Iterative Randomized Rounding
Thomas Rothvoß
Department of Mathematics, M.I.T.
Cargèse 2011
Joint work with Jaroslaw Byrka,
Fabrizio Grandoni and Laura Sanità
What is Iterative Randomized Rounding?
Set Cover:
◮
◮
Input: Sets S1 , . . . , Sm over elements 1, . . . , n; cost c(Si )
P
S
Goal: minI⊆[m] { i∈I ci | i∈I Si = [n]}
What is Iterative Randomized Rounding?
Set Cover:
Input: Sets S1 , . . . , Sm over elements 1, . . . , n; cost c(Si )
P
S
◮ Goal: minI⊆[m] {
i∈I ci |
i∈I Si = [n]}
Standard LP:
◮
min
m
X
c(Si ) · xi
i=1
X
xi ≥ 1 ∀j ∈ [n]
i:j∈Si
xi ≥ 0 ∀i ∈ [m]
What is Iterative Randomized Rounding?
Set Cover:
Input: Sets S1 , . . . , Sm over elements 1, . . . , n; cost c(Si )
P
S
◮ Goal: minI⊆[m] {
i∈I ci |
i∈I Si = [n]}
Standard LP:
◮
min
m
X
c(Si ) · xi
i=1
X
xi ≥ 1 ∀j ∈ [n]
i:j∈Si
xi ≥ 0 ∀i ∈ [m]
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
What is Iterative Randomized Rounding?
Set Cover:
Input: Sets S1 , . . . , Sm over elements 1, . . . , n; cost c(Si )
P
S
◮ Goal: minI⊆[m] {
i∈I ci |
i∈I Si = [n]}
Standard LP:
◮
min
m
X
c(Si ) · xi
i=1
X
xi ≥ 1 ∀j ∈ [n]
i:j∈Si
xi ≥ 0 ∀i ∈ [m]
Known:
◮
Integrality gap is Θ(ln n)
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
What is Iterative Randomized Rounding?
Set Cover:
Input: Sets S1 , . . . , Sm over elements 1, . . . , n; cost c(Si )
P
S
◮ Goal: minI⊆[m] {
i∈I ci |
i∈I Si = [n]}
Standard LP:
◮
min
m
X
c(Si ) · xi
i=1
X
xi ≥ 1 ∀j ∈ [n]
i:j∈Si
1/2
1/2
1/2
1/2
1/2
xi ≥ 0 ∀i ∈ [m]
Known:
◮
Integrality gap is Θ(ln n)
◮
Suppose |Si | ≤ k. Then gap is Θ(ln k).
1/2
1/2
1/2
1/2
Iterative randomized rounding algorithm:
(1) FOR t = 1 TO ∞
(2) Solve LP → x
(3) FOR ALL i: Buy Si with prob. xi (remove covered el.)
(4) IF no elemente left THEN RETURN all bought sets
Iterative randomized rounding algorithm:
(1) FOR t = 1 TO ∞
(2) Solve LP → x
(3) FOR ALL i: Buy Si with prob. xi (remove covered el.)
(4) IF no elemente left THEN RETURN all bought sets
iteration t = 1
1
2
x
1
2
1
2
1
2
1
2
bought
sets
1
2
1
2
1
2
1
2
iteration t = 2
Iterative randomized rounding algorithm:
(1) FOR t = 1 TO ∞
(2) Solve LP → x
(3) FOR ALL i: Buy Si with prob. xi (remove covered el.)
(4) IF no elemente left THEN RETURN all bought sets
iteration t = 1
1
2
x
1
2
1
2
1
2
1
2
bought
sets
1
2
1
2
1
2
1
2
iteration t = 2
Iterative randomized rounding algorithm:
(1) FOR t = 1 TO ∞
(2) Solve LP → x
(3) FOR ALL i: Buy Si with prob. xi (remove covered el.)
(4) IF no elemente left THEN RETURN all bought sets
iteration t = 1
1
2
x
iteration t = 2
1
1
1
2
1
2
1
2
1
2
bought
sets
1
2
1
2
1
2
1
2
1
Iterative randomized rounding algorithm:
(1) FOR t = 1 TO ∞
(2) Solve LP → x
(3) FOR ALL i: Buy Si with prob. xi (remove covered el.)
