Simulating draws from a Uniform Distribution using draws from a Normal Distribution I recently purchased a data science interview resource in which one of the probability questions was as follows:

Given draws from a normal distribution with known parameters, how can you simulate draws from a uniform distribution?

My original thought process was that, for a discrete random variable, we could break the normal distribution into K unique subsections where each subsection has an equal area under the normal curve. Then we could determine which of K values the variable takes by recognizing which area of the normal curve the variable ends up falling into. 
But this would only work for discrete random variables. I did some research into how we might do the same for continuous random variables, but unfortunately I could only find techniques like inverse transform sampling that would use as input a uniform random variable, and could output random variables from some other distribution. I was thinking that perhaps we could do this process in reverse to get uniform random variables?
I also thought about possibly using the Normal random variables as inputs into a linear congruential generator, but I'm not sure if this would work.
Any thoughts on how I might approach this question?
 A: Adding on to 5:
The trick of transforming random variables to bits works for any independent pair of absolutely continuous random variables X and Y, even if X and Y are dependent on each other or the two variables are not identically distributed, as long as the two variables are statistically indifferent (Montes Gutiérrez 2014, De Schuymer et al. 2003); equivalently, their probabilistic index (Acion et al. 2006) is 1/2 or their net benefit is 0. This means in our case that P(X < Y) = P(X > Y). In particular, two independent normal random variables X and Y are statistically indifferent if and only if they have the same mean (they remain so even if their standard deviations differ) (Montes Gutiérrez 2014).
In this case, to generate unbiased random bits this way, sample an independent pair of statistically indifferent, absolutely continuous random variates, X and Y, and compare them. If X is less than Y, output 1; otherwise, output 0 (Morina et al. 2019). Because they are statistically indifferent, this method will output 1 or 0 with equal probability (see the appendix in my Note on Randomness Extraction).
(Note that statistical indifference also makes sense for discrete and singular random variables, but in this case an additional rejection step will be necessary if X and Y turn out to be equal [Morina et al. 2019].)
REFERENCES:

*

*Montes Gutiérrez, I., "Comparison of alternatives under uncertainty and imprecision", doctoral thesis, Universidad de Oviedo, 2014.

*De Schuymer, Bart, Hans De Meyer, and Bernard De Baets. "A fuzzy approach to stochastic dominance of random variables", in International Fuzzy Systems Association World Congress 2003.

*Morina, G., Łatuszyński, K., et al., "From the Bernoulli Factory to a Dice Enterprise via Perfect Sampling of Markov Chains", arXiv:1912.09229 [math.PR], 2019.

*Acion, Laura, John J. Peterson, Scott Temple, and Stephan Arndt. "Probabilistic index: an intuitive non‐parametric approach to measuring the size of treatment effects." Statistics in medicine 25, no. 4 (2006): 591-602.

A: You can use a trick very similar to what you mention. Let's say that $X \sim N(\mu, \sigma^2)$ is a normal random variable with known parameters. Then we know its distribution function, $\Phi_{\mu,\sigma^2}$, and $\Phi_{\mu,\sigma^2}(X)$ will be uniformly distributed on $(0,1)$. To prove this, note that for $d \in (0,1)$ we see that
$P(\Phi_{\mu,\sigma^2}(X) \leq d) = P(X \leq \Phi_{\mu,\sigma^2}^{-1}(d)) = d$.
The above probability is clearly zero for non-positive $d$ and $1$ for $d \geq 1$. This is enough to show that $\Phi_{\mu,\sigma^2}(X)$ has a uniform distribution on $(0,1)$ as we have shown that the corresponding measures are equal for a generator of the Borel $\sigma$-algebra on $\mathbb{R}$. Thus, you can just tranform the normally distributed data by the distribution function and you'll get uniformly distributed data.
