=Paper=
{{Paper
|id=Vol-3010/PAPER_03
|storemode=property
|title=Duality and Outermost Boundaries in Generalized Percolation Lattices
|pdfUrl=https://ceur-ws.org/Vol-3010/PAPER_03.pdf
|volume=Vol-3010
|authors=Ghurumuruhan Ganesan
}}
==Duality and Outermost Boundaries in Generalized Percolation Lattices==
Duality and Outermost Boundaries in Generalized Percolation Lattices Ghurumuruhan Ganesan1 1 Institute of Mathematical Sciences, HBNI, Chennai Abstract In this paper we consider a connected planar graph πΊ and impose conditions that results in πΊ having a percolation lattice-like cellular structure. Assigning each cell of πΊ to be either occupied or vacant, we describe the outermost boundaries of star and plus connected components in πΊ. We then consider the dual graph of πΊ and impose conditions under which the dual is also a percolation lattice. Finally, using πΊ and its dual, we construct vacant cell cycles surrounding occupied components and study left right crossings and bond percolation in rectangles. Keywords Percolation lattices, Star and plus connected components, Outermost boundaries, Duality, Left-right crossings, 1. Introduction The structure of the outermost boundary of finite components is crucial for contour analysis problems of percolation [3] and random graphs [5]. For the square lattice, self-duality plays a crucial role in determining the properties of star and plus connected components and we refer to Chapter 3 [2] for a detailed discussion of combinatorial properties of percolation in regular lattices. For general graphs, [6] uses separating sets in equivalence class of infinite paths to study duality in locally finite graphs and [5] uses unicoherence and topological arguments to investigate plus connected components. In many applications, it might happen that the lattice on which percolation occurs is not necessarily regular, like for example percolation in Voronoi tessellations [2]. It would therefore be interesting to study the duality properties of such irregular lattices and determine conditions under these lattices have behaviour similar to the regular lattices. In this paper we study outermost boundaries in generalized percolation lattices and prove duality properties analogous to regular lattices. We first consider an arbitrary planar graph πΊ and impose certain cyclic conditions on πΊ that results in a cellular structure analogous to regular lattices. We then define the dual graph of πΊ and determine necessary and sufficient conditions for the dual to be a percolation lattice, analogous to πΊ. Using πΊ and its dual, we study outermost boundaries, occupied components and rectangular left-right crossings in generalized percolation lattices. The paper is organized as follows: In Section 2, we define generalized percolation lattices and describe the cellular structure of such lattices in Theorem 1 and in Section 3 we study outermost Algorithms, Computing and Mathematics Conference (ACM 2021), Chennai, India " gganesan82@gmail.com (G. Ganesan) ORCID: 0000-0001-5857-5411 (G. Ganesan) Β© 2021 Copyright for this paper by its authors. Use permitted under Creative Commons License Attribution 4.0 International (CC BY 4.0). CEUR Workshop Proceedings http://ceur-ws.org ISSN 1613-0073 CEUR Workshop Proceedings (CEUR-WS.org) 36 boundaries of star and plus connected components in generalized percolation lattices. Next, in Section 4, we define the dual graph to a percolation lattice and describe conditions under which the dual graph has the properties of a percolation lattice. Following this, we use dual lattices to study vacant cycles of cells surrounding plus and star connected components. Finally in Section 5, we study left right crossings of generalized percolation lattices in rectangles. 2. Percolation lattices Let πΊ = (π, πΈ) be any connected finite planar graph in R2 where each edge is a straight line segment. Two vertices π£1 and π£2 are said to be adjacent if they share an edge in common. Two edges π1 and π2 are said to be adjacent if they share a vertex in common. A subgraph π = (π£1 , . . . , π£π ) β πΊ is said to be a walk if π£π is adjacent to π£π+1 for 1 β€ π β€ π β 1. If ππ = (π£π , π£π+1 ) is the edge containing end-vertices π£π and π£π+1 , we also represent π = (π1 , . . . , ππβ1 ) and say that π£1 and π£π are end-vertices of π. We say that π is a circuit if π is a walk and π£1 = π£π . We say that π is a path if π is a walk and all the π vertices in π are distinct. Finally, we say that π is a cycle if π is a path and π£1 = π£π . For a cycle πΆ β πΊ, let π΄ = π΄(πΆ) be the bounded open set whose boundary is πΆ. We define the interior of πΆ to be π΄ and the closed interior of πΆ to be π΄ βͺ πΆ. We also define the exterior of πΆ to be (πΆ βͺ π΄)π and the closed exterior of πΆ to be π΄π . We say that the graph πΊ is a percolation lattice if every edge in πΊ belongs to a cycle. We say that a cycle πΆ in πΊ is a cell if there exists no point of an edge of πΊ in the interior of πΆ. By definition any two distinct cells πΆ1 and πΆ2 have mutually disjoint interiors and the intersection πΆ1 β© πΆ2 is either empty or a union of vertices and edges in πΊ. Any edge of π belongs to at most two cells and we say that π is unicellular if there is at most one cell containing π as an edge. The following intuitive result captures the main features of percolation lattices as studied in [2]. Theorem 1. If πΊ is a percolation lattice, then there are distinct cells π1 , π2 , . . . , ππ such that π βοΈ πΊ= ππ . (2.1) π=1 Moreover, the representation (2.1) is unique in the sense that if π1 , . . . , ππ are cells such that πΊ = βοΈ π π=1 ππ , then π = π and {ππ }1β€πβ€π = {ππ }1β€πβ€π . The following additional properties hold: (π₯1) For every edge π, there are at most two cells containing π as an edge. (π₯2) If π is contained in the closed interior of a cycle πΆ β πΊ, then there are two cells containing π as an edge and both cells lie in the closed interior of πΆ. If π is contained in the closed exterior of πΆ, then all cells containing π lie in the closed exterior of πΆ. (π₯3) There are cycles Ξ1 , Ξ2 , . . . , Ξπ΅ with mutually disjoint interiors such that every cell in πΊ is contained in the closed interior of one of these cycles. For any π ΜΈ= π, the cycles Ξπ and Ξπ have at most one vertex in common and an edge π β πΊ is unicellular if and only if π belongs to some cycle in {Ξπ }. 