rightward direction, like the route drawn in Fig. \[fig:latt\](c), we first have to move one step to the right with probability $p$ and the result becomes $$Q_{\rm right}(l)=p~b(p)\frac{1-a^{N(l-1)}(p)}{1-a(p)}.$$ The probability for a tagged route with no backsteps is then given by $\prod_{l=1}^{L}Q_{d(l)}(l)$ where $d(l)=\pm 1$ defines a route to the left or right on that particular $l$-level. The average number of such tagged routes is hence $$\begin{aligned}
r^{\rm tag} &=& \prod_{l=1}^{L} \left[ Q_{\rm left}(l)+Q_{\rm right}(l)
\right] \nonumber\\
&=& \left[\frac{(1+p)b(p)}{1-a(p)} \right]^{L}\prod_{l=1}^{L} \left[
1-a^{N(l-1)}(p) \right] =
\left[ \frac{p(1+p)}{1-p(1-p)} \right]^{L}\prod_{l=1}^{L} \left[
1-p^{2}(1-p)^{2^{l}} \right],
\label{rtag}\end{aligned}$$ which means that $$c \left[ \frac{p(1+p)}{1-p(1-p)} \right]^{L}\leq r^{\rm tag}\leq
\left[ \frac{p(1+p)}{1-p(1-p)} \right]^{L},
\label{size}$$ where $0<c\leq 1$ is a constant independent of $L$. Thus for $\frac{p(1+p)}{1-p(1-p)}<1$, the average number of tagged routes without backsteps vanishes in the limit $L\rightarrow \infty $. This happens for $p<1/2$ which gives the estimate $p_{c2}=1/2$: at this threshold the probability for having a tagged route from an arbitrary surface point to the midpoint vanishes. In contrast, for $p>1/2$ there is a nonzero probability that a finite fraction of the surface points are connected to the midpoint. Since the number of surface points is $B=2^{L}$, this also means that, when $Br^{\rm tag} $ becomes finite, there 