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| Branch-Based Local Capture in Tree-Ball Geometry: Sharp Positive and Negative Results | 2026-05-06 | We study a local team-chase problem in a $d$-regular graph whose ball of large radius around the robber is a tree. We isolate the right local invariant — the deep load along a nonbacktracking path tube — and prove a coordinated package of positive and negative results: (i) sharp counts of geodesic cones and their truncations, (ii) a one-round universal branch-persistence lemma, (iii) a $t$-round generalization with depth budget $2t+1$, (iv) a sampling barrier showing that any proof relying on the path-tube certificate requires $\Omega((d-1)^{t-1} \cdot t)$ cops, and (v) a finite-order potential-degeneration barrier showing that any local invariant depending only on order-$r$ tube data is strictly insufficient to certify even $(r+1)$-round persistence. Together, these results form a double pincer: certifying $t$-round persistence by an order-$r$ local invariant requires $r \geq t$, and order-$t$ resolution requires $\Omega((d-1)^{t-1})$ cops to occupy by uniform sampling. This rules out order-$r$ branch-aggregated potentials (for any fixed $r$) as a route to $\Theta(\log n)$-round chase from polylogarithmic cops, and identifies the precise structural reason why the natural iteration of the one-round argument fails. |
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data/branch-capture-paper.bib | /papers/branch-capture-paper.pdf | true |
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Introduction
Motivation
The cops-and-robbers game is a pursuit-evasion game on a graph in which k cops attempt to capture a single robber. The minimum k for which the cops have a winning strategy is the cop number c(G). Meyniel's conjecture (Frankl, 1987) asserts that c(G) = O(\sqrt{n}) for every connected graph on n vertices, and remains the central open problem in the area. The current best upper bound is c(G) \leq n / 2^{(1-o(1))\sqrt{\log n}} (Lu–Peng; Scott–Sudakov; Frieze–Krivelevich–Loh).
A natural approach toward Meyniel's conjecture, in the spirit of Prałat–Wormald's resolution for random graphs and recent partial progress on expanders and high-girth graphs[@BradshawHosseiniMoharStacho; @HMG], is to combine random cop placement with a deterministic local chase. The bottleneck of this approach is the local team-chase problem: assuming several cops have been placed within a small distance r of the robber, can they coordinate their gradient-descent moves to capture the robber within O(r \log n) rounds?
In high-girth regular graphs, where the local geometry is tree-like, the difficulty of this problem is structural: a single cop is always evaded (Aigner–Fromme), so capture requires multi-cop coordination, and the right coordination invariant is not obvious. In this paper, we work in the cleanest possible local geometry — a $d$-regular tree-ball — and identify the right invariant, prove that it works for one round, prove that it generalizes to t rounds, and prove a matching barrier.
Setup
Throughout, G denotes a $d$-regular graph with d \geq 3. We fix a vertex v_0 \in V(G) and a radius R \geq 1. The fundamental hypothesis is:
::: {.annotation .annotation--static #def-tree-ball}
In a tree-ball, v_0 has d children (its neighbors), and every interior vertex has exactly one parent and d-1 children. We write S_j(v) = \{x : \operatorname{dist}(x, v) = j\} for the depth-j shell at v.
::: {.annotation .annotation--static #def-cone}
C_u(v, r) := \{x \in B_r(v) \setminus \{v\} : \text{the unique geodesic from } x \text{ to } v \text{ has penultimate vertex } u\}.
The truncated cone at depth $\geq k$ is
C_u^{\geq k}(v, r) := \{x \in C_u(v, r) : \operatorname{dist}(x, v) \geq k\}.
::: {.annotation .annotation--static #def-tube}
\begin{gathered}
T_\sigma^{\geq k}(R) := \bigl\{x \in B_R(v_0) : \operatorname{dist}(x, v_0) \geq k, \\
\text{the unique geodesic from } x \text{ to } v_0 \text{ visits } v_t, v_{t-1}, \ldots, v_1 \text{ in order}\bigr\}.
\end{gathered}$$
</div>
:::
The cops execute *gradient descent*: at each cop turn, every cop moves to a neighbor minimizing distance to the current robber position, breaking ties by an arbitrary rule (deterministic or randomized). The order of play is: cops move, then robber moves. A robber move to $w \in N(v(t))$ is *safe* if no cop occupies $w$ at the start of the robber's turn (i.e., after the cops' move).
For the robber's position $v$ and a neighbor $u' \in N(v)$, we say a cop $c$ *contributes to branch $u'$ at $v$* if $c \neq v$ and the geodesic from $c$ to $v$ has penultimate vertex $u'$. The *branch-load support* at $v$ is the set of $u' \in N(v)$ that receive at least one contributing cop.
### Results
The paper has four main results, prefaced by a sharp counting lemma.
::: {.annotation .annotation--static #lem-shell}
<div class="annotation-header">
<span class="annotation-label">Lemma 1.4</span>
<span class="annotation-name">Sharp shell and cone counts</span>
</div>
<div class="annotation-body">
Let $G$ be $d$-regular with $d \geq 3$, and suppose $B_r(v)$ is a tree-ball. Then for every $u \in N(v)$ and $1 \leq j \leq r$,
$$|C_u(v, r) \cap S_j(v)| = (d-1)^{j-1}, \qquad |S_j(v)| = d(d-1)^{j-1}.$$
Hence
$$|C_u(v, r)| = \sum_{j=1}^{r} (d-1)^{j-1} = \frac{(d-1)^r - 1}{d-2}, \quad |B_r(v)| = 1 + d \cdot \frac{(d-1)^r - 1}{d - 2}.$$
For the truncated cone,
$$|C_u^{\geq k}(v, r)| = \sum_{j=k}^{r} (d-1)^{j-1}, \quad 1 \leq k \leq r.$$
</div>
:::
The first main result is a probabilistic deep-load corollary.
