Proof. At a lower endpoint whose triangle center is nearest its lower side, append the vertical half-edge to that side’s port. At the other triangle type, first append its fixed downward edge and then the vertical half-edge. The reflected operation applies at the upper endpoint. Every newly added vertex lies strictly outside the old height range, so it was unvisited. The two additions are disjoint for positive macroscopic span. Bounded spans have only finitely many endpoint types and use the same construction; a zero-span piece is trivial. At most three extra vertex factors occur relative to the original edge count. Removing the recorded one- or two-edge endpoint patterns recovers the original path, proving all bounds. Rotations and reflections give the other lattice normal directions. ◻

Proof. Take real parts in the boundary winding identity before and after making the cut. Contributions of paths contained in the near subdomain cancel. For every other boundary exit the cosine is between the same positive constant and one, since both domains are convex. Thus the lost old exit mass and the new-cut exit mass are comparable. For a parallel cut, its exits are a subset of the bridges across the intervening strip. This also explains the lower comparison used later: a family of paths in the old domain which necessarily passes the cut gives a lower bound, up to a constant, for the mass of exits on that cut. No bound on the continuation of an individual prefix is required.

For localization truncate the strip to a parallelogram whose width is a small fixed multiple of \(M\), and average the winding comparison over bottom sources in its central half. There are \(\asymp M\) such sources. Reverse paths from a side exit to one of these sources. For each side port they reach a cut parallel to that side at distance at least \(cM\), so their total mass is at most \(C B_{\lfloor cM\rfloor}\). There are \(O(h)\) side ports. The average loss is therefore at most \(C h M^{-1}B_{\lfloor cM\rfloor}\le C hM^{-5/4}\). If \(M/h\) is sufficiently large, the parallelogram lies within lateral distance \(M\) of every central source. For each such source, exceeding lateral distance \(M\) requires leaving this parallelogram. The former mass is the same for all these sources by translation invariance, so it is bounded by the average exit loss. Infinite strips are obtained by increasing finite truncations. ◻

Proof. Put \(\lambda=\pi/8\). Add the two adjacent rhombi shown in Figure 2 above ports \(a=x\) and \(b=x+1\). Each rhombus consists of two ordinary triangles. Their four new dual vertices form the chain \(u_0,v_0,u_1,v_1\), with \(a\) incident to \(u_0\) and \(b\) incident to \(u_1\). Denote the left and right step exits by \(L,R\). The horizontal top exits have winding zero, while \(L\) and \(R\) have windings \(+\pi/3\) and \(-2\pi/3\), respectively: closing to the source along the exterior boundary determines these turns.

A path to \(L\) or \(R\) crosses the two-port interface an odd number of times, hence exactly once. The four possible continuations have weights \[\begin{array}{c|cc} & L & R\\ \hline a & \rho & \rho^4\\ b & \rho^3 & \rho^2 \end{array}\] according to the number of new vertices they visit. Thus their combined contribution to the imaginary identity is \[\sin\lambda\bigl(\rho b_h(x)+\rho^3b_h(x+1)\bigr) -\sin(2\lambda)\bigl(\rho^4b_h(x)+\rho^2b_h(x+1)\bigr).\] Using \(\sin\lambda\,\rho=\sin(2\lambda)\rho^2\), this equals \[\sin\lambda\,\rho(1-\rho^2) \bigl(b_h(x)-b_h(x+1)\bigr).\] There can be no other occupied segment in the raised part: another visit would require at least three interface crossings, while its horizontal top ports are vacant for a path ending at a step exit.

Subtract the old imaginary winding identity. The absolute value of the remaining bottom contribution is bounded by a constant times the mass of new arches using the bump. For distinct bump positions these arch families are disjoint: a member enters and leaves the last band through that specific adjacent pair, and uses no other port there. Their sum is bounded by \(A_{h+1}-A_h\). The real strip identity \(\cos(3\lambda)A_h+B_h=1\), with \(\lambda=\pi/8\), gives the assertion. To justify the infinite-strip identities, first fix a finite set of bump positions and perform each calculation in a wide finite parallelogram. Lemma 12 bounds every side-error mass by \(C_h M^{-5/4}\), which tends to zero as the lateral cutoff \(M\) increases with \(h\) fixed. After this limit, increase the finite set of bump positions. The nonnegative, disjoint new-arch masses sum monotonically to at most \(A_{h+1}-A_h\), whose finiteness is part of Theorem 6. ◻

Proof. Expand \(N^j\) and order the selected levels, allowing repetitions. Each internal gap \(g\) requires one bridge in the first case and two in the second. Removing mutual avoidance can only increase the resulting upper bound. The gap factors are therefore at most \(C(1+g)^{-1/4}\) and \(C(1+g)^{-1/2}\), respectively. Summing the first level in \(R\) ways and each following gap gives the displayed bounds, with the ordering and repetitions absorbed in \(C_j\). For a threshold \(Q\), use \[\sum\rho^{|\gamma|}N^a\mathbf1_{\{N>Q\}} \le Q^{-(j-a)}\sum\rho^{|\gamma|}N^j\] for a fixed \(a\) and then choose \(j\) sufficiently large. All paths in a box of diameter \(O(R_0)\) have length \(O(R_0^2)\), so length powers and raw geometric multiplicities can be included in the polynomial factor. ◻

Proof. We first allow a sufficiently large fixed tube radius. Let \(H\) be even, choose \(|k-H/4|\le H/40\), and write \(l=H/2-k\). Lemma 13, summed over these heights, implies that a positive fraction of the choices have both kernel variations at most \(C H^{-5/4}\). Their suprema have the same bound, since each summable kernel tends to zero at infinity. Reflection gives \(b_k(x)=b_k(-x)\), and likewise for \(l\).

For either good kernel \(f\), localization leaves mass \(\ge cH^{-1/4}\) on \(O(H)\) endpoint positions. Thus \(\sum f(x)^2\ge cH^{-3/2}\). Moreover \[\left|\sum_x f(x)f(x+t)-\sum_x f(x)^2\right| \le \|f\|_\infty\,\|f(\cdot+t)-f\|_1 \le C|t|H^{-5/2}.\] For \(|t|\le v_0H\), with \(v_0>0\) sufficiently small, the correlation is at least \(c'H^{-3/2}\). Symmetry turns this into the same lower bound for the self-convolution. If both constituent bridges are confined to a lateral box of radius \(DH\), the convolution loss is at most twice their lost mass times \(\|f\|_\infty\), hence at most \(CD^{-5/4}H^{-3/2}\). Fix \(D\) large enough.

Concatenate bridges of successive heights \(k,k,l,l\). Convolving the two pair bounds over \(\asymp H\) intermediate displacements gives mass \(\ge cH^{-2}\) to each prescribed \(|x|\le v_0H/2\), for every good \(k\), inside a fixed large tube. Let \(F_*\) count admissible indices \(k\) for which all three levels \(k,2k,H/2+k\) are crossed once. Then \[\sum_{\gamma\text{ bridge of height }H} \rho^{|\gamma|}F_*(\gamma)^j \le C_j H^{(j-1)/4}.\] To verify this stronger version of the seam bound, order the chosen indices. Their levels lie in three disjoint macroscopic bands. A gap between successive indices creates three bridge gaps, with combined factor \((1+g)^{-3/4}\). The four outer and interband gaps all have height \(\asymp H\), contributing \(H^{-1}\). Thus the level sum is \(H^{-1}H H^{(j-1)/4}\), as asserted, also when some indices coincide. The sum over good \(k\) of representations of a fixed endpoint has mass \(\ge cH^{-1}\). By the high moment bound its part with \(F_*>H^{1/4+\varepsilon}\) is negligible compared with \(H^{-1}\). Dividing the remaining representation mass by this threshold gives \(H^{-5/4-\varepsilon}\). An odd height is handled by appending a one-band bridge of fixed allowable displacement and positive weight; retain slack in the permitted slope.

We finish with the quantitative iteration. Suppose the endpoint estimate, with arbitrary exponent slack, holds at radius \(D_jh\) and all slopes at most \(v_j\). Set \(D_{j+1}=3D_j/4\). Choose \[0<\eta_j\le\min\left\{\frac1{16},\frac{D_j}{8(1+D_j)}, \frac{v_0'2^{-j}}{64}\right\}, \qquad v_{j+1}=v_j-8\eta_j,\] where \(v_0'>0\) is the slope range obtained in the preceding construction. Start with \(v_0=v_0'\) in this iteration. Since \(\sum_{j\ge0}8\eta_j\le v_0'/4\), all \(v_j\) remain at least \(3v_0'/4\).

Take \(|x|\le v_{j+1}h\). Choose a joining level \(r\) with \(|r-h/2|\le\eta_jh\), and an attainable port \(y\in r/2+\mathbb Z\) with \(|y-xr/h|\le\eta_jh\). There are at least \(c_jh^2\) such pairs for all large \(h\). The two slopes differ from \(x/h\) by at most \(3\eta_j\), so both lie in the previously allowed range. The difference between either piece’s straight join and the whole join is at most \(\eta_jh\). Its tube therefore lies within distance \[D_j(1/2+\eta_j)h+\eta_jh\le(3D_j/4)h\] of the whole join. The second displacement is compatible because \(x-y\in(h-r)/2+\mathbb Z\).

For any \(\tau>0\), the product of the two endpoint mass bounds is at least \(c_{j,\tau}h^{-5/2-2\tau}\). After summing over the joining choices, representation mass is at least \(c_{j,\tau}h^{-1/2-2\tau}\). A representation is recovered from its joining seam alone: that seam determines the joining port and both pieces. Its multiplicity is thus bounded by the count \(N\) of once-crossed joining levels. Lemma 14, with \(a=1\), removes the part where \(N>h^{3/4+\tau}\) at a cost smaller than any prescribed negative power of \(h\). Dividing the retained representation mass by this threshold gives \[c'_{j,\tau}h^{-5/4-3\tau}.\] As \(\tau\) is arbitrary, this proves the endpoint estimate at radius \(D_{j+1}\) with arbitrary exponent slack. Each step uses finitely many previous constants; they may depend on \(j\).

For fixed \(\delta>0\), a finite number of steps gives \(D_j\le\delta\). The uniform lower bound on all \(v_j\) permits a single choice, for example \(v=v_0'/2\), for every \(\delta\). Choose the exponent slacks successively small enough for the desired final \(\varepsilon\). This proves the assertion. ◻

Proof. First construct bridges in a large fixed box with terminal displacement \(|x|\le v'h\), where \(v'>0\) can be any fixed small number. Use the four-piece construction above without restricting to good heights. The localized kernels can be chosen symmetric and retain mass \(\ge ch^{-1/4}\). Partition their \(O(h)\) endpoint positions into intervals of width a small multiple of \(v'h\). Pair an interval with its reflection. Cauchy–Schwarz on the finitely many interval masses shows that an equal-height pair with small total displacement has mass \(\ge c(v')h^{-1/2}\). The two pairs therefore give \(ch^{-1}\) per \(k\), and their representation mass summed over \(\asymp h\) indices is bounded below by a positive constant. The previous estimate \(\sum\rho^{|\gamma|}F_*^2\le Ch^{1/4}\) and weighted Cauchy–Schwarz give mass \(\ge ch^{-1/4}\). Odd heights again cost only a fixed factor.

For a tube of radius \(\delta h\), divide the height into a sufficiently large fixed number \(m\) of pieces. Vary the \(m-1\) joining levels in disjoint small macroscopic intervals near equal division. Apply the large-box construction to each piece, taking its permitted terminal slope sufficiently small that cumulative endpoint shifts remain below \(\delta h/2\), and then \(m\) sufficiently large that each piece’s box fits the remaining clearance. The representation mass is at least \[c_m h^{-1/4}h^{3(m-1)/4}.\] The representation count is bounded by the product of the seam counts in the \(m-1\) intervals. Its second moment is at most \[C_m h^{-1/4}h^{3(m-1)/2}.\] Indeed in each interval the two selected levels introduce one bridge gap, whose ordered sum is \(O(h^{7/4})\), while the \(m\) other gaps are macroscopic and contribute \(h^{-m/4}\). Cauchy–Schwarz proves the fixed-proportion tube assertion.

For the widening tube use seam intervals \([r_i,5r_i/4]\), where \(r_i=4^is\), stopping before \(h/2\). The proportional width for each piece is chosen small enough that the geometric sum of all previous endpoint shifts and current fluctuations is at most \(\delta(s+y)\). If there are no intervals, one fixed-proportion tube suffices. With \(k\) intervals the representation mass and second moment are bounded respectively below and above by \[c^{k+1}h^{-1/4}\prod_i r_i^{3/4},\qquad C^{k+1}h^{-1/4}\prod_i r_i^{3/2}.\] The moment estimate follows by selecting two seams in each interval: its gap factor sums to \(O(r_i^{7/4})\), and the outer gaps contribute \(h^{-1/4}\prod_i r_i^{-1/4}\). Constants are uniform in the number of intervals because the scales are separated by four. Thus Cauchy–Schwarz leaves \((c^2/C)^{k+1}h^{-1/4}\). Since \(k=O(1+\log(h/s))\), this is the asserted bound for some fixed finite \(P\). ◻

Proof. There are constants \(0<a<b<\infty\) such that height \(h\) arches, localized to a fixed box of size \(h\), have rightward displacement in \([ah,bh]\) with mass at least \(ch^{-1/4}\). To see this, start from \(\sum_{s>0}sK_h(s)\asymp h^{3/4}\). Localization and dyadic summation bound the first moment at displacements above \(bh\) by \(Ch(bh)^{-1/4}\). For \(1\le q\le h\), arches of displacement at least \(q\) have total mass \(O(q^{-1/4})\): those leaving the height-\(\lfloor q\rfloor\) strip cost \(O(B_{\lfloor q\rfloor})\) by the strip winding identity, and those remaining have this bound by the first displacement moment in that smaller strip. Integrating this tail up to \(ah\) bounds its first moment by \(O((ah)^{3/4})\). Choose \(a\) small and \(b\) large. Finally, removal of paths leaving a sufficiently wide box changes the first moment on \([ah,bh]\) by an arbitrarily small multiple of \(h^{3/4}\), by Lemma 12. Division by \(bh\) proves the claim.

From the prescribed ports run two bridges into the chosen half-plane to a line at height \(k\in[Dr,2Dr]\), with \(D\) a large fixed constant. Join their terminal ports by one of the just-constructed arches beyond that line, of height \(r\) and displacement in \([ar,br]\). There are \(\asymp r\) translations of this arch in a small macroscopic interval near the starting ports. For each translation and arch endpoint separation, use Lemma 15 for both bridges. Choose \(D\) large enough for their slopes and their tube widths small enough that they are disjoint. This is possible because the two straight segments remain in the same order and their separation is at least \(\min(1,a)r\). Translation along the terminal line preserves the port parities. For each \(k\) the product mass is therefore at least \(r\,r^{-5/2-o(1)}r^{-1/4}=r^{-7/4-o(1)}\).

Summing \(k\) gives representation mass \(r^{-3/4-o(1)}\). A representation is recovered at its two-crossing seam. The end pieces are two bridges from the prescribed ports, the internal gaps require two bridges each, and the last piece is an arch; all their endpoint sums are bounded. Lemma 14 discards multiplicity above \(r^{1/2+\varepsilon}\), proving the assertion. The construction has diameter \(O(r)\). The finitely many bounded nonzero distances have positive weight by fixed ordinary arches and are absorbed in the constants. ◻

Proof. Figure 4 shows the two closures and their recovery lines. An arch needs one movable line; a bridge needs two lines that move together at fixed separation. This coupling improves the bridge’s inverse multiplicity and compensates for its more costly closure.

Close an arch by an exterior arch supplied by Lemma 17, at mass at least \(h^{-5/4-o(1)}\). The result is a simple polygon of diameter \(O(h)\), and its original boundary is a two-crossing seam. Modulo translation, fixing this line recovers the two arcs and the input start up to bounded choices. Its placement count has moments at most \(C_jh^{C+j/2}\), with \(C\) independent of \(j\): choose an initial line and crossing port in polynomially many ways, order the other levels, use two bridges per internal gap and bounded terminal arch sums. Thus multiplicities above \(h^{1/2+\varepsilon}\) have negligible mass, even with squared length and raw multiplicity included. The polygon second moment gives \(h^{5/4+1/2+2/3+o(1)}=h^{29/12+o(1)}\).

For a bridge, attach exterior arches at its bottom and top, ending far to one common side of the original path. Each new endpoint ranges over \(ch\) positions at distance \(\asymp h\), with their intervals chosen so that their mutual displacement satisfies the slope condition of Lemma 15. Join them by a thin bridge inside the original strip, separated from the original path. The three added kernels and two endpoint sums have total mass at least \(h^2h^{-15/4-o(1)}=h^{-7/4-o(1)}\).

