Add a print(N) line after all the N+= lines, and try various Z arrays.
For example
Define a small z with a block of 1s in the middle:
In [29]: z = np.zeros((10,10),int)
In [31]: z[4:6,4:6]=1
In [34]: z[4:8,5]=1
In [35]: z
Out[35]:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
pass it to the function:
In [36]: iterate(z)
[[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 1. 2. 2. 1. 0. 0. 0.]
[ 0. 0. 0. 2. 3. 3. 2. 0. 0. 0.]
[ 0. 0. 0. 2. 4. 4. 3. 0. 0. 0.]
[ 0. 0. 0. 1. 4. 3. 3. 0. 0. 0.]
[ 0. 0. 0. 0. 2. 1. 2. 0. 0. 0.]
[ 0. 0. 0. 0. 1. 1. 1. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]
N has counted the number of neighbors that are 1. Check the counts yourself.
Out[36]:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 1, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
Try various patterns, repeat the iterate and watch the pattern change. Some die away, some move in an order fashion, some 'blink', etc.
In a line like:
N[1:, 1:] += Z[:-1, :-1]
the RHS is an upper left portion (here 9x9); LHS is a bottom right, again 9x9. There are 8 N+= expressions, calculating the 8 neighbors (in a 3x3 block, minus the center). With this offset slicing it can do the count for all points in Z at once.
For a start, a 1 row array might be easier to visualize
In [47]: z = np.zeros((1,10),int)
In [49]: z[0,4:7]=1
In [50]: z
Out[50]: array([[0, 0, 0, 0, 1, 1, 1, 0, 0, 0]])
In [51]: iterate(z)
[[ 0. 0. 0. 1. 1. 2. 1. 1. 0. 0.]]
Out[51]: array([[0, 0, 0, 0, 0, 1, 0, 0, 0, 0]])
I think this works best if all the edge values of z are 0.
This array creates the glider that is animated on https://en.wikipedia.org/wiki/Glider_(Conway%27s_Life)
In [64]: z = np.zeros((10,10),int)
In [65]: z[1,2]=1;z[2,3]=1;z[3,1:4]=1
