Please read carefully the documentation on backward() to better understand it.
By default, pytorch expects backward() to be called for the last output of the network – the loss function. The loss function always outputs a scalar and therefore, the gradients of the scalar loss w.r.t all other variables/parameters is well defined (using the chain rule).
Thus, by default, backward() is called on a scalar tensor and expects no arguments.
For example:
a = torch.tensor([[1,2,3],[4,5,6]], dtype=torch.float, requires_grad=True)
for i in range(2):
for j in range(3):
out = a[i,j] * a[i,j]
out.backward()
print(a.grad)
yields
tensor([[ 2., 4., 6.],
[ 8., 10., 12.]])
As expected: d(a^2)/da = 2a.
However, when you call backward on the 2-by-3 out tensor (no longer a scalar function) – what do you expects a.grad to be? You’ll actually need a 2-by-3-by-2-by-3 output: d out[i,j] / d a[k,l](!)
Pytorch does not support this non-scalar function derivatives. Instead, pytorch assumes out is only an intermediate tensor and somewhere “upstream” there is a scalar loss function, that through chain rule provides d loss/ d out[i,j]. This “upstream” gradient is of size 2-by-3 and this is actually the argument you provide backward in this case: out.backward(g) where g_ij = d loss/ d out_ij.
The gradients are then calculated by chain rule d loss / d a[i,j] = (d loss/d out[i,j]) * (d out[i,j] / d a[i,j])
Since you provided a as the “upstream” gradients you got
a.grad[i,j] = 2 * a[i,j] * a[i,j]
If you were to provide the “upstream” gradients to be all ones
out.backward(torch.ones(2,3)) print(a.grad)
yields
tensor([[ 2., 4., 6.],
[ 8., 10., 12.]])
As expected.
It’s all in the chain rule.