Superbubbles, Ultrabubbles, and Cacti

Abstract

A superbubble is a type of directed acyclic subgraph with single distinct source and sink vertices. In genome assembly and genetics, the possible paths through a superbubble can be considered to represent the set of possible sequences at a location in a genome. Bidirected and biedged graphs are a generalization of digraphs that are increasingly being used to more fully represent genome assembly and variation problems. In this study, we define snarls and ultrabubbles, generalizations of superbubbles for bidirected and biedged graphs, and give an efficient algorithm for the detection of these more general structures. Key to this algorithm is the cactus graph, which, we show, encodes the nested decomposition of a graph into snarls and ultrabubbles within its structure. We propose and demonstrate empirically that this decomposition on bidirected and biedged graphs solves a fundamental problem by defining genetic sites for any collection of genomic variations, including complex structural variations, without need for any single reference genome coordinate system. Further, the nesting of the decomposition gives a natural way to describe and model variations contained within large variations, a case not currently dealt with by existing formats [e.g., variant cell format (VCF)].

1. Introduction

Graphs are used extensively in biological sequence analysis, where they are often used to represent uncertainty about, or ensembles of, potential nucleotide sequences. Several subtypes have become especially prominent for sequence representation, in particular the de Bruijn graph (de Bruijn, 1946; Pevzner et al., 2001), the string graph (Myers, 2005), the breakpoint graph (Pevzner, 2000; Alekseyev and Pevzner, 2009), and the bidirected graph (aka sequence graph; Edmonds and Johnson, 1970; Medvedev and Brudno, 2009).

In the context of de novo sequence assembly, several characteristic types of subgraph are recognized, in particular the bubble (Zerbino and Birney, 2008), a pair of paths that start and end at common source and sink nodes but are otherwise disjoint. In the context of sequence analysis, a bubble can represent a potential sequencing error or a genetic variation within a set of homologous molecules. An efficient algorithm for bubble detection was proposed by Birmelé et al. (2012).

A generalization of the notion of a bubble, the superbubble is a more complex subgraph type in which a set of (not necessarily disjoint) paths start and end at common source and sink nodes. This problem was initially proposed by Onodera et al. (2013), who gave a quadratic solution. Brankovic et al. (2015) recently provided a linear time algorithm for superbubbles on directed acyclic graphs (DAGs). This result, when paired with a previous linear time transformation of the problem of superbubbles on directed graphs to superbubbles on DAGs (Sung et al., 2015), yields a linear cost solution for computing superbubbles on digraphs. For a review of superbubbles and their use in sequence analysis, refer Iliopoulos et al. (2016). In this article, we generalize the idea of superbubble to the more general case of a bidirected graph, connect a slight generalization of the superbubble, which we call the ultrabubble, and show how it relates to the decomposition of the graph into 2- and 3-edge connected (2-EC and 3-EC) components.

2. Methods

2.1. Directed, bidirected, and biedged graphs

A bidirected graph is a graph in which each endpoint of every edge has an independent orientation (denoted either “left” or “right”), indicating whether the endpoint is incident with the left or right side of the given vertex. The sides of D are, therefore, the set , and each edge in E_D is a pair set of two sides (Fig. 1). We say for all , and are opposite sides.

FIG. 1.

(A) A digraph. (B) A bidirected graph. Each node is drawn as a box, and the orientation for each edge endpoint is indicated by the connection to either the left or right side of the node. The graph excluding the dotted edges is the equivalent bidirected graph for the digraph in (A); the dotted edges encode an inversion that cannot be expressed in the digraph representation. (C) A biedged graph equivalent to the bidirected graph shown in (B).

Any digraph is a special case of a bidirected graph in which each edge connects a left and a right side (by convention, we here consider the right side to be the outgoing side and the left side the incoming side, so that the conversion from a digraph to a bidirected graph is determined; Fig. 1).

A biedged graph is a graph with two types of edges: black edges and gray edges, such that each vertex is incident with at most one black edge (Fig. 1C).

For any bidirected graph D, there exists an equivalent biedged graph , where:

• • , the sides of V_D,

• • , where E_D are the gray edges,

• • and are the black edges.

For a vertex , we use the notation to denote its opposite side.

Clearly, the bidirected and biedged representations are essentially equivalent, and the choice to use either one is largely a stylistic consideration. For the remainder of this article, we will mostly use the biedged representation. As any digraph is a special case of a bidirected graph and any bidirected graph has an equivalent biedged graph, so any digraph has an equivalent biedged graph.

2.2. Directed walks on biedged and bidirected graphs

A directed walk on a bidirected graph is a walk that at each visited vertex exits the opposite side to that which it enters. On a biedged graph, a directed walk is equivalent to a walk that alternates between black and gray edges. A directed cycle is a closed directed walk that starts and ends either on the same side (e.g., a self-loop edge) or on opposites sides of a vertex (in which case the start and end is arbitrary due to symmetry). A bidirected or biedged graph is acyclic if it contains no directed cycles.

