LP-Based Parameterized Algorithms for Separation Problems D. Lokshtanov , N.S. Narayanaswamy V. Raman, M.S. Ramanujan S. Saurabh

Message of this talk

It was open for quite a while whether Odd Cycle Transversal and Almost 2-Sat are FPT .

A simple branching algorithm for Vertex Cover known since the mid 90’s solves both problems in time 𝑂 ∗ (4 𝑘 ) .

Some more work gives 𝑂 ∗ (2.32

𝑘 ) .

Results

( Above LP ) Multiway Cut 4 𝑘−𝐿𝑃

[CPPW11]

2.32

𝑘−𝐿𝑃 ( Above LP ) Vertex Cover Almost 2 -SAT 2.32

𝑘 2.32

𝑘 Odd Cycle Transversal

How does one get a 4 k-LP algorithm?

Branching: on both sides k-LP least ½ .

decreases by at How to improve ? Decrease k-LP more.

Multiway Cut

In:

Graph G , set T of vertices, integer k .

Question:

of G\S ∃𝑆 ⊆ 𝑉 𝐺 such that no component has at least two vertices of T ?

FPT by Marx , 04 Faster FPT by Chen et al, 07 Fastest FPT and FPT/ k-LP by Cygan et al, 11

Vertex Cover

In:

G , t

Question:

edge in G ∃𝑆 ⊆ 𝑉 𝐺 , 𝑆 ≤ t such that every has an endpoint in S ?

Long story...

Here: 𝑂 ∗ (2.32

𝑡−𝐿𝑃 ) .

Almost 2-SAT

In:

2 SAT formula 𝜓 , integer k

Question:

Can we remove k variables from 𝜓 and make it satisfiable?

FPT by Razgon and O’Sullivan , 08 Here: 𝑂 ∗ (2.32

𝑘 ) .

Odd Cycle Transversal

In:

G , k

Question:

bipartite?

∃𝑆 ⊆ 𝑉 𝐺 , 𝑆 ≤ k such that G\S is FPT: 𝑂 ∗ (3 𝑘 ) by Reed et al.

Here: 𝑂 ∗ (2.32

𝑘 ) .

Vertex Cover

In:

G , t

Question:

edge in G ∃𝑆 ⊆ 𝑉 𝐺 , 𝑆 ≤ t such that every has an endpoint in S ?

Minimize ∑𝑥𝑖 ∀𝑢𝑣 ∈ 𝐸 𝐺 : 𝑥𝑢 + 𝑥𝑣 ≥ 1 𝑥 𝑖 ≥ 0 𝑥 𝑖 ∈

Z

Vertex Cover Above LP

In:

G , t

Question:

edge in G ∃𝑆 ⊆ 𝑉 𝐺 , 𝑆 ≤ t such that every has an endpoint in S ?

Running Time:

𝑂 ∗ (𝑓 𝑡 − 𝐿𝑃 ) , where LP value of the optimum LP solution.

is the 𝜇 = 𝑡 − 𝐿𝑃

x z Odd Cycle Transversal  Almost 2-Sat y 𝑥 ∨ ¬𝑦 ¬𝑥 ∨ 𝑦 x 𝑥 ∨ ¬𝑧 ¬𝑥 ∨ 𝑧 𝑦 ∨ ¬𝑧 z y ¬𝑦 ∨ 𝑧

Almost 2-SAT 

Vertex Cover/

t-LP 𝑥 𝑥 ∨ 𝑦 𝑦 𝑦 ∨ ¬𝑧 𝑧 ¬𝑥 ¬𝑦 ¬𝑧

Nemhauser Trotter Theorem

(a) There is always an optimal solution to Vertex 1 Cover LP that sets variables to {0 , , 1} .

2 1 (b) For any {0 , , 1} –solution there is a matching 2 from the 1 -vertices to the 0 -vertices, saturating the 1 -vertices.

Nemhauser Trotter Proof

+ ϵ + ϵ ϵ ϵ ϵ < 𝟏 𝟐 > 𝟏 𝟐 𝟏 𝟐

Reduction Rule

If exists optimal LP solution that sets x v to 1 , then exists optimal vertex cover that selects v .

