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Minimum Routing Cost Tree. Definition For two nodes u and v on a tree , there is a path between them. The sum of all edge weights on this path is called the routing cost of this pair of nodes. MRCT: Minimize the sum of routing costs between all pairs of nodes. - PowerPoint PPT Presentation
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1
Minimum Routing Cost Tree
• Definition– For two nodes u and v on a tree, there is a path
between them.– The sum of all edge weights on this path is
called the routing cost of this pair of nodes.– MRCT: Minimize the sum of routing costs
between all pairs of nodes.
• cf. Minimum Spanning Tree
2
Example 1
• Total sum of all pairs of routing costs of this tree:– RC(a,b)+RC(a,c)+RC(a,d)+RC(b,a)+RC(b,c)+RC(b,d)+RC(c,a)+RC(c,b)
+RC(c,d)+RC(d,a)+RC(d,b)+RC(d,c) = 1+2+2+1+1+1+2+1+2+2+1+2 = 18
a b
dc
2
2
1
1 1
1
b
a c d
1 1 1
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Example 2
• Total sum of all pairs of routing costs of this tree:– RC(a,b)+RC(a,c)+RC(a,d)+RC(b,a)+RC(b,c)+RC(b,d)+RC(c,a)
+RC(c,b)+RC(c,d)+RC(d,a)+RC(d,b)+RC(d,c) = 30
a b
dc
2
2
1
1 1
1
a
b c d
1 2 2
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Centroid of A Tree
• Deleting a centroid of a tree will produce subgraphs such that each subgraph contains no more than n/2 nodes.
a
b m
g
c d
h i
f
e
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Centroid (2)
• Let T be rooted by a centroid m of T.– Every subtree of m contains no more than n/2 nodes.
• Consider any node v in T. The subtree containing v contains no more than n/2 nodes.– At least n/2 paths between some node u (in other subtre
e) and v will pass through m.
m
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Centroid (3)
• Note: we count every pair of nodes twice.– The path from u to v and the path from v to u
will both be counted.
• In the routing cost of this tree, the length of the path from any node u to m will be counted at least 2(n/2) = n times.
u
muRCnTC ),()(
b
a c d
1 1 11218
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1-Star
• A tree with only one internal node.– All other nodes are leaf nodes.
…
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Approximation Algorithm for MRCT
1 For each node i
1-1 Form a 1-star Si rooted at i
1-2 Calculate the routing cost C(Si)
2 Return as the approximate solution
)(min ii SC
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Time Complexity
• Step 1: n iterations– Step 1-1: O(n)– Step 1-2: O(n)
• Step 2: O(n)
• Total: O(n2)
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Performance Ratio• Let C(S) denote the cost of 1-star S.
• Assume T is a minimum routing cost spanning tree and m is a centroid of T.
• Because of the triangular inequality, w(v,m) is less than RC(v,m) of T.
v
mvwnSC ),()22()(
v
mvRCnSC ),(2)(
)(2)( TCSC
…
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Can we find a better ratio?
• Yes. There is a PTAS.
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PTAS for MRCT
• We shall use a k-star to approximate the optimal solution, and show
• When k=1, this is exact the 1-star case we described in the previous section.
• We shall use the case k=3 to illustrate the basic concept of this PTAS.
)(1
3)( TCk
kStarkC
)(2)( TCSC
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k-Star
• A k-star is a tree with exactly k internal nodes.
a bj
i
ha
c
d
e
f
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-separator
• For 0 < ½, a -separator of a graph G is a minimum subgraph of G whose deletion will result in subgraphs, each of which contains no more than n nodes.– For =1/2, the -separator contains only one po
int, i.e. the controid.– We shall choose = 2/(k+3)– For k=3, = 1/3
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Analysis of Routing Cost
• Assume T is a minimum routing cost spanning tree.• Assume T is rooted at its centroid m.• Then at most two subtrees of m contain more than n/3 nodes.• Let a and b be the lowest nodes whose descendants have at least n/3 n
odes.• P: the path from node a to node b.
