A Polynomial Time Algorithm for Obtaining Minimum Edge

8/14/2019 A Polynomial Time Algorithm for Obtaining Minimum Edge

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A Polynomial Time Algorithm for

Obtaining Minimum Edge Ranking onTwo-Connected Outerplanar Graphs

Shin-ichi Nakayama, Shigeru MasuyamaInformation Processing Letters 103 (2007) 216-221

Adviser: Yue-Li Wang

Speaker: Ying-Jhih Chen


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Outerplanar Graph

Embedding in the plane

Such that every vertex appears on theboundary of the exterior face.

Lie on a fixed circle


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Outerplanar Graph

Example:


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Edge Ranking

An edge ranking of G is a labeling r from the edges ofG to the positive integers such that foreachpath

between any two edges eu andev, eu ev, with r(eu)= r(

ev), there exists at least one edge

ew on the

path with r(ew) > r(eu) = r(ev).

The value r(ev) of an edge evis called the rank of edgeev.


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Minimum Edge Ranking

An edge ranking ofGis minimum if thelargest rankk assigned is the smallestamong all rankings ofG. Such rankk is

called the edge ranking numberofG.Denoted bye(G).


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Minimum Edge Ranking Graph

Example:


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Abstract

Minimum edge ranking problem is NP-hard and nopolynomial time algorithm for solving it is knownfor non-trivial classes of graphs other than the classof trees.

On a general graph G, a relation between a minimumedge ranking ofG and its minimal cuts, whichensures that we can obtain apolynomial timealgorithm for obtaining minimum edge ranking ofa given graph G ifminimal cutsfor each subgraphof G can be found in polynomial time and thenumber of those is polynomial.


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Minimal Cut

G(E\E) denotes the subgraph obtained from G bydeleting the edges inE from E.

A set C E is said to be a cut of G if G(E\C) isdisconnected.

A cut C E of G is said to be a minimal cut of G, ifnoproper subsetof C is a cutofG.

AsetS2 is a proper subset of another set S1 if every

element in S2 is in S1 and S1 has some elementswhich are not in S2.


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Lemma 1

Suppose that edges of G are labeled with an edgeranking. Let k be the largest rank assigned to morethan one edge. After removing edges whose rank islarger than k from G, the resulting graph G is not

connected.


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Lemma 2

Let G = (V ,E) be a connected graph and be theclass of all minimal cuts of G. The following

formula holds with respect to a minimum edgeranking.


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Proof Lemma 2

let be k. We assign a unique rankchosen from {k+1, k+2,. . . , k +|C|} to each edge ofC.

Example:( |C|=2 )

Thus

K+1

K+2


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Proof Lemma 2

Lete(G) l be the largest rank among ranks assigned to morethan one edge. Let C be a set of edges with rankse(G),e(G) 1, . . . , e(G) l + 1. Then, |C| = l.By Lemma 1, G(E\C) is dividedinto at least two connectedsubgraphs, but C may not be a minimal cut.

Fortunately, however, a minimal cut C is obviously included inC. Let |C| = i ( l). We reassign labels e(G),e(G) 1, . . . ,e(G) i+ 1 to edges in C ande(G) i,e(G) i 1, . . . , e(G) l + 1 to edges inC\C. Note that, by the above process, the labels of edges inE\Cassigned by the minimum edge ranking ofG are not changed. As labels in

Care mutually different, even if we reassign labels to edges in C by the above manner, the resulting labeling is a minimum edgeranking ofG.


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Proof Lemma 2

As C is a minimalcut, G(E\C) is divided into twoconnected subgraphs , .

Each edge of and has a rank among 1, . . . ,

e(G) |C|. Hence, the largest rank ofedges of

and is not greater thane(G) |C|.Consequently,

Thus


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Example

|C|=i=3 , |C*|=l=5

By lemma 1, G(E,C*) divide into at least two connected subgraphs.

C*

e(G)

e(G)-1

e(G)-2

e(G)-3

e(G)-5

e(G)-5

e(G)-4

C


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Half-cycle Subgraph


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Anti-half-cycle subgraph


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Find Minimum Edge Ranking

Takes timesExample:

PS. 46 = 4096

1 2

34


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Find All Minimal Cuts That Divide G into Two MaximalConnected Subgraphs

We first find all minimal cuts: c(k,l,G[1,1;1,4])c(1,1,G[1,1;1,4])=2

12

34


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Find c(1,2,G[1,1;1,4])

c(1,2,G[1,1;1,4])=3

1 2

34


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Find All Minimal Cuts: c(k,l,G[1,1;1,4])

Thus:c(1,3,G[1,1;1,4])=3 , c(1,4,G[1,1;1,4])= c(2,1,G[1,1;1,4])= , c[2,2,G[1,1;1,4])=3c(2,3,G[1,1;1,4])=3, c[2,4,G[1,1;1,4])=2c(3,1,G[1,1;1,4])=3, c(3,2,G[1,1;1,4])= c(3,3,G[1,1;1,4])=2, c(3,4,G[1,1;1,4])=3c(4,1,G[1,1;1,4])=3, c(4,2,G[1,1;1,4])=2c(4,3,G[1,1;1,4])= , c(4,4,G[1,1;1,4])=3The number ofc(k, l,G), k, l = 1, . . . ,n, is O(n^2). Then, as

each c(k, l,G) can be found in O(n) time, this processtakes O(n^3) time in total.


