Lecture 11 Intersection

1
Lecture 11 Intersection
2
The goal of the chapter

• To study intersection problems of a set of
  geometric objects
• Geometric objects polygons, segments, polyhedra,
  etc.
• To study four kinds of intersection problems

3
Intersection Problems 4

• Intersection Construction
• Given two geometric objects, compute (or
  construct) their intersection.

4
Intersection Problem 3

• Intersection Detection
• Given two geometric objects, do they
  intersect?
• Intersection detection is frequently easier than
  the corresponding construction problem

5
Intersection Problem 2

• Pairwise Intersection
• Given n geometric objects,
• determine whether any two intersect?
• report all intersecting pairs.
• For testing problem we look for an algorithm that
  avoids testing each object against every other
  object.

6
Intersection Problem 1

• Common Intersection
• Given n geometric objects, find the common
  intersection of all geometric objects.

7
Linear Programming 2

• LP can be viewed as a common intersection
  problem.
• The feasible region of an LP problem is the
  intersection of the half-spaces determined by its
  constraint set.
• The objective function is optimized at some
  vertex (extreme point) of the feasible region
  (convex polyhedron or polytope).

8
Linear Programming 1

• By constructing the common intersection of these
  n half-spaces, we can obtain a solution to LP by
  examining each extreme point.
• However, all we need is the identification of one
  vertex for which the objective function is
  optimized and there is no reason to think that
  the complete construction of the feasible region
  is necessary.
• The two problems are not equivalent !

9
Intersection of convex polygons

• Problem Given two convex polygons, P with L
  vertices and Q with M vertices, compute their
  intersection
• Brute force method O(LM)
• Theorem The intersection of a convex L-gon and a
  convex M-gon can be computed in ?(LM) time.

10
Trivial method for computing intersection of
convex polygons 2

• The boundary of P?Q consists of alternate chains
  of vertices of the two polygons, interspersed
  with points at which the boundaries intersect.

Red, Black, Blue, Black,
11
Trivial method for computing intersection of
convex polygons 1

• To proceed around P, edge by edge, finding all of
  the boundary intersection point involving Q and
  keeping a list of the intersection points and
  vertices along the way.
• This takes O(LM) time because each edge of P will
  be checked against every edge of Q to see if they
  intersect.
• Can this method generalize to solve the
  intersection of two general polygons ? YES

12
?(LM) algorithm for computing intersection of
convex polygons 2

• The approach consists in subdividing the plane
  into regions, in each of which the intersection
  of the two polygons can be easily computed.

Inside a slab each polygon forms a trapezoid
13
?(LM) algorithm for computing intersection of
convex polygons 1

• To partition plane into regions. O(LM)
  merge-sort
• The boundary of each polygon is divided into two
  monotone chains by the left and right extreme
  points.
• Perform a plane-sweep algorithm (merge-sort) from
  left to right to partition the interior of each
  polygon into trapezoids.
• The key observation is that in each slab the
  intersection of P and Q is an intersection of two
  quadrilaterals, which can be found in O(1) time.
  The resulting pieces can be fitted together in
  O(LM) time. O(LM)
• Question Can this method generalize to solve the
  intersection of two star-shaped polygons or two
  general polygons ? NO

14
Intersection of star-shaped polygons 3

• Problem Given two star-shaped polygons, P and Q,
  compute their intersection

Figure 1
15
Intersection of star-shaped polygons 2

• The intersection of two star-shaped polygons P
  Q is not a polygon, but is a union of many
  polygons.
• In figure 1, P Q both have n vertices, and
  every edge of P intersects every edge of Q, so
  the intersection has order of n2 vertices.

16
Intersection of star-shaped polygons 1

• Theorem Finding the intersection of two
  star-shaped polygons requires ?(n2) time in the
  worst case.

17
Intersection of general polygons

• Problem Given two general polygons, P and Q,
  compute their intersection

18
Intersection computation

• The intersection of two convex polygons ?(n)
• The intersection of two star-shaped polygons
  ?(n2).
• The intersection of two general polygons ?(n2).
• But it may not be necessary to spend this much
  time if we only want to know whether two polygons
  P Q intersect at all.

19
Intersection of line segments

• Problem (Line Segment Intersection Test) Given n
  line segments in the plane, determine whether any
  two intersect.
• Brute-force Take each pair of segments and
  determine whether they intersect. This takes
  O(n2) time.
• Theorem Whether any two of n line segments in
  the plane intersect can be determined in ?(n log
  n) time.

