= Solution
On the <circular cylinder>, $\rho=R$ is constant, so the given <line element> reduces to $ds^2=R^2d\phi^2+dz^2$. Unroll the <circular cylinder> with $u=R\phi$. The <line element> is then $ds^2=du^2+dz^2$, so the usual <Euclidean geometry> formula measures <arc length> in the $(u,z)$ <plane>.
The endpoint has infinitely many lifts $(R(\phi_1+2\pi k),z_1)$, with $k\in\mathbb Z$. For each lift, the <straight line> segment is shortest; its <length> is $\sqrt{R^2(\phi_1+2\pi k)^2+z_1^2}$. Comparing the lifts gives the unrestricted <shortest path on a circular cylinder>:
$$
\boxed{L_{\min}=\min_{k\in\mathbb Z}\sqrt{R^2(\phi_1+2\pi k)^2+z_1^2}.}
$$
If $\delta$ is a minimizing angular increment, a shortest <geodesic> is parametrized by $(\rho,\phi,z)=(R,\delta t,z_1t)$, $0\leq t\leq1$. This is a <helix> when both increments are nonzero, and a circular arc or vertical generator in the limiting cases. A nonminimal lift also gives a <geodesic>, but it is only shortest within its fixed winding class.
For $0\leq\phi_1\leq2\pi$, put $m=\min(\phi_1,2\pi-\phi_1)$. Then $\boxed{L_{\min}=\sqrt{R^2m^2+z_1^2}}$. At $\phi_1=\pi$, both winding directions give distinct shortest paths of equal <length>.
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