Minicourse on parameterized algorithms and complexity Part 2: Iterative compression

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Minicourse on parameterized algorithms and complexity Part 2: Iterative compression

D´aniel Marx (slides by Marek Cygan)

November 3, 2016

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What iterative compression is?

Iterative compression — main idea

Recursive approach exploiting instance structure exposed by a bit oversized solution.

Solution compression:

1 First, apply some simple trick so that you can assume that a slightly too large solution is available.

2 Then exploit the structure it imposes on the input graph to construct an optimal solution.

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What iterative compression is?

Iterative compression — main idea

Solution compression:

1 First, apply some simple trick so that you can assume that a slightly too large solution is available.

2 Then exploit the structure it imposes on the input graph to construct an optimal solution.

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Vertex Cover

Input: undirected G, integer k

Question: is there a subsetX ⊆V(G)of size at most k such that for eachuv ∈E(G) we have{u,v} ∩X 6=∅.

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Vertex Cover

Input: undirected G, integer k

Question: is there a subsetX ⊆V(G)of size at most k such that for eachuv ∈E(G) we have{u,v} ∩X 6=∅.

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Additional input: oversized solution

We exemplify the iterative compression technique by showing 2^kn^O(1) algorithm for Vertex Cover.

Vertex CoverCompression Input: undirected G, integer k,

vertex cover Z ⊆V(G) of size at most 2k Question: is there a vertex cover of size at most k?

Where do we get Z from? How do we useZ?

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Additional input: oversized solution

vertex cover Z ⊆V(G) of size at most 2k Question: is there a vertex cover of size at mostk?

Where do we get Z from? How do we useZ?

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Additional input: oversized solution

vertex cover Z ⊆V(G) of size at most 2k Question: is there a vertex cover of size at mostk?

Where do we get Z from?

How do we useZ?

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Additional input: oversized solution

Where do we getZ from?

Use polynomial time2-approximation: Find any inclusionwise maximal matching M. If|M|>k, then no VC of size ¬k exists. Otherwise, set Z =V(M), we have|Z| ¬2k.

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Additional input: oversized solution

Use polynomial time2-approximation:

Find any inclusionwise maximal matching M. If|M|>k, then no VC of size ¬k exists. Otherwise, set Z =V(M), we have|Z| ¬2k.

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Additional input: oversized solution

Find any inclusionwise maximal matching M.

If|M|>k, then no VC of size ¬k exists. Otherwise, set Z =V(M), we have|Z| ¬2k.

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Additional input: oversized solution

Find any inclusionwise maximal matching M. If|M|>k, then no VC of size ¬k exists.

Otherwise, set Z =V(M), we have|Z| ¬2k.

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Additional input: oversized solution

Find any inclusionwise maximal matching M. If|M|>k, then no VC of size ¬k exists.

Otherwise, set Z =V(M), we have|Z| ¬2k.

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Additional input: oversized solution

How do we useZ?

Guess X ∩Z =X_Z (by branching into 2^|Z|¬4^k cases). Check if Z\X_Z is independent and|X_Z∪N(Z \X_Z)| ¬k.

Z

X_Z Z \XZ

V \Z

N(Z\XZ)∩(V \Z)

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Additional input: oversized solution

How do we useZ?

Guess X ∩Z =X_Z (by branching into 2^|Z|¬4^k cases).

Check if Z\X_Z is independent and|X_Z∪N(Z \X_Z)| ¬k.

Z X_Z

Z \XZ

V \Z

N(Z\XZ)∩(V \Z)

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Additional input: oversized solution

How do we useZ?

Z X_Z Z \XZ

V \Z

N(Z\XZ)∩(V \Z)

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Additional input: oversized solution

How do we useZ?

Z X_Z Z \XZ

V \Z

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Additional input: oversized solution

How do we useZ?

Z X_Z Z \XZ

V \Z N(Z\XZ)∩(V \Z)

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Additional input: oversized solution

How do we useZ?

