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Approximation Algorithms Greedy Strategies Max and Min min f is equivalent to max –f. However, a good approximation for min f may not be a good approximation for max –f. For example, consider a graph G=(V,E). C is a minimum vertex cover of G iff V \ C is a maximum independent set of G. The minimum vertex cover has a polynomial-time 2-approximation, but the maximum independent set has no constant-bounded approximation unless NP=P. Another example: Minimum Connected Dominating Set and Minimum Spanning Tree with Maximum Number of Leaves I hear, I forget. I learn, I remember. I do, I 2 understand! Greedy for Max and Min Max --- independent system Min --- submodular potential function I hear, I forget. I learn, I 3 remember. I do, I understand! Independent System I hear, I forget. I learn, I 4 remember. I do, I understand! Independent System Consider a set E and a collection C of subsets of E. (E,C) is called an independent system if A B, B C A C The elements of C are called independent sets I hear, I forget. I learn, I 5 remember. I do, I understand! Maximization Problem Consider a nonnegative cost function : c : E R Consider the following maximation problem on (E, C) : max c( A) subject to A C where c( A) c( x) xA I hear, I forget. I learn, I 6 remember. I do, I understand! Greedy Approximation MAX Sort all elements in E into ordering c( x1) c ( x 2) c ( xn ). A ; for i 1 to n do if A {xi} C hen A A {xi}; t output A. I hear, I forget. I learn, I 7 remember. I do, I understand! Theorem Let AG be obtained solution by the Greedy Algorithm MAX. Let A * be an optimal solution. Then c( A*) u(F ) max F E c( AG ) v( F ) whereu ( F ) is the maximum size of independent set in F and v( F )is the minimum size of maximal independent set in F . I hear, I forget. I learn, I 8 remember. I do, I understand! Proof Denote Ei {x1, ..., xi}. n c( AG ) c( x1) | E1 AG | c( xi )(| Ei AG | | Ei 1 AG |) i 2 n 1 | Ei AG | (c( xi ) c( xi 1)) | En AG | c( xn ) i 1 Similarly, c( A*) n 1 | Ei A* | (c( xi ) c( xi 1)) | En A* | c( xn ) i 1 I hear, I forget. I learn, I 9 remember. I do, I understand! Therefore, c( A*) | Ei A* | max . c( AG ) 1i n | Ei AG | Note that | Ei A* | u ( Ei ) since Ei A* is independent. We need only to show that | Ei AG | v( Ei ), that is, to show Ei AG is a maximal independent set of Ei. I hear, I forget. I learn, I 10 remember. I do, I understand! Supposeindependent set Ei AG is not maximal in Ei. Then thereis a xj Ei \ AG such that( Ei AG ) {xj} is independent. So, ( Ej 1 AG ) {xj} is independent, contradicting the rule of not choosing xj. I hear, I forget. I learn, I 11 remember. I do, I understand! Maximum Weight Hamiltonian Cycle Given an edge-weighted complete graph, find a Hamiltonian cycle with maximum total weight. I hear, I forget. I learn, I 12 remember. I do, I understand! Independent sets E = {all edges} A subset of edges is independent if it is a Hamiltonian cycle or a vertex-disjoint union of paths. C = a collection of such subsets I hear, I forget. I learn, I 13 remember. I do, I understand! Maximal Independent Sets Consider a subset F of edges. For any two maximal independent sets I and J of F, |J| < 2|I| I hear, I forget. I learn, I 14 remember. I do, I understand! I hear, I forget. I learn, I 15 remember. I do, I understand! Theorem: For the maximum Hamiltonian cycle problem, the greedy algorithm MAX produces a polynomial time approximation with performance ratio at most 2. I hear, I forget. I learn, I 16 remember. I do, I understand! Maximum Weight Directed Hamiltonian Cycle Given an edge-weighted complete digraph, find a Hamiltonian cycle with maximum total weight. I hear, I forget. I learn, I 17 remember. I do, I understand! Independent sets E = {all edges} A subset of edges is independent if it is a directed Hamiltonian cycle or a vertex-disjoint union of directed paths. I hear, I forget. I learn, I 18 