Important separators and parameterized algorithms

Important separators and parameterized algorithms

Dániel Marx¹

1Institute for Computer Science and Control, Hungarian Academy of Sciences (MTA SZTAKI)

Budapest, Hungary

ICERM Providence, RI April 27, 2014

1

Overview

Main message

Small separators in graphs have interesting extremal properties that can be exploited in combinatorial and algorithmic results.

Bounding the number of “important” cuts.

Edge/vertex versions, directed/undirected versions.

Algorithmic applications: FPT algorithm for Multiway cut,

Directed Feedback Vertex Set, and (p,q)-Clustering.

Important cuts

Definition: δ(R)is the set of edges with exactly one endpoint inR. Definition: A setS of edges is aminimal (X,Y)-cutif there is no X−Y path inG\S and no proper subset ofS breaks everyX−Y path.

Observation: Every minimal(X,Y)-cutS can be expressed asS = δ(R)for some X ⊆R andR∩Y =∅.

R δ(R) X Y

3

Important cuts

Definition: A minimal (X,Y)-cut δ(R) isimportant if there is no (X,Y)-cutδ(R⁰)with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|.

Note: Can be checked in polynomial time if a cut is important.

R δ(R) X Y

Important cuts

Definition: A minimal (X,Y)-cut δ(R) isimportant if there is no (X,Y)-cutδ(R⁰)with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|.

Note: Can be checked in polynomial time if a cut is important.

R⁰ δ(R)

R

δ(R⁰)

X Y

3

Important cuts

Definition: A minimal (X,Y)-cut δ(R) isimportant if there is no (X,Y)-cutδ(R⁰)with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|.

Note: Can be checked in polynomial time if a cut is important.

R δ(R)

X Y

Important cuts

The number of important cuts can be exponentially large.

Example:

X Y

1 2 k/2

This graph has2^k/2 important(X,Y)-cuts of size at mostk.

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk. (Proof is implicit in[Chen, Liu, Lu 2007], worse bound in[M. 2004].)

4

Important cuts

The number of important cuts can be exponentially large.

Example:

X Y

1 2 k/2

This graph has2^k/2 important(X,Y)-cuts of size at mostk.

Theorem

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

Proof: Determine separately the contribution of the different types of edges.

5

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

Proof: Determine separately the contribution of the different types of edges.

A B

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

0 1 1 0

Proof: Determine separately the contribution of the different types of edges.

B A

5

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

1 0 1 0

Proof: Determine separately the contribution of the different types of edges.

A B

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

0 1 0 1

Proof: Determine separately the contribution of the different types of edges.

A B

5

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

1 0 0 1

Proof: Determine separately the contribution of the different types of edges.

B A

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

1 1 1 1

Proof: Determine separately the contribution of the different types of edges.

B A

5

Submodularity

Fact: The functionδ issubmodular: for arbitrary sets A,B,

|δ(A)| + |δ(B)| ≥ |δ(A∩B)| + |δ(A∪B)|

1 1 0 0

Proof: Determine separately the contribution of the different types of edges.

B A

Submodularity

Lemma

Letλbe the minimum (X,Y)-cut size. There is a unique maximal Rmax⊇X such that δ(Rmax) is an(X,Y)-cut of sizeλ.

Proof: Let R₁,R₂ ⊇X be two sets such that δ(R₁), δ(R₂)are (X,Y)-cuts of size λ.

|δ(R₁)|+|δ(R₂)| ≥ |δ(R₁∩R₂)|+|δ(R₁∪R₂)|

λ λ ≥λ

⇒ |δ(R₁∪R₂)| ≤λ

R₂ R₁

Y

X Note: Analogous result holds for a unique minimal R_min.

6

Submodularity

Lemma

Letλbe the minimum (X,Y)-cut size. There is a unique maximal Rmax⊇X such that δ(Rmax) is an(X,Y)-cut of sizeλ.

Proof: Let R₁,R₂ ⊇X be two sets such that δ(R₁), δ(R₂)are (X,Y)-cuts of size λ.

|δ(R₁)|+|δ(R₂)| ≥ |δ(R₁∩R₂)|+|δ(R₁∪R₂)|

λ λ ≥λ

⇒ |δ(R₁∪R₂)| ≤λ

R₂ R₁

Y

X Note: Analogous result holds for a unique minimal R_min.

Important cuts

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk.

Proof: Let λbe the minimum(X,Y)-cut size and letδ(R_max) be the unique important cut of sizeλsuch that R_max is maximal.

(1) We show thatR_max⊆R for every important cutδ(R).

By the submodularity ofδ:

|δ(R_max)|+|δ(R)| ≥ |δ(R_max∩R)|+|δ(R_max∪R)|

λ ≥λ

⇓

|δ(R_max∪R)| ≤ |δ(R)|

⇓

IfR 6=R_max∪R, then δ(R)is not important. Thus the important(X,Y)- and(R_max,Y)-cuts are the same.

