Common intersection of half planes in r 2 2
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Common Intersection of Half-Planes in R 2 2. PROBLEM (Common Intersection of half-planes in R 2 ) Given n half-planes H 1 , H 2 ,..., H n in R 2 compute their intersection H 1  H 2  ...  H n .

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Common Intersection of Half-Planes in R 2 2

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Common Intersection of Half-Planes in R22

  • 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.


Common Intersection of Half-Planes in R21

  • 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.


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.


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).


Linear Programming in R212

  • The 2-variable LP problem

    Minimize ax + by

    subject to aix + biy + ci 0, i= 1,...,n. (LP1)

    can be transformed by setting Y=ax+by & X=x 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.


Y

P

X

Optimum vertex

Linear Programming in R211

  • 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.


Y

P

F(X)

F+(X)

X

u1

u2

Linear Programming in R210

  • 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+.


Linear Programming in R29

  • I0: All constraints in I0 are vertical lines and determine the feasible interval for X

    u1X u2

    u1 = max{-ci/i: iI0, i<0}

    u2 = min{-ci/i: iI0, i>0}

  • 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)


Linear Programming in R28

  • 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’) > 0, then X’ infeasible

    • if H(X’) =F-(X’) - F+(X’)  0, then X’ feasible


Linear Programming in R27

  • 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);


Linear Programming in R26

  • 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-1n<kn/=O(n)


Linear Programming in R25

  • 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 < u1 or Xij > 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)


Eliminated

Eliminated

Eliminated

Y=iX+i

Y=jX+j

u1

u2 < Xij

u2

Xij < u1

Fig c

Fig a

Fig b

Linear Programming in R24


Linear Programming in R23

  • 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 Xij‘s 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.


Y

P

F(X)

F+(X)

X

Xij

u1

u2

X1/2

Eliminated

Fig d: optimal lies on the left side of X1/2

Linear Programming in R22


Linear Programming in R21

  • Prune & Search Algorithm for 2-variable LP problem

  • Transform (LP1) to (LP2) & (LP3) O(M)

  • For each pair of constraints

    if (i= i or Xij<u1 or Xij>u2),

    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 Xij‘s 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)


Common Intersection

  • Common Intersection of half-planes in R2: (n log n)

  • 2-varialbe Linear Programming: (n)


  • 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.


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