A practical O(n log^{2} n) time algorithm for computing the triplet distance on binary trees
 Andreas Sand^{1, 2},
 Gerth Stølting Brodal^{2, 3},
 Rolf Fagerberg^{4},
 Christian NS Pedersen^{1, 2, 5} and
 Thomas Mailund^{1}Email author
https://doi.org/10.1186/1471210514S2S18
© Sand et al.; licensee BioMed Central Ltd. 2013
Published: 21 January 2013
Abstract
The triplet distance is a distance measure that compares two rooted trees on the same set of leaves by enumerating all subsets of three leaves and counting how often the induced topologies of the tree are equal or different. We present an algorithm that computes the triplet distance between two rooted binary trees in time O (n log^{2} n). The algorithm is related to an algorithm for computing the quartet distance between two unrooted binary trees in time O (n log n). While the quartet distance algorithm has a very severe overhead in the asymptotic time complexity that makes it impractical compared to O (n^{2}) time algorithms, we show through experiments that the triplet distance algorithm can be implemented to give a competitive walltime running time.
Background
Using trees to represent relationships is widespread in many scientific fields, in particular in biology where trees are used e.g. to represent species relationships, so called phylogenies, the relationship between genes in gene families or for hierarchical clustering of highthroughput experimental data. Common for these applications is that differences in the data used for constructing the trees, or differences in the computational approach for constructing the trees, can lead to slightly different trees on the same set of leaf IDs.
To compare such trees, distance measures are often used. Common distance measures include the RobinsonFoulds distance [1], the triplet distance [2], and the quartet distance [3]. Common for these three distance measures is that they all enumerate certain features of the trees they compare and count how often the features differ between the two trees. The RobinsonFoulds distance enumerates all edges in the trees and tests if the bipartition they induce is found in both trees. The triplet distance (for rooted trees) and quartet distance (for unrooted trees) enumerate all subsets of leaves of size three and four, respectively, and test if the induced topology of the leaves is the same in the two trees.
Efficient algorithms to compute these three distance measures exist. The RobinsonFoulds distance can be computed in time O (n) [4] for trees with n leaves, which is optimal. The quartet distance can be computed in time O (n log n) for binary trees [5], in time O (d^{9} n log n) for trees where all nodes have degree less than d [6], and in subcubic time for general trees [7]. See also Christiansen et al. [8] for a number of algorithms for general trees with different tradeoffs depending on the degree of inner nodes. For the triplet distance, O (n^{2}) time algorithms exist for both binary and general trees [2, 9].
Brodal et al. [5] present two algorithms for computing the quartet distance for binary trees; one running in time O (n log n), and one running in time O (n log^{2} n). The latter is the most practical, and was implemented in [10, 11], where it was shown to be slower in practice compared to a simple O (n^{2}) time algorithm [12] unless n is above 2000. In this paper we focus on the triplet distance and develop an O (n log^{2} n) time algorithm for computing this distance between two rooted binary trees. The algorithm is related to the O (n log^{2} n) time algorithm for quartet distance, but its core accounting system is completely changed. As we demonstrate by experiments, the resulting algorithm is not just theoretically efficient but also Efficient in practice, as it is faster than a simple O (n^{2}) time algorithm based on [12] already for n larger than 12, and is e.g. faster by a factor of 50 when n is 2900.
Methods
The triplet distance is the number of triplets whose topology differ in the two trees. It can naïvely be computed by enumerating all O (n^{3}) sets of three leafs and comparing the induced topologies in the two trees, counting how often the trees agree or disagree on the topology. Triplet topologies in a tree, however, are not independent, and faster algorithms can be constructed exploiting this, comparing sets of triplet topologies faster. Critchlow et al. [2] for example exploit information about the depth of shared ancestors of leaves in a tree to achieve an O (n^{2}) time algorithm for binary trees while Bansal et al. [9] construct a table of shared leafsets and achieve an O (n^{2}) time algorithm for general trees.
For the quartet distance, the analogue to the triplet distance for unrooted trees, Brodal et al. [5] construct an even faster algorithm by identifying sets of four different leaves in one tree through coloring all leaves with one of three different colors and then counting the number of topologies compatible with the coloring in the other tree. Using a variant of the socalled "smaller half trick" for keeping the number of different relevant colorings low, the algorithm manages to construct all relevant colorings with O (n log n) color changes. The number of topologies compatible with the coloring can then be counted in the other tree using a data structure called a "hierarchical decomposition tree". Maintaining the hierarchical decomposition tree, however, involves a number of polynomial manipulations that, while theoretically can be done in constant time per polynomial, are quite time consuming in practice [10, 11], making the algorithm slow in practice.