37 βοΈ4 1: Example of a percolation lattice πΊ along with the corresponding cellular decomposi- Figure tion π=1 ππ . In Figure 1(π), we illustrate the above result using a percolation lattice containing 4 cells π1 , π2 , π3 and π4 . For completeness, we prove Theorem 1 using the following auxiliary result regarding merging of two cycles that is of independent interest and used throughout the paper. Proposition 1. Let πΆ and π· be cycles in the graph πΊ that have more than one vertex in common. There exists a unique cycle πΈ consisting only of edges of πΆ and π· with the following properties: (π) The closed interior of πΈ contains the closed interior of both πΆ and π·. (ππ) If an edge π belongs to πΆ or π·, then either π belongs to πΈ or is contained in its closed interior. Moreover, if π· contains at least one edge in the closed exterior of πΆ, then the cycle πΈ also contains an edge of π· that lies in the closed exterior of πΆ. The above result essentially says that if two cycles intersect at more than one point, there is an innermost cycle containing both of them in its interior. For illustration, we refer to Figure 2(π) where two cycles πππππ ππ and ππβπππ have the edge ππ and the vertex π in common. The merged cycle πππππ ππ contains both the smaller cycles in its closed interior. Analogous to [4], we use an iterative piecewise algorithmic construction for obtaining the merged cycle. Proof of Proposition 1: Let π β π· be any path that has its end-vertices in the cycle π·0 := πΆ and lies in the exterior of π·0 (for illustration see Figure 2(π) where π = ππ π and πΆ = ππ ππ π). Letting π = ππ π β πΆ the cycle π·1 := π βͺ π then contains the cycle πΆ = π·0 in its interior. We then repeat the above procedure with the cycle π·1 and look for another path π1 β π· that lies in the exterior of π·1 and has end-vertices in π·1 . Arguing as before, we get a cycle π·2 that contains π1 as a subpath and has the cycle πΆ in its closed interior. This procedure continues until we obtain a cycle π·π that does not contain any edge of π· in its exterior. We argue that π·π is the desired cycle πΈ. By construction both πΆ and π· are contained in the closed interior of π·π and π·π consists of only edges of πΆ and π· so (π) and (ππ) are true. Moreover, each cycle π·π , 1 β€ π β€ π contains an edge of π· that lies in the closed exterior of πΆ. The uniqueness of πΈ is also true since if πΈ1 ΜΈ= πΈ is any edge satisfying (π) β (ππ) and πΈ1 contains an edge of πΈ in its closed exterior, then some edge of πΆ or π· is present in the closed exterior of πΈ1 , a contradiction. 38 (b) (a) Figure 2: (π) The cycle πππππ ππ formed by merging the cycles πππππ ππ and ππβπππ. (π) The path π = ππ π β π· lies in the exterior cycle πΆ = ππ ππ π. Proof of Theorem 1: Let πΆ1 , . . . , πΆπ be the set of all cycles containing π = (π’, π£) as an edge. We shrink the cycles in an iterative manner as follows. Let π·0 := πΆ1 and suppose there exists some point of an edge of the cycle πΆπ , π β₯ 2 in the interior of π·0 . Because the graph πΊ is planar, there exists a path π1 β πΆπ present in the closed interior of π·0 . We shrink the cycle π·0 to the cycle π·1 containing the edge π and the path π1 . For illustration we again use Figure 2(π) with π·0 = πΆ1 = ππ ππ π and πΆ2 = ππ ππ π. In this case π1 = ππ π and π·0 = ππ ππ π. The βshrinked" cycle π·1 = ππ ππ π contains every edge π1 . We now repeat the above procedure with the cycle βοΈππ·1 and proceed iteratively to finally obtain a cycle π·π ππ that does not contain any point of π=1 πΆπ in its interior. The cycle π·π ππ =: ππ is a cell containing the edge π. If there exists a cycle πΆπ such that πΆπ and ππ have mutually disjoint interiors, then we repeat the above procedure starting with the cycle πΆπ and obtain another cell π π , also containing π as an edge. By construction ππ and π π have mutually disjoint interiors and for any cycle πΆπ we have that either βοΈ ππ or π π is contained in the closed interior of πΆπ . The set of all distinct cells in {ππ }πβπΊ {π π }πβπΊ is the desired cellular decomposition (2.1) of πΊ. The decomposition (2.1) is unique since every ππ must necessarily be one of π1 , . . . , ππ . The proof of (π₯1) and the proof of (π₯2) for π present in the closed exterior of πΆ, follow from the above construction. To prove the remaining part of (π₯2), suppose that π is present in the closed interior of πΆ. Since πΊ is a percolation lattice, there exists a cycle πΆπ β πΊ containing π as an edge. This cycle πΆπ contains an edge π in the closed interior of πΆ and therefore a path ππ β πΆπ with end-vertices π, π β πΆ contained in the closed interior of πΆ. Let πΉ1 and πΉ2 be the two paths with end-vertices π, π that form the cycle πΆ. For reference, in Figure 2(π), we have π = π, π = π, πΉ1 = ππ π, πΉ2 = ππ π and ππ = ππ π. The cycles ππ βͺ πΉ1 and ππ βͺ πΉ2 have mutually disjoint interiors and so arguing as before, we obtain two cells ππ and π π containing π as an edge. By construction both ππ and π π are contained in the interior of πΆ. The cycles in (π₯3) are obtained by repeatedly merging cells in πΊ as follows: We first pick one cell π1 and using Proposition 1, merge π1 with another cell, say π2 , that shares an edge with π1 to get a new cycle π12 . We then pick another cell, say π3 , that shares an edge with π12 and lies in the exterior of π12 and merge these together to get a new cycle π123 . Continuing 39 this way, we get a cycle Ξ1 satisfying the property that no cell in {ππ } lying in the exterior of Ξ1 shares an edge with Ξ1 . If there still exists cells in the exterior of Ξ1 , then because πΊ is connected, one of these exterior cells (call it π21 ) necessarily shares a vertex with Ξ1 . We then repeat the above procedure starting with the cell π21 . Continuing this way until all cells are exhausted, we get the desired cycles Ξπ , 1 β€ π β€ π΅. Finally, if π is unicellular and is contained within Ξπ , then the cell ππ necessarily shares the edge π with Ξπ . This completes the proof of (π₯3). 3. Outermost boundaries βοΈπ Let πΊ be a percolation lattice with cellular decomposition πΊ = π=1 ππ as in (2.1). We have the following definition of star and plus adjacency. Definition 1. We say that two cells π1 and π2 in πΊ are star adjacent if π1 β© π2 contains a vertex in πΊ and plus adjacent if π1 β© π2 contains an edge in πΊ. We assign every cell ππ , 1 β€ π β€ π, one of the two states, occupied or vacant and assume that there exists an occupied cell π0 containing the origin. We say that the cell ππ is connected to the cell ππ by a star connected πβpath if there is a sequence of distinct cells (π1 , π2 , ..., ππ‘ ), ππ β {ππ }, 1 β€ π β€ π‘ such that ππ is star adjacent to ππ+1 for all 1 β€ π β€ π‘ β 1 and π1 = ππ and ππ‘ = ππ . If all the cells in {ππ }1β€πβ€π‘ are occupied, we say that ππ is connected to ππ by an occupied star connected πβpath. Let πΆ(0) be the collection of all occupied cells in {ππ }1β€πβ€π each of which is connected to the cell π0 by an occupied star connected πβpath. We say that πΆ(0) is the star connected occupied component containing the origin and let {π½π }1β€πβ€π β {ππ } be the set of all the occupied cells belonging to the component πΆ(0). To