Problem Set 9 Solutions


 Cynthia Harrison
 4 years ago
 Views:
Transcription
1 Desgn and Analyss of Algorthms May 4, 2015 Massachusetts Insttute of Technology 6.046J/18.410J Profs. Erk Demane, Srn Devadas, and Nancy Lynch Problem Set 9 Solutons Problem Set 9 Solutons Ths problem set s due at 11:59pm on Thursday, Aprl 30, Exercse 91. Read CLRS, Chapter 35. Exercse 92. Exercse Exercse 93. Exercse Exercse 94. Read Prof. Demane s notes on fxedparameter algorthms. Problem 91. Knapsack [25 ponts] In the Knapsack problem, we are gven a set A = {a 1,..., a n } of tems, where each a has a specfed postve nteger sze s and a specfed postve nteger value v. We are also gven a postve nteger knapsack capacty B. Assume that s B for every. The problem s to fnd a subset of A whose total sze s at most B and for whch the total value s maxmzed. In ths problem, we wll consder approxmaton algorthms to solve the Knapsack problem. Notaton: For any subset S of A, we wrte s S for the total of all the szes n S and v S for the total of all the values n S. Let Opt denote an optmal soluton to the problem. (a) [5 ponts] Consder the followng greedy algorthm Alg 1 to solve the Knapsack problem: Order all the tems a n nonncreasng order of ther densty, whch s the rato of value to sze, v s. Make a sngle pass through the lst, from hghest to lowest densty. For each tem encountered, f t stll fts, nclude t, otherwse exclude t. Prove that algorthm Alg 1 does not guarantee any constant approxmaton rato. That s, for any postve nteger k, there s an nput to the algorthm for whch the total value of the set of tems returned by the algorthm s at most v Opt. k Soluton: Fx any k. Consder the twotem nput lst a 1, a 2, where v 1 = 2, s 1 = 1, v 1 v v 2 2 = 2k, and s k = 2k, wth knapsack capacty B = 2k. Snce = 2 and = 1, s 1 s 2 Alg 1 consders the tems n order a 1, a 2, and ncludes a 1 and excludes a 2. The total value acheved s only 2, but the largest achevable total value s 2k.
2 2 Problem Set 9 Solutons (b) [7 ponts] Consder the followng algorthm Alg 2. If the total sze of all the tems s B, then nclude all the tems. If not, then order all the tems n nonncreasng order of ther denstes. Wthout loss of generalty, assume that ths orderng s the same as the orderng of the tem ndces. Fnd the smallest ndex n the ordered lst such that the total sze of the frst tems exceeds B (.e., 1 1 s j > B, but s j B). If v > v j, then return {a }. Otherwse, return {a 1,..., a 1 }. Prove that Alg 2 always yelds a 2approxmaton to the optmal soluton. Soluton: We know that fractonal knapsack problem can be solved va greedy algorthm; the optmal soluton for fractonal knapsack problem takes frst 1 tems, and takes some fracton α of tem. That s, fractonal knapsack optmal soluton v fopt s 1 v fopt = v j + αv. Snce every knapsack problem s a vald fractonal knapsack problem, we know that v Opt v fopt. Now consder the soluton output by Alg v fopt max( v j, v ) max( v j, αv ), 2 because at least one of the two terms needs to be larger than half of v fopt. Therefore, we have 1 v Opt v fopt 2 max( v j, v ), whch shows that Alg 2 s a 2approxmaton algorthm. Soluton: Ths s an alternate soluton. Assume for contradcton that there s some nput nstance for whch Alg 2 does not acheve a 2approxmaton to the optmal soluton. Then for as specfed n the 1 algorthm, the maxmum of v and v j s strctly less than v Opt. That s, both 2 1 v Opt v and v j are strctly less than. Then v j < v Opt. We also know that 2 s j > B. Let C = {a 1,..., a } Opt, D = Opt {a 1,..., a }, and E = {a 1,..., a } Opt. We know that v C = v j v E, and v D = v Opt v E. Snce v j < v Opt, we now have that v C < v D. We also know that s C > s D, snce s j > B and s D B. < v D Therefore, the densty of C s strctly less than the densty of D (.e., v C ). s C s D
3 Problem Set 9 Solutons 3 Now recall that the denstes of {a 1,..., a n } are nonncreasng. Consder j v s j j j:v C (v j s v s j ) = j. We know that v j j:v j C s s j s j s v C v v D v j v C v v D s C s s D s j s C s s D j j:v j C v j j:v j C s j v for a j C, and thus v j s v s j > 0. Ths mples that, and smlar argument holds for. Therefore,. Ths contradcts our result from the prevous paragraph that says v C < v D, and Alg 2 must be a 2approxmaton. s C s D (c) [5 ponts] Let A = {a 1,..., a n } be the nput ordered arbtrarly, and V be the largest value for any tem; then nv s an upper bound on the total value that could be acheved by any soluton. For every {1,..., n} and v {1,..., nv }, defne S,v to be the smallest total sze of a subset of {a 1,..., a } whose total value s exactly v; S,v = f no such subset exsts. Gve a dynamc programmng algorthm Alg 3 to solve Knapsack exactly. Specfcally, gve a recurrence to compute all values of S,v, and explan how to use ths to solve the Knapsack problem. Analyze ts tme complexty. Soluton: S 1,v = s 1 f v 1 = v, and = otherwse. For 1 n 1: S +1,v = mn{s,v, s +1 + S,v v+1 } f v +1 v, S,v otherwse. The computaton of all of these values takes tme O(n 2 V ). As usual for dynamc programmng algorthms, we can add bookkeepng to calculate the actual subsets as we go along. The maxmum total value achevable s max vls n,v B. The fnal output s a subset of A whose total value s ths maxmum. Snce the tme complexty O(n 2 V ) depends lnearly on V, whch s represented n bnary n the nput to the problem, Alg 3 s a pseudopolynomaltme algorthm. (d) [8 ponts] Fnally, we develop Alg 4, whch s a Fully Polynomal Tme Approxmaton Scheme (FPTAS) for Knapsack. The dea s to use an exact dynamc programmng algorthm lke the one n Part (c), but nstead of usng the gven (possbly large) values for the tems, we use versons of the gven values that are sutably scaled and rounded down. As n Part (c), order A = {a 1,..., a n } arbtrarly, and let V be the largest value for any tem. For any ε, 0 < ε < 1, Alg 4 behaves as follows: = l( v For each tem a wth value v, defne a scaled value v )( n)j. V ε Usng these scaled values (and the gven szes), run Alg 3 and output the set C of tems that t returns. Prove that Alg 4 s a FPTAS for Knapsack. Soluton:
