, b] and y = [c, d] (The Cartesian product of two intervals), where: 0 ∈ f ([a, b], [c, d]) 0 ∈ fx([a, b], [c, d]) 0 ∈ fy([a, b], [c, d])

Eliminate Loops

So we want to find a rectangular region (x, y) where x = [a, b] and y = [c, d] (The Cartesian product of two intervals), where: 0 ∈ f ([a, b], [c, d]) 0 ∈ fx([a, b], [c, d]) 0 ∈ fy([a, b], [c, d]) Start with the original bounding region and subdivide until you can verify that 0 ∈ f (x, y) or [0 ∈ fx(x, y) or 0 ∈ fy(x, y)]

Eliminate Loops

(x, y)

Eliminate Loops

Eliminate Loops

Eliminate Loops

Eliminate Loops

Eliminate Loops

Eliminate Loops

Overview

Refinement

Refinement

Refinement

Overview

Demo

Overview

Conclusion

There is no formal analysis of the run-time of either curve tracing or encasement.

Conclusion

There is no formal analysis of the run-time of either curve tracing or encasement. Encasement appears to match the order of the running time

• f curve tracing.

Conclusion

There is no formal analysis of the run-time of either curve tracing or encasement. Encasement appears to match the order of the running time

• f curve tracing.

Separating two curves takes an equal number of tracing steps

• vs. cell splits.

Conclusion

There is no formal analysis of the run-time of either curve tracing or encasement. Encasement appears to match the order of the running time

• f curve tracing.

Separating two curves takes an equal number of tracing steps

• vs. cell splits.

But splitting a cell also adds to the space

Conclusion

There is no formal analysis of the run-time of either curve tracing or encasement. Encasement appears to match the order of the running time

• f curve tracing.

Separating two curves takes an equal number of tracing steps

• vs. cell splits.

But splitting a cell also adds to the space Future work needs to address space cost.

Future Work

Relax the constraint on number of non-intersecting curves If you can construct a polygonal chain between two curves, don’t split

Future Work

Allow splits with non-linear geometry (i.e. edges that are degree 2 or 3 curves).

Questions?

Any questions? Thank you!