Given a knot diagram $K$, we get a corresponding set $\mathrm{C}(K)$ of crossings of $K$. Also, to each subset $x$ of $\mathrm{C}(K)$, we can assign a natural number $f(x)$ by defining $f(x)$ to be the number of copies of the unknot we get when we take the overstrand left at every crossing that appears in $x$, and take it right for every crossing that doesn't appear in $x$. This is extremely vague, but if you know how to compute the bracket polynomial you probably know what I mean. Please comment if my meaning is unclear. In summary, each knot diagram $K$ yields a set $\mathrm{C}(K)$ and a function $f : \mathcal{P}(\mathrm{C}(K)) \rightarrow \mathbb{N}$, where $\mathcal{P}$ is the powerset operator.
Question. Does the pair $(\mathrm{C}(K),f)$ allow us to recover the knot diagram $K$?
The pair $(C(K),f)$ is not enough to recover the knot diagram $K$. Mutation on a knot diagram will preserve the function $f$, but not the knot diagram (or even the knot type).
As a specific example, consider the two-component links below.
The second of these links is obtained from the first by performing a mutation inside of the red circle. These two diagrams are different even up to planar isotopy because the first contains a component with no self-crossings and the second does not. It turns out that these are diagrams of the same link, but in the interest of keeping the number of crossings small, I chose this example.
The set $C(K)$ here is of size $2^6=64$. If my back of the envelope computations are correct, then the following chart displays the behavior of $f$ for both diagrams: $$\begin{array}{c | c} \#~\text{subsets of crossings} & \#~\text{components}\\ \hline 12 & 1\\ 27 & 2\\ 20 & 3\\ 5 & 4 \end{array}.$$