1. Data Visualization Lecture 4 Two Dimensional Scalar Visualization
Part 2: Further Contouring and Other 2D Techniques

2. Return to Example Consider this data: -5 10 1 -2
Where does the zero level contour go?
Can we draw a straight line which will
approximate the contour line?

3. Joining Intersections Directly
It is tempting to approximate by joining intersections with straight lines: 10 -5 -2 1

4. Ambiguity But this does not always work - look at this data: 10 -5 -3
Try it - it is ambiguous!

5. What is the Problem?
The contour of the bilinear interpolant is NOT a straight line – it is a curve - 10 -5 -2 1
Joining intersections
with straight lines
was only an
approximation
This is curve of: f(x,y) = (1-x)(1-y)f00+x(1-y)f10+(1-x)yf01+ xyf11
= 0
BUT how can we draw it?

6. Tracking Contours
We can track the contour in small steps through the grid rectangle - starting from the intersection with the edges
take a step, probe at an equal distance to either side, then predict next point; and so on
-0.3
0.9
-0.3
0.9
Next point on
contour
Current point
on contour
Probes
BUT THIS IS SLOW!!

7. Solution by Decomposing Cell
One technique – more efficient than tracking - is to split cell into four triangles
Within a triangle, we can fit a linear model
F(x,y) = a + bx +cy
How do we split?
How do we calculate a,b,c?
What is the gain? f1 f2 f3

8. Cell Decomposition
Problem of drawing the curved lines has been circumvented by decomposing cell into four pieces within which the contours are well defined straight lines -5 10
0.75 -3 1

9. Avoiding Cell Decomposition
As an alternative to decomposing the cell, we can try to understand how the curved contours of a bilinear interpolant behave
The difficult case is: + -
Opposite vertices: two +
and two -
Where do the contour
lines go?

10. Bilinear Interpolant
The bilinear interpolant has contours which are hyperbolas, and can be one of two forms: + - A + - B
In each case there is a saddle point - if saddle point is
-ve, then we have case A; if +ve, case B.
Saddle point: fx=fy=0 .. Max in one direction, min in other

11. A Possible Method This suggests following method:
in ambiguous case, calculate saddle point and join intersection points according to how bilinear behaves: + - A B - + - +
saddle negative
saddle positive
Try it on the example earlier

12. Saddle Point
It is possible to calculate the saddle point in terms of vertex values (from fx = fy =0 ):
And the corresponding value:
x = (f00 - f01 ) / D;
y = (f00 - f10 ) / D
D = f00 + f11 - f01 -f10
Saddle value = (f00f11 - f01f10)/D

13. Problem
In the ambiguous case, will the earlier four triangle method always give the same result as the saddle point method? 5
-150
100 -1
Try this
with both
approaches

14. Greater Accuracy
The greater understanding of the bilinear leads to a more accurate method
A single straight line approximation can be made more accurate by using two straight line pieces
see Lopes and Brodlie paper on Web site B + - + - - + A

15. 2D Interpolation - Scattered Data
Often the data will be given, not on a regular grid, but at scattered locations:
f given at each
marked point
Approach:
(i) triangulate
(ii) build interpolant
in each triangle
(iii) draw contours

16. Triangulation
Triangulation is the process of forming a grid of triangles from the data points
How can we construct the triangulation?

17. Tesselation We solve the DUAL problem:
Suppose a wolf is stationed at each data point. Each wolf is equally powerful and dominates the territory closest to its own base
What are the territories dominated by each wolf?

18. Tesselation - Two or Three Wolves

19. Tesselation - Two or Three Wolves

20. Dirichlet Tesselation
The resulting tesselation is known as the Dirichlet or Voronoi tesselation
Given the Dirichlet tesselation for N points
P1, P2, ... PN
there is an algorithm for constructing the tesselation when an extra point is added

21. Dirichlet TesselationP2
Tesselation for
P1, P2, P3 P3 P1 Q
Point Q
added

22. Dirichlet TesselationP2 P3 P1 Q

23. Dirichlet TesselationP2 P3 P1 Q

24. Dirichlet Tesselation
Determine polygon containing Q - here D3, surrounding P3
Construct perpendicular bisector of P3Q and find intersection with D3 - this becomes point of modified tesselation
Determine adjacent polygon - here D2
Repeat the above two steps until D3 is reached again, or there is no intersection
Remove all vertices and edges interior to the new polygon

25. Delaunay TriangulationP2 P3 P1 Q

26. Delaunay Triangulation
Triangulation formed by joining points whose ‘territories’ share a common boundary in the tesselation
This has the nice property that it avoids long skinny triangles
See the nice applets at:
Delaunay.html
Note the ‘empty circle’ property of the Delaunay triangulation

27. Contouring from Triangulated Data
The final step is to contour from the triangulated data
Easy – because contours of linear interpolant are straight lines – see earlier

28. Implementing Triangle-based Contouringf1 f2 f3
Each vertex can be positive or negative (ignore zero for now)
This gives 23 = 8 possible cases…
… but there are only 2 distinct configurations
No contour (all same sign)
Contour (2 of one sign, 1 of the other)
Implementation:
Determine which of 8 cases
Select code for the appropriate configuration
All same sign f1 f2 f3
Two same sign

29. Implementing Rectangle-based Contouring+
For a rectangle, there will be 24 = 16 cases
There are 4 configurations
All same sign (no contour)
3 same sign (one contour piece)
2 adjacent with same sign (one contour piece)
2 opposite with same sign (two pieces, but ambiguous) + - + - + -

30. Surface Views
A different mapping technique for 2D scalar data is the surface view.
Here a surface is created in 3D space, the height representing the scalar value
Construction is quite easy - suppose we have a rectangular grid

31. Constructing a Surface View - 1

32. Constructing a Surface View - 2
Surface created as pair of triangles per grid rectangle.
Rendering step is then display of triangles.

33. Examples - with added contours

34. Image Plots A further mapping technique for 2D data is the image plot
There are three variants:
dot array : draw a dot at each data point, coloured according to the value (very fast, but low quality)

35. Image Plots
Grid lines:

36. Image Plots
Areas:

37. Cross Sections
Another option is to look at a cross-section through the data
For example, if x and y are the independent variables, we could fix y and look at f in terms of just x
then repeat for different y
this reduces the ES2 problem to a sequence of ES1 problems