This document discusses brute force and divide-and-conquer algorithms. It covers various brute force algorithms like computing an, string matching, closest pair problem, convex hull problems, and exhaustive search algorithms like the traveling salesman problem and knapsack problem. It then discusses divide-and-conquer techniques like merge sort, quicksort, and multiplication of large integers. Examples and analysis of algorithms like selection sort, bubble sort, and mergesort are provided.

1. Unit II
BRUTE FORCE AND DIVIDE-AND-
CONQUER
PART 1
Dr.S.Gunasundari
Associate Professor
CSE Dept

2. Syllabus
Brute Force
 Computing an
 String matching
 Closest-Pair
 Convex Hull Problems
 Exhaustive Search
Traveling Salesman Problem
 Knapsack Problem
Assignment problem.
Divide and conquer
• Merge sort
• Quick sort
• Binary search-
• Closest Pair
• Multiplication of Large Integers
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3. Brute Force
A straightforward approach, usually based directly on the problem’s statement and definitions
of the concepts involved
Examples:
1. Computing an (a > 0, n a nonnegative integer)
2. Computing n!
3. Multiplying two matrices
4. Searching for a key of a given value in a list
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4. Brute-Force Sorting Algorithm
Selection Sort
Scan the array to find its smallest element and swap it with the first element.
Then, starting with the second element, scan the elements to the right of it
to find the smallest among them and swap it with the second element.
Generally, on pass i (0  i  n-2), find the smallest element in A[i..n-1] and
swap it with A[i]:
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5. Example
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6. Analysis of Selection Sort
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7. 09-03-2022

8. Time Efficiency
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9. Bubble sort
Consecutive adjacent pairs of elements in the array
are compared with each other.
If the element at the lower index is greater than the
element at the higher index
the two elements are interchanged so that the
element is placed before the bigger one.
This process will continue till the list of unsorted
elements exhausts
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10. Example
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11. Code and Example
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12. Example
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13. Time Efficiency
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14. Computing an
 an = a * a * a …..(n times)
 a – non-zero number
 n- non negative integer
 Compute an by multiplying a by n times
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15. Brute-Force String Matching
 pattern: a string of m characters to search for
 text: a (longer) string of n characters to search in
 problem: find a substring in the text that matches the pattern
Brute-force algorithm
Step 1 Align pattern at beginning of text
Step 2 Moving from left to right, compare each character of pattern to the corresponding
character in text until
 all characters are found to match (successful search); or
 a mismatch is detected
Step 3 While pattern is not found and the text is not yet exhausted, realign pattern one
position to the right and repeat Step 2
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16. Examples of Brute-Force String Matching
1. Pattern: 001011
Text: 10010101101001100101111010
1. Pattern: happy
Text: It is never too late to have a happy childhood.
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17. Example
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18. Pseudocode and Efficiency Efficiency:
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19. Closest-Pair Problem
Find the two closest points in a set of n points (in the two-dimensional Cartesian plane).
Brute-force algorithm
Compute the distance between every pair of distinct points n(n-1)/2 pairs and return the
indexes of the points for which the distance is the smallest.
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20. Closest-Pair Brute-Force Algorithm (cont.)
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21. Analysis
 Basic Operation: Computing Euclidean distance
 Avoid finding square roots
 Now basic operation is squaring the number
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22. Convex Hull
 The convex hull problem is the problem of
constructing the convex hull for a given set S
of n points
 Definition A set of points (finite or infinite) in
the plane is called convex if for any two points
P and Q in the set, the entire line segment with
the endpoints at P and Q belongs to the set
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23. 09-03-2022

24. Convex Hull Brute Force Algorithm
 A line segment connecting two points Pi and Pj of a set of n points is a part of its
convex hull’s boundary
 if and only if all the other points of the set lie on the same side of the straight line through
these two points
 Repeating this test for every pair of points yields a list of line segments that make up
the convex hull’s boundary
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25. 09-03-2022

26. Analysis of Convex Hull Brute Force Algorithm
 The time efficiency of this algorithm is in O(n3)
 for each of n(n-1)/ 2 pairs of distinct points, we may need to find the sign of ax +
by - c for each of the other n - 2 points
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27. Brute-Force Strength and Weakness
 Strengths
 wide applicability
 simplicity
 yields reasonable algorithms for some important problems
(e.g., matrix multiplication, sorting, searching, string matching)
 Weaknesses
 rarely yields efficient algorithms
 some brute-force algorithms are unacceptably slow
 not as constructive as some other design techniques
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28. Exhaustive Search
A brute force solution to a problem involving search for an element with a special
property, usually among combinatorial objects such as permutations, combinations,
or subsets of a set.
Method:
 generate a list of all potential solutions to the problem in a systematic manner
 evaluate potential solutions one by one, disqualifying infeasible ones and, for an
optimization problem, keeping track of the best one found so far
 when search ends, announce the solution(s) found
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29. Example 1: Traveling Salesman Problem
 Given n cities with known distances between each pair, find the shortest tour that passes
through all the cities exactly once before returning to the starting city
 Alternatively: Find shortest Hamiltonian circuit in a weighted connected graph
 Hamiltonian circuit is defined as a cycle that passes through all the vertices of the graph exactly once.
 Graph Vertices represent cities
 Edge weight represent distance
 Example:
a b
c d
8
2
7
5 3
4
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30. TSP by Exhaustive Search
Tour Cost
a→b→c→d→a 2+3+7+5 = 17
a→b→d→c→a 2+4+7+8 = 21
a→c→b→d→a 8+3+4+5 = 20
a→c→d→b→a 8+7+4+2 = 21
a→d→b→c→a 5+4+3+8 = 20
a→d→c→b→a 5+7+3+2 = 17
a b
c d
8
2
7
5 3
4
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we can get all the tours by generating all the permutations of n − 1 intermediate
cities, compute the tour lengths, and find the shortest among them
Efficiency: (n-1)!
• Some pair of tours differ by direction
• a→b→c→d→a AND a→d→c→b→a
• we could cut the number of vertex permutations by half ((n-1)!)/2

31. Example
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32. Example 2: Knapsack Problem
Given n items:
 weights: w1 w2 … wn
 values: v1 v2 … vn
 a knapsack of capacity W
Find most valuable subset of the items that fit into the knapsack
Example: Knapsack capacity W=16
item weight value
1 2 $20
2 5 $30
3 10 $50
4 5 $10
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33. Knapsack Problem by Exhaustive Search
Subset Total weight Total value
{1} 2 $20
{2} 5 $30
{3} 10 $50
{4} 5 $10
{1,2} 7 $50
{1,3} 12 $70
{1,4} 7 $30
{2,3} 15 $80
{2,4} 10 $40
{3,4} 15 $60
{1,2,3} 17 not feasible
{1,2,4} 12 $60
{1,3,4} 17 not feasible
{2,3,4} 20 not feasible
{1,2,3,4} 22 not feasible
Efficiency:
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Since
the number of subsets of an n-element set is 2n, the
exhaustive search leads to a (2n) algorithm, no
matter how efficiently individual subsets are
generated.

34. Example
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35. Example 3: The Assignment Problem
There are n people who need to be assigned to n jobs, one person per job. The
cost of assigning person i to job j is C[i,j]. Find an assignment that minimizes the
total cost.
Job 0 Job 1 Job 2 Job 3
Person 0 9 2 7 8
Person 1 6 4 3 7
Person 2 5 8 1 8
Person 3 7 6 9 4
Algorithmic Plan: Generate all legitimate assignments, compute
their costs, and select the cheapest one.
How many assignments are there?
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36. Assignment Problem by Exhaustive Search
9 2 7 8
6 4 3 7
5 8 1 8
7 6 9 4
Assignment (col.#s) Total Cost
1, 2, 3, 4 9+4+1+4=18
1, 2, 4, 3 9+4+8+9=30
1, 3, 2, 4 9+3+8+4=24
1, 3, 4, 2 9+3+8+6=26
1, 4, 2, 3 9+7+8+9=33
1, 4, 3, 2 9+7+1+6=23
etc.
C =
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Since the number of permutations to be
considered for the general case of the
assignment problem is n!,

