Approximation Algorithms Greedy Strategies. I hear, I forget. I learn, I remember. I do, I understand! 2 Max and Min  min f is equivalent to max –f.

Approximation Algorithms Greedy Strategies

I hear, I forget. I learn, I remember. I do, I understand! 2 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 understand! 3 Greedy for Max and Min  Max --- independent system  Min --- submodular potential function

I hear, I forget. I learn, I remember. I do, I understand! 4 Independent System

I hear, I forget. I learn, I remember. I do, I understand! 5 Independent System  Consider a set E and a collection C of subsets of E. (E,C) is called an independent system if The elements of C are called independent sets

I hear, I forget. I learn, I remember. I do, I understand! 6 Maximization Problem

I hear, I forget. I learn, I remember. I do, I understand! 7 Greedy Approximation MAX

I hear, I forget. I learn, I remember. I do, I understand! 8 Theorem

I hear, I forget. I learn, I remember. I do, I understand! 9 Proof

I hear, I forget. I learn, I remember. I do, I understand! 10

I hear, I forget. I learn, I remember. I do, I understand! 12 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 remember. I do, I understand! 13 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 remember. I do, I understand! 14 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 remember. I do, I understand! 16  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 remember. I do, I understand! 17 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 remember. I do, I understand! 18 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 remember. I do, I understand! 20 Tightness 1 1 1 1+ε ε The rest of all edges have a cost ε

I hear, I forget. I learn, I remember. I do, I understand! 21 A Special Case  If c satisfies the following quadrilateral condition: For any 4 vertices u, v, u’, v’ in V, Then the greedy approximation for maximum weight Hamiltonian cycle has the performance ratio 2.

I hear, I forget. I learn, I remember. I do, I understand! 26 Superstring  Given n strings s 1, s 2, …, s n, find a shortest string s containing all s 1, s 2, …, s n as substrings.  No s i is a substring of another s j.

I hear, I forget. I learn, I remember. I do, I understand! 27 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 remember. I do, I understand! 28 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 remember. I do, I understand! 29 Relationship to SC (cont) ||| Let M be the set that consists of the strings σ ijk sisi sjsj σ ijk k

I hear, I forget. I learn, I remember. I do, I understand! 30 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 Relationship to SC (cont) Now, define S Define cost of

I hear, I forget. I learn, I remember. I do, I understand! 31 Algorithm 1 for SS

I hear, I forget. I learn, I remember. I do, I understand! 32 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:

I hear, I forget. I learn, I remember. I do, I understand! 33 Proof of Lemma 1 (cont)

I hear, I forget. I learn, I remember. I do, I understand! 34 Approximation Ratio  Theorem1: Algorithm 1 has an approximation ratio within a factor of 2H n  Proof: We know that the approximation ratio of Set Cover is H n. From Lemma 1, it follows directly that Algorithm 1 is a 2H n factor algorithm for SS

I hear, I forget. I learn, I remember. I do, I understand! 35 Prefix and Overlap  For two string s 1 and s 2, we have:  Overlap(s 1,s 2 ) = the maximum between the suffix of s 1 and the prefix of s 2.  pref(s 1,s 2 ) = the prefix of s 1 that remains after chopping off overlap(s 1,s 2 )  Example:  s 1 = abcbcaa and s 2 = bcaaca, then  overlap(s 1,s 2 ) = bcaa  pref(s 1,s 2 ) = abc  Note: overlap(s 1,s 2 ) ≠ overlap(s 2, s 1 )

I hear, I forget. I learn, I remember. I do, I understand! 36 Is there any better approach?  Now, suppose that in the optimal solution, the strings appear from the left to right in the order: s 1, s 2, …, s n  Define: opt = |pref(s 1,s 2 )| + …+|pref(s n-1,s n )| + |pref(s n,s 1 )| + |overlap(s n,s 1 )| Why overlap(s n,s 1 )? 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 remember. I do, I understand! 37 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 s i )  For each edge i→j, i ≠ j, assign a weight of |pref(s i, s j )|  Example: S={abc, bcd, dab} 1( ) ( )3 2( )

I hear, I forget. I learn, I remember. I do, I understand! 38 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 remember. I do, I understand! 39 How to find a min. weight cycle cover  Corresponding to the prefix graph, construct a bipartite graph H=(X,Y;E) such that:  X = {x 1, x 2, …, x n } and Y = {y 1, y 2, …, y n }  For each i, j (in 1…n), add edge (x i, y j ) of weight |pref(s i,s j )|  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 remember. I do, I understand! 40 How to break the cycle s 11 s 12 s 13