(4) IF no elemente left THEN RETURN all bought sets
iteration t = 1
1
2
x
iteration t = 2
1
1
1
2
1
2
1
2
1
2
bought
sets
1
2
1
2
1
2
1
2
1
Iterative randomized rounding algorithm:
(1) FOR t = 1 TO ∞
(2) Solve LP → x
(3) FOR ALL i: Buy Si with prob. xi (remove covered el.)
(4) IF no elemente left THEN RETURN all bought sets
iteration t = 1
1
2
x
iteration t = 2
1
1
1
2
1
2
1
2
1
2
bought
sets
1
2
1
2
1
2
1
2
1
SOLUTION
Analysis
E[AP X]
=
X
E[OP Tf in step t]
t≥1
I∗
Analysis
◮
Let I ∗ be optimal Set Cover solution.
E[AP X]
=
X
E[OP Tf in step t]
t≥1
≤
X
t≥1
E[c(I t )]
I∗
Analysis
◮
Let I ∗ be optimal Set Cover solution.
E[AP X]
=
X
E[OP Tf in step t]
t≥1
≤
X
E[c(I t )]
b
b
t≥1
I∗
b
b
b
covered after t it
Analysis
◮
◮
Let I ∗ be optimal Set Cover solution.
I t := {i ∈ I ∗ | not yet all elements in Si covered} (←
feasible in step t)
E[AP X]
=
X
E[OP Tf in step t]
t≥1
≤
X
E[c(I t )]
b
b
t≥1
It
b
b
b
covered after t it
Analysis
◮
◮
Let I ∗ be optimal Set Cover solution.
I t := {i ∈ I ∗ | not yet all elements in Si covered} (←
feasible in step t)
E[AP X]
=
X
E[OP Tf in step t]
t≥1
≤
X
E[c(I t )]
=
X
E[# iterations Si is in I t ] · c(Si ) b
b
t≥1
i∈I ∗
b
It
b
b
covered after t it
Analysis
◮
◮
Let I ∗ be optimal Set Cover solution.
I t := {i ∈ I ∗ | not yet all elements in Si covered} (←
feasible in step t)
◮
Pr[element j not covered after ln(2k) it.] ≤ e− ln(2k) =
◮
Pr[not all el. in Si covered after ln(2k) it.] ≤ k ·
E[AP X]
=
X
1
2k
=
1
2k
1
2
E[OP Tf in step t]
t≥1
≤
X
E[c(I t )]
=
X
E[# iterations Si is in I t ] ·c(Si )
|
{z
}
b
t≥1
i∈I ∗
≤O(ln(k))
b
It
b
b
b
covered after t it
Analysis
◮
◮
Let I ∗ be optimal Set Cover solution.
I t := {i ∈ I ∗ | not yet all elements in Si covered} (←
feasible in step t)
◮
Pr[element j not covered after ln(2k) it.] ≤ e− ln(2k) =
◮
Pr[not all el. in Si covered after ln(2k) it.] ≤ k ·
E[AP X]
=
X
1
2k
=
1
2k
1
2
E[OP Tf in step t]
t≥1
≤
X
E[c(I t )]
=
X
E[# iterations Si is in I t ] ·c(Si )
|
{z
}
b
t≥1
i∈I ∗
=
≤O(ln(k))
b
It
b
b
b
O(ln k) · OP T
covered after t it
Steiner Tree
Given:
◮
undirected graph G = (V, E)
◮
cost c : E → Q+
◮
terminals R ⊆ V
Find: Min-cost Steiner tree, spanning R.
OP T := min{c(S) | S spans R}
terminals
W.l.o.g.: c is metric.
Steiner Tree
Given:
◮
undirected graph G = (V, E)
◮
cost c : E → Q+
◮
terminals R ⊆ V
Find: Min-cost Steiner tree, spanning R.
OP T := min{c(S) | S spans R}
Steiner node
W.l.o.g.: c is metric.
Steiner tree
Spanning tree
terminal spanning tree
Min-cost terminal spanning tree (MST):
Spanning tree
terminal spanning tree
Min-cost terminal spanning tree (MST):
◮ Can be computed in poly-time.