A naïve algorithm that computes the quartet distance between two unrooted trees by explicitly inspecting each of the O (n^{4}) quartets can be modified to compute the triplet distance between two rooted trees without loss of time by adding a new leaf x above the two root nodes and limit the inspection of quartets to the quartets containing this new leaf. However, the efficient algorithms for computing the quartet distance presented in [5, 7] do not explicitly inspect every quartet and therefore cannot be modified to compute the triplet distance following this simple approach.
In the following, we develop an efficient algorithm for computing the triplet distance between two rooted binary trees T_{1} and T_{2} with the same set of leaf IDs. Our key contribution is to show how all triplets in one tree, say T_{1}, can be captured by coloring the leaves with colors, and how the smaller half trick lets us enumerate all such colorings in time O (n log n). We will then construct a hierarchical decomposition tree (HDT) for T_{2} that counts its number of compatible triplets. Unlike the algorithms for computing the quartet distance [5], where the counting involves manipulations of polynomials, the HDT for triplets involves simple arithmetic computations that are efficient in practice.
Counting shared triplets through leaf colorings
where Shared(S) on a set S of triplets is its number of triplets having the same topology in T_{1} and T_{2}. The triplet distance of T_{1} and T_{2} is then $\left(\begin{array}{c}\hfill n\hfill \\ \hfill 3\hfill \end{array}\right)\mathsf{\text{Shared}}\left(\mathcal{T}\right)$.
For such a coloring according to a node v ∈ T_{1}, and for a node u ∈ T_{2}, the number Shared(τ_{ v } ∩ τ_{ u } ) could be found as follows: let x and y be the two subtrees of u, let x(r) and y(r) be the number of leaves colored red in the subtrees x and y, respectively, and x(b) and y(b) the number of leaves colored blue. The number Shared(τ_{ v } ∩ τ_{ u } ) is then $\left(\begin{array}{c}\hfill x\left(b\right)\hfill \\ \hfill 2\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}\cdot \phantom{\rule{2.77695pt}{0ex}}y\left(r\right)\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\left(\begin{array}{c}\hfill y\left(b\right)\hfill \\ \hfill 2\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}\cdot \phantom{\rule{2.77695pt}{0ex}}x\left(r\right)+\left(\begin{array}{c}\hfill y\left(r\right)\hfill \\ \hfill 2\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}\cdot \phantom{\rule{2.77695pt}{0ex}}x\left(b\right)\phantom{\rule{2.77695pt}{0ex}}+\phantom{\rule{2.77695pt}{0ex}}\left(\begin{array}{c}\hfill x\left(r\right)\hfill \\ \hfill 2\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}\cdot \phantom{\rule{2.77695pt}{0ex}}y\left(b\right).$ We call these the triples of u compatible with the coloring.
Explicitly going through T_{1} and coloring for each node v would take time O (n) per node, for a total time of O (n^{2}). We reduce this to O (n log n) by using the smaller half trick. Going through T_{2} for each coloring and counting the number of compatible triplets would also take time O (n^{2}). Using a HDT we find this count in O(1) time, while updating the structure takes time O (log n) after each leaf color change. The result is O(n log^{2} n) total running time. In essence, the HDT performs the inner sum of $\mathsf{\text{Shared}}\left(\mathcal{T}\right)={\sum}_{v\in {T}_{1}}{\sum}_{u\in {T}_{2}}\mathsf{\text{Shared}}\left({\tau}_{v}\cap {\tau}_{u}\right)$, while the coloring algorithm performs the outer sum.
Smaller half trick
We go through nodes v in a depth first order while maintaining two invariants of the algorithm: 1) Before we process v, the entire subtree of v is colored "red" and the rest of the tree has no color; 2) When we return from the depth first recursion, the entire tree has no color.
 1.
Color S(v) "blue". Now v has the coloring that enable us to count the triplets for v.
 2.
Remove the color for S(v). Now we can call recursively on L(v) while satisfying the invariant.
 3.
Returning from the recursive call, the entire tree is colorless by invariant 2.
 4.
Color S(v) "red". Now we satisfy invariant 1 for calling recursively on S(v).
 5.
Call recursively on S(v). When we return we satisfy invariant 2 for returning from the recursive call.
Using this recursive algorithm, we go through all colorings of the tree. In each instance (not counting recursive calls), we only color leaves in S(v), and only a constant number of times. Thus, a leaf ℓ is only colored when visiting an ancestor node v where ℓ S(v), i.e. ℓ is in the smaller subtree of v. Since ℓ can have at most O (log n) such ancestors, each leaf will only be colored at most O (log n) times, implying a total of O (n log n) color changes.
Hierarchical decomposition tree
We build a data structure, the hierarchical decomposition tree (HDT), on top of the second tree T_{2} in order to count the triplets in T_{2} compatible with the coloring of leaves in the first tree T_{1}. The HDT is a balanced binary tree where each node corresponds to a connected part of T_{2}. Each node in the HDT, or component, keeps a count of the number of compatible triplets the correspondent part of T_{2} contains, plus some additional bookkeeping that makes it possible to compute this count in each component in constant time using the information stored in the component's children in the HDT.
The HDT contains three different kinds of components:

