Convex Hulls - Department of Computer Sciencehacamero/ConvexHulls.pdf · Convex Hull: Informally...

Convex Hulls

Helen Cameron

Helen Cameron Convex Hulls 1/101

What Is a

Convex Hull?

Starting Point: Points in 2D

x

y


Convex Hull: Informally

Imagine that the x , y -plane is a board and the points are nailssticking out of the board. If you stretch an elastic so that all thenails are inside it and then let go of the elastic, the elastic willtighten into the boundary of the convex hull of the points.

x

y


Starting Point: Simple Polygons

x

yPolygon: A sequenceof points connectedby edges into a closed“curve”.

Simple: Only consec-utive edges intersect(at their common end-points).


Starting Point: Simple Polygons

x

y

For comparison, here’s anon-simple polygon.


Inside and Outside

x

y

A simple polygon has aninside and an outside.

The inside is always onyour left if you walk theboundary in a counter-clockwise direction


Interior Angles

x

yThe interior angle at apoint is the angle insidethe polygon between thepoint’s adjacent edges.


Convex Angles

x

y

A convex angle is anangle < 180 degrees (πin radians).

convex

not convex


Convex Polygon

x

y

A convex polygon hasonly convex interior an-gles.


Convex Polygon

x

y Equivalent definition: A polygon isconvex if, for any two points in oron the polygon, the line segmentconnecting the two points is entirelycontained in the polygon.


Convex Hull: Formally

The convex hull of a set of 2D points is the smallest convexpolygon that contains the points.

x

y


Why Care About

Convex Hulls?(Some applications!)

Convex Hulls Are Easier

Collision avoidance is faster and easier using the convex hull of arobot or a car than working with its exact (more complex) shape.

SevenTransparent.jpg By Brian

Snelson (originally posted to Flickr

as My car photoshopped) [CC BY

2.0 (https://creativecommons.

org/licenses/by/2.0)], via

Wikimedia Commons.


https://commons.wikimedia.org/wiki/File:SevenTransparent.jpg

https://creativecommons.org/licenses/by/2.0


Removing Outliers

Often, we want to estimate someparameter from a small,randomly-chosen sample of apopulation of points.

This sort of estimation is used,for example, in medicaldiagnostics and network intrusiondetection.

File:Pankreas-Ca im CT -

Coeliacusblockade.png By Hellerhoff

[CC BY 2.0

(https://creativecommons.org/

licenses/by/2.0)], via Wikimedia

Commons.


https://commons.wikimedia.org/wiki/File:File:Pankreas-Ca im CT - Coeliacusblockade.png

https://commons.wikimedia.org/wiki/File:File:Pankreas-Ca im CT - Coeliacusblockade.png



Removing Outliers

Some parameter estimations arevery sensitive to “outliers”(points that are far from most ofthe other points in thepopulation).

Example outliers


Removing Outliers

Get rid of outlier points in 2D:Repeatedly find and then “peel”off the convex hull, until no morethan some chosen fraction of theoriginal points remains.

Then you can safely estimate thedesired parameter.

Originalpoints

Outermosthull

Remainingpoints

Next hull

Points to estimatefrom:


Finding the Convex Hull:

Input and Output

The Input (If You’re Writing a Program)

Input: A list of 2D points (px , py ), in no particular order.

Example input:

(9.8,10.6) (7,13) (12.8,8.7)(10.35,12.1) (11.2, 14.5) (8,11)(7.5,9.8) (9.1,12.2) (7.8,7.3)(10.2,9.5) (11.7,9.6) (9.6,9.1)(9.5,7.8) (8.55,10.2) (10.7,10.1)(12.55,8.3) (8.85,9.1) (8.4,13.8)(10,8.7) (12.2,10.7) (11, 10.4)(10.55,8.3)


The Output (If You’re Writing a Program)

Output: The points on the boundary of the convex hull of theinput points, listed in the order of a counter-clockwise traversal ofthe boundary.

