Using a characterisation of $\min\{\mathfrak{d},\mathfrak{r}\}$ that comes from dualizing a result in Kamburelis' and Węglorz's paper called "Splittings":
A. Kamburelis, B. Węglorz, Splittings, Archive for Mathematical Logic 35, Issue 4 (1996) 263-277, doi:10.1007/s001530050044,
the cardinal invariant $\mathfrak{bidi}$ is equal to that minimum.
To introduce their result, consider the following notion of splitting: given $a\in[\omega]^\omega$ and $\bar{I}=\langle I_n\rangle_{n<\omega}$ an interval partition of $\omega$, $a$ splits $\bar{I}$ iff $a$ contains infinitely many $I_n$'s and also its complement $\omega\smallsetminus a$ contains infinitely many $I_n$'s.
A family $\mathcal{I}$ of interval partitions of $\omega$ is unreapable if no $a\in[\omega]^\omega$ splits all the members of $\mathcal{I}$. On the other hand, say that $A\subseteq[\omega]^\omega$ is interval-splitting if any interval partition of $\omega$ is splitted by some member of $A$.
Kamburelis and Weglorz proved that $\max\{\mathfrak{b},\mathfrak{s}\}$ is equal to the smallest size of an interval-splitting family. On the other hand, the dual proof leads to see that $\min\{\mathfrak{d},\mathfrak{r}\}$ is the smallest size of an unreapable family of interval partitions of $\omega$.
We use this result to obtain $\min\{\mathfrak{d},\mathfrak{r}\}\leq\mathfrak{bidi}$ (the other inequality was pointed in Tsaban's question). Let $Y\subseteq\omega^\omega$ of size $<\min\{\mathfrak{d},\mathfrak{r}\}$, wlog, we may assume that, for any $h\in Y$, $h$ is increasing and $h(0)>0$. Now, consider $h^*\in\omega^\omega$ defined as $h^*(0)=0$ and $h^*(k+1)=h(h^*(k))$ and let $\bar{I}^h$ be the interval partition where $I^h_k=[h^*(2k),h^*(2k+2))$. Therefore (by Kamburelis and Weglorz), there exists an $a\in[\omega]^\omega$ that splits $\bar{I}^h$ for all $h\in Y$. It is very easy to conclude that $Y$ is bi-nondominating (as defined by Tsaban) by $a$, to be more specific, if $I^h_k\subseteq a$ then the $h^*(2k)$-th element of $\omega\smallsetminus a$ is bigger than $h^*(2k+1)=h(h^*(2k))$ and, likewise, if $I^h_k\subseteq \omega\smallsetminus a$ then the $h^*(2k)$-th element of $a$ is bigger than $h(h^*(2k))$.
Naturally, the previous argument would lead to a Tukey embedding that also gives $\max\{\mathfrak{b},\mathfrak{s}\}$ bigger than or equal (actually, just equal) to the dual of $\mathfrak{bidi}$, i.e., the minimal size of a family $A\subseteq[\omega]^\omega$ with the property that, for any $y\in\omega^\omega$ there is an $a\in A$ such that $y$ doesn't dominate both the increasing enumerations of $a$ and $\omega\smallsetminus a$.
A lot of very good observations have already been put into the comments. I'll add one more observation that's too long for a comment:
It is consistent that $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$.
The basic idea is to start with a model of MA + $\neg$CH (where $\mathrm{non}(\mathcal L)$ is already big), and then to force over this model to make $\acute{\mathfrak{m}}$ smaller while leaving $\mathrm{non}(\mathcal L)$ large. The proof uses two facts:
Fact 1: There is a $\sigma$-centered notion of forcing $P$ that does not change the value of $\mathfrak{c}$ and that forces $\acute{\mathfrak{m}} = \aleph_1$.
Fact 2: Suppose $V$ is a model of MA, and $P$ is a $\sigma$-centered notion of forcing (in $V$). Then $P$ does not lower the value of $\mathrm{non}(\mathcal L)$. In particular, if $V \not\models$ CH and if $P$ does not change the value of $\mathfrak{c}$, then $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ in the extension.
My claim follows easily from these two facts. To get a model of $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$, begin with a model of MA + $\mathfrak{c} > \aleph_1$ and force with the notion of forcing $P$ described in Fact 1. The resulting model has $\acute{\mathfrak{m}} = \aleph_1$ (because $P$ makes this true) and $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ (by the "in particular" part of Fact 2). QED
Fact 2 is possibly "well known" but I don't know the standard reference. It was first explained to me by Andreas Blass, who exposits it nicely in the proof of Corollary 49 in this paper.
