 Hirzebruch–Jung resolution

For the cyclic quotient surface singularity $\frac{1}{r}(1,q)$, expand $r/q=[b_1,\dots ,b_s{]}^{-}$ with $b_i\ge 2$. In the quotient lattice, start with $v_0=(0,1)$, $v_1=(1,q)/r$, and set $v_{i+1}=b_i{v}_i-v_{i-1}$. The last vector is $(1,0)$, each consecutive pair is a lattice basis, and subdivision by these rays is a resolution. Its exceptional chain has self-intersections $-b_i$. None is a minus-one curve, so the resolution is minimal. All coefficients are two exactly for $q=r-1$.

// Target: geometry-and-topology.bigb