CS133 Computational Geometry Intersection Problems 1 Riddle: Fair - - PowerPoint PPT Presentation

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CS133 Computational Geometry Intersection Problems 1 Riddle: Fair - - PowerPoint PPT Presentation

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CS133 Computational Geometry Intersection Problems 1 Riddle: Fair Cake-cutting Using only one straight-line cut, how to split the cake into two equal pieces (by area)? Cake 2 Riddle: Fair cake-cutting Mixed cake! Still one cut Cake Cake

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CS133

Computational Geometry

Intersection Problems

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Riddle: Fair Cake-cutting

Using only one straight-line cut, how to split the cake into two equal pieces (by area)?

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Cake

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Riddle: Fair cake-cutting

Mixed cake! Still one cut

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Cake Cake

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Line Segment Intersections

Given a set of line segments, each defined by two end points, find all intersecting line segments

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Line Segment Intersection

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Line Segment Intersection

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NaΓ―ve Algorithm

Enumerate all possible pairs of lines Test for intersection Running time O(n2) What is the lower bound of the running time? Worst case: O(n2) Is this optimal?

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Plane-sweep Algorithm

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Elements of Plane-sweep

The sweep line: Sweeps the plane in specific direction, e.g., top-down The state of the sweep line 𝑇: A set of all line segments that intersect the sweep line at any

  • position. The state changes as the line move.

The event points 𝐹: Is the set of points where the state 𝑇 changes. In this case, the end points of the line segments comprise the event points.

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Example: Event Points

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π‘š1 π‘š2 π‘š3

(4,10) (1,7) (2,5) (3,5) (5,6) (3,3) 𝒛 π’Žπ’‹ Start/End 10 π‘š1 Start 7 π‘š2 Start 6 π‘š3 Start 5 π‘š1 End 5 π‘š2 End 3 π‘š3 End

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Plane-sweep Simple Impl.

Input 𝑀 = π‘šπ‘— π‘šπ‘— = π‘šπ‘—. π‘ž1, π‘šπ‘—. π‘ž2 𝐹= the list of 𝑧-coordinates sorted in decreasing

  • rder

𝑇 = {} For each event with a corresponding line π‘šπ‘—

If top point

Compare π‘šπ‘— to each 𝑑 ∈ 𝑇 Insert π‘šπ‘— to S

If end point

Remove π‘šπ‘— from S

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Plane-sweep Poor Behavior

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𝑃 π‘œ2

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Bentley-Ottmann Algorithm

An improved scan-line algorithm Maintains the state S in a sorted order to speed up checking a line segment against segments in S

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Example

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Sweep line state (in sorted order)

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Algorithm Pseudo code

Create a list of event points 𝑄

𝑄 is always sorted by the 𝑧 coordinate

Initialize the state 𝑇 to an empty list Initialize 𝑄 with the first point (top point) of each line segment While 𝑄 is not empty

π‘ž  𝑄.pop 𝑧𝑑 = π‘ž. 𝑧 processEvent(π‘ž)

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Process Event Point (p)

// p is the top (starting) point If π‘ž is the top point Add π‘ž. π‘š to 𝑇 at the order π‘ž. 𝑦 (π‘ž. π‘š = 𝑇𝑗) checkIntersection(π‘‡π‘—βˆ’1, 𝑇𝑗) checkIntersection(𝑇𝑗, 𝑇𝑗+1) Add the end point of π‘ž. π‘š to 𝑄

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Process Event Point (p)

// p is the bottom (ending) point If π‘ž is the bottom point // let π‘ž. π‘š be at position 𝑇𝑗 before removal Remove π‘ž. π‘š from 𝑇 checkIntersection(π‘‡π‘—βˆ’1, 𝑇𝑗)

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Process Event Point (p)

If π‘ž is an interior point

Report π‘ž as an intersection Find π‘ž. π‘š in 𝑇 (π‘ž. π‘š = 𝑇𝑗, 𝑇𝑗+1 ) Swap(𝑇𝑗, 𝑇𝑗+1) checkIntersection(π‘‡π‘—βˆ’1, 𝑇𝑗) checkIntersection(𝑇𝑗+1, 𝑇𝑗+2)

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Check Intersection(π‘š1, π‘š2)

If π‘š1 does not intersect π‘š2 then return Compute the intersection π‘žπ‘— of π‘š1 and π‘š2 If π‘žπ‘—. 𝑧 is above the sweep line then return If π‘žπ‘— ∈ 𝑄 then return Insert π‘žπ‘— into 𝑄

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Sweep Line State (𝑇)

A list of line segments [π‘šπ‘—] Sorted by the 𝑦-coordinate of the intersections between π‘šπ‘— and the sweep line

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π‘šπ‘—[0] π‘šπ‘—[1]

𝑧𝑑

π‘šπ‘— 0 . 𝑦 + Δ𝑦 Δ𝑧 𝑧𝑑 βˆ’ π‘šπ‘— 0 . 𝑧 , 𝑧𝑑 Δ𝑦 Δ𝑧

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Analysis

Initial sort of starting points 𝑃 π‘œ β‹… log π‘œ Number of processed event points 2π‘œ + 𝑙 For each event point

Remove from P: 𝑃 log 𝑄 Insert or remove from S: 𝑃 log 𝑇 Check intersection with at most two lines: 𝑃 1 Insert a new event points: 𝑃 log 𝑄

Upper limit of 𝑄 = 2π‘œ Upper limit of 𝑇 = π‘œ Overall running time 𝑃 π‘œ + 𝑙 log π‘œ

