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Differentiable Stencil computations in JAX

Installation |Description |Quick example |More Examples |Benchmarking

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πŸ› οΈ Installation

pip install kernex

πŸ“– Description

Kernex extends jax.vmap/jax.lax.map/jax.pmap with kmap and jax.lax.scan with kscan for general stencil computations.

⏩ Quick Example

kmap kscan
import kernex as kex
import jax.numpy as jnp

@kex.kmap(kernel_size=(3,))
def sum_all(x):
    return jnp.sum(x)

x = jnp.array([1,2,3,4,5])
print(sum_all(x))
# [ 6  9 12]
import kernex as kex 
import jax.numpy as jnp

@kex.kscan(kernel_size=(3,))
def sum_all(x):
return jnp.sum(x)

x = jnp.array([1,2,3,4,5])
print(sum_all(x))
# [ 6 13 22]
`jax.vmap` is used to sum each window content. `lax.scan` is used to update the array and the window sum is calculated sequentially. the first three rows represents the three sequential steps used to get the solution in the last row.

πŸ”’ More examples

1️⃣ Convolution operation
import jax
import jax.numpy as jnp
import kernex as kex

@jax.jit
@kex.kmap(
    kernel_size= (3,3,3),
    padding = ('valid','same','same'))
def kernex_conv2d(x,w):
    # JAX channel first conv2d with 3x3x3 kernel_size
    return jnp.sum(x*w)
2️⃣ Laplacian operation
# see also
# https://numba.pydata.org/numba-doc/latest/user/stencil.html#basic-usage
import jax
import jax.numpy as jnp
import kernex as kex

@kex.kmap(
    kernel_size=(3,3),
    padding= 'valid',
    relative=True) # `relative`= True enables relative indexing
def laplacian(x):
    return ( 0*x[1,-1]  + 1*x[1,0]   + 0*x[1,1] +
             1*x[0,-1]  +-4*x[0,0]   + 1*x[0,1] +
             0*x[-1,-1] + 1*x[-1,0]  + 0*x[-1,1] )

print(laplacian(jnp.ones([10,10])))
# [[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., 0., 0., 0., 0.],
#  [0., 0., 0., 0., 0., 0., 0., 0.],
#  [0., 0., 0., 0., 0., 0., 0., 0.]]
3️⃣ Get Patches of an array
import jax
import jax.numpy as jnp
import kernex as kex

@kex.kmap(kernel_size=(3,3),relative=True)
def identity(x):
    # similar to numba.stencil
    # this function returns the top left cell in the padded/unpadded kernel view
    # or center cell if `relative`=True
    return x[0,0]

# unlike numba.stencil , vector output is allowed in kernex
# this function is similar to
# `jax.lax.conv_general_dilated_patches(x,(3,),(1,),padding='same')`
@jax.jit
@kex.kmap(kernel_size=(3,3),padding='same')
def get_3x3_patches(x):
    # returns 5x5x3x3 array
    return x

mat = jnp.arange(1,26).reshape(5,5)
print(mat)
# [[ 1  2  3  4  5]
#  [ 6  7  8  9 10]
#  [11 12 13 14 15]
#  [16 17 18 19 20]
#  [21 22 23 24 25]]


# get the view at array index = (0,0)
print(get_3x3_patches(mat)[0,0])
# [[0 0 0]
#  [0 1 2]
#  [0 6 7]]
4️⃣ Linear convection
Problem setup Stencil view
import jax
import jax.numpy as jnp
import kernex as kex
import matplotlib.pyplot as plt

# see https://nbviewer.org/github/barbagroup/CFDPython/blob/master/lessons/01_Step_1.ipynb

tmax,xmax = 0.5,2.0
nt,nx = 151,51
dt,dx = tmax/(nt-1) , xmax/(nx-1)
u = jnp.ones([nt,nx])
c = 0.5

# kscan moves sequentially in row-major order and updates in-place using lax.scan.

F = kernex.kscan(
        kernel_size = (3,3),
        padding = ((1,1),(1,1)),
        # n for time axis , i for spatial axis (optional naming)
        named_axis={0:'n',1:'i'},  
        relative=True
    )


# boundary condtion as a function
def bc(u):
    return 1

# initial condtion as a function
def ic1(u):
    return 1

def ic2(u):
    return 2

def linear_convection(u):
    return ( u['i','n-1'] - (c*dt/dx) * (u['i','n-1'] - u['i-1','n-1']) )


