Advanced Algorithms (II) Chihao Zhang Shanghai Jiao Tong University - - PowerPoint PPT Presentation

Advanced Algorithms (II)

Chihao Zhang

Shanghai Jiao Tong University

• Mar. 4, 2019

Advanced Algorithms (II) 1/11

MaxSAT

Recall that we can obtain a

• approximation by choosing the betuer
• f the two.

We show that this ratio can also be achieved via direct rounding. Recall that we have the following linear programming relaxation. max

m j

zj subject to

i Pj

yi

k Nj

yk zj Cj

i Pj

xi

k Nj

xk zj j m yi i n

Advanced Algorithms (II) 2/11

MaxSAT

Recall that we can obtain a 3

4-approximation by choosing the betuer

• f the two.

We show that this ratio can also be achieved via direct rounding. Recall that we have the following linear programming relaxation. max

m j

zj subject to

i Pj

yi

k Nj

yk zj Cj

i Pj

xi

k Nj

xk zj j m yi i n

Advanced Algorithms (II) 2/11

MaxSAT

Recall that we can obtain a 3

4-approximation by choosing the betuer

• f the two.

We show that this ratio can also be achieved via direct rounding. Recall that we have the following linear programming relaxation. max

m j

zj subject to

i Pj

yi

k Nj

yk zj Cj

i Pj

xi

k Nj

xk zj j m yi i n

Advanced Algorithms (II) 2/11

MaxSAT

Recall that we can obtain a 3

4-approximation by choosing the betuer

• f the two.

We show that this ratio can also be achieved via direct rounding. Recall that we have the following linear programming relaxation. max

m

∑

j=1

zj subject to ∑

i∈Pj

yi + ∑

k∈Nj

(1 − yk) ≥ zj, ∀Cj = ∨

i∈Pj

xi ∨ ∨

k∈Nj

¯ xk zj ∈ [0, 1], ∀j ∈ [m] yi ∈ [0, 1], ∀i ∈ [n]

Advanced Algorithms (II) 2/11

Let { y∗

i

}

i∈[n] ,

{ z∗

j

}

j∈[m] be an optimal solution of the LP.

Instead of tossing yi -biased coins, we toss f yi -biased coins for some function f . For Cj

i Pj xi k Nj xk,

Pr Cj is not satisfied

i Pj

f yi

k Nj

f yk We can choose a suitable f to get approximation.

Advanced Algorithms (II) 3/11

Let { y∗

i

}

i∈[n] ,

{ z∗

j

}

j∈[m] be an optimal solution of the LP.

Instead of tossing y∗

i -biased coins, we toss f(y∗ i )-biased coins for

some function f(·) ∈ [0, 1]. For Cj

i Pj xi k Nj xk,

Pr Cj is not satisfied

i Pj

f yi

k Nj

f yk We can choose a suitable f to get approximation.

Advanced Algorithms (II) 3/11

Let { y∗

i

}

i∈[n] ,

{ z∗

j

}

j∈[m] be an optimal solution of the LP.

Instead of tossing y∗

i -biased coins, we toss f(y∗ i )-biased coins for

some function f(·) ∈ [0, 1]. For Cj = ∨

i∈Pj xi ∨ ∨ k∈Nj ¯

xk, Pr [ Cj is not satisfied ] = ∏

i∈Pj

(1 − f(y∗

i ))

∏

k∈Nj

f(y∗

k).

We can choose a suitable f to get approximation.

Advanced Algorithms (II) 3/11

Let { y∗

i

}

i∈[n] ,

{ z∗

j

}

j∈[m] be an optimal solution of the LP.

Instead of tossing y∗

i -biased coins, we toss f(y∗ i )-biased coins for

some function f(·) ∈ [0, 1]. For Cj = ∨

i∈Pj xi ∨ ∨ k∈Nj ¯

xk, Pr [ Cj is not satisfied ] = ∏

i∈Pj

(1 − f(y∗

i ))

∏

k∈Nj

f(y∗

k).