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Set Default Value

e(G[1,1;1,1])=0,e(G[2,2;2,2])=0e(G[3,3;3,3])=0,e(G[4,4;4,4])=0e(G[1,1;1,2])=1,e(G[2,2;2,3])=1

e(G[3,3;3,4])=1,e(G[4,4;4,1])=1


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Bottom Up

e(G[1,1;2,2])=e(G[1,1;1,2])=1e(G[1,1;3,3])= (its disconnected graph itself)

e(G[4,4;1,1])=e(G[4,4;4,1])=1

e(G[1,1;4,4])=e(G[4,4;1,1])=1e(G[2,2;4,4]): (C=1)=min{|C|+max{G[2,2;2,2],G[4,4;4,4]}}=min{1+max{0,0}}=min{1+0}=1

e(G[4,4;2,2])=e(G[2,2;4,4])=1

1 2

2

4


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Finde(G[1,1;1,3])

e(G[1,1;1,3])=min{|C|+max{G[3,3;3,3],G[1,1;1,2]},|C|+max{G[1,1;1,1],G[2,2;2,3]}}=min{1+max{0,1} ,1+max{0,1} }=min{1+1,1+1}=min{2,2}=2

1 2

3

1 2

3

1 2

3


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Finde(G[2,2;2,4])

e(G[2,2;2,4])=min{|C|+max{G[2,2;2,2],G[3,3;3,4]} ,|C|+max{G[3,3;3,3],G[4,4;2,2]} , |C|+max{G[4,4;4,4],G[2,2;2,3]}}=min{2+max{0,1},2+max{0,1},2+max{0,1}}=min{2+1,2+1,2+1}=min{3,3,

3}=3

3

2

4 2

344

2

3

43

2


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e(G[x,y;z,w])

e(G[4,4;4,2])=3e(G[1,2;4,4])=min{|C|+max{G[1,1;1,1],G[2,2;4,4]} , |

C|+max{G[2,2;2,2],G[4,4;4,1]} , |C|+max{G[4,4;4,4],G[1,1;1,2]}}=min{2+1,2+1,2+1}=3


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Sorte(G[x,y;z,w])

Sorte(G[x,y;z,w]) according to increasing numberof vertices ofG[x,y;z,w].

Can be executed using thebucket sort and thenumber of data is O(n^4), thus its execution takesO(n^4) time.

Bucket sort, orbin sort, is a sorting algorithm thatworks by partitioning an array into a finite number

of buckets. bucket sort is not a comparison sort.Under certain conditions for input data the bucketsort may run in linear time ((n)).


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Finds a Minimum Edge Ranking on a 2-connected outerplanargraph

Formula:

A graph is said to be K-connected if there does not

exist a set of K-1 vertices whose removal disconnectsthe graph.


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Add Dummy Edge Case (i)

Case (i) The case where the degree of either vertex xor y of a half-cycle subgraph G[x, x;x, y] is 1

Example:

y

x

Dummy Edge


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Add Dummy Edge Case(ii)

Case (ii) The case where the degree of at least one ofx, y, z, w of an anti-half-cycle subgraphG[x,y;z,w] is 1

x y

zw

Dummy Edge


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Time Complexity

For O(n^4) subgraphs G[x,y;z,w], x, y, z,w = 1, . . . ,n, each subgraph is calculated for at mostO(n^2)cuts.

This step can be executed in O(n^6) time.


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Final Step

Final step: Outpute(G[1, 1;1,n]) that is a minimumedge ranking of G.

This step can be done in O(1) time.

As a whole, it takes time.


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Finde(G[1,1;1,4])

e(G[1,1;1,4])=min{|c(4,2,G[1,1;1,4])|+max{e(G[1,2;4,4]), e(G[3,3;3,3])} , |c(2,3,G[1,1;1,4])|+max{e(G[4,4;4,1]),e(G[2,2;2,3])} , }=min{2+max{3,0},3+max{1,1},

}=min{2+3,3+1,}=min{5,4,}=4


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Figure

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1

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Minimum Edge Ranking Graph

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(1)

(1)

(4)

(3)

(2)

A Polynomial Time Algorithm for Obtaining Minimum Edge

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