# Re: [Meep-discuss] Pymeep geometries using epsilon_func

```Hi,

If you return a `Medium` in your epsilon function, then you have to pass it
to the `Simulation` with the `material_function` keyword argument instead
of `epsilon_func`. See the section titled "material functions" under the
`Medium` documentation here:
Additionally, I think you want `epsilon_function` to check `if
vector.norm() < 1` instead of `vector.unit()`?```
```
Chris

On Wed, Jul 11, 2018 at 4:19 AM, Karl-Johan Olofsson <
karol...@student.liu.se> wrote:

> Hello. I would like to draw a hexagonal pyramid in MEEP using the python
> interface. My first idea was to use a hexagonal prism and then use blocks
> to "cut" the pyramid shape. That proved to be quite the nuisance so I
> decided to use the epsilon function to draw the shape instead. Any ideas
> what the best way might be?
>
>
> Anyway, attempting to draw a simple circle using the epsilon function
> gives me segmentation fault, please see the code below:
>
> import meep as mp
> import numpy as np
> import matplotlib.pyplot as plt
> import math as math
>
> sxy=10
> resolution=10
> cell=mp.Vector3(sxy,sxy,0)
>
> def epsilon_function(vector):
>
>         if vector.unit() < 1:
>
>             return mp.Medium(epsilon=2)
>
>         else:
>
>             return mp.Medium(epsilon=1)
>
> sim=mp.Simulation(cell_size=cell,
>         #geometry=geometry,
>         epsilon_func=epsilon_function,
>         resolution=resolution)
>
> sim.run(until=1)
>
> eps_data = sim.get_array(center=mp.Vector3(), size=cell,
> component=mp.Dielectric)
> plt.figure(dpi=100)
> plt.imshow(eps_data.transpose(), interpolation='spline36',  cmap='binary')
> plt.axis('on')
> plt.show()
>
> Any help would be greatly appreciated
>
> Karl-Johan
>
>
>
> _______________________________________________
> meep-discuss mailing list
> meep-discuss@ab-initio.mit.edu
> http://ab-initio.mit.edu/cgi-bin/mailman/listinfo/meep-discuss
>
```
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