Now there are two clean two-crossing lines at a fixed separation \(h\). Their joint placement count on the polygon has moments at most \(C_jh^C(1+\log h)^j\). Indeed restrict the lower line to an interval of width at most \(h/3\); only boundedly many intervals are needed after anchoring a polygon of diameter \(O(h)\). Each successive gap \(g\) then forces two bridges in the lower band and two in the upper, giving \((1+g)^{-1}\); its sum is \(O(1+\log h)\). The excursions are nested and all remaining pieces have bounded strip sums, with only polynomial placement costs. For fixed lines the two bridges and exterior arches are recovered up to bounded orientation choices. High moments therefore discard multiplicity above \(h^\varepsilon\). The polygon estimate now gives \(h^{7/4+2/3+o(1)}=h^{29/12+o(1)}\). In both cases the input length is at most the polygon length. Exceptional representation mass is harmless because every path stays in an \(O(h)\) box and hence has length \(O(h^2)\). ◻

Proof. Group the excluded chords by dyadic endpoint scale \(s\), taking \(s\asymp1\) for bounded separations. For a common side, fix a source and close outside \(D\) with an arch of mass \(s^{-5/4-o(1)}\). The two-crossing seam argument in Proposition 18, now using diameter \(O(H)\), gives the squared-length bound, summed over sources, \[\sum_{\substack{\gamma\text{ with same-side endpoints}\\ \text{at scale }s}} \rho^{|\gamma|}|\gamma|^2 \le H^{1+2/3+1/2+o(1)}s^{5/4}.\] The factor \(H\) counts sources; the factor \(H^{1/2+o(1)}\) bounds seam multiplicity. All small exponent losses can be assigned to \(H\). For distinct sides we claim \[ \sum_{\substack{\gamma\text{ with distinct-side endpoints}\\ \text{at corner scale }s}} \rho^{|\gamma|}|\gamma|^2 \le H^{2/3+o(1)}s^{11/4}. \tag{10}\] Here the sum includes every source and endpoint pair at that scale.

By symmetry put the sides at \[A:\ y=0,\qquad B:\ x=y/\sqrt3,\] with \(D\) above \(A\) and to the right of \(B\). Write the endpoints as \(a\in A,b\in B\), choosing \(b\) farther from the shared corner. Then \(y_b\asymp s\). Put \(w=y/2-d_0x\); the half-plane exterior to \(B\) is \(w>0\). We first construct a closing arc outside \(D\) with mass \(s^{-3/2-o(1)}\) and diameter \(O(s)\). Join \(a\), below \(A\), to a port \((X,0)\) with \(X\in[-8s,-7s]\). The arch kernel costs \(s^{-5/4-o(1)}\). We next construct a join from \((X,0)\) to \(b\), within \(y>0,w>0\), at cost \(s^{-5/4-o(1)}\). Summing over the \(\asymp s\) choices of \(X\) then gives the required closing mass.

For this two-face join choose small fixed \(\alpha>0\). From \(b\) first reach a horizontal line \(y=t\), where \(t-y_b\in[\alpha s,2\alpha s]\). The comparison domain lies between \(B\) and a parallel line at \(w\)-depth approximately \(10\alpha s\), with the additional inequalities \[y\ge0,\qquad x\le x_b+(\alpha/4)s-(y-y_b)/\sqrt3.\] A thin bridge tube in the inward normal direction \((-d_0,1/2)\) from \(B\) to the far parallel line fits these inequalities and passes beyond \(y=t\). Its mass is \(\ge cs^{-1/4}\) by Lemma 16. Convex cut comparison gives the same lower bound for exits on \(y=t\). Every such exit \(z\) satisfies \(w(z)\ge c\alpha s\), by the oblique upper bound on \(x\). The lower bound on \(x\), from \(y\ge0,w=O(\alpha s)\), keeps all these paths well to the right of \(X\).

Continue from \(z\) vertically to \(y=L\), where \(L\in[D_2s,2D_2s]\) and \(D_2\) is a large fixed constant. A tube of width a sufficiently small multiple of \(s\) stays clear of \(B\) and of the future arm from \(X\); its mass is \(\ge c_{D_2}s^{-1/4}\). For each \(t\) the product mass is therefore \(\ge c_{D_2}s^{-1/2}\). Sum over \(\asymp s\) choices of \(t\). Each is recovered as a once-crossed seam. The initial segment has bounded convex-boundary mass in the two-face wedge with distant walls, the final segment has bounded strip mass, and internal gaps are bridges. Lemma 14 discards counts above \(s^{3/4+\varepsilon}\). Thus the distinct \(b\)-arms to \(L\) have mass at least \(s^{-1/4-o(1)}\).

Independently take a vertical arm from \((X,0)\) to \(L\) in a disjoint small-width tube, with mass \(\ge cs^{-1/4}\). Connect the two new ports by an arch beyond \(L\) of mass \(s^{-5/4-o(1)}\). Their lateral separation is \(O(s)\) independently of \(D_2\); hence the arch diameter constant is independent of \(D_2\), and increasing \(D_2\) makes this arch lie wholly in \(w>0\). Vary \(L\) over its interval. These are exactly two-crossing seams; the two initial arms have bounded convex boundary sums, the intermediate gaps give pairs of bridges, and the terminal arch has bounded mass. Removing multiplicity above \(s^{1/2+\varepsilon}\) gives \[s\,s^{-1/4-o(1)}s^{-1/4}s^{-5/4-o(1)} /s^{1/2+\varepsilon}=s^{-5/4-o(1)}\] for the two-face join. All walls and cuts are lattice lines rounded with proportional slack; distant walls can be added to make every comparison domain finite. For bounded scales, first move \(b\) a fixed number of bands into \(w>0\) by elementary bridges with displacement \((-1/2,d_0)\), which remain in \(y>0\). Perform the construction from that translated parallel boundary. Keep the auxiliary interval on \(y=0\), shifting it tangentially relative to the new boundary’s intersection with \(y=0\). This costs a positive constant and preserves the required original-side crossings.

We now count inverse images of the resulting polygon. It crosses \(A\) exactly twice, and its arc above \(A\) crosses \(B\) exactly once; other crossings of \(B\) do not matter. Given both lines, these properties recover the original chord and closing arc up to bounded choices. Each line position is within \(O(s)\) levels of the corresponding directional extremum of the polygon. For \(A\), the two-crossing moment bound gives at most \(s^{1/2}H^\varepsilon\) choices, outside negligible weighted exceptions. For each \(A\), apply the one-crossing moment bound to its positive-side arc to obtain at most \(s^{3/4}H^\varepsilon\) choices of \(B\). To justify the end factors in this second application, retain the \(A\) half-plane and the relevant extreme \(B\) half-plane for each end piece, with distant walls. Convex winding bounds its boundary sum; the opposite \(A\)-arc has bounded arch mass. Initial roots, outermost levels, and the union over \(A\) cost only a fixed polynomial in \(H\), independent of the moment order. Thus the total inverse multiplicity is at most \(s^{5/4}H^{2\varepsilon}\). Two nonparallel lines determine the placement of the fixed triangle. Squared lengths and all discarded raw multiplicities are polynomial in \(H\), so their exceptional contribution is negligible. Dividing the polygon second moment by the added mass \(s^{-3/2-o(1)}\) proves (10).

Finally apply Cauchy–Schwarz with the central-visit mass \(CH^{3/4}\). The same-side contribution at scale \(s\) is at most \(H^{35/24+o(1)}s^{5/8}\); the distinct-side contribution is at most \(H^{17/24+o(1)}s^{11/8}\). At \(s\le CH^{1-\eta}\) these are respectively at most \(H^{25/12-5\eta/8+o(1)}\) and \(H^{25/12-11\eta/8+o(1)}\). Summing the \(O(\log H)\) scales is power-smaller than the original first-length lower bound. ◻

Proof. For distinct sides use the coordinates \(A,B,a,b,w\) from the preceding proof and a horizontal outer strip. Connect \(a\) to the lower boundary by a localized bridge below \(A\). For \(b\), take bridges from \(B\) in direction \((-d_0,1/2)\) to a sufficiently distant parallel line, so that their straight tubes pass beyond the upper target boundary. Because \(y_b\asymp s\), the widening tubes of Lemma 16 with initial clearance \(s\) fit inside \(y>0,w>0\), on reducing their fixed proportional width. Include these half-plane constraints and distant bounding walls in a convex comparison domain. Cut at the upper target line. Lemma 12 gives mass \(\ge cH^{-1/4}(s/H)^P\) for the \(b\)-arm; the \(a\)-arm has mass \(\ge cH^{-1/4}\). The two arms lie on opposite sides of \(A\), and avoid \(D\). On the resulting bridge, \(A\) is crossed once and \(B\) is crossed once on the tail following \(A\).

For a common horizontal side \(A\), write \(x_a<x_b\) with separation \(\asymp s\). Orient the outer strip in crossing direction \(p=(d_0,-1/2)\). Separate the arms by a lattice wall parallel to \[x=(x_a+x_b)/2+y/\sqrt3.\] The separator is a \(p\)-level, since \(p\cdot(x,y)=d_0x-y/2\) is constant on it. Both arms lie below \(A\), the arm from \(b\) to the upper \(p\)-boundary on the right of the wall. Construct this arm by taking downward bridges in direction \(n=(0,-1)\) in a longer convex domain and cutting at the target \(p\)-line. Since \(n\cdot p=1/2\), they reach that target. Their widening tubes, with initial scale \(s\), remain right of the separator and inside the opposite outer strip bound. They give mass \(\ge cH^{-1/4}(s/H)^P\).

The arm from \(a\) must turn toward the lower outer boundary. Put \(q=(-d_0,-1/2)\). First reach a \(q\)-normal line at depth in \([\alpha s,2\alpha s]\) from \(a\), with small fixed \(\alpha>0\). Writing \(t=-y\), use the convex domain with \[0\le t\le10\alpha s, \qquad x\ge x_a-(\alpha/4)s-t/\sqrt3,\] also restricted to the left of the separator and to the outer strip where needed. A thin downward bridge tube fits and passes beyond the chosen \(q\)-line, so cutting gives mass \(\ge cs^{-1/4}\). Its exits \(z\) satisfy \(t(z)\ge c\alpha s\) and \(|x(z)-x_a|=O(\alpha s)\), using \[q\cdot(z-a)=-d_0(x(z)-x_a)+t(z)/2.\] From \(z\), continue in positive \(q\) direction in the half-plane beyond this switching line. A widening tube of initial scale \(s\) remains below \(A\), left of the separator, and clear of the opposite outer boundary: its depth below \(A\) and clearance from the separator both grow linearly along \(q\). Since \(q\cdot(-p)=1/2\), long enough bridges in this direction cross the lower target line. Convex truncation there gives mass \(\ge cH^{-1/4}(s/H)^P\).

Sum the switching line over \(\asymp s\) choices. It is recovered as a once-crossed seam; the extreme end pieces have bounded convex-boundary mass after retaining the original and target boundaries and the extreme seam half-planes. The seam moment bound, allowing a polynomial factor in \(H\), removes counts above \(s^{3/4}H^\varepsilon\). The line count \(s\) cancels the short-arm factor \(s^{-1/4}\) and this seam factor. The resulting left arm has mass \(H^{-1/4-o(1)}(s/H)^P\), proving (11) after combining with the right arm and, if necessary, increasing \(P\). The entire output crosses \(A\) exactly twice, with its endpoints on its exterior side.

The widening-tube construction above requires \(s\) to exceed a fixed threshold. To include bounded \(s\), choose a fixed integer \(K\) above that threshold and first add the following elementary stems. On a common side, take \(K\) downward one-band bridges from \(a\), each with displacement \((-1/2,-d_0)\), and from \(b\), each with displacement \((1/2,-d_0)\). The stems are disjoint, remain respectively in \(x\le x_a\) and \(x\ge x_b\), and end on \(A'=\{y=-Kd_0\}\) with separation \((x_b-x_a)+K\asymp s+K\). Apply the preceding common-side construction below \(A'\). The new arms avoid the stems; the completed output still crosses the original \(A\) exactly twice. For distinct sides, extend \(b\) by \(K\) elementary \(B\)-normal bridges of displacement \((-1/2,d_0)\). Their endpoint \(b'\) lies on \(B'=\{w=Kd_0\}\) with \(y_{b'}\ge Kd_0\). Run the \(b\)-arm construction in \(y>0,w>Kd_0\). The new arm avoids the stem, the triangle, and the downward \(a\)-arm; the completed \(b\)-arm retains the required single crossing of the original \(B\). The enlarged scales are fixed positive constants for bounded \(s\), so their powers of \(s/H\) are comparable up to constants. The stems have fixed positive weights, and change all outer-strip clearances by \(O(K)\) only.

All comparison domains here are intersections of the specified lattice half-planes, a starting and a far terminal line normal to the comparison bridge, and distant lattice walls. The fitted tube stays inside with proportional slack away from its endpoints. This verifies that every cut comparison is between convex domains and avoids assuming uniform continuation weights for individual prefixes. The allowed normal strip placements form an interval of \(\asymp M\) choices; all distances and tube diameters are fixed multiples of \(M\). ◻

Proof. Fix \(0<\eta<1\). Retain the chords from Lemma 19 and attach the two connectors for each outer strip placement. Their endpoint scale obeys \(s\ge H^{1-\eta}\), so the added mass is at least \(H^{-1/2-P\eta-o(1)}\). Translate every output bridge to a fixed bottom source. We must count how many inputs yield the same translated output, including the original placement of \(D\).

For distinct sides, once-crossed \(A\)-levels have at most \(H^{3/4+\varepsilon}\) choices outside negligible mass. For each such \(A\), apply the same bound to candidate \(B\)-levels on the specified tail above \(A\), obtaining another \(H^{3/4+\varepsilon}\). The initial piece up to \(A\) has a bounded bridge sum. The end pieces for the \(B\)-counts have bounded convex-boundary mass with the outer strip, \(A\) half-plane, and relevant extreme \(B\) half-plane retained. The union over the first level or over \(A\) costs only a polynomial in \(H\) before taking high moments. These two nonparallel lines determine \(D\) and the junctions.

For a common side, the permissible \(A\)-levels are crossed twice, with both output endpoints on the exterior side. Their nested excursions give at most \(H^{1/2+\varepsilon}\) choices. The end arms lie in the intersection of the outer strip and the exterior half-plane of the first level, and the last piece is an arch, so their sequential boundary sums are bounded. Once the seam and its two crossing ports are fixed, at most \(O(H)\) tangential placements of \(D\) remain. Thus both endpoint types have total inverse multiplicity \(H^{3/2+o(1)}\). This also recovers the outer strip placement by translating back relative to \(D\). The auxiliary construction parameters within a connector were already removed in Lemma 20; they are not counted again. Exceptions remain negligible with length and raw placement factors, since all output paths lie in an \(O(M)\) box.

The input first-length mass, the \(\asymp M\) outer placements, the connector mass, and the inverse multiplicity therefore give \[H^{25/12-o(1)}\,cM\,H^{-1/2-P\eta-o(1)} /H^{3/2+o(1)} =M^{13/12-P\eta-o(1)}.\] There are only boundedly many orientations and endpoint orders; apply a lattice symmetry to the selected class. The width constant is independent of \(\eta\). Since \(\eta>0\) is arbitrary, this proves the asserted lower exponent.

For the localized moments defined at the start of the section, we now have \(m_1\ge M^{13/12-o(1)}\), \(m_2\le M^{29/12+o(1)}\), and \(m_0\le B_M\asymp M^{-1/4}\). For \(a=M^{4/3-\xi}\), the first-length mass below \(a\) is at most \(aB_M=O(M^{13/12-\xi})\), hence negligible. Weighted Cauchy–Schwarz gives mass at least \(m_1^2/(4m_2)\ge M^{-1/4-o(1)}\) above the threshold, for large \(M\). Division by \(B_M\) proves (12). This step alone does not give a high-probability lower bound. ◻

Proof. We first construct paths along a finite chain of normal tubes and turning regions, keeping their terminal port free. Consecutive directions will have positive scalar product, and nonadjacent stages will have fixed positive clearance. After proving the mass bound for this construction, we fit its chain around the forbidden set.

We begin with a turn and its two recovery lines. Write \(n_1,n_2\) for the incoming and outgoing unit normals. It suffices to use turns by \(60^\circ\), so \(n_1\cdot n_2=1/2\); a longer turn is a sequence of these. Let \(C\) be the corner and put \(F=C-\ell n_1\), where \(\ell\) is a small fixed multiple of \(H\). The line through \(F\) is normal to \(n_1\), and the line through \(C\) is normal to \(n_2\). Each line will vary in its own band of width \(\varepsilon\ell\), with \(\varepsilon>0\) fixed and small. The incoming long tube ends at the \(F\)-line. A short piece turns from that line to the \(C\)-line, and the next long tube begins there.

Here is an explicit region for the short piece. Before rounding its sides to lattice lines, take the intersection of the half-planes between the \(F\)-line and the parallel far line \(n_1\cdot(p-C)=\ell/100\), and impose \[ -(.5+.02)\ell<(n_1-n_2)\cdot(p-C)<.02\ell. \tag{13}\] For a source within \(\varepsilon\ell\) of \(F\), this convex region contains a thin normal tube extending past the chosen \(C\)-line. The straight center starts with the side-wall coordinate near \(-\ell/2\) and finishes near \(\ell/200\), leaving positive margins in (13). The free-tube lemma therefore supplies \(c\ell^{-1/4}\) mass reaching the far wall. Cutting at the chosen \(C\)-line and using the convex cut comparison supplies that much mass ending on the cut. It is a lower bound on the whole exit sum, rather than a lower bound on the continuation of any individual prefix.

At an exit \(p\) on the unshifted \(C\)-line, \(n_2\cdot(p-C)=0\), so (13) gives \(n_1\cdot(p-F)>.48\ell\). Small source errors, cut shifts and rounding leave the uniform bound \[ n_1\cdot(p-F)>.4\ell. \tag{14}\] Along the following normal direction \(n_2\) the left-hand side increases. Choosing the next tube narrow enough makes it greater than \(.3\ell\) throughout that tube. The preceding tube, in contrast, ends near \(F\). Thus the two cuts have disjoint port regions and the outgoing long tube cannot revisit the earlier one. Figure 5 shows the two separated port regions.

