1. Fast Retrieval of Subparts Windowing Queries
D7013E! Lecture 10! Project! Geometric ! Data Structures II!
Interval ! Trees!
Efficient windowing queries!
Priority Search! Trees!
Segment! Trees!
D7013E Computational Geometry - Håkan Jonsson!
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Windowing queries vs range queries
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2. Interval Trees • Let S be a set of n axis-parallel line segments.!
• Range trees: !
– A restriction we will eventually abandon.!
– Points, often i higher dimensions.!
• Windowing Queries: ! – Line segments, polygons, curves, etc usually in low dimensions (2D, 3D).!
• Task: Build a data structure based on S that can report which line segments of S intersect a given (query-) window. ! – The window is a rectangle and the queryalgorithm that traverses the data structure gets its vertices as input. !
• This data structure will be based on a binary tree called an interval tree.!
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An easy special case:
Interval trees… trees on intervals • Question:!
• Line segments with at least one end point in the window:!
– Which, out of a bunch of one-dimensional intervals, contain a given number?!
– Use a range tree (Ch 5).! • O(n log n) storage to answer queries in O(log2n + k) time, where k is the number of line segments reported.! • (Fractional cascading lowers the query time to O(log n + k).)!
• Assume we represent each interval by a horizontal line segment on the x-axis and the number by a vertical line.! • We then get the question: ! – What line segments are ”stabbed” by a given vertical line?!
• Now, since we ask for what a vertical line intersects, an answer also works for any set of horizontal line segments.!
• However, what about line segments that cross over the window?!
– Regradless of y-coordinates. ! – Project segments onto the x-axis.!
– They will not be accounted for.!
• To also include such, let’s first look at a slightly simpler problem…! D7013E Computational Geometry - Håkan Jonsson!
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Building an interval tree
(Building cont.)
• Find the median end point xmid of a line segment in the x-direction and divide segments in 3 subsets Imid, Ileft, and Iright:" " " " " " " " ! • Recur on Ileft, and Iright to get the left and right subtrees respectively. !
• Store the segments intersected by xmid, i.e. those in each root, in sorted x-order on each side of xmid:!
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An example of an interval tree
Performing a query
• Query: ! – Check on which side of xmid the query line lies and walk through the list from outmost while reporting segments whose end point lie further from xmid than qx. ! – Recur into one of the subtrees.! D7013E Computational Geometry - Håkan Jonsson!
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Properties of interval trees
Stabbing with line segments
• Theorem 10.4:!
• In our original windowing problem we stab with a line segment, one side of the query rectangle (q below), and not a line. !
– An interval tree can be built in O(n log n) time and occupies O(n) storage, where n is the number of intervals.! – A query (to report all intervals that contain a given number) takes O(log n + k) time, where k is the number of intervals reported. ! • Since we split on the median the tree ends up balanced with depth O(log n).! • T(n) = 2T(n/2) + O(n) solves to O(n log n).! D7013E Computational Geometry - Håkan Jonsson!
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– There’s an interval also in the y-direction to take into account.! – We are now considering line segments that can lie anywhere in the plane:!
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Solution
Conclusion
• We need to sort out exactly those end points that share y-coordinates with the line segment q.! • Use range trees in the nodes of the interval tree!!
• Build interval trees with range trees in the nodes. !
• Theorem 10.6:!
Build a left range tree on these end points…
A range query as a rectangle
– We can then also detect line segments that pass straight over our query window by two “stabbing queries”. ! • In the x and y directions.!
…and a right one on these.
– Let S be a set of n axis-parallel line segments in the plane. Then…! – The line segments intersected by a query window (rectangle) can be answered in O(log2n + k) time with a data structure that uses O(n log n) storage and can be built in O(n log n) time. !
• Here, k is the number of reported line segments.! A query line segment q D7013E Computational Geometry - Håkan Jonsson!
– So, once again we have a case where the query time is outputsensitive.!
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3. Priority Search Trees
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Heaps • A heap is a complete binary tree.!
• A storage efficient variant of the range tree for queries that are unbounded on one side:!
– Only the down-most level might contain free space.!
• A heap has the following property, called the heap property:!
"
– The root [of the entire heap but also of a subtree] is smaller than any element in the subtrees below the root.!
! • Implemented using a heap. ! D7013E Computational Geometry - Håkan Jonsson!
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Unbounded queries in Heaps
Idea
• Example query: (-!,5]!