::: {.annotation .annotation--static #cor-occupancy}
<div class="annotation-header">
<span class="annotation-label">Corollary 1.5</span>
<span class="annotation-name">Deep-load occupancy</span>
</div>
<div class="annotation-body">
Let $G$ be $d$-regular with $B_r(v)$ a tree-ball, and let $X_1, \ldots, X_m$ be i.i.d. uniform samples from $B_r(v)$. Set
$$p_{d,r}^{\geq k} := \frac{|C_u^{\geq k}(v, r)|}{|B_r(v)|},$$
which by [Lemma 1.4](#lem-shell) does not depend on the choice of $u \in N(v)$, since all branches in a tree-ball have equal size. Then
$$\Pr[\exists u \in N(v): C_u^{\geq k}(v, r) \text{ receives no sample}] \leq d \cdot (1 - p_{d,r}^{\geq k})^m.$$
</div>
:::
The second is the universal $t$-round persistence theorem with the correct depth budget.
::: {.annotation .annotation--static #thm-persistence}
<div class="annotation-header">
<span class="annotation-label">Theorem 1.6</span>
<span class="annotation-name">Universal $t$-round deep-load persistence</span>
</div>
<div class="annotation-body">
Let $G$ be $d$-regular with $d \geq 3$, $v_0 \in V(G)$, and assume $B_R(v_0)$ is a tree-ball for some $R \geq 2t+1$. Suppose that for every nonbacktracking path $\sigma = (v_0, v_1, \ldots, v_t)$ of length $t$ from $v_0$, there exists at least one cop in $T_\sigma^{\geq 2t+1}(R)$.
Then for every nonbacktracking robber path $(v_0, v_1, \ldots, v_t)$ and every $s \in \{1, \ldots, t\}$, after the cops' gradient-descent step at round $s$, exactly one of the following holds:
- **(a)** a cop occupies $v_s$, so the robber's intended move to $v_s$ is illegal; or
- **(b)** the move to $v_s$ is legal, and after the robber moves to $v_s$, the branch-load support at $v_s$ has size at least $2$.
</div>
:::
The third is the sampling barrier.
::: {.annotation .annotation--static #thm-barrier}
<div class="annotation-header">
<span class="annotation-label">Theorem 1.7</span>
<span class="annotation-name">Barrier on branch-tube certificates</span>
</div>
<div class="annotation-body">
Fix $d \geq 3$. In the $d$-regular tree-ball model, any argument that places $m$ cops by uniform sampling from $B_R(v_0)$ and relies on occupancy of every length-$t$ deep tube $T_\sigma^{\geq 2t+1}(R)$ requires
$$m = \Omega\!\left((d-1)^{t-1} \cdot t\right)$$
cops. In particular, for $m = \mathrm{polylog}(n)$ and fixed $d$, this method certifies only $t = O(\log\log n)$ rounds, not $t = \Theta(\log n)$.
</div>
:::
The fourth is a complementary degeneration barrier showing that bounded-order compressions of the path-tube data are themselves insufficient. [Theorem 4.6](#thm-finite-order-barrier) (stated and proved in [Section 4.5](#a-finite-order-potential-degeneration-barrier)) constructs, for every $r \geq 1$, two cop configurations $X$ and $Y$ that agree on all order-$r$ tube data but differ in their post-$(r+1)$-round branch-load support: $X$ has support $\geq 2$ while $Y$ has support exactly $1$. Hence no certificate depending only on order-$r$ data can certify universal $(r+1)$-round persistence.
### Strategic significance
[Theorems 1.6](#thm-persistence), [1.7](#thm-barrier), and [4.6](#thm-finite-order-barrier) together capture the precise reach of branch-tube methods. The natural iteration of the one-round argument extends to $t$ rounds with the correct depth budget $2t+1$ ([Theorem 1.6](#thm-persistence)), but the path-entropy cost is exponential in $t$ ([Theorem 1.7](#thm-barrier)), and any compression to bounded-order data fails dynamically ([Theorem 4.6](#thm-finite-order-barrier)). Together, the two barriers tighten this trade-off: certifying $t$-round persistence by an order-$r$ local invariant requires $r \geq t$, and order-$t$ resolution requires $\Omega((d-1)^{t-1})$ cops to occupy.
The barriers rule out a substantial family of proof techniques but are silent on what could replace them. The pivots in [Section 5](#discussion-pivots-and-open-directions) sketch the most natural escape routes: epoch refresh (which probably needs a global progress measure to succeed), soft potentials that necessarily break locality or symmetry, and a non-local spectral approach via nonbacktracking-walk concentration.
## Sharp Counts
### Proof of Lemma 1.4
::: {.exhibit .exhibit--proof data-exhibit-name="Sharp shell and cone counts" data-exhibit-type="proof" data-exhibit-caption="Counting tree-ball shells, cones, and truncated cones via the (d−1)-ary branch structure."}
:::: exhibit-body
Since $B_r(v)$ is a tree-ball and $G$ is $d$-regular, $v$ has exactly $d$ children in the rooted tree (one per neighbor in $N(v)$), and every non-root vertex at depth $< r$ has exactly $d-1$ children. Fix $u \in N(v)$. The branch rooted at $u$ is a $(d-1)$-ary rooted tree of depth $r - 1$; its level $j-1$ has $(d-1)^{j-1}$ vertices, so the depth-$j$ shell of the branch (relative to $v$) has $(d-1)^{j-1}$ vertices.
