Why take the minimum in the acceptance ratio in the Metropolis-Hastings algorithm? The Metropolis-Hastings ratio is defined as

$$
\alpha(x'|x) = \min\left(1, \frac{P(x')g(x|x')}{P(x)g(x'|x)}\right)
$$

and the state $x'$ is accepted if $u \leq \alpha(x'|x)$, where $u$ is uniformly distributed in $[0,1]$. Given that $u$ can never be larger than $1$, it seems odd to impose the minimum condition on $\alpha(x'|x)$. Is there any logic behind this choice?
 A: $\alpha$ is the acceptance probability.  You don't have to clip it in your code since as you point out $u \le \alpha_{\textrm{clipped}}$ iff $u \le \alpha_{\textrm{unclipped}}$.
A: One very useful fact of the standard uniform distribution is for any $r \in [0,1]$ $P(u \leq r)=  r$ for $u \sim U(0,1)$. We're doing a stochastic hill-climbing procedure which means we have two cases to consider: (1) if it's more likely to move $x\to x'$ than $x'\to x$, i.e. the proposed point is "better", we'll always move there; (2) if $x'$ is not as good, we'll move there with some probability that depends on how "good" it is.
We can make the description of this process simpler by forcing $\alpha(x'|x)\in[0,1]$ so that $\alpha(x'|x)$ itself becomes the probability of accepting.
In general $\alpha(x'|x) \geq 0$ but it does not have to be bounded. But if the proposed point $x'$ is better, as measured by $\frac{P(x')g(x|x')}{P(x)g(x'|x)}\geq 1$, we want to move there with probability $1$. If the proposed point $x'$ is not better we still want a chance to move there, and want to do so with probability $\frac{P(x')g(x|x')}{P(x)g(x'|x)}$. A simple way to do this in one line is
$$
\alpha(x'|x) = \begin{cases} 1 & \frac{P(x')g(x|x')}{P(x)g(x'|x)} \geq 1 \\ \frac{P(x')g(x|x')}{P(x)g(x'|x)} & \text{o.w.}\end{cases} = \min\left\{1, \frac{P(x')g(x|x')}{P(x)g(x'|x)}\right\}
$$
which makes it so
$$
P(\text{move to } x' \text{ from } x) = P(u \leq \alpha(x'|x)) = \alpha(x'|x).
$$