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Corner Case 1: Horizontal Line

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l1 l2 l3 If two points have the same y-coordinate, sort them by the x-coordinate Starting point

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Corner Case 2: Three Intersecting Lines

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l1 l2 l3 Allow the event point to store a list of all intersecting line segments When processed, reverse the

  • rder of all the lines
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Rectangle Intersection

Given a set of orthogonal rectangles (𝑆), find the set of all intersections between pairs of rectangles 𝑠

1 ∩ 𝑠2: 𝑠 1, 𝑠2 ∈ 𝑆

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Example

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r1 r2 r3 r4

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Example

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r1 r2 r3 r4

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Rectangle Primitives

An orthogonal rectangle is represented by its two corner points, lower and upper Test if two rectangles overlap

Two rectangles overlap if both their x intervals and y-intervals overlap Intervals overlap [x1,x2], [x3,x4]: x4>=x1 and x2>=x3

R1(x1,y1,x2,y2)Γ—R2(x3,y3,x4,y4) βž” R3(Max(x1,x3), Max(y1,y3), Min(x2,x4), Min(y2,y4))

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NaΓ―ve Algorithm

Test all pairs of rectangles and report the intersections Running time O(n2) Is it optimal?

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Simple Plane-sweep Algo.

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r1 r2 r3 r4

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Simple Plane-sweep Algo.

What is the state of the sweep line? What is an event? What processing should be done at each event?

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Improved Plane-sweep Algo.

Keep the sweep line state sorted

But how?

Interval tree A variation of BST Stores intervals Supports two operation

Find all intervals that overlap a query point π‘ž Find all intervals that overlap a query interval π‘Ÿ

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A Simple Interval Tree

Store the intervals in a BST ordered by 𝑗. π‘¦π‘›π‘—π‘œ Augment the BST with the value 𝑦𝑛𝑏𝑦 which stores the maximum value of all the intervals in the subtree

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Augmented BST

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Credit: https://en.wikipedia.org/wiki/Interval_tree

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Polygon Intersection

Given a set of polygons, find all intersecting polygons

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Polygon Representation

A polygon is represented as a sequence of points that form its boundary A general polygon might also contain holes, e.g., a grass area with a lake inside For simplicity, we will only deal with solid polygons with no holes

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p1 p2 p3 p4 p5 p6 p7 p8 Corners Edge or Segment

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Filter-and-refine Approach

Convert all polygons to rectangles

For each polygon, compute its minimum bounding rectangle (MBR)

Filter: Find all overlapping rectangles Refine: Test the polygons that have

  • verlapping MBBs

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Filter-and-refine Approach

Filter step: Already studied Refine: How to find the intersection of two polygons? For any two polygons, there are three cases:

1.

Polygons are disjoint

2.

One polygon is contained in the other polygon

3.

Polygon boundaries intersect

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Case 1: Disjoint

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P Q No intersection points Neither 𝑄 βŠ‚ 𝑅 nor 𝑅 βŠ‚ 𝑄 The intersection is empty

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Case 2: Contained

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P Q No intersection points If 𝑄 βŠ‚ 𝑅, then any corner

  • f P is ∈ 𝑅

If 𝑅 βŠ‚ 𝑄, then any corner

  • f Q is ∈ 𝑄

The intersection is the contained polygon

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Case 3: Intersecting

If the boundaries of the two polygons overlap, then there is at least two polygon edges that

  • verlap

NaΓ―ve solution: Compute all intersections between every pair of edges 𝑃 𝑛 β‹… π‘œ

Where 𝑛 and π‘œ are the sizes of the two polygons

We can also use the line-segment sweep-line algorithm

Run in 𝑃 𝑙 + 𝐽 log 𝑙 where 𝑙 = 𝑛 + π‘œ and 𝐽 is the number of intersections If we only need to test, we can stop at the first intersection

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Computing the Intersection

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P Q

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Special Case: Convex Polygons

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P Q

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Convex Polygon Overlaps

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Right-left Right-right left-left Left-right

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Left-right Split

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Left-right Overlap Test

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Observation: the points of each half are monotone along the 𝑧-axis

R L

Since the segments that form each hull are monotone (i.e., going down), there is no chance for the top red segment to intersect the green half-hull Start with the top two segments from each half-hull Do they intersect? No!

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Left-right Overlap Test

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The bottom point of the green segment is higher than all remaining red segments

R L

Skip to the next green line segment Do the current segments intersect? No!

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Left-right Overlap Test

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R L

The bottom point of the red segment is higher. Skip to the next. Do the current segments intersect? No!

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Left-right Overlap Test

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Report the intersection

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Bottom green point is higher. Skip to the next green segment. Do the current segments intersect? Yes!

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Left-right Overlap Test

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R L

Report the intersection Do the current segments intersect? Yes! Bottom red point is higher. Skip to the next red segment.

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Left-right Overlap Test

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R L

Do the current segments intersect? No! Both bottom points are at equal 𝑧-coordinate. Skip any of them

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Right-right Overlap Test

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Right-right Overlap Test

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Right-right Overlap Test

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Right-right Overlap Test

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Right-right Overlap Test

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Right-right Overlap Test

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Convex Polygon Intersection

If only testing is required, the algorithm can terminate as soon as the first intersection is found If the polygon intersection is needed, the algorithm reports all intersections The algorithm terminates when all segments are inspected Running time: 𝑃 𝑛 + π‘œ , each iteration skips

  • ne segment

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