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 Będlewo, Poland

August 19, 2014 1

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.

Random selection of important separators: a new tool with many applications.

Overview

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

Minimum cuts

Theorem

A minimum(X,Y)-cut can be found in polynomial time.

Theorem

The size of a minimum(X,Y)-cut equals the maximum size of a pairwise edge-disjoint collection ofX −Y paths.

R δ(R) X Y

Minimum cuts

There is a long list of algorithms for finding disjoint paths and minimum cuts.

Edmonds-Karp: O(|V(G)| · |E(G)|²) Dinitz: O(|V(G)|²· |E(G)|)

Push-relabel: O(|V(G)|³)

Orlin-King-Rao-Tarjan: O(|V(G)| · |E(G)|) . . .

But we need only the following result:

Theorem

An(X,Y)-cut of size at mostk (if exists) can be found in time O(k·(|V(G)|+|E(G)|)).

Finding minimum cuts

Theorem

An(X,Y)-cut of size at mostk (if exists) can be found in time O(k·(|V(G)|+|E(G)|)).

We try to grow a collectionP of edge-disjointX −Y paths.

Residual graph:

not used byP: bidirected,

used by P: directed in the opposite direction.

X Y X Y

original graph residual graph

If we cannot find an augmenting path, we can find a (minimum) cut of size|P|.

Finding minimum cuts

Theorem

An(X,Y)-cut of size at mostk (if exists) can be found in time O(k·(|V(G)|+|E(G)|)).

We try to grow a collectionP of edge-disjointX −Y paths.

Residual graph:

not used byP: bidirected,

used by P: directed in the opposite direction.

X Y X Y

original graph residual graph

If we cannot find an augmenting path, we can find a (minimum) cut of size|P|.

Finding minimum cuts

Theorem

An(X,Y)-cut of size at mostk (if exists) can be found in time O(k·(|V(G)|+|E(G)|)).

We try to grow a collectionP of edge-disjointX −Y paths.

Residual graph:

not used byP: bidirected,

used by P: directed in the opposite direction.

X Y X Y

original graph residual graph

If we cannot find an augmenting path, we can find a (minimum) cut of size|P|.

Finding minimum cuts

Theorem

An(X,Y)-cut of size at mostk (if exists) can be found in time O(k·(|V(G)|+|E(G)|)).

We try to grow a collectionP of edge-disjointX −Y paths.

Residual graph:

not used byP: bidirected,

used by P: directed in the opposite direction.

X Y X Y

original graph residual graph

If we cannot find an augmenting path, we can find a (minimum) cut of size|P|.

Finding minimum cuts

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.

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

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

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

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 R_max⊇X such that δ(R_max) 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.

Submodularity

Lemma

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

Submodularity

Lemma

Given a graphG and sets X,Y ⊆V(G), the sets R_min andR_max can be found in polynomial time.

Proof: Iteratively add vertices to X if they do not increase the minimumX −Y cut size. When the process stops, X =R_max. Similar forR_min.

But we can do better!

Finding R

and R

Lemma

Given a graphG and sets X,Y ⊆V(G), the sets R_min andR_max can be found inO(λ·(|V(G)|+|E(G)|))time, whereλis the minimumX −Y cut size.

Proof: Look at the residual graph.

X Y X Y

original graph residual graph

Rmin Rmax Rmin Rmax

R_min: vertices reachable from X.

Rmax: vertices from whichY isnotreachable.

Finding R

and R

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

Important cuts

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

Important cuts

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

Important cuts

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

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.

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.

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.

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.

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.

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.

Important cuts

We are using the following two statements:

Branch 1: If uv ∈S, then Sis an important(X,Y)-cut inG

S \ uv is an important (X,Y)-cut in G\uv

Converse is not true:

Set {ab,ay} is important (X,Y)-cut in G\xb, but {xb,ab,ay} is not an impor- tant(X,Y)-cut in G.

Branch 2: If S is an(X ∪v,Y)-cut, then Sis an important(X,Y)-cut

inG

S is an important(X∪v,Y)- cut in G

Converse is true!

Important cuts — some details

We are using the following two statements:

Branch 1: If uv ∈S, then Sis an important(X,Y)-cut inG

S \ uv is an important (X,Y)-cut in G\uv

Converse is not true:

Set {ab,ay} is important (X,Y)-cut in G\xb, but {xb,ab,ay} is not an impor-

tant(X,Y)-cut in G. X Y

a

c

x b y

Branch 2: If S is an(X ∪v,Y)-cut, then Sis an important(X,Y)-cut

inG

S is an important(X∪v,Y)- cut in G

Converse is true!

