# Lecture 12_ 13 - PowerPoint Presentation

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```					  Advanced Iso-Surfacing
Algorithms
Jian Huang, CS594, Spring 2002

This set of slides are developed and used by Prof.
Han-Wei Shen at Ohio State University.
Iso-contour/surface Extractions

2D Iso-contour   3D Iso-surface
Iso-contour (0)

Remember bi-linear interpolation

p2              p3
To know the value of P,
P =?               we can first compute p4 and
p4                p5       P5 and then linearly interpolate
P
p0              p1
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    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

p0               p1
Iso-contour (2)

We can solve the problem based on linear interpolation

p2              p3     (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
p0              p1
Iso-contouring – Step 1

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

If v1 < C < v2 then the edge contains such a point
v1            v2
Iso-contouring – Step 2

(2) Calculate the position of P

p1     P                 Use linear interpolation:
p2
P = P1 + (C-v1)/(v2-v1) * (P2 – P1)
v1     C      v2
Iso-contouring – Step 3

p2      p3     Connect the points with line(s)

Based on the principle of linear variation, all
the points on the line have values equal C
p0      p1
Inside or Outside?
Just a naming convention

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

p2       p3               p2        p3

-        +    p0       p1         -     p0        p1
outside cell          inside cell
Iso-surface Extraction

Extend the same divide-and-conquer algorithm to
three dimension

• 3D cells

• Look at one cell at a time

• Let‟s only focus on voxel
Divide and Conquer
_
+
_
+               _
+
_
+
+       _
_
(2 triangles)    +               _
+
_
+
How many cases?

Now we have 8 vertices

So it is: 2 8= 256

How many unique topological cases?
Case Reduction (1)

Value Symmetry
_                _          +           +
_
+                           _       +
_                     +
_                       +
_          _
+                                       +
Case Reduction (2)

Rotation Symmetry
_               _              _            _
_                               +
+                             _
_              _
_                                       _
_             _
+
+

By inspection, we can reduce 256       14
Iso-surface Cases
Total number of cases: 14 + 3
Marching Cubes Algorithm

A Divide-and-Conquer Algorithm

Vi is „1‟ or „0‟ (one bit)
v8          v7    1: > C; 0: <C (C= iso-value)
v4         v3
v6   Each cell has an index mapped to a
v5          value ranged [0,255]

v1         v2

Index = v8 v7 v6 v5 v4 v3 v2 v1
Marching Cubes (2)

Given the index for each cell, a table lookup is performed
to identify the edges that has intersections with the
iso-surface                     Index intersection edges
0    e1, e3, e5
e7
1     …
e11        e8    e12      e6   2
e3
e5        3
e4           e2
e9              e10
e1
14
Marching Cubes (3)

+       _
• Perform linear interpolations
_             at the edges to calculate the
+               _     intersection points
+
_       • Connect the points
+
Why is it called marching cubes?

Linear search through cells

•Row by row, layer by layer
•Reuse the interpolated points
Iso-surface cell search

• Iso-surface cells: cells that contain iso-
surface.
min < iso-value < max

• Marching cubes algorithm performs a linear
search to locate the iso-surface cells – not
very efficient for large-scale data sets.
Iso-surface Cells

• For a given iso-value, only a smaller portion of
cells are iso-surface cell.

• For a volume with
n x n x n cells, the
n
average number of the                         n
iso-surface cells is O(n x n)       n
(ratio of surface v.s. volume)
Efficient iso-surface cell search
• Problem statement:
Given a scalar field with N cells, c1, c2, …,
cn, with min-max ranges (a1,b1), (a2,b2), …,
(an, bn)

Find {Ck | ak < C < bk; C=iso-value}
Efficient search methods

1. Spatial subdivision (domain search)
2. Value subdivision (range search)
3. Contour propagation
Domain search
•   Subdivide the space into several sub-domains, check the
min/max values for each sub-domain
•   If the min/max values (extreme values) do not contain the
iso-value, we skip the entire region
Min/max

Complexity = O(Klog(n/k))
Range Search (1)

Subdivide the cells based on their min/max ranges
Global minimum                                        Global maximum

Hierarchically subdivide the cells based on
their min/max ranges
Isovalue
Range Search (2)

Within each subinterval, there are more than one cells
To further improve the search speed, we sort them.

Sort by what ?          Min and Max values
G1
Max M5    M2    M6    M4     M1 M3 M7     M8 M11 M10 M9

G2
Min m5    m1   m6     m3    m8 m7 m2      m9 m11 m4 m10

Isosurface cells = G1      G2
Range Search (3)
A clean range subdivision is difficult …

?

Difficult to get an optimal speed
Range Search: Span Space

Span Space: Instead of treating each cell as a range,
we can treat it as a 2D point at (min, max)

This space consists of min and max axes is called span space

Any problem here?
Span Space

What are the iso-surface cells?

max
How to search them?

min
C
Span Space Search (1)

With the point representation, subdividing the space is
much easier now.

Search method 1:    K-D tree subdivision (NOISE algorithm)
K-d tree:
• A multi-dimensional version of binary tree
• Partition the data by alternating between each
each of the dimensions at each level of the tree
NOISE Algorithm (K-d tree)
Median point
Min
Construction
left
Max
right

?
max
up        down

…              … …
* One node per cell

min
NOISE Algorithm (Query)
Median point
Min   If ( iso-value < root.min )
• check the ?? Sub-tree
left
Max
right
If (iso-value > root.min)
?   • Check the ?? Sub-tree
• Don‟t forget to check the
up        down                       root„s interval as well.
…              … …
Complexity = O( N + k)
Span Space Search (2)
Search Method (2): ISSUE, discretized span space
O(log(N/L))
O(1)
Complexity = ?

