CIS 068 / 2: Lab-Assignment #4: Shape Abstraction

Score: 10 points + 3 Bonus points
Due date: friday, march 3rd (2 weeks !)

A 2-dimensional shape can be defined by its boundary-polygon,which is an ordered list of all coordinates of its outline points. See the following figure for an example:

The left picture shows the original shape, the middle picture the outline of the shape. We want to create a LIST of all points of the outline. In order to do so, we start at some arbitrarily chosen point and traverse the outline clockwise. We store the coordinates of each point in a node, and therefore create an ordered list of x,y-pairs of coordinates of all boundary vertices, e.g.
10 10
11 10
11 11
...

Now take a look at the rightmost picture in the figure above. It shurely shows a very similar shape, it seems to be an abstracted version of the original boundary. In fact it is: the rightmost picture shows a certain subset of the original boundary vertices (points), just connected by lines.
In image processing it is a very important task to simplify shapes without changing their general visual appearance. This can be done, like in the example figure, by detecting a subset of boundary vertices containing visually significant information (in terms of human visual perception).

A very simple way to achieve this is an algorithm called Discrete Curve Evolution (latecki/lakaemper 1999): It keeps the important points, and dismisses unimportant ones. The importance is measured by the amount of visual information in the following way:

The figure above shows a cutout of the boundary, L,P and R a vertices, the lengths of the segments connecting these vertices are: LP = l1, PR = l2, RL = l3.
(Since someone from Greece long ago defined the length between to points as the 'Euclidean distance', it is given by sqrt((x1-x2)^2 + (y1-y2)^2), as you should know.)
The line LR shown in the figure is not part of the boundary, but we need it's length for the significance measure:
Define the visual significance S of vertex P simply by S(P) = l1 + l2 - l3.
Just for better understanding, here are some properties of this significance measure S(P) of point P:

• It is zero if L,P,R are collinear (certainly P does not contain any visual information then !)
• It increases with P moving away from the baseline LR
• It is never less than zero

Now comes the important part: the abstraction. Remember we want to create a subset of the original boundary, i.e. want to pick out certain 'important' vertices, dropping 'unimportant' ones.
A natural way to do this is the following:

1. Assign the signifcance value S() to every vertex except the first and last one (they don't have 2 neighbors)
2. set S for the first and last vertex to INFINITY (they'll never be removed)
3. LOOPSTART: Find the vertex containing the minimum significance
4. Drop the vertex by removing it from the list
5. compute the new values S for the 2 points, that were the neighbors of the removed vertex
6. loop (goto 'find the vertex...') until the list contains a predefined number of vertices

• No sorting is necessary in this implementation, we search the minimum by traversing the list (O(n))
• We need the next and previous node of the removed node. This can be done either by using a double linked list, or, if using a single linked list, memorizing the previous node.

And here's your assignment:
- Read the boundary-list (DOWNLOAD HERE!) , given as a textfile of (x,y)-coordinates of boundary vertices, into a single linked list. You can either use what JAVA offers or define your own list-structure
- Visualize it, connecting the vertices given in the file by lines
- Implement the abstraction-algorithm
- Show the simplified polygon

Your program should take the filename and the desired number of remaining vertices as input, e.g.
'java abstraction shapelist.txt 38' for reading the file 'shapelist.txt' and simplification down to 38 vertices.

Remark:

• Implementing the boundary polygon as a linked list is mandatory. No arrays this time (zero points for arrays).
• The abstraction in the first figure above, rightmost picture, is done by exactly this algorithm, but taking a different measure for the significance value. The simplified image shown contains 38 vertices. Your results will look a bit different.

Good luck !