MCMC Metropolis Hastings - Normalised distribution

Question

I am reading about MCMC from this PDF Murphy's MCMC (1), on page 4, above equation (21) the author states:

"Note that when evaluating α (acceptance probability), we only need to know the target density p up to a normalization constant. In particular, suppose $π(x) = 1/Z (π′(x))$, where $π′(x)$ is an unnormalized distribution and $Z$ is the normalization constant."

Can someone please, especially graphically, explain to me why is this important? The fact that we only need to know the unnormalised target distribution? Later on in the text, he states that often is difficult to calculate $Z$, which I suppose $Z$ might be the integral of $π(x)$? No? Thanks!

(1) Kevin P. Murphy (2006),
"Markov chain Monte Carlo (MCMC)",
Last updated November 3, 2006

Answer 1

$Z$ would be the integral of $\pi^\prime$ (i.e. the integral of the unnormalized density). The integral of $\pi$ would be 1.

Evaluating the integral accurately would often be difficult (or where it's not difficult, at least time consuming).