Divide and Conquer

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Closest Points Problem

• Problem:

  Given N points in a plane. Find the closest pair of points. (If two points have the same position, then that pair is the closest with distance 0.)

• Solution:

Note

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Recursive Time Complexity Analysis

• Substitution Method:

  Guess, then prove by induction

Example

• Recursion-tree method:

  Draw the recursion tree, then sum the cost of each level.

Example

Thus, there is no need to draw the whole tree, just make a thorough guess and proove by substitution

• Master Method:

• \(T(N)=aT(N/b)+f(N)\)

  If \(f(N)=O(N^{\log_b{a}-\epsilon})\), for \epsilon>0, then \(T(N)=\Theta(N^{\log_b{a}})\)
  If \(f(N)=\Theta(N^{\log_b{a}})\), then \(T(N)=\Theta(N^{\log_b{a}}\log{N})\)
  If \(f(N)=\Omega(N^{\log_b{a}+\epsilon})\), for \epsilon>0, and \(af(N/b)\le kf(N)\), for k<1, thus N sufficiently large then \(T(N)=\Theta(f(N))\)

• Another Form:

  If \(af(N/b)=Kf(N)\), for K<1, then \(T(N)=\Theta(f(N))\)
  If \(af(N/b)=Kf(N)\), for K=1, then \(T(N)=\Theta(f(N)\log{N})\)
  If \(af(N/b)=Kf(N)\), for K>1, then \(T(N)=\Theta(N^{\log_b{a}})\)

• Another Theorem:

  the solution to the equation \(T(N)=aT(N/b)+\Theta(N^k\log^p{N})\) is \(T(N)= \begin{cases} O(N^k\log^{p+1}{N}) & \text{if} k=\log_b{a} \\ O(N^{\log_b{a}}\log{N}) & \text{if} k<\log_b{a} \\ O(N^k\log^pN) & \text{if} k>\log_b{a} \end{cases}\)