DLS Course 4 Week 2 Exercise 1: 1x1 convolution with strides=2

[Huang_Erdong]

Under the function convolutional_block(), in the first component of the main path, we are doing 1x1 convolution with strides=2.
This really confuses (and annoys) me.
Why are we looking at only 1/2 * 1/2 = 1/4 of activations of the input layers?
Why are we discarding 3/4 of the activations (neurons) from the previous layer, without examining them?
If they are really unnecessary, then we can just make the input layer smaller in size.
Otherwise, we can have used a max pooling to reduce the size, instead of just discarding them anyway.

[TMosh]

Were you able to find an answer to this question?

[Huang_Erdong]

Not yet. I am waiting for an answer here.

[Jerome_Griffon]

You can find some answers in this discussion:

Course 4 week 2: Residual Networks - kernel size=1 stride = 2

In the residual network, for the first filter of the convolutional block we use: kernel size = (1,1) and stride = (s,s), where s is usually 2. Doesn’t that mean that we just throw away 75% of the input information? That doesn’t seem to make sense to me. What I would expect: the first and third filter to have kernel size(1,1) and stride (1,1), and the second filter to have kernel size (f,f) and stride(s,s). So that we don’t throw away information. Could someone help me to understand?