SOFSEM 2007 Weighted Nearest Neighbor Algorithms for the
SOFSEM 2007 Weighted Nearest Neighbor Algorithms for the Graph Exploration Problem on Cycles Eiji Miyano Kyushu Institute of Technology, Japan Joint work with Yuichi Asahiro, Shuichi Miyazaki, and Takuro Yoshimuta 1
Overview o o o Problems n Related problem: Traveling Salesperson Problem (TSP) n Our problem: Online Graph Exploration Problem Simple online algorithms and their performances Our algorithm: Weighted nearest neighbor algorithm Main results n 1. 5 competitive algorithm for special graphs, ie. , cycles n 1. 5 tight example n 1. 25 lower bound for deterministic online algorithms Conclusions 2
One of the most popular problems Traveling Salesperson Problem, TSP (more precisely, the shortest tour problem) Input: We in advance have the complete map of an input graph, i. e. , 8 n the origin node o, n the number of nodes, n the number of edges, n the length of every edge, and n the topology. Goal: o Minimize the total distance of the traveled tour, beginning at the origin o, visiting all the nodes at least once, and finally returning to the origin o. origin 3 o 5 5 4 3 3 7 2 1 3 4 3
One of the most popular problems Traveling Salesperson Problem, TSP (more precisely, the shortest tour problem) Input: We in advance have the complete map of an input graph, i. e. , 8 n the origin node o, n the number of nodes, n the number of edges, n the length of every edge, n the topology. Goal: o Minimize the total distance of the traveled tour, beginning at the origin o, visiting all the nodes at least once, and finally returning to the origin o. origin 3 o 5 5 4 3 3 7 2 1 4 3 3 Total distance = 36 4
Our problem Online Graph Exploration Problem Initial Information: o Only the partial information of the map is given, i. e. , the searcher knows only n n n o searcher 3 origin 5 u the origin o its neighbor nodes the length of edge (o, u) for each neighbor node u ? ? ? By using the partial information, the searcher has to select the next node and visit it. 5
Our problem Online Graph Exploration Problem Initial Information: o Only the partial information of the map is given n the origin o n its neighbor nodes n the length of edge (o, u) for each neighbor node u 3 u origin 5 4 8 Online New Information: o When the searcher visits a node u, its neighbor nodes v’s, and the length of edges (u, v)’s are obtained. ? ? ? 6
Our problem Online Graph Exploration Problem Initial Information: o Only the partial information of map is given n the origin o n its neighbor nodes n the length of edge (o, u) for each neighbor node u Online Information: o When the searcher visits a node u, its neighbor nodes v’s, and the length of edges (u, v)’s are obtained. 3 5 4 8 3 Goal: o 5 3 7 ? ? ? Minimize the total distance of the traveled tour, beginning at the origin o, visiting all the nodes at least once, and returning to the origin o. 7
Quality of an online algorithm Competitive analysis o o Let OPT(G) denote the total length of the shortest exploring tour of G taken by an optimal offline algorithm OPT. Let ALG(G) denote the total length of the exploring tour of G taken by algorithm ALG. We say that ALG is -competitive for a class of graphs G if for all graphs G G. o The ratio is called the competitive o ALG is competitive if ALG is c-competitive for some constant c. ratio. 8
Strategy 1: Nearest Neighbor Algorithm (NN) o always chooses the unvisited new node nearest to the current position. 10 1 origin 2 4 a 2 8 1 b origin o a 3 b NN is Bad [RSL 97] n There is a planar graph G (with n nodes) for which 9
NN is bad There is a planar graph G for which o 2 [RSL 97] origin 1 OPT(G) = 32 NN(G) = 66 2 2 1 1 2 1 3 3 10 3 6 1 2 1 3 2 2 1 2 Generally, OPT(G) = n NN(G) = (nlog n) 10