We can choose a suitable f to get 3

4 approximation.

Advanced Algorithms (II) 3/11

Integrality Gap

In most LP based approximation algorithms, the upper bound for the OPT is OPT OPT LP Then we establish OPT OPT LP OPT If we alreadly know OPT OPT LP then we cannot have ! The ratio is called the integrality gap of the LP relaxation.

Advanced Algorithms (II) 4/11

Integrality Gap

In most LP based approximation algorithms, the upper bound for the OPT is OPT ≤ OPT(LP). Then we establish OPT OPT LP OPT If we alreadly know OPT OPT LP then we cannot have ! The ratio is called the integrality gap of the LP relaxation.

Advanced Algorithms (II) 4/11

Integrality Gap

In most LP based approximation algorithms, the upper bound for the OPT is OPT ≤ OPT(LP). Then we establish OPT∗ ≥ α · OPT(LP) ≥ α · OPT. If we alreadly know OPT OPT LP then we cannot have ! The ratio is called the integrality gap of the LP relaxation.

Advanced Algorithms (II) 4/11

Integrality Gap

In most LP based approximation algorithms, the upper bound for the OPT is OPT ≤ OPT(LP). Then we establish OPT∗ ≥ α · OPT(LP) ≥ α · OPT. If we alreadly know OPT ≤ β · OPT(LP), then we cannot have ! The ratio is called the integrality gap of the LP relaxation.

Advanced Algorithms (II) 4/11

Integrality Gap

In most LP based approximation algorithms, the upper bound for the OPT is OPT ≤ OPT(LP). Then we establish OPT∗ ≥ α · OPT(LP) ≥ α · OPT. If we alreadly know OPT ≤ β · OPT(LP), then we cannot have α > β! The ratio is called the integrality gap of the LP relaxation.

Advanced Algorithms (II) 4/11

Integrality Gap

In most LP based approximation algorithms, the upper bound for the OPT is OPT ≤ OPT(LP). Then we establish OPT∗ ≥ α · OPT(LP) ≥ α · OPT. If we alreadly know OPT ≤ β · OPT(LP), then we cannot have α > β! The ratio β is called the integrality gap of the LP relaxation.

Advanced Algorithms (II) 4/11

Integrality Gap for MaxSAT

Consider the instance, x x x x x x x x OPT and OPT LP . The integrality gap of our LP is .

• Corollary. We cannot beat

if we use OPT OPT LP upper bound.

Advanced Algorithms (II) 5/11

Integrality Gap for MaxSAT

Consider the instance, (x1 ∨ x2) ∧ (x1 ∨ ¯ x2) ∧ (¯ x1 ∨ x2) ∧ (¯ x1 ∨ ¯ x2) OPT and OPT LP . The integrality gap of our LP is .

• Corollary. We cannot beat

if we use OPT OPT LP upper bound.

Advanced Algorithms (II) 5/11

Integrality Gap for MaxSAT

Consider the instance, (x1 ∨ x2) ∧ (x1 ∨ ¯ x2) ∧ (¯ x1 ∨ x2) ∧ (¯ x1 ∨ ¯ x2) OPT = 3 and OPT(LP) = 4. The integrality gap of our LP is .

• Corollary. We cannot beat

if we use OPT OPT LP upper bound.

Advanced Algorithms (II) 5/11

Integrality Gap for MaxSAT

Consider the instance, (x1 ∨ x2) ∧ (x1 ∨ ¯ x2) ∧ (¯ x1 ∨ x2) ∧ (¯ x1 ∨ ¯ x2) OPT = 3 and OPT(LP) = 4. The integrality gap of our LP is 3

4.

• Corollary. We cannot beat

if we use OPT OPT LP upper bound.

