Contouring - The Problem More Formallycrawfis.3/cis694L/Slides/... · 2003. 4. 17. · (2)...

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Iso-Contouring and Level-Sets

Roger Crawfis

Contributors: Roger Crawfis, Han-Wei Shen, Raghu Machiraju, Torsten Moeller, Fan Ding, Charles Dyer, and Huang Zhiyong

4/17/2003 R. Crawfis, Ohio State Univ. 2

Iso-contour/surface Extractions

2D Iso-contour 3D Iso-surface


Contouring - The Problem

Extracting an isosurface from an implicit function, that is, Extracting a surface from volume data (discrete implicit function), f (x ,y ,z ) = T


More Formally

A scalar visualization technique that creates curves (in 2D) or surfaces (in 3D) representing a constant scalar value across a scalar field.Contour lines are called isovalue lines or isolines.Contour surfaces are called isovaluesurfaces or isosurfaces


Contouring a 2D structured grid with contour line value = 5

0 1 1 3 2

1 3 6 6 3

3 7 9 7 3

2 7 8 6 2

1 2 3 4 3

1. Using interpolation to generatepoints along edges with the constantvalue

2. Connect these points into contoursusing a few different approaches. Oneof the approaches:

. Detects edge intersection

. Tracks this contour as it moves acrosscell boundary

. Repeat for all contours4/17/2003 R. Crawfis, Ohio State Univ. 6

2D Contouring – Not so easy

AnnotationsSmooth Curves

Topographic Map of Jerusalem (Contour interval 10 meters)

North is at the top of the map. The Mount of Olives is on the far right, Mount Zion on the left. Mount Moriah rises as a long ridge at the south end of the City of David and continues on past the present Temple Mount, and reaches its highest point outside the Northern walls of the Old City, at the top of the map.

http://www.skullandcrossbones.org/articles/solomontemple/solomontemple2.htm

2


Quantitative and Qualitative

In this figure, we can actually read off any value to about three significant digits.Hard and tedious problem to determine where to place labels. Usually done as vector graphics, rather than raster graphics for precision.

http://omnimap.com/catalog/cats/fish/fish-contour.htm


Divide and Conquer

Partition space into rectangular sub-regions.


Dividing Cubes/Squares

Generates Points and renders themRendering points is faster than polygonsPrinciple - Divide a square until contour passes thru the cell

1 2 3 4 3

2 7 8 6 2

3 7 9 7 3

1 3 6 6 3

0 1 1 3 2

Iso-value=54/17/2003 R. Crawfis, Ohio State Univ. 10

Dividing Cubes

1. Create a cell2. Determine if the cell is a surface cell3. Subdivide surface cells to image resolution4. Determine intensity or color of refined cell5. Output a point for each remaining surface cell


Step 3: Subdivide

Subdivide each cell such that the resolution is higher than the image resolution. For example, when a 1282

volume is rendered to a 5122 image, cells containing the contour will be divided to 4x4x4 smaller cubes.

classify subdivide output points


Surface TrackingGiven function S(t,x,y,z) that returns 1 if (x,y,z) is on the surface t and 0 otherwiseNeeded: a seed voxel p that is on the surface.

1 2 3 4 3

2 7 8 6 2

3 7 9 7 3

1 3 6 6 3

0 1 1 3 2

Iso-value=5Seed cell

3


Surface TrackingLet Q be a queue of voxels.

Push p onto QFlag p as “visited”While Q is not empty do

Pop q from Q and output it.For each voxel v in the neighborhood of q doIf S(t, v) = 1 then

If v was not visited thenPush v onto QFlag it as “visited”

end {while}


Tracking and Smooth Curves

Recall that the gradient to a scalar field is normal to the isolines.Therefore, the curve will be towards the direction perpendicular to the gradient.Can also use the gradient to calculate tangents for each point on the curve to generate smooth curves.


Restricted Topological Cases

Filled circle vertices indicate scalar value is above the contour valueUnfilled ones which scalar values are below the contour values

•A closed curve can not be contained in the cell.•The curve can only enter and exit the cell once per edge.


