• Nem Talált Eredményt

The Complexity of Tree Multicolorings

N/A
N/A
Protected

Academic year: 2022

Ossza meg "The Complexity of Tree Multicolorings"

Copied!
23
0
0

Teljes szövegt

(1)

The Complexity of Tree Multicolorings

Dániel Marx

Budapest University of Technology and Economics dmarx@cs.bme.hu

(2)

Minimum sum multicoloring

Given: a graph G(V, E), and demand function x : V → N

Find: an assignment Ψ of x(v) colors (integers) to every vertex v, such that neighbors receive disjoint sets

Goal: The finish time f(v) of vertex v is the largest color (integer) assigned to it in the coloring. Minimize P

vV f(v), the sum of the coloring.

(3)

Minimum sum multicoloring

Given: a graph G(V, E), and demand function x : V → N

Find: an assignment Ψ of x(v) colors (integers) to every vertex v, such that neighbors receive disjoint sets

Goal: The finish time f(v) of vertex v is the largest color (integer) assigned to it in the coloring. Minimize P

vV f(v), the sum of the coloring.

1

3 1

1 2

2

(4)

Minimum sum multicoloring

Given: a graph G(V, E), and demand function x : V → N

Find: an assignment Ψ of x(v) colors (integers) to every vertex v, such that neighbors receive disjoint sets

Goal: The finish time f(v) of vertex v is the largest color (integer) assigned to it in the coloring. Minimize P

vV f(v), the sum of the coloring.

1

3 1

1 2

2

1,4

2,5 1 2

1,4,5

3

(5)

Minimum sum multicoloring

Given: a graph G(V, E), and demand function x : V → N

Find: an assignment Ψ of x(v) colors (integers) to every vertex v, such that neighbors receive disjoint sets

Goal: The finish time f(v) of vertex v is the largest color (integer) assigned to it in the coloring. Minimize P

vV f(v), the sum of the coloring.

1

3 1

1 2

2

1,4

2,5 1 2

1,4,5 3

,5

,4 3 ,5

1 2

Sum of the coloring:

5 + 1 + 2 + 4 + 3 + 5 = 20

(6)

Application in scheduling

Scheduling of interfering jobs, minimizing the sum of completion times (same as minimizing the average completion times)

vertices ⇐⇒ jobs

demands ⇐⇒ days required

edges ⇐⇒ interfering pairs of jobs

colors ⇐⇒ days

assignment of colors ⇐⇒ assignment of days

finish time of a vertex ⇐⇒ day when the job is finished sum of the coloring ⇐⇒ sum of the completion times

(7)

Example

E: 3 B: 1 C: 2 A: 2 5

D: 1 4 F: 1 4 5

Day 1 Day 2 Day 3 Day 4 Day 5

A B C D E F ,

, ,

,

Finish time

5 1 2 4 3 5 Sum of the coloring: 20 1 1

1 2

3 4

4 5

2 5

1

3 1

1 2

2

(8)

Example

E: 3 B: 1 C: 2 A: 2 5

D: 1 4 F: 1 4 5

Day 1 Day 2 Day 3 Day 4 Day 5

A B C D E F ,

, ,

,

Finish time

5 1 2 4 3 5 Sum of the coloring: 20 1 1

1 2

3 4

4 5

2 5

1

3 1

1 2

2

Preemptive scheduling: the jobs can be interrupted

(9)

Known results

Special case, the chromatic sum problem: x(v) = 1, ∀v ∈ V

General graphs:

? cannot be approximated within n1 even if every demand is 1 (unless NP = ZPP) [Bar-Noy et al., 1998],

? O(n/log2n) approximation for sum multicoloring [Bar-Noy et al., 2000]

Bipartite graphs:

? 1.5-approximation for sum multicoloring [Bar-Noy and Kortsarz, 1998]

? APX-hard, even if every demand is 1

(10)

Known results

Planar graphs:

? (1 + )-approximation for sum multicoloring [Halldórsson and Kortsarz, 1999]

? NP-complete even if every demand is 1

Trees:

? (1 + )-approximation for sum multicoloring [Halldórsson et al., 1999]

? polynomial time solvable if every demand is 1 [Kubicka, 1989],

(11)

Known results

Planar graphs:

? (1 + )-approximation for sum multicoloring [Halldórsson and Kortsarz, 1999]

? NP-complete even if every demand is 1

Trees:

? (1 + )-approximation for sum multicoloring [Halldórsson et al., 1999]

? polynomial time solvable if every demand is 1 [Kubicka, 1989],

New result: Minimum sum multicoloring is NP-hard on binary trees, even if every demand is polynomially bounded (in the size of the tree)

(12)

List multicoloring

As a first step of the proof, we introduce another problem where trees are difficult to color:

List multicoloring

Given: a graph G(V, E), a demand function x : V → N, and a set of avail- able colors L(v) for every vertex

Find: an assignment Ψ of x(v) colors to every vertex v, such that

? neighbors receive disjoint sets and

? Ψ(v) ⊆ L(v)

(13)

List multicoloring

As a first step of the proof, we introduce another problem where trees are difficult to color:

List multicoloring

Given: a graph G(V, E), a demand function x : V → N, and a set of avail- able colors L(v) for every vertex

Find: an assignment Ψ of x(v) colors to every vertex v, such that

? neighbors receive disjoint sets and

? Ψ(v) ⊆ L(v)

New result: List multicoloring is NP-complete in binary trees.

(14)

Theorem: List multicoloring is NP-complete in trees.

(Sketch of proof) Reduction from the maximum independent set problem (“Is there an independent set of size k?”)

The tree is a star with one leaf for each edge.

For every edge vxvy, let {x, y} be the list of the corresponding leaf.