Example output: (7,13) → (7.8,7.3) → (12.55,8.3) → (12.8,8.7)→ (11.2, 14.5) → (8.4,13.8) → (7,13).


Left, Right, or Collinear?

Leftness Testing for a Line Segment

Useful test when finding the convex hull: Is point q to the left,

right or collinear with a directed line segment−→a, b — that is, if

we’re standing on endpoint a and facing the other endpoint b, is qto our left, to our right or somewhere on the infinite line through aand b?

a

bq0

q1

q2


Leftness Testing from a Line Segment

A method that returns true if q is to the left of−→a, b (returns false

otherwise):

boolean isLeft( Point q, Point a, Point b ) {return 0 < (b.x-a.x)*(q.y-a.y) - (b.y-a.y)*(q.x-a.x);

}

A method that returns true if q is to the right of−→a, b (returns false

otherwise):

boolean isRight( Point q, Point a, Point b ) {return 0 < (b.x-a.x)*(q.y-a.y) - (b.y-a.y)*(q.x-a.x);

}

A method that returns true if q is collinear with−→a, b (returns false

otherwise):

boolean isCollinear( Point q, Point a, Point b ) {return 0 == (b.x-a.x)*(q.y-a.y) - (b.y-a.y)*(q.x-a.x);

}Helen Cameron Convex Hulls 23/101

Leftness Testing from a Line Segment

The leftness test is using the “right-hand rule”:

Lay the outer edge of your right hand along−→a, b (fingers

pointing from a to b) with your thumb sticking up.

If your fingers naturally curl towards q, then q is to the left.

The right-hand rule has to do with the “cross product” of twovectors . . .


A Vector

A vector ~p is a directed line segment starting at the origin andending at point p.

Note: We’ll work with a directed line segment starting at somepoint a and ending at point b — we’ll have to translate a to theorigin by subtracting a from all points. (Really: subtract ax fromall points’ x-coordinates and subtract ay from all points’y -coordinates).


Area of a Triangle: Vector Cross Product

The cross product p × q of two vectors ~p = (px , py , pz) and~q = (qx , qy , qz) only makes sense in 3D: it is the vectorperpendicular to both p and q given by the “right-hand rule” andwhose magnitude is given by the absolute value of∣∣∣∣∣∣~x ~y ~zpx py pzqx qy qz

∣∣∣∣∣∣ = (pyqz−pzqy )~x+(pzqx−pxqz)~y+(pxqy−pyqx)~z .

Note: ~x is the unit vector in the positive x-direction, ~y is the unitvector in the positive y -direction, and ~z is the unit vector in thepositive z-direction.


Area of a Triangle: Vector Cross Product

For 2D vectors, the z-coordinates pz = qz = 0 and thecross-product formula reduces to

(pxqy − pyqx)~z .

If point q is to the left of vector ~p, then pxqy − pyqx ispositive (follows right-hand rule).

If point q is to the right, then pxqy − pyqx is negative(perpendicular in the opposite direction).

If point q is collinear with vector ~p, then pxqy − pyqx is zero.


Examples

Example 1: If ~p =−−−→(3, 1) and q = (2, 4), then the calculation gives

(pxqy − pyqx)~z = (3× 4− 1× 2)~z = 10~z .

Conclusion: q is to the left of ~p.

(0, 0)

p

q


Examples

Example 2: If ~p =−−−−→(−1, 4) and q = (2, 4), then the calculation is

(pxqy − pyqx)~z = (−1× 4− 4× 2)~z = −12~z .

Conclusion: q is to the right of ~p.

(0, 0)

p q


Examples

Example 3: If ~p =−−−→(1, 1) and q = (2, 2), then the calculation gives

(pxqy − pyqx)~z = (1× 2− 1× 2)~z = 0~z .

Conclusion: q is collinear with ~p.

(0, 0)

p

q


Leftness Testing and Line Segments

We’re usually working not with a vector (tail at the origin andhead at some point p), but with a directed line segment (tail atsome point a and head at some other pont b).

