6.854 Advanced Algorithms Petar Maymounkov Problem Set 11 (November 23, 2005) With: Benjamin Rossman, Oren Weimann, and Pouya Kheradpour


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1 6.854 Advanced Algorthms Petar Maymounkov Problem Set 11 (November 23, 2005) Wth: Benjamn Rossman, Oren Wemann, and Pouya Kheradpour Problem 1. We reduce vertex cover to MAXSAT wth weghts, such that the correspondng MAXSAT nstance has the same dependancy graph as the vertex cover problem (and hence the same treewdth). Then we apply the fact (proven n the prevous problem set) that MAXSAT wth wegths runs n tme 2 T O(n), where T s the tree wdth. For each vertex v n the vertex set nstance, ntroduce a varable x v n the MAXSAT nstance. For each edge vw n vertex cover, ntroduce an edge clause (x v x w ) n MAX SAT wth weght n. Lastly, for each vertex v, ntroduce a vertex clause (x v ) wth weght 1 n MAXSAT. Observe that the dependancy graph of the MAXSAT problem (where an edge corresponds to the relaton share a clause ) s dentcal to the orgnal vertex cover graph, and hence they have the same treewdth. By constructon, every vertex cover corresponds to an assgnment that satsfes all edge clauses and some of the vertex clauses. We now prove that a MAXSAT assgnment corresponds to a smallest vertex cover. Frst we show that a MAXSAT assgnment s a vald vertex cover. Assume not. Then there s at least one edge clause that s not satsfed, and hence the total satsfed weght s mn (where m s the number of edges n the orgnal graph). On the other hand, any vald vertex cover assgnment that satsfes all edge clauses satsfes > mn weght, because t must also satsfy at least one vertex clause. So we get a contradcton. Fnally, argue that the MAXSAT assgnment s ndeed the mnmal vertex cover. If t sn t, then the mnmal vertex cover nduces an assgnment wth hgher satsfed weght (because all vertex clauses have the same weght), and hence we get a contradcton. Problem 2a. We use the same potental functon as n the lnear case: Φ = km + <j d( ) s DC, s DC j. Notce t s always nonnegatve snce all terms are ndvdually nonnegatve. 111
2 Much lke n the lnear case, when OPT moves dstance d, t ncreases the current mncost matchng by at most d, and hence t ncreases the optmal (after the move) mncost matchng by at most d. The overall change n Φ s then kd, snce DC doesn t change. Consder the case when DC moves. Subdvde t nto phases, where wthn a phase the set of neghbors movng towards the request does not change. Let the number of neghbors wthn a phase be n. WLOG one of the neghbors s matched wth an OPT server at the request pont (otherwse we could uncross the mncost matchng wthout ncreasng ts cost). If the total dstance traveled by DC n ths phase s d, each neghbor ndvdually travels d/n. In the worst case (when there s only one OPT server between the movng neghbors) the current matchng s (a) decreased by d/n due to the neghbor matched wth the OPT server, and (b) ncreased by (n 1)d/n by the remanng neghbors movng away from ther matches. In total M (n 2)d/n. Let s consder the change to DC. Let b = B, where B s the set of DC servers behnd neghbor. Then: j B d ( s DC ( ), s DC d j = b ) b k n k Further, the dstance change among all pars j of neghbors s: d ( ( ) ) s DC, s DC d n j = 2 = d n(n 1) n 2 n j Summng over all changes n dstances among DC s servers, we get: <j ( s DC ), s DC d j = (n 2)(k n) n n(n 1) d n And combnng ths wth M, we get: Φ d In concluson, every d move of OPT makes room for at most kd move of DC, hence DC s kcompettve. 112
3 Problem 2b. In the pagng problem, recallng a new page nto cache costs 1 unt. We model the pagng algorthm by a starshaped tree. Wth leaves correspondng to the pages, and k servers correspondng to the cache entres. Each edge n the tree has length 1/2. Assume that the kserver problem starts off wth all servers on the leaves (f not, the extra ncurred cost s absorbed by the asymptotcs). The star model establshes a mnmum cost of 1 unt for a server to move from one page to another. (Of course, the kserver algorthm s allowed to be more wasteful, f t so desres.) Ths shows that the model s correct, because the mnmum cost for page swtch s the same n both problem statements. We convert any gven kserver algorthm nto a pagng algorthm as follows. If a server leaves a page, the correspondng pagng algorthm does no evct that page untl that server arrves at a dfferent page. It s obvous that under ths constructon the pagng algorthm ncurs no more cost than the underlyng kserver algorthm. Problem 2c. We get Fat s markng algorthm. To argue ths, we need to dsambguate the case when the DC algorthm reaches a phase when multple neghbors (of the request) reach the same tree vertex, because we need to specfy whch sngle one contnues towards the request. It s clear that t doesn t matter whch one does that, but for the sake of ths problem, we