Multiway Cut and important cuts 19

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

19

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

19

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

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

19

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

19

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

²⁰

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

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

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

S∈S₂

4^−|S|≤2^−(λ+1)=2^−λ/2.

A refined bound

21

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

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

S∈S2

4^−|S|≤2^−(λ+1)=2^−λ/2.

A refined bound

21

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

²²

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

23

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

23

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

23

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

²⁴

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

25

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

25

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

25

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

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

26

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

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

26

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

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

26

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

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

26

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

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

26

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

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

26

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

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

26

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

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

26

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

26

s

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

26

s

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

26

In document Important separators and parameterized algorithms (Pldal 49-75)