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Common Intersection of Half-Planes in R 2 2

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- PROBLEM (Common Intersection of half-planes in R2)
Given n half-planes H1, H2,..., Hn in R2 compute their intersection H1H2 ...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.

- 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 = H1H2 ...Hn is formed, we may read off the x.'s in sorted Order by reading the slope of successive edges of P.

- 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.

- 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).

- 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

- 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

- 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+.

- I0: All constraints in I0 are vertical lines and determine the feasible interval for X
u1X u2

u1 = max{-ci/i: iI0, i<0}

u2 = min{-ci/i: iI0, i>0}

- I+: All constraints in I+ define a piecewise upward-convex function F+ = miniI+(i X+i), where i = - (i /i) & i = - (ci /i)
- I-: All constraints in I- define a piecewise downward-convex function F- = miniI-(i X+i), where i = - (i /i) & i = - (ci /i)

- Our problem so becomes: O(n)
Minimize F-(X)

subject to F-(X) F+(X) (LP3)

u1Xu2

- 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

- 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);

- 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)

- 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)

- If i = j (i.e. the corresponding straight lines are parallel)
- 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

- 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

- 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. u1Xij 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 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.