On percolation and the bunkbed conjecture
We study a problem on edge percolation on product graphs . Here is any finite graph and consists of two vertices connected by an edge. Every edge in is present with probability independent of other edges. The Bunkbed conjecture states that for all and the probability that is in the same component as is greater than or equal to the probability that is in the same component as for every pair of vertices .
We generalize this conjecture and formulate and prove similar statements for randomly directed graphs. The methods lead to a proof of the original conjecture for special classes of graphs , in particular outerplanar graphs.
This note is concerned with discussing a property of edge-percolation on finite graphs that should be intuitively clear, but more difficult to prove rigorously. To the best of my knowledge, the conjecture was first formulated in a slightly different form (equivalent to model below) by P.W. Kasteleyn in 1985, see Remark 5 in [vdBK]. In the form stated above, the conjecture has been presented in [OH1] and [OH2], by Olle Häggström (who claimed it to be folklore).
For any graph we consider the bunkbed graph , where is the graph with two vertices and one edge. A vertex will have two images and one edge between them. Such edges will be called vertical edges. Every edge has also two images that will be called horizontal edges. We will use the terms downstairs and upstairs to denote all vertices and edges in the 0-layer and 1-layer respectively. Let .
Model (Edge percolation): Every edge in is present with probability independently of the other edges. We call the corresponding random graph .
Of course this definition is not restricted to bunkbed graphs. For the theory of percolation in general we refer the reader to [GG2]. For any bunkbed graph , any vertices and , we define
We will often omit if it is clear from the context what graph we are considering. We will only be interested in connected graphs .
The bunkbed conjecture may now be defined as follows.
Conjecture 1.1 (Bunkbed conjecture [Oh2]).
Let be any graph and the corresponding bunkbed graph. For any and any we have
One motivation for formulating this problem is that we would like the probability to be a measure of how close the vertices and are in the graph . For this to be a good concept we would like to make sure that ”intuitive obvious” properties of closeness are true. To this end, the bunkbed graphs are natural testing candidates and we certainly want to be true. In [OH2] Häggström coins the term bunkbed graph and proves the corresponding statement for a related model called random cluster model (also known as Fortuin-Kasteleyn model) with a certain parameter . In that model graphs with a large number of non-connected components occur with higher probability and there is thus dependence between edges. In [BB], Bollobás and Brightwell consider random walks on Bunkbed graphs and more general product graphs. They prove a number of interesting intuitively pleasing statements, but they also have a warning example that intuition sometimes might go wrong. In [OH1] Häggström study continuous random walks on bunkbed graphs and proves a conjecture by Bollobás and Brightwell. The interesting papers [vdBK] and [vdBHK] have been inspired by the conjecture.
The rest of this paper is organized as follows. In Section 2 we generalize the model in several steps to be able to use the combinatorial tools we want. In Section 3 we prove (generalization of) for outerplanar graphs. In Section 4 we present the corresponding problem for randomly directed graphs. Lemma 4.1 states that probabilities of existence of paths in are equal to existence of directed paths in the randomly directed case thus giving a direct connection to directed graphs. In Theorem 4.2 the corresponding bunkbed property is proved for a related model. Finally, in Section 5 we define a critical probability for finite graphs.
Acknowledgement: I thank Olle Häggström for inspiring discussions on the conjecture. I also thank Jörgen Backelin, Madeleine Leander and a very helpful anonymous referee for comments on an earlier version of the manuscript.
2. Generalizations and tools
We start with generalizing the model in three steps. The first step actually consists of two. First we condition on which vertical edges are present in . Second we replace with a vector for every , giving a probability for each edge of . We will call such a vector a probability vector on . For given and we define the following model.
Model : Vertical edges in are present exactly at positions in . For each , the horizontal edges in are present with probability . All events that different edges are present are independent.
The vertices in will be called transversal. The natural generalization of the bunkbed conjecture, let us call it seems also very likely to be true.
Conjecture 2.1 (Kasteleyn, [vdBK]).
Let be any graph and the corresponding bunkbed graph. For any , any and any probability vector we have
This conjecture is the original conjecture as formulated by P.W. Kasteleyn, see Remark 5 of [vdBK]. In fact, that beautiful paper was inspired by Kasteleyn’s conjecture.
Given a graph and a probability , let for every . If is true for all , then is true for the same graph and same .
For any vertices we have
The proposition follows. ∎
With this formulation one may start to prove the Bunkbed Conjecture for certain sets and we will present two easy examples. Recall that a set is called a cutset for if is disconnected. If are in different components of , then is said to separate and .
