View
220
Download
2
Category
Preview:
Citation preview
Computing Convex HullsCLRS 33.3
Computational Geometry
• Studies algorithms for solving geometric problems• Applications in many fields
– Computer graphics
– Robotics
– VLSI design
– Computer-aided design, etc.
Geometric algorithms
• The input (typically): a description of a set of geometric objects– A set of points
– A set of line segments
– Vertices of a polygon
– Etc.
• We will look at sample problems in two dimensions, that is, in the plane
Convex hulls in two dimensions
• What is a convex hull?
• What is convexity?– A subset S of the plane is called convex if and only if for
any pair of points p, q in S, the line segment pq is completely contained in S
p
q
convex
p q
not convex
Convex Hull
• The convex hull CH(S) of a set s is the smallest convex set that contains S.
• Consider the problem of computing the convex hull of a finite set P of n points in the plane.
},...,,{ 21 npppP
Convex Hull
Convex Hull
Convex Hull
• Alternative Definition: It is the unique convex polygon whose vertices are points from P and contains all the points of P
Problem
• Given P, compute a list of points that contains the vertices of CH(P), listed in clockwise (CW) order
• Expected output:
10p
2p
1p6p
4p
3p
5p
7p
9p
8p
},,,,{ 781042 ppppp
Brute-Force Solution
• Consider an edge of the convex hull between vertices p and q
• Consider the directed line passing through p and q
• Observe that CH(P) lies to the right of this line
p
q
Brute-Force Solution
• Reverse holds too:• If all points in lie to the right of the directed
line through p and q, the pq is an edge of CH(P)
},{ qpP
p
q
BruteForce_ConvexHull(P) {
E=empty set
for all ordered pairs (p,q) in PxP, and p!= q do {
isEdge = true
for all points r in P, r!=p and r!=q do
if r lies to the left of directed line from p to q
isEdge= false
if isEdge the add pq to E
}
from the set of edges in E, construct a list L of vertices of CH(P), in CW order
return L
}
How to test?
if r lies to the left of directed line from p to q
Orientation test
• Given an ordered triple of points <p,q,r> in the plane, we say that they have – positive orientation if they define a CCW oriented triangle
– negative orientation if they define a CW oriented triangle
– Zero orientation if they are collinear
p
q
r
Positive orientation
p
q
r
r is to the left of pq
Orientation test
• Given an ordered triple of points <p,q,r> in the plane, we say that they have – positive orientation if they define a CCW oriented triangle
– negative orientation if they define a CW oriented triangle
– Zero orientation if they are collinear
p
q
r
negative orientation
p
q
r
r is to the right of pq
Orientation test
• Given an ordered triple of points <p,q,r> in the plane, we say that they have – positive orientation if they define a CCW oriented triangle
– negative orientation if they define a CW oriented triangle
– Zero orientation if they are collinear
zero orientation
p
q
r
r is on pq
p
q
r
Orientation
)
1
1
1
det ( ),,(
yx
yx
yx
rr
pp
signrqpOrient
How about the last step?
• Given E, construct the list of vertices of CH(P) in CW order?
How about the last step?
• Given E, construct the list L of vertices of CH(P) in CW order?
• Remove an arbitrary edge pq from E, put p and q in L• Now find in E the edge with origin q (say qr), remove that edge from
E and add r to L• Keep doing this until E has only one edge left
• time to contsruct L – why?
)( 2nO
Overall,
• BruteForce takes
– How many ordered pairs? • n (n-1)
– For each we compare n-2 other points (r)
)( 3nO
Can we do better?
• A number of algorithms that can compute CH(P) in O(n log n)
• We will study the Incremental Algorithm– Points are added one at a time and the hull is updated at each
insertion
Incremental Algorithm
• At any intermediate step of the algorithm– There is a current hull of points – Add point– Update the hull
• Add points in order of increasing x-coordinate– To guarantee that the new point is outside the
current hull– If same x-coordinate, then in increasing order of y
11,..., ipp
ip
Upper Hull and Lower Hull
We will construct the hull in two pieces:
construct upper hull from left to right
construct lower hull from right to left
1p np
Construction of the Upper Hull
• already processed11,..., ipp
1p
4p7p
9p
12p
13p
ip
Construction of the Upper Hull
• Observation: for a convex polygon, if we walk through the boundary in CW order, we always make a right turn– i.e. if we consider any three consecutive points: p, q, r
Orient(p,q,r) should be negative
• How can we use this?
1p
4p7p
9p
12p
13p
ip
Construction of the Upper Hull
• Check whether the last two points on the current hull and
make a right turn
1p
4p 7p9p
12p
13p
ip
ip
Left turn
Construction of the Upper Hull
• If a left turn, remove the last point from the current hull and try again
1p
4p 7p9p
12p
13p
ipLeft turn
Construction of the Upper Hull
• If a left turn, remove the last point from the current hull and try again
1p
4p 7p9p
12p
ipLeft turn
Construction of the Upper Hull
• If a left turn, remove the last point from the current hull and try again
1p
4p 7p9p
ipLeft turn
Construction of the Upper Hull
• If a right turn, add to the current hull
right turn
1p
4p7p
ip
ip
Construction of the Upper Hull
• If a right turn, add to the current hull
right turn
1p
4p7p
ip
ip
• Store current hull on a stack U– U.top() returns the last point on the current hull– U.second() returns the second to last point on the
current hull
– while ( Orient(U.second(), U.top() ) >= 0)
U.push( )
U.pop()
ip
IncrementalConvexHull(P) {
sort points in increasing order of their x-coordinate
U.push(p[1])
U.push(p[2])
for i=3 to n do
while U.size() >= 2 and orient(U.second(), U.top(), p[i]) >= 0) do
U.pop()
U.push(p[i])
L.push(p[n])
L.push(p[n-1])
for i=n-2 to 1 do
while L.size() >= 2 and Orient(L.second(), L.top(), p[i]) >= 0) do
L.pop()
L.push(p[i])
Combine U and L into a single list of points in CW direction
Incremental Convex Hull
• Running time?– Sort: O(n log n)
– Each point is removed from the stack U at most once: O(n)
– Each point is entered into the stack U once: O(n)
– Thus, O(n) for upper hull
– Similarly O(n) for lower hull
– Total: O(n log n)
Recommended