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Preliminaries Hardness of Ind - Max Approximability of Ind - Max References Movement problems on graphs Stefano Leucci stefano.leucci@graduate.univaq.it Dipartimento di Ingegneria e Scienze dellInformazione e Matematica, University of


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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Movement problems on graphs

Stefano Leucci

stefano.leucci@graduate.univaq.it

Dipartimento di Ingegneria e Scienze dell’Informazione e Matematica, University

  • f L’Aquila, Italy.

Movement problems on graphs Stefano Leucci

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Scenario

A central entity needs to plan the motion of a set P of agents (or pebbles) in a complex environment in order to reach a specific goal.

  • The environment is modelled as an undirected graph G.
  • Agents are placed on the vertices of G.
  • We want to move the agents in order to reach a certain goal

configuration (e.g. they must be on a clique of G).

  • Moving an agent trough an edge costs 1 to the agent (e.g.
  • ne unit of energy, one unit of time, . . . ).
  • Amongst all feasible movements we want the one that

minimizes a certain cost function, e.g. the sum of the agents’ costs.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Assumptions

  • Devices do not choose their trajectory autonomously: rather,

their overall movement is planned by a central authority, and hence our focus is on the computational complexity of such a centralized task.

  • Quite naturally, the pebbles should follow a shortest path in G.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example (Connectivity)

Movement problems on graphs Stefano Leucci

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example (Connectivity)

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example (Connectivity)

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Motivation

  • Robot motion planning:
  • Minimizing energy consumption.
  • Minimizing completion time.
  • Radio-equipped agents: form a connected ad-hoc network

(either single-hop or multi-hop).

  • Moving antennas: build an interference-free networks.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Definition

An instance of the problem is defined as follows: Input:

  • An undirected, unweighed graph G = (V , E) on n vertices.
  • A set of k pebbles P.
  • A function σ : P → V that assigns each pebble to its starting

position. Output:

  • A function µ : P → V that assigns each pebble to its final

position, such that the set of final pebble positions achieves a certain goal. Measure:

  • A non-negative function that maps each feasible solution to

its cost.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Goals

Let U be the set of the final position of the pebbles. We consider the following goals: Connectivity(Con): the subgraph of G induced by the set U must be connected. Independency (Ind): U must be an independent set of size k (|U| = k) for G. (Here we are not allowed to place more than one pebble on the same vertex). Clique (Clique): U must a clique of G. (We are allowed to place more than on pebble on the same vertex).

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Measures

Every pebble p ∈ P is moved from its starting vertex σ(p) to its end vertex µ(p) by using a shortest path on G. Overall movement: sum of the distances travelled by pebbles. Sum(µ) =

  • p∈P

dG(σ(p), µ(p)) Maximum movement: maximum distance travelled by a pebble. Max(µ) = max

p∈P dG(σ(p), µ(p))

Number of moved pebbles: number of pebbles that moved from their starting positions. Num(µ) = |{p ∈ P : σ(p) = µ(p)}|

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example

Ind-Max.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example

Ind-Max.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example

Ind-Max. Cost=1

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example

Ind-Sum.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Example

Ind-Sum.

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Example

Ind-Sum. Cost=2

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Example

Ind-Num. Cost=1

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Complexity results

All the movement problems defined here are NP-hard. Some are known to admit a polynomial-time algorithms for special classes of graphs:

  • All connectivity problems (Sum, Max, Num) on trees.
  • Ind-Sum and Ind-Num on trees.
  • Ind-Max on paths.
  • Clique-Num on graphs where a maximum weight clique

can be computed in polynomial time.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Independent set

Definition (Independent set)

An independent set of a graph G = (V , E) is a set of vertices U ⊆ V that are pairwise non-adjacent, i.e. such that ∀u, v ∈ U, (u, v) ∈ E.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Independent set

Definition (Independent set)

An independent set of a graph G = (V , E) is a set of vertices U ⊆ V that are pairwise non-adjacent, i.e. such that ∀u, v ∈ U, (u, v) ∈ E.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Maximum independent set

Definition (Maximum independent set)

A maximum independent set of a graph G = (V , E) is an independent set U∗ of maximum cardinality, i.e. such that for every other independent set U we have |U∗| ≥ |U|.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Maximum independent set

Definition (Maximum independent set)

A maximum independent set of a graph G = (V , E) is an independent set U∗ of maximum cardinality, i.e. such that for every other independent set U we have |U∗| ≥ |U|.

