# Capacity bits in SHA-3

I am trying to understand the sponge construction, and in the official web site I found this sentence:

The last c bits of the state are never directly affected by the input blocks and are never output during the squeezing phase.

Now I understood that the capacity is used to hide the internal state to an attacker. But I can not understand two aspects:

1. How is the capacity filled?
2. If it is part of the internal state, how can be possible that is not affected by the input blocks?

The capacity $$c$$, together with the rate $$r$$, is the state of the algorithm between the applications of the sponge.
It is of course affected by the message (there is no other input after all) but not directly: the input first has to go through the sponge function, $$f$$, after it is XOR'red with the rate, $$r$$. Similarly, only the rate $$r$$ is output during the squeezing. So the previous capacity $$c$$ is similarly protected by a sponge. Note that there is always an $$f$$ between the input and output (of course).