Suppose $X_1,\ldots,X_n$ is an i.i.d. sample from a probability distribution with continuous c.d.f. $F.$ Let $F_n$ be the empirical c.d.f.
$$
F_n(x) = \frac 1 n \sum_{k=1}^n \mathbf 1_{X_n\le x} = \frac{\text{how many of }X_1,\ldots,X_n \text{ are} \le x} n.
$$
Let
\begin{align}
D_n & = \sup \left\{ |F(x) - F_n(x)| : x\in\mathbb R \right\} \tag 1 \\[4pt]
& = \text{the maximum discrepancy statistic.}
\end{align}
The probability distribution of $D_n$ does not depend on which continuous c.d.f. $F$ is involved. This is the **Kolmogorov–Smirnov distribution.**

Now suppose $X_1\sim N(\mu,\sigma^2)$ and they're still i.i.d. and instead of $F$ in line $(1)$ above, you put
\begin{align}
\widehat F_n = {} & \text{the c.d.f. of the normal} \\[4pt] & \text{distribution with expectation} \\[6pt]
& \overline X = (X_1+\cdots+X_n)/n \\[6pt]
& \text{and variance } \frac 1 {n-1} \sum_{k=1}^n (X_i -\overline X)^2.
\end{align}
The maximum discrepancy statistic then becomes stochastically smaller. Its distribution becomes the **Lilliefors distribution.**

Wikipedia says that values of the c.d.f. of the Lilliefors distribution have been computed only by Monte Carlo methods.

Is there some proof that no deterministic numerical method can do this efficiently?