SlowGuess/NeurIPS_2023_OCR huggingface dataset:
This establishes that if  $D_{1}, D_{2}$  are  $k$ -Markov equivalent then  $\mathcal{C}_k(D_1), \mathcal{C}_k(D_2)$  must have the same skeleton and the same unshielded colliders. By Corollary 3.13,  $\mathcal{C}_k(D_1), \mathcal{C}_k(D_2)$  are Markov equivalent.
(Left)
Suppose that  $\mathcal{C}_k(D_1)$  and  $\mathcal{C}_k(D_2)$  are Markov equivalent. Then they impose the same d-separation statements. Therefore they impose the same d-separation statements when the conditioning set is restricted to size at most  $k$ . By Lemma 3.6, this means that  $D_{1}, D_{2}$  must also impose the same d-separation statements for conditioning sets of size of at most  $k$ . This establishes that  $D_{1}, D_{2}$  are  $k$ -Markov equivalent.
A.7 Proof of Lemma 4.1
Suppose in the  $k$ -closure graph  $\mathcal{C}_k(D)$  for some DAG  $D$  and integer  $k$ , we have a discriminating path for  $Y$  between the nodes  $a, v$  of the form
$$
a * \rightarrow \leftrightarrow \dots \leftrightarrow u \leftrightarrow Y \leftrightarrow v.
$$
By definition of discriminating path,  $u$  must be a collider along the path, and  $u \to v$ . If  $Y \gets v$ , then we would have an almost directed cycle  $u \to v \to Y \leftrightarrow u$ . Thus, we have  $u \leftrightarrow Y \leftrightarrow v$ .
First, we show that the arrowhead at  $Y$  of the edge  $Y \leftrightarrow v$  can be learned by first orienting unshielded colliders and then applying  $\mathcal{R}1$  and  $\mathcal{R}2$ . Consider the bidirected edge  $u \leftrightarrow Y$ . By definition of discriminating path  $a, v$  must be non-adjacent and thus separable by a set of size at most  $k$  by Lemma A.18. Therefore, we have a set  $S: |S| \leq k$  such that  $a \perp v |S$ . By the discriminating path definition, every collider along the path must be a parent of  $v$ , and therefore it must be the case that every collider along the discriminating path including  $u$  must be in  $S$ , since otherwise there would be a d-connecting path. From Lemma A.2, conditioned on any set of size at most  $k$ , we have a d-connecting path that starts with an edge into  $Y$ . Consider the shortest such path  $q$ . Let  $X$  be the node immediately before  $Y$  along  $q$ . Since this path exists in the DAG by the lemma, we have  $X \rightarrow Y$ . If  $X, v$  are non-adjacent, then  $X \rightarrow Y \leftrightarrow v$  would be an unshielded collider and we are done.
Suppose  $X, v$  are adjacent. Note that conditioned on  $S$ ,  $a$  and  $X$  are d-connected. If  $X$  is a non-collider along the path obtained by concatenating the subpath of  $q$  between  $a, X$  and the edge between  $X, v$ , then  $a, v$  would be d-connected given  $S$ , which is a contradiction. Therefore,  $X$  must be a collider along this path. Thus we have  $X \leftrightarrow v$ . Note that we cannot have  $X \leftarrow v$  since this would create an almost directed cycle in  $C_k(D)$ . Let  $V$  be the node immediately before  $X$  along  $q$ . Thus we have  $V \rightarrow X \leftrightarrow v$  (not  $V \leftrightarrow X$  since this edge exists in  $D$ ).
Suppose  $V, v$  are non-adjacent. Thus, the collider  $V \to X \leftarrow *v$  is unshielded, and therefore can be learned. Furthermore,  $V, Y$  must be non-adjacent since otherwise, we must have  $V \to Y$  to avoid a cycle and there would be a path that is shorter than  $q$ , which "jumps over" the node  $X$  along  $q$ . Thus, we can learn that  $X \to Y$  from  $\mathcal{R}1$ . Finally, since now we have learned  $v * \rightarrow X \rightarrow Y$  and that  $Y, v$  are adjacent, by  $\mathcal{R}2$ , we must have  $Y \leftarrow *v$ . Thus the arrowhead mark at  $Y$  of the edge  $Y \leftrightarrow v$  can be learned, and we are done.