For Fact 1, the notion of forcing from Theorem 4 in this paper of Arnie Miller does the job. This forcing -- let us call it $P$ -- is $\sigma$-centered and does not change the value of $\mathfrak c$.$^{(*)}$ $P$ is designed to add a partition of $2^\omega$ into $\aleph_1$ closed sets, and it is easy to show (using a genericity argument) that each set in this partition has measure $0$; Miller even points this out in a comment after the proof of his Theorem 4.
There may be a small issue here about partitioning $2^\omega$ instead of $[0,1]$, but we can get around it. Given our partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, first observe that none of them has interior in $2^\omega$ (because then it would fail to have measure $0$). Thus there is a countable, dense $D \subset 2^\omega$ that contains no more than one point of any member of our partition. Recall that if $C$ is a closed measure-zero subset of $2^\omega$, then $C$ minus one point is homeomorphic to a countable disjoint union of Cantor sets, so it can be partitioned into countably many closed measure-zero sets. Thus, we may modify our partition by adding in the sets of the form $\{d\}$ with $d \in D$, and then dividing some other sets into countably many pieces. In this way we obtain a partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, including all sets of the form $\{d\}$ for $d \in D$. Once we have done this, we observe that there is a measure-preserving homeomorphism from $2^\omega \setminus D$ onto $[0,1] \setminus \mathbb Q$. When we push our partition through this homeomorphism, we obtain a partition of $[0,1]$ into $\aleph_1$ closed measure-zero sets.
$(*)$ Neither of these facts is stated explicitly in the linked paper, but neither is difficult to prove either. (Using Miller's notation from the linked paper:) Each forcing of the form $P(X)$ is $\sigma$-centered, because if two conditions agree on the part that asserts sentences of the form "$[s] \cap C_n = \emptyset$'' then they are compatible (take the union of the other part). It is clear that no $P(X)$ increases $\mathfrak c$, because it is too small ($|P(X)| \leq \mathfrak c$ and it has the c.c.c., so there are only $\mathfrak c$ nice names for reals). Thus $P$, which is a length-$\omega_1$, finite support iteration of forcings of the form $P(X)$, also is $\sigma$-centered and does not increase $\mathfrak c$.
Best Answer
The answer to this problem is negative: For the compact Polish group $X=S_3^\omega$ we have $Sn(X)\le\mathfrak r$ where $$\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq [\omega]^\omega\;\wedge\;\forall f\in 2^\omega\;\;\exists R\in\mathcal R\;\;|f[R]|=1\}.$$ By induction it can be shown $\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq[\omega]^\omega\;\wedge\;\forall f\in n^\omega\;\exists R\in\mathcal R\;\;|f[R]|=1\}$ for every $n\ge 2$. So, we can find a family $\mathcal R\subseteq[\omega]^\omega$ of cardinality $|\mathcal R|=\mathfrak r$ such that for every function $f:\omega\to S_3$ there exists $R\in\mathcal R$ such that $|f[R]|=1$. Then $$S_3^\omega=\bigcup_{R\in\mathcal R}\bigcup_{g\in S_3}A_{R,g},\quad\mbox{where $A_{R,g}=\{f\in S_3^\omega:\{g\}=f[R]\}$}.$$ Observe that for every $R\in\mathcal R$ and $g\in S_3$ the set $A_{R,g}A_{R,g}^{-1}=A_{R,e}$ is nowhere dense in $S_3^\omega$ and hence $FA_{R,g}A_{R,g}^{-1}F\ne S_3^\omega$ for any finite set $F\subseteq S_3^\omega$.
It remains to show that for every $R\in\mathcal R$ and $g\in S_3$, the set $A_{R,g}$ is algebraic in $S_3^\omega$.
Let $d$ and $t$ be elements of order $2$ and $3$ in the symmetric group $S_3$. It is easy to check that $S_3=\{x\in S_3:dxxdxx=e\}$ where $e$ is the identity of the group $S_3$. On the other hand, $\{x\in S_3:xtxxtx=t^2\}=\{e\}$ and hence $\{x\in S_3:g^{-1}xtg^{-1}xg^{-1}xtg^{-1}x=t^2\}=\{g\}$.
Then $A_{R,g}=\{x\in S_3^\omega:a_0xa_1xa_2xa_3x=b\}$ where $$a_0(i)=a_2(i)=\begin{cases}g^{-1}&\mbox{if $i\in R$};\\ d&\mbox{if $i\in\omega\setminus R$}, \end{cases}\quad a_1(i)=a_3(i)=\begin{cases}tg^{-1}&\mbox{if $i\in R$};\\ e&\mbox{if $i\in\omega\setminus R$}, \end{cases} $$ and $$ b(i)=\begin{cases}t^2&\mbox{if $i\in R$};\\ e&\mbox{if $i\in\omega\setminus R$}, \end{cases} $$