I am currently reading about Trapdoor Permutations (TDP). While I can totally understand and think of examples of TDP. I cannot think of any examples of Enhanced TDP. The definition of both TDP and Enhanced TDP is given below :

**Standard Trapdoor Permutations Collection** : It is a collection of finite permutations, denoted $\left\{f_{\alpha}: D_{\alpha} \rightarrow\right.$ $\left.D_{\alpha}\right\}$, accompanied by four probabilistic polynomial-time algorithms, denoted $I, S, F$ and $B$ (for index, sample, forward and backward), such that the following (syntactic) conditions hold:

- On input $1^{n}$, algorithm $I$ selects at random an $n$-bit long index $\alpha$ (not necessarily uniformly) of a permutation $f_{\alpha}$, along with a corresponding trapdoor $\tau$;
- On input $\alpha$, algorithm $S$ samples the domain of $f_{\alpha}$, returning an almost uniformly distributed element in it;
- For any $x$ in the domain of $f_{\alpha}$, given $\alpha$ and $x$, algorithm $F$ returns $f_{\alpha}(x)$ (i.e., $\left.F(\alpha, x)=f_{\alpha}(x)\right)$;
- For any $y$ in the range of $f_{\alpha}$ if $(\alpha, \tau)$ is a possible output of $I\left(1^{n}\right)$, then, given $\tau$ and $y$, algorithm $B$ returns $f_{\alpha}^{-1}(y)$ (i.e., $\left.B(\tau, y)=f_{\alpha}^{-1}(y)\right)$.

Let $I_{1}\left(1^{n}\right)$ denote the first element in the output of $I\left(1^{n}\right)$ (i.e., the index), it is required that, for every probabilistic polynomial-time algorithm $A$ (resp., every non-uniform family of polynomial-size circuits $A=\left\{A_{n}\right\}_{n}$ ), we have
$$
\underset{\substack{\alpha \leftarrow I_{1}\left(1^{n}\right) \\ x \leftarrow S(\alpha)}}{\operatorname{Pr}}\left[A\left(\alpha, f_{\alpha}(x)\right)=x\right]=\mu(n),
$$

or equivalently

$$
\underset{\substack{\alpha \leftarrow I_{1}\left(1^{n}\right) \\ r \leftarrow R_{n}}}{\operatorname{Pr}}\left[A(\alpha, S(\alpha ; r))=f_{\alpha}^{-1}(S(\alpha ; r))\right]=\mu(n)
$$

**Enchanced Trapdoor Permutations Collection** : The only thing that changes it the last condition, where the adversary has access to the random coins of $S$
$$
\underset{\substack{\alpha \leftarrow I_{1}\left(1^{n}\right) \\ r \leftarrow R_n}}{\operatorname{Pr}}\left[A(\alpha, r)=f_{\alpha}^{-1}(S(\alpha ; r))\right]=\mu(n),
$$