Suppose  $V, v$  are adjacent. Following a similar argument, we either have some unshielded collider that can be propagated using the argument above to orient  $Y \leftarrow *v$ , or we can continue covering unshielded colliders, which would imply the previous nodes are always parents along  $q$ . But this implies that  $u$  has a directed path to  $Y$ , which cannot happen since we have  $u \leftrightarrow Y$  in  $\mathcal{C}_k(D)$ . This establishes that  $Y \leftarrow *v$  can be learned by orienting unshielded colliders and applying the rules  $\mathcal{R}1$  and  $\mathcal{R}2$ .
For the arrowhead mark at  $Y$  of the edge  $u \leftrightarrow Y$ , similarly consider the shortest d-connecting path  $q$  between  $Y, v$  in  $D$  given  $S$  that starts with an arrow into  $Y$ . The argument follows similarly that either there would be directed path from  $v$  to  $Y$ , which is a contradiction with the existence of the edge  $Y \leftrightarrow v$  in the  $k$ -closure graph, or that there exists an unshielded collider along  $q$  that can be learned by orienting unshielded colliders, which can be propagated to learn  $u * \rightarrow Y$  using rules  $\mathcal{R}1$ , and  $\mathcal{R}2$ . This establishes the lemma.
A.8 Proof of Theorem 4.4
We show soundness of the two new rules with the following two lemmas:
Lemma A.19. Let  $K$  be a mixed graph that is sandwiched between  $\varepsilon_{k}(D)$  and  $PAG(\mathcal{C}_k(D))$ , i.e.,  $\varepsilon_{k}(D) \subseteq K \subseteq PAG(\mathcal{C}_{k}(D))$ .  $\mathcal{R}11$  is sound on  $K$  for learning the  $k$ -essential graph, i.e., if  $K' = \mathcal{R}11(K)$ , then  $\varepsilon_{k}(D) \subseteq K' \subseteq K$ .
Proof. For the sake of contradiction, suppose otherwise:  $\mathcal{R}11$  orients an edge  $a o \rightarrow b$  in  $K$  as  $a \rightarrow b$ , and there is a DAG  $D'$  with a  $k$ -closure graph  $\mathcal{C}_k(D')$  that is Markov equivalent to  $\mathcal{C}_k(D)$  and is consistent with  $K$  where  $a \leftrightarrow b$ . This means  $a, b$  are  $k$ -covered in  $D'$ . Then from Lemma A.2, conditioned on any subset  $S$  of size at most  $k$ , there must be a d-connecting path that starts with an arrow into both  $a$  and  $b$  in  $D$ . By construction of the  $k$ -closure graph, this path must also exist in  $\mathcal{C}_k(D')$ . Therefore, there must be some node  $w$  such that  $a \leftarrow w$ . Since  $a$  has no incoming edges, it must be the case that  $w \in C$ . However,  $b$  is chosen so that  $b$  is non-adjacent to any node in  $C$ . Therefore,  $b$  must be non-adjacent to  $w$  in  $K$ . However, this creates the unshielded collider  $w \rightarrow a \leftrightarrow b$ . However, note that  $w o - o a o \rightarrow b$  in  $\mathrm{PAG}(\mathcal{C}_k(D))$ , and thus  $\langle w, a, b \rangle$  is a non-collider in  $\mathcal{C}_k(D)$ . Therefore,  $\mathcal{C}_k(D')$  cannot be Markov equivalent to  $\mathcal{C}_k(D)$ , which is a contradiction.