L: A leaf in T_{2},

I: An inner node in T_{2},

C: A connected subpart of T_{2},
where for type C we require that at most two edges in T_{2} crosses the boundary of the component; at most one going up towards the root and at most one going down to a subtree.
The construction algorithm operates on components and edges. Each component is of one of the types L, I, or C. It has a parent pointer, pointing to its parent component in the HDT, and a bit, down_closed, indicating if the component corresponds to a complete subtree in the original tree. When a component does not yet have a parent in the HDT, we make the parent pointer point to the component itself. We use this pointer to test if an edge has been contracted in the HDT construction. Edges consist of two pointers, an up pointer and a down pointer that points to the components above and below the edge in T_{2}.
In case 1, one of e's neighbors already has a parent, thus this neighbor has already been contracted into a C component in this iteration and should not be contracted again. Case 2 is the situation in Figure 5a, and case 3 is the situation in Figure 5b. In case 4, e.down is an I component or e.up is an I component and e.down does not correspond to a complete subtree, hence none of the constructions in Figure 5 apply. After removing all edges from es, the up and down pointers of the remaining edges in next are updated in lines 2122, such that they point to either a newly created component or still point to the same component. The edges in next are finally moved back to es in line 23, and a new iteration is started. The algorithm finishes when es is empty, and the root of the HDT is the C component resulting from joining the ends of the last edge. We now argue that height of the HDT is O(log n), and that the construction time is O(n). Each iteration of the code in Figure 6 takes time linear in the number E = es of edges remaining to be contracted at the beginning of the iteration. The height and time bounds follow if we can argue that the number of edges decreases geometrically for each iteration. In the following we argue that the number of edges after one iteration is at most 11/12. E.
We first argue that the number of contractible edges at the beginning of the iteration is at least E/4. Note that only edges incident to I components might not be contractible, and that the number of downclosed components is at least one larger than the number of I components. If the number of I components is at most E/4, then at most 3 · E/4 incident edges might not be contractible, i.e. at least E/4 edges are contractible. Otherwise the number of I components is more than E/4, and therefore the number of downclosed components is more than E/4 + 1. Since the parent edges from all downclosed components are contractible (for E ≥ 1), the number of contactable edges is again at least E/4.
Since each contracted edge can prevent at most two other edges incident to the two merged components (see Figure 5) from being contracted in the same iteration, each iteration will contract at least 1/3 of the at least E/4 contractible edges. It follows that an iteration reduces the number of contractible edges by at least E/12.
Counting triplets in the hierarchical decomposition tree
In each component we keep track of N, the number of triplets contained within the component (i.e., where the three leaves of the triplet are within the component) that are compatible with the coloring. When we change the coloring, we update the HDT to reflect this, so we can always read o the total number of compatible triplets from the root of the HDT in constant time.
By adding a little bookkeeping information to each component we make it possible to compute N (and the bookkeeping information itself) for a component in constant time from the information in the component's children in the HDT. This has two consequences: it makes it possible to add the bookkeeping and N to all components after the construction in linear time, and it makes it possible to update the information when a leaf changes color by updating only components on the path from the leafcomponent to the root in the HDT, a path that is bounded in length by O (log n). Since we only change the color of a leaf O (n log n) times in the triplet distance algorithm, it is this property of the HDT that keeps the total running time at O (n log^{2} n). For the bookkeeping we store six numbers in each component in addition to N:

R: The number of leaves colored red in the component.

