Proof. Unique factorization and multiplication of vertex weights give the geometric series with nonnegative terms. The finite strip estimate makes \(b(s,0)\) finite for \(s>0\), so \(\sum p(\gamma)e^{-sH(\gamma)}<1\). Monotone convergence as \(s\downarrow0\) gives \(\sum p\le1\). On the other hand, summing \(B_h\asymp(1+h)^{-1/4}\) gives \(b(s,0)\asymp s^{-3/4}\to\infty\), forcing \(\sum p=1\). Positive heights make the events that the \(k\)th renewal has height \(h\) disjoint as \(k\) varies. Summing their probabilities yields the last assertion. ◻

Proof. Mark one piece in a bridge and then cut immediately before and after it. The pieces before the mark form an arbitrary bridge, the mark is one irreducible, and the pieces after it form another arbitrary bridge. This is a bijection, with multiplicative weights. Tonelli’s theorem gives both formulas. Since length is additive across ports, summing the lengths of all pieces gives the full visited-vertex length. ◻

Proof. The first assertion follows directly from Proposition 2. Since \(1-e^{-H/x}\ge1-e^{-1}\) on \(H>x\), it gives the height tail.

We first prove the diameter tail, which is not a consequence of the height tail alone. Set \(s=A/x\) with a fixed large \(A\). Give all finite bridges their normalized weight proportional to \(\rho^L e^{-sH}\); its normalizer is \(b(s,0)\). The mass of bridges of diameter greater than \(x\) is at most \[Cx^{-1/4}\sum_{h\ge1}e^{-sh}\le Cx^{-1/4}s^{-1}.\] Relative to \(b(s,0)\ge c s^{-3/4}\) this is at most \(C A^{-1/4}\). Choose \(A\) so that this is at most \(1/2\). Put \(q=\mathbb E_p[e^{-sH};D>x]\) and \(d=1-\mathbb E_p e^{-sH}\). The relative mass of bridge lists containing at least one such piece is exactly \[1-\frac{d}{d+q}=\frac{q}{d+q}.\] A piece of diameter greater than \(x\) forces the whole bridge to have that diameter. Thus \(q\le d\le C_Ax^{-3/4}\). On \(H\le x\), \(e^{-sH}\ge e^{-A}\); on \(H>x\) use the height tail. This proves \(p(D>x)\le Cx^{-3/4}\).

For the reward bound, retain in Lemma 3 only the marked pieces of height at most \(x\) and outer bridges of heights at most \(x\). Their total height is at most \(3x\), whence \[\left(\sum_{h\le x}B_h\right)^2\mathbb E_p[L;H\le x] \le\sum_{h\le3x}M_h\le Cx^{25/12}.\] Since the squared factor on the left is at least \(cx^{3/2}\), the reward is at most \(Cx^{7/12}\). Taking \(x=n^{3/4}\) and splitting according to \(H>x\) gives \[p(L>n)\le p(H>x)+n^{-1}\mathbb E_p[L;H\le x] \le Cn^{-9/16}.\] Integration of this tail yields \(\mathbb E_p(1-e^{-uL})\le Cu^{9/16}\).

For the reverse bound take \(s=R^{-1}\) and \(u=R^{-4/3}\). On every height \(R\le h\le2R\), Theorem 1 retains critical mass at least \(cR^{-1/4}\) with \(L\ge cR^{4/3}\). Consequently \[b(s,0)-b(s,u)\ge cR^{3/4}\ge c' b(s,0).\] Writing \(d_s=1-\mathbb E_p e^{-sH}\) and \(e_{s,u}=\mathbb E_p[e^{-sH}(1-e^{-uL})]\), the renewal identity makes the relative difference equal to \(e_{s,u}/(d_s+e_{s,u})\). The displayed lower bound therefore implies \(e_{s,u}\ge c d_s\ge cR^{-3/4}\). Finally \(\mathbb E_p(1-e^{-uL})\ge e_{s,u}\) and \(R^{-3/4}=u^{9/16}\). ◻

Proof. Fix \(\epsilon>0\) and let \(k=2^j\). Put \(t=k^{1/a+\epsilon}\). The probability of a summand above \(t\) is at most \(k t^{-a+o(1)}\). Moreover \(\mathbb E[Y\mathbf 1_{Y\le t}]\le t^{1-a+o(1)}\), by integrating the tail, so Markov’s inequality bounds the probability that the truncated sum exceeds \(t\) by the same expression. This is a summable geometric sequence in \(j\). For the lower bound put \(t=k^{1/a-\epsilon}\) and \(u=t^{-1}\). Exponential Markov gives \[\mathbb P\{\textstyle\sum_{i\le k}Y_i\le t\} \le e\big(\mathbb Ee^{-Y/t}\big)^k \le\exp\{1-k t^{-a+o(1)}\},\] also summable. Borel–Cantelli, monotonicity between consecutive dyadic \(k\), and a countable sequence of \(\epsilon\downarrow0\) prove the claims. ◻

Proof. Lemma 6 and Proposition 4 give \[\sum_{i\le k}L_i=k^{16/9+o(1)},\qquad \sum_{i\le k}H_i=k^{4/3+o(1)},\qquad \sum_{i\le k}D_i\le k^{4/3+o(1)}\] on one event of probability one. If vertex \(n\) lies in piece \(k+1\), it is above the last completed renewal boundary and within the total diameter of the first \(k+1\) pieces. The first of the three bounds implies \(k=n^{9/16+o(1)}\): both consecutive partial sums have the same exponent, so an exceptionally long current piece creates no missing interpolation interval. The height lower bound and diameter upper bound give the first conclusion. For each fixed \(\epsilon>0\), all sufficiently large \(n\) then obey \(n^{3/4-\epsilon}\le |\gamma_n-\gamma_1|\le n^{3/4+\epsilon}\). Inverting these two inequalities gives the asserted count of vertices in a ball. ◻

Proof. First \(a_n^{1/n}\to\rho^{-1}\). The upper bound is the honeycomb connective-constant theorem [4]. For the lower bound, let \(b_m\) count strict bridges of length \(2m\) over all heights. Concatenation at a prescribed length is injective, so \(b_{m+j}\ge b_mb_j\); also \(b_1>0\). The divergence \(\sum_m b_m\rho^{2m}=\sum_hB_h=\infty\) and supermultiplicativity imply the required root bound along even lengths. Length-two bridges fill the bounded remainders after repeating a fixed long block. Extending the terminal port by one vertex gives the odd-length bound.

A path is specified, up to bounded initial choices, by its left/right turn word. Runs have length at most four, since five equal turns close an elementary hexagon. Except for an exponentially negligible fraction at length \(n\), there are at least \(\delta n\) maximal runs of length three, for some fixed \(\delta>0\). Indeed, compositions into parts \(1,2,4\) have growth less than \(\rho^{-1}\) because \(\rho+\rho^2+\rho^4<1\). Assign a sufficiently small positive weight \(v<1\) to every part of length three. The geometric composition series still has growth at most \(\lambda<\rho^{-1}\); the words with at most \(\delta n\) such parts number at most \(C(v^{-\delta}\lambda)^n\). Choose \(\delta\) small enough and use the root bound above.

Such a run follows a four-edge arc of a hexagonal face, with exterior entrance and exit edges. The one omitted face vertex cannot be used internally elsewhere, because two of its neighbors are already saturated. Discard faces incident to either endpoint and the bounded number of runs at the ends of the word. The remaining faces are wholly in the strict half-plane: a honeycomb face intersecting its boundary has at most three vertices strictly above it. Each retained four-edge arc may therefore be replaced by the complementary two-edge arc.

Call a wholly allowed face a slot if it avoids the endpoints, is traversed in one consecutive two- or four-edge arc using exterior entrance and exit edges, and has all its remaining vertices unused. A flip changes the length by two. Color faces with finitely many colors so that faces of one color have disjoint vertex sets. For a fixed color the slot set is unchanged by its flips. Each resulting equivalence class with \(k\) slots has \(\binom{k}{j}\) paths of lengths \(n_0+2j\), where \(j\) is the number of four-edge arcs.

Put \(\theta=\rho^2/(1+\rho^2)\). For each color and each fixed \(\epsilon>0\), paths of length \(n\) for which \(k\ge\epsilon n\) but \(|j/k-\theta|\ge\epsilon\) have exponentially negligible proportion. To see this, weight every path by \(\rho^{\text{length}}\). In each class shift \(j\) towards \(k\theta\) by \(\lfloor\eta n\rfloor\) steps, with a sufficiently small fixed \(\eta>0\), using the same sign for each of the two deviation classes. The binomial successive ratios show that the new weighted class count exceeds the old by \(e^{c n}\). The target lengths are \(n\pm2\lfloor\eta n\rfloor\), whose total critical counts are \(e^{o(n)}\) by the root limit; the original denominator is also \(e^{o(n)}\). The claim follows. Summing over colors, the fraction of four-edge arcs among all slots therefore tends in probability to \(\theta\). Colors with fewer than \(\epsilon n\) slots contribute at most a fixed multiple of \(\epsilon n\), while the total slot count is at least \(\delta n-O(1)\) with probability tending to one.

Double-count the edges representing one flip between lengths \(n\) and \(n+2\), weighting an edge by the reciprocal of the larger endpoint slot count. One flip changes that count by at most a fixed constant, since only neighboring faces can change their status. The mean incident weight tends to \(1-\theta\) at length \(n\) and to \(\theta\) at length \(n+2\). The incident sums are at most one, so the negligible exceptional paths do not affect the limits. Hence \(a_n(1-\theta+o(1))=a_{n+2}(\theta+o(1))\), proving the ratio limit.

Fix an ordered list of \(k\) irreducibles, of total length \(m\). Paths which start with this list and then remain strictly above its top boundary have exactly \(a_{n-m}\) possible continuations. Here \(m\) is even. The probability of this event tends to \(\rho^m\), the product of the specified irreducible masses. For fixed \(k\), the events over all such lists are disjoint and their limiting probabilities sum to one. Truncation to a finite set of lists therefore proves convergence on every fixed initial path. ◻

Proof. Evaluate the Laplace deficit at \(A/x\). Its contribution from \(X>x\) is at most \(Cx^{-a}\), while its contribution from \(X\le x/M\) is at most \[\frac A x\mathbb E[X;X\le x/M]\le C A M^{a-1}x^{-a}.\] Choose \(A\) large enough that the lower deficit \(cA^ax^{-a}\) dominates the first term, and then \(M\) large enough to dominate the second. The remaining contribution is at most \(\mathbb P(x/M<X\le x)\), proving the claim.

Let \(\phi(\theta)=\mathbb Ee^{i\theta X}\). Use two independent copies in \(1-|\phi(\theta)|^2=\mathbb E[1-\cos(\theta(X-X'))]\). Keep \(X'\) in a fixed bounded set of positive probability and \(X\) in the annulus \([\epsilon/(M|\theta|),\epsilon/|\theta|]\), with a fixed small \(\epsilon>0\). For sufficiently small \(|\theta|\), the phase is bounded above by a small constant and below in magnitude by a fixed positive constant depending only on \(\epsilon,M\). The annular bound therefore gives \(|\phi(\theta)|\le e^{-c|\theta|^a}\). Away from zero, span one gives a uniform bound strictly below one. Fourier inversion yields \[\sup_j\mathbb P(S_k=j)\le(2\pi)^{-1}\int_{-\pi}^{\pi}|\phi(\theta)|^k\,d\theta \le C k^{-1/a}.\] Finally, either some \(X_i>t\) or \(\sum_iX_i\mathbf 1_{X_i\le t}>t\). The union bound and Markov’s inequality, using \(\mathbb E[X;X\le t]\le Ct^{1-a}\), give the sum-tail estimate. ◻

Proof. For \(u>0\) tilt the law by \[p_u(j)=\frac{e^{-uj}\mathbb P(X=j)}{Z(u)},\qquad Z(u)=\mathbb Ee^{-uX}.\] Write \(\mu_u,v_u\) for its mean and variance. Tail integration gives the upper bounds for its first three moments. The annular mass at scale \(u^{-1}\) gives the lower bounds for the first two moments; for the variance, apply the identity \(2v_u=\mathbb E_u(X-X')^2\), keeping one variable bounded and the other in that annulus. Thus \[\mu_u\asymp u^{a-1},\quad v_u\asymp u^{a-2},\quad \mathbb E_u|X-\mu_u|^3\le Cu^{a-3}.\] Let \(\phi_u\) be the tilted characteristic function. The same two-copy argument, now using the annulus of size \(u^{-1}\) for \(|\theta|\le u\) and of size \(|\theta|^{-1}\) for \(u\le|\theta|\le\theta_0\), gives uniform bounds \[ |\phi_u(\theta)|\le \begin{cases} e^{-cv_u\theta^2},& |\theta|\le u,\\ e^{-c|\theta|^a},& u\le|\theta|\le\theta_0,\\ 1-c,&\theta_0\le|\theta|\le\pi. \end{cases} \tag{1}\] In the first annulus the tilt is bounded below by a positive constant, and the phase is comparable to \(|\theta|/u\). In the second it is again bounded below because \(u/|\theta|\le1\). The final line follows from span one and the total-variation convergence \(p_u\to p\).

Fix a large constant \(A\) and put \(u=A/m\). Consider integers \(k\) with \[|m-k\mu_u|\le\tfrac1{10}\sqrt{k v_u}.\] There are at least \(c_A m^a\) such integers for all large \(m\). Indeed their center is \(m/\mu_u\asymp_A m^a\) and their interval width is comparable to \(\sqrt{(m/\mu_u)v_u}/\mu_u\asymp_A m^a\). On this interval \(ku^a\asymp A\), and \(\sqrt{kv_u}\asymp_A m\).

We verify a uniform local lower bound under the tilted law. Write \(\sigma=\sqrt{kv_u}\). On \(|\theta|\le\omega/\sigma\), Taylor expansion of the centered characteristic function gives its Gaussian approximation with error at most \(C_\omega A^{-1/2}\), since \[\frac{k\mathbb E_u|X-\mu_u|^3}{(kv_u)^{3/2}} \le \frac C{\sqrt{ku^a}}\le\frac C{\sqrt A}.\] After the change of variables \(t=\sigma\theta\), Fourier inversion over this interval therefore differs by at most \(C_\omega A^{-1/2}/\sigma\) from \[\frac1{2\pi\sigma}\int_{-\omega}^{\omega} e^{-t^2/2}e^{-it(m-k\mu_u)/\sigma}\,dt.\] The latter has real part at least \(c/\sigma\), uniformly for the allowed \(k\), once \(\omega\) is large enough. The portion \(\omega/\sigma<|\theta|\le u\) is at most \(C e^{-c\omega^2}/\sigma\) by the first line of (1). The portion \(u<|\theta|\le\theta_0\) is at most \(Cu e^{-c'A}\), by substituting \(|\theta|=u t\) into the second line; relative to \(1/\sigma\) this is at most \(C\sqrt A e^{-c'A}\). The final portion is exponentially small in \(k\). Choose first \(\omega\) and then \(A\) large enough. We obtain \[\mathbb P_u(S_k=m)\ge c_A/m.\] Removing the tilt multiplies this by \(e^{um}Z(u)^k\). Since \(um=A\) and \(k(1-Z(u))=O(A)\), this factor is bounded below by a positive constant depending only on \(A\). Summing over the \(c_A m^a\) choices of \(k\) proves the theorem. ◻

Proof. Port bridge lengths are even: their first and last triangles have opposite types. The irreducible law admits two- and four-vertex zigzags in one row, so \(X=L/2\) has atoms at \(1\) and \(2\) and has span one. Proposition 4 verifies the hypotheses of Theorem 10 with \(a=\alpha=9/16\). Thus, writing \(m=n/2\), the total critical bridge mass at that length is \[V_m=\sum_{k\ge1}\mathbb P_p\{\textstyle\sum_{i\le k}L_i=2m\} \ge c m^{\alpha-1}.\] All bridges of this length have the same critical weight, so normalization by \(V_m\) gives the uniform law.

We first restrict the number \(k\) of pieces. For \(k\ge2\) split the list into two comparable halves. If the total \(X\)-sum is \(m\), one half is at least \(m/2\). Lemma 9 bounds that probability by \(Ckm^{-\alpha}\); the other independent half pays at most \(Ck^{-1/\alpha}\) for the required residual length. Therefore, for any fixed \(\delta>0\), \[\sum_{2\le k\le m^{\alpha-\delta}}\mathbb P(S_k=m) \le C m^{-\alpha}\sum_{k\le m^{\alpha-\delta}}k^{1-1/\alpha} \le C m^{\alpha-1-\delta(2-1/\alpha)}.\] The exponent \(2-1/\alpha=2/9\) is positive. The \(k=1\) term is at most \(Cm^{-\alpha}=o(m^{\alpha-1})\). At the other end, exponential Markov at discount \(1/m\) gives \[\mathbb P(S_k=m)\le e\exp(-c k m^{-\alpha}),\] so the sum over \(k>m^{\alpha+\delta}\) is negligible. Here \(k\le m\) because \(X\ge1\).

For the remaining \(k\), total height below \(t=m^{3/4-\xi}\) costs at most \[\mathbb P\{\textstyle\sum H_i\le t\} \le e\exp(-c k t^{-\beta}) \le e\exp(-c m^{\beta\xi-\delta}),\] which is negligible if \(\delta<\beta\xi\). This step does not impose a length condition, so it is a valid upper bound for the exceptional length mass as well.

For the diameter upper bound put \(r=m^{3/4+\xi}\). If \(\sum_{i\le k}D_i>r\), one of the two halves has diameter sum exceeding \(r/2\). The diameter tail and the same truncation argument as in Lemma 9 bound this by \(Ckr^{-\beta}\). Pay for the exact length using the independent other half. Summing over \(k\le m^{\alpha+\delta}\) gives \[C r^{-\beta}\sum_{k\le m^{\alpha+\delta}}k^{1-1/\alpha} \le C m^{\alpha-1-\beta\xi+\delta(2-1/\alpha)} =o(V_m)\] for sufficiently small \(\delta\). Independence was used between disjoint sets of pieces, never between the length and diameter of a piece. Finally \(d_0\sum H_i\le A\le D\le\sum D_i+O(1)\). Apply the argument with slightly smaller tolerances to absorb fixed geometric constants. ◻

Proof. The mean length is \(M_h/B_h\le Ch^{4/3}\), proving the upper length tail at every positive exponent slack. The height variable has Laplace deficit comparable to \(s^\beta\) and tail \(O(x^{-\beta})\). Its support is an unbounded subset of the integers. The modulus-one characters therefore form a finite set on the Fourier torus; the same near-origin estimate translates to each of them. Lemma 9’s Fourier proof gives \(\sup_j\mathbb P(\sum_{i\le k}H_i=j)\le Ck^{-1/\beta}\). Splitting into halves as above shows \[\sum_{k\le h^{\beta-\delta}}\mathbb P\{\textstyle\sum_{i\le k}H_i=h\} \le Ch^{-\beta}\left(1+\sum_{2\le k\le h^{\beta-\delta}}k^{1-1/\beta}\right) =o(h^{\beta-1})=o(B_h).\] For \(k>h^{\beta-\delta}\), length below \(h^{4/3-\epsilon}\) costs at most \(\exp\{1-c h^{\alpha\epsilon-\delta}\}\) by the length deficit. There are at most \(h\) possible counts, so choosing \(\delta<\alpha\epsilon\) proves the lower length tail. The diameter is at least \(d_0h\); its upper tail at \(h^{1+\epsilon}\) has relative mass at most \(Ch^{-\epsilon/4}\) by Theorem 1. A restriction retaining a fixed fraction of the denominator multiplies all these vanishing upper bounds by only a constant. ◻

Proof. Take \(t=h^{-4/3}\). On the length window in (2), \(1-e^{-tL}\ge c(\log h)^{-1/8}\). Summing this damping loss over the \(O(h(\log h)^{1/64})\) allowed heights shows that some height \(1\le h_*\le2R_0\) has loss at least \(h^{-1/4}(\log h)^{-C'}\). For a bridge with irreducible lengths \(L_i\), \(1-e^{-t\sum_iL_i}\le\sum_i(1-e^{-tL_i})\). The marked reward identity of Lemma 3 therefore bounds this loss above by \[\sum_{j\le h_*}\mathbb E_p[1-e^{-tL};H=j] \sum_{b+d=h_*-j}B_bB_d \le C h_*^{1/2}q_L(t).\] Here \(B_j\asymp(1+j)^{-1/4}\) gives the convolution bound. Since \(h_*\le2h(\log h)^{1/64}\), we obtain \[q_L(h^{-4/3})\ge c h^{-3/4}(\log h)^{-P}\] for some finite \(P\), proving the lower estimate.

