Symbolic computing for ML

!wget --no-cache -O init.py -q https://raw.githubusercontent.com/jdariasl/OTBD/main/content/init.py
import init; init.init(force_download=False)

import sys
import sympy as sy
from IPython.display import Image
%load_ext tensorboard
sy.init_printing(use_latex=True)    
import tensorflow as tf
tf.__version__

The tensorboard extension is already loaded. To reload it, use:
  %reload_ext tensorboard

'2.14.0'

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from scipy.optimize import minimize
%matplotlib inline

d = pd.read_csv("local/datasets/trilotropicos.csv")
print(d.shape)
plt.scatter(d.longitud, d.densidad_escamas)
plt.xlabel(d.columns[0])
plt.ylabel(d.columns[1]);

(150, 2)

Let’s define a simple loss function

\[\mathcal{L}({\bf{w}}) = \frac{1}{n}\sum_{i=1}^{n} ({\bf{w}}^T{\bf{x}}_i- y_i)^2 \]

The gradient of \(\mathcal{L}(\cdot)\) is:

\[\nabla \mathcal{L}({\bf{w}}) = \frac{1}{n}\sum_{i=1}^{n} ({\bf{w}}^T{\bf{x}}_i -y_i ) {\bf{x}}_i\]

y = d.densidad_escamas.values
X = np.r_[[[1]*len(d), d.longitud.values]].T


def n_cost(t):
    return np.mean((X.dot(t)-y)**2)

def n_grad(t):
    return 2*X.T.dot(X.dot(t)-y)/len(X)

init_t = np.random.random()*40-5, np.random.random()*20-10
r = minimize(n_cost, init_t, method="BFGS", jac=n_grad)
r

      fun: 2.74476625708012
 hess_inv: array([[ 5.57513613, -1.09143751],
       [-1.09143751,  0.23468758]])
      jac: array([-1.19252074e-07, -6.09487721e-07])
  message: 'Optimization terminated successfully.'
     nfev: 7
      nit: 6
     njev: 7
   status: 0
  success: True
        x: array([12.68999806, -0.71805909])

Using sympy computer algebra system (CAS)

x,y = sy.symbols("x y")

z = x**2 + x*sy.cos(y)
z

we can evaluate the expresion by providing concrete values for the symbolic variables

z.subs({x: 2, y: sy.pi/4})

and obtain numerical approximations of these values

sy.N(z.subs({x: 2, y: sy.pi/4}))

a derivative can be seen as a function that inputs and expression and outputs another expression

observe how we compute \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\)

z.diff(x)

z.diff(y)

r = z.diff(x).subs({x: 2, y: sy.pi/4})
r, sy.N(r)

r = z.diff(y).subs({x: 2, y: sy.pi/4})
r, sy.N(r)

How is this done?

Computational graph

Image("local/imgs/ComputationalGraph.png")

Image("local/imgs/ComputationalGraph2.png")

A symbolic computational package can do a lot of things:

sy.expand((x+2)**2)

sy.factor( x**2-2*x-8 )

sy.solve( x**2 + 2*x - 8, x)

a = sy.symbols("alpha")
sy.solve( a*x**2 + 2*x - 8, x)

differential equations, solving \(\frac{df}{dt}=f(t)+t\)

t, C1 = sy.symbols("t C1")
f = sy.symbols("f", cls=sy.Function)
dydt = f(t)+t
eq = dydt-sy.diff(f(t),t)
eq

yt = sy.dsolve(eq, f(t))
yt

systems of equations

sy.solve ([x**2+y, 3*y-x])

Sympy to Python and Numpy

See Sympy Numeric Computation

f = (sy.sin(x) + x**2)/2
f

f.subs({x:10})

sy.N(f.subs({x:10}))

f1 = sy.lambdify(x, f)
f1(10)

and a vectorized version

f2 = sy.lambdify(x, f, "numpy")

f2(10)

f2(np.array([10,2,3]))

array([49.72798944,  2.45464871,  4.57056   ])

the lambdified version is faster, and the vectorized one is even faster

%timeit sy.N(f.subs({x:10}))

112 µs ± 884 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

%timeit f1(10)

1.14 µs ± 7 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)

%timeit [f1(i) for i in range(1000)]

1.18 ms ± 14.6 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

%timeit f2(np.arange(1000))

13.2 µs ± 93.1 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

Let’s remember the loss function we defined before:

\[\mathcal{L}({\bf{w}}) = \frac{1}{n}\sum_{i=1}^{n} ({\bf{w}}^T{\bf{x}}_i- y_i)^2 \]

The gradient of \(\mathcal{L}(\cdot)\) is:

\[\nabla \mathcal{L}({\bf{w}}) = \frac{1}{n}\sum_{i=1}^{n} ({\bf{w}}^T{\bf{x}}_i -y_i ) {\bf{x}}_i\]

Using sympy to obtain the gradient.

y = d.densidad_escamas.values
X = np.r_[[[1]*len(d), d.longitud.values]].T

w0,w1 = sy.symbols("w_0 w_1")
w0,w1

we first obtain the cost expression for a few summation terms, so that we can print it out and understand it

L = 0
for i in range(10):
    L += (X[i,0]*w0+X[i,1]*w1-y[i])**2
L = L/len(X)
L

we can now simplify the expression, using sympy mechanics

L = L.simplify()
L

we now build the full expression

def build_regression_cost_expression(X,y):
    expr_cost = 0
    for i in range(len(X)):
        expr_cost += (X[i,0]*w0+X[i,1]*w1-y[i])**2/len(X)
    expr_cost = expr_cost.simplify()
    return expr_cost

y = d.densidad_escamas.values
X = np.r_[[[1]*len(d), d.longitud.values]].T

expr_cost = build_regression_cost_expression(X,y)
expr_cost

obtain derivatives symbolically

expr_dw0 = expr_cost.diff(w0)
expr_dw1 = expr_cost.diff(w1)
expr_dw0, expr_dw1

and obtain regular Python so that we can use them in optimization

s_cost = sy.lambdify([[w0,w1]], expr_cost, "numpy")

d0 = sy.lambdify([[w0,w1]], expr_dw0, "numpy")
d1 = sy.lambdify([[w0,w1]], expr_dw1, "numpy")
s_grad = lambda x: np.array([d0(x), d1(x)])

and now we can minimize

r = minimize(s_cost, [0,0], jac=s_grad, method="BFGS")
r

      fun: 2.7447662570799594
 hess_inv: array([[ 5.57425906, -1.09086733],
       [-1.09086733,  0.23434848]])
      jac: array([-1.73778293e-07, -9.26928138e-07])
  message: 'Optimization terminated successfully.'
     nfev: 10
      nit: 9
     njev: 10
   status: 0
  success: True
        x: array([12.6899981, -0.7180591])

observe that hand derived functions and the ones obtained by sympy evaluate to the same values

w0 = np.random.random()*5+10
w1 = np.random.random()*4-3
w = np.r_[w0,w1]

print ("theta:",w)
print ("cost analytic:", n_cost(w))
print ("cost symbolic:", s_cost(w))

print ("gradient analytic:", n_grad(w))
print ("gradient symbolic:", s_grad(w))

theta: [14.35986113 -2.46039072]
cost analytic: 50.580711344683486
cost symbolic: 50.58071134468287
gradient analytic: [-12.86378994 -67.23901186]
gradient symbolic: [-12.86378994 -67.23901186]