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How do I traverse a KDTree to find k nearest neighbors?

This question concerns the implementation of KNN searching of KDTrees. Traversal of a KDTree to find a single best match (nearest neighbor) is straightforward, akin to a modified binary search.

How is the traversal modified to exhaustively and efficiently find k-best matches (KNN)?

Edit for clarification: After finding the nearest node M to the input query I, how does the traversal algorithm continue to find the remaining K-1 closest matches to the query? Is there a traversal pattern which guarantees that nodes are visited in order of best to worst match to the query?

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user2647513 Avatar asked Jan 09 '16 02:01

user2647513


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2 Answers

You can maintain a max heap of size k (k is the count of nearest neighbors which we wanted to find).

Start from the root node and insert the distance value in the max heap node. Keep on searching in k-d tree using dimensional splitting , criteria and keep updating Max Heap tree.

https://gopalcdas.wordpress.com/2017/05/24/construction-of-k-d-tree-and-using-it-for-nearest-neighbour-search/

~Ashish

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Ashish Mittal Avatar answered Nov 01 '22 23:11

Ashish Mittal


Adding to @Ashish's answer, you can use a max-heap in the following manner:

1) Build a max-heap of the first k elements (arr[0] to arr[k-1]) of the given array. 

This step is O(k). Then

2) For each element, after the kth element (arr[k] to arr[n-1]), compare it with 
   root of the max-heap.
    a) If the element is smaller than the root then make it root 
       and call heapify for max-heap.
    b) Else ignore it.

The step 2 is O((n-k)*log(k)).

3) Finally, the max-heap has k smallest elements and root of the heap 
   is the kth smallest element.

Time Complexity: O(k + (n-k)*log(k)) without sorted output. If sorted output is needed then O(k + (n-k)*log(k) + k*log(k)).

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sks-15 Avatar answered Nov 01 '22 22:11

sks-15



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