(Background: For ranked pairs voting when all voters rank all candidates, pairs of candidates $(A, B)$ are sorted by the margin of victory between the two candidates (that is, if $x$ voters prefer $A$ to $B$, and $y$ voters prefer $B$ to $A$, then we sort by $x-y$), and then these sorted pairs are added one by one to a directed graph, drawing an arrow from the winner to the loser in each pair, skipping any pairs that would create a cycle. This produces a ranking of candidates from the source to the sink of the graph.)
I'm working on a variation of this method with two additional complications: firstly, voters are not required to rank all candidates, and secondly, it cannot be assumed that a voter prefers a candidate that they ranked over a candidate that they did not rank. Instead of using $x-y$ as the metric by which the pairs should be sorted, I need a different function with the following properties (which are hopefully not contradictory). "$(A, B) > (C, D)$" means that pair $(A, B)$ comes before pair $(C, D)$ in the final sorting (where $A$, $B$, $C$, and $D$ are not necessarily distinct).
Multiplying by a constant greater than $1$ results in a more decisive victory or defeat, that is, a $10-5$ victory would be ranked above a $2-1$ victory, but a $1-2$ defeat would be ranked above a $5-10$ defeat.
Adding a constant greater than $0$ results in a less decisive victory or defeat, that is, a $10-1$ victory would be ranked above a $100-91$ victory, but a $91-100$ defeat would be ranked above a $1-10$ defeat.
A unanimous winning pair (that is, a pair $(A, B)$ where all candidates who ranked both $A$ and $B$ preferred $A$ to $B$) must be ranked above all other non-unanimous pairs, that is, a $1-0$ victory would be ranked above a $100-1$ victory.
If $(A, B) > (C, D)$, then $(D, C) > (B, A)$. That is, the sorted list of pairs should read the same backwards and forwards excluding ties between pairs, so if a $7-3$ victory is ranked above a $5-2$ victory, then a $2-5$ defeat must be ranked above a $3-7$ defeat.
All pairs involving tied candidates are equal. A $1-1$ tie is the exact same as a $5-5$ tie.
All other seemingly intuitive properties about how the pairs should be ranked.
I know this is a bizarre question and not at all like the usual questions asked on this site, so even if you can't give a definite answer, any insight at all would be hugely appreciated.