These definitions are a generalization of a directed walk on a digraph. In a bidirected representation of a digraph, all edges in a directed walk are left-to-right or all are right-to-left. A directed walk on a general bidirected (or biedged) graph can mix these two types and additionally include edges that do not alternate the orientation of their endpoints (e.g., left-right, right-right, and left-left edges).

Given these generalizing relationships, clearly a digraph D is acyclic if B(D) is acyclic. Note that any acyclic biedged graph can also be converted into an equivalent DAG:

Lemma 1. For any acyclic biedged graph B(D), there exists an isomorphic biedged graph B(D) such that D is a DAG.

Proof. For each connected component in B(D), use a depth first search (DFS) beginning at side x to label the sides either “red” or “white”: If x is not already labeled, then label x red and white. For each gray edge incident with , if the connected side is not labeled, label the connected side red and continue recursively via DFS. In this way, all the sides in the connected component containing x will be labeled in a single DFS. If during the recursion the connected side encountered is already labeled, then it must be labeled red, else there would exist a directed cycle, a contradiction. Use the labeling to create B(D), isomorphic to B(D) but replacing the orientation of the sides so that each side labeled white is a left side and each side labeled red is a right side. All edges in B(D) connect a left and a right side.

2.3. Superbubbles, snarls, and ultrabubbles

Repeating the definition from Onodera et al. (2013), any pair of distinct vertices in a digraph D is called a superbubble (Fig. 2A) if:

FIG. 2.

(A) Superbubbles in a digraph. The superbubbles are indicated by pairs of numbered arrows, numbered consistently with (B). (B) A biedged graph representation of the digraph in (A). The snarls are illustrated by numbered arrows; the ultrabubbles are those numbered 1, 4, 9, and 12. Note, a side incident with a black bridge edge may be in multiple snarls (see snarls numbered 10).

• • reachability: y is reachable from x.

• • matching: The set of vertices, X, reachable from x without passing through y is equal to the set of vertices from which y is reachable without passing through x (passing through here means to enter and then exit a vertex on the path).

• • acyclicity: The subgraph induced by X is acyclic.

• • minimality: No vertex in X other than y forms a pair with x that satisfies the criteria previously defined, and similarly for y.

We call the subgraph induced by X the superbubble subgraph.

To generalize superbubbles for biedged graphs, we introduce the notion of a snarl, a minimal subgraph in a biedged graph whose vertices are at most 2-black-edge-connected (2-BEC) to the remainder of the graph (two vertices in a biedged graph are k-BEC if it takes the deletion of at least k black edges to disconnect them). In a biedged graph B(D), a pair set of distinct, non-opposite vertices are a snarl (Fig. 2B) if:

• • separable: The removal of the black edges incident with x and y disconnects the graph, creating a separated component X containing x and y and not and .

• • minimality: No pair of opposites in X exists such that and fulfills the criteria described earlier.

We call a vertex not incident with a gray edge a tip (Zerbino and Birney, 2008). In a biedged graph B(D), a snarl is an ultrabubble if its separated component is acyclic and contains no tips.

The following shows that a superbubble in a digraph is an ultrabubble in the equivalent biedged graph.

Lemma 2. For any superbubble (x, y) in a digraph D, the pair set is an ultrabubble in B(D).

Proof. Let d and e be the black edges incident with and , respectively, and let X be the superbubble subgraph of .

We start by proving that satisfies the separable criteria. As y is reachable from x by definition, there exists a directed path in B(D) between (the right side of x) and (the left side of y) that excludes d and e. After the deletion of these black edges, and , therefore, remain connected. If the separable criteria are not satisfied, the deletion of d and e must, therefore, not disconnect and from either or both and , without loss of generality assume (and therefore ) remains connected to .

If is on a directed walk from that excludes d, then the addition of d to this walk defines a directed cycle in B(D). As all nodes reachable from x are in the separated component X, the existence of this cycle in B(D) implies the existence of a corresponding directed cycle in X, a contradiction.

If there exists a non-directed walk from to , then let be the last node on the walk from such that the subwalk between and is a directed walk. By definition, there exists a directed walk from to . The next node on the walk from to after is, by definition, not reachable from but must be reachable from this node. This implies a contradiction of the matching criteria for the corresponding nodes in X.

We have, therefore, established that fulfills the separable criteria. We have already established that if a digraph is acyclic, its equivalent biedged graph is acyclic, therefore the separated component of is acyclic. As every node in X is reachable from both x and on a path from y, the separated component clearly contains no tips.

It remains to prove that fulfills the minimality criteria. If do not satisfy the minimality criteria without loss of generality, there exists a node in the separated component of such that are separable. It follows that all directed paths from to that exclude d and e visit , and for the node z in D contained in , fulfills (clearly) all the superbubble criteria, a contradiction.

