= Solution
Lift the angular coordinate continuously to a real coordinate and write $\Delta z=z_B-z_A$ and $\Delta\phi_m=\phi_B-\phi_A+2\pi m$ for a prescribed winding number $m\in\mathbb Z$. Assume $z_A<z_B$ for the calculation; reversing the endpoints changes no length. The <arc length> functional on the <circular cylinder> is
$$
\mathcal L[\phi]=\int_{z_A}^{z_B}\sqrt{1+R^2\phi'(z)^2}\,dz.
$$
The <Euler-Lagrange equation> says $R^2\phi'/\sqrt{1+R^2\phi'^2}$ is constant. This expression is strictly increasing in $\phi'$, so $\phi'$ is constant. The candidate is the <helix>
$$
\boxed{\phi(z)=\phi_A+\frac{\Delta\phi_m}{\Delta z}(z-z_A),\qquad
\mathcal L_m=\sqrt{\Delta z^2+R^2\Delta\phi_m^2}.}
$$
It is a global minimum within the chosen winding class: the integrand is strictly <convex>, and <Jensen inequality> bounds the functional below by $(z_B-z_A)\sqrt{1+R^2(\Delta\phi_m/\Delta z)^2}$, with equality only for constant slope. Equivalently, unrolling the cylinder makes the minimizing path a straight segment.
The signed <pitch of a helix>, measured for an increase of $2\pi$ in angle, is
$$
\boxed{P_m=\frac{2\pi\Delta z}{\Delta\phi_m}.}
$$
Its geometric magnitude is $|P_m|$. If only the physical endpoints are fixed, choose the integer $m$ minimizing $|\Delta\phi_m|$; opposite points may give two equally short helices. If the lifted endpoint angles are prescribed, use that winding class. A zero angular increment gives a straight generator, the limiting infinite-pitch case.
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