4 4 Problem Set 9 Solutons By Part (c), the runnng tme of the algorthm s O(n 2 ( V )( n ) ) = V ε O(n2 n ), whch ε s polynomal n n and 1. ε It remans to show that the set C returned by the algorthm has total value v C (1 ε)v Opt. Let K denote εv, so each v = v v n. It follows that K K v K. Consderng all the elements of O, we get v O pt K v Opt nk. Now consder the set C that s returned by the dynamc programmng step. We have that v C Kv C. Also, snce the set C s optmal n terms of the scaled values, we have v C v O pt. Therefore, we have: v C Kv C Kv O pt v Opt nk = v Opt εv v Opt εv Opt = (1 ε)v Opt. Ths s as needed. Problem 92. FxedParameter Algorthms [25 ponts] We consder the Tournament Edge Reversal problem. Defne a tournament to be a drected graph T = (V, E) such that, for every par of vertces u, v V, exactly one of (u, v) and (v, u) s n E. Furthermore, defne a cycle cover to be a set A E of drected edges of a tournament T such that, every drected cycle of T contans at least one edge from A. (a) [5 ponts] Let a mnmal cycle cover of T be a cycle cover wth the least number of edges. Prove that reversng all the edges of a mnmal cycle cover A turns T nto an acyclc tournament. (Hnt: Any edge e A must be the only edge n A on some drected cycle of T.) Soluton: Let T be the new tournament. Suppose for contradcton that T contans a cycle C. Let F be the set of edges of C that were obtaned by reversng the edges of A, and let F be ther reversals, whch are edges n T. For each e F, let C e be a drected cycle of T that e covers. From the hnt, we know e s the only edge n A that s also n C e. Now we can construct a cycle C n T, consstng entrely of edges of T that dd not get reversed: Include all reverse of edges n C F (whch are vald edges n T ), and for each e C F, nclude all the edges of C e except for e. Ths yelds a cycle C n T that contans no edges n A, whch s a contradcton. In the Tournament Edge Reversal problem, we are gven a tournament T and a postve nteger k, and the objectve s to decde whether T has a cycle cover of sze at most k. (b) [15 ponts] Show that ths problem has a kernel wth at most k 2 + 2k vertces. (Hnt: Defne a trangle to be a drected cycle of length 3. Consder the number of tmes that a node or an edge appears n dfferent trangles.) Soluton:
5 Problem Set 9 Solutons 5 Part (a) mples that a tournament T has a cycle cover of sze at most k f and only f t can be turned nto an acyclc tournament by reversng drectons of at most k edges. We wll use ths characterzaton for the kernel. We gve two smple reducton rules: Rule 1: If an edge e s contaned n k + 1 trangles, then reverse e and reduce k by 1. Ths rule s safe because f we do not reverse e, we must reverse at least one edge from each of k + 1 trangles contanng e. Thus e belongs to every cycle cover of sze k. Rule 2: If a vertex v s not contaned n any trangle, then delete v from T. To see why ths s safe, let X be the set of vertces u such that T contans the edge (v, u) (the outgong neghbors of v), and let Y be the set of vertces u such that T contans the edge (u, v) (the ncomng neghbors of v). X and Y partton the vertces n V {v}. Snce v s not contaned n any trangle, there s no edge from X to Y. Thus, every drected cycle n T s ether wholly contaned wthn the subgraph nduced by X or the subgraph nduced by Y. Therefore, removng v and ts ncdent edges from T does not affect the sze of the mnmum cycle cover. Thus, startng wth nstance (T ; k) = (T 0 ; k 0 ), we apply our reducton rules repeatedly untl they cannot be appled any longer, obtanng a sequence of equvalent nstances (T 1 ; k 1 ), (T 2 ; k 2 ),..., (T m ; k m ) = (T ; k ), where nether Rule s applcable to the fnal nstance (T ; k ). Clam: If T has a cycle cover A of sze k, then T has at most k (k + 2) vertces. Proof of clam: Snce Rule 2 s not applcable, every vertex of T s n a trangle, whch must contan some edge n A. Snce Rule 1 s not applcable, for every edge e A, there are at most k vertces other than e s endponts that are n trangles contanng e. Snce A k, t follows that T has at most k (k + 2) vertces. Thus, after reducng, we consder the fnal nstance (T, k ). If T has more than k (k + 2) vertces, then t cannot have a cycle cover of sze k, and s a nonstance. By equvalence, (T, k) s also a nonstance. We return no n ths case. Otherwse, we get the desred kernel wth at most k 2 + 2k vertces. (c) [5 ponts] Obtan a FPT algorthm for the Tournament Edge Reversal problem. Soluton: Gven an nstance (T, k) for the Tournament Edge Reversal problem, use the algorthm from Part (b) to ether determne that (T, k) s a nonstance or obtan a kernel (T, k ) where k k and T has at most k (k + 2) vertces. In the former case, answer no. In the latter case, run any algorthm wth runnng tme g(k ), for any functon g, to solve the Tournament Edge Reversal problem on (T, k ). Return the result as the answer. To reduce the problem, we have to check for all trangles, and check membershp of all edges and vertces n the trangles. For vertces, to check all trangles, t takes O( ( ) V 2 ) per vertex. For edges, t takes V per edge. Each reducton step thus takes
6 6 Problem Set 9 Solutons O( V 3 + V E )). There are at most V +E reducton steps snce each step removes an edge or a vertex. In total, reducton takes O( V 4 + V 3 E + V 2 E + V E 2 ). Snce O( E ) = O( V 2 ) n tournaments, ths s O( V 5 ). To fnd the mnmum cycle cover, one soluton s to check all possble subsets of edges n T. There are O(2 k/2 (k / +2) 2 ) subsets, and for each subset we need to run a cycle fndng algorthm, such as DFS, to ensure all cycles are covered. Therefore, the fnal tme complexty s O( V 5 )+2 O(k4).