2.4. Cactus graphs

A cactus graph is a graph in which any two vertices are at most 2-EC (Harary and Uhlenbeck, 1953). In a cactus graph, each edge is part of at most one simple cycle, and, therefore, any two simple cycles intersect at most one vertex.

For a graph , let be a multigraph created by merging subsets of the vertices, such that:

• •  is a partition of V_G,

• •  is a multiset,

where is a graph homomorphism that maps each vertex in V_G to the set in that contains it.

Merging all equivalence classes of 3-EC vertices in a graph results in a cactus graph (Paten et al., 2011).

For a biedged graph B(D), let C(D) be the cactus graph created by first contracting all the gray edges in B(D); then for each equivalence class of 3-EC vertices in the resulting graph merging together the vertices within the equivalence class (Fig. 3A–C). As with and G, is a partition of the vertices of , and is a multiset.

FIG. 3.

(A) A biedged graph B(D) with the snarls indicated by pairs of numbered arrows. (B) The graph in (A) after contracting the gray edges. (C) The cactus graph C(D) for B(D), constructed by merging the vertices in each 3-EC in (B). (D) The bridge forest D(D), constructed by contracting the edges in simple cycles in (C). 3-EC, 3-edge connected.

For a vertex , we call its projection in C(D). Similarly for a set of vertices , we call the projection of X in C(D). Let , which is the projection of the black edge incident with x in C(D).

Appendix 1 gives lemmas that make explicit the relationship between the edge connectivity of vertices in B(D) and C(D), and that we use to prove the relationship between the snarls of B(D) and C(D).

2.5. Snarls and cacti

A pair set of distinct vertices in B(D) are a chain pair if they project to the same vertex in C(D) and their incident black edges project to the same simple cycle in C(D) (e.g., pairs of arrows in simple cycles in Fig. 3C). A cyclic sequence of chain pairs within the same simple cycle in C(D) and ordered according to the ordering of this simple cycle is a (cyclic) chain. Contiguous chain pairs in a chain share two opposite sides of a black edge in B(D).

For a cactus graph C(D), the graph D(D) resulting from contracting all the edges in simple cycles in C(D) is a called a bridge forest (Fig. 3D).

A pair set of distinct vertices in B(D) are a bridge pair if they project to the same vertex in D(D) and both their incident black edges are bridges (e.g., pairs of arrows numbered 1 and 2 in Fig. 3D). A maximum sequence of bridge pairs within D(D) connected by incident nodes with degree 2 is an (acyclic) chain. As with chain pairs, contiguous bridge pairs in a chain share two opposite sides of a black (bridge) edge in B(D).

Theorem 1. The set of snarls in B(D) is equal to the union of chain pairs and bridge pairs.

Proof. Follows from Lemmas 10 and 11 given in Appendix 2.

Given Theorem 1 to calculate the set of snarls for a given biedged graph, it is sufficient to calculate the cactus graph to give the set of snarls that map to chain pairs and the bridge forest to calculate the set of snarls that map to bridge pairs. Constructing a cactus graph of the type described for a biedged graph is linear in the size of the biedged graph [using the algorithm described in Paten et al. (2011)], and clearly the cost of then calculating the bridge forest from the cactus graph is similarly linear. The number of chain pairs is clearly linear in the size of the biedged graph; however, the number of bridge pairs is potentially quadratic in the number of bridge pairs, so enumerating these latter snarls has potentially worst case quadratic cost in terms of the size of the biedged graph. Next, we consider ways to prune the set of snarls by using their natural nesting relationships to create a hierarchy of snarls that is at most linear in the size of the biedged graph.

2.6. Compatible snarl families

One particularly attractive feature of superbubbles is that they have nested containment relationships. That is, superbubbles have subgraphs that are either strictly nested or disjoint. Accordingly, a digraph is partitioned into a set of top-level superbubble subgraphs and other graph members not contained in a superbubble subgraph, and each top-level superbubble component then contains one or more child superbubbles, forming a tree structure. The situation is more complex for snarls. The separated component of snarls can overlap (Fig. 4) such that each partially contains the other. To create a properly nested hierarchy of snarls, it is, therefore, necessary to exclude some snarls.

FIG. 4.

Overlapping snarls. (A) A bidirected graph, its corresponding (B) cactus graph. The snarl numbered 2 contains the snarl numbered 4; similarly, the snarl numbered 3 contains the snarl numbered 1. The snarls numbered 2 and 3 overlap.

We will call a family of snarls compatible if all pairs of distinct snarls in the family have snarl subgraphs that are either disjoint or nested. A compatible family of snarls has a nesting structure that is a forest, similar to superbubbles. The following theorem provides a sufficient condition for constructing such a family in many bidirected graphs.

Theorem 2. In a connected biedged graph with at least one black bridge edge, the family of snarls whose subgraphs have no black bridge edges is compatible.