To test if point q is to the left of directed line segment−→a, b, we

take the cross product of vectors−−−→b − a and

−−−→q − a.


Leftness Testing and Line Segments

Example: If we want to know if point q = (4, 5) is to the left of

line segment−→a, b, where a = (2, 1) and b = (5, 2), we translate to

asking if point q − a = (2, 4) is to the left of directed line segment−−−−−−−−→a− a, b − a = vector

−−−→b − a =

−−−→(3, 1) — this translation is

Example 1 above:

(0, 0)

b − a

q − a

a

b

q


Graham’s Scan

Graham’s Scan: Overview

Graham’s scan: Finds a convex hull of a set S of two-dimensionalpoints.

Example input: Find the convex hull of (8, 8), (2, 4), (0, 11),(5, 6), (6, 10), (10, 5), (1, 6), (5, 12),(2, 8), and (9, 13):

(0, 11)

(1, 6)

(2, 8)

(2, 4)

(5, 12)

(6, 10)

(8, 8)

(9, 13)

(10, 5)

(5, 6)



Graham’s scan uses a list:

Each point is pushed onto the end of the list once, andeventually popped off the end of the list if the point is not onthe convex hull.

At the end of the algorithm, the list contains thecounterclockwise ordering of points on the hull, from thebeginning to the end of the list.



Example Output: For the previous example points, Graham’sscan returns (2, 4), (10, 5), (9, 13), (0, 11), and (1, 6) (in thatorder), because that’s what its list contains (from beginning to endof the list) at the end of Graham’s scan:

(0, 11)

(1, 6)

(2, 8)

(2, 4)

(5, 12)

(6, 10)

(8, 8)

(9, 13)

(10, 5)

(5, 6)


Graham’s Scan Has Four Steps

Step 1: Find the point p0 with the lowest y -coordinate.

(0, 11)

(1, 6)

(2, 8)

(2, 4)p0

(5, 12)

(6, 10)

(8, 8)

(9, 13)

(10, 5)

(5, 6)



Step 2: Sort the points by angle around p0 (that is, by “polarcoordinates”).

p7

p8

p6

p0

p5

p4

p2

p3

p1

discarded



Step 3: Initialize the list L to contain p0, p1, and p2 (in thatorder):

p7

p8

p6

p0

p5

p4

p2

p3

p1

discarded



Step 4: Process each of the other points in order, p3, p4, . . . ,pm−1.

When we’re processing point pi :(4a) Loop:

Suppose points p2nd last and plast are the last two points onthe end our list L.

Remove plast from the end of L if pi (the point we’reprocessing) is NOT to the left of directed line segment−−−−−−−→p2nd lastplast.

Repeat this until pi is to the left of −−−−−−−→p2nd lastplast.(4b) Add pi to the end of L.


Graham-Scan( S )

1. Find p0, the point in S with the lowest y-coordinate(break ties by choosing the leftmost such point).

2. Sort the other points in S by polar angle around p0

(if two or more points have the same polar angle,

remove all but the one farthest from p0).

Let the sorted order be p1, p2, . . . , pm−1.

3. L = empty list.

Add p0 to the end of L.



4. For each point pi in order from p3 to pm−1 do

4a. While pi is not left of directed

line segment −−−−−−−−−−−→p 2nd lastp last

Remove the point p last from the end of L.

4b. Add pi to the end of L.

Step 1: Finding Lowest Point p0

Question: How do we find the point p0 with the lowesty -coordinate?

(0, 11)

(1, 6)

(2, 8)

(2, 4)p0

(5, 12)

(6, 10)

(8, 8)

(9, 13)

(10, 5)

(5, 6)

Answer: Simply loop through the points, looking only at they -coordinates of each point (do a findLowest).


Step 1: Finding Lowest Point p0

Question: How do we find the point p0 with the lowesty -coordinate?

findLowest( Point[] S ) returns int {int minIndex = 0; // Assume S[0] is p0

for ( int i = 1; i < S.length; i++ )

{if (S[i].y < S[minIndex].y )

minIndex = i;

else if ((S[i].y == S[minIndex].y)

&& (S[i].x < S[minIndex].x))

minIndex = i;

} // end for

return minIndex;

} // end findLowest

At the end, S[minIndex] is the point with the lowest y -coordinate(in case of ties, it’s the leftmost of the lowest points).