Spanning tree
terminal spanning tree
Min-cost terminal spanning tree (MST):
◮ Can be computed in poly-time.
◮
Costs ≤ 2 · OP T .
Known results for Steiner tree:
Approximations:
◮
2-apx (minimum spanning tree heuristic)
◮
1.83-apx [Zelikovsky ’93]
◮
1.667-apx [Prömel & Steger ’97]
◮
1.644-apx [Karpinski & Zelikovsky ’97]
◮
1.598-apx [Hougardy & Prömel ’99]
◮
1.55-apx [Robins & Zelikovsky ’00]
Known results for Steiner tree:
Approximations:
◮
2-apx (minimum spanning tree heuristic)
◮
1.83-apx [Zelikovsky ’93]
◮
1.667-apx [Prömel & Steger ’97]
◮
1.644-apx [Karpinski & Zelikovsky ’97]
◮
1.598-apx [Hougardy & Prömel ’99]
◮
1.55-apx [Robins & Zelikovsky ’00]
Hardness:
◮
NP-hard even if edge costs ∈ {1, 2} [Bern & Plassmann ’89]
◮
no <
96
95 -apx
unless NP = P [Chlebik & Chlebikova ’02]
Our results:
Theorem
There is a polynomial time 1.39-approximation.
◮
LP-based! (Directed-Component Cut Relaxation)
◮
Algorithmic framework: Iterative Randomized Rounding
Our results:
Theorem
There is a polynomial time 1.39-approximation.
◮
LP-based! (Directed-Component Cut Relaxation)
◮
Algorithmic framework: Iterative Randomized Rounding
Theorem
The Directed-Component Cut Relaxation has an integrality gap
of at most 1.55.
◮
First < 2 bound for any LP-relaxation.
Bi-directed cut relaxation
Bi-directed cut relaxation
◮
Pick a root r ∈ R
root r
Bi-directed cut relaxation
◮
◮
Pick a root r ∈ R
Bi-direct edges
root r
Bi-directed cut relaxation
◮
◮
Pick a root r ∈ R
Bi-direct edges
min
X
c(e)ze
(BCR)
ze ≥ 1
∀U ⊆ V \ {r} : U ∩ R 6= ∅
ze ≥ 0
∀e ∈ E.
root r
e∈E
X
e∈δ+ (U )
U
Bi-directed cut relaxation
◮
◮
Pick a root r ∈ R
Bi-direct edges
min
X
c(e)ze
(BCR)
ze ≥ 1
∀U ⊆ V \ {r} : U ∩ R 6= ∅
ze ≥ 0
∀e ∈ E.
root r
ze =
e∈E
X
e∈δ+ (U )
U
1
2
Bi-directed cut relaxation
◮
◮
Pick a root r ∈ R
Bi-direct edges
min
X
c(e)ze
(BCR)
ze ≥ 1
∀U ⊆ V \ {r} : U ∩ R 6= ∅
ze ≥ 0
∀e ∈ E.
root r
ze =
e∈E
X
e∈δ+ (U )
Theorem (Edmonds ’67)
R = V ⇒ BCR integral
◮
◮
Integrality gap ≤ 4/3 for quasi-bipartite graphs
[Chakrabarty, Devanur, Vazirani ’08]
Integrality gap ∈ [1.16, 2]
U
1
2
Components
directed component C
sink(C)
◮
C = set of directed components
Directed component cut relaxation
min
X
c(C) · xC
(DCR)
C∈C
X
xC
≥ 1
∀∅ ⊂ U ⊆ R \ {r}
xC
≥ 0
∀C ∈ C
C ∈ C : R(C) ∩ U 6= ∅,
sink(C) ∈
/U
root r
U
Directed component cut relaxation
min
X
c(C) · xC
(DCR)
C∈C
X
xC
≥ 1
∀∅ ⊂ U ⊆ R \ {r}
xC
≥ 0
∀C ∈ C
C ∈ C : R(C) ∩ U 6= ∅,
sink(C) ∈
/U
root r
xC1 =
1
2
b
b
xC2 =
U
b
1
2
xC3 =
1
2
Directed component cut relaxation
min
X
c(C) · xC
(DCR)
C∈C
X
xC
≥ 1
∀∅ ⊂ U ⊆ R \ {r}
xC
≥ 0
∀C ∈ C
C ∈ C : R(C) ∩ U 6= ∅,
sink(C) ∈
/U
Properties:
◮
Number of variables: exponential
◮
Number of constraints: exponential
◮
Approximable within 1 + ε (we ignore the ε here).
