Let's consider the given directed graph of clearings and trails to be divided into ~M = \text{ceil}(N/4)~ rows, with nodes ~0 \dots 3~ in row ~1~, nodes ~4 \dots 7~ in row ~2~, and so on. The constraint that an edge from node ~a~ to node ~b~ may only exist if ~0 \le \text{floor}(b/4)-\text{floor}(a/4) \le 1~ can then be restated as follows: an edge from node ~a~ to node ~b~ may only exist if ~a~ and ~b~ are in the same row, or if ~b~ is in the row just after ~a~'s row.

When taking a stroll starting from a node in row ~r~, note that Mr. Fox will walk amongst nodes in row ~r~, walk to a node in row ~r+1~, walk amongst nodes in row ~r+1~, walk to a node in row ~r+2~, and so on until he reaches a closed set of nodes in some row which have no outgoing edges to the following row, which he'll walk amongst forever (or until he reaches a node with no outgoing edges, at which point the stroll will end).

For each row, we'd like to calculate a ~4 \times 4~ table of values ~P~, where ~P_{i,j}~ is the probability that, given that Mr. Fox is currently in the ~i^\text{th}~ node in row ~r~, he will eventually reach the ~j^\text{th}~ node in row ~r+1~ without having reached any other nodes in row ~r+1~ previously. Note that ~\sum_{j=1}^4 P_{i,j}~ may not be equal to ~1~, if there's a chance that Mr. Fox will never reach row ~r+1~ from the ~i^\text{th}~ node in row ~r~.

This table can be calculated by modeling the ~8~ nodes in rows ~r~ and ~r+1~ (which we can re-number as nodes ~1 \dots 4~ and ~5 \dots 8~ respectively) as a Markov chain, with nodes ~5 \dots 8~ (and any nodes with no outgoing edges) being absorbing states. In particular, we can construct an ~8 \times 8~ one-step transition matrix ~A~, where ~A_{i,i} = 1~ if node ~i~ is an absorbing state, ~A_{i,j} = 0~ if node ~i~ is an absorbing state and ~i \ne j~, and ~A_{i,j}~ is the weighed probability of Mr. Fox walking to node ~j~ while at node ~i~ (based on the relative rock counts on outgoing edges from node ~i~) if ~i~ is not an absorbing state. The long-run transition matrix ~B~ can then be computed by using fast matrix exponentiation to take ~A~ to a high enough power (for example, ~B = A^{2^{16}}~). ~P~ then consists of the values ~B_{i,j}~ for ~i = 1 \dots 4~ and ~j = 5 \dots 8~. This computation takes ~16 \times 8^3~ time per table. ~P~ can be computed more efficiently (with ~8^3~ time) with Gaussian elimination and some additional operations, but this is not required.

The above ~P~ table calculations will need to be done for each row initially (once the starting edges have been inputted), and then each time an edge from some node ~a~ is added or removed (due to an event of type 1 or 2), the ~P~ table for node ~a~'s row will have to be recomputed. Therefore, the total time spent on them will be ~(M+D) \times 16 \times 8^3~, which is expensive but acceptable.

Now, to handle events of type 3, let's say that Mr. Fox starts at the ~{c_1}^\text{th}~ node in row ~r_1~, and we want the probability that he'll eventually visit the ~{c_2}^\text{th}~ node in row ~r_2~. If ~r_2 < r_1~, then the answer is trivially ~0~. Otherwise, the first step is to determine a list of ~4~ values ~S~ similar to the ~P~ table, where ~S_i~ is the probability that Mr. Fox will eventually reach the ~i^\text{th}~ node in row ~r_2~ without having reached any other nodes in row ~r_2~ previously. In fact, if we consider the product of matrices ~P_{r_1} \times P_{r_1+1} \times \dots \times P_{r_2-1}~ (or the identity matrix if ~r_1 = r_2~), then ~S~ is just the ~{c_1}^\text{th}~ row of this resulting matrix.

We can't afford to loop over ~\mathcal O(M)~ rows to compute ~S~ for each type 3 operation, but the standard concept of using a segment tree to perform range queries will cut that factor down to ~\mathcal O(\log M)~. In this case, each node in the segment tree will store a ~4 \times 4~ matrix of probabilities – the leaf nodes of the tree will contain the computed ~P~ tables, and each interior node will store the product of its left child's matrix with its right child's matrix. Building the tree initially takes ~M \times 4^3~ time, updating it when a ~P~ table changes takes ~\log(M) \times 4^3~ time, and querying it to compute ~S~ also takes ~\log(M) \times 4^3~ time. Therefore, the total time from these calculations is ~(M + D \log M) \times 4^3~.

Finally, given ~S~, we still need to consider Mr. Fox walking amongst the nodes in row ~r_2~ to determine the probability that he'll ever visit its ~{c_2}^\text{th}~ node. Just like the ~P~ table calculation, this may be modeled as a Markov chain, with an ~8 \times 8~ one-step transition matrix ~A~ calculated exactly as before but with node ~c_2~ also being made into an absorbing state (by ignoring all of its outgoing edges and giving it a self loop). The long-run transition matrix ~B = A^{2^{16}}~ may also be calculated as before, in ~16 \times 8^3~ time, and then the final answer will be ~\sum_{i=1}^4 S_i \times B_{i,c_2}~, considering each of the ~4~ possible points of entry into row ~r_2~.