Overlay Stitch Meshing

5/2/2007 Don SheehyOverlay Stitch Meshing

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Overlay Stitch MeshingOverlay Stitch Meshing


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A competitive algorithm for A competitive algorithm for no-large-angle triangulationno-large-angle triangulation

Don SheehyDon Sheehy

Joint work with Gary Miller Joint work with Gary Miller and Todd Phillipsand Todd Phillips

To appear at ICALP 2007To appear at ICALP 2007


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The ProblemThe ProblemInput: A Planar Straight Line Input: A Planar Straight Line

GraphGraph


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GraphGraph

Output: A Conforming TriangulationOutput: A Conforming Triangulation


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GraphGraph

Output: A Conforming TriangulationOutput: A Conforming Triangulation


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Why would you want to do that?Why would you want to do that?


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What went wrong?What went wrong?


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What if you don’t know the function?What if you don’t know the function?

?? ??


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2 Definitions of Quality2 Definitions of Quality

1. No Large Angles [Babuska, Aziz 1. No Large Angles [Babuska, Aziz 1976]1976]


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2 Definitions of Quality2 Definitions of Quality

1. No Large Angles [Babuska, Aziz 1. No Large Angles [Babuska, Aziz 1976]1976]

2. No Small Angles2. No Small Angles


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No Small AnglesNo Small Angles


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No Small AnglesNo Small Angles

You may have heard of these You may have heard of these before.before.

Delaunay RefinementDelaunay Refinement

Sparse Voronoi RefinementSparse Voronoi Refinement

QuadtreeQuadtree


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Paying for the spreadPaying for the spread

LL

ss

Spread = L/sSpread = L/s


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Optimal No-Large-Angle TriangulationOptimal No-Large-Angle Triangulation


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Paying for the spreadPaying for the spreadWhat if we don’t allow small angles?What if we don’t allow small angles?


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O(L/s)O(L/s) triangles! triangles!

What if we don’t allow small angles?What if we don’t allow small angles?


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O(L/s)O(L/s) triangles! triangles!

What if we don’t allow small angles?What if we don’t allow small angles?Fact: For inputs with Fact: For inputs with NO NO edgesedges, no-small-angle , no-small-angle

meshing algorithms produce meshing algorithms produce output with output with O(n log L/s)O(n log L/s) size size and angles between 30and angles between 30oo and and

120120oo


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What to do?What to do?

Small input angles can force even smaller Small input angles can force even smaller ouput angles. [Shewchuk ’02]ouput angles. [Shewchuk ’02]


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No Large AnglesNo Large Angles


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Polygons with HolesPolygons with Holes[Bern, Mitchell, Ruppert 95] – [Bern, Mitchell, Ruppert 95] –

- All triangles are - All triangles are nonobtuse. nonobtuse.

- Output has O(n) triangles.- Output has O(n) triangles.


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Polygons with HolesPolygons with Holes[Bern, Mitchell, Ruppert 95] – [Bern, Mitchell, Ruppert 95] –

- All triangles are - All triangles are nonobtuse. nonobtuse.

- Output has O(n) triangles.- Output has O(n) triangles.

Does not work for arbitrary PSLGsDoes not work for arbitrary PSLGs


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Paterson’s ExamplePaterson’s ExampleRequires Requires (n(n22) points.) points.

O(n) O(n) pointpoint

ss

O(n) O(n) lineline

ss


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ss

O(n) O(n) lineline

ss


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ss

O(n) O(n) lineline

ss


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ss

O(n) O(n) lineline

ss


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ss

O(n) O(n) lineline

ss


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ss

O(n) O(n) lineline

ss


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Propagating PathsPropagating Paths


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First introduced by Mitchell [93]First introduced by Mitchell [93]

Later Improved by Tan [96]Later Improved by Tan [96]

Worst Case Optimal Size O(nWorst Case Optimal Size O(n22))

Angle bounds: 132Angle bounds: 132oo


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The OSM AlgorithmThe OSM Algorithm(Overlay Stitch Meshing)(Overlay Stitch Meshing)


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Results:Results:1.1. Angles bounded by 170Angles bounded by 170oo

2.2. O(log L/s) competitive sizeO(log L/s) competitive size


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• An Overlay Edge is An Overlay Edge is keptkept if if1. It does not intersect the input,1. It does not intersect the input,

OROR

2. It forms a 2. It forms a goodgood intersection with the intersection with the input.input.

at least 30at least 30oo


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Angle GuaranteesAngle Guarantees


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We want to show that angles at We want to show that angles at stitchstitch vertices are bounded by vertices are bounded by

170170oo


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An Overlay TriangleAn Overlay Triangle


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An Overlay EdgeAn Overlay Edge


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An Overlay EdgeAn Overlay Edge

Gap BallGap Ball


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A Good IntersectionA Good Intersection


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A Bad IntersectionA Bad Intersection


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How Bad can it How Bad can it be?be?


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Angle GuaranteesAngle GuaranteesHow Bad can it How Bad can it

be?be?

About 10About 10oo

About 170About 170oo


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Angle GuaranteesAngle GuaranteesHow Bad can it How Bad can it

be?be?

About 10About 10oo

Theorem: If any input Theorem: If any input edge makes a edge makes a goodgood intersection with an intersection with an overlay edge then any overlay edge then any other intersection on other intersection on that edge is that edge is not too not too badbad..

