Is there a better-than-brute-force algorithm to generate a graph whose relation is string edit distance=1?

It's possible to use BK-Trees to speed this up. Inserting $n$ elements into the tree takes $O(n log n)$ time. After this you can query the tree for all strings whose edit distance is exactly one away from your input. Doing this for all strings again takes $O(n^2)$ complexity, however with a very small constant factor (this page mentions only 5-8% of the tree needs to be inspected per query).

Here's a short description of how it works:

Data structure

A BK-Tree is either

• Empty
• A node with a root value $r$ and a set of indexed children $c_i$, each a BK-Tree (implemented using hash map, dynamic array, or similar)

A metric (important!) distance function $d(x,y)$ is used for insertion and queries.

Inserting string $s$

• If the tree is empty, make it a new node with $s$ as the root value and no children
• If the tree is a node with root $r$ and children $c_i$, let $k=d(r,s)$.
  • If $k=0$, skip insertion as $s$ is already at the root
  • Otherwise recursively insert $s$ into the $k$-th child tree $c_k$.

The last step makes sure that all nodes in $c_i$ have distance $i$ to the root

Querying string $s$

• If the tree is empty, return no results
• If the tree is a node with root $r$ and children $c_i$, let $k=d(r,s)$.
  • If $k=1$, add $r$ as a result (Note: This step is different from usual BK-Tree queries)
  • In addition, recursively query $s$ from the children $c_{d-1}$, $c_d$ and $c_{d+1}$. Return all results from those queries as well

The last step there is the magic of BK-Trees, as it doesn't need to check the other children due to the distance being metric. Here's a partial proof of why others don't need to be checked:

Let's assume there was some string $x$ which has distance one from our query, so $d(s, x)=1$, but it's in the child tree $c_{k+2}$. We know that $d(r, x)=k+2$ from the insertion procedure. This however (with the triangle inequality for metric spaces) gives

$$ k+2=d(r, x)leq d(r, s)+d(s,x)=k+1 $$

But this is a contradiction! Similar can be done for all children with $i>k+1$ and $i<k-1$. This means that all strings in other children won't have distance one by construction, so we don't need to even check them, which saves processing time.

Edit: Another explanation: https://signal-to-noise.xyz/post/bk-tree/

In the worst case any such algorithm will work $Omega(n^2)$ because your graph can have $Omega(n^2)$ edges.

By the way, are you interested in some particular string metric?