1

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

2

Overview

Problems Related problem: Traveling Salesperson Problem (TSP) Our problem: Online Graph Exploration Problem

Simple online algorithms and their performances Our algorithm: Weighted nearest neighbor algorithm Main results

1.5 competitive algorithm for special graphs, ie., cycles 1.5 tight example 1.25 lower bound for deterministic online algorithms

Conclusions

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., the origin node o, the number of nodes, the number of edges, the length of every edge, and the topology.

Goal: Minimize the total distance of the traveled

tour, beginning at the origin o, visiting allthe nodes at least once, and finally returningto the origin o.

origin3

3

4

2

3

7

4

3

5

5

1

8

4

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., the origin node o, the number of nodes, the number of edges, the length of every edge, the topology.

Goal: Minimize the total distance of the traveled

tour, beginning at the origin o, visiting allthe nodes at least once, and finally returningto the origin o.

origin3

3

4

2

3

7

4

3

5

5

1

Total distance = 36

8

3

5

Our problem

Online Graph Exploration ProblemInitial Information: Only the partial information of

the map is given, i.e.,the searcher knows only 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 nodeand visit it.

origin

u

35

???

searcher

6

Our problem

Online Graph Exploration ProblemInitial Information: Only the partial information of

the map is given the origin o its neighbor nodes the length of edge (o, u) for

each neighbor node u

Online New Information: When the searcher visits a node u,

its neighbor nodes v’s, and the lengthof edges (u, v)’s are obtained.

origin

u

35

4

8

???

7

Our problem

Online Graph Exploration ProblemInitial Information: Only the partial information of

map is given the origin o its neighbor nodes the length of edge (o, u) for

each neighbor node uOnline Information: When the searcher visits a node u,

its neighbor nodes v’s, and the lengthof edges (u, v)’s are obtained.

Goal: Minimize the total distance of the traveled tour,

beginning at the origin o, visiting all the nodesat least once, and returning to the origin o.

35

4

8 3

7

3

5

???

8

Quality of an online algorithm

Competitive analysis 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.

The ratio is called the competitive ratio.

ALG is competitive if ALG is c-competitive for some constant c.

)(OPT)(ALG GG

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Strategy 1:

Nearest Neighbor Algorithm (NN) always chooses the unvisited new node nearest to the

current position.

NN is Bad [RSL97] There is a planar graph G (with n nodes) for which

origin

1

24

origin3

2

81

10

ab

a

b

)(log)(

)(n

GOPT

GNN

10

NN is bad

There is a planar graph G for which

[RSL97]

)(log)(

)(n

GOPT

GNN

origin2

2

2

2

2

2

2

11

11

1

1

1

1

6

6

2

103

3

3

3

OPT(G) = 32NN(G) = 66

Generally,OPT(G) = nNN(G) = (nlog n)

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Strategy 2:

Depth-First algorithm (DF) 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.

a

b

o

d

1origin

c

782

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DF is bad

Basically, if the current node has new edges, thenDF always chooses one of them.

Otherwise, DF chooses the nearest new edge.

Proposition 1’

DF is not competitive even for cycles.

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DF is bad

Basically, if the current node has new edges, thenDF always chooses one of them.

Otherwise, DF chooses the nearest new edge.

Proposition 1’

DF is not competitive even for cycles.

Proof.

If the graph includes cycles, the problem would become non-trivial.

1

1 100

1

origin

OPT(G) = 6DF(G) = 103

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Best previous algorithm

ShortCut algorithm: proposed by Kalyanasundarama and Pruhs (TCS 130, 199

4). It is shown that ShortCut achieves the competitive ratio of 1

6 for planar graphs.

Basic strategy of ShortCut: the weighted nearest neighbor strategy (WNN).

What is WNN?

16)graphplanar(OPT

)graphplanar(ShortCut

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Weighted Nearest Neighbor Algorithm

Suppose that the searcher is currently on u.

Weighted Nearest Neighbor WNN If d1 + d2 < d3, then

the searcher visits y; Otherwise,

the searcher visits v.

Nearest Neighbor NN sets = 1 WNN1

x y

u v

explored area

d3

d1

d2

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Weighted Nearest Neighbor Algorithm

Suppose that the searcher is currently on u.

Weighted Nearest Neighbor WNN If d1 + d2 < d3, then

the searcher visits y; Otherwise,

the searcher visits v.

Nearest Neighbor NN sets = 1, i.e., WNN1

visits v since 99 +50 > 55 worst case: its competitive ratio is (log n)

x y

u v

explored area

55

99

50

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Weighted Nearest Neighbor Algorithm

Suppose that the searcher is currently on u.

Weighted Nearest Neighbor WNN If d1 + d2 < d3, then

the searcher visits y; Otherwise,

the searcher visits v.

ShortCut [KP94] sets = 3, WNN3

visits y since 99 + 50 < 3 50 its competitive ratio is 16 for planar graphs

x y

u v

explored area

55

99

50

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This paper

Focuses the WNN strategy, Applies WNN to cycles, Investigates its ability more intensively.