Hence $|C_u(v, r) \cap S_j(v)| = (d-1)^{j-1}$ for $1 \leq j \leq r$. Summing over the $d$ branches gives $|S_j(v)| = d(d-1)^{j-1}$, and summing $j$ from 1 to $r$ gives the cone size. Including $v$ itself yields the ball size, and restricting $j \geq k$ yields the truncated cone count. [□]{.proof-qed}
::::
:::
::: {.annotation .annotation--static #rem-ratio}
<div class="annotation-header">
<span class="annotation-label">Remark 2.1</span>
<span class="annotation-name">Asymptotic branch ratio</span>
</div>
<div class="annotation-body">
The ratio $|C_u^{\geq k}(v, r)| / |B_r(v)|$ tends to $1/d$ as $r \to \infty$ for fixed $k$. Thus each branch asymptotically owns a $1/d$ fraction of the ball, and depth-$k$ truncation costs only $O(d^k)$ vertices out of $\Theta(d^r)$, which is negligible at large $r$.
</div>
:::
### Proof of Corollary 1.5
::: {.exhibit .exhibit--proof data-exhibit-name="Deep-load occupancy" data-exhibit-type="proof" data-exhibit-caption="Union bound over the d branches at v."}
:::: exhibit-body
For a fixed branch $u \in N(v)$, the probability that no $X_i$ lies in $C_u^{\geq k}(v, r)$ is $(1 - p_{d,r}^{\geq k})^m$. Union bounding over the $d$ choices of $u$ gives the claim. [□]{.proof-qed}
::::
:::
For $k = 3$ and $r$ large, $p_{d,r}^{\geq 3} \to 1/d - O(d^{-r})$, so $m = \Theta(d \log d)$ samples suffice to hit every truncated branch with high probability. For fixed $d$, $m = O(\log n)$ drives the failure probability polynomially small in $n$.
## One-Round and $t$-Round Persistence
### The one-round case
We first prove the case $t = 1$ separately, both for clarity and because the argument is the model for the multi-round induction. [Lemma 3.1](#lem-oneround) below is the $t = 1$ specialization of [Theorem 1.6](#thm-persistence), presented here in self-contained form.
::: {.annotation .annotation--static #lem-oneround}
<div class="annotation-header">
<span class="annotation-label">Lemma 3.1</span>
<span class="annotation-name">One-round persistence</span>
</div>
<div class="annotation-body">
Let $G$ be $d$-regular with $d \geq 3$, $v_0 \in V(G)$, and assume $B_3(v_0)$ is a tree-ball. Suppose that for every $u \in N(v_0)$, there is at least one cop in $C_u^{\geq 3}(v_0, R)$ for some $R \geq 3$.
Then for every $w \in N(v_0)$, after the cop gradient-descent step, exactly one of the following holds:
- **(a)** a cop occupies $w$, so the robber's intended move to $w$ is illegal; or
- **(b)** the move to $w$ is legal, and after the robber moves to $w$, the branch-load support at $w$ has size at least $2$.
</div>
:::
::: {.exhibit .exhibit--proof data-exhibit-name="One-round persistence" data-exhibit-type="proof" data-exhibit-caption="Descendant witness from the cop in C_w; parent witness from a cop in C_u with u ≠ w."}
:::: exhibit-body
Fix $w \in N(v_0)$. After the cop gradient-descent step, if some cop occupies $w$ then case (a) holds and we are done. Otherwise we exhibit both a descendant witness and a parent witness at $w$, establishing case (b).
*Descendant witness.* By hypothesis there is a cop $c^\downarrow \in C_w^{\geq 3}(v_0, R)$. Then $\operatorname{dist}(c^\downarrow, v_0) = j \geq 3$ with the geodesic from $c^\downarrow$ to $v_0$ passing through $w$. Gradient descent toward $v_0$ moves $c^\downarrow$ to its parent in the tree rooted at $v_0$, which is at depth $j - 1 \geq 2$, still in the $w$-subtree. After this move, $c^\downarrow$ is at distance $j - 1 - 1 = j - 2 \geq 1$ from $w$. The geodesic from this new position to $w$ passes through some child of $w$ in the tree, so $c^\downarrow$ contributes to a descendant branch at $w$.
*Parent witness.* Pick $u \in N(v_0)$ with $u \neq w$ (possible since $d \geq 3 \geq 2$). By hypothesis there is a cop $c^\uparrow \in C_u^{\geq 3}(v_0, R)$. After gradient descent, $c^\uparrow$ remains in the $u$-subtree at depth $\geq 2$ from $v_0$. Relative to $w$, the geodesic from $c^\uparrow$ to $w$ passes through $v_0$ (the unique tree path between distinct subtrees), so $c^\uparrow$ contributes to the parent branch at $w$ through $v_0$.
*Combining witnesses.* The descendant and parent witnesses lie in distinct branches at $w$, so the branch-load support at $w$ has size at least $2$, establishing case (b). [□]{.proof-qed}
::::
:::
### Proof of Theorem 1.6
We now extend the one-round argument to general $t$, with depth budget $2t+1$. The reason this budget is necessary, not merely sufficient, is illustrated in the sharpness remark of [Section 3.3](#the-depth-budget-is-sharp).
