Added a Tutte Embedding Layout for Graphs

[nturillo]

Adds enhancement requested in #38410.

For 3-connected and planar graphs (https://en.wikipedia.org/wiki/Polyhedral_graph), there exists an embedding into
the plane called the Tutte embedding. In the Tutte embedding, an outer face is set, and all other vertices are placed inside
the outer face such that they are at the mean position of their neighbors.
(https://en.wikipedia.org/wiki/Tutte_embedding)

📝 Checklist

• [x ] The title is concise and informative.
• [x ] The description explains in detail what this PR is about.
• [x ] I have linked a relevant issue or discussion.
• I have created tests covering the changes.
• I have updated the documentation and checked the documentation preview.

⌛ Dependencies

[dcoudert]

The coding style will have to be improved, but let's start with the most important.

[dcoudert]

Why do you need a copy of the graph?

[nturillo]

I was copying another layout function, but you're right I don't need this.

[dcoudert]

you check if the vertex connectivity is more than 3. What if the graph is only 2 or 1 connected ? what if the graph is not connected ?

[nturillo]

Whoops, I was getting my connectivity rules reversed. It should be fixed now. If the connectivity is more than three, that's actually okay; it's the reverse situation which induces a problem.

[dcoudert]

a call to V.index(u) takes time $O(n)$. You should better first define a mapping from vertices to index in V like vertex_to_int = {v: I for I, v in enumerate(V)} and then use it. Each query will then be constant time.

[nturillo]

Okay good catch fixed!

[dcoudert]

• Is the proposed method working on directed graphs ? if not, it should be in graph.py and not in generic_graph.py
• you should add some tests showing the behavior of the method when the connectivity requirement is not fulfilled, or when the graph is not planar
• For all comments, please use 80 columns mode (lines of length at most 80).

[dcoudert]

Be more precise on the connectivity requirement. Is it exactly or at least 3 ?

[nturillo]

Any graph that is k-connected for k $\geq$ 3 is 3-connected. For example, a 5-connected graph is 3-connected.

[dcoudert]

isn't it possible to find an embedding without specifying the external face ?

[nturillo]

In order to calculate the embedding, an external face must be specified. Any subgraph of the graph that is a cycle will work as an external face. It would be possible to find an arbitrary external face automatically if none is specified, but it's unclear to me how one would be chosen from the available options. Also, that functionality is outside the scope of my original intention.

[dcoudert]

Consider tha following:

H = Graph(self) # take a (undirected) copy H of the graph
u, v = next(H.edge_iterator(labels=False)  # take any edge (u, v) of H
H.delete_edge(u, v)  # remove edge (u, v) from H
C = H.shortest_path(v, u) # Compute a shortest path from v to u in H minus (u, v)

You get a short cycle C, and this cycle is a face.

This can of course be done in a follow up PR, but this seems quite simple.

[dcoudert]

This doctest is unstable. I obtain different results on my laptop (macOS...)

File "src/sage/graphs/generic_graph.py", line 21041, in sage.graphs.generic_graph.GenericGraph.layout_tutte
Failed example:
    g.layout_tutte(external_face=[1, 2, 3, 4, 5, 6])
Expected:
    {1: (-0.499999999999999, 0.866025403784438),
    2: (0.500000000000000, 0.866025403784439),
    3: (1.00000000000000, 8.85026869717109e-17),
    4: (0.500000000000000, -0.866025403784439),
    5: (-0.500000000000001, -0.866025403784438),
    6: (-1.00000000000000, 4.55896726715916e-16),
    0: (-5.55111512312578e-17, 1.86944233679563e-16)}
Got:
    {0: (4.16333634234434e-16, 7.42055745992141e-16),
     1: (-0.499999999999999, 0.866025403784439),
     2: (0.500000000000000, 0.866025403784440),
     3: (1.00000000000000, 7.31549942167514e-16),
     4: (0.500000000000000, -0.866025403784438),
     5: (-0.500000000000000, -0.866025403784437),
     6: (-1.00000000000000, 1.29124025055007e-15)}

You should find another test that is more stable. Printing vertex positions is never a good idea.

[nturillo]

In many of the other layout examples, they include this sort of example. However, I've removed it from this function.

[nturillo]

The function is intended to work on digraphs as well. Since it's an embedding, it just embeds the digraph as if it were a simple graph. I've added more examples for when one of the requirements is not met.

Can you be more specific about the 80 column mode? For what comments? These on GitHub or in the actual code? Sorry, this is my first time contributing.