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

School on Parameterized Algorithms and Complexity

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

Definition

A minimal(X,Y)-cutδ(R)is importantif there is no(X,Y)-cut δ(R⁰) withR ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|.

Note: Can be checked in polynomial time if a cut is important (δ(R) is important ifR =R_max).

R δ(R) X Y

Definition

A minimal(X,Y)-cutδ(R)is importantif there is no(X,Y)-cut δ(R⁰) withR ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|.

Note: Can be checked in polynomial time if a cut is important (δ(R) is important ifR =R_max).

R⁰ δ(R)

R

δ(R⁰)

X Y

Definition

A minimal(X,Y)-cutδ(R)is importantif there is no(X,Y)-cut δ(R⁰) withR ⊂R⁰ and|δ(R⁰)|≤ |δ(R)|.

Note: Can be checked in polynomial time if a cut is important (δ(R) is important ifR =R_max).

R δ(R)

X Y

Theorem

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

R δ(R)

X Y

Lemma:

At mostk·4^k edges incident to t can be part of an inclusionwise minimals−t cut of size at most k.

Proof: We show that every such edge is contained in an important (s,t)-cut of size at mostk.

Suppose thatvt ∈δ(R) and|δ(R)|=k.

There is an important(s,t)-cutδ(R⁰)withR ⊆R⁰ and|δ(R⁰)|≤k. Clearly,vt ∈δ(R⁰): v ∈R, hencev ∈R⁰.

Lemma:

At mostk·4^k edges incident to t can be part of an inclusionwise minimals−t cut of size at most k.

Proof: We show that every such edge is contained in an important (s,t)-cut of size at mostk.

v

R t

s

Suppose thatvt ∈δ(R) and|δ(R)|=k.

There is an important(s,t)-cutδ(R⁰)withR ⊆R⁰ and|δ(R⁰)|≤k. Clearly,vt ∈δ(R⁰): v ∈R, hencev ∈R⁰.

Lemma:

At mostk·4^k edges incident to t can be part of an inclusionwise minimals−t cut of size at most k.

Proof: We show that every such edge is contained in an important (s,t)-cut of size at mostk.

v R

R⁰

s t

Suppose thatvt ∈δ(R) and|δ(R)|=k.

There is an important(s,t)-cutδ(R⁰)withR ⊆R⁰ and|δ(R⁰)|≤k.

Clearly,vt ∈δ(R⁰): v ∈R, hencev ∈R⁰.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

Is the opposite possible, i.e.,S_i separates everyt_j except t_i? Lemma

IfS_i separatest_j froms if and only j 6=i and every S_i has size at mostk, then n≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

S1

Is the opposite possible, i.e.,S_i separates everyt_j except t_i? Lemma

IfS_i separatest_j froms if and only j 6=i and every S_i has size at mostk, then n≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

S2

Is the opposite possible, i.e.,S_i separates everyt_j except t_i? Lemma

IfS_i separatest_j froms if and only j 6=i and every S_i has size at mostk, then n≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

S3

Is the opposite possible, i.e.,S_i separates everyt_j except t_i? Lemma

IfS_i separatest_j froms if and only j 6=i and every S_i has size at mostk, then n≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

S1

Is the opposite possible, i.e.,S_i separates everyt_j exceptt_i?

Lemma

IfS_i separatest_j froms if and only j 6=i and every S_i has size at mostk, then n≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

S2

Is the opposite possible, i.e.,S_i separates everyt_j exceptt_i?

Lemma

IfS_i separatest_j froms if and only j 6=i and every S_i has size at mostk, then n≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

S3

Is the opposite possible, i.e.,S_i separates everyt_j exceptt_i?

Lemma

IfS_i separatest_j froms if and only j 6=i and every S_i has size at mostk, then n≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lets,t₁, . . . ,t_n be vertices andS₁, . . . ,S_n be sets of at mostk edges such thatS_i separates t_i froms, butS_i does notseparatet_j froms for any j 6=i.