If contains a cutset of separating from , or if , or if then , in particular is true.
This is easily proved with a mirror argument. Let be the cutset. Let be the edges in the same component as in together with the edges with one endpoint in that component and the other in . For any configuration of present edges , let . Define as the mirror image of , i.e. an edge is present upstairs in if and only if it is present downstairs in and vice versa. Define and it is clear that the two configurations and have the same probability. Also if there is a path from some vertex in to in then there is a path from the same vertex of to in and vice versa. Since any path from to some vertex in may continue both upstairs and downstairs we receive a matching between cases with paths to and to respectively. The lemma follows. ∎
If then is true for any graph and any probability vector .
If then it is clear. Assume . Given two complementary events , it will suffice to prove for . Here, we let be the event that there is no path from to . Then the probability of a path to is zero, so the inequality follows. Let be the event that there exists a path from to . The probability of a path from to upstairs is at most as large as a the probability of a path from to downstairs, because conditioning on the existence of a path from to downstairs can only affect the probability positively (Harris’ inequality on increasing events [H, GG2]). ∎
To introduce our next generalization, let us fix any particular edge . If there are four possibilities in model which we group into three cases as follows.
are both present
is present and is absent, or
is absent and is present
are both absent
The intuitive idea is to condition on which case we belong to and to use that case (1) can be thought of as contracting and (3) as removing . This is made precise in Proposition 2.6. The remaining case (2) leads to defining the following new model for a given set .
Model : Vertical edges exist exactly at positions in . Every horizontal edge upstairs in is present independently with probability and otherwise the corresponding edge exists downstairs. But no horizontal edge exists both upstairs and downstairs.
The natural generalization of the bunkbed conjecture is the following.
Conjecture 2.5 ().
Let be any graph and the corresponding bunkbed graph. For any and any we have
Recall that is a minor of if it can be obtained by deleting and contracting edges of . For we use and for the graph obtained when deleting and contracting the edge . When we say minor in the proposition below we mean the usual notion in graph theory where multiple edges have been removed. However, it will later sometimes be convenient to allow multiple edges and then it will be explicitly stated.
If is true for any minor of and all , then is true for any and thus also for any .
Assume that and are given and that is true for any minor of and all We will now condition on the edges of one at a time and prove the proposition by induction over the number of non-conditioned edges. For a given edge , we will in case (1) contract , in case (3) delete and in case (2) leave in the graph and remember that it now appears either upstairs or downstairs in the corresponding bunkbed graph. When we contract an edge we will in this proof allow the creation of multiple edges, but loops are irrelevant and may be deleted. Also when we contract an edge we let the new vertex be in if at least one of are in . This way the probabilities for existence of paths will be preserved. Note that we have no assumption on in the bunkbed conjectures.
Let and let be any graph where we have conditioned on the edges in . So in for we have that will occur independently with probability . For an edge exactly one of is present in each with probability . The inductive hypothesis is that the corresponding bunkbed conjecture is true for any such graph . With slight abuse of notation we will talk of such a graph also when we mean the entire model with probabilities for all possible configurations in .
The base case, when , is a graph as in Model with the difference that there might be multiple edges in . If there are no multiple edges in , then is a minor of and we are done. Assume are multiple (parallel) edges, then we consider the following complementary events: is the event that both and are present or that both and are present, is the event that or are present. If we condition on being in case then we let . Since edge is irrelevant in this situation the conditional connection probabilities in are the same as the connection probabilities in . If we condition on then we contract and and call it (possibly creating new multiple edges). Again, the conditional connection probabilities in are the same as the connection probabilities in . As in the proof of Lemma 2.4 it suffices to prove the bunkbed inequality in the cases or equivalently for and . Since and have strictly fewer edges we can perform another induction, this time over the number of edges, and it follows that they satisfy the bunkbed inequality. As base case for this induction over we have the graphs with no multiple edges.
For the inductive step, let be any graph obtained by conditioning on the edges in and . Let be the graphs obtained by contracting and deleting the edge respectively and note that connection probabilities (resp. ) are equal to the conditional connection probabilities in case (1) for (case (3) respectively). Let also be the graph such that exactly one of and is present in , which similarly correspond to the case (2). Thus, for any vertices we have that
For and the non-conditioned edges are , for they form a subset of (edges parallel to become loops and thus removed). In any case we have by induction that all three graphs satisfy the bunkbed inequality. It follows that the bunkbed conjecture is true also for . ∎
It might seem more difficult to prove a conjecture not only for the graph but also for all its minors, but if the line of reasoning is to show that a minimal counterexample cannot exist then it is no more difficult. Model has the great advantage that we no longer have the parameter .