In addition, the next theorem shows that this family of snarls is a generalization of ultrabubbles.

Theorem 3. No ultrabubble contains a black bridge edge in its subgraph.

Proofs of these theorems are included in Appendix 3.

The bridge edge condition can also be used to construct a compatible family of snarls in a graph with no black bridge edges. To do so, we break one black edge into two tips. Each of these tips is then a bridge edge, so the family of snarls we construct from the modified graph is compatible. However, the family of snarls we obtain will depend on our choice of a black edge to break. Heuristically, an edge corresponding to a highly conserved genomic element should be chosen, since by construction it will not occur in any snarl's subgraph.

Given a snarl decomposition, the following algorithm will filter them down to the compatible family we have described:

• • Iterate over the black bridge edges of the graph [i.e., the edges of D(D)].

• • For a bridge edge , if either u or is the boundary of a snarl, mark that snarl as not containing .

• • Initialize a queue with u and , and traverse outward in breadth-first order, ignoring restrictions on directed biedged walks.

• • On reaching a node x that is a boundary for a snarl , if y, , or has not been traversed, mark the snarl as containing .

• • On reaching a node whose opposite is boundary for a snarl , if , x, or y has not been traversed, mark the snarl as not containing .

• • After completing every traversal, retain only snarls that were never marked as containing a black bridge edge.

The validity of this algorithm is proved by Lemma 21. Naively, this algorithm requires time for the traversals, and to mark all snarls. However, we can implement optimizations that improve on this behavior. First, we can also stop the breadth first search (BFS) traversals whenever they encounter a bridge edge. Lemmas 22 and 23 demonstrate that the portion of the BFS traversal after a bridge edge is redundant. This reduces the time required for the traversal to , where . In general, , so this does not improve over the worst case asymptotic bound. However, in many practical cases, M is approximately constant.

We can also reduce the total number of snarls we need to filter by neglecting to produce some snarls a priori. The quadratic bound on the number snarls is due to the fact that there is a bridge pair for all pairs of edges incident on a node in D(D). However, Lemma 17 shows that none of these bridge pairs will pass the filter. Accordingly, we can reduce the set of snarls we consider to only chain pairs and bridge pairs that project to nodes of degree 2 in D(D), which we call simple bridge pairs. This reduces the total number of snarls to .

2.7. Ultrabubbles and cacti

Given Theorem 1, to determine the ultrabubbles in B(D), it is sufficient to check for each chain and bridge pair if the separated component is acyclic and contains no tips.

Using Theorem 3, we can restrict the search to snarls whose separated component does not contain a black bridge edge. This implies that we need only consider bridge pairs whose projection in D(D) is a node whose degree is two, and we call such bridge pairs simple. The number of simple bridge pairs must be less than the cardinality of D(D), and therefore the total number of chain pairs and simple bridge pairs is less than or equal to . Using D(D) and C(D), which both can be constructed in time, we can clearly enumerate the set of simple chain pairs and bridge pairs in time.

A simple algorithm to find the set of ultrabubbles enumerates all chain pairs and simple bridge pairs and checks for each the acyclicity and tipless requirement by using a DFS, and is therefore worst case time.

3. Results

We implemented algorithms to create the cactus graph and bridge forest for an arbitrary bidirected graph in the vg software package (http://github.com/vgteam/vg), where these structures are used to decompose graphs into sites for variant calling.

Families of compatible snarls are created by picking the longest path in the bridge graph of simple bridge pairs, making this the top-level chain. The subset of ultrabubbles can also be computed by running vg stats -u.

In this study, we present the results of running this decomposition on a graph for human chromosome 1 constructed from the (∼6.5 million) variant calls from phase 3 of the 1000 Genomes Project (Consortium et al., 2015). The graph contained 19,917,881 nodes and 26,782,661 edges, and the runtime was 23 minutes by using a maximum of 49G RAM on a single 2.27 GHz Intel Xeon core (4 minutes and 30G of RAM were spent loading the graph into memory, a process that can be made an order of magnitude more efficient by switching the implementation to use xg, vg's succinct representation).

Table 1 shows the relative proportion of each of these structures. The first three rows describe the top-level ultrabubble decomposition, which covers exactly every base in the input graph. The second three rows display the same statistics but for structures that are entirely contained within top-level ultrabubbles or snarls. The remaining rows describe the third and deepest nesting level, which is contained within second-level ultrabubbles or snarls. Every base within the graph is part of either a top-level chain, ultrabubble, or snarl in this decomposition.

Table 1.

