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

Is the opposite possible, i.e.,Si separates everytj except ti? Lemma

IfSi separatestj froms if and onlyj 6=i and every Si 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

Is the opposite possible, i.e.,Si separates everytj except ti? Lemma

IfSi separatestj froms if and onlyj 6=i and every Si 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.

26

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

Is the opposite possible, i.e.,Si separates everytj except ti? Lemma

IfSi separatestj froms if and onlyj 6=i and every Si 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

Is the opposite possible, i.e.,Si separates everytj except ti? Lemma

IfSi separatestj froms if and onlyj 6=i and every Si 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.

26

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

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

Lemma

IfSi separatestj froms if and onlyj 6=i and every Si 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

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

Lemma

IfSi separatestj froms if and onlyj 6=i and every Si 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.

26

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

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

Lemma

IfSi separatestj froms if and onlyj 6=i and every Si 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

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

IfSi separatestj froms if and onlyj 6=i and every Si 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.

26

Anti isolation

t1 t2 t3 t4 t5 t6

s t

S3

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

IfSi separatestj froms if and onlyj 6=i and every Si has size at mostk, then n≤(k+1)·4^k+1.

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

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

Anti isolation

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

IfSi separatestj froms if and onlyj 6=i and every Si has size at mostk, then n≤(k+1)·4^k+1.

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

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

26

Anti isolation

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

IfSi separatestj froms if and onlyj 6=i and every Si has size at mostk, then n≤(k+1)·4^k+1.

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

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

Multicut

Multicut

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

Find: A set S of edges such that G\S has no si-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

Multicutis FPT parameterized by the sizek of the solution.

27

Multicut

Multicut

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

Find: A set S of edges such that G\S has no si-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

Multicutis FPT parameterized by the sizek of the solution.

In document Minicourse on parameterized algorithms and complexity Part 6: Important Separators (Pldal 65-78)