Another advantage is that we may reformulate it in terms of edge colorings of the original graph as follows.
Model reformulated: Let . Every edge in is colored either red or blue with equal probability. A walk in may change color only at a vertex in .
Here we think of a blue edge as existing upstairs (blue as in heaven) and a red edge being downstairs. Arriving to or is the same as arriving to along a red and blue edge respectively. Recall that a walk in a graph is more general than a path since it is allowed to revisit a vertex. It is an elementary fact from graph theory that there exist a walk between two vertices if and only if there exists a path between the same vertices. We need to use the term walk in this model since we could use a vertex both going along red edges and later along blue edges or vice versa. In the non-colored models the probability of existence of a path and of a walk is of course the same. We will from now on use mostly this second formulation of , but for notational convenience we use for a walk entering along a red edge (if blue edge also legal) and for a walk entering along a blue edge (if red edge also legal).
Our proof in Section 3 requires in fact yet one more level of generalization. In this next model we assume that some edges forming a connected subgraph are required to have the same color. To this end we think of the edges as partitioned into disjoint subsets , i.e. and if . Let be such a partition into connected subgraphs and .
Model : (Hypergraph) All edges in a set are given the same color red or blue with equal probability independent of the other sets. A walk in may change color only at a vertex in .
Note that model is the special case, where all sets contains one edge. It is helpful to think of a set as a hyperedge having a color, which enables passage between any two of the vertices in the hyperedge. Thus model is a generalization to hypergraphs.
Conjecture 2.7 ().
Let be any graph and and as in model . For any we have
It seems me that also this more general conjecture is likely to be true.
It is worth noting however that one may not in general assume that two edges have different colors without violating the bunkbed condition. As an example, let be the path of length two from to with as middle vertex and let . If we now assumed that the two edges have to have different colors then we would have
contrary to what we conjecture in the other models.
3. Outerplanar graphs
In this section we will prove the Bunkbed conjectures and for outerplanar graphs . A connected planar graph is called outerplanar if it is has a drawing such that every vertex lies on the boundary of the outer region. This is equivalent to the graph not having or as minors.
Our line of proof is to recursively prove that a minimal counterexample may not exist. To this end we will present a number of recursive operations. We will often need to work in model . In each case we have a triple , a graph , a set of transversal vertices and a partition of . We say that the triple reduces to a set of triples if whenever is true for all then also is true. The operations below will be constructed by conditioning on mutually exclusive events and thus every probability for a walk in is a linear combination of the probability of the corresponding walks in which implies that reduces to . When we are interested in only we take to be the partition into singletons.
Whenever we contract an edge we let the new vertex be in if at least one of are in . To avoid technicalities we will allow multigraphs and keep multiple edges that are formed after a contraction. Loops will always be removed. Note that we have not assumed that in general. This might be the case after a contraction and it is no problem.
T-operation: If and , then we contract the edge to the graph , with and the partition restricted to the new edge set.
Any walk can always run freely between the vertices and and assume any color when leaving . Thus every probability is preserved when contracting the edge . When this is the case we will call the graphs equivalent. Thus reduces to .
V2-operation: Assume and . Let be the neighbors of and assume that at least one of and form a singleton set in . Then we define two subgraphs of as follows. and with and the natural restrictions on .
If the edges have different colors we are in a situation equivalent to . If have the same color, then we are in a situation equivalent to . Thus reduces to , .
-operation: Assume are any vertices (possibly including or ) of the graph that form a triangle, i.e. . Assume further that no other edge is dependent on the color of or , i.e. each of them form a singleton set . Then we form the following four cases: , , and finally is the same graph as , but we require and to have the same color so they form on block in the partition . In this particular situation we do not want the graph to have double edges between and so we remove one of them and similarly for . Other multiple edges could have been created as usual.
Assume first that . Consider the eight possible colorings of , see Figure 1. Let case be the two leftmost figures, where has a different color than . Then we see that any walk can run freely between vertices and along either of the two different colors. This case is thus equivalent to . Since we have assumed that have the same color they have this also in and this is the reason we removed one of them in the definition of . Similarly let be the case where has a different color than and let be the case where has a different color than . Then these case are for the same reasons equivalent to and respectively. In the two rightmost figures the colors are equal for all three edges which is . Thus reduces to .