Coverage Statistics for the Ultrabubble Decomposition of the Human Chromosome 1 Variant Graph

Structure	Nesting level	Count	Coverage (bp)	Coverage (pct)
Chains	Top	1	221,715,143	86.60
Ultrabubbles	Top	5,554,903	12,539,619	4.90
Snarls	Top	75	21,775,387	8.50
Chains	Second	919	20,594,450	8.04
Ultrabubbles	Second	533,252	1,199,777	0.47
Snarls	Second	0	0	0
Chains	Third	67	495	0.00
Ultrabubbles	Third	694	1623	0.00
Snarls	Third	0	0	0

bp, base pairs; pct, percent.

Figure 5 shows the size distribution of the top-level ultrabubble and snarl sizes. All but 22 top-level ultrabubbles (totaling 3251 bases) are 100 bases long or shorter. If we consider such sites “easy” to call, along with top-level chains, then we can assign roughly 91.5% of chromosome 1 into this category. Figure 6 displays three examples of such small ultrabubbles. The remaining 9.5% of cases are found in a small number of relatively large snarls.

FIG. 5.

Histograms of top-level ultrabubble and snarl sizes in number of bases, as found in the 1000 Genomes graph for chromosome 1.

FIG. 6.

Ultrabubbles found in the 1000 Genomes-derived graph for chromosome 1. (A) Two adjacent SNPs inside a deletion (chr1:209,887,366). (B) A more complex combination of SNP and indel events (chr1:237,977,845). (C) Copy number changes in a GT repeat (chr1:1,200,943). SNP, single nucleotide polymorphism.

4. Discussion and Conclusion

We have presented a partial decomposition of a bidirected graph into a set of nested snarls and ultrabubbles. We believe this solves an important problem in using graphs for representing arbitrary genetic variations by defining a decomposition that determines sites and alleles.

As the decomposition is only partial, not all elements in a graph will necessarily fit into one of the ultrabubbles. However, we demonstrate that for an existing large library of variation (1000 Genomes), the large majority of sites are either invariant or described by simple, top-level ultrabubbles.

For bases outside of these easy sites, it is possible to imagine further subclassification. For example, classifying snarls that contain tips but are acyclic might define a useful class of subgraph that is common in some subproblems (e.g., sequence assembly). Some structures representing dense or overlapping collections of sequence polymorphisms, insertions, and deletions cannot be fully described by using nested ultrabubbles. We have previously shown that a generalization of the separability criterion for ultrabubbles can describe sites in these cases (Rosen et al., 2017). Similarly, characteristic structures representing genomic phenomena, such as inversions and translocations, are imaginable. Beyond our initial investigation, a more thorough evaluation of how much of a graph fits within a snarl, ultrabubble, or one of these more complex structures would be a useful exercise. We propose that the compatible family of snarls we constructed provides one path forward in this endeavor.

We can also envision that the nesting structure of snarls could play a powerful role in decomposing genotyping problems. Nested graph structures often arise from nested indels and substitutions.

In the context of assembly, various error correction algorithms have been proposed to remove graph elements and reduce the complexity of the graph. This increases the fraction of the graph that is contained within an ultrabubble structure. We foresee the cactus graph structure providing a useful basis for exploring such algorithms.

5. Appendix

5.1. Appendix 1

Lemma 3. A pair of vertices x, y are in the same component of B(D) if their projections are in the same component of C(D).

Proof. IF: Follows given that, by definition, no pair of vertices not connected in B(D) project to the same vertex in C(D). ONLY IF: Follows given that is a graph homomorphism from B(D) to C(D) and graph homomorphisms preverse connectedness.

Lemma 4. For a subset of edges , if the removal of the projection of X disconnects C(D), then the removal of X disconnects B(D).

Proof. Follows given that graph homomorphisms preverse connectedness.

Lemma 5. The vertices in C(D) are the equivalence classes of 3-BEC in B(D).

Proof. Each pair of vertices B(D) that project to the same vertex in C(D) are either/or-both connected by a path of gray edges (and hence 3-BEC) or connected by at least three black-edge-disjoint paths (using Menger's theorem).

Lemma 6. A black edge in B(D) is a bridge edge if its projection in C(D) is a bridge edge.

Proof. Let .

ONLY IF: Suppose e is a bridge. As e is a bridge, the vertices X reachable from x without visiting are black-edge connected only by e to the vertices reachable from without visiting x. Given Lemma 5, it follows that the projection of X and the projection of are disjoint, therefore the projection of e is a bridge.

IF: Suppose e is not a bridge but its projection is. By definition, there exists a path in from x to that does not include e. As is a homomorphism, the projection of that path connects and without traversing , implying that it is not a bridge, a contradiction.

Lemma 7. A maximal set of vertices in C(D) is 2-EC if the union of its members is a 2-BEC equivalence class of vertices in B(D).

Proof. Delete the black bridge edges in B(D) and the bridge edges in C(D) to create and , respectively. Each component is is, by definition, 2-BEC, and similarly each component in is 2-EC. The proof follows from Lemmas 3 and 6, by showing there exists a bijection between components in and such that for each component X in all the vertices in X project to vertices in the same component in .