For the upper estimate fix a sufficiently large confinement constant \(T\) in Proposition 13. At each height \(d\le h\), its confined bridges have mass at least \(c_Td^{-1/4}\) and first-length mass at most \(C_Td^{13/12}\). Choosing a large fixed \(C'\) and restricting to \(L\le C'h^{4/3}\) preserves at least half the mass: the discarded fraction is at most a constant times \(d^{4/3}/(C'h^{4/3})\). On the retained paths \(e^{-tL}\ge e^{-C'}\). Summing over \(d\le h\) gives damped bridge mass at least \(c h^{3/4}\). By renewal the total damped bridge mass over all heights is \(1/q_L(t)\), so \(q_L(h^{-4/3})\le Ch^{-3/4}\). ◻

Proof. A block law from the finite window. The height deficit gives \[p\{S_H(k)\le2R_0\}\le e\exp(-c kR_0^{-a}).\] Consequently, the contribution of counts \(k>h^a\log^2h\) to (2) is negligible. Dividing the remaining window mass by this number of counts yields a block whose probability of lying in the length window is bounded below by a negative power of \(\log h\). Since \(l_-/l_+=(\log h)^{-5/8}\), fix any \(b\in(5/8,1)\), say \(b=3/4\). For each sufficiently large \(r\), choose \(h\) so that \(l_+\) is comparable to \(r\) and \(2l_+\le r/2\). Then there is a deterministic integer \(k(r)\) such that \[ p\{r(\log r)^{-b}\le S_L(k(r))\le r/2\} \ge(\log r)^{-C}. \tag{3}\] Fix one such choice for every \(r\). We will use these blocks to advance through a large remaining gap without passing its target.

Proof. Put \(\vartheta_r=1-p(e^{-H/r})\). The renewal identity gives \[\vartheta_r^{-1}=\sum_{h\ge0}e^{-h/r}B_h\asymp r^{3/4}.\] After normalization, this bridge sum consists of a geometric number of increments with one-increment law proportional to \(e^{-H/r}p\). Its number of increments has mean of order \(r^{3/4}\). The inequality \(1-e^{-H/r}\ge(1-e^{-1})\mathbf1_{\{H\ge r\}}\) first gives the height tail.

The diameter of the whole bridge divided by \(r\) is tight in this normalized sum, by (6), summation over heights, and the exponentially small contribution of heights much larger than \(r\). An increment of diameter greater than \(Mr\) forces the same diameter event for the whole bridge. The elementary formula for a geometrically stopped list therefore gives \(p(e^{-H/r};D>Mr)\le C_M\vartheta_r\) for one sufficiently large fixed \(M\). On \(H\le r\) the damping is bounded below; the already proved height tail handles \(H>r\). Rescaling by the fixed \(M\) proves (9), and integration of that tail proves the truncated first-moment bounds.

The lower bound for \(\vartheta_r\) cannot be supplied by very small or very large heights. Indeed their contributions to \(p(1-e^{-H/r})\) are at most \(C\delta^{1/4}r^{-3/4}\) for \(H\le\delta r\) and \(C M^{-3/4}r^{-3/4}\) for \(H>Mr\). Choosing \(\delta\) small and \(M\) large proves (10).

In the normalized bridge sum, heights in \([r,2r]\) have probability bounded below. At any such height, the pointwise bound in (6) shows that terminal positions in an interval of width \(\delta r\) have probability at most \(C\delta\). Thus the total horizontal displacement has magnitude at least \(c r\) with probability bounded below. The expected sum of \(\min(|X_i|,\delta r)\) over the geometric list is at most \(C\delta^{1/4}r\); increments of diameter greater than \(Mr\) have arbitrarily small probability when \(M\) is large. Hence, with probability bounded below, the list contains an increment with \(c' r<|X_i|<C'r\) and \(H_i<C'r\). Its expected number is an upper bound for that probability. The geometric mean is of order \(r^{3/4}\) and the tilted probability of this event is at most a constant times its \(p\) probability. This proves (11). ◻

Proof. Use independent increments \(Y,Y'\) in \(1-|\varphi(\theta)|^2=p\otimes p[1-\cos(\theta\cdot(Y-Y'))]\). Keep \(Y'\) in a fixed bounded set of positive probability. At scale \(r=\epsilon/|\theta|\), Lemma 16 supplies mass \(c r^{-3/4}\) with the coordinate whose coefficient has larger absolute value between fixed positive multiples of \(r\) in magnitude. For the height shell, discard \(D>Mr\) with fixed \(M\) sufficiently large: the diameter tail removes at most \(CM^{-3/4}r^{-3/4}\), leaving a fixed fraction of that shell mass. Both coordinates are now bounded by a fixed multiple of \(r\); the horizontal shell already has this property. Reflection replaces \(X\) by \(-X\) without changing \(H\). One of these two signs avoids cancellation between the two coordinates in the scalar product. Choose the fixed \(\epsilon>0\) small enough to avoid wrapping the resulting phase around \(2\pi\). The integrand is then bounded below by a positive constant on a fixed fraction of this mass. This gives the claimed loss near zero. The same argument using only \(H\) proves the one-dimensional loss.

The support differences span two dimensions: otherwise reflection would first force an affine relation involving only \(H\), and (10) rules that out. A finite subset of the differences therefore generates a rank-two sublattice. Thus \(\Gamma_1\) has finite index in \(\Gamma\). Equality in the triangle inequality for \(p(e^{i\theta\cdot Y})\) occurs exactly when that character is constant on the support, which gives precisely the stated finite set. At each of these characters its modulus is a translate of the modulus near zero. Away from their neighborhoods compactness gives a bound strictly below one. Fourier inversion, with \(\int_{\mathbb R^d}e^{-c k|\theta|^{3/4}}d\theta=O(k^{-4d/3})\), proves both atom estimates. No span-one assumption in two dimensions is being made. ◻

Proof. For a polygon of diameter in \([c r,C'r]\), let \(K\) count horizontal row cuts crossed exactly twice. Mark \(s\) of these cuts, allowing repeated marks, and order their distinct levels. Outside the extreme cuts there are two arches. Between successive cuts there are two bridges. Anchor the first cut and its left crossing port, sum its arch gap using \(\sum_b a_b<\infty\), and drop all further disjointness restrictions. Each internal gap contributes at most \(\sum_{t\le C''r}B_t^2\le C\sqrt r\). Repeated marks and the choices of the extreme cuts contribute a fixed polynomial in \(r\) whose degree does not depend on \(s\). Consequently \[\sum_{D(P)\le C'r}w(P)K(P)^s\le C_s r^{C_0+s/2}.\] Apply Hölder with arbitrarily large fixed \(s\) and the first two bounds in (7). For the length-weighted measure use also \(|P|\le C r^2\) in the high-moment bound. We obtain \[\sum_{c r\le D(P)\le C'r}w(P)K(P)\le r^{-3/2+o(1)},\qquad \sum_{c r\le D(P)\le C'r}|P|w(P)K(P)^2\le r^{1/3+o(1)}.\]

Close a pair counted by \(D_h^*\) with a fixed bounded lower cap and an independent upper arch. The latter has mass at least \(c r^{-5/4}\) and diameter at most \(C'r\); the two exterior half-planes keep the closures disjoint from the bridges. The lower cut is at bounded distance from the bottom of the resulting polygon. An output polygon has at most \(C K\) preimages, since choosing its upper cut recovers both bridges and both caps. This proves (12).

For (13), place a second bridge in a disjoint box to the right. Its starting port has order \(r\) choices and its confined mass is at least \(c r^{-1/4}\). Close with exterior arches above and below, each of mass at least \(c r^{-5/4}\). The added mass per input is at least \(c r^{-7/4}\). Both separating cuts recover the construction with multiplicity at most \(C K^2\), and the input length is bounded by the output polygon length. The preceding weighted bound gives \(r^{7/4+1/3+o(1)}=r^{25/12+o(1)}\).

Finally surround an irreducible of diameter at most \(r\) by confined bridges whose heights each range over \([r,2r]\). Each side has mass at least \(c r^{3/4}\). In the output bridge the sum of the lengths of all possible marked irreducibles is at most its full length. Applying (13) on a bounded number of height bins therefore gives \(p(L;D\le r)\le r^{25/12-3/2+o(1)}\). In \(q(t)\le t^{-1}p(L;D\le r)+p(D>r)\) take \(r=t^{3/4}\). ◻

Proof. We prove this bound by induction, for \(0<\delta<1\). Set \(r=k^{4/3}\) and expose \(k_0=\lfloor k/2\rfloor\) increments on each side. Call their endpoint coordinates \((X_i,t_i)\). If the sum of their diameters exceeds \(Mr\), the sum of the diameters truncated at \(Mr\) also does. On \(E_k\) every earlier paired prefix satisfies \(E_{a-1}\). Markov’s inequality and independence of the next increment give an upper bound \[\frac1{Mr}\sum_{i=1}^2\sum_{a\le k_0} \Pr(E_{a-1})p(D\wedge Mr) \le C_\delta M^{-3/4} A_\delta k^{-1+\delta},\] where \(A_\delta\) is the induction constant. Choose \(M\) large enough that this is at most one quarter of the proposed bound.

Label the lower endpoint path 1, so \(t_1\le t_2\), and reflect horizontally if necessary so that it starts at the left port. For small fixed \(d>0\), let \(m\) be the leftmost coordinate of the second history between heights \(t_1\) and \(t_1+dr\), with \(m=+\infty\) if it has no point there. We discard \(|m-X_1|\le ur\), where \(u>0\) will be small. Condition on both histories through \(\lfloor k/4\rfloor\) increments and then on the rest of the second history. For each possible \(t_1=O(Mr)\), the prohibited interval for \(X_1\) has width \(2ur+O(1)\). The two-dimensional atom bound on the still independent part of the first history shows that this conditional probability is at most \(C_M(u+r^{-1})\). The induction bound for \(E_{\lfloor k/4\rfloor}\) pays for the earlier histories.

Next discard the case \(m<X_1-ur\). Were a disjoint common-height extension possible, the left crosscut would separate the second path from the far left of the slab. Thus its continuation above \(t_1\) would have to pass to the left of a point of the second history in this band; at height \(t_1\) such a point is already excluded by the single crossing there. In the next \(\lfloor sk\rfloor\) increments, however, the first path, with probability as close to one as desired, gains height greater than \(dr\) while its total diameter sum is less than \(ur\). Indeed the latter exceptional probability is at most \(C s u^{-3/4}\) by truncation and the diameter tail. The probability of no height increment exceeding \(dr\) is at most \(\exp(-c s d^{-3/4})\), by the height shell lower bound. Choose \(s\) small after \(u\), and \(d\) small after \(s\). These fresh increments fit among the first \(k\) and are independent of the exposed histories. They contradict the required leftward passage. Conditional on those histories their failure probability is uniformly small. The induction bound at \(\lfloor k/4\rfloor\), with \(u,s,d\) chosen in that order, makes the total of these two discarded cases, including reflected cases, at most another quarter of \(A_\delta k^{-1+\delta}\).

For each remaining pair of histories the strip just above \(t_1\) of height \(dr\), to the left of \(X_1+ur\), is free of the second history. Extend the first history through this strip to the left of the box containing both histories, then upward. Extend the second upward in a separate corridor. If \(t_2<t_1+dr\), its tip lies to the right of the cleared strip; if \(t_2\ge t_1+dr\), the first path has reached its separate corridor before the second extension begins. These are piecewise linear guides with fixed positive separation. The corridor input supplies joint mass at least \(c r^{-1/2}\) to every common terminal level in \([C_Mr,2C_Mr]\). The diameter stays at most \(C'_Mr\).

In an output pair the original histories are recovered by cutting after exactly \(k_0\) irreducibles on both paths. Hence there is no multiplicity factor for those cuts. Summing the order \(r\) terminal levels and using (12), the mass of the remaining histories is at most \(r^{-3/4+o(1)}=k^{-1+o(1)}\). For large \(k\) this fits in the remaining half of the induction bound. Increasing \(A_\delta\) handles the finitely many smaller \(k\). ◻

Proof. Fix a small \(\eta>0\). Lists with more than \(R^{3/4+\eta}\) increments have superpolynomially small mass: after removing the marked slot, exponential Markov with \(e^{-H/R}\) gives \(\exp(-cR^\eta)\). There are at most \(2R\) increments on either list, so summing all counts and possible marks still bounds this part by \(p(f)R^{-M}\) for any fixed \(M\). We henceforth impose the count cutoff.

Bin the numbers of increments before and after the marked slot on the first path, each increased by one, at dyadic scales \(a_-,a_+\). There are \(O(a_-a_+)\) choices of these two numbers. Fix them, and put \(K=\max(a_-,a_+)\). Near each end of the second path choose a grid of more than \(2/\eta\) height thresholds in a fixed small fraction of \(R\). The counts to their first reaching renewals are positive and at most \(2R\). At some consecutive pair the counts differ by a factor at most \(R^\eta\). Bin the lower count at scale \(c_e\), where \(e=-,+\) specifies the end. Sum over these finitely many threshold choices and logarithmically many bins at the end.

At each end expose \(m_e=\lfloor R^{-2\eta}\min(a_e,c_e)\rfloor\) increments on both paths. If necessary reduce this by a fixed factor to account for the added one in \(a_e\); zero histories cost one. For large \(R\) these are disjoint coordinate blocks and do not include the mark. On the second path they lie before the selected lower height thresholds. On the first path that extra restriction is unnecessary: the actual full disjoint bridges supply the extensions required by the prefix event. At the upper end read the lists backwards in relative coordinates from adjacent ports. Their still unknown common translation is immaterial to this test. Lemma 19 supplies the independent cost \[ R^{O(\eta)}\prod_{e=-,+}\min(a_e,c_e)^{-1}. \tag{16}\]

The choices of counts and of the marked slot cost \(O(a_-a_+)\). To cancel that cost against the two history factors, we will obtain a matching gain \(\min(1,c_e/K)\) at each end. We first bound the central part of the second path, then match its displacement using an independent batch of increments on the first path.

Fix the histories. Sum over the remaining count on the second path and let \(F(y)\) be its central bridge kernel, where \(y\) is its two-dimensional displacement. Its height is comparable to \(R\). From either end, within \(O(c_eR^\eta)\) central increments it must gain at least \(c_\eta R\) height, by the gap between the two selected thresholds. Retain these requirements and drop the other restrictions. The pointwise slab estimate gives \(F(y)\le CR^{-5/4}\).

We need the following improvement on a small displacement ball \(U_b\), valid for \(1\ll b\ll c_\eta R\) and either subset \(S\) of the two ends: \[ \sum_{y\in U_b}F(y)\le CR^{-5/4}b^2 \prod_{e\in S}(C_\eta c_eR^\eta b^{-3/4}). \tag{17}\] To prove it, append at each indicated end an independently chosen bridge of height in \([b,2b]\) and diameter at most \(Cb\). Its mass, with terminal position and height summed, is at least \(c b^{3/4}\). The enlarged displacement lies in a ball of radius \(C'b\), hence has at most \(Cb^2\) possible values, each with mass \(CR^{-5/4}\). In any output, a possible removed attachment ends within \(O(b)\) height of the new endpoint. Moreover that cut precedes the first renewal at or beyond height \(c_\eta R/2\) from the new endpoint by at most \(O(c_eR^\eta)\) renewal steps. Thus there are at most that many possible cuts at each indicated end. Dividing by the attachment masses proves (17). Positivity is used both in adjoining these bridges and in dropping all further restrictions.

The first path still has order \(K\) unexposed increments outside the mark and the histories. If \(K\le R^{5\eta}\), the pointwise bound on \(F\) suffices, since all additional factors below cost only \(R^{O(\eta)}\). Otherwise choose \(b_0=\lfloor KR^{-4\eta}\rfloor\) slots in each of \(\lceil30/\eta\rceil\) disjoint batches. Put \(b=R^\eta b_0^{4/3}\). The count cutoff implies \(b\le R^{1-3\eta}\), up to fixed factors. By the diameter tail and truncation, a batch has displacement norm greater than \(b\) with probability at most \(C R^{-3\eta/4}\). The batches are independent, so the chance that all of them fail is \(O(R^{-10})\).

On that exception use the pointwise bound on \(F\). Otherwise take a union over the fixed finite set of possible successful batch labels. For each label, fix everything outside that batch and use its two-dimensional atom bound \(C b_0^{-8/3}\) together with (17). At the end \(e\) use the improvement when \(c_e<K\). Since \(b_0^{-8/3}b^2=R^{2\eta}\) and \(c_eR^\eta b^{-3/4}\le R^{O(\eta)}c_e/K\), the resulting matching cost is \[ C_\eta R^{-5/4+O(\eta)} \prod_{e=-,+}\min(1,c_e/K). \tag{18}\] The estimate is uniform in the value of the marked increment. The mark can therefore be integrated separately at cost \(p(f)\).

Multiply (18) by (16) and the \(O(a_-a_+)\) count and mark choices. Each end contributes at most one because \[a_e\min(1,c_e/K)\le\min(a_e,c_e).\] Summing the threshold and dyadic labels costs only logarithmic powers. Taking \(\eta\) sufficiently small in terms of \(\epsilon\) proves (15).

For prescribed separated bottom ports, retain exactly the same central kernel and matching proof. The displacement to be matched is merely shifted by that prescribed separation. The bottom avoidance test is now absent, so its factor in (16) is lost. The remaining count factor is bounded by \[\frac{a_-a_+\prod_e\min(1,c_e/K)}{\min(a_+,c_+)} \le a_-\le R^{3/4+\eta}.\] This proves the \(R^{3/4+\epsilon}\) loss. If the marked path has at most \(R^{3/4-\delta}\) increments, then \(a_-\le C R^{3/4-\delta}\), which gives the stated improvement. This is the only change in the argument; no bottom avoidance probability is retained implicitly. ◻

Proof. The upper bound is (14). For a lower bound fix \(\eta>0\). On polygon diameters in \([r,r^{1+\eta}]\), the length-weighted mass is \(r^{-2/3+o(1)}\). Lengths less than \(r^{4/3-\eta}\) contribute at most \(r^{-2/3-\eta+o(1)}\) by the unweighted tail. Lengths greater than \(r^{4/3+2\eta}\) contribute at most \[r^{-4/3-2\eta}\sum_{D(P)\le r^{1+\eta}}|P|^2w(P) \le r^{-2/3-4\eta/3+o(1)}.\] Thus the unweighted mass with both diameter and length in these windows is at least \(r^{-2-2\eta+o(1)}\).

Choose a symmetry-related row normal retaining a fixed fraction with vertical span comparable to diameter. A bounded hexagonal-face modification at its lowest and highest vertices turns the polygon into two disjoint bridges with adjacent ports at each end. Here is the local operation. In coordinates \(P_{ij}=iu+jv\), \(R_{ij}=P_{ij}+(u+v)/3\), with \(u=(1,0)\) and \(v=(1/2,\sqrt3/2)\), the neighbors of \(P_{ij}\) are \(R_{ij},R_{i-1,j},R_{i,j-1}\). A lowest polygon vertex is a \(P\) and uses both ascending bonds. At the leftmost lowest such vertex, take the adjacent face below and to its right. Its overlap with the polygon is a path of one or two edges and includes every occupied vertex of that face. Symmetric difference replaces it by the complementary face path, creating a bounded exterior lower cap. The reflected operation creates the upper cap. Remove the two caps. Simplicity forces the remainder to be exactly two through bridges; otherwise one cap would bound a separate cycle. The faces are disjoint for large \(r\), and length, diameter, and weight change by bounded amounts. Anchoring the lower left port, the cuts and caps recover the input with bounded multiplicity. This suffices for the mass lower bound, whether or not a tie convention makes the map injective.

The output heights lie between \(cr\) and \(Cr^{1+\eta}\). Set \(t=r^{4/3-2\eta}\). Across both output lists, \(\sum_i(1-e^{-L_i/t})\ge1-e^{-\sum_iL_i/t}\) is bounded below by a positive constant. Apply Proposition 20 with \(f=1-e^{-L/t}\) to both paths and sum the height bins. It gives \(r^{-2-2\eta+o(1)}\le r^{-5/4+o(1)}q(t)\), hence \(q(t)\ge r^{-3/4-2\eta+o(1)}\). Letting \(\eta\) decrease proves the lower exponent. ◻

Proof. For counts in \([k,2k)\), split each list into two groups of comparable size. If their total height is \(h\), one group gains at least \(h/2\), at cost \(C\min(1,k h^{-3/4})\); the other matches the remaining height at cost \(C(1+k)^{-4/3}\). There are \(O(k)\) count choices. Thus counts at most \(h^{3/4-\delta}\) have total mass at most \(h^{-1/4-2\delta/3+o(1)}\), negligible compared with \(B_h\). Bounded counts follow directly from the height tail. Counts exceeding \(h^{3/4+\delta}\) have superpolynomially small mass by exponential Markov with \(e^{-H/h}\).