::: {.exhibit .exhibit--proof data-exhibit-name="Universal $t$-round deep-load persistence" data-exhibit-type="proof" data-exhibit-caption="Induction on rounds: each cop in a deep tube descends one step per round, and the depth budget 2t+1 keeps the descendant witness strictly below v_s."}
:::: exhibit-body
Fix any nonbacktracking robber path $(v_0, v_1, \ldots, v_t)$ and any $s \in \{1, \ldots, t\}$. We analyze round $s$ on the assumption that no cop landed on $v_{s'}$ during the round-$s'$ gradient-descent step, for each $1 \leq s' < s$; otherwise case (a) holds at the first such $s'$, establishing the dichotomy there. Under this assumption the robber is at $v_{s-1}$ at the start of round $s$. After the cops' gradient-descent step, if some cop occupies $v_s$ then case (a) holds and we are done. Otherwise we exhibit both a descendant-branch witness and a parent-branch witness at $v_s$, establishing case (b).
*Descendant witness at $v_s$.* Let $\sigma_s = (v_0, v_1, \ldots, v_s)$ be the robber's trajectory through time $s$. Extend $\sigma_s$ to any nonbacktracking length-$t$ path $\sigma$ (e.g., by appending $t - s$ steps in any nonbacktracking way). By hypothesis there is a cop $c_\sigma^\downarrow \in T_\sigma^{\geq 2t+1}(R)$ with initial depth $j_0 \geq 2t+1$, and the unique geodesic from $c_\sigma^\downarrow$ to $v_0$ visits $v_t, v_{t-1}, \ldots, v_1$ in order.
In a tree-ball, gradient descent toward any vertex on the cop's geodesic to $v_0$ is forced: the unique distance-decreasing neighbor is the cop's parent in the $v_0$-rooted tree. Therefore each round the cop moves exactly one step upward along its geodesic to $v_0$. After $s$ rounds, the cop has moved $s$ steps up its geodesic to $v_0$, so its current geodesic to $v_0$ visits $v_s, v_{s-1}, \ldots, v_1$ in order, and its depth from $v_0$ is $j_0 - s$.
Since $j_0 \geq 2t+1$ and $s \leq t$, we have $j_0 - s \geq t + 1 > s$, so the cop remains *strictly below* $v_s$: it is in the $v_s$-subtree at depth $\geq s + 1$ from $v_0$, equivalently at distance $\geq 1$ from $v_s$ within that subtree. Hence the geodesic from the cop's current position to $v_s$ has penultimate vertex equal to some child of $v_s$, and the cop contributes to a descendant branch at $v_s$.
*Parent witness at $v_s$.* Choose any neighbor $u_0 \in N(v_0)$ with $u_0 \neq v_1$ (possible since $d \geq 3$). Extend $u_0$ to any nonbacktracking length-$t$ path $\sigma' = (v_0, u_0, \ldots)$. By hypothesis there is a cop $c^\uparrow \in T_{\sigma'}^{\geq 2t+1}(R)$.
Initially $c^\uparrow$ is in the $u_0$-subtree at depth $\geq 2t+1$. At round $1$, gradient descent toward $v_0$ moves $c^\uparrow$ to depth $2t$, still in the $u_0$-subtree. At round $2$, gradient descent toward $v_1$ has target in a different subtree from $c^\uparrow$, so the cop's geodesic to $v_1$ goes up through $v_0$. The unique distance-decreasing neighbor is the cop's parent in the $v_0$-rooted tree, so the cop moves to depth $2t - 1$, still in the $u_0$-subtree. The same argument applies at every round $k$ with $1 \leq k \leq s$: from the cop's position in the $u_0$-subtree at depth $2t + 2 - k$, the robber's current position $v_{k-1}$ is reached only via $v_0$ (since $u_0 \neq v_1$), so the unique distance-decreasing direction is up to the parent.
By induction, after $s$ rounds, $c^\uparrow$ is at depth $2t + 1 - s$ from $v_0$, still in the $u_0$-subtree. Since $u_0 \neq v_1$, the entire $u_0$-subtree is disjoint from the $v_1$-subtree, and since each $v_j$ for $j \geq 1$ lies in the $v_1$-subtree, the cop $c^\uparrow$ at depth $2t + 1 - s \geq t + 1 \geq 1$ from $v_0$ in the $u_0$-subtree is on the parent side of $v_s$. For $s \geq 1$, the geodesic from $c^\uparrow$ to $v_s$ passes through $v_0$, so the penultimate vertex of that geodesic (the one adjacent to $v_s$) is $v_{s-1}$. So $c^\uparrow$ contributes to the parent branch of $v_s$ through $v_{s-1}$.
*Combining witnesses.* The descendant witness contributes to a non-parent branch at $v_s$, and the parent witness contributes to the parent branch at $v_s$. These are distinct branches, so the branch-load support at $v_s$ is at least $2$, establishing case (b). [□]{.proof-qed}
::::
:::
### The depth budget is sharp
The hypothesis $j_0 \geq 2t + 1$ in [Theorem 1.6](#thm-persistence) is necessary, not merely sufficient. We illustrate.