⇒We can assume X =R_max.

7

Important cuts

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk.

Proof: Let λbe the minimum(X,Y)-cut size and letδ(R_max) be the unique important cut of sizeλsuch that R_max is maximal.

(1) We show thatR_max⊆R for every important cutδ(R).

By the submodularity ofδ:

|δ(R_max)|+|δ(R)| ≥ |δ(R_max∩R)|+|δ(R_max∪R)|

λ ≥λ

⇓

|δ(R_max∪R)| ≤ |δ(R)|

⇓

If R 6=R_max∪R, thenδ(R)is not important.

Thus the important(X,Y)- and(R_max,Y)-cuts are the same.

⇒We can assume X =R_max.

Important cuts

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk.

Proof: Let λbe the minimum(X,Y)-cut size and letδ(R_max) be the unique important cut of sizeλsuch that R_max is maximal.

(1) We show thatR_max⊆R for every important cutδ(R).

By the submodularity ofδ:

|δ(R_max)|+|δ(R)| ≥ |δ(R_max∩R)|+|δ(R_max∪R)|

λ ≥λ

⇓

|δ(R_max∪R)| ≤ |δ(R)|

⇓

If R 6=R_max∪R, thenδ(R)is not important.

Thus the important(X,Y)- and (R_max,Y)-cuts are the same.

⇒We can assume X =R_max.

7

Important cuts

(2) Search tree algorithm for enumerating all these cuts:

An (arbitrary) edgeuv leavingX =R_max is either in the cut or not.

v Y X =R_maxu

Branch 1: Ifuv ∈S, thenS\uv is an important (X,Y)-cut of size at mostk−1 inG \uv.

⇒ k decreases by one, λdecreases by at most 1. Branch 2: If uv 6∈S, thenS is an important

(X ∪v,Y)-cut of size at mostk in G.

⇒ k remains the same,λincreases by 1. The measure2k−λdecreases in each step.

⇒ Height of the search tree ≤2k

⇒ ≤2^2k =4^k important cuts of size at mostk.

Important cuts

(2) Search tree algorithm for enumerating all these cuts:

An (arbitrary) edgeuv leavingX =R_max is either in the cut or not.

v Y X =R_maxu

Branch 1: Ifuv ∈S, thenS\uv is an important (X,Y)-cut of size at mostk−1 inG \uv.

⇒ k decreases by one, λdecreases by at most 1.

Branch 2: If uv 6∈S, thenS is an important (X∪v,Y)-cut of size at most k in G.

⇒ k remains the same,λincreases by 1. The measure2k−λdecreases in each step.

⇒ Height of the search tree ≤2k

⇒ ≤2^2k =4^k important cuts of size at mostk.

8

Important cuts

(2) Search tree algorithm for enumerating all these cuts:

An (arbitrary) edgeuv leavingX =R_max is either in the cut or not.

v Y X =R_maxu

Branch 1: Ifuv ∈S, thenS\uv is an important (X,Y)-cut of size at mostk−1 inG \uv.

⇒ k decreases by one, λdecreases by at most 1.

Branch 2: If uv 6∈S, thenS is an important (X∪v,Y)-cut of size at most k in G.

⇒ k remains the same,λincreases by 1.

The measure2k−λdecreases in each step.

⇒ Height of the search tree ≤2k

⇒ ≤2^2k =4^k important cuts of size at mostk.

Important cuts

(2) Search tree algorithm for enumerating all these cuts:

An (arbitrary) edgeuv leavingX =R_max is either in the cut or not.

v Y X =R_maxu

Branch 1: Ifuv ∈S, thenS\uv is an important (X,Y)-cut of size at mostk−1 inG \uv.

⇒ k decreases by one, λdecreases by at most 1.

Branch 2: If uv 6∈S, thenS is an important (X∪v,Y)-cut of size at most k in G.

⇒ k remains the same,λincreases by 1.

The measure2k−λdecreases in each step.

⇒ Height of the search tree ≤2k

⇒ ≤2^2k =4^k important cuts of size at mostk.

8

Important cuts

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk.

Example: The bound 4^k is essentially tight.

Y X

Any subtree withk leaves gives an important(X,Y)-cut of sizek. The number of subtrees withk leaves is the Catalan number

C_k−1= 1 k

2k−2 k−1

≥4^k/poly(k).

Important cuts

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk.

Example: The bound 4^k is essentially tight.

X

Y

Any subtree withk leaves gives an important(X,Y)-cut of sizek.

The number of subtrees withk leaves is the Catalan number C_k−1= 1

k

2k−2 k−1

≥4^k/poly(k).

