Design and Analysis of Algorithms (Divide & Conquer: Convex Hull, Median Finding)

Paradigm

Given a problem of size n divide it into subproblems of size n, a ≥ 1, b > 1. Solve each subproblem recursively. Combine solutions of subproblems to get the overall solution.

T(n) = aT(n/b ) + [work for merge]

Convex Hull

Given n points in plane

S = {(xi, yi) | i = 1, 2,...,n}

assume no two have the same x coodinates, no two have sa`me y coordinates, no three in a line.

Convex Hull: smallest polygon contain all points in the plane.

CH(S) represented by the sequence of points on the boundary in clockwise order as a doubly linked list.

Brute force for Convex Hull

Test each line segment to see if it makes up an edge of the convex hull

If the rest of the points are on one side of the segment, the segment is on the convex hull.
else the segment is not.
O(n2) edges, O(n) tests ⇒ O(n3) complexity

Divide and Conquer Convex Hull

Sort points by x coordinates

divide into left half A and right half B by x coords
compute CH(A) and CH(B)
combine CH's of two halves (merge step)

How to merge?

Find the upper tangent (ai, bj).
Find the lower tangent (ak, bm).
Cut and paste in time Θ(n).

Firstly, link ai to bj, go down b list till you see bm and link bm to ak, continue along the a list until you return to ai.

Find Tangents

Assume ai maximizes x within CH(A) (a1, a2,...,ap). b1 minimizes x within CH(B) (b1, b2,..., bq)

L is the vertical line separating A and B. Define y(i, j) as y-coordinate of intersection between L and segment (ai, bj ).

Claim: (ai, bj ) is upper tangent if it maximizes y(i, j)

If y(i, j) is not maximum, there will be points on both sides of (ai, bj ) and it cannot be a tangent.

Algorithm: Obvious O(n2) algorithm looks at all ai, bj pairs. T(n)=2T(n/2)+ Θ(n^2) = Θ(n^2).

i = 1
j = 1
while(y(i, j+1) > y(i, j) or y(i − 1, j) > y(i, j))
    if (y(i, j + 1) > y(i, j)) -> move right finger clockwise
        j = j + 1( mod q)
    else
        i = i - 1( mod p) -> move left finger anti-cloclwise
    return (ai, bj ) as upper tangent

T(n)=2T(n/2) + Θ(n) = Θ(n log n)

Median Finding

Given a set of n numbers, define rank(x) as the number of numbers in the set that are ≤ x. Find an element of rank $\left \lfloor \frac{ n+1 }{2} \right \rfloor$ (lower median) and $\left \lceil \frac{n + 1}{2}\right \rceil$ (upper median).

Select(S, i)
    Pick x in S
    Compute k = rank(x)
    B = {y ∈ S|y<x}
    C = {y ∈ S|y>x}
    if k = i
        return x
    else if k > i
        return Select(B, i)
    else if k < i
        return Select(C, i - k)

Picking x Cleverly

Need to pick x so rank(x) is not extreme.

Arrange S into columns of size 5 (n/5 cols)
Sort each column (bigger elements on top) (linear time)
Find “median of medians” as x