::: {.annotation .annotation--static #rem-sharpness}
<div class="annotation-header">
<span class="annotation-label">Remark 3.2</span>
<span class="annotation-name">Sharpness of the depth budget</span>
</div>
<div class="annotation-body">
Consider $j_0 = 2t$. After $s = t$ rounds, the cop is at depth $j_0 - t = t$ from $v_0$, with the robber at $v_t$ (also depth $t$). The cop's geodesic to $v_0$ contains $v_t, v_{t-1}, \ldots, v_1$, so the cop is at $v_t$ itself, not at a strict descendant. This blocks the robber's move (case (a)) but does not provide a descendant branch at $v_t$ for the post-move analysis. Reducing further to $j_0 < 2t$ would place the cop on an ancestor of $v_t$, and the cop would contribute to the parent branch only — destroying the support-2 claim.
</div>
:::
## The Barrier
### Tube density
::: {.annotation .annotation--static #lem-density}
<div class="annotation-header">
<span class="annotation-label">Lemma 4.1</span>
<span class="annotation-name">Asymptotic tube density</span>
</div>
<div class="annotation-body">
For fixed $d \geq 3$ and a length-$t$ nonbacktracking prefix $\sigma$ in a $d$-regular tree-ball $B_R(v_0)$ with $R \geq 2t+1$,
$$|T_\sigma^{\geq 2t+1}(R)| = \sum_{j = 2t+1}^{R} (d-1)^{j-t} = \sum_{\ell = t+1}^{R-t} (d-1)^\ell = (d-1)^{t+1} \cdot \frac{(d-1)^{R-2t} - 1}{d - 2}.$$
The tube density
$$q_t(R) := \frac{|T_\sigma^{\geq 2t+1}(R)|}{|B_R(v_0)|}$$
satisfies, as $R \to \infty$,
$$q_t(R) \to \frac{1}{d (d-1)^{t-1}}.$$
</div>
:::
::: {.exhibit .exhibit--proof data-exhibit-name="Asymptotic tube density" data-exhibit-type="proof" data-exhibit-caption="Both numerator and denominator are geometric sums dominated by their top shells; the ratio of top shells gives the limit."}
:::: exhibit-body
Each vertex in $T_\sigma^{\geq 2t+1}(R) \cap S_j(v_0)$ is a descendant of $v_t$ at depth $j$ from $v_0$, hence at depth $j - t$ within the $v_t$-rooted subtree. The latter is a $(d-1)$-ary tree (since $v_t$ has $d - 1$ children in the tree-ball), so its depth-$(j - t)$ shell has $(d-1)^{j-t}$ vertices for $j \geq t + 1$. Restricting to $j \geq 2t + 1$, summing, and substituting $\ell = j - t$ gives the closed form above.
For the asymptotic density: both $|T_\sigma^{\geq 2t+1}(R)|$ and $|B_R(v_0)|$ are geometric sums dominated by their top shells. The top shell of the tube at $j = R$ has $(d-1)^{R-t}$ vertices; the top shell of the ball at $j = R$ has $d(d-1)^{R-1}$ vertices. Hence the ratio tends to
$$\frac{(d-1)^{R-t}}{d(d-1)^{R-1}} = \frac{1}{d(d-1)^{t-1}},$$
as claimed. [□]{.proof-qed}
::::
:::
### Number of length-$t$ prefixes
::: {.annotation .annotation--static #lem-prefixes}
<div class="annotation-header">
<span class="annotation-label">Lemma 4.2</span>
<span class="annotation-name">Number of length-$t$ prefixes</span>
</div>
<div class="annotation-body">
The number of nonbacktracking length-$t$ paths from $v_0$ in a $d$-regular tree-ball is
$$N_t = d(d-1)^{t-1}.$$
</div>
:::
::: {.exhibit .exhibit--proof data-exhibit-name="Number of length-$t$ prefixes" data-exhibit-type="proof" data-exhibit-caption="d choices for the first step, d−1 choices thereafter."}
:::: exhibit-body
The first step has $d$ choices (any neighbor of $v_0$). Each subsequent step has $d - 1$ choices (any neighbor of the current vertex except the immediately previous one). So $N_t = d(d-1)^{t-1}$. [□]{.proof-qed}
::::
:::
### Proof of Theorem 1.7
::: {.exhibit .exhibit--proof data-exhibit-name="Barrier on branch-tube certificates" data-exhibit-type="proof" data-exhibit-caption="Coupon-collector-style union bound: covering N_t = d(d−1)^{t−1} tubes each with density 1/[d(d−1)^{t−1}] requires Ω((d−1)^{t−1} · t) cops."}
:::: exhibit-body
Suppose $m$ cops are sampled i.i.d. uniformly from $B_R(v_0)$, and the proof relies on every length-$t$ tube being occupied. By [Lemma 4.1](#lem-density), for each fixed prefix $\sigma$,
$$\Pr\!\left[T_\sigma^{\geq 2t+1}(R) \text{ is empty}\right] = (1 - q_t(R))^m \leq \exp(-m \cdot q_t(R)).$$
Union bounding over the $N_t$ prefixes,
$$\Pr[\exists \sigma : T_\sigma \text{ empty}] \leq N_t \cdot \exp(-m \cdot q_t(R)).$$
For this to be $o(1)$, we require
$$m \geq \frac{\log N_t + \omega(1)}{q_t(R)}.$$
Using $N_t = d(d-1)^{t-1}$ and $q_t(R) \sim 1/[d(d-1)^{t-1}]$ from Lemmas [4.2](#lem-prefixes) and [4.1](#lem-density),
$$m \gtrsim d(d-1)^{t-1} \cdot \log\!\bigl[d(d-1)^{t-1}\bigr].$$
For fixed $d$, this is $m = \Omega((d-1)^{t-1} \cdot t)$.
