1. Computing Convex Hulls CLRS 33.3

2. Computational Geometry
Studies algorithms for solving geometric problems
Applications in many fields
Computer graphics
Robotics
VLSI design
Computer-aided design, etc.

3. Geometric algorithms
The input (typically): a description of a set of geometric objects
A set of points
A set of line segments
Vertices of a polygon
Etc.
We will look at sample problems in two dimensions, that is, in the plane

4. Convex hulls in two dimensions
What is a convex hull?
What is convexity?
A subset S of the plane is called convex if and only if for any pair of points p, q in S, the line segment pq is completely contained in S p q p q
not convex
convex

5. Convex Hull
The convex hull CH(S) of a set s is the smallest convex set that contains S.
Consider the problem of computing the convex hull of a finite set P of n points in the plane.

6. Convex Hull

7. Convex Hull

8. Convex Hull
Alternative Definition: It is the unique convex polygon whose vertices are points from P and contains all the points of P

9. Problem
Given P, compute a list of points that contains the vertices of CH(P), listed in clockwise (CW) order
Expected output:

10. Brute-Force Solution
Consider an edge of the convex hull between vertices p and q
Consider the directed line passing through p and q
Observe that CH(P) lies to the right of this line q p

11. Brute-Force Solution Reverse holds too:
If all points in lie to the right of the directed line through p and q, the pq is an edge of CH(P) q p

12. BruteForce_ConvexHull(P) {
E=empty set
for all ordered pairs (p,q) in PxP, and p!= q do {
isEdge = true
for all points r in P, r!=p and r!=q do
if r lies to the left of directed line from p to q
isEdge= false
if isEdge the add pq to E }
from the set of edges in E, construct a list L of vertices of CH(P), in CW order
return L

13. How to test?
if r lies to the left of directed line from p to q

14. Orientation test
Given an ordered triple of points <p,q,r> in the plane, we say that they have
positive orientation if they define a CCW oriented triangle
negative orientation if they define a CW oriented triangle
Zero orientation if they are collinear p q r
Positive orientation p q r
r is to the left of pq

15. Orientation test
Given an ordered triple of points <p,q,r> in the plane, we say that they have
positive orientation if they define a CCW oriented triangle
negative orientation if they define a CW oriented triangle
Zero orientation if they are collinear q q r r p p
negative orientation
r is to the right of pq

16. Orientation test
Given an ordered triple of points <p,q,r> in the plane, we say that they have
positive orientation if they define a CCW oriented triangle
negative orientation if they define a CW oriented triangle
Zero orientation if they are collinear q q r r p p
zero orientation
r is on pq

17. Orientation

18. How about the last step?
Given E, construct the list of vertices of CH(P) in CW order?

19. How about the last step?
Given E, construct the list L of vertices of CH(P) in CW order?
Remove an arbitrary edge pq from E, put p and q in L
Now find in E the edge with origin q (say qr), remove that edge from E and add r to L
Keep doing this until E has only one edge left
time to contsruct L
why?

20. Overall, BruteForce takes How many ordered pairs?
n (n-1)
For each we compare n-2 other points (r)

21. Can we do better?
A number of algorithms that can compute CH(P) in O(n log n)
We will study the Incremental Algorithm
Points are added one at a time and the hull is updated at each insertion

22. Incremental Algorithm
At any intermediate step of the algorithm
There is a current hull of points
Add point
Update the hull
Add points in order of increasing x-coordinate
To guarantee that the new point is outside the current hull
If same x-coordinate, then in increasing order of y

23. Upper Hull and Lower Hull
We will construct the hull in two pieces:
construct upper hull from left to right
construct lower hull from right to left

24. Construction of the Upper Hull
already processed

25. Construction of the Upper Hull
Observation: for a convex polygon, if we walk through the boundary in CW order, we always make a right turn
i.e. if we consider any three consecutive points: p, q, r
Orient(p,q,r) should be negative
How can we use this?

26. Construction of the Upper Hull
Check whether the last two points on the current hull and
make a right turn
Left turn

27. Construction of the Upper Hull
If a left turn, remove the last point from the current hull and try again
Left turn

28. Construction of the Upper Hull
If a left turn, remove the last point from the current hull and try again
Left turn

29. Construction of the Upper Hull
If a left turn, remove the last point from the current hull and try again
Left turn

30. Construction of the Upper Hull
If a right turn, add to the current hull
right turn

31. Construction of the Upper Hull
If a right turn, add to the current hull
right turn

32. Store current hull on a stack U
U.top() returns the last point on the current hull
U.second() returns the second to last point on the current hull
while ( Orient(U.second(), U.top() ) >= 0)
U.push( )
U.pop()

33. IncrementalConvexHull(P) {
sort points in increasing order of their x-coordinate
U.push(p[1])
U.push(p[2])
for i=3 to n do
while U.size() >= 2 and orient(U.second(), U.top(), p[i]) >= 0) do
U.pop()
U.push(p[i])
L.push(p[n])
L.push(p[n-1])
for i=n-2 to 1 do
while L.size() >= 2 and Orient(L.second(), L.top(), p[i]) >= 0) do
L.pop()
L.push(p[i])
Combine U and L into a single list of points in CW direction

34. Incremental Convex Hull
Running time?
Sort: O(n log n)
Each point is removed from the stack U at most once: O(n)
Each point is entered into the stack U once: O(n)
Thus, O(n) for upper hull
Similarly O(n) for lower hull
Total: O(n log n)