Choose the finitely many corner lengths successively, with each much larger than the accumulated endpoint errors from the preceding stages. Choose the first length so small that the last is still a small fraction of every original segment. Then choose all cut ranges and tube widths smaller still. These are choices of fixed constants: every allowed cut band contains at least \(cH\) lattice levels when \(H\) is large. The endpoint of one stage may move over its actual lattice ports; the next stage is started at that actual endpoint. Equation (14) and the positive scalar products preserve all clearances uniformly over these choices.

Each long stage has mass at least \(cH^{-1/4}\) by Lemma 16; the preceding convex construction gives that bound for each short corner stage. Adjacent stages occupy opposite strict half-planes of their joining line. Nonadjacent stage regions are disjoint, after making the widths sufficiently small. Consequently their concatenation is self-avoiding and its weights multiply. If there are \(m\) stages and \(F(\gamma)\) counts their representations of an output path, summing the \(m-1\) joining levels and all the intermediate ports gives \[ \sum_\gamma \rho^{|\gamma|}F(\gamma) \ge cH^{-m/4}H^{m-1}. \tag{15}\] No separate endpoint-position factor occurs: it is already included in each stage’s free-terminal mass.

We next give the second-moment argument, including the upper bounds for the pieces that remain after two representations are cut. Each joining port region is visited only by its two adjacent stages. At the first cut of a corner this region is close to \(F\); at the second it is within \(O(\ell)\) of \(C\) and has the positive separation (14) from \(F\). Hence, in any two representations, both splits at one joint precede both splits at the next joint. Otherwise the intervening portion would belong to nonadjacent stage regions, which are disjoint. At a single joint, the portion between its two split times stays strictly between the two parallel joining lines, by the opposite half-plane restrictions in the two representations. It is therefore a bridge. Equal levels give the same split and port; the corresponding bridge has weight \(B_0=1\).

Delete these \(m-1\) short bridge gaps. Exactly \(m\) portions remain: one before the first pair of splits, one between every consecutive pair, and one after the last. Each starts on its incoming supporting line and stays strictly on its forward side. It ends on its outgoing supporting line at normal distance at least \(\kappa H\) into that forward half-plane, where \(\kappa>0\) depends only on the fixed geometry. For a long stage this is the segment length less the two narrow cut bands. For a corner it is (14), less the cut shifts. These conclusions hold for a mixed pair of representations: retaining the later incoming split and the earlier outgoing split only moves the relevant lines within their already allowed bands.

For completeness, this depth gives an upper mass \(CB_{\lfloor\kappa H\rfloor}\) for each remaining portion. Enclose it in the convex intersection of its endpoint half-planes and distant lattice walls. Its source is on the incoming wall. Cut this domain by a line parallel to that wall at depth \(\kappa H/2\). Every allowed terminal port is lost in the cut. Lemma 12 bounds their total mass by the new-cut mass, which is at most \(CB_{\lfloor\kappa H/2\rfloor}\le CH^{-1/4}\). This estimate is uniform in the actual starting port. We can thus sum the portions in their temporal order, dropping mutual avoidance for an upper bound, without paying for another translation or for a new root.

For each joint the sum over its two levels and their bridge gap is at most \[CH\sum_{g=0}^{CH}B_g\le CH^{7/4}.\] The \(m\) macroscopic portions therefore give \[ \sum_\gamma\rho^{|\gamma|}F(\gamma)^2 \le CH^{-m/4}H^{7(m-1)/4}. \tag{16}\] The finitely many choices of the order of two levels are included in \(C\). Weighted Cauchy–Schwarz, applied to (15) and (16), yields \[\sum_{F(\gamma)>0}\rho^{|\gamma|} \ge cH^{-m/4+2(m-1)-7(m-1)/4}=cH^{-1/4}.\] This proves the asserted finite-chain estimate with no exponent loss.

We now fit such a chain to the obstacle. We may decrease \(\delta\) below one and increase \(D\) above one. If \(v-u\le\delta H/2\), a sufficiently narrow vertical free tube stays to the right of \(x-\delta H\) and has mass at least \(c(1+v-u)^{-1/4}\ge c'H^{-1/4}\); zero and bounded gaps cost only fixed elementary paths. Otherwise rise a small fixed fraction of \(\delta H\), then take one right-up segment and a fixed number of right-down/right-up pairs, with normals \((\sqrt3/2,-1/2)\) and \((\sqrt3/2,1/2)\). Choose equal segment lengths in each pair, much smaller than \(\delta H\), and use sufficiently many pairs that the horizontal coordinate exceeds \(x+(D+1)H\). The last oblique direction is right-up, so it has scalar product \(1/2\) with the final upward direction. Keep this part in the first \(\delta H/2\) bands, and finish by rising vertically to \(y=v\). There is fixed proportional clearance from the obstacle and from both horizontal walls except at the required endpoints. Adjacent normals have scalar product \(1/2\); the stages are simple and separated at all nonadjacent segments. The two-cut construction just proved fits inside this clearance. All lengths and the enclosing box are fixed multiples of \(H\), depending only on \(D,\delta\). This proves the lemma. ◻

Proof. Use the same normal-direction chain as in Lemma 22: rise first, make equal right-up and right-down pairs inside a small fraction of the \(\sigma h\)-high slab, and end with a downward segment beyond \(x_A+(D'+1)h\). Choose the last oblique direction right-down, so its scalar product with the final downward direction is \(1/2\). All intermediate regions lie strictly above the starting line, and the final downward tube ends on that line. They remain right of \(x_A-\sigma h\) with fixed proportional clearance. The finite-chain estimate gives the lower bound. The terminal displacement forces lateral extent at least \(D'h\). For \(D'\ge C_1\sigma\), Lemma 12 bounds the total such arch mass in the height-\(\lceil\sigma h\rceil\) strip by \(C\sigma h(D'h)^{-5/4}\le C'h^{-1/4}\). Lattice rounding changes only fixed constants. ◻

Proof. Prepend and append bridges with heights in \([a,2a]\) to a marked irreducible. Their product mass is at least \(ca^{2\alpha}\). Output heights are \(O(a)\), and a marked irreducible of height at least \(a\) has bounded multiplicity in such an output. Its lateral excursion forces comparable output range. The strip lateral estimate bounds output mass by \(Ca^\alpha\Phi(d)\), proving the first assertion. Summing dyadic \(a\) and width bins gives \(\mathbf E\min\{1,(J+W)/x\}\le Cx^{-\alpha}\). The event \(U_{k-1}<b\) is decided before jump \(k\); multiplying by the fresh-jump expectation and summing gives the charge estimate, because \(\sum_{j<b}B_j\le C(1+b)^\alpha\). This includes the last, possibly overshooting jump. For the moment estimate, order the marked renewals, restart at each, and sum their nonnegative height increments using \(\sum_{j\le v}B_j\le C(1+v)^\alpha\). Diagonal repetitions cost only a constant depending on \(p\). A delayed interval is handled by first stopping on entrance; overshoot can only shorten its remaining length. ◻

Proof. Constructing the hairpin. Apply Corollary 23 at the worker endpoint. It supplies an arch contained in the next \(\sigma h\) bands, to the right of \(x_A-\sigma h\), whose far endpoint lies beyond \(D'h\), for any sufficiently large fixed \(D'\). To pass from the relative displacement in that lemma to this absolute bound, increase its fixed displacement constant by \(D\), using \(|x_A|\le Dh\). Its total critical mass is between \(c h^{-1/4}\) and \(C h^{-1/4}\). The lower bound has no exponent loss; the fixed-endpoint tube bound would not suffice here. The corridor lemma also establishes the endpoint half-plane bounds needed when this arch is cut in the second-moment calculation below.

Choose \(z\in[eh/5,eh/3]\). From the far end of the long arch descend by a narrow bridge to level \(z\), with mass at least \(ch^{-1/4}\). Continue downward by a bridge of free height in \([2r,4r]\); use \(r=h\) for \(Y\). Require its portion through depth \(z+1\) from its starting port to stay within horizontal distance \(D''h\) of that port. This restriction retains mass at least \(cr^\alpha\). Indeed its failure probability in the renewal law is small for large \(D''\), by Lemma 26. If \(N\) counts renewals in the desired terminal interval, then \(\mathbf EN\ge cr^\alpha\) and \(\mathbf EN^2\le Cr^{2\alpha}\); consequently \(\mathbf E[N\mathbf1_{\rm failure}]\le C r^\alpha\mathbf P(\mathrm{failure})^{1/2}\). Take \(D''\) large and then \(D'\) much larger.

Join the two near launches by a fixed short arch in the one band on the opposite side. The resulting path rises from its new lower end, goes round the long arch, descends along the worker to this short turn, then rises along the obstacle. Append an upper continuation with height in \([2r,4r]\). For \(X_{12}\) also make the reflected hairpin at the other end. Qualification gives all lateral clearances. The two near-end structures protrude by at most \((2/5+\sigma)h_i\); since \(h_i\le r\) and the giant has height at least \(r\), they are separated in height. These operations produce strict bridges at a fixed translated root. Local connections change weights by fixed factors. Figure 7 shows the one-giant construction; there \(\tau\) is the absolute height of the near launch.

Representation mass and nearby turn lines. For the one-giant construction, let \(m(\Gamma)\) count the possible input tuples and joining choices that produce the rooted output bridge \(\Gamma\). Define \[Z_1=\sum_\Gamma\rho^{|\Gamma|}m(\Gamma),\qquad V_1=\sum_{m(\Gamma)>0}\rho^{|\Gamma|},\qquad M_2=\sum_\Gamma\rho^{|\Gamma|}m(\Gamma)^2.\] The weight of an input tuple is comparable to the output weight, because its pieces partition \(\Gamma\) apart from the bounded local port changes. Thus \(Z_1\) is comparable to the summed construction weight. Weighted Cauchy–Schwarz gives \(Z_1^2\le V_1M_2\). For \(X_d=X_d(h,r)\), the preceding construction and the strip lateral estimate therefore give \[Z_1\ge cX_d r^{2\alpha}h^{1/2},\qquad V_1\le Cr^\alpha\Phi(d).\] The factor \(h^{1/2}\) is the product of the \(\asymp h\) choices of \(z\) and the two \(h^{-1/4}\) connector masses. The marked giant’s width forces comparable output range, giving \(\Phi(d)\).

We estimate the second moment by pairs of representations of the same output. First bin their turn lines \(\tau\), in absolute output coordinates, into intervals of width \(r/50\). There are only boundedly many such bins, because output heights are \(O(r)\). Cauchy–Schwarz permits us to sum same-bin second moments at bounded cost. For turn-line difference at most \(Nh\), the contribution is \[ C_Nr^{2\alpha}X_d(h,r)h^{1/2}h^{5/4}. \tag{20}\] Here is the full cost allocation. Subdivide into intervals of width \(ceh\), with a sufficiently small fixed \(c\). For nearby cells, \(ab\le(a^2+b^2)/2\) reduces pair counting to same-cell pairs with a factor depending on \(N,e\), but no factor for the absolute cell index. A line strictly between all \(\tau+z\) and all \(\tau+u\) in the cell has the following property: after its first visit, the lowest point of the path is the short turn. Before the descending worker reaches that turn, the initial rising branch stays above \(\tau+z\); afterwards everything other than the fixed short arch stays above the turn line. Thus the short turn, its line, and its two bounding visits coincide in the two representations. The upward branch determines the giant mark up to bounded choices, since its prefix is at most \(h/8\). Retain the marked input associated to the smaller worker height \(u\).

Between the two worker heights there are two bridges, one on either side of the long arch. Their jointly summed cost is \(\sum_{j\le Ch}B_j^2\le Ch^{1/2}\). The higher long arch costs \(Ch^{-1/4}\). The far rising bridge from the higher \(z\) to the smaller \(u\) has height comparable to \(h\), hence costs \(Ch^{-1/4}\). The pair of \(z\)-levels and their gap cost \(Ch\sum_{j\le Ch}B_j\le Ch^{1+\alpha}\). Multiplying these four costs gives \(Ch^{1/2+5/4}\), as in (20). The two exterior continuations supply \(r^{2\alpha}\). Ties use \(B_0=1\). Every cut is determined by its level: each selected \(u\)-level is crossed twice on the pre-turn part, while the \(z\)-cuts are first visits before these crossings. A rooted tuple of these relative shapes has one translation to the prescribed joining port, also when attached by its far endpoint. Thus no horizontal multiplicity is omitted.

Distant turn lines. For distant turn lines in the same coarse bin, write \(\Delta=\tau_2-\tau_1>Nh\), and let \(p_i=\tau_i+f_i\) be their giant junctions. Everything before \(p_1\) has height at most \(\tau_1+(2/5+\sigma)h<\tau_2+z_2\), since \(\Delta>Nh\). Thus the first visit to \(\tau_2+z_2\) occurs after \(p_1\). Moreover, \(\Delta<r/50\) and \(z_2=O(h)\), with \(h<\Delta/N\), put this cut below the terminal level of the first giant. The portion from \(p_1\) to this cut is therefore contained in that giant. It is strict above its lower level by the giant’s lower wall, and strict below its upper level by the first-visit property. After the cut the second representation never falls below \(\tau_2-O(1)\). If this subbridge had a single-crossing level below \(\tau_2-O(1)\), it would also be a seam of the first giant. Its first irreducible \(I'\) consequently has height \[J(I')\in[\Delta-f_1-O(1),\Delta-f_1+O(h)],\] and the rest is a bridge of free height \(O(h)\). Replacing the first giant by \(I'\) preserves the earlier qualification, which concerns only its near range. Retain this reduced first construction and then the second construction with its original giant. The very first and last exterior continuations are retained as well. The removed endpoint of the first giant is recoverable as its first ensuing renewal in the upward continuation; it has no unbounded marking multiplicity.

The second construction costs \(h^{1/2}X_d(h,r)\); the reduced first one at dyad \(a\) costs \(h^{1/2}X_0(h,a)\); the intervening bridge costs \(h^\alpha\). Hence the distant contribution is at most \[Cr^{2\alpha}h^{1/2}X_d(h,r)\, h^{\alpha+1/2}\sum_{Nh/C\le a<r}X_0(h,a).\] Relative offsets of the two turn lines are already determined by the reduced shapes. For \(d=0\), induct on the dyadic \(r\) with the bound \(X_0(h,a)\le Kh^\alpha a^{-\alpha}\). After division by \(r^{2\alpha}h^{1/2}X_0(h,r)h^{5/4}\), the distant contribution is at most \(CKN^{-\alpha}\). Cauchy–Schwarz, \(Z_1^2\le V_1\) times the second moment, therefore gives \[X_0(h,r)\le h^\alpha r^{-\alpha}(C_N+CKN^{-\alpha}).\] Choose \(N\) first to make the coefficient of \(K\) smaller than one half, then \(K\) to absorb \(C_N\) and the bounded initial scales. For \(d>0\) use this proved \(X_0\) bound in the distant sum and retain the factor \(\Phi(d)\) in \(V_1\). This proves the one-giant bound.

For \(Y\), only nearby bins are needed, with \(r=h\). The obstacle’s terminal cut \(v\) is now movable on the common upward branch. Retain the smaller worker height and smaller obstacle height, even if they come from different representations. Shortening preserves qualification and \(v-u\ge2\sigma h\). The larger \(v\) costs one additional summed bridge gap, at most \(Ch^\alpha\). The same Cauchy–Schwarz calculation gives \(Y\le Ch^\alpha\).

Two ends of a shared giant. It remains to bound both ends of one giant simultaneously. Use the analogous representation count for the paired construction. Its first moment is at least \(cX_{12}r^{2\alpha}\prod_i h_i^{1/2}\), and output mass is at most \(Cr^\alpha\). Bin both turn lines into intervals of width \(r/50\). On side \(i\), color the fine cells of width \(ceh_i\) periodically with a sufficiently large fixed number of colors. Matching colors mean either the same cell or separation exceeding \(100h_i\). Splitting into these finitely many classes and applying Cauchy–Schwarz lets us use only matching-color pairs.

Call a same-cell side local. The preceding recovery of the short turn also recovers its giant junction: all seams within the first \(h_i/8\) bands are unaffected by the other-end hairpin, which lies far beyond that range. Distinct junctions would put an internal seam in one of the two giants. On a distant side the outer junction precedes the inner one in the inward orientation, and the preceding \(I'\)-replacement applies.

We describe the retained subpaths before counting them. Write \(a_j,b_j\) for the lower and upper giant junctions of representation \(j\), and write \(\prec\) for order along the oriented output path. At the lower end denote the two junctions by \(a_{\rm out}\preceq a_{\rm in}\), and at the upper end by \(b_{\rm in}\preceq b_{\rm out}\); equality means that side is local. The subpath between \(a_{\rm in}\) and \(b_{\rm in}\) will be the central giant. When both sides are distant, the two possible orders, after interchanging representation names or reflecting, are \[a_1\prec a_2\prec b_1\prec b_2 \quad\hbox{and}\quad a_1\prec a_2\prec b_2\prec b_1.\] Thus the central subpath is \([a_2,b_1]\) in the crossed case and \([a_2,b_2]\) in the nested case. Figure 8 shows these path orders; it does not depict the spatial positions of the junctions.

The central subpath is indeed an irreducible bridge of scale \(r\) or \(r/2\). When both its junctions come from one representation, it is that representation’s giant. In the crossed case it is contained in both giants: the giant starting at \(a_2\) keeps it strictly above \(a_2\), and the giant ending at \(b_1\) keeps it strictly below \(b_1\). Divide its height range into lower and upper halves. A seam in the lower half would extend to a seam of the giant \([a_2,b_2]\): between \(b_1\) and \(b_2\), the upper hairpin can descend below \(b_1\) by at most \((2/5+\sigma)h_{\rm up}+O(1)\), where \(h_{\rm up}\) is the upper-side test scale. A seam in its upper half is excluded by the reflected argument using \([a_1,b_1]\). Each turn-line difference is less than \(r/50\), and a distant difference exceeds \(100h_i\), so the near-end excursions cannot invalidate either extension. The central height can fall below \(r\) only in this two-distant case, where necessarily \(h_i\le r/2\). On a distant side retain the inner representation’s worker and prefix; on a local side retain the smaller worker height. They qualify against the central giant by containment, and so give a central paired input.

