“Searching” is the best way to solve common questions like this.
Do you really want to learn? Or you only want to ask to act like a learner.
CNN: TODO:for you
RNN: http://preview.d2l.ai/d2l-en/master/chapter_recurrent-neural-networks/bptt.html
How about:
x = np.linspace(- np.pi,np.pi,100)
x = torch.tensor(x, requires_grad=True)
y = torch.sin(x)
z = y.sum()
z.backward()
d2l.plot(x.detach(),(y.detach(),x.grad),legend = (('sin(x)','grad w.s.t x')))My solution to the last exercise
import torch
import matplotlib.pyplot as plt
import numpy as np
import math
%matplotlib inline
x = torch.arange(0, 2*math.pi, 0.1)
x.requires_grad_(True)
y = torch.sin(x).sum().backward() #Derivando y respecto a x
dy_dx = x.grad
x = x.detach().numpy()
y = np.sin(x)
dy_dx = dy_dx.numpy()
fig = plt.figure()
plt.plot(x, y, ‘-’, label = ‘y = sin(x)’)
plt.plot(x, dy_dx, ‘–’, label =’ dy_dx = cos(x)’)
plt.axis(‘equal’)
leg = plt.legend();
x = torch.arange(-np.pi, np.pi, 0.1, requires_grad=True)
y = torch.sin(x)
y.sum().backward()
x.grad
Exercise 1. The second derivative is much more expensive because, for a function f with n variables, the Hessian Matrix has n² elements while the gradient vector has n elements. In other words, there are n² possible second order derivatives ( ∂²f/∂x1², ∂²f/∂x1∂x2, …) while there are n possible first order derivatives (∂f/∂x1, ∂f/∂x2, …, ∂f/∂xn).
Backward for Non-Scalar Variables:
Seems, in the text, by calculating y.backward(torch.ones(len(x))), we contract the Jacobian over the 4 components of y(x). The Jacobian is a 4x4 matrix. I can extract each column via y.backward(torch.tensor([1,0,0,0])), #where i shift the 1 to each of the 4 column slots.
Is this best way to calculate the Jacobian?
My anwser for ex 2 3 4，welcom to tell me if you find any error
'''
ex.2
Aafter running the function for backpropagation, immediately run it again and see what hap-
pens. Why?
reference:
[https://blog.csdn.net/rothschild666/article/details/124170794](https://blog.csdn.net/rothschild666/article/details/124170794)
'''
import torch
x = torch.arange(4.0, requires_grad=True)
x.requires_grad_(True)
y = 2 * torch.dot(x, x)
y.backward(retain_graph=True)#need to set para = Ture
y.backward()
x.grad
out for ex2:
tensor([ 0., 8., 16., 24.])
'''
ex.3
In the control flow example where we calculate the derivative of d with respect to a, what
would happen if we changed the variable a to a random vector or a matrix? At this point, the
result of the calculation f(a) is no longer a scalar. What happens to the result? How do we
analyze this?
reference:
[https://www.jb51.net/article/211983.htm](https://www.jb51.net/article/211983.htm)
'''
import torch
def f(a):
b = a * 2
while b.norm() < 1000:
b = b * 2
if b.sum() > 0:
c = b
else:
c = 100 * b
return c
a = torch.randn(size=(2,3), requires_grad=True)
#a = torch.tensor([1.0,2.0], requires_grad=True)
print(a)
d = f(a)
d.backward(d.detach())#This is neccesary, I don't know why, is it because there are b and c in the function?
#d.backward()#wrong
print(a.grad)
out for ex3:
tensor([[-0.6848, 0.2447, 1.5633],
[-0.1291, 0.2607, 0.9181]], requires_grad=True)
tensor([[-179512.9062, 64147.3203, 409814.3750],
[ -33844.5430, 68346.7969, 240686.7344]])
'''
ex.4
Let f (x) = sin(x). Plot the graph of f and of its derivative f ′ . Do not exploit the fact that
f ′ (x) = cos(x) but rather use automatic differentiation to get the result.
