The number of constraints is then equal to the number of rows in the matrix $P$ which is the number of rows in $D$ + the number of rows in $B$ + 3 times the number of rows in $G$. That is $P$ has $5n = 270$ which is near to $2^8$ which would lead to proofs of around 47 kiB.
In order to fully understand Greyhound, we need to understand deeply LaBRADOR because it’s the heart of the PCS. LaBRADOR stands for:
Labrador consists of proving this relation:
For greyhound, the first sum isn’t interesting but we can represent our last required check with what remains:
By picking the $\varphi_i$ such that they represent the matrix $P$ of Greyhound and the vector $b$ with the vector $z$ of Greyhound, we could prove the Greyhound relation $Pz=h$ without communicating $z$, drastically reducing the proof size.
So let’s keep the same notation as in Greyhound! (The LaBRADOR papers recomment tu set $r = N^{1/3}$ and $n = N^{2/3}$.
First the prover commit to $z$ with an Ajtai commitment (because $z$ was made of small norm using $G$).
The prover knows witness that we want to prove. In Greyhound, it’s the single column vector $z$ composed of elements of $R_q$ It is plit in $r$ (not the same as in Greyhound) vectors each composed of $n$ elements. On the right side, the verifier knows the $\varphi$ that represent the matrix $P$ in Greyhound, the matrix composed of $a_{i,j}$ that are all equal to $0$ in Greyhound and the vector $b$ corresponding to $-h$ from Greyhound.