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e. Minimum Routing Cost Spanning Trees (MRCTs). Dean L. Zeller Kent State University October 4 th , 2005. 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

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### Minimum Routing Cost Spanning Trees (MRCTs)

Dean L. Zeller

Kent State University

October 4th, 2005

• (T,e)

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

Minimum Routing Cost Spanning Trees (MRCTs)

• Let T be a tree

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

Minimum Routing Cost Spanning Trees (MRCTs)

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

Minimum Routing Cost Spanning Trees (MRCTs)

• 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.

Minimum Routing Cost Spanning Trees (MRCTs)

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

Minimum Routing Cost Spanning Trees (MRCTs)

Routing Cost and Sum of Distances

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

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

Claim:

Minimum Routing Cost Spanning Trees (MRCTs)

sum table columns first

sum table rows first

Minimum Routing Cost Spanning Trees (MRCTs)

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 Formula, con’t

Minimum Routing Cost Spanning Trees (MRCTs)

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

Proof by Example

Minimum Routing Cost Spanning Trees (MRCTs)

l(T,a) = 2·1·4 = 8

l(T,b) = 2·2·3 = 12

l(T,c) = 2·1·4 = 8

l(T,d) = 2·1·4 = 8

C(T) = 8·5 + 12·10 + 8·3 + 8·1 = 192

Minimum Routing Cost Spanning Trees (MRCTs)

• 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.

Minimum Routing Cost Spanning Trees (MRCTs)

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

Minimum Routing Cost Spanning Trees (MRCTs)

• G = (V,E,w)

• r = median of graph G

• Y = shortest-paths tree rooted at r

Minimum Routing Cost Spanning Trees (MRCTs)

d(x,y) is a metric…

• d(x,y)=0 iff x=y

• d(x,y) = d(y,x)

• d(x,y)  d(x,z) + d(z,y)

Minimum Routing Cost Spanning Trees (MRCTs)

• 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.

Minimum Routing Cost Spanning Trees (MRCTs)

• 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

Minimum Routing Cost Spanning Trees (MRCTs)

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

• Suppose r is also the centroid of the optimal MRCT

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

Minimum Routing Cost Spanning Trees (MRCTs)

• 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)

Minimum Routing Cost Spanning Trees (MRCTs)

• Since…

• And…

• It follows that…

• Thus, Y is a 2-approximation of

Minimum Routing Cost Spanning Trees (MRCTs)

• 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.

Minimum Routing Cost Spanning Trees (MRCTs)

• 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.

Minimum Routing Cost Spanning Trees (MRCTs)

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).

Minimum Routing Cost Spanning Trees (MRCTs)

• 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 matches

2 matches

Minimum Routing Cost Spanning Trees (MRCTs)

• 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 matches

5 matches

Minimum Routing Cost Spanning Trees (MRCTs)

• 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.

Minimum Routing Cost Spanning Trees (MRCTs)

Definition: 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.

Minimum Routing Cost Spanning Trees (MRCTs)

• A X X _ Z

• A X _ _ Z

• A _ X _ Z

• A Y _ _ Z

• A Y X X Z

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

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

Minimum Routing Cost Spanning Trees (MRCTs)

• 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.

Minimum Routing Cost Spanning Trees (MRCTs)

• 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)