I hear, I forget. I learn, I remember. I do, I understand! 41 A constant factor algorithm Algorithm 2:

I hear, I forget. I learn, I remember. I do, I understand! 42 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 remember. I do, I understand! 43 Approximation Ratio (cont)  Theorem 2: Algorithm 2 has an approximation ratio of 4.  Proof: (see next slide)

I hear, I forget. I learn, I remember. I do, I understand! 45 Modification to 3-Approximation

I hear, I forget. I learn, I remember. I do, I understand! 46 3-Approximation Algorithm  Algorithm 3:

I hear, I forget. I learn, I remember. I do, I understand! 47 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 = {s 1, s 2, …, s n } |ov(u,v)| is edge weight.  Suppose s* is the shortest supper string. Let s 1, …, s n be the strings in the order of appearance from left to right. Then s i, s i+1 must have maximum overlap in s*. Hence s 1, …, s n form a directed Hamiltonian path in G.

I hear, I forget. I learn, I remember. I do, I understand! 50 The Algorithm (via Hamiltonian)

I hear, I forget. I learn, I remember. I do, I understand! 51 A special property v u v’ u’

I hear, I forget. I learn, I remember. I do, I understand! 52 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={ab k, b k+1, b k a}. s* = ab k+1 a. Our obtained solution: ab k ab k+1.

I hear, I forget. I learn, I remember. I do, I understand! 53 Submodular Function

I hear, I forget. I learn, I remember. I do, I understand! 54 What is a submodular function? Consider a finite set E, (called ground set), and a function f : 2 E →Z. The function f is said to be submodular if for any two subsets A and B in 2 E : Example: f(A) = |A| is submodular.

I hear, I forget. I learn, I remember. I do, I understand! 55 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 remember. I do, I understand! 56 Greedy Algorithm 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 remember. I do, I understand! 57 Analyze the Approximation Ratio

I hear, I forget. I learn, I remember. I do, I understand! 60 Alternative Analysis

I hear, I forget. I learn, I remember. I do, I understand! 61 What do we need?

I hear, I forget. I learn, I remember. I do, I understand! 62 What ’ s we need?

I hear, I forget. I learn, I remember. I do, I understand! 63 Actually, this inequality holds if and only if f is submodular and (monotone increasing)

Proof of (1) I hear, I forget. I learn, I remember. I do, I understand! 66

Proof of (2) I hear, I forget. I learn, I remember. I do, I understand! 67

I hear, I forget. I learn, I remember. I do, I understand! 68 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 remember. I do, I understand! 69 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 remember. I do, I understand! 70 Greedy Algorithm

I hear, I forget. I learn, I remember. I do, I understand! 71 A General Problem

I hear, I forget. I learn, I remember. I do, I understand! 72 Greedy Algorithm

I hear, I forget. I learn, I remember. I do, I understand! 73 A General Theorem Remark (Normalized):

I hear, I forget. I learn, I remember. I do, I understand! 75 Proof (cont)  We will prove these following claims:

I hear, I forget. I learn, I remember. I do, I understand! 76 Show the First Claim

I hear, I forget. I learn, I remember. I do, I understand! 79 Show the Second Claim For any integers p > q > 0, we have: $(p – q)/p = \sum_{j=q+1}^p 1/p \le \sum_{j=q+1}^p 1/j$

I hear, I forget. I learn, I remember. I do, I understand! 80 Connected Vertex-Cover  Given a connected graph, find a minimum vertex-cover which induces a connected subgraph.

I hear, I forget. I learn, I remember. I do, I understand! 81  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 remember. I do, I understand! 82 -p-q  Define f(A)= -p(A) –q(A). Then f(A) is submodular and monotone increasing

I hear, I forget. I learn, I remember. I do, I understand! 83 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

I hear, I forget. I learn, I remember. I do, I understand! 84 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.

Approximation Algorithms Greedy Strategies. I hear, I forget. I learn, I remember. I do, I understand! 2 Max and Min  min f is equivalent to max –f.

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Approximation Algorithms Greedy Strategies. I hear, I forget. I learn, I remember. I do, I understand! 2 Max and Min  min f is equivalent to max –f.

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Presentation on theme: "Approximation Algorithms Greedy Strategies. I hear, I forget. I learn, I remember. I do, I understand! 2 Max and Min  min f is equivalent to max –f."— Presentation transcript:

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