A cut pair is a pair of edges whose deletion disconnects the graph.

Lemma 8. A pair of edges in a 2-EC component of a cactus graph is a cut pair if both edges are contained within the same simple cycle.

Proof. By definition, a 2-EC component of a cactus graph is a set of simple cycles connected by articulation (cut) vertices. It is easily verified that such a graph is and can only be disconnected by a pair of edges if they occur within one such simple cycle.

Lemma 9. A pair of black edges in a 2-BEC component X of B(D) is a cut pair if its projection is a cut pair in C(D).

Proof. Let be a vertex-induced subgraph of the projection of X. By Lemma 7, is a 2-EC component in C(D).

IF: If the deletion of the projection of d and e disconnects , then, using Lemma 4, the deletion of d and e disconnects X.

ONLY IF: If the projections of d and e are not a cut pair, by the definition of a cactus graph and Lemma 8, the projections of d and e in are each members of two distinct simple cycles. If the projection of d (similarly e) were a self-loop, then its endpoints are 3-BEC, implying that after the deletion of d and e its endpoints remain connected. This is impossible if the deletions of d and e disconnect the 2-EC component, hence each simple cycle containing the projection of d or e has length >1. For any pair of distinct vertices x, y in B(D) that project to the same vertex in C(D), there exists a path in B(D) that connects them that excludes their incident black edges, because by Lemma 5 they are 3-BEC, and are therefore connected either by a path of gray edges or by Menger's theorem, connected by at least three edge disjoint paths containing black edges. From this observation, it is easily verified that the endpoints of d (and similarly e) must be connected by a path Y in B(D) that includes the black edges that project to the simple cycle containing d, in the order of the cycle, and that excludes both d and e. This implies that the endpoints of d (similarly e) remain connected after the deletion of d and e, contradicting the claim that they are a cut pair.

5.2. Appendix 2

Lemma 10. Each snarl in B(D) is either a chain pair or a bridge pair.

Proof. Using Lemma 3, both x and y must project to a vertex in the same component of C(D) as they are connected in B(D).

Let d and e be the black edges incident with x and y, respectively. If d is a bridge, then e must be a bridge, or else, by definition, e connects two vertices in a 2-EC component X, the removal of d and e cannot therefore disconnect X, and therefore y and , violating the snarl separation criteria. Using Lemma 6, in this case the projections of d and e must, therefore, also be bridges. If both d and e are bridge edges but x and y do not project to the same vertex in D(D) (and are, therefore, not a bridge pair), there exists an intermediate bridge edge on the path between and . The deletion d, e and for B(D) disconnects B(D) into distinct components: One contains x and z, one contains and y, one contains , and one contains . This implies that and each fulfill the separation criteria, contradicting the minimality of .

If d and e are not bridges, both must be in the same 2-BEC component or contradict the separation criteria, by the same reasoning as earlier. In this case, Lemma 7 implies that both d and e must project edges in the same 2-EC component in C(D). Lemmas 8 and 9 further imply that they must project to edges in the same simple cycle. If x and y do not project to the same vertex in C(D) (and are, therefore, not a chain pair), then there exists an intermediate black edge on the path between and that excludes and . As with the case that both d and e were bridge edges, this similarly contradicts the minimality of .

Lemma 11. Each chain pair or bridge pair in B(D) is a snarl.

Proof. Lemmas 4 and 8 imply that meet the separation criteria. It remains to prove that is minimal. If is not minimal, then there must exist an intermediate edge on a path between and that excludes and , and that, using Lemma 10, forms chain or bridge pairs with and . As if is a chain pair, or if is a bridge pair, this is clearly impossible.

5.3. Appendix 3

In this section, we prove Theorems 2 and 3, which characterize a sufficient condition to produce a family of compatible snarls. We begin with two useful lemmas.

Lemma 12. Let be a snarl with snarl subgraph X. If u is a node in X and v is a node that is not in X, then any path from u to v includes the black edge incident on x or the black edge incident on y.

Proof. Suppose a path exists that does not include either of the black edges incident on x and on y. Then u is not disconnected from v after deleting these edges, which contradicts the separability of .

Lemma 13. Let be a snarl with subgraph X. Then there exists a path from u to either x or y that includes neither the black edge incident on x nor the black edge incident on y if u is in X.

Proof. First assume u is in X. Some path exists from u to either x or y, else u is not in the same connected component as x and y. Consider the shortest such path. Without loss of generality, assume this path is between u and x. Suppose the black edge incident on x or the black edge incident on y occurs somewhere along the path. Without loss of generality, assume it is the black edge incident on x. By Lemma 1, x or y must occur in the prefix of the path between u and . This implies that the path was not the shortest, which is a contradiction. Therefore, there exists a path from u that contains neither the black edge incident on x nor the black edge incident on y.