*Inverting.* Suppose $m = \mathrm{polylog}(n)$. From $(d-1)^{t-1} \leq m$, taking logarithms gives
$$t - 1 \leq \frac{\log m}{\log(d-1)} = \frac{O(\log\log n)}{\log(d-1)},$$
so $t = O(\log\log n)$.
In particular, the certificate cannot certify $t = \Theta(\log n)$ rounds with $m = \mathrm{polylog}(n)$ cops. To certify $t = \Theta(\log n)$ rounds, the certificate requires $m = (d-1)^{\Theta(\log n)} \cdot \Theta(\log n) = n^{\Theta(\log(d-1))} \cdot \Theta(\log n)$ cops, which is polynomial in $n$ with exponent depending on $d$. [□]{.proof-qed}
::::
:::
### What the barrier says, precisely
The barrier rules out a specific proof technique: it does not say that the $\Theta(\log n)$-round local team-chase is impossible, only that it cannot be obtained by occupying every length-$t$ deep tube uniformly. The barrier identifies the cost as exponential in $t$, with the exponent base $d - 1$ — the local branching factor of the tree-ball.
::: {.annotation .annotation--static #rem-robust}
<div class="annotation-header">
<span class="annotation-label">Remark 4.3</span>
<span class="annotation-name">Robustness of the barrier</span>
</div>
<div class="annotation-body">
The barrier is robust to mild relaxations of the certificate. For example, requiring *most* (rather than all) tubes to be occupied still requires $\Omega((d-1)^{t-1})$ cops in expectation: the expected number of empty tubes is $N_t (1 - q_t)^m$, and bounding this even by $1$ gives $m \geq \log N_t / q_t$, which is the same order. Similarly, weakening the depth threshold from $2t + 1$ to any $\Theta(t)$ does not change the exponential dependence on $t$, only the constant in the exponent.
</div>
:::
### A finite-order potential-degeneration barrier
[Theorem 1.7](#thm-barrier) shows that occupying every length-$t$ deep tube is exponentially expensive in $t$ under uniform local sampling. A natural response is to compress the local state: instead of tracking every length-$t$ tube, one might try to use an invariant that remembers only bounded-order tube data, aggregating all information below a fixed depth. The next theorem shows that this cannot certify persistence beyond one additional round.
::: {.annotation .annotation--static #def-order-r-profile}
<div class="annotation-header">
<span class="annotation-label">Definition 4.4</span>
<span class="annotation-name">Order-$r$ tube profile</span>
</div>
<div class="annotation-body">
Fix a rooted tree-ball $B_R(v_0)$ in a $d$-regular graph. For a cop configuration $X$ and an integer $r \geq 1$, the *order-$r$ tube profile* of $X$ is the family
$$\Pi_r(X) := \bigl\{N_X(\sigma, j)\bigr\}_{\sigma, j},$$
where $\sigma = (v_0, v_1, \ldots, v_r)$ ranges over all nonbacktracking paths of length $r$ from $v_0$, and $j \geq r + 1$, and
$$N_X(\sigma, j) := \#\bigl\{x \in X : \operatorname{dist}(x, v_0) = j, \text{ and the unique geodesic from } x \text{ to } v_0 \text{ begins with } \sigma\bigr\}.$$
Thus $\Pi_r(X)$ records the exact depth histogram inside every length-$r$ tube, but forgets how the mass in that tube splits below level $r$.
</div>
:::
::: {.annotation .annotation--static #def-order-r-invariant}
<div class="annotation-header">
<span class="annotation-label">Definition 4.5</span>
<span class="annotation-name">Order-$r$ invariant</span>
</div>
<div class="annotation-body">
A local invariant $F$ on cop configurations in a rooted tree-ball is *order-$r$* if it factors through $\Pi_r$, i.e. if $F(X) = F(Y)$ whenever $\Pi_r(X) = \Pi_r(Y)$. It is *neighbor-symmetric* if it is invariant under rooted automorphisms of the tree-ball fixing $v_0$.
</div>
:::
::: {.annotation .annotation--static #thm-finite-order-barrier}
<div class="annotation-header">
<span class="annotation-label">Theorem 4.6</span>
<span class="annotation-name">Finite-order potential-degeneration barrier</span>
</div>
<div class="annotation-body">
Fix $d \geq 3$ and $r \geq 1$. Let $B_R(v_0)$ be a $d$-regular tree-ball with $R \geq 2r + 3$. Then there exist two cop configurations $X$ and $Y$ in $B_R(v_0)$ such that:
- **(i)** $\Pi_r(X) = \Pi_r(Y)$; in particular, every order-$r$ invariant takes the same value on $X$ and $Y$.
- **(ii)** There exists a nonbacktracking path
$$(v_0, v_1, \ldots, v_{r+1})$$
such that this robber path is safe in both configurations, but after $r+1$ rounds of cop gradient descent followed by robber motion along that path:
- in configuration $X$, the branch-load support at $v_{r+1}$ has size at least $2$;
- in configuration $Y$, the branch-load support at $v_{r+1}$ has size exactly $1$.
Consequently, no certificate whose hypotheses depend only on order-$r$ data can correctly distinguish $\geq 2$-branch support from $1$-branch support after $r+1$ rounds.
