Fall 204 Randomized Algoithms Oct 8, 204 Lectue 3 Pof. Fiedich Eisenband Scibes: Floian Tamè In this lectue we will be concened with linea pogamming, in paticula Clakson s Las Vegas algoithm []. The main esult is an expected polytime eduction fom an LP in dimension d with n constaints to O(n) LPs in dimension d with O(d 2 ) constaints. In a second pat, we intoduce bounds on the diamete of a polyheda, a topic which we will view in moe detail in the next lectue. Finally, we discuss the smallest enclosing disk poblem as an application of Clakson s algoithm. Linea Pogamming We conside the taditional fomulation of linea pogamming in d dimensions with n constaints as max c > x, x2r d subject to Ax apple b. Thoughout this lectue, we will make the following assumptions: A 2 R n d has full ank. Thee exists a basis B {,...,n}, B = d such that A B is non singula and x = A B b B is optimal (x has d tight constaints). The optimal solution x is unique (note that this can always be achieved by slightly petubing the objective function). Clakson s Algoithm We begin with a vey useful lemma. We emind that a multiset is a set in which membes may appea moe than once. We denote the cadinality of a multiset H by. By H, we denote the collection of all sub-multisets of H of cadinality. Fo a set (o multiset) U of constaints, let x U be the optimal solution (if it exists) of the LP esticted to the constaints in U. Lemma (Sampling Lemma) Let G be a set of constaints and H be a multiset of constaints of dimension d. Let R 2 H be picked unifomly at andom. Let V R = {h 2 H : x G[R violates h}. Then, E [ V R ] apple d +.
Poof We have E [ V R ]= P (h, R) = ( R2( H ) ) V R. Define the chaacteistic function ( if x G[R violates h 0 othewise. We get E [ V R ]= R2( H ) = h2h\r Q2( +) H h2q (h, R) (h, Q h), whee the last equality holds since picking constaints and then an additional constaint fom H is equivalent to picking + constaints fom H and then one constaint out of these. Fix a basis B of G [ Q. Suppose that the chosen constaint h is violated by x G[(Q h) (and thus = ). Then, it follows that h has to be pat of the basis B. Since B = d, wehave = fo at most d constaints. Thus, E [ V R ]= Q2( H +) d = d + = d. + We can now pesent the Clakson(2) algoithm. The Clakson() algoithm will be discussed in the next lectue. Algoithm Clakson(2) Input: A multiset H containing the n constaints of the LP Output: The solution x to the LP : Set =6 d 2 2: epeat 3: Pick R 2 H unifomly at andom 4: Compute x R and V R = {h 2 H : x R violates h} 5: if V R apple then 6: H = H + V R (double the occuence of each h 2 V R in H. 7: end if 8: until V R = ; 9: etun x R 2
Analysis of Clakson(2) Fist of all, note that we educe the poblem of solving ou LP with n constaints to the poblem of solving LPs with =6 d 2 = O(d 2 ) constaints (line 4). We thus still need to show that the whole pocedue is epeated O(n) times. We analyse the pobability that the if statement (line 5) is satisfied. By the Sampling Lemma, we have E[ V R ] apple d 6d2 6d 2 + Thus, using Makov s inequality, we get apple P V R apple apple = P V R > apple 6d. = P [ V R > 2 E[ V R ]] {z } apple 2 2. We now bound the numbe of successful iteations, whee the if statement is satisfied (and thus V R apple ). Note that fom the pevious esult, we may then bound the expected numbe of iteations (and thus the expected numbe of LPs to solve) by twice the numbe of successful iteations. Lemma 2 The numbe of successful iteations is O(d log(n)). Poof Let B H be the basis fo the optimal solution x of the LP (note that B is a sub-multiset of H so it may contain a constaint moe than once). Let µ i (B) denote the cadinality of B afte iteation i. By definition of the basis, we have µ 0 (B) =d. Note that if x R doesn t violate any constaint of the basis, then it is an optimal solution to the LP. Fo all but the last iteation, we have V R 6= ;, meaning that x R is not the oveall-optimal solution, and thus V R contains at least one constaint of B. Thus, at each iteation, at least one basis constaint has its numbe of occuences doubled in H. We thus get that µ i d (B) 2 i d 2 i. Now conside how the cadinality of H gows. We have µ 0 (H) =. On a successful iteation, we add V