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Alice and Bob fix a largish $n$, say $n = 1000$, and they publish a simple, undirected graph $G$ on $n$ vertices (more precisely, its adjacency matrix, or a similar form of representation). Moreover they publicly agree on some hash function $h:\{0,1\}^*\to \{0,1\}^{1024}$ (or some other constant number of output bits).

They also publicly agree on a large commutative subgroup of $U\subseteq S_n$, the symmetric group of permutations $\varphi:\{1,\ldots,n\}\to \{1,\ldots,n\}$. Then:

  1. Alice and Bob privately choose members of $U$, say $\varphi$ and $\psi$, respectively.

  2. Alice publishes the adjacency graph of $\varphi(G)$, and Bob does the same with $\psi(G)$. It is computationally infeasible to compute $\varphi,\psi$ given $G, \varphi(G), \psi(G)$.

  3. Alice secretly computes $G_0=\varphi(\psi(G))$, Bob secretly computes $G_0 = \psi(\varphi(G))$ -- and they end up with the adjacency matrix $M_0$ of the same graph, since $\varphi,\psi\in U$ commute.

  4. Each computes $h(M_0)$, and so they end up with a common $1024$-bit key.

Problems with this approach:

  1. Identifying a large and good commutative subgroup $U\subseteq S_n$.

  2. The object exchanged, that is $\varphi(G), \psi(G)$ have the size ${\cal O}(n^2)$ if $n$ is the number of vertices used for the graph.

Question. Has such a key exchange system been tried / implemented?

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    $\begingroup$ What are you trying to do is this; take a large array (of size $1000^2$) randomly fill to 0s and 1s. and generate a permutation on this array as your public key. A Graph is redundant here. Instead, we choose a group where discrete logarithm is hard ( $g^x$ is a permutation in some settings). $\endgroup$
    – kelalaka
    Nov 7, 2023 at 12:36

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