A Belated Valentine’s Day Post

This is romantic!  So listen up!

A 3D heart shape may be drawn using the following implicit function:


Or, in Python:

def  heart_3d(x,y,z):
    return (x**2+(9/4)*y**2+z**2-1)**3-x**2*z**3-(9/80)*y**2*z**3

Trouble is, there is no direct way of graphing implicit functions in Python. But anything can be found on Stack Overflow.

Putting it all together:

#!/usr/bin/env python
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import matplotlib.pyplot as plt
import numpy as np
def heart_3d(x,y,z):
   return (x**2+(9/4)*y**2+z**2-1)**3-x**2*z**3-(9/80)*y**2*z**3

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, 40) # 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',colors=('r',))
        # [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',colors=('red',))

    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',colors=('red',))

    # 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.


if __name__ == '__main__':

Show this to your date on the next Valentine’s Day, because it is too late for this one. Trust me, results are guranteed. Not sure what kind of results though.

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2 Responses to “A Belated Valentine’s Day Post”

  1. Julien says:

    What about this?
    Copy-paste this in Google:
    5 + (sqrt(1-x^2(y-abs(x))^2))cos(30((1-x^2-(y-abs(x))^2))), x is from -1 to 1, y is from -1 to 1.5, z is from 1 to 6

  2. Iddo says:

    Wicked! Thanks, Julien.