On the remaining counts, length at most \(t=h^{4/3-\epsilon}\) costs at most \(e\exp(-k q(t))\). Choose \(\delta<9\epsilon/32\); the deficit estimate makes this superpolynomially small. For length greater than \(t=h^{4/3+\epsilon}\), mark \(1-e^{-L_i/t}\), whose sum is bounded below on that event. The other increments match height at cost \(Ck^{-4/3}\). Summing marks and count choices gives \(Ck^{2/3}q(t)\) in a dyadic bin. For sufficiently small \(\delta\) this is \(o(h^{-1/4})\). Finally \(D\ge c h\) deterministically, and (6) gives \(B_h[D>h^{1+\epsilon}]=o(B_h)\). Division by any retained fixed fraction of \(B_h\) preserves every conclusion. ◻

Proof. The denominator is \(a_r\asymp r^{-5/4}\). First discard arches of height at most \(r^{1-\epsilon}\). Glue two independent such arches, one reflected into the lower half-plane, to make a marked polygon. This squares their total mass, up to bounded weight factors. Choose a symmetry-related period of order \(r^{1-\epsilon}\) whose vertical component exceeds the polygon’s span. Its projection is simple, and anchoring the first marked edge recovers the lift. The transverse span is of order at least \(r\), because of the original boundary gap and the nonparallel period. The possible marks cost at most a length factor. The cylinder estimate (8), with arbitrarily large \(M\), makes the squared mass and hence the mass itself smaller than every power of \(r\).

For every remaining arch perform only the upper extremal-face operation described in the proof of Corollary 21. Removing the new upper cap leaves two disjoint bridges from the original prescribed bottom ports to adjacent top ports. Weights and lengths change boundedly and the inverse multiplicity is bounded. At height \(h\in[R,2R]\), Proposition 20 with the bottom test omitted gives \(R^{-1/2+o(1)}p(f)\) for a mark on either path. Take \(f=\min(1,H/R)\): its sum on either path is bounded below, while \(p(f)\le CR^{-3/4}\). The total mass of the bin is therefore \(R^{-5/4+o(1)}\). Summing \(R>r^{1+\epsilon}\) is negligible relative to \(a_r\).

In the remaining height range \([r^{1-\epsilon},r^{1+\epsilon}]\), requiring one path to have count at most \(R^{3/4-\delta}\) adds the factor \(R^{-\delta+o(1)}\). Choose \(\epsilon\) small after \(\delta\); these pairs are negligible. For larger counts, a total length at most \(r^{4/3-\delta'}\) is superpolynomially unlikely under the free increment law, by \(\exp(-k q(r^{4/3-\delta'}))\). One may drop all pair constraints here; each count is at most \(2R\), so their choices cost only a polynomial. Taking \(\epsilon,\delta\) small enough in terms of \(\delta'\) proves the lower length bound.

For the upper length bound mark \(f=1-e^{-L/t}\) with \(t=r^{4/3+\delta'}\). Summing the two paths in the typical height bins gives \(R^{-1/2+o(1)}q(t)\), which is \(o(r^{-5/4})\) when \(\epsilon\) is sufficiently small. For diameter greater than \(r^{1+\delta'}\), use instead \(f=\min(1,D/r^{1+\delta'})\). The sum of increment diameters dominates each path diameter, and Lemma 16 bounds its expectation by \(Cr^{-3(1+\delta')/4}\). The same calculation applies. The lower diameter bound follows from the prescribed boundary gap. Bounded offsets in the cap construction are absorbed by decreasing exponent slacks. A retained positive fraction of the denominator changes none of the estimates. ◻

Proof. Use the \(P,R\) coordinates in the proof of Corollary 21. At a top endpoint an \(R\) extends by its upward bond to the next cut; a top \(P\) must have been reached along its unique descending bond and can be trimmed to the port immediately below. No other top \(P\) is internal to the segment, since it has only one descending bond. Reflect at bottom endpoints.

At a common top turn the pivot is an \(R_{ij}\) using both descending bonds. Use the face immediately above and to its right, with lower path \(R_{ij},P_{i+1,j},R_{i+1,j}\). If the other uppermost \(R_{i+1,j}\) is occupied, it too uses both descending bonds. The union’s overlap with this face is therefore a path of one or two edges containing all its occupied face vertices. Symmetric difference with the face, followed by removal of the new upper cap, creates disjoint ends at adjacent ports. Reflect at a bottom turn. Only boundedly many bonds change. For large spans the far-end modification cannot affect disjointness within, say, one quarter of the smaller original span from this turn.

If individual trimming makes the endpoint levels meet or cross, the segment was confined to one row. Its induced graph has degree at most two, so there are at most constantly many paths of each length from its endpoint. Their critical weights have an exponential tail because \(\rho<1\). Otherwise the result is a positive-height bridge. Record the case, each bounded local change, and the offsets between consecutive ports. There are constantly many such records per piece, and reversing them recovers the original path. The next anchor is determined by the preceding piece and its record; no free translation factor is added. ◻

Proof. Fix every actual count. At \(e=ij\) put \(d_e=\min(R_i,R_j)\) and \(s_e=s_{ij}\). If \(d_e\) is bounded in terms of \(\eta\), drop this test at a bounded cost. Otherwise place more than \(3/\eta\) consecutive thresholds within a small fixed fraction of \(d_e\) from each incident end, with gaps at least \(c_\eta d_e\). The counts to their first reaching renewals lie in \([1,C_0h]\). At some consecutive pair their ratio is at most \(h^\eta\), since otherwise their product would exceed the count cap. Bin the lower crossing counts dyadically at \(c_{i,e}\) and \(c_{j,e}\), and put \(c_e=\min(c_{i,e},c_{j,e})\).

Once these labels are fixed, the first \(\ell_e=\lfloor h^{-2\eta}c_e\rfloor\) slots at both ends are deterministic coordinate sets. On the actual event their heights are below the lower thresholds, and the actual disjoint paths extend them to a common cut within the tested range. Opposite-end histories are disjoint coordinate blocks: each uses a vanishing fraction of the node’s count. Lemma 19 supplies the independent cost \[ C_\eta h^{3\eta}/c_e. \tag{22}\] This includes \(\ell_e=0\). If \(c_e\ge h^{-4\eta}s_e\), it already costs at most \(C_\eta h^{7\eta}/s_e\).

If \(c_e<h^{-4\eta}s_e\), designate an incident node attaining this smaller bin. On it take the deterministic group of slots \[\ell_e+1,\ldots,\lceil2h^\eta c_e\rceil\] read from that end. It contains the later threshold renewal, so its height sum is at least \(c_\eta d_e\). It has at most \(C c_eh^\eta\le C h^{-3\eta}s_e\le C h^{-3\eta}k_i\) slots. Call it a rapid group. Such groups are disjoint from the histories and from any opposite-end rapid group. The height tail and its truncated first moment imply \[ \Pr(H_1+\cdots+H_n\ge u)\le C\min(1,n u^{-3/4}). \tag{23}\] The rapid group therefore costs at most \(C_\eta c_eh^\eta/d_e^{3/4}\le C_\eta h^\eta c_e/s_e\). Multiplying by (22) pays \(s_e^{-1}\). We now discard the original crossing-index restrictions: only necessary events on the deterministic independent coordinate blocks are retained.

The histories removed at one node have total height less than a small fixed multiple of \(R_i\). Their complement must gain at least \(c_0R_i\). It consists of at most two rapid groups and the remaining slots. Partition those remaining slots into eight groups of nearly equal size. For sufficiently large fixed counts, each has order \(k_i\) slots. Some one of these at most ten groups gains height at least \(c'_0R_i\). Take a finite union over a label selecting that height-carrying group. After fixing the label, reserve three different free groups for count damping, concentration, and the length test. None is the height-carrying group.

Write \(b_i=\min(1,k_i/K_i)\). If a free group carries the height, (23) charges it by \(Cb_i\), independently of the rapid groups. If the carrier is the rapid group for \(e\) on node \(i\), charge its larger height directly at scale \(R_i\). The cost \(C_\eta c_eh^\eta/K_i\) supplies both required factors, because \[ c_e/K_i\le \min(1,k_i/K_i)c_e/s_e. \tag{24}\] This follows from \(s_e\le\kappa_i\) in either regime for \(k_i/K_i\). The other rapid group, if any, remains independent. Also \(d_e^{3/4}=\min(K_i,K_j)\ge s_e\); thus this calculation allows unrelated height scales.

The reserved damping group has height at most \(CR_i\). Exponential Markov and \(1-p(e^{-H/R_i})\ge c/K_i\) give \(C e^{-c k_i/K_i}\). All previously charged events use other slots. Conditional on coordinates outside the concentration group, its product law is unchanged. Lemma 17 bounds its height atom by \(Ck_i^{-4/3}\le C\kappa_i^{-4/3}\), proving (20). The same statement holds with any proved concentration function for an additive statistic of these slots. On the separate length group, \[\Pr\{\textstyle\sum L_j\le l\} \le e(1-q(l))^{c k_i}\le C e^{-c k_i q(l)}.\] These bounds multiply because their coordinate sets are disjoint. For several height constraints, condition outside all their groups, bound the joint atom by the product, and sum the allowed height vectors. Successive interval constraints follow by successive integration with the same uniform bounds.

For bounded counts the incident turn factors are bounded. Drop them and apply (23) to the whole list. The height factor is then bounded below by a positive constant, so its insertion changes only the constant; no large-count length test is needed. There are \(O(k_i)\) actual counts at each node, and \[k_i\min(1,k_i/K_i)e^{-c k_i/K_i}\le C\kappa_i^2/K_i.\] For \(k_i\le K_i\) this is immediate; otherwise use \(x e^{-cx}\le C\). Thresholds, crossing-count bins, height-carrier labels, and local records number at most \((C_\eta\log^C h)^{C(m+1)}\). These choices and the power losses above are at most \(h^{C_1\eta(m+1)}\) for large \(h\). This proves the proposition. ◻

Proof. Take \(h=C'N^2\) in Proposition 25 and choose \(\eta\) sufficiently small in terms of \(\epsilon\). Once terminal pieces and local records are fixed, their final-height constraint reads \(H_i\in[H_j+b-r-C_0,H_j+b+r+C_0]\). The interval depends on the other bridge and fixed external data, hence is measurable outside the selected group. No horizontal displacement of that group has been fixed. The one-node assertion is the proposition with no turn test. ◻

Proof. Split a path at a global minimum of its height and read both parts outward from that minimum. In either part, end the first piece at the last maximum of the remaining path, the next at its last minimum, and continue. Each piece stays between its endpoint heights. The spans decrease strictly after a possible initial equality, since the selected extreme cannot occur again in the remaining path. Every honeycomb edge has nonzero vertical projection in these coordinates.

In a dyadic span bin \([R,2R)\) there are at most \(CR\) pieces in a chain. Convert their endpoints by Lemma 24. Test turns only between consecutive pieces in the same bin. An omitted piece splits that bin into runs; no test crosses an omission. At a run of \(m\) pieces put \(K=R^{3/4}\) and \(x_i=\kappa_i/K\le1\), absorbing fixed comparability constants into the per-node constants. The product in (19), before its exponent loss, is \[ Kx_1x_m\prod_{i<m}\max(x_i,x_{i+1}). \tag{26}\] There are only \(O(\log R)\) possible count bins per piece.

Fix a small \(\zeta>0\). If fewer than \(m/4\) nodes have \(x_i\ge R^{-\zeta}\), at least \(m/2-O(1)\) neighboring pairs have both labels smaller than this threshold. They supply the factor \(R^{-\zeta(m/2-O(1))}\). If instead \(b\ge m/4\) nodes have large labels, use the reserved height atom at those nodes. Their atom bounds are at most \(C R^{-1+4\zeta/3}\). Their original spans are ordered and lie in a range of \(CR\) levels. Modified heights differ from them by specified bounded offsets. Hence, conditional on the other coordinates, the number of allowed vectors for these \(b\) heights is at most \[C^b\binom{CR+b}{b}.\] The product atom bound from Proposition 25 therefore costs at most \(C^mR^{4\zeta b/3}/b!\), since \(b\le CR\). This is a joint atom calculation followed by a count of ordered vectors, not a repeated conditioning on already fixed bridge heights.

Choose \(\eta\) sufficiently small after \(\zeta\). The factors \(R^{C_1\eta(m+1)}\), the count bins, and the local records are absorbed by the preceding decay whenever \(m>R^{C_2\zeta}\), for one sufficiently large absolute \(C_2\). For example, in the second case \(b!\ge(b/e)^b\) beats \(R^{(4\zeta/3+C\eta)m}\); in the first case choose \(C\eta<\zeta/8\). These long-run contributions are summable. For \(m\le R^{C_2\zeta}\), even bounding every displayed factor and choice separately costs at most \(\exp(R^{C_3\zeta})\), with a slightly larger absolute \(C_3\). Thus that bound covers the whole bin.

The bins occur in decreasing order. Multiplying over dyadic scales up to \(Cn\) gives \(\exp(Cn^{C_3\zeta})\), because the powers sum geometrically. Choose \(\zeta\) small enough for the requested \(\delta\), and absorb bounded bins and both halves into the constant. If a fixed number of pieces is omitted, the number of runs per bin increases by only a fixed amount. Choosing their bins and positions costs a polynomial in \(n\), and the same argument covers the remainder. Root choice or allowing both vertex types changes a constant only. ◻

Proof. First consider a port bridge of length at most \(n+O(1)\) and height at least \(d\ge c n^{3/4+\epsilon}\). Decreasing \(\epsilon\) slightly absorbs the fixed \(c\). Put \(h=\lfloor n^{3/4-\gamma}\rfloor\), where \(\gamma=\epsilon/10\). Starting upward, track the running maximum until the first drop of \(h\) from it, if such a drop occurs. Place a turn at the last vertex attaining that maximum before the drop. From there track the running minimum until the first rise of \(h\), and repeat with directions alternating. Consecutive pieces stay between their endpoint levels, have spans at least \(h\), and have no move against their direction of more than \(h+O(1)\). The first and last pieces point upward since the whole path is a bridge. Choosing the last extremum makes the same statements valid in the portion preceding each trigger.

Within every such run place transverse cuts at deterministic spacings of order \(10h\), leaving margins of at least \(5h\) from the ends and no residual gap greater than a fixed multiple of \(h\). A run too short for a cut is kept as one piece. At a cut crossed exactly once, split at that port. At a repeatedly crossed cut of an upward run, let \(M\) be the largest height before its last crossing, attained at a selected pivot before that crossing. Let \(m\) be the smallest height from that pivot to the end of the run. Then \[M>q>m,\qquad M-m\le h+O(1),\] where \(q\) is the cut level. The minimum occurs before the last crossing; after that crossing all heights are above \(q\). Split at the two pivots. The intervening descending segment is a repair piece. Before the upper pivot the path never exceeds it, and after the lower pivot it never goes below that lower level. Thus the neighboring upward pieces are again slab pieces. Reflect this construction in a downward run. The cut spacing and drawdown bound put different repairs in disjoint ordered zones.

Figure 1 shows the local reason for the reverse piece: cutting only at the last crossing would leave the preceding piece above its terminal level, so it would not be a strict bridge.

Call the pieces in the original direction primary pieces. Their spans are between fixed positive multiples of \(h\); repair spans are at most \(Ch\). After the local port changes, test every turn with sufficiently large smaller span. If \(P\) is the number of primary pieces, then \[ c d/h\le P\le Cn/h. \tag{29}\] There are at most \(P-1\) repairs, and the total modified length is at most \(n+CP\).

Apply Proposition 25 with \(K=h^{3/4}\). Write a primary label as \(\kappa_i=Kx_i\). A repair of height scale \(a\le Ch\) has \(K_{\rm rep}=a^{3/4}\); put \(t=K_{\rm rep}/K\) and \(y=\kappa_{\rm rep}/K\le t\). Fixed comparability constants can be absorbed into all bounds below, so the normalized labels are at most one. For a single-crossing cut without a repair use the formal values \(a=K_{\rm rep}=\kappa_{\rm rep}=1\). This virtual node has unit mass and no avoidance test; its algebraic node and seam factors are also one, since every actual \(\kappa_i\ge1\).

At either kind of cut, smooth the height of the preceding primary piece. Its endpoint lies within \(O(a)\) of the cut, where the cut level is specified relative to the run’s starting height by the discrete cut pattern. We claim that this supplies the factor \[ C\min(1,(t/x_i)^{4/3}). \tag{30}\] Bounded-count nodes use the bounded-factor height estimate in Proposition 25. At the other nodes, use this constraint only when \(a\kappa_i^{-4/3}<1\), and fix all coordinates outside the selected reserved groups. Their height sums have a joint atom bound equal to the product of their individual bounds \(C\kappa_i^{-4/3}\) from Proposition 25. The allowed height vectors can be counted in traversal order: once the earlier selected sums in the run are given, the current endpoint constraint allows at most \(Ca\) values for the next sum. The run’s starting height cancels from this relative constraint. Thus the number of allowed vectors is at most the product of these \(Ca\) factors. Multiplying by the joint atom bound, and dropping the constraints at the unselected nodes, proves the product of (30), since \(a\kappa_i^{-4/3}=(t/x_i)^{4/3}\). Each primary receives at most one factor. This count already includes its possible endpoint heights. The turn tests use only their separate history and rapid groups in adjacent-port coordinates; none fixes the displacement of a selected reserved group.

For completeness, the product calculation is as follows. A direct turn between primaries contributes \(\max(x_i,x_{i+1})\) after the primary factors have been collected. A cut with a repair contributes \[Q=\frac{\max(x_i,y)\max(x_{i+1},y)}{t} \min(1,(t/x_i)^{4/3}) \le\max(x_{i+1},y).\] If \(x_i\le t\), use \(y\le t\); if \(x_i>t\), the remaining factor is \((t/x_i)^{1/3}\le1\). This proves the inequality in both cases, including the virtual-node convention. The full product is bounded by \(Kx_1x_P\) times these factors and constant bases. Every connecting factor is bounded, and is \(O(h^{-s})\) if all its incident labels are less than \(h^{-s}\). The patterns of runs and cuts, repair scales, count bins, and local records cost at most \(h^{C\eta(P+1)}\).

Take \(s'=\epsilon/2\) and \(s=9\epsilon/64\). For small \(\epsilon\) these choices satisfy \(9s'/16>s\) and \[\frac{n}{h^{4/3-s'}}=o\bigl(n^{3/4+\epsilon}/h\bigr).\] Set \(l=h^{4/3-s'}\). Since the total length is at most \(n+CP\), only \(o(P)\) actual nodes have length greater than \(l\). Designate a superset of these nodes; all possible designations cost at most \(2^{2P}\). On every other actual node with normalized label at least \(h^{-s}\), the separate short-length group supplies \[C e^{-c k_i q(l)}\le\exp(-h^{c_s}),\] for some \(c_s>0\): indeed \(k_i\ge K h^{-s}\) and \(K h^{-s}q(l)=h^{9s'/16-s+o(1)}\). Virtual and bounded one-row nodes have small labels.

If a positive fraction of labels are large, this gives stronger than exponential decay in \(P\log h\). Otherwise a positive fraction of the connections have all incident labels small. Each large or designated long node can spoil at most two such connections, so those connections supply \(h^{-s}\) each. Choose \(\eta\) sufficiently small after \(s\). After summing all the records and designations, the contribution with \(P\) primary pieces is at most \(\exp(-cP\log h)\). Equation (29) gives the tall-bridge bound \[ \exp(-c d\log h/h). \tag{31}\] All constants are uniform for \(d\) in the stated range.