Next assume without loss of generality that a path exists from u to x that includes neither the black edge incident on x nor the black edge incident on y. This path is preserved after removing these two edges. This implies that u is in the same connected component as x (and hence also y) in the resulting graph, so u is in X.

Let and be two snarls with snarl subgraphs X₁ and X₂ respectively. We will say that splits if either a) x₂ is in X₁ but y₂ is not in X₁ or b) y₂ is in X₁ but x₂ is not in X₁. This condition clearly violates compatibility.

Lemma 14. Let and be snarls with snarl subgraphs X₁ and X₂. If splits , then x₁ and y₁ are in X₂.

Proof. We will proceed by showing that all other cases lead to contradictions. Without loss of generality, assume x₂ is in X₁ and y₂ is not in X₁.

Case I: x₁ and y₁ are not in X₂

Consider the set of paths from x₂ to y₂ that do not pass through or . This set is nonempty, else X₂ is disconnected. By Lemma 12, any such path must include x₁ or y₁, which would imply that x₁ is in X₂ or y₁ is in X₂, respectively, by Lemma 13. This violates the assumption of the case, so this case is contradictory.

Case II: x₁ is in X₂, and y₁ is not in X₂

Any path from x₁ to y₁ that does not include the black edges incident on x and y cannot include y₂, else y₂ is in X₁ by Lemma 13. Therefore, it must contain the black edge incident on x₂ by Lemma 12. Without loss of generality, this implies that and are separable, which violates the minimality of . Thus, this case is contradictory as well.

Case III: y₁ is in X₂, and x₁ is not in X₂

Same as Case II.

This proves the lemma.

Lemma 15. Let and be snarls with snarl subgraphs X₁ and X₂. If splits , then X₁ contains a black bridge edge.

Proof. Without loss of generality, assume x₂ is in X₁ and y₂ is not in X₁. Then, there exists at least one path from to either x₁ or y₁, else X₁ is disconnected. By Lemma 14, x₁ and y₁ are in X₂, so all such paths must include the black edge incident on x₂ or the black edge incident on y₂ by Lemma 12. Since y₂ is not in X₁, all paths from x₁ or y₁ to in X₁ must include the black edge incident on x₂. Therefore, the black edge incident on x₂ is a bridge edge by Menger's theorem.

There are also cases that violate compatibility without splitting a snarl. The following lemmas characterize these cases.

Lemma 16. Let and be snarls with distinct boundaries in a connected graph whose snarl subgraphs are X₁ and X₂. If x₁ and y₁ are in X₂, and x₂ and y₂ are in X₁, then .

Proof. Let u be an arbitrary node that is not in X₁ be arbitrary. There exists at least one path from u to a node in X₁, else B(D) is not connected. Let p₁ be the shortest such path. Clearly, no node from X₁ occurs in p₁ except at its terminus. In particular, p₁ does not contain either or . By Lemma 12, p₁ includes x₁ or y₁, so one of these must be the terminal node. Without loss of generality, assume it is x₁. Since x₁ is in X₂, there also exists a path p₂ from to either x₂ or y₂ that does not include or by Lemma 13. Note that is a path from u to either x₂ or y₂ that does not include the black edges incident on x₂ and y₂. Thus, u is in X₂ by Lemma 13. This implies .

Lemma 17. Let and be snarls with snarl subgraphs X₁ and X₂. If , then both X₁ and X₂ contain a black bridge edge.

Proof. Suppose y₂ is not in X₁. Then, all paths from y₂ to x must include the black edge incident on x or the black edge incident on y₁ by Lemma 12. There exists at least one path between x and y₂ in X₂, which cannot include the black edge incident on x. Therefore, all paths between y₂ and x must include the black edge incident on y₁. This implies without loss of generality that and are separable, which violates the minimality of Thus, y₂ is in X₁. Note that the black edge incident on x is not in X₁. Therefore, removing the black edge incident on y₂ from X₁ disconnects x from because of the separability of . Thus, the black edge incident on y₂ is a bridge edge in X₁. Similarly, the black edge incident on y₁ is a bridge edge in X₂.

Finally, we establish the relationship between pairs of snarls that allow for compatibility.

Lemma 18. Let and be snarls with snarl subgraphs X₁ and X₂. If both x₂ and y₂ are in X₁, and both x₁ and y₁ are not in X₂, then .

Proof. Let u be an arbitrary node in X₂. There exists a path p₁ from u to x₁ or y₁ that consists of only nodes in X₂, else X₂ is not connected. In particular, , else x₁ or y₁ would be in X₂. There also exists a path p₂ from x₂ to x₁ that includes neither nor by Lemma 13. The path connects u to x₁ and includes neither nor . Thus, u is in X₁ by Lemma 13.

Lemma 19. Let and be snarls with snarl subgraphs X₁ and X₂. If x₂ and y₂ are not in X₁, and x₁ and y₁ are not in X₂, then X₁ and X₂ are disjoint.