F[:,0]  = F[:,-1] = bc # assign 1 for left and right boundary for all t

# square wave initial condition
F[:,:int((nx-1)/4)+1] = F[:,int((nx-1)/2):] = ic1
F[0:1, int((nx-1)/4)+1 : int((nx-1)/2)] = ic2

# assign linear convection function for
# interior spatial location [1:-1]
# and start from t>0  [1:]
F[1:,1:-1] = linear_convection

kx_solution = F(jnp.array(u))

plt.figure(figsize=(20,7))
for line in kx_solution[::20]:
    plt.plot(jnp.linspace(0,xmax,nx),line)
5️⃣ Gaussian blur
import jax
import jax.numpy as jnp
import kernex as kex

def gaussian_blur(image, sigma, kernel_size):
    x = jnp.linspace(-(kernel_size - 1) / 2.0, (kernel_size- 1) / 2.0, kernel_size)
    w = jnp.exp(-0.5 * jnp.square(x) * jax.lax.rsqrt(sigma))
    w = jnp.outer(w, w)
    w = w / w.sum()

    @kex.kmap(kernel_size=(kernel_size, kernel_size), padding="same")
    def conv(x):
        return jnp.sum(x * w)

    return conv(image)
6️⃣ Depthwise convolution
import jax
import jax.numpy as jnp
import kernex as kex

@jax.jit
@jax.vmap
@kex.kmap(
    kernel_size= (3,3),
    padding = ('same','same'))
def kernex_depthwise_conv2d(x,w): 
    return jnp.sum(x*w)

h,w,c = 5,5,2
k=3

x = jnp.arange(1,h*w*c+1).reshape(c,h,w)
w = jnp.arange(1,k*k*c+1).reshape(c,k,k)
print(kernex_depthwise_conv2d(x,w))
7️⃣ Average pooling 2D
@jax.vmap # vectorize over the channel dimension
@kex.kmap(kernel_size=(3,3), strides=(2,2))
def avgpool_2d(x):
    # define the kernel for the Average pool operation over the spatial dimensions
    return jnp.mean(x)
8️⃣ Runge-Kutta integration
# lets solve dydt = y, where y0 = 1 and y(t)=e^t
# using Runge-Kutta 4th order method
# f(t,y) = y
import jax.numpy as jnp
import matplotlib.pyplot as plt
import kernex as kex


t = jnp.linspace(0, 1, 5)
y = jnp.zeros(5)
x = jnp.stack([y, t], axis=0)
dt = t[1] - t[0]  # 0.1
f = lambda tn, yn: yn


def ic(x):
    """ initial condition y0 = 1 """
    return 1.


def rk4(x):
    """ runge kutta 4th order integration step """
    # β”Œβ”€β”€β”€β”€β”¬β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”      β”Œβ”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”
    # β”‚ y0 β”‚*y1*β”‚ y2 β”‚      β”‚[0,-1]β”‚[0, 0]β”‚[0, 1]β”‚
    # β”œβ”€β”€β”€β”€β”Όβ”€β”€β”€β”€β”Όβ”€β”€β”€β”€β”€ ==>  β”œβ”€β”€β”€β”€β”€β”€β”Όβ”€β”€β”€β”€β”€β”€β”Όβ”€β”€β”€β”€β”€β”€β”€
    # β”‚ t0 β”‚ t1 β”‚ t2 β”‚      β”‚[1,-1]β”‚[1, 0]β”‚[1, 1]β”‚
    # β””β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”˜      β””β”€β”€β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”€β”€β”˜
    t0 = x[1, -1]
    y0 = x[0, -1]
    k1 = dt * f(t0, y0)
    k2 = dt * f(t0 + dt / 2, y0 + 1 / 2 * k1)
    k3 = dt * f(t0 + dt / 2, y0 + 1 / 2 * k2)
    k4 = dt * f(t0 + dt, y0 + k3)
    yn_1 = y0 + 1 / 6 * (k1 + 2 * k2 + 2 * k3 + k4)
    return yn_1


F = kex.kscan(kernel_size=(2, 3), relative=True, padding=((0, 1)))  # kernel size = 3

F[0:1, 1:] = rk4
F[0, 0] = ic
# compile the solver
solver = jax.jit(F.__call__)
y = solver(x)[0, :]

plt.plot(t, y, '-o', label='rk4')
plt.plot(t, jnp.exp(t), '-o', label='analytical')
plt.legend()

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