Advanced Algorithms (II) 5/11

Integrality Gap for MaxSAT

Consider the instance, (x1 ∨ x2) ∧ (x1 ∨ ¯ x2) ∧ (¯ x1 ∨ x2) ∧ (¯ x1 ∨ ¯ x2) OPT = 3 and OPT(LP) = 4. The integrality gap of our LP is 3

4.

• Corollary. We cannot beat 3

4 if we use OPT ≤ OPT(LP) upper

bound.

Advanced Algorithms (II) 5/11

Minimum Label Cut

Minimum Label s-t Cut Input: A graph G V E ; a set of labels L L such that each e E is labelled with

• ne

e L ; two vertices s t V. Problem: Compute a minimum set of labels L L such that the removal of all edges with label in L disconnects s and t. NP-hard, and even hard to approximate with any constant ratio (unless NP P).

Advanced Algorithms (II) 6/11

Minimum Label Cut

Minimum Label s-t Cut Input: A graph G = (V, E); a set of labels [L] = {1, 2, . . . , L} such that each e ∈ E is labelled with

• ne ℓ(e) ∈ [L]; two vertices s, t ∈ V.

Problem: Compute a minimum set of labels L′ ⊆ [L] such that the removal of all edges with label in L′ disconnects s and t. NP-hard, and even hard to approximate with any constant ratio (unless NP P).

Advanced Algorithms (II) 6/11

Minimum Label Cut

Minimum Label s-t Cut Input: A graph G = (V, E); a set of labels [L] = {1, 2, . . . , L} such that each e ∈ E is labelled with

• ne ℓ(e) ∈ [L]; two vertices s, t ∈ V.

Problem: Compute a minimum set of labels L′ ⊆ [L] such that the removal of all edges with label in L′ disconnects s and t. NP-hard, and even hard to approximate with any constant ratio (unless NP = P).

Advanced Algorithms (II) 6/11

LP Relaxation

We introduce a variable zj for each label j L . Let

s t be the

collection of paths between s and t. min

L j

zj subject to

e P

z

e

P

s t

zj j L Q1: How to solve this LP efgiciently? Q2: What is the integrality gap of this LP? m .

Advanced Algorithms (II) 7/11

LP Relaxation

We introduce a variable zj for each label j ∈ [L]. Let Ps,t be the collection of paths between s and t. min

L j

zj subject to

e P

z

e

P

s t

zj j L Q1: How to solve this LP efgiciently? Q2: What is the integrality gap of this LP? m .

Advanced Algorithms (II) 7/11

LP Relaxation

We introduce a variable zj for each label j ∈ [L]. Let Ps,t be the collection of paths between s and t. min

L

∑

j=1

zj subject to ∑

e∈P

zℓ(e) ≥ 1, ∀P ∈ Ps,t zj ∈ [0, 1], ∀j ∈ [L] Q1: How to solve this LP efgiciently? Q2: What is the integrality gap of this LP? m .

Advanced Algorithms (II) 7/11

LP Relaxation

We introduce a variable zj for each label j ∈ [L]. Let Ps,t be the collection of paths between s and t. min

L

∑

j=1

zj subject to ∑

e∈P

zℓ(e) ≥ 1, ∀P ∈ Ps,t zj ∈ [0, 1], ∀j ∈ [L] Q1: How to solve this LP efgiciently? Q2: What is the integrality gap of this LP? m .

Advanced Algorithms (II) 7/11

LP Relaxation

We introduce a variable zj for each label j ∈ [L]. Let Ps,t be the collection of paths between s and t. min

L

∑

j=1

zj subject to ∑

e∈P

zℓ(e) ≥ 1, ∀P ∈ Ps,t zj ∈ [0, 1], ∀j ∈ [L] Q1: How to solve this LP efgiciently? Q2: What is the integrality gap of this LP? m .

Advanced Algorithms (II) 7/11

LP Relaxation

We introduce a variable zj for each label j ∈ [L]. Let Ps,t be the collection of paths between s and t. min

L

∑

j=1

zj subject to ∑

e∈P

zℓ(e) ≥ 1, ∀P ∈ Ps,t zj ∈ [0, 1], ∀j ∈ [L] Q1: How to solve this LP efgiciently? Q2: What is the integrality gap of this LP? Ω(m).

