variables:

$n =$ spatial dimension (3 for 3+1 simulations)
$Q =$ lapse function ($f K$ for Bona Masso, $f=1$ for harmonic slicing)
$\alpha = $ lapse
$\gamma_{ij} = $ spatial metric
$a_i = ln(\alpha)_{,i} = \alpha_{,i} / \alpha$
$\gamma = det(\gamma_{ij})$
$\phi = \frac{1}{c n} (ln \gamma) $ for $c = $ constant, typically 4
$\tilde{\gamma}_{ij} = e^{-c \phi} \gamma_{ij} = e^{-\frac{1}{n} ln \gamma} \gamma_{ij} = \gamma_{ij} / \gamma^{1/n}$
such that $det(\tilde{\gamma}_{ij}) = 1$
$\tilde{\gamma}^{ij} = e^{c \phi} \gamma^{ij}$
such that $\tilde{\gamma}^{ik} \tilde{\gamma}_{kj} = \delta^i_j$
$\tilde{A}_{ij} = K_{ij} - \frac{1}{n} \gamma_{ij} K$
such that $tr(\tilde{A}_{ij}) = \gamma^{ij} \tilde{A}_{ij} = \gamma^{ij} K_{ij} - \gamma^{ij} \gamma_{ij} \frac{1}{n} K = K - \frac{n}{n} K = 0$
$\Phi_i = \phi_{,i}$
$\tilde{d}_{ijk} = \frac{1}{2} \tilde{\gamma}_{jk,i}$
${\tilde{\Lambda}^k}_{ij} = ({\tilde{d}^k}_{ij} + \delta^k_{(i} (a_{j)} - \tilde{\Gamma}_{j)} + 2 \Phi_{j)}))^{TF}$
...what does "trace-free" mean for a degree-3 tensor?
$\tilde{\Gamma}^i = \tilde{\gamma}^{jk} {\tilde{\Gamma}^i}_{jk}$
$ = {\tilde{\gamma}^{ij}}_{,j} = e^{4\phi} \Gamma^i + 2 \tilde{\gamma}^{ij} \phi_{,j}$

BSSNOK formalism

$\alpha_{,t} = -\alpha^2 Q$
$\phi_{,t} = -\frac{1}{6} \alpha K$
$\tilde{\gamma}_{ij,t} = -2 \alpha \tilde{A}_{ij}$
$a_{i,t} = -\alpha Q_{,i}$
$\Phi_{i,t} = -\frac{1}{6} \alpha K_{,i}$
$\tilde{d}_{ijk,t} = -\alpha \tilde{A}_{jk,i}$
$K_{,t} = -\alpha e^{-c \phi} \tilde{\gamma}^{mn} a_{n,m} = -\alpha \gamma^{mn} a_{n,m}$
$\tilde{A}_{ij,t} = -\alpha e^{-c \phi} {\tilde{\Lambda}^k}_{ij,k}$
${\tilde{\Gamma}^i}_{,t} = \alpha ((\xi - 2) \tilde{A}^{ik} - \frac{2}{3} \xi \tilde{\gamma}^{ik} K)_{,k}$
standard BSSNOK: $\xi = 2$, so
${\tilde{\Gamma}^i}_{,t} = -\alpha (\frac{4}{3} \tilde{\gamma}^{ik} K)_{,k}$

BSSNOK 1D

This is my attempt at recasting the BSSNOK equations in 1D.
Note that you can either choose a spatial dimension of $n=1$, or keep the spatial dimension at 3 and only consider the 1st dimension.
If you choose $n=1$ then wouldn't you need to revisit the original GR equations? And in doing so, wouldn't the Riemann curvature tensor always be 1?
Therefore I am going to treat $n=3$ and consider a metric where:

$\gamma_{ij} = \delta_{ij} \delta_{ix}$
$\gamma_{yy} = \gamma_{zz} = 1$
$\gamma_{pq} = 0$ for $p \ne q$

$\gamma_{ij} = \left[\matrix{ \gamma_{xx} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 }\right]$