</div>
:::
::: {.exhibit .exhibit--proof data-exhibit-name="Finite-order potential-degeneration barrier" data-exhibit-type="proof" data-exhibit-caption="Two two-cop configurations agree on order-r tube data but differ on which children of v_{r+1} they cover after r+1 forced upward steps."}
:::: exhibit-body
Fix any nonbacktracking path
$$(v_0, v_1, \ldots, v_{r+1})$$
inside the tree-ball. Since $d \geq 3$, the vertex $v_{r+1}$ has at least two distinct children in the rooted tree. Choose two such children and call them $a$ and $b$.
We now define two cop configurations, each consisting of exactly two cops, both at depth $2r + 3$ from $v_0$.
*Configuration $X$.* Place one cop in the subtree rooted at $a$ and one cop in the subtree rooted at $b$, both at total depth $2r + 3$ from $v_0$.
*Configuration $Y$.* Place both cops in the subtree rooted at $a$, at distinct vertices, again at total depth $2r + 3$ from $v_0$. (Since the depth-$(r+1)$ shell of the $a$-subtree has $(d-1)^{r+1} \geq 2$ vertices, this is possible.)
In both configurations, every cop lies in the same length-$r$ tube determined by the prefix
$$\sigma_0 = (v_0, v_1, \ldots, v_r),$$
and every cop lies at the same total depth $2r + 3$ from $v_0$. Therefore the order-$r$ tube counts are identical:
$$N_X(\sigma_0, 2r+3) = N_Y(\sigma_0, 2r+3) = 2,$$
and $N_X(\sigma, j) = N_Y(\sigma, j) = 0$ for all other $(\sigma, j)$. Hence $\Pi_r(X) = \Pi_r(Y)$. The order-$r$ invariant cannot see how the two cops in $\sigma_0$ split between the children $a$ and $b$ of $v_{r+1}$.
Now consider the robber path
$$(v_0, v_1, \ldots, v_{r+1}).$$
We claim it is safe in both configurations. Each cop starts at depth $2r + 3$ from $v_0$. In a tree-ball, gradient descent toward any robber position on the cop's geodesic to $v_0$ is forced: each round the cop moves exactly one step upward toward $v_0$. Hence after $s$ rounds, each cop is at depth $2r + 3 - s$ from $v_0$. After the cop's move at round $s$, the robber moves $v_{s-1} \to v_s$. The cop's distance to the robber's destination $v_s$ is
$$(2r + 3 - s) - s = 2r + 3 - 2s.$$
For every $1 \leq s \leq r + 1$ this is at least $1$, so no cop ever lands on $v_s$ during these $r + 1$ rounds. Thus the robber path is safe in both configurations.
Finally, inspect the cop locations after $r + 1$ rounds. Each cop has moved upward by exactly $r + 1$ steps, so each is now at depth
$$2r + 3 - (r + 1) = r + 2$$
from $v_0$, while $v_{r+1}$ has depth $r + 1$. Hence each cop is now a child of $v_{r+1}$.
In configuration $X$, the two cops lie on two distinct children of $v_{r+1}$, namely $a$ and $b$. Therefore the branch-load support at $v_{r+1}$ has size at least $2$.
In configuration $Y$, both cops climbed within the $a$-subtree and now sit on its root, namely $a$ itself (since gradient descent in a tree-ball is forced upward). Therefore both cops contribute to the same branch at $v_{r+1}$, and the branch-load support has size exactly $1$.
This proves (ii). Since $\Pi_r(X) = \Pi_r(Y)$ but the outcomes differ, no certificate depending only on order-$r$ data can correctly classify both configurations. [□]{.proof-qed}
::::
:::
::: {.annotation .annotation--static #cor-first-order-insufficient}
<div class="annotation-header">
<span class="annotation-label">Corollary 4.7</span>
<span class="annotation-name">First-order data is insufficient</span>
</div>
<div class="annotation-body">
Any invariant depending only on first-level branch data, even with exact depth histograms inside each first-level branch, cannot certify universal $2$-round persistence in the tree-ball model.
</div>
:::
::: {.exhibit .exhibit--proof data-exhibit-name="First-order data is insufficient" data-exhibit-type="proof" data-exhibit-caption="Apply the finite-order barrier with r = 1."}
:::: exhibit-body
Take $r = 1$ in [Theorem 4.6](#thm-finite-order-barrier). [□]{.proof-qed}
::::
:::
::: {.annotation .annotation--static #rem-complementary}
<div class="annotation-header">
<span class="annotation-label">Remark 4.8</span>
<span class="annotation-name">Complementarity of the two barriers</span>
</div>
<div class="annotation-body">
Theorems [1.7](#thm-barrier) and [4.6](#thm-finite-order-barrier) are complementary. [Theorem 1.7](#thm-barrier) shows that full order-$t$ path information is sufficient but exponentially expensive to certify by uniform random local sampling. [Theorem 4.6](#thm-finite-order-barrier) shows that any bounded-order compression of that information is dynamically insufficient beyond one additional round. Together they imply that, in the tree-ball model, there is no bounded-order shortcut around path entropy: certifying $t$-round persistence by a local invariant requires at least order-$t$ resolution, and order-$t$ resolution requires $\Omega((d-1)^{t-1})$ cops to occupy. Together they rule out order-$r$ branch-aggregated potentials for every fixed $r$ as a route to $\Theta(\log n)$-round chase from polylogarithmic cops.
</div>
:::
## Discussion: Pivots and Open Directions
The barriers of Theorems [1.7](#thm-barrier) and [4.6](#thm-finite-order-barrier) together rule out a substantial family of approaches but leave several escape routes open.