9

Important cuts

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk.

Example: The bound 4^k is essentially tight.

Y X

Any subtree withk leaves gives an important(X,Y)-cut of sizek.

The number of subtrees withk leaves is the Catalan number C_k−1= 1

k

2k−2 k−1

≥4^k/poly(k).

Important cuts

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk.

Example: The bound 4^k is essentially tight.

Y X

Any subtree withk leaves gives an important(X,Y)-cut of sizek.

The number of subtrees withk leaves is the Catalan number C_k−1= 1

k

2k−2 k−1

≥4^k/poly(k).

9

Multiway Cut

Definition: Amultiway cut of a set of terminalsT is a setS of edges such that each component ofG\S contains at most one vertex ofT.

Multiway Cut

Input: GraphG, setT of vertices, inte- gerk

Find: Amultiway cutS of at most k

edges. ^t4

t5

t4

t3

t2

t1

Polynomial for|T|=2, but NP-hard for any fixed|T| ≥3 [Dalhaus et al. 1994].

Multiway Cut

Definition: Amultiway cut of a set of terminalsT is a setS of edges such that each component ofG\S contains at most one vertex ofT.

Multiway Cut

Input: GraphG, setT of vertices, inte- gerk

Find: Amultiway cutS of at most k

edges. ^t4

t5

t4

t3

t2

t1

Trivial to solve in polynomial time for fixedk (in time n^O(k)).

Theorem

Multiway cutcan be solved in time4^k ·n^O(1), i.e., it is

fixed-parameter tractable (FPT) parameterized by the sizek of the solution.

10

Multiway Cut

Intuition: Consider a t∈T. A subset of the solutionS is a (t,T \t)-cut.

t

There are many such cuts.

But a cut farther fromt and closer toT \t seems to be more useful.

Multiway Cut

Intuition: Consider a t∈T. A subset of the solutionS is a (t,T \t)-cut.

t

There are many such cuts.

But a cut farther fromt and closer toT \t seems to be more useful.

11

Multiway Cut

Intuition: Consider a t∈T. A subset of the solutionS is a (t,T \t)-cut.

t

There are many such cuts.

But a cut farther fromt and closer toT \t seems to be more useful.

Multiway Cut

Intuition: Consider a t∈T. A subset of the solutionS is a (t,T \t)-cut.

t

There are many such cuts.

But a cut farther fromt and closer toT \t seems to be more useful.

11

Multiway Cut and important cuts

Pushing Lemma

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Proof: Let R be the vertices reachable fromt inG \S for a solutionS.

δ(R)is not important, then there is an important cut δ(R⁰) with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|. ReplaceS with

S⁰ := (S\δ(R))∪δ(R⁰) ⇒ |S⁰| ≤ |S|

S⁰ is a multiway cut: (1) There is not-u path inG\S⁰ and (2) a u-v path inG \S⁰ implies at-u path, a contradiction.

Multiway Cut and important cuts

Pushing Lemma

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Proof: Let R be the vertices reachable fromt inG \S for a solutionS.

R t

δ(R)is not important, then there is an important cut δ(R⁰) with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|. ReplaceS with

S⁰ := (S\δ(R))∪δ(R⁰) ⇒ |S⁰| ≤ |S|

S⁰ is a multiway cut: (1) There is not-u path inG\S⁰ and (2) a u-v path inG \S⁰ implies at-u path, a contradiction.

12

Multiway Cut and important cuts

Pushing Lemma

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Proof: Let R be the vertices reachable fromt inG \S for a solutionS.

R⁰ R t

δ(R)is not important, then there is an important cut δ(R⁰) with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|. ReplaceS with

S⁰ := (S\δ(R))∪δ(R⁰) ⇒ |S⁰| ≤ |S|

S⁰ is a multiway cut: (1) There is not-u path inG\S⁰ and (2) a u-v path inG \S⁰ implies at-u path, a contradiction.

Multiway Cut and important cuts

Pushing Lemma

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Proof: Let R be the vertices reachable fromt inG \S for a solutionS.

R⁰ R

t u

v

δ(R)is not important, then there is an important cut δ(R⁰) with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|. ReplaceS with

S⁰ := (S\δ(R))∪δ(R⁰) ⇒ |S⁰| ≤ |S|

S⁰ is a multiway cut: (1) There is not-u path inG\S⁰ and (2) a u-v path inG \S⁰ implies a t-u path, a contradiction.

