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generalised eden growth model and random planar trees
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Generalised Eden growth model and random planar trees Marco - - PowerPoint PPT Presentation

Generalised Eden model Shape result Flake model Open problems Generalised Eden growth model and random planar trees Marco Longfils Sergei Zuyev Chalmers University of Technology, Gothenburg, Sweden CG Week 2015, Eindhoven Sergei Zuyev


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Generalised Eden model Shape result Flake model Open problems

Generalised Eden growth model and random planar trees

Marco Longfils Sergei Zuyev

Chalmers University of Technology, Gothenburg, Sweden

CG Week 2015, Eindhoven

Sergei Zuyev Generalised Eden growth model and random planar trees

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Generalised Eden model Shape result Flake model Open problems

Notation

Given a finite subset C of Z2 which we call a crystal, its (external) boundary ∂C are these nodes of Z2 \ C which have at least one neighbour in C: ∂C = {y ∈ Z2 \ C : ∃x ∈ C such that x − y = 1}.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Notation

Given a finite subset C of Z2 which we call a crystal, its (external) boundary ∂C are these nodes of Z2 \ C which have at least one neighbour in C: ∂C = {y ∈ Z2 \ C : ∃x ∈ C such that x − y = 1}. Four types of nodes: ∂C = ∂1C ∪ ∂2C ∪ ∂3C ∪ ∂4C, where ∂iC = {y ∈ Z2\C : exactly i neighbours of y lie in C}, i = 1, 2, 3, 4.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Crystal and its boundary

Sergei Zuyev Generalised Eden growth model and random planar trees

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Growth model

At time t = 0 we start with a fixed connected set C0 ⊂ Z2 – the initial crystal.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Growth model

At time t = 0 we start with a fixed connected set C0 ⊂ Z2 – the initial crystal. Let Cn is the crystal at time t = n. At time t = n + 1

  • ne of the external boundary nodes z ∈ ∂Cn will

become crystallised, i.e. a new crystal is Cn+1 = Cn ∪ {z}, where z is chosen randomly with probability depending on the number of neighbouring crystallised nodes, i.e. their type.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Generalised Eden model

We consider the following Generalised Eden model: given 4 non-negative parameters r1, . . . , r4 not all equal 0, the probability that z ∈ ∂iCn, i = 1, 2, 3, 4 is crystallised at time n + 1 is given by ri 4

i=1 ri|∂iCn|

Once crystallised, nodes stay crystallised forever.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Continuous time version

At time t = 0, each boundary node z ∈ ∂iC0 is given independently an exponentially Exp(ri) distributed clock and the one z1 with the minimal time t1 is

  • crystallised. Neighbours of z1 have their clocks reset

depending on their new type.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Continuous time version

At time t = 0, each boundary node z ∈ ∂iC0 is given independently an exponentially Exp(ri) distributed clock and the one z1 with the minimal time t1 is

  • crystallised. Neighbours of z1 have their clocks reset

depending on their new type. Classical Eden model is the one with parameters ri = i. Equivalently, every node retains its Exp(1) clock. It is equivalent to first-passage percolation model: the crystal Ct at time t are the nodes which are “wet” at time t when the water source is C0 and the water speed along each edge is independent 1/Exp(1) r.v.’s.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Infinite growth

If r1 > 0, the crystal cannot stop growing. Let z(Cn) be the leftmost among the lowest nodes of Cn and |C0| = n0. Then |∂Cn| ≤ 4(n + n0), probability that the node f(Cn) ∈ ∂Cn just below z(Cn) crystallise is at least 1/(4(n + n0)) and by the Borel-Cantelli lemma, this would happen infinitely often. We consider only the case r1 > 1 and, without loss of generality, assume r1 = 1.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Infinite growth

If r1 > 0, the crystal cannot stop growing. Let z(Cn) be the leftmost among the lowest nodes of Cn and |C0| = n0. Then |∂Cn| ≤ 4(n + n0), probability that the node f(Cn) ∈ ∂Cn just below z(Cn) crystallise is at least 1/(4(n + n0)) and by the Borel-Cantelli lemma, this would happen infinitely often. If r1 = 0, the crystal can got stuck (e.g., when r2 = 1 and C0 = {0, 1}2). We consider only the case r1 > 1 and, without loss of generality, assume r1 = 1.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Shape result