The list of the central node v contains every color.

1,2 1,3

2,4

4,5 3,4

v

3

v

5

v

4

v

2

v

1

e

e

1,4 1,2,3,4,5

k

1 1

1 1

1 1

(15)

Theorem: List multicoloring is NP-complete in trees.

(Sketch of proof) Reduction from the maximum independent set problem (“Is there an independent set of size k?”)

The tree is a star with one leaf for each edge.

For every edge vxvy, let {x, y} be the list of the corresponding leaf.

The list of the central node v contains every color.

1,2 1,3

2,4

4,5 3,4

v

3

v

5

v

4

v

2

v

1

e

e

1,4 1,2,3,4,5

k

1 1

1 1

1 1

Claim: In every list coloring of the star, the colors assigned to the central node form an independent set.

(16)

Theorem: List multicoloring is NP-complete in trees.

(Sketch of proof) Reduction from the maximum independent set problem (“Is there an independent set of size k?”)

The tree is a star with one leaf for each edge.

For every edge vxvy, let {x, y} be the list of the corresponding leaf.

The list of the central node v contains every color.

1,2 1,3

2,4

4,5 3,4

v

3

v

5

v

4

v

2

v

1

e

e

1,4 1,2,3,4,5

k

1 1

1 1

1 1

4,

,4

,4

1, 1,

1, 1,2,3,4,5

3 1,2,3,41,2,31,

Claim: In every list coloring of the star, the colors assigned to the central node form an independent set.

(17)

Theorem: List multicoloring is NP-complete in trees.

(Sketch of proof) Reduction from the maximum independent set problem (“Is there an independent set of size k?”)

The tree is a star with one leaf for each edge.

For every edge vxvy, let {x, y} be the list of the corresponding leaf.

The list of the central node v contains every color.

1,2 1,3

2,4

4,5 3,4

v

3

v

5

v

4

v

2

v

1

e

e

1,4 1,2,3,4,5

k

1 1

1 1

1 1

4,

,4

,4

1, 1,

1, 1,2,3,4,5

3 1,2,3,41,2,31,

Claim: In every list coloring of the star, the colors assigned to the central node form an independent set.

(18)

Returning to minimum sum multicoloring. (There are no lists, the goal is to minimize P

vV f(v), where f(v) is the largest color assigned to v.)

The NP-hardness of minimum sum coloring in trees is proved by a similar reduction. The lists are simulated by “penalty gadgets”.

(19)

Returning to minimum sum multicoloring. (There are no lists, the goal is to minimize P

vV f(v), where f(v) is the largest color assigned to v.)

The NP-hardness of minimum sum coloring in trees is proved by a similar reduction. The lists are simulated by “penalty gadgets”.

Illustration: forcing vertex v to use only colors greater than a

v

(20)

Returning to minimum sum multicoloring. (There are no lists, the goal is to minimize P

vV f(v), where f(v) is the largest color assigned to v.)

The NP-hardness of minimum sum coloring in trees is proved by a similar reduction. The lists are simulated by “penalty gadgets”.

Illustration: forcing vertex v to use only colors greater than a

v

a a a

x

1

a

x

2

. . .

x

C

(21)

v

a a a

x

1

a

x

2

. . . x

C

Every vertex xi has demand a ⇒ the sum of vertices xi is at least aC. If C is

“very large”, then this forces v to have only colors greater than a:

• If v has only colors greater than a ⇒ every vertex xi can receive colors {1, . . . , a} ⇒ their total sum is aC.

• If v has a color ≤ a ⇒ every xi has a color greater than a ⇒ their total sum is at least aC + C.

(22)

Remaining steps

• A similar gadget can force v to have only colors less than b

• Using these two gadgets, we can force v to have colors from a given set

we can prove that minimum sum multicoloring is NP-complete in trees

• With a more complicated construction, we can make penalty gadgets with maximum degree 3

we can prove that minimum sum multicoloring is NP-complete in binary trees

(23)

Summary

Coloring problem: Minimum sum multicoloring (minimize the sum of the finish times)

Previous positive result: Minimum sum multicoloring is polynomial in trees if every demand is 1 (or bounded by a constant)

More general result: if every demand is at most p, then the problem can be solved in O(n · (p log n)p) time ⇒ polynomial time, if every demand is O(log n/log log n)

Previous positive result: (1 + )-approximation algorithm for minimum sum multicoloring in trees.

New negative result: Minimum sum multicoloring is NP-complete in binary trees.

List multicoloring is NP-complete in binary trees.

Hivatkozások

KAPCSOLÓDÓ DOKUMENTUMOK

We have proved that computing a geometric minimum-dilation graph on a given set of points in the plane, using not more than a given number of edges, is an NP-hard problem, no matter

These gadgets are trees with a single pendant edge, and have the following general property: if a coloring is “cheap,” meaning that it has as small error on the internal vertices

Bounded degree trees: In edge coloring bounded degree trees, a constant number of color sets is sufficient for each edge ⇒ linear time PTAS... Bounded

Clearly, every small and frequent edge becomes a leaf edge in its subtree, thus if every node has at most D large child edges in the tree, then in every subtree each node has at most

[This paper] PTAS for the minimum sum edge multicoloring of partial k-trees and planar graphs....

The NP-hardness for the minimum-cost popular assignment problem (see Section 7) implies that given a set of desired edges, it is NP-hard (even for strict preferences) to find a

For more than four tree degree sequences on a small number of vertices, it is hard to prove the existence of a rainbow matching of size k − 1 within an arbitrary k − 1 of

The compact suffix tree is a modified version of the suffix tree, and it can be stored in linear space of the length of the string, while the non-compact suffix tree is quadratic