If any of belong to , the reduction still works to the same four triples. For instance, if , , let be the same case as above. If, say, is red and are blue, then any walk entering blue can leave or in any color. Entering in either color the walk can leave at the other vertex in either color, or at in blue. Similarly with the colors reversed. Thus again, conditioning on case is equivalent to . We leave to the reader to verify all other possibilities, which are not more difficult.
is true for all outerplanar graphs and all possible . Thus the bunkbed conjectures are true for any outerplanar graph and any .
We will in fact prove the theorem for outerplanar multigraphs. Assume the contrary and let be a minimal counterexample, for some set . Minimal here means that all graphs obtained by deleting or contracting an edge are not counterexamples for any set . Note that if then we get equality by a mirror argument changing the color of every edge, similar to the proof of Lemma 2.3. Hence, we may assume that . We may also assume that is connected, since otherwise we could reduce to the component containing . Similarly is connected. In fact we may assume that is 2-connected. If not there would be a cutpoint such that is disconnected. If there is a component s.t. then we can condition on the colors of the edges in which will imply a situation where we may or may not change color using a tour into . This is equivalent to conditioning on if or not which means that is not a minimal counterexample. If and are in different components , then let and . In this situation every walk from to passes through so
Here we use that by symmetry , and . The notation means the probability that there are walks in from both and to .
So we may assume that is 2-connected and there are therefore two independent paths from to along the outer region, call them the outer paths. A chord is any edge not in the boundary of the outer region.
Claim 1: All chords in separates and , that is and are in different components of the graph obtained by removing vertices from .
Note that this implies in particular there are no chords with or as an endvertex. To prove the claim we assume the opposite, that contains a chord between two vertices on the same outer path from to . Then there is one such chord , with as few vertices as possible between and along the outer path. By construction and . If any then we can use operation to reduce to smaller graphs for which the bunkbed conjecture is true by assumption, which gives a contradiction. Similarly if we get a contradiction from the T-operation. This gives that the only possible configuration is a triangle , where is of degree 2 between and along the outer path and . The -operation reduces to subgraphs , for which the conjecture is true by assumption and where the three edges of the triangle have the same color. In the latter case one may remove and its two edges without altering any probability. This is again a subgraph of with no color assumptions and this contradicts being a minimal counterexample. The claim follows.
The claim has the direct consequence . Let be the neighbors of . We now condition on the color of . If it was blue (corresponding to upstairs) we can never use that edge for any walk containing (downstairs) since . In that case we could remove to obtain a smaller graph for which the theorem is true by assumption. We may thus assume that is red and is a minimal counterexample when and is red. Similarly we can argue that is red.
If then Claim 1 and the outerplanarity of implies that one of say has degree two and we may contract the red edge to obtain a minor . This graph is a smaller graph with and the condition that edge is red, which by assumption is not a counterexample.
If , then we condition on the color of . Again, because of outerplanarity and Claim 1 one of , say has degree at most 3. If is blue, then and are connected both with a red path and a blue edge. We may thus contract without changing any probabilities for walks. Since and are both red no path can ever enter and every path starting in must first go to . We may thus contract also and the resulting minor must satisfy the bunkbed conjecture. If is red, then we may contract and remove one of the parallel red edges to get a new graph . Probabilities for walks starting in in will be the same as walks starting in in . Since we get and has exactly one red edge . But was a minimal such counterexample so the bunkbed conjecture is true for and we get the desired contradiction. ∎
Note that there are other operations that one possibly may use to prove the conjectures for larger classes of graphs. We end this section with two examples.
Restricted -operation: Assume form a triangle, i.e. . Assume further that , , whereas the color of and is not dependent on the color of any other edge. Then we form the following three cases: , and finally is the same graph as , but we require and to have the same color. As in the -operation we remove the multiple edge in and the edge from . The set such that do not change. The same reasoning as for -operation shows that reduces to .
The reason we cannot use the ordinary -operation is that if we contract this would form a situation where the edges are forced to have different color than , which is not legal in model . See also Remark 2.8.
Y-operation: Assume and . Let be the neighbors of and assume that the color of no other edge is dependent on the color of . Then we form four subgraphs of as follows. , , and is the same graph as but the edges must have the same color.
If the edges have the same color but different from we are in a situation equivalent to . If the color of is different from then we are in a situation equivalent to and similarly for . The remaining cases are when all three edges have the same color which gives . As for previous operations we see that reduces to .
There is also a restricted Y-operation, whose formulation is left to the reader.
Note that a unicolored and a unicolored give the same hypergraph. This opens the possibility to perform transformations of graphs. It is well-known that every planar graph is reducible to . I have however not been able to use this fact to prove the for planar graphs. One obstacle is that one may perform the Y-operation only if .