Advanced Algorithms (II) 7/11

Separation Oracle

We can solve the LP in poly-time using ellipsoid method provided a separation oracle: Given a point, in PTIME either confirm it is a feasible solution; or find a violated constraint. Oracle here: shortest s-t path

Advanced Algorithms (II) 8/11

Separation Oracle

We can solve the LP in poly-time using ellipsoid method provided a separation oracle: Given a point, in PTIME either confirm it is a feasible solution; or find a violated constraint. Oracle here: shortest s-t path

Advanced Algorithms (II) 8/11

Separation Oracle

We can solve the LP in poly-time using ellipsoid method provided a separation oracle: Given a point, in PTIME either ▶ confirm it is a feasible solution; or ▶ find a violated constraint. Oracle here: shortest s-t path

Advanced Algorithms (II) 8/11

Separation Oracle

We can solve the LP in poly-time using ellipsoid method provided a separation oracle: Given a point, in PTIME either ▶ confirm it is a feasible solution; or ▶ find a violated constraint. Oracle here: shortest s-t path

Advanced Algorithms (II) 8/11

Beat the Interality Gap

Obtain a partial solution via rounding; Complement the solution by combinatorial construction. Let zj

j L be an optimal solution of the LP. Let

be a parameter.

• 1. Let L

j L zj .

• 2. Let G be the graph obtained from G by removing

edges with label in L .

• 3. Let F be the minimum s-t cut of G , L be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; Complement the solution by combinatorial construction. Let zj

j L be an optimal solution of the LP. Let

be a parameter.

• 1. Let L

j L zj .

• 2. Let G be the graph obtained from G by removing

edges with label in L .

• 3. Let F be the minimum s-t cut of G , L be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; ▶ Complement the solution by combinatorial construction. Let zj

j L be an optimal solution of the LP. Let

be a parameter.

• 1. Let L

j L zj .

• 2. Let G be the graph obtained from G by removing

edges with label in L .

• 3. Let F be the minimum s-t cut of G , L be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; ▶ Complement the solution by combinatorial construction. Let zj

j L be an optimal solution of the LP. Let

be a parameter.

• 1. Let L

j L zj .

• 2. Let G be the graph obtained from G by removing

edges with label in L .

• 3. Let F be the minimum s-t cut of G , L be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; ▶ Complement the solution by combinatorial construction. Let { z∗

j

}

j∈[L] be an optimal solution of the LP. Let β > 0

be a parameter.

• 1. Let L

j L zj .

• 2. Let G be the graph obtained from G by removing

edges with label in L .

• 3. Let F be the minimum s-t cut of G , L be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; ▶ Complement the solution by combinatorial construction. Let { z∗

j

}

j∈[L] be an optimal solution of the LP. Let β > 0

be a parameter.

• 1. Let L1 ≜

{ j ∈ L : z∗

j ≥ β

} .

• 2. Let G be the graph obtained from G by removing

edges with label in L .

• 3. Let F be the minimum s-t cut of G , L be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; ▶ Complement the solution by combinatorial construction. Let { z∗

j

}

j∈[L] be an optimal solution of the LP. Let β > 0

be a parameter.

• 1. Let L1 ≜

{ j ∈ L : z∗

j ≥ β

} .

• 2. Let G′ be the graph obtained from G by removing

edges with label in L1.

• 3. Let F be the minimum s-t cut of G , L be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; ▶ Complement the solution by combinatorial construction. Let { z∗

j

}

j∈[L] be an optimal solution of the LP. Let β > 0

be a parameter.

• 1. Let L1 ≜

{ j ∈ L : z∗

j ≥ β

} .

• 2. Let G′ be the graph obtained from G by removing

edges with label in L1.