Recovering the paired representations. In the crossed case, the lower reduced construction retains the first representation’s worker and prefix, followed by its first irreducible \(I'_-\); thus \(a_1\) is still the prefix–irreducible junction of that counted input. On the upper side, read inward from the upper end, the reduced construction retains the second representation’s worker and prefix, followed by its first irreducible \(I'_+\); its prefix–irreducible junction is \(b_2\). The remaining endpoints \(a_2,b_1\) are the two endpoints of the central giant. These are existing input junctions, rather than additional marks placed on an unmarked bridge.

Here is the order in which the lower pieces are joined. For \(j=1,2\), write \(A_j,P_j,S_j\) for the lower worker, obstacle prefix, and fixed short turn, respectively. Let \(K_j\) be the connector from the cut at \(\tau_j+z_j\), through the narrow bridge and long arch, to the worker endpoint, retaining their counted joining ports and local connection data; here \(\tau_j\) denotes the turn-line height. Let \(R_-\) be the bridge remainder after \(I'_-\), and let \(E_-\) be the first exterior continuation. With a bar denoting path reversal, the lower part of the common output is assembled in the order \[E_-\,K_1\,\overline{A_1}\,S_1\,P_1\,I'_-\,R_-\, K_2\,\overline{A_2}\,S_2\,P_2,\] and then the central giant is attached. The upper part is assembled by the reflected reverse procedure, with representation 2 outer and representation 1 inner. In the forward cutting operation, \(I'_-\) ends at the first positive renewal of the strict subbridge from \(a_1\) to the second connector’s first-visit cut; \(R_-\) is the rest of that subbridge. The upper cut uses the same rule in the reversed orientation. Consequently the reduced-irreducible cuts require no further renewal index. The connector cuts themselves are the first-visit and ordered worker cuts already specified above.

Each item in this list is a rooted relative shape: part of the central or a reduced outer input, a connector, a bridge remainder, or an exterior continuation. Once one joining port is placed, the next shape has a unique translation to it, including when the shape is attached by its terminal port. Its height and horizontal displacement are then fixed. Starting at the prescribed output root therefore places every piece, and in particular all four retained junctions, without a separate sum over turn-line differences or horizontal offsets. The second lower exterior continuation is the assembled prefix \(E_-K_1\overline{A_1}S_1P_1I'_-R_-\) ending at the exterior port of \(K_2\); the remaining upper exterior is the corresponding recovered suffix. Thus each original exterior is recovered by cutting at its retained connector port. Finally, extract the ordered subpaths \([a_1,b_1]\) and \([a_2,b_2]\), and assign the retained workers, prefixes and connectors to their recorded representation. This recovers both original constructions. Interchanging the representation names, choosing the reflected orientation, and the bounded local connection choices require only finitely many records.

For the nested two-distant case, the order is \(a_1\prec a_2\prec b_2\prec b_1\). The central paired input is bounded by \(a_2,b_2\), both from representation 2, and the two reduced inputs retain \(a_1,b_1\), both from representation 1. A finite matching record assigns the four retained boundaries \(a_{\rm out},a_{\rm in},b_{\rm in},b_{\rm out}\) to the two representations; the same ordered assembly and exterior-port cuts recover the nested case. On a local side the two junctions coincide as proved above. The tuple pieces partition the common output, apart from bounded port modifications. Their product critical weight is therefore comparable to the single critical weight of that output in the second moment. In particular the central giant is charged once; the two original giants are recovered subpaths of the assembled output and do not contribute additional weight factors.

The reconstruction has bounded ambiguity, uniformly in all scales. We can therefore integrate these pieces. Each exterior continuation costs \(Cr^\alpha\). A local side has the four gap and arch costs already computed, namely \(Ch_i^{1/2+5/4}\). On a distant side the inner connector costs \(Ch_i^{1/2}\), the reduced outer construction costs \(Ch_i^{1/2}X_0(h_i,a)\), and the bridge remainder costs \(Ch_i^\alpha\). The proved one-sided estimate gives \[h_i^{1/2}h_i^{1/2}h_i^\alpha \sum_{a\ge c h_i}X_0(h_i,a) \le Ch_i^{1+\alpha}=Ch_i^{1/2+5/4}.\] The central paired input has scale \(r\) or \(r/2\), and its giant weight is charged once. Combining these costs with the first moment and output mass yields \[\begin{split} &\left(cX_{12}(h_1,h_2,r)r^{2\alpha}\prod_i h_i^{1/2}\right)^2\\ &\qquad\le Cr^{3\alpha}\prod_i h_i^{1/2+5/4} \{X_{12}(h_1,h_2,r)+X_{12}(h_1,h_2,r/2)\}, \end{split}\] omitting an out-of-domain last term. Writing \(V_r=X_{12}(h_1,h_2,r)/(h_1^\alpha h_2^\alpha r^{-\alpha})\), this becomes \(V_r^2\le C(V_r+2^\alpha V_{r/2})\). A sufficiently large constant majorant closes the dyadic induction, proving the paired bound. ◻

Proof. Write \((t,x_0)\) for the worker stop. A regular history has \(t<h/16\) and horizontal range less than \(h/16\). For \(m=\lceil3c'h^\alpha\rceil\), the probability that the next \(m\) jumps advance more than \(h/4\) is \(O(c')\): bound jumps larger than \(h/4\) by a union bound and the remaining sum by its truncated mean. Choose \(c'\) small. There are then at least \(2c'h^\alpha\) future renewals before \(t+h/4<2h/5\), except with arbitrarily small probability. Renewals within \(eh\) of the stop have expected count \(O((1+eh)^\alpha)\), so remove them by taking \(e\) small.

No short bin contains a substantial fraction of these renewals with high probability. Indeed partition the candidate interval into \(O(\theta^{-1})\) bins of length \(\theta h\). By Lemma 26, for any fixed \(p>1/\alpha\), the probability that some bin contains at least \(\varepsilon h^\alpha\) renewals is at most \(C_{p,\varepsilon}\theta^{p\alpha-1}\), once \(\theta h\) is large. Use neighboring bins as well to control every interval of length \(2\theta h\). The charge bound also makes the entire worker’s lateral range through its test stop at \(h\) at most \(Dh\), with failure probability \(O(D^{-\alpha})\).

Freeze the opposite path. For a right worker let \(R(u)\) be the rightmost opposite abscissa in \([u,u+\sigma h]\). From \((t,x_0)\) the renewal-occurrence density at \((u,x)\) is exactly \(b_{u-t}(x-x_0)\). It is not divided by \(B_{u-t}\), since we count renewal occurrences. Thus the expected number of renewals, at distance at least \(eh\) above the stop, with \(|x-R(u)|\le2e'h\), is at most \[\begin{align*} C(e'h+1)\sum_{j\ge eh}\|b_j\|_\infty &\le C(e'h+1)\sum_{j\ge eh}(B_j-B_{j+1})\\ &\le C_e(e'+h^{-1})h^\alpha. \end{align*}\] The middle inequality follows from Lemma 13: a summable nonnegative kernel tends to zero at infinity, so its supremum is at most its total variation. The sum of the resulting differences telescopes. No height-by-height derivative bound on \(B_j\) is needed. At each renewal \(u\), restart and discard it if the path through its first renewal strictly above \(u+\sigma h\) moves more than \(e'h\) laterally. Its conditional probability is at most \(C((1+\sigma h)/(e'h))^\alpha\); multiplying by the expected candidate count bounds the expected number discarded by this quantity times \(Ch^\alpha\).

On continued order, every opposite point in this slab lies to the left of the worker’s rightmost point at that level. For a retained candidate that point is at most \(x+e'h\), because strict renewals exclude earlier and later pieces from the slab. Hence \(R(u)\le x+e'h\). Since \(|x-R(u)|>2e'h\), necessarily \(R(u)<x-2e'h\). Choose \(\sigma<e'\). This is the required clearance; the obstacle’s global one-sided bound below \(u+\sigma h\) follows from order and the worker’s bounded range. Reflection gives the other orientation.

Choose the constants successively: the desired failure tolerance, \(c'\), then \(e,\theta\), then \(D\) large and \(e'\) small, and finally \(\sigma\) small relative to \(e',e,\theta\). The expectation bounds and Markov’s inequality leave \(ch^\alpha\) candidates. For two workers, apply this uniform estimate first with the opposite future frozen and then with roles reversed, integrating by Tonelli. A union bound suffices; independence of the two good events is unnecessary. Bin nonconcentration ensures that only a small fraction of all cross-pairs have heights within \(2\sigma h\). The remaining pairs have the claimed lower/higher qualification. ◻

Proof of Theorem 27. We first prove the \(E\) and \(F\) bounds simultaneously by strong induction on their test height, with \(r,d\) unrestricted. Floors below one mean the trivial test, with constant cost. All uses of nontrivial smaller tests have strictly smaller test height.

For \(F(h,r,d)\), stop the worker at height \(s=h/L\). If its history is irregular, then \[\mathbf1_{\rm irregular} \le16\sum_{k:U_{k-1}<s}\min\{1,(J_k+W_k)/h\}.\] Integrate this inequality in (17), retaining the large test only where it is needed to imply a smaller test. Let \(f,g\) be the dyads of one plus the birth depths of the main giant and a charged worker jump; let \(a,l\) be that worker jump’s height and width bins. Its charge is at most \(CT_h(a,l)\). Both birth dyads are at most \(C(1+s)\). Constants in what follows do not grow with \(L\).

When \(f<g/(16L)\), retain the whole smaller event \(F(\lfloor g/2\rfloor,r,d)\). Do not also sum separately over \(f\). On contributing paths the worker’s stop for this smaller test is no later than the charged birth, which itself is a renewal. Restarting there, the expected total charges born in the \(g\)-bin are at most \(Cg^\alpha h^{-\alpha}\), by Lemma 26. After dropping the large test, this is a uniform bound on the conditional future integral. Hence the contribution, divided by \(r^{-\alpha}\Phi(d)\), is at most \[CK_2h^{-\alpha}\sum_{g\le C(1+s)}g^\alpha \le CK_2L^{-\alpha}.\]

When \(g<\min(f,a)/(16L)\), retain instead \(F(\lfloor\min(f,a)/2\rfloor,a,l)\), with the workers’ roles exchanged. The charged jump is now the unique marked jump of that smaller test; there is no separate \(g\)-birth factor. The original giant is born later than its arm’s smaller test stop. Its subsequent birth count in the \(f\)-bin is at most \(Cf^\alpha\), uniformly in the stop, including a possible birth at the stop itself. The ensuing fresh jump costs \(Cr^{-\alpha}\Phi(d)\). Thus the relative contribution is bounded by \[CK_2\sum_{f\le C(1+s)}f^\alpha \sum_{a,l}a^{-\alpha}\Phi(l)T_h(a,l) \le CK_2L^{-\alpha}.\] This formula is a product of a smaller tested integral, a renewal count, and a fresh-jump integral; it is not a conditional probability estimate under the larger \(F\) event.

In the remaining case retain \(E(\lfloor\min(f,g)/2\rfloor)\). Both subsequent births are renewals at or beyond their respective smaller stops. Once their earlier histories are fixed, their count integrals are bounded by \(Cf^\alpha\) and \(Cg^\alpha\), and the new jump integrals are independent. The smaller test costs \(CK_1\min(f,g)^{-\alpha}\), leaving \(CK_1\max(f,g)^\alpha\). The restrictions defining this case are \(f\ge g/(16L)\) and \(g\ge\min(f,a)/(16L)\). Their dyadic sum obeys \[\sum_{f,g}\max(f,g)^\alpha \le C(1+s)^\alpha [\log(2L)+\log^+((1+s)/a)].\] Indeed for \(f\le a\) the two birth dyads are within factor \(16L\), so at each larger dyad there are \(O(\log(2L))\) partners. For \(f>a\), the permitted smaller partner is at least \(a/(16L)\), adding at most \(O(\log^+((1+s)/a))\) choices. Geometric summation of the larger dyad proves the displayed bound. Equation (23) makes the relative contribution at most \(CK_1L^{-\alpha}\log(2L)\).

Consider regular histories next. To apply Lemma 29 without paying the unrestricted prefix mass, retain a smaller-test envelope. If the main birth is before \(s/(4L)\), use \(F(\lfloor s/2\rfloor,r,d)\). Otherwise use \(E(\lfloor s/(8L)\rfloor)\), followed by the main birth count \(O(s^\alpha)\) and its jump weight. In the later-birth branch the obstacle’s smaller stop occurs no later than the counted birth. In both branches the worker’s smaller stop occurs no later than its stop at \(s\). The total envelope mass is therefore at most \[C_L(K_1+K_2)r^{-\alpha}\Phi(d).\] Condition on the worker history to \(s\) and on the full opposite obstacle before invoking the lemma. Its failure probability is uniformly at most \(\delta\), without conditioning on continued \(E(h)\). The envelope is measurable with respect to this exposed information; integrating it bounds the failure contribution by \(\delta\) times the last display.

On success there are at least \(ch^\alpha\) qualified renewal choices. Count them with \(\mathcal R\), using its occurrence identity. Each choice leaves a qualified worker bridge and the original marked obstacle; all unused worker futures have probability at most one. Tonelli therefore bounds the total choice mass by \(X_d(h,r)\). Lemma 28, divided by the guaranteed number of choices, bounds this success contribution by \(Cr^{-\alpha}\Phi(d)\). Altogether, \[ \frac{F(h,r,d)}{r^{-\alpha}\Phi(d)} \le C+CL^{-\alpha}[K_2+K_1\log(2L)] +\delta C_L(K_1+K_2). \tag{24}\]

For \(E(h)\), stop both workers at \(s'=h/L'\), where \(L'\) will be large compared to \(L/\kappa\). If either early history has a jump of height dyad \(a\ge\kappa h\), retain the smaller event \(F(\lfloor\kappa h\rfloor,a,0)\); its birth is early enough since \(s'<\kappa h/L\). Summing over \(a\ge c\kappa h\) costs at most \(CK_2\kappa^{-\alpha}h^{-\alpha}\). For all other charged jumps, \(a<\kappa h\). If their birth is before \(a/L\), use \(F(a,a,l)\), with no further birth-count factor. Otherwise retain a smaller \(E\)-test below the birth dyad and then count subsequent births and the new jump as above. The number of possible late birth dyads contributes \(1+\log^+(s'L/a)\). Consequently the irregular-history contribution, relative to \(h^{-\alpha}\), is at most \[\begin{align*} Ch^\alpha\sum_{a<\kappa h,l} &[K_2+K_1\{1+\log^+(s'L/a)\}] a^{-\alpha}\Phi(l)T_h(a,l)\\ &\le C(K_1+K_2)\kappa^{1-\alpha}, \end{align*}\] by (22) and \(s'L\le\kappa h\).

For regular histories retain the envelope \(E(\lfloor s'\rfloor)\), whose mass is at most \(CK_1(s')^{-\alpha}\). Apply the usable-renewal lemma to each worker while freezing the other path; integrate each failure estimate and take a union bound. Their total cost is \(C\delta K_1(s')^{-\alpha}\). On success count the \(ch^{2\alpha}\) separated cross-pairs. Each pair, according to which renewal is lower, contributes to \(Y(h)\) in one of two reflected orientations. The occurrence identity integrates unused futures to at most one, so the total success probability is at most \(CY(h)/h^{2\alpha}\le Ch^{-\alpha}\). Thus \[ \frac{\mathbf P(E(h))}{h^{-\alpha}} \le C+CK_2\kappa^{-\alpha} +C(K_1+K_2)\kappa^{1-\alpha} +C\delta K_1(L')^\alpha. \tag{25}\]

Choose \(\kappa\) so small that the coefficient of \((K_1+K_2)\) in the third term of (25) is small. Choose the ratio \(K_1/K_2\) sufficiently large compared to \(\kappa^{-\alpha}\). Next choose \(L\) so that the charge terms of (24) are a small fraction of \(K_2\), and then \(L'\gg L/\kappa\). Choose the two failure tolerances small enough for these fixed constants. Finally increase \(K_1,K_2\) together in their fixed ratio to absorb the success constants and bounded test heights. At bounded heights, the prefix sum is bounded and the giant tail supplies \(r^{-\alpha}\Phi(d)\). Both recursions close.

It remains to prove the paired bound. Regard the two established constants as fixed and induct on \(h_1+h_2\), using \(s_i=h_i/L_2\). If either test height is bounded, integrate its prefix and worker at bounded cost and use the single \(F\) bound on the other side; since \(L_2\ge L\), its prefix restriction is stronger than that of \(F\). This gives the initial cases.

