Divide and Conquer¶

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Simple Example¶

Divide and conquer.

Let \(\delta = \) \(min\{d_l, d_r\}\)

For a point lies in the strip, we only search for the near by \(\delta * \delta\) square.

At the beginning, the points are preprocessed to sorted by x distance.

During conquer, we merge them with sorted y distance. (Like merge sort).

Each point at most inspect 8 neighbors.

In the conquer part, we only scan the point list linearly. \(O(NlogN)\)

Guess first, proof by induction

\(\Large{T(N) = \sum_{j=0}^{log_b(N-1)} a^j \mathbf{f(N/b^j)} + \Theta(N^{log_b a})}\)

The last layer has \(N^{log_b a}\) nodes

compare \(f(N)\) with \(N^{log_b a}\) in \(\textbf{Polynomial}\) way.

\(f(N) = NlogN > N^{log_b a + \epsilon}\) \(\textbf{WRONG}\)

Why comparing with \(N^{log_b a}\) ?

This determines the result of \((\frac{N}{b^j})^{log_b a + ?} = 1 \) or \(b^{\epsilon j}\) or ...

if \(f(N) = \Theta(N^{log_ba})\)

then we have

\(\sum_{j=0}^{log_b(N-1)} a^j \mathbf{f(N/b^j)}= \Theta(\sum_{j=0}^{log_b(N-1)} a^j \mathbf{(N/b^j)^{log_b a}}) = \)

\(\Theta(N^{log_ba}\sum_{j=0}^{log_bN-1} (\frac{a}{b^{log_ba}})^j) = \)\(\Theta(N^{log_ba} * \sum_{j=0}^{log_bN-1} 1) = \Theta(N^{log_ba}log_bN)\)

\(\Theta(N^{log_ba}log_bN) + \Theta(N^{log_b a}) = \Theta(N^{log_ba}log_bN))\)

Compare \(af(N/b)\) with \(f(N)\)