On the Manhattan pinball problem

We consider the periodic Manhattan lattice with alternating orientations going north-south and east-west. Place obstructions on vertices independently with probability $0<p<1$. A particle is moving on the edges with unit speed following the orientation of the lattice and it will turn only when encountering an obstruction. The problem is that for which value of $p$ is the trajectory of the particle closed almost surely. We prove this for $p>\frac{1}{2}-\varepsilon$ with some $\varepsilon>0$.


Introduction
We consider the Manhattan pinball problem which was previously proposed and analysed in [BOC03]. See also [Spe12] and [Car10]. There are equivalent ways to state the problem formally and we state it by using bond percolation. Consider the Z 2 lattice embedded into the plane R 2 . Denote the tilted lattice by x + 1 2 , y + 1 2 : x, y are integers and x − y is even , and for any a, b ∈ Z 2 , a and b are connected by an edge if and only if |a−b| = √ 2. Here, |·| is the Euclidean distance on R 2 . See Figure 1 for an illustration. Given 0 < p < 1, we consider the Bernoulli bond percolation on Z 2 . We declare each edge of Z 2 to be closed with probability p and open with probability 1 − p, independent of all other edges. We use ω to denote a configuration of open and closed edges. We place a two-sided plane mirror on each closed edge. Suppose a ray of light starts from the origin and initially moves towards east. When the light reaches an open edge, it passes through without deflection. When the light reaches a closed edge, it is deflected through a right angle by the mirror on the edge. Let L(ω) denote the trajectory of the light.
Our main result is the following Theorem 1.1. Let E n denote the event that the diameter of L(ω) is at most n.
Then there exists ε 0 > 0 such that following holds. For each p > 1 2 − ε 0 , there are α, c > 0 such that for n > α, we have (1)  In the right hand side, after placing mirrors on closed edges of G 0 , the light comes from the west of the red edge will travel in the blue path before leaving.

Enhancement
The proof of Theorem 1.1 uses an enhancement procedure which we define now. Denote by G 0 the finite configuration illustrated in the left hand side of Figure  2 and its caption. Given any configuration ω, we enhance it by the following procedure. For any translated copy of G 0 which appears in ω, we close the open edge which coincides with the red edge (illustrated in Figure 3). We denote the Figure 3: The picture contains a doubly-infinite closed path if and only if the enhancement is activated so that the red edge is turned to be closed. This means the enhancement is essential.
configuration after this procedure byω.
The procedure above is a special case of the general enhancement (see e.g. [Gri99, Section 3.3]). Intuitively, we have added closed edges to ω in a systematic way and we expect that the resulting configurationω is more "percolative" than ω. This is the content of [Gri99, Theorem (3.16)]. For [Gri99, Theorem (3.16)] to hold, the enhancement is required to be essential (a concept defined in [Gri99, Page 64]) which is true for our case (see Figure 3 and its caption).
Remark 2.1. There are other choices of the enhancement G 0 for our proof of Theorem 1.1. We will see in Section 3 that, our choice of G 0 makes the proof easier because of an important feature contained in Lemma 2.2 below.
The following lemma can be directly checked from Figure 3. Denote Q n = (x, y) ∈ Z 2 : |x + y − 1| ≤ n and |x − y| ≤ n to be the tilted box centered at ( 1 2 , 1 2 ). For any edge e of Z 2 , we say e is inside Q n if e connects two vertices in Q n , otherwise we say it is outside Q n . The following proposition follows from the proof of [Gri99, Theorem (3.16)] (more precisely, the argument near [Gri99, Equation (3.8)]) and the fact that the enhancement which we constructed is essential. Proposition 2.3 (Theorem (3.16) in [Gri99]). There exists ε 1 > 0 such that, for each p > 1 2 − ε 1 , (2) Figure 4: The picture illustrates a closed path in T n which joins some vertex on northwest side to southeast side.
for large enough n. Here, A n denotes the event that there exists a path of closed edges joining ( 1 2 , 1 2 ) to some vertex in Z 2 \ Q n . In fact, by directly adapting the proof of [Gri99, Theorem (3.16)], we can substitute A n in Proposition 2.3 by A ′ n which denotes the crossing event in a 4n × n tilted rectangle. Thus we have Proposition 2.4. There exists ε 1 > 0 such that, for each p > 1 2 − ε 1 , for large enough n. Here, A ′ n denotes the event that there is a path of closed edges in the tilted rectangle joining some vertex on northwest side {(x, y) ∈ T n : x − y = −2n} to some vertex on southeast side {(x, y) ∈ T n : x − y = 2n}. See Figure 4 for an illustration.
The following well-known lemma states that in the supercritical phase (i.e. p > 1 2 ), the crossing event A ′ n happens with high probability. Lemma 2.5. For any p > 1 2 , there is constant c > 0 such that for large enough n.
Together with Proposition 2.4 and Lemma 2.5, we have Proposition 2.6. There exists ε 1 > 0 such that, for each p > 1 2 − ε 1 , for a constant c p > 0 and large enough n. Here, A ′′ n denotes the event that Q n lies in the interior of a circuit which consists of closed edges in Q 2n .
By Lemma 2.2, if G 0 is a translated copy of G 0 which appears in ω, then only the red edge will be closed under the enhancement and other edges in G 0 will remain unchanged. We denote the set of edges (outside Q 100 ) which are closed under the enhancement by E. We now start from ω 0 and open edges in E to reach the initial configuration ω and we keep track of the trajectory. Only those edges in E which intersects with L(ω 0 ) will affect the trajectory by adding translated copies of the blue path (or its inverse, see Figure 2) to L(ω 0 ). Here, we used the fact that the edges in E are outside Q 100 and thus if the light comes to some e ∈ E, it must come from west or south (otherwise, L(ω 0 ) is a translated copy of the blue loop and is outside Q 90 which contradicts with the fact that L(ω 0 ) starts from the origin). Hence, L(ω) is "trapped" inside Q 2n+10 (see Figure 6) and our theorem follows.