We have obtained 2^|Z|n^O(1)¬4^kn^O(1) time algorithm.

Can we improve the dependency on k to2^k?

Notice that it would be enough to have |Z| ¬k+1, but so far we only have|Z| ¬2k.

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Additional input: oversized solution

How do we useZ?

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Additional input: oversized solution

How do we useZ?

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Vertex Cover

vertex cover Z ⊆V(G) of size at most k+1 Question: is there a vertex cover of size at mostk?

Idea: get Z from recursion!

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Vertex Cover

vertex cover Z ⊆V(G) of size at most k+1 Question: is there a vertex cover of size at mostk?

Idea: get Z from recursion!

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Bootstrapping

How to getZ of size at most k+1?

Assume that an instance I = (G,k)without Z is given.

Pick any v ∈V(G) and solveI⁰ = (G\ {v},k) recursively. IfI⁰ is a NO-instance thenI is a NO-instance.

Otherwise set Z =X ∪ {v}, where X is a solution for I⁰. (G,k,Z) is VC Compression instance to solve.

Lemma

f(k)n^c time algorithm for VC Compression impliesf(k)n^c+1 time algorithm for VC.

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Bootstrapping

Pick any v ∈V(G) and solveI⁰ = (G\ {v},k) recursively.

IfI⁰ is a NO-instance thenI is a NO-instance.

Lemma

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Bootstrapping

Lemma

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Bootstrapping

Otherwise set Z =X ∪ {v}, whereX is a solution forI⁰.

(G,k,Z) is VC Compression instance to solve. Lemma

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Bootstrapping

Otherwise set Z =X ∪ {v}, whereX is a solution forI⁰. (G,k,Z) is VC Compression instance to solve.

Lemma

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Bootstrapping

Otherwise set Z =X ∪ {v}, whereX is a solution forI⁰. (G,k,Z) is VC Compression instance to solve.

Lemma

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Vertex Cover - summary

Lemma

Reduction: Vertex Cover→ Vertex Cover Compression.

Vertex Cover Compression can be solved in time2^|Z|n^O(1), which leads to2^kn^O(1) algorithm for VC.

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Vertex Cover - summary

Lemma

Reduction: Vertex Cover→ Vertex Cover Compression.

Vertex Cover Compression can be solved in time2^|Z|n^O(1), which leads to2^kn^O(1) algorithm for VC.

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Outline

1 Iterative compression - introduction.

2 Learning by example - vertex cover.

3 Learning by example - FVS in tournament.

4 Generic steps of the method.

5 5^kn^O(1) algorithm for FVS.

6 3^kn^O(1) algorithm for OCT - sketch.

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FVS in tournaments

Feedback Vertex Set (FVS) in Tournaments Input: a tournament (oriented clique)T, integerk Question: is there a subsetX ⊆V(T) of size at most k,

such that T \X is acyclic

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FVS in tournaments

Feedback Vertex Set (FVS) in Tournaments Input: a tournament (oriented clique)T, integerk Question: is there a subsetX ⊆V(T) of size at most k,

such that T \X is acyclic

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FVS in tournaments

Lemma

If a tournament contains a cycle, then it contains a3-cycle.

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FVS in tournaments

Lemma

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FVS in tournaments

Lemma

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FVS in tournaments

Lemma

This lemma implies a simple 3^kn^O(1) branching algorithm.

By using iterative compression we will see how to improve the running time to2^kn^O(1).

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FVS in tournaments

Lemma

This lemma implies a simple 3^kn^O(1) branching algorithm.

By using iterative compression we will see how to improve the running time to2^kn^O(1).

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FVS in tournaments

Start with the recursive trick, reducing the problem to its compression version.