Step 2: Sorting by Polar Coordinate

Question: How do we sort by polar coordinate?

p7

p8

p6

p0

p5

p4

p2

p3

p1

discarded

Answer: Use any sorting algorithm such as merge sort or quicksort.



In the sorting algorithm, when comparing two points pi and pj todecide which comes first in the sorted order (“Is pi before pj in thesorted order?”):

Test if pj is to the left of directed line segment −−−→p0, pi usingthe cross product method.

If pj is to the left, then pi comes before pj in the sorted order.

If pj is to the right of −−−→p0, pi , then pi comes after pj in thesorted order.

If the cross product gives 0, then p0, pi and pj are collinear.Compute the distance of pi and pj from p0 and discard thecloser one.



Further note: If the cross product gives 0, do not perform thesquare root in the calculation of the distance: Just discard pi if

(pi .x − p0.x)2 + (pi .y − p0.y)2

is smaller than

(pj .x − p0.x)2 + (pj .y − p0.y)2;

otherwise, discard pj .

Computing square roots is too expensive and we do not need toknow the exact distance measurement, we just need to know whichis closer to p0.


Graham’s Scan Example

Example: Given the following set of points, compute the convexhull.



Example (continued): First choose p0 to be the leftmost pointamong the two points with the same lowest y -coordinate. Thensort the remaining points by polar angle:

0p

1p

2p

3p

4p

5p

6p

7p

8p



Example (continued): Now add p0, p1, p2 on the list L.

0p

1p

2p

3p

4p

5p

6p

7p

8p

Next, consider points p3, . . . , p8 in turn.



Example (continued): Consider point p3: Because p2, p3 is nota left turn from p1, p2, remove p2 from the list L.

0p

1p

2p

3p

4p

5p

6p

7p

8p



Example (continued): Still considering point p3: Becausep1, p3 is a left turn from p0, p1, add p3 onto the list L.

0p

1p

2p

3p

4p

5p

6p

7p

8p



Example (continued): Consider point p4: Because p3, p4 is aleft turn from p1, p3, add p4 onto the list L.

0p

1p

2p

3p

4p

5p

6p

7p

8p




0p

1p

2p

3p

4p

5p

6p

7p

8p




0p 1

p

2p

3p

4p

5p

6p

7p

8p




0p

1p

2p

3p

4p

5p

6p

7p

8p




0p 1

p

2p

3p

4p

5p

6p

7p

8p

The convex hull is represented by the counterclockwise ordering

p0, p1, p3, p5, p7, p8.


What Is Step 4(a) Doing?

Idea: When Graham’s scan is about to process pi , list L containsthe convex hull of the points previously processed p0, . . . , pi−1.Graham’s scan tries to add the next point pi by simply addingtriangle p0, pi−1, pi to the convex hull.

p0 p1

pi−1

pi

CH(p0 − pi−1)


What Is Step 4(a) Doing?

Idea: But the result is not a convex hull if pi is not a left turnfrom the previous edge in CH(p0, . . . , pi−1)). Graham’s Scan fixesthe convexity problem by adding more triangles on the outside ofthe polygon (“filling in” non-left turns to pi with another triangle):

p0 p1

pk

pj

pi−1

pi

CH(p0 − pi−1)


How Fast Is Graham’s Scan?

To find the convex hull of n points using Graham’s Scan:

Running time = Time to find p0 + Time to sort by polar angle

+ Time for adds, removes, and left-turn tests

= O(n) + O(n log n) + O(n)

= O(n log n).


Divide and Conquer

Convex Hull

Divide and Conquer Algorithm

This Preparata and Hong (1977) algorithm can be extended tothree or higher dimensions.