Lemma A.20. Let  $K$  be a mixed graph that is sandwiched between  $\varepsilon_{k}(D)$  and  $PAG(\mathcal{C}_k(D))$ , i.e.,  $\varepsilon_{k}(D) \subseteq K \subseteq PAG(\mathcal{C}_{k}(D))$ .  $\mathcal{R}12$  is sound on  $K$  for learning the  $k$ -essential graph, i.e., if  $K' = \mathcal{R}12(K)$ , then  $\varepsilon_{k}(D) \subseteq K' \subseteq K$ .
Proof. For the sake of contradiction, suppose otherwise:  $\mathcal{R}12$  orients an edge  $ao - oc$  in  $K$  as  $a - c$ , and there is a DAG  $D'$  with a  $k$ -closure graph  $\mathcal{C}_k(D')$  that is Markov equivalent to  $\mathcal{C}_k(D)$  and is consistent with  $K$  where  $a \leftrightarrow c$ . This means  $a, c$  are  $k$ -covered in  $D'$ . Then from Lemma A.2, conditioned on any subset  $S$  of size at most  $k$ , there must be a d-connecting path that starts with an arrow into both  $a$  and  $c$  in  $D$ . By construction of the  $k$ -closure graph, this path must also exist in  $\mathcal{C}_k(D')$ . Therefore, there must be some node  $w$  such that  $a \gets w$ . Since  $a$  has no incoming edges, it must be the case that  $w \in C$ . However,  $c$  is chosen so that  $c$  is non-adjacent to any other node in  $C$ . Therefore,  $c$  must be non-adjacent to  $w$  in  $K$ . However, note that  $wo - oao - oc$  in  $\mathrm{PAG}(\mathcal{C}_k(D))$ , and thus  $\langle w, a, c \rangle$  is a non-collider in  $\mathcal{C}_k(D)$ . Therefore,  $\mathcal{C}_k(D')$  cannot be Markov equivalent to  $\mathcal{C}_k(D)$ , which is a contradiction.
Now consider the execution of the algorithm  $k$ -PC. When the algorithm completes Step 4, from Corollary 4.2 we have that  $K = \mathrm{PAG}(\mathcal{C}_k(D))$ . Since we start Step 5 with  $K = \mathrm{PAG}(\mathcal{C}_k(D))$ , from Lemma A.19 and A.20, any arrowhead and tail orientation of the  $K$  obtained at the end of step 5 must be consistent with the  $k$ -essential graph of  $D$ . Therefore, we have that  $\varepsilon_k(D) \subseteq K$ .
B FCI Orientation Rules
Algorithm 2 FCI_Orient
Input: Mixed graph  $K$
Apply the orientation rules of  $\mathcal{R}1, \mathcal{R}2, \mathcal{R}3$  of [25] to  $K$  until none applies.
Apply the orientation rules of  $\mathcal{R}8, \mathcal{R}9, \mathcal{R}10$  of [25].
We restate the FCI orientation rules in detail and demonstrate how they are applicable for learning  $k$ -closure graphs. The following definitions are from [25].
Definition B.1 (Partial Mixed Graph (PMG)). Any graph that contains the edge marks arrowhead, tail, circle is called a partial mixed graph (PMG).
Definition B.2 (Uncovered path). In a PMG, a path  $\langle u_1, u_2 \dots u_m \rangle$  is called an uncovered path if  $u_i, u_{i+2}$  are non-adjacent for all  $i \in \{1, 2, \dots, m-2\}$ .
Definition B.3 (Potentially directed path). In a PMG, a path  $\langle u_1, u_2 \ldots u_m \rangle$  is called a potentially directed (p.d.) path if the edge between  $u_i$  and  $u_{i+1}$  does not have an arrowhead at  $u_i$  for all  $i \in \{1, 2, \ldots, m-1\}$ .
Definition B.4 (Circle path). In a PMG, a path  $\langle u_1,\dots u_m\rangle$  is called a circle path if  $u_{i}o - ou_{i + 1}$  for all  $i\in \{1,2,\ldots m - 1\}$ .
Note that circle paths are special cases of p.d. paths.