Feedback Vertex Set (FVS) in TournamentsCompression Input: a tournament (oriented clique)T, integerk

a FVSZ ⊆V(T) of size at mostk+1

Question: is there a subsetX ⊆V(T) of size at most k, such that T \X is acyclic

Lemma

f(k)n^c time algorithm for FVST Compression impliesf(k)n^c+1 time algorithm for FVST.

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FVS in tournaments

Lemma

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FVS in tournaments

Lemma

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FVS in tournaments

Pf: this time we use induction (loop) - alternative to recursion.

Let V(T) ={v₁, . . . ,v_n}.

We want to solve FVST(T[V_i],k) fori =1, . . . ,n, where Vi ={v₁, . . . ,vi}.

Set X₁ =∅, which is a solution forFVST(T[v₁],k). For 2¬i ¬n do

Zi =Xi−1∪ {vi},

letXi be a solution toFVST Compression(T[Vi],k,Zi). if no solution found forT[Vi], then return NO.

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FVS in tournaments

Let V(T) ={v₁, . . . ,v_n}.

We want to solve FVST(T[V_i],k) fori =1, . . . ,n, whereVi ={v₁, . . . ,vi}.

Set X₁ =∅, which is a solution forFVST(T[v₁],k). For 2¬i ¬n do

Zi =Xi−1∪ {vi},

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FVS in tournaments

Let V(T) ={v₁, . . . ,v_n}.

Set X₁ =∅, which is a solution forFVST(T[v₁],k).

For 2¬i ¬n do Zi =Xi−1∪ {vi},

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FVS in tournaments

Let V(T) ={v₁, . . . ,v_n}.

Set X₁ =∅, which is a solution forFVST(T[v₁],k).

For 2¬i ¬n do Zi =Xi−1∪ {vi},

letXi be a solution toFVST Compression(T[Vi],k,Zi).

if no solution found forT[Vi], then return NO.

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FVS in tournaments

Feedback Vertex Set (FVS) in Tournaments Compression Input: a tournament (oriented clique)T, integerk

a FVSZ ⊆V(T) of size at most k+1

By guessing a partitionZ =X_Z]W, we get to the disjoint version.

Disjoint FVS in Tournaments Compression Input: a tournament (oriented clique)T, integerk

a FVSW ⊆V(T) of size at mostk+1

Question: is there a subsetX ⊆V(T) of size at most k, disjoint with W, such thatT \X is acyclic Lemma

Poly time algorithm for Disjoint FVST Compression implies 2^kn^O(1) time algorithm for FVST Compression.

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FVS in tournaments

Feedback Vertex Set (FVS) in Tournaments Compression Input: a tournament (oriented clique)T, integerk

a FVSZ ⊆V(T) of size at most k+1

By guessing a partitionZ =X_Z]W, we get to the disjoint version.

Question: is there a subsetX ⊆V(T) of size at most k, disjoint with W, such thatT \X is acyclic

Lemma

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FVS in tournaments

Question: is there a subsetX ⊆V(T) of size at most k, disjoint with W, such thatT \X is acyclic Lemma

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Disjoint FVS in tournaments

Observation

For an acyclic tournament, there is a single topological ordering.

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Disjoint FVS in tournaments

Simple reduction rules:

Reduction 1

IfT[W]is not acyclic, then answer NO.

Let A=V(T)\W (removable set). Reduction 2

If forv ∈Athe graphT[W ∪ {v}]contains a cycle, then removev and reducek by one.

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Disjoint FVS in tournaments

Reduction 1

Let A=V(T)\W (removable set). Reduction 2

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Disjoint FVS in tournaments

Reduction 1

Let A=V(T)\W (removable set).

Reduction 2

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 3 1 1 2 3 1 5 2 5

3

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 3 1 1 2 3 1 5 2 5

3

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 3 1 1 2 3 1 5 2 5

3

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 3 1 1 2 3 1 5 2 5

3

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

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Disjoint FVS in tournaments

W

A=V(T)\W

0 1 2 3 4 5

2

0 0 5 33 1 1 2 3 1 5 2 5

0 0 1 1 2 3 5 5

Consequently Disjoint FVST Compression may be reduced to

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General framework

Iterative compression schema:

By using induction we can assume that a solution Z ⊆V(G),

|Z| ¬k+1 is given as part of input.