To keep things simple, assume:

No three points lie on a line. (No three points are collinear.)

No two points lie on a vertical line.

These assumptions can be removed, but they make theexplanations simpler, so we will use them.


Divide and Conquer Algorithm: The Idea

The algorithm sorts the points by x-coordinate.

The algorithm then builds many small convex hulls of small groupsof adjacent points. These small convex hulls will not intersect eachother — they’re disjoint.

Then the algorithm merges (joins together) adjacent small convexhulls to make bigger convex hulls of bigger groups of adjacentpoints.

The algorithm keeps merging adjacent convex hulls until it hasONE convex hull of all the points.


Step 1: Sort the Points

Step 1 gets as its input an array p of points:

(8, 3) (−1, 4) (0, 1) (5, 2) (4, 4) (6, 0) (3, 6)

0 1 2 3 4 5 6

Here’s the input points plotted:

p[0]

p[1]

p[2]

p[3]

p[4]

p[5]

p[6]

You can see that the points are not in any particular order.Helen Cameron Convex Hulls 65/101


Step 1 sorts the points into ascending order by x-coordinate:

(−1, 4) (0, 1) (3, 6) (4, 4) (5, 2) (6, 0) (8, 3)

0 1 2 3 4 5 6

Here’s the points plotted, so you can see that they’re now sorted.

p[6]

p[0]

p[1]

p[4]

p[3]

p[5]

p[2]



Step 1 can use any sorting method (e.g., quick sort) — when thesorting algorithm wants to compare two items (points), it simplycompares their x-coordinates.


Step 2: Pair Up the Points

Step 2 pairs adjacent points to form edges. (If there’s an oddnumber of points, group the last three points into a triangle.)

p[6]

p[0]

p[1]

p[4]

p[3]

p[5]

p[2]



An edge is a “degenerate” convex hull of the pair of points.

Consequence: Step 2 is making the list of points into a list ofconvex hulls of small groups of points.



Note: Make sure you get a counter-clockwise ordering of thepoints of a triangle.

When p[i+2] is left of−−−−−−−−−→p[i],p[i+1]

p[i]

p[i+1]

p[i+2]

Order: p[i],p[i+1],p[i+2]

p[i]

p[i+1]

p[i+2]

When p[i+2] is right of−−−−−−−−−→p[i],p[i+1]

p[i]

p[i+1]

p[i+2]

Order: p[i],p[i+2],p[i+1]

p[i]

p[i+1]

p[i+2]


Step 3: Merge Pairs of Hulls

Step 3 does the following Pair-and-Merge step over and overagain until we have ONE convex hull:

Pair up adjacent small convex hulls and merge each pair into oneconvex hull (more on merging in a minute).

Before Pair-and-Merge 1:

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Step 3 pairs up adjacent small convex hulls and merges each pairinto one convex hull (more on merging in a minute).

After Pair-and-Merge 1:

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Step 3 Each time we pair and merge, we are halving the numberof convex hulls we have and doubling the number of points thateach convex hull contains.

If we have n points, we can do that at most log2 n times.


Step 3: Finding Tangent Lines to Merge Hulls

The merge step: How do we merge two non-overlapping convexhulls separated by an imaginary vertical line?

We find two “tangent lines”, one supporting the two hulls belowand the other resting on the two hulls above.

0

12

3

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0

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34

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12

LR


Finding the Lower Tangent Line


What is the lower tangent line T? The two points on L’s hullimmediately before and after T ’s endpoint in L are both to the leftof T (when T is directed from L to R).The two points on R’s hull immediately before and after T ’sendpoint in R are both to the left of T (when T is directed from Lto R).

0

12

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0

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34

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LR



To find the lower tangent line: Start by connecting therightmost point of L with the leftmost point of R with a linesegment T .

L R

T



To find the lower tangent line: (continued) Then “walk” Tdownward . . .

Repeat while T is not tangent to both hulls at the same time:

Move T ’s endpoint down L’s hull while T is not tangent toL’s hull, then

Move T ’s endpoint down R’s hull while T is not tangent toR’s hull.