Solvability of the LP
Lemma
For any ε > 0, a solution x of cost ≤ (1 + ε)OP Tf can be
computed in polynomial time.
Solvability of the LP
Lemma
For any ε > 0, a solution x of cost ≤ (1 + ε)OP Tf can be
computed in polynomial time.
◮
Use only components of size 2⌈1/ε⌉ = O(1)
[Borchers & Du ’97]: Increases cost by ≤ 1 + ε
→ # variables polynomial
Solvability of the LP
Lemma
For any ε > 0, a solution x of cost ≤ (1 + ε)OP Tf can be
computed in polynomial time.
≤ 2⌈1/ε⌉
◮
Use only components of size 2⌈1/ε⌉ = O(1)
[Borchers & Du ’97]: Increases cost by ≤ 1 + ε
→ # variables polynomial
Solvability of the LP
Lemma
For any ε > 0, a solution x of cost ≤ (1 + ε)OP Tf can be
computed in polynomial time.
≤ 2⌈1/ε⌉
◮
◮
Use only components of size 2⌈1/ε⌉ = O(1)
[Borchers & Du ’97]: Increases cost by ≤ 1 + ε
→ # variables polynomial
Compact flow formulation → # constraints polynomial
(or solve with ellipsoid method).
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
xC1 =
1
2
b
b
xC2 =
b
1
2
xC3 =
1
2
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
C1 b
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
b
C2
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
C1 b
b
C2
An iterative randomized rounding algo
(1) FOR t = 1, . . . , ∞ DO
(2) Compute opt. LP solution x
(3) Sample a component:
Pr[sample C] =
xC
1T x
and contract it.
(4) IF all terminals connected THEN output sampled
components
◮
W.l.o.g. M := 1T x invariant
Roadmap
◮
In one iteration t:
E[c(comp. sampled in it. t)] =
X xC
C
M
·c(C) ≤
1
·OP T in it t
M
2 · OP T
1 · OP T
1·M
2·M
t = #iterations
Roadmap
◮
In one iteration t:
E[c(comp. sampled in it. t)] =
X xC
C
◮
M
·c(C) ≤
1
·OP T in it t
M
In total
X
X 1
E[c(comp. sampled in it. t)] ≤
·E[OP T in iteration t]
M
t≥1
t≥1
2 · OP T
1 · OP T
1·M
2·M
t = #iterations
Roadmap
◮
In one iteration t:
E[c(comp. sampled in it. t)] =
X xC
C
◮
M
·c(C) ≤
1
·OP T in it t
M
In total
X
X 1
E[c(comp. sampled in it. t)] ≤
·E[OP T in iteration t]
M
t≥1
t≥1
2 · OP T
E[OP T after t it] ≤ (1 −
1 · OP T
1·M
2·M
1 t
2M )
· OP T
t = #iterations
Roadmap
◮
In one iteration t:
E[c(comp. sampled in it. t)] =
X xC
C
◮
1
·OP T in it t
M
In total
X
X 1
E[c(comp. sampled in it. t)] ≤
·E[OP T in iteration t]
M
t≥1
2 · OP T
M
·c(C) ≤
t≥1
E[OP T after t it] ≤ (1 −
1 t
M)
· 2 · OP T
E[OP T after t it] ≤ (1 −
1 · OP T
1·M
2·M
1 t
2M )
· OP T
t = #iterations
Bridges
◮
Let S be Steiner tree
C
Bridges
◮
Let S be Steiner tree, C a component
C
Bridges
◮
Let S be Steiner tree, C a component
C
◮
Bridges:
BrS (C) = argmax{c(B) | B ⊆ S, S\B ∪ C is connected}
Bridges
◮
Let S be Steiner tree, C a component
C
◮
Bridges:
BrS (C) = argmax{c(B) | B ⊆ S, S\B ∪ C is connected}
The saving function
Definition
For a Steiner tree S, the saving function w : E → Q+ is
defined as
w(u, v) := max{c(e) | e on u − v path in S}.
