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

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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 - PowerPoint PPT Presentation

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• Minimum Routing Cost Spanning Trees (MRCTs)Dean L. ZellerKent State UniversityOctober 4th, 2005

Minimum Routing Cost Spanning Trees (MRCTs)

• Routing Load(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)

• Routing Load, contLet T be a tree

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

Minimum Routing Cost Spanning Trees (MRCTs)

• Routing Load, contRouting load on eX 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)

3

2

7

6

4

1

5

X

Y

e

A

• Routing CostThe 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)

• MRCT ProblemThe 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 DistancesSum of distances between all pairs of vertices in TSum of the routing load for each edge in T times its weightClaim:

Minimum Routing Cost Spanning Trees (MRCTs)

• Proof by Formulasum table columns firstsum table rows first

Minimum Routing Cost Spanning Trees (MRCTs)

• Proof by Formula, cont

Minimum Routing Cost Spanning Trees (MRCTs)

• Proof by Example

Minimum Routing Cost Spanning Trees (MRCTs)

a 5

v1

B

v2

v3

v4

v5

c 3

d 1

b

10

• Proof by Example, contl(T,a) = 214 = 8l(T,b) = 223 = 12l(T,c) = 214 = 8l(T,d) = 214 = 8C(T) = 85 + 1210 + 83 + 81 = 192

Minimum Routing Cost Spanning Trees (MRCTs)

a 5

v1

B

v2

v3

v4

v5

c 3

d 1

b

10

• ApproximationsNave algorithms will find a solution to any problem, given enough timeTime 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)

• TheoremA 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)

• ProofG = (V,E,w)r = median of graph GY = shortest-paths tree rooted at r

Minimum Routing Cost Spanning Trees (MRCTs)

u

r

v

G

• Proof, contd(x,y) is a metricd(x,y)=0 iff x=yd(x,y) = d(y,x)d(x,y) d(x,z) + d(z,y)

Minimum Routing Cost Spanning Trees (MRCTs)

u

r

v

G

• Proof, contSince 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)

• Solution DecompositionAnalysis technique used widely in approximation algorithmsServes as a justification as to why the researchers used this method to find solutionSuppose X is the optimal solutionDecompose X to compose a feasible solution YAccuracy 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)

• Solution Decomposition, contCut a tree at the centroid r. All subtrees will have no more than half of the trees verticesSuppose 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)

• Solution Decomposition, contIf u and v are nodes not in the same branch, thenIn 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)

• Solution Decomposition, contSince

And

It follows that

Thus, Y is a 2-approximation of

Minimum Routing Cost Spanning Trees (MRCTs)

• Time RequirementsIt 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)

• Application Computational BiologyDNA 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)

• Computational Biology, contThe 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)

• Computational Biology, contConsider the following three strings:T C C G A T GC C G G A C GT C G A C G+ + + + +One column and five pairwise matchesTotal of eight matches4 matches2 matches2 matches

Minimum Routing Cost Spanning Trees (MRCTs)

• Computational Biology, contMutations can be added within strings to create more sequence alignmentsT C C - G A T - G- C C G G A - C GT C - - G A - C G+ + + Four column and three pairwise matches15 total matchesGoal: find a minimum-cost mutation path to maximize multiple alignments5 matches5 matches5 matches

Minimum Routing Cost Spanning Trees (MRCTs)

• Computational Biology, contThe nave solution to this problem is O(2nln).Infeasible for all but the smallest of computational biology problems.5 strings, 500 characters in length255005 = 1000000000000000 = 1015By creating a mutation decision tree, an approximation can be created in a fraction of the time.

Minimum Routing Cost Spanning Trees (MRCTs)

• Computational Biology, contDefinition: 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)

• Computational Biology, contA X X _ ZA X _ _ ZA _ X _ ZA Y _ _ ZA Y X X ZA tree with its nodes labeled by a (multi)set of strings.A multiple alignment of those strings that is consistent with the tree.

Minimum Routing Cost Spanning Trees (MRCTs)

5 AYXYZ

AXZ

1

4 AYZ

3 AXXZ

AXZ

2

• Computational Biology, contThe 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)

• BibliographyGusfield, 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)