Coding Example: Implement the behavior of Boids (NumPy approach)

NumPy Implementation

As you might expect, the NumPy implementation takes a different approach and we’ll gather all our boids into a position array and a velocity array:

n = 500
velocity = np.zeros((n, 2), dtype=np.float32)
position = np.zeros((n, 2), dtype=np.float32)

The first step is to compute the local neighborhood for all boids, and for this, we need to compute all paired distances:

#np.subtract.outer  apply the ufunc op to all pairs (a, b) with a in A and b in B.
dx = np.subtract.outer(position[:, 0], position[:, 0])
dy = np.subtract.outer(position[:, 1], position[:, 1])
distance = np.hypot(dx, dy)

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We could have used the scipy cdist but we’ll need the dx and dy arrays later. Once those have been computed, it is faster to use the hypot method. Note that distance shape is (n, n) and each line relates to one boid, i.e. each line gives the distance to all other boids (including self).

From these distances, we can now compute the local neighborhood for each of the three rules, taking advantage of the fact that we can mix them together. We can actually compute a mask for distances that are strictly positive (i.e. have no self-interaction) and multiply it with other distance masks.

Note: If we suppose that boids cannot occupy the same position, how can you compute mask_0 more efficiently?

$$$$

mask_0 = (distance > 0)
mask_1 = (distance < 25)
mask_2 = (distance < 50)
mask_1 *= mask_0
mask_2 *= mask_0
mask_3 = mask_2

Then, we compute the number of neighbors within the given radius and we ensure it is at least 1 to avoid division by zero.

mask_1_count = np.maximum(mask_1.sum(axis=1), 1)
mask_2_count = np.maximum(mask_2.sum(axis=1), 1)
mask_3_count = mask_2_count

We’re ready to write our three rules:

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Alignment