Strategy 2: Depth-First algorithm (DF) o o Basically, if the current node has new edges, then DF always chooses one of them. Otherwise, DF chooses the nearest new edge. Proposition 1 DF is 1 -competitive (optimal) for trees, i. e. , graphs without loops. origin 2 d o 1 a 7 b 8 c 11
DF is bad o o Basically, if the current node has new edges, then DF always chooses one of them. Otherwise, DF chooses the nearest new edge. Proposition 1’ DF is not competitive even for cycles. 12
DF is bad o o Basically, if the current node has new edges, then DF always chooses one of them. Otherwise, DF chooses the nearest new edge. Proposition 1’ DF is not competitive even for cycles. Proof. origin 1 100 OPT(G) = 6 DF(G) = 103 1 1 o If the graph includes cycles, the problem would become non-trivial. 13
Best previous algorithm Short. Cut algorithm: o proposed by Kalyanasundarama and Pruhs (TCS 130, 1994). o It is shown that Short. Cut achieves the competitive ratio of 16 for planar graphs. Basic strategy of Short. Cut: o the weighted nearest neighbor strategy (WNN). n What is WNN? 14
Weighted Nearest Neighbor Algorithm o Suppose that the searcher is currently on u. Weighted Nearest Neighbor WNN o If d 1 + d 2 < d 3, then the searcher visits y; o Otherwise, the searcher visits v. d 1 y x d 2 explored area Nearest Neighbor NN o sets = 1 o WNN 1 u v d 3 15
Weighted Nearest Neighbor Algorithm o Suppose that the searcher is currently on u. Weighted Nearest Neighbor WNN o If d 1 + d 2 < d 3, then the searcher visits y; o Otherwise, the searcher visits v. 99 y x 50 explored area Nearest Neighbor NN o sets = 1, i. e. , WNN 1 o visits v since 99 +50 > 55 o worst case: its competitive ratio is (log n) u v 55 16
Weighted Nearest Neighbor Algorithm o Suppose that the searcher is currently on u. Weighted Nearest Neighbor WNN o If d 1 + d 2 < d 3, then the searcher visits y; o Otherwise, the searcher visits v. 99 y x 50 explored area Short. Cut [KP 94] o sets = 3, WNN 3 o visits y since 99 + 50 < 3 50 o its competitive ratio is 16 for planar graphs u v 55 17
This paper o o o Focuses the WNN strategy, Applies WNN to cycles, Investigates its ability more intensively. d 1 y x Weighted Nearest Neighbor WNN o If d 1 + d 2 < d 3, then the searcher visits y; o Otherwise, the searcher visits v. d 2 explored area u v d 3 18
Our problem Online Graph Exploration Problem for Cycles Initial Information: o Only the partial information of map is given n the origin o n its neighbor nodes n the length of edge (o, u) for each neighbor node u Online Information: o When the searcher visits a node u, its neighbor nodes v’s, and the length of edges (u, v)’s are obtained. Goal: o Minimize the total distance of the tour, beginning at o, visiting all the nodes at least once, and returning to o. ? ? ? The gray area is a simple path, but the searcher does not know its shape. 19
Summary of Our Results Upper bounds for cycles: o NN (i. e. , WNN 1) achieves the competitive ratio of 1. 5 for cycles. o Our analysis of the 1. 5 -competitive ratio is tight since we can provide an instance for which the bound of 1. 5 is attained. Lower bounds for WNN: o Setting = 1 for WNN is the best for cycles, i. e. , n if 1, the competitive ratio of WNN is at least 1. 5. Lower bound for general algorithms. o No deterministic online algorithm has a competitive ratio less than 1. 25. 20
Cycles o o o Let C = (V, E, l) be a cycle with |V| = n, |E| = n, and edge-length l(e). Let L be the sum of the length of all edges in E. Let lmax be the maximum edge length 1 2 1 4 3 2 21
Optimal Tour o If the graph is a cycle, an optimal tour forms either n a simple cycle including all n edges, or n a U-turn tour including n – 1 different edges. 2 5 4 2 2 3 1 5 4 200 3 1 22
Our Result (1) Theorem 1: WNN 1 is 1. 5 -competitive for cycles. Proof. There are two cases: (Case 1) lmax < L/2 o OPT(G) = L OPT 23
