The Metric DimensionProblem.

J. Dıaz

Monash U., May 2018

The Metric Dimension problem

Given G (V ,E ) its metric dimension, β(G ) is the cardinality of thesmallest L ⊂ V s.t. ∀x , y ∈ V , ∃z ∈ L with dG (x , z) 6= dG (y , z).The set L is called a resolving set.

Harary, Melter, (1976), Slater, (1974)

(4,3,0)

(2,1,2)

(4,1,4)

(3,2,1)

(1,2,3)(0,3,4)

(3,0,3)

The Metric Dimension problem

Given G (V ,E ) its metric dimension, β(G ) is the cardinality of thesmallest L ⊂ V s.t. ∀x , y ∈ V , ∃z ∈ V with dG (x , z) 6= dG (y , z).The set L is called a resolving set.

Harary, Melter, (1976), Slater, (1974)

(4,0)

(1,3) (2,2)

(3,3)

(3,1)

(4,4)

(0,4)

Characterizations of MD for some particular graphs

• β(G ) = 1 iff G is a path.

• β(G ) = n − 1 iff G is a n-clique.

• If β(G ) = 2 ⇒ G does not contain K3,3 or K5

Khuller,Raghavachari,Rosenfeld (1996)

• If T a tree, L the set of leaves and F the set offathers of L with degree ≥ 3⇒β(T ) = |L| − |F |. (Slater 1975) F

L

MD and graph properties

• Metric dimension of certain Cartesian product of graphs:For different examples of G and H produce UB and LB to theMD of G2H. They gave an example of a G with bounded MD,where G2G has unbounded MD.Caceres,Hernandez,Mora,Pelayo,Puertas,Sera,D.Wood (2007)

• If G has diameter D, n ≤ Dβ(G)−1 + β(G ).Khuller,Raghavachari,Rosenfeld (1996)

• Let Gβ,D be the class of graphs with MD= β and diameter=D,the authors determine the max. number of vertices for G ∈ Gβ,D.

Hernando,Mora,Pelayo,Seara,Wood (2010)

Complexity of Metric Dimension

• NPC for general graphs, Garey,Johnson (1979)

• P for trees, Khuller,Raghavachari,Rosenfeld (1996)

• NPC for bounded degree planar graphsDıaz, Pottonen, Serna, Van Leeuwen, (2012)

• NPC for Gabriel graphs Hoffman, Wanke (2012)G is Gabriel ∀u, v ∈ V (G ) are adjacent if the closed disc of

which line segment uv is diameter contains no w ∈ V (G ).⇒ Unit Disks Graphs are NPC

• NPC for weighted MD for a variety of graphsEpstein, Levin, Woeginger (2012)

NPC for bounded degree planar graphs:Sketch

Consider the 1-Negative Planar 3-SAT problem: Given a satformula φ s.t.

I every variable occurs exactly once negatively and once ortwice positively,

I every clause contains two or three distinct variables,

I every clause with three distinct variables contains at least onenegative literal,

I the clause-variable graph Gφ is planar.

decide if it is SAT.1-Negative Planar 3-SAT problem is NPC: reduction fromPlanar-SAT.

1-Negative Planar 3-SAT problem ≤p decisional MD boundeddegree planar graphs.

Aproximability to MD

• There is a 2 log n-approximation for general graphs, Khuller*

• If P 6= NP, there is not a o(2 log n)-approximation,Beliova,Eberhard,Erlebach,Hall,Hoffmann,Mihalak,Ram (2006)

• ∀ε > 0, There is no (1− ε) log n for general graphs, unlessNP⊆ DTIME (nlog log n), Hauptmann,Scmhied,Viehmann(12)

• If P 6= NP, not o(log n)-approximation for general graphs withmaximum degree 3, Hartung,Nichterlein (2013)

MD is in P for outerplanar graphs

An undirected G is said to be anouterplanar graph if it can be drawn in theplane without crossings in such a way thatall of the vertices belong to the unboundedface of the drawing.

For k > 1, G is said to be an k-outerplanargraph if removing the vertices on the outerface results in a (k − 1)-outerplanarembedding.

MD ∈ P for outer-planar graphs

1. Characterize the resolving sets by giving 2 conditions: oneover the vertices and another over the faces

2. Define a T where the vertices are the cut vertices and faces ofG and the edges in T correspond to inner edges and bridges(separators) of G . Notice as size of an inner face could bearbitrarily large, the width of T could be arbitrary.Explore T in bottom-up fashion using two data structures:

2.1 Boundary conditions2.2 Configurations

Algorithm for outerplanar

Even the number of vertices in G represented by v ∈ V (T ) couldbe unbounded, the total number of configurations is polynomial.