Proof. Let u be an arbitrary node in X₁, and let p be any path from u to x₂ or y₂. By Lemma 12, p includes x₁ or y₁. Thus, by Lemma 12, p includes or . Since p was chosen arbitrarily, this implies u is not in X₂ by Lemma 13. Therefore, X₁ and X₂ are disjoint.

Taken together, these results yield the sufficient condition for compatibility that we set out to prove.

Theorem 4. In a connected biedged graph with at least one black bridge edge, the family of snarls whose subgraphs have no black bridge edges is compatible.

Proof. Let and be arbitrary snarls with snarl subgraphs X₁ and X₂ such that neither subgraph contains a black bridge edge. By Lemma 15, neither snarl splits the other. By Lemma 17, the two snarls cannot share a boundary node. Therefore, we also cannot have both x₁ and y₁ in X₂, and x₂ and y₂ in X₁, else either X₁ or X₂ must contain the graph's black bridge edge by Lemma 16. This leaves three cases:

• 1. x₁ and y₁ are in X₂, and x₂ and y₂ are not in X₁

• 2. x₁ and y₁ are not in X₂, and x₂ and y₂ are in X₁

• 3. x₁ and y₁ are not in X₂, and x₂ and y₂ are not in X₁

In the first two cases, one subgraph is nested in the other by Lemma 18. In the last case, the subgraphs are disjoint by Lemma 19. Therefore, and are compatible.

We will now move on to proving that ultrabubbles are included in the family of snarls with no black bridge edges.

Lemma 20. Let u be a terminal of a black bridge edge whose removal separates a graph B(D) into connected components B₁ and B₂ with u in B₁ and in B₂. Then, all snarls have either both x and y in B₁ or both x and y in B₂.

Proof. Suppose without loss of generality that x is in B₁ and y is in B₂. All paths between x and y include the black edge incident on u. Therefore, and are separable. This contradicts the minimality of .

Theorem 3. No ultrabubble contains a black bridge edge in its subgraph.

Proof. Let be an ultrabubble with subgraph X. Suppose X contains a black bridge edge with terminals u and . Removing this edge separates X into connected components X₁ and X₂ with u in X₁ and in X₂. By Lemma 20, we may assume without loss of generality that x and y are in X₁.

Since , there are no cyclic walks in X₂. Moreover, there is at least one edge in X₂, else the black bridge edge is a tip. Let w be the longest biedged walk starting from in X₂. This walk must exist, since walks of unbounded length could only exist if there is a cyclic biedged walk. Moreover, w is not empty since X₂ contains at least one edge.

Suppose the final edge in w is gray. Since x and y are not in X₂, one endpoint of this gray edge must have no black edge incident on it in the full graph, else w could be lengthened. This violates the definition of a bidirected graph. Therefore, the final edge in w must be black. However, this implies that one endpoint of this black edge has no gray edges incident on it, else w could be lengthened. That is, the black edge is a tip, which contradicts the definition of ultrabubble. Therefore, X does not contain a black bridge edge.

Lemma 21. Let be a snarl with subgraph X. Further, let u be any node, and let p be the shortest path from u to either x or y and q be the shortest path from u to either or . If u is in X, then , and if u is not in X, then .

Proof. First assume that u is in X. Then, q contains x or y by Lemma 12. The subpath up to this point is a path between u and either x or y that is strictly shorter than q. Therefore, . Similarly, if u is in X, then .

Lemma 22. Let be a black bridge edge that separates a graph into connected components B₁ and B₂ with u in B₁ and in B₂. Further, let be a snarl subgraph X such that (1) x is in B₂ and (2) . Then, X contains if X contains w.

Proof. By Lemma 20, y is in B₂ as well as x. Note that if or , then the claim is verified trivially, so we may focus on the case where and . In this case, the black edges and must be in B₂.

First, assume is in X. There exists a path p₁ from w to u in B₁. Note that this implies that p₁ contains neither nor . There also exists a path p₂ from to x or y that includes neither nor by Lemma 13. Thus, is a path from w to x or y that includes neither nor , which implies that w is in X by Lemma 13.

Next, assume w is in X. There exists a path p from w to x or y that includes neither nor by Lemma 13. Since x and y are in B₂, this path must include , which means that it includes a subpath from to x or y. Therefore, is in X by Lemma 13.

Lemma 23. Let be a black bridge edge whose removal separates a graph B(D) into connected components B₁ and B₂ with x in B₁ and in B₂. If is a snarl with subgraph X, then .

Proof. X consists of only nodes that can be reached from x without crossing by Lemma 13. Therefore, .

Acknowledgments

This work was supported by the National Human Genome Research Institute of the National Institutes of Health under Award Number 5U54HG007990 and grants from the W.M. Keck Foundation and the Simons Foundation. This work benefited from numerous conversations with David Haussler and Daniel Zerbino.

Author Disclosure Statement

No competing financial interests exist.