define the outermost boundary of πΆ(0), we begin with a few preliminary definitions. Let πΊ0 be the graph with vertex set and edge set, respectively being the vertex set and edge set of the cells {π½π }1β€πβ€π = πΆ(0). An edge π β πΊ0 is a said to be boundary edge if π is unicellular or π is adjacent to a vacant cell. (By definition, π is already adjacent to an occupied cell of the component πΆ(0)). We have the following definition. Definition 2. We say that the edge π in the graph πΊ0 is an outermost boundary edge of the component πΆ(0) if the following holds true for every cycle πΆ in πΊ0 : either π is an edge of πΆ or π is in the closed exterior of πΆ. We define the outermost boundary π0 of πΆ(0) to be the set of all outermost boundary edges of πΊ0 . Thus outermost boundary edges cannot be contained in the interior of any cycle in the graph πΊ0 . We have the following result regarding the outermost boundary of the star compo- nent πΆ(0). Theorem 2. There are cycles πΆ1 , πΆ2βοΈ , . . . , πΆπ β πΊ satisfying the following properties: (π) An edge π ββοΈπ0 if and only if π β ππ=1 πΆπ . (ππ) The graph ππ=1 πΆπ is a connected subgraph of πΊ0 . 40 (πππ) If π ΜΈ= π, the cycles πΆπ and πΆπ have disjoint interiors and have at most one vertex in common. (ππ£) Every occupied cell π½π β πΆ(0) is contained in the interior of some cycle πΆπ . (π£) If π β πΆπ for some π, then π is a boundary edge belonging to an occupied square of πΆ(0) contained in the interior of πΆπ . If π is not unicellular, then π also belongs to a vacant cell lying in the exterior of all the cycles in π0 . Moreover, there exists a circuit πΆππ’π‘ containing every edge of βͺ1β€πβ€π πΆπ . The outermost boundary π0 is a connected union of cycles satisfying properties (π) β (π£) and is therefore an Eulerian graph with πΆππ’π‘ denoting the corresponding Eulerian circuit (see Chapter 1, [1]). As an illustration, Figure 3(π) describes a percolation lattice with six cells. The cells with circle inside them are occupied and the rest are vacant. The occupied cells form a star connected component and the outermost boundary consists of two cycles πΆ1 = ππππππ ππ and πΆ2 = π βππ. (a) (b) Figure 3: (π) A star connected component formed by the cells with circle inside them. The outermost boundary consists of two cycles πΆ1 = ππππππ ππ and πΆ2 = π βππ. (π) Replacing an interior path π = πππππ in the cycle π·π (denoted by the wavy curve) with a path π = ππ₯π β π·π . To prove Theorem 2, we use the following Proposition also of independent interest. Proposition 2. For every occupied cell π½π β πΆ(0), 1 β€ π β€ π, there exists a unique cycle π·π in πΊ0 satisfying the following properties. (π) The cell π½π is contained in the closed interior of π·π . (π) Every edge π β π·π is a boundary edge adjacent to an occupied cell of πΆ(0) present in the closed interior of π·π . If π is not unicellular, then π is also adjacent to a vacant cell present in the closed exterior of π·π . (π) If πΆ is any cycle in πΊ0 that contains π½π in the interior, then every edge in πΆ either belongs to π·π or is contained in the interior. Every edge of π·π is also an outermost boundary edge in the graph πΊ0 and so we denote π·π to be the outermost boundary cycle containing the cell π½π β πΆ(0). For example in Figure 3(π), the outermost boundary cycle containing the cell ππ ππ is the cycle πΆ1 = ππππππ ππ. 41 Below we prove Proposition 2 and Theorem 2 in that order. Proof of Proposition 2: Let β° = ΜΈ β be the set of all cycles in the graph πΊ0 satisfying prop- erty (π); i.e., if πΆ β β° then π½π in present in the closed interior of πΆ. The set β° is not empty since π½π is itself a cycle containing π½π in its closed interior and so belongs to β°. Moreover if πΆ1 and πΆ2 are any two cycles in β°, then it cannot be the case that πΆ1 and πΆ2 have mutually disjoint interiors since both πΆ1 and πΆ2 contain the cell π½π in its closed interior. Thus it is possible to merge πΆ1 and πΆ2 using Proposition 1 to get a new cycle πΆ3 β β° that contains both πΆ1 and πΆ2 in its closed interior. Continuing this way, we obtain an βoutermost cycle" πΆπ ππ that contains all the cycles of β° in its closed interior. By construction, the cycle πΆπ ππ satisfies properties (π) and (π). To see that (π) also holds, suppose there exists an edge π β πΆπ ππ that is not a boundary edge. Since π belongs to the graph πΊ0 , the edge π is adjacent to an occupied cell π΄1 β {π½π }. But since π is not a boundary edge, there exists one other cell π΄2 β {π½π } β π΄1 containing π as an edge and moreover π΄2 is also occupied. One of these cells, say π΄1 , is contained in the interior of πΆπ ππ and the other cell π΄2 , is contained in the exterior. The cell π΄2 and the cycle πΆπ ππ have the edge π in common and thus more than one vertex in common. We then use Proposition 1 to obtain a larger cycle πΆπππ ΜΈ= πΆπ ππ containing both πΆπ ππ and π΄2 in its closed interior. This is a contradiction to the fact that πΆπ ππ satisfies property (π). Thus every edge π of πΆπ ππ is a boundary edge. By the same argument above, we also get that the edge π cannot be adjacent to an occupied cell in the exterior of the cycle πΆπ ππ . Thus π is adjacent to an occupied cell in the interior of πΆπ ππ and a vacant cell (if it exists) in the exterior. Proof of Theorem 2: We argue that the set of distinct cycles in the set π := βͺ1β€πβ€π {π·π } obtained in Proposition 2 is the desired outermost boundary π0 and satisfies the properties (π) β (π£) mentioned in the statement of the theorem. To prove (π), it suffices to see that every edge in the union of the cycles π = βͺ1β€πβ€π {π·π } is an outermost boundary edge. This is because, by definition, no edge with an end-vertex present in the interior of some cycle in π can be an outermost boundary edge. Now, suppose some edge π β π·π has an end-vertex in the interior of some cycle πΆ β πΊ0 . This necessarily implies that at least one edge of πΆ is present in the exterior of π·π and moreover πΆ and π·π cannot have a single common vertex. Therefore it is possible to merge πΆ and π·π using Proposition 1 to get a bigger cycle πΉπ β πΊ containing π·π in its closed interior, a contradiction to the construction of π·π . This completes the proof of (π). To prove (ππ), we first see that the graph πΊ0 formed by the vertices and edges of the com- ponent πΆ(0), is connected. Indeed, let π’1 and π’2 be vertices in πΊ0 so that each π’π , π = 1, 2 is a corner of an occupied cell π½π β πΆ(0). By definition, there is a star connected cell path connecting π½1 and π½2 , consisting only of cells in πΆ(0) and consequently, there exists a path in πΊ0 from π’1 to π’2 . Now let π£1 and π£2 be vertices in π belonging to cycles π·π1 and π·π2 , respectively, for some 1 β€ π1 , π2 β€ π. By previous discussion, there exists