w(u, v) := max{c(e) | e on u − v path in S}
u
v
A saving lemma
Lemma
For any component C, ∃ saving tree spanning the terminals of
C with
c(BrS (C)) = w(saving tree)
C
r
r
1
r
1
2
1
5
S 1
A saving lemma
Lemma
For any component C, ∃ saving tree spanning the terminals of
C with
c(BrS (C)) = w(saving tree)
b
saving tree
r
5
1
r
2
1
2
1
5
S 1
r
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
1
terminal spanning tree
· c(T )
E c
≤ 1−
after 1 sampling step
M
root r
1/2
1/2
1/2
T
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
X
xC · c(BrT (C)) ≥ c(T )
C∈C
root r
1/2
1/2
1/2
T
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
X
xC · c(BrT (C)) ≥ c(T )
C∈C
◮
For any C, ∃ saving tree:
c(BrT (C)) = w(saving tree of C)
root r
1/2
1/2
1/2
T
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
X
xC · c(BrT (C)) ≥ c(T )
C∈C
◮
For any C, ∃ saving tree:
c(BrT (C)) = w(saving tree of C)
◮
Transfer capacity from component to
its saving tree
1/2
b
→ capacity reservation y : E → Q+
root r
1/2
1/2
T
1/2
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
X
xC · c(BrT (C)) ≥ c(T )
C∈C
◮
For any C, ∃ saving tree:
c(BrT (C)) = w(saving tree of C)
◮
Transfer capacity from component to
its saving tree
1/2
b
→ capacity reservation y : E → Q+
root r
1/2
1/2
T
1/2
1/2
1/2
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
X
xC · c(BrT (C)) ≥ c(T )
C∈C
◮
For any C, ∃ saving tree:
c(BrT (C)) = w(saving tree of C)
◮
Transfer capacity from component to
its saving tree
1/2
b
→ capacity reservation y : E → Q+
root r
b
1/2
T
1/2
1/2
1/2
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
X
xC · c(BrT (C)) ≥ c(T )
C∈C
◮
For any C, ∃ saving tree:
c(BrT (C)) = w(saving tree of C)
◮
Transfer capacity from component to
its saving tree
1/2
b
→ capacity reservation y : E → Q+
root r
b
1/2
T
1/2
1/2
1/2
1/21/2
The Bridge Lemma (1)
Lemma (Bridge Lemma)
For T terminal spanning tree, x LP solution:
X
xC · c(BrT (C)) ≥ c(T )
C∈C
◮
For any C, ∃ saving tree:
c(BrT (C)) = w(saving tree of C)
◮
Transfer capacity from component to
its saving tree
1/2
b
→ capacity reservation y : E → Q+
X
xC · c(BrT (C)) = w(y)
root r
b
C∈C
1/2
T
1/2
1/2
1/2
1/2b
The Bridge Lemma (2)
X
xC · c(BrT (C)) = w(y)
C∈C
root r
1/2
1/2
1/2
T
1/2
1/2
1/2
The Bridge Lemma (2)
Edmonds Thm
X
xC · c(BrT (C)) = w(y) ≥ w(F )
C∈C
root r
F
T
The Bridge Lemma (2)
Cycle rule
Edmonds Thm
X
xC · c(BrT (C)) = w(y) ≥ w(F ) ≥ c(T )
C∈C
root r
F
T
A 1st bound on OPT
Lemma
E[OP T after it. t] ≤ 1 −
1 t
M
· 2 · OP T .
A 1st bound on OPT
Lemma
E[OP T after it. t] ≤ 1 −
◮
1 t
M
· 2 · OP T .
Initially c(MST) ≤ 2 · OP T
A 1st bound on OPT
Lemma
E[OP T after it. t] ≤ 1 −
1 t
M
· 2 · OP T .