wll assume that a random one does. The equvalence between Fat and the DC algorthm s then gven n the followng smple terms. A cache entry s marked, f the correspondng server s at a leaf, and t s unmarked f t s at the center of the star. Ths equvalence llumnates an nterestng fact. In partcular, t shows that the random choce of an unmarked page n Fat s algorthm s unncessary. And that choosng an arbtrary unmakred page s suffcent to acheve kcompettveness. Problem 2d. We are gong to model the weghted pagng algorthm as a star as well. Where the length of the edge to page p wll be equal to w p /2, where w p s the cost of recallng p nto cache. To prove equvalence, we have to show that swtchng pages ncurs the same cost n both the pagng and the server models. We make the followng observaton. In any sequence of events 113
4 each recall of a page s pared wth an eventual evcton. We spread the cost of the recall equally over the recall and the evcton. Ths gves us drect correspondance to the server model, where the recall corresponds to traversng the edge towards the leaf, and the evcton corresponds to traversng the edge away from the leaf. In summary, a par of recall/evct operatons for a page p n the server model s forced to ncur cost at least w p. We use the same nterpretaton of the kserver algorthm as a pagng algorthm as the one descrbed n part 2b. Problem 3a. At step, the onlne algorthm makes a mstake when a fracton F > 1/2 of the total weght w s n mstake. The updated total weght w +1 s then: w +1 = (1/2)F w + (1 F )w = (1 F/2)w (3/4)w Let x be the total number of mstakes, the onlne algorthm makes. Then after the completon of the experment for the fnal total faculty weght w, we have that: w (3/4) x n Problem 3b. On the other hand, f we examne the weght of the wsest faculty member, t begns at 1, and s shrnked by a factor of 2 exactly m tmes. Hence at the end of the algorthm t s equal to 2 m. Problem 3c. The weght of the wsest faculty member s a lower bound on the total fnal weght. Combnng ths wth the upper bound from 3a, we get: Solvng the nequalty for x gves: x 2 m w (3/4) x n 1 (m + log n) log(4/3) 2.41(m + log n) 114
5 Problem 3d. At queston, regardless of whether the onlne algorthm guesses (probablstcally) the answer correctly, the total weght decreses as follows: w +1 = F w β + (1 F )w = ( 1 F (1 β) ) w Further, f X s the ndcator random varable that the onlne algorthm errs on the th queston, and X = X, then E[X] = E[X ] = F. And hence our goal wll be to bound F. Problem 3e. Much lke n the determnstc case, the total fnal weght w s lowerbounded by the weght of the wsest faculty β m : β m w = n ( 1 F (1 β) ) Solve ths nequalty, usng that ln(1 y) y for y 0: ( 1 F (1 β) ) β m n m ln β ln n + ln ( 1 F (1 β) ) ln n (1 β) F F m ln(1/β) + ln n 1 β ( ) And snce E[X] = F, we get the needed result. Problem 3f. Set: β = And drectly substtute nto ( ) to get: ln n/m E[X] m + ln n + 2 m ln n 115
6 Problem 4a. Transform each axesalgned rectangle n R 2 defned by ts dagonal corners (a, b) and (a, b ) nto the pont (a, b, a, b ) n R 4. Also, transform each query rectange (x, y), (x, y ) n R 2 nto the query rectangle (x, y, x, y), (x, y, x, y ) n R 4. By defnton a < a and b < b. In R 2, the rectangle (a, b), (a, b ) s contaned n the query (x, y), (x, y ) ff x a < a x and y b < b y (Condton ). In R 4, the pont (a, b, a, b ) s contaned n the query rectangle (x, y, x, y), (x, y, x, y ) ff x a x, x a x, y b y, and y b y (Condton ). Condtons ( ) and ( ) are equvalent. We prove the desred space and tme requrements by nducton. For space, nductve hypothess s that ddmensonal structure requres O(n log d 1 n) space. Base case s trval. A d + 1dmensonal structure requres O(n) space for the hghest level BST as well some space for the ddmensonal substructures for each nternal tree node of the hghest level BST (by nductve hypothess): log n 2 n n 2 logd 1 2 n logd n =1 For tme, agan by nducton. Hypothess s that n d dmensons, dentfyng the last dmenson s BST subtrees that contan the query ponts takes O(log d n) tme. Base step s trval. In d + 1 dmensons, n O(log n) tme we dentfy O(log n) subtrees of the hghest level subtree, and for each of them, we need to recursvely run the algorthm, so total tme s O(log n log d n) = O(log d+1 n). Once we ve dentfed all subtrees (O(log d+1 n count) n the lowest dmenson that contan the query answers, we enumerate the answers n another k steps n total. Problem 4b. Observe that f a polygon s contaned n an axesalgned query rectange, so s ts boundng axesalgned query rectangle. Therefore, preprocess (n lnear tme) all polygons and replace them by ther boundng rectangles. (Smply fnd the smallest x, largest x, smallest y and largest y coordnates for each polygon across all ts vertces). And apply the algorthm from 4a to the boundng rectangles. 116
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