Branch into 2^|Z|cases, guessing what part ofZ should be in a solution.

Solve a disjoint version of the problem, where given a solution W ⊆V(G) we look forX ⊆V(G)\W of size at most k. c^kn^O(1) time algorithm for the disjoint version implies (2c)^kn^O(1) time algorithm for the general problem.

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General framework

Solve a disjoint version of the problem, where given a solution W ⊆V(G) we look forX ⊆V(G)\W of size at most k. c^kn^O(1) time algorithm for the disjoint version implies (2c)^kn^O(1) time algorithm for the general problem.

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General framework

Solve a disjoint version of the problem, where given a solution W ⊆V(G) we look forX ⊆V(G)\W of size at most k.

c^kn^O(1) time algorithm for the disjoint version implies (2c)^kn^O(1) time algorithm for the general problem.

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General framework

Solve a disjoint version of the problem, where given a solution W ⊆V(G) we look forX ⊆V(G)\W of size at most k.

c^kn^O(1) time algorithm for the disjoint version implies (2c)^kn^O(1) time algorithm for the general problem.

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General framework

Lemma

c^kn^O(1) time algorithm for the disjoint version implies (c+1)^kn^O(1) time algorithm for the general problem.

X

X⊆Z

c^k−|X^|=

k+1

X

i=0

k+1 i

!

c^k−i1ⁱ = (c +1)^k+1/c

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General framework

Lemma

c^kn^O(1) time algorithm for the disjoint version implies (c+1)^kn^O(1) time algorithm for the general problem.

X

X⊆Z

c^k−|X^|=

k+1

X

i=0

k+1 i

!

c^k−i1ⁱ = (c +1)^k+1/c

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General framework

Remarks:

To make induction work, we need to find a solution (answering YES/NO is not enough).

By default iterative compression adds n factor to the running time.

Ex: show that for VC and FVST this factor can be reduced to O(k) (hint: use O(1)-approximation).

Some natural problems are not vertex deletion closed. Ex: reduce Connected Vertex Cover (CVC) to

CVC-Compression.

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General framework

Remarks:

By default iterative compression addsn factor to the running time.

CVC-Compression.

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General framework

Remarks:

CVC-Compression.

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General framework

Remarks:

Some natural problems are not vertex deletion closed.

Ex: reduce Connected Vertex Cover (CVC) to CVC-Compression.

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General framework

Remarks:

Some natural problems are not vertex deletion closed.

Ex: reduce Connected Vertex Cover (CVC) to CVC-Compression.

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FVS

Feedback Vertex Set (FVS) Input: undirected G, integer k

Question: is there a subsetX ⊆V(G)of size at most k, such that G\X is a forest

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FVS

FVS is vertex deletion closed, so we can apply iterative compression schema and solving the following problem in time c^kn^O(1) leads to(c+1)^kn^O(1) time algorithm for FVS.

Disjoint FVS Compression Input: undirected G, integer k

a FVSW ⊆V(G) of size at most k+1

Question: is there a subsetX ⊆V(G) of size at mostk, disjoint with W, such that G \X is a forest

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FVS

FVS is vertex deletion closed, so we can apply iterative compression schema and solving the following problem in time c^kn^O(1) leads to(c+1)^kn^O(1) time algorithm for FVS.

Disjoint FVS Compression Input: undirected G, integer k

a FVSW ⊆V(G) of size at most k+1

Question: is there a subsetX ⊆V(G)of size at most k, disjoint with W, such that G \X is a forest

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FVS - reduction rules

Reduction 0

IfG[W]contains a cycle, return NO.

W (forest)

A (forest) v

We wantv to have2 incident edges going to W.