Finally take an unrestricted path of diameter at least \(d\). One of the finitely many symmetry-related row normals has vertical span at least \(c d\). In its extremum decomposition from Proposition 27, an initial piece on one of the two halves reaches the global maximum from the global minimum. Omit that piece and its neighboring turn tests. Its individual port conversion has length at most \(n+O(1)\) and height at least \(c'd\); its mass is bounded by (31). The remaining pieces, including their choices and records, cost at most \(C_\delta e^{n^\delta}\) for every fixed \(\delta>0\). Choose \(\delta\) smaller than the power gain in \(d/h\ge c n^{\epsilon+\gamma}\). That cost is absorbed by (31). This proves (27). It also proves the uniform bound (28), since \(h\le n^{3/4}\) and the absorbed exponent is a fixed fraction of \(d\log h/h\). ◻

Proof. Factor at all once-crossed positive row boundaries. The factors before the last remainder are strict irreducible bridges, and this factorization is unique. Conversely, adjoining any finite list of irreducibles and a no-renewal end gives a half-plane end. The interiors of successive pieces lie in disjoint height slabs, and the remainder lies above the last joining line. Thus weights multiply and no avoidance correction is needed. A list of \(j\) unspecified irreducibles has \(z\)-mass \((1-q_N)^j\), whence \[H_N=U_N\sum_{j\ge0}(1-q_N)^j=U_N/q_N.\] For a prescribed tuple, all possible later pieces together form an arbitrary half-plane end. Its numerator is \(z^\ell H_N\), and division by \(H_N\) gives the claimed probability. ◻

Proof. Take a nonempty bridge from a fixed lower port to an arbitrary upper port. At the lower port attach the reverse of a no-renewal end in the lower exterior half-plane. At the upper port attach a no-renewal end in the upper exterior half-plane. The three pieces have disjoint vertex sets, their weights multiply, and the result is an oriented plane walk.

The construction is injective in the translation convention. In the resulting path, the lowest and highest horizontal row boundaries crossed exactly once are the two attachment lines. There are no single-crossing lines farther outward because both attached ends have no renewals. Recover those two lines from the output, cut at their crossing ports, and translate the lower port to the fixed anchor. This recovers the central bridge and both ends. Internal renewals of the central bridge do not affect this recovery.

The total \(z\)-mass of all nonempty bridges is \[\sum_{j\ge1}(1-q_N)^j=q_N^{-1}-1.\] Multiplying by the two end masses proves the first bound. Since \(q_N\to0\), the second follows. ◻

Proof. If either endpoint attains the minimum height, use that endpoint as the minimum, reversing orientation if necessary. Otherwise cut at the first internal vertex attaining the minimum. This gives either one half-plane end or two disjoint half-plane ends. The passage between vertex endpoints and ports uses bounded local modifications, with bounded inverse multiplicity and bounded weight factors, uniformly for \(\rho/2\le z<\rho\).

Here is the local modification, illustrated in Figure 2. Write \[u=(1,0),\quad v=(1/2,\sqrt3/2),\qquad P_{ij}=iu+jv,\quad R_{ij}=P_{ij}+(u+v)/3.\] The neighbors of \(P_{ij}\) are \(R_{ij},R_{i-1,j},R_{i,j-1}\). An internal lowest vertex must be of type \(P\), because a vertex of type \(R\) has only one ascending bond. Translate the minimum to \(P_{00}\). Keep the branch toward \(R_{-1,0}\) and move the start of the branch toward \(R_{00}\) to \(P_{10}\). If \(P_{10}\) is absent, append its free bond to \(R_{00}\). If it is present, it cannot be an endpoint, by our initial choice of case. Minimality then forces it to use both ascending bonds, so this branch already reaches it through \(R_{00}\); trim that bounded initial segment. Attach the two incoming half-edges from below. The resulting ends start at adjacent ports and remain disjoint. The modification changes only boundedly many vertices, preserves their far endpoints and is undone from a bounded local record.

If a minimum is an endpoint of type \(P\), its incoming bottom port gives the one-end representation directly. If it has type \(R\) and the path is nontrivial, its first bond is the unique ascending bond; trim that bond and start at its midpoint. No internal vertex can be a lowest \(R\). If the other endpoint is also a lowest \(R\), trim its last descending bond as well. All remaining vertices then lie strictly above the new starting cut. A singleton walk is handled separately at bounded cost. These operations change the endpoint heights by at most a fixed constant.

First suppose there is one end. On the event \(V\le r\), its last renewal is at height at most \(C(1+r)\). After removing its no-renewal remainder and dropping the restriction on that remainder’s final vertex, the mass is at most \[C U_N\sum_{h\le C(1+r)}B_h \le C'(1+r)^{3/4}U_N \le C''(1+r)^{3/4}U_N^2,\] using (32) and \(U_N\ge\rho/2\).

Now suppose there are two ends. Remove their no-renewal remainders and fix those remainders up to translation. Their \(z\)-weights will be summed at the end. Discard all avoidance constraints involving a remainder and the remaining total-length restriction, but retain the resulting \(O(N^2)\) cap on bridge heights and piece counts. Drop the bridge damping factors \(e^{-L_i/N}\le1\), so their weights are critical and the third input applies. The remaining two bridges have adjacent starts. If both are nonempty, they remain disjoint throughout the common height range. Put their heights and counts into dyadic bins and use the input’s notation. If \(a=\min(\kappa_1,\kappa_2)\) and \(M=\max(\kappa_1,\kappa_2)\), then \(K_i\ge\kappa_i\) gives \[F=\frac{\kappa_1^2\kappa_2^2}{K_1K_2a}\le M.\] If \(M\) is bounded by the fixed count threshold of Proposition 25, the bound without a concentration test is already \(O_\varepsilon(N^\varepsilon)\) and suffices. Otherwise choose a bridge \(i\) attaining \(M\), with a fixed tie rule; its count is large enough to supply the reserved group. In the integration for (35), fix the other bridge, the local records and the increments of bridge \(i\) outside its reserved group; the two remainder shapes are already fixed. If \(h_j\) is the other bridge’s height, the endpoint-height condition restricts the total height of bridge \(i\) to \[[h_j+b-r-C_0,\ h_j+b+r+C_0],\] where \(C_0\) bounds the endpoint-height changes and \(b\) is determined by the two remainder endpoint heights and the fixed local records. Subtracting the fixed height of the other increments gives an interval of the same size for the reserved group’s height. Its horizontal displacement is still summed over, so this is precisely the permitted concentration test. Thus (35) bounds the mass in this pair of bins by \[C_\varepsilon N^\varepsilon M \min\{1,(1+r)M^{-4/3}\} \le C_\varepsilon N^\varepsilon(1+r)^{3/4}.\] The last inequality follows by considering \(M\le(1+r)^{3/4}\) and its complement. If only one bridge is nonempty, its factor \(\kappa^2/K\le\kappa\) gives the same estimate: bounded \(\kappa\) needs no concentration test, and larger \(\kappa\) permits the calculation above. If neither is nonempty, the mass is at most one before the remainder weights are restored.

Because \(L\le N^2\), there are only \(O((\log N)^4)\) pairs of height/count bins. Choose the exponent slack in the geometric input smaller than the desired \(\varepsilon\) and absorb this factor. Summing the two fixed remainder weights gives \(U_N^2\). Their translations were determined by the bridge tips; no free placement factor is introduced. This proves the bound. ◻

Proof. We use the equivalent vertex-weighted translation-class law. First record a consequence of (33). For each \(\epsilon>0\), \(p\ge0\) and \(B>0\), \[ \sum_{m>N^{1+\epsilon}}m^p a_m e^{-m/N}=O_{\epsilon,p,B}(N^{-B}). \tag{36}\] Indeed choose \(0<\delta<\epsilon/(1+\epsilon)\). Uniformly for \(m\ge N^{1+\epsilon}\), one has \(m^\delta\le m/(2N)\) for all large \(N\). The summands are then at most \(C_\delta m^p e^{-m/(2N)}\), whose sum has faster than polynomial decay. In particular, paths with \(L>N^2\) have negligible absolute mass and negligible mass after multiplication by any fixed length power. Since Lemma 30 gives \(Z_N\ge c/q_N\), this remains true after normalization.

Take \(r=N^{3/4-\delta}\) with fixed \(0<\delta<3/4\). Lemmas 30 and 31 give \[\mathbb P_N\{V\le r,\ L\le N^2\} \le C_\varepsilon N^\varepsilon q_N(1+r)^{3/4} =N^{\varepsilon-3\delta/4+o(1)}.\] Choose \(\varepsilon<3\delta/4\) and add the tail just discarded. It follows that \(V\ge N^{3/4-\delta}\) with probability tending to one. After slightly decreasing the exponent tolerance to absorb \(d_0\), the same lower bound holds for \(A\).

We next obtain the upper spatial estimate from the absolute fast-path bound. Fix \(\xi>0\). Choose small positive \(\epsilon,\eta\) such that \[(1+\epsilon)(3/4+\eta)<3/4+\xi.\] For lengths \(m\le N^{1/2}\), the deterministic estimate \(D\le Cm\) already gives the desired upper bound for large \(N\). For \(N^{1/2}<m\le N^{1+\epsilon}\), any violation of \(D\le N^{3/4+\xi}\) implies \(D>m^{3/4+\eta}\). Summing (34) over this range gives a faster than polynomially small absolute mass. Lengths above \(N^{1+\epsilon}\) are handled by (36). Division by \(Z_N\) proves the upper spatial assertion.

The same argument gives the lower length estimate, rather than assuming it from the discount scale. Fix \(0<\xi<1\), choose \(\eta,\delta>0\) so small that \[(1-\xi)(3/4+\eta)<3/4-\delta, \qquad \delta<1/4.\] On the typical event \(A\ge N^{3/4-\delta}\), a walk of length \(m\le N^{1/2}\) is impossible for large \(N\). If instead \(N^{1/2}<m<N^{1-\xi}\), that endpoint event forces \(D>m^{3/4+\eta}\), up to fixed unit factors absorbed by the strict inequality above. The summed exceptional mass is again negligible by (34). Hence \(L\ge N^{1-\xi}\) with high probability. The upper length bound follows from (36). Larger tolerances follow from smaller ones.

For moments, the lower probability estimates give the respective lower exponents. For the upper length moment, restrict to \(L\le N^{1+\epsilon}\) and use (36) on its complement: \[\mathbb E_N L^p\le N^{p(1+\epsilon)}+o(1).\] For the spatial upper moments, repeat the fast-path split with the additional factor \(D^p\le C_pL^p\). On the exceptional paths of polynomially large length, the factor is harmless in the sum of (34); on the damped long-length tail use (36) with this \(p\). The remaining paths obey \(D\le N^{3/4+\xi}\), apart from the deterministic short-length class. Thus \[\mathbb E_N A^p\le\mathbb E_N D^p\le N^{p(3/4+\xi)}+o(1).\] Let the positive slacks decrease to zero. This completes the proof. ◻

Proof. There are two estimates to combine. A test with substantial transverse clearance is rare by a positive flux comparison. After any candidate for the last such test, each later test costs a power through either a large height increment or a small transverse interval. Those later costs use fresh coordinate blocks.

Proof of Theorem 33. An input of (37) contains a weak bridge between a minimum and maximum in some signed normal direction of span \(H\asymp R\), since the three lattice directions control diameter; take the sign so it is oriented from low to high. We call this coordinate \(y\) again. The exterior pieces cost at most \(V_{2R}^2\) up to constants; it suffices to estimate such weak bridges from their fixed first vertex. Pay \(O(R)\) for specifying the last height and normalize the first height by translation.

Take \(\delta=\xi/2\), small \(0<\sigma\ll\xi\), integer \(S\asymp R^{1-\sigma}\), so \(R^{d_*-\xi}=o(S^{d_*-\delta})\). First divide the bridge into swings with drawdown/up threshold \(s_0=S/10\). Track running maxima from the beginning; on first detecting a fall by at least \(s_0\) from one, mark the last corresponding peak before detection and switch to tracking running minima starting at detection, with next switch symmetrically after a rise, marking the trough, and so on. Use the marked extrema (not the detection points) and the two endpoints to cut. Every resulting swing is a weak bridge of span \(\ge s_0\), with adverse increments between any earlier and later point of that swing strictly less than \(s_0\). Indeed, e.g. on an upswing the initial trough-to-rise-detection part lies above the trough, and strictly within \(s_0\) above it before detection. From detection to the peak the running maximum rule gives the bounds; all heights stay above the trough and at most the peak on both parts. The reversed argument works going down, and the global endpoint extrema handle first/last swings (after detecting a fall there must still be a rise by threshold before or at the terminal maximum). Without any switch the span itself is \(H\).

Inside each swing use all absolute grid levels spaced by \(S\) (multiples of \(S\) in wall coordinates) with interior margin \(\ge 2s_0\) from both extrema. Here is the subdivision in upward orientation, symmetrically otherwise. For each such level \(l\), if there is just one crossing, cut there at the port. Otherwise take a highest vertex \(A_l\) before the last upward crossing of \(l\), and a lowest \(B_l\) after \(A_l\) in this swing. Then \(y(A_l)>l>y(B_l)\), both are between the first and last crossings (a return occurs after \(A_l\), all vertices after last crossing are above). The path between them is a descending weak bridge of span \(<s_0\); mark it as an inserted piece. These crossing zones for successive levels occur in order since a backtrack of size \(S\) is impossible. All intermediate pieces are upward weak bridges: e.g. from \(B_l\) to \(A_{l+S}\) all vertices respect those extremal bounds, with the analogous inequalities at single crossings and swing endpoints. These main pieces have spans between \(c S\) and \(C S\), by the margins, offsets \(<s_0\) and the minimum full swing span if no cut. Thus throughout the original bridge there are \(k\ge c H/S\) main pieces and at most \(k\) inserts.

Use strict virtual pieces with the bounded endpoint extensions described above. Specify their types/order, signs, and the following boundary data, at cost \(R^{O(1)}(C(1+\log R))^{C k}\) before any fine coordinates at turns:

• At a turn between swings, label the coarse height interval (grid of size \(S\)) and assign boundary scale \(D_j=S\).

• At the two turns for an insert, label the grid level and a common dyad \(D_j\asymp 1+|y(A_l)-y(B_l)|\) (same at both). Each height has offset \(O(D_j)\) from that level.

• At the single-cut ports label the grid level; take \(D_j=1\) there and at the prescribed endpoint heights.

Coarse coordinates cost boundedly many choices per piece from those before, since increments are \(O(S)\). We can include any finite endpoint/port adjustment types as labels. Thus each virtual boundary height \(w_j\) still to sum ranges over \(O(D_j)\) possibilities (already fixed at external endpoints and single cuts), and for fixed next \(w_{j+1}\) has bounded multiplicity per intervening strict-piece height \(h\), their signed difference. We restrict \(h\asymp P\), where \(P=S\) on main pieces, \(P=D_j=D_{j+1}\) on inserts; other fine height ordering restrictions will not be needed. Every strict piece length is \(\le S^{d_*-\delta}\) by the global input bound, including its bounded adjustments.

Use (39), and fix \(0<u\ll v\ll \min(v_0,\sigma_0)\). Call a boundary active if \(D_j\ge S^{1-v}\). At scale \(D'=D_j\) take the threshold \[g(D')=\max(1,\lfloor c_*D'S^{-u\,{\bf1}_{\rm active}}\rfloor)\] for small fixed \(c_*\). Since each endpoint scale is \(O(P)\) this ensures \(h-b-d\asymp P\) at the censored kernel, and pair tests at turns are either vacuous or well below the adjacent far adjustments. Write \(F(D')=C_\eta g(D')^{-a+\eta}\) as a pair bound, or \(1\) at unpaired boundaries with \(D'=1\).

The subdivision has reduced rapid travel to many main pieces of scale \(S\). It remains to extract a power saving from each such piece without spending the same independent end history twice.

We sum each kernel over its left boundary height at fixed right height and histories from both ends (left/right here meaning first/last in traversal). Include in its numerical budget also the factor \(F(D_L)\) at the first boundary of that piece. We claim a bound \(S^{-c_1}\) for this product on main pieces and active-scale inserts, for some \(c_1>0\) uniformly at all sufficiently small \(\eta\); on inactive-scale inserts at most \(C_\eta S^{2\eta}\) suffices.

Indeed in the coincident-jump case the summed kernel is at most \(C g_L^a g_R^a P^{-a}\), using bounded multiplicity per height. With \(F(D_L)\) this leaves \(O_\eta(S^\eta (g_R/P)^a)\), giving the claim (on main pieces \(g_R/P\le C\max(S^{-v},S^{-u})\), and on active inserts \(\le CS^{-u}\)). In the distinct case sort \(1+m\) in dyads of scale \(M\). For fixed crossing increments each \(m\) can occur boundedly often as the left height varies, with \(O(D_L)\) possible heights total. Thus by \(B_m\le C M^{-\vartheta}\) we have kernel bound \[C\min(D_L M^{-\vartheta},M^a) = C D_L^a\min((D_L/M)^\vartheta,(M/D_L)^a).\] Here \(D_L^a F(D_L)\le C_\eta S^\eta\) times \(S^{a u}\) if active. When \(M\ge S^{1-v_0}\) (39) allows multiplying by \(S^{-\sigma_0}\), enough. For smaller \(M\) on a main piece or active insert, there must be a giant crossing jump \(y_i\ge cP\) on some side \(G\). Accordingly we can multiply the bound, after splitting into the two possible sides, by \(C (g_G/P)^a\) by its conditional tail. On a main piece this saves \(S^{-av}\) if \(G\) inactive, otherwise \(S^{-au}\); on an active insert (both sides active) also \(S^{-au}\). In the latter savings cases, if left itself is active and pays \(S^{au}\), the factor \((M/D_L)^a\le S^{-a(v_0-v)}\) supplies the gain. Inactive inserts just use the unconstrained dyad estimate with both boundary scales inactive. For clarity, all these cases have a common strictly positive saving before the small \(\eta\) loss: \[\Lambda=\min\{au,\ a(v-u),\ a(v_0-v),\ \sigma_0-au\}>0.\] Every dyad on a main piece or active insert costs at most \(C_\eta S^{\eta-\Lambda}\). Choose \(\eta\) sufficiently small, then sum \(O(\log S)\) dyads. This proves the claimed negative power; inactive inserts cost at most \(C_\eta S^{2\eta}\).

To combine, first hold all histories fixed and sum out the fine variables in sequence from first to last using the uniform summed-kernel bounds (the last height fixed). Applying the pair constraints (40) in the product of censored-history probabilities contributes exactly the indicated upper costs allocated to the left boundaries. Taking \(\eta\) small, all labeling costs, inactive inserts, and endpoint weight factors are absorbed: the resulting weak-bridge sum is at most \(R^{O(1)}\sum_{k\ge cH/S} S^{-c_2 k}\) for some \(c_2>0\). This is stretched-exponentially small; (42) at the two exterior pieces preserves this, proving (37). ◻

Proof. Absolute upper tails and moment bounds. For the infinite walk, consider the first \(n\) whole irreducibles; their concatenation contains the first \(n\) centers. Except with probability \(Cn^{1-aA}\), every one of these pieces has diameter at most \(n^A\), where \(A\) is any sufficiently large fixed constant. The concatenation has exactly its critical bridge weight as probability, by unique factorization. Split it at its \(n\)th center. If this prefix reaches radius \(n^{1/d_*+\epsilon}\), summing its critical mass over diameter dyads costs stretched-exponentially little by (37), with its slack \(\xi\) sufficiently small. The remaining suffix has diameter at most \(Cn^{A+1}\) and costs at most \(CV_{Cn^{A+1}}\), including bounded endpoint weight corrections. Choose the exponent \(\nu\) in (42) sufficiently small after \(A\); this suffix factor is absorbed by the stretched-exponential bound. Since \(A\) can be arbitrarily large, the upper tail for the first \(n\) centers is smaller than every fixed inverse power of \(n\).

The almost-sure lower exponent of Theorem 7, this upper tail, and the deterministic linear distance bound give the stated moments for endpoint distance and maximum radius. For uniform strict or vertex bridges, divide the absolute fast-path bound by their previously proved polynomial lower mass bound and use their typical lower cutoff. The same argument gives the upper tail for uniform endpoint-free walks. For plane walks the total edge-weighted mass is at least one, by the connective constant and submultiplicativity. For half-plane walks it has a polynomial lower bound by inclusion of strict bridges, with one outward center appended when required by parity.

Proof. Write \(\alpha=9/16\). We first record the length atom estimate under the exponent-precision hypotheses of the turning-chain route. If the length \(L\) of one irreducible satisfies \(1-\mathbb E e^{-L/r}=r^{-\alpha+o(1)}\), then independent sums satisfy \[ \sup_m p(S_L(j)=m)\le j^{-1/\alpha+o(1)}. \tag{44}\] Here is a direct verification, without logarithmic precision. Fix a small \(\zeta>0\) and put \(z=r^{1-\zeta/2}\). The Laplace estimate implies \(p(L>s)\le s^{-\alpha+o(1)}\) and \(\mathbb E[L;L\le s]\le s^{1-\alpha+o(1)}\). Hence the contributions of \(L<r^{1-\zeta}\) and \(L>r/10\) to \(\mathbb E(1-e^{-L/z})\) are negligible compared with \(z^{-\alpha+o(1)}\). The intermediate interval therefore has probability at least \(r^{-(1-\zeta/2)\alpha+o(1)}\). Subtracting an independent bounded-length copy in the characteristic-function symmetrization gives phases between \(c r^{-\zeta}\) and a fixed constant below \(\pi\) at frequency \(1/r\). Thus the squared characteristic-function modulus is at most \(1-c r^{-\alpha-O(\zeta)+o(1)}\). On the actual length lattice all modulus-one characters form a finite set and translate the same bound. Fourier inversion, followed by \(\zeta\downarrow0\), proves (44).