12

Multiway Cut and important cuts

Pushing Lemma

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Proof: Let R be the vertices reachable fromt inG \S for a solutionS.

t u

R v R⁰

δ(R)is not important, then there is an important cut δ(R⁰) with R ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|. ReplaceS with

S⁰ := (S\δ(R))∪δ(R⁰) ⇒ |S⁰| ≤ |S|

Algorithm for Multiway Cut

1 If every vertex of T is in a different component, then we are done.

2 Let t∈T be a vertex that is not separated from everyT \t.

3 Branch on a choice of an important(t,T \t) cut S of size at most k.

4 Set G :=G\S andk :=k− |S|.

5 Go to step 1.

We branch into at most4^k directions at most k times.

(Better analysis gives4^k bound on the size of the search tree.)

13

Multicut

Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integerk

Find: A set S of edges such that G\S has no s_i-t_i path for any i.

Theorem

Multicutcan be solved in timef(k, `)·n<sup>O(1)</sup> (FPT parameterized by combined parametersk and`).

Proof: The solution partitions{s₁,t₁, . . . ,s_{`</sub>,t<sub>`}}into components. Guess this partition, contract the vertices in a class, and solve Multiway Cut.

Theorem[Bousquet, Daligault, Thomassé 2011] [M., Razgon 2011] Multicutis FPT parameterized by the sizek of the solution.

Multicut

Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integerk

Find: A set S of edges such that G\S has no s_i-t_i path for any i.

Theorem

Multicutcan be solved in timef(k, `)·n<sup>O(1)</sup> (FPT parameterized by combined parametersk and`).

Proof: The solution partitions{s₁,t₁, . . . ,s_{`</sub>,t<sub>`}}into components.

Guess this partition, contract the vertices in a class, and solve Multiway Cut.

Theorem[Bousquet, Daligault, Thomassé 2011] [M., Razgon 2011]

Multicutis FPT parameterized by the sizek of the solution.

14

Directed graphs

Definition: ~δ(R) is the set of edgesleavingR.

Observation: Every inclusionwise-minimal directed (X,Y)-cutS can be expressed asS =~δ(R) for some X ⊆R andR∩Y =∅.

Definition: A minimal(X,Y)-cut~δ(R) isimportant if there is no (X,Y)-cut~δ(R⁰)with R ⊂R⁰ and|~δ(R⁰)|≤ |~δ(R)|.

R

~δ(R)

Y X

Directed graphs

Definition: ~δ(R) is the set of edgesleavingR.

Observation: Every inclusionwise-minimal directed (X,Y)-cutS can be expressed asS =~δ(R) for some X ⊆R andR∩Y =∅.

Definition: A minimal(X,Y)-cut~δ(R) isimportant if there is no (X,Y)-cut~δ(R⁰)with R ⊂R⁰ and|~δ(R⁰)|≤ |~δ(R)|.

R⁰

~δ(R⁰)

R

~δ(R)

Y X

15

Directed graphs

Definition: ~δ(R) is the set of edgesleavingR.

Observation: Every inclusionwise-minimal directed (X,Y)-cutS can be expressed asS =~δ(R) for some X ⊆R andR∩Y =∅.

Definition: A minimal(X,Y)-cut~δ(R) isimportant if there is no (X,Y)-cut~δ(R⁰)with R ⊂R⁰ and|~δ(R⁰)|≤ |~δ(R)|.

The proof for the undirected case goes through for the directed case:

Theorem

There are at most4^k importantdirected(X,Y)-cuts of size at mostk.

Directed Multiway Cut

The undirected approach does not work: the pushing lemma is not true.

Pushing Lemma (for undirected graphs)

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Directed counterexample:

s t

a

b

Unique solution with k = 1 edges, but it is not an important cut (boundary of{s,a}, but the boundary of{s,a,b} has same size).

16

Directed Multiway Cut

The undirected approach does not work: the pushing lemma is not true.

Pushing Lemma (for undirected graphs)

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Directed counterexample:

s t

a

b

Directed Multiway Cut

The undirected approach does not work: the pushing lemma is not true.

Pushing Lemma (for undirected graphs)

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Directed counterexample:

b a

t s

Unique solution with k = 1 edges, but it is not an important cut (boundary of{s,a}, but the boundary of{s,a,b} has same size).

16

Directed Multiway Cut

The undirected approach does not work: the pushing lemma is not true.

Pushing Lemma (for undirected graphs)

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Problem in the undirected proof:

v t u

R R⁰

Directed Multiway Cut

The undirected approach does not work: the pushing lemma is not true.

Pushing Lemma (for undirected graphs)

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-cut.

Using additional techniques, one can show:

Theorem[Chitnis, Hajiaghayi, M. 2011]

Directed Multiway Cutis FPT parameterized by the sizek of the solution.

16

Directed Multicut

Directed Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integerk

Find: A setS of edges such thatG\S has nos_i →t_i path for any i.

Theorem[M. and Razgon 2011]

Directed Multicutis W[1]-hard parameterized by k.

Directed Multicut

Directed Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integerk

Find: A setS of edges such thatG\S has nos_i →t_i path for any i.