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FVS - reduction rules

Reduction 0

W (forest)

A (forest)

v

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FVS - reduction rules

Reduction 0

W (forest)

A (forest) v

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FVS - reduction rules

Reduction 1

Remove all degree at most1 vertices from G.

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FVS - reduction rules

Reduction 2

If there isv ∈A with deg(v) =2and at least one neighbor in A, then add an edge between neighbours ofv (even if there was one) and removev.

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FVS - reduction rules

W (forest)

A (forest) v (deg_A(v)¬1)

Any leaf v inA has now at least two edges toW.

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FVS - one more reduction rule

W (forest)

A (forest) v

Reduction 3

If forv ∈A=V(G)\W the graphG[W ∪ {v}]contains a cycle, then removev and decrease k by one.

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FVS - one more reduction rule

W (forest)

A (forest) v

Reduction 3

If forv ∈A=V(G)\W the graph G[W ∪ {v}]contains a cycle, then removev and decrease k by one.

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FVS - one more reduction rule

W (forest)

A (forest) v

Reduction 3

If forv ∈A=V(G)\W the graph G[W ∪ {v}]contains a cycle, then removev and decrease k by one.

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FVS branching

W (forest)

A(forest) v

v

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FVS branching

Formally, we branch into instances:

(G\ {v},k−1,W), (G,k,W ∪ {v}).

Observation

A potentialπ(I) =k+ #cc(G[W])decreases in each branch.

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FVS branching

(G\ {v},k−1,W), (G,k,W ∪ {v}).

Observation

W (forest) A(forest) v

v

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FVS branching

(G\ {v},k−1,W), (G,k,W ∪ {v}).

Observation

Lemma

Disjoint FVS Compression can be solved in time4^kn^O(1), consequently there is5^kn^O(1) time algorithm for FVS.

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OCT

Odd Cycle Transversal (OCT) Input: undirected G, integer k

Question: is there a subsetX ⊆V(G)of size at most k, such that G\X is bipartite

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OCT

The heart of the solution for OCT by iterative compression is the following problem, which can be solved in polynomial time!

Annotated Bipartite Coloring

Input: bipartiteG = (V₁,V₂,E), integerk, a partial coloringf₀ :V(G)→ {1,2,?}

Question: is there a subsetX ⊆V(G)of size at most k, and a proper coloring f ofG \X consistent withf0.

V₁

V2

1 2 ?

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OCT

Question: is there a subsetX ⊆V(G)of size at most k, and a proper coloring f ofG \X consistent withf₀.

V1

V2

1 2 ?

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OCT

Question: is there a subsetX ⊆V(G)of size at most k, and a proper coloring f ofG \X consistent withf₀.

V1

V2

1 2 ?

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OCT

V1

V2

1 2 ?

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OCT

V₁

V₂

1 2 ?

each blue vertex is either removed or recolored wrt V1]V2,

each green vertex is removed or maintains color wrt V1]V2, for eache ∈E(G\X) either both vertices are recolored, or none,

algorithm: find min cut between green and blue!

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OCT

V₁

V₂

1 2 ?

each blue vertex is either removed or recolored wrt V1]V2, each green vertex is removed or maintains color wrt V1]V2,

for eache ∈E(G\X) either both vertices are recolored, or none,

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OCT

V₁

V₂

1 2 ?

each blue vertex is either removed or recolored wrt V1]V2, each green vertex is removed or maintains color wrt V1]V2, for eache ∈E(G\X) either both vertices are recolored, or none,

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OCT

V₁

V₂

1 2 ?

each blue vertex is either removed or recolored wrt V1]V2, each green vertex is removed or maintains color wrt V1]V2, for eache ∈E(G\X) either both vertices are recolored, or none,

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Summary

Iterative compression

We have seen it applied to:

Vertex Cover,

FVS in Tournaments, FVS,

OCT (sketch).

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