Now by eq. (2) the denominator of the rightmost expression must be unity, so that we finally arrive at:

$$ \begin{equation} R = \sum_{j=1}^N d_{ij}(\mathbf{r}^{\mathrm{new}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}}) = \sum_{j=1}^N \phi_j(\mathbf{r}_i^{\mathrm{new}})\, d_{ji}^{-1}(\mathbf{r}^{\mathrm{old}}) \tag{6} \end{equation} $$

What this means is that in order to get the ratio when only the i-th particle has been moved, we only need to calculate the dot product of the vector \( \left(\phi_1(\mathbf{r}_i^\mathrm{new}),\,\dots,\,\phi_N(\mathbf{r}_i^\mathrm{new})\right) \) of single particle wave functions evaluated at this new position with the i-th column of the inverse matrix \( \hat{D}^{-1} \) evaluated at the original position. Such an operation has a time scaling of \( O(N) \). The only extra thing we need to do is to maintain the inverse matrix \( \hat{D}^{-1}(\mathbf{x}^{\mathrm{old}}) \).