The algorithm works in O(n8) (plenty of room for possibleimprovement)

Open probelms on the complexity of MD

Prob. 1: Find if MD for K -outerplanar graphs is in P or in NPC.

Baker’s Technique (1994):The technique aims to produce FPTAS for problems that areknown to be NPC on planar graphs. They decompose the planarrealization into k-outerplanar, get an exact solution for eachk-outerplanar slice and combine them. Solving for eachk-outerplanar using DP on a tree decomposition, that for eachvertex separator of size at most 2k .Prob. 2: We know that unless NP⊆ DTIME (nlog log n), MD hastor PTAS in planar graphs ∈ PTAS for planar graphs. Is it inAPX-hard?

Why MD is difficult? 1

• Strongly non-local. A vertex in L can resolve vertices very faraway.

• Non-closed under vertex addition, subtraction, or subdivision.

e

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fg

ac

d

e

b

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ac

d

Why MD is difficult?

• MD does not have the bidimensionality behavior.A problem is bidimensional if it does not increase when performingcertain operations as contraction of edges, and the solution valuefor the problem on a n × n-grid is Ω(n2) Demaine, Fomin,Hajiaghayi, Thilikos (2005)

Bidimensionality has been used as a tool to find PTAS forbidimensional problems that are NPC on planar graphs. Demaine,Hajiaghayi (2005).

Examples: feedback vertex set, minimum maximal matching, facecover, edge dominating set . . ..

Background on parametrized complexity

The Tree-width of G = (V ,E ) is a tree (Xi,T):

I ∪Xi = V

I ∀e ∈ E , ∃i : e ∈ Xi

I If v ∈ Xi ∩ Xj then ∀Xk ∈ Xi ; Xj we have v ∈ Xk

The tree width of a graph G is the size of its largest set |Xi | − 1.

Treewidth = 2

b

a c d

gfe

abc

cde

efd

fg

Parametrized complexity

Classify the problems according to their difficulty with respect tothe input size n an input parameter k of the problem.Downey, Fellows (1999)

Fixed parameter tractable: FPT is the class of problems solvable intime f (k)poly(n) (where f (k) = 2k)

• Ex. (k-vertex cover) Given (G , k), does G have a VC ≤ k?Time of k-VC = (kn + 1.2k). ∴ k-VC ∈ FPT.

• Another ex. SAT with m clauses and k variables it can bechecked in time O(m2k).

P ⊆ FPT ⊆W[1] ⊆W[2] ⊆ · · · ⊆ XP

Metric Dimension and parametrized complexity

• W[2]-complete for general graphs, Hartung, Nichterlein (2013)

Courcelle’s Theorem Any problem definable by Monadic SecondOrder Logic is FPT when parametrized by tree width and thelength of the formula.

So far, it seems to be difficult to formulate MD as anMSOL-formula ⇒ Courcelle’s Theorem can’t apply.

Prob. 3: Prove formally that MD can not be expressed as anMSOL formula.

Prob. 4: Show if MD ∈ P (or not) for bounded tree-width graphs.

Prob. 5: Study the parametrized complexity of MD on planargraphs.

Binomial Graphs G (n, p)

G ∈ G (n, p) if given nvertices V (G ), eachpossible edge e isincluded independentlywith probabilityp = p(n).

Whp |E (G )| = p(n2

)and the expected degree of a vertex: d = np.

Giant component threshold: pt = (1 + ε) 1n .

Connectivity threshold: pc = (1 + ε) log nn .

Expected β(G ) in G (n, p)

Bollobas, Mitsche, Pralat (2013)Given G ∈ G (n, p), choose randomly the resolving set L ⊆ V andbound Pr [∃u, v not separated by L].

d = np

β

log5 n

Θ(n)

n1/2 log n

n1/3 log n

n1/4 log n

log n

log nΘ(1) logc n n1/5 n1/3n1/4 n1/2 n(1− ǫ)

Prob. 6: Find if there is a E [β(G )] for Θ(1/n) < p < log5 n/n

Random t-regular Graphs G(n, t)

G ∈ G(n, t) if it is uniformlysampled from the set of allgraphs with n vertices anddegree t. Assume t = Θ(1).

Let G ∈ G(n, t):

I For t ≥ 3 aas G is strongly connected Cooper (93).

I For t ≥ 3 aas G is Hamiltonian Robinson,Wormald (92,93),Cooper, Frieze (94).

I For t ≥ 3 aas the diameter of G = logt−1 +o(log n) Bollobas,Fernandez de la Vega (81)

I For t ≥ 3, G is an expander, i.e. ∃c > 1 s.t. ∀S ⊂ V (G ) with1 ≤ |S | ≤ n

2 , N (S) ≥ c |S |.