remember. I do, I understand! Consider F E. Suppose I and J are two maximal independent sets of F . Denote J 1 {e J | e shares the head with an edge in I } J 2 {e J | e shares the tail with an edge in I } J 3 {e J | e connectsa maximal path of I from its head to its tail} | Ji || I | for i 1, 2, 3. | J || J 1 | | J 2 | | J 3 | 3 | I | . I hear, I forget. I learn, I 19 remember. I do, I understand! Tightness The rest of all edges have a cost ε ε 1 1 1 1+ε I hear, I forget. I learn, I 20 remember. I do, I understand! A Special Case If c satisfies the following quadrilateral condition: For any 4 vertices u, v, u’, v’ in V, c(u, v) max{c(u, v' ), c(u' , v)} c(u, v) c(u' , v' ) c(u, v' ) c(u' , v) Then the greedy approximation for maximum weight Hamiltonian cycle has the performance ratio 2. I hear, I forget. I learn, I 21 remember. I do, I understand! Suppose e1, e2, ..., en are selected by the Greedy algorithm in the order of appearance. Let Qi Pi {e1, e2, ..., ei} where Pi is the maximum weight Hamiltonia n cycle containing e1, e2, ..., ei. We claim that for i 1, ..., n, c(Qi 1) c(Qi ) 2c(ei ). If this is proved, then wehave c(Q 0) c(Q0 ) 2c(e1 ) c(Q2 ) 2c(e1 ) 2c(e2 ) c(Qn) 2(c(e1) c(en)). I hear, I forget. I learn, I 22 remember. I do, I understand! Let ei (u, v). By the Greedy rule, c(ei ) c(e) for all e Qi 1. After ei is selected, at most threeedges in Qi 1 become not permissible : (1) one shares the head with ei, (2) one shares the tail with ei, (3) the third one is on the path from v to u. I hear, I forget. I learn, I 23 remember. I do, I understand! Case 1. At most twoedges, say e and e' , of Qi 1 become not permissible after ei is selected.That is, we may form a new path passing through e1 , e 2 , ..., ei by removing at most two edges e and e'. Then c((Pi 1 \ {e, e'}) {ei }) c( Pi ). Hence, c(Qi 1) c(Qi ) c(e) c(e' ) c(Qi ) 2c(ei ) I hear, I forget. I learn, I 24 remember. I do, I understand! Case 2. We must remove3 edges from Pi-1 to form a new path. They are (u, v' ), (u ' , v) and an edge e on the subpath from v to u in Pi 1. This means that Pi-1 has a subpath from u' to v', which contains these three edges.Thus after deleting (u, v' ), (u', v), and e, we can add edge (u', v' ) to form a new path. Therefore c(Qi ) c(Qi 1) c(e) c(u , v' ) c(u ' , v) c(u ' , v' ) c(Qi 1) c(e) c(u, v) c(Qi 1) 2c(ei ) the secondinequality follows from the Quadrilateral condition u, v, u' , v', and the fact thatc(u,v) c(e') for all e' Q i-1 I hear, I forget. I learn, I 25 remember. I do, I understand! Superstring Given n strings s1, s2, …, sn, find a shortest string s containing all s1, s2, …, sn as substrings. No si is a substring of another sj. I hear, I forget. I learn, I 26 remember. I do, I understand! An Example Given S = {abcc, efaab, bccef} Some possible solutions: Concatenate all substrings = abccefaabbccef (14 chars) A shortest superstring is abccefaab (9 chars) I hear, I forget. I learn, I 27 remember. I do, I understand! Relationship to Set Cover? How to “transform” the shortest superstring (SS) to the Set Cover (SC) problem? Need to identify U Need to identify S Need to define the cost function The SC instance is an SS instance Let U = S (a set of n strings). How to define S ? I hear, I forget. I learn, I 28 remember. I do, I understand! Relationship to SC (cont) k si | | sj | | | | σijk Let M be the set that consists of the strings σijk I hear, I forget. I learn, I 29 remember. I do, I understand! Relationship to SC (cont) For a string S M , define set( ) {s S | s is a substring of } Now, define S {set ( ) | S M } Define cost of set ( ) | |, i.e., the length of string Let C is the set cover of this constructed SC, then the concatenation of all strings in C is a solution of SS. Note that C is a collection of set ( ) I hear, I forget. I learn, I 30 remember. I do, I understand! Algorithm 1 for SS 1 : Construct the SC as described 2 : Use the SC algorithm to find a set cover C Let