Now take the full-plane thermal law with edge weights \((\rho e^{-1/N})^m\) and its deterministic minimum-split decomposition. Use the notation of the two-end estimate in Corollary 26: for a height scale \(H_i\) and count scale \(k_i\), let \(K_i=H_i^{3/4}\) and \(\kappa_i=\min(k_i,K_i)\). Put \(M=\max_i\kappa_i\) on the nonempty bridges, and \(M=1\) if there are none. The endpoint-minimum case is included using its one-end decomposition. Restrict to length at most \(N^2\), whose complement has negligible thermal probability by the absolute critical mass bound.

The turning-chain estimate without smoothing bounds the critical mass of each dyadic family, including count choices, by \(MN^{o(1)}\). The two no-renewal ends cost at most a constant times \(U_N^2\). Lemma [R:thermal:denominator] bounds the plane normalization below by \((q_N^{-1}-1)U_N^2\), where \(q_N=N^{-\alpha+o(1)}\). Consequently all families with \(M<N^{\alpha-\eta}\) have vanishing normalized mass, for every fixed \(\eta>0\).

In the remaining families fix the end shapes, all witness labels, and all pieces except the unused batch belonging to an index attaining \(M\). Proposition 25 reserves this batch independently of the avoidance tests and the large-height and short-length tests. Exact total length is therefore one scalar equality in that batch. Applying (44) costs an additional \(M^{-1/\alpha+o(1)}\), uniformly in the fixed outside data. The bounded endpoint length corrections merely change the target integer. Damping may be dropped on these bridges before using the critical estimate. After normalization, for every integer \(m\) the mass of length \(m\) in the retained families is at most \[N^{-\alpha+o(1)}M^{1-1/\alpha} \le N^{-1+(1/\alpha-1)\eta+o(1)}.\] No independence between the length and height of an individual irreducible is used: its entire batch remains unexposed until this scalar test.

Theorem [R:thermal:law] gives probability tending to one to \(N^{1-s}\le m\le N^{1+s}\) and to the corresponding spatial exponent window with fixed slack \(s>0\). Conditional on \(m\), the thermal law is the uniform law. Taking \(s\) sufficiently small relative to \(\epsilon\) and using Markov’s inequality on the conditional failure probability, the lengths satisfying the proposition carry probability tending to one. Their intersection with the retained families still does so. The uniform atom bound forces their number to be at least \(N^{1-O(\eta)+o(1)}\).

Set \(N=R^{1/(1+2s)}\). For small enough \(s\) the resulting length interval lies in \([R^{1-\epsilon},R]\). Since \(s\) and \(\eta\) are arbitrarily small, the count is at least \(R^{1-o(1)}\), and the trivial bound by \(R\) proves the claim. The limsup assertions combine these favorable lengths with the all-length upper spatial tail. This argument gives power-cardinality; the separate refolding argument establishes the density-one conclusion. ◻

Proof. We first establish the estimate without the length and diameter restrictions. Although subsequential laws of the rescaled sums may differ, each has full forward support and a strictly positive density in the open forward half-plane. Fourier inversion will then give a uniform lattice lower bound. Finally we mark individual increments to impose the two restrictions.

Subsequential jump limits. Lemma 16 gives \[ p(|Y|>r)\le Cr^{-\beta},\qquad p(|Y|;|Y|\le r)\le Cr^{1-\beta}. \tag{50}\] Consider any sequence \(H\to\infty\). On a subsequence the measures \[\mu_H=H^\beta p(Y/H\in\cdot)\] converge vaguely on \(\mathbb R^2\setminus\{0\}\) to a measure \(\mu\). This follows by compactness on each closed annulus and a diagonal extraction. The bounds \[\mu_H(|y|>r)\le Cr^{-\beta},\qquad \int_{|y|\le r}|y|\,d\mu_H(y)\le Cr^{1-\beta}\] give the corresponding bounds for \(\mu\), by approximation away from zero. In particular \(\int(1\wedge|y|)\,d\mu(y)<\infty\). The measure is supported in \(\{T\ge0\}\).

If \(k/H^\beta\to u>0\), expansion of one characteristic function gives \[ \mathbb E\exp\left(is\cdot H^{-1}\sum_{i=1}^kY_i\right) \longrightarrow F_u(s):=\exp\left(u\int(e^{is\cdot y}-1)\,d\mu(y)\right). \tag{51}\] For completeness, truncate the integral to \(\delta\le|y|\le M\). Vague convergence gives the limit there at continuity radii. The small-jump error is at most \(C|s|\delta^{1-\beta}\); the large-jump error is at most \(CM^{-\beta}\). Send first \(H\) to infinity, then \(\delta\) to zero and \(M\) to infinity. The one-factor error is \(O(H^{-\beta})\), so its squared term disappears after multiplication by \(k\).

The right side is a characteristic function. Construct independent Poisson point collections on disjoint annuli, with intensity \(u\mu\), and sum their points. There are finitely many points outside the unit disk, and the sum of the norms inside it has finite expectation. Thus the small-jump series converges absolutely almost surely. This is the finite-variation form of the Poisson construction of a pure-jump infinitely divisible law; see Schilling [26] for the general construction. Tightness of the rescaled sums also follows directly from (50): remove jumps above \(M\), and use the truncated first moment for the others. Fourier uniqueness therefore proves weak convergence to the constructed law. We prove the support and uniform lattice lower bound needed here next.

Full forward support. For any \(u>0\) the support of the limiting law is the closed additive semigroup generated by the support of \(\mu\), including zero, and is independent of \(u\). One inclusion follows by truncating the Poisson sum. Conversely, a specified finite list of jumps can be approximated with positive probability, excluding every other jump above a small cutoff. The remaining small jumps have arbitrarily small total norm with positive probability, by their first-moment bound. Hence each finite sum of support points belongs to the support.

This semigroup contains the entire open forward half-plane. To prove it, fix an open ball \(B\) whose closure lies in \(\{T>0\}\) and then a smaller closed ball \(B'\) in \(B\). The free-terminal guide estimate in Section 5.1, applied over a fixed positive interval of target heights, gives \[ \sum_{k\ge0}p^{*k}(HB')\ge c_B H^\beta. \tag{52}\] Indeed a narrow straight guide places the terminal port in \(HB'\), uniformly over order \(H\) heights, and contributes \(cH^{-1/4}\) at each height. Counts \(k\le\xi H^\beta\) have total contribution at most \(C_B\xi^2H^\beta\): to reach \(HB'\) their height sum must exceed \(c_BH\), and the truncated-sum bound from (50) gives probability at most \(C_B kH^{-\beta}\). Counts \(k\ge S H^\beta\) contribute at most \(C_BH^\beta e^{-c_BS}\), because \[p\left(\sum_{i=1}^kT_i\le C_BH\right) \le e^{C_B}\bigl(p(e^{-T/H})\bigr)^k \le C_Be^{-c_B k/H^\beta}.\] Choose \(\xi\) small and \(S\) large. The remaining mass in (52) is at least \(c'_BH^\beta\), spread over at most \(SH^\beta+1\) counts. Thus for some \(k_H\in[\xi H^\beta,SH^\beta]\), the probability of \(H^{-1}\sum_{i\le k_H}Y_i\in B'\) is bounded below independently of \(H\). On a further subsequence \(k_H/H^\beta\to u\in[\xi,S]\). Weak convergence and the closedness of \(B'\) show that the limiting law gives positive mass to \(B'\). Because all positive \(u\) have the same support, every such ball meets the support of every limiting law.

Fourier inversion on the displacement lattice. We have obtained full support in the open forward half-plane. To turn this into a positive lattice lower bound, we need a density and Fourier domination. Write \(\varphi(\theta)=p(e^{i\theta\cdot Y})\) for the increment characteristic function. Lemma 17 gives, near each of its modulus-one characters, the uniform estimate \[ |\varphi(\theta)|\le 1-c\operatorname{dist}(\theta,\mathcal P)^\beta, \tag{53}\] where \(\mathcal P\) is the finite character group annihilating \(\Gamma_1\) on the dual torus of \(\Gamma\). Away from fixed neighborhoods of \(\mathcal P\) the modulus is at most a constant strictly smaller than one. Passing to the limit near zero in (53) gives \(|F_u(s)|\le e^{-cu|s|^\beta}\). Thus the limiting law has a continuous density \(g_u\).

In fact \(g_u(v)>0\) whenever the first coordinate of \(v\) is positive. The convolution identity \(g_u=g_{u/2}*g_{u/2}\) follows from (51). Choose a small ball of heights strictly between zero and the first coordinate of \(v\). Full support and continuity provide a point in that ball where \(g_{u/2}>0\), and a smaller ball on which it is bounded below. The reflected difference ball about \(v\) remains in the forward half-plane and has positive mass under the other half-time law. The convolution is therefore positive at \(v\).

Take compatible \(k\) and \(y\) with \(k/H^\beta\to u\in[1,2]\) and \(y/H\to v\). Fourier inversion on \(\Gamma\) yields \[ H^2p^{*k}(y)\longrightarrow c_\Gamma g_u(v), \qquad c_\Gamma>0. \tag{54}\] Here \(c_\Gamma\) is the covolume of \(\Gamma_1\) in the chosen coordinates. To check the periodicity factor, for \(\theta_0\in\mathcal P\) all increments have the same phase \(e^{i\theta_0\cdot a_*}\), so \(\varphi(\theta_0+\theta)=e^{i\theta_0\cdot a_*}\varphi(\theta)\). Compatibility makes the phase in the inversion integrand equal to one. Thus all the finitely many neighborhoods contribute identically. In each neighborhood rescale \(\theta\) by \(H^{-1}\) and dominate by \(e^{-c'|s|^\beta}\) using (53); the complement is exponentially small in \(k\). This proves (54) with its stated constant.

The same subsequence argument proves a uniform positive lower bound when \(|x|/H\le C\) and \(k/H^\beta\in[1,2]\) is compatible. If it failed, choose offending targets and counts, extract \(x/H\to v_2\), \(k/H^\beta\to u\), and then a vague limit \(\mu\) as above. Equation (54) would contradict \(g_u(1,v_2)>0\). A fixed positive fraction of the integers in \([K,2K]\) are compatible with each \(y\in\Gamma\), because \(\Gamma/\Gamma_1\) is finite cyclic. Consequently \[ \sum_{K\le k\le2K}p^{*k}(y)\ge c_CKH^{-2}. \tag{55}\]

Length and diameter restrictions. We now remove paths whose length or total increment diameter is too large. If \(\sum_iL_i>M\), then \[\sum_i(1-e^{-L_i/M})\ge1-e^{-\sum_iL_i/M}\ge1-e^{-1}.\] Mark one increment by \(1-e^{-L/M}\). Its total marked mass is \(q_M:=p(1-e^{-L/M})\). The other \(k-1\) increments match the remaining displacement with probability at most \(C(k-1)^{-8/3}\le CH^{-2}\) by Lemma 17, independently of the value of the marked increment. Summing the mark and then the count gives the excluded mass bound \[C K^2H^{-2}q_M.\] For \(M=H^{4/3+\eta}\), the weak deficit bound \(q_M=M^{-9/16+o(1)}\) makes this \(o(KH^{-2})\).

The same argument controls spatial excursions without a new independence assumption. If \(\sum_i\Delta_i>BH\), then \(\sum_i\min(1,\Delta_i/(BH))\ge1\). The diameter tail implies \[p\bigl(\min(1,\Delta/(BH))\bigr)\le C(BH)^{-\beta}.\] Marking by this function excludes mass at most \(CB^{-\beta}KH^{-2}\). Choose a fixed \(B=M_0\) large enough to leave at least half the lower bound in (55), and then take \(H\) large enough for the length exclusion. The entire concatenated path is within its sum of increment diameters of its first port. This proves (49). ◻

Proof. Fix a small \(\delta>0\), and put \(H=\lfloor N^{3/4-\delta}\rfloor\). Choose one of the three inward port normals at \(o\) whose projection of \(b-o\) is at least \(c|b-o|\). Call the vertex type of \(o\) type \(P\), and the other type \(R\). Start the port path on the cut immediately below \(o\). If \(b\) has type \(R\), end at its incident port on the cut immediately above \(b\); delete the initial and terminal half-edges to obtain the ordinary walk. If \(b\) has type \(P\), end at its incident port immediately below \(b\); delete the initial half-edge and append the final half-edge from that port to \(b\). The appended vertex lies beyond the final open slab and is therefore unvisited. If the port path has \(L\) internal vertices, the resulting edge count is \(L-1\) in the first case and \(L\) in the second. These operations are recoverable from the fixed orientation and terminal vertex. The associated port displacement differs from \(b-o\) by a bounded amount, has positive height comparable to \(|b-o|\), and has horizontal component at most a constant times that height.

When \(|b-o|\gg H\), divide this displacement along its straight guide into \(q=O(|b-o|/H)\) exact port displacements, each with integer height between \(H\) and \(2H\) and bounded horizontal-to-height ratio. Round intermediate positions to ports on the chosen cuts; this changes each horizontal displacement by a bounded amount and leaves the final port exact. Proposition 45, with a fixed length slack \(\eta>0\) sufficiently small after \(\delta\), gives critical mass at least \(cH^{-5/4}\) for every segment, using only lengths at most \((2H)^{4/3+\eta}\) and diameter at most \(C H\). Since \[H^{4/3+\eta}=o(N),\] the thermal discount on each segment is bounded below by a fixed positive constant.

Concatenate the segments in their consecutive slabs. They cannot share vertices across slabs. The predetermined intermediate cuts are each crossed once, so the pieces are recoverable from their concatenation. They remain in a corridor of width \(CH\) about the straight guide. After the bounded endpoint conversion, we obtain \[ G_{1/N}(b)\ge \exp\left[-C_\delta\left(1+\frac{|b-o|}{H}\right)\log N\right] \qquad(|b-o|\gg H). \tag{56}\]

For fixed \(N\) and \(s<m_{\mathrm{rad}}(1/N)\), \[G_{1/N}(b)\le \left(\sum_v e^{s|v-o|}G_{1/N}(v)\right)e^{-s|b-o|}.\] The parenthesized quantity is finite and independent of \(b\). Hence the lower spatial rate is at least \(m_{\mathrm{rad}}(1/N)\), which is at least \(N^{-3/4-\varepsilon}\) for every fixed \(\varepsilon>0\) and sufficiently large \(N\). On the other hand (56) gives the upper spatial rate at most \(C_\delta H^{-1}\log N\). Letting the fixed exponent slacks tend to zero proves both asserted rate scales. No equality or existence of a single directional spatial limit is required.

Now suppose \(|b_N-o|=N^{3/4+o(1)}\). For every fixed \(\zeta>0\), choose \(\delta<\zeta/3\) in (56). For all sufficiently large \(N\) this gives \[ G_{1/N}(b_N)\ge e^{-N^\zeta}. \tag{57}\] The choice of \(\zeta\) will be made after the desired tail slack, and stays fixed as \(N\) tends to infinity.

Let \(c_j\rho^j\) be the total critical mass at edge length \(j\). Proposition 27 gives \(c_j\rho^j\le\exp(C_\nu j^\nu)\) for every \(\nu>0\). Thus, for every \(\varepsilon>0\), the absolute thermal mass on \(j>N^{1+\varepsilon}\), even multiplied by \(j^r\) for any fixed \(r\), is at most \(\exp(-cN^\varepsilon)\) after decreasing \(\nu\). Dividing by (57) proves the length upper tail and its moment version.

For the spatial upper tail choose a small \(\varepsilon'>0\) relative to \(\varepsilon\). On lengths \(j\le N^{1+\varepsilon'}\), the event \(D>N^{3/4+\varepsilon}\) is covered by (48), with a sufficiently small fixed \(\eta\) there. Its absolute mass is at most \(\exp(-N^c)\) for some \(c>0\), after summing the polynomial number of lengths. The lengths above this cutoff have just been treated. Choose \(\zeta<c\) in (57). This proves the spatial upper tail. The same proof, using \(D\le a_{\mathrm{lat}}j\) and the weighted length tail, gives its positive-moment version.

Finally every path to \(b_N\) has \(D\ge|b_N-o|\). If \(j<N^{1-\varepsilon}\), this required displacement is at least \(N^{3/4-o(1)}\), so (48) again bounds the absolute mass by \(\exp(-N^{c'})\) for some \(c'>0\); lengths smaller than the graph distance contribute nothing. Division by (57) proves the length lower tail. The diameter lower bound is deterministic. Integrating the proved weighted upper tails and using these lower bounds gives both moment exponents. ◻

Proof. On a compact subinterval of \(t>0\), every differentiated series is dominated by a constant times the untilted susceptibility at a smaller positive value of \(t\). Hence \((-1)^j\partial_t^jG_t(b)/G_t(b)\) is exactly the \(j\)th edge-length moment under the normalized endpoint law. Apply Theorem 46. The logarithmic derivative is the first moment. Higher logarithmic derivatives are cumulants and are not identified by this argument. ◻

Proof. For \(s>0\), the renewal and reward identities give \[B(s):=\sum_{h\ge0}b_he^{-sh}=\frac1{1-\mathbb E_p e^{-sH}},\qquad \sum_{h\ge0}m_he^{-sh}=B(s)^2\mathbb E_p[Le^{-sH}].\] Positive coefficient summation of (58) yields \(B(1/r)=r^{\beta+o(1)}\) and \(\sum m_he^{-h/r}=r^{d+\beta+o(1)}\). This proves (61). The height deficit is \(r^{-\beta+o(1)}\), so \(p(H>r)\le r^{-\beta+o(1)}\) and \(\mathbb E[H;H\le r]\le r^{1-\beta+o(1)}\). To obtain the reverse height-tail bound, fix \(\delta>0\) and use the deficit at scale \(r^{1+\delta}\). Heights at most \(r\) contribute at most \(r^{-\beta-\delta+o(1)}\), whereas that deficit is \(r^{-\beta(1+\delta)+o(1)}\). The former is negligible because \(\beta<1\). Thus \(p(H>r)\ge r^{-\beta(1+\delta)-o(1)}\); let \(\delta\) decrease to zero.

Write \(j_h=\mathbb E_p[L;H=h]\). Coefficientwise the reward identity is \(m=b*j*b\). At height \(3r\), the convolution multiplier of each \(j_h\) with \(h\le r\) is \[(b*b)_{3r-h}\ge r^{2\beta-1-o(1)},\] because order \(r\) terms have both indices comparable to \(r\). Since \(m_{3r}\le r^{d+\beta-1+o(1)}\), this proves the upper bound in (60). Splitting \(p(L>n)\) according as \(H>n^{1/d}\) or not gives the first bound in (62).

We verify that the discounted reward is spatially retained. The elementary packing bound for a self-avoiding path in a strip is \(L\le C h(1+S)\) at height \(h\ge1\). Integrating (59) therefore gives, after decreasing \(c\), \[\mathbb E_p[L;H=h,S>t] \le C h(1+t+h)e^{-ct/(h+1)}.\] For \(h>r\log^2r\), use \(j_h\le m_h\) and the factor \(e^{-h/r}\); the resulting tail is smaller than every inverse power of \(r\). For \(h\le r\log^2r\) and \(S>r\log^4r\), the last display has the same property after summing over \(h\). Thus removing these two tails retains \(r^{d-\beta+o(1)}\). Evaluating this conclusion at \(r=h^{1-\delta}\) gives \(\sum_{j\le h}j_j\ge h^{d-\beta-o(1)}\) after letting \(\delta\downarrow0\). This proves the other half of (60).

On the retained set \(L=O(r^2\log^6r)\). Take \(r=n^{1/2-\delta}\), so eventually \(L\le n\) on that set. Since \(1-e^{-L/n}\ge(1-e^{-1})L/n\) for \(L\le n\), its retained reward gives \[1-\mathbb E_p e^{-L/n} \ge n^{-1}r^{7/12-o(1)} =n^{-17/24-(7/12)\delta-o(1)}.\] Let \(\delta\downarrow0\). Finally, split the event \(D>r\) at \(H=r/\log^2r\). The height tail costs \(r^{-\beta+o(1)}\); on the complement, diameter greater than \(r\) forces \(S\ge c_0r\). Summing (59) over the remaining heights is superpolynomially small. This proves the diameter bound. ◻

Proof. We give the estimates that distinguish these conclusions from the sharp spatial law. At a dyadic vertex time \(n\), let \(m=\lfloor n^{9/16-\varepsilon}\rfloor\). The length tail in (62), a union bound above \(n\), and Markov’s inequality for lengths truncated at \(n\) imply \[p\Bigl(\sum_{i\le m}L_i>n\Bigr)\le n^{-\varepsilon+o(1)}.\] The probability that all of those \(m\) heights are below \(n^{3/4-2\varepsilon/\beta}\) is stretched exponentially small, by the matching height tail. Once these pieces have been completed, every later vertex is above their slabs. Borel–Cantelli on dyadics, followed by monotonicity between dyadics, proves the asserted height lower bound.