Our Result (1) Theorem 1: WNN 1 is 1. 5 -competitive for cycles. Proof. There are two cases: (Case 1) lmax < L/2 o OPT(G) = L o WNN 1 needs at most L to visit all nodes, and at most L/2 to go back to the origin. WNN 1 1 2 10 1 5 24
Our Result (1) Theorem 1: WNN 1 is 1. 5 -competitive for cycles. Proof. Ttwo cases: (Case 2) lmax >= L/2 o OPT(G) = 2(L – lmax) o WNN 1 needs at most 2(L – lmax) to visit all nodes, and at most (L – lmax) to go back to the origin. 25
Our Result (2) Theorem 2: For any , the competitive ratio of WNN is at least 1. 5 for cycles. Theorem 1: WNN 1 is 1. 5 -competitive for cycles. From Theorems 1 and 2, it can be obtained that o Setting =1 for WNN is the best for cycles. 26
Our Result (2) Theorem 2: For any , the competitive ratio of WNN is at least 1. 5 for cycles. Proof. This theorem is shown by Lemmas 1 and 2: Lemma 1: For 0 < < 1, the competitive ratio of WNN exceeds 1. 5. Lemma 2: For 1 <= , the competitive ratio of WNN is at least 1. 5. 27
Lemma 1: For 0 < < 1, the competitive ratio of WNN exceeds 1. 5. Proof. o Construct a hard cycle C such that n the searcher of WNN goes through the longest edge, and n the searcher goes through the explored edges several times, but n OPT does not go through the longest edge. x o q p 1 p 2 pk y 28
Lemma 1: For 0 < < 1, the competitive ratio of WNN exceeds 1. 5. Proof. o Construct a hard cycle C such that n the searcher of WNN goes through the longest edge, and n the searcher goes through the explored edges several times, but n OPT does not go through the longest edge. x o q p 1 p 2 pk y 29
Lemma 1: For 0 < < 1, the competitive ratio of WNN exceeds 1. 5. Proof. o Construct a hard cycle C such that n the searcher of WNN goes through the longest edge, and n the searcher goes through the explored edges several times, but n OPT does not go through the longest edge. x o q p 1 p 2 pk y 30
Lemma 2: For 1 <= , the competitive ratio of WNN is at least 1. 5. Proof. o Construct a different hard cycle C such that n the searcher of WNN goes through the explored edges many times, but n OPT goes through every edge exactly once. am a 2 a 1 o b 1 b 2 bm 31
Theorems 1 and 2 Theorem 1: WNN 1 is 1. 5 -competitive for cycles. Theorem 2: For any , the competitive ratio of WNN is at least 1. 5. As a result, o Setting =1 for WNN is the best for cycles. 32
Our Result (3) Theorem 3: No online graph exploration algorithm has a competitive ratio less than 1. 25. Proof. Consider two cycles. 1 a d b 3 1 c 3 1 a b 3 1 d c ε 0 33
Summary and Future Work Result 1: Upper and tight bounds for cycles o o NN (i. e. , WNN 1) achieves the competitive ratio of 1. 5 for cycles. Our analysis of the 1. 5 -competitive ratio is tight since we can provide an instance for which the bound of 1. 5 is attained. Result 2: Lower bound for general algorithms o No deterministic online algorithm has a competitive ratio less than 1. 25. Future Work o o Different good online strategy for cycles. Smaller competitive ratio than 16 for planar graphs. Larger lower bound than 1. 25 for general algorithms. Competitive algorithm for general graphs. 34
Summary and Future Work Result 1: Upper and tight bounds for cycles o o NN (i. e. , WNN 1) achieves the competitive ratio of 1. 5 for cycles. Our analysis of the 1. 5 -competitive ratio is tight since we can provide an instance for which the bound of 1. 5 is attained. Result 2: Lower bound for general algorithms o No deterministic online algorithm has a competitive ratio less than 1. 25. Future Work o o Different good online strategy for cycles. Smaller competitive ratio than 16 for planar graphs. Larger lower bound than 1. 25 for general algorithms. Competitive algorithm for general graphs. Thank you. 35
- Slides: 35