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Maximum independent set

  • On general graphs the problem of finding a maximum

independent set is NP-hard.

  • The decision version of this problem requires determining if

there exists an independent set of at least a certain size.

  • In independency motion problems we need to find an

independent set of size at least |P|.

  • This means that it is NP-hard even to find a feasible solution.
  • Idea: We restrict to classes of graphs where a maximum

independent set can be computed in polynomial time.

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Maximum independent set

  • On general graphs the problem of finding a maximum

independent set is NP-hard.

  • The decision version of this problem requires determining if

there exists an independent set of at least a certain size.

  • In independency motion problems we need to find an

independent set of size at least |P|.

  • This means that it is NP-hard even to find a feasible solution.
  • Idea: We restrict to classes of graphs where a maximum

independent set can be computed in polynomial time.

Bad news: the problem is still hard!

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Preliminaries Hardness of Ind-Max Approximability of Ind-Max References

Special classes of graphs

A maximum independent set can be found in polynomial time on:

  • Paths
  • Trees
  • Bipartite graphs
  • Claw-free graphs (no induced

claws)

  • Perfect graphs
  • ...

A claw and an hole.

Definition (Perfect graph)

A graph G is perfect if neither G nor it’s complement have odd holes.

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Hardness of Ind-Max

Polynomial reduction from the 3-SAT problem to Ind-Max. Ingredients of 3-SAT:

  • A set X = {x1, x2, . . . } of boolean variables.
  • A literal is either an asserted or a negated variable.
  • A clause is a disjunction of three literals.
  • A formula f is a conjunction of clauses.

The 3-SAT problem: There exists a truth assignment to the variables so that f is true?

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Hardness of Ind-Max

Polynomial reduction from the 3-SAT problem to Ind-Max. Ingredients of 3-SAT:

  • A set X = {x1, x2, . . . } of boolean variables. E.g.

X = {x1, x2, x3, x4}.

  • A literal is either an asserted or a negated variable. E.g. x1,

¯ x3, ¯ x1, x2, . . . .

  • A clause is a disjunction of three literals. E.g. (x1 ∨ ¯

x2 ∨ x4), (¯ x1 ∨ ¯ x2 ∨ x3).

  • A formula f is a conjunction of clauses. E.g.

(x1 ∨ ¯ x2 ∨ x4) ∧ (¯ x1 ∨ ¯ x2 ∨ x3). The 3-SAT problem: There exists a truth assignment to the variables so that f is true?

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The variable gadget

For each variable xi of f we build the following “variable” gadget:

ui vi ¯ xi xi

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The clause gadget

For each clause cj = (ℓ1

j , ℓ2 j , ℓ3 j ) of f we build the following

“clause” gadget:

zj wj ℓ1 j ℓ2 j ℓ3 j

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Putting all together

For each clause cj = (ℓ1

j , ℓ2 j , ℓ3 j ) of f we connect each literal to the

  • pposite node of the corresponding variable gadget.

u1 v1 ¯ x1 x1 u2 v2 ¯ x2 x2 u3 v3 ¯ x3 x3 z2 w2 ℓ1 2 ℓ2 2 ℓ3 2 z1 w1 ℓ1 1 ℓ2 1 ℓ3 1

(¯ x1 ∨ ¯ x2 ∨ x3) ∧ (x1 ∨ x2 ∨ ¯ x3)

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Completing the proof (forward)

Claim

The formula f can be satisfied ⇐ ⇒ there exists a solution for the Ind-Max instance of cost 1.

Proof (forward).

  • Consider a truth assignment for f .
  • For each variable xi, if xi is asserted move the pebble starting
  • n vi to the vertex xi, otherwise move it to ¯

xi.

  • For each clause (ℓ1

j , ℓ2 j , ℓ3 j ) there must at least literal ℓk j that is

true.

  • This means that the vertex of the variable gadget that is

adjacent to ℓk

j does not contain pebble.

  • Move the pebble starting on zj to ℓk

j .