Theorem[M. and Razgon 2011]

Directed Multicutis W[1]-hard parameterized by k.

But the case`=2can be reduced toDirected Multiway Cut:

t₁ s₁

t₂ s₂

17

Directed Multicut

Directed Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integerk

Find: A setS of edges such thatG\S has nos_i →t_i path for any i.

Theorem[M. and Razgon 2011]

Directed Multicutis W[1]-hard parameterized by k.

But the case`=2can be reduced toDirected Multiway Cut:

x y

s t

s₁ t₁

Directed Multicut

Directed Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integerk

Find: A setS of edges such thatG\S has nos_i →t_i path for any i.

Theorem[M. and Razgon 2011]

Directed Multicutis W[1]-hard parameterized by k.

But the case`=2can be reduced toDirected Multiway Cut:

x y

s₂ t₂

s₁ t₁

17

Directed Multicut

Directed Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integerk

Find: A setS of edges such thatG\S has nos_i →t_i path for any i.

Theorem[M. and Razgon 2011]

Directed Multicutis W[1]-hard parameterized by k.

Corollary

Directed Multicutwith `=2is FPT parameterized by the sizek of the solution.

?

Skew Multicut

Skew Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integer k Find: A setS of k directed edges such that G\S con-

tains nos_i →t_j path for any i ≥j.

t₄ t₃ t₂ t₁

s₄ s₃ s₂ s₁

18

Skew Multicut

Skew Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integer k Find: A setS of k directed edges such that G\S con-

tains nos_i →t_j path for any i ≥j.

t₄ t₃ t₂ t₁

s₄ s₃ s₂ s₁

Pushing Lemma

Skew Multcutproblem has a solution S that contains an important(s_{`</sub>,{t<sub>1</sub>, . . . ,t<sub>`}})-cut.

Skew Multicut

Skew Multicut

Input: GraphG, pairs(s₁,t₁),. . .,(s_{`</sub>,t<sub>`}), integer k Find: A setS of k directed edges such that G\S con-

tains nos_i →t_j path for any i ≥j.

t₄ t₃ t₂ t₁

s₄ s₃ s₂ s₁

Pushing Lemma

Skew Multcutproblem has a solution S that contains an important(s_{`</sub>,{t<sub>1</sub>, . . . ,t<sub>`}})-cut.

Theorem[Chen, Liu, Lu, O’Sullivan, Razgon 2008]

Skew Multicutcan be solved in time 4^k ·n^O(1).

18

Directed Feedback Vertex Set

Directed Feedback Vertex/Edge Set Input: Directed graphG, integer k

Find: A setSofkvertices/edges such thatG\S is acyclic.

Note: Edge and vertex versions are equivalent, we will consider the edge version here.

Theorem[Chen, Liu, Lu, O’Sullivan, Razgon 2008]

Directed Feedback Edge Setis FPT parameterized by the sizek of the solution.

Solution uses the technique of

it er

The compression problem

Directed Feedback Edge Set Compression Input: Directed graphG, integer k,

a set W of k+1 edges such that G\W is acyclic

Find: A set S of k edges such that G \S is acyclic.

Easier than the original problem, as the extra inputW gives us useful structural information aboutG.

Lemma

The compression problem is FPT parameterized byk.

A useful trick for edge deletion problems: we define the

compression problem in a way that a solution ofk+1 vertices are given and we have to find a solution ofk edges.

20

The compression problem

Directed Feedback Edge Set Compression Input: Directed graphG, integer k,

a setW ofk+1verticessuch thatG\W is acyclic

Find: A set S of k edges such that G \S is acyclic.

Easier than the original problem, as the extra inputW gives us useful structural information aboutG.

Lemma

The compression problem is FPT parameterized byk.

A useful trick for edge deletion problems: we define the

The compression problem

Proof: Let W ={w₁, . . . ,w_k+1} Let us split eachw_i into an edge−→

t_is_i.

t₄ s₁

t₁ t₂s₂ t₃s₃ s₄

By guessing the order of {w₁, . . . ,w_k+1} in the acyclic ordering of G\S, we can assume thatw₁ <w₂ <· · ·<w_k+1 in G\S [(k+1)!possibilities].

⇒We can solve the compression problem by (k+1)!applications ofSkew Multicut.

21

The compression problem

Proof: Let W ={w₁, . . . ,w_k+1} Let us split eachw_i into an edge−→

t_is_i.

t₄ s₁

t₁ t₂s₂ t₃s₃ s₄ Claim:

G \S is acyclic and has an ordering with w1 <w2 <· · ·<w_k+1

⇓

S covers every s_i →t_j path for every i ≥j

⇓ G\S is acyclic

⇒We can solve the compression problem by (k+1)!applications ofSkew Multicut.

