Associativity in Braided Monoidal Categories with String Diagrams In A survey of graphical languages for monoidal categories, Peter Selinger gives a graphical language for braided monoidal categories by using string diagrams.
The following equation of braided monoidal categories :
$$(id_B \otimes c_{A,C}) \circ \alpha_{B,A,C} \circ (c_{A,B} \otimes id_C) = \alpha _{B,C,A} \circ (c_{B,C \otimes A}) \circ \alpha_{A,B,C}$$
where $$\alpha_{A,B,C} : (A \otimes B) \otimes C \rightarrow A \otimes (B \otimes C)$$
and the braiding is
$$c_{A,B} : A \otimes B \rightarrow B \otimes A$$
is represented as the following string diagram :

And moreover, Paul-André Meliès use these diagrams for (I think) the same thing in his Categorical Semantics of Linear Logic (Page 68) :

What I don't understand :
1) I see that the braiding is represented as an exchange/crossing of strings but I don't see how the associator $\alpha$ is represented. Is it invisible ?
If it is invisible, can string diagrams only be used for "structures with an associative operator" ?
2) In the diagram of Melliès, why are the strings for $B$ and $C$ "stucked to each other" on the left side of the $=$ and then "untangled" in the right side ? 
 A: 
1) I see that the braiding is represented as an exchange/crossing of strings but I don't see how the associator $\alpha$ is represented. Is it invisible ?

Yes, the associator is invisible. A justification I have heard for this is that every monoidal category is equivalent to one where the associator is the identity.

If it is invisible, can string diagrams only be used for "structures with an associative operator" ?

I would think so, yes.

2) In the diagram of Melliès, why are the strings for $B$ and $C$ "stucked to each other" on the left side of the $=$ and then "untangled" in the right side ? 

My impression is that the "stuck together" strings are like parentheses. They indicate that the braiding operator is swapping two strings together. The right-hand side of the equation does not have any stuck strings because the braiding operator is only interchanging single adjacent strands. Without those strands merged together, it would be harder (though probably not impossible? depends on how you draw it) to tell that the swap is moving both strands at the same time. But that's the difference between the two diagrams. On the right-hand side you swap one strand, then the other. On the left-hand side, you do two in one go.
