Minimum-Width Drawings of Phylogenetic Trees

Abstract

We show that finding a minimum-width orthogonal upward drawing of a phylogenetic tree is NP-hard for binary trees with unconstrained combinatorial order and provide a linear-time algorithm for ordered trees. We also study several heuristic algorithms for the unconstrained case and show their effectiveness through experimentation.

A Additional Proofs

1.1 A.1 Alignment Nodes

Alignment nodes only need to be able to fully realign satisfied clauses, therefore within the three structures at least one must be of width three and at least one of width one. Therefore if we consider the periodicity of the column at which the edge drops, the maximum possible phase difference in these periods is shown in Fig. 10. Considering this order as an extreme case is sufficient because after using both satisfied literal structures (after $x_3$ in $c_1$) the only remaining widths could be two (which maintains the phase difference) or one (which reduces the phase difference). The same argument can be made after using both unsatisfied literal structures after $x_2$.

Fig. 10.

Set of consecutive clauses, $c_1=x_2\vee x_3 \vee x_6$ and $c_2=x_1\vee \overline{x_2} \vee x_6$ requiring the largest realignment, with satisfying assignment $x=\{false,true,true, ...\}$ (Color figure online)

Without alignment nodes it would be impossible to connect $x_2$ and $x_3$ to the next clause, but adding the two row of alignment nodes (shown in blue) between clauses enable them to remain connected. This allows each clause to remain tight and assume width of at most $2n+1$ whenever they are satisfied, regardless of the previous clause.

1.2 A.2 Pyramidal Structure

Recall Lemma 1: At minimal width, the base and bridge can only assume a pyramidal embedding. We define a pyramidal embedding as the embedding of the base in which nodes closer to the root lie closer to the center and nodes further from the root approach the outer sides, as shown in Fig. 3b.

Proof

We define the filled area of a drawing as the sum of the space used by each node (equal to its width), and the space used by each edge (equal to its length). The length of the edges is fixed, but we can change the filled area by changing the layout to minimize the width of the non-leaf nodes. When the base is in the pyramidal embedding, it fills the least possible area, since every non-leaf node must have width equal to its number of children. We now show that this is the only configuration with minimal area.

We begin from the parents of the base gadget, which belong to a copy of $w_2$ (Fig. 3a). The two incoming edges from this structure must be next to one another, with no gap between them. The two nodes at the top level of the base each then have both a leaf and a large subtree attached.

The width of each of these top-level nodes must be two, and the only way to achieve this is that each node’s leaf must lie on the inside and its subtree on the outside. Similarly, for each subsequent node along the path, the same argument shows that its leaf must lie on the inside. This proves by induction that the base needs to be in a pyramidal embedding. The bridge then must fit against the base. The bridge nodes with the lowest leaves must be on the outside, with the next lowest leaves next to them, and so on by induction back to the center. This shows that the pyramidal embedding is the unique embedding that minimizes the filled area. Since the levels containing the base and bridge nodes are completely packed with no gaps, this also implies that the pyramidal embedding is the unique embedding that minimizes the width of these levels. $\blacksquare$

1.3 A.3 Constraint Graph is a DAG

Recall Lemma 4: The Constraint Graph D of the fixed order n-node phylogenetic tree $T = (V,E)$ is a DAG with $3n-1$ vertices and O(n) edges, where $n = |V|$.

Proof

Our objects are the left and right sides of each vertex in T, and the edges in T. This gives us two vertices in D for each vertex in T, and one for each edge. Since T is a tree, it must have $n - 1$ edges, so D has $3n - 1$ vertices. If two objects are horizontally visible, then there is a segment between them that crosses only those two objects. We will use these segments to build a planar embedding of D, which will imply that D has O(n) edges.

Let the collection of segments connecting our objects be S. We first construct a larger planar graph $D'$, in which the vertices are the endpoints of S. The edges of $D'$ include all of the segments in S. We also add edges connecting each vertex to the vertices immediately above and below it that represent the same object. By the definition of horizontally visible, each segment in S can only intersect two objects, so none of the segments in S can intersect. The additional edges also cannot intersect, since they are ordered by the height of the vertices. Therefore, $D'$ is planar.

We then contract all of the edges that connect two vertices in $D'$ corresponding to the same object. This produces the DAG D. Since D is a contraction of a planar graph, it must also be planar. $\blacksquare$

1.4 A.4 Approximation Guarantee for Minimum Area Heuristic

We first describe the running time of the heuristic. For each ordering of the children, the minimum width drawing is calculated using the algorithm from Theorem 2 and the bounding polygon is calculated by traversing the tree once to find the extreme-most branches, running in O(n). We repeat this O(1) times per vertex for a total running time of $O(n^2)$ for bounded degree trees.

Theorem 4

The minimum area heuristic has an approximation ratio of at least $\varOmega (\sqrt{n})$.

Proof

Recall, the structures for $w_k$ as defined in Theorem 3 and further constrain it to be a complete binary tree with all its k leaves in horizontal alignment. Recall the subtrees used in Theorem 3 and note the subtrees used in this tree instead increase the size of their $w_k$ by 3 each time (with the exception of the first two which have the same $w_k$). Furthermore each subtrees nodes end immediately before the first node in the subtree two subtrees away, the latter subtree also has a node aligned with the former’s $w_k$ leaves. The leaves in $w_k$ are horizontally aligned with the top node in the next subtree.

Fig. 11.

Tree structures causing worst case performance for minimum area heuristic.

Using these definitions Fig. 11 demonstrates a tree structure where a minimum area embedding of the two subtrees in yellow makes it impossible for the entire drawing to admit minimum width and minimum area. The heuristic achieves the right embedding for the subtree but fails to choose the right embedding for the two sibling subtrees. Although the optimal’s embedding uses a larger area for the combination of both siblings with $w_k$, it occupies an almost rectangular space resulting in a really small area (and width) increase when adding the next subtree. Define each subtree by the size of the $w_k$ structure inside it, consider the tree with 2k/3 subtrees $w_k, w_{k+3} ... w_{3k}$. This structure will have an optimal width of $3k+6$. The minimum area heuristic would instead have two subtrees with their $w_k$ on opposite sides and the $w_{k+3}$ overlapping the bottom-most $w_k$ and every next pair of subtrees overlapping in the same way. The total width achieved by the minimum area heuristic would therefore be $\sum _{i=0}^{2k/6} (k+6i+5)=2k^2/3+11k/3+5$. The total number of nodes is $n=\sum _{i=0}^{2k/3}(7+2(k-2i)+1)+6+k =4k^2/9+7k+14$, which we can use to find k in terms of n. We find that $k\approx \sqrt{n/2}$, and therefore the approximation ratio achieved by the greedy heuristic for this tree is $\frac{2k^2/3+11k/3+5}{3k+6}\approx \frac{2k}{9}=\varOmega (\sqrt{n})$. $\blacksquare$