'''
#suppose the functions in 2.4.2Visualization Utilities has been constructed in d2l
from d2l import torch as d2l
x = d2l.arange(0,10,0.1, requires_grad = True)
y = d2l.sin(x)
y.backward(gradient=d2l.ones(len(y)))#y.sum().backward() can do the same thing
#.detach.numpy() is needed because x is set to requires_grad = True
d2l.plot(x.detach().numpy(), [y.detach().numpy(),x.grad], 'x', 'f(x)', legend = ['sin(x)','sin\'(x)'])
out for ex4:
Can someone please explain the result of 2.5.1 to me. I understand how the first function is just y = 2x^2, but creating the new function that is y = x.sum(), just returns 6. Is this conceptually 6x or is it just 6 or is this just x (i’m assuming the third because that’s the only way we can get 1 as the gradient all the time) or am i missing something?
My understanding about 2.5.1’s pytorch code blocks:
y = x.sum() is just a way of defining a computational graph in PyTorch, which means evaluating each component of x and adding them up. Gradients are computed on each component of x, NOT on the y graph. Evaluating gradient on a component of x means computing for y_i = x_i (which yields 1).
The same principal applies to the y = 2 * x^Tx part, y is NOT 2x^2, it’s a computational graph for evaluating 2 * x^Tx where each component of it is actually y_i = 2 * x_i * x_i. So the graph y is in fact sum { 2 * x_i * x_i }.
HTH
Can someone help with Questions 5 and 6. Is this a pen-paper question, writing down backprop for the computation graph, or is this supposed to be implemented (if yes, then can someone elaborate more on it or provide a solution.)
Also problems 7, 8 are difficult ones, but at least there is a good resource provided by Denis Kazakov.
Hi! In Exercise 5, I don’t understand why my solution and the expected derivative do not correspond:
%matplotlib inline
import numpy as np
from matplotlib_inline import backend_inline
from d2l import torch as d2l
import math
x = torch.linspace(0.1, 10, 100)
x.requires_grad_(True)
def f(x):
return ((torch.log(x ** 2)) * torch.sin(x)) + (x ** -1)
y = f(x)
y.backward(torch.ones_like(y))
p = x.grad
Plot the function and its derivative
plt.plot(x.detach().numpy(), y.detach().numpy(), label=‘f(x)’)
plt.plot(x.detach().numpy(), p.detach().numpy(), label=“f’(x)”)
plt.xlabel(‘x’)
plt.ylabel(‘y’)
plt.legend()
plt.grid(True)
plt.show()
gives me this plot:
and when I compare
desired_derivative = -(2 / x) * torch.sin(x) + torch.log(x**2) * torch.cos(x) - (1/x **2)
p == desired_derivative
it evaluates to False.
Can you help me?
Ex2.
import torch
x = torch.arange(4.0)
x.requires_grad_(True)
y = x @ x
y.backward()
Output (second run):
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
/tmp/ipykernel_82/1055444503.py in <module>
----> 1 y.backward()
/opt/miniconda/lib/python3.7/site-packages/torch/_tensor.py in backward(self, gradient, retain_graph, create_graph, inputs)
394 create_graph=create_graph,
395 inputs=inputs)
--> 396 torch.autograd.backward(self, gradient, retain_graph, create_graph, inputs=inputs)
397
398 def register_hook(self, hook):
/opt/miniconda/lib/python3.7/site-packages/torch/autograd/__init__.py in backward(tensors, grad_tensors, retain_graph, create_graph, grad_variables, inputs)
173 Variable._execution_engine.run_backward( # Calls into the C++ engine to run the backward pass
174 tensors, grad_tensors_, retain_graph, create_graph, inputs,
--> 175 allow_unreachable=True, accumulate_grad=True) # Calls into the C++ engine to run the backward pass
176
177 def grad(
RuntimeError: Trying to backward through the graph a second time (or directly access saved tensors after they have already been freed). Saved intermediate values of the graph are freed when you call .backward() or autograd.grad(). Specify retain_graph=True if you need to backward through the graph a second time or if you need to access saved tensors after calling backward.
