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The Ferry Cover Problem

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The Ferry Cover Problem

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The Ferry Cover Problem

Michael Lampis - Valia Mitsou

National Technical University of Athens

Wolf

Goat

Cabbage

Guard

Boat

- “Propositiones ad acuendos iuvenes”, Alcuin of York, 8th century A.D (in latin).
- We propose a generalization of Alcuin’s puzzle

- We seek to transport n items, given their incompatibility graph.
- Objective: Minimize the size of the boat
- We call this the Ferry Cover Problem

OPTFC (G) ≥ OPTVC (G)

OPTFC (G) ≥ OPTVC (G)

OPTFC (G) ≥ OPTVC (G)

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

OPTFC (G) ≤ OPTVC (G) + 1

Lemma:

OPTVC (G) ≤ OPTFC (G) ≤ OPTVC (G) + 1

Graphs are divided into two categories:

- Type-0, ifOPTFC (G) = OPTVC (G)
- Type-1, if OPTFC (G) = OPTVC (G) + 1

- Ferry Cover is NP and APX-hard (like Vertex Cover [Håstad1997]).
- A ρ-approximation algorithm for Vertex Cover yields a (ρ+1/ OPTFC)-approximation algorithm for Ferry Cover.

Lemma:

For trees with OPTVC (G) > 1 (i.e. not stars)

OPTFC (G) = OPTVC (G) (Type-0)

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- For a star with three or more leaves
OPTFC (G) = OPTVC (G)+1 = 2 (Type-1)

- For any other tree
OPTFC (G) = OPTVC (G) (Type-0)

Fact:

The Vertex Cover Problem can be solved in Polynomial time on trees.

Theorem:

The Ferry Cover Problem can be solved in polynomial time on trees.

- Variation of Ferry Cover: we are also given a trip constraint. We seek to minimize the size of the boat s.t. there is a solution within this constraint.
- Definition: FCi→ determine the minimum boat size s.t. there is a solution with at most 2i+1 trips (i round-trips).

- An interesting special case: only one round-trip allowed.
- FC1 is NP-hard.
- 2-approximation for general graphs.
- A (4/3+ε)-approximation for trees.

FC1 is NP-hard

1

2

3

- A traditional 3-coloring of graph G:
Vertices of color 2 are connected with vertices of colors 1 or 3

H :

- A constrained 3-coloring of graph G:
Vertices of color 2 are only connected with vertices of color 3

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H :

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3

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3

- A loose 3-coloring of graph G:
Vertices of color 2 can be connected with any vertex.

H :

Vertices are partitioned into 3 groups:

- Those loaded and unloaded on the first trip
- Those remaining on the boat for all three trips
- Those loaded and unloaded on the third trip

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F1

3

- Boat size is |V2|+ max{|V1|, |V3|}
- FC1 is equivalent to finding an F1-coloring that minimizes the above function.

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F1

3

Via a reduction from NAE3SAT

Sketch:

- Given a NAE3SAT formula φ with m clauses, create a new formula φ’
- From φ’ create a graph G
- G has an F1-coloring of cost 7m iff φ is satisfiable.

In NAESAT

For example:

Then:

- For every clause construct a triangle.
- For every variable construct a complete bipartite graph.
- Connect each triangle vertex to one corresponding bipartite vertex.

Example:

Example:

Example:

Example:

false

true

true

false

...

...

F1

false

true

true

false

...

...

F1

false

true

true

false

...

...

F1

false

true

true

Cost = 7m = 2m + 2m + 3m

false

...

...

- Observe that 7m is the minimum possible cost.
- It is possible to show that a coloring of this cost is a coloring of the previous form.
- Therefore, φ is satisfiable.
- Bonus: This reduction is also gap-preserving. Therefore, FC1 is APX-hard.

Approximation algorithms for FC1

- The boat arrives to the destination bank twice.
- Therefore, its size must be at least n/2
- A boat of size n is a 2-approximation!

Fact:For a tree G OPTVC(G) ≤ n/2 (because tree is a bipartite graph)

ALGORITHM

- Load a vertex cover of size 2n/3.
- Unload n/3 vertices that form an Independent Set and return.
- Load the remaining vertices and transfer all of them to the destination.

i:

0

1

2

n-2

n-1

2n-1

NP-hard

≡FC

?

Trivial

NP-hard

- Is it NP-hard to determine whether a graph G is Type-0 or Type-1?
- Is FC equivalent to FCn?
- Is FCi for 1 < i < n-1 polynomially solved?
- Can we have an efficient approximation of FC1 in the general case?

The End!