jlcastrillon's blog

KD Tree Help
By jlcastrillon, 11 years ago, In English

Here is the problem Given a set of N points and some query points, What is the farthest point from the set to each of the query points. Here are a couple of questions: Is it the same time complexity when finding the nearest neighbor? Where can I find some theory about this?

If there is a better way (with the same or less time complexity) to solve this problem using a suitable amount of code please let me know...

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11 years ago, # |
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How about computing the convex hull for this as the farthest point will always lie on the convex hull.

Complexity O(nlogn + qh) = O(nlogn) (using graham scan) where h = no. of points on the convex hull ,and q = no. of queries

Edit: Worst Case complexity for the above is o(nq). But on the average it should be much faster.

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    11 years ago, # ^ |
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    Convex Hull will only work for small polygons with many points inside, and for two dimensional space, for bigger dimensional space it is a real pain. What I am looking for is an O(q logn + n logn) or O(q sqrt(n) + n sqrt(n)) worst case complexity algorithm

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11 years ago, # |
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Will be there any difference when the distance is euclidean or manhattan?

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    11 years ago, # ^ |
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    If we consider 2D points + Manhattan distance then Fenwick tree can be used to do offline queries.

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      11 years ago, # ^ |
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      Sorry, I forgot to add this to the blog, some points might be deleted on the course of the queries..

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        11 years ago, # ^ |
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        Could you explain your offline version of Fenwick Tree for this problem when distance is Manhattan?

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          11 years ago, # ^ |
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          For each point i, let it be the origin then we can divide other points into 4 quadrants. For the 3rd quadrant where x[k] <= x[i] and y[k] <= y[i], the Manhattan distance between i and j is equal to x[i] - x[k] + y[i] - y[k] = (x[i] + y[i]) - (x[k] + y[k]).

          Based on this observation, we first sort all points by x ascending, then by y ascending. Iterate in the sorted order and use a Fenwick tree to store min(x[k] + y[k]) for all k <= i and y[k] <= Y (don't forget to map the y values into [1..n]). For the other 3 quadrants, just perform a rotation and do the same.

          One similar problem with minimum distance requirement: http://www.spoj.com/problems/GANNHAT/

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