Rules 1, 2, 3 are straightforward extensions of the orientation rules for constraint-based learning to mixed graphs. For completeness, we restate them below. The star marks that appear both before and after the application of the rules are edge marks that remain unchanged by the rule.
$\underline{\mathcal{R}} 1$  : If  $a * \rightarrow bo * c$ , and  $a, c$  are not adjacent, then orient the triple as  $a * \rightarrow b \rightarrow c$ .
$\underline{\mathcal{R}} 2$  : If  $a\to b*\to c$  or  $a*\to b\to c$  and  $a* - o c$  , then orient  $a*$  -oc as  $a*\rightarrow c$
$\underline{\mathcal{R}} 3$  : If  $a*\rightarrow b\gets *c,a*$  -ODO-  $*c,a,c$  are non-adjacent, and  $d^{*}$  -ob then orient  $d^{*}$  -ob are  $d*\to b$
We now restate  $\mathrm{FCI + }$  rules3  $\mathcal{R}8$ $\mathcal{R}9$ $\mathcal{R}10$  and explain their relevance for learning  $k$  -closure graphs. Note that the rules are simplified since we do not have undirected edges that represent selection bias, and our undirected edges are treated as if they are circle edges.
$\underline{\mathcal{R}} 8$  : If  $a\to b\to c$  and  $ao\to c$  , orient  $ao\rightarrow c$  as  $a\to c$
$\underline{\mathcal{R}} 9$  : If  $a o \rightarrow c$ , and  $p = \langle a, b, u_1, u_2 \dots u_m, c \rangle$  is an uncovered p.d. path from  $a$  to  $c$  such that  $b, c$  are non-adjacent, then orient  $a o \rightarrow c$  as  $a \rightarrow c$ .
$\mathcal{R}10$ : Suppose  $a o \to c, b \to c, d \to c$ ,  $p_1$  is an uncovered p.d. path from  $a$  to  $d$  and  $p_2$  is an uncovered p.d. path from  $a$  to  $b$ . Let  $t_d$  be the node adjacent to  $a$  on  $p_1$  ( $t_d$  can be  $d$ ) and  $t_c$  be the node adjacent to  $a$  on  $p_2$  ( $t_c$  can be  $c$ ). If  $t_d, t_c$  are distinct and non-adjacent, then orient  $a o \to c$  as  $a \to c$ .
$\mathcal{R}11$  and  $\mathcal{R}12$  cannot replace any of the above rules. For example, consider Figure 5. None of the FCI rules apply to the output of Step 3, thus we can only learn of the unshielded colliders at the end of Step 4 of the algorithm. The completeness of FCI implies that any edge  $x o \rightarrow y$  can be oriented as  $x \leftrightarrow y$  or  $x \rightarrow y$  and give a MAG consistent with the PAG. However, not all such MAGs are valid  $k$ -closure graphs.  $\mathcal{R}11$  can be applied to orient several tails, which gives the graph in  $(d)$ . Similarly in Figure 6,  $\mathcal{R}12$  helps orient the tail edges between  $a, b$  which cannot be learned by FCI rules.
This version was originally called A-FCI, short for augmented FCI rules by [25]. Augmented graphs are recently used in a different context in the causality literature, which is why in this work we are calling this version  $\mathrm{FCI + }$  to avoid confusion.
C Sample Runs of  $k$ -PC Algorithm
Consider the figures below for two sample runs of  $k$ -PC algorithm. Note that  $k$ -PC outputs the  $k$ -essential graph in these examples, i.e., it can orient every invariant arrowhead and tail mark in the  $k$ -closure graph of  $D$ .
(a)  $D$
(b)  $\mathcal{C}_k(D)$
(c)  $K$  After Step 4
(d)  $K$  after Step 5.  $\mathcal{R}11$  applied on  $a,u,v$
(e)  $D^{\prime}\sim_{k}D$
(f)  $\mathcal{C}_k(D^{\prime})$  is consistent with the output of  $k$ -PC.