Expected β(G ) for G(n, t)

Given G ∈ G(n, t), |V | = n and 2 < t = Θ(1), then whp

E [β(G )] = Θ(log n)

Given G ∈ G(n, t), v ∈ V (G ), let Si = u ∈ V (G ) | dG (v , u) = i

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(with 0 < α < 1)

S1(v) S2(v)S3(v) Si(v)

v

t

αin αi+1nt(t− 1)3

i = logt−1 n2

t(t− 1)

Θ(√n)

t(t− 1)2

Given v ∈ V (G ), for any pair (u,w) ∈ V 2:v does not separate u and w if u,w ∈ Si , andv separates u and w if u ∈ Si & w ∈ Si+1 (or vice versa).

Expected β(G ) for G(n, t)

Therefore, Pr [v separates u&w ] ≥ 2αiαi+1, andPr [v does not separate u&v ] ≥ α2

i + α2i+1,

where αi and αi+1 are constants between 0 and 1.(1− αiαi+1)︸ ︷︷ ︸

α

≥ Pr [v separates u&w ] ≥ 2αiαi+1︸ ︷︷ ︸α′

Upper Bound

Randomly choose a resolving L ⊂ V (G ) with |L| = C log n, forlarge constant C > 0.

Then for a particular pair of vertices u,wPr [L does not separate u&w ] < αC log n ∼ o( 1

n2) (union bound)

Let XC = be the number of pairs not separated by L,

E [XC ] < n2αC log n → 0⇒ Pr [XC > 0]→ 0

Expected β(G ) for G(n, t): Lower Bound

Randomly choose a resolving set L ⊂ V (G ) with |L| = c log n, forsmall constant c > 0.

Pr [L does not separate u&w ] ≥ α′c log n ∼ ω( 1n2

)

⇒ If Xc = number pairs not separated by L, then

E [Xc ] > n2α′c log n →∞⇒ Pr [Xc > 0] = 1− o(1)

Therefore, β(G ) = Θ(log n) .

Prob. 7: Find the constant in E [β(G )] = Θ(log n)For t = 3, empirically β(G ) = 1.13 log n.

Random Geometric Graphs G(n, r(n))

Given a square Q = [0,√n]2 and a

real r(n) > 0 define a randomgeometric graph G ∈ G (n, r) byscattering n expected vertices V onQ according to a Poisson distributionwith intensity 1, and for anyu, v ∈ V , (u, v) ∈ E iff dE (u, v) ≤ r .

It is known: (1) The giant component appears at rt = Θ(1).(2) There is a sharp connectivity threshold at rc = Θ

√log n.

(3) For v ∈ V , the expected degree d(v) = π log n.

M. Penrose: Random Graphs. Oxford (2002)

Expected metric dimension on G(n, r(n))

Given G ∈ G(n, r(n)) what can we say about E [β]?

If rt = O(1) ⇒ β(G ) = Θ(n)

Given v , u ∈ V (G ) how can they be separated?

E [β(G )] for rc = c√

log n

Let G ∈ G(n, rc) and let u, v ∈ V (G )2 with dE (u, v) = xDefine the crowns:Ci (u, v) := w ∈ V (G ) : dE (u,w) = i and dE (v ,w) = i + 1

2r

xr

C1

C1

C2

C2

v

u

LB: Compute the number of pairs for which C1 = ∅.Area of C1 = 4xrc ⇒ Pr [C1 = ∅] = e−4xrcn

Number of (u, v) with C1 = ∅ is 2πn2∫ r0 xe−4xrcndx = n

log n

E [β(G )] for rc = c√

log n

UB: Let x0 = c√n(log n)1/3

Divide the pairs (u, v) in two groups : those with x ≤ x0 and theremaining ones.For the first group, E [|(u, v) ≤ x0|] = O( n

(log n)1/3)

For the second group choose a random resolving L ⊆ V (G ), with|L| = n

(log n)1/3,

If d(u, v) > x there are sufficiently large numbers of crowns eachwith enough vertices assure us each Ci (u, v) intersects L.

Therefore at rc = Θ(√

log nn ):

n

log n≤ β(G ) ≤ n

log1/3 n

Expected metric dimension on G(n, r(n))What we know and don’t know:

I If r = O(1) ⇒ β(G ) = Θ(n) XI If 1 << r <<

√log log n ⇒ β(G ) = Θ(ne−πr

2) X

I If r = C√

log n ⇒ nlog n ≤ β(G ) ≤ n

log1/3 n?

I If log n ≤ r ≤ (n log3 n)1/4 ⇒ nr2≤ β(G ) ≤ n log2 n

r2?

I If (n log1/3 n)1/4 ≤ r ≤√n4 ⇒ β(G ) = Θ(r2/3n1/3) ?

I If r ≥√n√2⇒ β(G ) = Θ(n) ?

Thank you for your attention