Conversely take \(m=\lceil n^{17/24+\varepsilon}\rceil\). The last estimate in (62) and exponential Markov give \[p\Bigl(\sum_{i\le m}L_i\le n\Bigr) \le e\bigl(\mathbb E_p e^{-L/n}\bigr)^m \le \exp(-n^{\varepsilon/2})\] for large \(n\). The diameter-tail bound, a union bound and truncated first moments show that \(\sum_{i\le m}D_i\le m^{1/\beta+\varepsilon}\) except with a power-decaying probability. These bounds are summable on dyadics, and the union of the first \(m\) blocks contains every vertex up to time \(n\). Sending \(\varepsilon\) to zero gives \((17/24)/\beta=17/18\).

Including the empty bridge, the height-summed tilted law has normalizer \[Z_n=(1-\mathbb E_pe^{-L/n})^{-1}\ge n^{9/16-o(1)}.\] The bound uses the upper length tail. Excluding the empty bridge replaces \(Z_n\) by \(Z_n-1\) and does not change the conclusion. Its part of height at most \(h\) is bounded by \(\sum_{j\le h}b_j=h^{\beta+o(1)}\). Take \(h=n^{3/4-\varepsilon}\). For arches, split at height \(h=n^{3/4}\): the part leaving that strip has mass \(O(b_h)\); the remaining long arches have mass at most \(n^{-1}h^{13/12+o(1)}\). Both contributions have exponent \(-3/16\).

For the completed prefix, put \(j_h=\mathbb E_p[L;H=h]\) and mark the start height \(l\) of each counted block. The expected count is exactly \[\mathbb E V_h=\sum_{l+j\le h}b_l j_j.\] The cumulative bounds in Proposition 48 prove the upper bound. For the lower restrict \(l\le h/2\) and apply the retained reward at scale \(r=h^{1-\delta}\), so eventually \(j\le h/2\). The prefix masses with \(l\le h/2\) total \(h^{\beta+o(1)}\). Their contribution with lateral excursion greater than \(h\log^2h\) is superpolynomially small by (59); the retained last block has excursion at most \(r\log^4r=o(h)\). For each marked block, its preceding prefix stays within \(h\log^2h\) of the source and the marked block stays within \(o(h)\) of its own start. Hence the whole prefix through that block is confined to \([-C h\log^2h,C h\log^2h]\) for one fixed \(C\). These are precisely contributions to \(V_h^{(C h\log^2h)}\), of expected total mass at least \(h^{\beta+o(1)}r^{d-\beta-o(1)}\). Let \(\delta\downarrow0\) and use \(V_h^{(w)}\le V_h\) to obtain both parts of (63). Taking \(h\) a sufficiently small multiple of \(R/\log^3R\) puts every retained prefix inside the disk. All its counted vertices occur before first exit, which proves the last assertion. ◻

Proof. Again put \(j_h=\mathbb E_p[L;H=h]\) and \(\beta=3/4\). The renewal and reward identities with (64) give \[B(1/R)\asymp R^\beta,\quad \sum_hm_he^{-h/R}=R^{25/12+o(1)},\quad \sum_hj_he^{-h/R}=R^{7/12+o(1)}.\] They imply \(\sum_{h\le R}j_h=R^{7/12+o(1)}\). For the lower bound, evaluate at \(R^{1-\varepsilon}\): the tail beyond \(R\) is exponentially negligible because \(j_h\le m_h\) has polynomial growth; then let \(\varepsilon\downarrow0\). Consequently \(\sum_{l+h\le R}b_lj_h=R^{4/3+o(1)}\), proving the first expectation and the upper bound for the second.

The fixed-ball lower bound needs the confined bridge estimate, rather than only these transforms. Choose \(c_0>0\) so small that each height-\(m\) bridge with \(c_0R\le m\le2c_0R\) and lateral wandering at most \(Cm\) lies in \(B(0,R)\) and has height at most \(R\). The sum of their length masses is \[ \sum_{c_0R\le m\le2c_0R} \sum_{\substack{\gamma:H(\gamma)=m\\\gamma\subset B(0,R)}} L(\gamma)\rho^{L(\gamma)} \ge R^{25/12-o(1)}. \tag{65}\] Under the iid irreducible law, the left side counts visits in prefixes ending at such renewal heights, with a visit repeated at later permissible prefix endings. For a vertex in a block ending at height \(s\le R\), condition on that block and its past. Dropping every future confinement condition, the expected number of repetitions at later renewal endings is at most \(\sum_{h\le R}b_h=O(R^\beta)\), by independence of the future pieces. Every counted vertex belongs to \(\mathcal V_R\cap B(0,R)\). Tonelli’s theorem therefore bounds the left side of (65) by \(CR^\beta\mathbb E|\mathcal V_R\cap B(0,R)|\). Since \(25/12-\beta=4/3\), this proves the lower bound. The same choice of a smaller \(c_0\) handles any fixed positive multiple of the ball radius.

For the almost-sure statement, the bounded-factor height deficit gives \(p(H>R)\asymp R^{-\beta}\): its upper bound is immediate, and after using that upper bound the contribution of \(H\le\eta R\) to the deficit at \(R\) is at most \(C\eta^{1-\beta}R^{-\beta}\). Choose a fixed small \(\eta\); the remaining contribution gives \(p(H>\eta R)\ge cR^{-\beta}\). Replacing \(R\) by \(R/\eta\) proves the lower bound at \(R\). Splitting according to height and using the cumulative reward gives \(p(L>t)\le t^{-9/16+o(1)}\). The first \(k=\lceil R^{\beta+\varepsilon}\rceil\) blocks reach a height larger than the greatest layer meeting \(B(0,R)\) except with probability \(\exp(-cR^\varepsilon)\). Their total length is at most \(R^{4/3+3\varepsilon(4/3)/\beta}\) except with probability \(R^{-2\varepsilon+o(1)}\), by a union bound for a length above that threshold and Markov’s inequality for truncated lengths. After those blocks, the path remains above the ball. The estimates are summable on dyadic radii; monotonicity and \(\varepsilon\downarrow0\) complete the proof. ◻

Proof. First exhaust a tangent half-plane by aligned tessellation hexagons. Boundary saturation in (66) and positive real parts show that the artificial exits tend to zero. Artificial exits in any smaller convex domain are a subset of these exits, so they too vanish. This justifies its limiting boundary balance.

For the quantitative assertion, cut the domain by a tessellation hexagon about a nearby lattice vertex, whose inner and outer radii about the source lie between \(cS\) and \(S/2\). Subtract the balances before and after the cut. A chord valid in both domains has the same terminal phase and cancels; all remaining old chords are precisely those lost on imposing the cut. Every chord reaching distance \(S\) is among these lost chords. The new exits lie on the cap, and convex positivity compares their mass with the lost mass in both directions, with absolute constants. Their paths are also cap exits in the cut tangent half-plane. Applying the same subtraction there bounds their mass by the lost half-plane arch mass, at most \(W_{cS}\). The exhaustion argument above applies if the original domain is infinite. ◻

Proof. Consider first the \(60^\circ\) wedge between the positive horizontal ray and the ray at angle \(\pi/3\), with source at half-integer coordinate \(a>0\) on the horizontal ray. Relative to the full upper half-plane, let \(L_<(a)\) and \(L_>(a)\) be the lost masses to the left and right of the source, including endpoints beyond the apex. Let \(D(a,b)\) be the lost mass when both endpoint coordinates are positive. Reversal gives \(D(a,b)=D(b,a)\).

The phases of a left base exit, a right base exit and an exit on the other ray are respectively \(e^{3i\vartheta}\), \(e^{-3i\vartheta}\) and \(e^{i\vartheta}\). If \(K_{\rm w}(a)\) is the mass to the other ray, subtraction of the balances gives \[e^{3i\vartheta}L_<(a)+e^{-3i\vartheta}L_>(a) =e^{i\vartheta}K_{\rm w}(a).\] Multiplication by \(e^{-i\vartheta}\) and imaginary parts yield \(L_<(a)=\sqrt2 L_>(a)\). Summing this relation over \(a\le R\) and using symmetry of \(D\) gives \[(\sqrt2-1)\sum_{0<a<b\le R}D(a,b) +\sqrt2\sum_{0<a\le R<b}D(a,b) =\sum_{0<a\le R}A(\lceil a\rceil).\] The total loss is at most \(CW_{ca}\) by Lemma 52. For a lower bound, take also the reflected wedge with its apex on the other side of the source. Their intersection is a triangle of diameter \(O(a)\). An arch escaping this triangle is lost in at least one wedge; reflection makes their loss masses equal. Thus the total loss in either wedge is at least \(cW_{Ca}\), and has exponent \(-1/4\).

Since \(L_>(a)=\sum_{b>a}D(a,b)\), the preceding identity is bounded above and below by fixed multiples of \(\sum_{a\le R}L_>(a)\). Consequently \[\sum_{a\le R} A(\lceil a\rceil)=R^{3/4+o(1)}.\] Monotonicity gives the upper bound for \(A(R)\). For the lower bound, subtract the cumulative sum at \(R\) from that at \(R^{1+\delta}\) and bound the difference above by \(R^{1+\delta}A(R)\). Letting \(\delta\downarrow0\) proves the first assertion in (69).

Half-plane–strip subtraction gives the exact relation \[b_h=\sin\vartheta\sum_{j>h}a_j.\] An arch in this sum reaches distance comparable to \(h\), so the cap bound gives \(b_h\le h^{-1/4+o(1)}\). To obtain the reverse bound, let \(K_{\rm w}(a,v)\) be the other-ray wedge kernel. Reversal and reflection make it symmetric in \(a,v\), and the balance above shows \(\sum_vK_{\rm w}(a,v)=a^{-1/4+o(1)}\). Cut the wedge at height \(h\). All exits with \(v>Ch\) are lost, and the new exits have mass at most \(b_h\) per source. By symmetry, \[\sum_{a\le R}\sum_{v>Ch}K_{\rm w}(a,v) \ge R^{3/4-o(1)}-h^{3/4+o(1)}.\] The subtracted term bounds the sum with \(v\le Ch\) by summing first over \(v\) and then over all \(a\). Therefore \[Rb_h\ge c\bigl(R^{3/4-o(1)}-h^{3/4+o(1)}\bigr).\] Set \(R=h^{1+\epsilon}\) and let \(\epsilon\downarrow0\).

For either stated wedge angle, the missing same-side endpoints beyond the apex have mass at least \(A(CH)=H^{-1/4+o(1)}\). Its boundary balance compares the total same-side loss with the other-ray mass, with fixed positive constants. Lemma 52 removes at most \(H^{-(1+\epsilon)/4+o(1)}\) outside the prescribed radius, which is negligible. This proves the confined assertion. ◻

Proof. The interior visit identity bounds the sum of positive chord visits to any non-boundary-adjacent vertex by a constant times its three incident \(+2\) nesting partitions. Each such one-face partition in the full strip has square at most \(G_M(2)\), for a physical period \(M\asymp Ch\) divisible by four in a transverse \(60^\circ\) direction. To see the square, translate one strip and its marked face to each of the two cylinder marks. For large fixed \(C\), the vertical separation of the half-period translates exceeds the strip width. The two strips are disjoint, their cycles embed without intersections, and every cycle separates the marks. Their independent union is therefore counted by \(G_M(2)\).

It follows from (67) that the raw visit mass at each such vertex is at most \(h^{1/12+o(1)}\). This bound applies first in finite convex strip exhaustions and then, by positivity, to genuine strip chords. For a fixed source, sum visits along a horizontal row. Horizontal translation identifies this sum with the sum over the corresponding boundary sources at one representative vertex of that row; it is bounded by the raw all-source visit estimate. There are \(O(h)\) rows. Visits in a port-adjacent row can be charged, with bounded multiplicity, to slant neighbors on the path that are not adjacent to the boundary, when \(h\ge2\). This proves the first-length bounds. Smaller heights have finite masses and moments as well: append a fixed bridge to reach a larger height; for arches use monotonicity.

For (70), cut the strip on a transverse \(60^\circ\) tessellation line. Compare balances in the strip and in one remaining side of this cut, then sum over bottom sources on that side. Retain only same-side chords whose endpoints are on opposite sides of the cut; all are lost. A chord with endpoint separation \(d\) has exactly \(d\) such source placements, up to a fixed lattice factor. The total lost mass thus bounds the left side of (70). It is controlled by new cut exits. Reversing these paths gives, at each cut port of height \(y\), a mass of bottom exits at most \(O(1)\), and at most \(CW_{cy}\) when \(y\) is bounded away from zero, by Lemma 52. Summing over \(0\le y\le h\) gives \(h^{3/4+o(1)}\).

We finally record why arbitrary fixed-width walks, needed in the slit exhaustion below, are summable. Split such a walk at a global highest vertex. On each resulting chain, cut at the last occurrence of its lowest level, then the last highest level in the remainder, and continue. Vertex heights in thirds of a layer are integral and change at every step. The positive spans of these extremal pieces decrease strictly, except possibly for equality of the first two. Thus the number of pieces is bounded in terms of the strip height. Each piece can be extended at its extreme ends to horizontal ports by adding at most one new vertex per end, giving a bridge with bounded multiplicity and bounded weight change. Drop avoidance between different pieces and compensate for shared endpoints. There are finitely many possible lists of spans. For each list, the mass and its first-length insertion are bounded by products of finite bridge masses and moments. There are only finitely many possible rows of the split vertex; after fixing a representative in that row, the original initial vertex determines at most one horizontal translation. This proves both asserted finiteness statements. ◻

Proof. The slit is attached to the exterior. A stem from its boundary still approaches a first self-collision from outside the new loop, so the same local cancellation proves the signed balance. Figure 6 shows the domain. The following table gives the boundary-determined turns from an upward-entering source; a thin opening of the slit makes the concave tip turn \(-\pi\) explicit. Put \(\theta_0=3\pi/8\). \[\begin{array}{c|ccccc} \text{exit}&\text{upper left}&\text{upper right}&\text{top} &\text{bottom}&\text{lower shore}\\ \hline T&\pi&-\pi&0&\pi&2\pi\\ e^{3iT/8}&e^{i\theta_0}&e^{-i\theta_0}&1 &e^{i\theta_0}&e^{2i\theta_0} \end{array}\] Compare with the isolated upper strip of height \(L\). Let \(E_<(a)\) and \(E_>(a)\) be the nonnegative gains to other upper-shore ports on the left and right of the source, and let \(E_T(a)\) be the gain to the top. Let \(A_0(a)\) be the reference mass to same-side endpoints left of the tip, which are missing from the slit shore. Project the balance difference by \[\mathcal P(w)=-\frac{\Im(e^{-2i\theta_0}w)}{\sin\theta_0}.\] It kills every lower-shore term, including a return to the same cut location. The other coefficients give \[E_<+V+2\cos\theta_0\,E_T =A_0+\frac{\sin(\pi/8)}{\sin\theta_0}E_>.\]

These identities may first be written in remote finite truncations. Lemma 54 makes their exit errors vanish and allows absolute summation over sources. For the latter assertion, every gained or bottom-exiting path goes around the tip. Translate to a fixed source and ignore the slit: its possible offsets relative to the tip are at most a constant times its length. The fixed-width first-length sum therefore bounds the summed gains and bottom exits.

Reversal of an upper-shore gained chord gives \(\sum_a E_<(a)=\sum_a E_>(a)\). Since \(\sin(\pi/8)/\sin\theta_0=\sqrt2-1<1\), summation and positivity imply \[\sum_aV(a)\le\sum_aA_0(a)=\sum_{d>0}dH_L(d).\] Lemma 54 proves the desired bound. Its right side depends only on the upper height \(L\), which explains the uniformity in \(s_0\). ◻

Proof. Because \(a,b\) lie on the same straight side, \(T(a,e)=T(b,e)\); give both trunks the common phase \(e^{isT(a,e)}\) during the following exploration. Fix a trunk from \(a\) to \(e\) and grow a turning-weighted stem from \(b\), stopping at its first trunk contact as well as at boundary exits. Compare this with trunk \(b\) to \(e\) and stem from \(a\). Self-collisions cancel by the local rule. A contact term is an embedded Y tree with outward arms in cyclic order \(a,b,e\). At its junction, the turn from the \(a\) arm onto the \(e\) arm is \(+\pi/3\), whereas from the \(b\) arm it is \(-\pi/3\). The other turns and all vertex weights agree, so the ratio of the two contact phases is \(e^{is(2\pi/3)}=r\). This also covers immediate contact. After multiplying the second exploration by \(r\), all contact terms cancel. The remaining disjoint boundary exits have exactly the two cyclic ranges in (72); division by the common trunk phase gives that identity. In strip exhaustions, remote disjoint exits vanish by product domination and the cap bound.

If \(e\) lies beyond \(b\) on the same base segment, multiply the identity by \(e^{3i\vartheta}\) and take imaginary parts. The first sum has zero imaginary part. In the subtracted sum the sine phases are nonnegative, since \(5\vartheta+sT\in[2\vartheta,8\vartheta]\). As \(\sin3\vartheta=\sin5\vartheta>0\), this gives \(K(a,e)\le K(b,e)\). Reversal makes it outward decrease of the terminal from a fixed source; a tessellation reflection gives the other direction.

To prove the confined lower bound, use a tangent half-hexagonal cap whose radius is a small constant times \(H^{1+5\epsilon}\). Along its base, the sum of kernels to ports beyond distance \(H^{1+4\epsilon}\) is at least the half-plane tail \(A(H^{1+4\epsilon})\) minus the lost mass \(CW_{cH^{1+5\epsilon}}\). By Lemmas 52 and 53, the second term is negligible and the first has exponent \(-(1+4\epsilon)/4\). There are \(O(H^{1+5\epsilon})\) ports on the cap base. The outward monotonicity just proved bounds the kernel to every nearer port below by this sum divided by that number. The cap lies inside the required confinement radius, giving (73). ◻

Proof. Apply the imaginary-part projection in the preceding proof with \(e\) on the top wall. Bottom exits to the right of \(b\) and to the left of \(a\) vanish. The top coefficients agree because \(\sin3\vartheta=\sin5\vartheta\), giving the difference identity. Summing it from the left end of the strip gives the second identity; the kernel tends to zero at infinity by summability.

For the last bound, join the adjacent bottom ports below the strip by a three-center arch. Together with the two bridges this gives a downward half-plane arch from \(c\) to \(d\). Translate its first port to a fixed source. The connector has a unique deepest vertex, so the output recovers the connector, the translation and the original height. Thus the map is injective over all \(h\), with constant weight factor \(\rho^3\). Its image reaches distance at least a constant times \(H\), and (66) proves the bound. ◻

Proof. The height deficit is \(\mathbb E_p(1-e^{-\mathsf H/t})=t^{-3/4+o(1)}\), giving the height upper tail and \(\mathbb E[\mathsf H;\mathsf H\le t]\le t^{1/4+o(1)}\). Evaluate the deficit at \(t^{1+\epsilon}\). Heights at most \(t\) contribute only \(t^{-3/4-\epsilon+o(1)}\), which is smaller than the deficit \(t^{-3(1+\epsilon)/4+o(1)}\). This proves the lower tail as \(\epsilon\downarrow0\).

Call a block wide if \(S>t\), and put \(u=\mathbb E_p[z^{\mathsf H};S>t]\). A bridge containing such a block is itself wide, so at each height its mass is at most \(CW_{ct}\) by the cap bound. Under \(\mathbb Q_z\), the probability of at least one wide block is exactly \[\frac{B(z)u}{1+B(z)u}.\] Take \(z=e^{-1/H}\) with \(H=t^{1-\epsilon}\). This probability is at most \(CHW_{ct}/B(z)=o(1)\), and hence \(u\le H^{-1/2+o(1)}t^{-1/4+o(1)}\). On \(\mathsf H\le H\) the tilt is bounded below. Add the height-tail bound on the complement and let \(\epsilon\downarrow0\) to obtain the span bound.