• We like a substitute for a range tree…! • When building the heap, divide points into subtrees by y-coordinate!! – Each node contains the median in the y-direction in its subtree.! – A parent node has lower x-coordinate than its children.! Ordering among the points In x: p5, p3, p1, p6, p2, p4 In y: p5, p2, p1, p4, p6, p3
We start at “minus infinity” and go towards “plus infinity”, i.e. downwards in our traversal of the heap.
Stop nodes
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(The heap is deliberately tilted to show how points are stored depending on their x- and y-coordinates.)
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Performing a query in a priority search tree One of the priority search trees in a node
Query line segment
q
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Conclusion • Theorem 10.9:! – A priority search tree for a set of n points in the plane uses O(n) storage and can be built in O(n log n) time. ! • Sorting points and picking out medians.! – All k points in a query range (-!,qx]"[qy,q’y] can be reported in O(log n + k) time. ! • Just traverse the heap while not going down into subtrees whose roots contain points with ! – y-coordinates smaller/larger than those of q and ! – x-coordinates larger than q. ! D7013E Computational Geometry - Håkan Jonsson!
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4. Segment Trees
A simple solution
• A data structure suitable to answer questions like about a set of arbitrarily oriented but (still) non-intersecting line segments: !
• Replace each line segment by its bounding box.!
– Which line segments intersect a given query window?! – As before, segments with end points in the window are easily found using a range tree of all end points.! – But what about the other?!!
– First single out boxes that overlap with the window, then check the line segments in the boxes found.!
• Use our previous data structure for axisparallel line segments.! – ”Usually” good in practice.! – Bad news: In the worst case a query window could intersect all bounding boxes while not intersecting a single line segment(!)!
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A problem with interval trees • Why is an interval tree not that suitable?" " " " " " " " ! • Arbitrary oriented line segments could point outside the query region (thereby upsetting everything)!
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The locus approach • Divide parameter space into regions where the answer to a query is the same.! – A common technique to reduce “an infinite number of things to finite”.! – Plane sweep is an example where an infinite number of stops are reduced to a finite number.!
• In our original stabbing problem, there are infinitely many query lines but only a finite number of answers.! • The intervals in which the answer is the same is elementary. ! D7013E Computational Geometry - Håkan Jonsson!
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Supporting stabbing queries
Idea to reduce the storage
• To support stabbing queries we could build a (balanced) binary search tree with leaves that correspond to elementary intervals. !
• Instead of storing all pointers at leaves, “lift” pointers up to internal nodes when segments cover many elementary intervals in whole subtrees. " " " " " " " " ! • The segment s cover all 5 leaves but is just stored at node v and leaf µ5 !
– Store pointers to the actual line segments in the elementary intervals.!
• A query now amounts to locating the elementary interval that contains the query line.! • However, O(n2) storage in the worst case:!
– s is just one of many intervals.!
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An example of a segment tree
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About segment trees • Lemma 10.10: ! – A segment tree uses O(n log n) storage.! • Each segment is stored at most twice on each level in the tree, which is balanced.! – It can be built in O(n log n) time.!
• Lemma 10.11:! – Using a segment tree, the k intervals containing a query line (a point) can be reported in O(log n + k) time.!
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Solving the original windowing problem
Solving the original windowing problem
• In a segment tree, nodes contain those line segments that cover the set of elementary intervals in the subtree of the node only. ! – Canonical subset.!
• Having searched the segment tree for segments stabbed in the x direction by q, the vertical query line segment….! • …. left is to find those that also share ycoordinate with q.! – Only these are actually intersected.! D7013E Computational Geometry - Håkan Jonsson!
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Final touch
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Final conclusion
• So, each node corresponds to a horizontal slab… and the canonical subset of line segments of a node pass straight through the slab of the node… and it was assumed that line segments do not intersect…! • …build a balanced binary search tree in each slab and over the trapezoids in the slab!! • Searching in this tree using the end points of q (the vertical query segment) gives us all the correct segments.! D7013E Computational Geometry - Håkan Jonsson!
D7013E Computational Geometry - Håkan Jonsson!
• Corollary 10.14:! – Let S be a set of n line segments with arbitrary orientation and disjoint interiors. ! – The k line segments of S intersecting an axis -parallel rectangular query window can be reported in O(log2n + k) time with a data structure that uses O(n log n) storage (and can be built in O(n log n) time). ! • In each of O(log n) nodes we perform a search in a slab that takes O(log n) time. !
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