Ask Your Question
1

Does the set_embedding command have a bug?

asked 2019-07-09 01:35:26 +0100

KaylaG gravatar image

updated 2019-07-11 00:41:10 +0100

We created a bipartite graph and called the set_embedding command by using a dictionary with vertices as keys and a list of neighboring vertices in clockwise order. The graph did not show the correct embedding. We expected the vertices to be in the correct clockwise order, but the order is not consistent. We also tried using the on_embedding command. Each time we run the code our vertices are randomly placed in different places.

EDIT:

#S0 and S1 are lists of lists


def Graf(S0, S1):
    D={} #empty graph
    G={} #edited graph that we add to
    verts={}
    #opening an empty graph
    graph=Graph(D, multiedges=True)

    #adds edges and vertices to the graph
    #S0 and S1 are lists of lists
    def Dessin(S0, S1):
        numVertices = len(S0)+len(S1)

        #creating dictionary G of vertices with neighboring edges in counterclockwise order
        for i in range(numVertices):
            G[i]=(S0+S1)[i]

            verts[i]=[]
        #iterate over black vertices
        for i in range(len(S0)):
            #iterate over edges connected to black vertices
            for j in range(len(S0[i])):
                #iterate over white vertices
                for k in range(len(S1)):
                    #check if black and white vertices have an edge in common
                    if S0[i][j] in S1[k]:
                        #if they have an edge in common connect the vertices with the correct edge labeling
                        graph.add_edge(i,k+len(S0),S0[i][j])

        print('dictionary of neighboring edges:')
        print(G)

        #creating dictionary with vertices as keys and neighboring vertices in counterclockwise order
        for i in range(numVertices):
            for j in range(len(G[i])):
                for k in range(numVertices):
                    if i!=k:
                        for l in range(len(G[k])):
                            if G[i][j] in G[k]:
                                if k not in verts[i]:
                                    verts[i].append(k)
        print("verts before flipped order:")
        print(verts)

        #flip order of vertices so that we're counterclockwise.
        for i in range(len(verts)):
            L=[]
            for j in range(len(verts[i])):
                L.append(verts[i][(len(verts[i]))-j-1])
            verts[i]=L

    #all the vertices in S0 are black, all the vertices in S1 are white
    def color(S0, S1):
        grey=[]
        white=[]
        for i in range(len(S0)):
            grey.append(i)
        for i in range(len(S1)):
            white.append(i+len(S0))
        return [grey,white]

    #call and color the graph
    Dessin(S0, S1)
    color(S0, S1)
    print("verts after flipped order:")
    print(verts)

    #keep a certain ordering
    graph.set_embedding(verts)
edit retag flag offensive close merge delete

Comments

Please edit your question: paste the relevant code, select it and press the 101010 button to indent it by four spaces (to get the proper formatting).

rburing gravatar imagerburing ( 2019-07-09 13:35:31 +0100 )edit

I am not sure whether our planar plot function respects the embedding or not :

sage: G = graphs.RandomTriangulation(10)
sage: G.plot(layout='planar')

But at least it displays a planar graph.

FrédéricC gravatar imageFrédéricC ( 2019-07-09 15:04:16 +0100 )edit

Please include a full code sample, since G, verts and graphare currently not defined. Also include a full code sample of how you want to call this function, what you expect as output, and what you get instead. (It should be possible to run it e.g. on https://sagecell.sagemath.org/) It is very difficult to help you otherwise.

rburing gravatar imagerburing ( 2019-07-09 21:51:50 +0100 )edit

1 Answer

Sort by » oldest newest most voted
1

answered 2019-07-09 22:26:03 +0100

rburing gravatar image

updated 2019-07-10 11:45:39 +0100

Setting the combinatorial embedding does not immediately affect how the graph is plotted. However, you can call one of the layout algorithms that use this combinatorial embedding, such as planar, with save_pos=True.

G = Graph([[0,1,2,3,5,4],[[0,1],[0,2],[0,3],[0,4],[0,5]]])
G.set_embedding({0: [1,2,3,4,5], 1 : [0], 2 : [0], 3:[0], 4:[0], 5:[0]})
G.layout('planar', save_pos=True)
G.show()

graph with embedding

Indeed we get clockwise 1, 2, 3, 4, 5.

Edit: Obviously not clockwise but counterclockwise. I opened trac ticket #28152 for this issue.

edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

Stats

Asked: 2019-07-09 01:35:26 +0100

Seen: 1,628 times

Last updated: Jul 11 '19