C {set ( 1 ),..., set ( k )} 3 : Concatenate the strings 1 ,..., k in any order 4 : Return the resulting string s I hear, I forget. I learn, I 31 remember. I do, I understand! Approximation Ratio Lemma 1: Let opt be length of the optimal solution of SS and opt’ be the cost of the optimal solution of SC, we have: opt ≤ opt’ ≤ 2opt Proof: (a) Show opt opt ' Consider an optimal SC C* {set ( ij ) |1 j l}. Let s be a string concatenated from ij ,1 j l. Thus | s | opt '. Since each string of S is a substring of some ij ,1 j l , it is also a substring of s. Hence s is also a solution of SS. Therefore, opt | s | opt ' I hear, I forget. I learn, I 32 remember. I do, I understand! Proof of Lemma 1 (cont) (b) Show opt ' 2opt Note that each string in the set cover must start and end in different places from the rest of the strings since none of the strings are substrings of another string. Now, partition the strings of the set cover into group that are contained by 1 ,..., k . The strings are split such thatthe last string in group i will be the furthest overlapping string to the first string in i . After this partition, we claim that i cannot overlap with i 2 . (Use the contradiction method to prove it) Therefore,in the worstcase, every other group completely overlaps and we have opt ' 2opt I hear, I forget. I learn, I 33 remember. I do, I understand! Approximation Ratio Theorem1: Algorithm 1 has an approximation ratio within a factor of 2Hn Proof: We know that the approximation ratio of Set Cover is Hn. From Lemma 1, it follows directly that Algorithm 1 is a 2Hn factor algorithm for SS I hear, I forget. I learn, I 34 remember. I do, I understand! Prefix and Overlap For two string s1 and s2, we have: Overlap(s1,s2) = the maximum between the suffix of s1 and the prefix of s2. pref(s1,s2) = the prefix of s1 that remains after chopping off overlap(s1,s2) Example: s1 = abcbcaa and s2 = bcaaca, then overlap(s1,s2) = bcaa pref(s1,s2) = abc Note: overlap(s1,s2) ≠ overlap(s2, s1) I hear, I forget. I learn, I 35 remember. I do, I understand! Is there any better approach? Now, suppose that in the optimal solution, the strings appear from the left to right in the order: s1, s2, …, sn Define: opt = |pref(s1,s2)| + …+|pref(sn-1,sn)| + |pref(sn,s1)| + |overlap(sn,s1)| Why overlap(sn,s1)? Consider this example S={agagag, gagaga}. If we just consider the prefix only, the result would be ag whereas the correct result is agagaga I hear, I forget. I learn, I 36 remember. I do, I understand! Prefix Graph Define the prefix graph as follows: Complete weighted directed graph G=(V,E) V is a set of vertices, labeled from 1 to n (each vertex represents each string si) For each edge i→j, i ≠ j, assign a weight of 1( ) |pref(si, sj)| Example: S={abc, bcd, dab} 2( ) ( )3 I hear, I forget. I learn, I 37 remember. I do, I understand! Cycle Cover Cycle Cover: a collection of disjoint cycles covering all vertices (each vertex is in exactly one cycle) Note that the tour 1 → 2 → … → n → 1 is a cycle cover Minimum weight cycle cover: sum of weights is minimum over all covers Thus, we want to find a minimum weight cycle cover I hear, I forget. I learn, I 38 remember. I do, I understand! How to find a min. weight cycle cover Corresponding to the prefix graph, construct a bipartite graph H=(X,Y;E) such that: X = {x1, x2, …, xn} and Y = {y1, y2, …, yn} For each i, j (in 1…n), add edge (xi, yj) of weight |pref(si,sj)| Each cycle cover of the prefix graph ↔ a perfect matching of the same weight in H. (Perfect matching is a matching which covers all the vertices) Finding a minimum weight cycle cover = finding a minimum weight perfect matching (which can be found in poly-time) I hear, I forget. I learn, I 39 remember. I do, I understand! How to break the cycle If ci (i1 i2 il i1 ) is a cyclein the prefix graph, s12 let α(ci ) pref ( si1 , si2 ) pref ( sil 1 , sil ) pref ( sil , si1 ) Let w(ci ) | (ci ) | be the weight of cycleci