7 MIT OpenCourseWare J / J Desgn and Analyss of Algorthms Sprng 2015 For nformaton about ctng these materals or our Terms of Use, vst:
NPCompleteness : Proofs
NPCompleteness : Proofs Proof Methods A method to show a decson problem Π NPcomplete s as follows. (1) Show Π NP. (2) Choose an NPcomplete problem Π. (3) Show Π Π. A method to show an optmzaton problem
More informationCalculation of time complexity (3%)
Problem 1. (30%) Calculaton of tme complexty (3%) Gven n ctes, usng exhaust search to see every result takes O(n!). Calculaton of tme needed to solve the problem (2%) 40 ctes:40! dfferent tours 40 add
More informationComplete subgraphs in multipartite graphs
Complete subgraphs n multpartte graphs FLORIAN PFENDER Unverstät Rostock, Insttut für Mathematk D18057 Rostock, Germany Floran.Pfender@unrostock.de Abstract Turán s Theorem states that every graph G
More informationFinding Dense Subgraphs in G(n, 1/2)
Fndng Dense Subgraphs n Gn, 1/ Atsh Das Sarma 1, Amt Deshpande, and Rav Kannan 1 Georga Insttute of Technology,atsh@cc.gatech.edu Mcrosoft ResearchBangalore,amtdesh,annan@mcrosoft.com Abstract. Fndng
More informationOutline and Reading. Dynamic Programming. Dynamic Programming revealed. Computing Fibonacci. The General Dynamic Programming Technique
Outlne and Readng Dynamc Programmng The General Technque ( 5.3.2) 1 Knapsac Problem ( 5.3.3) Matrx ChanProduct ( 5.3.1) Dynamc Programmng verson 1.4 1 Dynamc Programmng verson 1.4 2 Dynamc Programmng
More informationMaximizing the number of nonnegative subsets
Maxmzng the number of nonnegatve subsets Noga Alon Hao Huang December 1, 213 Abstract Gven a set of n real numbers, f the sum of elements of every subset of sze larger than k s negatve, what s the maxmum
More informationAssortment Optimization under MNL
Assortment Optmzaton under MNL Haotan Song Aprl 30, 2017 1 Introducton The assortment optmzaton problem ams to fnd the revenuemaxmzng assortment of products to offer when the prces of products are fxed.
