# Background: Secret Sharing and Multiplication Triples

I know that a nearly identical question was asked in (Secret Sharing and Multiplication Triples), but I am a few points shy of asking follow up questions in the comments, so I understand if this is closed soon afterward.

First and foremost, just to clarify, the use of a multiplicative triple in this secret sharing problem is solely as a vehicle for the sharing no? I cannot see any other motivation for the triple other than facilitating the protocol.

Second, in @Guut boy's comment he mentions that the previous asker was mistaking the "notation for XOR'ing with a public value", but I don't see how this is true. The previous asker seems to imply that: $d_a = x_a \oplus a$ and $d_b = x_b$, similarly for $e$. If this is so then the XORing with the public value $a$ was done correctly and I still get the extra $ed$ at the end of the calculation because: $$z_a = c_a \oplus ex_a \oplus dy_a \oplus ed$$ $$z_b = c_b \oplus ex_b \oplus dy_b \oplus ed$$ because $ed$ is not defined in $[z]$ with XOR notation brackets it is given as $$[z] = [c] \oplus e[x] \oplus d[y] \oplus ed$$

Basically, I don't understand why the last $ed$ still exists any help is appreciated and I apologize in advance for the follow up to a previously asked question.

For the second question. The mistake was exactly that the asker did the XOR as $$d_a = x_a \oplus a \\ d_b = x_b \oplus a$$ But the correct way is $$d_a = x_a \oplus a\\ d_b = x_b$$ Thus to compute the shares for $$[z] = [c]\oplus e[x]\oplus d[y]\oplus ed$$ You would set $$z_a = c_a \oplus ex_A \oplus dy_a \oplus ed\\ z_b = c_b \oplus ex_b \oplus dy_b$$ You can check that this gives the expected result.
I am not sure what you first question is about. The purpose of the having secret shared a random multiplication triple $[a],[b],[c]$ is solely to use it later in this protocol to perform a multiplication operation (in this case AND). Not sure if this answers you question.