and let σ (ci ) (ci ) si1 From now on, we will call si1 the representa string for ci tive s13 s11 Informally, we can obtain the string σ(ci ) by " opening" ci at arbitrary string si1 . (c1 ) pref ( s11 , s12 ) pref ( s12 , s13 ) pref ( s13 , s11 ) bc a a bcaa (c1 ) (c1 ) s11 bcaabca I hear, I forget. I learn, I 40 remember. I do, I understand! A constant factor algorithm Algorithm 2: ing 1 : Constructthe prefix graph correspond to strings in S 2 : Find a minimum weight cyclecover of the prefix graph, C {c1, , ck } 3 : Return σ(c1 ) σ(ck ) I hear, I forget. I learn, I 41 remember. I do, I understand! Approximation Ratio Lemma 2: Let C be the minimum weight cycle cover of S. Let c and c’ be two cycles in C, and let r, r’ be representative strings from these cycles. Then |overlap(r, r’)| < w(c) + w(c’) Proof: Exercise I hear, I forget. I learn, I 42 remember. I do, I understand! Approximation Ratio (cont) Theorem 2: Algorithm 2 has an approximation ratio of 4. Proof: (see next slide) I hear, I forget. I learn, I 43 remember. I do, I understand! Proof The output of Alg 2 has length k k k | (c ) | w(c ) | r | i 1 i i 1 i i 1 i tive where ri is the representa string of ci . k k Now, weshow that | ri | opt 2 w(ci ) whereopt is i 1 i 1 the size of an optimal solution. Let l be the shortestsuperstring of all ri . Then | l | opt k k k k and opt | ri | | overlap(ri , ri 1 ) | | ri | 2 w(ci ) (Lemma 2) i 1 i 1 i 1 i 1 k k k Hence, | (ci ) | w(ci ) opt 2 w(ci ) i 1 i 1 i 1 k 4opt (since w(ci ) opt ) i 1 I hear, I forget. I learn, I 44 remember. I do, I understand! Modification to 3-Approximation Note that in Alg 2, after we find all strings i , we simply concatenate them to obtain the resuting string s. However, we can merge these i to obtain a shorter string by applying the shortest supperstring algorithm on i With this modificati on, we can obtain a 3 - approximat ion I hear, I forget. I learn, I 45 remember. I do, I understand! 3-Approximation Algorithm Algorithm 3: 1: Construct the prefix graph corresponding to strings in S 2: Find a minimum weight cycle cover of the prefix graph, C {c1 , , ck } 3: Return the greedy superstring algorithm on σ(c1 ) σ(ck ) to obtain the maximum compression 4: Output the resulting string, I hear, I forget. I learn, I 46 remember. I do, I understand! Superstring via Hamiltonian path |ov(u,v)| = max{|w| | there exist x and y such that u=xw and v=wy} Overlap graph G is a complete digraph: V = {s1, s2, …, sn} |ov(u,v)| is edge weight. Suppose s* is the shortest supper string. Let s1, …, sn be the strings in the order of appearance from left to right. Then si, si+1 must have maximum overlap in s*. Hence s1, …, sn form a directed Hamiltonian path in G. I hear, I forget. I learn, I 47 remember. I do, I understand! I hear, I forget. I learn, I 48 remember. I do, I understand! Suppose s1, s 2, ..., sn appear in minimum superstring s* in this natural ordering. Then n n 1 | s* | | si | | ov( si, si 1) | . i 1 i 1 Hence, (s1, s 2, ..., sn) is a maximum Hamiltonian path of the overlap graph. The longest Hamiltonian path would give a shortest supperstring I hear, I forget. I learn, I 49 remember. I do, I understand! The Algorithm (via Hamiltonian) I hear, I forget. I learn, I 50 remember. I do, I understand! A special property | ov(u, v) | max(| ov(u, v' ) |, | ov(u ' , v) |) | ov(u, v) | | ov(u ' , v' ) || ov(u, v' ) | | ov(u ' , v) | u’ v u v’ If | ov(u, v) | | ov(u, v ') | | ov(u ', v) |, proof is trivial. Else, write ov(u,v)=xwy such that ov(u,v')=xw and ov(u',v)=wy. Then w is an overlap between u' and v'. Therefore, |ov(u',v')| |w| = |ov(u,v')|+|ov(u',v)|-|ov(u,v)| I hear, I forget. I learn, I 51 remember. I do, I understand! Theorem The Greedy approximation MAX for maximum Hamiltonian path in overlapping graph has performance ratio 2. Conjecture: This greedy approximation also give the minimum superstring an approximation solution within a factor of 2 from