More informationCS 331 DESIGN AND ANALYSIS OF ALGORITHMS DYNAMIC PROGRAMMING. Dr. Daisy Tang
CS DESIGN ND NLYSIS OF LGORITHMS DYNMIC PROGRMMING Dr. Dasy Tang Dynamc Programmng Idea: Problems can be dvded nto stages Soluton s a sequence o decsons and the decson at the current stage s based on the
More informationCollege of Computer & Information Science Fall 2009 Northeastern University 20 October 2009
College of Computer & Informaton Scence Fall 2009 Northeastern Unversty 20 October 2009 CS7880: Algorthmc Power Tools Scrbe: Jan Wen and Laura Poplawsk Lecture Outlne: Prmaldual schema Network Desgn:
More informationU.C. Berkeley CS294: Beyond WorstCase Analysis Luca Trevisan September 5, 2017
U.C. Berkeley CS94: Beyond WorstCase Analyss Handout 4s Luca Trevsan September 5, 07 Summary of Lecture 4 In whch we ntroduce semdefnte programmng and apply t to Max Cut. Semdefnte Programmng Recall that
More informationA 2D Bounded Linear Program (H,c) 2D Linear Programming
A 2D Bounded Lnear Program (H,c) h 3 v h 8 h 5 c h 4 h h 6 h 7 h 2 2D Lnear Programmng C s a polygonal regon, the ntersecton of n halfplanes. (H, c) s nfeasble, as C s empty. Feasble regon C s unbounded
More informationfind (x): given element x, return the canonical element of the set containing x;
COS 43 Sprng, 009 Dsjont Set Unon Problem: Mantan a collecton of dsjont sets. Two operatons: fnd the set contanng a gven element; unte two sets nto one (destructvely). Approach: Canoncal element method:
More informationEdge Isoperimetric Inequalities
November 7, 2005 Ross M. Rchardson Edge Isopermetrc Inequaltes 1 Four Questons Recall that n the last lecture we looked at the problem of sopermetrc nequaltes n the hypercube, Q n. Our noton of boundary
More informationLecture Notes on Linear Regression
Lecture Notes on Lnear Regresson Feng L fl@sdueducn Shandong Unversty, Chna Lnear Regresson Problem In regresson problem, we am at predct a contnuous target value gven an nput feature vector We assume
More informationLecture 10: May 6, 2013
TTIC/CMSC 31150 Mathematcal Toolkt Sprng 013 Madhur Tulsan Lecture 10: May 6, 013 Scrbe: Wenje Luo In today s lecture, we manly talked about random walk on graphs and ntroduce the concept of graph expander,
More informationAffine transformations and convexity
Affne transformatons and convexty The purpose of ths document s to prove some basc propertes of affne transformatons nvolvng convex sets. Here are a few onlne references for background nformaton: http://math.ucr.edu/
More informationExample: (13320, 22140) =? Solution #1: The divisors of are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 41,
The greatest common dvsor of two ntegers a and b (not both zero) s the largest nteger whch s a common factor of both a and b. We denote ths number by gcd(a, b), or smply (a, b) when there s no confuson
More informationDynamic Programming. Preview. Dynamic Programming. Dynamic Programming. Dynamic Programming (Example: Fibonacci Sequence)
/24/27 Prevew Fbonacc Sequence Longest Common Subsequence Dynamc programmng s a method for solvng complex problems by breakng them down nto smpler subproblems. It s applcable to problems exhbtng the propertes
More informationarxiv: v2 [cs.ds] 1 Feb 2017
Polynomaltme Algorthms for the Subset Feedback Vertex Set Problem on Interval Graphs and Permutaton Graphs Chars Papadopoulos Spyrdon Tzmas arxv:170104634v2 [csds] 1 Feb 2017 Abstract Gven a vertexweghted
More informationMin Cut, Fast Cut, Polynomial Identities
Randomzed Algorthms, Summer 016 Mn Cut, Fast Cut, Polynomal Identtes Instructor: Thomas Kesselhem and Kurt Mehlhorn 1 Mn Cuts n Graphs Lecture (5 pages) Throughout ths secton, G = (V, E) s a multgraph.
More informationThe Minimum Universal Cost Flow in an Infeasible Flow Network
Journal of Scences, Islamc Republc of Iran 17(2): 175180 (2006) Unversty of Tehran, ISSN 10161104 http://jscencesutacr The Mnmum Unversal Cost Flow n an Infeasble Flow Network H Saleh Fathabad * M Bagheran
More informationFirst day August 1, Problems and Solutions
FOURTH INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS July 30 August 4, 997, Plovdv, BULGARIA Frst day August, 997 Problems and Solutons Problem. Let {ε n } n= be a sequence of postve
More informationStructure and Drive Paul A. Jensen Copyright July 20, 2003
Structure and Drve Paul A. Jensen Copyrght July 20, 2003 A system s made up of several operatons wth flow passng between them. The structure of the system descrbes the flow paths from nputs to outputs.
More informationHMMT February 2016 February 20, 2016
HMMT February 016 February 0, 016 Combnatorcs 1. For postve ntegers n, let S n be the set of ntegers x such that n dstnct lnes, no three concurrent, can dvde a plane nto x regons (for example, S = {3,
More informationStanford University CS359G: Graph Partitioning and Expanders Handout 4 Luca Trevisan January 13, 2011
Stanford Unversty CS359G: Graph Parttonng and Expanders Handout 4 Luca Trevsan January 3, 0 Lecture 4 In whch we prove the dffcult drecton of Cheeger s nequalty. As n the past lectures, consder an undrected
More information2.3 Nilpotent endomorphisms
s a block dagonal matrx, wth A Mat dm U (C) In fact, we can assume that B = B 1 B k, wth B an ordered bass of U, and that A = [f U ] B, where f U : U U s the restrcton of f to U 40 23 Nlpotent endomorphsms
More informationU.C. Berkeley CS294: Spectral Methods and Expanders Handout 8 Luca Trevisan February 17, 2016
U.C. Berkeley CS94: Spectral Methods and Expanders Handout 8 Luca Trevsan February 7, 06 Lecture 8: Spectral Algorthms Wrapup In whch we talk about even more generalzatons of Cheeger s nequaltes, and
More informationEvery planar graph is 4colourable a proof without computer
Peter Dörre Department of Informatcs and Natural Scences Fachhochschule Südwestfalen (Unversty of Appled Scences) Frauenstuhlweg 31, D58644 Iserlohn, Germany Emal: doerre(at)fhswf.de Mathematcs Subject
More informationLectures  Week 4 Matrix norms, Conditioning, Vector Spaces, Linear Independence, Spanning sets and Basis, Null space and Range of a Matrix
Lectures  Week 4 Matrx norms, Condtonng, Vector Spaces, Lnear Independence, Spannng sets and Bass, Null space and Range of a Matrx Matrx Norms Now we turn to assocatng a number to each matrx. We could
More informationCHAPTER 17 Amortized Analysis
CHAPTER 7 Amortzed Analyss In an amortzed analyss, the tme requred to perform a sequence of data structure operatons s averaged over all the operatons performed. It can be used to show that the average
More informationKernel Methods and SVMs Extension
Kernel Methods and SVMs Extenson The purpose of ths document s to revew materal covered n Machne Learnng 1 Supervsed Learnng regardng support vector machnes (SVMs). Ths document also provdes a general
More informationFor now, let us focus on a specific model of neurons. These are simplified from reality but can achieve remarkable results.
