Capacity bits in SHA-3

Question

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?

Answer 1

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).

The capacity and the rate are first initialized to zero.