Not Allowed


Linear Topology

The restricted set is topologically equivalent to linear segments.Topologically equivalent if we restrict the edge intersections to lie at the mid-points of each edge.


Using Symmetry to Reduce to 5

Contour ambiguity: either solidor dash lines look OK

4


2D Iso-contour (0)

Remember bi-linear interpolation

p2 p3

p0 p1

P =?p4 p5

To know the value of P, we can first compute p4 andP5 and then linearly interpolateP


2D Iso-contour (1)

Consider a simple case: one cell data set

The problem of extracting an iso-contour is an inverse of value interpolation. That is:

p2 p3

p0 p1

Given f(p0)=v0, f(p1)=v1, f(p2)=v2, f(p3)=v3

Find the point(s) P within the cell that have values

F(p) = C


2D Iso-contour (2)

p2 p3

p0 p1

We can solve the problem based on linear interpolation

(1) Identify edges that contain points P that have value f(P) = C

(2) Calculate the positions of P

(3) Connect the points with lines


2D Iso-contouring – Step 1

(1) Identify edges that contain points P that have value f(P) = C

v1 v2

If v1 < C < v2 then the edge contains such a point



(2) Calculate the position of P

Use linear interpolation:

P = P1 + (C-v1)/(v2-v1) * (P2 – P1)v1 v2

Pp1 p2

C



p2 p3

p0 p1

Connect the points with line(s)

Based on the principle of linear variation, all the points on the line have values equal C

5


Inside or Outside?

Just a naming convention

1. If a value is smaller than the iso-value, we call it “Outside”2. If a value is greater than the iso-value, we call it “Inside”

p2 p3

p0 p1- +inside cell

p2 p3

p0 p1-

outside cell4/17/2003 R. Crawfis, Ohio State Univ. 26

Contouring in 3D

Treat volume as a set of 2D slicesApply 2D Contouring algorithm on each slice.Or given as a set of hand-drawn contours

Stitch the slices together.


Contour Stitching••Problem:Problem:

Given: 2 twoGiven: 2 two--dimensional dimensional closedclosed curvescurvesCurve #1 has Curve #1 has mm pointspointsCurve #2 has Curve #2 has nn pointspoints

Which point(s) does vertex Which point(s) does vertex ii on curve one correspond to on curve one correspond to on curve two?on curve two? ?

?

i


A Solution

Fuchs, et. al.Optimization problem1 stitch consists of:⌧2 spans between curves⌧1 contour segment

Triangles of {Pi,Qj,Pi+1} or {Qj+1,Pi,Qj}⌧Consistent normal directions

Pi Pi+1

Qj


Fuchs, et. al.

Left spanPiQj => go up

Right span (either)Pi+1Qj => go downPiQj+1 => go down


Fuchs, et. al.

ConstraintsEach contour segment is used once and only once.If a span appears as a left span, then it must also appear as a right span.If a span appears as a right span, then it must also appear as a left span.

6


Fuchs, et. al.

This produces an acceptable surface (from a topological point of view)

No holes

We would like an optimal one in some sense.


Fuchs et. al.

Graph problemVertices Vij = span between Pi and QjEdges are constructed from a left span to a right span.

Only two valid right spans for a left span.

PPii

QQjj QQj+1j+1

PPi+1i+1


Fuchs, et. al.

Organize these edges as a grid or matrix.

PP

QQ

i

j

PiQjPi+1

QjPiQj+1


Fuchs, et. al.

Acceptable graphsExactly one vertical arc between Pi and Pi+1

Exactly one horiz. arc between Qj and Qj+1

Either⌧indegree(Vij) = outdegree(Vij) = 0⌧both > 0

• (if a right, also has to be a left)


Fuchs, et. al.

Claim:An acceptable graph, S, is weakly connected.

Lemma 2Only 0 or 1 vertex of S has an indegree = 2.⌧E.g., Two cones touching in the center.