About 170About 170oo


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Size BoundsSize Bounds


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Size BoundsSize BoundsHow big is the resulting triangulation?How big is the resulting triangulation?


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Size BoundsSize BoundsHow big is the resulting triangulation?How big is the resulting triangulation?

Goal: log(L/s)-competitive with optimalGoal: log(L/s)-competitive with optimal


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Local Feature SizeLocal Feature Size

lfs(x) = distance to second nearest input vertex.lfs(x) = distance to second nearest input vertex.

xxlfs(x)lfs(x)

Note: lfs is defined on Note: lfs is defined on the whole plane before the whole plane before

we do any meshing.we do any meshing.


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Ruppert’s IdeaRuppert’s Idea

rr

cc

r = r = (lfs(c))(lfs(c))

area(area() = ) = (lfs(c)(lfs(c)22))

# of triangles = # of triangles = ((ssss 1/lfs(x,y) 1/lfs(x,y)22 dxdy) dxdy)


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rr

cc



# of triangles = # of triangles = ((ssss 1/lfs(x,y) 1/lfs(x,y)22 dxdy) dxdy)

dydy

dxdxThis represents a This represents a (dxdy/lfs(x,y)(dxdy/lfs(x,y)22)) fraction of the fraction of the

triangle.triangle.


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rr

cc



# of triangles = # of triangles = ((ssss lfs(x,y) lfs(x,y)-2-2 dxdy) dxdy)

CaveatCaveat: Only Works for meshes : Only Works for meshes

with bounded with bounded smallestsmallest angle. angle.


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ExtendingExtending Ruppert’s Idea Ruppert’s Idea

An input edge An input edge e e intersecting intersecting

the overlay meshthe overlay mesh

# of triangles along # of triangles along ee = = ((sszz22 e e lfs(z) lfs(z)-1-1 dz) dz)



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ExtendingExtending Ruppert’s Idea Ruppert’s Idea

Now we can compute the Now we can compute the size of our output mesh. We size of our output mesh. We just need to compute these just need to compute these

integrals.integrals.

# of triangles along # of triangles along ee = = ((sszz22 e e lfs(z) lfs(z)-1-1 dz) dz)



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Output SizeOutput Size

Stitch Vertices Added = Stitch Vertices Added = ((sszz22 E E lfs(z) lfs(z)-1-1 dz) dz)

Overlay Mesh Size = Overlay Mesh Size = ((ssss lfs(x,y) lfs(x,y)-2-2 dxdy) dxdy) O(n log L/s)O(n log L/s)

O(log L/s)O(log L/s)££ |OPT| |OPT|

We’ll Prove this nowWe’ll Prove this now


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What can we say about the What can we say about the optimal optimal no-large-angle no-large-angle

triangulation?triangulation?


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Competitive AnalysisCompetitive AnalysisWarm-upWarm-up

9090oo


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??oo


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Suppose we have a triangulation with all angles at least 170Suppose we have a triangulation with all angles at least 170oo..

the optimalthe optimal


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Suppose we have a triangulation with all angles at least 170Suppose we have a triangulation with all angles at least 170oo..

the optimalthe optimal Every input edge is Every input edge is covered by empty covered by empty

lenses.lenses.


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Competitive AnalysisCompetitive AnalysisThis is what we have to integrate over.This is what we have to integrate over.

We show that in each lens, we We show that in each lens, we put O(log (L/s)) points on the put O(log (L/s)) points on the

edge.edge.

In other words:In other words:

sse’e’ 1/lfs(z) dz = O(log (L/s)) 1/lfs(z) dz = O(log (L/s))

ee’’


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Competitive AnalysisCompetitive Analysis

ee’’

sszz22 e’ e’ 1/lfs(z) dz = O(log (L/s)) 1/lfs(z) dz = O(log (L/s))


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Competitive AnalysisCompetitive Analysissszz22 e’ e’ 1/lfs(z) dz = O(log (L/s)) 1/lfs(z) dz = O(log (L/s))

ee’’

11stst trick: lfs trick: lfs ¸̧ s everywhere s everywhere

22ndnd trick: lfs trick: lfs ¸̧ ct for t ct for t22 [0,[0,ll/2]/2]

tt

sszz22 e’ e’ 1/lfs(z) dz 1/lfs(z) dz ·· 2 2ss00ss 1/s + 2 1/s + 2ssss

ll/2/2 1/x 1/x dx dx

= O(1) + O(log = O(1) + O(log ll/s)/s)

= O(log L/s)= O(log L/s)

Parametrize e’ as [0, Parametrize e’ as [0, ll]]


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ConclusionsConclusions


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ConclusionConclusion• A new algorithm for no-large-angle A new algorithm for no-large-angle

triangulation.triangulation.• The output has bounded degree The output has bounded degree

triangles.triangles.• The first log-competitive analysis for The first log-competitive analysis for

such an algorithm.such an algorithm.• We used some of the high school We used some of the high school

calculus you thought you forgot.calculus you thought you forgot.


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Where to go from here?Where to go from here?• 3D?3D?• Better angle bounds?Better angle bounds?• Leverage lower bound technology for Leverage lower bound technology for

constant competitive algorithm.constant competitive algorithm.


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Thank you.Thank you.

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Overlay Stitch Meshing