a path π12 from π£1 to π£2 containing only edges of πΊ0 and by construction, each such edge lies in the closed interior of some cycle in {π·π }. This is true since all the occupied cells are present in some cycle in {π·π }. For each sub-path π β π12 that contains a point in the interior of some cycle π·π and has end-vertices in π·π , we replace π with a path π β π·π (see Figure 3(π)). Iteratively replacing all such interior paths, we get a path from π£1 to π£2 containing only edges of {π·π }. Thus the 42 union of cycles π is connected and this proves (ππ). Property (πππ) is true since otherwise we could merge the cycles π·π and π·π obtained in Proposition 2 to get a larger cycle π·π‘ππ‘ containing both π·π and π·π in its closed interior, a contradiction to the fact that π·π satisfies property (π) in Proposition 2. Indeed for any occupied cell π½π β πΆ(0) the corresponding outermost boundary cycle π·π satisfies property (π) of Proposition 2 and so (ππ£) is true. Moreover if π β π·π is any edge, then using the fact that π·π satisfies property (π) of Proposition 2, we get that the edge π satisfies property (π£). Finally, to obtain the circuit πΆππ’π‘ containing the outermost boundary, we first compute the cycle graph πππ¦π as follows. Let πΈ1 , πΈ2 , ..., πΈπ be the distinct outermost boundary cycles in π. Represent πΈπ by a vertex π in πππ¦π . If πΈπ and πΈπ share a corner, we draw an edge π(π, π) between π and π. Since the union of cycles πΈπ , 1 β€ π β€ π is connected, we get that πππ¦π is connected as well. Let π»ππ¦π be any spanning tree of πππ¦π and consider an increasing sequence of tree sub- graphs {1} = π»1 β π»2 β . . . π»π = π»ππ¦π . The graph π»1 contains a single vertex {1} and so we set Ξ 1 = πΈ1 to be the circuit obtained at the end of the first iteration. Having obtained the circuit Ξ π , let ππ+1 β π»π+1 β π»π be adjacent to some leaf π£π β π»π . This implies that the cycle πΈππ+1 shares a vertex π€π with the cycle πΈπ£π . Since the circuit Ξ π contains π€π , we assume that π€π is the starting and ending vertex of Ξ π and also of πΈππ+1 . The concatenation of Ξ π and πΈππ+1 is the desired circuit Ξ π+1 . Continuing this way, the final circuit Ξ π obtained is the desired circuit πΆππ’π‘ . Plus connected components The techniques used in the previous sections also allows us to obtain the outermost boundary for plus connected components. We recall that cells ππ and ππ are said to be plus adjacent if they share an edge between them. We say that the cell ππ is connected to the cell ππ by a plus connected πβpath if there is a sequence of distinct cells (ππ = π1 , π2 , ..., ππ‘ = ππ ) β {ππ }1β€πβ€π such that ππ is plus adjacent to ππ+1 for all 1 β€ π β€ π‘ β 1. If all the cells in {ππ }1β€πβ€π‘ are occupied, we say that ππ is connected to ππ by an occupied plus connected πβpath. Let πΆ + (0) be the collection of all occupied cells in {ππ }1β€πβ€π each of which is connected to the occupied cell π0 containing the origin, by an occupied plus connected πβpath. We say that πΆ + (0) is the plus connected occupied component containing the origin. Further we also define πΊ+ 0 to be the graph with vertex set being the set of all vertices of the cells of {ππ } present in πΆ + (0) and edge set consisting of the edges of the cells of {ππ } present in πΆ + (0). Every plus connected component is also a star connected component and so the definition of outermost boundary edge in Definition 2 holds for the component πΆ + (0) with πΊ0 replaced by πΊ+ 0 . We have the following result. Theorem 3. The outermost boundary π0+ of πΆ + (0) is unique cycle in πΊ+ 0 with the following properties: (π) All cells of πΆ + (0) are contained in the interior of π0+ . (ππ) Every edge in π0+ is a boundary edge adjacent to an occupied cell of πΆ + (0) contained in the interior of π0+ and a vacant cell in the exterior. This is in contrast to star connected components which may contain multiple cycles in the 43 outermost boundary. In Figure 3(π) for example, the union of the cells ππππππ and ππ πππ forms a plus connected component whose outermost boundary is the cycle πππππ πππ. Proof of Theorem 3: Proposition 2 holds with πΆ(0) replaced by πΆ + (0). Since πΆ + (0) is plus connected, the outermost boundary cycle π·0 the cell π0 in its interior also contains all the cells of πΆ + (0) in its interior. Therefore cycle π·0 satisfies the conditions (π) and (ππ) in the statement of the theorem, is unique and so π0+ = π·0 . 4. Vacant cell cycles surrounding occupied components In this section, we study vacant cycles of cells surrounding occupied star and plus components. We therefore βοΈπ begin with a discussion of the dual lattice. Let πΊ be any percolation lattice and let πΊ = π=0 ππ be the cellular decomposition of πΊ. Definition 3. We say that a graph πΊπ is dual to πΊ if the following conditions hold: (π1) Every cell in πΊ contains exactly one vertex of πΊπ in its interior. Suppose vertex π€ β πΊπ is the present in the interior of the cell π(π€) of πΊ. (π2) Vertices π€1 , π€2 β πΊπ are adjacent if and only if the cells π(π€1 ) and π(π€2 ) are plus adjacent. To study the similarity between πΊ and πΊπ , we would like to first ensure that the dual graph πΊπ is also a percolation lattice, i.e., we prefer that πΊπ is connected, planar and each edge of πΊπ belongs to a cycle. This is because, there are in fact dual graphs that satisfy exactly two of these three properties. For example, consider two plus adjacent squares π1 and π2 of same side length, to be the graph πΊ. If we let the centres of π1 and π2 be the vertex set of πΊπ , then the edge set of πΊπ is the single edge joining the centres of π1 and π2 . The graph πΊπ is connected and planar but acyclic. In Figure 4 (π), we have an example of a graph πΊ (denoted by solid lines) and the corresponding dual graph πΊπ (denoted by the dotted lines). The graph πΊπ is connected and every edge of πΊπ belongs to a cycle but πΊπ is not planar. In Figure 4 (π), the dual graph πΊπ is planar and every edge of πΊπ belongs to a cycle but πΊπ is not connected. Throughout the paper, we assume that the graph πΊ admits a dual graph πΊπ satisfying the following properties: (π1) (Niceness property) The percolation lattice πΊ is nice in the sense that any two plus adjacent cells in πΊ share exactly one edge in common and no other vertex. (π2) (Interior edge property) Any edge (π€1 , π€2 ) β πΊπ is present in the interior of the cycle formed by merging the plus adjacent cells π(π€1 ) and π(π€2 ). (π3) The dual graph πΊπ is a connected and planar percolation lattice and πΊ is dual to πΊπ . (π4) The dual graph πΊπ satisfies the niceness and interior edge property. (π5) If π§ is a vertex of the dual cell π (π£) and π(π§) is the cell in πΊ containing π§, then π£ is a vertex of π(π§). Thus πΊ and πΊπ have similar cellular structure. For example, the usual square lattice on the plane satisfies properties (π1) β (π5). Let πΊ be a percolation lattice with cellular decomposition π π=0 ππ and let πΊπβοΈbe a lattice dual βοΈ to πΊ satisfying properties (π1)β(π5) and having cellular decomposition πΊπ = πΏ π=0 ππ . We say that the sequence πΏπ = (π1 , ..., ππ ) is a plus connected cell path in πΊ if for each 1 β€ π β€ π β 1, the cell ππ is plus adjacent with the cell ππ+1 . We say that πΏπ is a plus connected cell cycle if ππ 44 (a) (b) Figure 4: (π) An example where the dual graph πΊπ is non-planar but connected and every edge of πΊπ belongs to a cycle. (π) An example where πΊπ is planar and every edge of πΊπ belongs to a cycle but πΊπ is not connected. is adjacent to ππβ1 and ππ+1 for each 1 β€ π β€ π with the notation that ππ+1 = π1 . Analogous definition holdsβοΈfor star connected paths and cycles. Let πΆ(0) = ππ=0 π½π be the star connected occupied cell component containing the cell π0 with origin in its interior. By definition, every cell in πΆ(0) is connected to π0 by a star connected cell path. We have the following result. Theorem 4. Suppose properties (π1) β (π5) hold and suppose every vertex in the component πΆ(0) is present in the interior of some dual cell of πΊπ . There exists a unique cycle π«ππ’π‘ = (π€1 , . . . , π€π ) β πΊπ such that each vertex π€π is present in the interior of a vacant cell ππ and satisfies the following properties: (π) For every π, 1 β€ π β€ π , the cell ππ is vacant and star adjacent to some occupied cell in πΆ(0). (ππ) All occupied cells in πΆ(0) are contained in the interior of π«ππ’π‘ . (πππ) If πΉππ’π‘ ΜΈ= π«ππ’π‘ is any other cycle in πΊπ that satisfies (π) β (ππ) above, then πΉππ’π‘ is contained in the closed interior of π«ππ’π‘ . The sequence of cells in (π1 , . . . , ππ ) form a plus connected cycle of vacant cells surrounding the star connected component πΆ(0). In Figure 5, the two cells containing the circles form the star connected occupied component. Every other cell is vacant. The two dual cycles 12345671 and 1234598671 both satisfy (π) β (ππ) and π«ππ’π‘ = 12345671. Proof of Theorem 4: Let π0 denote the outermost boundary of the star connected compo- nent πΆ(0) in the percolation lattice πΊ. From Theorem 2 we have that π0 = βͺ1β€πβ€π πΆπ is a connected union of cycles {πΆπ } with mutually disjoint interiors and moreover, for π ΜΈ= π, the cycles πΆπ and πΆπ have at most one common vertex. If vertices π£1 , π£2 β πΊ are adjacent in the outermost boundary π0 of the component πΆ(0), then the corresponding dual cells π (π£1 ) and π (π£2 ) are plus adjacent (property (π4). Also, because π0 is connected (see property (ππ) Theorem 2), the union of dual cells βοΈ πΆπ (π0 ) := π (π£) (4.1) π£βπ0 45 Figure 5: Cycle of vacant cells surrounding a star connected occupied component. is a plus connected component in the dual graph πΊπ . Moreover, each edge of π0 is present in the interior of closed interior of the union of cells of πΆπ (π0 ). Using Theorem 3 we therefore have that the outermost boundary ππ (π0 ) of πΆπ (π0 ) is a single cycle in πΊπ containing all cells of πΆπ (π0 ) in its closed interior and all edges of π0 in its interior. Here the dual outermost boundary ππ (π0 ) is obtained as follows. Every dual cell belonging to πΆπ (π0 ) is labelled 1 and every dual cell sharing an edge with a cell in πΆπ (π0 ) and not belonging to πΆπ (π0 ), is labelled 0. We then apply Theorem 3 with label 1 cells as occupied and label 0 cells as vacant. Suppose π§1 , π§2 , . . . , π§π‘ are the vertices of the dual cycle ππ (π0 ) encountered in that order; i.e., the vertex π§1 is adjacent to π§2 , the vertex π§2 is adjacent to π§3 and so on. Each vertex π§π is a vertex of the dual cell π (π£) for some π£ β π0 . Therefore if π(π§π ) is the cell in πΊ containing π§π in its interior, then π£ is a vertex of π(π§π ), by property (π5). Moreover π(π§π ) lies in the exterior of π0 and is adjacent to the vertex π£ of πΆ(0). This implies that π(π§π ) must necessarily be vacant. This proves that the cycle ππ (π0 ) satisfies properties (π) β (ππ). To get a unique dual cycle satisfying properties (π) β (πππ), we merge all dual cycles satisfying properties (π) β (ππ). This is possible since any two dual cycles satisfying (π) β (ππ) both contain the cell π0 in their respective interiors and therefore cannot have mutually disjoint interiors. Plus connected components In this subsection we let πΆ + (0) = ππ=0 π½π be the plus connected occupied cell component βοΈ containing the cell π0 with origin in its interior. By definition, every cell in πΆ + (0) is connected to π0 by a plus connected cell path and the outermost boundary π0+ of πΆ + (0) is a single cycle containing all the cells of πΆ + (0) in its interior. We have the following result. Theorem 5. There exists a star connected cell cycle β³ππ’π‘ = (π1 , . . . , ππ‘ ) β πΊ such that: (π) Each cell ππ is vacant, lies in the exterior of π0+ and is plus adjacent to some occupied cell of πΆ + (0). (ππ) The outermost boundary of β³ππ’π‘ is a single cycle containing all the cells of β³ππ’π‘ βͺ πΆ + (0) in its interior. The sequence of cells in (π1 , . . . , ππ‘ ) form a star connected cell cycle of vacant cells surround- ing the plus connected component πΆ + (0). Proof of Theorem 5: From Theorem 3, we have that the outermost boundary π0+ = (π1 , . . . , ππ‘ ) of πΆ + (0) is a single cycle containing all cells of πΆ + (0) in its interior. Moreover every edge ππ 46 of π + (0) belongs to a occupied cell of πΆ + (0) and also to a vacant cell ππ lying in the ex- terior of π0+ . It is possible that multiple edges in π0+ belong to the same cell ππ and so the sequence (π1 , π2 , . . . , ππ ) could have repetitions. In such a case, we remove recurring entries and assume without loss of generality that ππ is star adjacent to ππ+1 for 1 β€ π β€ π with the notation that ππ+1 = π1 . The set of cells in Ξ := (π1 , . . . , ππ ) form a star connected component and we suppose that the sequence of cells (π1 , . . . , ππ ) form a star connected cycle πΏπ for some π β€ π. The outermost boundary ππ of πΏπ is a connected union of cycles 1β€πβ€π π·π and contains every βοΈ cell of πΏπ in the interior of some π·π . If some cell of πΆ + (0) is contained in the interior of π·π , then because πΆ + (0) is plus connected, every cell of πΆ + (0) is contained in the interior of π·π . Moreover, since π·π and π·π share at most one vertex in common, it must the case that π = 1 and so the outermost boundary of the star cycle (π1 , . . . , ππ ) is the single cycle π·1 . If on the other hand, every cell of πΆ + (0) is contained in the exterior of every cycle of ππ , then every edge of ππ is the edge of some occupied cell of πΆ + (0) that lies in the exterior of ππ . This implies that the outermost boundary π0+ of πΆ + (0) is contained in the strict exterior of every cycle in ππ . But this contradicts the fact that each ππ contains at least one edge of π0+ and so there is at least one edge of π0+ contained in the closed interior of some cycle of ππ . 