 Remove v from G and decrease t by 1 .

Correctness follows from Nemhauser Trotter Polynomial time by LP solving.

Branching

Pick an edge uv . Solve (G\u, t-1) and (G\v, t-1) .

LP(G\u) > LP(G) – 1 since otherwise there is an optimal LP solution for G that sets u to 1 .

Then LP(G\u) ≥ LP(G) − 1 2

Branching - Analysis

LP – t drops by ½ ... in both branches!

𝑇 𝜇 ≤ 2𝑇 𝜇 − 2 ≤ 4 𝜇 Total time: 𝑂 ∗ (4 𝑡−𝐿𝑃 )

Caveat: The reduction does not increase the measure!

Moral

Nemhauser Trotter

reduction + classic «

branch on an edge

» gives 𝑂 ∗ 4 𝐿𝑃−𝑡 time algorithm for Vertex Cover and 𝑂 ∗ 4 𝑘 time algorithm for Odd Cycle Transversal and Almost 2-Sat .

Can we do better?

Surplus

The

surplus

be negative!

of a set I is |N(I)| – |I| . The surplus can 1 In any {0 , , 1} -LP solution, the total weight is 2 n/2 + surplus(V 0 )/2 .

Solving the Vertex Cover LP an independent set I is equivalent to finding of minimum surplus.

Surplus and Reductions

If « all ½ » is the unique LP optimum then surplus(I) > 0 for all independent sets.

Can we say anything meaningful for independent sets of surplus 1? 2? k?

Surplus Branching Lemma

Let I be an independent set in G with minimum surplus . There exists an optimal vertex cover C that either contains I or avoids I .

Surplus Branching Lemma Proof

𝐶 ∩ 𝐼 𝐶\𝐼 I N(I) R

Branching Rule

Find an independent set Solve (G\I, t-|I|) I of minimum surplus. and (G\N(I), t-|N(I)|) .

LP(G\I) > LP(G) - |I| , since otherwise LP(G) optimal solution that sets I to 1 .

has an So 𝐿𝑃 𝐺\I ≥ 𝐿𝑃 𝐺 − 𝐼 + 1 2 t-LP drops by at least ½ .

Branching Rule Analysis Cont’d

Analyzing the (G\N(I), t-N(I)) side: LP(G\N(I)) + |N(I)| = n/2 + surplus(I)/2 ≥ LP G + surplus(I)/2 So LP(G\N(I)) ≥ LP(G) − |N(I)| + surplus(I)/2 t-LP drops by at least surplus(I)/2

Branching Summary

The measure k-LP drops by (½, surplus(I)/2) .

Will see that independent sets of surplus 1 be reduced in polynomial time !

can Measure drops by (½,1) giving a 𝑂 ∗ time algorithm for Vertex Cover 2.618

𝑡−𝐿𝑃

Reducing Surplus 1 sets.

Lemma:

and N(I) If surplus(I) = 1 , I has minimum surplus is not independent then there exists an optimum vertex cover containing N(I) .

I N(I) R

Reducing Surplus 1 sets.

Reduction Rule:

surplus and N(I) If surplus(I) = 1 , I has minimum is independent then solve (G’,t-|I|) where single vertex v .

G’ is G with N[I] contracted to a I N(I) R

Summary

Nemhauser Trotter lets us assume surplus > 0 More rules let us assume surplus > 1 ( ≥ 2 ) * If surplus 𝑂 ∗ 2.618

≥ 2 𝑡−𝐿𝑃 then branching yields time for Vertex Cover The correctness of these rules were also proved by NT!

Can we do better?

Can get down to 𝑂 ∗ (2.32

𝑡−𝐿𝑃 ) by more clever branching rules. Yields 𝑂 ∗ (2.32

𝑡−𝐿𝑃 ) for Almost 2-SAT and Odd Cycle Transversal .

Should not be the end of the story.

Better OCT?

Can we get down to 𝑂 ∗ (2 𝑘 ) Transversal ?

for Odd Cycle

LP Branching in other cases

I believe many more problems should have FPT algorithms by LP -guided branching.

What about ... ( Directed ) Feedback Vertex Set , parameterized by solution size k ?