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I have to write a report on "Classical Interaction Cannot Replace a Quantum Message" by Gavinsky (arxiv link).

The author gave a lecture, and at 14:55 explains that a communication problem exists where Alice has input $x$, Bob has input $y$ and together they produce an output $z$. This problem is relational so every $x,y$ there are multiple output $z$. In the video he explains that a table(galois field) exists in the communication channel which stores the $x$ and $y$ and then calculates $z$ by taking the bitwise XOR and giving orthogonal bits to the resultant.

After watching that part of the video and reading through some of the paper,

I also want to know if Alice and Bob have a fixed number of rows and columns dedicated to them for calculating the output, and are the number of elements in each cell fixed or variable?

How are the elements placed in the table?

These definitions are provided in the paper on page 3:

Definition 1. Let $x, y⊂[n^2]$, such that $|x|=n/2$ and $|y|=n$. Let $z \in \mathcal{G} \mathcal{F}_{2}^{2 \log n} \backslash\{\overline{0}\}$. Let $\Sigma=\left\{\sigma_{2^{2 i}}\right\}_{i=1}^{\infty}$ be a set of reversible mappings from [2$^{2i}$] to GF$^{2i}_2$. Then $(x, y, z) ∈ P^Σ_{1×1}$ if either $|x \cap y| \neq 2$ or〈z, a+b〉= 0, where $\sigma_{n^{2}}(x \cap y)=\{a, b\}$.

Let $\Sigma_{0}$ be the set of reversible mappings from $\left[2^{2 i}\right] \text { to } \mathcal{G} \mathcal{F}_{2}^{2 i}$, preserving the lexicographic ordering of the elements.

Definition 2. Let $x \subset\left[2 n^{2}\right],|x|=n$. Let $y = (y_1, . . . , y_{n/4})$ be a tuple of disjoint subsets of [n$^{2}$], each of size $n$, such that $\left|x \cap y_{j}\right|=2$ for all $1 \leq j \leq n / 4$. Let $z \in \mathcal{G} \mathcal{F}_{2}^{2 \log n+1} \backslash\{\overline{0}\}$ and $1 \leq i \leq n / 4$, then $(x, y,(i, z)) \in P^{(n)} \text { if }\langle z, a+b\rangle=0$, where $\sigma_{0}\left(x \cap y_{i}\right)=\{a, b\}$ for some $\sigma_{0} \in \Sigma_{0}$.

In the rest of the paper we will implicitly assume equivalence between the arguments and the corresponding values of every $\sigma_{0} \in \Sigma_{0}$. We will show that P is easy to solve in Q$^{1}$ and is hard for R. In order to prove the lower bound we will use the following modification of $P_{1 \times 1}^{\Sigma}$.

Definition 3. Let $x$, $y⊂[n^2]$, such that $|x|=n/2$ and $|y|=n$. Let $z\subset[n^2]$. Then $(x, y, z) ∈ P^{search}_{1×1}$ if $x∩y=z$.

1Please consider editing the question to be as self-contained as possible. Please summarize the paper, and also the video, to make it easier on the readers. Also, please link to the abstract, and not the PDF. The title of the question looks interesting, but it’s difficult to have to hunt down and download PDFs and watch videos. – Mark S – 2021-02-03T03:14:46.227

1I have made the changes. But since I can not include all the things from the PDF I have included the definitions from it. And I have explained the basic communication problem from the video as well. so it is not necessary to watch it, but leaving the link for those who would want to see it. – Asjad Sohail – 2021-02-03T13:45:11.870

@AsjadSohail you might also want to have a look at https://quantumcomputing.meta.stackexchange.com/q/49/55 for how to write math in the site

– glS – 2021-02-03T14:54:26.1932I've made all the required changes. I hope it is easier to understand now. – Asjad Sohail – 2021-02-03T16:13:47.590