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. There is a simple O(n 2 ) algorithm for computing the intersection of n half-planes in R 2. Theorem: The intersection of n half-planes in R 2 can be found in  (n log n) time, and this is optimal.

Common Intersection of Half-Planes in R 2 1 Theorem: The intersection of n half-planes in R 2 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 x 1,..., x n Let Hi: y  2x i x – x i 2 Once P = H 1  H 2 ...  H n is formed, we may read off the x.'s in sorted Order by reading the slope of successive edges of P.

Linear Programming in R 2 14 PROBLEM (2-variable LP) Minimize ax + by, subject to a i x + b i y + c i  0, i= 1,...,n. 2-variable LP  O(n) Common intersection of half-planes in R 2 Theorem: A linear program in two variables and n constraints can be solved in O(n log n) time.

Linear Programming in R 2 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).

The 2-variable LP problem Minimize ax + by subject to a i x + b i y + c i  0, i= 1,...,n. (LP1) can be transformed by setting Y=ax+by & X=x as follows: O(n) Minimize Y subject to  i X +  i Y + c i  0, i= 1,...,n. (LP2) where  i =(a i -(a/b)b i ) &  i = b i /b. Linear Programming in R 2 12

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 X P Optimum vertex Linear Programming in R 2 11

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 I 0, I , I +. Y X P u1u1 u2u2 F  (X) F + (X) Linear Programming in R 2 10

I 0 : All constraints in I 0 are vertical lines and determine the feasible interval for X u 1  X  u 2 u 1 = max{-c i /  i : i  I 0,  i <0} u 2 = min{-c i /  i : i  I 0,  i >0} I + : All constraints in I + define a piecewise upward- convex function F + = min i  I+ (  i X+  i ), where  i = - (  i /  i ) &  i = - (c i /  i ) I - : All constraints in I - define a piecewise downward- convex function F - = min i  I- (  i X+  i ), where  i = - (  i /  i ) &  i = - (c i /  i ) Linear Programming in R 2 9

Our problem so becomes: O(n) Minimize F - (X) subject to F - (X)  F + (X) (LP3) u 1  X  u 2 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 R 2 8

Given X ’ of X in [u 1, u 2 ], 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 R 2 7

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-1 n<kn/  =O(n) Linear Programming in R 2 6

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 X ij denote the abscissa of their intersection If (X ij u 1 ) then one can be eliminated. (Fig b) If (u 1  X ij  u 2 ) then we retain X ij with no elimination. (Fig c) For each pair i, j of I -, it is similar to I + O(M) Linear Programming in R 2 5

Y=  i X+  i Y=  j X+  j Eliminated Fig a X ij < u 1 u 2 Fig b Eliminated u1u1 u 2 < X ij Fig c Eliminated Linear Programming in R 2 4

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 X 1/2 of S O(M) If X 1/2 is not the extreminzing abscissa, then We test which side of X 1/2 the optimum lies. O(M) So half of the X ij ‘ s lie in the region which are known not to contain the optimum. For each X ij 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. Linear Programming in R 2 3

Y X P u1u1 u2u2 F  (X) F + (X) X 1/2 X ij Eliminated Fig d: optimal lies on the left side of X 1/2 Linear Programming in R 2 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 X ij u 2 ), then eliminate one constraint O(M) Let S be all the pairs of constraints s.t. u 1  X ij  u 2, Find the median X 1/2 of S & test which side of X 1/2 the optimum lies O(M) Half of the X ij ‘ s lie in the region which are known not to contain the optimum. For each X ij in the region, one constraint can be eliminated. O(M) Linear Programming in R 2 1

Common Intersection Common Intersection of half-planes in R 2 :  (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 2 n vertices. The size of Common Intersection of half-spaces in R k 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.