Lecture The Ellipsoid Algorithm

Transcription

1 8.433 Combinatorial Optimization November 4,9 Lecture The Ellipsoid Algorithm November 4,9 Lecturer: Santosh Vempala The Algorithm for Linear rograms roblem. Given a polyhedron, written as Ax b, find a point in. Before tackling this problem, we begin with some definitions. A real symmetric matrix A with the property that x T Ax > for all x is called positive definite. If A is positive definite, then there exists an invertible matrix, such that A T. Let D be a positive definite matrix and consider the ellipsoid Ell(D, z) {x : (x z) T D (x z) }. Let ν be the maximum number of bits required to describe a vertex of and set R ν. To solve roblem we apply the following algorithm: The Ellipsoid Algorithm Start with the ellipsoid E Ell(R I, ). At the i th iteration, check whether z i is in. - YES. Output z i as the feasible point. - NO. Find a constraint for, a k x b k, violated by z i. Recurse on E i+, the minimum volume ellipsoid containing E i {x a k x a k z i }. E i+ E i Figure : One cycle of the algorithm Observe that the algorithm halts when a point z i is found to be within. It must halt,

2 since at any step i, is a subset of E i, and we will see that after each step, the volume of E i+ has decreased by an appreciable amount. For some value of i, the volume of E i will be smaller than the volume of, so the algorithm must halt before reaching this point. In the next section we show that the algorithm can actually be implemented in polynomial time. The Time Bound Lemma. The minimum volume ellipsoid containing Ell(D, z) {x a x a z} is exactly E Ell(D, z ), where and and z z n + n Da a T Da ( D n + Daa T ) D a T Da () () vol(e ) vol(e) e n+ (3) Sketch of proof : First, note that Ell(A, ) can be obtained from Ell(I, ) (the unit ball) using the transformation y Bx, where A B T B. To see this, consider the following: x T x y T (B ) T (B )y x T x y T (B ) T (B )y y T A y where the first and last equations define the unit ball and Ell(A, ), respectively. Now, first we will prove the results () and () for the special case of the unit ball, E Ell(I, ). In this case, () reduces to z z n + a a T a and () reduces to n (I aa T n + a T a ) Since E is a ball, we can rotate a without affecting anything. So, assume a [,,, ] T. Then, z [ /(n + ),,, ] T.

3 E E / yd x E E Figure : Transform space to take E to a ball n I n + n n+. The simplified statements for z and D can be proved by calculus. The general case is then proved by applying a transformation of A to the unit ball (B above; the transformation scales the volume of a convex set by the factor det(b)). Assuming () and (), we can now prove (3). observe that: vol(ell(d, z )) vol(ell(d, z)) vol(ell(i, )) vol(ell(i, )) det(d ) det(d) We transform space to take E to the unit ball. The transformed E is still the minimum ellipsoid containing half of E. 3

4 Then, assuming D I, we have det(d ) vol(e )/vol(e). Now we can use (). n I n + The determinant of this matrix is ( ) n det(d n ( ) n ) n + Hence, (using e x + x). vol(e ) vol(e) ( n n ) n/ ( ) n / n + ( n ) n n (n ) / n (n ) / (n + ) / (n + ) / ( + ) n ( n ) n + e (n )(n+) (n ) e n+ e (n+) We need to calculate how small can be in order to obtain a bound on the number of times we shrink E. We will see that vol( ) nν by finding a simplex inside. Clearly the volume of the simplex will be less than or equal to the volume of. Now there exist n + affinely independent verticies of, say x, x,..., x n. vol(conv(x,..., x n )) det x x x n A vertex x i is a solution to a subset C i of rows of Ax b. We can solve for it using Cramer s Rule, x ij det(c ij ) det(c j ), where C ij is the matrix C i with the i th column replaced by b restricted to the relevant rows for C i. So, det C det C det C det C vol(conv(x,..., x n )) det 4 det C det C det C det C.

5 ulling out the denominators, we see that det C det C det C det C det det C nn det C det C det C n det(c ) det(c ) det(c n ) As det C i ν, we have vol(conv(x,..., x n )) n n ( ν ) n nν. After i steps, vol(e i ) (R) n e i n+. We stop before vol(ei ) < vol( ). Thus, (ν+)n e i n+ < nν which means we stop when i O(n ν). Recall that ν was less than the number of bits required to write down any n n subset of {A, b}, plus log n bits. So, the the number of iterations is O(n C, d ). If we use L-bit numbers, then C, d O(n L). To check the validity of a point, we must check each constraint of, taking O(mn) time. This dominates the time required to calculate the minimum ellipsoid. So the total time required to complete the algorithm is at most O(mn 5 L). 5