minimum routing cost spanning trees (mrcts)

Minimum Routing Cost Spanning Trees (MRCTs)

Dean L. Zeller

Kent State University

October 4th, 2005

e

Minimum Routing Cost Spanning Trees (MRCTs) Slide 2 of 30

Routing Load

(T,e)

• Serves as a measurement of the number of vertices that will use an edge for routing purposes.


Routing Load, con’t

• Let T be a tree

• Let X and Y be the subtrees created by removing edge e

)(TEe

)(XVx )(YVy

xyeT 2),(


Routing Load, con’t

Routing load on e

X has 3 vertices.

Y has 4 vertices.

Vertex pairs using e for routing:

14, 15, 16, 17, 24, 25, 26, 27, 34, 35, 36, 37, 41, 42, 43, 51, 52, 53, 61, 62, 63, 71, 72, 73 24)4)(3(2),( eA

3

2 6

4

1 5

X Y

e

A

7


Routing Cost

• The routing cost of a tree is the sum of the routing load of each edge times its weight.

• For a tree T with edge length w, C(T) can be computed in O(n) time.

)(

)(),()(TEe

eweTTC


MRCT Problem

• The goal of the MRCT problem is to minimize the overall cost C(T) for all spanning trees among G.


Routing Cost and Sum of Distances

)(,

),(TVvu

T vud “Sum of distances between all pairs of vertices in T”

“Sum of the routing load for each edge in T times its weight”

Claim:)(),(),(

)()(,

ewvuvudTEeTVvu

T

)(

)(),(TEe

eweT


Proof by Formula

)(TC

)(,

),(TVvu

T vud

)(, ),(

)(TVvu vuSPe T

ew

)(),(|)( )(

ewvuSPevTEe TVu

T

)(

)(),(TEe

eweT

sum table columns first

sum table rows first


Proof by Formula, con’tVertex Pairs

v1,v2 v1,v3 v1,v4 … vn-1,vn

e1 0 3 0 0

e2 2 0 2 0

e3 5 0 5 0

…

em 0 0 0 3


Proof by Example

a 5

v1

B v2

v3

v4

v5 c 3

d 1

b 10

dB(v1,v1) = 0

dB(v2,v1) = 10

dB(v3,v1) = 5

dB(v4,v1) = 13

dB(v5,v2) = 11

dB(v1,v2) = 10

dB(v2,v2) = 0

dB(v3,v2) = 15

dB(v4,v2) = 3

dB(v5,v2) = 1

dB(v1,v3) = 5

dB(v2,v3) = 15

dB(v3,v3) = 0

dB(v4,v3) = 18

dB(v5,v3) = 16

dB(v1,v4) = 13

dB(v2,v4) = 3

dB(v3,v4) = 18

dB(v4,v4) = 0

dB(v5,v4) = 4

dB(v1,v5) = 11

dB(v2,v5) = 1

dB(v3,v5) = 16

dB(v4,v5) = 4

dB(v5,v5) = 0

192),(,

vu

B vud


Proof by Example, con’t

a 5

v1

B v2

v3

v4

v5 c 3

d 1

b 10

192)(),()()(

TEe

eweTlTC

l(T,a) = 2·1·4 = 8l(T,b) = 2·2·3 = 12l(T,c) = 2·1·4 = 8l(T,d) = 2·1·4 = 8 C(T) = 8·5 + 12·10 + 8·3 + 8·1 = 192


Approximations

• Naïve algorithms will find a solution to any problem, given enough time

• Time is money – solutions must come quickly.

• Oftentimes a “good enough” solution will serve just as well.

• By choosing a good starting point, one can obtain a good MRCT approximation.


Theorem

• A shortest-paths tree rooted at the median of a graph is a 2-approximation of an MRCT of the graph.