All other vertices have indegree=1(recall indegree = outdegree)


Fuchs, et. al.

Thereom 1: S is an acceptable surface if and only if:

S has one and only one horizontal arc between adjacent columns.S has one and only one vertical arc between adjacent rows.S is Eulerian (closed walk with every arc only once).

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Fuchs, et. al.

Number of arcs is thus m+n.Many possible solutions still!!!Associate costs with each edge

Area of resulting triangleAspect ratio of resulting triangle


Fuchs, et. al.

Note that Vi0 is in S for some i.Note that V0j is in S for some j. With the weights (costs), we can compute the minimum path from a starting node Vi0.

Since we do not know which Vi0, we compute the paths for all of them and take the minimum.⌧Some cost saving are achievable.


Marching Cubes

Predominant method used today.Efficient and simpleIt was independently reported by Wyvilland McPeeters in 1986, Lorenson and Cline in 1987


Marching Cubes

Treat each cube individualNo 2D contour curves

Allow intersections only on the edges or at vertices.Pre-calculate all of the necessary information to construct a surface.


Marching Cubes

Consider a single cubeAll vertices above the contour thresholdAll vertices belowMixed above and below

+ +

+

+ +

+

-+


Marching Cubes

Binary label each node => (above/below) Examine all possible cases of above or below for each vertex.8 vertices implies 256 possible cases.

+ +

+

+ +

+

-+

8


How many cases?

Now we have 8 vertices

So it is: 2 = 2568

How many unique topological cases?


Case Reduction

Value Symmetry

+

+

_ _

_

__

_+

+

_

_

+

+

+

+


Case Reduction

Rotation Symmetry

+

+

_ _

_

__

__

_

+

+

_ _

__


Case Reduction

Mirror Symmetry

+

+

_ +

_

+_

__

+

+

+

+ _

__

By inspection, we can reduce 256 14


Marching Cube - Summary

Create a cubeClassify each voxelBuild an indexLookup edge listInterpolate triangle verticesCalculate and interpolate normals


Step 1: Create a Cube

Consider a cube defined by eight data values: four from slice K, and four from slice K+ 1

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Step 2: Classify Each Voxel

Binary classification of each vertex of the cube as to whether it lies

outside the surface (voxel value < isosurface value)or inside the surface (voxel value <= isosurfacevalue).


Step 3: Build an Index

Use the binary labeling of each voxel to create an 8-bit index:

8 vertices - 256 casesRequires a canonical ordering of the cube.


Step 4: Look-up the Topology

Given the index for each cell, a table lookup is performed to identify the edges that has intersections with the iso-surface.Also requires a canonical ordering of the edges: e1, e2, …, e12

0123

14

e1, e3, e5…

Index intersection edges

e1

e2

e3

e4e5

e6

e7e8

e9 e10

e11 e12


Step 4: Look-up the Topology

all 256 cases could be derived from 15 base cases.Difficult to determine rotational symmetry and map it correctly to the table.


Step 4: Example

Index = 10001101triangle 1 = e4,e7,e12triangle 2 = e1, e7, e4triangle 3 = e1, e6, e7triangle 4 = e1, e10, e6

e1e10

e6

e7e12

e4

v1 v2

v7

e2


Step 5: Interpolate Triangle Vertices

For each edge, find the vertex location along the edge using linear interpolation of the voxel values.

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Step 6: Compute Normals

Calculate the normal at each cube vertexUse linear interpolation to interpolate the polygon vertex normal


Why is it called marching cubes?

Linear search through cells

•Row by row, layer by layer•Reuse the interpolated pointsfor adjacent cells

No neighborhood informationNo neighborhood informationis required!!!is required!!!


Problems

Ambiguities: Are the curves below right or wrong?


Problems

Inconsistencies: Does the result produce the correct topology?An iso-contour surface should always be a manifold.

We know whether these cases are right or wrong.

wrong


2D Contouring Ambiguity

Break contour Join contour

or

Either choices areacceptable becausecontour lines are continuousand closed

Independent of otherchoices


Marching Cubes

Topological inconsistencies in the 15 cases

Turns out positive and negative are not symmetric.