optimal. Example: S={abk, bk+1, bka}. s* = abk+1a. Our obtained solution: abkabk+1. I hear, I forget. I learn, I 52 remember. I do, I understand! Submodular Function I hear, I forget. I learn, I 53 remember. I do, I understand! What is a submodular function? Consider a finite set E, (called ground set), and a function f : 2E →Z. The function f is said to be submodular if for any two subsets A and B in 2E : f ( A) f ( B) f ( A B) f ( A B) Example: f(A) = |A| is submodular. I hear, I forget. I learn, I 54 remember. I do, I understand! Set-Cover Given a collection C of subsets of a set E, find a minimum subcollection C’ of C such that every element of E appears in a subset in C’ . I hear, I forget. I learn, I 55 remember. I do, I understand! Greedy Algorithm C ' ; while | E | f (C ') do choose S C to maximize f (C ' S ) and C ' C ' S ; Return C’ Here: f(C) = # of elements in C Basically, the algorithm pick up the set that cover the most uncovered elements at each step I hear, I forget. I learn, I 56 remember. I do, I understand! Analyze the Approximation Ratio Suppose S 1, S 2, ..., Sk are selected by Greedy Algorithm. Denote Ci {S 1, ..., Si}. Then f (Ci 1) f (Ci ) (| E | f (Ci )) / opt I hear, I forget. I learn, I 57 remember. I do, I understand! (| E | f (Ci ))(1 1/ opt ) |E | f (Ci 1) | E | f (Ci 1) (| E | f (Ci ))(1 1/ opt ) (| E | f (Ci 1))(1 1/ opt ) 2 | E | (1 1/ opt ) i 1 (call this *) Choose i to be the largest one satisfying opt | E | f (Ci) (call this **) Then k i opt opt | E | (1 1/ opt ) i (subtitute ** to *) I hear, I forget. I learn, I 58 remember. I do, I understand! opt | E | (1 1 / opt ) i i / opt | E |e (note : 1 x e ) x So i opt ln (| E | / opt ) Thus, k opt i opt (1 ln (| E | / opt )) I hear, I forget. I learn, I 59 remember. I do, I understand! Alternative Analysis Suppose S 1, S 2, ..., Sk are selected by Greedy Algorithm. Denote Ci {S 1, ..., Si}. Then f (Ci 1) f (Ci ) (| E | f (Ci )) / opt I hear, I forget. I learn, I 60 remember. I do, I understand! Denote Xf ( A) f ( A X ) f ( A). Let C* { X 1, X 2,..., Xopt} be an optimal solution. Denote Cj* { X 1,..., Xj}. What By greedy rule, Si 1 f (Ci ) Xj 1 f (Ci ) do we need? for all 0 j opt 1 Thus, Si 1 f (Ci ) ( 0 j opt 1 Xj 1 f (Ci )) / opt ( 0 j opt 1 Xj 1 f (Ci Cj*)) / opt (f (Ci C*) f (Ci )) / opt ( | E | f (Ci)) / opt I hear, I forget. I learn, I 61 remember. I do, I understand! What’s we need? A B Xf ( A) Xf ( B) I hear, I forget. I learn, I 62 remember. I do, I understand! Actually, this inequality holds if and only if f is submodular and A B f ( A) f ( B) (monotone increasing) I hear, I forget. I learn, I 63 remember. I do, I understand! Let f be a function on all subsets of a set E. Then f is submodular and monotone increasing iff for any two subsets A B and any element x E : A B xf ( A) xf ( B ) I hear, I forget. I learn, I 64 remember. I do, I understand! Proof (1) f is submodular iff A B xf ( A) xf ( B) for x B (2) f is monotoneincreasing iff A B xf ( A) xf ( B) for x B I hear, I forget. I learn, I 65 remember. I do, I understand! Proof of (1) I hear, I forget. I learn, I 66 remember. I do, I understand! Proof of (2) This is very simple. Note that f is monotoneincreasing iff for any subset A E and x E , x f ( A) 0. For x B, by monotonicity of f, x f ( A) 0 x f ( B) I hear, I forget. I learn, I 67 remember. I do, I understand! Theorem Greedy Algorithm produces an approximation within ln n +1 from optimal for the set cover problem The same result holds for weighted set-cover. I hear, I forget. I learn, I 68 remember. I do, I understand! Weighted Set Cover Given a collection C of subsets of a set E and a weight function w on C, find a minimum total- weight subcollection C’ of C such that every element of E appears in a subset in C’ . I hear, I forget. I learn, I 69 remember. I do, I understand! Greedy Algorithm C ' ; while | E | f (C ' ) do choose S C to maximize Sf (C ' ) / w( S ) and C ' C '{S }; output C '. I hear, I forget. I learn, I 70 remember. I do, I understand! A General Problem Consider a monotoneincreasing, submodular function f defined on all subsetsof a set E. Define ( f ) { A | x E , xf ( A) 0}. min c( A) xA c( x) s.t. A ( f ) I hear, I forget. I learn, I 71 remember. I do, I understand! Greedy Algorithm A ; while there exists x E st xf ( A) 0 do choose x E to maximize xf ( A) / c( x) and A A {x}; output A. I hear, I forget. I learn, I 72 remember. I do, I understand! A General Theorem If f () 0, f is an integer function,and c( x) 0 for x E , then Greedy gives a (1 ln ) - approximation where max xE f ({x}). Remark (Normalized): If f () 0, we may consider g ( A) f ( A) f (). Therefore max xE ( f ({x}) f ()). I hear, I forget. I learn, I 73 remember. I do, I understand! Proof Suppose x1, x 2, ..., xk are selected by Greedy Algorithm in the order of their appearance. Denote Ai {x1, ..., xi} and ri xif ( Ai 1). Suppose A* { y1, ..., yh} is an optimal solution. For each ye A*, denote zei ye f ( Ai 1), ci c( xi ) and k 1 ci ck w(e) ( zei ze , i 1) zek i 1 ri rk k c1 ci ci 1 ze1 ( ) zei r1 i 2 ri ri 1 I hear, I forget. I learn, I 74 remember. I do, I understand! Proof (cont) We will prove these following claims: c( A) w(e) ye A* 1 w(e) c( ye ) for every ye i 1 i Then by the two claims, we have: 1 c( A) ye A* w(e) c( A*) i 1 i I hear, I forget. I learn, I 75 remember. I do, I understand! Show the First Claim By Greedy rule and submodular property of f , r1 r 2 rk c1 c 2 ck ************************** r1 x 2 f ( A0) by Greedy rule, where A0 c1 c2 x 2 f ( A1) r 2 since f is submodular c2 c2 I hear, I forget. I learn, I 76 remember. I do, I understand! k Now, show r z j i j yeA* ei for i 1, ..., k k r j i j f ( Ak ) f ( Ai 1) f ( Ai 1 A*) f ( Ai 1) A* f ( Ai 1) ye A* ye f ( Ai 1) by submodular and monotone increasing properties of f ye A* zei I hear, I forget. I learn, I 77 remember. I do, I understand! k ri c( A) ci i 1 ri k 1 ci k k ck k ( rj 1 rj ) rk rj i 1 ri j i j i j k c1 k k ci ci 1 k rj ( ) rj r1 j 1 i 2 ri ri 1 j i k c1 ci ci 1 * ze1 ( ri ri 1 ) y* zei r 1 yeA i2 e A w(e) ye A* I hear, I forget. I learn, I 78 remember. I do, I understand! Show the Second Claim ci c ( ye ) By greedy rule, . ri zei By the submodular and monotone increasing property of f , zei ze , i 1. Therefore, k 1 ci ck w(e) ( zei ze, i 1) i 1 ri zek rk For any integers p > q > k 1 0, we have: c ( ye ) c ( ye ) ( zei ze, i 1) i 1 zei zek zek $(p – q)/p = \sum_{j=q+1}^p 1/p \le \sum_{j=q+1}^p 1/j$ ze 1 1 c ( ye ) (since zei are integers) i 1 i 1 c ( ye) since ze1 for all ye A* i 1 i I hear, I forget. I learn, I 79 remember. I do, I understand! Connected Vertex-Cover Given a connected graph, find a minimum vertex-cover which induces a connected subgraph. I hear, I forget. I learn, I 80 remember. I do, I understand! For any vertex subset A, p(A) is the number of edges not covered by A. For any vertex subset A, q(A) is the number of connected component of the subgraph induced by A. -p is submodular. -q is not submodular. Note that when A is a connected vertex cover, the q(A) = 1 and p(A) = 0. I hear, I forget. I learn, I 81 remember. I do, I understand! -p-q Define f(A)= -p(A) –q(A). Then f(A) is submodular and monotone increasing Proof: Homework! I hear, I forget. I learn, I 82 remember. I do, I understand! Theorem Connected Vertex-Cover has a (1+ln Δ)- approximation where Δ is the maximum degree. -p(Ø)=-|E|, -q(Ø)=0. |E|-p(x)-q(x) < Δ-1 The maximum value of - x p(emptyset) , and that of - x g (emptyset) 1 for all x V I hear, I forget. I learn, I 83 remember. I do, I understand! Weighted Connected Vertex-Cover Given a vertex-weighted connected graph, find a connected vertex-cover with minimum total weight. Theorem Weighted Connected Vertex-Cover has a (1+ln Δ)-approximation. I hear, I forget. I learn, I 84 remember. I do, I understand!

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