For $k1,$ if the path has $k$ as an intermediate vertex, so that it is $ikj$

where $\frac{?}{\text{b}\text{a}\text{r}}(k)j,$ then choose as the predecessor of $j$ on this path the same vertex

as the predecessor of $j$ chosen on a shortest path from $k$ with all intermediate

vertices in the set $\{1,2,$dots,$k-1\}.$ Otherwise, when the path from $i$ to $j$ does not

have $k$ as an intermediate vertex, choose the same predecessor of $j$ as on a shortest

path from i with all intermediate vertices in the set $\{1,2,$dots,$k-1\}.$ Formally, for

$k1,$

${}_{ij}^{(k)}=\{\begin{array}{ccc}{}_{kj}^{(k-1)}& if{d}_{ij}^{(k-1)}>d_{ik}^{(k-1)}+d_{kj}^{(k-1)}(kisan\text{i}\text{n}\text{t}\text{e}\text{r}\text{m}\text{e}\text{d}\text{i}\text{a}\text{t}\text{e}\text{v}\text{e}\text{r}\text{t}\text{e}\text{x}& ),\\ {}_{ij}^{(k-1)}& if{d}_{ij}^{(k-1)}{d}_{ik}^{(k-1)}+d_{kj}^{(k-1)}(kis\text{n}\text{o}\text{t}an\text{i}\text{n}\text{t}\text{e}\text{r}\text{m}\text{e}\text{d}\text{i}\text{a}\text{t}\text{e}\text{v}\text{e}\text{r}\text{t}\text{e}\text{x}& ).\end{array}$

$23\mathrm{.}2-5$

Consider the following change to how equation $(23\mathrm{.}8)$ handles equality:

$$