37. Assignment Total Cost
2,1,3,4 2+6+1+4=13
2,1,4,3 2+6+8+9=25
2,3,1,4 2+3+5+4=14
2,3,4,1 2+3+8+7=20
2,4,1,3 2+7+5+9=23
2,4,3,1 2+7+1+7=17
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38. AssignmentTotal Cost
3,1,2,4 7+6+8+4=25
3,1,4,2 7+6+8+6=27
3,2,1,4 7+4+5+4=20
3,2,4,1 7+4+8+7=26
3,4,1,2 7+7+5+6=25
3,4,2,1 7+7+8+7=29
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39. AssignmentTotal Cost
4,1,2,3 8+6+8+9=31
4,1,3,2 8+6+1+6=21
4,2,1,3 8+4+5+9=26
4,2,3,1 8+4+1+7=20
4,3,1,2 8+3+5+6=22
4,3,2,1 8+3+8+7=26
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40. Solution
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41. Final Comments on Exhaustive Search
 Exhaustive-search algorithms run in a realistic amount of time only on very small
instances
 In some cases, there are much better alternatives!
 Euler circuits
 shortest paths
 minimum spanning tree
 assignment problem
 In many cases, exhaustive search or its variation is the only known way to get
exact solution
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42. Divide-and-Conquer
The most-well known algorithm design strategy:
1. Divide instance of problem into two or more smaller instances
2. Solve smaller instances recursively
3. Obtain solution to original (larger) instance by combining these solutions
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43. Divide-and-Conquer Technique (cont.)
subproblem 2
of size n/2
subproblem 1
of size n/2
a solution to
subproblem 1
a solution to
the original problem
a solution to
subproblem 2
a problem of size n
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44. Divide-and-Conquer Examples
 Sorting: mergesort and quicksort
 Binary tree traversals
 Binary search (?)
 Multiplication of large integers
 Matrix multiplication: Strassen’s algorithm
 Closest-pair and convex-hull algorithms
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45. General Divide-and-Conquer Recurrence
T(n) = aT(n/b) + f (n) where f(n)  (nd), d  0
Master Theorem: If a < bd, T(n)  (nd)
If a = bd, T(n)  (nd log n)
If a > bd, T(n)  (nlog b a )
Note: The same results hold with O instead of .
Examples: T(n) = 4T(n/2) + n  T(n)  ?
T(n) = 4T(n/2) + n2  T(n)  ?
T(n) = 4T(n/2) + n3  T(n)  ?
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46. Mergesort
 Split array A[0..n-1] in two about equal halves and make copies of each half in arrays B and C
 Sort arrays B and C recursively
 Merge sorted arrays B and C into array A as follows:
Repeat the following until no elements remain in one of the arrays:
compare the first elements in the remaining unprocessed portions of
the arrays
copy the smaller of the two into A, while incrementing the index
indicating the unprocessed portion of that array
Once all elements in one of the arrays are processed, copy the remaining
unprocessed elements from the other array into A.
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47. Pseudocode of Mergesort
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48. Pseudocode of Merge
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49. Mergesort Example
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
3 8 2 9 1 7 4 5
2 3 8 9 1 4 5 7
1 2 3 4 5 7 8 9
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50. Analysis of Mergesort
 All cases have same efficiency: Θ(n log n)
 Number of comparisons in the worst case is close to theoretical minimum for
comparison-based sorting:
log2 n! ≈ n log2 n - 1.44n
 Space requirement: Θ(n) (not in-place)
 Can be implemented without recursion (bottom-up)
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51. Quicksort
 Select a pivot (partitioning element) – here, the first element
 Rearrange the list so that all the elements in the first s positions are smaller than or equal to the pivot
and all the elements in the remaining n-s positions are larger than or equal to the pivot (see next slide
for an algorithm)
 Exchange the pivot with the last element in the first (i.e., ) subarray — the pivot is now in its final
position
 Sort the two subarrays recursively
p
A[i]p A[i]p
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52. Partitioning Algorithm
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53. Quicksort Example
5 3 1 9 8 2 4 7
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54. Analysis of Quicksort
 Best case: split in the middle — Θ(n log n)
 Worst case: sorted array! — Θ(n2)
 Average case: random arrays — Θ(n log n)
 Improvements:
better pivot selection: median of three partitioning
switch to insertion sort on small subfiles
elimination of recursion
These combine to 20-25% improvement
 Considered the method of choice for internal sorting of large files (n ≥ 10000)
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55. Binary Search
Very efficient algorithm for searching in sorted array:
K
vs
A[0] . . . A[m] . . . A[n-1]
If K = A[m], stop (successful search); otherwise, continue
searching by the same method in A[0..m-1] if K < A[m]
and in A[m+1..n-1] if K > A[m]
l  0; r  n-1
while l  r do
m  (l+r)/2
if K = A[m] return m
else if K < A[m] r  m-1
else l  m+1
return -1
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56. Analysis of Binary Search
 Time efficiency
worst-case recurrence: Cw (n) = 1 + Cw( n/2 ), Cw (1) = 1
solution: Cw(n) = log2(n+1)
This is VERY fast: e.g., Cw(106) = 20
 Optimal for searching a sorted array
 Limitations: must be a sorted array (not linked list)
 Bad (degenerate) example of divide-and-conquer
 Has a continuous counterpart called bisection method for solving equations in one unknown f(x) = 0 (see
