 Solution

Choose the variable ordering of a binary decision diagram $y<x<z$. The Boolean cofactor with $y=1$ is the constant one, while the Boolean cofactor with $y=0$ is $x\wedge z$. Consequently the right panel in part (d) has the following arrows:Here an entry such as $0\to x$ denotes the zero arrow to the $x$ decision node. There are $\boxed{3+2=5}$ vertices: three decision nodes and two terminal nodes, saving one decision node relative to either of the previous orderings. Testing $y$ first is effective because its one branch settles the entire Boolean function immediately. The ordering $y<z<x$ works equally well.

This is also minimal among ordinary binary decision diagrams for this Boolean function. Each Boolean variable is essential: setting $y=0,z=1$ leaves $f=x$; setting $y=0,x=1$ leaves $f=z$; and setting $x=z=0$ leaves $f=y$. If a Boolean variable were never tested, the represented Boolean function could not depend on it. Thus at least three decision nodes are necessary, and the construction attains that bound.