Assume C0 = 0. One speaks of a Shape result if there exist a compact set D containing the origin, such that lim

n→∞ distH(n−1/2Cn, D) = 0 a.s.,

where distH(A, B) = supx∈A infy∈B x − y is the Hausdorff distance between sets.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Non-decreasing rates

For the case r1 ≤ r2 ≤ r3 ≤ r4 the main tool is Kingman’s subadditivity theorem for time t(x, y) when y crystallises from initial crystal C0 = {x}: Show that for co-linear 0, x, y along each rational direction θ ∈ [0, 2π) t(0, y) ≤ t(0, x) + t(x, y). (1) This is proved by coupling two crystallisation processes, starting from {0} and from {x}. Eq. (1) implies existence of an a.s. limit lim

y→∞ y−1t(0, y) = ρ(θ)

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Then show continuity of ρ(θ) using subadditivity again: t(0, y) ≤ t(0, x) + t(x, y) and t(0, x) ≤ t(0, y) + t(x, y) for x = y = n and x − y = nε. Theorem When r1 ≤ r2 ≤ r3 ≤ r4 the Shape result holds.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Flake model

Consider now an extreme case r1 = 1, r2 = r3 = r4 = 0: a node can crystallise if only one of its neighbour is crystallised. The crystal is a tree: a node which would close a cycle has at least two crystallised neighbours and so will never crystallise.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Types of nodes

One may distinguish

1

The crystallised nodes: Cn – the crystal

2

The nodes ∂1Cn which can be crystallised at the next step (their clocks are set)

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Types of nodes

One may distinguish

1

The crystallised nodes: Cn – the crystal

2

The nodes ∂1Cn which can be crystallised at the next step (their clocks are set)

3

forbidden nodes: Fn = ∂2Cn ∪ ∂3Cn ∪ ∂4Cn

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Types of nodes

One may distinguish

1

The crystallised nodes: Cn – the crystal

2

The nodes ∂1Cn which can be crystallised at the next step (their clocks are set)

3

forbidden nodes: Fn = ∂2Cn ∪ ∂3Cn ∪ ∂4Cn

4

All the rest: Z2 \ (Cn ∪ ∂Cn) among which are the nodes which will never get crystallised since they belong to holes. Definition A hole is a finite connected set H ⊂ Z2 \ (Cn ∪ ∂Cn) such that ∂H ⊂ Fn.

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Holes

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Holes

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Geometry of a hole

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Consider F = ∂H of a hole H and let f(H) be the leftmost

  • f its lowest nodes.

F contains no more than 2 neighbouring horizontally

  • r vertically aligned nodes. If there are 3, the central
  • ne cannot be forbidden, since its neighbours are 2

forbidden and 1 node from the hole.

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Consider F = ∂H of a hole H and let f(H) be the leftmost

  • f its lowest nodes.

F contains no more than 2 neighbouring horizontally

  • r vertically aligned nodes. If there are 3, the central
  • ne cannot be forbidden, since its neighbours are 2

forbidden and 1 node from the hole. Connect the consecutive nodes from F by arrows going counter-clockwise starting from f(H) so that the hole stays “on the left”. The angle these arrows form with the abscissa cannot decrease and can increase

  • nly by π/4 or π/2.

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Impossible turns

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Geometry of hole’s boundary

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Big holes

There are O(n) possible configurations of holes with perimeter |∂H| = n with a fixed f(H).

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Big holes

There are O(n) possible configurations of holes with perimeter |∂H| = n with a fixed f(H). Probability to observe a hole with diameter at least n with a fixed f(H) is at most exp{−βn} for some 0 < β < log 2.

