the shortest horizontal distance $(l,d,s_{l}^{(1)})$. By choosing the shortest route-marker distance on every level one obtains a class of tagged routes from the point on the surface to the midpoint. The probability for one such tagged route, denoted by $j$, is $P_{j}^{\rm (tag)}=\prod_{l=0}^{L}
P^{\rm (tag)}(l,d_{l},s_{l}^{(1)})$ and the average number of such tagged routes, which reach the midpoint, is $r^{\rm (tag)}=\sum_{j}P_{j}^{\rm (tag)}$. Here $j=1...2^L$ enumerates the possible tagged routes which correspond to the possible sequences $d(1),d(2),...d(L)$. The point is that $r^{\rm (tag)}$ can be explicitly calculated: starting from the first route-marker on level $l$ in Fig. \[fig:latt\](c), the chance for getting to the closest point on the level $l-1$ above, when moving to the left, is $b(p)=p+(1-p)p^{2},$ because there are two routes leading to the closest point. In the same way, the chance to end at the second closest is $a(p)b(p)$ where $a(p)=p^{2}(1-p)^{2}$, the third $a^{2}(p)b(p)$ and to point number $i$ is $a^{i-1}(p)b(p)$. This also means that the chance of ending at one of the points on the level above, when moving in the left direction starting from a right triangle corner is $$Q_{\rm left}(l)=\sum_{\nu =1}^{N(l-1)}a^{\nu -1}(p)b(p)=b(p)\frac{
1-a^{N(l-1)}(p)}{1-a(p)},$$ where $N(l-1)=2^{l-1}$ is the number of points on the $l-1$ level. In the 