Counting topologies#

Yan Wong

This tutorial is intended to be a gentle introduction to the combinatorial treatment of tree topologies in tskit. For a more formal introduction, see the Identifying and counting topologies section of the official tskit documentation.

The topology of a single tree is the term used to describe the branching pattern, regardless of the lengths of the branches. For example, both trees below have the same topology, although the branch lengths differ:

import tskit
node_labels = {0: "a", 1: "b", 2: "c"}  # avoid confusion by using letters to label tips
tree = tskit.Tree.generate_comb(3)
display(tree.draw_svg(node_labels=node_labels, y_axis=True))

deep_tree = tskit.Tree.generate_comb(10).tree_sequence.simplify([0, 1, 2]).first()
display(deep_tree.draw_svg(node_labels=node_labels, y_axis=True))


The treatment of topologies in tskit is restricted to trees with a single defined root, without nodes with a single child (i.e. trees must consist of nodes that are either leaves, or internal nodes with two or more children). For convenience in the examples below, trees are drawn with the tips flagged as samples, although whether a node is a sample or not does not change the topology of the tree.

Tree labellings and shapes#

The topology of a tree also takes into account the labelling of tips, so that the trees below, although they have the same shape, count as three different topologies:

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from string import ascii_lowercase
from IPython.display import SVG

def str_none(s, prefix=None):
    if s is not None:
        if prefix is None:
            return str(s)
            return prefix + " = " + str(s)
    return None

def draw_svg_trees(trees, node_labels={}, x_lab_attr=None, width=100, height=150, space=10):
    w = width + space
    h = height + space
    trees = list(trees)
    s = f'<svg height="{h}" width="{w * len(trees)}" xmlns="">'
    s += f'<style>.x-axis {{transform: translateY({space}px)}}</style>'
    for i, tree in enumerate(trees):
        s += tree.draw_svg(
            size=(width, height),
            canvas_size=(w, h),
            root_svg_attributes={"x": i * w},
            x_label=str_none(getattr(tree.rank(), x_lab_attr or "", None), x_lab_attr)
    s += '</svg>'
    return SVG(s)

draw_svg_trees(tskit.all_tree_labellings(tree), node_labels={u: ascii_lowercase[u] for u in tree.samples()})

These are, in fact, the only possible three labellings for a three-tip tree of that shape. There is only one other possible shape for a three-tip tree, and for this shape, all labelling orders are equivalent (in other words, there is only one possible labelling):

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A 3-tip tree therefore has only four possible topologies. These can be generated with the all_trees() function.

generated_trees = tskit.all_trees(3)
print("For a three-tip tree there are", len(list(generated_trees)), "labelled topologies.")
For a three-tip tree there are 4 labelled topologies.

Here they are, plotted out with their shapes enumerated from zero:

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    node_labels={u: ascii_lowercase[u] for u in tree.samples()},

Enumerating shapes and labellings#

For a tree with four tips, more topologies and shapes are possible. As before, we can generate the topologies using all_trees(). Alternatively, if we only want the (unlabelled) shapes, we can use the all_tree_shapes() function:

print("For a four-tip tree there are", len(list(tskit.all_trees(4))), "labelled topologies.")

generated_trees = tskit.all_tree_shapes(4)
print("These can be categorised into", len(list(generated_trees)), "shapes.")
For a four-tip tree there are 26 labelled topologies.
These can be categorised into 5 shapes.

Again, we can give each shape a number or index, starting from zero:

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draw_svg_trees(tskit.all_tree_shapes(4), x_lab_attr="shape")

Each of these shapes will have a separate number of possible labellings, and trees with these labellings can be created using all_tree_labellings():

for shape_index, tree in enumerate(tskit.all_tree_shapes(4)):
    labellings = tskit.all_tree_labellings(tree)
    num_labellings = len(list(labellings))
        f"Tree shape {shape_index} for a four-tip tree has "
        f"{num_labellings} labelling{'' if num_labellings==1 else 's'}."
Tree shape 0 for a four-tip tree has 1 labelling.
Tree shape 1 for a four-tip tree has 6 labellings.
Tree shape 2 for a four-tip tree has 4 labellings.
Tree shape 3 for a four-tip tree has 12 labellings.
Tree shape 4 for a four-tip tree has 3 labellings.

