1. Data Structures and Algorithms (AT70. 02) Comp. Sc. and Inf. Mgmt
Data Structures and Algorithms (AT70.02) Comp. Sc. and Inf. Mgmt. Asian Institute of Technology
Instructor: Prof. Sumanta Guha
Slide Sources: CLRS “Intro. To Algorithms” book website
(copyright McGraw Hill)
adapted and supplemented

2. CLRS “Intro. To Algorithms” Ch. 33: Computational Geometry

3. If p1 = (x1, y1) and p2 = (x2, y2) then p1  p2 = x1 x2 = x1y2 – x2y1*
y1 y2
If p1  p2 > 0, then p1 is clockwise from p2 (turning around the origin);
if p1  p2 < 0, then p1 is counterclockwise from p2;
If p1  p2 = 0, then p1 and p2 are collinear (pointing in either same or
opposite directions)
* The cross product is actually a 3D vector whose magnitude is |x1y2 – x2y1|. Here we simplify for 2D.

5. If DIRECTION(pi, pj, pk) > 0, then pj – pi is a left turn from pk – pi around pi
< 0, then pj – pi is a right turn from pk – pi around pi
= 0, then pj – pi and pk – pi are collinear

7. How?!

10. Eliminate the polar co-ordinate sort part of Graham’s scan by
modifying it to scan the vertices sorted from left to right.
Otherwise, the underlying principle, data structures, etc. remain the same.
Comment : This modification of Graham’s scan was made by Andrew (1979) and is often called the Monotone Chain algorithm.

11. Jarvis’s march is output-sensitive!

12. Closest Pair by Divide-and-Conquer
Following slides are copied from Dariusz Kowalski (

13. Lecture 5: Divide and Conquer, cont. (Dariusz Kowalski)
Solution
Preprocessing:
Sort points according to the first coordinate (obtain horizontal list H) and according the second coordinate (obtain vertical list V)
Main Algorithm:
Partition list H input into halves in linear time (draw vertical line L containing medium point)
Solve the problem for each half of the input separately, obtaining two pairs of closest points; let  be the smallest distance from the obtained ones
Find if there is a pair which has distance smaller than  - if yes then find the smallest distance pair
Lecture 5: Divide and Conquer, cont. (Dariusz Kowalski)

14. Lecture 5: Divide and Conquer, cont. (Dariusz Kowalski)
Main difficulty
Find if there is a pair which has distance smaller than  -
if yes then find the smallest distance pair
How to do it in linear time?   
Lecture 5: Divide and Conquer, cont. (Dariusz Kowalski)

15. Closest pair in the strip
1. Find a sublist P of V containing all points with a distance at most  from the line L - in linear time
2. For each element in P check its distance to the next 8 elements and remember the closest pair ever found   
Lecture 5: Divide and Conquer, cont. (Dariusz Kowalski) L

16. Why to check only 8 points ahead?
1. Partition the strip into squares of length /2 as shown in the picture.
2. Each square contains at most 1 point - by definition of .
3. If there are at least 2 squares between points then they can not be the closest points.
4. There are at most 8 squares to check. 
/2
/2
/2
/2
/2
/2
Lecture 5: Divide and Conquer, cont. (Dariusz Kowalski) L

17. Problems Ex. 33.3-2 Ex. 33.3-4 Ex. 33.3-5 Ex. 33.3-6 Ex. 33.4-3