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This is from Justin Thaler's book - Proofs, Arguments & Zero Knowledge

Page 43

For it to make sense to talk about multilinear extensions, we need to view the adjacency matrix $A$ not as a matrix, but rather as a function $f_A$ mapping $\lbrace0,1\rbrace^ {log n} \times \lbrace0,1\rbrace^ {log n}$ to $\lbrace0,1\rbrace$. The natural way to do this is to define $f_A(x,y)$ so that it interprets $x$ and $y$ as the binary representations of some integers $i$ and $j$ between $1$ and $n$, and outputs $A_{i, j}$. See Figure $4.5$ for an example.

This is figure 4.5

enter image description here

(I have shaded in yellow the element which I am going use below as an example).

The 4 parameters of function $f_A$ seem to be the bit representation of the element number of the matrix - i.e if I consider the matrix elements having numbers $0$ to $15$ (Considering element numbers of first row as $0, 1, 2$ & $3$ & so on)

For e.g. if I consider the element $6$ in row number $4$ & column number $2$- it's the $13$th element of the Matrix. The bit pattern of $13$ is $1101$.

So $f_A$ takes 4 parameters same as the bit pattern - i.e. $f_A(1,1,0,1) = 6$

I hope my understanding above is correct.

However, right after, on Page 44, he says counts triangles using the following equation

$\Delta = \frac {1}{6} \sum_{{x,y,z \in \lbrace 0, 1 \rbrace}^{log\space n}} f_A(x,y)\cdot f_A(y,z) \cdot fA(x,z)$

I assume above that the parameters to $f_A$ are row & column numbers.

In figure $4.5$, the parameters of $f_A$ is the bit pattern of the element number of the matrix. However, above it changes to row & column numbers. How does this match with figure $4.5$?

In the particular case of a $4 \times 4$ matrix the element number bit pattern is the same as bit pattern of row number joined with bit pattern of column number. But this isn't true for others - i.e. not true even for $5\times 5$?

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1 Answer 1

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In figure 4.5, the parameters of $f_A$ is the bit pattern of the element number of the matrix. However, above it changes to row & column numbers. How does this match with figure 4.5?

The inputs to $f_A$ are always a pair consisting of a row and column. If the book wanted to be extremely precise, they would have used the following notation, for example $$f_A((1,1),(0,1)) = 6.$$ But that can hurt readability.

But this isn't true for others - i.e. not true even for 5×5 ?

True. The example simplifies to make it work nicely. But for the general case, I believe the "fix" is to arbitrarily define the values of $f_A$ at inputs that aren't valid indices. Then, one can use something akin to a selector polynomial in the specific application (that evaluates to zero, for example, in a counting application, etc..).

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  • $\begingroup$ fA((1,1),(0,1))=6. with such a definition, how exactly would you use Lemma 3.6 (Page 29) to interpolate & find the polynomial which would represent the matrix? $\endgroup$
    – user93353
    Commented Nov 13, 2023 at 14:08
  • $\begingroup$ There's a 1-1 mapping from $((w,x),(y,z))$ to $(w, x ,y ,z)$. So you can define an intermediary function that "flattens" the tuple before doing the interpolation. At this point, it's mainly a question of presentation. $\endgroup$ Commented Nov 13, 2023 at 14:13
  • $\begingroup$ Thank you. So the sumcheck also instead of $\sum_{{x,y,z \in \lbrace 0, 1 \rbrace}^{log\space n}} f_A(x,y)\cdot f_A(y,z) \cdot fA(x,z)$ would also become something like $\sum_{x_1 \in \lbrace 0, 1 \rbrace} ... \sum_{y_1 \in \lbrace 0, 1 \rbrace} ... \sum_{z_1 \in \lbrace 0, 1 \rbrace}... f_A(x_1, x_2, ..., y_1, y_2, ..., z_1, z_2, ...)$ $\endgroup$
    – user93353
    Commented Nov 14, 2023 at 10:40
  • $\begingroup$ You can certainly define things such that the expression you give works. But I suppose the "ultimate" implementation will still be something like $f_A(x_1,\cdots, y_1, \cdots) f_A(y_1,\cdots, z_1, \cdots) f_A(z_1,\cdots, x_1, \cdots) $. The reason being that the running time for the MLE evaluation is basically exponential in the number of variables in $f_A$. But surely, the long expression you can be used either directly or as "syntactic sugar" for the more efficient implementation. $\endgroup$ Commented Nov 14, 2023 at 13:34

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