Neural Networks : Dervaton compled by Alvn Wan from Professor Jtendra Malk s lecture Ths type of computaton s called deep learnng and s the most popular method for many problems, such as computer vson
More informationn ). This is tight for all admissible values of t, k and n. k t + + n t
MAXIMIZING THE NUMBER OF NONNEGATIVE SUBSETS NOGA ALON, HAROUT AYDINIAN, AND HAO HUANG Abstract. Gven a set of n real numbers, f the sum of elements of every subset of sze larger than k s negatve, what
More informationAPPENDIX A Some Linear Algebra
APPENDIX A Some Lnear Algebra The collecton of m, n matrces A.1 Matrces a 1,1,..., a 1,n A = a m,1,..., a m,n wth real elements a,j s denoted by R m,n. If n = 1 then A s called a column vector. Smlarly,
More informationand problem sheet 2
8 and 55 problem sheet Solutons to the followng seven exercses and optonal bonus problem are to be submtted through gradescope by :0PM on Wednesday th September 08. There are also some practce problems,
More informationModule 9. Lecture 6. Duality in Assignment Problems
Module 9 1 Lecture 6 Dualty n Assgnment Problems In ths lecture we attempt to answer few other mportant questons posed n earler lecture for (AP) and see how some of them can be explaned through the concept
More informationModule 3 LOSSY IMAGE COMPRESSION SYSTEMS. Version 2 ECE IIT, Kharagpur
Module 3 LOSSY IMAGE COMPRESSION SYSTEMS Verson ECE IIT, Kharagpur Lesson 6 Theory of Quantzaton Verson ECE IIT, Kharagpur Instructonal Objectves At the end of ths lesson, the students should be able to:
More informationCommon loop optimizations. Example to improve locality. Why Dependence Analysis. Data Dependence in Loops. Goal is to find best schedule:
15745 Lecture 6 Data Dependence n Loops Copyrght Seth Goldsten, 2008 Based on sldes from Allen&Kennedy Lecture 6 15745 20058 1 Common loop optmzatons Hostng of loopnvarant computatons precompute before
More informationProblem Set 6: Trees Spring 2018
Problem Set 6: Trees 129 Sprng 2018 A Average dstance Gven a tree, calculate the average dstance between two vertces n the tree. For example, the average dstance between two vertces n the followng tree
More informationCluster Validation Determining Number of Clusters. Umut ORHAN, PhD.
Cluster Analyss Cluster Valdaton Determnng Number of Clusters 1 Cluster Valdaton The procedure of evaluatng the results of a clusterng algorthm s known under the term cluster valdty. How do we evaluate
More information= z 20 z n. (k 20) + 4 z k = 4
Problem Set #7 solutons 7.2.. (a Fnd the coeffcent of z k n (z + z 5 + z 6 + z 7 + 5, k 20. We use the known seres expanson ( n+l ( z l l z n below: (z + z 5 + z 6 + z 7 + 5 (z 5 ( + z + z 2 + z + 5 5
More informationExercises. 18 Algorithms
18 Algorthms Exercses 0.1. In each of the followng stuatons, ndcate whether f = O(g), or f = Ω(g), or both (n whch case f = Θ(g)). f(n) g(n) (a) n 100 n 200 (b) n 1/2 n 2/3 (c) 100n + log n n + (log n)
More informationprinceton univ. F 17 cos 521: Advanced Algorithm Design Lecture 7: LP Duality Lecturer: Matt Weinberg
prnceton unv. F 17 cos 521: Advanced Algorthm Desgn Lecture 7: LP Dualty Lecturer: Matt Wenberg Scrbe: LP Dualty s an extremely useful tool for analyzng structural propertes of lnear programs. Whle there
More information= = = (a) Use the MATLAB command rref to solve the system. (b) Let A be the coefficient matrix and B be the righthand side of the system.
Chapter Matlab Exercses Chapter Matlab Exercses. Consder the lnear system of Example n Secton.. x x x y z y y z (a) Use the MATLAB command rref to solve the system. (b) Let A be the coeffcent matrx and
More informationMath 261 Exercise sheet 2
Math 261 Exercse sheet 2 http://staff.aub.edu.lb/~nm116/teachng/2017/math261/ndex.html Verson: September 25, 2017 Answers are due for Monday 25 September, 11AM. The use of calculators s allowed. Exercse
More informationCapacity Constraints Across Nests in Assortment Optimization Under the Nested Logit Model
Capacty Constrants Across Nests n Assortment Optmzaton Under the Nested Logt Model Jacob B. Feldman School of Operatons Research and Informaton Engneerng, Cornell Unversty, Ithaca, New York 14853, USA
More informationarxiv: v3 [cs.dm] 7 Jul 2012
Perfect matchng n unform hypergraphs wth large vertex degree arxv:1101.580v [cs.dm] 7 Jul 01 Imdadullah Khan Department of Computer Scence College of Computng and Informaton Systems Umm AlQura Unversty
More informationOn the Multicriteria Integer Network Flow Problem
BULGARIAN ACADEMY OF SCIENCES CYBERNETICS AND INFORMATION TECHNOLOGIES Volume 5, No 2 Sofa 2005 On the Multcrtera Integer Network Flow Problem Vassl Vasslev, Marana Nkolova, Maryana Vassleva Insttute of
More informationGraph Reconstruction by Permutations
Graph Reconstructon by Permutatons Perre Ille and Wllam Kocay* Insttut de Mathémathques de Lumny CNRS UMR 6206 163 avenue de Lumny, Case 907 13288 Marselle Cedex 9, France emal: lle@ml.unvmrs.fr Computer
More informationNUMERICAL DIFFERENTIATION
NUMERICAL DIFFERENTIATION 1 Introducton Dfferentaton s a method to compute the rate at whch a dependent output y changes wth respect to the change n the ndependent nput x. Ths rate of change s called the
More informationU.C. Berkeley CS294: Beyond WorstCase Analysis Handout 6 Luca Trevisan September 12, 2017
U.C. Berkeley CS94: Beyond WorstCase Analyss Handout 6 Luca Trevsan September, 07 Scrbed by Theo McKenze Lecture 6 In whch we study the spectrum of random graphs. Overvew When attemptng to fnd n polynomal
More informationWeek 2. This week, we covered operations on sets and cardinality.
