Here are three slightly different notions of the “One More Discrete Logarithm Problem” (OMDL), formulated as games.

First, the classic notion:

$$ \boxed{ \begin{aligned} &\colorbox{#FBCFE8}{\large $\text{OMDL-Adaptive}_b$ }\cr \cr &x_0, \ldots, x_{n + 1} \xleftarrow{R} \mathbb{Z}/(q)\cr &X_0, \ldots, X_{n + 1} \gets x_i \cdot G\cr \cr &\underline{\texttt{Public()}:}\cr &\ \texttt{return } (X_0, \ldots, X_{n + 1})\cr \cr &\text{count} \gets 1\cr &\underline{\texttt{DLog(}i\texttt{)}:}\cr &\ \texttt{assert } \text{count++} < n + 1\cr &\ \texttt{return } x_i\cr \cr &\underline{\texttt{Challenge(}\hat{x}_0, \ldots, \hat{x}_{n + 1}\texttt{)}:}\cr &\ \texttt{return } b = 0 \land \forall i.\ \hat{x}_i = x_i\cr \end{aligned} } $$

In this game, the adversary is given $n + 1$ group elements, and can query for $n$ of the discrete logarithms, adaptively.

Another variant allows queries for elements of the adversary’s choice, but not adaptively. This game is identical, with the $\texttt{DLog}$ being the only change:

$$ \boxed{ \begin{aligned} &\colorbox{#FBCFE8}{\large $\text{OMDL-Dynamic}_b$ }\cr &\ldots \cr &\text{count} \gets 1\cr &\underline{\texttt{DLog(}\hat{i}\texttt{)}:}\cr &\ \texttt{assert } \text{count++} = 1\cr &\ \texttt{return } (x_i\ |\ i \neq \hat{i})\cr \end{aligned} } $$

Finally, you can also make it so that the adversary doesn’t get to choose which group elements they get the logarithm for:

$$ \boxed{ \begin{aligned} &\colorbox{#FBCFE8}{\large $\text{OMDL-Static}_b$ }\cr &\ldots \cr &\underline{\texttt{DLog(}\texttt{)}:}\cr &\ \texttt{return } (x_i\ |\ i \neq n + 1)\cr \end{aligned} } $$

Now, because each of these gives progressively weaker capabilities, we have:

$$ \text{OMDL-Static}_b \leq \text{OMDL-Dynamic}_b \leq \text{OMDL-Adapative}_b $$

This can be proved via straight-line reductions.

The other direction is harder.

$\text{OMDL-Dynamic}_b \leq \text{OMDL-Static}_b$

The idea behind this reduction is that if the adversary $\mathcal{A}$ for $\text{OMDL-Dynamic}$ happens to ask for $\hat{i} = n + 1$, you can satisfy their query immediately. At first, this seems like it would reduce your advantage by a factor of $n + 1$, since you need to get lucky.

The trick is that you can reset the adversary, and try again. The adversary may not ever pick $n + 1$ in their strategy, so instead you need to rearrange the group elements yourself, picking where you place the “hole” randomly, so that there’s always a $1 / (n + 1)$ chance of the adversary picking it.

Proof:

More formally, given this adversary $\mathcal{A}$ for $\text{OMDL-Dynamic}$, we construct an adversary $\mathcal{B}$ for $\text{OMDL-Static}$, as follows:

1. Query $\texttt{Public}$ and $\texttt{DLog}$ to learn $X_0, \ldots, X_{n + 1}$ and associated $x_0, \ldots, x_n$. Let $x_{n + 1} = \bot$.
2. Retry the following steps until they succeed (resetting any state modifications):
3. Choose random $r_0, \ldots, r_{n + 1} \xleftarrow{R} \mathbb{Z}/(q)$, and set $x_j \gets r_j \cdot x_j$, as well as $X_j \gets r_j \cdot X_j$.
4. Choose a random $j \xleftarrow{R} [1, \ldots, n + 1]$, and swap $X_j$ and $X_{n + 1}$, as well as $x_j$ and $x_{n + 1}$.
5. Run $\mathcal{A}$ using these group elements and scalars to answer their queries, dividing by $r_j$ before forwarding their answers to $\texttt{Challenge}$, and abort if they call $\texttt{DLog}$ with $\hat{i} \neq j$.
6. Return what $\mathcal{A}$ returned.

The expected number of retries is $n + 1$. This is because the behavior of $\mathcal{A}$ is completely independent of $j$, particularly because we randomize the group elements at each iteration, so that they’re indistinguishable from freshly sampled group elements. Thus, at each try, there’s at most an $n / (n + 1)$ chance that we have to abort when running $\mathcal{A}$.

$\square$

$\text{OMDL-Adaptive}_b \leq \text{OMDL-Static}_b$

It’s actually not any easier to prove $\text{OMDL-Adaptive} \leq \text{OMDL-Dynamic}$, so we’ll prove this one instead.

The basic idea of the reduction is the same, it’s just that the analysis is more complicated. We re-randomize and shuffle the group elements as in the previous reduction, and we abort the adversary $\mathcal{A}$ if it ever queries the group element we have no scalar for. Analyzing this is trickier because of the adaptivity.

The idea is that given our choices of group elements $X_0, \ldots, X_{n + 1}$, we can look at the sequence of queries made by $\mathcal{A}$, if they get the answer each time. This defines a sequence of random variables $i_0, \ldots, i_n$, which depends solely on $X_i$. We can also define a related random variable, $\hat{i}$, represented the index which isn’t queried. This is a function of $i_0, \ldots, i_n$, and so depends solely on our choice of $X_i$.

In particular, $\hat{i}$ is independent from the random $j$ we choose at each iteration, because of the re-randomization, and so the analysis from the previous reduction applies exactly.

$\square$