For an irregular worker on side 1, use the same three birth-order regimes, now with \(L_2\). In the very early main-birth regime, retain the smaller paired test on side 1 and the original side-2 test. Restart only the side-1 worker after its smaller stop. This gives relative charge \(CK_3L_2^{-\alpha}\). In the very early worker-birth regime, the smaller single \(F\) marks the charged worker jump and only uses side 1’s obstacle before its later shared-giant birth. Once this tested prefix and the subsequent side-1 birth count have been integrated, the remaining shared-giant and side-2 integral is \[\sum_{r\le J(I)<2r}\nu(I) \int_{f_2<h_2/L_2}\!d\mathcal R(P_2) \int\mathbf1_{E(h_2)}(P_2I^{(2)},A_2)\,d\mathcal Q_{h_2}(A_2) \le K_2r^{-\alpha}.\] It is bounded by the single \(F\) mass because \(L_2\ge L\). The pre-giant side-1 factors are independent of this remaining integral: translating \(I\) to the endpoint of the side-1 prefix adds no weight and no choice. This yields relative charge \(CK_2^2L_2^{-\alpha}\). In the comparable-birth regime the smaller side-1 test is \(E\), followed by its two birth counts and fresh charged jump, then exactly the same remaining integral. Its charge is \(CK_1K_2L_2^{-\alpha}\log(2L_2)\). The giant has been charged once in all three regimes. The full side configurations are not independent; their only dependence left after peeling the pre-giant factors is precisely the one shared giant in the last integral. Repeating the argument on side 2 gives total relative bad-history cost \[CL_2^{-\alpha} [K_3+(K_2^2+K_1K_2)\log(2L_2)].\]

For regular histories keep measurable smaller-test envelopes before integrating failure probabilities. If both main births are very early, retain a paired test with heights \(\lfloor s_i/2\rfloor\). On each later-birth side retain instead \(E(\lfloor s_i/(8L_2)\rfloor)\) and its subsequent prefix count. If one early side remains, keep its smaller single \(F\) test. If neither remains, charge the giant by its one-jump tail. Thus the total exposed envelope mass is at most \[C_{L_2}(K_3+K_1K_2+K_1^2)r^{-\alpha}.\] All retained worker test stops occur no later than the corresponding worker stops at \(s_i\); retained obstacle data belong to the exposed paths. Hence these envelopes are measurable before either tested worker future is integrated. Apply the conditional failure bound separately to the two independent worker futures and integrate by Tonelli; a union bound makes its cost at most \(\delta\) times a fixed multiple of this display. This does not assert independence of the usability events or condition either future on continued survival.

On success each side has at least \(ch_i^\alpha\) usable renewals. Their simultaneous choices number at least \(c h_1^\alpha h_2^\alpha\), since there is no condition between the two side tests. In (18), sum both renewal occurrences before integrating. The shared giant remains a single factor \(\nu(I)\); each worker prefix becomes its critical bridge measure and its unused future has mass at most one. The resulting integral is bounded by \(X_{12}(h_1,h_2,r)\), so the paired hairpin bound makes the success contribution at most \(Cr^{-\alpha}\). We have proved \[\frac{F_{12}(h_1,h_2,r)}{r^{-\alpha}} \le C+CL_2^{-\alpha} [K_3+(K_2^2+K_1K_2)\log(2L_2)] +\delta C_{L_2}(K_3+K_1K_2+K_1^2).\] Choose \(L_2\ge L\) large, then \(\delta\) small, and finally \(K_3\) large. This closes the induction and proves (19). ◻

Proof. Stop both renewal strings at their first renewals at or above \(h/4\). If both stops are below \(h\), disjoint first-hit arcs to \(h\) imply \(E(\lfloor h/4\rfloor)\). Indeed those renewal stops are single crossings, so later portions cannot revisit the tested lower region; the two first-hit crosscuts have their launch order throughout it. Equation (19) bounds this case.

Otherwise mark each jump born below \(h/4\) that reaches or exceeds \(h\). There are one or two such marked jumps, each of height at least \(3h/4\). Bin their birth depths plus one into dyads \(f_i\). Retain the ordered test at half the smallest birth dyad, or the trivial test at bounded depth. Its stops precede the corresponding births, and in the unmarked arm precede its stop below \(h\), so they precede both first hits at \(h\). The same crosscut argument makes this test valid. It costs \(Cf_{\min}^{-\alpha}\). Each subsequent marked birth count costs \(Cf_i^\alpha\) by restarting, and each ensuing jump costs \(Ch^{-\alpha}\) by the irreducible tail. Thus one marked bin costs at most \(Ch^{-\alpha}\), and a pair of marked bins costs at most \[C f_{\min}^{-\alpha}(f_1/h)^\alpha(f_2/h)^\alpha =C f_{\max}^\alpha h^{-2\alpha}\le Ch^{-\alpha}.\] There are \(O(1+\log h)\) bins per marked jump. Summing proves (26). ◻

Proof. Corollary 30 gives \(S_2(H)\le CH^{-3/4}(1+\log H)^2\). For fixed \(\epsilon>0\), the logarithmic factor is bounded by \(C_\epsilon H^\epsilon\). ◻

Proof. The polynomial lower-length failure bound in Theorem 25 gives, for some \(a>0\), \(B_j[|\gamma|\le j^{4/3-\varepsilon/2}] \le C B_jj^{-a}\). Choose \(\kappa>0\) so small that \((1-2\kappa)(4/3-\varepsilon/2)>4/3-\varepsilon\). For large \(H\), if \(j\ge H^{1-2\kappa}\), the length cap in Equation (27) is below that threshold. Such heights contribute at most \(Cr^{3/4}H^{-a(1-2\kappa)}\). Smaller heights, including zero, have total mass \(O(H^{3(1-2\kappa)/4})\), at most \(Cr^{3/4}H^{-3\kappa/4}\) in the stated range. Taking \(a_0\le\min\{a(1-2\kappa),3\kappa/4\}\) proves the lemma. ◻

Proof. Average a cut level \(m\) over lattice lines in the middle third of the bridge. If the line is crossed once, cut there and insert an empty retreat. Otherwise, after the first hit of \(m\), let \(i\) be the last point attaining the minimum height of the remaining path, and let \(j\) be a point attaining the maximum height before \(i\), with a fixed tie convention. The multiple crossing implies \(i<m<j\), and \(j\) occurs after the first hit. The three pieces are the up bridge to \(j\), the down retreat to \(i\), and the remaining up bridge. They are weak bridges: the defining extrema bound all their intermediate heights. Repeat on both up pieces through depth \(d\). Each up span loses at most a fixed factor per generation, so its final span is at least \(c_dH\).

For a fixed final decomposition, the cut at an internal tree node must lie in its retreat’s height interval, or at the single-crossing cut if the retreat is empty. Its number of choices is \(O(s_\ell)\). The averaging denominator at that node is \(\asymp H\), with constants depending on \(d\). The tree order is fixed, so multiplying these bounds proves the asserted placement factor.

We describe the local recoding because neighboring starts are needed for Corollary 31. A nonempty peak is a down-pointing triangle, two thirds up its band, visited via its two lower neighbors. Assign that triangle to the retreat, entering through its top port. The reversed up piece can start at the neighboring top port on its branch side and pass through the adjacent down-pointing triangle to the same lower neighbor of the peak. If it already used that triangle, the visit must be immediately next to this lower neighbor: it cannot cross above its extremum, and the piece continues macroscopically below. In that case start directly there and remove the short peak portion. The adjacent triangle cannot be in the relevant initial retreat arc, because any such visit would use that same lower neighbor, contradicting the original disjointness.

Reflect this construction at minima. If an original endpoint is a vertex, extend it to the nearest exterior band line by the bounded construction in the proof of Lemma 5. The added vertices lie outside the old height range; the extension changes length and height by \(O(1)\), has bounded multiplicity, and changes weight by a bounded factor. Original port endpoints and single-crossing port cuts already lie on boundary lines. These local operations produce strict port bridges within the outer band lines of their extrema. At a macroscopic turn, modifications at remote ends are outside the first-hit portions of distance \(o(s_\ell)\) and \(o(H)\); hence those neighboring arms remain disjoint. A minimum attained again at the starting height is still a local minimum between upper neighbors, so the same construction applies there. Each cut has only finitely many possible local records and weight ratios. Relative shapes and these records recover the original pieces, their translations being fixed by concatenation up to bounded lattice choices. In upper bounds all ending heights may therefore be summed freely over their indicated scale ranges. ◻

Proof. For unbounded scales choose \(t\) less than, say, one tenth of both probe scales, reducing the constant to handle fixed scale conventions. Expose each infinite end walk through its first renewal at or above \(t\). The first-visit avoidance event is measurable with respect to these two stopped strings. A good string has conditional remaining mass at most 1. For a marked string, condition on its exposed endpoint height \(a\). If \(a\ge2f\) there is no admissible marker, except for harmless bounded boundary cases. Otherwise its conditional expected number of possible marked renewals is at most \[1+\sum_{j\le2f}B_j\le Cf^{3/4}.\] This includes the stopping renewal itself and is uniform in overshoot. The infinite renewal law up to a marked renewal has exactly the unnormalized bridge weights, so the argument bounds the required finite-string measure. After multiplying the two conditional bounds, the remaining event has probability \(S_2(t)\). This proves all good/marked combinations without conditioning a terminal height in advance.

For simultaneous turns, first express every end in relative coordinates. The local records at a turn fix its neighboring starting ports up to a bounded choice. Once middle bridges and large jumps have been summed, discard consistency of their translations and heights with these relative strings. This enlarges the set being counted. Each end then appears in exactly one turn, so the pairs of independent string variables used by different avoidance tests are disjoint. The preceding conditional integration can be performed at every turn, and its factors multiply. In particular, no avoidance estimate is being applied to a string already biased by a different turn. ◻

Proof. It suffices to take small \(\varepsilon>0\). Use Lemma 33 with \(k=2^d\), where \(d\) is fixed now and chosen large at the end. Local recodings increase the total length by \(O(k)\), so every resulting portion has length at most \(2H^{4/3-\varepsilon}\) for large \(H\). Let \(\kappa,a_0\) come from Lemma 32, and choose \[0<\Delta<\min\{1/40,\kappa/10,a_0/10\}.\] An up piece has scale \(r=H\), while a retreat has scale \(r=s_\ell\). The unrestricted summed-height mass of either is \(O(r^{3/4})\). Retreats with \(s_\ell\le H^{1/20}\) will use only this bound.

For any other piece inspect from both ends its first renewal beyond \(q=\lfloor rH^{-\Delta}\rfloor\), a threshold less than half its span. The irreducible decomposition is unchanged by reversal. If the two stopped strings do not overlap, allowing a shared cut level, call the piece type \(\mathrm M\). Sum its two end strings separately and give its middle bridge the full mass bound \(Cr^{3/4}\). If \(r\ge H^{1-\kappa}\), the middle also has the length cap, so Equation (27) supplies the extra factor \(H^{-a_0}\), regardless of its actual remaining span.

An \(\mathrm M\) end is good, denoted \(\mathrm G\), if its stopping jump lands within distance \(cr\) of that end, for a small fixed \(c>0\). Ignore this extra constraint and use its stopped renewal probability law. Otherwise the end is bad. Group \(1+\)the height before its stopping jump in a dyadic bin of scale \(f\le Cq\). Its preceding marked-renewal prefix has mass \(O(f^{3/4})\), and its next independent jump has height at least \(cr/2\), with probability \(O(r^{-3/4})\). Thus the bad end costs \(C(f/r)^{3/4}\) in addition to the middle’s baseline.

If the end strings overlap, call the piece type \(\mathrm Z\). They contain the same irreducible jump between the two near-end ranges. There cannot be two such jumps, since the ranges are disjoint and the irreducibles are linearly ordered in height. Mark both ends bad. Summing their preceding prefixes and this single shared jump gives \[ C f_1^{3/4}f_2^{3/4}r^{-3/4} =Cr^{3/4}(f_1/r)^{3/4}(f_2/r)^{3/4}. \tag{29}\] The shared jump has height comparable to \(r\). There is no length gain for this piece, and its tail is charged only once.

For each turn of a retreat with \(s>H^{1/20}\), impose adjacent-arm avoidance at a small multiple of the minimum probe scale: \(q\) at a good end, \(f\) at a bad end. These distances are \(o(s)\) and \(o(H)\), as required by the local recoding. When the minimum is bounded use the constant bound instead. By Lemma 34, the marked terminal renewal sum after exposing this avoidance still costs at most \(Cf^{3/4}\). Large-jump tail factors follow these prefix portions and do not consume them. In the \(\mathrm Z\) case the one shared jump has already been integrated as in Equation (29). The remaining end variables are distinct at distinct turns, so their avoidance gains multiply. Corollary 31 therefore supplies the minimum probe scale to power \(-3/4+o(1)\) at every such turn.

Each turn adjoins an up piece of scale \(H\) to a retreat of scale \(s\). Combine its avoidance factor with its attached bad-end costs. Apart from \(H^{o(1)}\), it supplies \(s^{-3/4}\) and the following additional factors: \[\begin{array}{c|c} \text{end types}&\text{factor after extracting }s^{-3/4}\\ \hline \mathrm G,\mathrm G&H^{3\Delta/4}\\ \text{one bad}&1\\ \text{both bad}&H^{-3\Delta/4}. \end{array}\] For two good ends, the smaller threshold is comparable to \(sH^{-\Delta}\). With one bad end of scale \(r\), if its probe \(f\) is smaller, its \(f^{3/4}\) cost cancels the avoidance factor, leaving \(r^{-3/4}\le Cs^{-3/4}\). If the good probe \(r'H^{-\Delta}\) is smaller, use \(f/r\le CH^{-\Delta}\) and \(s\le Cr'\). With two bad ends, writing their probes as \(f_1,f_2\), the factor is exactly \[(f_1/H)^{3/4}(f_2/s)^{3/4} \min(f_1,f_2)^{-3/4} =s^{-3/4}\bigl(\max(f_1,f_2)/H\bigr)^{3/4},\] which is at most \(Cs^{-3/4}H^{-3\Delta/4}\).

Nominally extract \(s_\ell^{-3/2}\) from the two turns of every retreat. Multiplying all baseline masses and placement factors gives \[ H^{3k/4}\prod_{\ell=1}^{k-1} \left[s_\ell^{3/4}\frac{s_\ell}{H}s_\ell^{-3/2}\right] =H^{3/4}\prod_{\ell=1}^{k-1}(s_\ell/H)^{1/4}. \tag{30}\] For an untreated retreat \(s\le H^{1/20}\), restore the nominal factor \(s^{3/2}\). Its net factor is at most \(H^{-1/4}s^{7/4}\le H^{-13/80}\). A treated retreat with \(s<H^{1-\kappa}\) pays at least \(H^{-\kappa/4}\) through its scale factor in Equation (30).

Every both-good loss can be charged to its incident \(\mathrm M\) up piece, since a \(\mathrm Z\) piece has both ends bad. Each up piece has at most two turns, so its own \(H^{-a_0}\) gain leaves at least \(H^{-a_0/2}\), by the choice of \(\Delta\). Assign each \(\mathrm Z\) up piece to one incident retreat; an endpoint up piece still has one such retreat because \(k\ge2\). A retreat receives at most two assignments. If its scale is below \(H^{1-\kappa}\), its scale saving pays those assignments. If it is larger and type \(\mathrm M\), its own \(H^{-a_0}\) length gain pays them. If it is larger and type \(\mathrm Z\), the assigned turn is both bad and pays \(H^{-3\Delta/4}\). Thus there is a constant \(b=b(\varepsilon)>0\), independent of fixed \(k\), such that all up pieces together save \(H^{-bk}\). For example one can choose any sufficiently small positive number below \(\min\{a_0/2,\kappa/8,13/160,3\Delta/8\}\).

For fixed \(k\), there are finitely many type choices and only logarithmically many dyadic scales per piece. All avoidance losses are \(H^{o(1)}\) for this fixed number of tests. The total mass is consequently at most \(C_kH^{3/4-bk+o(1)}\). Taking a fixed power of two \(k\) sufficiently large in terms of \(A\), and then enlarging the constant for bounded \(H\), proves Equation (28). ◻

Proof. Decreasing \(c\) only enlarges the event, so assume that \(c\) is a small fixed constant. The middle-third extremal decomposition used for (28), now at the line of height \(\lfloor cr\rfloor\), gives an initial rising bridge \(A_1\), a retreat \(T\), and a final rising bridge \(A_2\). Place the initial endpoint at height zero. Let \(y\) denote the height of the last minimum after the first hit of this line, and let \(z\) denote the height of a chosen maximum before that minimum, with a fixed tie rule for its location along the path. Before recoding, their heights are \(z\), \(z-y\), and \(J-y\). For large \(r\), the return assumption gives \(y=o(r)\) while \(z\ge cr+O(1)\) and \(r\le J<2r\), so all three spans are comparable to \(r\). After recoding at the two turns, write their heights as \(m_1,m_T,m_2\). Choose dyads \(ar\) and \(br\) for \(1+y\) and \(1+J-z\), respectively. Thus \(1/r\le a,b\le C\), \(a\le C(1+\log r)^{-D}\), and \[|m_1-m_T|\le Car,\qquad |m_2-m_T|\le Cbr.\] The first irreducible of \(A_1\) has height at least a constant times \(ar\), and the last of \(A_2\) has height at least a constant times \(br\). A single-crossing seam below \(y\), or above \(z\), would otherwise be a seam of \(I\) itself: the bounded changes at the remote turn lie on the far side of that seam. This would contradict irreducibility.

At each turn inspect the incident end strings to their first renewals at depth at least \(q\), where \(q\asymp r(1+\log r)^{-Q}\), with \(Q\) fixed. Separate the required outer irreducible at each unprobed end of \(A_1,A_2\). For an outer bridge, type \(\mathrm M\) means that this outer jump and the string inspected from the turn do not overlap; type \(\mathrm Z\) means that they share a jump, which must be the required outer jump. For \(T\), inspect from both ends: type \(\mathrm M\) means the two stopped strings are disjoint, allowing a common cut, and type \(\mathrm Z\) means they share one irreducible. The three baseline masses are \[r^\alpha(ar)^{-\alpha},\qquad r^\alpha, \qquad r^\alpha(br)^{-\alpha}.\] Each \(\mathrm M\) piece leaves an independent residual bridge with freely summed height at most \(Cr\), supplying the factor \(r^\alpha\) in its baseline.