◮
Initially c(MST) ≤ 2 · OP T
◮
In any iteration
E[c(new MST)]
≤
c(old MST) − E[c(Brold MST (C))]
A 1st bound on OPT
Lemma
E[OP T after it. t] ≤ 1 −
1 t
M
· 2 · OP T .
◮
Initially c(MST) ≤ 2 · OP T
◮
In any iteration
E[c(new MST)]
≤
=
c(old MST) − E[c(Brold MST (C))]
1 X
xC · c(Brold MST (C))
c(old MST) −
M
C∈C
A 1st bound on OPT
Lemma
E[OP T after it. t] ≤ 1 −
1 t
M
· 2 · OP T .
◮
Initially c(MST) ≤ 2 · OP T
◮
In any iteration
E[c(new MST)]
≤
=
c(old MST) − E[c(Brold MST (C))]
1 X
xC · c(Brold MST (C))
c(old MST) −
M
C∈C
|
{z
}
≥c(old MST)
A 1st bound on OPT
Lemma
E[OP T after it. t] ≤ 1 −
1 t
M
· 2 · OP T .
◮
Initially c(MST) ≤ 2 · OP T
◮
In any iteration
E[c(new MST)]
≤
=
c(old MST) − E[c(Brold MST (C))]
1 X
xC · c(Brold MST (C))
c(old MST) −
M
C∈C
|
{z
}
≥c(old MST)
≤
1−
1
M
· c(old MST)
A 2nd bound on OP T
Theorem
In any iteration
1 · old OPT
E[new OPT] ≤ 1 −
2M
◮
Let S be opt. Steiner tree
S
A 2nd bound on OP T
Theorem
In any iteration
1 · old OPT
E[new OPT] ≤ 1 −
2M
◮
Let S be opt. Steiner tree
◮
From each inner node in S: Contract
the cheapest edge going to a child
S
A 2nd bound on OP T
Theorem
In any iteration
1 · old OPT
E[new OPT] ≤ 1 −
2M
b
◮
Let S be opt. Steiner tree
◮
From each inner node in S: Contract
the cheapest edge going to a child
◮
A terminal spanning tree T
remains
b
T
b
A 2nd bound on OP T
Theorem
In any iteration
1 · old OPT
E[new OPT] ≤ 1 −
2M
b
◮
Let S be opt. Steiner tree
◮
From each inner node in S: Contract
the cheapest edge going to a child
◮
b
T
A terminal spanning tree T
remains
E[save on S] ≥ E[save on T ]
Bridge Lem
≥
1
1
· c(T ) ≥
· c(S)
|{z}
M
2M
1
≥ c(S)
2
b
The approximation guarantee
Theorem
E[AP X] ≤ (1.5+ε) · OP T .
2 · OP T
1 · OP T
◮
1·M
2 · M t = #iterations
Cost of sampled components:
∞
X
1
· E[OP T in it. t]
M
t=1
The approximation guarantee
Theorem
E[AP X] ≤ (1.5+ε) · OP T .
1 t
E[OP T after t it] ≤ (1 − M
) · 2 · OP T
−t/M
≤ 2e
· OP T
2 · OP T
1 t
E[OP T after t it] ≤ (1 − 2M
) · OP T
−t/(2M
)
≤ e
· OP T
1 · OP T
◮
1·M
2 · M t = #iterations
Cost of sampled components:
∞
X
1
· E[OP T in it. t]
M
t=1
The approximation guarantee
Theorem
E[AP X] ≤ (1.5+ε) · OP T .
2 · OP T
1 · OP T
◮
1 t
E[OP T after t it] ≤ (1 − M
) · 2 · OP T
−t/M
≤ 2e
· OP T
1 t
E[OP T after t it] ≤ (1 − 2M
) · OP T
−t/(2M
)
≤ e
· OP T
1·M
2 · M t = #iterations
Cost of sampled components:
∞
X
1
· E[OP T in it. t]
M
t=1
Z ∞
M →∞
min{2e−x , e−x/2 } dx
→
OP T ·
0
The approximation guarantee
Theorem
E[AP X] ≤ (1.5+ε) · OP T .
