CODEWITHSUNDEEP
c++computational-geometryintervalsskip-lists
ramino
ramino

Reputation: 498

Finding All Intervals That Overlap a Point

Consider a large set of floating-point intervals in 1-dimension, e.g. 

 [1.0, 2.5],                1.0 |---------------|2.5

 [1.5, 3.6],                      1.5|---------------------|3.6

 .....

It is desired to find all intervals that contain a given point. For example given point = 1.2, algorithm should return the first interval, and if given point = 2.0, it should return the first two interval in the above example.

In the problem I am dealing, this operation needs to be repeated for a large number of times for a large number of intervals. Therefore a brute-force search is not desired and performance is an important factor.

After searching about it, I saw this problem is addressed using interval skip list in the context of computational geometry. I was wondering if there is any simple, efficient C++ implementation available.

EDIT: To be more precise about the problem, there are N intervals and for M points, it should be determined which intervals contain each point. N and M are large numbers where M is larger than N.

Upvotes: 1

Views: 2353

Answers (2)

user1196549
user1196549

Reputation:

If your distribution of intervals allows it, it may be worth to consider a gridding approach: choose some grid size s and create an array of lists. Every k-th list enumerates the intervals that overlap with the "cell" [k.s, (k+1).s[.

Then a query amounts to finding the cell that contains the query point (in O(1)) and reporting all intervals in the list that effectively contain it (in O(K)).

Both preprocessing time and storage are O(I.L+G) where I is the number of intervals and L the average interval length in terms of the grid size and G the total number of grid cells. s must be chosen carefully.

Upvotes: 1

Jason S
Jason S

Reputation: 189646

Suggest using CGAL range trees:

Wikipedia says interval trees (1-dimensional range trees) can "efficiently find all intervals that overlap with any given interval or point".

Upvotes: 1

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