The marked-reward identity gives \[\sum_hm_B(h)z^h=B(z)^2\mathbb E_p[\mathsf Lz^{\mathsf H}], \qquad \mathbb E_p[\mathsf L e^{-\mathsf H/H}] \le H^{7/12+o(1)}.\] Splitting according to \(\mathsf H>n^{3/4}\) and applying Markov to the remaining length gives the upper length tail in (75).

It remains to prove that the signed displacement has a matching lower tail. By Lemma 57, bridge weights to any endpoint windows of size \(G\) per row, summed over \(R\le h\le2R\), are at most \(C(G+1)R^{-1/4+o(1)}\). For \(z=e^{-1/R}\), the whole height bin has \(\mathbb Q_z\)-probability \(R^{-o(1)}\), whereas the part with \(|\mathsf X_{\rm total}|\le R^{1-\epsilon}\) has probability at most \(R^{-\epsilon+o(1)}\). Thus displacement larger than this threshold has at least subpower probability.

If every piece had displacement at most \(t_*=R^{1-\delta}\), with \(\epsilon\ll\delta\), such a total displacement would be superpolynomially unlikely. Here are the estimates. Reflection preserves height and reverses displacement, so the truncated increments \(\mathsf X_{\rm trunc}=\mathsf X\mathbf1_{\{|\mathsf X|\le t_*\}}\) are centered under \(p_z\). Their variance is at most \(t_*^{5/4+o(1)}\) by the span tail. The geometric number of pieces exceeds \(R^{3/4+c\delta}\) with superpolynomially small probability for small fixed \(c>0\). For \(|v|\le1/t_*\), \[\log\mathbb E_{p_z}e^{v\mathsf X_{\rm trunc}} \le C v^2\mathbb E_{p_z}\mathsf X_{\rm trunc}^2.\] Both the displacement threshold divided by \(t_*\) and its square divided by the resulting total variance grow as positive powers of \(R\). Exponential Markov therefore gives the claimed negligible probability. A union bound, using the geometric mean \(R^{3/4+o(1)}\), now forces \[\mathbb E_p[z^{\mathsf H};|\mathsf X|>R^{1-\delta}] \ge R^{-3/4-o(1)}.\] Let \(\delta\downarrow0\) and combine with the span upper tail. ◻

Proof. Fix a small \(\delta>0\), and set \[L=\lceil m^{4/3+\delta}\rceil,\qquad G=m^{4/3-\delta}.\] We first show that, outside a superpolynomially small event, on each arm at most \(m/4\) renewal endpoints lie within horizontal distance \(G\) of the pertinent extreme crossing of the other full infinite path at that height. The pertinent crossing is the leftmost on the right arm or the rightmost on the left arm; it exists because the path reaches every height.

Condition on the other path. Divide the renewal indices into residue classes modulo \(m_0=\lfloor m^{1-c\delta}\rfloor\), with a small fixed \(c>0\). Successive sampled endpoints in one class are separated by a fresh chunk of \(m_0\) independent pieces. By the displacement lower tail in Lemma 58, such a chunk contains a positive power of \(m\) increments with magnitude greater than \(2G\), except with probability tending to zero. Conditional on all chunk heights and absolute displacements, their nonzero signs are independent fair signs. The target crossing position depends on the heights and the other path, but not these signs.

If there are \(k\) increments of magnitude greater than \(2G\), their signed sum lands in an interval of width \(2G\) with probability at most \(C/\sqrt{k}\). Indeed, after fixing the other signs, the subsets of positive large signs that yield a sum in that interval form an antichain: adding one such sign changes the sum by more than \(4G\). Counting maximal chains bounds its size by the middle binomial coefficient. Thus the conditional window probability at each sampled endpoint tends uniformly to zero. Successive conditioning makes a positive fraction of hits in any residue class exponentially unlikely in \(m/m_0\). Union over the classes proves the assertion about at most \(m/4\) close endpoints; the initial incomplete chunk costs at most \(m_0=o(m)\) indices.

Suppose now that \(P_m\) holds, both total heights are at most \(L\), and the exception above does not occur. There are at least \(cm^2\) pairs of renewal indices for which the two prefixes are disjoint and their gap at the shorter height exceeds \(G\). Crossings there by the longer prefix equal those by its full path: later pieces lie strictly above its endpoint. We claim that the expected number of all such height-bounded pairs is \[ O(1)+O(L^{7/4+o(1)}/G). \tag{77}\] Summing over renewal prefixes gives ordinary bridge weights. Equal-height pairs have bounded total mass by (74). For unequal heights, fix their difference \(s_0=p-n>0\) and which arm is shorter. Join their adjacent bottom starts by the three-center connector used in Lemma 57.

At row \(n\), choose a tessellation vertex between the shorter endpoint and the extreme crossing of the longer prefix. The gap gives at least \(cG\) choices. From that tip cut the horizontal ray pointing away from the longer path’s crossings, through the shorter endpoint. The joined path avoids the slit: the shorter bridge is strict below its endpoint, and all crossings of the longer bridge at this row are on the other side. It starts at the slit shore, goes toward the connector, and terminates on the line of height \(p\).

The map is illustrated in Figure 7. Send the tip \((t,n)\) to the origin. If the shorter endpoint is on the right, use \((x',y')=(x-t,n-y)\); if it is on the left, use \((x',y')=(t-x,n-y)\). These tessellation symmetries make the slit rightward and the initial shore upper. The endpoint is on \(y'=-s_0\), while the unique highest vertex, on the connector, has \(y'=n+2/3\). The path is therefore counted by the slit estimate with upper wall \(L+1\). Conversely, its unique highest vertex recovers \(n\), the connector and the original translation, and hence both bridges and the chosen tip. For the fixed assignment this map is injective, with a fixed weight factor. Summing (71) over the slit sources and dividing by the \(cG\) tip choices bounds the pair mass, already summed over \(n\le L\), by \(CL^{3/4+o(1)}/G\). Sum over the at most \(L\) height differences to obtain (77).

Division by \(cm^2\) now bounds \(p_m\) in the restricted height range by \(m^{-1+C\delta+o(1)}\). To remove that restriction, set \(t=m^{4/3+\delta/2}\). Exponential Markov at \(1/t\), together with \(\mathbb E[\mathsf H;\mathsf H\le t]\le t^{1/4+o(1)}\), shows that a sum of \(m\) heights exceeding \(L\) with no increment above \(t\) has superpolynomially small probability. A height above \(t\) at index \(j\) on either arm is independent of \(P_{j-1}\). Thus \[p_m\le Cm^{-1+C'\delta+o(1)} +Cm^{-1-c'\delta}\sum_{j<m}p_j, \qquad p_0=1,\] for some \(c'>0\). Induction gives \(p_m=O_\eta(m^{-1+\eta})\) whenever \(\eta>C'\delta\): under this bound the second term is \(O(m^{-1+\eta-c'\delta})\). Choose \(\delta\) arbitrarily small to obtain (76). ◻

Proof. First count valid segments wholly outside \(\xi\). Marking such a segment in either exterior bridge gives two ordinary bridge factors and the folded arch factor. One bridge normalization cancels, so this contribution is at most \[2\rho B(z)\sum_{k\text{ in range}}a_kz^{2k} \le 2\rho B(z)\sum_{k\ge c_1H}a_k \le H^{1/2+O(\epsilon)+o(1)}.\] The last inequality uses \(B(z)=T^{3/4+o(1)}\) and the arch radius tail.

Every other valid segment contains the whole defect. Its middle renewal cannot lie strictly inside an irreducible. Suppose first that this renewal is \(l\ge0\) levels below the start of \(\xi\); the other case follows by reversal and height reflection, which preserve the leftmost rule and replace \(\xi\) by a fixed transformed defect if necessary.

Here is the normalization cancellation for these segments. Cut at their two endpoint renewals. The ordinary pieces outside the segment are free and contribute \(B(z)^2\), canceling the two exterior normalizations. Thus for this orientation their contribution to \(\mathbb E^*_\xi K\) is the sum, over valid internal segments containing the fixed defect, of \[\rho^{L(\text{segment})-L(\xi)}z^{2k-d}.\] There is no weight for the prescribed defect in this expression and no sum over an independently chosen joining port. Unfolding the segment at its middle gives two arms of height \(k\): one ordinary bridge, and one ordinary bridge of height \(l\), followed by \(\xi\), followed by an ordinary bridge of height \(k-l-d\). The pieces have total ordinary height \(2k-d\), as in the displayed weight. Reversed or reflected ordinary pieces can be rooted at their first port by translation. Figure 8 shows these heights.

If \(l=0\), dropping avoidance already bounds the aligned arm sum by \(T^{1/2+o(1)}\), using the squared-sum estimate below. Suppose \(l>0\). The unfolded arms initially meet only at their common port and first down triangle, and take its distinct slant branches. They must be turned into disjoint arms from adjacent ports before Lemma 59 can be used.

Orient the unfolding downward and put its common port at \(P=(0,0)\). Write \(h_0=\sqrt3/2\). The first down center is \(D=(0,-h_0/3)\); the left branch next visits \(U=(-1/2,-2h_0/3)\). Replace the initial portion \(P,D,U\) of this branch by \[P_-=(-1,0),\qquad D_-=(-1,-h_0/3),\qquad U.\] These are actual neighboring ports and centers, as shown in Figure 9. The center \(D_-\) is unused by both arms, because \(D\) was chosen as the leftmost visited highest down triangle of the arch. From \(U\) the next edge cannot return to \(D\) or visit \(D_-\), so it continues downward. The replacement therefore preserves self-avoidance and makes the two arms disjoint. It substitutes one vertex for one vertex, preserving each length and weight. All changed edges lie at depths less than one; every positive integral-depth crossing and its order is unchanged. Since \(l\ge1\), the fixed defect is untouched.

The inverse restores the left initial portion using the unchanged right port and first center. Keep the finitely many choices of arm and symmetry as separate assignments. After reanchoring the adjacent starts at fixed ports, the unchanged far endpoint and the original anchor determine the translation back. This gives bounded inverse multiplicity with no height-dependent factor.

A simple strip crosscut meeting a level just once separates its two horizontal rays. Consequently, whenever either repaired full bridge renews, all crossings of the other at that level lie on the prescribed side of its endpoint; the endpoints on the two exterior lines are ordered as well. This supplies the ordering condition used next.

Let the ordinary prefix on the defect arm contain \(m\) blocks and end at height \(l\). Expose the other arm through its first renewal at height \(p\ge l\). Sum the remaining ordinary portions, discarding their avoidance constraints. Cauchy–Schwarz applied to the shifted sequence \((b_jz^j)_{j\ge0}\) bounds the alignment cost uniformly in \(l,p,d\) by \[\sum_{k\ge\max(p,l+d)} b_{k-p}b_{k-l-d} z^{(k-p)+(k-l-d)} \le\sum_{j\ge0}b_j^2z^{2j}\le T^{1/2+o(1)}.\]

After this summation the exposed portions can be counted as independent \(p\)-sequences: the first \(m\) blocks on the defect arm and the other arm stopped at its first renewal at or above \(l\). Their necessary disjointness and ordering conditions imply \(P_m\). If at least \(m\) blocks of the other arm have been exposed, restrict to its first \(m\). If fewer have been exposed, extend that arm by ordinary blocks: they stay beyond \(p\ge l\) and neither meet the defect arm’s ordinary prefix nor alter crossings at its renewal levels. Thus every such extension has the required conditions for the first \(m\) blocks of both arms. This explains why a probability bound for two equally indexed prefixes applies to the unequal stopping rule.

Retain the tilt \(z^l\) on the \(m\) ordinary blocks of the defect arm, and drop the other exposed tilt factors. For \(m\le T^{3/4+\delta'}\), Lemma 59 bounds the sum of exposed-prefix weights by \[\sum_{m\le T^{3/4+\delta'}}p_m=T^{o(1)}.\] For larger \(m\), discard ordering instead and use \(I(z)^m\); since \(1-I(z)=T^{-3/4+o(1)}\), that tail is negligible for each fixed \(\delta'>0\). Multiplying by the alignment bound proves (78) for segments containing the defect, and hence for all valid segments. ◻

Proof. If a polygon has many exact-two levels in one direction, cut at their two extremes. There is an outer arch on each side, and between the extremes there are two through-bridges: the other pairing would disconnect the single cycle. Every intermediate exact-two level is a common renewal of these bridges. Fixing the extreme levels and ports modulo translation costs a polynomial in \(Q\). The outer arches have bounded mass. Drop avoidance between the through-bridges and count their prefixes as two independent \(p\)-sequences.

Let \(N_Q\) count their common renewal heights up to \(CQ\). For every fixed positive integer \(r\), the renewal property gives \[\mathbb E\binom{N_Q}{r} \le\left(\sum_{j\le CQ}b_j^2\right)^r \le Q^{r/2+o(1)}.\] One obtains the first inequality by listing the \(r\) common hits in order and summing the independent height increments between consecutive hits. Therefore \[\mathbb P\{N_Q\ge Q^{1/2+\epsilon}\} \le \frac{Q^{r/2+o(1)}}{\binom{\lfloor Q^{1/2+\epsilon}\rfloor}{r}} =Q^{-r\epsilon+o(1)}.\] Taking \(r\) as large as needed absorbs the polynomial specification cost and any fixed polynomial length factor.

For the second assertion, fix the original cut level and one arc. Its endpoints are ports on that line, so neither lies on a tessellation line in a different direction. Take the first and last, in level order, among the alleged many exact-one crossings in the second direction. The portion between them is a strict through-bridge and all intermediate exact-one levels are its renewals. For a single sequence the same argument uses \[\sum_{j\le CQ}b_j\le Q^{3/4+o(1)}\] in place of the squared sum, and the threshold is \(Q^{3/4+\epsilon}\). The end portions lie between the cutting-line constraint and the corresponding extreme-level constraint, so they are boundary chords in convex wedges and have bounded mass. The other polygon arc is a bounded-mass half-plane arch. Fixing their cut ports and levels costs only a polynomial. For a fixed-source arch use its base line in place of the polygon’s first cutting line; the same decomposition applies. Finally every path of diameter at most \(Q\) has \(O(Q^2)\) vertices, so fixed polynomial length factors are harmless. Crossings throughout are at bond midpoints. ◻

Proof. Use the regular hexagons \(D_R\) in the finite input, summed over integers \(H\le R\le2H\). Consider ordered chords that visit the middle region (some center within \(R/10\) of the middle). Their mass is at most \(H^{7/4+o(1)}\) by the boundary estimate. Their length sum is at least \(H^{37/12-o(1)}\) by the imported bulk-visit lower bound (at least \(R^{2+1/12-o(1)}\) per hexagon). Close chords with additional paths, using a small \(\epsilon>0\) and polygons of diameter \(\le Q=H^{1+6\epsilon}\). All pieces joined at ports below have disjoint center sets and weights just multiply. Fix endpoint side choices, finitely many possibilities.

• If both endpoints are on the same side, join them by an independent arch in its exterior half-plane. By (73) the confinement toll (total available factor) is at least \(H^{-5/4-C\epsilon-o(1)}\). That side level has exactly two crossings in the polygon. For a given nonexceptional polygon up to translation and fixed \(R\), there are at most \(Q^{1/2+\epsilon}\) choices for the cut level, and \(O(H)\) more for translation along it (place a port on the side); the pieces can then be read from the polygon up to bounded choices.

• If the sides are nonparallel, denote their lines by \(v=0\) (source side) and \(u=0\) (destination side), signed oblique level coordinates chosen so the interior lies at \(u<0,v>0\). These tessellation lines meet at a lattice vertex at distances \(O(H)\) from the two endpoints \(a,b\). Join \(a\) into the exterior wedge \(u<0,v<0\) to an exit port \(c\) on \(u=0\). Both possible wedge angles are covered by the confined other-ray bound after (69); use radii \(\le H^{1+\epsilon}\). Then join \(c\) to \(b\) in the half-plane \(u>0\). By (73) the joint toll is at least \(H^{-3/2-C\epsilon-o(1)}\). Now \(u=0\) has exactly two crossings, and in the arc in \(u<0\) the level \(v=0\) has exactly one. Thus the translation choices at each \(R\) cost at most \(Q^{1/2+\epsilon}Q^{3/4+\epsilon}\) for nonexceptional polygons; the two level placements determine the translation, and the three pieces are recovered from the cuts.

• For opposite parallel sides orient the strip vertically in height, the chord a bridge from \(a\) to \(b\) of height \(h=2R\). Use an independent second height-\(h\) bridge, started to the right of \(a\) at an integer gap in \([H^{1+2\epsilon},2H^{1+2\epsilon}]\), confined within radius \(H^{1+\epsilon}\) of its start. This has mass \(h^{-1/4+o(1)}\) by (69) and the cap comparison, and is separate from the chord. Join the pairs of ends by arches below and above the strip using (73). Including the gap choices, the toll is at least \(H^{-7/4-C\epsilon-o(1)}\). Across all \(R\), the two cut levels both have exact-two crossings, and determine \(h\), hence \(R\); thus multiplicity for a nonexceptional polygon is at most \(C Q^{1+2\epsilon}H\), including translation along the source side. The original pieces and the gap are then read off up to bounded choices.

In each case the number of inverse specifications even in the exceptional set is only polynomial (port assignments and translations in bounded ranges). Thus by (68) and the observations above, counting polygon lengths squared (which majorize the chord lengths squared) costs \(Q^{2/3+o(1)}\) times the stated multiplicities, plus negligible errors. To display the losses, suppress the already specified \(O(\epsilon)+o(1)\) corrections. The leading exponents from the inverse toll and the inverse count (including the \(R\) sum) are \[\begin{aligned} \text{same side:}&\quad \tfrac54+\tfrac52=\tfrac{15}4,\\ \text{nonparallel sides:}&\quad \tfrac32+\tfrac94=\tfrac{15}4,\\ \text{opposite sides:}&\quad \tfrac74+2=\tfrac{15}4. \end{aligned}\] In the first two cases one sums over \(R\); in the opposite-side case the two recovered cut levels already determine \(R\). Including the polygon second-moment exponent gives \(15/4+2/3=53/12\). Let \(\epsilon\downarrow0\) to obtain (79). ◻

Proof. The shorter chords carry first-length mass at most \(H^{4/3-\delta}M_0(H)\le H^{37/12-\delta+o(1)}\), negligible relative to \(M_1(H)\ge H^{37/12-o(1)}\). Thus the long chords retain first-length mass \(H^{37/12-o(1)}\). Cauchy–Schwarz and Lemma 62 give their mass at least \[\frac{H^{2(37/12)-o(1)}}{H^{53/12+o(1)}}=H^{7/4-o(1)}.\] ◻

Proof of Theorem 51. Fix a small \(\eta>0\), choose \(0<\epsilon\ll\delta\ll\eta\), and put \[n=H^{4/3-\eta},\qquad T=H^{1+3\epsilon},\qquad z=e^{-1/T}, \qquad u=\mathbb E_p[z^{\mathsf H};\mathsf L\ge n].\] Call the chords supplied by Corollary 63 long; their lengths are at least \(H^{4/3-\delta}\).

From long chords to bridges or arches. There are finitely many assignments of endpoint sides. At least one assignment carries mass \(H^{7/4-o(1)}\), though the assignment may vary with \(H\). Its conversion is as follows.

• For opposite sides, the chord is already a bridge of height \(h=2R\). There are \(O(H)\) possible source placements at each \(R\), and \(h\) determines \(R\). Translating to a fixed source gives long bridge mass, summed over \(2H\le h\le4H\), at least \(H^{3/4-o(1)}\).

• For the same side, regard the chord as an arch in its inward half-plane. Visiting the middle forces its maximum layer between \(c_0H\) and \(CH\). There are \(O(H^2)\) choices of \(R\) and source placement, so after translation its long arch mass is at least \(H^{-1/4-o(1)}\).

• For nonparallel sides, use the oblique coordinates from the closure proof: the source side is \(v=0\), the destination side is \(u=0\), and the hexagon lies in \(u<0,v>0\). Append at the destination \(b\) a chord in the wedge \(u>0,v>0\) ending on \(v=0\). By Lemma 53, it has mass at least \(H^{-1/4-o(1)}\) within radius \(H^{1+\epsilon}\). The result is a long arch above \(v=0\), with maximum layer in \([c_0H,H^{1+2\epsilon}]\) and diameter at most \(Q=H^{1+6\epsilon}\). Its crossing of \(u=0\) is unique.

  Anchor the arch’s initial port. At each \(R\), different source placements would give different single-crossing cut levels \(u=0\). The arch version of Lemma 61 thus bounds inverse multiplicity, summed over \(R\), by \(CHQ^{3/4+\epsilon}\) outside a negligible set. Consequently the long arch mass in this height range is at least \(H^{-1/4-C\epsilon-o(1)}\).