Proof

• G = (V,E,w)

• r = median of graph G

• Y = shortest-paths tree rooted at r

u

r

v

G


Proof, con’t

d(x,y) is a metric…

1) d(x,y)=0 iff x=y

2) d(x,y) = d(y,x)

3) d(x,y) d(x,z) + d(z,y)

u

r

v

G

),(),(),( vrdrudvud YYY

u v

Yu v

Yu v

Y vrdrudvudYC ),(),(),()(

v

Yu

Y vrdnrudn ),(),(

u

Y rudn ),(2


Proof, con’t• Since r is the median of G…

• Dividing by n…

• In a shortest-paths tree…

• And thus…

• Y is a 2-approximation of an MRCT of G.

vu

Gv

G vudvrdn,

),(),(

vu

Gnv

G vudvrd,

1 ),(),(

),(),( vrdvrd GY

vu

Gv

G vudvrdnYC,

),(2),(2)(


Solution Decomposition

• Analysis technique used widely in approximation algorithms

• Serves as a justification as to why the researchers used this method to find solution

• Suppose X is the optimal solution• Decompose X to compose a feasible solution Y• Accuracy of Y depends on how “feasible” is

defined.• Y is a good approximation of X by belonging to a

restricted subset of feasible solutions


Solution Decomposition, con’t

1. Cut a tree at the centroid r. All subtrees will have no more than half of the tree’s vertices

2. Suppose r is also the centroid of the optimal MRCT

3. Construct a shortest-paths tree Y rooted at r. The routing cost of Y will be at most twice that of

T̂

T̂



• If u and v are nodes not in the same branch, then

• In calculating the total distance for all pairs of nodes on will be counted at least n times ( times from v to others, times from others to v)

),(),(),( ˆˆˆ vrdrudvudTTT

T̂ ),(ˆ rvdT

2n

2n



• Since…

• And…

• It follows that…

• Thus, Y is a 2-approximation of

v

TrvdnTC ),()ˆ( ˆ

v

G rvdnYC ),(2)(

)ˆ(2)( TCYC

T̂


Time Requirements

• It was recently discovered that a shortest-paths tree can be constructed in O(nlogn+m) time.

• The routing cost of a tree can be completed in O(n) time.

• Thus, completing the 2-approximation algorithm can be done in O(n2logn+mn) time.


Application – Computational Biology

• DNA is represented by a string of characters involving only A, C, G, and T.

• Similarity of DNA strands is determined through multiple sequence alignments.

• Mutations can be introduced by inserting gaps into the string.


Computational Biology, con’t

The second fact of biological sequence comparison Evolutionarily and functionally related molecular strings can differ significantly throughout much of the string and yet preserve the same three-dimensional structure(s).



• Consider the following three strings:

T C C G A T G

C C G G A C G

T C G A C G

+ ▲ + + + +• One column and five pairwise matches

• Total of eight matches

4 matches

2 matches2 matches



• Mutations can be added within strings to create more sequence alignments

T C C - G A T - G

- C C G G A - C G

T C - - G A - C G

+ ▲ + ▲ ▲ + ▲• Four column and three pairwise matches• 15 total matches• Goal: find a minimum-cost mutation path to

maximize multiple alignments

5 matches

5 matches5 matches


Computational Biology, con’t• The naïve solution to this problem is

O(2nln).

• Infeasible for all but the smallest of computational biology problems. 5 strings, 500 characters in length 255005 = 1000000000000000 = 1015

• By creating a mutation decision tree, an approximation can be created in a fraction of the time.


Computational Biology, con’tDefinition: Let S be a set of strings, and let T be a tree where each node is labeled with a distinct string from S. Then, a multiple alignment M of S is called consistent with T if the induced pairwise alignment of Si and Sj has score D(Si,Sj) for each pair of strings (Si,Sj) that label adjacent nodes in T.

Theorem: For any set of strings S and for any tree T whose nodes are labeled by distinct strings of S, we can efficiently find a multiple alignment M(T) of S that is consistent with T.



3 A X X _ Z

4 A X _ _ Z

5 A _ X _ Z

4 A Y _ _ Z

5 A Y X X Z

A tree with its nodes labeled by a (multi)set of strings.

A multiple alignment of those strings that is consistent with the tree.

AXZ 1

AXZ 2

3 AXXZ

4 AYZ

5 AYXYZ



• The time needed to compute M(T) is dominated by the time to compute k-1 pairwise alignments. If each string has length n, then each pairwise alignment takes time O(n2) and the time to construct M(T) is O(kn2).

• Theorem is considered “folklore” in the algorithm research community.


Bibliography

• Gusfield, Dan. Algorithms on Strings,Trees, and Sequences. Cambridge, 1997.

• Wu, B.Y. & K.M. Chao. Spanning Trees and Optimization Problems. Chapman & Hall/CRC, 2004.

minimum routing cost spanning trees (mrcts)

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