+ +

+

+ +

+- -

+ -

-

+ +

---

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A Bad Example

Resolution:Favor positive pathways thru the voxel(or negative, but be consistent).

Case 3 Case 3

Case 3Case 6

(compliment symmetry) 4/17/2003 R. Crawfis, Ohio State Univ. 62

Ambiguous CasesAmbiguous cases:3, 4, 6, 7, 10, 12, 13Adjacent vertices: different statesDiagonal vertices: same stateCase 4 is not usually considered ambiguous, but we could favor a tunnel through the long diagonal.

or

or


Ambiguous Cases

The ambiguous cases lead to a continuous manifold, but due to improper sampling of the volume, a (somewhat arbitrary) decision can be made on whether to close two curves or connect them.

Asymptotic Decider - by Nielson and Hamann(IEEE Vis’91) examines neighboring cells to determine this choice.


Asymptotic Decider

Based on bilinear interpolation over faces

B01

B00 B10

B11

(s,t)B(s,t) = (1-s, s) B00 B01

B10 B111-tt

The contour curves of B:

{(s,t) | B(s,t) = α } are hyperbolas


Asymptotic Decider (2)

(0,0)

(1,1)

Where the hyperbolasgo through the cell depends on the valuesat the corners, I.e., B00, B01, B10, B11



(0,0)

(1,1)

Asymptote

(Sα, Tα)

If α < B(Sα, Tα)

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(1,1)

Asymptote

(Sα, Tα)

(0,0)

If α > B(Sα, Tα)



(1,1) (Sα, Tα)

(0,0)

Sα = B00 - B01 B00 + B11 – B01 – B10

Tα= B00 – B10 B00 + B11 – B01 – B10

B(Sα,Tα) = B00 B11 + B10 B01 B00 + B11 – B01 – B10



Based on the result of asymptotic decider, we expand the marching cube case 3, 6, 12, 10, 7, 13(These are the cases with at least one ambiguiousfaces).


Marching Cubes Demo

Animating the contour valueSpecial functions for contouringVarying speeds and numbers of triangles


Marching Cubes

Data Structures/Tablesstatic int const HexaEdges[12][2] = { {0,1}, {1,2}, {2,3}, {3,0},

{4,5}, {5,6}, {6,7}, {7,4},{0,4}, {1,5}, {3,7}, {2,6}};

typedef struct {EDGE_LIST HexaEdges[16];

} HEXA_TRIANGLE_CASES;

/* Edges to intersect. Three at a time form a triangle. */

static const HEXA_TRIANGLE_CASES HexaTriCases[] = {{-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1}, /* 0 */{ 0, 8, 3, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1}, /* 1 */{ 0, 1, 9, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1}, /* 2 */{ 1, 8, 3, 9, 8, 1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1}, /* 3 */

...


Marching Cubes - How simple

/* Determine the marching cubes index */for ( i=0, index = 0; i < 8; i++)

if (val1[nodes[i]] >= thresh) /* If the nodal value is above the */index |= CASE_MASK[i]; /* threshold, set the appropriate bit. */

triCase = HexaTriCases[index]; /* triCase indexes into the MC table. */edge = triCase->HexaEdges; /* edge points to the list of intersected edges */

for ( ; edge[0] > -1; edge += 3 ) /* stop if we hit the -1 flag */{for (i=0; i<3; i++) /* Calculate and store the three edge intersections */{

vert = HexaEdges[edge[i]];n0 = nodes[vert[0]];n1 = nodes[vert[1]];t = (thresh - val1[n0]) / (val1[n1] - val1[n0]);tri_ptr[i] = add_intersection( n0, n1, t ); /* Save an index to the pt. */

}add_triangle( tri_ptr[0], tri_ptr[1], tri_ptr[2], zoneID); /* Store the triangle */

}}

Contouring - The Problem More Formallycrawfis.3/cis694L/Slides/... · 2003. 4. 17. · (2)...

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