- The
.backward()call will automatically free all the intermediate tensors in a computational graph after the backprop. - In that sense, “double backward” can not be realized unless one specifies
retain_graph=True.
Ex3.
def f(a):
b = a * 2
while b.norm() < 1000:
b = b * 2
if b.sum() > 0:
c = b
else:
c = 100 * b
return c
a = torch.randn(size=(4, ), requires_grad=True)
d = f(a)
d.backward()
Output:
---------------------------------------------------------------------------
RuntimeError Traceback (most recent call last)
/tmp/ipykernel_82/535593037.py in <module>
1 a = torch.randn(size=(4, ), requires_grad=True)
2 d = f(a)
----> 3 d.backward()
/opt/miniconda/lib/python3.7/site-packages/torch/_tensor.py in backward(self, gradient, retain_graph, create_graph, inputs)
394 create_graph=create_graph,
395 inputs=inputs)
--> 396 torch.autograd.backward(self, gradient, retain_graph, create_graph, inputs=inputs)
397
398 def register_hook(self, hook):
/opt/miniconda/lib/python3.7/site-packages/torch/autograd/__init__.py in backward(tensors, grad_tensors, retain_graph, create_graph, grad_variables, inputs)
164
165 grad_tensors_ = _tensor_or_tensors_to_tuple(grad_tensors, len(tensors))
--> 166 grad_tensors_ = _make_grads(tensors, grad_tensors_, is_grads_batched=False)
167 if retain_graph is None:
168 retain_graph = create_graph
/opt/miniconda/lib/python3.7/site-packages/torch/autograd/__init__.py in _make_grads(outputs, grads, is_grads_batched)
65 if out.requires_grad:
66 if out.numel() != 1:
---> 67 raise RuntimeError("grad can be implicitly created only for scalar outputs")
68 new_grads.append(torch.ones_like(out, memory_format=torch.preserve_format))
69 else:
RuntimeError: grad can be implicitly created only for scalar outputs
- Since
aas well asdis no longer a scalar, we need to reducedto make backprop feasible.
d.sum().backward()
a.grad == d / a
Output:
tensor([True, True, True, True])
Ex4-5.
- The dependency graph (computational graph):
- Applying the backprop algorithm according to the chain rule…
- The
+operationy = n1 + n2:n1.grad = 1, n2.grad = 1 - The inverse operation
n1 = 1 / x:x.grad1 = (- 1 / x ** 2) * n1.grad = - 1 / x ** 2 - The
*operationn2 = n3 * n4:n3.grad = n4 * n2.grad = log(x ** 2), n4.grad = n3 * n2.grad = sin(x) - The
sinoperationn3 = sin(x):x.grad2 = cos(x) * n3.grad = cos(x) * log(x ** 2) - The
logoperationn4 = log(n5):n5.grad = 1 / n5 * n4.grad = 1 / x ** 2 * sin(x) - The square operation
n5 = x ** 2:x.grad3 = 2 * x * n5.grad = 2 / x * sin(x) - Final result:
x.grad = x.grad1 + x.grad2 + x.grad3 = - 1 / x ** 2 + cos(x) * log(x ** 2) + 2 / x * sin(x)
- The
import torch
torch.manual_seed(713)
x = torch.randn(size=(), requires_grad=True)
n1, n3, n5 = 1 / x, torch.sin(x), x ** 2
n4 = torch.log(n5)
n2 = n3 * n4
y = n1 + n2
y.backward()
x.grad, - 1 / x ** 2 + torch.cos(x) * torch.log(x ** 2) + 2 / x * torch.sin(x)
Output:
(tensor(-1.7580), tensor(-1.7580, grad_fn=<AddBackward0>))
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