1. Convex Sets & Concave Sets
A planar region 𝑅 is called convex if and only if for any pair
of points 𝑝, 𝑞 in 𝑅, the line segment 𝑝𝑞 lies completely in 𝑅.
Otherwise, it is called concave. 𝑝 𝑞
𝑅 2
Concave
Convex 𝑝 𝑞
𝑅 1

2. An Example 4 1 2 3
Regions 1 & 2: convex
Regions 3 & 4: concave

3. Convex Hull
The convex hull CH(𝑄) of a set 𝑄 is the smallest convex region
that contains 𝑄.
Rubber band
When 𝑄 is finite, its convex hull is the unique convex polygon whose vertices
are from 𝑄 and that contains all points of 𝑄.

4. Degenerate Cases ♦ The convex hull of a single point is itself.
♦ The convex hull of several collinear points is the line segment
joining the two extreme points.

5. The Convex Hull Problem
Input: a set 𝑃 = { 𝑝 1 , 𝑝 2 , …, 𝑝 𝑛 } of points
Output: a list of vertices of CH(𝑃) in counterclockwise order.
(direction of traversal about the
outward axis with the interior
on the left)
Example
𝑝 1
𝑝 6
𝑝 10
𝑝 7
𝑝 8
𝑝 2
𝑝 5
𝑝 9
𝑝 4
𝑝 3
Output: 𝑝 5 , 𝑝 9 , 𝑝 2 , 𝑝 8 , 𝑝 10 , 𝑝 7 .

6. Edges of a Convex Hull For every edge with endpoints 𝑝, 𝑞 ∈ 𝑃.
All other points in 𝑃 lie
to the same side of the line passing through 𝑝 and 𝑞 q p

7. A Slow Convex Hull Algorithm
𝐸 ≠ {} // set of directed edges of CH(𝑃) that bounds the
// points of P on the right.
for every ordered pair (𝑝, 𝑞), where 𝑝, 𝑞 ∈ 𝑃 and 𝑝 ≠ 𝑞 // ( 𝑛 2 ) pairs
do valid  true
for every point 𝑟 ≠ 𝑝 or 𝑞 // 𝑛 −2 such points
do if 𝑟 lies to the right of 𝑝𝑞 or
collinear with 𝑝 and 𝑞 but not on 𝑝𝑞
then valid  false
if valid
then 𝐸 ≠ 𝐸 ∪ { 𝑝𝑞 } // 𝑝𝑞 and 𝑞𝑝 cannot be both in 𝐸
From 𝐸 construct a list 𝐿 of vertices of CH(𝑃), sorted in
counterclockwise order. // 𝑂( 𝑛 2 ) improvable to 𝑂(𝑛 log 𝑛 )
return L 𝑝 𝑞 𝑟
Running time ( 𝒏 𝟑 )

8. Polar Angle 𝑦 𝑞
polar angle p 𝑥 𝑟

9. Graham Scan 1) Select the node with the smallest 𝑦 coordinate. 𝑝 0
This node will be a vertex of the convex hull.

10. Tie Breaking (1)
When more than one point has the smallest 𝑦 coordinate, pick
the leftmost (or rightmost) one.
𝑝 0

11. Sorting by Polar Angle 2) Sort by polar angle with respect to 𝑝 0 .
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5
𝑝 0
Labels are in the polar angle order.

12. No Polar Angle Evaluation
𝑝 0 is the lowest (and leftmost)
all polar angles ∈ 0, 𝜋 .
𝐔𝐬𝐞 𝐜𝐫𝐨𝐬𝐬 𝐩𝐫𝐨𝐝𝐮𝐜𝐭!
𝑝 𝑖 < 𝑝 𝑗 if
𝑝 0 𝑝 𝑖 × 𝑝 0 𝑝 𝑗 >0
𝑝 𝑖
𝑝 𝑗
𝑝 0

13. Tie Breaking (2) What if 𝑝 0 , 𝑝 𝑖 , 𝑝 𝑗 are on the same line?
Order them by distance from 𝑝 0 .
𝑝 𝑖 < 𝑝 𝑗
if and
𝑝 0 𝑝 𝑖 × 𝑝 0 𝑝 𝑗 =0
𝑝 0 𝑝 𝑖 <| 𝑝 0 𝑝 𝑗 |
𝑝 0
𝑝 𝑗
𝑝 𝑖
No square roots.
Use dot product!
𝑝 0 𝑝 𝑖 ∙ 𝑝 0 𝑝 𝑖 < 𝑝 0 𝑝 𝑗 ∙ 𝑝 0 𝑝 𝑗

14. Point Elimination
When multiple points have the same polar angle, keep the one
furthest from 𝑝 0 .
Remove the rest since they cannot possibly be the hull vertices.
𝑝 0
furtherest
remove
𝑝 𝑗
𝑝 𝑖

15. Stack Initialization
3) Scan points in the increasing order of polar angle,
maintaining a stack.
𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5
𝑝 0
𝑝 2
𝑝 1 𝑆