Any tree topology for a tree of \(N\) tips can therefore be described by a shape index combined with a labelling index. This is known as the rank of a tree, and it can be obtained using the Tree.rank() method. For instance, here is the rank of a simulated tree of 10 tips:

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import msprime
num_tips = 10
simulated_ts = msprime.sim_ancestry(10, ploidy=1, random_seed=123)
simulated_tree = simulated_ts.first()
print("The topology of the simulated tree below can be described as", simulated_tree.rank())
ascii_node_labels = {u: ascii_lowercase[u] for u in simulated_tree.samples()}
The topology of the simulated tree below can be described as Rank(shape=1270, label=205540)

A tree with the same topology (i.e. the same shape and labelling, but ignoring the branch lengths) can be generated using the Tree.unrank() method, by specifying the number of tips and the appropriate (shape, labelling) tuple:

new_tree = tskit.Tree.unrank(num_tips, (1270, 21580))

Note that this method generates a single tree in a new tree sequence whose a default sequence length is 1.0.

Methods for large trees#

The number of possible topologies for a tree with \(N\) tips grows very rapidly with \(N\). For instance, with 10 tips, there are 282,137,824 possible topologies.

For this reason, the all_trees(), all_tree_shapes() and all_tree_labellings() methods do not return a list of trees but an iterator over the trees. This means it is perfectly possible to start iterating over (say) all tree shapes for a tree of 100 leaves, but the iterator will not finish before the death of our galaxy.

for num_trees, tree in enumerate(tskit.all_tree_shapes(100)):
    shape = tree.rank().shape
    b2 = tree.b2_index()
    print(f"A 100-tip tree with shape index {shape} has a b2 balance index of {b2}")
    if num_trees > 5:
      break  # better not let this run too long!
A 100-tip tree with shape index 0 has a b2 balance index of 2.0000000000000013
A 100-tip tree with shape index 1 has a b2 balance index of 1.9986759016244608
A 100-tip tree with shape index 2 has a b2 balance index of 1.9960946599243323
A 100-tip tree with shape index 3 has a b2 balance index of 1.9958336776669476
A 100-tip tree with shape index 4 has a b2 balance index of 1.9973695449917637
A 100-tip tree with shape index 5 has a b2 balance index of 1.992978538300552
A 100-tip tree with shape index 6 has a b2 balance index of 1.9927249774271467

For similar combinatorial reasons, the Tree.rank() method can be inefficient for large trees. To compare the topology of two trees, you are therefore recommended to use e.g. the Tree.kc_distance() method rather than comparing ranks directly.

simulated_tree = simulated_ts.first(sample_lists=True)  # kc_distance requires sample lists
if simulated_ts.first(sample_lists=True).kc_distance(simulated_tree) == 0:
    print("Trees are identical")
    # To compare to the new_tree we need to fix 
    # print("The simulated and topology-constructed trees have the same topology")
Trees are identical

Despite the combinatorial explosion associated with topologies of many-tip trees, it is still possible to efficiently count the number of embedded topologies in a large tree.

Embedded topologies#

An embedded topology is a a topology involving a subset of the tips of a tree. If the tips are classified into (say) three groups, red, green, and blue, we can efficiently count all the embedded three-tip trees which have one tip from each group using the Tree.count_topologies() method.