The compression problem

Proof: Let W ={w₁, . . . ,w_k+1} Let us split eachw_i into an edge−→

t_is_i.

s₄ t₃s₃ t₂s₂

t₁s₁ t₄

Claim:

G \S is acyclic and has an ordering with w1 <w2 <· · ·<w_k+1

⇓

S covers every s_i →t_j path for every i ≥j

⇓ G\S is acyclic

⇒We can solve the compression problem by (k+1)!applications ofSkew Multicut.

21

The compression problem

Proof: Let W ={w₁, . . . ,w_k+1} Let us split eachw_i into an edge−→

t_is_i.

s₄ t₃s₃ t₂s₂

t₁s₁ t₄

Claim:

G \S is acyclic and has an ordering with w1 <w2 <· · ·<w_k+1

⇓

S covers every s_i →t_j path for every i ≥j

⇓ G\S is acyclic

Iterative compression

We have given af(k)n^O(1) algorithm for the following problem:

Directed Feedback Edge Set Compression Input: Directed graphG, integer k,

a setW ofk+1 vertices such thatG\W is acyclic

Find: A set S of k edges such that G \S is acyclic.

Nice, but how do we get a solutionW of size k+1?

We get it for free!

Powerful technique:

it er

22

Iterative compression

We have given af(k)n^O(1) algorithm for the following problem:

Directed Feedback Edge Set Compression Input: Directed graphG, integer k,

a setW ofk+1 vertices such thatG\W is acyclic

Find: A set S of k edges such that G \S is acyclic.

Nice, but how do we get a solutionW of size k+1?

We get it for free!

Powerful technique:

it er

Iterative compression

Letv₁,. . .,v_nbe the edges of G and letG_i be the subgraph induced by{v₁, . . . ,v_i}.

For everyi =1, . . . ,n, we find a setS_i of at mostk edges such thatG_i\S_i is acyclic.

For i =1, we have the trivial solutionS_i =∅.

Suppose we have a solutionS_i for G_i. LetW_i contain the head of each edge in S_i. ThenW_i∪ {v_i+1} is a set of at mostk+1 vertices whose removal makes Gi+1 acyclic.

Use the compression algorithm for G_i+1 with the set W_i∪ {v_i+1}.

If there is no solution of sizek forGi+1, then we can stop. Otherwise the compression algorithm gives a solutionSi+1of sizek forGi+1.

We call the compression algorithmn times, everything else is polynomial.

⇒Directed Feedback Edge Set is FPT.

23

Iterative compression

Letv₁,. . .,v_nbe the edges of G and letG_i be the subgraph induced by{v₁, . . . ,v_i}.

For everyi =1, . . . ,n, we find a setS_i of at mostk edges such thatG_i\S_i is acyclic.

For i =1, we have the trivial solutionS_i =∅.

Suppose we have a solutionS_i for G_i. LetW_i contain the head of each edge in S_i. ThenW_i∪ {v_i+1} is a set of at mostk+1 vertices whose removal makes G_i+1 acyclic.

Use the compression algorithm for G_i+1 with the set W_i∪ {v_i+1}.

If there is no solution of sizek forGi+1, then we can stop.

Otherwise the compression algorithm gives a solutionSi+1of sizek forGi+1.

We call the compression algorithmn times, everything else is

Outline

So far we have seen:

Definition of important cuts.

Combinatorial bound on the number of important cuts.

Pushing argument: we can assume that the solution contains an important cut. SolvesMultiway Cut,Skew

Multiway Cut.

Iterative compression reduces Directed Feedback Vertex Set toSkew Multiway Cut.

Next:

Randomized sampling of important separators.

24

Randomized sampling of important cuts

A new technique used by several results:

Multicut[M. and Razgon STOC 2011]

Clustering problems [Lokshtanov and M. ICALP 2011]

Directed Multiway Cut [Chitnis, Hajiaghayi, M. SODA 2012]

Directed Multicut in DAGs [Kratsch, Pilipczuk, Pilipczuk, Wahlström ICALP 2012]

Directed Subset Feedback Vertex Set ^[Chitnis, Cygan, Hajiaghayi, M. ICALP 2012]

Parity Multiway Cut [Lokshtanov, Ramanujan ICALP 2012]

List homomorphism removal problems [Chitnis, Egri, and M.

ESA 2013]

Clustering

We want to partition objects into clusters subject to certain requirements (typically: related objects are clustered together, bounds on the number or size of the clusters etc.)

(p,q)-clustering

Input: A graphG, integers p,q.

Find:

A partition(V₁, . . . ,V_m)of V(G)such that for every i

|V_i| ≤p and δ(V_i)≤q.

δ(V_i): number of edges leavingV_i. Theorem[Lokshtanov and M. 2011]

(p,q)-clusteringcan be solved in time 2^O(q)·n^O(1).