Weighted Nearest Neighbor WNN If d1 + d2 < d3, then

the searcher visits y; Otherwise,

the searcher visits v.

x y

u v

explored area

d3

d1

d2

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Our problem

Online Graph Exploration Problem for CyclesInitial Information: Only the partial information of

map is given the origin o its neighbor nodes the length of edge (o, u) for

each neighbor node uOnline Information: When the searcher visits a node u,

its neighbor nodes v’s, and the lengthof edges (u, v)’s are obtained.

Goal: Minimize the total distance of the tour,

beginning at o, visiting all the nodesat least once, and returning to o.

???

The gray area is a simple path,but the searcher does not knowits shape.

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Summary of Our Results

Upper bounds for cycles: NN (i.e., WNN1) 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.

Lower bounds for WNN: Setting = 1 for WNN is the best for cycles, i.e.,

if 1, the competitive ratio of WNN is at least 1.5.

Lower bound for general algorithms. No deterministic online algorithm has a competitive ratio less

than 1.25.

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Cycles

Let C = (V, E, l) be a cycle with |V| = n, |E| = n, andedge-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

3

4

2

13

242311

L

4max l

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Optimal Tour

If the graph is a cycle, an optimal tour forms either a simple cycle including all n edges, or a U-turn tour including n – 1 different edges.

4

3 1

2

52

Ll

L

l

21

max

max

17

5

17

)(

LGOPT

4

3 1

200

52

Ll

L

l

21

max

max

215

200

30

)(2)( max

lLGOPT

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Our Result (1)

Theorem 1:

WNN1 is 1.5-competitive for cycles.

Proof. There are two cases:

(Case 1) lmax < L/2 OPT(G) = L

OPT

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Our Result (1)

Theorem 1:

WNN1 is 1.5-competitive for cycles.

Proof. There are two cases:

(Case 1) lmax < L/2 OPT(G) = L WNN1 needs at most L to visit all nodes,

and at most L/2 to go back to the origin.

L

LLWNN

23

21

1

5.12

323

L

L

WNN1

21

1

5

10

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Our Result (1)

Theorem 1:

WNN1 is 1.5-competitive for cycles.

Proof. Ttwo cases:

(Case 2) lmax >= L/2

OPT(G) = 2(L – lmax)

WNN1 needs at most 2(L – lmax) to visitall nodes, and at most (L – lmax) to goback to the origin.

)(3

)()()(

max

maxmaxmax1

lL

lLlLlLWNN

5.12

3

)(2

)(3

max

max

lL

lL

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Our Result (2)

Theorem 2:

For any , the competitive ratio of WNN is at least 1.5 for cycles.

Theorem 1:

WNN1 is 1.5-competitive for cycles.

From Theorems 1 and 2, it can be obtained that Setting =1 for WNN is the best for cycles.

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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.

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Lemma 1

Lemma 1:

For 0 < < 1, the competitive ratio of WNN exceeds 1.5.

Proof. Construct a hard cycle C such that

the searcher of WNN goes through the longest edge, and

the searcher goes through the explored edges several times, but

OPT does not go through thelongest edge.

ox q p1 p2 y

maxl

pk

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Lemma 1

Lemma 1:

For 0 < < 1, the competitive ratio of WNN exceeds 1.5.

Proof. Construct a hard cycle C such that

the searcher of WNN goes through the longest edge, and

the searcher goes through the explored edges several times, but

OPT does not go through thelongest edge.

ox q p1 p2 y

maxl

pk

30

Lemma 1

Lemma 1:

For 0 < < 1, the competitive ratio of WNN exceeds 1.5.

Proof. Construct a hard cycle C such that

the searcher of WNN goes through the longest edge, and

the searcher goes through the explored edges several times, but

OPT does not go through thelongest edge.

ox q p1 p2 y

maxl

pk

5.11

2

8)1(

8)2(

L

L

OPT

WNN

Ll 21

max

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Lemma 2

Lemma 2:

For 1 <= , the competitive ratio of WNN is at least 1.5.

Proof. Construct a different hard cycle C such that

the searcher of WNN goes through the explored edges

many times, but OPT goes through every edge

exactly once.

o b1a1a2 b2am bm

)1(5.1

)12(2

))1(222(32

2

o

m

OPT

WNNm

m

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Theorems 1 and 2

Theorem 1:

WNN1 is 1.5-competitive for cycles.

Theorem 2:

For any , the competitive ratio of WNN is at least 1.5.

As a result, Setting =1 for WNN is the best for cycles.

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Our Result (3)

Theorem 3:

No online graph exploration algorithm has a competitive ratio less than 1.25.

Proof. Consider two cycles.

1

3 3

1 1

3 ε0

1a a

b

c

d b

c

d

25.18

10

OPT

ALG

25.124

5

0

0

OPT

ALG

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Summary and Future Work

Result 1: Upper and tight bounds for cycles NN (i.e.,WNN1) 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 No deterministic online algorithm has a competitive ratio less than 1.25.

Future Work 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.

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Summary and Future Work

Result 1: Upper and tight bounds for cycles NN (i.e.,WNN1) 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 No deterministic online algorithm has a competitive ratio less than 1.25.

Future Work 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.