Week 2 Ths week, we covered operatons on sets and cardnalty. Defnton 0.1 (Correspondence). A correspondence between two sets A and B s a set S contaned n A B = {(a, b) a A, b B}. A correspondence from
More informationTHE ARIMOTOBLAHUT ALGORITHM FOR COMPUTATION OF CHANNEL CAPACITY. William A. Pearlman. References: S. Arimoto  IEEE Trans. Inform. Thy., Jan.
THE ARIMOTOBLAHUT ALGORITHM FOR COMPUTATION OF CHANNEL CAPACITY Wllam A. Pearlman 2002 References: S. Armoto  IEEE Trans. Inform. Thy., Jan. 1972 R. Blahut  IEEE Trans. Inform. Thy., July 1972 Recall
More informationDynamic Programming 4/5/12. Dynamic programming. Fibonacci numbers. Fibonacci: a first attempt. David Kauchak cs302 Spring 2012
Dynamc Programmng Davd Kauchak cs32 Sprng 212 Dynamc programmng l One of the most mportant algorthm tools! l Very common ntervew queston l Method for solvng problems where optmal solutons can be defned
More informationChapter 5. Solution of System of Linear Equations. Module No. 6. Solution of Inconsistent and Ill Conditioned Systems
Numercal Analyss by Dr. Anta Pal Assstant Professor Department of Mathematcs Natonal Insttute of Technology Durgapur Durgapur713209 emal: anta.bue@gmal.com 1 . Chapter 5 Soluton of System of Lnear Equatons
More informationAnnexes. EC.1. Cyclebase move illustration. EC.2. Problem Instances
ec Annexes Ths Annex frst llustrates a cyclebased move n the dynamcblock generaton tabu search. It then dsplays the characterstcs of the nstance sets, followed by detaled results of the parametercalbraton
More information20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The first idea is connectedness.
20. Mon, Oct. 13 What we have done so far corresponds roughly to Chapters 2 & 3 of Lee. Now we turn to Chapter 4. The frst dea s connectedness. Essentally, we want to say that a space cannot be decomposed
More informationErrors for Linear Systems
Errors for Lnear Systems When we solve a lnear system Ax b we often do not know A and b exactly, but have only approxmatons Â and ˆb avalable. Then the best thng we can do s to solve Âˆx ˆb exactly whch
More informationComplex Numbers Alpha, Round 1 Test #123
Complex Numbers Alpha, Round Test #3. Wrte your 6dgt ID# n the I.D. NUMBER grd, leftjustfed, and bubble. Check that each column has only one number darkened.. In the EXAM NO. grd, wrte the 3dgt Test
More informationVickrey Auction VCG Combinatorial Auctions. Mechanism Design. Algorithms and Data Structures. Winter 2016
Mechansm Desgn Algorthms and Data Structures Wnter 2016 1 / 39 Vckrey Aucton VckreyClarkeGroves Mechansms SngleMnded Combnatoral Auctons 2 / 39 Mechansm Desgn (wth Money) Set A of outcomes to choose
More informationFinding Primitive Roots PseudoDeterministically
Electronc Colloquum on Computatonal Complexty, Report No 207 (205) Fndng Prmtve Roots PseudoDetermnstcally Ofer Grossman December 22, 205 Abstract Pseudodetermnstc algorthms are randomzed search algorthms
More informationIntroduction to Vapor/Liquid Equilibrium, part 2. Raoult s Law:
CE304, Sprng 2004 Lecture 4 Introducton to Vapor/Lqud Equlbrum, part 2 Raoult s Law: The smplest model that allows us do VLE calculatons s obtaned when we assume that the vapor phase s an deal gas, and
More informationNotes on Frequency Estimation in Data Streams
Notes on Frequency Estmaton n Data Streams In (one of) the data streamng model(s), the data s a sequence of arrvals a 1, a 2,..., a m of the form a j = (, v) where s the dentty of the tem and belongs to
More informationSingular Value Decomposition: Theory and Applications
Sngular Value Decomposton: Theory and Applcatons Danel Khashab Sprng 2015 Last Update: March 2, 2015 1 Introducton A = UDV where columns of U and V are orthonormal and matrx D s dagonal wth postve real
More informationLecture 20: Lift and Project, SDP Duality. Today we will study the Lift and Project method. Then we will prove the SDP duality theorem.