An inspected end is good, denoted \(\mathrm G\), if its stopping renewal lands below \(c'r\), where \(c'\) is sufficiently small depending on \(c\). Otherwise it is bad. A type \(\mathrm Z\) retreat has two bad ends: the two birth depths are below \(q=o(r)\), so the jump they share spans a fixed proportion of the retreat. If \(f\) is the dyadic scale of one plus the birth depth of its stopping jump, the prefix mass is at most \(Cf^\alpha\) and the jump has mass at most \(Cr^{-\alpha}\). Consequently a bad end costs \((f/r)^\alpha\) beyond baseline. There is only one jump variable when the two inspections overlap. For a type \(\mathrm Z\) retreat its actual mass factor is \[f_1^\alpha f_2^\alpha r^{-\alpha} =r^\alpha(f_1/r)^\alpha(f_2/r)^\alpha.\] For the first outer piece of type \(\mathrm Z\), the shared jump is macroscopic and its mass factor is instead \[f^\alpha r^{-\alpha} =\bigl[r^\alpha(ar)^{-\alpha}\bigr](f/r)^\alpha a^\alpha.\] Thus the first outer piece has an extra factor \(a^\alpha\) beyond its baseline and bad-end factor; the second has the corresponding \(b^\alpha\). These equalities allocate the weight of one shared jump, without replacing it by independent copies. The same additional \(a^\alpha\) is available when \(A_1\) is \(\mathrm M\) and its required first jump has height at least \(c'r\).

Use the logarithmic first-hit estimate (26) at each turn, at a sufficiently small fraction of the smaller available depth: \(q\) on a good end and \(f\) on a bad end. At bounded depth use the trivial bound instead. The test precedes every counted bad jump; after its stopping renewal, the remaining prefix occurrences still cost at most \(Cf^\alpha\) by the renewal property. The two end prefixes at a turn are distinct, and the recoding preserves their first-hit disjointness. The two turn tests therefore give \(r^{-\alpha}\) each, with the following additional factors: \((r/q)^\alpha\) for two good ends, a constant for one good and one bad end, and \((q/r)^\alpha\) for two bad ends. All other logarithmic losses, including birth dyads and the choices of \(a,b\), are bounded by \((1+\log r)^{C_1}\), where \(C_1\) does not depend on \(D,Q\).

It remains to use the height constraints without conditioning the probe laws. Put \[K_r(j)=r^{-\alpha}B_j\mathbf 1_{\{0\le j\le Cr\}}.\] Uniformly under arbitrary translations, one kernel restricted to an interval of width \(w\ge1\) has mass at most \(C(w/r)^\alpha\). For \(m=2,3\), Hölder’s inequality gives \[\sum_j\prod_{\nu=1}^m K_r(j+u_\nu) \le \prod_{\nu=1}^m\left(\sum_jK_r(j+u_\nu)^m\right)^{1/m} \le C r^{1-m}.\] Here \(m(\alpha-1)>-1\), so the last estimate follows by summing the power kernel. Summing the allowed height differences now shows that two residual kernels matching within \(Car\) cost \(Ca\), and three kernels with the two displayed height constraints cost \(Cab\). These bounds hold after every outer jump and every probe has been fixed. If a residual height is itself at least \(c''r\), its normalized kernel is at most \(C/r\), so restriction to width \(Car\) costs \(Ca\).

We apply these estimates in the following order. Fix all end strings and distinct jump shapes, and first sum the residual bridge heights with their translated restrictions. The kernel estimates just proved are uniform in the fixed data. Then integrate every distinct separated or shared jump once. Finally apply the first-hit estimates to the remaining pre-jump end strings. Different turns use distinct such strings. Their untested continuations contribute the uniform renewal occurrence bounds already stated. This order proves simultaneous gains without conditioning a probe law on a completed bridge height.

After the turn factors the baseline is \(r^{-\alpha}(ab)^{-\alpha}\). It remains to gain either a positive power of the small return scale \(a\) or a positive power of \(q/r\). In the all-\(\mathrm M\) case, \[(ab)^{-\alpha}\,ab=a^{1-\alpha}b^{1-\alpha}\le C a^{1/4}.\] The other cases distribute the same kernel restrictions differently. If \(T\) and both outer pieces are \(\mathrm M\), the triple-kernel bound leaves \(a^{1-\alpha}b^{1-\alpha}\). If \(T,A_1\) are \(\mathrm M\) and \(A_2\) is \(\mathrm Z\), use the two-kernel bound \(a\) and its extra \(b^\alpha\). If \(T\) is \(\mathrm M\) and \(A_1\) is \(\mathrm Z\), use its extra \(a^\alpha\), restrict \(T\) to width \(Car\), and obtain \(b^\alpha\) from either a single restriction on \(A_2\) or its \(\mathrm Z\) improvement. In these three cases the remaining factor is bounded by \(C a^{\kappa}\), where \(\kappa=\min(\alpha,1-\alpha)\), before possible good–good losses.

If \(T\) is \(\mathrm Z\), neither turn has a good–good loss. Obtain \(b^\alpha\) from \(A_2\) as above. When \(A_1\) is \(\mathrm Z\), its extra \(a^\alpha\) comes from the macroscopic shared-jump tail. When \(A_1\) is \(\mathrm M\) with a bad turn probe, fix the \(\mathrm Z\) retreat’s height and use \(|m_1-m_T|\le Car\) to restrict the free residual kernel of \(A_1\), obtaining \(a^\alpha\). In both cases the tail and restriction factors cancel \((ab)^{-\alpha}\); the remaining first-turn factor is explicitly \((q/r)^\alpha\). When \(A_1\) has a good turn probe and a macroscopic first jump, the tail improvement and the single residual restriction give \(a^{2\alpha}\). In the remaining case its first jump and good turn string both have height below \(c'r\), whereas \(A_1\) has macroscopic height. Its residual height is therefore macroscopic; the sharper single-kernel restriction gives \(a\). Thus, with \(\kappa=\min(\alpha,1-\alpha)=1/4\), the total mass is at most \[C r^{-\alpha}(1+\log r)^{C_1} \left[(1+\log r)^{-D\kappa+2Q\alpha} +(1+\log r)^{-Q\alpha}\right].\] Choose \(Q\) first so that \(Q\alpha>P+C_1\), and then \(D\) so that \(D\kappa>P+C_1+2Q\alpha\). Bounded scales are absorbed by increasing \(C\). ◻

Proof. If such an arc exists, simplicity and the lattice area bound force horizontal range at least \(cH^{3\lambda/4}\). One of the oblique lattice normal projections consequently has range \(R\ge cH^{3\lambda/4}\). Between suitable extreme vertices of that projection take a weak bridge, and extend its ends at bounded cost. Its height is comparable to \(R\) and its vertical range is still \(O(H^{\lambda/4})\). This is confinement in the direction tangent to an oblique cut, rather than confinement in the oblique height.

Use the turning-extremum amplification in the proof of Theorem 35, replacing its length cap by this confinement. That argument turns a fixed power saving for every free \(\mathrm M\) residual bridge into arbitrary polynomial decay by increasing a fixed iteration depth. We verify the only changed input, namely suppression of a free \(\mathrm M\) residual bridge. Let \(V=O(H^{\lambda/4})=O(R^{1/3})\), and suppose \(R^{1-\kappa}\le s\le CR\), for a small fixed \(\kappa>0\). Residual heights \(j\le R^{1-2\kappa}\) contribute at most \(CR^{\alpha(1-2\kappa)}\). For larger heights, the permitted vertical endpoint difference gives at most \(C(1+V)\) lateral endpoint offsets at each height, since the tangent to the oblique cut has a nonzero vertical component. If \(b_j\) is the endpoint kernel, the variation bound therefore gives \[\sum_{R^{1-2\kappa}<j\le Cs} \sum_{\text{permitted offsets}}b_j \le C(1+V)\sum_{j\ge R^{1-2\kappa}}\|b_j\|_\infty \le C(1+V)B_{\lfloor R^{1-2\kappa}\rfloor} \le C R^{1/12+\kappa/2}.\] Both contributions are \(O(s^\alpha R^{-a_0})\) for sufficiently small \(\kappa,a_0>0\). Every other factor in the turning-extrema amplification is unchanged, so its arbitrarily large fixed iteration depth gives mass \(O_A(R^{-A})\). Sum over dyadic \(R\), placements, and the two exterior walk portions. Their cost is a fixed polynomial in \(H\) by the full-box bound; since \(R\) is a positive power of \(H\), it is absorbed by choosing the preceding exponent large enough. Trimming port endpoints costs only a bounded factor. ◻

Proof. Cuts and bridge weights. We first record the cuts and relative bridge shapes. The calculation then separates end tests from the remaining height variables. The height restrictions are summed by bounds uniform in the end data, so the tests can be integrated with their original renewal laws.

Every outer child of an expansion has span at least one third of its parent’s span. Choose \(0<c_0\le1/(12U)\). Every expanded parent has span at least \(s/U\ge r/U\), so all leaves have spans between \(c_0r\) and \(2r\), except possibly retreat leaves. There are only \(O_{p,U}(1)\) small retreat leaves, and they are isolated between macroscopic leaves. To see the bound, the minimal witness has \(O(p)\) branching or terminal expansions and otherwise consists of \(O(p)\) unary chains. If a parent has height \(h\) and a small retreat has height \(d<c_0r\), the middle-third cut gives both forward heights at most \(2h/3+d\le3h/4\). The retreat itself has height below \(s/U\) and cannot continue. Along a continuing chain heights never increase and stay at least \(s/U\) until the last expansion; there can consequently be only \(O_U(1)\) such reductions on each chain. The first and last leaves of an expanded subtree descend through outer children and are macroscopic, so small retreats are isolated in the chronological leaf order.

Bin each small span plus one dyadically with scale \(r_i\), and put \(r_i=r\) on macroscopic leaves. Fix the complete leaf pieces and all cut records. At the parent of a small retreat, the chosen middle-third line must belong to that retreat’s height interval (one specified line when it is empty); there are \(O(r_i)\) possible choices out of at least \(cr\). Each small leaf therefore contributes \(Cr_i/r\) to the probability of these witness data. All other line-choice probabilities sum to at most one. This uses only choices recorded by the witness, without specifying any descendants outside it.

At a turn whose incident spans are both larger than \(Cr^{1/20}\), recode its incident leaves to strict bridges from neighboring ports, preserving disjoint first-hit portions to distance \(c_A\min(r_i,r_{i+1})\). Here \(c_A\) is small depending on \(c_0\). For completeness, at a peak both lower neighbors of its triangle occur on the path. One arm uses the central top port. Launch the other from the adjacent top port through the neighboring downward triangle to its original lower neighbor. If that neighboring triangle already occurs on this arm, it must occur immediately next to that lower neighbor, so shorten the initial segment instead. An internal visit would have to use the same lower neighbor. The other arm cannot use this triangle in the retained first-hit portion for the same reason. Reflect this operation at minima. Remote-end changes lie beyond the retained first-hit portions. Where no turn test is required, extend to bounding port lines; if an extreme triangle does not border that line, one extra triangle nearer the line is unused by extremality. Empty leaves have bounded bookkeeping. This gives a finite record per cut and changes lengths and spans by \(O(1)\).

After fixing those records, normalized bridge shapes have their translations fixed successively by incidence. There is no additional sum over horizontal positions at each turn. A turn test involves only two relative shapes launched at fixed neighboring ports; changing absolute translations does not change the test. The root location, the two exterior portions, and endpoints cost \(O(H^C)\) altogether by the full-box bound, with \(C\) independent of the number of leaves.

Baseline leaf \(i\) by \(r_i^\alpha\) and initially seek \(\min(r_i,r_{i+1})^{-\alpha}\) at every turn. Because small leaves are isolated, multiplying the baselines, these turn factors, and the placement probabilities gives at most \[C^k r^\alpha\prod_{i\text{ small}}(r_i/r)^{1-\alpha}.\] If \(r_i\le Cr^{1/20}\), omit both adjacent turn tests. Restoring their nominal factors multiplies this expression by at most \(Cr_i^{2\alpha}\). The resulting contribution from that leaf is bounded, since \[(r_i/r)^{1-\alpha}r_i^{2\alpha} \le C r^{-(1-\alpha)+(1+\alpha)/20}.\] No probes are needed at its own ends. Call turns touching small or marked leaves rough; there are \(O_{p,U}(1)\) of these. All other turns are ordinary.

End probes at ordinary turns. The factors at turns will cancel all but one of the bridge mass scales. Their dependence on \(k\) must remain explicit, because \(k\) can be of order \(\log H\).

For a non-tiny leaf inspect its two strings of irreducibles through the first renewals reaching \[q_i=\left\lfloor c_qr_i(k+1)^{-\delta}\right\rfloor,\] with a sufficiently small fixed \(\delta>0\). The coefficient \(c_q\) is small, and common on macroscopic leaves. For large \(H\) these are positive. The strings are of type \(\mathrm M\) when disjoint, allowing a common terminal cut, and then leave a free middle bridge. Otherwise they share exactly one irreducible and have type \(\mathrm Z\). Call an end good when its stop lands below \(c'r_i\), and bad otherwise, with \(c_q\ll c'\ll c_Ac_0\). In type \(\mathrm Z\) both ends are bad. A bad string is a prefix with endpoint depth below \(q_i\), followed by a jump of height comparable to \(r_i\). For a birth dyad \(f\), its prefix cost is \(Cf^\alpha\) and its jump cost is \(Cr_i^{-\alpha}\). On an \(\mathrm M\) leaf the residual middle supplies baseline \(r_i^\alpha\); on a \(\mathrm Z\) leaf its single shared jump supplies \(r_i^\alpha(r_i^{-\alpha})^2=r_i^{-\alpha}\). This equality is only bookkeeping: every distinct shared jump is integrated once. For example, among \(n_0=m_0+z_0\) macroscopic pieces with \(b_0\) bad ends there are \(d_{\mathrm{bad}}=b_0-z_0\) distinct bad jumps, and \[r^{\alpha n_0}r^{-\alpha b_0} =r^{\alpha m_0}r^{-\alpha d_{\mathrm{bad}}}.\] The right-hand side records the genuine independent middle and jump variables; the left-hand side assigns their costs to leaves and turns.

At an ordinary turn the desired estimates, beyond these leaf baselines, are \[ \begin{array}{c|c} \text{incident ends}&\text{turn factor}\\ \hline \mathrm{GG}&Cr^{-\alpha}(k+1)^{\delta\alpha}\\ \text{one good, one bad}&Cr^{-\alpha}\\ \text{two bad}&Cr^{-\alpha}(k+1)^{-\delta\alpha}. \end{array} \tag{35}\] The total additional loss over all ordinary turns will be at most \(C^k(1+\log\log H)^{Ck}\).

First fix the birth dyads. Apply the ordered test \(E\) from (19) at a small fraction of the minimum of the available depths, using \(q_i\) at a good end and \(f\) at a bad end. At bounded minimum use the trivial test. Its stopping renewals occur before the first-hit portions terminate: on a bad end they precede the specified birth, and a good stop is below \(c'r\). Renewal cuts prevent later visits below the tested level. Moreover two disjoint crosscuts from the neighboring launches to the common first-hit line have the required left–right extremal order at each lower level; a point past the opposite extremum would lie on the wrong exterior side of that crosscut. Hence \(E\) is a necessary condition. After these stops the remaining birth counts are at most \(Cf^\alpha\), including the possibility that the birth occurs at the stop itself. The three costs for fixed dyads are respectively \[Cq^{-\alpha},\qquad Cr^{-\alpha}f^\alpha\min(f,q)^{-\alpha},\qquad Cr^{-2\alpha}f^\alpha g^\alpha\min(f,g)^{-\alpha}.\] Since \(f,g\le Cq\) these give the table; the last expression is more precisely \(Cr^{-\alpha}(\max(f,g)/r)^\alpha\).

Summing birth depths. We must sum the birth dyads without a power of \(\log H\) at every turn. Set \[t=\lfloor r(1+\log H)^{-D_*}\rfloor,\qquad T_0=r(1+\log H)^{-10},\] where \(D_*>12\) is a sufficiently large fixed constant. Tag an unmarked macroscopic bad jump if, from either of its relevant ends, it returns to depth at most \(2t\) after reaching \(c''r\), where \(c''>0\) is small. By Lemma 37, choosing \(D_*\) larger if necessary, the mass of tagged jumps is \[Cr^{-\alpha}(1+\log H)^{-5}.\] The constants are uniform for \(H^\beta/2\le r\le C_0H\), since \(\log r\asymp_\beta\log H\). The union over two ends costs only a constant, and no designated-jump test below will be assigned to a tagged jump.

If both bad birth depths are at most \(T_0\), sum their dyads using the improved maximum factor. Summing over the larger dyad first gives \[\sum_{f,g\le CT_0}\left(\frac{\max(f,g)}r\right)^\alpha \le C(1+\log H)(T_0/r)^\alpha \le C(k+1)^{-\delta\alpha},\] for small \(\delta\), since \(k\le pL\). Birth dyads between \(t/L_2\) and \(Cq\) have only \(O(1+\log\log H)\) possible values. Here \(L_2\) is the fixed paired-test constant in (19), unrelated to the trial depth \(L\). The remaining situation has a birth below \(t/L_2\) opposite a good end or a bad birth larger than \(T_0\). If its jump is tagged, sum this early birth dyad directly. A distinct jump has at most two ends; its factor \((1+\log H)^{-5}\) pays both possible lost dyad sums, each of size \(O(1+\log H)\).