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big_tree = tskit.load("data/topologies_sim_speciestree.trees").first()
# Check all observed topologies have the same counts
assert list(big_tree.count_topologies()[0, 1, 2].values()) == [32, 32]
styles = [
    f".node.sample.p{} > .sym " + "{" + f"fill: {colour}" + "}"
    for colour, p in zip(['red', 'green', 'blue'], big_tree.tree_sequence.populations())
big_tree.draw_svg(style="".join(styles), node_labels={}, time_scale="rank", x_label="big_tree")

In this tree, it is clear that the green and blue tips never cluster together. The Tree.count_topologies() method exhaustively looks at all combinations of one red, one blue, and one green tip, and confirms that we never see the topology grouping green and blue. However, as might be expected from examination of the plot above, a red tip is equally likely to be a sister to a green tip as to a blue tip:

# By default `count_topologies` chooses one tip from each population, like setting
# sample_sets=[ts.samples( for p in ts.populations() if len(ts.samples( > 0]

topology_counter = big_tree.count_topologies()

colours = ['red', 'green', 'blue']
styles = [f".n{u}>.sym {{fill: {c} }}" for u, c in enumerate(colours)]

embedded_counts = topology_counter[0, 1, 2]
for embedded_tree in tskit.all_trees(3):
    rank = embedded_tree.rank()
    number_of_instances = embedded_counts[rank]
    label = f"{number_of_instances} instances embedded in big_tree"
    display(embedded_tree.draw_svg(style="".join(styles), node_labels={}, x_label=label))

Methods over tree sequences#

It can be useful to count embedded topologies over an entire tree sequence. For instance, we might want to know the number of embedded topologies that support Neanderthals as a sister group to europeans versus africans. Tskit provides the efficient TreeSequence.count_topologies() method to do this incrementally, without having to re-count the topologies independently in each tree.

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from myst_nb import glue
ts = tskit.load("data/topologies_sim_stdpopsim.trees")
print(f"Loaded a stdpopsim of {ts.num_trees} African+European+Chinese trees, each with {ts.num_samples} tips")
glue("seq_len", int(ts.sequence_length/1000), display=False)
Loaded a stdpopsim of 465 African+European+Chinese trees, each with 6000 tips

Although the trees in this tree sequence are very large, counting the embedded topologies is quite doable (for speed in this demo we are only simulating 49 kilobases, but calculating the average over an entire chromosome simply takes a little longer)

from datetime import datetime
names = {"YRI": "African", "CEU": "European", "CHB": "Chinese"}
colours = {"YRI": "yellow", "CEU": "green", "CHB": "blue"}

population_map = {p.metadata["id"]: for p in ts.populations()}
sample_populations = list(sorted({ts.node(u).population for u in ts.samples()}))
topology_span = {tree.rank(): 0 for tree in tskit.all_trees(len(sample_populations))}

start =
total = 0
for topology_counter, tree in zip(ts.count_topologies(), ts.trees()):
    embedded_topologies = topology_counter[sample_populations]
    weight = tree.span / ts.sequence_length
    for rank, count in embedded_topologies.items():
        topology_span[rank] += count * weight
        total += count
print(f"Counted {total} embedded topologies in { - start} seconds")
Counted 3720000000000 embedded topologies in 0:00:07.953739 seconds
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ntips = len(sample_populations)
styles = ".sample text.lab {baseline-shift: super; font-size: 0.7em;}"
node_labels = {}

for p in range(ntips):
    name = ts.population(sample_populations[p]).metadata["id"]
    node_labels[p] = names[name]
    styles += f".n{p}>.sym {{fill: {colours[name]} }}"

total = sum(topology_span.values())
for rank, weight in topology_span.items():
    label = f"{weight/total *100:.1f}% of genome"
    embedded_tree = tskit.Tree.unrank(ntips, rank)
    display(embedded_tree.draw_svg(size=(160, 150), style="".join(styles), node_labels=node_labels, x_label=label))

Perhaps unsurprisingly, the most common topology is the one that groups the non-African populations together (although there are many trees of the other two topologies, mostly reflecting genetic divergence prior to the emergence of humans out of Africa).

For an example with real data, see Topological analysis in the Tskit for population genetics tutorial.