prnceton u. sp 02 cos 598B: algorthms and complexty Lecture 20: Lft and Project, SDP Dualty Lecturer: Sanjeev Arora Scrbe:Yury Makarychev Today we wll study the Lft and Project method. Then we wll prove
More informationChapter Newton s Method
Chapter 9. Newton s Method After readng ths chapter, you should be able to:. Understand how Newton s method s dfferent from the Golden Secton Search method. Understand how Newton s method works 3. Solve
More informationLecture 4. Instructor: Haipeng Luo
Lecture 4 Instructor: Hapeng Luo In the followng lectures, we focus on the expert problem and study more adaptve algorthms. Although Hedge s proven to be worstcase optmal, one may wonder how well t would
More informationLecture 2: GramSchmidt Vectors and the LLL Algorithm
NYU, Fall 2016 Lattces Mn Course Lecture 2: GramSchmdt Vectors and the LLL Algorthm Lecturer: Noah StephensDavdowtz 2.1 The Shortest Vector Problem In our last lecture, we consdered short solutons to
More information8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS
SECTION 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS 493 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS All the vector spaces you have studed thus far n the text are real vector spaces because the scalars
More informationMore metrics on cartesian products
More metrcs on cartesan products If (X, d ) are metrc spaces for 1 n, then n Secton II4 of the lecture notes we defned three metrcs on X whose underlyng topologes are the product topology The purpose of
More informationCase A. P k = Ni ( 2L i k 1 ) + (# big cells) 10d 2 P k.
THE CELLULAR METHOD In ths lecture, we ntroduce the cellular method as an approach to ncdence geometry theorems lke the SzemerédTrotter theorem. The method was ntroduced n the paper Combnatoral complexty
More informationLinear, affine, and convex sets and hulls In the sequel, unless otherwise specified, X will denote a real vector space.
Lnear, affne, and convex sets and hulls In the sequel, unless otherwse specfed, X wll denote a real vector space. Lnes and segments. Gven two ponts x, y X, we defne xy = {x + t(y x) : t R} = {(1 t)x +
More informationLecture 4: November 17, Part 1 Single Buffer Management
Lecturer: Ad Rosén Algorthms for the anagement of Networs Fall 20032004 Lecture 4: November 7, 2003 Scrbe: Guy Grebla Part Sngle Buffer anagement In the prevous lecture we taled about the Combned Input
More informationLecture 5 Decoding Binary BCH Codes
Lecture 5 Decodng Bnary BCH Codes In ths class, we wll ntroduce dfferent methods for decodng BCH codes 51 Decodng the [15, 7, 5] 2 BCH Code Consder the [15, 7, 5] 2 code C we ntroduced n the last lecture
More informationTHE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens
THE CHINESE REMAINDER THEOREM KEITH CONRAD We should thank the Chnese for ther wonderful remander theorem. Glenn Stevens 1. Introducton The Chnese remander theorem says we can unquely solve any par of
More informationBasic Regular Expressions. Introduction. Introduction to Computability. Theory. Motivation. Lecture4: Regular Expressions
Introducton to Computablty Theory Lecture: egular Expressons Prof Amos Israel Motvaton If one wants to descrbe a regular language, La, she can use the a DFA, Dor an NFA N, such L ( D = La that that Ths
More informationPartitioning Graphs of Supply and Demand
Parttonng Graphs of Supply and Demand Takehro Ito Graduate School of Informaton Scences, Tohoku Unersty, Aobayama 6605, Senda, 9808579, Japan. take@nshzek.ece.tohoku.ac.p Abstract. Suppose that each
More informationSpectral Graph Theory and its Applications September 16, Lecture 5
Spectral Graph Theory and ts Applcatons September 16, 2004 Lecturer: Danel A. Spelman Lecture 5 5.1 Introducton In ths lecture, we wll prove the followng theorem: Theorem 5.1.1. Let G be a planar graph
More informationAn Interactive Optimisation Tool for Allocation Problems
An Interactve Optmsaton ool for Allocaton Problems Fredr Bonäs, Joam Westerlund and apo Westerlund Process Desgn Laboratory, Faculty of echnology, Åbo Aadem Unversty, uru 20500, Fnland hs paper presents
More informationLecture 12: Discrete Laplacian
Lecture 12: Dscrete Laplacan Scrbe: Tanye Lu Our goal s to come up wth a dscrete verson of Laplacan operator for trangulated surfaces, so that we can use t n practce to solve related problems We are mostly
More informationDesign and Analysis of Algorithms
Desgn and Analyss of Algorthms CSE 53 Lecture 4 Dynamc Programmng Junzhou Huang, Ph.D. Department of Computer Scence and Engneerng CSE53 Desgn and Analyss of Algorthms The General Dynamc Programmng Technque
More informationFoundations of Arithmetic
Foundatons of Arthmetc Notaton We shall denote the sum and product of numbers n the usual notaton as a 2 + a 2 + a 3 + + a = a, a 1 a 2 a 3 a = a The notaton a b means a dvdes b,.e. ac = b where c s an
More informationMath1110 (Spring 2009) Prelim 3  Solutions
Math 1110 (Sprng 2009) Solutons to Prelm 3 (04/21/2009) 1 Queston 1. (16 ponts) Short answer. Math1110 (Sprng 2009) Prelm 3  Solutons x a 1 (a) (4 ponts) Please evaluate lm, where a and b are postve numbers.
More informationProblem Do any of the following determine homomorphisms from GL n (C) to GL n (C)?
