# We are nerds, we are cheesy

In [1]:

```
%matplotlib inline
import matplotlib
matplotlib.rcParams['figure.figsize'] = (16,13)
```

In [2]:

```
# Source: http://stackoverflow.com/a/4687582
# Thanks!
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
def plot_implicit(fn, bbox=(-1.5,1.5)):
'''
create a plot of an implicit function
fn ...implicit function (plot where fn==0)
bbox ..the x,y,and z limits of plotted interval
'''
xmin, xmax, ymin, ymax, zmin, zmax = bbox*3
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
A = np.linspace(xmin, xmax, 100) # resolution of the contour
B = np.linspace(xmin, xmax, 15) # number of slices
A1,A2 = np.meshgrid(A,A) # grid on which the contour is plotted
for z in B: # plot contours in the XY plane
X,Y = A1,A2
Z = fn(X,Y,z)
cset = ax.contour(X, Y, Z+z, [z], zdir='z')
# [z] defines the only level to plot for this contour for this value of z
for y in B: # plot contours in the XZ plane
X,Z = A1,A2
Y = fn(X,y,Z)
cset = ax.contour(X, Y+y, Z, [y], zdir='y')
for x in B: # plot contours in the YZ plane
Y,Z = A1,A2
X = fn(x,Y,Z)
cset = ax.contour(X+x, Y, Z, [x], zdir='x')
# must set plot limits because the contour will likely extend
# way beyond the displayed level. Otherwise matplotlib extends the plot limits
# to encompass all values in the contour.
ax.set_zlim3d(zmin,zmax)
ax.set_xlim3d(xmin,xmax)
ax.set_ylim3d(ymin,ymax)
plt.show()
```

In [3]:

```
def happy_saint_valentine_ipython(x,y,z):
return -x**2*z**3 - 9*y**2*z**3/80 + (4*x**2 + 9*y**2 + 4*z**2 - 4)**3/64
plot_implicit(happy_saint_valentine_ipython)
```

In [3]:

```
```