The scheme of representation that we will use is slightly different from the formalism of completed PDAGs. It is not necessary for each configuration of our search space (which we call restricted PDAG or RPDAG) to correspond to a different equivalence class; two different RPDAGs may correspond to the same equivalence class. The main reason for this is efficiency: by allowing an equivalence class to be represented (only in some cases) by different RPDAGs, we will gain in efficiency to explore the space. Before explaining this in greater detail, let us define the concept of RPDAG:
This condition states that an arc exists in only if it is either part of an h-h pattern or there is another arc (originated by an h-h pattern) going to .
As an RPDAG is a PDAG, it could be considered to be a representation of a set of (equivalent) DAGs. We therefore must define which the set of DAGs is represented by a given RPDAG , i.e. how direction may be given to the links in in order to extend it to a DAG. The following definition formalizes this idea.
We will use to denote the set of DAGs that are extensions of a given PDAG .
(a) As has no directed cycle (condition 2 in Definition 1), then either is already a DAG or it has some links. Let us consider an arbitrary link . Using condition 1 in Definition 1, neither nor can have a parent. We can then direct the link in either direction without creating an h-h pattern. If we direct the link as and is part of another link , then we direct it as (in order to avoid a new h-h pattern). We can continue directing the links in a chain in this way, and this process cannot generate a directed cycle because, according to condition 3 in Definition 1, has no completely undirected cycle.
(b) The extension process of does not modify the skeleton and does not create new h-h patterns. Therefore, all the extensions of have the same skeleton and the same v-structures (a v-structure is a particular case of h-h pattern), hence they are equivalent.
It should be noted that condition 4 in Definition 1 is not necessary to prove the results in Proposition 1. This condition is included to ensure that the type of PDAG used to represent subsets of equivalent DAGs is as general as possible. In other words, condition 4 guarantees that an RPDAG is a representation of the greatest number of equivalent DAGs, subject to the restrictions imposed by conditions 1-3 in Definition 1. As we will see in the next proposition, this is achieved by directing the minimum number of edges. For example, would not be a valid RPDAG. The RPDAG that we would use in this case is .
The proof is constructive. We shall build the RPDAG as follows: the skeleton and the h-h patterns of are the same as those in . An arc in shall now be considered such that and (if such an arc does not exist, then itself would be an RPDAG): we convert the arc into the link . This process is then repeated. Obviously, the PDAG obtained in this way has no directed cycle and verifies condition 4 in Definition 1. Moreover, we cannot obtain a configuration as a subgraph of because (we only remove the direction of arcs whose initial nodes have no parent). In addition, cannot have any completely undirected cycle because either the arc is not part of any cycle in or it is part of a cycle in that must contain at least one h-h pattern (and the directions of the arcs in this pattern will never be removed). is therefore an RPDAG.
Let us now prove that : if then and have the same skeleton and h-h patterns, hence and also have the same skeleton and h-h patterns. Moreover, as all the arcs in are also arcs in , if then , which in turn implies that . Therefore, according to Definition 2, .
Finally, let us prove the uniqueness of : we already know that any other RPDAG verifying that has the same skeleton and h-h patterns as . According to condition 1 in Definition 1, the edges that are not part of any of these h-h patterns but are incident to the middle node in any h-h pattern must be directed away from (in order to avoid new h-h patterns). The remaining edges that are not part of any h-h pattern must be undirected, in order to satisfy condition 4 in Definition 1. There is therefore only one RPDAG that matches a given skeleton and a set of h-h patterns, so is the only RPDAG verifying that . Figure 6 shows an example of the construction process.
The following proposition ensures that the concept of RPDAG allows us to define a partition in the space of DAGs.
Let be any DAG. Then itself is a PDAG and obviously . By applying the result in Proposition 2, we can assert that there is a single RPDAG such that .
In the proposition below, we show the properties which are common to all the DAGs belonging to the same extension of an RPDAG.
The necessary condition is obvious. Let us prove the sufficient condition: let and be two DAGs with common skeleton and h-h patterns. We shall construct a PDAG as follows: the skeleton and the h-h patterns of are the same as those in and ; the edges that have the same orientation in and are directed in in the same way; the other edges in remain undirected. From Definition 2, it is clear that .
has no directed cycles because and are DAGs. has no completely undirected cycles, since all the cycles in and share at least the h-h patterns. In addition, cannot be a subgraph of because this would imply the existence of the subgraphs and in and , respectively, and therefore these two DAGs would not have the same h-h patterns.
Therefore, the PDAG satisfies conditions 1-3 in Definition 1. By applying Proposition 2, we can then build a single RPDAG such that , hence .
A characterization of the extension of an RPDAG that will be useful later is:
- : Let , i.e., . Then, from condition 2 in Definition 2, , i.e., . Moreover, is an h-h pattern in . From condition 3 in Definition 2, this occurs if and only if is an h-h pattern in , which is equivalent to . Therefore, .
- and : Let . If there is another node , then is an h-h pattern in and therefore it is also an h-h pattern in , which contradicts the fact that . So, cannot have more than one parent in , hence .
- If then . From condition 2 we have , hence .
- If is an h-h pattern in , once again from condition 2, and therefore is an h-h pattern in .
- If is an h-h pattern in , then and . So, from condition 3, we obtain and, from condition 2, . Therefore is an h-h pattern in .