Why does tf.keras.layers.Conv2DTranspose need no output_shape compared to tf.nn.conv2d_transpose?

Question

I am missing something basic here. But I always used the tf.nn API for transpose convolution, where I have to specify the output shape, because it is ambiguous(Link). However, for TF 2.0 I switch to Keras sub-classing and now I wonder why I do not have to specify the output shape for transpose convolution in the higher level API. How to they compute it exactly?

Answer 1

tf.keras.layers.Conv2DTranpose backends to tf.nn.conv2d_transpose via tf.keras.backend.conv2d_transpose.

To compute the output_shape argument for tf.nn.conv2d_transpose it utilizes the function deconv_output_length (defined here):

def deconv_output_length(input_length, filter_size, padding, stride):
  """Determines output length of a transposed convolution given input length.
  Arguments:
      input_length: integer.
      filter_size: integer.
      padding: one of "same", "valid", "full".
      stride: integer.
  Returns:
      The output length (integer).
  """
  if input_length is None:
    return None
  input_length *= stride
  if padding == 'valid':
    input_length += max(filter_size - stride, 0)
  elif padding == 'full':
    input_length -= (stride + filter_size - 2)
  return input_length