Movatterモバイル変換


[0]ホーム

URL:


Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly

Sign up
Appearance settings

Commit9b2611c

Browse files
authored
Update nearest_points.md - block equation spacing for correct rendering
1 parente09c042 commit9b2611c

File tree

1 file changed

+1
-0
lines changed

1 file changed

+1
-0
lines changed

‎src/geometry/nearest_points.md‎

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -315,6 +315,7 @@ While this algorithm may look slow, because of recomputing everything multiple t
315315
**Proof.** Let $X_i$ the random variable that is $1$ when point $p_i$ causes a change of $\delta$ and a recomputation of the data structures, and $0$ if not. It is easy to show that the cost is $O(n + \sum_{i=1}^{n} i X_i)$, since on the $i$-th step we are considering only the first $i$ points. However, turns out that $\Pr(X_i = 1) \le \frac{2}{i}$. This is because on the $i$-th step, $\delta$ is the distance of the closest pair in $\{p_1,\dots,p_i\}$, and $\Pr(X_i = 1)$ is the probability of $p_i$ belonging to the closest pair, which only happens in $2(i-1)$ pairs out of the $i(i-1)$ possible pairs (assuming all distances are different), so the probability is at most $\frac{2(i-1)}{i(i-1)} = \frac{2}{i}$, since we previously shuffled the points uniformly.
316316

317317
We can therefore see that the expected cost is
318+
318319
$$O(n + \sum_{i=1}^{n} i \Pr(X_i = 1)) \le O(n + \sum_{i=1}^{n} i \frac{2}{i}) = O(3n) = O(n) \quad \quad \blacksquare$$
319320

320321

0 commit comments

Comments
 (0)

[8]ページ先頭

©2009-2025 Movatter.jp