For an untagged jump in that remaining situation, use \(F(t,r',0)\) from (19), designating the early-birth end; \(r'\) is the jump’s height dyad, with only a bounded number of choices comparable to \(r\). The opposite end supplies an independent worker string. Its test stop precedes any specified bad birth there, since that birth is above \(T_0\gg t\). On the designated arc, every visit below \(t\) up through the jump occurs before the common first-hit line. Otherwise, as the designated birth is below \(t/L_2\), the jump itself would reach \(c''r\) and subsequently return below \(2t\), forcing its tag. The crosscut argument thus gives the required \(E(t)\) event. If the same jump is designated from both ends, use the single paired bound \(F_{12}(t,t,r')\) instead. It integrates that one jump, both early prefixes, and both opposite workers’ tests. In either case the cost is \(Cr^{-\alpha}\) for the distinct jump and all its designated prefixes and tests. A worker with a later bad birth still contributes \(Cg^\alpha\) after its test, so a two-bad turn retains its factor \((g/r)^\alpha\le C(k+1)^{-\delta\alpha}\). The single-test birth condition holds because \(L_2\) is at least the single-test ratio in (19).

Simultaneous integration. We will combine the turn gains with restrictions on residual middle heights. The needed restriction can be stated now. For an unmarked macroscopic \(\mathrm M\) leaf, let \(z\) be the combined height of its two end strings and \(j\) its middle height. Its total height is \(z+j\). The normalized middle kernel \[K_r(j)=r^{-\alpha}B_j\mathbf1_{\{0\le j\le Cr\}}\] obeys, for every interval \(I\) of width \(w\ge1\) and every shift \(z\), \[ \sum_{j:\,z+j\in I}K_r(j)\le C(w/r)^\alpha. \tag{36}\] Indeed \(B_j\le C(1+j)^{\alpha-1}\), so the sum over any \(w+1\) consecutive heights is at most \(Cw^\alpha\). Since \(k\le pL\) and \(r\ge H^\beta/2\), requiring two leaf heights to differ by \(O(1+r/\sqrt{k})\) costs \(O((k+1)^{-\alpha/2})\) if one leaf is of type \(\mathrm M\): fix the partner’s height, sum that middle in its translated window, and then sum the partner’s middle freely if it has one. The bound is uniform in the partner’s end strings and jumps, leaving those variables available for their later probe estimates. For disjoint leaf pairs these middle integrations can be performed together. The construction of sufficiently many such pairs is given below.

We now integrate the turn tests while retaining these uniform middle-height gains. Work with nonnegative critical-weight sums and fix only discrete records: the chain shape, scales, piece types, good/bad labels, tag labels, and assignments of turn tests. Do not condition any bridge law on its total height or on the completed chain. For an \(\mathrm M\) leaf its atomic variables are its two oriented end prefixes, any distinct bad stopping jumps, and its middle bridge. For a \(\mathrm Z\) leaf they are the two pre-jump prefixes and one shared jump. Concatenation identifies a tuple of such variables with at most a constant number of original records. Dropping unused self-avoidance and incidence conditions therefore bounds the original sum by a product of their critical measures. The restrictions retained in this product are local turn tests, individual jump restrictions, and the residual-height constraints just described.

Each end prefix occurs at exactly one turn, or at a free chain end. Assign every distinct bad jump to exactly one of three blocks: its individual tail bound (with its tag restriction if present), a single \(F\) block, or a paired \(F_{12}\) block. A designated block contains the designated pre-jump prefix or prefixes and the opposite worker prefixes through their test stops. An ordinary \(E\) block contains its two prefixes only through their test stops. These sets of variables are disjoint. In particular a worker at a designated turn has no early designated birth there: it is either good or its birth is above \(T_0\). The tested worker portion therefore excludes its own separately charged bad jump. If a shared jump has just one assigned \(F\) test, the test at its other end uses only the prefix before that jump; it belongs to a different block and never uses the shared jump a second time.

First sum the residual middles with the prescribed height restrictions, using (36) for one middle in each disjoint pair and the unrestricted bound for the others. These bounds are uniform in all the end variables. Leave every designated pre-jump prefix intact inside its assigned \(F\) or \(F_{12}\) block. For an ordinary \(E\) prefix or a worker prefix, next sum only its untested continuation. Given its test stopping history, the mass of subsequent renewal occurrences in a birth dyad \(g\) is at most \(Cg^\alpha\), uniformly in the stopping location and overshoot: an overshoot beyond the permitted endpoint contributes zero, and otherwise the remaining range has length at most \(Cg\). A good completion is a probability at most one. Length insertions are handled by the same conditional argument described below. Replacing these continuation sums by their uniform bounds removes them without conditioning the preceding tests. Now integrate each disjoint \(E\), \(F\), or \(F_{12}\) block using (19); integrate every remaining jump using its individual tail bound. Tonelli’s theorem justifies this order. Relative launches in each block are fixed neighboring ports, so absolute translations depending on variables outside the block introduce no parameter into its estimate. In a paired block this is exactly the two-end convention in the definition of \(F_{12}\). Finally sum the discrete labels and dyads as above. Thus no estimate has been applied to a renewal law conditioned on an eventual height, a tag, another turn’s survival, or full-chain avoidance. This proves (35) with the stated total loss.

Long marked leaves. The ordinary-turn estimates control the unmarked chain. A marked leaf must additionally pay \(H^{-\epsilon+o(1)}\) for exceeding its length threshold. Only a fixed number of turns touch such leaves, so the logarithmic first-hit estimate suffices there.

We next include rough turns and the marked length factors. Uniformly for \(1\le u\le C H\), the first-length moment gives \[\sum_{j\le Cu}\sum_{B:\,\operatorname{ht}(B)=j} \rho^{|B|}(1+|B|) \le u^{\alpha+4/3}H^{o(1)}.\] The analogous summed moment for an irreducible of height at most \(Cu\) is at most \(u^{4/3-\alpha}H^{o(1)}\). Indeed, in a height dyad \(b\), prepend and append bridges of heights in \([b,2b]\), whose summed masses each are at least \(cb^\alpha\). The resulting bridge has height \(O(b)\) and has only boundedly many irreducibles of height at least \(b\). Its length moment bounds the marked jump’s length moment, giving \(b^{4/3-\alpha}H^{o(1)}\) after division by \(b^{2\alpha}\). Sum the dyads, using \(4/3>\alpha\). These moment statements mean that for every fixed \(\eta>0\) the displayed factors \(H^{o(1)}\) are bounded by \(C_\eta H^\eta\), uniformly over all indicated scales.

Expand the product of length-plus-one factors on marked leaves by assigning each insertion to its middle, a bad jump, a good end string, or a pre-bad-jump prefix. There are only boundedly many categories per marked leaf; for an end string sum over its individual jumps. A middle or tiny-piece insertion costs at most \(r_i^{4/3}H^{o(1)}\) beyond its unweighted baseline. A bad-jump insertion has the same relative cost by the irreducible moment. No tag saving or designated \(F\) test uses a marked leaf’s own jump.

For an insertion in an end string, let \(d\) be the dyadic scale of one plus the inserted jump’s birth depth. Up to that birth the occurrence mass is at most \(Cd^\alpha\). The jump’s length moment, after extracting \(r_i^{4/3}H^{o(1)}\), costs \(Cr_i^{-\alpha}\). Thus this insertion gives an additional factor \((d/r_i)^\alpha\). Any subsequent required bad birth of dyad \(f\) costs at most \(Cf^\alpha\) by restarting after the inserted jump, with its bad-jump price already separately budgeted; completion to a good stop has probability at most one. This includes the terminal stopping jump of a good string. Additive constants in the length weight use the unweighted estimate.

At a non-omitted rough turn apply the first-hit bound (26) at a small fraction of the smaller available depth. Use \(d\) on an end with an insertion in its prefix, \(f\) on another bad end, and \(q_i\) on another good end. The test precedes the counted births and lies within the preserved disjointness range. Subsequent occurrence counts have the same conditional bounds as before. A side of available depth \(v\) has, after extraction of its baseline and any length price, cost at most \(H^{o(1)}(v/r_i)^\alpha\). For a good side the factor is supplied at polylogarithmic cost, since \(q_i/r_i\) is a negative power of \(k+1\). For two incident sides the test and these factors give \[H^{o(1)}\frac{v_1^\alpha v_2^\alpha} {r_i^\alpha r_{i+1}^\alpha} \min(v_1,v_2)^{-\alpha} \le H^{o(1)}\min(r_i,r_{i+1})^{-\alpha},\] because \(v_j\le Cr_j\). Free ends and omitted turns cost at most \(H^{o(1)}\). There are only \(O_{p,U}(1)\) rough turns and small leaves. Their scale choices cost \((1+\log H)^{O_{p,U}(1)}\), so all these losses remain \(H^{o_{p,U}(1)}\).

This rough calculation is compatible with the preceding block integration. At a rough turn an unmarked neighboring leaf supplies only its good prefix or its prefix before a bad jump; no marked insertion or rough test consumes that unmarked bad jump. Stop the rough tests before the inserted jumps, then use the conditional moment and renewal bounds for the continuations. Distinct marked leaves have distinct inserted jumps. Their length prices hence multiply. On a long marked leaf, \[\mathbf 1_{\{|B|>H^\epsilon(1+\operatorname{ht}(B))^{4/3}\}} \le \frac{1+|B|}{H^\epsilon(1+\operatorname{ht}(B))^{4/3}}.\] Its scale is comparable to \(1+\operatorname{ht}(B)\), so division by this threshold gives \(H^{-\epsilon+o(1)}\) per mark.

The gain from many leaves. We have obtained the turn baselines and the cost of each marked length. It remains to obtain the factor \((k+1)^{-bk}\) from the many unmarked macroscopic leaves. We now construct the close-height pairs to which (36) applies.

There are disjoint sibling pairs covering all but \(O_{p,U}(\sqrt{k}+1)\) leaves for which the two leaf heights differ by \(O(1+r/\sqrt{k})\). Indeed all but \(O(p)\) expansions lie on unary chains. The total drop of continuing heights along each such chain is at most \(2r\), so at most \(O(\sqrt{k})\) expansions on that chain have a drop larger than \(r/\sqrt{k}\). At every other unary expansion the two non-continuing children are leaves. Write \(h\) for the parent height, \(a,c\) for the forward child heights, and \(d\) for the retreat height. The signed-height identity is \(h=a-d+c\), and all three child heights are at most \(h\). If the first child continues, the drop \(h-a=c-d\) is exactly the difference of the other two heights; the last-child case is symmetric. If the retreat continues, then \(a+c=h+d\), with \(a,c\le h\), so \(a,c\in[d,h]\) and \(|a-c|\le h-d\). Consequently a continuing-height drop at most \(r/\sqrt{k}\) always supplies a sibling pair whose heights differ by at most that amount, up to the bounded port recodings. Remove the boundedly many pairs touching marked or small leaves. Specifying which of these pairs meet the constraint and their types costs at most exponentially many choices in \(k\).

In each retained pair with at least one \(\mathrm M\) leaf, integrate one such middle kernel with the height constraint, holding the partner’s height fixed. If the partner also has a middle, sum that middle without the constraint. The pairs are disjoint, so these integrations are simultaneous and uniform in all probe variables. These are the residual integrations used in the block argument above. There are \(N=2k+1\) leaves. If at least \(N/10\) are \(\mathrm M\), then, after removing the \(O_{p,U}(\sqrt{k}+1)\) exceptions, at least \(N/40\) retained pairs contain an \(\mathrm M\) leaf for large \(k\). Their total gain is at most \((k+1)^{-\alpha N/80}\). The at most \(N\) good–good turns cost at most \((k+1)^{\delta\alpha N}\), so a fixed choice \(0<\delta<1/160\) leaves a negative multiple of \(k\log(k+1)\). If fewer than \(N/10\) leaves are \(\mathrm M\), at least \(4N/5-1-O_{p,U}(1)\) adjacencies join two \(\mathrm Z\) leaves, allowing for the bounded number of small leaves and rough exceptions. At most \(N/10\) adjacencies can be good–good. For large \(k\) the excess of bad–bad over good–good ordinary turns is therefore at least \(N/2\), giving the factor \((k+1)^{-\delta\alpha N/2}\). Thus in either case the total factor is at most \((k+1)^{-bk}\) for an absolute \(b>0\), apart from constants to the power \(k\) and the stated logarithmic losses. Bounded \(k\) are absorbed by changing the constants.

Ordered ternary tree shapes, choices of the continuing children, piece labels, and the finite recoding records have at most exponential multiplicity in \(k\). Choices of the fixed number of marks contribute \(H^{o_p(1)}\). Small-scale dyads do so as well. The factor \(r^\alpha\), root placements and exterior paths are all covered by a fixed polynomial \(H^C\) independent of \(p,U\). This proves the bound in (34).

For the final assertion of the lemma, start instead with the given alternating chain of \(k\le pL\) comparable, nonincreasing spans. This range preserves the probe and dyad estimates above. There are no small leaves or placement weights. Pair successive pieces, leaving at most one unpaired. Since the total height drop is at most \(2r\), all but \(O(\sqrt{k})\) pairs have difference at most \(Cr/\sqrt{k}\). The same recoding, tests, variable partition, and middle-kernel integrations apply, and give the stated estimate. ◻

Proof of Theorem 36. Choose \(\epsilon>0\) much smaller than \(\tau\). In the notation of Lemma 39, when \(k=L\) the logarithm of the last two factors of (34) is \[-bL\log L+O\bigl(L\log(2+\log\log H)\bigr),\] which is smaller than \(-A\log H\) for every fixed \(A\) and all sufficiently large \(H\). Thus spine witnesses have superpolynomially small mass. For arbitrary \(k\le pL\), let \(V_H=C'(1+\log\log H)^{C'}\). The maximum of \(k\log V_H-bk\log(k+1)\) is \(O(V_H^{1/b})=o(\log H)\). Consequently the sum over \(k\) costs \(H^{o(1)}\). Choose \(p\) sufficiently large that the factor \(H^{-p\epsilon}\) absorbs the fixed polynomial and the desired failure power. Next choose \(U\) large enough that \(pU^{-4/3}<1/4\). Constants depending on these now fixed parameters do not affect either conclusion. Summing the \(O(\log H)\) root dyads makes all unwanted witnesses negligible at any prescribed power.

Sum the witness bounds over their sizes and use the kernel \(K_\gamma\) defined above. Since each witness count dominates its existence indicator, the resulting bound controls the \(\mu_H(d\gamma)K_\gamma(d\omega)\)-mass of pairs containing any unwanted witness. The root-subarc choices are already included in that count. Add the path exceptional set from Lemma 38, with parameter \(\lambda=\epsilon\).

Every remaining path-and-trial configuration satisfies the deterministic implication (33). That weak-bridge statement is a property of \(\gamma\) alone. If a path violates it and is not in the thin exceptional set, every trial configuration over that path must contain an unwanted witness. Therefore, writing \(\mathcal B\) for the bad paths and \(\mathcal W\) for the unwanted configurations, \[\mu_H(\mathcal B) =\int_{\mathcal B}K_\gamma(\Omega_\gamma)\,d\mu_H(\gamma) \le (\mu_H K)(\mathcal W) +\mu_H(\text{thin-excursion exception}).\] There is no loss from projecting away the trial variables.

To pass to arbitrary subwalks, split at a lowest vertex and read both halves away from that vertex. Successively cut each half at the last alternating extremum of the remaining tail. This gives weak bridges of nonincreasing spans. Outside another superpolynomial exceptional set, there are fewer than \(\lceil\log H\rceil\) consecutive pieces in any one span dyad above \(H^\beta/2\): otherwise apply the final part of Lemma 39 to \(k=L=\lceil\log H\rceil\) of those pieces. Root locations and exterior pieces again have only fixed polynomial cost. Summing the weak-bridge length bound over decreasing dyads gives at most \(CH^{2\epsilon}\log H(1+s)^{4/3}\) per half. Once the remaining span is below \(H^\beta\), the whole remaining tail is thin and has length at most \(H^\epsilon\). Choose \(\epsilon\) so that these bounds fit \(H^\tau(1+s)^{4/3}\) for large \(H\). Enlarging the exceptional-set constant handles bounded \(H\). A finite union over lattice symmetries proves the simultaneous directional statement. ◻

Proof. Put \(d_0=\sqrt3/2\) and use triangular coordinates with generators \((1,0)\) and \((1/2,d_0)\). Choose a fixed integer \(b=b(D)\) sufficiently large and put \(m=bj\). In the physical cylinder the successive half-period translations are \[m(-1/2,d_0),\qquad m(1/2,d_0).\] Thus the total period is \((0,2md_0)\) and the marker abscissas differ by \(m/2\). For \(b\) sufficiently large, an integer horizontal translation places the entire width-\(Dj\) slab strictly between those abscissas, with a fixed positive margin. Align its bottom with a marker band level.

Label the fixed lower ports \(s_1,\ldots,s_a\) and the fixed upper ports \(t_1,\ldots,t_a\), and write \(\pi\) for the fixed matching: strand \(i\) joins \(s_i\) to \(t_{\pi(i)}\). Choose independently \(2b\) tuples from the ensemble. Put the first in bands \(0,\ldots,j\) and reflect the second across the line at height \(jd_0\), traversing it upward. Reflection across any horizontal band boundary is a lattice symmetry. The second strip takes \(t_{\pi(i)}\) to the translate of \(s_i\) at height \(2jd_0\). Repeat this construction in successive double strips. Each double strip therefore induces the identity on the lower port labels. After all \(2b\) strips, the vertical quotient closes each strand at its own initial port.

The strip interiors are disjoint. Strict bridges meet their boundary cuts only at the fixed ports, and those ports are distinct within each tuple. Hence the resulting family contains exactly \(a\) disjoint simple polygons. Each lift travels through one full vertical period, so each polygon is essential. Since the whole family stays between the marker abscissas, each polygon separates the two markers.