20
Polygon Intersection Test

• Problem (Polygon Intersection Test)
• Given two simple polygons P and Q, determine
  if they intersect.
• Polygon Intersection Test ?O(n) Line Segment
  Intersection Test
• Detect any edge intersection between P and Q by
  using the Line Segment Intersection Test.
• If no intersection is found, we must test whether
  P?Q or Q?P. If P?Q, then every vertex of P is
  internal to Q, so we may apply the single-shot
  point inclusion test in O(n) time using any
  vertex of P.

21
Simple Polygon Test

• Problem (Simplicity Test)
• Given a polygon, determine if it is simple.
• Simplicity Test ?O(n) Line Segment Intersection
  Test

22
Line segment Intersection Problem in R1

• Suppose we are given n intervals on the real line
  and wish to know whether any two overlap.
• This can be answered in O(n2) time by inspecting
  all pairs of intervals
• If we sort the 2n endpoints of the intervals and
  designate them as either right or left, then the
  intervals themselves are disjoint if and only if
  the endpoints occur in alternating order L R L
  R... R L R. This test can be performed in O(n log
  n) time.

23

• Two questions
• Can this algorithm be improved?
• Can it generalize to two dimensions, i.e. to the
  case when the line segments are arbitrarily
  oriented?

24
Lower Bound

• Element Uniqueness (Element Distinctness) Problem
  (Given n real numbers, are they all distinct ?)
  requires ?(n log n) time in algebraic computation
  tree computation model
• Element Uniqueness ?O(n) Interval Overlap.
• Given a collection of n real numbers xi, they can
  be converted in linear time to n intervals xi,
  xi. These intervals overlap if and only if the
  original points are not distinct .
• We conclude that ?(n log n) comparisons are
  necessary and sufficient to determine whether n
  intervals in R1 are disjoint.

25
Generalize Interval Overlap Problem in R1 to R2

• The idea behind Interval Overlap Algorithm in R1
  If any two intervals contain a point of the same
  x-coord, then the two intervals overlap.
• The idea behind Line Segment Intersection
  Algorithm in R2 Even though two line-segments
  contain a point of the same x-coord., i.e. both
  are intersected by a vertical line, it doesnt
  mean that these two line-segments intersect.
• For each vertical line L, the line segments
  intersected by L are ordered in y-coord. by their
  intersections.
• When two line segment cross each other, they must
  be adjacent in y-coord. for some vertical line L.

26
Line-segment Intersection Algorithm 10

• Line-segment Intersection problem can be solved
  by plane-sweep technique.
• plane-sweep technique makes use of two basic data
  structures the sweep-line status and the
  event-point schedule.

27
Line-segment Intersection Algorithm 9

• sweep-line status a data structure maintaining
  the ordering of line segments intersecting with a
  give sweep line L and must support the following
  operations (a height-balanced tree will do)
• INSERT(s, L). Insert segment s into the total
  order maintained by L. O(log n)
• DELETE(s, L). Delete segment s from L. O(log n)
• ABOVE(s, L). Return the name of the segment
  immediately above s in the ordering. O(1)
• BELOW(s, L). Return the name of the segment
  immediately below s in the ordering. O(1)

28
Line-segment Intersection Algorithm 8

• event-point schedule a data structure recording
  the ordering of events at which the sweep-line
  status changes and it must support the following
  operation (A min-priority queue will do)
• MIN(E). Determine the smallest element in E and
  delete it O(log n)
• INSERT(x, E). Insert abscissa x into the total
  order maintained by E. O(log n)
• MEMBER(x, E). Determine if abscissa x is a member
  of E. O(log n)

29
Line-segment Intersection Algorithm 7

• The ordering of line segments can change in only
  three ways for each sweep-line
• 1. At the left endpoint of segment s. In this
  case s must be added to the ordering.
• 2. At the right endpoint of segment s. In this
  case s must be removed from the ordering.
• 3. At an intersection point of two segments s1
  and s2. Here s1 and s2 exchange places in the
  ordering.

30
Line-segment Intersection Algorithm 6

• We now outline the algorithm as the vertical
  line sweeps the plane, at each event point
  sweep-line status L is updated and all pairs of
  segments that become adjacent in this update are
  checked for intersection.

S1 ABOVE(s2, L) S3 BELOW(s2, L)
31
Line-segment Intersection Algorithm 5

• procedure LINE SEGMENT INTERSECTION
• sort the 2n endpoints lexicographically by x and
  y and place them into priority queue E
• A ?
• while (E ? ?) do
• p MIN(E)
• if (p is left endpoint)
• then HANDLE LEFT EVENT(p)
• else if (p is a right endpoint)
• then HANDLE RIGHT EVENT(p)
• else HANDLE INTERSECTION EVENT(p) /P is an
  intersection/
• while (A ? ?) do
• (s, s') ? A
• x common abscissa of s and s'
• if (MEMBER(x, E) FALSE)
• then output (s, s')
• INSERT(x, E)

32
Line-segment Intersection Algorithm 4

• HANDLE LEFT EVENT(p)
• s segment of which p is an endpoint
• INSERT(s, L)
• s1 ABOVE(s, L)
• s2 BELOW(s, L)
• if (s1 intersects s) then A ? (s1, s)
• if (s2 intersects s) then A ? (s2, s)

33
Line-segment Intersection Algorithm 3

• HANDLE RIGHT EVENT(p)
• s segment of which p is an endpoint
• s1 ABOVE(s, L)
• s2 BELOW(s, L)
• if (s1 intersects s2 to the right of p)
• then A ? (s1, s2)
• DELETE(s, L)

34
Line-segment Intersection Algorithm 2

• HANDLE INTERSECTION EVENT(p)
• (s1, s2) segment of which p is intersection
• / with s1 ABOVE(s2) to the left of p /
• s3 ABOVE(s1, L)
• s4 BELOW(s2, L)
• if (s3 intersects s2) then A ? (s3, s2)
• if (s1 intersects s4) then A ? (s1, s4)

35
Line-segment Intersection Algorithm 1

• That no intersection is missed by this algorithm
  follows from the observation that only adjacent
  segments may intersect and that all adjacencies
  are correctly examined at least once.

36
Example
37

• Problem (Line Segment Intersection)
• Given n line segments in the plane, find all
  intersecting pairs.
• TheoremBentley-Ottmann(1979) The K
  intersecting pairs of a set of n line segments
  can be reported in time O((nK) log n).
• TheoremChazelle-Edelsbrunner(1988) The K
  intersecting pairs of a set of n line segments
  can be reported in time ?(K n log n).

38
Intersection Detection

• Polygon Intersection Test ?(n log n)
• Simplicity Test ?(n log n)

39
Pairwise Intersection

• Pairwise Line Segment Intersection
• ?(K n log n)
• Line Segment Intersection Test
• ?(n log n)

40
Common Intersection of Half-Planes in R2 2

• PROBLEM (Common Intersection of half-planes in
  R2)
• Given n half-planes H1, H2,..., Hn in R2
  compute their intersection H1?H2 ?...?Hn.
• There is a simple O(n2) algorithm for computing
  the intersection of n half-planes in R2.
• Theorem The intersection of n half-planes in R2
  can be found in ?(n log n) time, and this is
  optimal.

41
Common Intersection of Half-Planes in R2 1

• Theorem The intersection of n half-planes in R2
  can be found in ?(n log n) time, and this
  is optimal.
• Proof.
• (1)To show upper bound, we solve it by
  Divide-and-Conquer
• T(n) 2T(n/2) O(n) O(n log n)
• Merge the solutions to sub-problems solutions
  by finding the intersection of two resulting
  convex polygons.
• (2)To prove the lower bound we show that
• Sorting ?O(n) Common intersection of
  half-planes.
• Given n real numbers x1,..., xn
• Let Hi y ? 2xix xi2
• Once P H1?H2 ?...?Hn is formed, we may read
  off the x.'s in sorted Order by reading the slope
  of successive edges of P.

42
Linear Programming in R2 14

• PROBLEM (2-variable LP)
• Minimize ax by, subject to aix biy ci ?
  0, i 1,...,n.
• 2-variable LP ?O(n) Common intersection of
  half-planes in R2
• Theorem A linear program in two variables and n
  constraints can be solved in O(n log n) time.

43
Linear Programming in R2 13

• Theorem A linear program in two variables and n
  constraints can be solved in ?(n).
• It can be solved by Prune-and-Search technique.
  This technique not only discards redundant
  constraints (i.e. those that are also irrelevant
  to the half-plane intersection task) but also
  those constraints that are guaranteed not to
  contain a vertex extremizing the objective
  function (referred to as the optimum vertex).

44
Linear Programming in R2 12

• The 2-variable LP problem
• Minimize ax by
• subject to aix biy ci ? 0, i 1,...,n.
  (LP1)
• can be transformed by setting Yaxby Xx as
  follows O(n)
• Minimize Y
• subject to ?iX ?iY ci ? 0, i 1,...,n.
  (LP2)
• where ?i(ai-(a/b)bi) ?i bi/b.

45
Linear Programming in R2 11

• In the new form we have to compute the smallest Y
  of the vertices of the convex polygon P (feasible
  region) determined by the constraints.

46
Linear Programming in R2 10

• To avoid the construction to the entire boundary
  of P, we proceed as follows. Depending upon
  whether ?i is zero, negative, or positive we
  partition the index set 1, 2, , n into sets
  I0, I?, I.

47
Linear Programming in R2 9

• I0 All constraints in I0 are vertical lines and
  determine the feasible interval for X
• u1?X ?u2
• u1 max-ci/?i i?I0, ?ilt0
• u2 min-ci/?i i?I0, ?igt0
• I All constraints in I define a piecewise
  upward-convex function F mini?I(?i X ?i),
  where ?i - (?i / ?i) ?i - (ci / ?i)
• I- All constraints in I- define a piecewise
  downward-convex function F- maxi?I-(?i X ?i),
  where ?i - (?i / ?i) ?i - (ci / ?i)

48
Linear Programming in R2 8

• Our problem so becomes O(n)
• Minimize F-(X)
• subject to F-(X) ? F(X) (LP3)
• u1?X?u2
• Given X of X, the primitive, called evaluation,
  F(X) F-(X) can be executed in O(n)
• if H(X) F-(X) - F(X) gt 0, then X
  infeasible
• if H(X) F-(X) - F(X) ? 0, then X feasible

49
Linear Programming in R2 7

• Given X of X in u1, u2 , we are able to reach
  one of the following conclusions in time O(n)
• X infeasible no solution to the problem
• X infeasible we know on which side of X
  (right or left) any feasible value of X may lies
• X feasible we know on which side of X (right
  or left) the minimum of F-(X) lies
• X achieves the minimum of F-(X)

50
Linear Programming in R2 6

• We should try to choose abscissa X where
  evaluation takes place s.t. if the algorithm does
  not immediately terminate, at least a fixed
  fraction ? of currently active constraints can be
  pruned. We get the overall running time T(n) ? ?i
  k(1-?)i-1nltkn/?O(n)

51
Linear Programming in R2 5

• We show that the value ?1/4 as follows
• At a generic stage assume the stage has M active
  constraints
• let I I- be the index set as defined earlier,
  with I I-M.
• We partition each of I I- into pairs of
  constraints.
• For each pair i, j of I , O(M)
• If ?i ?j (i.e. the corresponding straight lines
  are parallel)
• then one can be eliminated. (Fig a)
• Otherwise, let Xij denote the abscissa of their
  intersection
• If (Xij lt u1 or Xij gt u2)
• then one can be eliminated. (Fig b)
• If (u1 ? Xij ? u2)
• then we retain Xij with no elimination. (Fig
  c)
• For each pair i, j of I- , it is similar to I
  O(M)

52
Linear Programming in R2 4
53
Linear Programming in R2 3

• For all pairs, neither member of which has been
  eliminated, we compute the abscissa of their
  abscissa of their intersection. Thus, if k
  constraints have been eliminated, we have
  obtained a set S of (M-k)/2 intersection
  abscissae. O(M)
• Find the median X1/2 of S O(M)
• If X1/2 is not the extreminzing abscissa, then We
  test which side of X1/2 the optimum lies. O(M)
• So half of the Xijs lie in the region which are
  known not to contain the optimum. For each Xij in
  the region, one constraint can be eliminated O(M)
  (Fig d)
• This concludes the stage, with the result that at
  least k (M-k)/2/2 ?M/4 constraints have been
  eliminated.

54
Linear Programming in R2 2
55
Linear Programming in R2 1

• Prune Search Algorithm for 2-variable LP
  problem
• Transform (LP1) to (LP2) (LP3) O(M)
• For each pair of constraints
• if (?i ?i or Xijltu1 or Xijgtu2),
• then eliminate one constraint O(M)
• Let S be all the pairs of constraints s.t. u1?Xij
  ?u2,
• Find the median X1/2 of S test which side of
  X1/2 the optimum lies O(M)
• Half of the Xijs lie in the region which are
  known not to contain the optimum. For each Xij in
  the region, one constraint can be eliminated.
  O(M)

56
Common Intersection

• Common Intersection of half-planes in R2 ?(n log
  n)
• 2-varialbe Linear Programming ?(n)

57

• We must point out that explicit construction of
  the feasible polytope is not a viable approach to
  linear programming in higher dimensions because
  the number of vertices can grow exponentially
  with dimension. For example, n-dim hypercube has
  2n vertices.
• The size of Common Intersection of half-spaces in
  Rk is exponential in k, but the time complexity
  for k-variable linear programming is polynomial
  in k.
• These two problems are not equivalent in higher
  dimensions.