Important cuts — some details

We are using the following two statements:

Branch 1: If uv ∈S, then Sis an important(X,Y)-cut inG

S \ uv is an important (X,Y)-cut in G\uv

Converse is not true:

Set {ab,ay} is important (X,Y)-cut in G\xb, but {xb,ab,ay} is not an impor-

tant(X,Y)-cut in G. X Y

a

c

x b y

Branch 2: If S is an(X ∪v,Y)-cut, then Sis an important(X,Y)-cut

inG

S is an important(X∪v,Y)- cut in G

Converse is true!

Important cuts — some details

Theorem

There are at most4^k important(X,Y)-cuts of size at mostk and they can be enumerated in timeO(4^k·k·(|V(G)|+|E(G)|)).

Algorithm for enumerating important cuts:

1 Handle trivial cases (k =0,λ=0,k < λ)

2 FindR_max.

3 Choose an edge uv ofδ(Rmax).

Recurse on(G−uv,Rmax,Y,k−1).

Recurse on(G,Rmax∪v,Y,k).

4 Check if the returned cuts are important and throw away those that are not.

Important cuts — algorithm

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

Ck−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 Ck−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 Ck−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 Ck−1= 1

k

2k−2 k−1

≥4^k/poly(k).

Important cuts

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 ·k³·(|V(G)|+|E(G)|).

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.

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.

Multiway Cut

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.

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.

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|

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.

Multiway Cut and important cuts

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: 4^k2·n^O(1) running time.

Next: Better analysis gives 4^k bound on the size of the search tree.

Algorithm for Multiway Cut

We have seen: at most4^k important cut of size at most k.

Better bound:

Lemma

IfS is the set of all important(X,Y)-cuts, thenP

S∈S4^−|S|≤1 holds.

Proof: We show the stronger statement P

S∈S4^−|S|≤2^−λ, where λis the minimum(X,Y)-cut size.

Branch 1: removinguv.

λincreases by at most one and we add the edgeuv to each separator, increasing the cut by one. Thus the total contribution is

X

S∈S1

4^−(|S|+1)= X

S∈S1

4^−|S|/4≤2^−(λ−1)/4=2^−λ/2.

Branch 2: replacing X with X ∪v.

λincreases by at least one. Thus the total contribution is X

S∈S₂

4^−|S|≤2^−(λ+1)=2^−λ/2.

A refined bound

Lemma

IfS is the set of all important(X,Y)-cuts, thenP

S∈S4^−|S|≤1 holds.

Proof: We show the stronger statement P

S∈S4^−|S|≤2^−λ, where λis the minimum (X,Y)-cut size.

Branch 1: removinguv.

λincreases by at most one and we add the edgeuv to each separator, increasing the cut by one. Thus the total contribution is

X

S∈S₁

4^−(|S|+1)= X

S∈S₁

4^−|S|/4≤2^−(λ−1)/4=2^−λ/2.

Branch 2: replacing X with X ∪v.

λincreases by at least one. Thus the total contribution is X

S∈S2

4^−|S|≤2^−(λ+1)=2^−λ/2.

A refined bound

Lemma

The search tree for theMultiway Cutalgorithm has4^k leaves.

Proof: Let L_k be the maximum number of leaves with parameter k. We proveL_k ≤4^k by induction. After enumerating the setS_k of important separators of size≤k, we branch into|S_k|directions.

X

S∈Sk

4^k−|S|=4^k· X

S∈Sk

4^−|S|≤4^k

Still need: bound the work at each node.

Refined analysis for Multiway Cut

We have seen:

Lemma

We can enumerate every important(X,Y)-cut of size at mostk in timeO(4^k ·k·(|V(G)|+|E(G)|)).

Problem: running time at a node of the recursion tree is not linear in the number children.

Refined enumeration algorithms

We have seen:

Lemma

We can enumerate every important(X,Y)-cut of size at mostk in timeO(4^k ·k·(|V(G)|+|E(G)|)).

Problem: running time at a node of the recursion tree is not linear in the number children.

Easily follows:

Lemma

We can enumerate a supersetS_k⁰ of every important (X,Y)-cut of size at mostk in timeO(|S_k⁰| ·k²·(|V(G)|+|E(G)|))such that P

S∈S_k⁰ 4^−|S|≤1holds.

Refined enumeration algorithms

We have seen:

Lemma

We can enumerate every important(X,Y)-cut of size at mostk in timeO(4^k ·k·(|V(G)|+|E(G)|)).

Problem: running time at a node of the recursion tree is not linear in the number children.

Needs more work:

Lemma

We can enumerate the setS_k of every important (X,Y)-cut of size at mostk in time O(|S_k| ·k²·(|V(G)|+|E(G)|)).

Refined enumeration algorithms

Theorem

Multiway Cutcan be solved in time O(4^k ·k³·(|V(G)|+|E(G)|)).

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.

Algorithm for Multiway Cut

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.

Simple application

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.

Simple application

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.

Simple application

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation

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.

Anti isolation