4. Randomly oriented graphs
In this section we present a connection to randomly directed graphs. First the basic model.
Model : Every edge in is given one of the two possible directions with equal probability independently of the other edges.
We call the corresponding random directed graph .
By analogy with the undirected case we define Probability that there exist a directed path from vertex to in under model . This model is a natural candidate to define a random orientation of a given graph. It was for example studied for the -lattice in [GG1] and for questions of correlation of directed paths in [AL1, AL2].
The following lemma gives a direct connection between model and . It gives an interesting non-trivial reformulation of the problem. It is, to the best of my knowledge, first published by McDiarmid [CM] and seemingly independently and with an elegant proof by Karp [K] (My thanks to Jeff Kahn and the anonymous referee for pointing out these two references.) The lemma might seem surprising at first sight but once discovered it is not so difficult to prove. A third proof can be found in [SL].
For any graph and any vertices we have
This means that for the special case , we may study randomly oriented graphs instead. A different model of directed graphs that also is applicable to other values of is discussed in [SL]. Note that is a truly different model than . We may for instance not generalize by conditioning on the direction of vertical edges as we have conditioned on the presence of vertical edges in . We have however not been able to prove the bunkbed conjecture using these directed graphs either.
The reformulation of model inspired the following two models replacing red and blue with directions.
Model : Let . Every edge in is given one of the two possible directions with equal probability. A walk in may change direction at a vertex in , i.e. switch from following the direction of the edges to going against them and vice versa.
The corresponding question for this model is to start a walk from following the direction of the edges and compare the probabilities for arriving at going with or against the direction of the last edge into . For this model we may in fact prove the corresponding bunkbed theorem. Let and denote starting at following the direction of the edges (resp. going against the direction of the edges). If , then we can for both symbols start with or against the direction. Also let and denote entering going with (resp. against) the directions of the edge. Again, if then it is in both cases legal to enter either going forward or reverse direction.
Let be any graph and . For any we have
First we consider all orientations of such that there is no walk from to any vertex in . In this case the right hand side is zero so the inequality is clear.
In the remaining cases we condition on the existence of a walk from to some vertex in . In this case we will construct a involution on the set of orientations which will show that the probability is equal arriving to and to .
To this end fix an orientation of and define as all vertices to which there exists walks from to both and . For instance every vertex on a directed path from to a vertex in belongs to . This is because we may follow the path to the transversal vertex and then go backwards along the same path. Hence . If then we do nothing. If , let be all edges between two vertices in . Now we define a new orientation by reversing the direction of all edges not in . By construction . If there were a vertex , this would mean that there were two shortest paths in with orientation starting at some, possibly different, vertices in , using only edges in and ending in and respectively. But every edge on has the reverse orientation in than in and every vertex in can be reached either way. Thus is a legal path also in orientation of but ending in instead of , and the other way around for . This gives a contradiction and we can conclude that and thus and . There is a walk in from to if and only if there is a walk from to in and vice versa. The theorem follows. ∎
Model : Let . Every edge in is given one of the two possible directions with equal probability. A walk in may change direction at a vertex in , i.e. switch from following the direction of the edges to going against them and vice versa. A walk must not use an edge in both directions.
The model seems closer to than , but unfortunately they are not equivalent in general. Figure 2 shows an example with four vertices and five edges, , where , whereas .
We end with the corresponding bunkbed conjecture for model .
Conjecture 4.3 ().
Let be any graph and . For any we have
5. A critical probability for finite graphs
We end this note with the definition of a critical probability for finite graphs that could be interesting to study further. Consider the following modification of Model .
Model : Given a graph and , let . Every edge in is colored red with probability and otherwise colored blue. A walk in may change color only at a vertex in .
Recall that we think of red edges as being downstairs (in the 0-layer) and blue as being upstairs. Now we define the average probability that there is a walk from to . That is, the walk must start from along a red edge (unless then we can switch to a blue edge at once) and arrive to along a red edge (again unless ).
Similarly we define
Intuitively it is clear that if is large (close to 1) the first quantity should be larger and vice versa if is close to 0. We conjecture that for any connected graph and any there is a critical probability such that
If this and the conjecture are true, then for . The inequality is strict because of the case .
Example: Let be the path with edges and let be the endpoints. It is easy to compute that for and for . Defining an appropriate recursion one may also prove that the conjectured properties of holds for any path and that is increasing, monotone and converging to for . This may be interpreted as the endpoints of long paths being further apart. Does this make some sense also for other graphs?
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