Sergei Zuyev Generalised Eden growth model and random planar trees

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Idea of the proof

Consider continuous time version and let F = ∪t≥0F(Ct) – all the forbidden nodes. Consider a hole H with breadth along (1, 1) and (1, −1) directions n and centroid f(H) at a fixed node

  • x1. For definitiveness, let the longest boundary is

along (1, −1) direction. Enumerate each second node going clockwise from x1: x1, x2, . . . , x[n/2] and on the opposite side x[n/2]+1, . . . , xn. Let yi, zi be their boundary nodes at the left and below.

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x3 H z1 H z y w x x1 y1 z2 x2 y2

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Let τi = min{t(yi), t(zi)} be the time the first neighbour of xi crystallises, τ(1) ≤ τ(2) . . . and x(i) the i-th among x’s whose neighbour crystallises. P{∂H ∈ F x1 = f(H)} ≤ P{x1, x2, . . . , xn ∈ F x1 = f(H)} = E P{x(1) ∈ F τ(1), x1 = f(H)} × P{x(2) ∈ F τ(1), τ(2), x(1) ∈ F, x1 = f(H)} . . . × P{x(n) ∈ F τ(1), . . . , τ(n), x(1), . . . , x(n−1) ∈ F, x1 = f(H)}

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By the strong Markov property, since {τ(i)} are stopping times, P{x(i) ∈ F x(1), . . . , x(i−1) ∈ F, τ(1), . . . , τ(i), x1 = f(H)} = P{x(i) ∈ F x(1), . . . , x(i−1) ∈ F, τ(i), x1 = f(H)} Moreover, x(i) ∈ F depends on the clocks at nodes y(i), z(i) which are reset at time τ(i) by memoryless of the Exponential distribution. Thus = P{x(i) ∈ F τ(i)}

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Omitting index (i), let εx be the clock started at node x at time t(z) so that x is set to crystallise at time t(z) + εx. Let εy is the clock started at node y when the first of its neighbours, say w crystallised. We have two cases:

1

at time t(z), both neighbours of y was not yet crystallised so y did not have clock set yet: t(z) < t(w);

2

at time t(z), y was not crystallised, but had a clock εy already ticking: t(w) < t(z).

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εx t(z) t(w) εy εx εy t(w) t(z)

x ∈ F if t(z) + εx > t(w) + εy. By memoryless of the exponential r.v.’s, this is equivalent εx > εy so that P{x(i) ∈ F τ(i)} = 1/2.

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Thus for a given configuration of H with diameter n, P{∂H ∈ F x1 = f(H)} ≤ 2−n P{there is a hole H with f(H) = x1 with diameter ≥ n} ≤ C

  • m=n

m 2m so by Borel-Cantelli, the probability that there is always a hole of diameter α √ N in CN is 0. Thus If the shape result still holds, D is 1-connected.

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Open problems

Does the shape result still hold for non-monotonely growing rates? Problem: coupling argument does not work – crystal at 0 inhibits growth of crystal at x!

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105 steps Courtesy of Arvind Singh (Orsay)

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107 steps

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109 steps

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Limiting shape is not a ball

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Square root law for boundary size

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Long memory

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Connctivity

Let C0 = {0} and grow the crystal to infinity. Let L(x, y) be the length of the path from x to y given they are

  • crystallised. Will an a.s. limit exist:

lim

n→∞(2n)−1L

  • (−n, 0), (n, 0)
  • ?

If yes, will it be different from lim

n→∞(2n)−1L

  • (−n, N), (n, N)
  • ?

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Random forest

If C0 = {x, y} with x − y > 1 then the trees connected to x and y are disjoint. How does ‘interface’ looks like? If the shape result then like a bisector up to o(r) at distance r?

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Random forest

If C0 = {x, y} with x − y > 1 then the trees connected to x and y are disjoint. How does ‘interface’ looks like? If the shape result then like a bisector up to o(r) at distance r? Choose nodes to C0 independently with prob. p. When p ↓ 0, would the trees converge to the Voronoi cells when the grid size is √p?

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References

1

Klaus Sch¨ urger On the asymptotic geometrical behavior of a class of contact interaction process with a monotone infection rate, Z. Wahrsch. verw Gebiete, 48, 35–48, 1979

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Thank you! Questions?

Sergei Zuyev Generalised Eden growth model and random planar trees