An demonstration of autograd. Guess the linear recursive relation of fibnonssi number. i.e. Given f(0) = a0, f(1) = a1, f(x) = w0*f(x-1) + w1*f(x-2), find a0,a1,w that best approximate fibonacci sequence fib(i) where i = 5..9

pytorch_play.py

import torch as th
from torch.autograd import Variable

## helpers
def var(t):
    return Variable(t, requires_grad=True)

## training data
fibs = [1, 1]
for i in range(8):
    fibs.append(fibs[i] + fibs[i - 1])

## variables
a0 = var(th.randn(1))  # optimal = 1
a1 = var(th.randn(1))  # optimal = 1
w = var(th.randn(2))   # optimal = [1,1]

## model
def model(x):
    if x == 0: return a0
    if x == 1: return a1
    a = model(x - 1)
    b = model(x - 2)
    return w[0] * a + w[1] * b

def report(name):
    print(f'==== {name} ====')
    print(f'a0={a0.data[0]}')
    print(f'a1={a1.data[0]}')
    print(f'w0={w.data[0]}')
    print(f'w1={w.data[1]}')
    loss = [(model(i) - fibs[i]).abs().data[0] for i in range(len(fibs))]
    print(f'loss={loss}')

## training
report('initial')
rate = 0.001
for i in range(10000):
    print(f'running iteration {i}...', end='\r')
    ## approximate last 5 values
    for j in range(len(fibs) - 5, len(fibs)):
        ## absolute diff. as loss
        loss = (model(j) - fibs[j]).abs()
        ## calculate gradient
        loss.backward()
        ## update variables
        for v in [a0, a1, w]:
            v.data -= rate * v.grad.data
            v.grad.data.zero_()
report('final')