Cutting this family at the fixed \(j\)-spaced levels recovers every input tuple. The strip index specifies its reflection, and the prescribed ports recover the strand labels and matching. Recovery is therefore injective even though the polygon family itself is unlabeled and unoriented. Ports carry no vertex weight, so the critical weights multiply exactly at every concatenation. If \(M\) is the input tuple mass, positivity and (37) give \[M^{2b}\le w_{2bj}(a) \le\exp\!\left(C\log(2bj)-\frac{ca^2}{\log(2bj)}\right).\] Take the \(2b\)th root and use \(\log(2bj)\asymp_b\log(2+j)\). For bounded \(j\), the fixed-width slab contains only boundedly many distinct ports, so \(a\) is bounded; the finite bridge masses cover these cases by changing the constants. ◻

Proof. Use Theorem 36 with a smaller slack \(\tau'>0\). It suffices to prove the statement at dyadic radii with further slack; rounding a radius upwards then absorbs the constant for all radii. There are polynomially many possible centers and \(O(\log H)\) scales. Small fixed radii are harmless by the lattice area bound.

For fixed \(v,s\), surround \(v\) by a lattice hexagonal box whose faces are at perpendicular distances between \(10s\) and \(20s\). For each face fix a parallel inner lattice line about \(4s\) inward. Starting at the next visit to \(\overline B(v,s)\), follow the walk until its first subsequent hit of a box face, or its endpoint; repeat at the next visit to the ball. Each such visit batch is contained in the box. Its span is \(O(s)\), so the local length bound, with bounded trimming at face ports, allows at most \(CH^{\tau'}s^{4/3}\) visited vertices in that batch. If (39) fails, then for a fixed small \(0<\xi<\tau-\tau'\) and large \(H\), at least \(a=\lceil H^\xi\rceil\) completed passages reach the same face. There is at most one uncompleted terminal batch.

From each selected passage retain its segment from the last hit of that face’s inner line to the first hit of the face. These are mutually disjoint strict bridges, all of height \(j\asymp s\), in the common box and hence a slab of width \(Dj\), after a lattice symmetry. If there are fewer than \(a\) available ports, the event is empty. Otherwise the endpoint tuples, matchings, and chronological placements have at most \(\exp(O(a\log H+\log H))\) choices. The complementary path segments cost the same order: there are \(a+1\) of them, each with polynomial full-box susceptibility, and we may drop their mutual avoidance. Apply Lemma 40 to each fixed choice. Since \(s\le H\), the total mass is bounded by \[\exp\!\left(Ca\log H+C\log H-\frac{ca^2}{\log H}\right),\] which is smaller than every power of \(H^{-1}\) for \(a=\lceil H^\xi\rceil\). The polynomial union over centers and scales, and the exceptional set for the local length estimate, preserve the claim. ◻

Proof. A violation has diameter \(d>H^\tau(1+l)^{3/4}\). Some lattice normal projection has span at least \(cd\). Between its extreme levels take an intervening weak bridge and extend it to a strict bridge. Its height \(h\) is at least \(cd\) and at most \(CH\), and its length is at most \(l+O(1)\). Thus \(h\ge cH^\tau\) and, for a smaller fixed \(\tau'>0\) and large \(H\), \[l+O(1)\le C H^{-4\tau/3}h^{4/3}\le h^{4/3-\tau'}.\] Apply the fast-bridge estimate (28) dyadically in \(h\). The height lower bound is a positive power of \(H\), and all placements and exterior portions cost only a fixed polynomial by the full-box bound. Its arbitrary failure power therefore gives the assertion. ◻

Proof. The honeycomb lattice is vertex-transitive. Cutting a walk after \(m\) edges and forgetting the mutual avoidance of its two pieces gives \(c_{m+n}\le c_mc_n\). If \(Z_{n_0}=a<1\) for some positive \(n_0\), writing \(n=qn_0+r\), \(0\le r<n_0\), gives \[Z_n\le a^q\max_{0\le r<n_0}Z_r\le Ce^{-cn}.\] Trimming the two endpoint half-edges of a port bridge gives a vertex walk from one of finitely many possible neighbors of the source port; the weight ratio and inverse multiplicity are bounded. A height-\(h\) bridge has at least a constant multiple of \(h\) edges. The preceding bound would therefore imply \(B_h\le C\sum_{n\ge c'h}e^{-cn}\le C'e^{-c''h}\), contrary to \(B_h\asymp h^{-1/4}\). The zero-length case is immediate. ◻

Proof of Theorem 1. Fix \(\delta,k>0\). Use (41)–(43) with a positive \(\eta\) much smaller than \(\delta\) and with the prescribed failure exponent \(k\). All arguments below take place on their single simultaneous good event.

Applying (41) to the entire path gives \[1+D\ge c n^{(3/4)(1-\eta)}.\] Equation (42), again for the entire path, gives \(D\le Cn^{3/4+\eta}\). Taking \(\eta\) small enough, then \(n\) large enough, absorbs the constants and the additive one and proves (1).

For the upper local mass bound, use both the deterministic bound \(M_\gamma(z,s)\le n+1\) and (43). For sufficiently small \(\eta\) and large \(n\) this gives \[M_\gamma(z,s)\le n^\delta\min\{n,s^{4/3}\}.\] To prove the lower bound, fix \(z=\gamma_i\) and \(s\in[1,n]\), and put \(m_0=\min\{n,s^{4/3}\}\). If \(n^{-\delta}m_0\le1\), the vertex \(z\) itself suffices. Otherwise choose \[q=\left\lfloor n^{-\delta/2}m_0\right\rfloor.\] For large \(n\), one side of \(i\) has at least \(n/2\ge q\) edges available. Choose the contiguous block of \(q\) edges on that side. Since \(m_0>n^\delta\), we have \(q\ge\tfrac12n^{-\delta/2}m_0\) and \(q+1\le2n^{-\delta/2}m_0\) for large \(n\). Its diameter is at most \[C n^\eta(1+q)^{3/4} \le C' n^{\eta-3\delta/8}m_0^{3/4} \le C' n^{\eta-3\delta/8}s\le s,\] provided \(\eta<3\delta/8\) and \(n\) is large. The entire block lies in \(\overline B(z,s)\). Self-avoidance makes its \(q+1\) vertices distinct, and \(q+1\ge n^{-\delta}m_0\) for large \(n\). This proves the lower local mass bound simultaneously in \(i\) and \(s\), using only the already simultaneous modulus.

Proof of Corollary 2. The local-length theorem holds simultaneously in the finitely many lattice normal directions. Apply it to the whole walk after the same exact-length probability transfer. If \(s\) is any such band span, \(n\le Cn^\eta(1+s)^{4/3}\), so \(s\ge c n^{(3/4)(1-\eta)}-1\). Convert band height to Euclidean projection using \(d_0\), then choose \(\eta\) sufficiently small relative to \(\delta\). Constants and the additive one are absorbed for all sufficiently large \(n\). ◻

Proof of Corollary 3. Fix \(\epsilon\in(0,3/4)\). On the event of Theorem 1 with error \(\delta<\epsilon/2\), uniformly for \(s\le n^{3/4-\epsilon}\), \[\log(D/s)\ge(\epsilon-\delta)\log n, \qquad \log\mathcal N(s)=\log n-\frac43\log s+O(\delta\log n)+O(1).\] The second assertion follows because \(ns^{-4/3}\ge n^{4\epsilon/3}\); replacing \(1+ns^{-4/3}\) by \(ns^{-4/3}\) changes its logarithm by \(O(1)\). Also \[\frac43\log(D/s)=\log n-\frac43\log s+O(\delta\log n).\] The difference of the two quantities, divided by \(\log(D/s)\), is \(O(\delta/\epsilon)+o(1)\). Given a desired error, first choose \(\delta\) small enough and then let \(n\to\infty\). The bad event has probability tending to zero, proving the Corollary. ◻

Proof of Corollary 4. The exact identity \[R_g^2=\frac1{2(n+1)^2}\sum_{i,j=0}^n|\gamma_i-\gamma_j|^2\] shows \(R_g\le D\). To get the lower bound, fix \(0<\epsilon<3/4\) and put \(r=n^{3/4-\epsilon}\). Use the upper local mass bound with slack \(\eta<4\epsilon/3\). Uniformly in \(i\), the number of \(j\) within distance \(r\) of \(\gamma_i\) is at most \[n^\eta r^{4/3}=n^{1+\eta-4\epsilon/3}=o(n).\] For large \(n\), at least \((n+1)/2\) indices \(j\) are farther than \(r\) from each \(\gamma_i\). Hence \(R_g^2\ge r^2/4\). Arbitrarily small \(\epsilon\), together with the diameter upper bound and absorption of the constant, proves the claimed probability law.

Finally both \(D\) and \(R_g\) are deterministically at most a constant times \(n\). For fixed \(p>0\), choose the failure exponent \(k>p+1\). Their \(p\)th moments on the bad event are \(O(n^{p-k})\). On the good event, their \(p\)th powers lie between \(n^{3p/4-p\delta}\) and \(n^{3p/4+p\delta}\) after adjusting the exponent error. Taking expectations and then letting \(\delta\downarrow0\) proves both moment assertions. ◻

Proof. Choose a large fixed \(A\), to be specified below. We prove the bound for \(x\ge2A\); increasing the final constant covers \(1\le x<2A\). Put \(r=\lfloor x/A\rfloor\ge1\), and write \(p=\mathbb P(W>x,J\le r)\). Let \(Z\) count renewal levels \(u\le r\) whose next irreducible has \(W>x\) and \(J\le r\). Independence of the next increment gives \[\mathbb EZ=p\sum_{u=0}^r B_u\asymp r^{3/4}p.\] For an ordered pair of distinct marks, expose the first marked jump through its completion. Its terminal height is at most \(2r\). Conditional on this history, the expected number of later marks born at height at most \(r\) is zero if the endpoint exceeds \(r\), and otherwise is at most \(p\sum_{v=0}^r B_v\le Cr^{3/4}p\). Consequently \[\mathbb E[Z(Z-1)]\le 2Cr^{3/4}p\,\mathbb EZ, \qquad \mathbb EZ^2\le Cr^{3/4}p(1+r^{3/4}p).\] Cauchy–Schwarz therefore gives \(\mathbb P(Z>0)\ge c\min(1,r^{3/4}p)\).

On \(Z>0\), stop immediately after completing the first marked jump. This is a stopping index in the irreducible sequence, at height at most \(2r\); the fresh continuation is independent of the exposed history. It forces an excursion farther than \(x/2\) from the original root: either its start is already that far away, or its displacement forces this during the jump. After detection the expected number of further renewals within another \(r\) bands is \(\asymp r^{3/4}\). Every resulting bridge, ending by \(3r\), has that excursion. Summing the lateral loss in Lemma 12 gives total mass at most \(Cr^2x^{-5/4}\). Dividing by the continuation count bounds the detection probability by \(C(r/x)^{5/4}\). Take \(A\) large enough to exclude the value 1 in the preceding minimum. It follows that \(p\le C_Ax^{-3/4}\); adding \(\mathbb P(J>r)\) proves the first claim. The union bound for \(J+W\) and integration of its tail give the second: \[\mathbb E\min(1,(J+W)/x) =x^{-1}\int_0^x\mathbb P(J+W>u)\,du\le Cx^{-3/4}.\] ◻

Proof. Fix small \(\varepsilon>0\), put \(p_*=3/4-\varepsilon\), and use strong induction with bound \(KH^{-p_*}\). Choose a large fixed \(L\), then set \(H_0=\lceil H/L\rceil\). Stop each path at its first renewal at or above \(H_0\). On survival through \(H\), the event that either prefix has sum of \(J+W\) greater than \(H/8\) has probability at most \[ C_\varepsilon K L^{-\varepsilon}H^{-p_*}. \tag{46}\] Indeed, group jumps starting below \(H_0\) into dyadic height bins \([j,2j)\), with a separate bin for height zero. For one path expose the other path, and stop the first at its first renewal at or above \(j\). Survival through first visits to \(j\) is measurable then. The expected subsequent sum of charges \(\min(1,(J+W)/H)\) from starting heights in that bin is at most \(Cj^{3/4}H^{-3/4}\), uniformly in the overshoot; if it skips the bin the contribution is zero. Multiply by \(S_2(j)\) and sum: \[CH^{-3/4}\left[1+ K\sum_{j\le H_0}^{\rm dyadic}j^\varepsilon\right] \le C_\varepsilon K L^{-\varepsilon}H^{-p_*}.\] If a sum exceeds \(H/8\), its truncated charge sum is at least \(1/8\), so this proves Equation (46). Choose \(L\) so this costs less than one quarter of the induction budget.

Fix two stopped prefixes having these size bounds and surviving through \(H_0\). Their future irreducibles remain independent. Under the law of these two continuations, we shall exclude an event of arbitrarily small prescribed probability, uniformly over the fixed prefixes. Outside that event, continued survival through \(H\) supplies at least \(cH^{3/4}\) suitable renewals on each side, in \([\eta H,3H/4]\), such that every cross-pair can be extended to a disjoint equal-height pair at cost \(H^{-1/4-o(1)}\).

We first retain many renewals in this interior height window while controlling their horizontal range. Take the next \(\lceil c'H^{3/4}\rceil\) jumps after either stop. Lemma 45, applied to their truncated heights, shows that all these renewals lie within an additional \(H/2\) bands except with probability \(O(c')\). Renewals within \(\eta H\) of the stop have expected number \(O((1+\eta H)^{3/4})\), and can be discarded. The full first-hit paths to \(H\) have horizontal range at most \(D_0H\), apart from conditional probability \(O(D_0^{-3/4})\): sum the truncated charges of jumps starting below \(H\), whose expected number is \(O(H^{3/4})\).

For right renewals expose also the full left first-hit path. Its future is independent of the newly sampled right future. At height \(u\), let \(R(u)\) be its rightmost abscissa within \([u,u+\delta H]\). Lemma 13 implies \[\sum_{j\ge\eta H}\|b_j\|_\infty \le C\sum_{j\ge\eta H}(B_j-B_{j+1}) \le CB_{\lfloor\eta H\rfloor}.\] Thus the expected number of right renewals, at least \(\eta H\) above their stop, with \(|x-R(u)|\le2eH\), is at most \(C_\eta(e+H^{-1})H^{3/4}\). Discard also renewals whose path up to the next renewal more than \(\delta H\) above them travels horizontally by more than \(eH\). Restarting there and using the truncated-charge bound gives expected discarded count at most \(C(\delta/e)^{3/4}H^{3/4}\).

On survival, the disjoint first-hit arcs have fixed left-right order across the slab. At a given height no point of the left arc can lie to the right of the right arc’s rightmost point: that region is in the right component of its complement. At a retained right renewal the right arc in the next \(\delta H\) bands has abscissa at most \(x+eH\). Hence \(R(u)\le x+eH\); exclusion of \(|x-R(u)|\le2eH\) forces \(R(u)<x-2eH\). The reflected argument holds for left renewals. Choose first \(c'\), then \(\eta\), then \(e\), and then \(\delta<e\), sufficiently small, and finally \(D_0\) large. Markov’s inequality and a union bound, applied to the two tests without conditioning either on success of the other, leave \(cH^{3/4}\) renewals per side with any prescribed fixed small failure probability. Lemma 44 applies to every cross-pair: the shorter bridge is extended above its renewal level, so it cannot meet its own prefix, while the other prefix is the obstacle. This proves the assertion.

Choose the just-described failure probability less than \(\tfrac14L^{-p_*}\), with a harmless adjustment for the ceiling in \(H_0\). Its total contribution is then at most one quarter of \(KH^{-p_*}\), by the induction bound on \(S_2(H_0)\).

Finally, under the original product renewal law, let \(Z_H\) count pairs of disjoint renewal prefixes in the stated height and horizontal ranges for which the obstacle lemma’s clearance conditions allow the prefix of lesser height to be extended to the other’s height. Equal heights use the empty extension. Set \(Z_H=0\) unless the first-hit paths survive through \(H\). These geometric conditions depend only on the two finite prefixes, and their extension masses have a common lower bound \(a_H=H^{-1/4-o(1)}\). A pair of prefixes occurs with exactly the product of its bridge weights. Dropping the subsequent survival condition and appending every allowed extension therefore gives \[a_H\mathbb EZ_H\le\sum_{\Gamma}w(\Gamma)M(\Gamma),\] where \(\Gamma\) is an output pair of disjoint equal-height bridges, \(w(\Gamma)\) is the product weight, and \(M(\Gamma)\) counts its input representations. The common output height lies in \([\eta H,H]\). Equation (44) telescopes to total output mass \(O_\eta(H^{-1/4})\). For a fixed output, an input is recovered by choosing which bridge was extended and a single-crossing seam in that bridge; its unextended companion fixes the common height. The two end pieces in the seam decomposition are strip bridges. Dropping avoidance with the companion bounds its mass by \(B_j\le1\), and summing the common height costs only \(O(H)\), independently of the moment order. Lemma 14 bounds the multiplicity by \(H^{3/4+\zeta}\) outside arbitrarily small polynomial mass. Raw seam counts are \(O(H)\), so these exceptional contributions remain negligible. Dividing by the extension mass gives \[\mathbb EZ_H\le H^{3/4+o(1)}.\] On the retained survival event, \(Z_H\ge c^2H^{3/2}\). Its probability is therefore \(H^{-3/4+o(1)}\), less than the remaining half of the induction budget for sufficiently large \(H\). Increase \(K\) to cover the finitely many smaller heights. The induction proves the theorem. ◻