What is the difference between 'SAME' and 'VALID' padding in tf.nn.max_pool of tensorflow?

If you like ascii art:

• "VALID" = without padding:

     inputs:         1  2  3  4  5  6  7  8  9  10 11 (12 13)
                    |________________|                dropped
                                   |_________________|
  

• "SAME" = with zero padding:

                 pad|                                      |pad
     inputs:      0 |1  2  3  4  5  6  7  8  9  10 11 12 13|0  0
                 |________________|
                                |_________________|
                                               |________________|
  

In this example:

• Input width = 13
• Filter width = 6
• Stride = 5

Notes:

• "VALID" only ever drops the right-most columns (or bottom-most rows).
• "SAME" tries to pad evenly left and right, but if the amount of columns to be added is odd, it will add the extra column to the right, as is the case in this example (the same logic applies vertically: there may be an extra row of zeros at the bottom).

Edit:

About the name:

• With "SAME" padding, if you use a stride of 1, the layer's outputs will have the same spatial dimensions as its inputs.
• With "VALID" padding, there's no "made-up" padding inputs. The layer only uses valid input data.

When stride is 1 (more typical with convolution than pooling), we can think of the following distinction:

• "SAME": output size is the same as input size. This requires the filter window to slip outside input map, hence the need to pad.
• "VALID": Filter window stays at valid position inside input map, so output size shrinks by filter_size - 1. No padding occurs.

I'll give an example to make it clearer:

• x: input image of shape [2, 3], 1 channel
• valid_pad: max pool with 2x2 kernel, stride 2 and VALID padding.
• same_pad: max pool with 2x2 kernel, stride 2 and SAME padding (this is the classic way to go)

The output shapes are:

• valid_pad: here, no padding so the output shape is [1, 1]
• same_pad: here, we pad the image to the shape [2, 4] (with -inf and then apply max pool), so the output shape is [1, 2]

x = tf.constant([[1., 2., 3.],
                 [4., 5., 6.]])

x = tf.reshape(x, [1, 2, 3, 1])  # give a shape accepted by tf.nn.max_pool

valid_pad = tf.nn.max_pool(x, [1, 2, 2, 1], [1, 2, 2, 1], padding='VALID')
same_pad = tf.nn.max_pool(x, [1, 2, 2, 1], [1, 2, 2, 1], padding='SAME')

valid_pad.get_shape() == [1, 1, 1, 1]  # valid_pad is [5.]
same_pad.get_shape() == [1, 1, 2, 1]   # same_pad is  [5., 6.]