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I am reading about GKR protocol from Justin Thaler's book - Proofs, Arguments & Zero Knowledge

Section 4.6.5 - Page 64 - Description of GKR Protocol

$S_0$ is the number of gates in layer 0.

$k_0 = \log(S_0) $

$V$ picks a random $r_0 \in F^{k_0}$ and lets $m_0 = \tilde D(r_0)$. The remainder of the protocol is devoted to confirming that $m_0 = \tilde W_0(r_0)$

Now $k_0$ is the maximum number of bits in the gate label/gate number - $\tilde W_0$ is $k_0$-variate Polynomial.

So now when $r_0$ is picked, how does $P$ split it in order to evaluate $W_0$ - i.e. if say $k_0 = 4$, then $r_0$ would need to be decomposed into 4 values so that you can pass those as $x_1, x_2, x_3, x_4$. Or am I mistaken here? So how is this decomposition done - on what boundaries? Or is the technique different?

Likewise, in the Sum-check protocol in section 4.3, Counting Triangles.

Now though the book represents $f_A$ as taking 2 params $f_A(X,Y)$ as per this answer, this is only notational & based on the dimensions of the adjacency matrix, $f_A$ is actually $f_A(x_1, x_2, x_3, ...)$

So on Page 45, when 3 random values are picked & $r_1, r_2, r_3$, $\tilde f_A(r_3,r_3),\tilde f_A(r_1,r_3)$ evaluated, how is each of $r_1, r_2, r_3$ decomposed into individual $x_1, x_2 ... $ etc?

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$W$ is a function from $\{0,1\}^{k_0}$ to $\mathbb{F}$, but both $\tilde{W}$ and $\tilde{D}$ are functions from $\mathbb{F}^{k_0}$ to $\mathbb{F}$. The text says that $r_0$ belongs to $\mathbb{F}^{k_0}$, thus, it is in the domain of both $\tilde{W}$ and $\tilde{D}$, therefore, the evaluation $\tilde{D}(r_0)$ is already well-defined and no decomposition or extra technique is needed.

You would need a "decomposition" or some sort of "extension" if $r_0$ were an element of $\mathbb{F}$ instead of a $k_0$-dimensional vector of elements of this field.

As for the functions $f_A$ and $\tilde{f_A}$, you have to interpret $X$ and $Y$ as vectors as well.

Notice that saying that a function $F$ has as input two elements (vectors) of $\{0,1\}^k$ or that it has as input $2k$ elements of $\{0,1\}$ is basically the same.

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  • $\begingroup$ $F$ is a function takes input from $\lbrace 0, 1 \rbrace^{k_0}$. However, $\tilde F$ is an extension. It takes any value from $\mathbb F$, right? i.e. it's no longer limited to $\lbrace 0, 1 \rbrace^{k_0}$ - you are picking $r_1$ from $\mathbb F$ & not from $\lbrace 0, 1 \rbrace^{k_0}$. So after picking $r_1$, don't you have to decompose it into a vector? Or am I wrong? $\endgroup$
    – user93353
    Commented Apr 21 at 2:49
  • $\begingroup$ If $k_0 = 3$, then $\tilde W$ is 3-variate polynomial. When you pick $r_0$, how to get $x_1,x_2,x_3$ to pass to $\tilde W?$ Likewise for $\tilde{f_A}$, when you pick $r_1, r_2,r_3$, you need to do decompose $r_1$ into multiple elements to form the vector $X$ & likewise for $r_2, r_3$ & $Y$ & $Z$ respectively, right? $\endgroup$
    – user93353
    Commented Apr 21 at 2:55
  • $\begingroup$ The reason it's called a low degree extension is because the domain of the MLE is much bigger than $\lbrace 0,1\rbrace^{k_0}$ - i.e. though the map is from $\lbrace 0,1\rbrace^{k_0}$, the MLE's domain is $\mathbb F^v$ - i.e. each of the $v$ variables can take any value from $\mathbb F$ $\endgroup$
    – user93353
    Commented Apr 21 at 3:05
  • $\begingroup$ No, $\tilde{F}$ does not take a value from $\mathbb{F}$. It is a multivariate polynomial over $\mathbb{F}$, this means it takes $k_0$ variables from $\mathbb{F}$, in other words, its input is a vector from $\mathbb{F}^{k_0}$, that is why in page 64 it is written $r_0 \in \mathbb{F}^{k_0}$, instead of $r_0 \in \mathbb{F}$ $\endgroup$ Commented Apr 21 at 10:29
  • $\begingroup$ So you mean, when it says pick $r_0$, it means, pick $k_0$ random numbers $\in F$. And likewise in Counting Triangles, when it says pick Random $r_1, r_2, r_3$, it means pick $3*v$ random numbers? $\endgroup$
    – user93353
    Commented Apr 21 at 10:53

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