It is possible thatn is “large” even ifk is “small.”

s

t6

t5

t4

t3

t2

t1

S3

Is the opposite possible, i.e.,S_i separates everyt_j exceptt_i? Lemma

IfS_i separatest_j froms if and only j 6=i and everyS_i has size at mostk, thenn ≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

t1 t2 t3 t4 t5 t6

s t

S3

Is the opposite possible, i.e.,S_i separates everyt_j exceptt_i? Lemma

IfS_i separatest_j froms if and only j 6=i and everyS_i has size at mostk, thenn ≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

s

t6

t5

t4

t3

t2

t1

t

S2

Is the opposite possible, i.e.,S_i separates everyt_j exceptt_i? Lemma

IfS_i separatest_j froms if and only j 6=i and everyS_i has size at mostk, thenn ≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

s

t₆ t₅ t₄ t₃ t₂ t₁

t

S₁

Is the opposite possible, i.e.,S_i separates everyt_j exceptt_i? Lemma

IfS_i separatest_j froms if and only j 6=i and everyS_i has size at mostk, thenn ≤(k+1)·4^k+1.

Proof: Add a new vertex t. Every edge tt_i is part of an

(inclusionwise minimal)(s,t)-cut of size at mostk+1. Use the previous lemma.

Lemma

IfS_i separatest_j froms if and only j 6=i and everyS_i has size at mostk, thenn ≤(k+1)·4^k+1.

Lower bound: in a binary tree of heightk, any of the2^k leaves can be the only reachable leaf after removingk edges.

s

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 .

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

Theorem

Multiway Cuton planar graphs can be solved in time 2^O(|T|)·n^O(

√|T|).

Theorem

Multiway Cuton planar graphs is W[1]-hard parameterized by

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.

Pushing Lemma

Lett∈T. The Multiway Cutproblem has a solutionS that contains an important(t,T \t)-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 can give a4^k bound on the size of the search tree.

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.

Much more involved: Theorem

Multicutis FPT parameterized by the sizek of the solution.

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.

Much more involved: Theorem

Multicutis FPT parameterized by the sizek of the solution.

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.

Much more involved:

Theorem

Multicutis FPT parameterized by the sizek of the solution.

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

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

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.

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).

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).

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).

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

Replacing R by R⁰ cannot create a t → u path, but can create a u→t path.

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

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

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

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

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

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

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

t₁ s₁

t₂ s₂

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

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

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

x y

s₂ t₂

s₁ t₁

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

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

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

x y

s₂ t₂

s₁ t₁

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

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

Corollary

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

Open questions:

?

IsDirected Multicut FPT parameterized by k and`?

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

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

Theorem

Directed Multicutis NP-hard for `=2 on DAGs.

Theorem

Directed Multicutis FPT parameterized by k and`on DAGs.

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

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

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

Theorem

k O(1)

Pushing Lemma

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

Proof: Similar to the undirected pushing lemma. Let R be the vertices reachable fromt inG \S for a solutionS.

s` t`

t1

t2

. . . R

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 skew multicut: (1) There is nos_{`</sub>-t<sub>j</sub> path in G\S<sup>0</sup> for any j and (2) as<sub>i</sub>-t<sub>j</sub> path inG \S<sup>0</sup> implies a s<sub>`}-t_j path, a contradiction.

Pushing Lemma

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

Proof: Similar to the undirected pushing lemma. Let R be the vertices reachable fromt inG \S for a solutionS.

s` t`

t1

t2

. . . R

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|

S⁰ is a skew multicut: (1) There is nos_{`</sub>-t<sub>j</sub> path in G\S<sup>0</sup> for any j and (2) as<sub>i</sub>-t<sub>j</sub> path inG \S<sup>0</sup> implies a s<sub>`}-t_j path, a contradiction.

Pushing Lemma

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

Proof: Similar to the undirected pushing lemma. Let R be the vertices reachable fromt inG \S for a solutionS.

s` t`

t1

t2

. . . si

R

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|

S⁰ is a skew multicut: (1) There is nos_{`</sub>-t<sub>j</sub> path in G\S<sup>0</sup> for any j and (2) as<sub>i</sub>-t<sub>j</sub> path inG \S<sup>0</sup> implies a s<sub>`}-t_j path, a contradiction.

Pushing Lemma

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

Proof: Similar to the undirected pushing lemma. Let R be the vertices reachable fromt inG \S for a solutionS.

s` t`

t1

t2

. . . si

R

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|

S⁰ is a skew multicut: (1) There is nos_`-t_j path in G\S⁰ for any j

0

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

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

Solution uses the technique of

it er

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.

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

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

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.

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.

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.

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

of .

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

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

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 polynomial.

⇒Directed Feedback Edge Set is FPT.

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 polynomial.

⇒ is FPT.

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

Multicut.

Iterative compression reduces Directed Feedback Vertex Set toSkew Multicut.

Next:

Randomized sampling of important separators.