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*. The TSP : NP-Completeness Approximation and Hardness of Approximation. All exact science is dominated by the idea of approximation. -- Bertrand Russell (1872 - 1970). Based upon slides of Dana Moshkovitz, Kevin Wayne and others + some old slides.
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* The TSP : NP-Completeness Approximation and Hardness of Approximation • All exact science is dominated by the idea of approximation. • -- Bertrand Russell(1872 - 1970) Based upon slides of Dana Moshkovitz, Kevin Wayne and others + some old slides *TSP = Traveling Salesman Problem
YES: vertices and faces of a dodecahedron. A related problem: HC • HAM-CYCLE: given an undirected graph G = (V, E), does there exist a simple cycle that contains every node in V.
Hamiltonian Cycle • HAM-CYCLE: given an undirected graph G = (V, E), does there exist a simple cycle that contains every node in V. 1 1' 2 2' 3 3' 4 4' 5 NO: bipartite graph with odd number of nodes.
Directed Hamiltonian Cycle • DIR-HAM-CYCLE: given a digraph G = (V, E), does there exists a simple directed cycle that contains every node in V? • Claim. DIR-HAM-CYCLE P HAM-CYCLE. • Pf. Given a directed graph G = (V, E), construct an undirected graph G' with 3n nodes. Each node splits up into an output node, a regular node and an input node.
Example ain a aout a cin c cout c b bin G b bout
Example ain a aout a cin c cout c b bin G b bout
Directed Hamiltonian Cycle • Claim. G has a Hamiltonian cycle iff G' does. • Pf. • Suppose G has a directed Hamiltonian cycle . • Then G' has an undirected Hamiltonian cycle (same order). • Pf. • Suppose G' has an undirected Hamiltonian cycle '. • ' must visit nodes in G' using one of following two orders: …, B, G, R, B, G, R, B, G, R, B, … …, B, R, G, B, R, G, B, R, G, B, … • Blue nodes in ' make up directed Hamiltonian cycle in G, or reverse of one. ▪
3-SAT Reduces to Directed Hamiltonian Cycle • Claim. 3-SAT P DIR-HAM-CYCLE. • Pf. Given an instance of 3-SAT, we construct an instance of DIR-HAM-CYCLE that has a Hamiltonian cycle iff is satisfiable. • Construction. First, create graph that has 2n Hamiltonian cycles which correspond in a natural way to 2n possible truth assignments.
3-SAT Reduces to Directed Hamiltonian Cycle • Construction. Given 3-SAT instance with n variables xi and k clauses. • Construct G to have 2n Hamiltonian cycles. • Intuition: traverse path i from left to right set variable xi = 1. xi = 1 s x1 x2 x3 t 3k + 3
3-SAT Reduces to Directed Hamiltonian Cycle • Construction. Given 3-SAT instance with n variables xi and k clauses. • For each clause: add a node and 6 edges. clause node clause node s x1 x2 x3 t
3-SAT Reduces to Directed Hamiltonian Cycle • Claim. is satisfiable iff G has a Hamiltonian cycle. • Pf. • Suppose 3-SAT instance has satisfying assignment x*. • Then, define Hamiltonian cycle in G as follows: • if x*i = 1, traverse row ifrom left to right • if x*i = 0, traverse row i from right to left • for each clause Cj , there will be at least one row i in which we are going in "correct" direction to splice node Cj into tour
3-SAT Reduces to Directed Hamiltonian Cycle • Claim. is satisfiable iff G has a Hamiltonian cycle. • Pf. • Suppose G has a Hamiltonian cycle . • If enters clause node Cj , it must depart on mate edge. • thus, nodes immediately before and after Cj are connected by an edge e in G • removing Cj from cycle, and replacing it with edge e yields Hamiltonian cycle on G - {Cj } • Continuing in this way, we are left with Hamiltonian cycle ' inG - {C1 , C2 , . . . , Ck }. • Set x*i = 1 iff ' traverses row i left to right. • Since visits each clause node Cj , at least one of the paths is traversed in "correct" direction, and each clause is satisfied. ▪
3-SAT P Directed HC P HC • Objectives: • To explore the Traveling Salesman Problem. • Overview: • TSP: Examples and Defn. • Is TSPNP-complete? • Approximation algorithm for special cases • Hardness of Approximation in general.
