torch.autograd.functional.hvp¶
- torch.autograd.functional.hvp(func, inputs, v=None, create_graph=False, strict=False)[source]¶
Compute the dot product between the scalar function’s Hessian and a vector
v
at a specified point.- Parameters
func (function) – a Python function that takes Tensor inputs and returns a Tensor with a single element.
inputs (tuple of Tensors or Tensor) – inputs to the function
func
.v (tuple of Tensors or Tensor) – The vector for which the Hessian vector product is computed. Must be the same size as the input of
func
. This argument is optional whenfunc
’s input contains a single element and (if it is not provided) will be set as a Tensor containing a single1
.create_graph (bool, optional) – If
True
, both the output and result will be computed in a differentiable way. Note that whenstrict
isFalse
, the result can not require gradients or be disconnected from the inputs. Defaults toFalse
.strict (bool, optional) – If
True
, an error will be raised when we detect that there exists an input such that all the outputs are independent of it. IfFalse
, we return a Tensor of zeros as the hvp for said inputs, which is the expected mathematical value. Defaults toFalse
.
- Returns
- tuple with:
func_output (tuple of Tensors or Tensor): output of
func(inputs)
hvp (tuple of Tensors or Tensor): result of the dot product with the same shape as the inputs.
- Return type
output (tuple)
Example
>>> def pow_reducer(x): ... return x.pow(3).sum() >>> inputs = torch.rand(2, 2) >>> v = torch.ones(2, 2) >>> hvp(pow_reducer, inputs, v) (tensor(0.1448), tensor([[2.0239, 1.6456], [2.4988, 1.4310]]))
>>> hvp(pow_reducer, inputs, v, create_graph=True) (tensor(0.1448, grad_fn=<SumBackward0>), tensor([[2.0239, 1.6456], [2.4988, 1.4310]], grad_fn=<MulBackward0>))
>>> def pow_adder_reducer(x, y): ... return (2 * x.pow(2) + 3 * y.pow(2)).sum() >>> inputs = (torch.rand(2), torch.rand(2)) >>> v = (torch.zeros(2), torch.ones(2)) >>> hvp(pow_adder_reducer, inputs, v) (tensor(2.3030), (tensor([0., 0.]), tensor([6., 6.])))
Note
This function is significantly slower than vhp due to backward mode AD constraints. If your functions is twice continuously differentiable, then hvp = vhp.t(). So if you know that your function satisfies this condition, you should use vhp instead that is much faster with the current implementation.