Figure 5: An example where  $\mathcal{R}11$  helps orient tails. (a) A DAG  $D$ . (b)  $k$ -closure graph of  $D$  for  $k = 0$ . (c)  $K$  after Step 4 of  $k - PC$ , the same as  $PAG(\mathcal{C}_k(D))$ . (d)  $\mathcal{R}11$  helps orient several tail edges. (e) A DAG  $D'$  that is  $k$ -Markov to  $D$ . (f)  $k$ -closure graph of  $D'$ , which is Markov equivalent to  $\mathcal{C}_k(D)$ , showing that the circle at  $bo \to c$  is not an invariant tail. Thus  $k$ -PC outputs  $k$ -essential graph  $\varepsilon_k(D)$  in this case.
(a)  $D$
Figure 6: (a) A causal graph  $D$ . (b)  $k$ -closure of  $D$  for  $k = 0$ . (c)  $K$  at the end of Step 4. (d) Node  $a$  has one edge  $ao - o$ ,  $ao - ob$ . Thus we have  $\mathcal{C} = \mathcal{C}^*$  since  $b$  is non-adjacent to any other nodes in  $C$  since there are no other nodes in  $C$ . Thus it is oriented as  $a - b$  due to  $\mathcal{R}12$ .  $e \to c$  is oriented due to  $\mathcal{R}11$ , similarly since  $\mathcal{C} = \emptyset$  and the node  $c$  is trivially non-adjacent to all nodes in  $\mathcal{C}$ .  $(e, f) D'$ ,  $D''$  are  $k$ -Markov equivalent to  $D$  and their  $k$ -closure graphs contain  $a \leftrightarrow c$  and  $b \leftrightarrow c$ , respectively. This shows that the graph in (d) given by  $k$ -PC is the  $k$ -essential graph  $\varepsilon_k(D)$ .
(b)  $\mathcal{C}_k(D)$
(c)  $K$  after Step 4.
(d)  $K$  at the end of Step 5.
(e)  $D^{\prime}$
(f)  $D^{\prime \prime}$
D Discussions
D.1 No Local  $k$ -Markov Equivalence Characterization on DAG Space
(a)  $D_{1}$
Figure 7: As a collider,  $d$  blocks the path  $(a,c,d,e,b)$  in  $D_{3}$ , but not in  $D_{4}$ . Accordingly,  $(a\perp b)_{D_1}(a\not\perp b)_{D_2}$  and  $D_{1},D_{2}$  are not  $k$ -Markov for  $k = 0$ . However, this does not appear as a local condition since  $c,e$  are not separable by conditioning sets of size up to 0 in both graphs. Therefore, a local characterization of equivalence like Verma and Pearl is not possible when we are only allowed to check degree- $k$  d-separation tests.
(b)  $D_{2}$
Consider the graphs in Figure 7. The only difference is the orientation of the edge between  $e, d$ . Due to the collider  $c \to d \gets e$ , we have  $a \perp b$  in  $D_{1}$  but not in  $D_{2}$ . Therefore for  $k = 0$ , we have that  $D_{1}, D_{2}$  are not  $k$ -Markov equivalent. However, this is not detectable locally: The endpoints of the collider responsible for the change of the  $k$ -Markov equivalence class cannot be  $d$ -separated. The effect of the collider can be detected only farther out in the graph between  $a, b$ . This shows that a local characterization similar to Theorem 2.10 is not possible for  $k$ -Markov equivalence.
D.2  $k$ -closure Graphs vs. MAGs
(a)  $D$
Figure 8: The mixed graph on the right is a valid  $k$ -closure graph for  $k = 1$ : It is the  $k$ -closure graph of the DAG in (a). However, it is not a valid  $k$ -closure graph for  $k = 2$ . Because after removing  $c \leftrightarrow d$ , it is possible to d-separate  $c, d$  by conditioning on  $u_1, u_2$ .