– P is a (1/3)-separator
a p
q
m h b
rn
o
j
k si g
d f
ec
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Analysis of Routing Cost (2)
• Let dt(v,P) denote the path length from v to P.– This path length must be counted at least 2n/3 times because P is a (1/3)-s
eparator.
• For each edge of P, the edge is counted at least (n/3)(2n/3) times in routing cost.– Each edge on P is counted (n/3 + )(2n/3 - ) times.
– When n/3, the above formula is always greater than or equal to 2n2/9.
• Let w(P) denote the total path length of P, we have
Pv
Pwn
Pvdtn
TC )(9
2),(
3
2)(
2
a b3
n
3
n
9
n
9
n
9
n
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Partition The Nodes• Va consists of nodes whose lowest ancestors on P are a.• Vb consists of nodes whose lowest ancestors on P are b.• Vm consists of nodes whose lowest ancestors on P are m.• Vam consists of nodes whose lowest ancestors on P are bet
ween a and m.• Vbm consists of nodes whose lowest ancestors on P are bet
ween b and m.
a p
q
m h b
rn
o
j
k si g
d f
ec
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Transform to 3-Star (1)
• For each node v in Va, Vb, and Vm, connect v to a, b, and m respectively.
a p
q
m h b
rn
o
j
k si g
d f
ec
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Transform to 3-Star (1)
• For each node v in Va, Vb, and Vm, connect v to a, b, and m respectively.
a p
q
m h b
rn
o
j
k si g
d f
ec
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Transform to 3-Star (2)
a p
q
m h b
rn
o
j
k si g
d f
ec
• Replace P by a path (a,m) and (b,m).
• For each node v in Vam, connect v to a or m, depending on which one is nearer to v.
• For each node v in Vbm, connect v to b or m, depending on which one is nearer to v.
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Transform to 3-Star (2)
• Replace P by a path (a,m) and (b,m).
• For each node v in Vam, connect v to a or m, depending on which one is nearer to v.
• For each node v in Vbm, connect v to b or m, depending on which one is nearer to v.
a
p
q
m
h
b
rn
o
j
k si g
d f
ec
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Comparison with 3-Star• For each edge (v,a), (v,b), (v,m) in the 3-
star, it will be counted (n-1) times.
• For edge (a,m) or (b,m), it is counted no more than (n/2)(n/2)=(n2/4).
)(2
3
)(3
1),(
)(4
1),()1(
),3(
2
2
TC
PwnPvdtn
PwnPvdtn
TstarRC
v
v
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How to find the 3-Star
• Pick 3 nodes.• For i+j+k=n-3, connect i nodes to a, j nodes
to b, and k nodes to c.
• However, if we tries all combinations, it took exponential time.
a b c
… … …
i j k
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Minimum Perfect Bipartite Matching
• In a bipartite graph, there are two sets of nodes, denoted as X and Y.
• In our case, let X = V – {a,b,c}. Y contains i copies of a, j copies of node b, and k copies of c.– If a node v in X is matched with a node u in Y,
v will be connected to u in the 3-star.
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PTAS
• For all (a,b,c) where a, b and c are selected from V do– For all (i,j,k) where i+j+k = n-3 and I,j,j are all positive integers,
do• Perform a perfect minimal bipartite matching to create a 3-star.
• Compute the total routing cost C of this 3-star.
– Choose the 3-star with minimum cost to be our approximate solution.
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Time Complexity
• Total time complexity: O(n8)– O(n3) possible ways to select a,b and c.
– O(n2) possible ways to select I,j and k.
– The perfect minimum bipartite problem can be solved in O(n3) time.
• For k-star, it takes O(n2k+2) time to obtain a solution which is less than or equal to (k+3)/(k+1) of the optimal solution.– Error ratio: 2/(k+1)
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Mid-Term
• April 28th, Thursday– 9:00-12:00– No class on April 26th
• Closed Book
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Presentation
• Each student chooses a journal paper about approximation algorithm and gives a 40-minute presentation.