• 3. Let F be the minimum s-t cut of G′, L2 be the labels
• f edges in F.
• 4. Return L

L .

Advanced Algorithms (II) 9/11

Beat the Interality Gap

▶ Obtain a partial solution via rounding; ▶ Complement the solution by combinatorial construction. Let { z∗

j

}

j∈[L] be an optimal solution of the LP. Let β > 0

be a parameter.

• 1. Let L1 ≜

{ j ∈ L : z∗

j ≥ β

} .

• 2. Let G′ be the graph obtained from G by removing

edges with label in L1.

• 3. Let F be the minimum s-t cut of G′, L2 be the labels
• f edges in F.
• 4. Return L1 ∪ L2.

Advanced Algorithms (II) 9/11

Bounding L1 and L2

It is clear that L

j L

zj OPT LP OPT On the otherhand, there cannot be too many edge disjoint paths between s and t in G : at least edges on each s-t path; at most m

L

m L such paths; therefore L F m L (Menger’s theorem).

Advanced Algorithms (II) 10/11

Bounding L1 and L2

It is clear that |L1| ≤ ∑

j∈[L]

1 β · z∗

j = 1

β · OPT(LP) ≤ 1 β · OPT. On the otherhand, there cannot be too many edge disjoint paths between s and t in G : at least edges on each s-t path; at most m

L

m L such paths; therefore L F m L (Menger’s theorem).

Advanced Algorithms (II) 10/11

Bounding L1 and L2

It is clear that |L1| ≤ ∑

j∈[L]

1 β · z∗

j = 1

β · OPT(LP) ≤ 1 β · OPT. On the otherhand, there cannot be too many edge disjoint paths between s and t in G′: at least edges on each s-t path; at most m

L

m L such paths; therefore L F m L (Menger’s theorem).

Advanced Algorithms (II) 10/11

Bounding L1 and L2

It is clear that |L1| ≤ ∑

j∈[L]

1 β · z∗

j = 1

β · OPT(LP) ≤ 1 β · OPT. On the otherhand, there cannot be too many edge disjoint paths between s and t in G′: ▶ at least 1

β edges on each s-t path;

▶ at most m−|L1 |

1/β

= β(m − |L1|) such paths; ▶ therefore |L2| ≤ |F| ≤ β(m − |L1|) (Menger’s theorem).

Advanced Algorithms (II) 10/11

The Choice of β

We already have L L OPT m L OPT m Setuing

OPT m

yields an O m approximation.

Remark

Instead of using Menger’s theorem, we can find a small cut by BFS from s.

• Exercise. Find an O n
• approx algorithm via rounding + BFS.

Advanced Algorithms (II) 11/11

The Choice of β

We already have |L1| + |L2| ≤ 1 β · OPT + β(m − |L1|) ≤ 1 β · OPT + βm. Setuing β = √

OPT m

yields an O ( m

1 2

) approximation.

Remark

Instead of using Menger’s theorem, we can find a small cut by BFS from s.

• Exercise. Find an O n
• approx algorithm via rounding + BFS.

Advanced Algorithms (II) 11/11

The Choice of β

We already have |L1| + |L2| ≤ 1 β · OPT + β(m − |L1|) ≤ 1 β · OPT + βm. Setuing β = √

OPT m

yields an O ( m

1 2

) approximation.

Remark

Instead of using Menger’s theorem, we can find a small cut by BFS from s.

• Exercise. Find an O n
• approx algorithm via rounding + BFS.

Advanced Algorithms (II) 11/11

The Choice of β

We already have |L1| + |L2| ≤ 1 β · OPT + β(m − |L1|) ≤ 1 β · OPT + βm. Setuing β = √

OPT m

yields an O ( m

1 2

) approximation.

Remark

Instead of using Menger’s theorem, we can find a small cut by BFS from s.

• Exercise. Find an O

( n

2 3

)

• approx algorithm via rounding + BFS.

Advanced Algorithms (II) 11/11