16. 𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5 𝑆
𝑝 3
𝑝 1
𝑝 0

17. 𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5 𝑆
𝑝 4
𝑝 1
𝑝 0

18. 𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5 𝑆
𝑝 5
𝑝 4
𝑝 1
𝑝 0

19. 𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5 𝑆
𝑝 6
𝑝 4
𝑝 1
𝑝 0

20. 𝑆
𝑝 8
𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5
𝑝 7
𝑝 6
𝑝 4
𝑝 1
𝑝 0

21. 𝑆
𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5
𝑝 7
𝑝 6
𝑝 4
𝑝 1
𝑝 0

22. 𝑆
𝑝 10
𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5
𝑝 9
𝑝 6
𝑝 4
𝑝 1
𝑝 0

23. 𝑆
𝑝 11
𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5
𝑝 9
𝑝 6
𝑝 4
𝑝 1
𝑝 0

24. Finish S 𝑝 11 𝑝 6 𝑝 9 𝑝 4 𝑝 9 𝑝 11 𝑝 7 𝑝 6 𝑝 3 𝑝 8 𝑝 4 𝑝 5 𝑝 10 𝑝 1
𝑝 0
𝑝 8
𝑝 10
𝑝 9
𝑝 7
𝑝 6
𝑝 4
𝑝 3
𝑝 1
𝑝 2
𝑝 11
𝑝 5
𝑝 9
𝑝 6
𝑝 4
𝑝 1
𝑝 0

25. Graham’s Scan 𝑝 𝑗 𝑝 𝑖 𝑝 0 𝑝 𝑘 𝑝 𝑠 𝑝 0 finish 𝑝 𝑗 𝑝 𝑖 𝑝 0 𝑝 0 𝑝 𝑖 𝑝 𝑙
Non-vertices
of CH(𝑃)
determined
so far are
popped
candidates
for vertices
of CH(𝑃)
𝑝 𝑠
𝑝 0 
finish
𝑝 𝑗
𝑝 𝑖
𝑝 0 
𝑝 0
𝑝 𝑖
𝑝 𝑙 
𝑝 0
𝑝 2
𝑝 1
start
Every point
in 𝑃 is
pushed onto
𝑆 once
vertices of
CH(𝑃) in the
counter-
clockwise
order.

26. The Graham Scan Algorithm
let 𝑝 0 be the point in 𝑃 with minimum 𝑦-coordinate
let  𝑝 1 , 𝑝 2 , …, 𝑝 𝑛−1  be the remaining points in 𝑃
sorted in counterclockwise order by polar angle around 𝑝 0 .
Top[𝑆]  0
Push( 𝑝 0 , 𝑆)
Push( 𝑝 1 , 𝑆)
Push( 𝑝 2 , 𝑆)
for 𝑖 = 3 to 𝑛 − 1
do while 𝑝 𝑖 makes a nonleft turn from the line segment
determined by Top(𝑆) and Next-to-Top(𝑆)
do Pop(𝑆)
Push(𝑆, 𝑝 𝑖 )
return 𝑆

27. Running time #operations time / operation total Finding 1 (𝑛) (𝑛)
𝑝 0
(𝑛) (𝑛)
Sorting 𝑂(𝑛 log 𝑛 ) O(𝑛 log 𝑛 )
Push 𝑛 𝑂(1) (𝑛)
Pop  𝑛 − 𝑂(1) 𝑂(𝑛)
Why?
The running time of Graham’s Scan is 𝑶(𝒏 log 𝒏 ).

28. Proof of Correctness
Claim 1 Each point popped from stack S is not a vertex of CH(P).
Proof
Two cases when p is popped: j p i j k p i j k
In neither case can p become a vertex of CH(P). j

29. Claim 2 Graham-Scan maintains the invariant that the points
on stack S always form the vertices of a convex polygon
in counterclockwise order.
Proof
The claim holds right after initialization of S when
p , p , p form a triangle (which is obviously convex).
Popping a point from S preserves the invariant.
Consider a point p being pushed onto S. i p j i
The region
containing p i
The invariant still holds.

30. Correctness of Graham’s Scan
Theorem If Graham-Scan is run on a set P of at least three
points, then a point of P is on the stack S at termination
if and only if it is a vertex of CH(P).

31. Floating Arithmetic is not Exact!
Nearly colinear points p, q, r. q r p
p to the left of qr.
q to the left of rp.
All three accepted as edges!
r to the left of qp.
Not robust – the algorithm could fail with small numerical error.

32. Jarvis’ March A “package wrapping” technique p p p p Right chain Left3
(smallest polar
angle w.r.t. p ) 2 p 4 p 2
(smallest polar
angle w.r.t. p ) 1 p 5
(smallest polar
angle w.r.t. p ) 4
Right
chain
Left
chain p 1
(smallest polar
angle w.r.t. p ) p 6
(smallest polar
angle w.r.t. p ) 5 p
(lowest point)

33. Running Time of Jarvis’s March
Let ℎ be the number of vertices of the convex hull.
For each vertex, finding the point with the minimum
Polar angle, that is, the next vertex, takes time 𝑂(𝑛).
Comparison between two polar angles can be done
using cross product.
Thus 𝑶(𝒏𝒉) time in total.