$\gamma = det(\gamma_{ij}) = \gamma_{xx}$
$\phi = \frac{1}{12} ln(\gamma_{xx})$

$\tilde{\gamma}_{xx} = e^{-4 \phi} \gamma_{xx}$
$\tilde{\gamma}^{xx} = e^{4 \phi} \frac{1}{\gamma_{xx}}$
$\tilde{\gamma}_{yy} = \tilde{\gamma}_{zz} = e^{-4 \phi}$
$\tilde{\gamma}^{yy} = \tilde{\gamma}^{zz} = e^{4 \phi}$

${\Gamma^a}_{bc}:$
${\Gamma^t}_{xx} = \frac{1}{2} \gamma_{xx,t}$
${\Gamma^x}_{xa} = {\Gamma^x}_{ax} = \frac{\gamma_{xx,a}}{2 \gamma_{xx}}$
${\Gamma^p}_{xx} = -\frac{\gamma_{xx,p}}{2}$ for $p \ne t,x$

$\tilde{\Gamma}^x = (\frac{e^{4\phi}}{\gamma_{xx}})_{,x} $
$ = e^{4\phi} (\frac{4\phi_{,x}}{\gamma_{xx}} - \frac{\gamma_{xx,x}}{(\gamma_{xx})^2})$
$ = -\frac{2}{3} (\gamma_{xx})^{-5/3} \gamma_{xx,x}$
$\tilde{\Gamma}^p = {\tilde{\gamma}^{pp}}_{,p}$ for $p \ne x$
$ = {e^{4\phi} \gamma^{pp}}_{,p}$
$ = {e^{4\phi}}_{,p}$
$ = 4 \phi_{,p} e^{4\phi}$
$ = \frac{1}{3} ln(\gamma_{xx})_{,p} e^{\frac{1}{3} ln(\gamma_{xx})}$
$ = \frac{1}{3} \frac{\gamma_{xx,p}}{\gamma_{xx}} (\gamma_{xx})^{1/3}$
$ = \frac{1}{3} (\gamma_{xx})^{-2/3} \gamma_{xx,p}$


By $K_{ij} = -\alpha {\Gamma^t}_{ij}$ we know that only $K_{kk}$ exists ...
... but it is equal to $\frac{1}{2} \gamma_{xx,t}$, which is the time derivative of the spatial metric ...
... so do the other $K_{ij}$'s ever gain a value during simulation?
... that depends on $\tilde{A}_{ij}$
... which also determines $\tilde{\gamma}_{ij}$

$K = K_{ij} \gamma^{ij} = K_{xx} / \gamma_{xx} + K_{yy} + K_{zz}$

I am going to initially provide $K_{xx}$, and all else will be zero.
Therefore the need for $K_{yy}$ or $K_{zz}$ will depend on whether the $\tilde{A}_{pp}, p \ne x$ terms have a time derivative.
In fact, if $\tilde{A}_{pp,t} \ne 0$ then so will $\tilde{\gamma}_{pp}$ need to be simulated.

$\tilde{A}_{ij} = K_{ij} - \frac{1}{3} \gamma_{ij} K$
$\tilde{A}_{xx} = K_{xx} - \frac{1}{3} \gamma_{xx} (K_{xx} / \gamma_{xx} + K_{yy} + K_{zz}) = \frac{2}{3} K_{xx} - \frac{1}{3} \gamma_{xx} (K_{yy} + K_{zz})$
$\tilde{A}_{ij} = 0$ for $i \ne j$
$\tilde{A}_{pp} = K_{pp} - \frac{1}{3} \gamma_{pp} (K_{xx} / \gamma_{xx} + K_{pp} + K_{qq}) = \frac{2}{3} K_{pp} - \frac{1}{3} K_{qq} - \frac{1}{3} \frac{K_{xx}}{\gamma_{xx}}$ for $p \ne x$ and $q \ne p,x$
notice that, even if $K_{xx}$ is the only initial extrinsic curvature, $\tilde{A}_{yy}$ and $\tilde{A}_{zz}$ will still get nonzero values, therefore $\gamma_{yy,t}$ and $\gamma_{zz,t}$ will also be nonzero, therefore I have to consider those terms as well ...

By here I know I have to consider all diagonals of $\gamma_{ij}$ ... or heck, maybe even the whole thing.
Note that this wouldn't be the case if ${\tilde{\Lambda}^k}_{ij}$ did not have the TF over it. Removing the trace is what allows the $\gamma_{xx}$ terms to modify the $\tilde{A}_{pp}$ terms, for $p \ne x$.