?

O(log(N/L))
Range Search: Interval Tree
Interval Tree:         Sort all the data points
(x1,x2,x3,x4,…. , xn)
Let d = x n/2 (mid point)
Id
We use d to divide the cells into three
sets Id, I left, and I right
I left        I right

…             …            Id : cells that have
I left: cells that have
min < d
max < d
< max

I right: cells that have   min >   d
Interval Tree
Now, given an isovalue C
1)   If C < d
Id                            2)   If C > d
3) If C = d

I left          I right

…                         …                           Complexity = O(log(n)+k)

min < d     < max           Optimal!!
Id : cells that have
I left: cells that have    max < d
I right: cells that have   min >   d
Range Search Methods

In general, range search methods all are super fast –

two orders of magnitude faster than the marching cubes
algorithm in terms of cell search

But they all suffer a common problem …

Excessive extra memory requirement!!!
Contour Propagation
Basic Idea:

Given an initial cell that contains iso-surface, the remainder of
the iso-surface can be found by propagation
FIFO Queue
Initial cell: A    A

C      Enqueue: B, C      BC
E
A          Dequeue: B         C           Search
B
D      Enqueue: D         CD

…                  ….
Challenges
Need to know the initial cells!

For any given iso-value C, finding
the initial cells to start the propagation
is almost as hard as
finding the iso-surface cells.

You could do a global search, but …
Solutions

(1) Extrema Graph (Itoh vis‟95)
(2) Seed Sets (Bajaj volvis‟96)

Problem Statement:

Given a scalar field with a cell set G, find a subset S    G, such that
for any given iso-value C, the set S contains initial cells to start
the propagation.

We need search through S, but S is usually (hopefully) much smaller
than G.

We will only talk about extrema graph due to time constraint
Extrema Graph (1)

Basic Idea:

If we find all the local minimum and
maximum points (Extrema), and connect them
together by straight lines (Arcs), then any closed
Iso-contour is intersect by at least one of the arcs.
Extrema Graph (2)
Extrema Graph (3)
Extreme Graph:
E1         E2
{
E, A:
a1                                E: extrema points
a2
A: Arcs conneccts E
E3         a3                       }
E4
a5          An ‘arc’ consists of cells that connect
a4                               extrema points (we only store
E7
min/max of the arc though)
E5                      a7
E6        a6          E8
Extrema Graph (4)

Algorithm:

Given an iso-value
1) Search the arcs of the extrema graph (to find the arcs that have
min/max contains the iso-value
2) Walk through the cells along each of the arcs to find the seed
cells
3) Start to propagate from the seed cells
4) ….

There is something more needs to be done…
We are not done yet …

What ?!
We just mentioned that all the closed iso-contours will
intersect with the arcs connecting the extrema points

How about non-closed iso-contours? (or called open iso-
contours)
Extrema Graph (5)

Contours missed

These open iso-contours will
intersect with ?? cells

Boundary Cells!!
Extrema Graph (6)

Algorithm (continued)
Given an iso-value
1) Search the arcs of the extrema graph (to find the arcs that have
min/max contains the iso-value
2) Walk through the cells along each of the arcs to find the seed
cells
3) Start to propagate from the seed cells

4) Search the cells along the boundary and find seed cells from
there
5) Propagate open iso-contours
Extrema Graph
• Efficiency - Number of cells visisted:
– extrema graph - N0.33
– boundary - N 0.66
– Iso-surface - N 0.66
• based on tetrahedra - will create more surface triangles ...
• should extract the same number of cells/ triangulation as
Marching Cubes
Ambiguity Problem

Certain Marching Cube cases have more than one
possible triangulation

Mismatch!!!
Hole!

+                 +
Case 6                               Case 3
+                  +
The Problem

Ambiguous Face: a face that has two diagonally opposing
points with the same sign

+

+

Connecting either way is possible
To fix it …

Match!!!

+                     +
Case 6                                       Case 3 B
+                    +

The goal is to come up with a consistent triangulation
Solutions

There are many solutions available – we present a method called:

Asymptotic Decider by Nielson and Hamann (IEEE Vis‟91)
Asymptotic Decider

Based on bilinear interpolation over faces

B01           B11

(s,t)                             B00 B01      1-t
B(s,t) = (1-s, s)
B10 B11      t

B00           B10   The contour curves of B:

{(s,t) | B(s,t) = a } are hyperbolas
Asymptotic Decider (2)

(1,1)

(0,0)                    Where the hyperbolas
go through the cell
depends on the values
at the corners, I.e.,
B00, B01, B10, B11
Asymptotic Decider (3)

(1,1)           (Sa, Ta)

If a   < B(Sa, Ta)

(0,0)

Asymptote
Asymptotic Decider (4)

(1,1)           (Sa, Ta)

If a   >   B(Sa, Ta)

(0,0)

Asymptote
Asymptotic Decider (5)

(1,1)   (Sa, Ta)

Sa =      B00 - B01
B00 + B11 – B01 – B10
(0,0)

Ta=       B00 – B10
B00 + B11 – B01 – B10

B(Sa,Ta) = B00 B11 + B10 B01
B00 + B11 – B01 – B10
Asymptotic Decider (6)

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 ambiguious
faces).

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