2 · OP T
1 · OP T
◮
1 t
E[OP T after t it] ≤ (1 − M
) · 2 · OP T
−t/M
≤ 2e
· OP T
1 t
E[OP T after t it] ≤ (1 − 2M
) · OP T
−t/(2M
)
≤ e
· OP T
1·M
2 · M t = #iterations
Cost of sampled components:
∞
X
1
· E[OP T in it. t]
M
t=1
Z ∞
M →∞
min{2e−x , e−x/2 } dx = 1.5 · OP T
→
OP T ·
0
A generalized bridge lemma
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
A generalized bridge lemma
1/2 1/2 1/2
pr.
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1
3
pr.
1
3
pr.
1/2 1/2 1/2
1/2 1/2 1/2
1
3
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
A generalized bridge lemma
1/2 1/2 1/2
pr.
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
1
3
pr.
1
3
pr.
1/2 1/2 1/2
1/2 1/2 1/2
1
3
1/2 1/2 1/2
1/2 1/2 1/2
1/2 1/2 1/2
A generalized bridge lemma
1/2 1/2 1/2
pr.
1/2 1/2 1/2
pr. 1 pr. 0
1/2 1/2 1/2
◮
1/2 1/2 1/2
1
3
pr.
1
3
pr.
1
3
1/2 1/2 1/2
1/2 1/2 1/2
pr. 1 pr. 0
1/2 1/2 1/2
pr. 1
1/2 1/2 1/2
1/2 1/2 1/2
Observe: Each edge in T removed with prob 2 ·
1
3
=
1
M!
A generalized bridge lemma
1/2 1/2 1/2
pr.
1/2 1/2 1/2
pr. 1 pr. 0
1/2 1/2 1/2
◮
◮
1/2 1/2 1/2
1
3
pr.
1
3
pr.
1
3
1/2 1/2 1/2
1/2 1/2 1/2
pr. 1 pr. 0
1/2 1/2 1/2
pr. 1
1/2 1/2 1/2
1/2 1/2 1/2
Observe: Each edge in T removed with prob 2 · 31 =
To show: We can always find these probabilities!
1
M!
A generalized bridge lemma (2)
Lemma
Let T be terminal spanning tree. Sample C from LP solution x.
There are B ⊆ T (dep. on C) s.t.
◮
(T \B) ∪ C spans all terminals
◮
Pr[e ∈ B] ≥
1
M
∀e ∈ T .
A generalized bridge lemma (2)
Lemma
Let T be terminal spanning tree. Sample C from LP solution x.
There are B ⊆ T (dep. on C) s.t.
◮
(T \B) ∪ C spans all terminals
◮
Pr[e ∈ B] ≥
1
M
∀e ∈ T .
P
B:(T \B)∪C conn. Pr[rem. B | C] = 1 ∀C
P
1
B∋e,C Pr[C] · Pr[rem. B | C] ≥ M ∀e
◮
Suppose system (1) has no non-negative solution.
A generalized bridge lemma (2)
Lemma
Let T be terminal spanning tree. Sample C from LP solution x.
There are B ⊆ T (dep. on C) s.t.
◮
(T \B) ∪ C spans all terminals
◮
Pr[e ∈ B] ≥
1
M
∀e ∈ T .
P
B:(T \B)∪C conn. Pr[rem. B | C] = 1 ∀C
P
1
B∋e,C Pr[C] · Pr[rem. B | C] ≥ M ∀e
dual
◮
◮
P yC
C yC
≥ xC · c(B) ∀B : T \B ∪ C conn.
< c(T )
Suppose system (1) has no non-negative solution.
Farkas Lemma: System (2) has solution (y, c) ≥ 0
A generalized bridge lemma (2)
Lemma
Let T be terminal spanning tree. Sample C from LP solution x.
There are B ⊆ T (dep. on C) s.t.
◮
(T \B) ∪ C spans all terminals
◮
Pr[e ∈ B] ≥
1
M
∀e ∈ T .
P
B:(T \B)∪C conn. Pr[rem. B | C] = 1 ∀C
P
1
B∋e,C Pr[C] · Pr[rem. B | C] ≥ M ∀e
dual
◮
◮
◮
P yC
C yC
≥ xC · c(B) ∀B : T \B ∪ C conn.
< c(T )
Suppose system (1) has no non-negative solution.