Traveling Salesman Problem A Tour around USA • TSP. Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length D? All 13,509 cities in US with a population of at least 500Reference: http://www.tsp.gatech.edu
Traveling Salesperson Problem • TSP. Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length D? Optimal TSP tour Reference: http://www.tsp.gatech.edu
Traveling Salesperson Problem • TSP. Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length D? 11,849 holes to drill in a programmed logic array Reference: http://www.tsp.gatech.edu
Traveling Salesperson Problem • TSP. Given a set of n cities and a pairwise distance function d(u, v), is there a tour of length D? Optimal TSP tour Reference: http://www.tsp.gatech.edu
TSP • Given a weighted graph G=(V,E) V = Vertices = Cities E = Edges = Distances between cities • Find the shortest tour that visits all cities
TSP 3 2 1 10 3 4 • Instance: A complete weighted undirected graph G=(V,E) (all weights are non-negative). • Problem: To find a Hamiltonian cycle of minimal cost. 5
Naïve Solution • Try all possible tours and pick the minimum • Dynamic Programming Definitely we need something better
Approximation Algorithms • A “good” algorithm is one whose running time is polynomial in the size of the input. • Any hope of doing something in polynomial time for NP-Complete problems?
c-approximation algorithm • The algorithm runs in polynomial time • The algorithm always produces a solution which is within a factor of c of the value of the optimal solution c (1/c) A(x) ≤ Opt(x) ≤ A(x) For all inputs x. OPT(x) here denotes the optimal value of the minimization problem
c-approximation algorithm • The algorithm runs in polynomial time • The algorithm always produces a solution which is within a factor of c of the value of the optimal solution c c A(x) ≥ Opt(x) ≥ A(x) For all inputs x. OPT(x) here denotes the optimal value of the maximization problem
So why do we study Approximation Algorithms • As algorithms to solve problems which need a solution • As a mathematically rigorous way of studying heuristics • Because they are fun! • Because it tells us how hard problems are
Vertex Cover • Any guess on how to design approximation algorithms for vertex cover?
Vertex Cover: Greedy Greedy VC Approx = 8 Opt = 6 Factor 4/3 HW 2 Problem : Example can be extended to O(log n) approximation
A Simpler Approximation Algorithm • Choose an edge e in G • Add both endpoints to the Approximate VC • Remove e from G and all incident edges and repeat. Cover generated is at most twice the optimal cover! • Nothing better than 2-factor known. • If P <> NP, there is no poly-time algorithm that achieves • an approximation factor better than 1.1666 [Has97].
What Next? • We’ll show an approximation algorithm for TSP, with approximation factor 2 for cost functions that satisfy a certain property.
Polynomial Algorithm for TSP? What about the greedy strategy: At any point, choose the closest vertex not explored yet?
The Greedy StrategyFails 10 5 12 2 3 1 0
The Greedy StrategyFails 10 5 12 2 3 1 0
Don’t be greedy Always! Another ExampleGreedy strategy fails -11 -5 -1 0 1 3 7 Even monkeys can do better than this !!!
TSP is NP-hard The corresponding decision problem: • Instance: a complete weighted undirected graph G=(V,E) and a number k. • Problem: to find a Hamiltonian path whose cost is at most k.
TSP is NP-hard Theorem:HAM-CYCLE p TSP. Proof: By the straightforward efficient reduction illustrated below: 1 cn 1 1 cn 1 HAM-CYCLE TSP n = k = |V|
TSP • Is a minimization problem. • We want a 2-approximation algorithm • But only for the case when the cost function satisfies the triangle inequality.
The Triangle Inequality • Cost Function: Let c(x,y) be the cost of going from city x to city y. • Triangle Inequality: In most situations, going from x to y directly is no more expensive than going from x to y via an intermediate place z.
x y z The Triangle Inequality Definition: We’ll say the cost function c satisfies the triangle inequality, if x,y,zV : c(x,z)+c(z,y)c(x,y)
Approximation Algorithm 1. Grow a Minimum Spanning Tree(MST) for G. 2. Return the cycle resulting from a preorder walk on that tree.
Demonstration and Analysis The cost of a minimal Hamiltonian cycle the cost of a MST
Demonstration and Analysis The cost of a preorder walk is twice the cost of the tree
Demonstration and Analysis Due to the triangle inequality, the Hamiltonian cycle is not worse.
The Bottom Line optimal HAM cycle our HAM cycle preorder walk ½· = ½· MST
What About the General Case? • We’ll show TSP cannot be approximated within any constant factor 1 • By showing the corresponding gap version is NP-hard. Inapproximability
gap-TSP[] • Instance: a complete weighted undirected graph G=(V,E). • Problem: to distinguish between the following two cases: There exists a Hamiltonian cycle, whose cost is at most |V|. The cost of every Hamiltonian cycle is more than |V|. YES NO
1 1 1 0 1 |V| |V| 0 +1 0 Instances min cost