5. Left right and top bottom crossings in rectangles In this section, we study the mutual exclusivity of left right βοΈ and top down crossings in a rectangle. As before we assume that the percolation lattice πΊ = π π=0 ππ and the dual lattice πΊπ = βοΈπ π=0 π π satisfy properties (π1) β (π5) and the origin is the present in the interior of the cell π0 . For a fixed rectangle π , we assume that the sides of π are nicely covered by cells of πΊ as shown in Figure 6. Consider the edges π1 π1 and π2 π2 intersecting the left side of π . The vertices π1 and π2 are connected by a path π΄1 (shown by dotted line) in the interior of π . Similarly the vertices π1 and π2 are connected by a path π΅1 in the exterior of π . The union π΄1 βͺ π΅1 βͺ {π1 π1 , π2 π2 } forms the cell πΏ1 . We define πΏ1 , . . . , πΏπ to be the left cells. Similarly, the cells π1 , . . . , ππ‘ are top cells, the right cells are π 1 , . . . , π π and the bottom cells are π΅1 , . . . , π΅π . Any cell contained in the closed interior of π is called an interior cell. The cells πΆπ πΏ , πΆπ π , πΆπ΅πΏ and πΆπ΅π each contain a corner of the rectangle and have a single vertex contained in the interior of the rectangle. These four cells, called corner cells, are not plus adjacent to any interior cell. As in Figure 6, we assume that the cell πΏ1 is plus adjacent to πΏ2 and πΆπ πΏ and does not share a vertex with any other left, right,top or bottom cell. We make analogous assumptions for each of the left, right, top and bottom cells. Finally, we also assume that the cells are nicely padded in the following way: Let βπ‘ππ be the infinite line containing the top side of π and define βπππ‘π‘ππ , βπππ π‘ and βπππβπ‘ analogously. If π1 is any cell intersecting βπ‘ππ and star adjacent to a cell intersecting π and π2 is any cell intersecting βπππ‘π‘ππ and star adjacent to a cell intersecting π , then π1 and π2 are not star adjacent. A similar assumption holds for cells intersecting βπππ π‘ and βπππβπ‘ . Assuming that π is nicely covered and padded as described above, we have the following definition of left right crossings. 47 Figure 6: The sides of the rectangle are nicely covered by the cells of πΊ. Definition 4. A plus connected cell path πΏ = (π½1 , . . . , π½π ) β {ππ }, π β₯ 3 is said to be a plus connected left right crossing of a rectangle π if π½1 is a left cell, the cell π½π is a right cell and every π½π , 2 β€ π β€ π β 1 is an interior cell. By the nicely padded assumption, πΏ must contain at least one interior cell. Every interior cell is now assigned one of the following two states: occupied or vacant. If every interior cell in a left right crossing πΏ of π is occupied, we say that πΏ is an occupied plus connected left right crossing of the rectangle π . We denote πΏπ + (π , π) and πΏπ + (π , π ) to be the events that the rectangle π contains an occupied and vacant plus connected left right crossing, respectively. We have a similar definition of plus connected top down crossings and denote π π·+ (π , π) and π π·+ (π , π ) to be the events that the rectangle π contains an occupied and vacant plus connected top down crossing, respectively. Replacing plus adjacent with star adjacent, we obtain an analogous definition for star connected left right and top down crossings. We have the following result. Theorem 6. Suppose properties (π1) β (π5) hold and π is nicely covered and nicely padded by cells as in Figure 6. Also suppose that every vertex of a cell intersecting π is present in the interior of a dual cell and that there is at least one plus connected left right crossing and one plus connected top bottom crossing. We have the following. (π) One of the events πΏπ + (π , π) or π π·* (π , π ) always occurs but not both. (ππ) One of the events πΏπ * (π , π) or π π·+ (π , π ) always occurs but not both. The above result describes the mutual exclusivity of occupied and vacant left right and top down crossings in any rectangle. Proof of Theorem 6 We prove the following three statements. (πΌ) The events πΏπ * (π , π) and π π·+ (π , π ) cannot both occur simultaneously. 48 (πΌπΌ) If πΏπ * (π , π) does not occur, then π π·+ (π , π ) must necessarily occur. (πΌπΌπΌ) If πΏπ + (π , π) does not occur, then π π·* (π , π ) occurs. Using (πΌ) β (πΌπΌπΌ) and the fact that a top down crossing of π is a left right crossing of the rectangle π β² obtained by rotating π by ninety degrees around the centre, we then get Theorem 6. Proof of (πΌ): Suppose there exists a star connected occupied left right crossing πΏπ of π and let Ξ = (π1 , . . . , ππ‘ ) be a path in πΏπ crossing π from left to right so that π1 intersects the left edge of π , the edge ππ‘ intersects the right edge of π and each ππ , 2 β€ π β€ π‘ β 1 belongs to an interior occupied cell of πΏπ . The path Ξ divides the rectangles into two halves. Suppose π π·+ (π , π ) also occurs and let πΏ = (π½1 , . . . , π½π ) be a vacant plus connected top bottom crossing where π½1 is a top cell, π½π is a bottom cell and every other π½π is an interior cell. Let π€π be the vertex of πΊπ present in the cell π½π so that π := (π€1 , π€2 , . . . , π€π ) is a path in πΊπ . We claim that the vertex π€1 necessarily above Ξ. This is because if π€1 were to be present below Ξ, then the cell π½1 β πΊ containing π€1 in its interior also lies below Ξ. Let πΏπ€1 be the infinite line parallel to the left side of π and containing the point π€1 . Since π€1 is contained in the interior of π½1 , the some edge of π½1 contains a point π’ β πΏπ€1 lying above π€1 . Since Ξ lies above π½1 , there exists π£ β Ξ lying above π’ (see Figure 7 (π) for illustration). This would imply that some edge of Ξ crosses the top side of π , a contradiction. (a) (b) Figure 7: (π) If π½1 lies below Ξ, then the path Ξ would cross the top side of π . (π) The top cell π = ππ πππππ containing the dual vertex π€1 in its interior. The dual edge (π€1 , π€2 ) with π€2 present in the interior of π , must cross the top edge of π in the segment π π. From the above paragraph we therefore get that the dual vertex π€1 is necessarily above Ξ and an analogous analysis implies that π€π is below Ξ. Next, we argue that the βfirst" edge (π€1 , π€2 ) β π is either present in the interior of π or crosses the top edge of π . For illustration we consider a magnified top cell π = ππ πππππ in Figure 7(π) containing edges ππ and ππ that intersect the top edge of π (the dotted-dashed line). Because of the interior edge property (π2), the dual edge (π€1 , π€2 ) must cross the top edge of π in the