To find the lower tangent line: (continued) Move T down Lwhile T is not tangent to L:

L R



To find the lower tangent line: (continued) Now move T downR while T is not tangent to R:

L R



To find the lower tangent line: (continued) Move T down Lagain, while T is not tangent to L:

L R



To find the lower tangent line: (continued) When we try tomove T down R again, we see that T is already tangent to R.Now T is the lower tangent line because it is tangent to both Land R at the same time!

L R



To find the lower tangent line: (continued) Let l be the index ofthe endpoint of T in L and r be the endpoint of T in R.

L R

Tl

r



To find the lower tangent line: (continued) Assume that, ineach hull, the points are indexed from 0 and that arithmetic onindices is done using modular arithmetic so that we can say l − 1and r + 1 and get the index of the next point.

01

2

5

6

7

L

0

1

2

3

6

R

Tl = 4

l − 1 = 3

4 = r

5 = r + 1



To find the lower tangent line: (continued) Notice that movingfrom l to l − 1 is moving clockwise in L and moving from r tor + 1 is moving counterclockwise in R.

01

2

5

6

7

L

0

1

2

3

6

R

Tl = 4

l − 1 = 3

4 = r

5 = r + 1



To find the lower tangent line: (continued) T will be tangent toboth hulls when l − 1 and l + 1 both lie above T (are to the left oflr) and when r − 1 and r + 1 lie above T (to the left of lr).

2

3

4

5

6

L

2

3

4

5

R

Tl = 0

l − 1 = 7

1 = l + 10 = r

1 = r + 1

r − 1 = 6



To find the lower tangent line: (continued)

Algorithm LowerTangent

1. l = rightmost point of L /* maximum x-coordinate */

2. r = leftmost point of R /* minimum x-coordinate */

3. while (T=lr is not lower tangent to both L and R) {4. while (T is not lower tangent to L) {5. l=l-1;6. }7. while (T is not lower tangent to R) {8. r=r+1;9. }10. }



To find the lower tangent line:

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4

5

6

7

8

9

10

0

1

234567

8

9

10

L R


Finding the Upper Tangent Line


To find the upper tangent line: We find the upper tangentsimilarly — we walk T up until it is upper tangent to both hulls.

Note: Instead of considering T to be the directed line segmentfrom l to r , we consider T to be the directed line segment from rto l . T is tangent if the before and after points on L and R are to

the left of−→rl .



To find the upper tangent line: Again, start by connecting therightmost point of L with the leftmost point of R with a linesegment T .

L R

Tl

r



To find the upper tangent line: Move T = rl up L while T isnot tangent to L:

L R

T

l

r



To find the upper tangent line: Then move T = rl up R whileT is not tangent to R:

L R

Tl r



To find the upper tangent line: Since T is not yet tangent toboth hulls at the same time, move T = rl up L while T is nottangent to L:

L R

Tlr



To find the upper tangent line: When we try to move T up Ragain, we see that T is already tangent to R. Now T is the uppertangent line because it is tangent both hulls at the same time,

L R

Tlr



To find the upper tangent line:

Algorithm UpperTangent

1. l = rightmost point of L /* maximum x-coordinate */

2. r = leftmost point of R /* minimum x-coordinate */

3. while (T=rl is not upper tangent to both L and R) {4. while (T is not upper tangent to L) {5. l=l+1;6. }7. while (T is not upper tangent to R) {8. r=r-1;9. }10. }


Merging the Two Convex Hulls

Step 3: Merging Hulls with Tangent Lines

The merged convex hull: Once we’ve found the two tangents’endpoints, we simply copy the pieces of L’s boundary and R’sboundary and the tangent lines into the boundary of the convexhull of L and R merged.

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Convex Hulls - Department of Computer Sciencehacamero/ConvexHulls.pdf · Convex Hull: Informally...

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Transcript of Convex Hulls - Department of Computer Sciencehacamero/ConvexHulls.pdf · Convex Hull: Informally...