R apple constaints to H. Thus, µ i+ (H) apple µ i (H) + µ i(h) =(+ )µ i(h). Afte i d iteations, we have µ i d (H) apple ( + )i d apple e i 3. Finally, note that the multiset B is always a sub-multiset of H. Afte d i iteations, we thus must have 2 i apple n e i 3, and theefoe i = O(log(n)). We theefoe get that the numbe of successful iteations is O(d log(n)). Since the pobability of an iteation being successful is geate than 2 we expect to have at most 2 O(d log(n)) = O(d log(n)) iteations. Clakson s algoithm thus educes an LP in d dimensions with n constaints to an expected O(d log(n)) LPs with O(d 2 ) constaints in expected polynomial time. 3
A note about unbounded LPs To deal with LPs that might be unbounded, we can fix a box of 2 d constaints, thus constaining the optimal solutions inside this box. In the analysis, these 2 d constaints ae added to the oiginal set G defined in the Sampling Lemma, which we have consideed to be empty in the above analysis. Open poblem: bounding the diamete of a polyhedon Conside a polyhedon in d dimensions defined by P = x 2 R d : Ax apple b, whee A 2 R n d. We make the assumption that P is non-degeneate, meaning that each vetex of the polyhedon is defined by d tight constaints. Fo a vetex x of P, we define the basis B x as the set of d tight constaints defining x. Two vetices x and y ae neighbous if B x \ B y = d (the two vetices only di e in one tight constaint). Definition 3 Fo a polyhedon P, conside the gaph G on vetices of P obtained by inseting edges between neighbouing vetices. The diamete of P is the lagest shotest-path distance (in numbe of edges) between any two vetices in G. Definition 4 Let constaints. d,n be the lagest diamete of a polyhedon P R d with n Theoem 5 ([2, 3, 4]) d,n apple n log(d)+. Conjectue 6 Thee exists a function f such that which does not depend on n). d,n = O(n) f(d) (fo a f Smallest enclosing disk Conside a set of n points in R 2 and the task of finding the smallest disk enclosing all of these points. Fist of all, we discuss how we may convince ouselves that a given cicle is indeed the smallest enclosing disk. Fo a given cicle, take all the points appeaing on the bounday of the cicle. Now, if the cente of the cicle is not inside of the convex hull of these points, it is easy to see that we may shift the cente of the cicle towads the convex hull and shink it since it now englobes all the points which peviously wee on its bounday. One can actually find that a disk is the smallest enclosing disk if it contains all points and one of the following two conditions is satisfied In [4], Kalai and Kleitman povide an uppe bound of n log(d)+2.thetighteboundwas shown by Kalai [3] and Eisenband et al. [2]. 4
Thee exist 3 points on the bounday of the disk such that the cente of the disk lies in the convex hull of the thee points. Thee ae 2 points on the bounday such that the cente of the disk is thei midpoint. We can veify that the Clakson(2) algoithm we descibed can be used to find the smallest enclosing disk in expected O(n log n) time. Indeed, in each iteation we pick a set R of 6 d 2 = 24 points, find a smallest enclosing disk on R, check which emaining points fall outside of the disk and double thei weight. By a simila analysis as fo the LP, we can show that the numbe of iteations is O(log n) and that each iteation takes time O(n) (finding the optimal disk on R can be done in constant time since R has constant size; checking which points fall outside the disk can be done in O(n) time though some cleve manipulation of copies of points in the multiset). Refeences [] Kenneth L. Clakson. Las vegas algoithms fo linea and intege pogamming when the dimension is small. J. ACM, 42(2):488 499, Mach 995. [2] Fiedich Eisenband, Nicolai Hähnle, and Thomas Rothvoß. Diamete of polyheda: Limits of abstaction. In Poceedings of the Twenty-fifth Annual Symposium on Computational Geomety, SCG 09, pages 386 392, New Yok, NY, USA, 2009. ACM. [3] Gil Kalai. Linea pogamming, the simplex algoithm and simple polytopes. Math. Pogamming, 79:27 233, 997. [4] Gil Kalai and Daniel J Kleitman. A quasi-polynomial bound fo the diamete of gaphs of polyheda. Bulletin of the Ameican Mathematical Society, 26(2):35 36, 992. 5