26

A sufficient and necessary condition

Good cluster: size at most p and at most q edges leaving it.

Necessary condition:

Every vertex is contained in a good cluster.

But surprisingly, this is also asufficient condition! Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

A sufficient and necessary condition

Good cluster: size at most p and at most q edges leaving it.

Necessary condition:

Every vertex is contained in a good cluster.

But surprisingly, this is also asufficient condition!

Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

27

A sufficient and necessary condition

Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

Proof: Find a collection of good clusters covering every vertex and having minimum total size. Suppose two clusters intersect.

X Y

δ(X) +δ(Y)≥δ(X\Y) +δ(Y \X) (posimodularity)

A sufficient and necessary condition

Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

Proof: Find a collection of good clusters covering every vertex and having minimum total size. Suppose two clusters intersect.

X Y

δ(X) +δ(Y)≥δ(X\Y) +δ(Y \X) (posimodularity)

⇒either δ(X)≥δ(X \Y) or δ(Y)≥δ(Y \X) holds.

28

A sufficient and necessary condition

Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

Proof: Find a collection of good clusters covering every vertex and having minimum total size. Suppose two clusters intersect.

X \Y Y

δ(X) +δ(Y)≥δ(X\Y) +δ(Y \X) (posimodularity)

A sufficient and necessary condition

Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

Proof: Find a collection of good clusters covering every vertex and having minimum total size. Suppose two clusters intersect.

X Y \X

δ(X) +δ(Y)≥δ(X\Y) +δ(Y \X) (posimodularity)

Ifδ(Y)≥δ(Y \X), replace Y withY \X,

strictly decreasing the total size of the clusters. QED 28

Finding a good cluster

We have seen:

Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

All we have to do is to check if a given vertexv is in a good cluster. Trivial to do in timen^O(q).

We prove next: Lemma

We can check in time2^O(q)·n^O(1) ifv is in a good cluster.

Finding a good cluster

We have seen:

Lemma

GraphG has a(p,q)-clustering if and only if every vertex is in a good cluster.

All we have to do is to check if a given vertexv is in a good cluster. Trivial to do in timen^O(q).

We prove next:

Lemma

We can check in time2^O(q)·n^O(1) ifv is in a good cluster.

29

Important sets

Definition

Fix a distinguished vertexv in a graphG. A set X ⊆V(G) is an important setif

v 6∈X,

there is no set X ⊂X⁰ with v 6∈X andδ(X⁰)≤δ(X).

v

Observation: X is an important set if and only ifδ(X) is an important(x,v)-cut for everyx ∈X.

Consequence: Every vertex is contained in at most4^k important sets.

Important sets

Definition

Fix a distinguished vertexv in a graphG. A set X ⊆V(G) is an important setif

v 6∈X,

there is no set X ⊂X⁰ with v 6∈X andδ(X⁰)≤δ(X).

v

Observation: X is an important set if and only ifδ(X) is an important(x,v)-cut for everyx ∈X.

Consequence: Every vertex is contained in at most4^k important sets.

30

Important sets

Definition

Fix a distinguished vertexv in a graphG. A set X ⊆V(G) is an important setif

v 6∈X,

there is no set X ⊂X⁰ with v 6∈X andδ(X⁰)≤δ(X).

v

Observation: X is an important set if and only ifδ(X) is an important(x,v)-cut for everyx ∈X.

Consequence: Every vertex is contained in at most4^k important sets.

Important sets

Definition

Fix a distinguished vertexv in a graphG. A set X ⊆V(G) is an important setif

v 6∈X,

there is no set X ⊂X⁰ with v 6∈X andδ(X⁰)≤δ(X).

v

Observation: X is an important set if and only ifδ(X) is an important(x,v)-cut for everyx ∈X.

Consequence: Every vertex is contained in at most4^k important sets.

30

Important sets

Definition

Fix a distinguished vertexv in a graphG. A set X ⊆V(G) is an important setif

v 6∈X,

there is no set X ⊂X⁰ with v 6∈X andδ(X⁰)≤δ(X).

v

Observation: X is an important set if and only ifδ(X) is an important(x,v)-cut for everyx ∈X.

Consequence: Every vertex is contained in at most4^k important sets.

Important sets

Definition

Fix a distinguished vertexv in a graphG. A set X ⊆V(G) is an important setif

v 6∈X,

there is no set X ⊂X⁰ with v 6∈X andδ(X⁰)≤δ(X).

v

Observation: X is an important set if and only ifδ(X) is an important(x,v)-cut for everyx ∈X.

Consequence: Every vertex is contained in at most4^k important sets.

30

Pushing argument

Lemma

IfC is a good cluster of minimum size containing v, then every component ofG \C is an important set.

v

ThusC can be obtained by removing at mostq important sets from V(G) (but there aren^O(q) possibilities, we cannot try all of them).

