The Primal-Dual Method:
Steiner Forest
Network Design
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Input:
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undirected graph G=(V,E)
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cost function: c:E ! R+
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connectivity requirements
Output:
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minimum cost subgraph satisfying connectivity requirements
Example:
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minimum spanning tree: minimum cost subgraph connecting all
vertices
Steiner Tree
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Input:
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set of terminals T = {t1, …, tk} µ V
Output:
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minimum cost subgraph connecting T
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NP-complete
Algorithm:
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define complete graph H on T
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c(ti, tj) = shortest path in G from ti to tj
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find a minimum spanning tree F in H
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replace each edge in F by the corresponding path in G (delete
cycles)
Steiner Tree: Analysis
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F* - optimal Steiner tree in G
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duplicate each edge in F* - all degrees are even
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open up F* into a (Euler) cycle Z
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Z can be transformed into a spanning tree in H:
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c(Z) =2c(F*)
for any t,t’ 2 T: cZ(t,t’) ¸ cH(t,t’)
Approximation factor = 2
Constrained Steiner Forest Problem
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Input:
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undirected graph G=(V,E)
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cost function: c:E ! R+
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disjoint sets T1, … ,Tk ½ V
Output:
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minimum cost subgraph H such that each set Ti
belongs to a connected component of H
Requirement function f: S ! {0,1}
8S½V
f(S) =1 if there is a set Ti split between S and V-S
f(S) =0 otherwise
Note: f(S)=f(V-S), f is a function on cuts
S
Examples of Requirement Functions
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Steiner tree: for any cut S separating
terminals, f(S)=1
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shortest path between s and t: for any cut S
separating s and t, f(S)=1
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minimum weight perfect matching:
f(S) =1 iff |S| is odd
(At least one vertex is matched outside S)
Linear Programming Formulation
X
minimize
X
maximize
ce x e
S½V
e2 E
subject t o:
X
x(e) ¸ f (S)
S ½V
subject t o:
X
yS · ce
ej e2 ±( S)
Sj e2 ±( S)
xe ¸ 0
yS ¸ 0
Covering LP
f (S)yS
e2 E
Packing LP
S
δ(S)
Integrality Gap Example
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Minimum spanning tree: all cuts are covered
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Graph: cycle on n vertices, unit cost edges
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Optimal integral solution: cost is n-1
Optimal fractional solution:
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for each edge e, x(e)=1/2
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all cuts are covered
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cost is n/2
Gap is 2-1/n
Primal-Dual Algorithm
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yÃ0
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A à ; (x à 0)
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While A is not a feasible solution
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C - set of connected components S 2 (V,A) such
that f(S)=1
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increase yS for all S 2 C until dual constraint for
new e 2 E becomes tight
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A Ã A [ {e}
Remove unnecessary edges from A
Primal-Dual Algorithm
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yÃ0
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A à ; (x à 0)
•
While A is not a feasible solution
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C - set of connected components S 2 (V,A) such
that f(S)=1
•
increase yS for all S 2 C until dual constraint for
new e 2 E becomes tight
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A Ã A [ {e}
6
7
15
10
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Remove unnecessary edges from A
A’ – final solution
40
9
5
Demo of Primal-Dual Algorithm
Analysis (1)
Lemma: The primal solution generated by the algorithm
satisfies the requirement constraints
Proof: The `while’ loop continues till we get a feasible
solution
Lemma: The dual solution generated by the algorithm is
feasible
Proof: Whenever an edge becomes tight, no more cuts
are “packed” into it
Analysis (2)
Goal: charge the increase in primal cost in each
iteration to the increase in dual cost
X
)
X
ce · 2
e2 A 0
yS
S½V
Why is it important that we delete redundant
edges at the end?
Otherwise, we cannot charge tight primal edges to
a single dual variable (the ball) – too many such
edges!
Analysis (3):
X
X
ce · 2 ¢
e2 A 0
yS
S½V
Since every picked edge is t ight ,
0
X
X
ce =
e2 A 0
e2 A 0
Need t o show
S½V
@
e2 A 0
Changing order of summat ion,
0
X
X
@
ce =
1
X
yS A
Sj e2 ±( S)
1
X
X
yS A =
degA 0 (S) ¢yS
Sj e2 ±( S) \ A 0
S½V
X
X
degA 0 (S) ¢yS · 2 ¢
S½V
S½V
yS
X
Analysis (4):
X
degA 0 (S) ¢yS · 2 ¢
S½V
yS
S½V
consider an it erat ion in which act ive balls grew by ¢ . Need t o show:
Ã
!
X
¢ £
degA 0 (S)¢ · 2¢ £ (# of act ive set s)
S½V
t hat is,
Ã
!
X
degA 0 (S)
· 2 £ (# of act ive set s)
S½V
A 0 is a forest and each act ive set has non-zero degree.
in a t ree, average degree · 2 !!