Sec. 12.4)
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57. Binary Tree Algorithms
Binary tree is a divide-and-conquer ready structure!
Ex. 1: Classic traversals (preorder, inorder, postorder)
Algorithm Inorder(T)
if T   a a
Inorder(Tleft) b c b c
print(root of T) d e d e
Inorder(Tright)
Efficiency: Θ(n)
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58. Binary Tree Algorithms (cont.)
Ex. 2: Computing the height of a binary tree
T T
L R
h(T) = max{h(TL), h(TR)} + 1 if T   and h() = -1
Efficiency: Θ(n)
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59. Multiplication of Large Integers
Consider the problem of multiplying two (large) n-digit integers represented by arrays of their digits such as:
A = 12345678901357986429 B = 87654321284820912836
The grade-school algorithm:
a1 a2 … an
b1 b2 … bn
(d10) d11d12 … d1n
(d20) d21d22 … d2n
… … … … … … …
(dn0) dn1dn2 … dnn
Efficiency: n2 one-digit multiplications
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60. First Divide-and-Conquer Algorithm
A small example: A  B where A = 2135 and B = 4014
A = (21·102 + 35), B = (40 ·102 + 14)
So, A  B = (21 ·102 + 35)  (40 ·102 + 14)
= 21  40 ·104 + (21  14 + 35  40) ·102 + 35  14
In general, if A = A1A2 and B = B1B2 (where A and B are n-digit,
A1, A2, B1, B2 are n/2-digit numbers),
A  B = A1  B1·10n + (A1  B2 + A2  B1) ·10n/2 + A2  B2
Recurrence for the number of one-digit multiplications M(n):
M(n) = 4M(n/2), M(1) = 1
Solution: M(n) = n2
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61. Second Divide-and-Conquer Algorithm
A  B = A1  B1·10n + (A1  B2 + A2  B1) ·10n/2 + A2  B2
The idea is to decrease the number of multiplications from 4 to 3:
(A1 + A2 )  (B1 + B2 ) = A1  B1 + (A1  B2 + A2  B1) + A2  B2,
I.e., (A1  B2 + A2  B1) = (A1 + A2 )  (B1 + B2 ) - A1  B1 - A2  B2,
which requires only 3 multiplications at the expense of (4-1) extra add/sub.
Recurrence for the number of multiplications M(n):
M(n) = 3M(n/2), M(1) = 1
Solution: M(n) = 3log 2n = nlog 23 ≈ n1.585
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62. Example of Large-Integer Multiplication
2135  4014
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63. Strassen’s Matrix Multiplication
Strassen observed [1969] that the product of two matrices can be computed as follows:
C00 C01 A00 A01 B00 B01
= *
C10 C11 A10 A11 B10 B11
M1 + M4 - M5 + M7 M3 + M5
=
M2 + M4 M1 + M3 - M2 + M6
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64. Formulas for Strassen’s Algorithm
M1 = (A00 + A11)  (B00 + B11)
M2 = (A10 + A11)  B00
M3 = A00  (B01 - B11)
M4 = A11  (B10 - B00)
M5 = (A00 + A01)  B11
M6 = (A10 - A00)  (B00 + B01)
M7 = (A01 - A11)  (B10 + B11)
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65. Analysis of Strassen’s Algorithm
If n is not a power of 2, matrices can be padded with zeros.
Number of multiplications:
M(n) = 7M(n/2), M(1) = 1
Solution: M(n) = 7log 2n = nlog 27 ≈ n2.807 vs. n3 of brute-force alg.
Algorithms with better asymptotic efficiency are known but they
are even more complex.
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66. Closest-Pair Problem by Divide-and-Conquer
Step 1 Divide the points given into two subsets S1 and S2 by a vertical line x = c so that half the points lie to
the left or on the line and half the points lie to the right or on the line.
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67. Closest Pair by Divide-and-Conquer (cont.)
Step 2 Find recursively the closest pairs for the left and right
subsets.
Step 3 Set d = min{d1, d2}
We can limit our attention to the points in the symmetric
vertical strip of width 2d as possible closest pair. Let C1
and C2 be the subsets of points in the left subset S1 and of
the right subset S2, respectively, that lie in this vertical
strip. The points in C1 and C2 are stored in increasing
order of their y coordinates, which is maintained by
merging during the execution of the next step.
Step 4 For every point P(x,y) in C1, we inspect points in
C2 that may be closer to P than d. There can be no more
than 6 such points (because d ≤ d2)!
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68. Closest Pair by Divide-and-Conquer: Worst Case
The worst case scenario is depicted below:
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69. Efficiency of the Closest-Pair Algorithm
Running time of the algorithm is described by
T(n) = 2T(n/2) + M(n), where M(n)  O(n)
By the Master Theorem (with a = 2, b = 2, d = 1)
T(n)  O(n log n)
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70. Quickhull Algorithm
Convex hull: smallest convex set that includes given points
 Assume points are sorted by x-coordinate values
 Identify extreme points P1 and P2 (leftmost and rightmost)
 Compute upper hull recursively:
 find point Pmax that is farthest away from line P1P2
 compute the upper hull of the points to the left of line P1Pmax
 compute the upper hull of the points to the left of line PmaxP2
 Compute lower hull in a similar manner
P1
P2
Pmax
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71. Efficiency of Quickhull Algorithm
 Finding point farthest away from line P1P2 can be done in linear time
 Time efficiency:
worst case: Θ(n2) (as quicksort)
average case: Θ(n) (under reasonable assumptions about
distribution of points given)
 If points are not initially sorted by x-coordinate value, this can be accomplished in O(n log n) time
 Several O(n log n) algorithms for convex hull are known
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