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Completing the proof

u1 v1 ¯ x1 x1 u2 v2 ¯ x2 x2 u3 v3 ¯ x3 x3 z2 w2 ℓ1 2 ℓ2 2 ℓ3 2 z1 w1 ℓ1 1 ℓ2 1 ℓ3 1

(¯ x1 ∨ ¯ x2 ∨ x3) ∧ (x1 ∨ x2 ∨ ¯ x3)

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Completing the proof

u1 v1 ¯ x1 x1 u2 v2 ¯ x2 x2 u3 v3 ¯ x3 x3 z2 w2 ℓ1 2 ℓ2 2 ℓ3 2 z1 w1 ℓ1 1 ℓ2 1 ℓ3 1

(¯ x1 ∨ ¯ x2 ∨ x3) ∧ (x1 ∨ x2 ∨ ¯ x3) x1 = ⊤, x2 = ⊥, x3 = ⊤

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Completing the proof (backward)

Claim

The formula f can be satisfied ⇐ ⇒ there exists a solution for the Ind-Max instance of cost 1.

Proof (backward).

  • Consider a solution or the Ind-Max instance of cost 1.
  • Each pebble starting on vi must have been moved to either xi
  • r ¯
  • xi. Set the truth value of the variable xi accordingly.
  • For each clause, the pebble starting on zj must have been

moved to a vertex ℓk

j ∈ {ℓ1 j , ℓ2 j , ℓ3 j }.

  • This means that the vertex of the variable gadget that is

adjacent to ℓk

j does not contain a pebble.

  • Therefore ℓk

j , and the whole clause are satisfied.

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Completing the proof

Theorem

The problem Ind-Max is NP-hard.

u1 v1 ¯ x1 x1 u2 v2 ¯ x2 x2 u3 v3 ¯ x3 x3 z2 w2 ℓ1 2 ℓ2 2 ℓ3 2 z1 w1 ℓ1 1 ℓ2 1 ℓ3 1 Movement problems on graphs Stefano Leucci

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Completing the proof

Theorem

The problem Ind-Max is NP-hard. This holds even when G is a bipartite graph.

u1 v1 ¯ x1 x1 u2 v2 ¯ x2 x2 u3 v3 ¯ x3 x3 z2 w2 ℓ1 2 ℓ2 2 ℓ3 2 z1 w1 ℓ1 1 ℓ2 1 ℓ3 1 Movement problems on graphs Stefano Leucci

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Approximability of Ind-Max

Theorem

If a maximum independent set of G can be found in polynomial time (e.g. on perfect graphs), Ind-Max can be approximated within an additive error of 1.

That’s the best we could possibly do in polynomial time! (unless P = NP).

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Approximability of Ind-Max

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We will need...

Theorem (Hall’s Matching Theorem)

Let H = (V1 + V2, E) be a bipartite graph. There exists a matching of size |V1| on H iff |A| ≤ |NH(A)|, ∀A ⊆ V1.

NH(A) and NH[A] are the open and the closed neighborhood of A, respectively.

V1 V2 A N[A]

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We will need...

Theorem (Hall’s Matching Theorem)

Let H = (V1 + V2, E) be a bipartite graph. There exists a matching of size |V1| on H iff |A| ≤ |NH(A)|, ∀A ⊆ V1.

NH(A) and NH[A] are the open and the closed neighborhood of A, respectively.

V1 V2 A N[A]

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Approximability of Ind-Max

Lemma

Let U∗ be a maximum independent set of G. For each independent set U of G: |U∗ ∩ NG[U]| ≥ |U|.

N[U] G

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Approximability of Ind-Max

Lemma

Let U∗ be a maximum independent set of G. For each independent set U of G: |U∗ ∩ NG[U]| ≥ |U|.

Proof.

Suppose |U∗ ∩ NG[U]| < |U|.

N[U] G

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Approximability of Ind-Max

Lemma

Let U∗ be a maximum independent set of G. For each independent set U of G: |U∗ ∩ NG[U]| ≥ |U|.

Proof.

Suppose |U∗ ∩ NG[U]| < |U|. U′ = (U∗ \ NG[U]) ∪ U is an independent set.

N[U] G

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Approximability of Ind-Max

Lemma

Let U∗ be a maximum independent set of G. For each independent set U of G: |U∗ ∩ NG[U]| ≥ |U|.