Homework 8 solutons. Problem 16.1. Whch of the followng defne homomomorphsms from C\{0} to C\{0}? Answer. a) f 1 : z z Yes, f 1 s a homomorphsm. We have that z s the complex conjugate of z. If z 1,z 2
More informationPolynomial Regression Models
LINEAR REGRESSION ANALYSIS MODULE XII Lecture  6 Polynomal Regresson Models Dr. Shalabh Department of Mathematcs and Statstcs Indan Insttute of Technology Kanpur Test of sgnfcance To test the sgnfcance
More information2E Pattern Recognition Solutions to Introduction to Pattern Recognition, Chapter 2: Bayesian pattern classification
E395  Pattern Recognton Solutons to Introducton to Pattern Recognton, Chapter : Bayesan pattern classfcaton Preface Ths document s a soluton manual for selected exercses from Introducton to Pattern Recognton
More informationMATH 5707 HOMEWORK 4 SOLUTIONS 2. 2 i 2p i E(X i ) + E(Xi 2 ) ä i=1. i=1
MATH 5707 HOMEWORK 4 SOLUTIONS CİHAN BAHRAN 1. Let v 1,..., v n R m, all lengths v are not larger than 1. Let p 1,..., p n [0, 1] be arbtrary and set w = p 1 v 1 + + p n v n. Then there exst ε 1,..., ε
More informationDiscrete Mathematics
Dscrete Mathematcs 30 (00) 48 488 Contents lsts avalable at ScenceDrect Dscrete Mathematcs journal homepage: www.elsever.com/locate/dsc The number of C 3 free vertces on 3partte tournaments Ana Paulna
More informationA new construction of 3separable matrices via an improved decoding of Macula s construction
Dscrete Optmzaton 5 008 700 704 Contents lsts avalable at ScenceDrect Dscrete Optmzaton journal homepage: wwwelsevercom/locate/dsopt A new constructon of 3separable matrces va an mproved decodng of Macula
More informationREAL ANALYSIS I HOMEWORK 1
REAL ANALYSIS I HOMEWORK CİHAN BAHRAN The questons are from Tao s text. Exercse 0.0.. If (x α ) α A s a collecton of numbers x α [0, + ] such that x α
More informationGrover s Algorithm + Quantum Zeno Effect + Vaidman
Grover s Algorthm + Quantum Zeno Effect + Vadman CS 2942 Bomb 10/12/04 Fall 2004 Lecture 11 Grover s algorthm Recall that Grover s algorthm for searchng over a space of sze wors as follows: consder the
More informationMath 426: Probability MWF 1pm, Gasson 310 Homework 4 Selected Solutions
Exercses from Ross, 3, : Math 26: Probablty MWF pm, Gasson 30 Homework Selected Solutons 3, p. 05 Problems 76, 86 3, p. 06 Theoretcal exercses 3, 6, p. 63 Problems 5, 0, 20, p. 69 Theoretcal exercses 2,
More informationThe ExpectationMaximization Algorithm
The ExpectatonMaxmaton Algorthm Charles Elan elan@cs.ucsd.edu November 16, 2007 Ths chapter explans the EM algorthm at multple levels of generalty. Secton 1 gves the standard hghlevel verson of the algorthm.
More information18.781: Solution to Practice Questions for Final Exam
18.781: Soluton to Practce Questons for Fnal Exam 1. Fnd three solutons n postve ntegers of x 6y = 1 by frst calculatng the contnued fracton expanson of 6. Soluton: We have 1 6=[, ] 6 6+ =[, ] 1 =[,, ]=[,,
More informationFirst Year Examination Department of Statistics, University of Florida
Frst Year Examnaton Department of Statstcs, Unversty of Florda May 7, 010, 8:00 am  1:00 noon Instructons: 1. You have four hours to answer questons n ths examnaton.. You must show your work to receve
More informationCS : Algorithms and Uncertainty Lecture 17 Date: October 26, 2016
CS 29128: Algorthms and Uncertanty Lecture 17 Date: October 26, 2016 Instructor: Nkhl Bansal Scrbe: Mchael Denns 1 Introducton In ths lecture we wll be lookng nto the secretary problem, and an nterestng
More informationECE559VV Project Report
ECE559VV Project Report (Supplementary Notes Loc Xuan Bu I. MAX SUMRATE SCHEDULING: THE UPLINK CASE We have seen (n the presentaton that, for downlnk (broadcast channels, the strategy maxmzng the sumrate
More informationTornado and Luby Transform Codes. Ashish Khisti Presentation October 22, 2003
Tornado and Luby Transform Codes Ashsh Khst 6.454 Presentaton October 22, 2003 Background: Erasure Channel Elas[956] studed the Erasure Channel β x x β β x 2 m x 2 k? Capacty of Noseless Erasure Channel
More informationThe L(2, 1)Labeling on Product of Graphs
Annals of Pure and Appled Mathematcs Vol 0, No, 05, 939 ISSN: 79087X (P, 790888(onlne Publshed on 7 Aprl 05 wwwresearchmathscorg Annals of The L(, Labelng on Product of Graphs P Pradhan and Kamesh
More informationOline Temporary Tasks Assignment. Abstract. In this paper we consider the temporary tasks assignment
Olne Temporary Tasks Assgnment Yoss Azar and Oded Regev Dept. of Computer Scence, TelAvv Unversty, TelAvv, 69978, Israel. azar@math.tau.ac.l??? Dept. of Computer Scence, TelAvv Unversty, TelAvv, 69978,
More information