Add high-dimensional manifolds

[Filco306]

Hello!

First of all, thank you for a very nice python package! I think high-dimensional manifolds, that is, manifolds beyond 2 and 3 dimensions would be very interesting to add. d-dimensional spheres have already been added, but if somebody knows how generate other types of manifolds of higher dimensions, I would be happy to cooperate and contribute to create this for this package :)

Cheers,
Filip

[ctralie]

Yes, I would love to help with this. Uniformly sampling manifolds is tricky, but naively I would start by just doing furthest points sampling to make sure things are even

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On Thu, Nov 12, 2020 at 10:44 AM Filip Cornell ***@***.***> wrote: Hello! I think high-dimensional manifolds, that is, manifolds beyond 2 and 3 dimensions would be very interesting to add. d-dimensional spheres have already been added, but if somebody knows how generate other types of manifolds of higher dimensions, I would be happy to cooperate and contribute to create this for this package :) Cheers, Filip — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#10>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAJWDZRI7GVGYMASEY5VJ2LSPP7GXANCNFSM4TTOIQOA> .

[Filco306]

How great to hear! I think #11 is a start, but what other high-dimensional manifolds can we construct?

[Filco306]

Have a look here. He claims an n-dimensional Torus can simply be built by taking the cartesian of S^1, seen here. Can we use this perhaps?

[ctralie]

Yeah actually, a flat torus is very straightforward that way. Simply use meshgrid and have pairs of dimensions be a different circle, as you said

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On Thu, Nov 12, 2020 at 3:39 PM Filip Cornell ***@***.***> wrote: Have a look here <http://planning.cs.uiuc.edu/node137.html>. He claims an n-dimensional Torus can simply be built by taking the cartesian of S^1, seen here <http://planning.cs.uiuc.edu/node136.html>. Can we use this perhaps? — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#10 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAJWDZS6WHG6L5ICKFRRCWDSPRBY5ANCNFSM4TTOIQOA> .

[Filco306]

Hmm, I am not sure I understand how you mean. Do you perhaps have some sample code?

[ctralie]

For 10,000 samples on a flat 2-torus, for example, one could do this N = 100 t = np.linspace(0, 2*np.pi, N) theta, phi = np.meshgrid(t, t) theta = theta.flatten() phi = phi.flatten() X = np.zeros((N*N, 4)) X[:, 0] = np.cos(theta) X[:, 1] = np.sin(theta) X[:, 2] = np.cos(phi) X[:, 3] = np.sin(phi)

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On Fri, Nov 13, 2020 at 3:17 AM Filip Cornell ***@***.***> wrote: Hmm, I am not sure I understand how you mean. Do you perhaps have some sample code? — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#10 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAJWDZS3A5EO4OZHEL5QWPDSPTTQDANCNFSM4TTOIQOA> .

[Filco306]

Very sorry for such a late reply. That is very interesting, I suppose we can extend that to an arbitrary number of dimensions?

[Filco306]

Would something like this do as a first draft?

def sample_4d_torus(n_points, seed):
    assert np.sqrt(n_points) % 1 == 0, "Please pick a number of points with integer root. "
    np.random.seed(seed)
    N = int(np.sqrt())
    t = np.random.uniform(0,2*np.pi, N)
    t = np.linspace(0, 2*np.pi, N)
    theta, phi = np.meshgrid(t, t)
    theta = theta.flatten()
    phi = phi.flatten()
    X = np.zeros((N*N, 4))
    X[:, 0] = np.cos(theta)
    X[:, 1] = np.sin(theta)
    X[:, 2] = np.cos(phi)
    X[:, 3] = np.sin(phi)
    return X

[ctralie]

Yeah, for a flat torus in 4D, that should be uniform. Did you mean to say N = int(np.sqrt(n_points)) ?

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On Fri, Feb 12, 2021 at 3:50 AM Filip Cornell ***@***.***> wrote: Would something like this do as a first draft? def sample_4d_torus(n_points, seed): assert np.sqrt(n_points) % 1 == 0, "Please pick a number of points with integer root. " np.random.seed(seed) N = int(np.sqrt()) t = np.random.uniform(0,2*np.pi, N) t = np.linspace(0, 2*np.pi, N) theta, phi = np.meshgrid(t, t) theta = theta.flatten() phi = phi.flatten() X = np.zeros((N*N, 4)) X[:, 0] = np.cos(theta) X[:, 1] = np.sin(theta) X[:, 2] = np.cos(phi) X[:, 3] = np.sin(phi) return X — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#10 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAJWDZWS7TECUCE7PSM6RDDS6TTVNANCNFSM4TTOIQOA> .

[Filco306]

Oh yes, it should be int(np.sqrt(n_points)). Can we extend this to more than 4 dimensions?

[ctralie]

Yup, you can do this sort of thing in every even dimension at 4 and beyond

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On Fri, Feb 12, 2021 at 10:52 AM Filip Cornell ***@***.***> wrote: Oh yes, it should be int(np.sqrt(n_points)). Can we extend this to more than 4 dimensions? — You are receiving this because you commented. Reply to this email directly, view it on GitHub <#10 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAJWDZTS3GMDA5HQRJVGFBTS6VFDHANCNFSM4TTOIQOA> .