Proof.

Suppose |U∗ ∩ NG[U]| < |U|. U′ = (U∗ \ NG[U]) ∪ U is an independent set. We have |U′| > |U∗| ⇒⇐

N[U] G

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Approximability of Ind-Max

Lemma

For each independent set U of G, there exists an injective function f : U → U∗ such that dG(u, f (u)) ≤ 1.

Proof.

Construct the bipartite graph H = (U + U∗, E) and connect each vertex u ∈ U to U∗ ∩ N[{u}].

U U ∗

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Approximability of Ind-Max

Lemma

For each independent set U of G, there exists an injective function f : U → U∗ such that dG(u, f (u)) ≤ 1.

Proof.

Construct the bipartite graph H = (U + U∗, E) and connect each vertex u ∈ U to U∗ ∩ N[{u}]. ∀A ⊆ U, N(A) = |U∗ ∩ NG[A]| ≥ |A|. Claim follows using Hall’s Matching Theorem.

U U ∗ A N[A]

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Approximability of Ind-Max

There exists a solution of cost OPT+1 that places all the pebbles

  • n U∗.

Cost = 0

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Approximability of Ind-Max

There exists a solution of cost OPT+1 that places all the pebbles

  • n U∗.

Cost = 0

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Approximability of Ind-Max

There exists a solution of cost OPT+1 that places all the pebbles

  • n U∗.

Cost = OPT

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Approximability of Ind-Max

There exists a solution of cost OPT+1 that places all the pebbles

  • n U∗.

Cost = OPT

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Approximability of Ind-Max

There exists a solution of cost OPT+1 that places all the pebbles

  • n U∗.

Cost = OPT+1

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Approximability of Ind-Max

Theorem

If a maximum independent set of G can be found in polynomial time, there exists a polynomial-time algorithm that approximates Ind-Max within an additive error of 1. U∗ ← MaximumIndependentSet(G) if |U∗| < |P| then return No solution for k ← 0 to |V | − 1 do F ← {(p, u) ∈ P × U∗|d(σ(p), u) ≤ k} H ← (P + U∗, F)

k = 1

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Approximability of Ind-Max

Theorem

If a maximum independent set of G can be found in polynomial time, there exists a polynomial-time algorithm that approximates Ind-Max within an additive error of 1. U∗ ← MaximumIndependentSet(G) if |U∗| < |P| then return No solution for k ← 0 to |V | − 1 do F ← {(p, u) ∈ P × U∗|d(σ(p), u) ≤ k} H ← (P + U∗, F) M ← MaximumBipartiteMatching(H)

k = 1

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Approximability of Ind-Max

Theorem

If a maximum independent set of G can be found in polynomial time, there exists a polynomial-time algorithm that approximates Ind-Max within an additive error of 1. U∗ ← MaximumIndependentSet(G) if |U∗| < |P| then return No solution for k ← 0 to |V | − 1 do F ← {(p, u) ∈ P × U∗|d(σ(p), u) ≤ k} H ← (P + U∗, F) M ← MaximumBipartiteMatching(H) if |M| = |P| then return M

k = 2

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Approximability of Ind-Max

Theorem

If a maximum independent set of G can be found in polynomial time, there exists a polynomial-time algorithm that approximates Ind-Max within an additive error of 1. U∗ ← MaximumIndependentSet(G) if |U∗| < |P| then return No solution for k ← 0 to |V | − 1 do F ← {(p, u) ∈ P × U∗|d(σ(p), u) ≤ k} H ← (P + U∗, F) M ← MaximumBipartiteMatching(H) if |M| = |P| then return M

k = 2

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References

  • D. Bilò, L. Gualà, S. Leucci, and G. Proietti, Exact and

approximate algorithms for movement problems on (special classes of) graphs, SIROCCO’13. Further readings:

  • E.D. Demaine, M. Hajiaghayi, H. Mahini, A.S.

Sayedi-Roshkhar, S. Oveisgharan, and M. Zadimoghaddam, Minimizing movement, SODA’07.

  • P. Berman, E.D. Demaine, and M. Zadimoghaddam,

O(1)-approximations for maximum movement problems, APPROX-RANDOM’11.

Movement problems on graphs Stefano Leucci