(b)  $K = \mathcal{C}_k(D)$  for  $k = 1$
In this section, we give an example for a MAG that is not a valid  $k$ -closure graph. Consider the graph in Figure 8.  $K$  in (b) is a valid  $k$ -closure graph for  $k = 1$  since it is the  $k$ -closure graph of  $D$ . However,  $K$  is not a valid  $k$ -closure graph for  $k = 2$ . This is because the bidirected edge  $c \leftrightarrow d$  is added between a pair that is not  $k$ -covered for  $k = 2$ : We have  $c \perp d|u_1, u_2$  in  $D$ . In fact, using the if and only if characterization in Theorem 3.9, we can show that there does not exist any  $D'$  with the given  $k$ -closure graph where  $c, d$  have a bidirected edge.
D.3 Bidirected Edge in  $k$ -essential Graphs
In Figure 9, since every endpoint is an arrowhead and is part of an unshielded collider, there is no other Markov equivalent  $k$ -closure graphs, which implies that the  $k$ -essential graph is the same as the  $k$ -closure graph. Thus, the edge  $c \leftrightarrow e$  is in  $\varepsilon_k(D)$ . This example shows that we can learn that two nodes do not cause each other using conditional independence tests that are not even powerful enough to make them conditionally independent. It is worth noting that LOCI [23] can also infer this fact by removing this edge.
(a)  $D$
Figure 9: A DAG with a size-1  $k$ -Markov equivalence class for  $k = 1$ . Observe that  $\mathcal{C}_k(D)$  only has unshielded colliders and thus there is no other  $k$ -closure graph that is Markov equivalent. Thus, this  $k$ -closure is at the same time the  $k$ -essential graph of  $D$  and can be learned from data. In this case, we can learn  $c, e$  do not cause each other despite not being separable in the data.
D.4  $k$ -PC is Incomplete
One might hope that  $k$ -PC is complete and outputs the  $k$ -essential graph  $\varepsilon_{k}(D)$ . This, however, is not true. We discuss an example where  $k$ -PC cannot orient an invariant tail mark.
First, observe that  $k$ -PC does not leverage the value of  $k$ . If we had an efficient way to answer the question "Is there a  $k$ -closure graph that is consistent with  $K$  in which  $a, b$  are  $k$ -covered?" then we could leverage this to orient more  $o \rightarrow$  edges as  $\rightarrow$  edges. As an example, consider the causal graph in Figure 10.  $d$  may have an incoming edge that prevents us from eliminating the possibility that  $do \rightarrow b$  is a bidirected edge  $d \leftrightarrow b$ . However, we can only have a single  $d$ -connecting path between  $d, b$ . Since  $d$  is a non-collider along  $a$ — $d$ — $c$ , one of the edges must be out of  $d$ . Suppose  $d \leftarrow a$  and  $d \rightarrow c$ . For  $c$  to not block this path it has to be a non-collider, and thus  $c \rightarrow b$ . This is the only way we can have two  $d$ -connecting paths between  $d, b$  in the underlying DAG. However, now we have the path  $d \rightarrow c \rightarrow b$ , which makes  $d$  an ancestor of  $b$ . Therefore the edge  $d \leftrightarrow b$  is inconsistent. Thus in  $k$ -essential graph of  $D$ , we must have a tail at  $d$  as  $d \rightarrow b$ . This cannot be learned by  $k$ -PC.
(a)  $D_{1}$
Figure 10: A graph where  $k$ -essential graph contains an invariant tail that cannot be learned by  $k$ -PC algorithm.  $k = 1$ ; thus  $d, b$  are  $k$ -covered but not  $a, c$ , which gives the  $k$ -closure graph on the right.  $k$ -PC algorithm outputs the graph in (c). However, there is no  $k$ -closure in the Markov equivalence class where  $d \leftrightarrow b$ . Thus the edge in the  $k$ -essential graph should be  $d \rightarrow b$ . Similarly,  $ao \rightarrow b$  should be  $a \rightarrow b$ , however this requires reasoning about the number of d-connecting paths between  $a, b$  in  $K$  that is not captured in  $k$ -PC.
