Common Intersection of HalfPlanes in R2 2

1
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.

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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.

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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.

4
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).

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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.

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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.

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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.

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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- mini?I-(?i X ?i),
  where ?i - (?i / ?i) ?i - (ci / ?i)

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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

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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 in which side of X
  (right or left) any feasible value of X may lies
• X feasible we know in which side of X (right
  or left) the minimum of F-(X) lies
• X achieves the minimum of F-(X)

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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)

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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 u1)
• 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)

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Linear Programming in R2 4
14
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.

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Linear Programming in R2 2
16
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)

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Common Intersection

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

18

• 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.