B: The number of leaves colored blue in the component.

$\tilde{rr}$: The number of pairs of red leaves found in the same subtree of a C component. Let T_{ i }, i = 1,..., n, denote the subtrees in the component, see Figure 7a, and let r(i) denote the number of red leaves in tree T_{ i }. Then $\tilde{rr}={\sum}_{i=1}^{n}\left(\begin{array}{c}\hfill r\left(i\right)\hfill \\ \hfill 2\hfill \end{array}\right)$.

$\tilde{bb}$: The number of pairs of blue leaves found in the same subtree of a C component.

$\hat{rb}$: The number of pairs of leaves with a red leaf in one subtree and a blue in another subtree, where the red leaf is in a tree further down on the path in a C component. Let T_{ i }, i = 1,..., n, denote the subtrees in the component, see Figure 7, and let r(i) denote the number of red leaves in tree T_{ i }and b(i) the number of blue leaves in T_{ i }. Then $\hat{rb}={\sum}_{i=1}^{n}{\sum}_{j=i+1}^{n}r\left(i\right)\phantom{\rule{2.77695pt}{0ex}}\cdot \phantom{\rule{2.77695pt}{0ex}}b\left(j\right)$.

$\hat{br}$: The number of pairs of leaves with a red leaf in one subtree and a blue in another subtree, where the blue leaf is in a tree further down on the path in a C component.
We describe how the bookkeeping variables and N are computed through a caseanalysis on how the components are constructed. L and I components are constructed in only one way, while C components are constructed in one of two ways (see Figure 5).
L components: For a leaf component, R is 1 if the leaf is colored red and 0 otherwise, B is 1 if the leaf is colored blue and 0 otherwise, and all other counts are 0.
I components: All counts are 0.
C components, case Figure 5b: Let x(1) denote one of the counters listed above for component C_{1}. Then R = R(1), B = B(1), $\tilde{rr}=\left(\begin{array}{c}\hfill R\left(1\right)\hfill \\ \hfill 2\hfill \end{array}\right)$, $\tilde{bb}=\left(\underset{2}{B\left(1\right)}\right)$, $\hat{rb}=\hat{br}=0$. Since the inner node in the composition does not contain any leaves, the triplet count is simply N = N (1).
Results and discussion
We implemented the algorithm in C++ and a simple O(n^{2}) time algorithm to ensure that it computes the correct triplet distance.
When changing the color of leaves, we spent time O (log n) updating the bookkeeping in the HDT for each leaf. We only count after a complete subtree has changed color, however, so instead of updating the HDT for each colorchange we could just mark which leaves have changed color and then update the HDT bottomup, so each inner node would only be updated once when it is on a path from a changed leaf. We implemented this, but found that the extra bookkeeping from marking nodes and then updating increased the running time by 10%15% compared to just updating the HDT.
Conclusions
We have presented an O (n log^{2} n) time algorithm for computing the triplet distance between two binary trees and experimentally validated its correctness and time analysis.
The algorithm builds upon the ideas in the O (n log^{2} n) time algorithm for computing the quartet distance between binary trees [13], but where the bookkeeping in the quartet distance algorithm is rather involved, making it inefficient in practice, the bookkeeping in the triplet distance algorithm in this paper is entirely different, and significantly simpler and faster.
Compressing the HDT during the algorithm makes it possible to reduce the running time of the quartet distance algorithm to O (n log n) and the same approach can also reduce the running time of the triplet algorithm to O (n log n). We have left for future work to experimentally test whether this method incurs too much overhead to make it practically worthwhile.
Unlike the O (n^{2}) time algorithm of Bansal et al. [9] our algorithm does not generalize to nonbinary trees. It is possible to extend the algorithm to nonbinary trees by employing more colors, as was done for the quartet distance [6], but this makes the algorithm depend on the degree of nodes, and future work is needed to develop a subquadratic algorithm for general rooted trees.
Declarations
Declarations
The publication costs for this article were funded by PUMPKIN, Center for Membrane Pumps in Cells and Disease, a Center of the Danish National Research Foundation, Denmark.
This article has been published as part of BMC Bioinformatics Volume 14 Supplement 2, 2013: Selected articles from the Eleventh Asia Pacific Bioinformatics Conference (APBC 2013): Bioinformatics. The full contents of the supplement are available online at http://www.biomedcentral.com/bmcbioinformatics/supplements/14/S2.
Authors’ Affiliations
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