Farkas Lemma: System (2) has solution (y, c) ≥ 0
Contradiction to Bridge Lemma!
The 1.39 bound
◮
Let S ∗ optimum Steiner tree.
The 1.39 bound
e
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
The 1.39 bound
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
The 1.39 bound
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
The 1.39 bound
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
The 1.39 bound
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
The 1.39 bound
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
The 1.39 bound
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
The 1.39 bound
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
The 1.39 bound
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
The 1.39 bound
◮
◮
◮
◮
◮
Let S ∗ optimum Steiner tree.
Goal: Define Steiner tree S t ⊆ S ∗ after t iterations with
E[t : e ∈ S t ] ≤ 1.39 · M .
For every internal node in S ∗ : Mark a random outgoing
edge.
Consider cycles S ∗ ∪ {u, v} containing exactly one marked
edge.
Such edges {u, v} induce terminal spanning tree T
The 1.39 bound
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
The 1.39 bound
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
The 1.39 bound
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
The 1.39 bound
e
◮
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
1
per it.
Random process deletes an edge e ∈ T with pr. M
The 1.39 bound
e
◮
◮
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
1
per it.
Random process deletes an edge e ∈ T with pr. M
E[t : e deleted] ≤ M
The 1.39 bound
ek
e1
e2
◮
◮
◮
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
1
per it.
Random process deletes an edge e ∈ T with pr. M
E[t : e deleted] ≤ M
E[t : e1 , . . . , ek deleted] ≤ (1 + 12 + . . . + k1 ) · M = H(k) · M
→ Coupon Collector Theorem
The 1.39 bound
e
◮
◮
◮
◮
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
1
per it.
Random process deletes an edge e ∈ T with pr. M
E[t : e deleted] ≤ M
E[t : e1 , . . . , ek deleted] ≤ (1 + 12 + . . . + k1 ) · M = H(k) · M
→ Coupon Collector Theorem
E[t : e deleted] ≤ H(#cycles through e) · M
The 1.39 bound
e
◮
◮
◮
◮
◮
◮
Def S t : e ∈ T not deleted ⇒ keep edges in corr. cycle in S ∗
1
per it.
Random process deletes an edge e ∈ T with pr. M
E[t : e deleted] ≤ M
E[t : e1 , . . . , ek deleted] ≤ (1 + 12 + . . . + k1 ) · M = H(k) · M
→ Coupon Collector Theorem
E[t : e deleted] ≤ H(#cycles through e) · M
Pr[e in k cycles] =X
( 12 )k
1 k
E[t : e deleted] ≤
·H(k)·M = ln(4)·M ≈ 1.39·M.
2
k≥1
Open problems
Open Problem I
1.01 ≤ Steiner tree approximability ≤ 1.39
Open problems
Open Problem I
1.01 ≤ Steiner tree approximability ≤ 1.39
Open Problem II
Is there an iterative randomized rounding approach for
Facility Location or k-Median?
Open problems
Open Problem III
Is there an iterative randomized rounding approach for ATSP?
(1) Solve Held-Karp relaxation:
min cT x
x(δ+ (S)) ≥ 1 ∀∅ ⊂ S ⊂ V
x(δ+ (v)) = x(δ− (v)) = 1 ∀v ∈ V
xe ≥ 0 ∀e ∈ E
(2) Sample a collection of cycles C from x∗ .
(3) Show E[c(C)] ≤ 1000 · OP T
(4) Show E[OP T after contracting C] ≤ 0.999 · OP T .
This would yield a O(1)-apx.
Open problems
Open Problem III
Is there an iterative randomized rounding approach for ATSP?
(1) Solve Held-Karp relaxation:
min cT x
x(δ+ (S)) ≥ 1 ∀∅ ⊂ S ⊂ V
x(δ+ (v)) = x(δ− (v)) = 1 ∀v ∈ V
xe ≥ 0 ∀e ∈ E
(2) Sample a collection of cycles C from x∗ .
(3) Show E[c(C)] ≤ 1000 · OP T
(4) Show E[OP T after contracting C] ≤ 0.999 · OP T .
This would yield a O(1)-apx.
Thanks for your attention