Spanning Trees

This example shows how to generate a spanning tree from an input graph using igraph.Graph.spanning_tree(). For the related idea of finding a minimum spanning tree, see Minimum Spanning Trees.

import igraph as ig
import matplotlib.pyplot as plt
import random

First we create a two-dimensional, 6 by 6 lattice graph:

g = ig.Graph.Lattice([6, 6], circular=False)

We can compute the 2D layout of the graph:

layout = g.layout("grid")

To spice things up a little, we rearrange the vertex ids and compute a new layout. While not terribly useful in this context, it does make for a more interesting-looking spanning tree ;-)

random.seed(0)
permutation = list(range(g.vcount()))
random.shuffle(permutation)
g = g.permute_vertices(permutation)
new_layout = g.layout("grid")
for i in range(36):
    new_layout[permutation[i]] = layout[i]
layout = new_layout

We can now generate a spanning tree:

spanning_tree = g.spanning_tree(weights=None, return_tree=False)

Finally, we can plot the graph with a highlight color for the spanning tree. We follow the usual recipe: first we set a few aesthetic options and then we leverage igraph.plot() and matplotlib for the heavy lifting:

g.es["color"] = "lightgray"
g.es[spanning_tree]["color"] = "midnightblue"
g.es["width"] = 0.5
g.es[spanning_tree]["width"] = 3.0

fig, ax = plt.subplots()
ig.plot(
    g,
    target=ax,
    layout=layout,
    vertex_color="lightblue",
    edge_width=g.es["width"]
)
plt.show()
spanning trees

Note

To invert the y axis such that the root of the tree is on top of the plot, you can call ax.invert_yaxis() before plt.show().

Total running time of the script: (0 minutes 0.705 seconds)

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