segment π π. Summarizing, the edge (π€1 , π€2 ) is either present in the interior of π or crosses the top edge of π . Similarly the final edge (π€πβ1 , π€π ) is either contained in the interior of π or crosses the bottom edge of π . Every other vertex π€π , 2 β€ π β€ πβ1 is present in the interior of Ξ. Therefore the path π necessarily crosses Ξ in the sense that there are edges π β Ξ and π = (π€π , π€π+1 ) β π such π intersects π. The end-vertices of the dual edge π belong to cells π½π and π½π+1 and the 49 Figure 8: The edge π = (π£5 , π£6 ) belonging to the path Ξ = (π£1 , π£2 , . . . , π£7 ) and the dual edge π = (π€2 , π€3 ) belonging to the path π = (π€1 , . . . , π€5 ) intersect. edge π is common to π½π and π½π+1 . At least one of the two cells π½π or π½π+1 lies in the interior of π and so the edge π is necessarily contained within the rectangle π . Thus π is an edge of some occupied cell in the left right crossing πΏπ and so one of the cells π½π or π½π+1 must lie in the interior of π and also be occupied, a contradiction. We illustrate the above argument in Figure 8, where the edge π = (π£5 , π£6 ) belonging to the path Ξ = (π£1 , π£2 , π£3 , π£4 , π£5 , π£6 ) and the dual edge π = (π€2 , π€3 ) in the path π = (π€1 , π€2 , π€3 , π€4 , π€5 ) intersect. Proof of (πΌπΌ): The collection βπ‘ππ‘ of all left cells πΏ1 , . . . , πΏπ and the corner cells πΆπ πΏ and πΆπ΅πΏ in Figure 6 is a plus connected component. To each cell in βπ‘ππ‘ we now assign a label π. If some occupied cell π in the interior of π is connected to a left cell by a star connected occupied path, we assign the label π to π as well. The collection of all cells with the label π is a star connected component which we denote as β±π‘ππ‘ . If π is any cell star adjacent to some cell in β±π‘ππ‘ and not present in β±π‘ππ‘ , we assign the label πΏ to π . By assumption, any vertex of a cell π β β±π‘ππ‘ is contained in the interior of some dual cell. Therefore by Theorem 4, there exists a plus connected cell cycle Ξπ£ππ = (π1 , . . . , ππ‘ ) surround- ing β±π‘ππ‘ in such a way that the outermost boundary cycle ππ of Ξπ£ππ contains all the cells of β±π‘ππ‘ in its interior. The cell cycle Ξπ£ππ contains a plus connected cell sub-path Ξπ = (ππ’1 , . . . , ππ’2 ) that lies to the right of βπππ π‘ , with ππ’1 intersecting βπ‘ππ and ππ’2 intersecting βπππ‘π‘ππ . In Figure 9, we illustrate the part of the cycle Ξπ£ππ that intersects the rectangle π with the cell labelled π denoting ππ . As in Figure 9, the cycle Ξπ£ππ may intersect the top side of π multiple times but there exists a βlast" cell after which the cycle never intersects the top side of π . Formally, π’1 be the largest index π such that the cell ππ intersects the top side of π and ππ’2 is the βfirst" cell after ππ’1 that intersects the bottom side of π . In Figure 9, π’1 = 6 and π’2 = 10. By definition the cell ππ’1 intersects the line βπ‘ππ , the cell ππ’2 intersects the line βπππ‘π‘ππ and that every other ππ , π’1 < π < π’2 neither intersects βπ‘ππ nor intersects βπππ‘π‘ππ but lies between 50 Figure 9: The part of the vacant cycle Ξπ£ππ that intersects the rectangle π . these two lines. By the nicely padded assumption we have that π’2 > π’1 . No cell ππ , π’1 < π < π’2 can be a right cell because then there would exist an occupied cell contained in the interior of π which is star adjacent to ππ and is connected to some left cell πΏπ₯ by an occupied star connected cell path π. The concatenation (πΏπ₯ , π, ππ ) would then form an occupied star connected left right crossing of π , a contradiction. Thus each cell ππ , π’1 < π < π’2 is necessarily an interior cell. By the nicely covered assumption, this necessarily means that ππ’1 must be a top cell and not a corner cell. This is because, no corner cell is plus adjacent to an interior cell. Similarly ππ’2 must be a bottom cell and not a corner cell and so (ππ’1 , ππ’1 +1 , . . . , ππ’2 ) forms a vacant plus connected top down crossing of π . Proof of (πΌπΌπΌ): The proof is analogous to the proof of (πΌπΌ) with minor modifications. Here Ξπ£ππ = (π1 , . . . , ππ‘ ) is star connected and if the corner cell πΆπ π or the bottom cell πΆπ΅π in Figure 6 appear in Ξπ£ππ , we simply remove the corresponding entry from Ξπ£ππ . The resulting sequence of vacant cells is still star connected and we proceed as before to get the desired vacant star connected top bottom crossing of π . Bond Percolation In this section, we consider bond percolation in the lattice πΊ and the mutual exclusivity of left right and top down crossings of in a rectangle. We consider unoriented bond percolation and an analogous analysis βοΈπ holds for the oriented case asβοΈwell. As before we assume that the percolation π lattice πΊ = π=0 ππ and the dual lattice πΊπ = π=0 ππ satisfy properties (π1) β (π5) and the origin is the present in the interior of the cell π0 . Moreover, we also assume that for a fixed rectangle π , we assume that the sides of π nicely covered and nicely padded by cells of πΊ as described prior to Definition 4. Assuming that π is nicely covered as described above, we have the following definition of 51 left right crossings. Definition 5. A path π = (π£1 , . . . , π£π ) β πΊ, π β₯ 4 is said to be a left right crossing of a rectangle π if: (π1) The edge (π£1 , π£2 ) intersects the left side of π and π£1 lies in the exterior of π . (π2) The edge (π£πβ1 , π£π ) intersects the right side of π and π£π lies in the exterior of π . (π3) Every other edge (π£π , π£π+1 ), 2 β€ π β€ π β 2 is contained in the interior of π . By definition any left right crossing must contain at least one edge in the interior of π . Every edge in the closed interior of π is now assigned one of the following two states: open or closed. Moreover, if edge π β πΊ is open and if π is the unique dual edge intersecting πΊπ , then we assign π to be open as well. If every interior edge in a left right crossing π of π is open, we say that π is an open left right crossing of the rectangle π . An analogous definition holds for top down crossings. We denote πΏπ to be the event that the rectangle π contains an open left right crossing of πΊ. For the dual crossing we have a slightly different definition. We say that ππ := (π1 , . . . , ππ‘ ) is a dual top bottom crossing of π if the dual vertex π1 lies in a top cell, the dual vertex ππ‘ lies in a bottom cell and each edge (ππ , ππ+1 ), 1 β€ π β€ π‘ β 1 intersects an interior edge of πΊ; i.e., an edge of πΊ with both end-vertices present in the interior of π . We now see that every edge in a dual top bottom crossing has a state. By definition it suffices to see that the first edge (π1 , π2 ) and the last edge (ππ‘β1 , ππ‘ ) both have states. First consider the edge (π1 , π2 ) and let (π£1 , π£2 ) β πΊ intersect (π1 , π2 ). By the nicely covered