Pushing argument

Lemma

IfC is a good cluster of minimum size containing v, then every component ofG \C is an important set.

v

ThusC can be obtained by removing at mostq important sets from V(G) (but there aren^O(q) possibilities, we cannot try all of them).

31

Pushing argument

Lemma

IfC is a good cluster of minimum size containing v, then every component ofG \C is an important set.

v

ThusC can be obtained by removing at mostq important sets from V(G) (but there aren^O(q) possibilities, we cannot try all of them).

Pushing argument

Lemma

IfC is a good cluster of minimum size containing v, then every component ofG \C is an important set.

v

ThusC can be obtained by removing at mostq important sets from V(G) (but there aren^O(q) possibilities, we cannot try all of them).

31

Pushing argument

Lemma

IfC is a good cluster of minimum size containing v, then every component ofG \C is an important set.

v

ThusC can be obtained by removing at mostq important sets from V(G) (but there aren^O(q) possibilities, we cannot try all of them).

Pushing argument

Lemma

IfC is a good cluster of minimum size containing v, then every component ofG \C is an important set.

v

ThusC can be obtained by removing at mostq important sets from V(G) (but there aren^O(q) possibilities, we cannot try all of them).

31

Random sampling

Let X be the set of all important sets of boundary size at most q in G.

Let X⁰⊆ X contain each set with probability ¹₂ independently.

Let Z =S

X∈X⁰X.

Let B be the set of vertices inC with neighbors outside C. Lemma

LetC be a good cluster of minimum size containingv. With probability2^−2O(q),Z coversG \C and is disjoint from B.

B

Random sampling

Let X be the set of all important sets of boundary size at most q in G.

Let X⁰⊆ X contain each set with probability ¹₂ independently.

Let Z =S

X∈X⁰X.

Let B be the set of vertices inC with neighbors outside C. Lemma

LetC be a good cluster of minimum size containingv. With probability2^−2O(q),Z coversG \C and is disjoint from B.

v B

32

Random sampling

Lemma

LetC be a good cluster of minimum size containingv. With probability2^−2O(q),Z coversG \C and is disjoint from B.

Two events:

(E1) Z coversG\C.

Each of the at most q components is an important set

⇒ all of them are selected by probability at least2^−q. (E2) Z is disjoint fromB.

Each vertex ofB is in at most 4^q members of X

⇒ all of them are selected by probability at least2^−q4q. The two events are independent (involve different sets ofX), thus the claimed probability follows.

Finding good clusters

LetC be a good cluster of minimum size containingv and assume G \C is covered by Z, and

Z is disjoint fromB (hence no edge going out of C is contained inZ).

v Z

G\Z

Where is the good clusterC in the figure?

Observe: Components ofZ are either fully in the cluster or fully outside the cluster. What is this problem?

34

Finding good clusters

LetC be a good cluster of minimum size containingv and assume G \C is covered by Z, and

Z is disjoint fromB (hence no edge going out of C is contained inZ).

v Z

G\Z

Where is the good clusterC in the figure?

Finding good clusters

LetC be a good cluster of minimum size containingv and assume G \C is covered by Z, and

Z is disjoint fromB (hence no edge going out of C is contained inZ).

v Z

G\Z

KNAPSACK!

34

Finding good clusters by Knapsack

v Z

G\Z

We interpret the componenentsV₁,. . .,V_t ofG[Z]as items:

V_i has value δ(V_i) and V_i has weight |V_i|.

Finding good clusters by Knapsack

v Z

G\Z

Standard DP solves it in polynomial time: letT[i,j]be the maximum value of a subset of the firsti items having total weight at mostj. Recurrence:

T[i,j] =max{T[i−1,j],T[i−1,j− |V_i|] +δ(V_i)}

35

Summary of algorithm

(p,q)-clustering

Input: A graphG, integers p,q.

Find:

A partition(V1, . . . ,Vm)of V(G)such that for every i

|V_i| ≤p and δ(V_i)≤q.

It is sufficient to check for each vertex v if it is in a good cluster.

Enumerate all the important sets.

Let Z be the union of random important sets.

The solution is obtained by extendingG \Z with some of the components ofG[Z].

(p, q) -clustering

With a slightly different probability distribution, one can reduce the error probability to 2^−O(q).

Derandomization is possible using standard techniques, but nontrivial to obtain 2^O(q) running time.

Other variants: maximum degree in the cluster is at most p, etc.

37

Summary

A simple (but essentially tight) bound on the number of important cuts.

Algorithmic results: FPT algorithms for Multiway Cutin undirected graphs, Skew Multicutin directed graphs,

Directed Feedback Vertex/Edge Set, and (p,q)-Clustering.