### Pivot A: Epoch refresh
Run epochs of length $t_0 = O(\log\log n)$, for which the tube certificate is affordable with $\mathrm{polylog}(n)$ cops, then re-randomize and restart. The barrier does not preclude this: it constrains a single epoch but not a sequence of epochs.
The crucial open question is whether one $t_0$-round epoch produces *permanent* progress (a coarser robber freedom parameter shrinks by a constant factor). In a tree-ball, no such parameter is obvious: the robber's territory is unbounded by hypothesis, and after one epoch the robber can simply retreat to a fresh subtree. This suggests that epoch refresh in the pure tree-ball model is unlikely to succeed, and the program must combine the local tree-ball analysis with a global structural bound that limits the robber's total territory.
### Pivot B: Soft potentials must break locality or symmetry
[Theorem 4.6](#thm-finite-order-barrier) rules out any *order-$r$* potential as a route to $(r+1)$-round persistence. To escape the barrier via a soft potential, the potential must therefore depend on at least one of:
- *Unbounded local order:* information that depends on tube data of order growing with $t$. This re-introduces path-entropy costs and hits [Theorem 1.7](#thm-barrier).
- *Non-local data:* information that depends on the global graph structure beyond the local tree-ball, such as cycle structure at radius $> R$, or spectral data of the full graph.
- *Asymmetric data:* information that breaks the rooted-tree symmetry, such as a fixed orientation, an external labeling, or a privileged subset of vertices not invariant under tree automorphisms.
The most promising candidate among these is the second: *nonbacktracking-walk concentration* of cop mass on the full graph. If cops are placed by a global averaging that respects the nonbacktracking spectrum, the resulting cop distribution at time $t$ may concentrate on the robber's location at exponential rate $\rho(B)^{-1}$, where $B$ is the Hashimoto nonbacktracking transition matrix of the full graph. The local tree-ball view of such a placement is no longer order-$r$ for any fixed $r$; it inherits global spectral information that is invisible to any local invariant. This suggests that purely local tree-ball analysis is fundamentally insufficient.
### Pivot C: Hierarchical packets
Use cop packets at multiple depth scales. A packet at scale $j$ guarantees persistence for $2^j$ rounds inside a tube of depth $\Theta(2^j)$. Packets are activated sequentially, with later packets covering longer ranges. This is a multiscale variant of refresh.
A naive cost analysis is discouraging: each scale costs $(d-1)^{2^j}$ cops by the barrier of [Theorem 1.7](#thm-barrier), and at the top scale $j = \log\log n$ this is already $(d-1)^{\log n} = n^{\log_2(d-1)}$, which is polynomial in $n$. So the straightforward hierarchical-packet scheme does not deliver polylogarithmic cop count. This matches the cost of [Theorem 1.7](#thm-barrier) inverted at $t = \Theta(\log n)$: the multi-scale scheme inherits the same fundamental cost as the single-scale scheme it tries to circumvent. To salvage the idea one would need a sub-exponential variant in which packets at scale $j$ exploit structure (epoch refresh between scales, partial coverage of tubes, or shared cops across scales) to evade the barrier. We do not pursue this here.
### Beyond the tree-ball model
The deepest limitation of the present results is the tree-ball hypothesis. Real expanders — including Ramanujan graphs of girth $\Theta(\log n)$ — are tree-balls only up to radius $\Theta(\log n)$, but a chase argument typically needs to reason about radii $r \approx \log n / \gamma$ where $\gamma$ is the spectral gap. For $\gamma$ bounded below by a constant, these match up to constants; but for $\gamma$ small, the chase exits the tree-ball and must contend with cycles. In the cyclic regime, the geodesic cone $C_u(v, r)$ is no longer a clean object (vertices may have multiple shortest paths to $v$), and the entire branch decomposition becomes ill-defined.
A full local team-chase theorem on expanders likely requires either (i) a generalization of the present analysis to graphs with bounded but nontrivial girth, where "tubes" become equivalence classes of approximately-tree-like geodesics, or (ii) a fundamentally different invariant — spectral, fence-based, or potential-theoretic — that does not rely on tree branch decomposition at all.
## Conclusion
We have proved a sharp local persistence theorem in the tree-ball model and two matching barriers: a sampling barrier showing that the natural iteration of the one-round argument is exponentially expensive in time, and a finite-order potential-degeneration barrier showing that any bounded-order compression of the path-tube certificate is dynamically insufficient. Together they tighten the trade-off: certifying $t$-round persistence by an order-$r$ local invariant requires $r \geq t$, and order-$t$ resolution requires $\Omega((d-1)^{t-1})$ cops to occupy by uniform sampling. Together, they rule out order-$r$ branch-aggregated potentials (for any fixed $r$) as a route to $\Theta(\log n)$-round chase from polylogarithmic cops in the tree-ball model. A structural new ingredient — non-local information, broken symmetry, or escape from the tree-ball regime entirely — is required.
The four main results — deep-load occupancy, $t$-round persistence with depth budget $2t+1$, the sampling barrier, and the finite-order potential-degeneration barrier — together with the supporting tube-counting lemma form a self-contained module that we hope will be useful as one block in any future proof of Meyniel's conjecture via local chase arguments. The pivots in [Section 5](#discussion-pivots-and-open-directions) suggest several directions in which the program could be continued, though we leave their analysis to future work.
## Acknowledgments
I warmly thank Eric Jovinelly, who led me through Graph Theory at Brown and greatly assisted me in building both graph theoretic and broader mathematical skills and intuition. I also thank my peers in the Spring 2026 S02 of Graph Theory at Brown.