Partitioning antidirected trees such that the induced graph by the partition is also an antidirected tree of constant size

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Statement:

Define for an oriented graph $G$ and a partition $\mathcal{P}$ of $V(G)$, the contracted graph $G[\mathcal{P}]$, with vertex set $V(G[\mathcal{P}]) = \mathcal{P}$ and edge set $$E(G[\mathcal{P}]) = \{ (A,B) \in \mathcal{P}^2 : \exists a\in A, \exists b \in B \text{ such that }(a,b)\in E(G)\}$$

Let $\Delta,C\in \mathbb{N}$. Does there exist $n_{0} \in \mathbb{N}$ such that for every $n\geq n_{0}$ and every antidirected tree $T$ on $n$ vertices with bounded degree $\Delta(T)\leq \Delta$ there is a partition $\mathcal{P}$ of $V(T)$ such that $T[\mathcal{P}]$ is an antidirected tree with $|\mathcal{P}| \leq C$ and with maximum degree $\Delta(T[\mathcal{P}]) \leq 3$ (or some other small constant), and such that the partition is balanced, i.e: $\frac{1}{2C} \cdot n \leq |A| \leq \frac{2}{C}\cdot n$, for each $A\in \mathcal{P}$ ?

Questions :

  • Are there any "known" results that are similar to the statement I'm trying to prove, that I may be able to adapt to my scenario?

  • On the proof idea outlined below, given the partition $\mathcal{X}$, How could I ensure that after breaking the possible "big" sets, and coarsening the partition by joining small sets that I can still have a constant upper bound on the number of clusters, while also having that none is too big?

  • Slightly off topic: Is there an extremal result akin to Burr's result on embedding antidirected trees, that could make it easier embedding antidirected structures that are not necesarilly trees?

One idea:

Let $r$ be an arbitrary root of $V(T)$, and let $\mathcal{L} = \{L_{0},...,L_{t} \}$ be the levels partitioning of $T$ with respect to the root $r$, this is $L_{s} = N^{s}(r)\setminus \bigcup\limits_{i=0}^{s-1}N^{i}(r)$. Note that the graph $T[\mathcal{L}]$ is an antidirected path of edge-length $t$, and that each $L_{s}$ satisfies that $|L_{s}|\leq \Delta^{s}$. We have then that $n = \sum\limits_{s=0}^{t} |L_{s}|\leq \sum\limits_{s=0}^{t} \Delta^{s}$, therefore $\frac{\log(n)+\log(\Delta-1)}{\log(\Delta)}\leq t+1$.

This nets us with an antidirected tree $T[\mathcal{L}]$ (actually a path) of size close to $log(n)$, (i.e. bigger than $C$) and whose clusters can be too big in principle.

We can further "sieve" this structure in the following manner:

Define the following three functions, the parent function $\pi:V(T)\to V(T)$ which maps $v\in L_{s}$ to its parent with respect to the root $r$ (i.e. the only element of $N(v)\cap L_{s-1}$), the successor function $\sigma: V(T) \to \mathcal{P}(V(T))$ which maps $v\in L_{s}$ to its sons $N(v)\cap L_{s+1}$ and the brotherhood function $B(T): V(T) \to \mathcal{P}(V(T))$, which is equal to $\sigma \circ \pi$ (given $v\in L_{s}$, $B(v)$ are the vertices in $L_{s}$ that share $v$'s parent in $L_{s-1}$). Also define for a set $X\subseteq V(T)$ the "images" $B(X) := \bigcup\limits_{x\in X} B(x)$ and $\sigma(X) := \bigcup\limits_{x\in X} \sigma(x) $.

If we fix an arbitrary set $X\subseteq L_{t}$, then define the following partition:

$$\mathcal{X} = \bigcup\limits_{s=0}^{t} \bigcup\limits_{l=0}^{s} \{X^{t-s}_{l}\}$$

Where $X_{l}^{t-s} = \big( (\sigma \circ B)^{l-s} \circ (\pi\circ B)^{l} (B(X)) \big) \setminus X_{l-1}^{t-s}$, where $X_{-1}^{j}=\phi$ for all $j$ and $X_{j}^{s} \subseteq L_{s}$ for all $j,s$. This partition defines an antidirected tree $T[\mathcal{X}]$ of maximum degree $\Delta(T[\mathcal{X}]) = 3$, with at most $\frac{t(t-1)}{2}$ elements in the partition (which is at least $C'\cdot \log(n)^2$ for some constant $C'$). Also note that each element of the partition is at most $\Delta^{2}$ times bigger than any of its neighbours in $T[\mathcal{X}]$.

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This sets some precedent for constructing "small" antidirected trees of constant degree, but the size of each cluster is not properly bounded, and the amount of clusters is too big in principle.

Following each ray (the diagonal branches of the tree in the second diagram) and joining all the sink and source nodes from each ray until getting enough vertices in a cluster should in theory help, but i don't see a clever way of arguing that I can do so while preserving the upper and lower bounds on cluster sizes.

Further reference: I asked a really similar question (that I haven't been able to solve) which didn't get traction in mathoverflow Here, albeit with slightly different notation, and the problem is less 'striped' of the actual intent of solving it, in case someone wants more context.