The exceptional sets remain negligible after these conversions because all inverse specifications cost only fixed powers of \(Q\).

A long tilted bridge must contain a large block. Under \(\mathbb Q_z\), the geometric number of blocks exceeds \(H^{3/4+4\epsilon}\) with superpolynomially small probability, because \(1-I(z)=T^{-3/4+o(1)}\). Set larger lengths to zero and truncate at \(n\). Their mean under \(p_z\) is at most \(n^{7/16+o(1)}\) by (75). Exponential Markov at \(1/n\) therefore bounds the probability of length at least \(H^{4/3-\delta}\) with no block of length at least \(n\) by an exponential whose exponent is at most \[-H^{4/3-\delta}/n +CH^{3/4+4\epsilon}n^{-9/16+o(1)}.\] The negative term is \(-H^{\eta-\delta}\) and the positive term is \(H^{9\eta/16+4\epsilon+o(1)}\). Choose \(\delta+4\epsilon<7\eta/16\); the bound is then superpolynomially small. The same remains true with any fixed polynomial weight in the block count, by the geometric tail.

If the bridge alternative above holds, its tilt costs a constant factor in the relevant height bin. Marking a large block bounds its long mass above by \(B(z)^2u\) plus a negligible error. Thus it yields \(u\ge H^{-3/4-O(\epsilon)-o(1)}\).

In either arch alternative, fold the long arches. Their tilted mass is their original mass multiplied by \(\rho z^{2k}\), where \(c_1H\le k\le H^{1+2\epsilon}\). These extra factors are bounded below by a positive constant. Insert such a folded segment between two arbitrary ordinary bridges. Cutting at its endpoint renewals identifies the inserted segment, so the resulting weighted count is its folded mass times \(B(z)^2\). It is bounded by the full tilted bridge mass with insertion \(K\) and with total length at least \(H^{4/3-\delta}\).

Terms with no block of length at least \(n\) are negligible by the preceding estimate, since \(K\) is at most the cube of the block count plus one. In every remaining term, mark a large block \(\xi\). The weight sum of the two ordinary exterior bridges is \(B(z)^2\), and their conditional expected insertion is bounded by Lemma 60, uniformly in \(\xi\). Dividing by \(B(z)^2\) therefore bounds the folded mass by \[uH^{1/2+O(\epsilon)+o(1)}+\text{negligible}.\] Its lower bound is \(H^{-1/4-O(\epsilon)-o(1)}\). Thus every alternative gives \(u\ge H^{-3/4-O(\epsilon)-o(1)}\). Let \(\epsilon\) decrease to zero at fixed \(\eta\), and then take \(\eta\) arbitrarily small. Since \(u\) is bounded above by the untilted tail, the upper bound in (75) proves \[ \mathbb P_p(\mathsf L\ge n)=n^{-9/16+o(1)}. \tag{80}\] The height and diameter assertions follow from Lemma 58 and \(D\le C(\mathsf H+S)\).

Infinite concatenation. The matching length and height tails and the span upper tail now give, on one probability-one event, \[\sum_{i\le m}\mathsf L_i=m^{16/9+o(1)},\qquad \sum_{i\le m}\mathsf H_i=m^{4/3+o(1)},\qquad \sum_{i\le m}S_i\le m^{4/3+o(1)}.\] For upper bounds, moments just below \(9/16\) or \(3/4\), subadditivity of these powers, Markov’s inequality and dyadic Borel–Cantelli suffice. Independent maxima and the matching lower tails give the lower bounds. Equivalently one may apply Lemma 6 to the resulting Laplace deficits. Within the next block, every vertex lies above the last completed height and within the sum of the spans and heights. Consecutive length sums have the same exponent, so if vertex \(j\) is in block \(m+1\), then \(m=j^{9/16+o(1)}\). Hence \[|\gamma_j-\gamma_0|=j^{3/4+o(1)},\qquad \#\{j\ge1:|\gamma_j-\gamma_0|\le R\}=R^{4/3+o(1)}\] almost surely. The second assertion follows by inverting the two eventual power bounds in the first.

This infinite law is also the port-to-vertex fugacity prefix limit: Lemma [R:renew:halfplane] includes the final half-plane remainder, whose mass cancels from a prescribed-prefix probability. For this application, finiteness at each subcritical fugacity follows already from the connective-constant root bound of [4]; no stronger counting estimate or thermal spatial conclusion is required.

Conditioning on an exact strip height. Critical bridges of height \(h\) are iid prefixes conditioned on hitting \(h\), an event of probability \(b_h=h^{-1/4+o(1)}\). We first show that this conditioning leaves at least \(h^{3/4-\varepsilon}\) blocks with high probability. The arch identity makes \(b_h\) decreasing, including \(b_0=1\). Define the series with nonnegative coefficients \[C(z)=\sum_{h\ge1}(b_{h-1}-b_h)z^h,\qquad 1-C(z)=(1-z)B(z).\] Since \(B=(1-I)^{-1}\), one has \((1-I)(1-C)=1-z\). Taking negative logarithms as formal series gives \[[z^h]\{-\log(1-I(z))\}\le [z^h]\{-\log(1-z)\}=\frac1h.\] For any integer \(m\ge1\), coefficient positivity consequently yields \[[z^h]\sum_{r=1}^{m}I(z)^r \le m[z^h]\{-\log(1-I(z))\}\le\frac mh.\] Divide by \(b_h\) and take \(m=\lfloor h^{3/4-\varepsilon}\rfloor\).

Given \(\delta>0\), choose \(\varepsilon>0\) sufficiently smaller than \(\delta\). By (80), the probability that the first \(m\) iid blocks all have length below \(h^{4/3-\delta}\) is superpolynomially small; division by \(b_h\) leaves it negligible. This proves the typical length lower bound. The strip first moment gives \[\mathbb E_h\mathsf L=\frac{m_B(h)}{b_h}\le h^{4/3+o(1)},\] and Markov gives the typical upper bound. The typical lower bound also gives the matching lower mean. The bridge diameter is at least a constant times \(h\), and the cap comparison, divided by \(b_h\), makes diameter above \(h^{1+\delta}\) negligible. In particular the length and mean assertions hold at every sufficiently large integer height, not merely along a subsequence.

Summing heights with a length tilt. Under the bridge law with weight \(\rho^{\mathsf L}e^{-\mathsf L/t}\), the number of blocks is geometric with termination parameter \[q_t=\mathbb E_p(1-e^{-\mathsf L/t})=t^{-9/16+o(1)}.\] The upper bound follows by integrating the length tail, or comparing \(1-e^{-x}\) with \(\min(1,x)\); the lower bound follows from \(p(\mathsf L\ge t)\). Conditional on the count, block shapes are independent with law proportional to \(e^{-\mathsf L/t}p\). For each fixed small \(\delta>0\), the count lies between \(t^{9/16-\delta}\) and \(t^{9/16+\delta}\) with probability tending to one. Removing the empty bridge changes none of these assertions.

For any small fixed \(\eta>0\), each of the two events \[t^{1-\eta}\le\mathsf L\le t,\qquad \mathsf H\ge t^{3/4-\eta},\quad \mathsf L\le t\] has enough tilted mass that it occurs among \(t^{9/16-\delta}\) independent blocks with high probability, if \(\delta\) is sufficiently small. For the first event, subtract \(p(\mathsf L>t)\) from the matching length tail at \(t^{1-\eta}\). For the second, subtract that same upper tail from the height lower tail at \(t^{3/4-\eta}\). In each subtraction the first term has a strictly larger power, and on \(\mathsf L\le t\) the tilt costs at most a constant factor. These events give the lower length and diameter bounds.

The tilted block laws have uniformly bounded length moments at every power below \(9/16\), and height and span moments at every power below \(3/4\). Subadditivity at those powers and Markov’s inequality, applied to at most \(t^{9/16+\delta}\) samples, bound their summed length by \(t^{1+o(1)}\) and their summed height and span by \(t^{3/4+o(1)}\) in probability. These sums bound the bridge diameter. Taking the slacks arbitrarily small proves the claimed thermal bridge laws and completes the theorem. ◻

Proof. The factorization in Proposition 2 gives \(\mathbb P\{m\text{ is a renewal height}\}=B_m\). If \(I_j\) is the probability that one irreducible has height \(j\), the same factorization and (81) give \[\sum_{j\ge x}I_j\le Cx^{-\beta}.\] Indeed, \(\sum_m B_me^{-m/x}\asymp x^\beta\), its reciprocal is \(1-\sum_j I_je^{-j/x}\), and this deficit bounds \((1-e^{-1})\sum_{j\ge x}I_j\) from below. Decomposing according to the last renewal before \(h\), for \(x\ge2h\) we obtain \[ \mathbb P\{S_{\tau_h}>x\} =\sum_{i=0}^{\lfloor h\rfloor} B_i \sum_{j>x-i}I_j \le C(h/x)^\beta. \tag{85}\]

For the length lower bound choose small \(\theta,\xi>0\) and set \(a=\lfloor h^{1-\theta}\rfloor\), \(b=\lfloor h^{1-\theta/2}\rfloor\), \(k=\lfloor h^{\theta/16}\rfloor\). Run \(k\) successive trials of height \(a\), restarting at each terminal renewal. They are independent by the iid law after stopping renewal indices. If their height advances are \(\Delta_1,\ldots,\Delta_k\), then \[\mathbb P\Big\{\sum_{i=1}^k\Delta_i>b\Big\} \le Ck(ak/b)^\beta\le Ch^{-17\theta/64}.\] The renewal-trial lemma gives success probability at least \(h^{-\theta/32}\) for each trial when \(h\) is large. The probability that all fail is at most \(\exp(-c h^{\theta/32})\). Outside these two exceptional events a trial of length at least \(a^{4/3-\xi}\) is contained in \(\Gamma_b\), hence in \(\Gamma_h\). Choose \(\theta,\xi\) so that \((1-\theta)(4/3-\xi)>4/3-\varepsilon\). This proves the required polynomial lower-tail estimate.

For the upper bounds put \(s=\lfloor h^{1+\delta}\rfloor\), where \(\delta>0\) will be small, and let \(u=S_{\tau_h}\). The exceptional probability \(\mathbb P\{u>s\}\) is at most \(Ch^{-\beta\delta}\). Starting from \(u\), the expected number of continuation renewals in the next \(s\) bands, including the renewal at \(u\), is \[D_s=\sum_{j=0}^sB_j\asymp s^\beta.\] Conditionally on \(\Gamma_h\), this future renewal count has expectation \(D_s\). On \(u\le s\) every such continuation ends at a height at most \(2s\) and contains \(\Gamma_h\). Therefore \[\begin{align*} D_s\,\mathbb E[L(\Gamma_h)\mathbf1_{\{u\le s\}}] &\le\sum_{m\le2s} \sum_{\eta:\,H(\eta)=m}\rho^{L(\eta)}L(\eta)\\ &\le s^{25/12+o(1)}. \end{align*}\] The first inequality counts a subset of all renewal prefixes ending at heights at most \(2s\). Their distribution is the unnormalized bridge measure, by the pathwise factorization; it does not require independence between the length and the height of one irreducible. Markov’s inequality now gives \[\mathbb P\{L(\Gamma_h)>h^{4/3+\varepsilon},\ u\le s\} \le h^{4\delta/3-\varepsilon+o(1)}.\]

The same device controls horizontal excursions. If \(\Gamma_h\) reaches horizontal distance \(M=h^{1+\varepsilon}\), all its continuation prefixes do so. For \(\delta<\varepsilon\), summing (83) over their possible heights gives \[D_s\,\mathbb P\{\operatorname{lat}(\Gamma_h)>M,\ u\le s\} \le C\sum_{m\le2s}mM^{-5/4}\le Cs^2M^{-5/4}.\] After division by \(D_s\), this is at most \(Ch^{-5(\varepsilon-\delta)/4}\). On \(u\le s\) the vertical span is at most a fixed multiple of \(s\). Taking \(\delta=\varepsilon/4\), with a smaller slack first if necessary, proves the assertion.

Apply Borel–Cantelli on dyadic values of \(h\), simultaneously for a countable sequence of positive slacks tending to zero. For each fixed small \(\eta>0\), eventually \[h^{4/3-\eta}\le L(\Gamma_h)\le h^{4/3+\eta},\qquad \operatorname{diam}(\Gamma_h)\le Ch^{1+\eta}\] at every dyadic scale. A dyadic \(h\) a fixed factor below \(n^{1/(4/3+\eta)}\) has \(L(\Gamma_h)<n\); every subsequent vertex lies above its terminal renewal line, whose height exceeds \(h\). A dyadic \(h\) a fixed factor above \(n^{1/(4/3-\eta)}\) has \(L(\Gamma_h)>n\). Hence, eventually, \[c_\eta n^{1/(4/3+\eta)}\le|\omega_n-o| \le C_\eta n^{(1+\eta)/(4/3-\eta)}.\] Let \(\eta\downarrow0\). The resulting pointwise bounds also show that all sufficiently early indices, up to \(R^{4/3-o(1)}\), are in the ball, and all sufficiently late indices, beyond \(R^{4/3+o(1)}\), are outside. This proves the visit-count assertion. ◻

Proof. Write \(M=M_{\mathcal A}\) for either input class and expand \(M^k\) into ordered \(k\)-tuples of representations of the same output. We first locate their seams. Let \(z_0,z_f\) be the vertical coordinates of the output endpoints. Undoing the lower reflection sends the root to height \(2a-z_0\), while removing the inserted height sends the other endpoint to \(z_f-(b-a)\), up to bounded local changes. The original endpoint difference is therefore \(z_0+z_f-a-b+O(1)\). Both input classes satisfy \[ |a+b-z_0-z_f|\le C_2h,\qquad H\le b-a\le2H. \tag{88}\] All heights here are in row units, with fixed conversion factors absorbed in \(C_2\).

Put \(S=z_0+z_f\) and \(E=C_2h\). The possible seams lie in the intervals \[a\in\left[\frac{S-E-2H}{2},\frac{S+E-H}{2}\right],\qquad b\in\left[\frac{S-E+H}{2},\frac{S+E+2H}{2}\right].\] Every upper seam thus exceeds every lower seam by at least \(H-E>0\) for large \(H\). Sort the lower seams increasingly and the upper seams decreasingly: \[a_1\le\cdots\le a_k<b_k\le\cdots\le b_1.\] Record ties and the original representation pairing by a permutation \(\pi\): the representation with lower seam \(a_i\) has upper seam \(b_{\pi(i)}\). These records cost a factor depending only on \(k\).

Every positive gap between consecutive seams in either group is a strict bridge. The two exterior portions and the portion between \(a_k\) and \(b_k\) are the only three portions left over. In the diameter case the whole output lies in a box of side \(O(H)\); in the endpoint case it lies in a box of side \(O(H+H^K)\). Fixing the output endpoint and the four extreme seam ports costs a polynomial in \(H\) whose exponent is independent of \(k\). The full-box susceptibility bound then sums the three leftover portions at cost \(H^C\) or \(H^{C_K}\). Internal joining ports introduce no additional position sums: once the relative gap shapes are specified, each has a unique translation to its joining port.

For the remaining gaps, use heights \[s_1=t_1=0,\qquad s_i=a_i-a_{i-1},\quad t_i=b_{i-1}-b_i\quad(2\le i\le k).\] The zero slots account for the two extreme seams. Repeated seams give further zero gaps, with bounded weight. The lower gap of height \(s_i\) ends at \(a_i\), and the upper gap of height \(t_j\) starts at \(b_j\). Consequently the original representation \(i\) pairs the slots \((s_i,t_{\pi(i)})\). Every positive gap appears in exactly one such pair. Figure 11 distinguishes this pairing from the rank pairing used later to sum endpoint constraints.

Proof. Every \(m\)-edge walk lies in a box of side \(O(m)\), so full-box susceptibility gives \(c_m\rho^m\le C(1+m)^C\). In particular \(Z_T\) is finite, \(Z_T\ge1\), and for every \(\xi>0\), \[ \mathbb P_T^{\mathrm{th}}\{\ell(\gamma)>T^{1+\xi}\} \le\sum_{m>T^{1+\xi}}C(1+m)^Ce^{-m/T}=o(1). \tag{90}\] We first prove the endpoint lower bound on lengths at most \(T^2\). This restriction loses \(o(1)\) probability and permits use of the second multiplicity estimate.

Fix a small \(\zeta>0\) and set \(h=T^{3/4-\zeta}\), \(H=T^{3/4-\zeta/2}\), rounding heights to integers. The critical mass of bridges of height in \([H,2H]\) is comparable to \(H^\beta\). By (83), choosing the fixed confinement constant \(C_*\) sufficiently large retains a fixed fraction with diameter at most \(C_*H\). The mass of bridges in this height window with \(L(\eta)>T\) is at most \(T^{-1}H^{25/12+o(1)}\), by (82). Its ratio to \(H^\beta\) is \(H^{4/3+o(1)}/T=o(1)\). On the remainder, \(x_T^{L(\eta)}\ge e^{-1}\rho^{L(\eta)}\). Thus every input admits insertions of total \(x_T\)-mass at least \(cH^\beta\).

Let \(\mathcal A\) consist of inputs of length at most \(T^2\) whose absolute vertical endpoint difference is at most \(h\), and let \(V\) be their \(x_T\)-mass. Since \(T^2\le H^K\) for a fixed sufficiently large \(K\), Lemma 65 applies to \(M_{\mathcal A}\). Put \[A_h=h^\beta(H/h)^{2\beta-1}\le H^\beta, \qquad Q=A_hH^\delta,\] where \(\delta>0\) will be chosen small after \(\zeta\). For every fixed integer \(k\ge1\), the lemma gives \[\begin{align*} \sum_{M_{\mathcal A}(\omega)>Q} x_T^{\ell(\omega)}M_{\mathcal A}(\omega) &\le Q^{1-k}\sum_\omega \rho^{\ell(\omega)}M_{\mathcal A}(\omega)^k\\ &\le C_{k,K}A_hH^{C_K+o_k(1)-\delta(k-1)}. \end{align*}\] Choose \(k\) so that \(\delta(k-1)>C_K+\beta+1\), and then let \(T\to\infty\). The last expression is \(o(1)\). This order is possible because \(C_K\) is independent of \(k\).

Sum all insertions of inputs in \(\mathcal A\). For outputs with \(M_{\mathcal A}\le Q\), there are at most \(Q\) representations per output. Their unrestricted partition sum, rooted at \(o^*\), equals \(Z_T\) by the root symmetry established above. The preceding estimate handles the remaining outputs. Recoverability and the bounded weight factors therefore give \[ cH^\beta V\le C A_hH^\delta Z_T+o(1). \tag{91}\] Divide by \(H^\beta Z_T\) and use \[\frac{A_h}{H^\beta}=(h/H)^{1-\beta}.\] For sufficiently small \(\delta\), the bound \(V/Z_T\le C(h/H)^{1-\beta}H^\delta+o(1)\) tends to zero. Adding the discarded length tail proves that the absolute vertical endpoint difference exceeds \(T^{3/4-\zeta}\) with probability tending to one. The Euclidean endpoint distance has the same lower exponent after absorbing the fixed row-spacing factor into the slack.

It remains to obtain the diameter upper bound and the length lower bound. Conditioned on its length, the thermal law is uniform. Fix \(\varepsilon>0\) and choose \(\eta,\xi>0\) so that \[(1+\xi)(3/4+\eta)<3/4+\varepsilon.\] For \(m<T^{1/2}\) the diameter is deterministically at most \(CT^{1/2}\). For \(T^{1/2}\le m\le T^{1+\xi}\), the probability of violating \(D\le T^{3/4+\varepsilon}\) is, by (89), at most \(C_{\eta,A}T^{-A/2}\) under each conditional length law. The remaining lengths have probability \(o(1)\) by (90). This proves the upper diameter exponent, and hence the upper endpoint exponent.

For the lower length bound it suffices to take \(0<\xi<1/2\). Choose \(\eta,\zeta>0\) such that \[(1-\xi)(3/4+\eta)<3/4-\zeta, \qquad \zeta<1/4.\] The typical endpoint lower bound excludes lengths \(m<T^{1/2}\) deterministically. For \(T^{1/2}\le m<T^{1-\xi}\), it forces \(D>m^{3/4+\eta}\), up to a fixed factor absorbed by the strict exponent inequality. Equation (89) bounds the conditional probability of this event uniformly by \(C_{\eta,A}T^{-A/2}=o(1)\). Thus \(\ell\ge T^{1-\xi}\) with probability tending to one. Together with the length upper tail and \(A\le D\), this proves all three claims. ◻