(b)  $\mathcal{C}_k(D_1)$
(c)  $k$ -PC Output
We observe that a local algorithm such as  $k$ -PC cannot be used to assess if there is some  $k$ -closure graph that is consistent with the current graph in which two nodes are  $k$ -covered without the edge between them. One practical strategy would be to take the output of the  $k$ -PC algorithm and list all MAGs consistent with the circle edges, and then check if they are valid  $k$ -closure graphs by pruning every edge using Theorem 3.9. For small or sparse  $k$ -closure graphs,  $k$ -PC could be a practical way to reduce the search space efficiently, and then we can conduct an exhaustive search as the next step to obtain a sound and complete algorithm for learning the  $k$ -essential graph.
D.5 Heuristic Uses of Bounded Size Conditioning Sets
For large number of variables, and except for very sparse graphs, constraint-based algorithms take significant time to complete. This is because the progressive nature of such tests is not able to sufficiently sparsify the graph with low-degree conditional independence tests, and they have to perform many tests: If the neighbor size is  $\mathcal{O}(n)$ , where  $n$  is the number of nodes, algorithm needs to check exponentially many subsets of nodes in the conditioning set.
To prevent this issue, several implementations of these algorithms have the added functionality to restrict this search by limiting the size of the conditioning set. For example, causal-learn package has this functionality. However, this is a heuristic that simply prematurely stops the search algorithms.
The results of our paper build the theoretical understanding of what is learnable in this setting with a new equivalence class and its graphical representation.
D.6 Comparison of  $k$ -PC Output to LOCI Output
Lemma D.1. Consider three nodes  $a, b, c$ . Suppose  $a \not\perp b | S$ ,  $b \not\perp c | S$ ,  $a \perp c | S$  for some set  $S: |S| \leq k$  and  $b \notin S$ . If  $b, c$  are  $k$ -covered, then  $k-PC$  orients the edge between  $b, c$  as  $b \gets *c$ .
Proof. In the following, we show that given the CI pattern that LOCI uses to orient edges,  $k$ -PC also orients the same edges. Consider the CI pattern that LOCI uses between three nodes  $a, b, c$ .
Conditioned on  $S$ , there is a d-connecting path between  $a, b$ ; let us call this path  $p$ . Conditioned on  $S$ , there is a d-connecting path between  $b, c$ ; let us call this path  $q$ . Consider the path obtained by concatenating  $p, q$ . Since this path must be d-separating,  $b$  must be a collider on it and it must be the case that  $b \notin A n(S)$ . Let the node that is adjacent to  $b$  along  $p$  be  $a'$  and the node adjacent to  $b$  along  $q$  be  $c'$ . Thus we have  $a' \rightarrow b \leftarrow c'$ .
Case 1: Now suppose that  $a'$  and  $c$  are separable by some  $T: |T| \leq k$ .  $k$ -PC would then orient  $a' * \rightarrow b \leftarrow *c$  since  $b, c$  remains adjacent throughout the execution of  $k$ -PC. Thus  $b \leftarrow *c$  would be oriented in this case.
Case 2: Suppose that  $a', c$  are  $k$ -covered. Then, given  $S$ , there is a d-connecting path between  $a', c$ . Now consider the path obtained by concatenating the subpath of  $p$  between  $a, a'$  and this d-connecting path. Since  $a, c$  are d-separated given  $S$ ,  $a'$  must be a collider along this path and a non-ancestor of  $S$ . Thus we must have  $a'' \rightarrow a' \rightarrow b$  as the last three nodes of path  $p$ .
Case 2.a: Suppose  $a''$ ,  $c$  are separable by some  $T: |T| \leq k$ . Since  $a'$ ,  $c$  are  $k$ -covered, they remain adjacent throughout the execution of  $k$ -PC. Thus,  $k$ -PC would orient  $a'' * \rightarrow a' \leftarrow *c$ . Now we consider two sub-cases: