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1 ecture 10 Dyamic Programmig 10.1 Kapsack Problem November 1, 2004 ecturer: Kamal Jai Notes: Tobias Holgers We are give a set of items U = {a 1, a 2,..., a }. Each item has a weight w i Z + ad a utility u i Z +. Our task is to fid the most valuable set of items with respect to the utility fuctio uder the costrait that the kapsack has a fixed capacity of B. Fid S U satisfyig: coditio maximum w i B i S i S u i 10.2 Types of Polyomial Algorithms This sectio preset three types of polyomial algorithms. Each defiitio cotais a example ru time for the Kapsack problem if such a algorithm existed. Defiitio 1 Polyomial algorithm Ruig time Poly(Istace size) P oly( i log w i + i log u i ) Defiitio 2 Strogly polyomial Ruig time Poly(#umbers i istace) P oly() Defiitio 3 Pseudo polyomial Ruig time Poly(magitude of umbers). Where umbers are writte i uary ecodig. Uary ecodig for 0 is +1 cosecutive oes, e.g. 5 = P oly( i w i + i u i ) 10.3 Dyamic Programmig We have a istace I of a problem we like to solve usig dyamic programmig. Now create sub-istaces S = {I 1, I 2,..., I k }. The dyamic program the does: 1

2 1. It puts a total (or partial) order o S. 2. First sub-istace is easily solvable. (total order) All miimal istaces are easily solvable. (partial order) 3. The last istace is your origial istace. (total order) The maximum istace is your origial istace. (partial order) 4. Take ay istace I m S If you kow the solutios of the previous istaces the a solutio of I m ca be computed easily. For the partial orderig you must kow the solutios of the lower istaces Dyamic Programmig for Kapsack Problem We will here apply the previous four steps of a dyamic programmig algorithm o the kapsack problem. First defie a sub-istace A(i, w) which is the maximum utility usig oly items {a 1, a 2,..., a i } ad with a maximum capacity of w. The secod item is simply a questio of whether we are able to pick the oly item with regard to the maximum capacity, i.e. A(, B). { 0 w < a1 A(1, w) = u 1 w a 1 Our fial solutio, which correspods to item three, is the sub-istace over all items usig the origial capacity. We ca show item four by iductio. We ca determie the solutio to a sub-istace usig oly previous/lesser sub-istaces. A(i, w) A(i, w ) i i, w w where at least i < i or w < w The additio of oe more item is similar to the case where we oly had oe item. We either add it to the previous sub-istace with room for the ew item or we do t pick it at all. { ui + A(i 1, w w A(i, w) = max i ) w w i A(i 1, w) otherwise The ruig time for this dyamic programmig algorithm is B. It is also pseudo polyomial i B if all iputs are give i uary Kapsack Miimizatio Problem The ruig time of the previous algorithm depeds o B which might be large. We rather wat it to be pseudo polyomial i i u i. Rewrite the sub-istaces to be A(i, U) which is the miimum weight usig oly items {a 1, a 2,..., a i } ad a utility of at least U. The recursio the looks like 2

3 { wi + A(i 1, U u A(i, U) = mi i ) U u i A(i 1, U) otherwise This dyamic programmig algorithm has a ruig time of i U u i. The exact solutios of the two differet recursive formulatios of the kapsack problems are equivalet. I the approximatio case they are differet Approximatio Scheme Defiitio 4 A α-approximatio algorithm produces a solutio satisfyig. 1. Miimizatio problem, α 1 solutio α OPT 2. Maximizatio problem, α 1 solutio α OPT 10.7 A Approximatio Scheme 1. Miimizatio problem 1 + ɛ approximatio algorithm for every ɛ > 0 2. Maximizatio problem 1 ɛ approximatio algorithm for every ɛ > 0 A polyomial time approximatio scheme (PTAS) is a (1+ɛ)-approximatio with a rutime that is polyomial i the istace size. Here follows a couple of examples of ru times ad whether or ot they are algorithms polyomial i the istace size. ɛ, 2 ɛ, 1 ɛ, 2 1 ɛ ok ( 1 ɛ ) ot ok A fully polyomial time approximatio scheme (FPTAS) is a PTAS where the ru time is polyomial is 1 ɛ too. Proof 1 We have a iteger valued maximizatio problem such that OPT > 2 P oly(istace) 1 for all istaces. Now choose ɛ =. Assume there 2 P oly(istace)+1 exists a FPTAS, i.e. there exists a (1 ɛ)-approximatio algorithm with the ruig time P oly(istace, 1 ɛ ). (1 ɛ)op T > OP T ɛ2 P oly(istace) = OP T oly(istace) 2P > OP T 1 2P oly(istace)+1 Our approximatio is therefore a optimal solutio with the ruig time P oly(p oly(istace uity )). The algorithm is hece pseudo polyomial. 3

4 10.8 Approximatio of Kapsack A approximatio algorithm ca ofte be created by roudig off the least sigificat bits of the iput umbers. The amout of roudig is determied by 1 ɛ ad some error threshold. Here follows a (1 ɛ)-factor algorithm for the Kapsack problem. The solutio A (1 ɛ)opt ad the allowed ru time is P oly(, 1 ɛ ). Now assume that all u i B, if ot, throw the item away sice it caot be picked. The take the largest u i ɛ ad scale all weights by 1 ɛ ad tur ito a iteger. Ad the ew weight the becomes. f = ɛ u max u i = ui f Ad the ruig time is i U u i where each u i is bouded by so we get a pseudo polyomial algorithm. Proof 2 et OPT optimal solutio OPT value of OPT solutio with u i weight fuctio A OPTimum solutio to out rouded istace with utility fuctio u i A Value of out solutio with respect to origial utilities The followig holds A OPT A fa From which this follows u i f u i + 1 u i fu i + f OPT fopt + f fa + f A + f Which gives us the result OPT f A OPT u max A Ad sice u max OPT we get OPT ɛopt A which proves that our approximatio is withi a factor (1 ɛ). 4

5 10.9 Travelig Salesma Problem (TSP) For som dimesio d, we got vertices a 1, a 2,..., a R d. Fid the shortest legth tour visitig all the vertices. The distace fuctio is the vector orm defied as x, y R d : d (x i y i ) 2 i=1 emma 1 A optimum TSP tour is a o self-itersectig polygo o a 1, a 2,..., a. ets start by showig that the tour is a polygo. The example tour a 1, a 2, a 5, a 3, a 2, a 6,... visits the vertex a 2 multiple times ad sice the triagle iequality holds, i.e. distace(a 3, a 6 ) distace(a 3, a 2 ) + distace(a 2, a 6 ) We ca remove the vertex a 2 from the tour ad thereby decreasig the total legth which meas that the tour was t optimal to start with. This gives us that the tour is a polygo. What if we have crossig edges? Remove them ad recoect the two compoets. This will decrease the legth of the tour so the optimal solutio has o crossigs. The decrease i legth is give by the triagle iequality which ca be applied to the vertices ad the midpoit. a1 a3 a1 a3 a7 a5 a7 a5 a1 a3 a7 a5 Figure 1: Remove crossig ad recoect compoets Polyomial Time Approximatio Scheme for TSP Create a boudig box for all vertices. That is, create the smallest axis parallel box cotaiig all the poits. Now, without loss of geerality, assume the width ad height are = 4 2. Assume further, without loss of geerality, that = 2 k. Approximate by creatig a uit grid for the boudig box ad move poits to ceter of the square. 5

6 y x Figure 2: Approximatio of TSP tour. Gray tour is exact ad black approximated. emma 2 OPT is the origial optimal solutio ad let OPT be the optimum solutio, i.e. the same tour, for the rouded istace. ɛ > 0 : OPT (1 + ɛ)opt Which holds if is big eough, > 1 ɛ. Goig to the ceter ad goig back to the origial vertex adds a maximum distace of 4 for each square. OPT OPT + 4 OPT + 42 OPT + OPT = (1 + 1 )OPT Divide the boudig box ito four square areas ad place i a tree, see figure 3. Keep dividig ad addig square areas so that the leaf odes of the tree are uit squares. The height of the tree will become log(4). T y.. x Figure 3: Divisio ito square areas I each box we solve the problem of miimizig path legths with the give etry ad exit poits. The miimum path that covers all poits will be o self-itersectig due to the lemma. Add etry ad exit portals to each uit square, see figure 4. Edges will be rouded to eter/exit from the closest portal. ets create m portals o each 6

7 side of the square ad have double portals o the corers. Pick m = O( log ɛ ). Eumeratig all pairigs of portals the takes time Which is polyomial i. 2 m log O( 2 ɛ ) O( 1 ɛ ) m Figure 4: Etry ad exit portals o square Assumptio 1 There is a tour of cost at most (1 + ɛ )OPT such that the tour eter or exit a square from the portals. Proof will be preseted o ext lecture. 7