assumption in Figure 6, we see that the edge (π£1 , π£2 ) belongs to one of the interior paths represented by the dotted lines and so necessarily lies in the interior of π and consequently has a state. This implies that the dual edge (π1 , π2 ) has the same state as (π£1 , π£2 ). Similarly, the last edge (ππ‘β1 , ππ‘ ) also has a state. If every edge in ππ is closed, we say that ππ is a closed dual top bottom crossing and we let π π·π be the event that π contains a closed dual top bottom crossing consisting of edges of πΊπ . We have the following result. Theorem 7. Suppose properties (π1) β (π5) hold. Further suppose that the rectangle π is nicely covered and nicely padded by cells in πΊ as in Figure 6. One of the events πΏπ or π π·π always occurs, but not both. As before we need to prove three statements: (πΌ) Both πΏπ and π π·π cannot occur simultaneously. (πΌπΌ) If πΏπ does not occur, then π π·π occurs. (πΌπΌπΌ) If π π·π does not occur, then πΏπ occurs. The proof of (πΌ) is analogous to the proof of (πΌ) in Theorem 6. If Ξ is any open left right crossing and Ξ is any top bottom dual crossing, we obtain that one vertex of the dual crossing lies above Ξ and one vertex lies below Ξ and so these two paths must necessarily meet. The dual edge intersecting any edge π β πΊ has a state and in fact the same state as π and this leads to a contradiction. The proof of (πΌπΌπΌ) is analogous to (πΌπΌ) and we prove (πΌπΌ) below. Proof of (πΌπΌ): Let {ππ }1β€πβ€π‘ be the set of edges of πΊ intersecting the left side of π arranged in 52 Figure 10: The outermost boundary cycle ππΈ = (π§1 , . . . , π§9 , π§1 ) denoted by the dotted lines, intersects the top and bottom sides of the rectangle π . the decreasing order of the π¦βcoordinate of the intersection point and for 1 β€ π β€ π‘, let ππ and ππ be the end-vertices of ππ present in the interior and exterior of π , respectively. For example, in Figure 6, the edge π1 = π1 π1 , π2 = π2 π2 and so on. Let βπ‘ππ‘ be the set of all open edges lying in the interior of π and connected to some vertex in {ππ }1β€πβ€π‘ by an open path and for 1 β€ π β€ π‘ β 1, let π΄π be the path between ππ and ππ+1 lying in the interior of the rectangle π . The union β°π‘ππ‘ = βπ‘ππ‘ βͺ {π΄π }1β€πβ€π‘β1 is then a connected component and each vertex π£ β β°π‘ππ‘ is present in the interior of some dual cell π (π£) β πΊπ . The union of the dual cells {π (π£)}π£ββ°π‘ππ‘ forms a plus connected dual component whose outermost boundary ππ is a single cycle in πΊπ containing all edges of β°π‘ππ‘ in its interior. We now see that ππ contains at least one dual vertex present in the interior of a top cell. From Figure 6, the dual edge joining π₯ and π¦ intersects the edge (π1 , π) and so belongs to the dual cell π (π1 ) containing π1 in its interior. The dual vertex π¦ lying in the interior of the top cell π1 therefore belongs to the dual cell π (π1 ), by property (π5). The dual edge (π₯, π¦) is not present in any other dual cell π (π£), π£ β β°π‘ππ‘ and so belongs to the final cycle ππ as well. By an analogous argument, all the dual vertices present in the interior of the left cells πΏ1 , . . . , πΏπ and the corner cells πΆπ΅πΏ and πΆπ πΏ form a sub-path ππ of ππ and the dual edge (π₯0 , π¦0 ) β ππ as well, where π₯0 and π¦0 are the dual vertices is present in the interior of the bottom left corner cell πΆπ΅πΏ and the first bottom cell π΅1 , respectively (See Figure 6). The subpath Ξπ := ππ β ππ has end-vertices π¦ and π¦0 . For illustration, in Figure 10, the outermost boundary cycle ππΈ formed by the merging of the dual cells {π (ππ )}1β€πβ€π‘ is shown by the dotted line (π§1 , π§2 , . . . , π§9 , π§1 ). Here π§1 = π¦, π§9 = π₯, π§5 = π₯0 and π§4 = π¦0 . The sub-path ππ = (π§5 , π§6 , π§7 , π§8 , π§9 ) and Ξπ = (π§1 , π§2 , π§3 , π§4 ). The path Ξπ might contain many dual vertices present in the interior of some top cell and so we pick the βlast" such vertex and call it π¦1 . Similarly we pick the first dual vertex present in the interior of a bottom cell. Formally, we extract a sub-path Ξπ := (π¦1 , . . . , π¦π ) β Ξπ such 53 Figure 11: The edge (π¦π , π¦π+1 ) β Ξπ cuts an edge intersecting the right side of π . that the vertex π¦1 lies in the interior of a top cell, the vertex π¦π lies in the interior of a bottom cell and every other π¦π lies in the interior of a right, corner or interior cell. To prove that each edge of Ξπ has a state it suffices to see that there does not exist π such that π¦π belongs to a corner or a right cell and π¦π+1 belongs to a right cell. As in Figure 11, we assume that π¦π and π¦π+1 both belong to right cells and an analogous argument holds for the corner cells since no corner cell is plus adjacent with an interior cell. From Figure 11 we see that the edge (π¦π , π¦π+1 ) β Ξπ intersects some edge (π’, π£) that cuts the right side of π as in Figure 11. The vertex π’ β πΊ is therefore contained in the interior of the dual cell containing (π¦π , π¦π+1 ) (property (π5)) and so by definition of the component β°π‘ππ‘ , there exists an open path π from π’ to some vertex ππ of the edge ππ that intersects the left side of π . The concatenation (ππ , π, (π’, π£)) would then form an open left right crossing of π , a contradiction. Finally by the nicely padded assumption there must exist at least two edges in Ξπ and so the subpath (π¦1 , . . . , π¦π ) β ππ is a dual top bottom crossing of π . It remains to see that each such edge is closed. Suppose not and the edge (π¦π , π¦π+1 ) β Ξπ is open. By the interior edge property (π2), there exists exactly one edge (π£1 , π£2 ) β πΊ that intersects (π¦π , π¦π+1 ). By construction, one of the vertices, say π£1 belongs to the component β°π‘ππ‘ and so there is an open path from π£1 to some end-vertex ππ of the edge ππ intersecting the left side of π . Since (π¦π , π¦π+1 ) is open, the edge (π£1 , π£2 ) is open as well and so π£2 also belongs to β°π‘ππ‘ . But if π (π£1 ) and π (π£2 ) denote the dual cells containing π£1 and π£2 , respectively, then from property (π4) the cells π (π£1 ) and π (π£2 ) are plus adjacent and share the edge (π¦π , π¦π+1 ). This implies that (π¦π , π¦π+1 ) is present in the interior of the cycle formed by merging π (π£1 ) and π (π£2 ) and consequently (π¦π , π¦π+1 ) must be present in the interior of the outermost boundary cycle ππ as well, a contradiction. 54 Acknowledgments I thank Professors Rahul Roy, Thomas Mountford, Federico Camia, C. R. Subramanian and the referee for crucial comments that led to an improvement of the paper. I also thank IMSc for my fellowships. References [1] B. BollobΓ‘s, Modern Graph Theory, Springer, 1998. [2] B. BollobΓ‘s and O. Riordan, Percolation, Academic Press, 2006. [3] G. Grimmett, Percolation, Springer Verlag, 1999. [4] H. 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